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string | image_1
image | image_2
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image | image_4
image | image_5
image | modality
string | difficulty
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2,581
|
In the diagram, a straight, flat road joins $A$ to $B$.
<image_1>
Karuna runs from $A$ to $B$, turns around instantly, and runs back to $A$. Karuna runs at $6 \mathrm{~m} / \mathrm{s}$. Starting at the same time as Karuna, Jorge runs from $B$ to $A$, turns around instantly, and runs back to $B$. Jorge runs from $B$ to $A$ at $5 \mathrm{~m} / \mathrm{s}$ and from $A$ to $B$ at $7.5 \mathrm{~m} / \mathrm{s}$. The distance from $A$ to $B$ is $297 \mathrm{~m}$ and each runner takes exactly $99 \mathrm{~s}$ to run their route. Determine the two values of $t$ for which Karuna and Jorge are at the same place on the road after running for $t$ seconds.
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[
"Suppose that Karuna and Jorge meet for the first time after $t_{1}$ seconds and for the second time after $t_{2}$ seconds.\n\nWhen they meet for the first time, Karuna has run partway from $A$ to $B$ and Jorge has run partway from $B$ to $A$.\n\n<img_3496>\n\nAt this instant, the sum of the distances that they have run equals the total distance from $A$ to $B$.\n\nSince Karuna runs at $6 \\mathrm{~m} / \\mathrm{s}$ for these $t_{1}$ seconds, she has run $6 t_{1} \\mathrm{~m}$.\n\nSince Jorge runs at $5 \\mathrm{~m} / \\mathrm{s}$ for these $t_{1}$ seconds, he has run $5 t_{1} \\mathrm{~m}$.\n\nTherefore, $6 t_{1}+5 t_{1}=297$ and so $11 t_{1}=297$ or $t_{1}=27$.\n\nWhen they meet for the second time, Karuna has run from $A$ to $B$ and is running back to $A$ and Jorge has run from $B$ to $A$ and is running back to $B$. This is because Jorge gets to $A$ halfway through his run before Karuna gets back to $A$ at the end of her run.\n\n<img_3542>\n\nSince they each finish running after 99 seconds, then each has $99-t_{2}$ seconds left to run. At this instant, the sum of the distances that they have left to run equals the total distance from $A$ to $B$.\n\nSince Karuna runs at $6 \\mathrm{~m} / \\mathrm{s}$ for these $\\left(99-t_{2}\\right)$ seconds, she has to run $6\\left(99-t_{2}\\right) \\mathrm{m}$.\n\nSince Jorge runs at $7.5 \\mathrm{~m} / \\mathrm{s}$ for these $\\left(99-t_{2}\\right)$ seconds, he has to run $7.5\\left(99-t_{2}\\right) \\mathrm{m}$.\n\nTherefore, $6\\left(99-t_{2}\\right)+7.5\\left(99-t_{2}\\right)=297$ and so $13.5\\left(99-t_{2}\\right)=297$ or $99-t_{2}=22$ and so $t_{2}=77$.\n\nAlternatively, to calculate the value of $t_{2}$, we note that when Karuna and Jorge meet for the second time, they have each run the distance from $A$ to $B$ one full time and are on their return trips.\n\nThis means that they have each run the full distance from $A$ to $B$ once and the distances that they have run on their return trip add up to another full distance from $A$ to $B$, for a total distance of $3 \\cdot 297 \\mathrm{~m}=891 \\mathrm{~m}$.\n\nKaruna has run at $6 \\mathrm{~m} / \\mathrm{s}$ for $t_{2}$ seconds, for a total distance of $6 t_{2} \\mathrm{~m}$.\n\nJorge ran the first $297 \\mathrm{~m}$ at $5 \\mathrm{~m} / \\mathrm{s}$, which took $\\frac{297}{5} \\mathrm{~s}$ and ran the remaining $\\left(t_{2}-\\frac{297}{5}\\right)$ seconds at $7.5 \\mathrm{~m} / \\mathrm{s}$, for a total distance of $\\left(297+7.5\\left(t_{2}-\\frac{297}{5}\\right)\\right) \\mathrm{m}$.\n\nTherefore,\n\n$$\n\\begin{aligned}\n6 t_{2}+297+7.5\\left(t_{2}-\\frac{297}{5}\\right) & =891 \\\\\n13.5 t_{2} & =891-297+7.5 \\cdot \\frac{297}{5} \\\\\n13.5 t_{2} & =1039.5 \\\\\nt_{2} & =77\n\\end{aligned}\n$$\n\nTherefore, Karuna and Jorge meet after 27 seconds and after 77 seconds."
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"27, 77"
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2,585
|
In the diagram, rectangle $P Q R S$ is placed inside rectangle $A B C D$ in two different ways: first, with $Q$ at $B$ and $R$ at $C$; second, with $P$ on $A B, Q$ on $B C, R$ on $C D$, and $S$ on $D A$.
<image_1>
If $A B=718$ and $P Q=250$, determine the length of $B C$.
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[
"Let $B C=x, P B=b$, and $B Q=a$.\n\nSince $B C=x$, then $A D=P S=Q R=x$.\n\nSince $B C=x$ and $B Q=a$, then $Q C=x-a$.\n\nSince $A B=718$ and $P B=b$, then $A P=718-b$.\n\nNote that $P Q=S R=250$.\n\nLet $\\angle B Q P=\\theta$.\n\nSince $\\triangle P B Q$ is right-angled at $B$, then $\\angle B P Q=90^{\\circ}-\\theta$.\n\nSince $B Q C$ is a straight angle and $\\angle P Q R=90^{\\circ}$, then $\\angle R Q C=180^{\\circ}-90^{\\circ}-\\theta=90^{\\circ}-\\theta$.\n\nSince $A P B$ is a straight angle and $\\angle S P Q=90^{\\circ}$, then $\\angle A P S=180^{\\circ}-90^{\\circ}-\\left(90^{\\circ}-\\theta\\right)=\\theta$.\n\nSince $\\triangle S A P$ and $\\triangle Q C R$ are each right-angled and have another angle in common with $\\triangle P B Q$, then these\n\n<img_3870>\nthree triangles are similar.\n\nContinuing in the same way, we can show that $\\triangle R D S$ is also similar to these three triangles.\n\nSince $R S=P Q$, then $\\triangle R D S$ is actually congruent to $\\triangle P B Q$ (angle-side-angle).\n\nSimilarly, $\\triangle S A P$ is congruent to $\\triangle Q C R$.\n\nIn particular, this means that $A S=x-a, S D=a, D R=b$, and $R C=718-b$.\n\nSince $\\triangle S A P$ and $\\triangle P B Q$ are similar, then $\\frac{S A}{P B}=\\frac{A P}{B Q}=\\frac{S P}{P Q}$.\n\nThus, $\\frac{x-a}{b}=\\frac{718-b}{a}=\\frac{x}{250}$.\n\nAlso, by the Pythagorean Theorem in $\\triangle P B Q$, we obtain $a^{2}+b^{2}=250^{2}$.\n\nBy the Pythagorean Theorem in $\\triangle S A P$,\n\n$$\n\\begin{aligned}\nx^{2} & =(x-a)^{2}+(718-b)^{2} \\\\\nx^{2} & =x^{2}-2 a x+a^{2}+(718-b)^{2} \\\\\n0 & =-2 a x+a^{2}+(718-b)^{2}\n\\end{aligned}\n$$\n\nSince $a^{2}+b^{2}=250^{2}$, then $a^{2}=250^{2}-b^{2}$.\n\nSince $\\frac{718-b}{a}=\\frac{x}{250}$, then $a x=250(718-b)$.\n\nTherefore, substituting into $(*)$, we obtain\n\n$$\n\\begin{aligned}\n0 & =-2(250)(718-b)+250^{2}-b^{2}+(718-b)^{2} \\\\\nb^{2} & =250^{2}-2(250)(718-b)+(718-b)^{2} \\\\\nb^{2} & =((718-b)-250)^{2} \\quad\\left(\\text { since } y^{2}-2 y z+z^{2}=(y-z)^{2}\\right) \\\\\nb^{2} & =(468-b)^{2} \\\\\nb & =468-b \\quad(\\text { since } b \\neq b-468) \\\\\n2 b & =468 \\\\\nb & =234\n\\end{aligned}\n$$\n\nTherefore, $a^{2}=250^{2}-b^{2}=250^{2}-234^{2}=(250+234)(250-234)=484 \\cdot 16=22^{2} \\cdot 4^{2}=88^{2}$ and so $a=88$.\n\nFinally, $x=\\frac{250(718-b)}{a}=\\frac{250 \\cdot 484}{88}=1375$. Therefore, $B C=1375$."
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2,591
|
How many equilateral triangles of side $1 \mathrm{~cm}$, placed as shown in the diagram, are needed to completely cover the interior of an equilateral triangle of side $10 \mathrm{~cm}$ ?
<image_1>
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[
"If we proceed by pattern recognition, we find after row 1 we have a total of 1 triangle, after two rows we have $2^{2}$ or 4 triangles. After ten rows we have $10^{2}$ or 100 triangles.\n\n<img_3802>",
"This solution is based on the fact that the ratio of areas for similar triangles is the square of the ratio of corresponding sides. Thus the big triangle with side length ten times that of the smaller triangle has 100 times the area."
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2,593
|
A rectangle PQRS has side PQ on the x-axis and touches the graph of $y=k \cos x$ at the points $S$ and $R$ as shown. If the length of $P Q$ is $\frac{\pi}{3}$ and the area of the rectangle is $\frac{5 \pi}{3}$, what is the value of $k ?$
<image_1>
|
[
"If $P Q=\\frac{\\pi}{3}$, then by symmetry the coordinates of $R$\n\nare $\\left(\\frac{\\pi}{6}, k \\cos \\frac{\\pi}{6}\\right)$.\n\nArea of rectangle $P Q R S=\\frac{\\pi}{3}\\left(k \\cos \\frac{\\pi}{6}\\right)=\\frac{\\pi}{3}(k)\\left(\\frac{\\sqrt{3}}{2}\\right)$\n\nBut $\\frac{\\sqrt{3} k \\pi}{6}=\\frac{5 \\pi}{3} \\quad \\therefore k=\\frac{10}{\\sqrt{3}}$ or $\\frac{10}{3} \\sqrt{3}$.\n\n<img_3853>"
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[
"$\\frac{10}{\\sqrt{3}}$,$\\frac{10}{3} \\sqrt{3}$"
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2,594
|
In determining the height, $M N$, of a tower on an island, two points $A$ and $B, 100 \mathrm{~m}$ apart, are chosen on the same horizontal plane as $N$. If $\angle N A B=108^{\circ}$, $\angle A B N=47^{\circ}$ and $\angle M B N=32^{\circ}$, determine the height of the tower to the nearest metre.
<image_1>
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[
"In $\\triangle B A N, \\angle B N A=25^{\\circ}$\n\nUsing the Sine Law in $\\triangle B A N$,\n\n$\\frac{N B}{\\sin 108^{\\circ}}=\\frac{100}{\\sin 25^{\\circ}}$\n\nTherefore $N B=\\frac{100 \\sin 108^{\\circ}}{\\sin 25^{\\circ}} \\approx 225.04$,\n\n<img_3946>\n\nNow in $\\triangle M N B, \\frac{M N}{N B}=\\tan 32^{\\circ}$\n\n$$\nM N=\\frac{100 \\sin 108^{\\circ}}{\\sin 25^{\\circ}} \\times \\tan 32^{\\circ} \\doteq 140.6\n$$\n\nThe tower is approximately $141 \\mathrm{~m}$ high."
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2,595
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The points $A, P$ and a third point $Q$ (not shown) are the vertices of a triangle which is similar to triangle $A B C$. What are the coordinates of all possible positions for $Q$ ?
<image_1>
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[
"$Q(4,0), Q(0,4)$\n\n$Q(2,0), Q(0,2)$\n\n$Q(-2,2), Q(2,-2)$\n\n<img_3757>"
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[
"$(4,0),(0,4),(2,0),(0,2),(-2,2),(2,-2)$"
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2,605
|
In triangle $A B C, B C=2$. Point $D$ is on $\overline{A C}$ such that $A D=1$ and $C D=2$. If $\mathrm{m} \angle B D C=2 \mathrm{~m} \angle A$, compute $\sin A$.
<image_1>
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[
"Let $[A B C]=K$. Then $[B C D]=\\frac{2}{3} \\cdot K$. Let $\\overline{D E}$ be the bisector of $\\angle B D C$, as shown below.\n\n<img_3399>\n\nNotice that $\\mathrm{m} \\angle D B A=\\mathrm{m} \\angle B D C-\\mathrm{m} \\angle A=\\mathrm{m} \\angle A$, so triangle $A D B$ is isosceles, and $B D=1$. (Alternately, notice that $\\overline{D E} \\| \\overline{A B}$, and by similar triangles, $[C D E]=\\frac{4}{9} \\cdot K$, which means $[B D E]=\\frac{2}{9} \\cdot K$. Because $[C D E]:[B D E]=2$ and $\\angle B D E \\cong \\angle C D E$, conclude that $\\frac{C D}{B D}=2$, thus $B D=1$.) Because $B C D$ is isosceles, it follows that $\\cos \\angle B D C=\\frac{1}{2} B D / C D=\\frac{1}{4}$. By the half-angle formula,\n\n$$\n\\sin A=\\sqrt{\\frac{1-\\cos \\angle B D C}{2}}=\\sqrt{\\frac{3}{8}}=\\frac{\\sqrt{6}}{\\mathbf{4}}\n$$"
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[
"$\\frac{\\sqrt{6}}{4}$"
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2,716
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Two equilateral triangles of side length 1 and six isosceles triangles with legs of length $x$ and base of length 1 are joined as shown below; the net is folded to make a solid. If the volume of the solid is 6 , compute $x$.
<image_1>
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[
"First consider a regular octahedron of side length 1. To compute its volume, divide it into two square-based pyramids with edges of length 1 . Such a pyramid has slant height $\\frac{\\sqrt{3}}{2}$ and height $\\sqrt{\\left(\\frac{\\sqrt{3}}{2}\\right)^{2}-\\left(\\frac{1}{2}\\right)^{2}}=\\sqrt{\\frac{1}{2}}=\\frac{\\sqrt{2}}{2}$, so its volume is $\\frac{1}{3} \\cdot 1^{2} \\cdot \\frac{\\sqrt{2}}{2}=\\frac{\\sqrt{2}}{6}$. Thus the octahedron has volume twice that, or $\\frac{\\sqrt{2}}{3}$. The result of folding the net shown is actually the image of a regular octahedron after being stretched along an axis perpendicular to one face by a factor of $r$. Because the octahedron is only being stretched in one dimension, the volume changes by the same factor $r$. So the problem reduces to computing the factor $r$ and the edge length of the resulting octahedron.\n\nFor convenience, imagine that one face of the octahedron rests on a plane. Seen from above the plane, the octahedron appears as shown below.\n\n\n\n<img_3368>\n\nLet $P$ be the projection of $A$ onto the plane on which the octahedron rests, and let $Q$ be the foot of the perpendicular from $P$ to $\\overline{B C}$. Then $P Q=R_{C}-R_{I}$, where $R_{C}$ is the circumradius and $R_{I}$ the inradius of the equilateral triangle. Thus $P Q=\\frac{2}{3}\\left(\\frac{\\sqrt{3}}{2}\\right)-\\frac{1}{3}\\left(\\frac{\\sqrt{3}}{2}\\right)=\\frac{\\sqrt{3}}{6}$. Then $B P^{2}=P Q^{2}+B Q^{2}=\\frac{3}{36}+\\frac{1}{4}=\\frac{1}{3}$, so $A P^{2}=A B^{2}-B P^{2}=\\frac{2}{3}$, and $A P=\\frac{\\sqrt{6}}{3}$.\n\nNow let a vertical stretch take place along an axis parallel to $\\overleftrightarrow{A P}$. If the scale factor is $r$, then $A P=\\frac{r \\sqrt{6}}{3}$, and because the stretch occurs on an axis perpendicular to $\\overline{B P}$, the length $B P$ is unchanged, as can be seen below.\n\n<img_3791>\n\nThus $A B^{2}=\\frac{6 r^{2}}{9}+\\frac{1}{3}=\\frac{6 r^{2}+3}{9}$. It remains to compute $r$. But $r$ is simply the ratio of the new volume to the old volume:\n\n$$\nr=\\frac{6}{\\frac{\\sqrt{2}}{3}}=\\frac{18}{\\sqrt{2}}=9 \\sqrt{2}\n$$\n\nThus $A B^{2}=\\frac{6(9 \\sqrt{2})^{2}+3}{9}=\\frac{975}{9}=\\frac{325}{3}$, and $A B=\\frac{5 \\sqrt{39}}{3}$."
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"$\\frac{5 \\sqrt{39}}{3}$"
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2,719
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Let $T=5$. The diagram at right consists of $T$ congruent circles, each of radius 1 , whose centers are collinear, and each pair of adjacent circles are externally tangent to each other. Compute the length of the tangent segment $\overline{A B}$.
<image_1>
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[
"For each point of tangency of consecutive circles, drop a perpendicular from that point to $\\overline{A B}$. For each of the $T-2$ circles between the first and last circles, the distance between consecutive perpendiculars is $2 \\cdot 1=2$. Furthermore, the distance from $A$ to the first perpendicular equals 1 (i.e., the common radius of the circles), which also equals the distance from the last perpendicular to $B$. Thus $A B=1+(T-2) \\cdot 2+1=2(T-1)$. With $T=5$, it follows that $A B=2 \\cdot 4=8$."
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2,743
|
Square $A B C D$ has side length 22. Points $G$ and $H$ lie on $\overline{A B}$ so that $A H=B G=5$. Points $E$ and $F$ lie outside square $A B C D$ so that $E F G H$ is a square. Compute the area of hexagon $A E F B C D$.
<image_1>
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[
"Note that $G H=A B-A H-B G=22-5-5=12$. Thus\n\n$$\n\\begin{aligned}\n{[A E F B C D] } & =[A B C D]+[E F G H]+[A E H]+[B F G] \\\\\n& =22^{2}+12^{2}+\\frac{1}{2} \\cdot 5 \\cdot 12+\\frac{1}{2} \\cdot 5 \\cdot 12 \\\\\n& =484+144+30+30 \\\\\n& =\\mathbf{6 8 8} .\n\\end{aligned}\n$$"
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"688"
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2,769
|
Suppose that Xena traces a path along the segments in the figure shown, starting and ending at point $A$. The path passes through each of the eleven vertices besides $A$ exactly once, and only visits $A$ at the beginning and end of the path. Compute the number of possible paths Xena could trace.
<image_1>
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[
"Count the number of complete paths that pass through all vertices exactly once (such a path is called a Hamiltonian path). The set of vertices can be split into two rings:\n\n$$\n\\mathcal{I}=\\left\\{A_{1}, A_{2}, \\ldots, A_{6}\\right\\} \\text { (i.e., the inner ring), } \\quad \\mathcal{O}=\\left\\{B_{1}, B_{2}, \\ldots, B_{6}\\right\\} \\text { (i.e., the outer ring). }\n$$\n\nwhere $A_{1}=A$. The two rings are connected by the edges $E=\\left\\{A_{1} B_{1}, A_{2} B_{2}, \\ldots, A_{6} B_{6}\\right\\}$. Each vertex in the figure has exactly three edges joining it with the neighboring vertices. Also note that any closed loop must use exactly two edges (out of three) for each vertex.\n\nFurther note that a loop must use at least one edge from $E$ to move from one ring to the other. Consider two cases: the loop uses all six edges from $E$, or it uses some but not all of them.\n\nIf all edges from $E$ are used, there are two possible undirected loops. It is not possible to use both edges $A_{1} A_{2}$ and $B_{1} B_{2}$, so either $A_{1} A_{2}$ or $B_{1} B_{2}$ will be used. This choice determines how the entire loop is constructed.\n\nIf not all edges from $E$ are used, then there must be some $i$ for which the loop uses the edge $A_{i} B_{i}$ and does not use $A_{i+1} B_{i+1}$ (where $A_{j} B_{j}$ represents $A_{j-6} B_{j-6}$ if $7 \\leq j \\leq 12$ ). Because $A_{i+1}$ is only connected to three other vertices, the loop must use $A_{i} A_{i+1}$ and $A_{i+1} A_{i+2}$, and similarly must use $B_{i} B_{i+1}$ and $B_{i+1} B_{i+2}$. This also precludes using $A_{i+2} B_{i+2}$, because doing so would close the loop before it visits all 12 vertices. Therefore the loop must also use $A_{i+2} A_{i+3}$ and $B_{i+2} B_{i+3}$, which now precludes using $A_{i+3} B_{i+3}$. This continues to force the structure of the loop until it closes by using $A_{i+5} B_{i+5}=A_{i-1} B_{i-1}$. Hence the loop must use exactly two edges from $E$, and they must be consecutive: $A_{i-1} B_{i-1}$ and $A_{i} B_{i}$. There are 6 ways to choose those two consecutive edges, so there are 6 possible undirected loops in this case.\n\nThe forced path is a loop, and the only way the given conditions are satisfied if $A_{i+k}=A_{i-1}$ and $B_{i+k}=B_{i-1}$. Hence the loop must use (precisely) two consecutive edges from $E: A_{i-1} B_{i-1}$ and $A_{i} B_{i}$. There are 6 ways to choose two consecutive edges, so there are 6 possible undirected loops in this case.\n\nEach undirected loop can be traced in two ways, and thus the number of ways for Xena to trace the path is $(6+2) \\cdot 2=16$."
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2,774
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Let $T$ be a rational number. Two coplanar squares $\mathcal{S}_{1}$ and $\mathcal{S}_{2}$ each have area $T$ and are arranged as shown to form a nonconvex octagon. The center of $\mathcal{S}_{1}$ is a vertex of $\mathcal{S}_{2}$, and the center of $\mathcal{S}_{2}$ is a vertex of $\mathcal{S}_{1}$. Compute $\frac{\text { area of the union of } \mathcal{S}_{1} \text { and } \mathcal{S}_{2}}{\text { area of the intersection of } \mathcal{S}_{1} \text { and } \mathcal{S}_{2}}$.
<image_1>
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[
"Let $2 x$ be the side length of the squares. Then the intersection of $\\mathcal{S}_{1}$ and $\\mathcal{S}_{2}$ is a square of side length $x$, so its area is $x^{2}$. The area of the union of $\\mathcal{S}_{1}$ and $\\mathcal{S}_{2}$ is $(2 x)^{2}+(2 x)^{2}-x^{2}=7 x^{2}$. Thus the desired ratio of areas is $\\frac{7 x^{2}}{x^{2}}=7$ (independent of $T$ )."
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"7"
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2,779
|
In acute triangle $I L K$, shown in the figure, point $G$ lies on $\overline{L K}$ so that $\overline{I G} \perp \overline{L K}$. Given that $I L=\sqrt{41}$ and $L G=I K=5$, compute $G K$.
<image_1>
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[
"Using the Pythagorean Theorem, $I G=\\sqrt{(I L)^{2}-(L G)^{2}}=\\sqrt{41-25}=4$, and $G K=\\sqrt{(I K)^{2}-(I G)^{2}}=$ $\\sqrt{25-16}=3$."
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2,806
|
This Question involves one Robber and one or more Cops. After robbing a bank, the Robber retreats to a network of hideouts, represented by dots in the diagram below. Every day, the Robber stays holed up in a single hideout, and every night, the Robber moves to an adjacent hideout. Two hideouts are adjacent if and only if they are connected by an edge in the diagram, also called a hideout map (or map). For the purposes of this Power Question, the map must be connected; that is, given any two hideouts, there must be a path from one to the other. To clarify, the Robber may not stay in the same hideout for two consecutive days, although he may return to a hideout he has previously visited. For example, in the map below, if the Robber holes up in hideout $C$ for day 1 , then he would have to move to $B$ for day 2 , and would then have to move to either $A, C$, or $D$ on day 3.
<image_1>
Every day, each Cop searches one hideout: the Cops know the location of all hideouts and which hideouts are adjacent to which. Cops are thorough searchers, so if the Robber is present in the hideout searched, he is found and arrested. If the Robber is not present in the hideout searched, his location is not revealed. That is, the Cops only know that the Robber was not caught at any of the hideouts searched; they get no specific information (other than what they can derive by logic) about what hideout he was in. Cops are not constrained by edges on the map: a Cop may search any hideout on any day, regardless of whether it is adjacent to the hideout searched the previous day. A Cop may search the same hideout on consecutive days, and multiple Cops may search different hideouts on the same day. In the map above, a Cop could search $A$ on day 1 and day 2, and then search $C$ on day 3 .
The focus of this Power Question is to determine, given a hideout map and a fixed number of Cops, whether the Cops can be sure of catching the Robber within some time limit.
Map Notation: The following notation may be useful when writing your solutions. For a map $M$, let $h(M)$ be the number of hideouts and $e(M)$ be the number of edges in $M$. The safety of a hideout $H$ is the number of hideouts adjacent to $H$, and is denoted by $s(H)$.
The Cop number of a map $M$, denoted $C(M)$, is the minimum number of Cops required to guarantee that the Robber is caught.
Find $C(M)$ for the map below.
<image_2>
|
[
"First we prove that for all maps $M, C(M)<h(M)$.\n\nProve. Let $n=h(M)$. The following strategy will always catch a Robber within two days using $n-1$ Cops, which proves that $C(M) \\leq n-1$. Choose any subset $\\mathcal{S}$ of $n-1$ hideouts and position $n-1$ Cops at the hideouts of $\\mathcal{S}$ for 2 days. If the Robber is not caught on the first day, he must have been at the hideout not in $\\mathcal{S}$, and therefore must move to a hideout in $\\mathcal{S}$ on the following day.\n\nso $C(M) \\leq 6$. To see that 6 is minimal, note that it is possible for the Robber to go from any hideout to any other hideout in a single night. Suppose that on a given day, the Robber successfully evades capture. Without loss of generality, assume that the Robber was at hideout $A_{1}$. Then the only valid conclusion the Cops can draw is that on the next day, the Robber will not be at $A_{1}$. Thus the only hideout the Cops can afford to leave open is $A_{1}$ itself. If they leave another hideout open, the Robber might have chosen to hide there, and the process will repeat. So we get $C(M)=6$."
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2,810
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This Question involves one Robber and one or more Cops. After robbing a bank, the Robber retreats to a network of hideouts, represented by dots in the diagram below. Every day, the Robber stays holed up in a single hideout, and every night, the Robber moves to an adjacent hideout. Two hideouts are adjacent if and only if they are connected by an edge in the diagram, also called a hideout map (or map). For the purposes of this Power Question, the map must be connected; that is, given any two hideouts, there must be a path from one to the other. To clarify, the Robber may not stay in the same hideout for two consecutive days, although he may return to a hideout he has previously visited. For example, in the map below, if the Robber holes up in hideout $C$ for day 1 , then he would have to move to $B$ for day 2 , and would then have to move to either $A, C$, or $D$ on day 3.
<image_1>
Every day, each Cop searches one hideout: the Cops know the location of all hideouts and which hideouts are adjacent to which. Cops are thorough searchers, so if the Robber is present in the hideout searched, he is found and arrested. If the Robber is not present in the hideout searched, his location is not revealed. That is, the Cops only know that the Robber was not caught at any of the hideouts searched; they get no specific information (other than what they can derive by logic) about what hideout he was in. Cops are not constrained by edges on the map: a Cop may search any hideout on any day, regardless of whether it is adjacent to the hideout searched the previous day. A Cop may search the same hideout on consecutive days, and multiple Cops may search different hideouts on the same day. In the map above, a Cop could search $A$ on day 1 and day 2, and then search $C$ on day 3 .
The focus of this Power Question is to determine, given a hideout map and a fixed number of Cops, whether the Cops can be sure of catching the Robber within some time limit.
Map Notation: The following notation may be useful when writing your solutions. For a map $M$, let $h(M)$ be the number of hideouts and $e(M)$ be the number of edges in $M$. The safety of a hideout $H$ is the number of hideouts adjacent to $H$, and is denoted by $s(H)$.
The Cop number of a map $M$, denoted $C(M)$, is the minimum number of Cops required to guarantee that the Robber is caught.
The police want to catch the Robber with a minimum number of Cops, but time is of the essence. For a map $M$ and a fixed number of Cops $c \geq C(M)$, define the capture time, denoted $D(M, c)$, to be the minimum number of days required to guarantee a capture using $c$ Cops. For example, in the graph below, if three Cops are deployed, they might catch the Robber in the first day, but if they don't, there is a strategy that will guarantee they will capture the Robber within two days. Therefore the capture time is $D\left(\mathcal{C}_{6}, 3\right)=2$.
<image_2>
Definition: The workday number of $M$, denoted $W(M)$, is the minimum number of Cop workdays needed to guarantee the Robber's capture. For example, a strategy that guarantees capture within three days using 17 Cops on the first day, 11 Cops on the second day, and only 6 Cops on the third day would require a total of $17+11+6=34$ Cop workdays.
Determine $W(M)$ for each of the maps in figure a and figure b.
<image_3>
figure a
<image_4>
figure b
|
[
"or the map $M$ from figure a, $W(M)=7$. The most efficient strategy is to use 7 Cops to blanket all the hideouts on the first day. Any strategy using fewer than 7 Cops would require 6 Cops on each of two consecutive days: given that any hideout can be reached from any other hideout, leaving more than one hideout unsearched on one day makes it is impossible to eliminate any hideouts the following day. So any other strategy would require a minimum of 12 Cop workdays.\n\nFor the map $M$ from figure b, $W(M)=8$. The strategy outlined in $2 \\mathrm{~b}$ used three Cops for a maximum of four workdays, yielding 12 Cop workdays. The most efficient strategy is to use 4 Cops, positioned at $\\{B, E, H, K\\}$ for 2 days each. The following argument demonstrates that 8 Cop workdays is in fact minimal. First, notice that there is no advantage to searching one of the hideouts between vertices of the square (for example, $C$ ) without searching the other hideout between the same vertices (for example, $D$ ). The Robber can reach $C$ on day $n$ if and only if he is at either $B$ or $E$ on day $n-1$, and in either case he could just as well go to $D$ instead of $C$. So there is no situation in which the Robber is certain to be caught at $C$ rather than at $D$. Additionally, the Robber's possible locations on day $n+1$ are the same whether he is at $C$ or $D$ on day $n$, so searching one rather than the other fails to rule out any locations for future days. So any successful strategy that involves searching $C$ should also involve searching $D$ on the same day, and similarly for $F$ and $G, I$ and $J$, and $L$ and $A$. On the other hand, if the Robber must be at one of $C$ and $D$ on day $n$, then he must be at either $B$ or $E$ on day $n+1$, because those are the only adjacent hideouts. So any strategy that involves searching both hideouts of one of the off-the-square pairs on day $n$ is equivalent to a strategy that searches the adjacent on-the-square hideouts on day $n+1$; the two strategies use the same number of Cops for the same number of workdays. Thus the optimal number of Cop workdays can be achieved using strategies that only search the \"corner\" hideouts $B, E, H, K$.\n\nRestricting the search to only those strategies searching corner hideouts $B, E, H, K$, a total of 8 workdays can be achieved by searching all four hideouts on two consecutive days: if the Robber is at one of the other eight hideouts the first day, he must move to one of the two adjacent corner hideouts the second day. But each of these corner hideouts is adjacent to two other corner hideouts. So if only one hideout is searched, for no matter how many consecutive days, the following day, the Robber could either be back at the previously-searched hideout or be at any other hideout: no possibilities are ruled out. If two adjacent corners are searched, the Cops do no better, as the following argument shows. Suppose that $B$ and $E$ are both searched for two consecutive days. Then the Cops can rule out $B, E, C$, and $D$ as possible locations, but if the Cops then switch to searching either $H$ or $K$ instead of $B$ or $E$, the Robber can go back to $C$ or $D$ within two days. So searching two adjacent corner hideouts for two days is fruitless and costs four Cop workdays. Searching diagonally opposite corner hideouts is even less fruitful, because doing so rules out none of the other hideouts as possible Robber locations. Using three Cops each day, it is easy to imagine scenarios in which the Robber evades capture for three days before being caught: for example, if $B, E, H$ are searched for two consecutive days, the Robber goes from $I$ or $J$ to $K$ to $L$. Therefore if three Cops are used, four days are required for a total of 12 Cop workdays.\n\nIf there are more than four Cops, the preceding arguments show that the number of Cops must be even to produce optimal results (because there is no advantage to searching one hideout between vertices of the square without searching the other). Using six Cops with four at corner hideouts yields no improvement, because the following day, the Robber could get to any of the four corner hideouts, requiring at least four Cops the second day, for ten Cop workdays. If two or fewer Cops are at corner hideouts, the situation is even worse, because if the Robber is not caught that day, he has at least nine possible hideouts the following day (depending on whether the unsearched corners are adjacent or diagonally opposite to each other). Using eight Cops (with four at corner vertices) could eliminate one corner vertex as a possible location for the second day (if the non-corner hideouts searched are on adjacent sides of the square), but eight Cop workdays have already been used on the first day. So 8 Cop workdays is minimal."
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"12, 8"
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2,820
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In $\triangle A B C, \mathrm{~m} \angle A=\mathrm{m} \angle B=45^{\circ}$ and $A B=16$. Mutually tangent circular arcs are drawn centered at all three vertices; the arcs centered at $A$ and $B$ intersect at the midpoint of $\overline{A B}$. Compute the area of the region inside the triangle and outside of the three arcs.
<image_1>
|
[
"Because $A B=16, A C=B C=\\frac{16}{\\sqrt{2}}=8 \\sqrt{2}$. Then each of the large arcs has radius 8 , and the small arc has radius $8 \\sqrt{2}-8$. Each large arc has measure $45^{\\circ}$ and the small arc has measure $90^{\\circ}$. Therefore the area enclosed by each large arc is $\\frac{45}{360} \\cdot \\pi \\cdot 8^{2}=8 \\pi$, and the area enclosed by the small arc is $\\frac{90}{360} \\cdot \\pi \\cdot(8 \\sqrt{2}-8)^{2}=48 \\pi-32 \\pi \\sqrt{2}$. Thus the sum of the areas enclosed by the three arcs is $64 \\pi-32 \\pi \\sqrt{2}$. On the other hand, the area of the triangle is $\\frac{1}{2}(8 \\sqrt{2})^{2}=64$. So the area of the desired region is $64-64 \\pi+32 \\pi \\sqrt{2}$."
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"$\\quad 64-64 \\pi+32 \\pi \\sqrt{2}$"
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2,826
|
Given noncollinear points $A, B, C$, segment $\overline{A B}$ is trisected by points $D$ and $E$, and $F$ is the midpoint of segment $\overline{A C} . \overline{D F}$ and $\overline{B F}$ intersect $\overline{C E}$ at $G$ and $H$, respectively. If $[D E G]=18$, compute $[F G H]$.
<image_1>
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[
"Compute the desired area as $[E G F B]-[E H B]$. To compute the area of concave quadrilateral $E G F B$, draw segment $\\overline{B G}$, which divides the quadrilateral into three triangles, $\\triangle D E G, \\triangle B D G$, and $\\triangle B G F$. Then $[B D G]=[D E G]=18$ because the triangles have equal bases and heights. Because $D, G$, and $F$ are collinear, to compute $[B G F]$ it suffices to find the ratio $D G / G F$. Use Menelaus's Theorem on $\\triangle A D F$ with Menelaus Line $\\overline{E C}$ to obtain\n\n$$\n\\frac{A E}{E D} \\cdot \\frac{D G}{G F} \\cdot \\frac{F C}{C A}=1\n$$\n\n\n\nBecause $E$ and $F$ are the midpoints of $\\overline{A D}$ and $\\overline{C A}$ respectively, $A E / E D=1$ and $F C / C A=$ $1 / 2$. Therefore $D G / G F=2 / 1$, and $[B G F]=\\frac{1}{2}[B D G]=9$. Thus $[E G F B]=18+18+9=45$.\n\nTo compute $[E H B]$, consider that its base $\\overline{E B}$ is twice the base of $\\triangle D E G$. The ratio of their heights equals the ratio $E H / E G$ because the altitudes from $H$ and $G$ to $\\overleftrightarrow{B E}$ are parallel to each other. Use Menelaus's Theorem twice more on $\\triangle A E C$ to find these values:\n\n$$\n\\begin{gathered}\n\\frac{A D}{D E} \\cdot \\frac{E G}{G C} \\cdot \\frac{C F}{F A}=1 \\Rightarrow \\frac{E G}{G C}=\\frac{1}{2} \\Rightarrow E G=\\frac{1}{3} E C, \\text { and } \\\\\n\\frac{A B}{B E} \\cdot \\frac{E H}{H C} \\cdot \\frac{C F}{F A}=1 \\Rightarrow \\frac{E H}{H C}=\\frac{2}{3} \\Rightarrow E H=\\frac{2}{5} E C .\n\\end{gathered}\n$$\n\nTherefore $\\frac{E H}{E G}=\\frac{2 / 5}{1 / 3}=\\frac{6}{5}$. Thus $[E H B]=\\frac{6}{5} \\cdot 2 \\cdot[D E G]=\\frac{216}{5}$. Thus $[F G H]=45-\\frac{216}{5}=\\frac{9}{5}$.",
"The method of mass points leads to the same results as Menelaus's Theorem, but corresponds to the physical intuition that masses on opposite sides of a fulcrum balance if and only if the products of the masses and their distances from the fulcrum are equal (in physics-speak, the net torque is zero). If a mass of weight 1 is placed at vertex $B$ and masses of weight 2 are placed at vertices $A$ and $C$, then $\\triangle A B C$ balances on the line $\\overleftrightarrow{B F}$ and also on the line $\\overleftrightarrow{C E}$. Thus it balances on the point $H$ where these two lines intersect. Replacing the masses at $A$ and $C$ with a single mass of weight 4 at their center of mass $F$, the triangle still balances at $H$. Thus $B H / H F=4$.\n\nNext, consider $\\triangle B E F$. Placing masses of weight 1 at the vertices $B$ and $E$ and a mass of weight 4 at $F$, the triangle balances at $G$. A similar argument shows that $D G / G F=2$ and that $E G / G H=5$. Because $\\triangle D E G$ and $\\triangle F H G$ have congruent (vertical) angles at $G$, it follows that $[D E G] /[F H G]=(D G / F G) \\cdot(E G / H G)=2 \\cdot 5=10$. Thus $[F G H]=[D E G] / 10=$ $\\frac{18}{10}=\\frac{9}{5}$."
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"$\\frac{9}{5}$"
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2,833
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Let $T=6$. In the square $D E F G$ diagrammed at right, points $M$ and $N$ trisect $\overline{F G}$, points $A$ and $B$ are the midpoints of $\overline{E F}$ and $\overline{D G}$, respectively, and $\overline{E M} \cap \overline{A B}=S$ and $\overline{D N} \cap \overline{A B}=H$. If the side length of square $D E F G$ is $T$, compute $[D E S H]$.
<image_1>
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[
"Note that $D E S H$ is a trapezoid with height $\\frac{T}{2}$. Because $\\overline{A S}$ and $\\overline{B H}$ are midlines of triangles $E F M$ and $D G N$ respectively, it follows that $A S=B H=\\frac{T}{6}$. Thus $S H=T-2 \\cdot \\frac{T}{6}=\\frac{2 T}{3}$. Thus $[D E S H]=\\frac{1}{2}\\left(T+\\frac{2 T}{3}\\right) \\cdot \\frac{T}{2}=\\frac{5 T^{2}}{12}$. With $T=6$, the desired area is 15 ."
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2,837
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Let $R$ be the larger number you will receive, and let $r$ be the smaller number you will receive. In the diagram at right (not drawn to scale), circle $D$ has radius $R$, circle $K$ has radius $r$, and circles $D$ and $K$ are tangent at $C$. Line $\overleftrightarrow{Y P}$ is tangent to circles $D$ and $K$. Compute $Y P$.
<image_1>
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"Note that $\\overline{D Y}$ and $\\overline{K P}$ are both perpendicular to line $\\overleftrightarrow{Y P}$. Let $J$ be the foot of the perpendicular from $K$ to $\\overline{D Y}$. Then $P K J Y$ is a rectangle and $Y P=J K=\\sqrt{D K^{2}-D J^{2}}=$ $\\sqrt{(R+r)^{2}-(R-r)^{2}}=2 \\sqrt{R r}$. With $R=450$ and $r=\\frac{1}{3}$, the answer is $2 \\sqrt{150}=\\mathbf{1 0} \\sqrt{\\mathbf{6}}$."
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"$10 \\sqrt{6}$"
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2,868
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An $\boldsymbol{n}$-label is a permutation of the numbers 1 through $n$. For example, $J=35214$ is a 5 -label and $K=132$ is a 3 -label. For a fixed positive integer $p$, where $p \leq n$, consider consecutive blocks of $p$ numbers in an $n$-label. For example, when $p=3$ and $L=263415$, the blocks are 263,634,341, and 415. We can associate to each of these blocks a $p$-label that corresponds to the relative order of the numbers in that block. For $L=263415$, we get the following:
$$
\underline{263} 415 \rightarrow 132 ; \quad 2 \underline{63415} \rightarrow 312 ; \quad 26 \underline{341} 5 \rightarrow 231 ; \quad 263 \underline{415} \rightarrow 213
$$
Moving from left to right in the $n$-label, there are $n-p+1$ such blocks, which means we obtain an $(n-p+1)$-tuple of $p$-labels. For $L=263415$, we get the 4 -tuple $(132,312,231,213)$. We will call this $(n-p+1)$-tuple the $\boldsymbol{p}$-signature of $L$ (or signature, if $p$ is clear from the context) and denote it by $S_{p}[L]$; the $p$-labels in the signature are called windows. For $L=263415$, the windows are $132,312,231$, and 213 , and we write
$$
S_{3}[263415]=(132,312,231,213)
$$
More generally, we will call any $(n-p+1)$-tuple of $p$-labels a $p$-signature, even if we do not know of an $n$-label to which it corresponds (and even if no such label exists). A signature that occurs for exactly one $n$-label is called unique, and a signature that doesn't occur for any $n$-labels is called impossible. A possible signature is one that occurs for at least one $n$-label.
In this power question, you will be asked to analyze some of the properties of labels and signatures.
We can associate a shape to a given 2-signature: a diagram of up and down steps that indicates the relative order of adjacent numbers. For example, the following shape corresponds to the 2-signature $(12,12,12,21,12,21)$ :
<image_1>
A 7-label with this 2-signature corresponds to placing the numbers 1 through 7 at the nodes above so that numbers increase with each up step and decrease with each down step. The 7-label 2347165 is shown below:
<image_2>
Consider the shape below:
<image_3>
Find the 2-signature that corresponds to this shape.
|
[
"The first pair indicates an increase; the next three are decreases, and the last pair is an increase. So the 2-signature is $(12,21,21,21,12)$."
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"$(12,21,21,21,12)$"
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2,869
|
An $\boldsymbol{n}$-label is a permutation of the numbers 1 through $n$. For example, $J=35214$ is a 5 -label and $K=132$ is a 3 -label. For a fixed positive integer $p$, where $p \leq n$, consider consecutive blocks of $p$ numbers in an $n$-label. For example, when $p=3$ and $L=263415$, the blocks are 263,634,341, and 415. We can associate to each of these blocks a $p$-label that corresponds to the relative order of the numbers in that block. For $L=263415$, we get the following:
$$
\underline{263} 415 \rightarrow 132 ; \quad 2 \underline{63415} \rightarrow 312 ; \quad 26 \underline{341} 5 \rightarrow 231 ; \quad 263 \underline{415} \rightarrow 213
$$
Moving from left to right in the $n$-label, there are $n-p+1$ such blocks, which means we obtain an $(n-p+1)$-tuple of $p$-labels. For $L=263415$, we get the 4 -tuple $(132,312,231,213)$. We will call this $(n-p+1)$-tuple the $\boldsymbol{p}$-signature of $L$ (or signature, if $p$ is clear from the context) and denote it by $S_{p}[L]$; the $p$-labels in the signature are called windows. For $L=263415$, the windows are $132,312,231$, and 213 , and we write
$$
S_{3}[263415]=(132,312,231,213)
$$
More generally, we will call any $(n-p+1)$-tuple of $p$-labels a $p$-signature, even if we do not know of an $n$-label to which it corresponds (and even if no such label exists). A signature that occurs for exactly one $n$-label is called unique, and a signature that doesn't occur for any $n$-labels is called impossible. A possible signature is one that occurs for at least one $n$-label.
In this power question, you will be asked to analyze some of the properties of labels and signatures.
We can associate a shape to a given 2-signature: a diagram of up and down steps that indicates the relative order of adjacent numbers. For example, the following shape corresponds to the 2-signature $(12,12,12,21,12,21)$ :
<image_1>
A 7-label with this 2-signature corresponds to placing the numbers 1 through 7 at the nodes above so that numbers increase with each up step and decrease with each down step. The 7-label 2347165 is shown below:
<image_2>
List all 5-labels with 2-signature $(12,12,21,21)$.
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"12543,13542,14532,23541,24531,34521"
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"12543,13542,14532,23541,24531,34521"
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2,870
|
An $\boldsymbol{n}$-label is a permutation of the numbers 1 through $n$. For example, $J=35214$ is a 5 -label and $K=132$ is a 3 -label. For a fixed positive integer $p$, where $p \leq n$, consider consecutive blocks of $p$ numbers in an $n$-label. For example, when $p=3$ and $L=263415$, the blocks are 263,634,341, and 415. We can associate to each of these blocks a $p$-label that corresponds to the relative order of the numbers in that block. For $L=263415$, we get the following:
$$
\underline{263} 415 \rightarrow 132 ; \quad 2 \underline{63415} \rightarrow 312 ; \quad 26 \underline{341} 5 \rightarrow 231 ; \quad 263 \underline{415} \rightarrow 213
$$
Moving from left to right in the $n$-label, there are $n-p+1$ such blocks, which means we obtain an $(n-p+1)$-tuple of $p$-labels. For $L=263415$, we get the 4 -tuple $(132,312,231,213)$. We will call this $(n-p+1)$-tuple the $\boldsymbol{p}$-signature of $L$ (or signature, if $p$ is clear from the context) and denote it by $S_{p}[L]$; the $p$-labels in the signature are called windows. For $L=263415$, the windows are $132,312,231$, and 213 , and we write
$$
S_{3}[263415]=(132,312,231,213)
$$
More generally, we will call any $(n-p+1)$-tuple of $p$-labels a $p$-signature, even if we do not know of an $n$-label to which it corresponds (and even if no such label exists). A signature that occurs for exactly one $n$-label is called unique, and a signature that doesn't occur for any $n$-labels is called impossible. A possible signature is one that occurs for at least one $n$-label.
In this power question, you will be asked to analyze some of the properties of labels and signatures.
We can associate a shape to a given 2-signature: a diagram of up and down steps that indicates the relative order of adjacent numbers. For example, the following shape corresponds to the 2-signature $(12,12,12,21,12,21)$ :
<image_1>
A 7-label with this 2-signature corresponds to placing the numbers 1 through 7 at the nodes above so that numbers increase with each up step and decrease with each down step. The 7-label 2347165 is shown below:
<image_2>
Find a formula for the number of $(2 n+1)$-labels with the 2 -signature
$$
(\underbrace{12,12, \ldots, 12}_{n}, \underbrace{21,21, \ldots, 21}_{n})
$$
|
[
"The answer is $\\left(\\begin{array}{c}2 n \\\\ n\\end{array}\\right)$. The shape of this signature is a wedge: $n$ up steps followed by $n$ down steps. The wedge for $n=3$ is illustrated below:\n\n<img_3277>\n\nThe largest number in the label, $2 n+1$, must be placed at the peak in the center. If we choose the numbers to put in the first $n$ spaces, then they must be placed in increasing order. Likewise, the remaining $n$ numbers must be placed in decreasing order on the downward sloping piece of the shape. Thus there are exactly $\\left(\\begin{array}{c}2 n \\\\ n\\end{array}\\right)$ such labels.\n\n"
] |
[
"$\\binom{2n}{n}$"
] | null | Not supported with pagination yet
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2,871
|
An $\boldsymbol{n}$-label is a permutation of the numbers 1 through $n$. For example, $J=35214$ is a 5 -label and $K=132$ is a 3 -label. For a fixed positive integer $p$, where $p \leq n$, consider consecutive blocks of $p$ numbers in an $n$-label. For example, when $p=3$ and $L=263415$, the blocks are 263,634,341, and 415. We can associate to each of these blocks a $p$-label that corresponds to the relative order of the numbers in that block. For $L=263415$, we get the following:
$$
\underline{263} 415 \rightarrow 132 ; \quad 2 \underline{63415} \rightarrow 312 ; \quad 26 \underline{341} 5 \rightarrow 231 ; \quad 263 \underline{415} \rightarrow 213
$$
Moving from left to right in the $n$-label, there are $n-p+1$ such blocks, which means we obtain an $(n-p+1)$-tuple of $p$-labels. For $L=263415$, we get the 4 -tuple $(132,312,231,213)$. We will call this $(n-p+1)$-tuple the $\boldsymbol{p}$-signature of $L$ (or signature, if $p$ is clear from the context) and denote it by $S_{p}[L]$; the $p$-labels in the signature are called windows. For $L=263415$, the windows are $132,312,231$, and 213 , and we write
$$
S_{3}[263415]=(132,312,231,213)
$$
More generally, we will call any $(n-p+1)$-tuple of $p$-labels a $p$-signature, even if we do not know of an $n$-label to which it corresponds (and even if no such label exists). A signature that occurs for exactly one $n$-label is called unique, and a signature that doesn't occur for any $n$-labels is called impossible. A possible signature is one that occurs for at least one $n$-label.
In this power question, you will be asked to analyze some of the properties of labels and signatures.
We can associate a shape to a given 2-signature: a diagram of up and down steps that indicates the relative order of adjacent numbers. For example, the following shape corresponds to the 2-signature $(12,12,12,21,12,21)$ :
<image_1>
A 7-label with this 2-signature corresponds to placing the numbers 1 through 7 at the nodes above so that numbers increase with each up step and decrease with each down step. The 7-label 2347165 is shown below:
<image_2>
Compute the number of 5-labels with 2 -signature $(12,21,12,21)$.
|
[
"The answer is 16 . We have a shape with two peaks and a valley in the middle. The 5 must go on one of the two peaks, so we place it on the first peak. By the shape's symmetry, we will double our answer at the end to account for the 5 -labels where the 5 is on the other peak.\n\n<img_3879>\n\nThe 4 can go to the left of the 5 or at the other peak. In the first case, shown below left, the 3 must go at the other peak and the 1 and 2 can go in either order. In the latter case, shown below right, the 1,2 , and 3 can go in any of 3 ! arrangements.\n<img_3481>\n\n\n\nSo there are $2 !+3 !=8$ possibilities. In all, there are 165 -labels (including the ones where the 5 is at the other peak)."
] |
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2,872
|
An $\boldsymbol{n}$-label is a permutation of the numbers 1 through $n$. For example, $J=35214$ is a 5 -label and $K=132$ is a 3 -label. For a fixed positive integer $p$, where $p \leq n$, consider consecutive blocks of $p$ numbers in an $n$-label. For example, when $p=3$ and $L=263415$, the blocks are 263,634,341, and 415. We can associate to each of these blocks a $p$-label that corresponds to the relative order of the numbers in that block. For $L=263415$, we get the following:
$$
\underline{263} 415 \rightarrow 132 ; \quad 2 \underline{63415} \rightarrow 312 ; \quad 26 \underline{341} 5 \rightarrow 231 ; \quad 263 \underline{415} \rightarrow 213
$$
Moving from left to right in the $n$-label, there are $n-p+1$ such blocks, which means we obtain an $(n-p+1)$-tuple of $p$-labels. For $L=263415$, we get the 4 -tuple $(132,312,231,213)$. We will call this $(n-p+1)$-tuple the $\boldsymbol{p}$-signature of $L$ (or signature, if $p$ is clear from the context) and denote it by $S_{p}[L]$; the $p$-labels in the signature are called windows. For $L=263415$, the windows are $132,312,231$, and 213 , and we write
$$
S_{3}[263415]=(132,312,231,213)
$$
More generally, we will call any $(n-p+1)$-tuple of $p$-labels a $p$-signature, even if we do not know of an $n$-label to which it corresponds (and even if no such label exists). A signature that occurs for exactly one $n$-label is called unique, and a signature that doesn't occur for any $n$-labels is called impossible. A possible signature is one that occurs for at least one $n$-label.
In this power question, you will be asked to analyze some of the properties of labels and signatures.
We can associate a shape to a given 2-signature: a diagram of up and down steps that indicates the relative order of adjacent numbers. For example, the following shape corresponds to the 2-signature $(12,12,12,21,12,21)$ :
<image_1>
A 7-label with this 2-signature corresponds to placing the numbers 1 through 7 at the nodes above so that numbers increase with each up step and decrease with each down step. The 7-label 2347165 is shown below:
<image_2>
Determine the number of 9-labels with 2-signature
$$
(12,21,12,21,12,21,12,21) \text {. }
$$
Justify your answer.
|
[
"The answer is 7936. The shape of this 2-signature has four peaks and three intermediate valleys:\n\n<img_3473>\n\nWe will solve this problem by building up from smaller examples. Let $f_{n}$ equal the number of $(2 n+1)$-labels whose 2 -signature consists of $n$ peaks and $n-1$ intermediate valleys. In part (b) we showed that $f_{2}=16$. In the case where we have one peak, $f_{1}=2$. For the trivial case (no peaks), we get $f_{0}=1$. These cases are shown below.\n\n1\n\n<img_3650>\n\n<img_3227>\n\nSuppose we know the peak on which the largest number, $2 n+1$, is placed. Then that splits our picture into two shapes with fewer peaks. Once we choose which numbers from $1,2, \\ldots, 2 n$ to place each shape, we can compute the number of arrangements of the numbers on each shape, and then take the product. For example, if we place the 9 at the second peak, as shown below, we get a 1-peak shape on the left and a 2-peak shape on the right.\n\n<img_4030>\n\nFor the above shape, there are $\\left(\\begin{array}{l}8 \\\\ 3\\end{array}\\right)$ ways to pick the three numbers to place on the left-hand side, $f_{1}=2$ ways to place them, and $f_{2}=16$ ways to place the remaining five numbers on the right.\n\nThis argument works for any $n>1$, so we have shown the following:\n\n$$\nf_{n}=\\sum_{k=1}^{n}\\left(\\begin{array}{c}\n2 n \\\\\n2 k-1\n\\end{array}\\right) f_{k-1} f_{n-k}\n$$\n\n\n\nSo we have:\n\n$$\n\\begin{aligned}\n& f_{1}=\\left(\\begin{array}{l}\n2 \\\\\n1\n\\end{array}\\right) f_{0}^{2}=2 \\\\\n& f_{2}=\\left(\\begin{array}{l}\n4 \\\\\n1\n\\end{array}\\right) f_{0} f_{1}+\\left(\\begin{array}{l}\n4 \\\\\n3\n\\end{array}\\right) f_{1} f_{0}=16 \\\\\n& f_{3}=\\left(\\begin{array}{l}\n6 \\\\\n1\n\\end{array}\\right) f_{0} f_{2}+\\left(\\begin{array}{l}\n6 \\\\\n3\n\\end{array}\\right) f_{1}^{2}+\\left(\\begin{array}{l}\n6 \\\\\n5\n\\end{array}\\right) f_{2} f_{0}=272 \\\\\n& f_{4}=\\left(\\begin{array}{l}\n8 \\\\\n1\n\\end{array}\\right) f_{0} f_{3}+\\left(\\begin{array}{l}\n8 \\\\\n3\n\\end{array}\\right) f_{1} f_{2}+\\left(\\begin{array}{l}\n8 \\\\\n5\n\\end{array}\\right) f_{2} f_{1}+\\left(\\begin{array}{l}\n8 \\\\\n7\n\\end{array}\\right) f_{3} f_{0}=7936 .\n\\end{aligned}\n$$"
] |
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"7936"
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2,875
|
An $\boldsymbol{n}$-label is a permutation of the numbers 1 through $n$. For example, $J=35214$ is a 5 -label and $K=132$ is a 3 -label. For a fixed positive integer $p$, where $p \leq n$, consider consecutive blocks of $p$ numbers in an $n$-label. For example, when $p=3$ and $L=263415$, the blocks are 263,634,341, and 415. We can associate to each of these blocks a $p$-label that corresponds to the relative order of the numbers in that block. For $L=263415$, we get the following:
$$
\underline{263} 415 \rightarrow 132 ; \quad 2 \underline{63415} \rightarrow 312 ; \quad 26 \underline{341} 5 \rightarrow 231 ; \quad 263 \underline{415} \rightarrow 213
$$
Moving from left to right in the $n$-label, there are $n-p+1$ such blocks, which means we obtain an $(n-p+1)$-tuple of $p$-labels. For $L=263415$, we get the 4 -tuple $(132,312,231,213)$. We will call this $(n-p+1)$-tuple the $\boldsymbol{p}$-signature of $L$ (or signature, if $p$ is clear from the context) and denote it by $S_{p}[L]$; the $p$-labels in the signature are called windows. For $L=263415$, the windows are $132,312,231$, and 213 , and we write
$$
S_{3}[263415]=(132,312,231,213)
$$
More generally, we will call any $(n-p+1)$-tuple of $p$-labels a $p$-signature, even if we do not know of an $n$-label to which it corresponds (and even if no such label exists). A signature that occurs for exactly one $n$-label is called unique, and a signature that doesn't occur for any $n$-labels is called impossible. A possible signature is one that occurs for at least one $n$-label.
In this power question, you will be asked to analyze some of the properties of labels and signatures.
We can associate a shape to a given 2-signature: a diagram of up and down steps that indicates the relative order of adjacent numbers. For example, the following shape corresponds to the 2-signature $(12,12,12,21,12,21)$ :
<image_1>
A 7-label with this 2-signature corresponds to placing the numbers 1 through 7 at the nodes above so that numbers increase with each up step and decrease with each down step. The 7-label 2347165 is shown below:
<image_2>
For a general $n$, determine the number of distinct possible $p$-signatures.
|
[
"The answer is $p ! \\cdot p^{n-p}$.\n\nCall two consecutive windows in a $p$-signature compatible if the last $p-1$ numbers in the first label and the first $p-1$ numbers in the second label (their \"overlap\") describe the same ordering. For example, in the $p$-signature $(. ., 2143,2431, \\ldots), 2143$ and 2431 are compatible. Notice that the last three digits of 2143 and the first three digits of 2431 can be described by the same 3-label, 132 .\n\nTheorem: A signature $\\sigma$ is possible if and only if every pair of consecutive windows is compatible.\n\nProof: $(\\Rightarrow)$ Consider a signature $\\sigma$ describing a $p$-label $L$. If some pair in $\\sigma$ is not compatible, then there is some string of $p-1$ numbers in our label $L$ that has two different $(p-1)$-signatures. This is impossible, since the $p$-signature is well-defined.\n\n$(\\Leftarrow)$ Now suppose $\\sigma$ is a $p$-signature such that that every pair of consecutive windows is compatible. We need to show that there is at least one label $L$ with $S_{p}[L]=\\sigma$. We do so by induction on the number of windows in $\\sigma$, using the results from $5(\\mathrm{~b})$.\n\nLet $\\sigma=\\left\\{\\omega_{1}, \\omega_{2}, \\ldots, \\omega_{k+1}\\right\\}$, and suppose $\\omega_{1}=a_{1}, a_{2}, \\ldots, a_{p}$. Set $L_{1}=\\omega_{1}$.\n\nSuppose that $L_{k}$ is a $(p+k-1)$-label such that $S_{p}\\left[L_{k}\\right]=\\left\\{\\omega_{1}, \\ldots, \\omega_{k}\\right\\}$. We will construct $L_{k+1}$ for which $S_{p}\\left[L_{k+1}\\right]=\\left\\{\\omega_{1}, \\ldots, \\omega_{k+1}\\right\\}$.\n\nAs in $5(\\mathrm{~b})$, denote by $L_{k}^{(j)}$ the label $L_{k}$ with a $j+0.5$ appended; we will eventually renumber the elements in the label to make them all integers. Appending $j+0.5$ does not affect any of the non-terminal windows of $S_{p}\\left[L_{k}\\right]$, and as $j$ varies from 0 to $p-k+1$ the final window of $S_{p}\\left[L_{k}^{(j)}\\right]$ varies over each of the $p$ windows compatible with $\\omega_{k}$. Since $\\omega_{k+1}$ is compatible with $\\omega_{k}$, there exists some $j$ for which $S_{p}\\left[L_{k}^{(j)}\\right]=\\left\\{\\omega_{1}, \\ldots, \\omega_{k+1}\\right\\}$. Now we renumber as follows: set $L_{k+1}=S_{k+p}\\left[L_{k}^{(j)}\\right]$, which replaces $L_{k}^{(j)}$ with the integers 1 through $k+p$ and preserves the relative order of all integers in the label.\n\nBy continuing this process, we conclude that the $n$-label $L_{n-p+1}$ has $p$-signature $\\sigma$, so $\\sigma$ is possible.\n\nTo count the number of possible $p$-signatures, we choose the first window ( $p$ ! choices), then choose each of the remaining $n-p$ compatible windows ( $p$ choices each). In all, there are $p ! \\cdot p^{n-p}$ possible $p$-signatures."
] |
[
"$p ! \\cdot p^{n-p}$"
] | null | Not supported with pagination yet
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2,876
|
An $\boldsymbol{n}$-label is a permutation of the numbers 1 through $n$. For example, $J=35214$ is a 5 -label and $K=132$ is a 3 -label. For a fixed positive integer $p$, where $p \leq n$, consider consecutive blocks of $p$ numbers in an $n$-label. For example, when $p=3$ and $L=263415$, the blocks are 263,634,341, and 415. We can associate to each of these blocks a $p$-label that corresponds to the relative order of the numbers in that block. For $L=263415$, we get the following:
$$
\underline{263} 415 \rightarrow 132 ; \quad 2 \underline{63415} \rightarrow 312 ; \quad 26 \underline{341} 5 \rightarrow 231 ; \quad 263 \underline{415} \rightarrow 213
$$
Moving from left to right in the $n$-label, there are $n-p+1$ such blocks, which means we obtain an $(n-p+1)$-tuple of $p$-labels. For $L=263415$, we get the 4 -tuple $(132,312,231,213)$. We will call this $(n-p+1)$-tuple the $\boldsymbol{p}$-signature of $L$ (or signature, if $p$ is clear from the context) and denote it by $S_{p}[L]$; the $p$-labels in the signature are called windows. For $L=263415$, the windows are $132,312,231$, and 213 , and we write
$$
S_{3}[263415]=(132,312,231,213)
$$
More generally, we will call any $(n-p+1)$-tuple of $p$-labels a $p$-signature, even if we do not know of an $n$-label to which it corresponds (and even if no such label exists). A signature that occurs for exactly one $n$-label is called unique, and a signature that doesn't occur for any $n$-labels is called impossible. A possible signature is one that occurs for at least one $n$-label.
In this power question, you will be asked to analyze some of the properties of labels and signatures.
We can associate a shape to a given 2-signature: a diagram of up and down steps that indicates the relative order of adjacent numbers. For example, the following shape corresponds to the 2-signature $(12,12,12,21,12,21)$ :
<image_1>
A 7-label with this 2-signature corresponds to placing the numbers 1 through 7 at the nodes above so that numbers increase with each up step and decrease with each down step. The 7-label 2347165 is shown below:
<image_2>
If a randomly chosen $p$-signature is 575 times more likely of being impossible than possible, determine $p$ and $n$.
|
[
"The answer is $n=7, p=5$.\n\nLet $P$ denote the probability that a randomly chosen $p$-signature is possible. We are\n\n\n\ngiven that $1-P=575$, so $P=\\frac{1}{576}$. We want to find $p$ and $n$ for which\n\n$$\n\\begin{aligned}\n\\frac{p ! \\cdot p^{n-p}}{(p !)^{n-p+1}} & =\\frac{1}{576} \\\\\n\\frac{p^{n-p}}{(p !)^{n-p}} & =\\frac{1}{576} \\\\\n((p-1) !)^{n-p} & =576\n\\end{aligned}\n$$\n\nThe only factorial that has 576 as an integer power is $4 !=\\sqrt{576}$. Thus $p=5$ and $n-p=2 \\Rightarrow n=7$."
] |
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2,879
|
An $\boldsymbol{n}$-label is a permutation of the numbers 1 through $n$. For example, $J=35214$ is a 5 -label and $K=132$ is a 3 -label. For a fixed positive integer $p$, where $p \leq n$, consider consecutive blocks of $p$ numbers in an $n$-label. For example, when $p=3$ and $L=263415$, the blocks are 263,634,341, and 415. We can associate to each of these blocks a $p$-label that corresponds to the relative order of the numbers in that block. For $L=263415$, we get the following:
$$
\underline{263} 415 \rightarrow 132 ; \quad 2 \underline{63415} \rightarrow 312 ; \quad 26 \underline{341} 5 \rightarrow 231 ; \quad 263 \underline{415} \rightarrow 213
$$
Moving from left to right in the $n$-label, there are $n-p+1$ such blocks, which means we obtain an $(n-p+1)$-tuple of $p$-labels. For $L=263415$, we get the 4 -tuple $(132,312,231,213)$. We will call this $(n-p+1)$-tuple the $\boldsymbol{p}$-signature of $L$ (or signature, if $p$ is clear from the context) and denote it by $S_{p}[L]$; the $p$-labels in the signature are called windows. For $L=263415$, the windows are $132,312,231$, and 213 , and we write
$$
S_{3}[263415]=(132,312,231,213)
$$
More generally, we will call any $(n-p+1)$-tuple of $p$-labels a $p$-signature, even if we do not know of an $n$-label to which it corresponds (and even if no such label exists). A signature that occurs for exactly one $n$-label is called unique, and a signature that doesn't occur for any $n$-labels is called impossible. A possible signature is one that occurs for at least one $n$-label.
In this power question, you will be asked to analyze some of the properties of labels and signatures.
We can associate a shape to a given 2-signature: a diagram of up and down steps that indicates the relative order of adjacent numbers. For example, the following shape corresponds to the 2-signature $(12,12,12,21,12,21)$ :
<image_1>
A 7-label with this 2-signature corresponds to placing the numbers 1 through 7 at the nodes above so that numbers increase with each up step and decrease with each down step. The 7-label 2347165 is shown below:
<image_2>
Find two 5-labels with unique 2-signatures.
|
[
"12345 and 54321 are the only ones."
] |
[
"12345, 54321"
] | null | Not supported with pagination yet
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English
|
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2,882
|
An $\boldsymbol{n}$-label is a permutation of the numbers 1 through $n$. For example, $J=35214$ is a 5 -label and $K=132$ is a 3 -label. For a fixed positive integer $p$, where $p \leq n$, consider consecutive blocks of $p$ numbers in an $n$-label. For example, when $p=3$ and $L=263415$, the blocks are 263,634,341, and 415. We can associate to each of these blocks a $p$-label that corresponds to the relative order of the numbers in that block. For $L=263415$, we get the following:
$$
\underline{263} 415 \rightarrow 132 ; \quad 2 \underline{63415} \rightarrow 312 ; \quad 26 \underline{341} 5 \rightarrow 231 ; \quad 263 \underline{415} \rightarrow 213
$$
Moving from left to right in the $n$-label, there are $n-p+1$ such blocks, which means we obtain an $(n-p+1)$-tuple of $p$-labels. For $L=263415$, we get the 4 -tuple $(132,312,231,213)$. We will call this $(n-p+1)$-tuple the $\boldsymbol{p}$-signature of $L$ (or signature, if $p$ is clear from the context) and denote it by $S_{p}[L]$; the $p$-labels in the signature are called windows. For $L=263415$, the windows are $132,312,231$, and 213 , and we write
$$
S_{3}[263415]=(132,312,231,213)
$$
More generally, we will call any $(n-p+1)$-tuple of $p$-labels a $p$-signature, even if we do not know of an $n$-label to which it corresponds (and even if no such label exists). A signature that occurs for exactly one $n$-label is called unique, and a signature that doesn't occur for any $n$-labels is called impossible. A possible signature is one that occurs for at least one $n$-label.
In this power question, you will be asked to analyze some of the properties of labels and signatures.
We can associate a shape to a given 2-signature: a diagram of up and down steps that indicates the relative order of adjacent numbers. For example, the following shape corresponds to the 2-signature $(12,12,12,21,12,21)$ :
<image_1>
A 7-label with this 2-signature corresponds to placing the numbers 1 through 7 at the nodes above so that numbers increase with each up step and decrease with each down step. The 7-label 2347165 is shown below:
<image_2>
Determine the smallest $p$ for which the 20-label
$$
L=3,11,8,4,17,7,15,19,6,2,14,1,10,16,5,12,20,9,13,18
$$
has a unique $p$-signature.
|
[
"The answer is $p=16$. To show this fact we will need to extend the idea from part 8(b) about \"linking\" inequalities forced by the various windows:\n\nTheorem: A $p$-signature for an $n$-label $L$ is unique if and only if for every $k<n, k$ and $k+1$ are in at least one window together. That is, the distance between them in the $n$-label is less than $p$.\n\nProof: Suppose that for some $k$, the distance between $k$ and $k+1$ is $p$ or greater. Then the label $L^{\\prime}$ obtained by swapping $k$ and $k+1$ has the same $p$-signature, because there are no numbers between $k$ and $k+1$ in any window and because the two numbers never appear in the same window.\n\nIf the distance between all such pairs is less than $k$, we need to show that $S_{p}[L]$ is unique. For $i=1,2, \\ldots, n$, let $r_{i}$ denote the position where $i$ appears in $L$. For example, if $L=4123$, then $r_{1}=2, r_{2}=3, r_{3}=4$, and $r_{4}=1$.\n\nLet $L=a_{1}, a_{2}, \\ldots, a_{n}$. Since 1 and 2 are in some window together, $a_{r_{1}}<a_{r_{2}}$. Similarly, for any $k$, since $k$ and $k+1$ are in some window together, $a_{r_{k}}<a_{r_{k+1}}$. We then get a linked inequality $a_{r_{1}}<a_{r_{2}}<\\cdots<a_{r_{n}}$, which can only be satisfied if $a_{r_{1}}=1, a_{r_{2}}=$ $2, \\ldots, a_{r_{n}}=n$. Therefore, $S_{p}[L]$ is unique.\n\nFrom the proof above, we know that the signature is unique if and only if every pair of consecutive integers coexists in at least one window. Therefore, we seek the largest distance between consecutive integers in $L$. That distance is 15 (from 8 to 9 , and from 17 to 18). Thus the smallest $p$ is 16 ."
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2,886
|
In rectangle $M N P Q$, point $A$ lies on $\overline{Q N}$. Segments parallel to the rectangle's sides are drawn through point $A$, dividing the rectangle into four regions. The areas of regions I, II, and III are integers in geometric progression. If the area of $M N P Q$ is 2009 , compute the maximum possible area of region I.
<image_1>
|
[
"Because $A$ is on diagonal $\\overline{N Q}$, rectangles $N X A B$ and $A C Q Y$ are similar. Thus $\\frac{A B}{A X}=\\frac{Q Y}{Q C}=$ $\\frac{A C}{A Y} \\Rightarrow A B \\cdot A Y=A C \\cdot A X$. Therefore, we have $2009=[\\mathrm{I}]+2[\\mathrm{II}]+[\\mathrm{III}]$.\n\nLet the common ratio of the geometric progression be $\\frac{p}{q}$, where $p$ and $q$ are relatively prime positive integers ( $q$ may equal 1 ). Then [I] must be some integer multiple of $q^{2}$, which we will call $a q^{2}$. This gives $[\\mathrm{II}]=a p q$ and [III $=a p^{2}$. By factoring, we get\n\n$$\n2009=a q^{2}+2 a p q+a p^{2} \\Rightarrow 7^{2} \\cdot 41=a(p+q)^{2}\n$$\n\nThus we must have $p+q=7$ and $a=41$. Since $[\\mathrm{I}]=a q^{2}$ and $p, q>0$, the area is maximized when $\\frac{p}{q}=\\frac{1}{6}$, giving $[\\mathrm{I}]=41 \\cdot 36=\\mathbf{1 4 7 6}$. The areas of the other regions are 246,246, and 41 ."
] |
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"1476"
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Geometry
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2,889
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The numbers $1,2, \ldots, 8$ are placed in the $3 \times 3$ grid below, leaving exactly one blank square. Such a placement is called okay if in every pair of adjacent squares, either one square is blank or the difference between the two numbers is at most 2 (two squares are considered adjacent if they share a common side). If reflections, rotations, etc. of placements are considered distinct, compute the number of distinct okay placements.
<image_1>
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[
"We say that two numbers are neighbors if they occupy adjacent squares, and that $a$ is a friend of $b$ if $0<|a-b| \\leq 2$. Using this vocabulary, the problem's condition is that every pair of neighbors must be friends of each other. Each of the numbers 1 and 8 has two friends, and each number has at most four friends.\n\nIf there is no number written in the center square, then we must have one of the cycles in the figures below. For each cycle, there are 8 rotations. Thus there are 16 possible configurations with no number written in the center square.\n\n| 2 | 1 | 3 |\n| :--- | :--- | :--- |\n| 4 | - | 5 |\n| 6 | 8 | 7 |\n\n\n| 3 | 1 | 2 |\n| :--- | :--- | :--- |\n| 5 | - | 4 |\n| 7 | 8 | 6 |\n\nNow assume that the center square contains the number $n$. Because $n$ has at least three neighbors, $n \\neq 1$ and $n \\neq 8$. First we show that 1 must be in a corner. If 1 is a neighbor of $n$, then one of the corners neighboring 1 must be empty, because 1 has only two friends ( 2 and $3)$. If $c$ is in the other corner neighboring 1 , then $\\{n, c\\}=\\{2,3\\}$. But then $n$ must have three\n\n\n\nmore friends $\\left(n_{1}, n_{2}, n_{3}\\right)$ other than 1 and $c$, for a total of five friends, which is impossible, as illustrated below. Therefore 1 must be in a corner.\n\n| - | 1 | $c$ |\n| :--- | :--- | :--- |\n| $n_{1}$ | $n$ | $n_{2}$ |\n| | $n_{3}$ | |\n\nNow we show that 1 can only have one neighbor, i.e., one of the squares adjacent to 1 is empty. If 1 has two neighbors, then we have, up to a reflection and a rotation, the configuration shown below. Because 2 has only one more friend, the corner next to 2 is empty and $n=4$. Consequently, $m_{1}=5$ (refer to the figure below). Then 4 has one friend (namely 6) left with two neighbors $m_{2}$ and $m_{3}$, which is impossible. Thus 1 must have exactly one neighbor. An analogous argument shows that 8 must also be at a corner with exactly one neighbor.\n\n| 1 | 2 | - |\n| :--- | :--- | :--- |\n| 3 | $n$ | $m_{3}$ |\n| $m_{1}$ | $m_{2}$ | |\n\nTherefore, 8 and 1 must be in non-opposite corners, with the blank square between them. Thus, up to reflections and rotations, the only possible configuration is the one shown at left below.\n\n| 1 | - | 8 |\n| :--- | :--- | :--- |\n| $m$ | | |\n| | | |\n\n\n| 1 | - | 8 |\n| :---: | :---: | :---: |\n| $2 / 3$ | $4 / 5$ | $6 / 7$ |\n| $3 / 2$ | $5 / 4$ | $7 / 6$ |\n\nThere are two possible values for $m$, namely 2 and 3 . For each of the cases $m=2$ and $m=3$, the rest of the configuration is uniquely determined, as illustrated in the figure above right. We summarize our process: there are four corner positions for 1; two (non-opposite) corner positions for 8 (after 1 is placed); and two choices for the number in the square neighboring 1 but not neighboring 8 . This leads to $4 \\cdot 2 \\cdot 2=16$ distinct configurations with a number written in the center square.\n\nTherefore, there are 16 configurations in which the center square is blank and 16 configurations with a number in the center square, for a total of $\\mathbf{3 2}$ distinct configurations."
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2,906
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Let $T=80$. In circle $O$, diagrammed at right, minor arc $\widehat{A B}$ measures $\frac{T}{4}$ degrees. If $\mathrm{m} \angle O A C=10^{\circ}$ and $\mathrm{m} \angle O B D=5^{\circ}$, compute the degree measure of $\angle A E B$. Just pass the number without the units.
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[
"Note that $\\mathrm{m} \\angle A E B=\\frac{1}{2}(\\mathrm{~m} \\widehat{A B}-m \\widehat{C D})=\\frac{1}{2}(\\mathrm{~m} \\widehat{A B}-\\mathrm{m} \\angle C O D)$. Also note that $\\mathrm{m} \\angle C O D=$ $360^{\\circ}-(\\mathrm{m} \\angle A O C+\\mathrm{m} \\angle B O D+\\mathrm{m} \\angle A O B)=360^{\\circ}-\\left(180^{\\circ}-2 \\mathrm{~m} \\angle O A C\\right)-\\left(180^{\\circ}-2 \\mathrm{~m} \\angle O B D\\right)-$ $\\mathrm{m} \\widehat{A B}=2(\\mathrm{~m} \\angle O A C+\\mathrm{m} \\angle O B D)-\\mathrm{m} \\widehat{A B}$. Thus $\\mathrm{m} \\angle A E B=\\mathrm{m} \\widehat{A B}-\\mathrm{m} \\angle O A C-\\mathrm{m} \\angle O B D=$ $\\frac{T}{4}-10^{\\circ}-5^{\\circ}$, and with $T=80$, the answer is 5 ."
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"5"
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2,923
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Let $T=24$. A regular $n$-gon is inscribed in a circle; $P$ and $Q$ are consecutive vertices of the polygon, and $A$ is another vertex of the polygon as shown. If $\mathrm{m} \angle A P Q=\mathrm{m} \angle A Q P=T \cdot \mathrm{m} \angle Q A P$, compute the value of $n$.
<image_1>
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[
"Let $\\mathrm{m} \\angle A=x$. Then $\\mathrm{m} \\angle P=\\mathrm{m} \\angle Q=T x$, and $(2 T+1) x=180^{\\circ}$, so $x=\\frac{180^{\\circ}}{2 T+1}$. Let $O$ be the center of the circle, as shown below.\n\n<img_3423>\n\nThen $\\mathrm{m} \\angle P O Q=2 \\mathrm{~m} \\angle P A Q=2\\left(\\frac{180^{\\circ}}{2 T+1}\\right)=\\frac{360^{\\circ}}{2 T+1}$. Because $\\mathrm{m} \\angle P O Q=\\frac{360^{\\circ}}{n}$, the denominators must be equal: $n=2 T+1$. Substitute $T=24$ to find $n=\\mathbf{4 9}$."
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[
"49"
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2,929
|
A king strapped for cash is forced to sell off his kingdom $U=\left\{(x, y): x^{2}+y^{2} \leq 1\right\}$. He sells the two circular plots $C$ and $C^{\prime}$ centered at $\left( \pm \frac{1}{2}, 0\right)$ with radius $\frac{1}{2}$. The retained parts of the kingdom form two regions, each bordered by three arcs of circles; in what follows, we will call such regions curvilinear triangles, or $c$-triangles ( $\mathrm{c} \triangle$ ) for short.
This sad day marks day 0 of a new fiscal era. Unfortunately, these drastic measures are not enough, and so each day thereafter, court geometers mark off the largest possible circle contained in each c-triangle in the remaining property. This circle is tangent to all three arcs of the c-triangle, and will be referred to as the incircle of the c-triangle. At the end of the day, all incircles demarcated that day are sold off, and the following day, the remaining c-triangles are partitioned in the same manner.
Some notation: when discussing mutually tangent circles (or arcs), it is convenient to refer to the curvature of a circle rather than its radius. We define curvature as follows. Suppose that circle $A$ of radius $r_{a}$ is externally tangent to circle $B$ of radius $r_{b}$. Then the curvatures of the circles are simply the reciprocals of their radii, $\frac{1}{r_{a}}$ and $\frac{1}{r_{b}}$. If circle $A$ is internally tangent to circle $B$, however, as in the right diagram below, the curvature of circle $A$ is still $\frac{1}{r_{a}}$, while the curvature of circle $B$ is $-\frac{1}{r_{b}}$, the opposite of the reciprocal of its radius.
<image_1>
Circle $A$ has curvature 2; circle $B$ has curvature 1 .
<image_2>
Circle $A$ has curvature 2; circle $B$ has curvature -1 .
Using these conventions allows us to express a beautiful theorem of Descartes: when four circles $A, B, C, D$ are pairwise tangent, with respective curvatures $a, b, c, d$, then
$$
(a+b+c+d)^{2}=2\left(a^{2}+b^{2}+c^{2}+d^{2}\right),
$$
where (as before) $a$ is taken to be negative if $B, C, D$ are internally tangent to $A$, and correspondingly for $b, c$, or $d$.
Without using Descartes' Circle Formula, Find the combined area of the six remaining curvilinear territories after day 1.
|
[
"The four \"removed\" circles have radii $\\frac{1}{2}, \\frac{1}{2}, \\frac{1}{3}, \\frac{1}{3}$ so the combined area of the six remaining curvilinear territories is:\n\n$$\n\\pi\\left(1^{2}-\\left(\\frac{1}{2}\\right)^{2}-\\left(\\frac{1}{2}\\right)^{2}-\\left(\\frac{1}{3}\\right)^{2}-\\left(\\frac{1}{3}\\right)^{2}\\right)=\\frac{5 \\pi}{18}\n$$"
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"$\\frac{5 \\pi}{18}$"
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2,930
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A king strapped for cash is forced to sell off his kingdom $U=\left\{(x, y): x^{2}+y^{2} \leq 1\right\}$. He sells the two circular plots $C$ and $C^{\prime}$ centered at $\left( \pm \frac{1}{2}, 0\right)$ with radius $\frac{1}{2}$. The retained parts of the kingdom form two regions, each bordered by three arcs of circles; in what follows, we will call such regions curvilinear triangles, or $c$-triangles ( $\mathrm{c} \triangle$ ) for short.
This sad day marks day 0 of a new fiscal era. Unfortunately, these drastic measures are not enough, and so each day thereafter, court geometers mark off the largest possible circle contained in each c-triangle in the remaining property. This circle is tangent to all three arcs of the c-triangle, and will be referred to as the incircle of the c-triangle. At the end of the day, all incircles demarcated that day are sold off, and the following day, the remaining c-triangles are partitioned in the same manner.
Some notation: when discussing mutually tangent circles (or arcs), it is convenient to refer to the curvature of a circle rather than its radius. We define curvature as follows. Suppose that circle $A$ of radius $r_{a}$ is externally tangent to circle $B$ of radius $r_{b}$. Then the curvatures of the circles are simply the reciprocals of their radii, $\frac{1}{r_{a}}$ and $\frac{1}{r_{b}}$. If circle $A$ is internally tangent to circle $B$, however, as in the right diagram below, the curvature of circle $A$ is still $\frac{1}{r_{a}}$, while the curvature of circle $B$ is $-\frac{1}{r_{b}}$, the opposite of the reciprocal of its radius.
<image_1>
Circle $A$ has curvature 2; circle $B$ has curvature 1 .
<image_2>
Circle $A$ has curvature 2; circle $B$ has curvature -1 .
Using these conventions allows us to express a beautiful theorem of Descartes: when four circles $A, B, C, D$ are pairwise tangent, with respective curvatures $a, b, c, d$, then
$$
(a+b+c+d)^{2}=2\left(a^{2}+b^{2}+c^{2}+d^{2}\right),
$$
where (as before) $a$ is taken to be negative if $B, C, D$ are internally tangent to $A$, and correspondingly for $b, c$, or $d$.
Determine the number of curvilinear territories remaining at the end of day 3.
|
[
"At the beginning of day 2, there are six c-triangles, so six incircles are sold, dividing each of the six territories into three smaller curvilinear triangles. So a total of 18 curvilinear triangles exist at the start of day 3, each of which is itself divided into three pieces that day (by the sale of a total of 18 regions bounded by the territories' incircles). Therefore there are 54 regions at the end of day 3."
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"54"
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2,932
|
A king strapped for cash is forced to sell off his kingdom $U=\left\{(x, y): x^{2}+y^{2} \leq 1\right\}$. He sells the two circular plots $C$ and $C^{\prime}$ centered at $\left( \pm \frac{1}{2}, 0\right)$ with radius $\frac{1}{2}$. The retained parts of the kingdom form two regions, each bordered by three arcs of circles; in what follows, we will call such regions curvilinear triangles, or $c$-triangles ( $\mathrm{c} \triangle$ ) for short.
This sad day marks day 0 of a new fiscal era. Unfortunately, these drastic measures are not enough, and so each day thereafter, court geometers mark off the largest possible circle contained in each c-triangle in the remaining property. This circle is tangent to all three arcs of the c-triangle, and will be referred to as the incircle of the c-triangle. At the end of the day, all incircles demarcated that day are sold off, and the following day, the remaining c-triangles are partitioned in the same manner.
Some notation: when discussing mutually tangent circles (or arcs), it is convenient to refer to the curvature of a circle rather than its radius. We define curvature as follows. Suppose that circle $A$ of radius $r_{a}$ is externally tangent to circle $B$ of radius $r_{b}$. Then the curvatures of the circles are simply the reciprocals of their radii, $\frac{1}{r_{a}}$ and $\frac{1}{r_{b}}$. If circle $A$ is internally tangent to circle $B$, however, as in the right diagram below, the curvature of circle $A$ is still $\frac{1}{r_{a}}$, while the curvature of circle $B$ is $-\frac{1}{r_{b}}$, the opposite of the reciprocal of its radius.
<image_1>
Circle $A$ has curvature 2; circle $B$ has curvature 1 .
<image_2>
Circle $A$ has curvature 2; circle $B$ has curvature -1 .
Using these conventions allows us to express a beautiful theorem of Descartes: when four circles $A, B, C, D$ are pairwise tangent, with respective curvatures $a, b, c, d$, then
$$
(a+b+c+d)^{2}=2\left(a^{2}+b^{2}+c^{2}+d^{2}\right),
$$
where (as before) $a$ is taken to be negative if $B, C, D$ are internally tangent to $A$, and correspondingly for $b, c$, or $d$.
Determine the total number of plots sold up to and including day $n$.
|
[
"The total number of plots sold up to and including day $n$ is\n\n$$\n\\begin{aligned}\n2+\\sum_{k=1}^{n} X_{k} & =2+2 \\sum_{k=1}^{n} 3^{k-1} \\\\\n& =2+2 \\cdot\\left(1+3+3^{2}+\\ldots+3^{n-1}\\right) \\\\\n& =3^{n}+1\n\\end{aligned}\n$$\n\nAlternatively, proceed by induction: on day 0 , there are $2=3^{0}+1$ plots sold, and for $n \\geq 0$,\n\n$$\n\\begin{aligned}\n\\left(3^{n}+1\\right)+X_{n+1} & =\\left(3^{n}+1\\right)+2 \\cdot 3^{n} \\\\\n& =3 \\cdot 3^{n}+1 \\\\\n& =3^{n+1}+1 .\n\\end{aligned}\n$$"
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"$3^{n}+1$"
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2,933
|
A king strapped for cash is forced to sell off his kingdom $U=\left\{(x, y): x^{2}+y^{2} \leq 1\right\}$. He sells the two circular plots $C$ and $C^{\prime}$ centered at $\left( \pm \frac{1}{2}, 0\right)$ with radius $\frac{1}{2}$. The retained parts of the kingdom form two regions, each bordered by three arcs of circles; in what follows, we will call such regions curvilinear triangles, or $c$-triangles ( $\mathrm{c} \triangle$ ) for short.
This sad day marks day 0 of a new fiscal era. Unfortunately, these drastic measures are not enough, and so each day thereafter, court geometers mark off the largest possible circle contained in each c-triangle in the remaining property. This circle is tangent to all three arcs of the c-triangle, and will be referred to as the incircle of the c-triangle. At the end of the day, all incircles demarcated that day are sold off, and the following day, the remaining c-triangles are partitioned in the same manner.
Some notation: when discussing mutually tangent circles (or arcs), it is convenient to refer to the curvature of a circle rather than its radius. We define curvature as follows. Suppose that circle $A$ of radius $r_{a}$ is externally tangent to circle $B$ of radius $r_{b}$. Then the curvatures of the circles are simply the reciprocals of their radii, $\frac{1}{r_{a}}$ and $\frac{1}{r_{b}}$. If circle $A$ is internally tangent to circle $B$, however, as in the right diagram below, the curvature of circle $A$ is still $\frac{1}{r_{a}}$, while the curvature of circle $B$ is $-\frac{1}{r_{b}}$, the opposite of the reciprocal of its radius.
<image_1>
Circle $A$ has curvature 2; circle $B$ has curvature 1 .
<image_2>
Circle $A$ has curvature 2; circle $B$ has curvature -1 .
Using these conventions allows us to express a beautiful theorem of Descartes: when four circles $A, B, C, D$ are pairwise tangent, with respective curvatures $a, b, c, d$, then
$$
(a+b+c+d)^{2}=2\left(a^{2}+b^{2}+c^{2}+d^{2}\right),
$$
where (as before) $a$ is taken to be negative if $B, C, D$ are internally tangent to $A$, and correspondingly for $b, c$, or $d$.
Two unit circles and a circle of radius $\frac{2}{3}$ are mutually externally tangent. Compute all possible values of $r$ such that a circle of radius $r$ is tangent to all three circles.
|
[
"Use Descartes' Circle Formula with $a=b=1$ and $c=\\frac{3}{2}$ to solve for $d$ :\n\n$$\n\\begin{aligned}\n2 \\cdot\\left(1^{2}+1^{2}+\\left(\\frac{3}{2}\\right)^{2}+d^{2}\\right) & =\\left(1+1+\\frac{3}{2}+d\\right)^{2} \\\\\n\\frac{17}{2}+2 d^{2} & =\\frac{49}{4}+7 d+d^{2} \\\\\nd^{2}-7 d-\\frac{15}{4} & =0\n\\end{aligned}\n$$\n\nfrom which $d=\\frac{15}{2}$ or $d=-\\frac{1}{2}$. These values correspond to radii of $\\frac{2}{15}$, a small circle nestled between the other three, or 2 , a large circle enclosing the other three.\n\nAlternatively, start by scaling the kingdom with the first four circles removed to match the situation given. Thus the three given circles are internally tangent to a circle of radius $r=2$ and curvature $d=-\\frac{1}{2}$. Descartes' Circle Formula gives a quadratic equation for $d$, and the sum of the roots is $2 \\cdot\\left(1+1+\\frac{3}{2}\\right)=7$, so the second root is $7+\\frac{1}{2}=\\frac{15}{2}$, corresponding to a circle of radius $r=\\frac{2}{15}$."
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"$2$, $\\frac{2}{15}$"
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2,940
|
A king strapped for cash is forced to sell off his kingdom $U=\left\{(x, y): x^{2}+y^{2} \leq 1\right\}$. He sells the two circular plots $C$ and $C^{\prime}$ centered at $\left( \pm \frac{1}{2}, 0\right)$ with radius $\frac{1}{2}$. The retained parts of the kingdom form two regions, each bordered by three arcs of circles; in what follows, we will call such regions curvilinear triangles, or $c$-triangles ( $\mathrm{c} \triangle$ ) for short.
This sad day marks day 0 of a new fiscal era. Unfortunately, these drastic measures are not enough, and so each day thereafter, court geometers mark off the largest possible circle contained in each c-triangle in the remaining property. This circle is tangent to all three arcs of the c-triangle, and will be referred to as the incircle of the c-triangle. At the end of the day, all incircles demarcated that day are sold off, and the following day, the remaining c-triangles are partitioned in the same manner.
Some notation: when discussing mutually tangent circles (or arcs), it is convenient to refer to the curvature of a circle rather than its radius. We define curvature as follows. Suppose that circle $A$ of radius $r_{a}$ is externally tangent to circle $B$ of radius $r_{b}$. Then the curvatures of the circles are simply the reciprocals of their radii, $\frac{1}{r_{a}}$ and $\frac{1}{r_{b}}$. If circle $A$ is internally tangent to circle $B$, however, as in the right diagram below, the curvature of circle $A$ is still $\frac{1}{r_{a}}$, while the curvature of circle $B$ is $-\frac{1}{r_{b}}$, the opposite of the reciprocal of its radius.
<image_1>
Circle $A$ has curvature 2; circle $B$ has curvature 1 .
<image_2>
Circle $A$ has curvature 2; circle $B$ has curvature -1 .
Using these conventions allows us to express a beautiful theorem of Descartes: when four circles $A, B, C, D$ are pairwise tangent, with respective curvatures $a, b, c, d$, then
$$
(a+b+c+d)^{2}=2\left(a^{2}+b^{2}+c^{2}+d^{2}\right),
$$
where (as before) $a$ is taken to be negative if $B, C, D$ are internally tangent to $A$, and correspondingly for $b, c$, or $d$.
Find the areas of the circles removed on day 3.
|
[
"Day 3 begins with two circles of curvature 15 from the configuration $(2,2,3,15)$, and four circles of curvature 6 from the configuration $(-1,2,3,6)$. Consider the following two cases:\n\nCase 1: $(a, b, c, d)=(2,2,3,15), s=22$\n\n- $a=2: a^{\\prime}=2 s-3 a=\\mathbf{3 8}$\n- $b=2: b^{\\prime}=2 s-3 b=\\mathbf{3 8}$\n- $c=3: c^{\\prime}=2 s-3 c=\\mathbf{3 5}$\n- $d=15: d^{\\prime}=2 s-3 d=-1$, which is the configuration from day 1 .\n\nCase 2: $(a, b, c, d)=(-1,2,3,6), s=10$\n\n- $a=-1: a^{\\prime}=2 s-3 a=\\mathbf{2 3}$\n- $b=2: b^{\\prime}=2 s-3 b=\\mathbf{1 4}$\n- $c=3: c^{\\prime}=2 s-3 c=\\mathbf{1 1}$\n- $d=6: d^{\\prime}=2 s-3 d=2$, which is the configuration from day 1 .\n\n\n\nSo the areas of the plots removed on day 3 are:\n\n$$\n\\frac{\\pi}{38^{2}}, \\frac{\\pi}{35^{2}}, \\frac{\\pi}{23^{2}}, \\frac{\\pi}{14^{2}}, \\text { and } \\frac{\\pi}{11^{2}}\n$$\n\nThere are two circles with area $\\frac{\\pi}{35^{2}}$, and four circles with each of the other areas, for a total of 18 plots."
] |
[
"$\\frac{\\pi}{38^{2}}, \\frac{\\pi}{35^{2}}, \\frac{\\pi}{23^{2}}, \\frac{\\pi}{14^{2}}, \\frac{\\pi}{11^{2}}$"
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2,959
|
Points $A$ and $L$ lie outside circle $\omega$, whose center is $O$, and $\overline{A L}$ contains diameter $\overline{R M}$, as shown below. Circle $\omega$ is tangent to $\overline{L K}$ at $K$. Also, $\overline{A K}$ intersects $\omega$ at $Y$, which is between $A$ and $K$. If $K L=3, M L=2$, and $\mathrm{m} \angle A K L-\mathrm{m} \angle Y M K=90^{\circ}$, compute $[A K M]$ (i.e., the area of $\triangle A K M$ ).
<image_1>
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[
"Notice that $\\overline{O K} \\perp \\overline{K L}$, and let $r$ be the radius of $\\omega$.\n\n<img_3784>\n\nThen consider right triangle $O K L$. Because $M L=2, O K=r$, and $O L=r+2$, it follows that $r^{2}+3^{2}=(r+2)^{2}$, from which $r=\\frac{5}{4}$.\n\nBecause $\\mathrm{m} \\angle Y K L=\\frac{1}{2} \\mathrm{~m} \\widehat{Y R K}$ and $\\mathrm{m} \\angle Y M K=\\frac{1}{2} \\mathrm{~m} \\widehat{Y K}$, it follows that $\\mathrm{m} \\angle Y K L+\\mathrm{m} \\angle Y M K=$ $180^{\\circ}$. By the given condition, $\\mathrm{m} \\angle Y K L-\\mathrm{m} \\angle Y M K=90^{\\circ}$. It follows that $\\mathrm{m} \\angle Y M K=45^{\\circ}$ and $\\mathrm{m} \\angle Y K L=135^{\\circ}$. hence $\\mathrm{m} \\widehat{Y K}=90^{\\circ}$. Thus,\n\n$$\n\\overline{Y O} \\perp \\overline{O K} \\quad \\text { and } \\quad \\overline{Y O} \\| \\overline{K L}\n\\tag{*}\n$$\nCompute $[A K M]$ as $\\frac{1}{2}$ base $\\cdot$ height, using base $\\overline{A M}$.\n\n<img_3921>\n\nBecause of (*), $\\triangle A Y O \\sim \\triangle A K L$. To compute $A M$, notice that in $\\triangle A Y O, A O=A M-r$, while in $\\triangle A K L$, the corresponding side $A L=A M+M L=A M+2$. Therefore:\n\n$$\n\\begin{aligned}\n\\frac{A O}{A L} & =\\frac{Y O}{K L} \\\\\n\\frac{A M-\\frac{5}{4}}{A M+2} & =\\frac{5 / 4}{3}\n\\end{aligned}\n$$\n\nfrom which $A M=\\frac{25}{7}$. Draw the altitude of $\\triangle A K M$ from vertex $K$, and let $h$ be its length. In right triangle $O K L, h$ is the altitude to the hypotenuse, so $\\frac{h}{3}=\\sin (\\angle K L O)=\\frac{r}{r+2}$. Hence $h=\\frac{15}{13}$. Therefore $[A K M]=\\frac{1}{2} \\cdot \\frac{25}{7} \\cdot \\frac{15}{13}=\\frac{375}{182}$.",
"Notice that $\\overline{O K} \\perp \\overline{K L}$, and let $r$ be the radius of $\\omega$.\n\n<img_3784>\n\nThen consider right triangle $O K L$. Because $M L=2, O K=r$, and $O L=r+2$, it follows that $r^{2}+3^{2}=(r+2)^{2}$, from which $r=\\frac{5}{4}$.\n\nBecause $\\mathrm{m} \\angle Y K L=\\frac{1}{2} \\mathrm{~m} \\widehat{Y R K}$ and $\\mathrm{m} \\angle Y M K=\\frac{1}{2} \\mathrm{~m} \\widehat{Y K}$, it follows that $\\mathrm{m} \\angle Y K L+\\mathrm{m} \\angle Y M K=$ $180^{\\circ}$. By the given condition, $\\mathrm{m} \\angle Y K L-\\mathrm{m} \\angle Y M K=90^{\\circ}$. It follows that $\\mathrm{m} \\angle Y M K=45^{\\circ}$ and $\\mathrm{m} \\angle Y K L=135^{\\circ}$. hence $\\mathrm{m} \\widehat{Y K}=90^{\\circ}$. Thus,\n\n$$\n\\overline{Y O} \\perp \\overline{O K} \\quad \\text { and } \\quad \\overline{Y O} \\| \\overline{K L}\n\\tag{*}\n$$\nBy the Power of the Point Theorem, $L K^{2}=L M \\cdot L R$, so\n\n$$\n\\begin{aligned}\nL R & =\\frac{9}{2} \\\\\nR M & =L R-L M=\\frac{5}{2} \\\\\nO L & =r+M L=\\frac{13}{4}\n\\end{aligned}\n$$\n\nFrom (*), we know that $\\triangle A Y O \\sim \\triangle A K L$. Hence by $(\\dagger)$,\n\n$$\n\\frac{A L}{A O}=\\frac{A L}{A L-O L}=\\frac{K L}{Y O}=\\frac{3}{5 / 4}=\\frac{12}{5}, \\quad \\text { thus } \\quad A L=\\frac{12}{7} \\cdot O L=\\frac{12}{7} \\cdot \\frac{13}{4}=\\frac{39}{7}\n$$\n\nHence $A M=A L-2=\\frac{25}{7}$. The ratio between the areas of triangles $A K M$ and $R K M$ is equal to\n\n$$\n\\frac{[A K M]}{[R K M]}=\\frac{A M}{R M}=\\frac{25 / 7}{5 / 2}=\\frac{10}{7}\n$$\n\nThus $[A K M]=\\frac{10}{7} \\cdot[R K M]$.\n\nBecause $\\angle K R L$ and $\\angle M K L$ both subtend $\\widehat{K M}, \\triangle K R L \\sim \\triangle M K L$. Therefore $\\frac{K R}{M K}=\\frac{L K}{L M}=$ $\\frac{3}{2}$. Thus let $K R=3 x$ and $M K=2 x$ for some positive real number $x$. Because $R M$ is a diameter of $\\omega$ (see left diagram below), $\\mathrm{m} \\angle R K M=90^{\\circ}$. Thus triangle $R K M$ is a right triangle with hypotenuse $\\overline{R M}$. In particular, $13 x^{2}=K R^{2}+M K^{2}=R M^{2}=\\frac{25}{4}$, so $x^{2}=\\frac{25}{52}$ and $[R K M]=\\frac{R K \\cdot K M}{2}=3 x^{2}$. Therefore\n\n$$\n[A K M]=\\frac{10}{7} \\cdot[R K M]=\\frac{10}{7} \\cdot 3 \\cdot \\frac{25}{52}=\\frac{\\mathbf{3 7 5}}{\\mathbf{1 8 2}}\n$$\n\n<img_3387>",
"Notice that $\\overline{O K} \\perp \\overline{K L}$, and let $r$ be the radius of $\\omega$.\n\n<img_3784>\n\nThen consider right triangle $O K L$. Because $M L=2, O K=r$, and $O L=r+2$, it follows that $r^{2}+3^{2}=(r+2)^{2}$, from which $r=\\frac{5}{4}$.\n\nBecause $\\mathrm{m} \\angle Y K L=\\frac{1}{2} \\mathrm{~m} \\widehat{Y R K}$ and $\\mathrm{m} \\angle Y M K=\\frac{1}{2} \\mathrm{~m} \\widehat{Y K}$, it follows that $\\mathrm{m} \\angle Y K L+\\mathrm{m} \\angle Y M K=$ $180^{\\circ}$. By the given condition, $\\mathrm{m} \\angle Y K L-\\mathrm{m} \\angle Y M K=90^{\\circ}$. It follows that $\\mathrm{m} \\angle Y M K=45^{\\circ}$ and $\\mathrm{m} \\angle Y K L=135^{\\circ}$. hence $\\mathrm{m} \\widehat{Y K}=90^{\\circ}$. Thus,\n\n$$\n\\overline{Y O} \\perp \\overline{O K} \\quad \\text { and } \\quad \\overline{Y O} \\| \\overline{K L}\n\\tag{*}\n$$\n\nLet $U$ and $V$ be the respective feet of the perpendiculars dropped from $A$ and $M$ to $\\overleftrightarrow{K L}$. From (*), $\\triangle A K L$ can be dissected into two infinite progressions of triangles: one progression of triangles similar to $\\triangle O K L$ and the other similar to $\\triangle Y O K$, as shown in the right diagram above. In both progressions, the corresponding sides of the triangles have common ratio equal to\n\n$$\n\\frac{Y O}{K L}=\\frac{5 / 4}{3}=\\frac{5}{12}\n$$\n\n\n\nThus\n\n$$\nA U=\\frac{5}{4}\\left(1+\\frac{5}{12}+\\left(\\frac{5}{12}\\right)^{2}+\\cdots\\right)=\\frac{5}{4} \\cdot \\frac{12}{7}=\\frac{15}{7}\n$$\n\nBecause $\\triangle L M V \\sim \\triangle L O K$, and because $L O=\\frac{13}{4}$ by $(\\dagger)$,\n\n$$\n\\frac{M V}{O K}=\\frac{L M}{L O}, \\quad \\text { thus } \\quad M V=\\frac{O K \\cdot L M}{L O}=\\frac{\\frac{5}{4} \\cdot 2}{\\frac{13}{4}}=\\frac{10}{13}\n$$\n\nFinally, note that $[A K M]=[A K L]-[K L M]$. Because $\\triangle A K L$ and $\\triangle K L M$ share base $\\overline{K L}$,\n\n$$\n[A K M]=\\frac{1}{2} \\cdot 3 \\cdot\\left(\\frac{15}{7}-\\frac{10}{13}\\right)=\\frac{\\mathbf{3 7 5}}{\\mathbf{1 8 2}}\n$$"
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"$\\frac{375}{182}$"
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2,965
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Let $T=8 \sqrt{2}$. In the diagram at right, the smaller circle is internally tangent to the larger circle at point $O$, and $\overline{O P}$ is a diameter of the larger circle. Point $Q$ lies on $\overline{O P}$ such that $P Q=T$, and $\overline{P Q}$ does not intersect the smaller circle. If the larger circle's radius is three times the smaller circle's radius, find the least possible integral radius of the larger circle.
<image_1>
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[
"Let $r$ be the radius of the smaller circle. Then the conditions defining $Q$ imply that $P Q=$ $T<4 r$. With $T=8 \\sqrt{2}$, note that $r>2 \\sqrt{2} \\rightarrow 3 r>6 \\sqrt{2}=\\sqrt{72}$. The least integer greater than $\\sqrt{72}$ is 9 ."
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"9"
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3,050
|
The arrangement of numbers known as Pascal's Triangle has fascinated mathematicians for centuries. In fact, about 700 years before Pascal, the Indian mathematician Halayudha wrote about it in his commentaries to a then-1000-year-old treatise on verse structure by the Indian poet and mathematician Pingala, who called it the Meruprastāra, or "Mountain of Gems". In this Power Question, we'll explore some properties of Pingala's/Pascal's Triangle ("PT") and its variants.
Unless otherwise specified, the only definition, notation, and formulas you may use for PT are the definition, notation, and formulas given below.
PT consists of an infinite number of rows, numbered from 0 onwards. The $n^{\text {th }}$ row contains $n+1$ numbers, identified as $\mathrm{Pa}(n, k)$, where $0 \leq k \leq n$. For all $n$, define $\mathrm{Pa}(n, 0)=\operatorname{Pa}(n, n)=1$. Then for $n>1$ and $1 \leq k \leq n-1$, define $\mathrm{Pa}(n, k)=\mathrm{Pa}(n-1, k-1)+\mathrm{Pa}(n-1, k)$. It is convenient to define $\mathrm{Pa}(n, k)=0$ when $k<0$ or $k>n$. We write the nonzero values of $\mathrm{PT}$ in the familiar pyramid shown below.
<image_1>
As is well known, $\mathrm{Pa}(n, k)$ gives the number of ways of choosing a committee of $k$ people from a set of $n$ people, so a simple formula for $\mathrm{Pa}(n, k)$ is $\mathrm{Pa}(n, k)=\frac{n !}{k !(n-k) !}$. You may use this formula or the recursive definition above throughout this Power Question.
For $n=1,2,3,4$, and $k=4$, find $\mathrm{Pa}(n, n)+\mathrm{Pa}(n+1, n)+\cdots+\operatorname{Pa}(n+k, n)$.
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[
"$$\n\\begin{aligned}\n& \\mathrm{Pa}(1,1)+\\mathrm{Pa}(2,1)+\\mathrm{Pa}(3,1)+\\mathrm{Pa}(4,1)+\\mathrm{Pa}(5,1)=1+2+3+4+5=\\mathbf{1 5} \\\\\n& \\mathrm{Pa}(2,2)+\\mathrm{Pa}(3,2)+\\mathrm{Pa}(4,2)+\\mathrm{Pa}(5,2)+\\mathrm{Pa}(6,2)=1+3+6+10+15=\\mathbf{3 5} \\\\\n& \\mathrm{Pa}(3,3)+\\mathrm{Pa}(4,3)+\\mathrm{Pa}(5,3)+\\mathrm{Pa}(6,3)+\\mathrm{Pa}(7,3)=1+4+10+20+35=\\mathbf{7 0} \\\\\n& \\mathrm{Pa}(4,4)+\\mathrm{Pa}(5,4)+\\mathrm{Pa}(6,4)+\\mathrm{Pa}(7,4)+\\mathrm{Pa}(8,4)=1+5+15+35+70=\\mathbf{1 2 6}\n\\end{aligned}\n$$"
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"15, 35, 70, 126"
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3,051
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The arrangement of numbers known as Pascal's Triangle has fascinated mathematicians for centuries. In fact, about 700 years before Pascal, the Indian mathematician Halayudha wrote about it in his commentaries to a then-1000-year-old treatise on verse structure by the Indian poet and mathematician Pingala, who called it the Meruprastāra, or "Mountain of Gems". In this Power Question, we'll explore some properties of Pingala's/Pascal's Triangle ("PT") and its variants.
Unless otherwise specified, the only definition, notation, and formulas you may use for PT are the definition, notation, and formulas given below.
PT consists of an infinite number of rows, numbered from 0 onwards. The $n^{\text {th }}$ row contains $n+1$ numbers, identified as $\mathrm{Pa}(n, k)$, where $0 \leq k \leq n$. For all $n$, define $\mathrm{Pa}(n, 0)=\operatorname{Pa}(n, n)=1$. Then for $n>1$ and $1 \leq k \leq n-1$, define $\mathrm{Pa}(n, k)=\mathrm{Pa}(n-1, k-1)+\mathrm{Pa}(n-1, k)$. It is convenient to define $\mathrm{Pa}(n, k)=0$ when $k<0$ or $k>n$. We write the nonzero values of $\mathrm{PT}$ in the familiar pyramid shown below.
<image_1>
As is well known, $\mathrm{Pa}(n, k)$ gives the number of ways of choosing a committee of $k$ people from a set of $n$ people, so a simple formula for $\mathrm{Pa}(n, k)$ is $\mathrm{Pa}(n, k)=\frac{n !}{k !(n-k) !}$. You may use this formula or the recursive definition above throughout this Power Question.
If $\mathrm{Pa}(n, n)+\mathrm{Pa}(n+1, n)+\cdots+\mathrm{Pa}(n+k, n)=\mathrm{Pa}(m, j)$, find and justify formulas for $m$ and $j$ in terms of $n$ and $k$.
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[
"Notice that $\\mathrm{Pa}(n, n)+\\operatorname{Pa}(n+1, n)+\\cdots+\\operatorname{Pa}(n+k, n)=\\mathrm{Pa}(n+k+1, n+1)$, so $m=n+k+1$ and $j=n+1$. (By symmetry, $j=k$ is also correct.) The equation is true for all $n$ when $k=0$, because the sum is simply $\\mathrm{Pa}(n, n)$ and the right side is $\\mathrm{Pa}(n+1, n+1)$, both of which are 1 . Proceed by induction on $k$. If $\\mathrm{Pa}(n, n)+\\mathrm{Pa}(n+1, n)+\\cdots+\\mathrm{Pa}(n+k, n)=\\mathrm{Pa}(n+k+1, n+1)$, then adding $\\mathrm{Pa}(n+k+1, n)$ to both sides yields $\\mathrm{Pa}(n+k+1, n)+\\mathrm{Pa}(n+k+1, n+1)=$ $\\mathrm{Pa}(n+k+2, n+1)$ by the recursive rule for $\\mathrm{Pa}$."
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"$m=n+k+1$, $j=n+1$"
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3,058
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The arrangement of numbers known as Pascal's Triangle has fascinated mathematicians for centuries. In fact, about 700 years before Pascal, the Indian mathematician Halayudha wrote about it in his commentaries to a then-1000-year-old treatise on verse structure by the Indian poet and mathematician Pingala, who called it the Meruprastāra, or "Mountain of Gems". In this Power Question, we'll explore some properties of Pingala's/Pascal's Triangle ("PT") and its variants.
Unless otherwise specified, the only definition, notation, and formulas you may use for PT are the definition, notation, and formulas given below.
PT consists of an infinite number of rows, numbered from 0 onwards. The $n^{\text {th }}$ row contains $n+1$ numbers, identified as $\mathrm{Pa}(n, k)$, where $0 \leq k \leq n$. For all $n$, define $\mathrm{Pa}(n, 0)=\operatorname{Pa}(n, n)=1$. Then for $n>1$ and $1 \leq k \leq n-1$, define $\mathrm{Pa}(n, k)=\mathrm{Pa}(n-1, k-1)+\mathrm{Pa}(n-1, k)$. It is convenient to define $\mathrm{Pa}(n, k)=0$ when $k<0$ or $k>n$. We write the nonzero values of $\mathrm{PT}$ in the familiar pyramid shown below.
<image_1>
As is well known, $\mathrm{Pa}(n, k)$ gives the number of ways of choosing a committee of $k$ people from a set of $n$ people, so a simple formula for $\mathrm{Pa}(n, k)$ is $\mathrm{Pa}(n, k)=\frac{n !}{k !(n-k) !}$. You may use this formula or the recursive definition above throughout this Power Question.
Clark's Triangle: If the left side of PT is replaced with consecutive multiples of 6 , starting with 0 , but the right entries (except the first) and the generating rule are left unchanged, the result is called Clark's Triangle. If the $k^{\text {th }}$ entry of the $n^{\text {th }}$ row is denoted by $\mathrm{Cl}(n, k)$, then the formal rule is:
$$
\begin{cases}\mathrm{Cl}(n, 0)=6 n & \text { for all } n \\ \mathrm{Cl}(n, n)=1 & \text { for } n \geq 1 \\ \mathrm{Cl}(n, k)=\mathrm{Cl}(n-1, k-1)+\mathrm{Cl}(n-1, k) & \text { for } n \geq 1 \text { and } 1 \leq k \leq n-1\end{cases}
$$
The first four rows of Clark's Triangle are given below.
<image_2>
If $\mathrm{Cl}(n, 1)=a n^{2}+b n+c$, determine the values of $a, b$, and $c$.
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[
"Using the given values yields the system of equations below.\n\n$$\n\\left\\{\\begin{array}{l}\n\\mathrm{Cl}(1,1)=1=a(1)^{2}+b(1)+c \\\\\n\\mathrm{Cl}(2,1)=7=a(2)^{2}+b(2)+c \\\\\n\\mathrm{Cl}(3,1)=19=a(3)^{2}+b(3)+c\n\\end{array}\\right.\n$$\n\nSolving this system, $a=3, b=-3, c=1$."
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"$3,-3,1$"
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3,060
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The arrangement of numbers known as Pascal's Triangle has fascinated mathematicians for centuries. In fact, about 700 years before Pascal, the Indian mathematician Halayudha wrote about it in his commentaries to a then-1000-year-old treatise on verse structure by the Indian poet and mathematician Pingala, who called it the Meruprastāra, or "Mountain of Gems". In this Power Question, we'll explore some properties of Pingala's/Pascal's Triangle ("PT") and its variants.
Unless otherwise specified, the only definition, notation, and formulas you may use for PT are the definition, notation, and formulas given below.
PT consists of an infinite number of rows, numbered from 0 onwards. The $n^{\text {th }}$ row contains $n+1$ numbers, identified as $\mathrm{Pa}(n, k)$, where $0 \leq k \leq n$. For all $n$, define $\mathrm{Pa}(n, 0)=\operatorname{Pa}(n, n)=1$. Then for $n>1$ and $1 \leq k \leq n-1$, define $\mathrm{Pa}(n, k)=\mathrm{Pa}(n-1, k-1)+\mathrm{Pa}(n-1, k)$. It is convenient to define $\mathrm{Pa}(n, k)=0$ when $k<0$ or $k>n$. We write the nonzero values of $\mathrm{PT}$ in the familiar pyramid shown below.
<image_1>
As is well known, $\mathrm{Pa}(n, k)$ gives the number of ways of choosing a committee of $k$ people from a set of $n$ people, so a simple formula for $\mathrm{Pa}(n, k)$ is $\mathrm{Pa}(n, k)=\frac{n !}{k !(n-k) !}$. You may use this formula or the recursive definition above throughout this Power Question.
Clark's Triangle: If the left side of PT is replaced with consecutive multiples of 6 , starting with 0 , but the right entries (except the first) and the generating rule are left unchanged, the result is called Clark's Triangle. If the $k^{\text {th }}$ entry of the $n^{\text {th }}$ row is denoted by $\mathrm{Cl}(n, k)$, then the formal rule is:
$$
\begin{cases}\mathrm{Cl}(n, 0)=6 n & \text { for all } n \\ \mathrm{Cl}(n, n)=1 & \text { for } n \geq 1 \\ \mathrm{Cl}(n, k)=\mathrm{Cl}(n-1, k-1)+\mathrm{Cl}(n-1, k) & \text { for } n \geq 1 \text { and } 1 \leq k \leq n-1\end{cases}
$$
The first four rows of Clark's Triangle are given below.
<image_2>
Compute $\mathrm{Cl}(11,2)$.
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[
"$\\mathrm{Cl}(11,2)=1000$."
] |
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"1000"
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3,062
|
The arrangement of numbers known as Pascal's Triangle has fascinated mathematicians for centuries. In fact, about 700 years before Pascal, the Indian mathematician Halayudha wrote about it in his commentaries to a then-1000-year-old treatise on verse structure by the Indian poet and mathematician Pingala, who called it the Meruprastāra, or "Mountain of Gems". In this Power Question, we'll explore some properties of Pingala's/Pascal's Triangle ("PT") and its variants.
Unless otherwise specified, the only definition, notation, and formulas you may use for PT are the definition, notation, and formulas given below.
PT consists of an infinite number of rows, numbered from 0 onwards. The $n^{\text {th }}$ row contains $n+1$ numbers, identified as $\mathrm{Pa}(n, k)$, where $0 \leq k \leq n$. For all $n$, define $\mathrm{Pa}(n, 0)=\operatorname{Pa}(n, n)=1$. Then for $n>1$ and $1 \leq k \leq n-1$, define $\mathrm{Pa}(n, k)=\mathrm{Pa}(n-1, k-1)+\mathrm{Pa}(n-1, k)$. It is convenient to define $\mathrm{Pa}(n, k)=0$ when $k<0$ or $k>n$. We write the nonzero values of $\mathrm{PT}$ in the familiar pyramid shown below.
<image_1>
As is well known, $\mathrm{Pa}(n, k)$ gives the number of ways of choosing a committee of $k$ people from a set of $n$ people, so a simple formula for $\mathrm{Pa}(n, k)$ is $\mathrm{Pa}(n, k)=\frac{n !}{k !(n-k) !}$. You may use this formula or the recursive definition above throughout this Power Question.
Clark's Triangle: If the left side of PT is replaced with consecutive multiples of 6 , starting with 0 , but the right entries (except the first) and the generating rule are left unchanged, the result is called Clark's Triangle. If the $k^{\text {th }}$ entry of the $n^{\text {th }}$ row is denoted by $\mathrm{Cl}(n, k)$, then the formal rule is:
$$
\begin{cases}\mathrm{Cl}(n, 0)=6 n & \text { for all } n \\ \mathrm{Cl}(n, n)=1 & \text { for } n \geq 1 \\ \mathrm{Cl}(n, k)=\mathrm{Cl}(n-1, k-1)+\mathrm{Cl}(n-1, k) & \text { for } n \geq 1 \text { and } 1 \leq k \leq n-1\end{cases}
$$
The first four rows of Clark's Triangle are given below.
<image_2>
Compute $\mathrm{Cl}(11,3)$.
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[
"$\\mathrm{Cl}(11,3)=2025$."
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"2025"
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3,065
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Leibniz's Harmonic Triangle: Consider the triangle formed by the rule
$$
\begin{cases}\operatorname{Le}(n, 0)=\frac{1}{n+1} & \text { for all } n \\ \operatorname{Le}(n, n)=\frac{1}{n+1} & \text { for all } n \\ \operatorname{Le}(n, k)=\operatorname{Le}(n+1, k)+\operatorname{Le}(n+1, k+1) & \text { for all } n \text { and } 0 \leq k \leq n\end{cases}
$$
This triangle, discovered first by Leibniz, consists of reciprocals of integers as shown below.
<image_1>
For this contest, you may assume that $\operatorname{Le}(n, k)>0$ whenever $0 \leq k \leq n$, and that $\operatorname{Le}(n, k)$ is undefined if $k<0$ or $k>n$.
Compute Le(17,1).
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[
"$\\operatorname{Le}(17,1)=\\operatorname{Le}(16,0)-\\operatorname{Le}(17,0)=\\frac{1}{17}-\\frac{1}{18}=\\frac{1}{306}$."
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"$\\frac{1}{306}$"
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3,066
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Leibniz's Harmonic Triangle: Consider the triangle formed by the rule
$$
\begin{cases}\operatorname{Le}(n, 0)=\frac{1}{n+1} & \text { for all } n \\ \operatorname{Le}(n, n)=\frac{1}{n+1} & \text { for all } n \\ \operatorname{Le}(n, k)=\operatorname{Le}(n+1, k)+\operatorname{Le}(n+1, k+1) & \text { for all } n \text { and } 0 \leq k \leq n\end{cases}
$$
This triangle, discovered first by Leibniz, consists of reciprocals of integers as shown below.
<image_1>
For this contest, you may assume that $\operatorname{Le}(n, k)>0$ whenever $0 \leq k \leq n$, and that $\operatorname{Le}(n, k)$ is undefined if $k<0$ or $k>n$.
Compute $\operatorname{Le}(17,2)$.
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[
"$\\operatorname{Le}(17,2)=\\operatorname{Le}(16,1)-\\operatorname{Le}(17,1)=\\operatorname{Le}(15,0)-\\operatorname{Le}(16,0)-\\operatorname{Le}(17,1)=\\frac{1}{2448}$."
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"$\\frac{1}{2448}$"
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3,068
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Leibniz's Harmonic Triangle: Consider the triangle formed by the rule
$$
\begin{cases}\operatorname{Le}(n, 0)=\frac{1}{n+1} & \text { for all } n \\ \operatorname{Le}(n, n)=\frac{1}{n+1} & \text { for all } n \\ \operatorname{Le}(n, k)=\operatorname{Le}(n+1, k)+\operatorname{Le}(n+1, k+1) & \text { for all } n \text { and } 0 \leq k \leq n\end{cases}
$$
This triangle, discovered first by Leibniz, consists of reciprocals of integers as shown below.
<image_1>
For this contest, you may assume that $\operatorname{Le}(n, k)>0$ whenever $0 \leq k \leq n$, and that $\operatorname{Le}(n, k)$ is undefined if $k<0$ or $k>n$.
Compute $\sum_{n=1}^{2011} \operatorname{Le}(n, 1)$.
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"Because $\\operatorname{Le}(n, 1)=\\frac{1}{n}-\\frac{1}{n+1}$,\n\n$$\n\\begin{aligned}\n\\sum_{i=1}^{2011} \\operatorname{Le}(i, 1) & =\\sum_{i=1}^{2011}\\left(\\frac{1}{n}-\\frac{1}{n+1}\\right) \\\\\n& =\\left(\\frac{1}{1}-\\frac{1}{2}\\right)+\\left(\\frac{1}{2}-\\frac{1}{3}\\right)+\\cdots+\\left(\\frac{1}{2010}-\\frac{1}{2011}\\right)+\\left(\\frac{1}{2011}-\\frac{1}{2012}\\right) \\\\\n& =1-\\frac{1}{2012} \\\\\n& =\\frac{2011}{2012} .\n\\end{aligned}\n$$"
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"$\\frac{2011}{2012}$"
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3,070
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Leibniz's Harmonic Triangle: Consider the triangle formed by the rule
$$
\begin{cases}\operatorname{Le}(n, 0)=\frac{1}{n+1} & \text { for all } n \\ \operatorname{Le}(n, n)=\frac{1}{n+1} & \text { for all } n \\ \operatorname{Le}(n, k)=\operatorname{Le}(n+1, k)+\operatorname{Le}(n+1, k+1) & \text { for all } n \text { and } 0 \leq k \leq n\end{cases}
$$
This triangle, discovered first by Leibniz, consists of reciprocals of integers as shown below.
<image_1>
For this contest, you may assume that $\operatorname{Le}(n, k)>0$ whenever $0 \leq k \leq n$, and that $\operatorname{Le}(n, k)$ is undefined if $k<0$ or $k>n$.
If $\sum_{i=1}^{\infty} \operatorname{Le}(i, 1)=\operatorname{Le}(n, k)$, determine the values of $n$ and $k$.
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"Extending the result of $8 \\mathrm{~b}$ gives\n\n$$\n\\sum_{i=1}^{n} \\operatorname{Le}(i, 1)=\\frac{1}{1}-\\frac{1}{n}\n$$\n\nso as $n \\rightarrow \\infty, \\sum_{i=1}^{n} \\operatorname{Le}(i, 1) \\rightarrow 1$. This value appears as $\\operatorname{Le}(0,0)$, so $n=k=0$."
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"$0,0$"
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3,071
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Leibniz's Harmonic Triangle: Consider the triangle formed by the rule
$$
\begin{cases}\operatorname{Le}(n, 0)=\frac{1}{n+1} & \text { for all } n \\ \operatorname{Le}(n, n)=\frac{1}{n+1} & \text { for all } n \\ \operatorname{Le}(n, k)=\operatorname{Le}(n+1, k)+\operatorname{Le}(n+1, k+1) & \text { for all } n \text { and } 0 \leq k \leq n\end{cases}
$$
This triangle, discovered first by Leibniz, consists of reciprocals of integers as shown below.
<image_1>
For this contest, you may assume that $\operatorname{Le}(n, k)>0$ whenever $0 \leq k \leq n$, and that $\operatorname{Le}(n, k)$ is undefined if $k<0$ or $k>n$.
If $\sum_{i=m}^{\infty} \operatorname{Le}(i, m)=\operatorname{Le}(n, k)$, compute expressions for $n$ and $k$ in terms of $m$.
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"$n=k=m-1$."
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"$m-1,m-1$"
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3,093
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$\quad$ Let $T=12$. As shown, three circles are mutually externally tangent. The large circle has a radius of $T$, and the smaller two circles each have radius $\frac{T}{2}$. Compute the area of the triangle whose vertices are the centers of the three circles.
<image_1>
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"The desired triangle is an isosceles triangle whose base vertices are the centers of the two smaller circles. The congruent sides of the triangle have length $T+\\frac{T}{2}$. Thus the altitude to the base has length $\\sqrt{\\left(\\frac{3 T}{2}\\right)^{2}-\\left(\\frac{T}{2}\\right)^{2}}=T \\sqrt{2}$. Thus the area of the triangle is $\\frac{1}{2} \\cdot\\left(\\frac{T}{2}+\\frac{T}{2}\\right) \\cdot T \\sqrt{2}=\\frac{T^{2} \\sqrt{2}}{2}$. With $T=12$, the area is $\\mathbf{7 2} \\sqrt{\\mathbf{2}}$."
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"$72 \\sqrt{2}$"
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1,735
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Three circular arcs $\gamma_{1}, \gamma_{2}$, and $\gamma_{3}$ connect the points $A$ and $C$. These arcs lie in the same half-plane defined by line $A C$ in such a way that $\operatorname{arc} \gamma_{2}$ lies between the $\operatorname{arcs} \gamma_{1}$ and $\gamma_{3}$. Point $B$ lies on the segment $A C$. Let $h_{1}, h_{2}$, and $h_{3}$ be three rays starting at $B$, lying in the same half-plane, $h_{2}$ being between $h_{1}$ and $h_{3}$. For $i, j=1,2,3$, denote by $V_{i j}$ the point of intersection of $h_{i}$ and $\gamma_{j}$ (see the Figure below).
Denote by $\overparen{V_{i j} V_{k j}} \overparen{k_{k \ell} V_{i \ell}}$ the curved quadrilateral, whose sides are the segments $V_{i j} V_{i \ell}, V_{k j} V_{k \ell}$ and $\operatorname{arcs} V_{i j} V_{k j}$ and $V_{i \ell} V_{k \ell}$. We say that this quadrilateral is circumscribed if there exists a circle touching these two segments and two arcs.
Prove that if the curved quadrilaterals $\sqrt{V_{11} V_{21}} \sqrt{V_{22} V_{12}}, \sqrt{V_{12} V_{22}} \sqrt{V_{23} V_{13}}, \sqrt{V_{21} V_{31}} \sqrt{V_{32} V_{22}}$ are circumscribed, then the curved quadrilateral $\overparen{V_{22} V_{32}} \overparen{V_{33} V_{23}}$ is circumscribed, too.
<image_1>
Fig. 1
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"Denote by $O_{i}$ and $R_{i}$ the center and the radius of $\\gamma_{i}$, respectively. Denote also by $H$ the half-plane defined by $A C$ which contains the whole configuration. For every point $P$ in the half-plane $H$, denote by $d(P)$ the distance between $P$ and line $A C$. Furthermore, for any $r>0$, denote by $\\Omega(P, r)$ the circle with center $P$ and radius $r$.\n\nLemma 1. For every $1 \\leq i<j \\leq 3$, consider those circles $\\Omega(P, r)$ in the half-plane $H$ which are tangent to $h_{i}$ and $h_{j}$.\n\n(a) The locus of the centers of these circles is the angle bisector $\\beta_{i j}$ between $h_{i}$ and $h_{j}$.\n\n(b) There is a constant $u_{i j}$ such that $r=u_{i j} \\cdot d(P)$ for all such circles.\n\nProof. Part (a) is obvious. To prove part (b), notice that the circles which are tangent to $h_{i}$ and $h_{j}$ are homothetic with the common homothety center $B$ (see Fig. 2). Then part (b) also becomes trivial.\n\nLemma 2. For every $1 \\leq i<j \\leq 3$, consider those circles $\\Omega(P, r)$ in the half-plane $H$ which are externally tangent to $\\gamma_{i}$ and internally tangent to $\\gamma_{j}$.\n\n(a) The locus of the centers of these circles is an ellipse arc $\\varepsilon_{i j}$ with end-points $A$ and $C$.\n\n(b) There is a constant $v_{i j}$ such that $r=v_{i j} \\cdot d(P)$ for all such circles.\n\nProof. (a) Notice that the circle $\\Omega(P, r)$ is externally tangent to $\\gamma_{i}$ and internally tangent to $\\gamma_{j}$ if and only if $O_{i} P=R_{i}+r$ and $O_{j}=R_{j}-r$. Therefore, for each such circle we have\n\n$$\nO_{i} P+O_{j} P=O_{i} A+O_{j} A=O_{i} C+O_{j} C=R_{i}+R_{j} .\n$$\n\nSuch points lie on an ellipse with foci $O_{i}$ and $O_{j}$; the diameter of this ellipse is $R_{i}+R_{j}$, and it passes through the points $A$ and $C$. Let $\\varepsilon_{i j}$ be that arc $A C$ of the ellipse which runs inside the half plane $H$ (see Fig. 3.)\n\nThis ellipse arc lies between the $\\operatorname{arcs} \\gamma_{i}$ and $\\gamma_{j}$. Therefore, if some point $P$ lies on $\\varepsilon_{i j}$, then $O_{i} P>R_{i}$ and $O_{j} P<R_{j}$. Now, we choose $r=O_{i} P-R_{i}=R_{j}-O_{j} P>0$; then the\n\n\n\n<img_3334>\n\nFig. 2\n\n<img_3279>\n\nFig. 3\n\ncircle $\\Omega(P, r)$ touches $\\gamma_{i}$ externally and touches $\\gamma_{j}$ internally, so $P$ belongs to the locus under investigation.\n\n(b) Let $\\vec{\\rho}=\\overrightarrow{A P}, \\vec{\\rho}_{i}=\\overrightarrow{A O_{i}}$, and $\\vec{\\rho}_{j}=\\overrightarrow{A O_{j}}$; let $d_{i j}=O_{i} O_{j}$, and let $\\vec{v}$ be a unit vector orthogonal to $A C$ and directed toward $H$. Then we have $\\left|\\vec{\\rho}_{i}\\right|=R_{i},\\left|\\vec{\\rho}_{j}\\right|=R_{j},\\left|\\overrightarrow{O_{i} P}\\right|=$ $\\left|\\vec{\\rho}-\\vec{\\rho}_{i}\\right|=R_{i}+r,\\left|\\overrightarrow{O_{j} P}\\right|=\\left|\\vec{\\rho}-\\vec{\\rho}_{j}\\right|=R_{j}-r$, hence\n\n$$\n\\begin{gathered}\n\\left(\\vec{\\rho}-\\vec{\\rho}_{i}\\right)^{2}-\\left(\\vec{\\rho}-\\vec{\\rho}_{j}\\right)^{2}=\\left(R_{i}+r\\right)^{2}-\\left(R_{j}-r\\right)^{2} \\\\\n\\left(\\vec{\\rho}_{i}^{2}-\\vec{\\rho}_{j}^{2}\\right)+2 \\vec{\\rho} \\cdot\\left(\\vec{\\rho}_{j}-\\vec{\\rho}_{i}\\right)=\\left(R_{i}^{2}-R_{j}^{2}\\right)+2 r\\left(R_{i}+R_{j}\\right) \\\\\nd_{i j} \\cdot d(P)=d_{i j} \\vec{v} \\cdot \\vec{\\rho}=\\left(\\vec{\\rho}_{j}-\\vec{\\rho}_{i}\\right) \\cdot \\vec{\\rho}=r\\left(R_{i}+R_{j}\\right)\n\\end{gathered}\n$$\n\nTherefore,\n\n$$\nr=\\frac{d_{i j}}{R_{i}+R_{j}} \\cdot d(P)\n$$\n\nand the value $v_{i j}=\\frac{d_{i j}}{R_{i}+R_{j}}$ does not depend on $P$.\n\nLemma 3. The curved quadrilateral $\\mathcal{Q}_{i j}=\\overparen{V_{i, j} V_{i+1}, j} \\overparen{V_{i+1, j+1} V_{i, j+1}}$ is circumscribed if and only if $u_{i, i+1}=v_{j, j+1}$.\n\nProof. First suppose that the curved quadrilateral $\\mathcal{Q}_{i j}$ is circumscribed and $\\Omega(P, r)$ is its inscribed circle. By Lemma 1 and Lemma 2 we have $r=u_{i, i+1} \\cdot d(P)$ and $r=v_{j, j+1} \\cdot d(P)$ as well. Hence, $u_{i, i+1}=v_{j, j+1}$.\n\nTo prove the opposite direction, suppose $u_{i, i+1}=v_{j, j+1}$. Let $P$ be the intersection of the angle bisector $\\beta_{i, i+1}$ and the ellipse arc $\\varepsilon_{j, j+1}$. Choose $r=u_{i, i+1} \\cdot d(P)=v_{j, j+1} \\cdot d(P)$. Then the circle $\\Omega(P, r)$ is tangent to the half lines $h_{i}$ and $h_{i+1}$ by Lemma 1 , and it is tangent to the $\\operatorname{arcs} \\gamma_{j}$ and $\\gamma_{j+1}$ by Lemma 2. Hence, the curved quadrilateral $\\mathcal{Q}_{i j}$ is circumscribed.\n\nBy Lemma 3, the statement of the problem can be reformulated to an obvious fact: If the equalities $u_{12}=v_{12}, u_{12}=v_{23}$, and $u_{23}=v_{12}$ hold, then $u_{23}=v_{23}$ holds as well."
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Geometry
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1,975
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Construct a tetromino by attaching two $2 \times 1$ dominoes along their longer sides such that the midpoint of the longer side of one domino is a corner of the other domino. This construction yields two kinds of tetrominoes with opposite orientations. Let us call them Sand Z-tetrominoes, respectively.
<image_1>
S-tetrominoes
<image_2>
Z-tetrominoes
Assume that a lattice polygon $P$ can be tiled with $S$-tetrominoes. Prove than no matter how we tile $P$ using only $\mathrm{S}$ - and Z-tetrominoes, we always use an even number of Z-tetrominoes.
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"We may assume that polygon $P$ is the union of some squares of an infinite chessboard. Colour the squares of the chessboard with two colours as the figure below illustrates.\n\n<img_3847>\n\nObserve that no matter how we tile $P$, any S-tetromino covers an even number of black squares, whereas any Z-tetromino covers an odd number of them. As $P$ can be tiled exclusively by S-tetrominoes, it contains an even number of black squares. But if some S-tetrominoes and some Z-tetrominoes cover an even number of black squares, then the number of Z-tetrominoes must be even.",
"Let us assign coordinates to the squares of the infinite chessboard in such a way that the squares of $P$ have nonnegative coordinates only, and that the first coordinate increases as one moves to the right, while the second coordinate increases as one moves upwards. Write the integer $3^{i} \\cdot(-3)^{j}$ into the square with coordinates $(i, j)$, as in the following figure:\n\n| $\\vdots$ | | | | | |\n| :---: | :---: | :---: | :---: | :---: | :---: |\n| 81 | $\\vdots$ | | | | |\n| -27 | -81 | $\\vdots$ | | | |\n| 9 | 27 | 81 | $\\cdots$ | | |\n| -3 | -9 | -27 | -81 | $\\ldots$ | |\n| 1 | 3 | 9 | 27 | 81 | $\\ldots$ |\n\nThe sum of the numbers written into four squares that can be covered by an $S$-tetromino is either of the form\n\n$$\n3^{i} \\cdot(-3)^{j} \\cdot\\left(1+3+3 \\cdot(-3)+3^{2} \\cdot(-3)\\right)=-32 \\cdot 3^{i} \\cdot(-3)^{j}\n$$\n\n(for the first type of $S$-tetrominoes), or of the form\n\n$$\n3^{i} \\cdot(-3)^{j} \\cdot\\left(3+3 \\cdot(-3)+(-3)+(-3)^{2}\\right)=0\n$$\n\nand thus divisible by 32. For this reason, the sum of the numbers written into the squares of $P$, and thus also the sum of the numbers covered by $Z$-tetrominoes in the second covering, is likewise divisible by 32 . Now the sum of the entries of a $Z$-tetromino is either of the form\n\n$$\n3^{i} \\cdot(-3)^{j} \\cdot\\left(3+3^{2}+(-3)+3 \\cdot(-3)\\right)=0\n$$\n\n(for the first type of $Z$-tetrominoes), or of the form\n\n$$\n3^{i} \\cdot(-3)^{j} \\cdot\\left(1+(-3)+3 \\cdot(-3)+3 \\cdot(-3)^{2}\\right)=16 \\cdot 3^{i} \\cdot(-3)^{j}\n$$\n\ni.e., 16 times an odd number. Thus in order to obtain a total that is divisible by 32, an even number of the latter kind of $Z$-tetrominoes needs to be used. Rotating everything by $90^{\\circ}$, we find that the number of $Z$-tetrominoes of the first kind is even as well. So we have even proven slightly more than necessary."
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2,039
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An anti-Pascal pyramid is a finite set of numbers, placed in a triangle-shaped array so that the first row of the array contains one number, the second row contains two numbers, the third row contains three numbers and so on; and, except for the numbers in the bottom row, each number equals the absolute value of the difference of the two numbers below it. For instance, the triangle below is an anti-Pascal pyramid with four rows, in which every integer from 1 to $1+2+3+4=10$ occurs exactly once:
<image_1>
Prove that it is impossible to form an anti-Pascal pyramid with 2018 rows, using every integer from 1 to $1+2+\cdots+2018$ exactly once.
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"Let $T$ be an anti-Pascal pyramid with $n$ rows, containing every integer from 1 to $1+2+\\cdots+n$, and let $a_{1}$ be the topmost number in $T$ (Figure 1). The two numbers below $a_{1}$ are some $a_{2}$ and $b_{2}=a_{1}+a_{2}$, the two numbers below $b_{2}$ are some $a_{3}$ and $b_{3}=a_{1}+a_{2}+a_{3}$, and so on and so forth all the way down to the bottom row, where some $a_{n}$ and $b_{n}=a_{1}+a_{2}+\\cdots+a_{n}$ are the two neighbours below $b_{n-1}=a_{1}+a_{2}+\\cdots+a_{n-1}$. Since the $a_{k}$ are $n$ pairwise distinct positive integers whose sum does not exceed the largest number in $T$, which is $1+2+\\cdots+n$, it follows that they form a permutation of $1,2, \\ldots, n$.\n\n<img_3838>\n\nFigure 1\n\n<img_4049>\n\nFigure 2\n\nConsider now (Figure 2) the two 'equilateral' subtriangles of $T$ whose bottom rows contain the numbers to the left, respectively right, of the pair $a_{n}, b_{n}$. (One of these subtriangles may very well be empty.) At least one of these subtriangles, say $T^{\\prime}$, has side length $\\ell \\geqslant\\lceil(n-2) / 2\\rceil$. Since $T^{\\prime}$ obeys the anti-Pascal rule, it contains $\\ell$ pairwise distinct positive integers $a_{1}^{\\prime}, a_{2}^{\\prime}, \\ldots, a_{\\ell}^{\\prime}$, where $a_{1}^{\\prime}$ is at the apex, and $a_{k}^{\\prime}$ and $b_{k}^{\\prime}=a_{1}^{\\prime}+a_{2}^{\\prime}+\\cdots+a_{k}^{\\prime}$ are the two neighbours below $b_{k-1}^{\\prime}$ for each $k=2,3 \\ldots, \\ell$. Since the $a_{k}$ all lie outside $T^{\\prime}$, and they form a permutation of $1,2, \\ldots, n$, the $a_{k}^{\\prime}$ are all greater than $n$. Consequently,\n\n$$\n\\begin{array}{r}\nb_{\\ell}^{\\prime} \\geqslant(n+1)+(n+2)+\\cdots+(n+\\ell)=\\frac{\\ell(2 n+\\ell+1)}{2} \\\\\n\\geqslant \\frac{1}{2} \\cdot \\frac{n-2}{2}\\left(2 n+\\frac{n-2}{2}+1\\right)=\\frac{5 n(n-2)}{8},\n\\end{array}\n$$\n\nwhich is greater than $1+2+\\cdots+n=n(n+1) / 2$ for $n=2018$. A contradiction.\n\nSo it is not possible."
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2,184
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Let $A B C D$ be a cyclic quadrilateral, and let diagonals $A C$ and $B D$ intersect at $X$. Let $C_{1}, D_{1}$ and $M$ be the midpoints of segments $C X$, $D X$ and $C D$, respectively. Lines $A D_{1}$ and $B C_{1}$ intersect at $Y$, and line $M Y$ intersects diagonals $A C$ and $B D$ at different points $E$ and $F$, respectively. Prove that line $X Y$ is tangent to the circle through $E, F$ and $X$.
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"We are to prove that $\\angle E X Y=\\angle E F X$; alternatively, but equivalently, $\\angle A Y X+\\angle X A Y=\\angle B Y F+\\angle X B Y$.\n\nSince the quadrangle $A B C D$ is cyclic, the triangles $X A D$ and $X B C$ are similar, and since $A D_{1}$ and $B C_{1}$ are corresponding medians in these triangles, it follows that $\\angle X A Y=\\angle X A D_{1}=\\angle X B C_{1}=\\angle X B Y$.\n\nFinally, $\\angle A Y X=\\angle B Y F$, since $X$ and $M$ are corresponding points in the similar triangles $A B Y$ and $C_{1} D_{1} Y$ : indeed, $\\angle X A B=\\angle X D C=\\angle M C_{1} D_{1}$, and $\\angle X B A=\\angle X C D=\\angle M D_{1} C_{1}$."
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2,227
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Let $H$ be the orthocenter and $G$ be the centroid of acute-angled triangle $\triangle A B C$ with $A B \neq A C$. The line $A G$ intersects the circumcircle of $\triangle A B C$ at $A$ and $P$. Let $P^{\prime}$ be the reflection of $P$ in the line $B C$. Prove that $\angle C A B=60^{\circ}$ if and only if $H G=G P^{\prime}$.
<image_1>
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"Let $\\omega$ be the circumcircle of $\\triangle A B C$. Reflecting $\\omega$ in line $B C$, we obtain circle $\\omega^{\\prime}$ which, obviously, contains points $H$ and $P^{\\prime}$. Let $M$ be the midpoint of $B C$. As triangle $\\triangle A B C$ is acute-angled, then $H$ and $O$ lie inside this triangle.\n\nLet us assume that $\\angle C A B=60^{\\circ}$. Since\n\n$$\n\\angle C O B=2 \\angle C A B=120^{\\circ}=180^{\\circ}-60^{\\circ}=180^{\\circ}-\\angle C A B=\\angle C H B,\n$$\n\nhence $O$ lies on $\\omega^{\\prime}$. Reflecting $O$ in line $B C$, we obtain point $O^{\\prime}$ which lies on $\\omega$ and this point is the center of $\\omega^{\\prime}$. Then $O O^{\\prime}=2 O M=2 R \\cos \\angle C A B=A H$, so $A H=O O^{\\prime}=H O^{\\prime}=A O=R$, where $R$ is the radius of $\\omega$ and, naturally, of $\\omega^{\\prime}$. Then quadrilateral $A H O^{\\prime} O$ is a rhombus, so $A$ and $O^{\\prime}$ are symmetric to each other with respect to $H O$. As $H, G$ and $O$ are collinear (Euler line), then $\\angle G A H=\\angle H O^{\\prime} G$. Diagonals of quadrilateral $G O P O^{\\prime}$ intersects at $M$. Since $\\angle B O M=60^{\\circ}$, so\n\n$$\nO M=M O^{\\prime}=\\operatorname{ctg} 60^{\\circ} \\cdot M B=\\frac{M B}{\\sqrt{3}}\n$$\n\nAs $3 M O \\cdot M O^{\\prime}=M B^{2}=M B \\cdot M C=M P \\cdot M A=3 M G \\cdot M P$, then $G O P O^{\\prime}$ is a cyclic. Since $B C$ is a perpendicular bisector of $O O^{\\prime}$, so the circumcircle of quadrilateral $G O P O^{\\prime}$ is symmetrical with respect to $B C$. Thus $P^{\\prime}$ also belongs to the circumcircle of $G O P O^{\\prime}$, hence $\\angle G O^{\\prime} P^{\\prime}=\\angle G P P^{\\prime}$. Note that $\\angle G P P^{\\prime}=\\angle G A H$ since $A H \\| P P^{\\prime}$. And as it was proved $\\angle G A H=\\angle H O^{\\prime} G$, then $\\angle H O^{\\prime} G=\\angle G O^{\\prime} P^{\\prime}$. Thus triangles $\\triangle H O^{\\prime} G$ and $\\triangle G O^{\\prime} P^{\\prime}$ are equal and hence $H G=G P^{\\prime}$.\n\nNow we will prove that if $H G=G P^{\\prime}$ then $\\angle C A B=60^{\\circ}$. Reflecting $A$ with respect to $M$, we get $A^{\\prime}$. Then, as it was said in the first part of solution, points $B, C, H$ and $P^{\\prime}$ belong to $\\omega^{\\prime}$. Also it is clear that $A^{\\prime}$ belongs to $\\omega^{\\prime}$. Note that $H C \\perp C A^{\\prime}$ since $A B \\| C A^{\\prime}$ and hence $H A^{\\prime}$ is a diameter of $\\omega^{\\prime}$. Obviously, the center $O^{\\prime}$ of circle $\\omega^{\\prime}$ is midpoint of $H A^{\\prime}$. From $H G=G P^{\\prime}$ it follows that $\\triangle H G O^{\\prime}$ is equal to $\\triangle P^{\\prime} G O^{\\prime}$. Therefore $H$ and $P^{\\prime}$ are symmetric with respect to $G O^{\\prime}$. Hence $G O^{\\prime} \\perp H P^{\\prime}$ and $G O^{\\prime} \\| A^{\\prime} P^{\\prime}$. Let $H G$ intersect $A^{\\prime} P^{\\prime}$ at $K$ and $K \\not \\equiv O$ since $A B \\neq A C$. We conclude that $H G=G K$, because line $G O^{\\prime}$ is midline of the triangle $\\triangle H K A^{\\prime}$. Note that $2 G O=H G$. since $H O$ is Euler line of triangle $A B C$. So $O$ is midpoint of segment $G K$. Because of $\\angle C M P=\\angle C M P^{\\prime}$, then $\\angle G M O=\\angle O M P^{\\prime}$. Line $O M$, that passes through $O^{\\prime}$, is an external angle bisector of $\\angle P^{\\prime} M A^{\\prime}$. Also we know that $P^{\\prime} O^{\\prime}=O^{\\prime} A^{\\prime}$, then $O^{\\prime}$ is the midpoint of arc $P^{\\prime} M A^{\\prime}$ of the circumcircle of triangle $\\triangle P^{\\prime} M A^{\\prime}$. It\n\n\n\n<img_3844>\n\nfollows that quadrilateral $P^{\\prime} M O^{\\prime} A^{\\prime}$ is cyclic, then $\\angle O^{\\prime} M A^{\\prime}=\\angle O^{\\prime} P^{\\prime} A^{\\prime}=\\angle O^{\\prime} A^{\\prime} P^{\\prime}$. Let $O M$ and $P^{\\prime} A^{\\prime}$ intersect at $T$. Triangles $\\triangle T O^{\\prime} A^{\\prime}$ and $\\triangle A^{\\prime} O^{\\prime} M$ are similar, hence $O^{\\prime} A^{\\prime} / O^{\\prime} M=O^{\\prime} T / O^{\\prime} A^{\\prime}$. In the other words, $O^{\\prime} M \\cdot O^{\\prime} T=O^{\\prime} A^{\\prime 2}$. Using Menelaus' theorem for triangle $\\triangle H K A^{\\prime}$ and line $T O^{\\prime}$, we obtain that\n\n$$\n\\frac{A^{\\prime} O^{\\prime}}{O^{\\prime} H} \\cdot \\frac{H O}{O K} \\cdot \\frac{K T}{T A^{\\prime}}=3 \\cdot \\frac{K T}{T A^{\\prime}}=1\n$$\n\nIt follows that $K T / T A^{\\prime}=1 / 3$ and $K A^{\\prime}=2 K T$. Using Menelaus' theorem for triangle $T O^{\\prime} A^{\\prime}$ and line $H K$ we get that\n\n$$\n1=\\frac{O^{\\prime} H}{H A^{\\prime}} \\cdot \\frac{A^{\\prime} K}{K T} \\cdot \\frac{T O}{O O^{\\prime}}=\\frac{1}{2} \\cdot 2 \\cdot \\frac{T O}{O O^{\\prime}}=\\frac{T O}{O O^{\\prime}}\n$$\n\nIt means that $T O=O O^{\\prime}$, so $O^{\\prime} A^{\\prime 2}=O^{\\prime} M \\cdot O^{\\prime} T=O O^{\\prime 2}$. Hence $O^{\\prime} A^{\\prime}=O O^{\\prime}$ and, consequently, $O \\in \\omega^{\\prime}$. Finally we conclude that $2 \\angle C A B=\\angle B O C=180^{\\circ}-\\angle C A B$, so $\\angle C A B=60^{\\circ}$.\n<img_4029>",
"Let $O^{\\prime}$ and $G^{\\prime}$ denote the reflection of $O$ and $G$, respectively, with respect to the line $B C$. We then need to show $\\angle C A B=60^{\\circ}$ iff $G^{\\prime} H^{\\prime}=G^{\\prime} P$. Note that $\\triangle H^{\\prime} O P$ is isosceles and hence\n\n\n\n$G^{\\prime} H^{\\prime}=G^{\\prime} P$ is equivalent to $G^{\\prime}$ lying on the bisector $\\angle H^{\\prime} O P$. Let $\\angle H^{\\prime} A P=\\varepsilon$. By the assumption $A B \\neq A C$, we have $\\varepsilon \\neq 0$. Then $\\angle H^{\\prime} O P=2 \\angle H^{\\prime} A P=2 \\varepsilon$, hence $G^{\\prime} H^{\\prime}=G^{\\prime} P$ iff $\\angle G^{\\prime} O H^{\\prime}=\\varepsilon$. But $\\angle G O^{\\prime} H=\\angle G^{\\prime} O H^{\\prime}$. Let $D$ be the midpoint of $O O^{\\prime}$. It is known that $\\angle G D O=\\angle G A H=\\varepsilon$. Let $F$ be the midpoint of $H G$. Then $H G=F O$ (Euler line). Let $\\angle G O^{\\prime} H=\\delta$. We then have to show $\\delta=\\varepsilon$ iff $\\angle C A B=60^{\\circ}$. But by similarity $\\left(\\triangle G D O \\sim \\triangle F O^{\\prime} O\\right.$ ) we have $\\angle F O^{\\prime} O=\\varepsilon$. Consider the circumcircles of the triangles $F O^{\\prime} O$ and $G O^{\\prime} H$. By the sine law and since the segments $H G$ and $F O$ are of equal length we deduce that the circumcircles of the triangles $F O^{\\prime} O$ and $G O^{\\prime} H$ are symmetric with respect to the perpendicular bisector of the segment $F G$ iff $\\delta=\\varepsilon$. Obviously, $O^{\\prime}$ is the common point of these two circles. Hence $O^{\\prime}$ must be fixed after the symmetry about the perpendicular bisector of the segment $F G$ iff $\\delta=\\varepsilon$ so we have $\\varepsilon=\\delta$ iff $\\triangle H O O^{\\prime}$ is isosceles. But $H O^{\\prime}=H^{\\prime} O=R$, and so\n\n$$\n\\varepsilon=\\delta \\Longleftrightarrow O O^{\\prime}=R \\Longleftrightarrow O D=\\frac{R}{2} \\Longleftrightarrow \\cos \\angle C A B=\\frac{1}{2} \\Longleftrightarrow \\angle C A B=60^{\\circ} \\text {. }\n$$"
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Geometry
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English
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2,270
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In the diagram, two circles are tangent to each other at point $B$. A straight line is drawn through $B$ cutting the two circles at $A$ and $C$, as shown. Tangent lines are drawn to the circles at $A$ and $C$. Prove that these two tangent lines are parallel.
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"Let the centres of the two circles be $O_{1}$ and $O_{2}$.\n\nJoin $A$ and $B$ to $O_{1}$ and $B$ and $C$ to $O_{2}$.\n\nDesignate two points $W$ and $X$ on either side of $A$ on one tangent line, and two points $Y$ and $Z$ on either side of $C$ on the other tangent line.\n\n<img_3553>\n\nLet $\\angle X A B=\\theta$.\n\nSince $W X$ is tangent to the circle with centre $O_{1}$ at $A$, then $O_{1} A$ is perpendicular to $W X$, so $\\angle O_{1} A B=90^{\\circ}-\\theta$.\n\nSince $O_{1} A=O_{1} B$ because both are radii, then $\\triangle A O_{1} B$ is isosceles, so $\\angle O_{1} B A=$ $\\angle O_{1} A B=90^{\\circ}-\\theta$.\n\nSince the two circles are tangent at $B$, then the line segment joining $O_{1}$ and $O_{2}$ passes through $B$, ie. $O_{1} B O_{2}$ is a straight line segment.\n\nThus, $\\angle O_{2} B C=\\angle O_{1} B A=90^{\\circ}-\\theta$, by opposite angles.\n\nSince $O_{2} B=O_{2} C$, then similarly to above, $\\angle O_{2} C B=\\angle O_{2} B C=90^{\\circ}-\\theta$.\n\nSince $Y Z$ is tangent to the circle with centre $O_{2}$ at $C$, then $O_{2} C$ is perpendicular to $Y Z$. Thus, $\\angle Y C B=90^{\\circ}-\\angle O_{2} C B=\\theta$.\n\nSince $\\angle X A B=\\angle Y C B$, then $W X$ is parallel to $Y Z$, by alternate angles, as required.",
"Let the centres of the two circles be $O_{1}$ and $O_{2}$.\n\nJoin $A$ and $B$ to $O_{1}$ and $B$ and $C$ to $O_{2}$.\n\nSince $A O_{1}$ and $B O_{1}$ are radii of the same circle, $A O_{1}=B O_{1}$ so $\\triangle A O_{1} B$ is isosceles, so $\\angle O_{1} A B=\\angle O_{1} B A$.\n\n<img_3245>\n\nSince $\\mathrm{BO}_{2}$ and $\\mathrm{CO}_{2}$ are radii of the same circle, $B O_{2}=C_{2}$ so $\\triangle B O_{2} C$ is isosceles, so $\\angle O_{2} B C=\\angle O_{2} C B$.\n\nSince the two circles are tangent at $B$, then $O_{1} B O_{2}$ is a line segment (ie. the line segment joining $O_{1}$ and $O_{2}$ passes through the point of tangency of the two circles).\n\nSince $O_{1} B O_{2}$ is straight, then $\\angle O_{1} B A=\\angle O_{2} B C$, by opposite angles.\n\nThus, $\\angle O_{1} A B=\\angle O_{1} B A=\\angle O_{2} B C=\\angle O_{2} C B$.\n\nThis tells us that $\\triangle A O_{1} B$ is similar to $\\triangle B O_{2} C$, so $\\angle A O_{1} B=\\angle B O_{2} C$ or $\\angle A O_{1} O_{2}=$ $\\angle C O_{2} O_{1}$.\n\nTherefore, $A O_{1}$ is parallel to $C_{2}$, by alternate angles.\n\nBut $A$ and $C$ are points of tangency, $A O_{1}$ is perpendicular to the tangent line at $A$ and $\\mathrm{CO}_{2}$ is perpendicular to the tangent line at $C$.\n\nSince $A O_{1}$ and $C O_{2}$ are parallel, then the two tangent lines must be parallel."
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Theorem proof
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Geometry
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2,274
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A school has a row of $n$ open lockers, numbered 1 through $n$. After arriving at school one day, Josephine starts at the beginning of the row and closes every second locker until reaching the end of the row, as shown in the example below. Then on her way back, she closes every second locker that is still open. She continues in this manner along the row, until only one locker remains open. Define $f(n)$ to be the number of the last open locker. For example, if there are 15 lockers, then $f(15)=11$ as shown below:
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Prove that there is no positive integer $n$ such that $f(n)=2005$ and there are infinitely many positive integers $n$ such that $f(n)=f(2005)$.
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"First, we calculate $f(n)$ for $n$ from 1 to 32 , to get a feeling for what happens. We obtain $1,1,3,3,1,1,3,3,9,9,11,11,9,9,11,11,1,1,3,3,1,1,3,3,9,9,11,11,9,9,11,11$. This will help us to establish some patterns.\n\nNext, we establish two recursive formulas for $f(n)$.\n\nFirst, from our pattern, it looks like $f(2 m)=f(2 m-1)$.\n\nWhy is this true in general?\n\nConsider a row of $2 m$ lockers.\n\nOn the first pass through, Josephine shuts all of the even numbered lockers, leaving open lockers $1,3, \\ldots, 2 m-1$.\n\nThese are exactly the same open lockers as if she had started with $2 m-1$ lockers in total. Thus, as she starts her second pass from right to left, the process will be the same now whether she started with $2 m$ lockers or $2 m-1$ lockers.\n\nTherefore, $f(2 m)=f(2 m-1)$.\n\nThis tells us that we need only focus on the values of $f(n)$ where $n$ is odd.\n\nSecondly, we show that $f(2 m-1)=2 m+1-2 f(m)$.\n\n(It is helpful to connect $n=2 m-1$ to a smaller case.)\n\nWhy is this formula true?\n\nStarting with $2 m-1$ lockers, the lockers left open after the first pass are $1,3, \\ldots, 2 m-1$, ie. $m$ lockers in total.\n\nSuppose $f(m)=p$. As Josephine begins her second pass, which is from right to left, we can think of this as being like the first pass through a row of $m$ lockers.\n\nThus, the last open locker will be the $p$ th locker, counting from the right hand end, from the list $1,3, \\ldots, 2 m-1$.\n\nThe first locker from the right is $2 m-1=2 m+1-2(1)$, the second is $2 m-3=2 m+1-2(2)$, and so on, so the $p$ th locker is $2 m+1-2 p$.\n\nTherefore, the final open locker is $2 m+1-2 p$, ie. $f(2 m-1)=2 m+1-2 p=2 m+1-2 f(m)$.\n\nUsing these two formulae repeatedly,\n\n$$\n\\begin{aligned}\nf(4 k+1) & =f(2(2 k+1)-1) \\\\\n& =2(2 k+1)+1-2 f(2 k+1) \\\\\n& =4 k+3-2 f(2(k+1)-1) \\\\\n& =4 k+3-2(2(k+1)+1-2 f(k+1)) \\\\\n& =4 k+3-2(2 k+3-2 f(k+1)) \\\\\n& =4 f(k+1)-3\n\\end{aligned}\n$$\n\n\n\nand\n\n$$\n\\begin{aligned}\nf(4 k+3) & =f(2(2 k+2)-1) \\\\\n& =2(2 k+2)+1-2 f(2 k+2) \\\\\n& =4 k+5-2 f(2 k+1) \\\\\n& =4 k+5-2 f(2(k+1)-1) \\\\\n& =4 k+5-2(2(k+1)+1-2 f(k+1)) \\\\\n& =4 k+5-2(2 k+3-2 f(k+1)) \\\\\n& =4 f(k+1)-1\n\\end{aligned}\n$$\n\nFrom our initial list of values of $f(n)$, it appears as if $f(n)$ cannot leave a remainder of 5 or 7 when divided by 8 . So we use these recursive relations once more to try to establish this:\n\n$$\n\\begin{aligned}\nf(8 l+1) & =4 f(2 l+1)-3 \\quad(\\text { since } 8 l+1=4(2 l)+1) \\\\\n& =4(2 l+3-2 f(l+1))-3 \\\\\n& =8 l+9-8 f(l+1) \\\\\n& =8(l-f(l+1))+9 \\\\\nf(8 l+3) & =4 f(2 l+1)-1 \\quad(\\text { since } 8 l+3=4(2 l)+3) \\\\\n& =4(2 l+3-2 f(l+1))-1 \\\\\n& =8 l+11-8 f(l+1) \\\\\n& =8(l-f(l+1))+11\n\\end{aligned}\n$$\n\nSimilarly, $f(8 l+5)=8 l+9-8 f(l+1)$ and $f(8 l+7)=8 l+11-8 f(l+1)$.\n\nTherefore, since any odd positive integer $n$ can be written as $8 l+1,8 l+3,8 l+5$ or $8 l+7$, then for any odd positive integer $n, f(n)$ is either 9 more or 11 more than a multiple of 8 . Therefore, for any odd positive integer $n, f(n)$ cannot be 2005, since 2005 is not 9 more or 11 more than a multiple of 8 .\n\nThus, for every positive integer $n, f(n) \\neq 2005$, since we only need to consider odd values of $n$.\n\nNext, we show that there are infinitely many positive integers $n$ such that $f(n)=f(2005)$. We do this by looking at the pattern we initially created and conjecturing that\n\n$$\nf(2005)=f\\left(2005+2^{2 a}\\right)\n$$\n\nif $2^{2 a}>2005$. (We might guess this by looking at the connection between $f(1)$ and $f(3)$ with $f(5)$ and $f(7)$ and then $f(1)$ through $f(15)$ with $f(17)$ through $f(31)$. In fact, it appears to be true that $f\\left(m+2^{2 a}\\right)=f(m)$ if $2^{2 a}>m$.)\n\n\n\nUsing our formulae from above,\n\n$$\n\\begin{aligned}\n& f\\left(2005+2^{2 a}\\right)=4 f\\left(502+2^{2 a-2}\\right)-3 \\quad\\left(2005+2^{2 a}=4\\left(501+2^{2 a-2}\\right)+1\\right) \\\\\n& =4 f\\left(501+2^{2 a-2}\\right)-3 \\\\\n& =4\\left(4 f\\left(126+2^{2 a-4}\\right)-3\\right)-3 \\quad\\left(501+2^{2 a-2}=4\\left(125+2^{2 a-4}\\right)+1\\right) \\\\\n& =16 f\\left(126+2^{2 a-4}\\right)-15 \\\\\n& =16 f\\left(125+2^{2 a-4}\\right)-15 \\\\\n& =16\\left(4 f\\left(32+2^{2 a-6}\\right)-3\\right)-15 \\\\\n& =64 f\\left(32+2^{2 a-6}\\right)-63 \\\\\n& =64 f\\left(31+2^{2 a-6}\\right)-63 \\\\\n& =64\\left(4 f\\left(8+2^{2 a-8}\\right)-1\\right)-63 \\\\\n& =256 f\\left(8+2^{2 a-8}\\right)-127 \\\\\n& =256 f\\left(7+2^{2 a-8}\\right)-127 \\\\\n& =256\\left(4 f\\left(2+2^{2 a-10}\\right)-1\\right)-127 \\quad\\left(7+2^{2 a-8}=4\\left(1+2^{2 a-10}\\right)+3\\right) \\\\\n& =1024 f\\left(2+2^{2 a-10}\\right)-383 \\\\\n& =1024 f\\left(1+2^{2 a-10}\\right)-383 \\\\\n& \\left(125+2^{2 a-4}=4\\left(31+2^{2 a-6}\\right)+1\\right) \\\\\n& \\left(31+2^{2 a-6}=4\\left(7+2^{2 a-8}\\right)+3\\right)\n\\end{aligned}\n$$\n\n(Notice that we could have removed the powers of 2 from inside the functions and used this same approach to show that $f(2005)=1024 f(1)-383=641$.)\n\nBut, $f\\left(2^{2 b}+1\\right)=1$ for every positive integer $b$.\n\nWhy is this true? We can prove this quickly by induction.\n\nFor $b=1$, we know $f(5)=1$.\n\nAssume that the result is true for $b=B-1$, for some positive integer $B \\geq 2$.\n\nThen $f\\left(2^{2 B}+1\\right)=f\\left(4\\left(2^{2 B-2}\\right)+1\\right)=4 f\\left(2^{2 B-2}+1\\right)-3=4(1)-3=1$ by our induction hypothesis.\n\nTherefore, if $a \\geq 6$, then $f\\left(1+2^{2 a-10}\\right)=f\\left(1+2^{2(a-5)}\\right)=1$ so\n\n$$\nf\\left(2005+2^{2 a}\\right)=1024(1)-383=641=f(2005)\n$$\n\nso there are infinitely many integers $n$ for which $f(n)=f(2005)$."
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Combinatorics
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2,348
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In the diagram, $C$ lies on $B D$. Also, $\triangle A B C$ and $\triangle E C D$ are equilateral triangles. If $M$ is the midpoint of $B E$ and $N$ is the midpoint of $A D$, prove that $\triangle M N C$ is equilateral.
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"Consider $\\triangle B C E$ and $\\triangle A C D$.\n\n<img_3765>\n\nSince $\\triangle A B C$ is equilateral, then $B C=A C$.\n\nSince $\\triangle E C D$ is equilateral, then $C E=C D$.\n\nSince $B C D$ is a straight line and $\\angle E C D=60^{\\circ}$, then $\\angle B C E=180^{\\circ}-\\angle E C D=120^{\\circ}$.\n\nSince $B C D$ is a straight line and $\\angle B C A=60^{\\circ}$, then $\\angle A C D=180^{\\circ}-\\angle B C A=120^{\\circ}$.\n\nTherefore, $\\triangle B C E$ is congruent to $\\triangle A C D$ (\"side-angle-side\").\n\nSince $\\triangle B C E$ and $\\triangle A C D$ are congruent and $C M$ and $C N$ are line segments drawn from the corresponding vertex ( $C$ in both triangles) to the midpoint of the opposite side, then $C M=C N$.\n\nSince $\\angle E C D=60^{\\circ}$, then $\\triangle A C D$ can be obtained by rotating $\\triangle B C E$ through an angle of $60^{\\circ}$ clockwise about $C$.\n\nThis means that after this $60^{\\circ}$ rotation, $C M$ coincides with $C N$.\n\nIn other words, $\\angle M C N=60^{\\circ}$.\n\nBut since $C M=C N$ and $\\angle M C N=60^{\\circ}$, then\n\n$$\n\\angle C M N=\\angle C N M=\\frac{1}{2}\\left(180^{\\circ}-\\angle M C N\\right)=60^{\\circ}\n$$\n\nTherefore, $\\triangle M N C$ is equilateral, as required.",
"We prove that $\\triangle M N C$ is equilateral by introducing a coordinate system.\n\nSuppose that $C$ is at the origin $(0,0)$ with $B C D$ along the $x$-axis, with $B$ having coordinates $(-4 b, 0)$ and $D$ having coordinates $(4 d, 0)$ for some real numbers $b, d>0$.\n\nDrop a perpendicular from $E$ to $P$ on $C D$.\n\n<img_3660>\n\nSince $\\triangle E C D$ is equilateral, then $P$ is the midpoint of $C D$.\n\nSince $C$ has coordinates $(0,0)$ and $D$ has coordinates $(4 d, 0)$, then the coordinates of $P$ are $(2 d, 0)$.\n\nSince $\\triangle E C D$ is equilateral, then $\\angle E C D=60^{\\circ}$ and so $\\triangle E P C$ is a $30^{\\circ}-60^{\\circ}-90^{\\circ}$ triangle and so $E P=\\sqrt{3} C P=2 \\sqrt{3} d$.\n\nTherefore, the coordinates of $E$ are $(2 d, 2 \\sqrt{3} d)$.\n\nIn a similar way, we can show that the coordinates of $A$ are $(-2 b, 2 \\sqrt{3} b)$.\n\nNow $M$ is the midpoint of $B(-4 b, 0)$ and $E(2 d, 2 \\sqrt{3} d)$, so the coordinates of $M$ are $\\left(\\frac{1}{2}(-4 b+2 d), \\frac{1}{2}(0+2 \\sqrt{3} d)\\right)$ or $(-2 b+d, \\sqrt{3} d)$.\n\nAlso, $N$ is the midpoint of $A(-2 b, 2 \\sqrt{3} b)$ and $D(4 d, 0)$, so the coordinates of $N$ are $\\left(\\frac{1}{2}(-2 b+4 d), \\frac{1}{2}(2 \\sqrt{3} b+0)\\right)$ or $(-b+2 d, \\sqrt{3} b)$.\n\nTo show that $\\triangle M N C$ is equilateral, we show that $C M=C N=M N$ or equivalently that $C M^{2}=C N^{2}=M N^{2}$ :\n\n$$\n\\begin{aligned}\nC M^{2} & =(-2 b+d-0)^{2}+(\\sqrt{3} d-0)^{2} \\\\\n& =(-2 b+d)^{2}+(\\sqrt{3} d)^{2} \\\\\n& =4 b^{2}-4 b d+d^{2}+3 d^{2} \\\\\n& =4 b^{2}-4 b d+4 d^{2} \\\\\nC N^{2} & =(-b+2 d-0)^{2}+(\\sqrt{3} b-0)^{2} \\\\\n& =(-b+2 d)^{2}+(\\sqrt{3} b)^{2} \\\\\n& =b^{2}-4 b d+4 d^{2}+3 b^{2} \\\\\n& =4 b^{2}-4 b d+4 d^{2} \\\\\nM N^{2} & =((-2 b+d)-(-b+2 d))^{2}+(\\sqrt{3} d-\\sqrt{3} b)^{2} \\\\\n& =(-b-d)^{2}+3(d-b)^{2} \\\\\n& =b^{2}+2 b d+d^{2}+3 d^{2}-6 b d+3 b^{2} \\\\\n& =4 b^{2}-4 b d+4 d^{2}\n\\end{aligned}\n$$\n\nTherefore, $C M^{2}=C N^{2}=M N^{2}$ and so $\\triangle M N C$ is equilateral, as required.\n\n\n\n#"
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Geometry
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2,363
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In parallelogram $A B C D, A B=a$ and $B C=b$, where $a>b$. The points of intersection of the angle bisectors are the vertices of quadrilateral $P Q R S$.
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Prove that $P Q R S$ is a rectangle.
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"In a parallelogram opposite angles are equal.\n\nSince $D F$ and $B E$ bisect the two angles, let $\\angle A D F=\\angle C D F=\\angle A B E=\\angle C B E$\n\n$$\n=x \\text { (in degrees) }\n$$\n\nAlso $\\angle C D F=\\angle A F D=x$ (alternate angles)\n\nLet $\\angle D A M=\\angle B A M=\\angle D C N=\\angle B C N=y$ (in degrees)\n\n\n\nFor any parallelogram, any two consecutive angles add to $180^{\\circ}, \\therefore 2 x+2 y=180$ or, $x+y=90$.\n\n<img_3560>\n\nTherefore in $\\triangle P A F, \\angle A P F=90^{\\circ}$.\n\n<img_3544>\n\nUsing similar reasoning and properties of parallel lines we get right angles at $Q, R$ and $S$.\n\nThus $P Q R S$ is a rectangle."
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2,364
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In parallelogram $A B C D, A B=a$ and $B C=b$, where $a>b$. The points of intersection of the angle bisectors are the vertices of quadrilateral $P Q R S$.
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Prove that $P R=a-b$.
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"Since $A M$ is a bisector of $\\angle D A B$, let $\\angle D A M=\\angle B A M=y$.\n\nAlso, $\\angle D M A=y$ (alternate angles)\n\nThis implies that $\\triangle A D M$ is isosceles.\n\nUsing the same reasoning in $\\triangle C B N$, we see that it is also isosceles and so the diagram may now be labelled as:\n\n<img_3709>\n\n$A N=a-b$\n\nThus $\\triangle A D M$ and $\\triangle C B N$ are identical isosceles triangles.\n\nAlso, $A M \\| N C$ (corresponding angles)\n\nor, $A P \\| N R$.\n\nBy using properties of isosceles triangles (or congruency), $A P=N R$ implying that $A P R N$ is a parallelogram.\n\nThus $A N=P R$ and since $A N=a-b, P R=a-b$ (as required)"
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2,367
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An equilateral triangle $A B C$ has side length 2 . A square, $P Q R S$, is such that $P$ lies on $A B, Q$ lies on $B C$, and $R$ and $S$ lie on $A C$ as shown. The points $P, Q, R$, and $S$ move so that $P, Q$ and $R$ always remain on the sides of the triangle and $S$ moves from $A C$ to $A B$ through the interior of the triangle. If the points $P, Q, R$ and $S$ always form the vertices of a square, show that the path traced out by $S$ is a straight line parallel to $B C$.
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"Let $\\angle R Q C=\\theta$ and from $S$ draw a line perpendicular to the base at $P$.\n\nThen $\\angle T Q B=180-(90+\\theta)=90-\\theta$.\n\nLet $s$ be the length of the side of the square.\n\nFrom $R$ draw a line perpendicular to $B C$ at $D$ and then through $S$ draw a line parallel to $B C$. From $R$ draw a line perpendicular to this line at E.\n\n<img_3178>\n\nFrom $\\triangle R Q D, R D=s \\sin \\theta$.\n\nSince $\\angle Q R D=90-\\theta$ then $\\angle S R E=\\theta$.\n\nFrom $\\triangle S E R, E R=s \\cos \\theta$.\n\nThe perpendicular distance from $S$ to $B C$ is $R D+E R=s \\sin \\theta+s \\cos \\theta$ which we must now show is a constant.\n\nWe can now take each of the lengths $D C, D Q, P F, F B$ and express them in terms of $s$.\n\nFrom $\\triangle R D C$ which is a $30^{\\circ}-60^{\\circ}-90^{\\circ}$ triangle, $\\frac{D C}{R D}=\\frac{1}{\\sqrt{3}}$.\n\nSince $R D=s \\sin \\theta$ (from above)\n\n$$\nD C=\\frac{1}{\\sqrt{3}}(s \\sin \\theta)=\\frac{\\sqrt{3}}{3} s \\sin \\theta\n$$\n\n\n\nFrom $\\Delta R D Q, \\frac{Q D}{R Q}=\\cos \\theta, Q D=s \\cos \\theta$.\n\nFrom $\\triangle T F Q, \\sin \\theta=\\frac{F Q}{s}$ and $\\cos \\theta=\\frac{T F}{s}$. or, $F Q=s \\sin \\theta$ and $T F=s \\sin \\theta$.\n\nFrom $\\triangle T F B, \\frac{B F}{T F}=\\frac{1}{\\sqrt{3}}, B F=\\frac{1}{\\sqrt{3}} T F=\\frac{1}{\\sqrt{3}} s \\cos \\theta=\\frac{\\sqrt{3}}{3} s \\cos \\theta$.\n\nSince $D C+Q D+F Q+B F=2, \\frac{\\sqrt{3}}{3} s \\sin \\theta+s \\cos \\theta+s \\sin \\theta+\\frac{\\sqrt{3}}{3} s \\cos \\theta=2$.\n\n$$\n\\begin{aligned}\n\\frac{\\sqrt{3}}{3}(s \\cos \\theta+s \\sin \\theta)+(s \\cos \\theta+s \\sin \\theta) & =2 \\\\\ns \\cos \\theta+s \\sin \\theta & =\\frac{2}{\\left(\\frac{\\sqrt{3}}{3}+1\\right)}\n\\end{aligned}\n$$\n\nThus $s \\cos \\theta+s \\sin \\theta$ is a constant and the path traced out by $S$ is a straight line parallel to $B C$."
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Theorem proof
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2,398
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In the diagram, line segment $F C G$ passes through vertex $C$ of square $A B C D$, with $F$ lying on $A B$ extended and $G$ lying on $A D$ extended. Prove that $\frac{1}{A B}=\frac{1}{A F}+\frac{1}{A G}$.
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[
"Without loss of generality, suppose that square $A B C D$ has side length 1 .\n\nSuppose next that $B F=a$ and $\\angle C F B=\\theta$.\n\nSince $\\triangle C B F$ is right-angled at $B$, then $\\angle B C F=90^{\\circ}-\\theta$.\n\nSince $G C F$ is a straight line, then $\\angle G C D=180^{\\circ}-90^{\\circ}-\\left(90^{\\circ}-\\theta\\right)=\\theta$.\n\nTherefore, $\\triangle G D C$ is similar to $\\triangle C B F$, since $\\triangle G D C$ is right-angled at $D$.\n\nThus, $\\frac{G D}{D C}=\\frac{B C}{B F}$ or $\\frac{G D}{1}=\\frac{1}{a}$ or $G D=\\frac{1}{a}$.\n\nSo $A F=A B+B F=1+a$ and $A G=A D+D G=1+\\frac{1}{a}=\\frac{a+1}{a}$.\n\nThus, $\\frac{1}{A F}+\\frac{1}{A G}=\\frac{1}{1+a}+\\frac{a}{a+1}=\\frac{a+1}{a+1}=1=\\frac{1}{A B}$, as required.",
"We attach a set of coordinate axes to the diagram, with $A$ at the origin, $A G$ lying along the positive $y$-axis and $A F$ lying along the positive $x$-axis.\n\nWithout loss of generality, suppose that square $A B C D$ has side length 1 , so that $C$ has coordinates $(1,1)$. (We can make this assumption without loss of generality, because if the square had a different side length, then each of the lengths in the problem would be scaled by the same factor.)\n\n\n\nSuppose that the line through $G$ and $F$ has slope $m$.\n\nSince this line passes through $(1,1)$, its equation is $y-1=m(x-1)$ or $y=m x+(1-m)$. The $y$-intercept of this line is $1-m$, so $G$ has coordinates $(0,1-m)$.\n\nThe $x$-intercept of this line is $\\frac{m-1}{m}$, so $F$ has coordinates $\\left(\\frac{m-1}{m}, 0\\right)$. (Note that $m \\neq 0$ as the line cannot be horizontal.)\n\nTherefore,\n\n$$\n\\frac{1}{A F}+\\frac{1}{A G}=\\frac{m}{m-1}+\\frac{1}{1-m}=\\frac{m}{m-1}+\\frac{-1}{m-1}=\\frac{m-1}{m-1}=1=\\frac{1}{A B}\n$$\n\nas required.",
"Join $A$ to $C$.\n\nWe know that the sum of the areas of $\\triangle G C A$ and $\\triangle F C A$ equals the area of $\\triangle G A F$.\n\nThe area of $\\triangle G C A$ (thinking of $A G$ as the base) is $\\frac{1}{2}(A G)(D C)$, since $D C$ is perpendicular to $A G$.\n\nSimilarly, the area of $\\triangle F C A$ is $\\frac{1}{2}(A F)(C B)$.\n\nAlso, the area of $\\triangle G A F$ is $\\frac{1}{2}(A G)(A F)$.\n\nTherefore,\n\n$$\n\\begin{aligned}\n\\frac{1}{2}(A G)(D C)+\\frac{1}{2}(A F)(C B) & =\\frac{1}{2}(A G)(A F) \\\\\n\\frac{(A G)(D C)}{(A G)(A F)(A B)}+\\frac{(A F)(C B)}{(A G)(A F)(A B)} & =\\frac{(A G)(A F)}{(A G)(A F)(A B)} \\\\\n\\frac{1}{A F}+\\frac{1}{A G} & =\\frac{1}{A B}\n\\end{aligned}\n$$\n\nas required, since $A B=D C=C B$."
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2,448
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A circle with its centre on the $y$-axis intersects the graph of $y=|x|$ at the origin, $O$, and exactly two other distinct points, $A$ and $B$, as shown. Prove that the ratio of the area of triangle $A B O$ to the area of the circle is always $1: \pi$.
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"Since both the circle with its centre on the $y$-axis and the graph of $y=|x|$ are symmetric about the $y$-axis, then for each point of intersection between these two graphs, there should be a corresponding point of intersection symmetrically located across the $y$-axis. Thus, since there are exactly three points of intersection, then one of these points must be on the $y$-axis, ie. has $x$-coordinate 0 . Since this point is on the graph of $y=|x|$, then this point must be $(0,0)$.\n\nSince the circle has centre on the $y$-axis (say, has coordinates $(0, b)$ ), then its radius is equal to $b$ (and $b$ must be positive for there to be three points of intersection). So the circle has equation $x^{2}+(y-b)^{2}=b^{2}$. Where are the other two points of intersection? We consider the points with $x$ positive and use symmetry to get the other point of intersection.\n\n\n\nWhen $x \\geq 0$, then $y=|x|$ has equation $y=x$. Substituting into the equation of the circle,\n\n$$\n\\begin{aligned}\nx^{2}+(x-b)^{2} & =b^{2} \\\\\n2 x^{2}-2 b x & =0 \\\\\n2 x(x-b) & =0\n\\end{aligned}\n$$\n\nTherefore, the points of intersection are $(0,0)$ and $(b, b)$ on the positive side of the $y$-axis, and so at the point $(-b, b)$ on the negative side of the $y$ axis.\n\nThus the points $O, A$ and $B$ are the points $(0,0)$, $(b, b)$ and $(-b, b)$.\n\n<img_3620>\n\nSince the radius of the circle is $b$, then the area of the circle is $\\pi b^{2}$.\n\nTriangle $O A B$ has a base from $(-b, b)$ to $(b, b)$ of length $2 b$, and a height from the line $y=b$ to the point $(0,0)$ of length $b$, and so an area of $\\frac{1}{2} b(2 b)=b^{2}$.\n\nTherefore, the ratio of the area of the triangle to the area of the circle is $b^{2}: \\pi b^{2}=1: \\pi$."
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2,449
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In the diagram, triangle $A B C$ has a right angle at $B$ and $M$ is the midpoint of $B C$. A circle is drawn using $B C$ as its diameter. $P$ is the point of intersection of the circle with $A C$. The tangent to the circle at $P$ cuts $A B$ at $Q$. Prove that $Q M$ is parallel to $A C$.
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[
"Since $M$ is the midpoint of a diameter of the circle, $M$ is the centre of the circle.\n\nJoin $P$ to $M$. Since $Q P$ is tangent to the circle, $P M$ is perpendicular to $Q P$.\n\nSince $P M$ and $B M$ are both radii of the circle, then $P M=M B$.\n\n<img_3432>\n\nTherefore, $\\triangle Q P M$ and $\\triangle Q B M$ are congruent (Hypotenuse - Side).\n\nThus, let $\\angle M Q B=\\angle M Q P=\\theta$. So $\\angle Q M B=\\angle Q M P=90^{\\circ}-\\theta$\n\nThen $\\angle P M C=180^{\\circ}-\\angle P M Q-\\angle B M Q=180^{\\circ}-\\left(90^{\\circ}-\\theta\\right)-\\left(90^{\\circ}-\\theta\\right)=2 \\theta$.\n\nBut $\\triangle P M C$ is isosceles with $P M=M C$ since $P M$ and $M C$ are both radii.\n\nTherefore, $\\angle C P M=\\frac{1}{2}\\left(180^{\\circ}-\\angle P M C\\right)=90^{\\circ}-\\theta$.\n\nBut then $\\angle C P M=\\angle P M Q$, and since $P M$ is a transversal between $A C$ and $Q M$, then $Q M$ is parallel to $A C$ because of equal alternating angles.",
"Join $M$ to $P$ and $B$ to $P$.\n\nSince $Q P$ and $Q B$ are tangents to the circle coming from the same point, they have the same length. Since $Q M$ joins the point of intersection of the tangents to the centre of the circle, then by symmetry, $\\angle P Q M=\\angle B Q M$ and $\\angle P M Q=\\angle B M Q$. So let $\\angle P Q M=\\angle B Q M=x$ and $\\angle P M Q=\\angle B M Q=y$.\n\n<img_3687>\n\nLooking at $\\triangle Q M B$, we see that $x+y=90^{\\circ}$, since $\\triangle Q M B$ is right-angled.\n\nNow if we consider the chord $P B$, we see that its central angle is $2 y$, so any angle that it subtends on the circle (eg. $\\angle P C B$ ) is equal to $y$.\n\nThus, $\\angle A C B=\\angle Q M B$, so $Q M$ is parallel to $A C$.",
"Join $P B$.\n\nSince $Q P$ is tangent to the circle, then by the Tangent-Chord Theorem, $\\angle Q P B=\\angle P C B=x$ (ie. the inscribed angle of a chord is equal to the angle between the tangent and chord.\n\n<img_3658>\n\nSince $B C$ is a diameter of the circle, then $\\angle C P B=90^{\\circ}$ and so $\\angle A P B=90^{\\circ}$, whence\n\n$\\angle A P Q=90^{\\circ}-\\angle Q P B=90^{\\circ}-x$.\n\nLooking at $\\triangle A B C$, we see that $\\angle P A Q=90^{\\circ}-x$, so $\\angle P A Q=\\angle A P Q$, and so $A Q=Q P$.\n\nBut $Q P$ and $Q B$ are both tangents to the circle $(Q B$ is tangent since it is perpendicular to a radius), so $Q P=Q B$.\n\nBut then $A Q=Q B$ and $B M=M C$, so $Q$ is the midpoint of $A B$ and $M$ is the midpoint of $B C$. Thus we can conclude that $Q M$ is parallel to $A C$.\n\n(To justify this last statement, we can show very easily that $\\triangle Q B M$ is similar to $\\triangle A B C$, and so show that $\\angle C A B=\\angle M Q B$.)"
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2,462
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A large square $A B C D$ is drawn, with a second smaller square $P Q R S$ completely inside it so that the squares do not touch. Line segments $A P, B Q, C R$, and $D S$ are drawn, dividing the region between the squares into four nonoverlapping convex quadrilaterals, as shown. If the sides of $P Q R S$ are not parallel to the sides of $A B C D$, prove that the sum of the areas of quadrilaterals $A P S D$ and $B C R Q$ equals the sum of the areas of quadrilaterals $A B Q P$ and $C D S R$. (Note: A convex quadrilateral is a quadrilateral in which the measure of each of the four interior angles is less than $180^{\circ}$.)
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"We begin by \"boxing in\" square $P Q R S$ by drawing horizontal and vertical lines through its vertices to form rectangle $W X Y Z$, as shown. (Because the four quadrilaterals $A B Q P$, $B C R Q, C D S R$, and $D A P S$ are convex, there will not be any configurations that look substantially different from this the diagram below.) We also label the various areas.\n\n<img_3366>\n\nSince $W X$ is parallel to $A B$, then quadrilateral $A B X W$ is a trapezoid. Similarly, quadrilaterals $B C Y X, C D Z Y$, and $D A W Z$ are trapezoids.\n\nWe use the notation $|A B Q P|$ to denote the area of quadrilateral $A B Q P$, and similar notation for other areas.\n\nSuppose that the side length of square $A B C D$ is $x$ and the side length of square $P Q R S$ is $y$.\n\nAlso, we let $\\angle W P Q=\\theta$.\n\nSince each of $\\triangle W P Q, \\triangle X Q R, \\triangle Y R S$, and $\\triangle Z S P$ is right-angled and each of the four angles of square $P Q R S$ is $90^{\\circ}$, then $\\angle W P Q=\\angle X Q R=\\angle Y R S=\\angle Z S P=\\theta$. This is because, for example,\n\n$\\angle X Q R=180^{\\circ}-\\angle P Q R-\\angle W Q P=90^{\\circ}-\\left(180^{\\circ}-\\angle P W Q-\\angle W P Q\\right)=90^{\\circ}-\\left(90^{\\circ}-\\theta\\right)=\\theta$\n\nThis fact, together with the fact that $P Q=Q R=R S=S P=y$, allows us to conclude that the four triangles $\\triangle W P Q, \\triangle X Q R, \\triangle Y R S$, and $\\triangle Z S P$ are congruent.\n\nIn particular, this tells us\n\n* the four areas labelled $e, f, g$ and $h$ are equal (that is, $e=f=g=h$ ),\n* $P Z=Q W=R X=S Y=y \\sin \\theta$, and\n\n$* W P=X Q=Y R=Z S=y \\cos \\theta$.\n\n\n\nCombining these last two facts tells us that $W Z=X W=Y X=Z Y$, since, for example, $W Z=W P+P Z=Z S+S Y=Z Y$. In other words, $W X Y Z$ is a square, with side length $z$, say.\n\nNext, we show that $(a+r)+(c+n)$ is equal to $(b+m)+(d+s)$.\n\nNote that the sum of these two quantities is the total area between square $A B C D$ and square $W X Y Z$, so equals $x^{2}-z^{2}$.\n\nThus, to show that the quantities are equal, it is enough to show that $(a+r)+(c+n)$ equals $\\frac{1}{2}\\left(x^{2}-z^{2}\\right)$.\n\nLet the height of trapezoid $A B X W$ be $k$ and the height of trapezoid $Z Y C D$ be $l$.\n\nThen $|A B X W|=a+r=\\frac{1}{2} k(A B+W X)=\\frac{1}{2} k(x+z)$.\n\nAlso, $|Z Y C D|=c+n=\\frac{1}{2} l(D C+Z Y)=\\frac{1}{2} l(x+z)$.\n\nSince $A B, W X, Z Y$, and $D C$ are parallel, then the sum of the heights of trapezoid $A B X W$, square $W X Y Z$, and trapezoid $Z Y C D$ equals the height of square $A B C D$, so $k+z+l=x$, or $k+l=x-z$.\n\nTherefore,\n\n$$\n(a+r)+(c+n)=\\frac{1}{2} k(x+z)+\\frac{1}{2} l(x+z)=\\frac{1}{2}(x+z)(k+l)=\\frac{1}{2}(x+z)(x-z)=\\frac{1}{2}\\left(x^{2}-z^{2}\\right)\n$$\n\nas required.\n\nTherefore, $(a+r)+(c+n)=(b+m)+(d+s)$. We label this equation $(*)$.\n\nNext, we show that $r+n=m+s$.\n\nNote that $r=|\\triangle Q X B|$. This triangle can be viewed as having base $Q X$ and height equal to the height of trapezoid $A B X W$, or $k$.\n\nThus, $r=\\frac{1}{2}(y \\cos \\theta) k$.\n\nNote that $n=|\\triangle S Z D|$. This triangle can be viewed as having base $S Z$ and height equal to the height of trapezoid $Z Y C D$, or $l$.\n\nThus, $n=\\frac{1}{2}(y \\cos \\theta) l$.\n\nCombining these facts, we obtain\n\n$$\nn+r=\\frac{1}{2}(y \\cos \\theta) k+\\frac{1}{2}(y \\cos \\theta) l=\\frac{1}{2} y \\cos \\theta(k+l)=\\frac{1}{2} y \\cos \\theta(x-z)\n$$\n\nWe note that this sum depends only on the side lengths of the squares and the angle of rotation of the inner square, so is independent of the position of the inner square within the outer square.\n\nThis means that we can repeat this analysis to obtain the same expression for $m+s$.\n\nTherefore, $n+r=m+s$. We label this equation $(* *)$.\n\nWe subtract $(*)-(* *)$ to obtain $a+c=b+d$.\n\nFinally, we can combine all of this information:\n\n$$\n\\begin{aligned}\n& (|A B Q P|+|C D S R|)-(|B C R Q|+|A P S D|) \\\\\n& \\quad=(a+e+s+c+g+m)-(b+f+r+d+h+n) \\\\\n& \\quad=((a+c)-(b+d))+((m+s)-(n+r))+((e+g)-(f+h)) \\\\\n& \\quad=0+0+0\n\\end{aligned}\n$$\n\nsince $a+c=b+d$ and $n+r=m+s$ and $e=f=g=h$.\n\nTherefore, $|A B Q P|+|C D S R|=|B C R Q|+|A P S D|$, as required."
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2,474
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In triangle $A B C, \angle A B C=90^{\circ}$. Rectangle $D E F G$ is inscribed in $\triangle A B C$, as shown. Squares $J K G H$ and $M L F N$ are inscribed in $\triangle A G D$ and $\triangle C F E$, respectively. If the side length of $J H G K$ is $v$, the side length of $M L F N$ is $w$, and $D G=u$, prove that $u=v+w$.
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"Let $\\angle B A C=\\theta$. Then by parallel lines, $\\angle D J H=\\angle B D E=\\theta$.\n\nThus, $\\angle B E D=90^{\\circ}-\\theta$ and so\n\n$\\angle N E M=\\theta$ since\n\n$\\angle D E F=90^{\\circ}$.\n\nSince $D G=u$ and $H G=v$,\n\nthen $D H=u-v$.\n\n<img_3789>\n\nSimilarly, $E N=u-w$.\n\nLooking at $\\triangle D H J$ and $\\triangle M N E$, we see that $\\tan \\theta=\\frac{u-v}{v}$ and $\\tan \\theta=\\frac{w}{u-w}$.\n\nTherefore,\n\n$$\n\\begin{aligned}\n\\frac{u-v}{v} & =\\frac{w}{u-w} \\\\\n(u-v)(u-w) & =v w \\\\\nu^{2}-u v-u w+v w & =v w \\\\\nu(u-v-w) & =0\n\\end{aligned}\n$$\n\nand since $u \\neq 0$, we must have $u-v-w=0$ or $u=v+w$.\n\n[Note: If $u=0$, then the height of rectangle $D E F G$ is 0 , ie. $D$ coincides with point $A$ and $E$ coincides with point $C$, which says that we must also have $v=w=0$, ie. the squares have no place to go!]"
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2,488
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In the diagram, quadrilateral $A B C D$ has points $M$ and $N$ on $A B$ and $D C$, respectively, with $\frac{A M}{A B}=\frac{N C}{D C}$. Line segments $A N$ and $D M$ intersect at $P$, while $B N$ and $C M$ intersect at $Q$. Prove that the area of quadrilateral $P M Q N$ equals the sum of the areas of $\triangle A P D$ and $\triangle B Q C$.
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"We use the notation $|P M Q N|$ to represent the area of quadrilateral $|P M Q N|,|\\triangle A P D|$ to represent the area of $\\triangle A P D$, and so on.\n\nWe want to show that $|P M Q N|=|\\triangle A P D|+|\\triangle B Q C|$.\n\nThis is equivalent to showing\n\n$$\n|P M Q N|+|\\triangle D P N|+|\\triangle C Q N|=|\\triangle A P D|+|\\triangle D P N|+|\\triangle B Q C|+|\\triangle C Q N|\n$$\n\nwhich is equivalent to showing\n\n$$\n|\\triangle D M C|=|\\triangle D A N|+|\\triangle C B N|\n$$\n\nsince combining quadrilateral $P M Q N$ with $\\triangle D P N$ and $\\triangle C Q N$ gives $\\triangle D M C$, combining $\\triangle A P D$ with $\\triangle D P N$ gives $\\triangle D A N$, and combining $\\triangle B Q C$ with $\\triangle C Q N$ gives $\\triangle C B N$. Suppose that $D C$ has length $x$ and $D N$ has length $t x$ for some $t$ with $0<t<1$.\n\nThen $N C=D C-D N=x-t x=(1-t) x$.\n\nSuppose also that the height of $A$ above $D C$ is $a$, the height of $B$ above $D C$ is $b$ and the height of $M$ above $D C$ is $m$.\n\n<img_3814>\n\nFigure 1\n\nThen $|\\triangle D A N|=\\frac{1}{2}(t x)(a)$ and $|\\triangle C B N|=\\frac{1}{2}((1-t) x) b$ so\n\n$$\n|\\triangle D A N|+|\\triangle C B N|=\\frac{1}{2}(t x a+(1-t) x b)=\\frac{1}{2} x(t a+(1-t) b)\n$$\n\nAlso, $|\\triangle D M C|=\\frac{1}{2} x m$.\n\nIn order to prove that $|\\triangle D M C|=|\\triangle D A N|+|\\triangle C B N|$, we need to show that $\\frac{1}{2} x m$ equals\n\n\n\n$\\frac{1}{2} x(t a+(1-t) b)$ which is equivalent to showing that $m$ is equal to $t a+(1-t) b$.\n\nIn Figure 2, we draw a horizontal line from $A$ to $B G$, meeting $M F$ at $R$ and $B G$ at $S$.\n\nSince $M F$ and $B G$ are vertical and $A R S$ is horizontal, then these line segments are perpendicular.\n\nSince $A E=a, M F=m$ and $B G=b$, then $M R=m-a$ and $B S=b-a$.\n\n<img_3348>\n\nFigure 2\n\nNow $\\triangle A R M$ is similar to $\\triangle A S B$, since each is right-angled and they share a common angle at $A$.\n\nTherefore, $\\frac{M R}{B S}=\\frac{A M}{A B}=\\frac{N C}{D C}$.\n\nSince $M R=m-a$ and $B S=b-a$, then $\\frac{M R}{B S}=\\frac{m-a}{b-a}$.\n\nSince $\\frac{A M}{A B}=\\frac{N C}{D C}$, then $\\frac{M R}{B S}=\\frac{(1-t) x}{x}=1-t$.\n\nComparing these two expressions, we obtain $\\frac{m-a}{b-a}=(1-t)$ or $m-a=(b-a)(1-t)$ or $m=a+b(1-t)+(t-1) a=t a+(1-t) b$, as required.\n\nThis concludes the proof, and so $|P M Q N|=|\\triangle A P D|+|\\triangle B Q C|$, as required.",
"Let $A M=x$ and $M B=y$. Then $A B=x+y$ and so $\\frac{A M}{A B}=\\frac{x}{x+y}$.\n\nLet $N C=n x$ for some real number $n$.\n\nSince $\\frac{N C}{D C}=\\frac{A M}{A B}$, then $\\frac{n x}{D C}=\\frac{x}{x+y}$ and so $D C=n(x+y)$.\n\nThis tells us that $D N=D C-N C=n(x+y)-n x=n y$.\n\nJoin $M$ to $N$ and label the areas as shown in the diagram:\n\n<img_3372>\n\nWe repeatedly use the fact that triangles with a common height have areas in proportion to the lengths of their bases.\n\nFor example, $\\triangle M D N$ and $\\triangle M N C$ have a common height from line segment to $D C$ to $M$ and so the ratio of their areas equals the ratio of the lengths of their bases.\n\nIn other words, $\\frac{w+r}{u+v}=\\frac{n x}{n y}=\\frac{x}{y}$. Thus, $w+r=\\frac{x}{y}(u+v)$.\n\n\n\nAlso, the ratio of the area of $\\triangle N A M$ to the area of $\\triangle N M B$ equals the ratio of $A M$ to $M B$.\n\nThis gives $\\frac{k+v}{s+w}=\\frac{x}{y}$ or $k+v=\\frac{x}{y}(s+w)$.\n\nNext, we join $A$ to $C$ and relabel the areas divided by this new line segment as shown:\n\n<img_3488>\n\n(The unlabelled triangle adjacent to the one labelled $k_{1}$ has area $k_{2}$ and the unlabelled triangle adjacent to the one labelled $r_{2}$ has area $r_{1}$.)\n\nConsider $\\triangle A N C$ and $\\triangle A D N$.\n\nAs above, the ratio of their areas equals the ratio of their bases.\n\nThus, $\\frac{k_{2}+v_{2}+w_{2}+r_{2}}{z+u}=\\frac{n x}{n y}=\\frac{x}{y}$, and so $k_{2}+v_{2}+w_{2}+r_{2}=\\frac{x}{y}(z+u)$.\n\nConsider $\\triangle C A M$ and $\\triangle C M B$.\n\nAs above, the ratio of their areas equals the ratio of their bases.\n\nThus, $\\frac{k_{1}+v_{1}+w_{1}+r_{1}}{s+t}=\\frac{x}{y}$, and so $k_{1}+v_{1}+w_{1}+r_{1}=\\frac{x}{y}(s+t)$.\n\nAdding $k_{2}+v_{2}+w_{2}+r_{2}=\\frac{x}{y}(z+u)$ and $k_{1}+v_{1}+w_{1}+r_{1}=\\frac{x}{y}(s+t)$ gives\n\n$$\n\\left(k_{1}+k_{2}\\right)+\\left(v_{1}+v_{2}\\right)+\\left(w_{1}+w_{2}\\right)+\\left(r_{1}+r_{2}\\right)=\\frac{x}{y}(s+t+z+u)\n$$\n\nor\n\n$$\nk+v+w+r=\\frac{x}{y}(s+t+z+u)\n$$\n\nSince $k+v=\\frac{x}{y}(s+w)$ and $w+r=\\frac{x}{y}(u+v)$, then\n\n$$\n\\frac{x}{y}(s+w)+\\frac{x}{y}(u+v)=\\frac{x}{y}(s+t+z+u)\n$$\n\nwhich gives\n\n$$\ns+w+u+v=s+t+z+u\n$$\n\nor\n\n$$\nw+v=t+z\n$$\n\nBut $w+v$ is the area of quadrilateral $P M Q N, z$ is the area of $\\triangle A P D$ and $t$ is the area of $\\triangle B Q C$. In other words, the area of quadrilateral $P M Q N$ equals the sum of the areas of $\\triangle A P D$ and $\\triangle P Q C$, as required."
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2,499
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In the diagram, $A B$ and $B C$ are chords of the circle with $A B<B C$. If $D$ is the point on the circle such that $A D$ is perpendicular to $B C$ and $E$ is the point on the circle such that $D E$ is parallel to $B C$, carefully prove, explaining all steps, that $\angle E A C+\angle A B C=90^{\circ}$.
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"Join $A$ to $E$ and $C$, and $B$ to $E$.\n\n<img_3770>\n\nSince $D E$ is parallel to $B C$ and $A D$ is perpendicular to $B C$, then $A D$ is perpendicular to $D E$, ie. $\\angle A D E=90^{\\circ}$.\n\nTherefore, $A E$ is a diameter.\n\nNow $\\angle E A C=\\angle E B C$ since both are subtended by $E C$.\n\nTherefore, $\\angle E A C+\\angle A B C=\\angle E B C+\\angle A B C=\\angle E B A$ which is indeed equal to $90^{\\circ}$ as required, since $A E$ is a diameter.",
"Join $A$ to $E$ and $C$.\n\n<img_3932>\n\nSince $D E$ is parallel to $B C$ and $A D$ is perpendicular to $B C$, then $A D$ is perpendicular to $D E$, ie. $\\angle A D E=90^{\\circ}$.\n\nTherefore, $A E$ is a diameter.\n\nThus, $\\angle E C A=90^{\\circ}$.\n\nNow $\\angle A B C=\\angle A E C$ since both are subtended by $A C$.\n\nNow $\\angle E A C+\\angle A B C=\\angle E A C+\\angle A E C=180^{\\circ}-\\angle E C A$ using the sum of the angles in $\\triangle A E C$.\n\nBut $\\angle E C A=90^{\\circ}$, so $\\angle E A C+\\angle A E C=90^{\\circ}$.",
"Join $A$ to $E$ and $C$, and $C$ to $D$.\n\n<img_3930>\n\nSince $D E$ is parallel to $B C$ and $A D$ is perpendicular to $B C$, then $A D$ is perpendicular to $D E$, ie. $\\angle A D E=90^{\\circ}$.\n\nTherefore, $A E$ is a diameter.\n\nNow $\\angle A B C=\\angle A D C$ since both are subtended by $A C$.\n\nAlso $\\angle E A C=\\angle E D C$ since both are subtended by $E C$.\n\nSo $\\angle E A C+\\angle A B C=\\angle E D C+\\angle A D C=\\angle A D E=90^{\\circ}$."
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2,516
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Suppose that $m$ and $n$ are positive integers with $m \geq 2$. The $(m, n)$-sawtooth sequence is a sequence of consecutive integers that starts with 1 and has $n$ teeth, where each tooth starts with 2, goes up to $m$ and back down to 1 . For example, the $(3,4)$-sawtooth sequence is
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The $(3,4)$-sawtooth sequence includes 17 terms and the average of these terms is $\frac{33}{17}$.
Prove that, for all pairs of positive integers $(m, n)$ with $m \geq 2$, the average of the terms in the $(m, n)$-sawtooth sequence is not an integer.
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"In an $(m, n)$-sawtooth sequence, the sum of the terms is $n\\left(m^{2}-1\\right)+1$.\n\nIn each tooth, there are $(m-1)+(m-1)=2 m-2$ terms (from 2 to $m$, inclusive, and from $m-1$ to 1 , inclusive).\n\nThis means that there are $n(2 m-2)+1$ terms in the sequence.\n\nThus, the average of the terms in the sequence is $\\frac{n\\left(m^{2}-1\\right)+1}{n(2 m-2)+1}$.\n\nWe need to prove that this is not an integer for all pairs of positive integers $(m, n)$ with $m \\geq 2$.\n\nSuppose that $\\frac{n\\left(m^{2}-1\\right)+1}{n(2 m-2)+1}=k$ for some integer $k$. We will show, by contradiction, that this is not possible.\n\nSince $\\frac{n\\left(m^{2}-1\\right)+1}{n(2 m-2)+1}=k$, then\n\n$$\n\\begin{aligned}\n\\frac{m^{2} n-n+1}{2 m n-2 n+1} & =k \\\\\nm^{2} n-n+1 & =2 m n k-2 n k+k \\\\\nm^{2} n-2 m n k+(2 n k-n-k+1) & =0\n\\end{aligned}\n$$\n\nWe treat this as a quadratic equation in $m$.\n\nSince $m$ is an integer, then this equation has integer roots, and so its discriminant must be a perfect square.\n\nThe discriminant of this quadratic equation is\n\n$$\n\\begin{aligned}\n\\Delta & =(-2 n k)^{2}-4 n(2 n k-n-k+1) \\\\\n& =4 n^{2} k^{2}-8 n^{2} k+4 n^{2}+4 n k-4 n \\\\\n& =4 n^{2}\\left(k^{2}-2 k+1\\right)+4 n(k-1) \\\\\n& =4 n^{2}(k-1)^{2}+4 n(k-1) \\\\\n& =(2 n(k-1))^{2}+2(2 n(k-1))+1-1 \\\\\n& =(2 n(k-1)+1)^{2}-1\n\\end{aligned}\n$$\n\nWe note that $(2 n(k-1)+1)^{2}$ is a perfect square and $\\Delta$ is supposed to be a perfect square. But these perfect squares differ by 1 , and the only two perfect squares that differ by 1 are\n\n\n\n1 and 0.\n\n(To justify this last fact, we could look at the equation $a^{2}-b^{2}=1$ where $a$ and $b$ are non-negative integers, and factor this to obtain $(a+b)(a-b)=1$ which would give $a+b=a-b=1$ from which we get $a=1$ and $b=0$.)\n\nSince $(2 n(k-1)+1)^{2}=1$ and $2 n(k-1)+1$ is non-negative, then $2 n(k-1)+1=1$ and so $2 n(k-1)=0$.\n\nSince $n$ is positive, then $k-1=0$ or $k=1$.\n\nTherefore, the only possible way in which the average is an integer is if the average is 1.\n\nIn this case, we get\n\n$$\n\\begin{aligned}\nm^{2} n-2 m n+(2 n-n-1+1) & =0 \\\\\nm^{2} n-2 m n+n & =0 \\\\\nn\\left(m^{2}-2 m+1\\right) & =0 \\\\\nn(m-1)^{2} & =0\n\\end{aligned}\n$$\n\nSince $n$ and $m$ are positive integers with $m \\geq 2$, then $n(m-1)^{2} \\neq 0$, which is a contradiction.\n\nTherefore, the average of the terms in an $(m, n)$-sawtooth sequence cannot be an integer."
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Number Theory
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2,526
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In the diagram, $A B C D$ is a square. Points $E$ and $F$ are chosen on $A C$ so that $\angle E D F=45^{\circ}$. If $A E=x, E F=y$, and $F C=z$, prove that $y^{2}=x^{2}+z^{2}$.
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"Rotate $\\triangle D F C$ through an angle of $90^{\\circ}$ counterclockwise about $D$, so that $D C$ now lies along $D A$ and $F^{\\prime}$ is outside the square, as shown.\n\nJoin $F^{\\prime}$ to $E$.\n\n<img_3177>\n\nSince $A C$ is a diagonal of square $A B C D$, then $\\angle E A D=\\angle F C D=45^{\\circ}$.\n\nSince $\\angle E A D=45^{\\circ}$ and $\\angle F^{\\prime} A D=\\angle F C D=45^{\\circ}$, then $\\angle F^{\\prime} A E=45^{\\circ}+45^{\\circ}=90^{\\circ}$.\n\nBy the Pythagorean Theorem in $\\triangle F^{\\prime} A E$, we have $F^{\\prime} E^{2}=F^{\\prime} A^{2}+A E^{2}$.\n\nBut $F^{\\prime} A=F C=z$ and $A E=x$, so $F^{\\prime} E^{2}=z^{2}+x^{2}$.\n\nTo show that $y^{2}=x^{2}+z^{2}$, it is sufficient to show that $F^{\\prime} E=y$.\n\nConsider $\\triangle F^{\\prime} D E$ and $\\triangle F D E$.\n\nNote that $F^{\\prime} D=F D$ and $\\angle F^{\\prime} D A=\\angle F D C$ by construction and $E D=E D$.\n\nAlso, $\\angle F^{\\prime} D E=\\angle F^{\\prime} D A+\\angle E D A=\\angle F D C+\\angle E D A=90^{\\circ}-\\angle E D F=45^{\\circ}$, which tells us that $\\angle F^{\\prime} D E=\\angle F D E=45^{\\circ}$.\n\nTherefore, $\\triangle F^{\\prime} D E$ is congruent to $\\triangle F D E$ (side-angle-side), and so $F^{\\prime} E=F E=y$.\n\nThis gives us the desired conclusion that $y^{2}=x^{2}+z^{2}$.",
"Since $A C$ is a diagonal of square $A B C D$, then $\\angle E A D=\\angle F C D=45^{\\circ}$.\n\nLet $\\angle A D E=\\theta$.\n\nSince the angles in a triangle have a sum of $180^{\\circ}$, then\n\n$$\n\\angle A E D=180^{\\circ}-\\angle E A D-\\angle A D E=180^{\\circ}-45^{\\circ}-\\theta=135^{\\circ}-\\theta\n$$\n\nSince $A E F$ is a straight angle, then $\\angle D E F=180^{\\circ}-\\angle A E D=180^{\\circ}-\\left(135^{\\circ}-\\theta\\right)=45^{\\circ}+\\theta$. Continuing in this way, we find that $\\angle E F D=90^{\\circ}-\\theta, \\angle D F C=90^{\\circ}+\\theta$, and $\\angle F D C=45^{\\circ}-\\theta$.\n\n<img_3414>\n\nUsing the sine law in $\\triangle A E D$, we see that $\\frac{A E}{\\sin \\angle A D E}=\\frac{E D}{\\sin \\angle E A D}$ or $\\frac{x}{\\sin \\theta}=\\frac{E D}{\\sin 45^{\\circ}}$.\n\nUsing the sine law in $\\triangle D E F$, we see that $\\frac{E F}{\\sin \\angle E D F}=\\frac{E D}{\\sin \\angle E F D}$ or $\\frac{y}{\\sin 45^{\\circ}}=\\frac{E D}{\\sin \\left(90^{\\circ}-\\theta\\right)}$.\n\n\n\nUsing the sine law in $\\triangle D E F$, we see that $\\frac{E F}{\\sin \\angle E D F}=\\frac{F D}{\\sin \\angle D E F}$ or $\\frac{y}{\\sin 45^{\\circ}}=\\frac{F D}{\\sin \\left(45^{\\circ}+\\theta\\right)}$.\n\nUsing the sine law in $\\triangle D F C$, we get $\\frac{F C}{\\sin \\angle F D C}=\\frac{F D}{\\sin \\angle D C F}$ or $\\frac{z}{\\sin \\left(45^{\\circ}-\\theta\\right)}=\\frac{F D}{\\sin 45^{\\circ}}$.\n\nDividing the first of these equations by the second, we obtain $\\frac{x \\sin 45^{\\circ}}{y \\sin \\theta}=\\frac{\\sin \\left(90^{\\circ}-\\theta\\right)}{\\sin 45^{\\circ}}$ or $\\frac{x}{y}=\\frac{\\sin \\left(90^{\\circ}-\\theta\\right) \\sin \\theta}{\\sin ^{2} 45^{\\circ}}$.\n\nDividing the fourth of these equations by the third, we obtain $\\frac{z \\sin 45^{\\circ}}{y \\sin \\left(45^{\\circ}-\\theta\\right)}=\\frac{\\sin \\left(45^{\\circ}+\\theta\\right)}{\\sin 45^{\\circ}}$ or $\\frac{z}{y}=\\frac{\\sin \\left(45^{\\circ}+\\theta\\right) \\sin \\left(45^{\\circ}-\\theta\\right)}{\\sin ^{2} 45^{\\circ}}$.\n\nSince $\\sin \\left(90^{\\circ}-\\alpha\\right)=\\cos \\alpha$ for every angle $\\alpha$, then $\\sin \\left(90^{\\circ}-\\theta\\right)=\\cos \\theta$.\n\nAlso, $\\sin \\left(45^{\\circ}+\\theta\\right)=\\sin \\left(90^{\\circ}-\\left(45^{\\circ}-\\theta\\right)\\right)=\\cos \\left(45^{\\circ}-\\theta\\right)$.\n\nUsing this and the fact that $\\frac{1}{\\sin ^{2} 45^{\\circ}}=\\frac{1}{(1 / \\sqrt{2})^{2}}=2$, the expressions for $\\frac{x}{y}$ and $\\frac{z}{y}$ become\n\n$$\n\\frac{x}{y}=2 \\cos \\theta \\sin \\theta=\\sin 2 \\theta\n$$\n\nand\n\n$$\n\\frac{z}{y}=2 \\cos \\left(45^{\\circ}-\\theta\\right) \\sin \\left(45^{\\circ}-\\theta\\right)=\\sin \\left(2\\left(45^{\\circ}-\\theta\\right)\\right)=\\sin \\left(90^{\\circ}-2 \\theta\\right)=\\cos 2 \\theta\n$$\n\nFinally, this tells us that\n\n$$\n\\frac{x^{2}}{y^{2}}+\\frac{z^{2}}{y^{2}}=\\left(\\frac{x}{y}\\right)^{2}+\\left(\\frac{z}{y}\\right)^{2}=\\sin ^{2} 2 \\theta+\\cos ^{2} 2 \\theta=1\n$$\n\nSince $\\frac{x^{2}}{y^{2}}+\\frac{z^{2}}{y^{2}}=1$, then $x^{2}+z^{2}=y^{2}$, as required.\n\n\n\n\n#"
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Geometry
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2,538
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In the diagram, $A B$ is tangent to the circle with centre $O$ and radius $r$. The length of $A B$ is $p$. Point $C$ is on the circle and $D$ is inside the circle so that $B C D$ is a straight line, as shown. If $B C=C D=D O=q$, prove that $q^{2}+r^{2}=p^{2}$.
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"Join $O$ to $A, B$ and $C$.\n\n<img_3419>\n\nSince $A B$ is tangent to the circle at $A$, then $\\angle O A B=90^{\\circ}$.\n\nBy the Pythagorean Theorem in $\\triangle O A B$, we get $O A^{2}+A B^{2}=O B^{2}$ or $r^{2}+p^{2}=O B^{2}$.\n\nIn $\\triangle O D C$, we have $O D=D C=q$ and $O C=r$.\n\nBy the cosine law,\n\n$$\n\\begin{aligned}\nO C^{2} & =O D^{2}+D C^{2}-2(O D)(D C) \\cos (\\angle O D C) \\\\\nr^{2} & =q^{2}+q^{2}-2 q^{2} \\cos (\\angle O D C) \\\\\n\\cos (\\angle O D C) & =\\frac{2 q^{2}-r^{2}}{2 q^{2}}\n\\end{aligned}\n$$\n\nIn $\\triangle O D B$, we have $\\angle O D B=\\angle O D C$.\n\nThus, using the cosine law again,\n\n$$\n\\begin{aligned}\nO B^{2} & =O D^{2}+D B^{2}-2(O D)(D B) \\cos (\\angle O D B) \\\\\n& =q^{2}+(2 q)^{2}-2(q)(2 q)\\left(\\frac{2 q^{2}-r^{2}}{2 q^{2}}\\right) \\\\\n& =q^{2}+4 q^{2}-2\\left(2 q^{2}-r^{2}\\right) \\\\\n& =q^{2}+2 r^{2}\n\\end{aligned}\n$$\n\nSo $O B^{2}=r^{2}+p^{2}=q^{2}+2 r^{2}$, which gives $p^{2}=q^{2}+r^{2}$, as required."
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Geometry
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Math
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English
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2,542
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Suppose there are $n$ plates equally spaced around a circular table. Ross wishes to place an identical gift on each of $k$ plates, so that no two neighbouring plates have gifts. Let $f(n, k)$ represent the number of ways in which he can place the gifts. For example $f(6,3)=2$, as shown below.
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Throughout this problem, we represent the states of the $n$ plates as a string of 0's and 1's (called a binary string) of length $n$ of the form $p_{1} p_{2} \cdots p_{n}$, with the $r$ th digit from the left (namely $p_{r}$ ) equal to 1 if plate $r$ contains a gift and equal to 0 if plate $r$ does not. We call a binary string of length $n$ allowable if it satisfies the requirements - that is, if no two adjacent digits both equal 1. Note that digit $p_{n}$ is also "adjacent" to digit $p_{1}$, so we cannot have $p_{1}=p_{n}=1$.
Prove that $f(n, k)=f(n-1, k)+f(n-2, k-1)$ for all integers $n \geq 3$ and $k \geq 2$.
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"An allowable string $p_{1} p_{2} \\cdots p_{n-1} p_{n}$ has $\\left(p_{1}, p_{n}\\right)=(1,0),(0,1)$, or $(0,0)$.\n\nDefine $g(n, k, 1,0)$ to be the number of allowable strings of length $n$, containing $k 1$ 's, and with $\\left(p_{1}, p_{n}\\right)=(1,0)$.\n\nWe define $g(n, k, 0,1)$ and $g(n, k, 0,0)$ in a similar manner.\n\nNote that $f(n, k)=g(n, k, 1,0)+g(n, k, 0,1)+g(n, k, 0,0)$.\n\nConsider the strings counted by $g(n, k, 0,1)$.\n\nSince $p_{n}=1$, then $p_{n-1}=0$. Since $p_{1}=0$, then $p_{2}$ can equal 0 or 1 .\n\nWe remove the first and last digits of these strings.\n\nWe obtain strings $p_{2} p_{3} \\cdots p_{n-2} p_{n-1}$ that is strings of length $n-2$ containing $k-11$ 's.\n\nSince $p_{n-1}=0$, then the first and last digits of these strings are not both 1 . Also, since the original strings did not contain two consecutive 1's, then these new strings does not either.\n\nTherefore, $p_{2} p_{3} \\cdots p_{n-2} p_{n-1}$ are allowable strings of length $n-2$ containing $k-1$ 1's, with $p_{n-1}=0$ and $p_{2}=1$ or $p_{2}=0$.\n\nThe number of such strings with $p_{2}=1$ and $p_{n-1}=0$ is $g(n-2, k-1,1,0)$ and the number of such strings with $p_{2}=0$ and $p_{n-1}=0$ is $g(n-2, k-1,0,0)$.\n\nThus, $g(n, k, 0,1)=g(n-2, k-1,1,0)+g(n-2, k-1,0,0)$.\n\nConsider the strings counted by $g(n, k, 0,0)$.\n\nSince $p_{1}=0$ and $p_{n}=0$, then we can remove $p_{n}$ to obtain strings $p_{1} p_{2} \\cdots p_{n-1}$ of length $n-1$ containing $k 1$ 's. These strings are allowable since $p_{1}=0$ and the original strings were allowable.\n\nNote that we have $p_{1}=0$ and $p_{n-1}$ is either 0 or 1 .\n\nSo the strings $p_{1} p_{2} \\cdots p_{n-1}$ are allowable strings of length $n-1$ containing $k$ 1's, starting with 0 , and ending with 0 or 1 .\n\nThe number of such strings with $p_{1}=0$ and $p_{n-1}=0$ is $g(n-1, k, 0,0)$ and the number of such strings with $p_{1}=0$ and $p_{n-1}=1$ is $g(n-1, k, 0,1)$.\n\nThus, $g(n, k, 0,0)=g(n-1, k, 0,0)+g(n-1, k, 0,1)$.\n\n\n\nConsider the strings counted by $g(n, k, 1,0)$.\n\nHere, $p_{1}=1$ and $p_{n}=0$. Thus, $p_{n-1}$ can equal 0 or 1 . We consider these two sets separately.\n\nIf $p_{n-1}=0$, then the string $p_{1} p_{2} \\cdots p_{n-1}$ is an allowable string of length $n-1$, containing $k 1$ 's, beginning with 1 and ending with 0 .\n\nTherefore, the number of strings counted by $g(n, k, 1,0)$ with $p_{n-1}=0$ is equal to $g(n-1, k, 1,0)$.\n\nIf $p_{n-1}=1$, then the string $p_{2} p_{3} \\cdots p_{n-1}$ is of length $n-2$, begins with 0 and ends with 1 . Also, it contains $k-1$ 1's (having removed the original leading 1) and is allowable since the original string was.\n\nTherefore, the number of strings counted by $g(n, k, 1,0)$ with $p_{n-1}=1$ is equal to $g(n-2, k-1,0,1)$.\n\nTherefore,\n\n$$\n\\begin{aligned}\nf(n, k)= & g(n, k, 1,0)+g(n, k, 0,1)+g(n, k, 0,0) \\\\\n= & (g(n-1, k, 1,0)+g(n-2, k-1,0,1)) \\\\\n& \\quad+(g(n-2, k-1,1,0)+g(n-2, k-1,0,0)) \\\\\n& \\quad+(g(n-1, k, 0,0)+g(n-1, k, 0,1)) \\\\\n= & (g(n-1, k, 1,0)+g(n-1, k, 0,1)+g(n-1, k, 0,0)) \\\\\n& \\quad+(g(n-2, k-1,0,1)+g(n-2, k-1,1,0)+g(n-2, k-1,0,0)) \\\\\n= & f(n-1, k)+f(n-2, k-1)\n\\end{aligned}\n$$\n\nas required.",
"We develop an explicit formula for $f(n, k)$ by building these strings.\n\nConsider the allowable strings of length $n$ that include $k$ 1's. Either $p_{n}=0$ or $p_{n}=1$.\n\nConsider first the case when $p_{n}=0$. (Here, $p_{1}$ can equal 0 or 1.)\n\nThese strings are all of the form $p_{1} p_{2} p_{3} \\cdots p_{n-1} 0$.\n\nIn this case, since a 1 is always followed by a 0 and the strings end with 0 , we can build these strings using blocks of the form 10 and 0 . Any combination of these blocks will be an allowable string, as each 1 will always be both preceded and followed by a 0 .\n\nThus, these strings can all be built using $k 10$ blocks and $n-2 k 0$ blocks. This gives $k$ 1 's and $k+(n-2 k)=n-k 0$ 's. Note that any string built with these blocks will be allowable and will end with a 0 , and any such allowable string can be built in this way.\n\nThe number of ways of arranging $k$ blocks of one kind and $n-2 k$ blocks of another kind is $\\left(\\begin{array}{c}k+(n-2 k) \\\\ k\\end{array}\\right)$, which simplifies to $\\left(\\begin{array}{c}n-k \\\\ k\\end{array}\\right)$.\n\nConsider next the case when $p_{n}=1$.\n\nHere, we must have $p_{n-1}=p_{1}=0$, since these are the two digits adjacent to $p_{n}$.\n\nThus, these strings are all of the form $0 p_{2} p_{3} \\cdots 01$.\n\nConsider the strings formed by removing the first and last digits.\n\nThese strings are allowable, are of length $n-2$, include $k-11$ 's, end with 0 , and can begin with 0 or 1 .\n\nAgain, since a 1 is always followed by a 0 and the strings end with 0 , we can build these strings using blocks of the form 10 and 0 . Any combination of these blocks will be an allowable string, as each 1 will always be both preceded and followed by a 0 .\n\nTranslating our method of counting from the first case, there are $\\left(\\begin{array}{c}(n-2)-(k-1) \\\\ k-1\\end{array}\\right)$ or\n\n\n\n$\\left(\\begin{array}{c}n-k-1 \\\\ k-1\\end{array}\\right)$ such strings.\n\nThus, $f(n, k)=\\left(\\begin{array}{c}n-k \\\\ k\\end{array}\\right)+\\left(\\begin{array}{c}n-k-1 \\\\ k-1\\end{array}\\right)$ such strings.\n\nTo prove the desired fact, we will use the fact that $\\left(\\begin{array}{c}m \\\\ r\\end{array}\\right)=\\left(\\begin{array}{c}m-1 \\\\ r\\end{array}\\right)+\\left(\\begin{array}{c}m-1 \\\\ r-1\\end{array}\\right)$, which we prove at the end.\n\nNow\n\n$$\n\\begin{aligned}\nf & n-1, k)+f(n-2, k-1) \\\\\n& =\\left(\\begin{array}{c}\n(n-1)-k \\\\\nk\n\\end{array}\\right)+\\left(\\begin{array}{c}\n(n-1)-k-1 \\\\\nk-1\n\\end{array}\\right)+\\left(\\begin{array}{c}\n(n-2)-(k-1) \\\\\nk-1\n\\end{array}\\right)+\\left(\\begin{array}{c}\n(n-2)-(k-1)-1 \\\\\n(k-1)-1\n\\end{array}\\right) \\\\\n& =\\left(\\begin{array}{c}\nn-k-1 \\\\\nk\n\\end{array}\\right)+\\left(\\begin{array}{c}\nn-k-2 \\\\\nk-1\n\\end{array}\\right)+\\left(\\begin{array}{c}\nn-k-1 \\\\\nk-1\n\\end{array}\\right)+\\left(\\begin{array}{c}\nn-k-2 \\\\\nk-2\n\\end{array}\\right) \\\\\n& =\\left(\\begin{array}{c}\nn-k-1 \\\\\nk\n\\end{array}\\right)+\\left(\\begin{array}{c}\nn-k-1 \\\\\nk-1\n\\end{array}\\right)+\\left(\\begin{array}{c}\nn-k-2 \\\\\nk-1\n\\end{array}\\right)+\\left(\\begin{array}{c}\nn-k-2 \\\\\nk-2\n\\end{array}\\right) \\\\\n& =\\left(\\begin{array}{c}\nn-k \\\\\nk\n\\end{array}\\right)+\\left(\\begin{array}{c}\nn-k-1 \\\\\nk-1\n\\end{array}\\right) \\quad \\text { (using the identity above) } \\\\\n& =f(n, k)\n\\end{aligned}\n$$\n\nas required.\n\nTo prove the identity, we expand the terms on the right side:\n\n$$\n\\begin{aligned}\n\\left(\\begin{array}{c}\nm-1 \\\\\nr\n\\end{array}\\right)+\\left(\\begin{array}{c}\nm-1 \\\\\nr-1\n\\end{array}\\right) & =\\frac{(m-1) !}{r !(m-r-1) !}+\\frac{(m-1) !}{(r-1) !(m-r) !} \\\\\n& =\\frac{(m-1) !(m-r)}{r !(m-r-1) !(m-r)}+\\frac{r(m-1) !}{r(r-1) !(m-r) !} \\\\\n& =\\frac{(m-1) !(m-r)}{r !(m-r) !}+\\frac{r(m-1) !}{r !(m-r) !} \\\\\n& =\\frac{(m-1) !(m-r+r)}{r !(m-r) !} \\\\\n& =\\frac{(m-1) ! m}{r !(m-r) !} \\\\\n& =\\frac{m !}{r !(m-r) !} \\\\\n& =\\left(\\begin{array}{c}\nm \\\\\nr\n\\end{array}\\right)\n\\end{aligned}\n$$\n\nas required."
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Theorem proof
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Algebra
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Math
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English
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2,547
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In trapezoid $A B C D, B C$ is parallel to $A D$ and $B C$ is perpendicular to $A B$. Also, the lengths of $A D, A B$ and $B C$, in that order, form a geometric sequence. Prove that $A C$ is perpendicular to $B D$.
(A geometric sequence is a sequence in which each term after the first is obtained from the previous term by multiplying it by a non-zero constant.)
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"Since the lengths of $A D, A B$ and $B C$ form a geometric sequence, we suppose that these lengths are $a$, ar and $a r^{2}$, respectively, for some real numbers $a>0$ and $r>0$.\n\nSince the angles at $A$ and $B$ are both right angles, we assign coordinates to the diagram, putting $B$ at the origin ( 0,0$), C$ on the positive $x$-axis at $\\left(a r^{2}, 0\\right), A$ on the positive $y$-axis at $(0, a r)$, and $D$ at $(a, a r)$.\n\n<img_3911>\n\nTherefore, the slope of the line segment joining $B(0,0)$ and $D(a, a r)$ is $\\frac{a r-0}{a-0}=r$.\n\nAlso, the slope of the line segment joining $A(0, a r)$ and $C\\left(a r^{2}, 0\\right)$ is $\\frac{a r-0}{0-a r^{2}}=-\\frac{1}{r}$.\n\nSince the product of the slopes of these two line segments is -1 , then the segments are perpendicular, as required."
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Geometry
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Math
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English
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2,561
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In the graph, the parabola $y=x^{2}$ has been translated to the position shown.
Prove that $d e=f$.
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"Since the given graph is congruent to $y=x^{2}$ and has $x$-intercepts $-d$ and $e$, its general form is $y=(x+d)(x-e)$.\n\nTo find the $y$-intercept, let $x=0$. Therefore $y$-intercept $=-d e$.\n\nWe are given that the $y$-intercept is $-f$.\n\nTherefore $-f=-d e$ or $f=d e$."
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Geometry
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Math
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English
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2,562
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In quadrilateral $K W A D$, the midpoints of $K W$ and $A D$ are $M$ and $N$ respectively. If $M N=\frac{1}{2}(A W+D K)$, prove that $WA$ is parallel to $K D$.
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"Establish a coordinate system with $K(0,0), D(2 a, 0)$ on the $x$-axes. Let $W$ be $(2 b, 2 c)$ and $A$ be $(2 d, 2 e)$.\n\nThus $M$ is $(b, c)$ and $N$ is $(a+d, e)$.\n\n$K D$ has slope 0 and slope $W A=\\frac{e-c}{d-b}$.\n\nSince $M N=\\frac{1}{2}(A W+D K)$\n\n$$\n\\begin{aligned}\n& \\sqrt{(a+d-b)^{2}+(e-c)^{2}} \\\\\n= & \\frac{1}{2}\\left(2 a+\\sqrt{(2 d-2 b)^{2}+(2 e-2 c)^{2}}\\right) \\\\\n= & \\frac{1}{2}\\left(2 a+2 \\sqrt{(d-b)^{2}+(e-c)^{2}}\\right)\n\\end{aligned}\n$$\n\n<img_3380>\n\nSquaring both sides gives,\n\n$$\n\\begin{aligned}\n& (a+d-b)^{2}+(e-c)^{2}=a^{2}+2 a \\sqrt{(d-b)^{2}+(e-c)^{2}}+(d-b)^{2}+(e-c)^{2} \\\\\n& a^{2}+2 a(d-b)+(d-b)^{2}=a^{2}+2 a \\sqrt{(d-b)^{2}+(e-c)^{2}}+(d-b)^{2}\n\\end{aligned}\n$$\n\nSimplifying and dividing by $2 a$ we have, $d-b=\\sqrt{(d-b)^{2}+(e-c)^{2}}$.\n\nSquaring, $(d-b)^{2}=(d-b)^{2}+(e-c)^{2}$.\n\nTherefore $(e-c)^{2}=0$ or $e=c$.\n\nSince $e=c$ then slope of $W A$ is 0 and $K D \\| A W$.",
"Join $A$ to $K$ and call $P$ the mid-point of $A K$.\n\nJoin $M$ to $P, N$ to $P$ and $M$ to $N$.\n\nIn $\\triangle K A W, P$ and $M$ are the mid-points of $K A$ and $K W$.\n\nTherefore, $M P=\\frac{1}{2} W A$.\n\nSimilarly in $\\triangle K A D, P N=\\frac{1}{2} K D$.\n\nTherefore $M P+P N=M N$.\n\n<img_3772>\n\nAs a result $M, P$ and $N$ cannot form the vertices of a triangle but must form a straight line.\n\nSo if $M P N$ is a straight line with $M P \\| W A$ and $P N \\| K D$ then $W A \\| K D$ as required.",
"We are given that $\\overrightarrow{A N}=\\overrightarrow{N D}$ and $\\overrightarrow{W M}=\\overrightarrow{M K}$.\n\nUsing vectors,\n\n(1) $\\overrightarrow{M N}=\\overrightarrow{M W}+\\overrightarrow{W A}+\\overrightarrow{A N}$ (from quad. $M W A N$ )\n\n(2) $\\overrightarrow{M N}=\\overrightarrow{M K}+\\overrightarrow{K D}+\\overrightarrow{D N} \\quad$ (from quad. $K M N D$ )\n\nIt is also possible to write, $\\overrightarrow{M N}=-\\overrightarrow{M W}+\\overrightarrow{K D}-\\overrightarrow{A N}$,\n\n(3) (This comes from taking statement (2) and making appropriate substitutions.)\n\n\n\nIf we add (1) and (3) we find, $2 \\overrightarrow{M N}=\\overrightarrow{W A}+\\overrightarrow{K D}$.\n\nBut it is given that $2|\\overrightarrow{M N}|=|\\overrightarrow{A W}|+|\\overrightarrow{D K}|$.\n\nFrom these two previous statements, $\\overrightarrow{M N}$ must be parallel to $\\overrightarrow{W A}$ and $\\overrightarrow{K D}$ otherwise $2|\\overrightarrow{M N}|<|\\overrightarrow{A W}|+|\\overrightarrow{D K}|$.\n\nTherefore, $W A \\| K D$."
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Geometry
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Math
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English
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2,563
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Consider the first $2 n$ natural numbers. Pair off the numbers, as shown, and multiply the two members of each pair. Prove that there is no value of $n$ for which two of the $n$ products are equal.
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"The sequence is $1(2 n), 2(2 n-1), 3(2 n-2), \\ldots, k(2 n-k+1), \\ldots, p(2 n-p+1), \\ldots, n(n+1)$.\n\nIn essence we are asking the question, 'is it possible that $k(2 n-k+1)=p(2 n-p+1)$ where $p$ and $k$ are both less than or equal to $n$ ?'\n\n$$\n\\begin{aligned}\nk(2 n-k+1) & =p(2 n-p+1) \\\\\n2 n k-k^{2}+k=2 n p-p^{2}+p & \\text { (supposing them to be equal) } \\\\\np^{2}-k^{2}+2 n k-2 n p+k-p & =0 \\\\\n(p-k)(p+k)+2 n(k-p)+(k-p) & =0 \\\\\n(p-k)[(p+k)-2 n-1] & =0 \\\\\n(p-k)(p+k-2 n-1) & =0\n\\end{aligned}\n$$\n\nSince $p$ and $k$ are both less than or equal to $n$, it follows $p+k-2 n-1 \\neq 0$. Therefore $p=k$ and they represent the same pair. Thus the required is proven.",
"The products are $1(2 n+1-1), 2(2 n+1-2), 3(2 n+1-3), \\ldots, n(2 n+1-n)$.\n\nConsider the function, $y=x(2 n+1-x)=-x^{2}+(2 n+1) x=f(x)$.\n\n\n\nThe graph of this function is a parabola, opening down, with its vertex at $x=n+\\frac{1}{2}$.\n\nThe products are the $y$-coordinates of the points on the parabola corresponding to $x=1,2,3, \\ldots, n$. Since all the points are to the left of the vertex, no two have the same $y$ coordinate.\n\nThus the products are distinct.\n\n<img_3942>",
"The sum of these numbers is $\\frac{2 n(2 n+1)}{2}$ or $n(2 n+1)$.\n\nTheir average is $\\frac{n(2 n+1)}{2 n}=n+\\frac{1}{2}$.\n\nThe $2 n$ numbers can be rewritten as,\n\n$$\nn+\\frac{1}{2}-\\left(\\frac{2 n-1}{2}\\right), \\cdots, n+\\frac{1}{2}-\\frac{3}{2}, n+\\frac{1}{2}-\\frac{1}{2}, n+\\frac{1}{2}+\\frac{1}{2}, n+\\frac{1}{2}+\\frac{3}{2}, \\cdots, n+\\frac{1}{2}+\\left(\\frac{2 n-1}{2}\\right) .\n$$\n\nThe product pairs, starting from the middle and working outward are\n\n$$\n\\begin{gathered}\nP_{1}=\\left(n+\\frac{1}{2}\\right)^{2}-\\frac{1}{4} \\\\\nP_{2}=\\left(n+\\frac{1}{2}\\right)^{2}-\\frac{9}{4} \\\\\n\\vdots \\\\\nP_{n}=\\left(n+\\frac{1}{2}\\right)^{2}-\\left(\\frac{2 n-1}{2}\\right)^{2}\n\\end{gathered}\n$$\n\nEach of the numbers $\\left(\\frac{2 k-1}{2}\\right)^{2}$ is distinct for $k=1,2,3, \\ldots, n$ and hence no terms of $P_{k}$ are equal.",
"The sequence is $1(2 n), 2(2 n-1), 3(2 n-2), \\ldots, n[2 n-(n-1)]$.\n\nThis sequence has exactly $n$ terms.\n\nWhen the $k$ th term is subtracted from the $(k+1)$ th term the difference is $(k+1)[2 n-k]-k[2 n-(k-1)]=2(n-k)$. Since $n>k$, this is a positive difference.\n\nTherefore each term is greater than the term before, so no two terms are equal."
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Number Theory
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2,803
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This Question involves one Robber and one or more Cops. After robbing a bank, the Robber retreats to a network of hideouts, represented by dots in the diagram below. Every day, the Robber stays holed up in a single hideout, and every night, the Robber moves to an adjacent hideout. Two hideouts are adjacent if and only if they are connected by an edge in the diagram, also called a hideout map (or map). For the purposes of this Power Question, the map must be connected; that is, given any two hideouts, there must be a path from one to the other. To clarify, the Robber may not stay in the same hideout for two consecutive days, although he may return to a hideout he has previously visited. For example, in the map below, if the Robber holes up in hideout $C$ for day 1 , then he would have to move to $B$ for day 2 , and would then have to move to either $A, C$, or $D$ on day 3.
<image_1>
Every day, each Cop searches one hideout: the Cops know the location of all hideouts and which hideouts are adjacent to which. Cops are thorough searchers, so if the Robber is present in the hideout searched, he is found and arrested. If the Robber is not present in the hideout searched, his location is not revealed. That is, the Cops only know that the Robber was not caught at any of the hideouts searched; they get no specific information (other than what they can derive by logic) about what hideout he was in. Cops are not constrained by edges on the map: a Cop may search any hideout on any day, regardless of whether it is adjacent to the hideout searched the previous day. A Cop may search the same hideout on consecutive days, and multiple Cops may search different hideouts on the same day. In the map above, a Cop could search $A$ on day 1 and day 2, and then search $C$ on day 3 .
The focus of this Power Question is to determine, given a hideout map and a fixed number of Cops, whether the Cops can be sure of catching the Robber within some time limit.
Map Notation: The following notation may be useful when writing your solutions. For a map $M$, let $h(M)$ be the number of hideouts and $e(M)$ be the number of edges in $M$. The safety of a hideout $H$ is the number of hideouts adjacent to $H$, and is denoted by $s(H)$.
The Cop number of a map $M$, denoted $C(M)$, is the minimum number of Cops required to guarantee that the Robber is caught.
Consider the hideout map $M$ below.
<image_2>
Show that one Cop can always catch the Robber.
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"Have the Cop stay at $A$ for 2 days. If the Robber is not at $A$ the first day, he must be at one of $B_{1}-B_{6}$, and because the Robber must move along an edge every night, he will be forced to go to $A$ on day 2 ."
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2,804
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This Question involves one Robber and one or more Cops. After robbing a bank, the Robber retreats to a network of hideouts, represented by dots in the diagram below. Every day, the Robber stays holed up in a single hideout, and every night, the Robber moves to an adjacent hideout. Two hideouts are adjacent if and only if they are connected by an edge in the diagram, also called a hideout map (or map). For the purposes of this Power Question, the map must be connected; that is, given any two hideouts, there must be a path from one to the other. To clarify, the Robber may not stay in the same hideout for two consecutive days, although he may return to a hideout he has previously visited. For example, in the map below, if the Robber holes up in hideout $C$ for day 1 , then he would have to move to $B$ for day 2 , and would then have to move to either $A, C$, or $D$ on day 3.
<image_1>
Every day, each Cop searches one hideout: the Cops know the location of all hideouts and which hideouts are adjacent to which. Cops are thorough searchers, so if the Robber is present in the hideout searched, he is found and arrested. If the Robber is not present in the hideout searched, his location is not revealed. That is, the Cops only know that the Robber was not caught at any of the hideouts searched; they get no specific information (other than what they can derive by logic) about what hideout he was in. Cops are not constrained by edges on the map: a Cop may search any hideout on any day, regardless of whether it is adjacent to the hideout searched the previous day. A Cop may search the same hideout on consecutive days, and multiple Cops may search different hideouts on the same day. In the map above, a Cop could search $A$ on day 1 and day 2, and then search $C$ on day 3 .
The focus of this Power Question is to determine, given a hideout map and a fixed number of Cops, whether the Cops can be sure of catching the Robber within some time limit.
Map Notation: The following notation may be useful when writing your solutions. For a map $M$, let $h(M)$ be the number of hideouts and $e(M)$ be the number of edges in $M$. The safety of a hideout $H$ is the number of hideouts adjacent to $H$, and is denoted by $s(H)$.
The Cop number of a map $M$, denoted $C(M)$, is the minimum number of Cops required to guarantee that the Robber is caught.
The map shown below is $\mathcal{C}_{6}$, the cyclic graph with six hideouts. Show that three Cops are sufficient to catch the Robber on $\mathcal{C}_{6}$, so that $C\left(\mathcal{C}_{6}\right) \leq 3$.
<image_2>
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"The Cops should stay at $\\left\\{A_{1}, A_{3}, A_{5}\\right\\}$ for 2 days. If the Robber evades capture the first day, he must have been at an even-numbered hideout. Because he must move, he will be at an odd-numbered hideout the second day. Equivalently, the Cops could stay at $\\left\\{A_{2}, A_{4}, A_{6}\\right\\}$ for 2 days."
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2,805
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This Question involves one Robber and one or more Cops. After robbing a bank, the Robber retreats to a network of hideouts, represented by dots in the diagram below. Every day, the Robber stays holed up in a single hideout, and every night, the Robber moves to an adjacent hideout. Two hideouts are adjacent if and only if they are connected by an edge in the diagram, also called a hideout map (or map). For the purposes of this Power Question, the map must be connected; that is, given any two hideouts, there must be a path from one to the other. To clarify, the Robber may not stay in the same hideout for two consecutive days, although he may return to a hideout he has previously visited. For example, in the map below, if the Robber holes up in hideout $C$ for day 1 , then he would have to move to $B$ for day 2 , and would then have to move to either $A, C$, or $D$ on day 3.
<image_1>
Every day, each Cop searches one hideout: the Cops know the location of all hideouts and which hideouts are adjacent to which. Cops are thorough searchers, so if the Robber is present in the hideout searched, he is found and arrested. If the Robber is not present in the hideout searched, his location is not revealed. That is, the Cops only know that the Robber was not caught at any of the hideouts searched; they get no specific information (other than what they can derive by logic) about what hideout he was in. Cops are not constrained by edges on the map: a Cop may search any hideout on any day, regardless of whether it is adjacent to the hideout searched the previous day. A Cop may search the same hideout on consecutive days, and multiple Cops may search different hideouts on the same day. In the map above, a Cop could search $A$ on day 1 and day 2, and then search $C$ on day 3 .
The focus of this Power Question is to determine, given a hideout map and a fixed number of Cops, whether the Cops can be sure of catching the Robber within some time limit.
Map Notation: The following notation may be useful when writing your solutions. For a map $M$, let $h(M)$ be the number of hideouts and $e(M)$ be the number of edges in $M$. The safety of a hideout $H$ is the number of hideouts adjacent to $H$, and is denoted by $s(H)$.
The Cop number of a map $M$, denoted $C(M)$, is the minimum number of Cops required to guarantee that the Robber is caught.
Show that for all maps $M, C(M)<h(M)$.
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"Let $n=h(M)$. The following strategy will always catch a Robber within two days using $n-1$ Cops, which proves that $C(M) \\leq n-1$. Choose any subset $\\mathcal{S}$ of $n-1$ hideouts and position $n-1$ Cops at the hideouts of $\\mathcal{S}$ for 2 days. If the Robber is not caught on the first day, he must have been at the hideout not in $\\mathcal{S}$, and therefore must move to a hideout in $\\mathcal{S}$ on the following day."
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Combinatorics
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Math
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English
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2,807
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This Question involves one Robber and one or more Cops. After robbing a bank, the Robber retreats to a network of hideouts, represented by dots in the diagram below. Every day, the Robber stays holed up in a single hideout, and every night, the Robber moves to an adjacent hideout. Two hideouts are adjacent if and only if they are connected by an edge in the diagram, also called a hideout map (or map). For the purposes of this Power Question, the map must be connected; that is, given any two hideouts, there must be a path from one to the other. To clarify, the Robber may not stay in the same hideout for two consecutive days, although he may return to a hideout he has previously visited. For example, in the map below, if the Robber holes up in hideout $C$ for day 1 , then he would have to move to $B$ for day 2 , and would then have to move to either $A, C$, or $D$ on day 3.
<image_1>
Every day, each Cop searches one hideout: the Cops know the location of all hideouts and which hideouts are adjacent to which. Cops are thorough searchers, so if the Robber is present in the hideout searched, he is found and arrested. If the Robber is not present in the hideout searched, his location is not revealed. That is, the Cops only know that the Robber was not caught at any of the hideouts searched; they get no specific information (other than what they can derive by logic) about what hideout he was in. Cops are not constrained by edges on the map: a Cop may search any hideout on any day, regardless of whether it is adjacent to the hideout searched the previous day. A Cop may search the same hideout on consecutive days, and multiple Cops may search different hideouts on the same day. In the map above, a Cop could search $A$ on day 1 and day 2, and then search $C$ on day 3 .
The focus of this Power Question is to determine, given a hideout map and a fixed number of Cops, whether the Cops can be sure of catching the Robber within some time limit.
Map Notation: The following notation may be useful when writing your solutions. For a map $M$, let $h(M)$ be the number of hideouts and $e(M)$ be the number of edges in $M$. The safety of a hideout $H$ is the number of hideouts adjacent to $H$, and is denoted by $s(H)$.
The Cop number of a map $M$, denoted $C(M)$, is the minimum number of Cops required to guarantee that the Robber is caught.
Show that $C(M) \leq 3$ for the map below.
<image_2>
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[
"The following strategy guarantees capture using three Cops for four consecutive days, so $C(M) \\leq 3$. Position three Cops at $\\{B, E, H\\}$ for two days, which will catch any Robber who starts out at $B, C, D, E, F, G$, or $H$, because a Robber at $C$ or $D$ would have to move to either $B$ or $E$, and a Robber at $F$ or $G$ would have to move to either $E$ or $H$. So if the Robber is not yet caught, he must have been at $I, J, K, L$, or $A$ for those two days. In this case, an analogous argument shows that placing Cops at $B, H, K$ for two consecutive days will guarantee a capture."
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2,809
|
This Question involves one Robber and one or more Cops. After robbing a bank, the Robber retreats to a network of hideouts, represented by dots in the diagram below. Every day, the Robber stays holed up in a single hideout, and every night, the Robber moves to an adjacent hideout. Two hideouts are adjacent if and only if they are connected by an edge in the diagram, also called a hideout map (or map). For the purposes of this Power Question, the map must be connected; that is, given any two hideouts, there must be a path from one to the other. To clarify, the Robber may not stay in the same hideout for two consecutive days, although he may return to a hideout he has previously visited. For example, in the map below, if the Robber holes up in hideout $C$ for day 1 , then he would have to move to $B$ for day 2 , and would then have to move to either $A, C$, or $D$ on day 3.
<image_1>
Every day, each Cop searches one hideout: the Cops know the location of all hideouts and which hideouts are adjacent to which. Cops are thorough searchers, so if the Robber is present in the hideout searched, he is found and arrested. If the Robber is not present in the hideout searched, his location is not revealed. That is, the Cops only know that the Robber was not caught at any of the hideouts searched; they get no specific information (other than what they can derive by logic) about what hideout he was in. Cops are not constrained by edges on the map: a Cop may search any hideout on any day, regardless of whether it is adjacent to the hideout searched the previous day. A Cop may search the same hideout on consecutive days, and multiple Cops may search different hideouts on the same day. In the map above, a Cop could search $A$ on day 1 and day 2, and then search $C$ on day 3 .
The focus of this Power Question is to determine, given a hideout map and a fixed number of Cops, whether the Cops can be sure of catching the Robber within some time limit.
Map Notation: The following notation may be useful when writing your solutions. For a map $M$, let $h(M)$ be the number of hideouts and $e(M)$ be the number of edges in $M$. The safety of a hideout $H$ is the number of hideouts adjacent to $H$, and is denoted by $s(H)$.
The Cop number of a map $M$, denoted $C(M)$, is the minimum number of Cops required to guarantee that the Robber is caught.
The police want to catch the Robber with a minimum number of Cops, but time is of the essence. For a map $M$ and a fixed number of Cops $c \geq C(M)$, define the capture time, denoted $D(M, c)$, to be the minimum number of days required to guarantee a capture using $c$ Cops. For example, in the graph below, if three Cops are deployed, they might catch the Robber in the first day, but if they don't, there is a strategy that will guarantee they will capture the Robber within two days. Therefore the capture time is $D\left(\mathcal{C}_{6}, 3\right)=2$.
<image_2>
A path on $n$ hideouts is a map with $n$ hideouts, connected in one long string. (More formally, a map is a path if and only if two hideouts are adjacent to exactly one hideout each and all other hideouts are adjacent to exactly two hideouts each.) It is denoted by $\mathcal{P}_{n}$. The maps $\mathcal{P}_{3}$ through $\mathcal{P}_{6}$ are shown below.
<image_3>
Show that $D\left(\mathcal{P}_{n}, 1\right) \leq 2 n$ for $n \geq 3$.
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"The following argument shows that $C\\left(\\mathcal{P}_{n}\\right)=1$, and that capture occurs in at most $2 n-4$ days. It helps to draw the hideout map as in the following diagram, so that odd-numbered hideouts are all on one level and even-numbered hideouts are all on another level; the case where $n$ is odd is shown below. (A similar argument applies where $n$ is even.)\n\n<img_3525>\n\n\n\nEach night, the Robber has the choice of moving diagonally left or diagonally right on this map, but he is required to move from top to bottom or vice-versa.\n\nFor the first $n-2$ days, the Cop should search hideouts $A_{2}, A_{3}, \\ldots, A_{n-1}$, in that order. For the next $n-2$ days, the Cop should search hideouts $A_{n-1}, A_{n-2}, \\ldots, A_{2}$, in that order.\n\nThe Cop always moves from top to bottom, or vice-versa, except that he searches $A_{n-1}$ two days in a row. If the Cop is lucky, the Robber chose an even-numbered hideout on the first day. In this case, the Cop and the Robber are always on the same level (top or bottom) for the first half ( $n-2$ days) of the search. The Cop is searching from left to right, and the Robber started out to the right of the Cop (or at $A_{2}$, the first hideout searched), so eventually the Robber is caught (at $A_{n-1}$, if not earlier).\n\nIf the Robber chose an odd-numbered hideout on the first day, then the Cop and Robber will be on different levels for the first half of the search, but on the same level for the second half. For the second half of the search, the Robber is to the left of the Cop (or possibly at $A_{n-1}$, which would happen if the Robber moved unwisely or if he spent day $n-2$ at hideout $A_{n}$ ). But now the Cop is searching from right to left, so again, the Cop will eventually catch the Robber.\n\nThe same strategy can be justified without using the zig-zag diagram above. Suppose that the Robber is at hideout $A_{R}$ on a given day, and the Cop searches hideout $A_{C}$. Whenever the Cop moves to the next hideout or the preceding hideout, $C$ changes by \\pm 1 , and the Robber's constraint forces $R$ to change by \\pm 1 . Thus if the Cop uses the strategy above, on each of the first $n-2$ days, either the difference $R-C$ stays the same or it decreases by 2 . If on the first day $R-C$ is even, either $R-C$ is 0 (the Robber was at $A_{2}$ and is caught immediately) or is positive (because $C=2$ and $R \\geq 4$ ). Because the difference $R-C$ is even and decreases (if at all) by 2 each day, it cannot go from positive to negative without being zero. If on the first day $R-C$ is odd, then the Robber avoids capture through day $n-2$ (because $R-C$ is still odd), but then on day $n-1, R$ changes (by \\pm 1 ) while $C$ does not. So $R-C$ is now even (and is either 0 or negative), and henceforth either remains the same or increases by 2 each day, so again, $R-C$ must be zero at some point between day $n-1$ and day $2 n-4$, inclusive."
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2,811
|
This Question involves one Robber and one or more Cops. After robbing a bank, the Robber retreats to a network of hideouts, represented by dots in the diagram below. Every day, the Robber stays holed up in a single hideout, and every night, the Robber moves to an adjacent hideout. Two hideouts are adjacent if and only if they are connected by an edge in the diagram, also called a hideout map (or map). For the purposes of this Power Question, the map must be connected; that is, given any two hideouts, there must be a path from one to the other. To clarify, the Robber may not stay in the same hideout for two consecutive days, although he may return to a hideout he has previously visited. For example, in the map below, if the Robber holes up in hideout $C$ for day 1 , then he would have to move to $B$ for day 2 , and would then have to move to either $A, C$, or $D$ on day 3.
<image_1>
Every day, each Cop searches one hideout: the Cops know the location of all hideouts and which hideouts are adjacent to which. Cops are thorough searchers, so if the Robber is present in the hideout searched, he is found and arrested. If the Robber is not present in the hideout searched, his location is not revealed. That is, the Cops only know that the Robber was not caught at any of the hideouts searched; they get no specific information (other than what they can derive by logic) about what hideout he was in. Cops are not constrained by edges on the map: a Cop may search any hideout on any day, regardless of whether it is adjacent to the hideout searched the previous day. A Cop may search the same hideout on consecutive days, and multiple Cops may search different hideouts on the same day. In the map above, a Cop could search $A$ on day 1 and day 2, and then search $C$ on day 3 .
The focus of this Power Question is to determine, given a hideout map and a fixed number of Cops, whether the Cops can be sure of catching the Robber within some time limit.
Map Notation: The following notation may be useful when writing your solutions. For a map $M$, let $h(M)$ be the number of hideouts and $e(M)$ be the number of edges in $M$. The safety of a hideout $H$ is the number of hideouts adjacent to $H$, and is denoted by $s(H)$.
The Cop number of a map $M$, denoted $C(M)$, is the minimum number of Cops required to guarantee that the Robber is caught.
The police want to catch the Robber with a minimum number of Cops, but time is of the essence. For a map $M$ and a fixed number of Cops $c \geq C(M)$, define the capture time, denoted $D(M, c)$, to be the minimum number of days required to guarantee a capture using $c$ Cops. For example, in the graph below, if three Cops are deployed, they might catch the Robber in the first day, but if they don't, there is a strategy that will guarantee they will capture the Robber within two days. Therefore the capture time is $D\left(\mathcal{C}_{6}, 3\right)=2$.
<image_2>
Definition: The workday number of $M$, denoted $W(M)$, is the minimum number of Cop workdays needed to guarantee the Robber's capture. For example, a strategy that guarantees capture within three days using 17 Cops on the first day, 11 Cops on the second day, and only 6 Cops on the third day would require a total of $17+11+6=34$ Cop workdays.
Let $M$ be a map with $n \geq 3$ hideouts. Prove that $2 \leq W(M) \leq n$, and that these bounds cannot be improved. In other words, prove that for each $n \geq 3$, there exist maps $M_{1}$ and $M_{2}$ such that $W\left(M_{1}\right)=2$ and $W\left(M_{2}\right)=n$.
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"A single Cop can only search one hideout in a day, so as long as $M$ has two or more hideouts, there is no strategy that guarantees that a lone Cop captures the Robber the first day. Then either more than one Cop will have to search on the first day, or a lone Cop will have to search for at least 2 days; in either of these cases, $W(M) \\geq 2$. On the other hand, $n$ Cops can always guarantee a capture simply by searching all $n$ hideouts on the first day, so $W(M) \\leq n$.\n\nTo show that the lower bound cannot be improved, consider the star on $n$ hideouts; that is, the map $\\mathcal{S}_{n}$ with one central hideout connected to $n-1$ outer hideouts, none of which is connected to any other hideout. (The map in 1a is $\\mathcal{S}_{7}$.) Then $W\\left(\\mathcal{S}_{n}\\right)=2$, because a single Cop can search the central hideout for 2 days; if the Robber is at one of the outer hideouts on the first day, he must go to the central hideout on the second day. For the upper bound, the complete map $\\mathcal{K}_{n}$ is an example of a map with $W(M)=n$. As argued above, the minimum number of Cops needed to guarantee a catch is $n-1$, and using $n-1$ Cops requires two days, for a total of $2 n-2$ Cop workdays. So the optimal strategy for $\\mathcal{K}_{n}$ (see 2c) is to use $n$ Cops for a single day."
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2,812
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This Question involves one Robber and one or more Cops. After robbing a bank, the Robber retreats to a network of hideouts, represented by dots in the diagram below. Every day, the Robber stays holed up in a single hideout, and every night, the Robber moves to an adjacent hideout. Two hideouts are adjacent if and only if they are connected by an edge in the diagram, also called a hideout map (or map). For the purposes of this Power Question, the map must be connected; that is, given any two hideouts, there must be a path from one to the other. To clarify, the Robber may not stay in the same hideout for two consecutive days, although he may return to a hideout he has previously visited. For example, in the map below, if the Robber holes up in hideout $C$ for day 1 , then he would have to move to $B$ for day 2 , and would then have to move to either $A, C$, or $D$ on day 3.
<image_1>
Every day, each Cop searches one hideout: the Cops know the location of all hideouts and which hideouts are adjacent to which. Cops are thorough searchers, so if the Robber is present in the hideout searched, he is found and arrested. If the Robber is not present in the hideout searched, his location is not revealed. That is, the Cops only know that the Robber was not caught at any of the hideouts searched; they get no specific information (other than what they can derive by logic) about what hideout he was in. Cops are not constrained by edges on the map: a Cop may search any hideout on any day, regardless of whether it is adjacent to the hideout searched the previous day. A Cop may search the same hideout on consecutive days, and multiple Cops may search different hideouts on the same day. In the map above, a Cop could search $A$ on day 1 and day 2, and then search $C$ on day 3 .
The focus of this Power Question is to determine, given a hideout map and a fixed number of Cops, whether the Cops can be sure of catching the Robber within some time limit.
Map Notation: The following notation may be useful when writing your solutions. For a map $M$, let $h(M)$ be the number of hideouts and $e(M)$ be the number of edges in $M$. The safety of a hideout $H$ is the number of hideouts adjacent to $H$, and is denoted by $s(H)$.
The Cop number of a map $M$, denoted $C(M)$, is the minimum number of Cops required to guarantee that the Robber is caught.
The police want to catch the Robber with a minimum number of Cops, but time is of the essence. For a map $M$ and a fixed number of Cops $c \geq C(M)$, define the capture time, denoted $D(M, c)$, to be the minimum number of days required to guarantee a capture using $c$ Cops. For example, in the graph below, if three Cops are deployed, they might catch the Robber in the first day, but if they don't, there is a strategy that will guarantee they will capture the Robber within two days. Therefore the capture time is $D\left(\mathcal{C}_{6}, 3\right)=2$.
<image_2>
Definition: A map is bipartite if it can be partitioned into two sets of hideouts, $\mathcal{A}$ and $\mathcal{B}$, such that $\mathcal{A} \cap \mathcal{B}=\emptyset$, and each hideout in $\mathcal{A}$ is adjacent only to hideouts in $\mathcal{B}$, and each hideout in $\mathcal{B}$ is adjacent only to hideouts in $\mathcal{A}$.
Prove that if $M$ is bipartite, then $C(M) \leq n / 2$.
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"Suppose $M$ is bipartite, and let $\\mathcal{A}$ and $\\mathcal{B}$ be the sets of hideouts referenced in the definition. Because $\\mathcal{A}$ and $\\mathcal{B}$ are disjoint, either $|\\mathcal{A}| \\leq n / 2$ or $|\\mathcal{B}| \\leq n / 2$ or both. Without loss of generality, suppose that $|\\mathcal{A}| \\leq n / 2$. Then position Cops at each hideout in $\\mathcal{A}$ for two days. If the Robber was initially on a hideout in $\\mathcal{B}$, he must move the following day, and because no hideout in $\\mathcal{B}$ is connected to any other hideout in $\\mathcal{B}$, his new hideout must be a hideout in $\\mathcal{A}$."
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2,813
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This Question involves one Robber and one or more Cops. After robbing a bank, the Robber retreats to a network of hideouts, represented by dots in the diagram below. Every day, the Robber stays holed up in a single hideout, and every night, the Robber moves to an adjacent hideout. Two hideouts are adjacent if and only if they are connected by an edge in the diagram, also called a hideout map (or map). For the purposes of this Power Question, the map must be connected; that is, given any two hideouts, there must be a path from one to the other. To clarify, the Robber may not stay in the same hideout for two consecutive days, although he may return to a hideout he has previously visited. For example, in the map below, if the Robber holes up in hideout $C$ for day 1 , then he would have to move to $B$ for day 2 , and would then have to move to either $A, C$, or $D$ on day 3.
<image_1>
Every day, each Cop searches one hideout: the Cops know the location of all hideouts and which hideouts are adjacent to which. Cops are thorough searchers, so if the Robber is present in the hideout searched, he is found and arrested. If the Robber is not present in the hideout searched, his location is not revealed. That is, the Cops only know that the Robber was not caught at any of the hideouts searched; they get no specific information (other than what they can derive by logic) about what hideout he was in. Cops are not constrained by edges on the map: a Cop may search any hideout on any day, regardless of whether it is adjacent to the hideout searched the previous day. A Cop may search the same hideout on consecutive days, and multiple Cops may search different hideouts on the same day. In the map above, a Cop could search $A$ on day 1 and day 2, and then search $C$ on day 3 .
The focus of this Power Question is to determine, given a hideout map and a fixed number of Cops, whether the Cops can be sure of catching the Robber within some time limit.
Map Notation: The following notation may be useful when writing your solutions. For a map $M$, let $h(M)$ be the number of hideouts and $e(M)$ be the number of edges in $M$. The safety of a hideout $H$ is the number of hideouts adjacent to $H$, and is denoted by $s(H)$.
The Cop number of a map $M$, denoted $C(M)$, is the minimum number of Cops required to guarantee that the Robber is caught.
The police want to catch the Robber with a minimum number of Cops, but time is of the essence. For a map $M$ and a fixed number of Cops $c \geq C(M)$, define the capture time, denoted $D(M, c)$, to be the minimum number of days required to guarantee a capture using $c$ Cops. For example, in the graph below, if three Cops are deployed, they might catch the Robber in the first day, but if they don't, there is a strategy that will guarantee they will capture the Robber within two days. Therefore the capture time is $D\left(\mathcal{C}_{6}, 3\right)=2$.
<image_2>
Definition: A map is bipartite if it can be partitioned into two sets of hideouts, $\mathcal{A}$ and $\mathcal{B}$, such that $\mathcal{A} \cap \mathcal{B}=\emptyset$, and each hideout in $\mathcal{A}$ is adjacent only to hideouts in $\mathcal{B}$, and each hideout in $\mathcal{B}$ is adjacent only to hideouts in $\mathcal{A}$.
Prove that $C(M) \leq n / 2$ for any map $M$ with the property that, for all hideouts $H_{1}$ and $H_{2}$, either all paths from $H_{1}$ to $H_{2}$ contain an odd number of edges, or all paths from $H_{1}$ to $H_{2}$ contain an even number of edges.
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"The given condition actually implies that the graph is bipartite. Let $A_{1}$ be a hideout in $M$, and let $\\mathcal{A}$ be the set of all hideouts $V$ such that all paths from $A_{1}$ to $V$ have an even number of edges, as well as $A_{1}$ itself; let $\\mathcal{B}$ be the set of all other hideouts in $M$. Notice that there are no edges from $A_{1}$ to any other element $A_{i}$ in $\\mathcal{A}$, because if there were, that edge would create a path from $A_{1}$ to $A_{i}$ with an odd number of edges (namely 1). So $A_{1}$ has edges only to hideouts in $\\mathcal{B}$. In fact, there can be no edge from any hideout $A_{i}$ in $\\mathcal{A}$ to any other hideout $A_{j}$ in $\\mathcal{A}$, because if there were, there would be a path with an odd number of edges from $A_{1}$ to $A_{j}$ via $A_{i}$. Similarly, there can be no edge from any hideout $B_{i}$ in $\\mathcal{B}$ to any other hideout $B_{j}$ in $\\mathcal{B}$, because if there were such an edge, there would be a path from $A_{1}$ to $B_{j}$ with an even number of edges via $B_{i}$. Thus hideouts in $\\mathcal{B}$ are connected only to hideouts in $\\mathcal{A}$ and vice versa; hence $M$ is bipartite."
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2,814
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This Question involves one Robber and one or more Cops. After robbing a bank, the Robber retreats to a network of hideouts, represented by dots in the diagram below. Every day, the Robber stays holed up in a single hideout, and every night, the Robber moves to an adjacent hideout. Two hideouts are adjacent if and only if they are connected by an edge in the diagram, also called a hideout map (or map). For the purposes of this Power Question, the map must be connected; that is, given any two hideouts, there must be a path from one to the other. To clarify, the Robber may not stay in the same hideout for two consecutive days, although he may return to a hideout he has previously visited. For example, in the map below, if the Robber holes up in hideout $C$ for day 1 , then he would have to move to $B$ for day 2 , and would then have to move to either $A, C$, or $D$ on day 3.
<image_1>
Every day, each Cop searches one hideout: the Cops know the location of all hideouts and which hideouts are adjacent to which. Cops are thorough searchers, so if the Robber is present in the hideout searched, he is found and arrested. If the Robber is not present in the hideout searched, his location is not revealed. That is, the Cops only know that the Robber was not caught at any of the hideouts searched; they get no specific information (other than what they can derive by logic) about what hideout he was in. Cops are not constrained by edges on the map: a Cop may search any hideout on any day, regardless of whether it is adjacent to the hideout searched the previous day. A Cop may search the same hideout on consecutive days, and multiple Cops may search different hideouts on the same day. In the map above, a Cop could search $A$ on day 1 and day 2, and then search $C$ on day 3 .
The focus of this Power Question is to determine, given a hideout map and a fixed number of Cops, whether the Cops can be sure of catching the Robber within some time limit.
Map Notation: The following notation may be useful when writing your solutions. For a map $M$, let $h(M)$ be the number of hideouts and $e(M)$ be the number of edges in $M$. The safety of a hideout $H$ is the number of hideouts adjacent to $H$, and is denoted by $s(H)$.
The Cop number of a map $M$, denoted $C(M)$, is the minimum number of Cops required to guarantee that the Robber is caught.
The police want to catch the Robber with a minimum number of Cops, but time is of the essence. For a map $M$ and a fixed number of Cops $c \geq C(M)$, define the capture time, denoted $D(M, c)$, to be the minimum number of days required to guarantee a capture using $c$ Cops. For example, in the graph below, if three Cops are deployed, they might catch the Robber in the first day, but if they don't, there is a strategy that will guarantee they will capture the Robber within two days. Therefore the capture time is $D\left(\mathcal{C}_{6}, 3\right)=2$.
<image_2>
Definition: A map is bipartite if it can be partitioned into two sets of hideouts, $\mathcal{A}$ and $\mathcal{B}$, such that $\mathcal{A} \cap \mathcal{B}=\emptyset$, and each hideout in $\mathcal{A}$ is adjacent only to hideouts in $\mathcal{B}$, and each hideout in $\mathcal{B}$ is adjacent only to hideouts in $\mathcal{A}$.
A map $M$ is called $k$-perfect if its hideout set $H$ can be partitioned into equal-sized subsets $\mathcal{A}_{1}, \mathcal{A}_{2}, \ldots, \mathcal{A}_{k}$ such that for any $j$, the hideouts of $\mathcal{A}_{j}$ are only adjacent to hideouts in $\mathcal{A}_{j+1}$ or $\mathcal{A}_{j-1}$. (The indices are taken modulo $k$ : the hideouts of $\mathcal{A}_{1}$ may be adjacent to the hideouts of $\mathcal{A}_{k}$.) Show that if $M$ is $k$-perfect, then $C(M) \leq \frac{2 n}{k}$.
|
[
"Because a Robber in $\\mathcal{A}_{i}$ can only move to a hideout in $\\mathcal{A}_{i-1}$ or $\\mathcal{A}_{i+1}$, this map is essentially the same as the cyclic map $\\mathcal{C}_{k}$. So the Cops should apply a similar strategy. First, position $n / k$ Cops at the hideouts of set $\\mathcal{A}_{1}$ and $n / k$ Cops at the hideouts of set $\\mathcal{A}_{2}$. On day 2 , leave the first $n / k$ Cops at $\\mathcal{A}_{1}$, but move the second group of Cops to $\\mathcal{A}_{3}$. Continue until the second set of Cops is at $\\mathcal{A}_{k}$; then \"wait\" one turn (search $\\mathcal{A}_{k}$ again), and then search backward $\\mathcal{A}_{k-1}$, $\\mathcal{A}_{k-2}$, etc."
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Combinatorics
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Math
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2,815
|
This Question involves one Robber and one or more Cops. After robbing a bank, the Robber retreats to a network of hideouts, represented by dots in the diagram below. Every day, the Robber stays holed up in a single hideout, and every night, the Robber moves to an adjacent hideout. Two hideouts are adjacent if and only if they are connected by an edge in the diagram, also called a hideout map (or map). For the purposes of this Power Question, the map must be connected; that is, given any two hideouts, there must be a path from one to the other. To clarify, the Robber may not stay in the same hideout for two consecutive days, although he may return to a hideout he has previously visited. For example, in the map below, if the Robber holes up in hideout $C$ for day 1 , then he would have to move to $B$ for day 2 , and would then have to move to either $A, C$, or $D$ on day 3.
<image_1>
Every day, each Cop searches one hideout: the Cops know the location of all hideouts and which hideouts are adjacent to which. Cops are thorough searchers, so if the Robber is present in the hideout searched, he is found and arrested. If the Robber is not present in the hideout searched, his location is not revealed. That is, the Cops only know that the Robber was not caught at any of the hideouts searched; they get no specific information (other than what they can derive by logic) about what hideout he was in. Cops are not constrained by edges on the map: a Cop may search any hideout on any day, regardless of whether it is adjacent to the hideout searched the previous day. A Cop may search the same hideout on consecutive days, and multiple Cops may search different hideouts on the same day. In the map above, a Cop could search $A$ on day 1 and day 2, and then search $C$ on day 3 .
The focus of this Power Question is to determine, given a hideout map and a fixed number of Cops, whether the Cops can be sure of catching the Robber within some time limit.
Map Notation: The following notation may be useful when writing your solutions. For a map $M$, let $h(M)$ be the number of hideouts and $e(M)$ be the number of edges in $M$. The safety of a hideout $H$ is the number of hideouts adjacent to $H$, and is denoted by $s(H)$.
The Cop number of a map $M$, denoted $C(M)$, is the minimum number of Cops required to guarantee that the Robber is caught.
The police want to catch the Robber with a minimum number of Cops, but time is of the essence. For a map $M$ and a fixed number of Cops $c \geq C(M)$, define the capture time, denoted $D(M, c)$, to be the minimum number of days required to guarantee a capture using $c$ Cops. For example, in the graph below, if three Cops are deployed, they might catch the Robber in the first day, but if they don't, there is a strategy that will guarantee they will capture the Robber within two days. Therefore the capture time is $D\left(\mathcal{C}_{6}, 3\right)=2$.
<image_2>
Definition: The workday number of $M$, denoted $W(M)$, is the minimum number of Cop workdays needed to guarantee the Robber's capture. For example, a strategy that guarantees capture within three days using 17 Cops on the first day, 11 Cops on the second day, and only 6 Cops on the third day would require a total of $17+11+6=34$ Cop workdays.
Definition: A map is bipartite if it can be partitioned into two sets of hideouts, $\mathcal{A}$ and $\mathcal{B}$, such that $\mathcal{A} \cap \mathcal{B}=\emptyset$, and each hideout in $\mathcal{A}$ is adjacent only to hideouts in $\mathcal{B}$, and each hideout in $\mathcal{B}$ is adjacent only to hideouts in $\mathcal{A}$.
Find an example of a map $M$ with 2012 hideouts such that $C(M)=17$ and $W(M)=34$, or prove that no such map exists.
|
[
"There are many examples. Perhaps the simplest to describe is the complete bipartite map on the hideouts $\\mathcal{A}=\\left\\{A_{1}, A_{2}, \\ldots, A_{17}\\right\\}$ and $\\mathcal{B}=\\left\\{B_{1}, B_{2}, \\ldots, B_{1995}\\right\\}$, which is often denoted $M=\\mathcal{K}_{17,1995}$. That is, let $M$ be the map with hideouts $\\mathcal{A} \\cup \\mathcal{B}$ such that $A_{i}$ is adjacent to $B_{j}$ for all $i$ and $j$, and that $A_{i}$ is not adjacent to $A_{j}$, nor is $B_{i}$ adjacent to $B_{j}$, for any $i$ and $j$.\n\nIf 17 Cops search the 17 hideouts in $\\mathcal{A}$ for two consecutive days, then they are guaranteed to catch the Robber. This shows that $C(M) \\leq 17$ and that $W(M) \\leq 34$. If fewer than 17 Cops search on a given day, then the Robber has at least one safe hideout: he has exactly one choice if the Cops search 16 of the hideouts $A_{i}$ and the Robber was hiding at some $B_{j}$ the previous day, 1995 choices if the Robber was hiding at some $A_{i}$ the previous day. Thus $C(M)=17$.\n\nOn the other hand, fewer than 34 Cop workdays cannot guarantee catching the Robber. Unless the Cops search every $A_{i}$ (or every $B_{j}$ ) on a given day, they gain no information about where the Robber is (unless he is unlucky enough to be caught).\n\nThe complete bipartite map is not the simplest in terms of the number of edges. Try to find a map with 2012 hideouts and as few edges as possible that has a Cop number of 17 and a workday number of 34 ."
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2,816
|
This Question involves one Robber and one or more Cops. After robbing a bank, the Robber retreats to a network of hideouts, represented by dots in the diagram below. Every day, the Robber stays holed up in a single hideout, and every night, the Robber moves to an adjacent hideout. Two hideouts are adjacent if and only if they are connected by an edge in the diagram, also called a hideout map (or map). For the purposes of this Power Question, the map must be connected; that is, given any two hideouts, there must be a path from one to the other. To clarify, the Robber may not stay in the same hideout for two consecutive days, although he may return to a hideout he has previously visited. For example, in the map below, if the Robber holes up in hideout $C$ for day 1 , then he would have to move to $B$ for day 2 , and would then have to move to either $A, C$, or $D$ on day 3.
<image_1>
Every day, each Cop searches one hideout: the Cops know the location of all hideouts and which hideouts are adjacent to which. Cops are thorough searchers, so if the Robber is present in the hideout searched, he is found and arrested. If the Robber is not present in the hideout searched, his location is not revealed. That is, the Cops only know that the Robber was not caught at any of the hideouts searched; they get no specific information (other than what they can derive by logic) about what hideout he was in. Cops are not constrained by edges on the map: a Cop may search any hideout on any day, regardless of whether it is adjacent to the hideout searched the previous day. A Cop may search the same hideout on consecutive days, and multiple Cops may search different hideouts on the same day. In the map above, a Cop could search $A$ on day 1 and day 2, and then search $C$ on day 3 .
The focus of this Power Question is to determine, given a hideout map and a fixed number of Cops, whether the Cops can be sure of catching the Robber within some time limit.
Map Notation: The following notation may be useful when writing your solutions. For a map $M$, let $h(M)$ be the number of hideouts and $e(M)$ be the number of edges in $M$. The safety of a hideout $H$ is the number of hideouts adjacent to $H$, and is denoted by $s(H)$.
The Cop number of a map $M$, denoted $C(M)$, is the minimum number of Cops required to guarantee that the Robber is caught.
The police want to catch the Robber with a minimum number of Cops, but time is of the essence. For a map $M$ and a fixed number of Cops $c \geq C(M)$, define the capture time, denoted $D(M, c)$, to be the minimum number of days required to guarantee a capture using $c$ Cops. For example, in the graph below, if three Cops are deployed, they might catch the Robber in the first day, but if they don't, there is a strategy that will guarantee they will capture the Robber within two days. Therefore the capture time is $D\left(\mathcal{C}_{6}, 3\right)=2$.
<image_2>
Definition: The workday number of $M$, denoted $W(M)$, is the minimum number of Cop workdays needed to guarantee the Robber's capture. For example, a strategy that guarantees capture within three days using 17 Cops on the first day, 11 Cops on the second day, and only 6 Cops on the third day would require a total of $17+11+6=34$ Cop workdays.
Find an example of a map $M$ with 4 or more hideouts such that $W(M)=3$, or prove that no such map exists.
|
[
"We denote a map as star, that is, the map $\\mathcal{S}_{n}$ with one central hideout connected to $n-1$ outer hideouts, none of which is connected to any other hideout.\n\nNo such map exists. Let $M$ be a map with at least four hideouts. The proof below shows that either $W(M)>3$ or else $M$ is a star, in which case $W(M)=2$.\n\nFirst, suppose that there are four distinct hideouts $A_{1}, A_{2}, B_{1}$, and $B_{2}$ such that $A_{1}$ and $A_{2}$ are adjacent, as are $B_{1}$ and $B_{2}$. If the Cops make fewer than four searches, then they cannot search $\\left\\{A_{1}, A_{2}\\right\\}$ twice and also search $\\left\\{B_{1}, B_{2}\\right\\}$ twice. Without loss of generality, assume the Cops search $\\left\\{A_{1}, A_{2}\\right\\}$ at most once. Then the Robber can evade capture by moving from $A_{1}$ to $A_{2}$ and back again, provided that he is lucky enough to start off at the right one. Thus\n\n\n\n$W(M)>3$ in this case. For the second case, assume that whenever $A_{1}$ is adjacent to $A_{2}$ and $B_{1}$ is adjacent to $B_{2}$, the four hideouts are not distinct. Start with two adjacent hideouts, and call them $A_{1}$ and $A_{2}$. Consider any hideout, say $B$, that is not one of these two. Then $B$ must be adjacent to some hideout, say $C$. By assumption, $A_{1}, A_{2}, B$, and $C$ are not distinct, so $C=A_{1}$ or $C=A_{2}$. That is, every hideout $B$ is adjacent to $A_{1}$ or to $A_{2}$.\n\nIf there are hideouts $B_{1}$ and $B_{2}$, distinct from $A_{1}$ and $A_{2}$ and from each other, such that $B_{1}$ is adjacent to $A_{1}$ and $B_{2}$ is adjacent to $A_{2}$, then it clearly violates the assumption of this case. That is, either every hideout $B$, distinct from $A_{1}$ and $A_{2}$, is adjacent to $A_{1}$, or every such hideout is adjacent to $A_{2}$. Without loss of generality, assume the former.\n\nBecause every hideout is adjacent to $A_{1}$ (except for $A_{1}$ itself), the foregoing proves that the map $M$ is a star. It remains to show that there are no \"extra\" edges in the map. For the sake of contradiction, suppose that $B_{1}$ and $B_{2}$ are adjacent hideouts, both distinct from $A_{1}$. Because the map has at least four hideouts, choose one distinct from these three and call it $A_{2}$. Then these four hideouts violate the assumption of this case."
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Theorem proof
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Combinatorics
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Math
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English
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2,873
|
An $\boldsymbol{n}$-label is a permutation of the numbers 1 through $n$. For example, $J=35214$ is a 5 -label and $K=132$ is a 3 -label. For a fixed positive integer $p$, where $p \leq n$, consider consecutive blocks of $p$ numbers in an $n$-label. For example, when $p=3$ and $L=263415$, the blocks are 263,634,341, and 415. We can associate to each of these blocks a $p$-label that corresponds to the relative order of the numbers in that block. For $L=263415$, we get the following:
$$
\underline{263} 415 \rightarrow 132 ; \quad 2 \underline{63415} \rightarrow 312 ; \quad 26 \underline{341} 5 \rightarrow 231 ; \quad 263 \underline{415} \rightarrow 213
$$
Moving from left to right in the $n$-label, there are $n-p+1$ such blocks, which means we obtain an $(n-p+1)$-tuple of $p$-labels. For $L=263415$, we get the 4 -tuple $(132,312,231,213)$. We will call this $(n-p+1)$-tuple the $\boldsymbol{p}$-signature of $L$ (or signature, if $p$ is clear from the context) and denote it by $S_{p}[L]$; the $p$-labels in the signature are called windows. For $L=263415$, the windows are $132,312,231$, and 213 , and we write
$$
S_{3}[263415]=(132,312,231,213)
$$
More generally, we will call any $(n-p+1)$-tuple of $p$-labels a $p$-signature, even if we do not know of an $n$-label to which it corresponds (and even if no such label exists). A signature that occurs for exactly one $n$-label is called unique, and a signature that doesn't occur for any $n$-labels is called impossible. A possible signature is one that occurs for at least one $n$-label.
In this power question, you will be asked to analyze some of the properties of labels and signatures.
We can associate a shape to a given 2-signature: a diagram of up and down steps that indicates the relative order of adjacent numbers. For example, the following shape corresponds to the 2-signature $(12,12,12,21,12,21)$ :
<image_1>
A 7-label with this 2-signature corresponds to placing the numbers 1 through 7 at the nodes above so that numbers increase with each up step and decrease with each down step. The 7-label 2347165 is shown below:
<image_2>
Prove that the following signature is possible.
$(123,132,213)$,
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[
"The signature is possible, because it is the 3 -signature of 12435"
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Multimodal
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Theorem proof
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Combinatorics
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Math
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English
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2,874
|
An $\boldsymbol{n}$-label is a permutation of the numbers 1 through $n$. For example, $J=35214$ is a 5 -label and $K=132$ is a 3 -label. For a fixed positive integer $p$, where $p \leq n$, consider consecutive blocks of $p$ numbers in an $n$-label. For example, when $p=3$ and $L=263415$, the blocks are 263,634,341, and 415. We can associate to each of these blocks a $p$-label that corresponds to the relative order of the numbers in that block. For $L=263415$, we get the following:
$$
\underline{263} 415 \rightarrow 132 ; \quad 2 \underline{63415} \rightarrow 312 ; \quad 26 \underline{341} 5 \rightarrow 231 ; \quad 263 \underline{415} \rightarrow 213
$$
Moving from left to right in the $n$-label, there are $n-p+1$ such blocks, which means we obtain an $(n-p+1)$-tuple of $p$-labels. For $L=263415$, we get the 4 -tuple $(132,312,231,213)$. We will call this $(n-p+1)$-tuple the $\boldsymbol{p}$-signature of $L$ (or signature, if $p$ is clear from the context) and denote it by $S_{p}[L]$; the $p$-labels in the signature are called windows. For $L=263415$, the windows are $132,312,231$, and 213 , and we write
$$
S_{3}[263415]=(132,312,231,213)
$$
More generally, we will call any $(n-p+1)$-tuple of $p$-labels a $p$-signature, even if we do not know of an $n$-label to which it corresponds (and even if no such label exists). A signature that occurs for exactly one $n$-label is called unique, and a signature that doesn't occur for any $n$-labels is called impossible. A possible signature is one that occurs for at least one $n$-label.
In this power question, you will be asked to analyze some of the properties of labels and signatures.
We can associate a shape to a given 2-signature: a diagram of up and down steps that indicates the relative order of adjacent numbers. For example, the following shape corresponds to the 2-signature $(12,12,12,21,12,21)$ :
<image_1>
A 7-label with this 2-signature corresponds to placing the numbers 1 through 7 at the nodes above so that numbers increase with each up step and decrease with each down step. The 7-label 2347165 is shown below:
<image_2>
Prove that the following signature is impossible:
$(321,312,213)$.
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[
"The signature is impossible. Let a 5 -label be $a_{1}, a_{2}, a_{3}, a_{4}, a_{5}$. The second window of (ii) implies $a_{3}<a_{4}$, whereas the third window implies $a_{3}>a_{4}$, a contradiction."
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Multimodal
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Competition
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Theorem proof
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Combinatorics
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Math
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English
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2,877
|
An $\boldsymbol{n}$-label is a permutation of the numbers 1 through $n$. For example, $J=35214$ is a 5 -label and $K=132$ is a 3 -label. For a fixed positive integer $p$, where $p \leq n$, consider consecutive blocks of $p$ numbers in an $n$-label. For example, when $p=3$ and $L=263415$, the blocks are 263,634,341, and 415. We can associate to each of these blocks a $p$-label that corresponds to the relative order of the numbers in that block. For $L=263415$, we get the following:
$$
\underline{263} 415 \rightarrow 132 ; \quad 2 \underline{63415} \rightarrow 312 ; \quad 26 \underline{341} 5 \rightarrow 231 ; \quad 263 \underline{415} \rightarrow 213
$$
Moving from left to right in the $n$-label, there are $n-p+1$ such blocks, which means we obtain an $(n-p+1)$-tuple of $p$-labels. For $L=263415$, we get the 4 -tuple $(132,312,231,213)$. We will call this $(n-p+1)$-tuple the $\boldsymbol{p}$-signature of $L$ (or signature, if $p$ is clear from the context) and denote it by $S_{p}[L]$; the $p$-labels in the signature are called windows. For $L=263415$, the windows are $132,312,231$, and 213 , and we write
$$
S_{3}[263415]=(132,312,231,213)
$$
More generally, we will call any $(n-p+1)$-tuple of $p$-labels a $p$-signature, even if we do not know of an $n$-label to which it corresponds (and even if no such label exists). A signature that occurs for exactly one $n$-label is called unique, and a signature that doesn't occur for any $n$-labels is called impossible. A possible signature is one that occurs for at least one $n$-label.
In this power question, you will be asked to analyze some of the properties of labels and signatures.
We can associate a shape to a given 2-signature: a diagram of up and down steps that indicates the relative order of adjacent numbers. For example, the following shape corresponds to the 2-signature $(12,12,12,21,12,21)$ :
<image_1>
A 7-label with this 2-signature corresponds to placing the numbers 1 through 7 at the nodes above so that numbers increase with each up step and decrease with each down step. The 7-label 2347165 is shown below:
<image_2>
Show that $(312,231,312,132)$ is not a unique 3 -signature.
|
[
"The $p$-signature is not unique because it equals both $S_{3}[625143]$ and $S_{3}[635142]$."
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Multimodal
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Competition
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Theorem proof
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Combinatorics
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Math
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English
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2,878
|
An $\boldsymbol{n}$-label is a permutation of the numbers 1 through $n$. For example, $J=35214$ is a 5 -label and $K=132$ is a 3 -label. For a fixed positive integer $p$, where $p \leq n$, consider consecutive blocks of $p$ numbers in an $n$-label. For example, when $p=3$ and $L=263415$, the blocks are 263,634,341, and 415. We can associate to each of these blocks a $p$-label that corresponds to the relative order of the numbers in that block. For $L=263415$, we get the following:
$$
\underline{263} 415 \rightarrow 132 ; \quad 2 \underline{63415} \rightarrow 312 ; \quad 26 \underline{341} 5 \rightarrow 231 ; \quad 263 \underline{415} \rightarrow 213
$$
Moving from left to right in the $n$-label, there are $n-p+1$ such blocks, which means we obtain an $(n-p+1)$-tuple of $p$-labels. For $L=263415$, we get the 4 -tuple $(132,312,231,213)$. We will call this $(n-p+1)$-tuple the $\boldsymbol{p}$-signature of $L$ (or signature, if $p$ is clear from the context) and denote it by $S_{p}[L]$; the $p$-labels in the signature are called windows. For $L=263415$, the windows are $132,312,231$, and 213 , and we write
$$
S_{3}[263415]=(132,312,231,213)
$$
More generally, we will call any $(n-p+1)$-tuple of $p$-labels a $p$-signature, even if we do not know of an $n$-label to which it corresponds (and even if no such label exists). A signature that occurs for exactly one $n$-label is called unique, and a signature that doesn't occur for any $n$-labels is called impossible. A possible signature is one that occurs for at least one $n$-label.
In this power question, you will be asked to analyze some of the properties of labels and signatures.
We can associate a shape to a given 2-signature: a diagram of up and down steps that indicates the relative order of adjacent numbers. For example, the following shape corresponds to the 2-signature $(12,12,12,21,12,21)$ :
<image_1>
A 7-label with this 2-signature corresponds to placing the numbers 1 through 7 at the nodes above so that numbers increase with each up step and decrease with each down step. The 7-label 2347165 is shown below:
<image_2>
Show that $(231,213,123,132)$ is a unique 3 -signature.
|
[
"Let $L=a_{1}, a_{2}, a_{3}, a_{4}, a_{5}, a_{6}$. We have $a_{4}<a_{6}<a_{5}$ (from window \\#4), $a_{3}<a_{1}<a_{2}$ (from window \\#1), and $a_{2}<a_{4}$ (from window \\#2). Linking these inequalities, we get\n\n$$\na_{3}<a_{1}<a_{2}<a_{4}<a_{6}<a_{5} \\quad \\Rightarrow \\quad L=231465\n$$\n\nso $S_{3}[L]$ is unique."
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Multimodal
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Competition
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Theorem proof
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Combinatorics
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Math
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English
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||
2,881
|
An $\boldsymbol{n}$-label is a permutation of the numbers 1 through $n$. For example, $J=35214$ is a 5 -label and $K=132$ is a 3 -label. For a fixed positive integer $p$, where $p \leq n$, consider consecutive blocks of $p$ numbers in an $n$-label. For example, when $p=3$ and $L=263415$, the blocks are 263,634,341, and 415. We can associate to each of these blocks a $p$-label that corresponds to the relative order of the numbers in that block. For $L=263415$, we get the following:
$$
\underline{263} 415 \rightarrow 132 ; \quad 2 \underline{63415} \rightarrow 312 ; \quad 26 \underline{341} 5 \rightarrow 231 ; \quad 263 \underline{415} \rightarrow 213
$$
Moving from left to right in the $n$-label, there are $n-p+1$ such blocks, which means we obtain an $(n-p+1)$-tuple of $p$-labels. For $L=263415$, we get the 4 -tuple $(132,312,231,213)$. We will call this $(n-p+1)$-tuple the $\boldsymbol{p}$-signature of $L$ (or signature, if $p$ is clear from the context) and denote it by $S_{p}[L]$; the $p$-labels in the signature are called windows. For $L=263415$, the windows are $132,312,231$, and 213 , and we write
$$
S_{3}[263415]=(132,312,231,213)
$$
More generally, we will call any $(n-p+1)$-tuple of $p$-labels a $p$-signature, even if we do not know of an $n$-label to which it corresponds (and even if no such label exists). A signature that occurs for exactly one $n$-label is called unique, and a signature that doesn't occur for any $n$-labels is called impossible. A possible signature is one that occurs for at least one $n$-label.
In this power question, you will be asked to analyze some of the properties of labels and signatures.
We can associate a shape to a given 2-signature: a diagram of up and down steps that indicates the relative order of adjacent numbers. For example, the following shape corresponds to the 2-signature $(12,12,12,21,12,21)$ :
<image_1>
A 7-label with this 2-signature corresponds to placing the numbers 1 through 7 at the nodes above so that numbers increase with each up step and decrease with each down step. The 7-label 2347165 is shown below:
<image_2>
Prove that $S_{5}[495138627]$ is unique.
|
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"Let $L=a_{1}, \\ldots, a_{9}$ and suppose $S_{5}[L]=S_{5}[495138627]=\\left(\\omega_{1}, \\ldots, \\omega_{5}\\right)$. Then we get the following inequalities:\n\n| $a_{4}<a_{8}$ | $\\left[\\right.$ from $\\left.\\omega_{4}\\right]$ | $a_{8}<a_{5}$ | $\\left[\\right.$ from $\\left.\\omega_{4}\\right]$ |\n| :--- | :--- | :--- | :--- |\n| $a_{5}<a_{1}$ | $\\left[\\right.$ from $\\left.\\omega_{1}\\right]$ | $a_{1}<a_{3}$ | $\\left[\\right.$ from $\\left.\\omega_{1}\\right]$ |\n| $a_{3}<a_{7}$ | $\\left[\\right.$ from $\\left.\\omega_{3}\\right]$ | $a_{7}<a_{9}$ | $\\left[\\right.$ from $\\left.\\omega_{5}\\right]$ |\n| $a_{9}<a_{6}$ | $\\left[\\right.$ from $\\left.\\omega_{5}\\right]$ | $a_{6}<a_{2}$ | [from $\\left.\\omega_{2}\\right]$ |\n\nCombining, we get $a_{4}<a_{8}<a_{5}<a_{1}<a_{3}<a_{7}<a_{9}<a_{6}<a_{2}$, which forces $a_{4}=1, a_{8}=2, \\ldots, a_{2}=9$. So the label $L$ is forced and $S_{5}[495138627]$ is therefore unique."
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2,883
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An $\boldsymbol{n}$-label is a permutation of the numbers 1 through $n$. For example, $J=35214$ is a 5 -label and $K=132$ is a 3 -label. For a fixed positive integer $p$, where $p \leq n$, consider consecutive blocks of $p$ numbers in an $n$-label. For example, when $p=3$ and $L=263415$, the blocks are 263,634,341, and 415. We can associate to each of these blocks a $p$-label that corresponds to the relative order of the numbers in that block. For $L=263415$, we get the following:
$$
\underline{263} 415 \rightarrow 132 ; \quad 2 \underline{63415} \rightarrow 312 ; \quad 26 \underline{341} 5 \rightarrow 231 ; \quad 263 \underline{415} \rightarrow 213
$$
Moving from left to right in the $n$-label, there are $n-p+1$ such blocks, which means we obtain an $(n-p+1)$-tuple of $p$-labels. For $L=263415$, we get the 4 -tuple $(132,312,231,213)$. We will call this $(n-p+1)$-tuple the $\boldsymbol{p}$-signature of $L$ (or signature, if $p$ is clear from the context) and denote it by $S_{p}[L]$; the $p$-labels in the signature are called windows. For $L=263415$, the windows are $132,312,231$, and 213 , and we write
$$
S_{3}[263415]=(132,312,231,213)
$$
More generally, we will call any $(n-p+1)$-tuple of $p$-labels a $p$-signature, even if we do not know of an $n$-label to which it corresponds (and even if no such label exists). A signature that occurs for exactly one $n$-label is called unique, and a signature that doesn't occur for any $n$-labels is called impossible. A possible signature is one that occurs for at least one $n$-label.
In this power question, you will be asked to analyze some of the properties of labels and signatures.
We can associate a shape to a given 2-signature: a diagram of up and down steps that indicates the relative order of adjacent numbers. For example, the following shape corresponds to the 2-signature $(12,12,12,21,12,21)$ :
<image_1>
A 7-label with this 2-signature corresponds to placing the numbers 1 through 7 at the nodes above so that numbers increase with each up step and decrease with each down step. The 7-label 2347165 is shown below:
<image_2>
Show that for each $k \geq 2$, the number of unique $2^{k-1}$-signatures on the set of $2^{k}$-labels is at least $2^{2^{k}-3}$.
|
[
"Let $s_{k}$ denote the number of such unique signatures. We proceed by induction with base case $k=2$. From $8(\\mathrm{c})$, a 2-signature for a label $L$ is unique if and only if consecutive numbers in $L$ appear together in some window. Because $k=2$, the consecutive numbers must be adjacent\n\n\nin the label. The $2^{2}$-labels 1234 and 4321 satisfy this condition, ${ }^{1}$ so their $2^{1}$-signatures are unique. Thus we have shown that $s_{2} \\geq 2=2^{2^{2}-3}$, and the base case is established.\n\nNow suppose $s_{k} \\geq 2^{2^{k}-3}$ for some $k \\geq 2$. Let $L_{k}$ be a $2^{k}$-label with a unique $2^{k-1}$-signature. Write $L_{k}=\\left(a_{1}, a_{2}, \\ldots, a_{2^{k}}\\right)$. We will expand $L_{k}$ to form a $2^{k+1}$ label by replacing each $a_{i}$ above with the numbers $2 a_{i}-1$ and $2 a_{i}$ (in some order). This process produces a valid $2^{k+1}$-label, because the numbers produced are all the integers from 1 to $2^{k+1}$. Furthermore, different $L_{k}$ 's will produce different labels: if the starting labels differ at place $i$, then the new labels will differ at places $2 i-1$ and $2 i$. Therefore, each starting label produces $2^{2^{k}}$ distinct $2^{k+1}$ labels through this process. Summarizing, each valid $2^{k}$-label can be expanded to produce $2^{2^{k}}$ distinct $2^{k+1}$-labels, none of which could be obtained by expanding any other $2^{k}$-label.\n\nIt remains to be shown that the new label has a unique $2^{k-1}$-signature. Because $L_{k}$ has a unique $2^{k-1}$-signature, for all $i \\leq 2^{k}-1$, both $i$ and $i+1$ appeared in some $2^{k-1}$-window. Therefore, there were fewer than $2^{k-1}-1$ numbers between $i$ and $i+1$. When the label is expanded, $2 i$ and $2 i-1$ are adjacent, $2 i+1$ and $2 i+2$ are adjacent, and $2 i$ and $2 i+1$ are fewer than $2 \\cdot\\left(2^{k-1}-1\\right)+2=2^{k}$ places apart. Thus, every pair of adjacent integers is within some $2^{k}$-window.\n\nSince each pair of consecutive integers in our new $2^{k+1}$-label coexists in some $2^{k}$-window for every possible such expansion of $L_{k}$, that means all $2^{2^{k}}$ ways of expanding $L_{k}$ to a $2^{k+1}$-label result in labels with unique $2^{k}$-signatures. We then get\n\n$$\n\\begin{aligned}\ns_{k+1} & \\geq 2^{2^{k}} \\cdot s_{k} \\\\\n& \\geq 2^{2^{k}} \\cdot 2^{2^{k}-3} \\\\\n& =2^{2^{k+1}-3}\n\\end{aligned}\n$$\n\nwhich completes the induction."
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2,928
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A king strapped for cash is forced to sell off his kingdom $U=\left\{(x, y): x^{2}+y^{2} \leq 1\right\}$. He sells the two circular plots $C$ and $C^{\prime}$ centered at $\left( \pm \frac{1}{2}, 0\right)$ with radius $\frac{1}{2}$. The retained parts of the kingdom form two regions, each bordered by three arcs of circles; in what follows, we will call such regions curvilinear triangles, or $c$-triangles ( $\mathrm{c} \triangle$ ) for short.
This sad day marks day 0 of a new fiscal era. Unfortunately, these drastic measures are not enough, and so each day thereafter, court geometers mark off the largest possible circle contained in each c-triangle in the remaining property. This circle is tangent to all three arcs of the c-triangle, and will be referred to as the incircle of the c-triangle. At the end of the day, all incircles demarcated that day are sold off, and the following day, the remaining c-triangles are partitioned in the same manner.
Some notation: when discussing mutually tangent circles (or arcs), it is convenient to refer to the curvature of a circle rather than its radius. We define curvature as follows. Suppose that circle $A$ of radius $r_{a}$ is externally tangent to circle $B$ of radius $r_{b}$. Then the curvatures of the circles are simply the reciprocals of their radii, $\frac{1}{r_{a}}$ and $\frac{1}{r_{b}}$. If circle $A$ is internally tangent to circle $B$, however, as in the right diagram below, the curvature of circle $A$ is still $\frac{1}{r_{a}}$, while the curvature of circle $B$ is $-\frac{1}{r_{b}}$, the opposite of the reciprocal of its radius.
<image_1>
Circle $A$ has curvature 2; circle $B$ has curvature 1 .
<image_2>
Circle $A$ has curvature 2; circle $B$ has curvature -1 .
Using these conventions allows us to express a beautiful theorem of Descartes: when four circles $A, B, C, D$ are pairwise tangent, with respective curvatures $a, b, c, d$, then
$$
(a+b+c+d)^{2}=2\left(a^{2}+b^{2}+c^{2}+d^{2}\right),
$$
where (as before) $a$ is taken to be negative if $B, C, D$ are internally tangent to $A$, and correspondingly for $b, c$, or $d$.
Without using Descartes' Circle Formula, Show that the circles marked off and sold on day 1 are centered at $\left(0, \pm \frac{2}{3}\right)$ with radius $\frac{1}{3}$.
|
[
"By symmetry, $P_{1}, P_{2}$, the two plots sold on day 1 , are centered on the $y$-axis, say at $(0, \\pm y)$ with $y>0$. Let these plots have radius $r$. Because $P_{1}$ is tangent to $U, y+r=1$. Because $P_{1}$ is tangent to $C$, the distance from $(0,1-r)$ to $\\left(\\frac{1}{2}, 0\\right)$ is $r+\\frac{1}{2}$. Therefore\n\n$$\n\\begin{aligned}\n\\left(\\frac{1}{2}\\right)^{2}+(1-r)^{2} & =\\left(r+\\frac{1}{2}\\right)^{2} \\\\\n1-2 r & =r \\\\\nr & =\\frac{1}{3} \\\\\ny=1-r & =\\frac{2}{3} .\n\\end{aligned}\n$$\n\nThus the plots are centered at $\\left(0, \\pm \\frac{2}{3}\\right)$ and have radius $\\frac{1}{3}$."
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Math
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English
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2,934
|
A king strapped for cash is forced to sell off his kingdom $U=\left\{(x, y): x^{2}+y^{2} \leq 1\right\}$. He sells the two circular plots $C$ and $C^{\prime}$ centered at $\left( \pm \frac{1}{2}, 0\right)$ with radius $\frac{1}{2}$. The retained parts of the kingdom form two regions, each bordered by three arcs of circles; in what follows, we will call such regions curvilinear triangles, or $c$-triangles ( $\mathrm{c} \triangle$ ) for short.
This sad day marks day 0 of a new fiscal era. Unfortunately, these drastic measures are not enough, and so each day thereafter, court geometers mark off the largest possible circle contained in each c-triangle in the remaining property. This circle is tangent to all three arcs of the c-triangle, and will be referred to as the incircle of the c-triangle. At the end of the day, all incircles demarcated that day are sold off, and the following day, the remaining c-triangles are partitioned in the same manner.
Some notation: when discussing mutually tangent circles (or arcs), it is convenient to refer to the curvature of a circle rather than its radius. We define curvature as follows. Suppose that circle $A$ of radius $r_{a}$ is externally tangent to circle $B$ of radius $r_{b}$. Then the curvatures of the circles are simply the reciprocals of their radii, $\frac{1}{r_{a}}$ and $\frac{1}{r_{b}}$. If circle $A$ is internally tangent to circle $B$, however, as in the right diagram below, the curvature of circle $A$ is still $\frac{1}{r_{a}}$, while the curvature of circle $B$ is $-\frac{1}{r_{b}}$, the opposite of the reciprocal of its radius.
<image_1>
Circle $A$ has curvature 2; circle $B$ has curvature 1 .
<image_2>
Circle $A$ has curvature 2; circle $B$ has curvature -1 .
Using these conventions allows us to express a beautiful theorem of Descartes: when four circles $A, B, C, D$ are pairwise tangent, with respective curvatures $a, b, c, d$, then
$$
(a+b+c+d)^{2}=2\left(a^{2}+b^{2}+c^{2}+d^{2}\right),
$$
where (as before) $a$ is taken to be negative if $B, C, D$ are internally tangent to $A$, and correspondingly for $b, c$, or $d$.
Given three mutually tangent circles with curvatures $a, b, c>0$, suppose that $(a, b, c, 0)$ does not satisfy Descartes' Circle Formula. Show that there are two distinct values of $r$ such that there is a circle of radius $r$ tangent to the given circles.
|
[
"Apply Descartes' Circle Formula to yield\n\n$$\n(a+b+c+x)^{2}=2 \\cdot\\left(a^{2}+b^{2}+c^{2}+x^{2}\\right),\n$$\n\na quadratic equation in $x$. Expanding and rewriting in standard form yields the equation\n\n$$\nx^{2}-p x+q=0\n$$\n\nwhere $p=2(a+b+c)$ and $q=2\\left(a^{2}+b^{2}+c^{2}\\right)-(a+b+c)^{2}$.\n\nThe discriminant of this quadratic is\n\n$$\n\\begin{aligned}\np^{2}-4 q & =8(a+b+c)^{2}-8\\left(a^{2}+b^{2}+c^{2}\\right) \\\\\n& =16(a b+a c+b c)\n\\end{aligned}\n$$\n\nThis last expression is positive because it is given that $a, b, c>0$. Therefore the quadratic has two distinct real roots, say $d_{1}$ and $d_{2}$. These usually correspond to two distinct radii, $r_{1}=\\frac{1}{\\left|d_{1}\\right|}$ and $r_{2}=\\frac{1}{\\left|d_{2}\\right|}$.\n\nOne possible exception case to consider is if $r_{2}=r_{1}$, which would occur if $d_{2}=-d_{1}$. This case can be ruled out because $d_{1}+d_{2}=p=2(a+b+c)$, which must be positive if $a, b, c>0$. (Notice too that this inequality rules out the possibility that both circles have negative curvature, so that there cannot be two distinct circles to which the given circles are internally tangent.)\n\nWhen both roots $d_{1}$ and $d_{2}$ are positive, the three given circles are externally tangent to both fourth circles. When one is positive and one is negative, the three given circles are internally tangent to one circle and externally tangent to the other.\n\n\n\nWhile the foregoing answers the question posed, it is interesting to examine the result from a geometric perspective: why are there normally two possible fourth circles? Consider the case when one of $a, b$, and $c$ is negative (i.e., two circles are internally tangent to a third). Let $A$ and $B$ be circles internally tangent to $C$. Then $A$ and $B$ partition the remaining area of $C$ into two c-triangles, each of which has an incircle, providing the two solutions.\n\nIf, as in the given problem, $a, b, c>0$, then all three circles $A, B$, and $C$, are mutually externally tangent. In this case, the given circles bound a c-triangle, which has an incircle, corresponding to one of the two roots. The complementary arcs of the given circles bound an infinite region, and this region normally contains a second circle tangent to the given circles. To demonstrate this fact geometrically, consider shrink-wrapping the circles: the shrink-wrap is the border of the smallest convex region containing all three circles. (This region is called the convex hull of the circles). There are two cases to address. If only two circles are touched by the shrink-wrap, then one circle is wedged between two larger ones and completely enclosed by their common tangents. In such a case, a circle can be drawn so that it is tangent to all three circles as shown in the diagram below (shrink-wrap in bold; locations of fourth circle marked at $D_{1}$ and $D_{2}$ ).\n\n<img_3391>\n\nOn the other hand, if the shrink-wrap touches all three circles, then it can be expanded to make a circle tangent to and containing $A, B$, and $C$, as shown below.\n\n\n\n<img_3298>"
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2,935
|
A king strapped for cash is forced to sell off his kingdom $U=\left\{(x, y): x^{2}+y^{2} \leq 1\right\}$. He sells the two circular plots $C$ and $C^{\prime}$ centered at $\left( \pm \frac{1}{2}, 0\right)$ with radius $\frac{1}{2}$. The retained parts of the kingdom form two regions, each bordered by three arcs of circles; in what follows, we will call such regions curvilinear triangles, or $c$-triangles ( $\mathrm{c} \triangle$ ) for short.
This sad day marks day 0 of a new fiscal era. Unfortunately, these drastic measures are not enough, and so each day thereafter, court geometers mark off the largest possible circle contained in each c-triangle in the remaining property. This circle is tangent to all three arcs of the c-triangle, and will be referred to as the incircle of the c-triangle. At the end of the day, all incircles demarcated that day are sold off, and the following day, the remaining c-triangles are partitioned in the same manner.
Some notation: when discussing mutually tangent circles (or arcs), it is convenient to refer to the curvature of a circle rather than its radius. We define curvature as follows. Suppose that circle $A$ of radius $r_{a}$ is externally tangent to circle $B$ of radius $r_{b}$. Then the curvatures of the circles are simply the reciprocals of their radii, $\frac{1}{r_{a}}$ and $\frac{1}{r_{b}}$. If circle $A$ is internally tangent to circle $B$, however, as in the right diagram below, the curvature of circle $A$ is still $\frac{1}{r_{a}}$, while the curvature of circle $B$ is $-\frac{1}{r_{b}}$, the opposite of the reciprocal of its radius.
<image_1>
Circle $A$ has curvature 2; circle $B$ has curvature 1 .
<image_2>
Circle $A$ has curvature 2; circle $B$ has curvature -1 .
Using these conventions allows us to express a beautiful theorem of Descartes: when four circles $A, B, C, D$ are pairwise tangent, with respective curvatures $a, b, c, d$, then
$$
(a+b+c+d)^{2}=2\left(a^{2}+b^{2}+c^{2}+d^{2}\right),
$$
where (as before) $a$ is taken to be negative if $B, C, D$ are internally tangent to $A$, and correspondingly for $b, c$, or $d$.
Algebraically, it is possible for a quadruple $(a, b, c, 0)$ to satisfy Descartes' Circle Formula, as occurs when $a=b=1$ and $c=4$. Find a geometric interpretation for this situation.
|
[
"In this case, the fourth \"circle\" is actually a line tangent to all three circles, as shown in the diagram below.\n\n<img_3396>"
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2,936
|
A king strapped for cash is forced to sell off his kingdom $U=\left\{(x, y): x^{2}+y^{2} \leq 1\right\}$. He sells the two circular plots $C$ and $C^{\prime}$ centered at $\left( \pm \frac{1}{2}, 0\right)$ with radius $\frac{1}{2}$. The retained parts of the kingdom form two regions, each bordered by three arcs of circles; in what follows, we will call such regions curvilinear triangles, or $c$-triangles ( $\mathrm{c} \triangle$ ) for short.
This sad day marks day 0 of a new fiscal era. Unfortunately, these drastic measures are not enough, and so each day thereafter, court geometers mark off the largest possible circle contained in each c-triangle in the remaining property. This circle is tangent to all three arcs of the c-triangle, and will be referred to as the incircle of the c-triangle. At the end of the day, all incircles demarcated that day are sold off, and the following day, the remaining c-triangles are partitioned in the same manner.
Some notation: when discussing mutually tangent circles (or arcs), it is convenient to refer to the curvature of a circle rather than its radius. We define curvature as follows. Suppose that circle $A$ of radius $r_{a}$ is externally tangent to circle $B$ of radius $r_{b}$. Then the curvatures of the circles are simply the reciprocals of their radii, $\frac{1}{r_{a}}$ and $\frac{1}{r_{b}}$. If circle $A$ is internally tangent to circle $B$, however, as in the right diagram below, the curvature of circle $A$ is still $\frac{1}{r_{a}}$, while the curvature of circle $B$ is $-\frac{1}{r_{b}}$, the opposite of the reciprocal of its radius.
<image_1>
Circle $A$ has curvature 2; circle $B$ has curvature 1 .
<image_2>
Circle $A$ has curvature 2; circle $B$ has curvature -1 .
Using these conventions allows us to express a beautiful theorem of Descartes: when four circles $A, B, C, D$ are pairwise tangent, with respective curvatures $a, b, c, d$, then
$$
(a+b+c+d)^{2}=2\left(a^{2}+b^{2}+c^{2}+d^{2}\right),
$$
where (as before) $a$ is taken to be negative if $B, C, D$ are internally tangent to $A$, and correspondingly for $b, c$, or $d$.
Let $\phi=\frac{1+\sqrt{5}}{2}$, and let $\rho=\phi+\sqrt{\phi}$.
Prove that $\rho^{4}=2 \rho^{3}+2 \rho^{2}+2 \rho-1$.
|
[
"Note that\n\n$$\n\\begin{aligned}\n\\frac{1}{\\rho} & =\\frac{\\phi^{2}-\\phi}{\\phi+\\sqrt{\\phi}} \\\\\n& =\\phi-\\sqrt{\\phi}\n\\end{aligned}\n$$\n\n\n\nTherefore\n\n$$\n\\begin{aligned}\n\\left(\\rho-\\frac{1}{\\rho}\\right)^{2} & =(2 \\sqrt{\\phi})^{2}=4 \\phi \\\\\n& =2\\left(\\rho+\\frac{1}{\\rho}\\right) .\n\\end{aligned}\n$$\n\nMultiplying both sides of the equation by $\\rho^{2}$ gives $\\left(\\rho^{2}-1\\right)^{2}=2\\left(\\rho^{3}+\\rho\\right)$. Expand and isolate $\\rho^{4}$ to obtain $\\rho^{4}=2 \\rho^{3}+2 \\rho^{2}+2 \\rho-1$.\n\nAlternate Proof: Because $\\phi^{2}=\\phi+1$, any power of $\\rho$ can be expressed as an integer plus integer multiples of $\\sqrt{\\phi}, \\phi$, and $\\phi \\sqrt{\\phi}$. In particular,\n\n$$\n\\begin{aligned}\n\\rho^{2} & =\\phi^{2}+2 \\phi \\sqrt{\\phi}+\\phi \\\\\n& =2 \\phi \\sqrt{\\phi}+2 \\phi+1, \\\\\n\\rho^{3} & =4 \\phi \\sqrt{\\phi}+5 \\phi+3 \\sqrt{\\phi}+4, \\text { and } \\\\\n\\rho^{4} & =12 \\phi \\sqrt{\\phi}+16 \\phi+8 \\sqrt{\\phi}+9 .\n\\end{aligned}\n$$\n\nTherefore\n\n$$\n\\begin{aligned}\n2 \\rho^{3}+2 \\rho^{2}+2 \\rho-1 & =2(4 \\phi \\sqrt{\\phi}+5 \\phi+3 \\sqrt{\\phi}+4)+2(2 \\phi \\sqrt{\\phi}+2 \\phi+1)+2 \\rho-1 \\\\\n& =12 \\phi \\sqrt{\\phi}+16 \\phi+8 \\sqrt{\\phi}+9 \\\\\n& =\\rho^{4}\n\\end{aligned}\n$$"
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Multimodal
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Theorem proof
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Geometry
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Math
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English
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2,937
|
A king strapped for cash is forced to sell off his kingdom $U=\left\{(x, y): x^{2}+y^{2} \leq 1\right\}$. He sells the two circular plots $C$ and $C^{\prime}$ centered at $\left( \pm \frac{1}{2}, 0\right)$ with radius $\frac{1}{2}$. The retained parts of the kingdom form two regions, each bordered by three arcs of circles; in what follows, we will call such regions curvilinear triangles, or $c$-triangles ( $\mathrm{c} \triangle$ ) for short.
This sad day marks day 0 of a new fiscal era. Unfortunately, these drastic measures are not enough, and so each day thereafter, court geometers mark off the largest possible circle contained in each c-triangle in the remaining property. This circle is tangent to all three arcs of the c-triangle, and will be referred to as the incircle of the c-triangle. At the end of the day, all incircles demarcated that day are sold off, and the following day, the remaining c-triangles are partitioned in the same manner.
Some notation: when discussing mutually tangent circles (or arcs), it is convenient to refer to the curvature of a circle rather than its radius. We define curvature as follows. Suppose that circle $A$ of radius $r_{a}$ is externally tangent to circle $B$ of radius $r_{b}$. Then the curvatures of the circles are simply the reciprocals of their radii, $\frac{1}{r_{a}}$ and $\frac{1}{r_{b}}$. If circle $A$ is internally tangent to circle $B$, however, as in the right diagram below, the curvature of circle $A$ is still $\frac{1}{r_{a}}$, while the curvature of circle $B$ is $-\frac{1}{r_{b}}$, the opposite of the reciprocal of its radius.
<image_1>
Circle $A$ has curvature 2; circle $B$ has curvature 1 .
<image_2>
Circle $A$ has curvature 2; circle $B$ has curvature -1 .
Using these conventions allows us to express a beautiful theorem of Descartes: when four circles $A, B, C, D$ are pairwise tangent, with respective curvatures $a, b, c, d$, then
$$
(a+b+c+d)^{2}=2\left(a^{2}+b^{2}+c^{2}+d^{2}\right),
$$
where (as before) $a$ is taken to be negative if $B, C, D$ are internally tangent to $A$, and correspondingly for $b, c$, or $d$.
Let $\phi=\frac{1+\sqrt{5}}{2}$, and let $\rho=\phi+\sqrt{\phi}$.
Show that four pairwise externally tangent circles with nonequal radii in geometric progression must have common ratio $\rho$.
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[
"If the radii are in geometric progression, then so are their reciprocals (i.e., curvatures). Without loss of generality, let $(a, b, c, d)=\\left(a, a r, a r^{2}, a r^{3}\\right)$ for $r>1$. By Descartes' Circle Formula,\n\n$$\n\\left(a+a r+a r^{2}+a r^{3}\\right)^{2}=2\\left(a^{2}+a^{2} r^{2}+a^{2} r^{4}+a^{2} r^{6}\\right)\n$$\n\nCancel $a^{2}$ from both sides of the equation to obtain\n\n$$\n\\left(1+r+r^{2}+r^{3}\\right)^{2}=2\\left(1+r^{2}+r^{4}+r^{6}\\right) .\n$$\n\nBecause $1+r+r^{2}+r^{3}=(1+r)\\left(1+r^{2}\\right)$ and $1+r^{2}+r^{4}+r^{6}=\\left(1+r^{2}\\right)\\left(1+r^{4}\\right)$, the equation can be rewritten as follows:\n\n$$\n\\begin{aligned}\n(1+r)^{2}\\left(1+r^{2}\\right)^{2} & =2\\left(1+r^{2}\\right)\\left(1+r^{4}\\right) \\\\\n(1+r)^{2}\\left(1+r^{2}\\right) & =2\\left(1+r^{4}\\right) \\\\\nr^{4}-2 r^{3}-2 r^{2}-2 r+1 & =0 .\n\\end{aligned}\n$$\n\nUsing the identity from 4a, $r=\\rho$ is one solution; because the polynomial is palindromic, another real solution is $r=\\rho^{-1}=\\phi-\\sqrt{\\phi}$, but this value is less than 1 . The product of the corresponding linear factors is $r^{2}-2 \\phi r+\\phi^{2}-\\phi=r^{2}-2 \\phi r+1$. Division verifies that the other quadratic factor of the polynomial is $r^{2}+(2 \\phi-2) r+1$, which has no real roots because $(\\sqrt{5}-1)^{2}<1$."
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Competition
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Theorem proof
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Geometry
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Math
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2,938
|
A king strapped for cash is forced to sell off his kingdom $U=\left\{(x, y): x^{2}+y^{2} \leq 1\right\}$. He sells the two circular plots $C$ and $C^{\prime}$ centered at $\left( \pm \frac{1}{2}, 0\right)$ with radius $\frac{1}{2}$. The retained parts of the kingdom form two regions, each bordered by three arcs of circles; in what follows, we will call such regions curvilinear triangles, or $c$-triangles ( $\mathrm{c} \triangle$ ) for short.
This sad day marks day 0 of a new fiscal era. Unfortunately, these drastic measures are not enough, and so each day thereafter, court geometers mark off the largest possible circle contained in each c-triangle in the remaining property. This circle is tangent to all three arcs of the c-triangle, and will be referred to as the incircle of the c-triangle. At the end of the day, all incircles demarcated that day are sold off, and the following day, the remaining c-triangles are partitioned in the same manner.
Some notation: when discussing mutually tangent circles (or arcs), it is convenient to refer to the curvature of a circle rather than its radius. We define curvature as follows. Suppose that circle $A$ of radius $r_{a}$ is externally tangent to circle $B$ of radius $r_{b}$. Then the curvatures of the circles are simply the reciprocals of their radii, $\frac{1}{r_{a}}$ and $\frac{1}{r_{b}}$. If circle $A$ is internally tangent to circle $B$, however, as in the right diagram below, the curvature of circle $A$ is still $\frac{1}{r_{a}}$, while the curvature of circle $B$ is $-\frac{1}{r_{b}}$, the opposite of the reciprocal of its radius.
<image_1>
Circle $A$ has curvature 2; circle $B$ has curvature 1 .
<image_2>
Circle $A$ has curvature 2; circle $B$ has curvature -1 .
Using these conventions allows us to express a beautiful theorem of Descartes: when four circles $A, B, C, D$ are pairwise tangent, with respective curvatures $a, b, c, d$, then
$$
(a+b+c+d)^{2}=2\left(a^{2}+b^{2}+c^{2}+d^{2}\right),
$$
where (as before) $a$ is taken to be negative if $B, C, D$ are internally tangent to $A$, and correspondingly for $b, c$, or $d$.
Given $A, B, C, D$ as above with $s=a+b+c+d$, if there is a second circle $A^{\prime}$ with curvature $a^{\prime}$ also tangent to $B, C$, and $D$. We can describe $A$ and $A^{\prime}$ as conjugate circles. Use Descartes' Circle Formula to show that $a^{\prime}=2 s-3 a$ and therefore $s^{\prime}=a^{\prime}+b+c+d=$ $3 s-4 a$.
|
[
"The equation $(x+b+c+d)^{2}=2\\left(x^{2}+b^{2}+c^{2}+d^{2}\\right)$ is quadratic with two solutions. Call them $a$ and $a^{\\prime}$. These are the curvatures of the two circles which are tangent to circles with curvatures $b, c$, and $d$. Rewrite the equation in standard form to obtain $x^{2}-2(b+c+d) x+$ $\\ldots=0$. Using the sum of the roots formula, $a+a^{\\prime}=2(b+c+d)=2(s-a)$. So $a^{\\prime}=2 s-3 a$, and therefore\n\n$$\n\\begin{aligned}\ns^{\\prime} & =a^{\\prime}+b+c+d \\\\\n& =2 s-3 a+s-a \\\\\n& =3 s-4 a .\n\\end{aligned}\n$$"
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Multimodal
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Competition
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Theorem proof
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Geometry
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Math
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English
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2,939
|
A king strapped for cash is forced to sell off his kingdom $U=\left\{(x, y): x^{2}+y^{2} \leq 1\right\}$. He sells the two circular plots $C$ and $C^{\prime}$ centered at $\left( \pm \frac{1}{2}, 0\right)$ with radius $\frac{1}{2}$. The retained parts of the kingdom form two regions, each bordered by three arcs of circles; in what follows, we will call such regions curvilinear triangles, or $c$-triangles ( $\mathrm{c} \triangle$ ) for short.
This sad day marks day 0 of a new fiscal era. Unfortunately, these drastic measures are not enough, and so each day thereafter, court geometers mark off the largest possible circle contained in each c-triangle in the remaining property. This circle is tangent to all three arcs of the c-triangle, and will be referred to as the incircle of the c-triangle. At the end of the day, all incircles demarcated that day are sold off, and the following day, the remaining c-triangles are partitioned in the same manner.
Some notation: when discussing mutually tangent circles (or arcs), it is convenient to refer to the curvature of a circle rather than its radius. We define curvature as follows. Suppose that circle $A$ of radius $r_{a}$ is externally tangent to circle $B$ of radius $r_{b}$. Then the curvatures of the circles are simply the reciprocals of their radii, $\frac{1}{r_{a}}$ and $\frac{1}{r_{b}}$. If circle $A$ is internally tangent to circle $B$, however, as in the right diagram below, the curvature of circle $A$ is still $\frac{1}{r_{a}}$, while the curvature of circle $B$ is $-\frac{1}{r_{b}}$, the opposite of the reciprocal of its radius.
<image_1>
Circle $A$ has curvature 2; circle $B$ has curvature 1 .
<image_2>
Circle $A$ has curvature 2; circle $B$ has curvature -1 .
Using these conventions allows us to express a beautiful theorem of Descartes: when four circles $A, B, C, D$ are pairwise tangent, with respective curvatures $a, b, c, d$, then
$$
(a+b+c+d)^{2}=2\left(a^{2}+b^{2}+c^{2}+d^{2}\right),
$$
where (as before) $a$ is taken to be negative if $B, C, D$ are internally tangent to $A$, and correspondingly for $b, c$, or $d$.
Show that by area, $12 \%$ of the kingdom is sold on day 2 .
|
[
"On day 2, six plots are sold: two with curvature 15 from the configuration $(2,2,3,15)$, and four with curvature 6 from the configuration $(-1,2,3,6)$. The total area sold on day 2 is therefore\n\n$$\n2 \\cdot \\frac{\\pi}{15^{2}}+4 \\cdot \\frac{\\pi}{6^{2}}=\\frac{3}{25} \\pi\n$$\n\nwhich is exactly $12 \\%$ of the unit circle."
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Multimodal
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Competition
| true
| null | null | null |
Theorem proof
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Geometry
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Math
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English
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