id int64 | question string | solution list | final_answer list | context string | image_1 image | image_2 image | image_3 image | image_4 image | image_5 image | modality string | difficulty string | is_multiple_answer bool | unit string | answer_type string | error string | question_type string | subfield string | subject string | language string |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
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 $B$ t... | [
"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 hav... | [
"27, 77"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Multimodal | Competition | true | null | Numerical | null | Open-ended | Algebra | Math | English | |
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 $B C$. | [
"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}-\\... | [
"1375"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Multimodal | Competition | false | null | Numerical | null | Open-ended | Geometry | Math | English | |
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> | [
"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... | [
"100"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Multimodal | Competition | false | null | Numerical | null | Open-ended | Geometry | Math | English | |
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 $\\frac{\\sqrt{3} k \\pi}{6}=\\frac{5... | [
"$\\frac{10}{\\sqrt{3}}$,$\\frac{10}{3} \\sqrt{3}$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Multimodal | Competition | true | null | Numerical | null | Open-ended | Geometry | Math | English | |
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> | [
"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 3... | [
"141"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Multimodal | Competition | false | m | Numerical | null | Open-ended | Geometry | Math | English | |
2,595 | 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> | [
"$Q(4,0), Q(0,4)$\n\n$Q(2,0), Q(0,2)$\n\n$Q(-2,2), Q(2,-2)$\n\n<img_3757>"
] | [
"$(4,0),(0,4),(2,0),(0,2),(-2,2),(2,-2)$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Multimodal | Competition | true | null | Tuple | null | Open-ended | Geometry | Math | English | |
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> | [
"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 $B D=1$. (Alternatel... | [
"$\\frac{\\sqrt{6}}{4}$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Multimodal | Competition | false | null | Numerical | null | Open-ended | Geometry | Math | English | |
2,716 | 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> | [
"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 $\\sqrt{\\left(\\frac{\\sqrt{3}}{2}\\right)^{2}-\\left(\\frac{1}{2}\\right)^{2}}=\\sqrt{\\frac{1}{2}}=\\fr... | [
"$\\frac{5 \\sqrt{39}}{3}$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Multimodal | Competition | false | null | Numerical | null | Open-ended | Geometry | Math | English | |
2,719 | 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> | [
"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 (... | [
"8"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Multimodal | Competition | false | null | Numerical | null | Open-ended | Geometry | Math | English | |
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> | [
"Note that $G H=A B-A H-B G=22-5-5=12$. Thus\n\n$$\n\\begin{aligned}\n{[A E F B C D] } & =[A B C D]+[E F G H]+[A E H]+[B F G] \\\\\n& =22^{2}+12^{2}+\\frac{1}{2} \\cdot 5 \\cdot 12+\\frac{1}{2} \\cdot 5 \\cdot 12 \\\\\n& =484+144+30+30 \\\\\n& =\\mathbf{6 8 8} .\n\\end{aligned}\n$$"
] | [
"688"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Multimodal | Competition | false | null | Numerical | null | Open-ended | Geometry | Math | English | |
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> | [
"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... | [
"16"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Multimodal | Competition | false | null | Numerical | null | Open-ended | Geometry | Math | English | |
2,774 | 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{\tex... | [
"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 $\\fr... | [
"7"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Multimodal | Competition | false | null | Numerical | null | Open-ended | Geometry | Math | English | |
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> | [
"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$."
] | [
"3"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Multimodal | Competition | false | null | Numerical | null | Open-ended | Geometry | Math | English | |
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 ... | [
"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. I... | [
"6"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Multimodal | Competition | false | null | Numerical | null | Open-ended | Combinatorics | Math | English | ||
2,810 | 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 ... | [
"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 u... | [
"12, 8"
] | null | Not supported with pagination yet | Multimodal | Competition | true | null | Numerical | null | Open-ended | Combinatorics | Math | English | ||||
2,820 | 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 arc... | [
"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 \... | [
"$\\quad 64-64 \\pi+32 \\pi \\sqrt{2}$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Multimodal | Competition | false | null | Numerical | null | Open-ended | Geometry | Math | English | |
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> | [
"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 bas... | [
"$\\frac{9}{5}$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Multimodal | Competition | false | null | Numerical | null | Open-ended | Geometry | Math | English | |
2,833 | 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$ ... | [
"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}\\... | [
"15"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Multimodal | Competition | false | null | Numerical | null | Open-ended | Geometry | Math | English | |
2,837 | 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... | [
"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... | [
"$10 \\sqrt{6}$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Multimodal | Competition | false | null | Numerical | null | Open-ended | Geometry | Math | English | |
2,868 | 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... | [
"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)$."
] | [
"$(12,21,21,21,12)$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Multimodal | Competition | false | null | Tuple | null | Open-ended | Combinatorics | Math | English | |||
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... | [
"12543,13542,14532,23541,24531,34521"
] | [
"12543,13542,14532,23541,24531,34521"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Multimodal | Competition | true | null | Numerical | null | Open-ended | Combinatorics | Math | English | ||
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... | [
"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 nu... | [
"$\\binom{2n}{n}$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Multimodal | Competition | false | null | Expression | null | Open-ended | Combinatorics | Math | English | ||
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... | [
"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 le... | [
"16"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Multimodal | Competition | false | null | Numerical | null | Open-ended | Combinatorics | Math | English | ||
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... | [
"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... | [
"7936"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Multimodal | Competition | false | null | Numerical | null | Open-ended | Combinatorics | Math | English | ||
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... | [
"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... | [
"$p ! \\cdot p^{n-p}$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Multimodal | Competition | false | null | Expression | null | Open-ended | Combinatorics | Math | English | ||
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... | [
"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}}{(... | [
"7,5"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Multimodal | Competition | true | null | Numerical | null | Open-ended | Combinatorics | Math | English | ||
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... | [
"12345 and 54321 are the only ones."
] | [
"12345, 54321"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Multimodal | Competition | true | null | Numerical | null | Open-ended | Combinatorics | Math | English | ||
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... | [
"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 betw... | [
"16"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Multimodal | Competition | false | null | Numerical | null | Open-ended | Combinatorics | Math | English | ||
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... | [
"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 pr... | [
"1476"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Multimodal | Competition | false | null | Numerical | null | Open-ended | Geometry | Math | English | |
2,889 | 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 si... | [
"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 fr... | [
"32"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Multimodal | Competition | false | null | Numerical | null | Open-ended | Combinatorics | Math | English | |
2,906 | 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.
<image_1> | [
"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... | [
"5"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Multimodal | Competition | false | null | Numerical | null | Open-ended | Geometry | Math | English | |
2,923 | 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> | [
"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 ... | [
"49"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Multimodal | Competition | false | null | Numerical | null | Open-ended | Geometry | Math | English | |
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 circ... | [
"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$$\n\\pi\\left(1^{2}-\\left(\\frac{1}{2}\\right)^{2}-\\left(\\frac{1}{2}\\right)^{2}-\\left(\\frac{1}{3}\\right)^{2}-\\left(\\frac{1}{3}\\right)^... | [
"$\\frac{5 \\pi}{18}$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Multimodal | Competition | false | null | Numerical | null | Open-ended | Geometry | Math | English | ||
2,930 | A king strapped for cash is forced to sell off his kingdom $U=\left\{(x, y): x^{2}+y^{2} \leq 1\right\}$. He sells the two circular plots $C$ and $C^{\prime}$ centered at $\left( \pm \frac{1}{2}, 0\right)$ with radius $\frac{1}{2}$. The retained parts of the kingdom form two regions, each bordered by three arcs of circ... | [
"At the beginning of day 2, there are six c-triangles, so six incircles are sold, dividing each of the six territories into three smaller curvilinear triangles. So a total of 18 curvilinear triangles exist at the start of day 3, each of which is itself divided into three pieces that day (by the sale of a total of 1... | [
"54"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Multimodal | Competition | false | null | Numerical | null | Open-ended | Geometry | Math | English | ||
2,932 | A king strapped for cash is forced to sell off his kingdom $U=\left\{(x, y): x^{2}+y^{2} \leq 1\right\}$. He sells the two circular plots $C$ and $C^{\prime}$ centered at $\left( \pm \frac{1}{2}, 0\right)$ with radius $\frac{1}{2}$. The retained parts of the kingdom form two regions, each bordered by three arcs of circ... | [
"The total number of plots sold up to and including day $n$ is\n\n$$\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$$\n\nAlternatively, proceed by induction: on day 0 , there are $2=3^{0}+1$ plo... | [
"$3^{n}+1$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Multimodal | Competition | false | null | Expression | null | Open-ended | Geometry | Math | English | ||
2,933 | A king strapped for cash is forced to sell off his kingdom $U=\left\{(x, y): x^{2}+y^{2} \leq 1\right\}$. He sells the two circular plots $C$ and $C^{\prime}$ centered at $\left( \pm \frac{1}{2}, 0\right)$ with radius $\frac{1}{2}$. The retained parts of the kingdom form two regions, each bordered by three arcs of circ... | [
"Use Descartes' Circle Formula with $a=b=1$ and $c=\\frac{3}{2}$ to solve for $d$ :\n\n$$\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\\... | [
"$2$, $\\frac{2}{15}$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Multimodal | Competition | true | null | Numerical | null | Open-ended | Geometry | Math | English | ||
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 circ... | [
"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- $b=2: b^{\\prime}=2 s-3 b=\\mathbf{3 ... | [
"$\\frac{\\pi}{38^{2}}, \\frac{\\pi}{35^{2}}, \\frac{\\pi}{23^{2}}, \\frac{\\pi}{14^{2}}, \\frac{\\pi}{11^{2}}$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Multimodal | Competition | true | null | Numerical | null | Open-ended | Geometry | Math | English | ||
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 ... | [
"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 $\\mathrm{m} \\angle Y K L=\\frac{1}{2} \\math... | [
"$\\frac{375}{182}$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Multimodal | Competition | false | null | Numerical | null | Open-ended | Geometry | Math | English | |
2,965 | 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 radiu... | [
"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 ."
] | [
"9"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Multimodal | Competition | false | null | Numerical | null | Open-ended | Geometry | Math | English | |
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... | [
"$$\n\\begin{aligned}\n& \\mathrm{Pa}(1,1)+\\mathrm{Pa}(2,1)+\\mathrm{Pa}(3,1)+\\mathrm{Pa}(4,1)+\\mathrm{Pa}(5,1)=1+2+3+4+5=\\mathbf{1 5} \\\\\n& \\mathrm{Pa}(2,2)+\\mathrm{Pa}(3,2)+\\mathrm{Pa}(4,2)+\\mathrm{Pa}(5,2)+\\mathrm{Pa}(6,2)=1+3+6+10+15=\\mathbf{3 5} \\\\\n& \\mathrm{Pa}(3,3)+\\mathrm{Pa}(4,3)+\\mathrm{... | [
"15, 35, 70, 126"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Multimodal | Competition | true | null | Numerical | null | Open-ended | Combinatorics | Math | English | |
3,051 | The arrangement of numbers known as Pascal's Triangle has fascinated mathematicians for centuries. In fact, about 700 years before Pascal, the Indian mathematician Halayudha wrote about it in his commentaries to a then-1000-year-old treatise on verse structure by the Indian poet and mathematician Pingala, who called it... | [
"Notice that $\\mathrm{Pa}(n, n)+\\operatorname{Pa}(n+1, n)+\\cdots+\\operatorname{Pa}(n+k, n)=\\mathrm{Pa}(n+k+1, n+1)$, so $m=n+k+1$ and $j=n+1$. (By symmetry, $j=k$ is also correct.) The equation is true for all $n$ when $k=0$, because the sum is simply $\\mathrm{Pa}(n, n)$ and the right side is $\\mathrm{Pa}(n+... | [
"$m=n+k+1$, $j=n+1$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Multimodal | Competition | true | null | Expression | null | Open-ended | Combinatorics | Math | English | |
3,058 | The arrangement of numbers known as Pascal's Triangle has fascinated mathematicians for centuries. In fact, about 700 years before Pascal, the Indian mathematician Halayudha wrote about it in his commentaries to a then-1000-year-old treatise on verse structure by the Indian poet and mathematician Pingala, who called it... | [
"Using the given values yields the system of equations below.\n\n$$\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$$\n\nSolving this system, $a=3, b=-3, c=1$."
] | [
"$3,-3,1$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Multimodal | Competition | true | null | Numerical | null | Open-ended | Combinatorics | Math | English | ||
3,060 | The arrangement of numbers known as Pascal's Triangle has fascinated mathematicians for centuries. In fact, about 700 years before Pascal, the Indian mathematician Halayudha wrote about it in his commentaries to a then-1000-year-old treatise on verse structure by the Indian poet and mathematician Pingala, who called it... | [
"$\\mathrm{Cl}(11,2)=1000$."
] | [
"1000"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Multimodal | Competition | false | null | Numerical | null | Open-ended | Combinatorics | Math | English | ||
3,062 | The arrangement of numbers known as Pascal's Triangle has fascinated mathematicians for centuries. In fact, about 700 years before Pascal, the Indian mathematician Halayudha wrote about it in his commentaries to a then-1000-year-old treatise on verse structure by the Indian poet and mathematician Pingala, who called it... | [
"$\\mathrm{Cl}(11,3)=2025$."
] | [
"2025"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Multimodal | Competition | false | null | Numerical | null | Open-ended | Combinatorics | Math | English | ||
3,065 | Leibniz's Harmonic Triangle: Consider the triangle formed by the rule
$$
\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 {... | [
"$\\operatorname{Le}(17,1)=\\operatorname{Le}(16,0)-\\operatorname{Le}(17,0)=\\frac{1}{17}-\\frac{1}{18}=\\frac{1}{306}$."
] | [
"$\\frac{1}{306}$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Multimodal | Competition | false | null | Numerical | null | Open-ended | Combinatorics | Math | English | |
3,066 | Leibniz's Harmonic Triangle: Consider the triangle formed by the rule
$$
\begin{cases}\operatorname{Le}(n, 0)=\frac{1}{n+1} & \text { for all } n \\ \operatorname{Le}(n, n)=\frac{1}{n+1} & \text { for all } n \\ \operatorname{Le}(n, k)=\operatorname{Le}(n+1, k)+\operatorname{Le}(n+1, k+1) & \text { for all } n \text {... | [
"$\\operatorname{Le}(17,2)=\\operatorname{Le}(16,1)-\\operatorname{Le}(17,1)=\\operatorname{Le}(15,0)-\\operatorname{Le}(16,0)-\\operatorname{Le}(17,1)=\\frac{1}{2448}$."
] | [
"$\\frac{1}{2448}$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Multimodal | Competition | false | null | Numerical | null | Open-ended | Combinatorics | Math | English | |
3,068 | Leibniz's Harmonic Triangle: Consider the triangle formed by the rule
$$
\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 {... | [
"Because $\\operatorname{Le}(n, 1)=\\frac{1}{n}-\\frac{1}{n+1}$,\n\n$$\n\\begin{aligned}\n\\sum_{i=1}^{2011} \\operatorname{Le}(i, 1) & =\\sum_{i=1}^{2011}\\left(\\frac{1}{n}-\\frac{1}{n+1}\\right) \\\\\n& =\\left(\\frac{1}{1}-\\frac{1}{2}\\right)+\\left(\\frac{1}{2}-\\frac{1}{3}\\right)+\\cdots+\\left(\\frac{1}{20... | [
"$\\frac{2011}{2012}$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Multimodal | Competition | false | null | Numerical | null | Open-ended | Combinatorics | Math | English | |
3,070 | Leibniz's Harmonic Triangle: Consider the triangle formed by the rule
$$
\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 {... | [
"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 $n=k=0$."
] | [
"$0,0$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Multimodal | Competition | true | null | Numerical | null | Open-ended | Combinatorics | Math | English | |
3,071 | 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 {... | [
"$n=k=m-1$."
] | [
"$m-1,m-1$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Multimodal | Competition | true | null | Expression | null | Open-ended | Combinatorics | Math | English | |
3,093 | $\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> | [
"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 ... | [
"$72 \\sqrt{2}$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Multimodal | Competition | false | null | Numerical | null | Open-ended | Geometry | Math | English | |
1,735 | 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... | [
"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 ... | null | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Multimodal | Competition | true | null | null | null | Theorem proof | Geometry | Math | English | |
1,975 | 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-te... | [
"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 cove... | null | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Multimodal | Competition | true | null | null | null | Theorem proof | Combinatorics | Math | English | ||
2,039 | 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 di... | [
"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 ... | null | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Multimodal | Competition | true | null | null | null | Theorem proof | Combinatorics | Math | English | |
2,184 | 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$, ... | [
"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,... | null | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Multimodal | Competition | true | null | null | null | Theorem proof | Geometry | Math | English | |
2,227 | 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^{\p... | [
"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\n... | null | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Multimodal | Competition | true | null | null | null | Theorem proof | Geometry | Math | English | |
2,270 | 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.
<image_1> | [
"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=\\thet... | null | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Multimodal | Competition | true | null | null | null | Theorem proof | Geometry | Math | English | |
2,274 | 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 con... | [
"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 ... | null | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Multimodal | Competition | true | null | null | null | Theorem proof | Combinatorics | Math | English | |
2,348 | 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.
<image_1> | [
"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}$... | null | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Multimodal | Competition | true | null | null | null | Theorem proof | Geometry | Math | English | |
2,363 | 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$.
<image_1>
Prove that $P Q R S$ is a rectangle. | [
"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 ... | null | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Multimodal | Competition | true | null | null | null | Theorem proof | Geometry | Math | English | |
2,364 | 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$.
<image_1>
Prove that $P R=a-b$. | [
"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<i... | null | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Multimodal | Competition | true | null | null | null | Theorem proof | Geometry | Math | English | |
2,367 | 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 ... | [
"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 l... | null | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Multimodal | Competition | true | null | null | null | Theorem proof | Geometry | Math | English | |
2,398 | 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}$.
<image_1> | [
"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}-\\l... | null | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Multimodal | Competition | true | null | null | null | Theorem proof | Geometry | Math | English | |
2,448 | 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$.
<image_1> | [
"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 ... | null | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Multimodal | Competition | true | null | null | null | Theorem proof | Geometry | Math | English | |
2,449 | 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$.
<image_1> | [
"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 ... | null | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Multimodal | Competition | true | null | null | null | Theorem proof | Geometry | Math | English | |
2,462 | 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 paralle... | [
"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 ... | null | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Multimodal | Competition | true | null | null | null | Theorem proof | Geometry | Math | English | |
2,474 | 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... | [
"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 $\\t... | null | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Multimodal | Competition | true | null | null | null | Theorem proof | Geometry | Math | English | |
2,488 | 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$... | [
"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... | null | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Multimodal | Competition | true | null | null | null | Theorem proof | Geometry | Math | English | |
2,499 | 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 $\angle E A C+\angle A B C=90^{\circ}$.
<image_1> | [
"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\nTheref... | null | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Multimodal | Competition | true | null | null | null | Theorem proof | Geometry | Math | English | |
2,516 | 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
<image_1>
The $(3,4)$-sawtooth se... | [
"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 sequenc... | null | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Multimodal | Competition | true | null | null | null | Theorem proof | Number Theory | Math | English | |
2,526 | 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 $y^{2}=x^{2}+z^{2}$.
<image_1> | [
"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 $\\angle E A D=\\angle F C D=45^{\\circ}$.... | null | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Multimodal | Competition | true | null | null | null | Theorem proof | Geometry | Math | English | |
2,538 | 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 $q^{2}+r^{2}=p^{2}$.
<image_1> | [
"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... | null | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Multimodal | Competition | true | null | null | null | Theorem proof | Geometry | Math | English | |
2,542 | 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.
<image_1>
Throughout this... | [
"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 sim... | null | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Multimodal | Competition | true | null | null | null | Theorem proof | Algebra | Math | English | |
2,547 | 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 ter... | [
"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... | null | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Multimodal | Competition | true | null | null | null | Theorem proof | Geometry | Math | English | |
2,561 | In the graph, the parabola $y=x^{2}$ has been translated to the position shown.
Prove that $d e=f$.
<image_1> | [
"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 $f=d e$."
] | null | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Multimodal | Competition | true | null | null | null | Theorem proof | Geometry | Math | English | |
2,562 | 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 $K D$.
<image_1> | [
"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$$\n\\begin{aligned}\n& \\sqrt{(a+d-b)^{2}+(e-c)^{2}... | null | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Multimodal | Competition | true | null | null | null | Theorem proof | Geometry | Math | English | |
2,563 | Consider the first $2 n$ natural numbers. Pair off the numbers, as shown, and multiply the two members of each pair. Prove that there is no value of $n$ for which two of the $n$ products are equal.
<image_1> | [
"The sequence is $1(2 n), 2(2 n-1), 3(2 n-2), \\ldots, k(2 n-k+1), \\ldots, p(2 n-p+1), \\ldots, n(n+1)$.\n\nIn essence we are asking the question, 'is it possible that $k(2 n-k+1)=p(2 n-p+1)$ where $p$ and $k$ are both less than or equal to $n$ ?'\n\n$$\n\\begin{aligned}\nk(2 n-k+1) & =p(2 n-p+1) \\\\\n2 n k-k^{2}... | null | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Multimodal | Competition | true | null | null | null | Theorem proof | Number Theory | Math | English | |
2,803 | 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 ... | [
"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 ."
] | null | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Multimodal | Competition | true | null | null | null | Theorem proof | Combinatorics | Math | English | ||
2,804 | 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 ... | [
"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}\\rig... | null | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Multimodal | Competition | true | null | null | null | Theorem proof | Combinatorics | Math | English | ||
2,805 | 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 ... | [
"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 b... | null | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Multimodal | Competition | true | null | null | null | Theorem proof | Combinatorics | Math | English | |
2,807 | 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 ... | [
"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 Robbe... | null | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Multimodal | Competition | true | null | null | null | Theorem proof | Combinatorics | Math | English | ||
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 ... | [
"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... | null | null | Not supported with pagination yet | Not supported with pagination yet | Multimodal | Competition | true | null | null | null | Theorem proof | Combinatorics | Math | English | |||
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 ... | [
"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... | null | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Multimodal | Competition | true | null | null | null | Theorem proof | Combinatorics | Math | English | ||
2,812 | 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 ... | [
"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}| \\l... | null | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Multimodal | Competition | true | null | null | null | Theorem proof | Combinatorics | Math | English | ||
2,813 | 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 ... | [
"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 $M$. Notic... | null | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Multimodal | Competition | true | null | null | null | Theorem proof | Combinatorics | Math | English | ||
2,814 | 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 ... | [
"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 $n / k$ C... | null | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Multimodal | Competition | true | null | null | null | Theorem proof | Combinatorics | Math | English | ||
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 ... | [
"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 m... | null | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Multimodal | Competition | true | null | null | null | Theorem proof | Combinatorics | Math | English | ||
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 ... | [
"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 c... | null | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Multimodal | Competition | true | null | null | null | Theorem proof | Combinatorics | Math | English | ||
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... | [
"The signature is possible, because it is the 3 -signature of 12435"
] | null | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Multimodal | Competition | true | null | null | null | Theorem proof | Combinatorics | Math | English | ||
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... | [
"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."
] | null | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Multimodal | Competition | true | null | null | null | Theorem proof | Combinatorics | Math | English | ||
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... | [
"The $p$-signature is not unique because it equals both $S_{3}[625143]$ and $S_{3}[635142]$."
] | null | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Multimodal | Competition | true | null | null | null | Theorem proof | Combinatorics | Math | English | ||
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... | [
"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 u... | null | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Multimodal | Competition | true | null | null | null | Theorem proof | Combinatorics | Math | English | ||
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... | [
"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| :--- | ... | null | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Multimodal | Competition | true | null | null | null | Theorem proof | Combinatorics | Math | English | ||
2,883 | 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... | [
"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. ... | null | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Multimodal | Competition | true | null | null | null | Theorem proof | Combinatorics | Math | English | ||
2,928 | 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 circ... | [
"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 $r+\\frac{1}{2}$... | null | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Multimodal | Competition | true | null | null | null | Theorem proof | Geometry | Math | English | ||
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 circ... | [
"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... | null | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Multimodal | Competition | true | null | null | null | Theorem proof | Geometry | Math | English | ||
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 circ... | [
"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>"
] | null | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Multimodal | Competition | true | null | null | null | Theorem proof | Geometry | Math | English | ||
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 circ... | [
"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$$\n\\begin{aligned}\n\\left(\\rho-\\frac{1}{\\rho}\\right)^{2} & =(2 \\sqrt{\\phi})^{2}=4 \\phi \\\\\n& =2\\left(\\rho+\\frac{1}{\\rho}\\right) .... | null | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Multimodal | Competition | true | null | null | null | Theorem proof | Geometry | Math | English | ||
2,937 | A king strapped for cash is forced to sell off his kingdom $U=\left\{(x, y): x^{2}+y^{2} \leq 1\right\}$. He sells the two circular plots $C$ and $C^{\prime}$ centered at $\left( \pm \frac{1}{2}, 0\right)$ with radius $\frac{1}{2}$. The retained parts of the kingdom form two regions, each bordered by three arcs of circ... | [
"If the radii are in geometric progression, then so are their reciprocals (i.e., curvatures). Without loss of generality, let $(a, b, c, d)=\\left(a, a r, a r^{2}, a r^{3}\\right)$ for $r>1$. By Descartes' Circle Formula,\n\n$$\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^... | null | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Multimodal | Competition | true | null | null | null | Theorem proof | Geometry | Math | English | ||
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 circ... | [
"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 $x^{2}-2(b+c+d) x+$ $\\ldo... | null | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Multimodal | Competition | true | null | null | null | Theorem proof | Geometry | Math | English | ||
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 circ... | [
"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 ... | null | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Multimodal | Competition | true | null | null | null | Theorem proof | Geometry | Math | English |
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