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 |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
729 | 求方程 $3^{p}+4^{p}=n^{k}$ 的正整数解 $(p, n, k)$, 其中, $p$ 为质数, $k>1$. | [
"显然, $3^{2}+4^{2}=5^{2}$, 即 $p=2, n=5, k=2$是方程的一组解.\n\n以下不妨设 $p$ 为奇质数, $p=2 l+1$. 则 $n^{k}=3^{2 l+1}+4^{2 l+1}=(3+4)\\left(3^{2 l}-3^{2 l-1} \\times 4+3^{2 l-2} \\times 4^{2}-\\cdots+4^{2 l}\\right)$.\n\n于是, $7\\left|n^{k}, 7\\right| n$.\n\n由 $k>1$, 得 $49 \\mid n^{k}$, 即\n\n$3^{2 l+1}+4^{2 l+1} \\equiv 0(\\bmod 49)... | [
"$(2,5,2)$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Tuple | null | Open-ended | Algebra | Math | Chinese |
730 | 平面上满足任意三点不共线的 $n$ 个点 $P_{1}, P_{2}, \cdots, P_{n}$ 构成的集合为 $D$, 在任意两点之间连一条线段,且每条线段的长互不相等.在一个三角形的三条边中, 长度非最长、也非最短的边称为该三角形的“中边”; 若一个三角形的三条边都是中边 (不一定是这个三角形的中边), 则称这个三角形为集合 $D$ 中的一个“中边三角形”. 一条不过点 $P_{i}(i=1,2$, $\cdots, n)$ 的直线 $l$ 将集合 $D$ 分成两个子集 $D_{1}$ 、 $D_{2}$. 若无论这 $n$ 个点如何分布, 也无论 $l$ 如何选取, 总存在一个子集 $D_{k}(k \in\{1,2\})... | [
"当 $n \\geqslant 11$ 时,无论 $l$ 如何选取, 总存在一个子集, 不妨假设为 $D_{1}$, 满足 $D_{1}$ 中至少有六个点.\n\n考虑 $D_{1}$ 中的所有三角形的中边,并将其染为红色,然后将其他边染为蓝色.\n\n由拉姆塞定理知,一定存在一个同色三角形. 由于每个三角形都有中边, 因此, 这个同色三角形一定是红色的. 故在 $D_{1}$ 中存在中边三角形\n\n下面证明: 当 $n \\leqslant 10$ 时,存在点集 $D$ 和直线 $l$, 使得 $D_{1} 、 D_{2}$ 中均不存在中边三角形.\n\n若 $n=10$, 考虑在直线 $l$ 两侧的两个子集 $D_{... | [
"$11$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Combinatorics | Math | Chinese |
733 | 设 $a 、 b$ 为正整数, $a^{2}+b^{2}$ 除以 $a+b$ 的商为 $q$, 余数为 $r$, 且 $q^{2}+r=2010$. 求 $a b$的值. | [
"不妨设 $a \\leqslant b$.\n\n由 $a^{2}+b^{2}>b^{2}-a^{2}=(b-a)(b+a)$, 得 $q \\geqslant b-a$.\n\n又由 $a^{2}+b^{2} \\geqslant a^{2}+a b=a(a+b)$, 得\n\n$q \\geqslant a$.\n\n另一方面, 由带余除法的性质有\n\n$0 \\leqslant r<a+b$.\n\n故 $0 \\leqslant r<a+b=a+a+(b-a) \\leqslant 3 q$.\n\n由 $q^{2}+r=2010$, 得\n\n$q^{2} \\leqslant 2010<q^{2}+3 q$.... | [
"$1643$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Number Theory | Math | Chinese |
737 | 已知 $p$ 为质数, $\sqrt{p}$ 的小数部分为 $x, \frac{1}{x}$ 的小数部分为 $\frac{\sqrt{p}-31}{75}$. 求所有满足条件的质数 $p$ 的值. | [
"设 $p=k^{2}+r$, 其中, $k 、 r$ 是整数, 且满足 $0 \\leqslant r \\leqslant 2 k$.\n\n因为 $\\frac{\\sqrt{p}-31}{75}$ 是 $\\frac{1}{x}$ 的小数部分, 所以,\n\n$0 \\leqslant \\frac{\\sqrt{p}-31}{75}<1 \\Rightarrow 31 \\leqslant \\sqrt{p}<106$.\n\n由 $p$ 为质数,知 $\\sqrt{p}$ 为无理数,且\n\n$[\\sqrt{p}]=k$.\n\n于是, $x=\\sqrt{p}-k(0<x<1)$.\n\n由于 $\\frac... | [
"$2011$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Number Theory | Math | Chinese |
744 | 对于任意实数 $a_{1}, a_{2}, \cdots, a_{2016}$, 试求
$$
\frac{7+23 \sum_{1 \leqslant i<j \leqslant 2016} \sin ^{2}\left(a_{i}-a_{j}\right)}{7+24 \sum_{1 \leqslant i<j \leqslant 2016} \cos ^{2}\left(a_{i}-a_{j}\right)}
$$
的取值范围. | [
"设原式为 $S$,\n\n$\\sum_{1 \\leqslant i<j \\leqslant 2016} \\sin ^{2}\\left(a_{i}-a_{j}\\right)=x$.\n\n则 $\\sum_{1 \\leqslant i<j \\leqslant 2016} \\cos ^{2}\\left(a_{i}-a_{j}\\right)$\n\n$=\\sum_{1 \\leqslant i<j \\leqslant 2016}\\left[1-\\sin ^{2}\\left(a_{i}-a_{j}\\right)\\right]=\\mathrm{C}_{2016}^{2}-x$.\n\n于是, $... | [
"$\\left[\\frac{1}{6963841}, \\frac{3338497}{3480193}\\right]$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Interval | null | Open-ended | Algebra | Math | Chinese |
745 | 求所有的映射 $f: \mathbf{Z} \rightarrow \mathbf{Z}$, 使得对于任意的 $m 、 n \in \mathbf{Z}$,均有
$$
f(f(m+n))=f(m)+f(n)
$$ | [
"设 $f(0)=a \\in \\mathbf{Z}$, 令 $n=0$, 则\n\n$f(f(m))=f(m)+f(0)=f(m)+a \\quad \\quad (1)$.\n\n在式(1)中取 $m=x+y(x 、 y \\in \\mathbf{Z})$, 则\n\n$f(f(x+y))=f(x+y)+a$.\n\n结合原题条件得\n\n$f(x)+f(y)=f(x+y)+a \\quad \\quad (2)$.\n\n令 $g(x)=f(x)-a$. 由式(2)得\n\n$g(x+y)=f(x+y)-a$\n\n$=f(x)+f(y)-2 a$\n\n$=g(x)+g(y) \\quad \\quad (3)$... | [
"$f(n)=n+a, f(n)=0$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | true | null | Expression | null | Open-ended | Algebra | Math | Chinese |
746 | 已知 $n$ 个正整数 $x_{1}, x_{2}, \cdots, x_{n}$ 的和为 2016. 若这 $n$ 个数既可分为和相等的 32 个组,又可分为和相等的 63 个组,求 $n$ 的最小值. | [
"设分成的 32 个组为 $A_{1}, A_{2}, \\cdots, A_{32}$, 每组中的各数之和均为 63 , 称这种组为 $A$ 类组; 而分成的 63 个组为 $B_{1}, B_{2}, \\cdots, B_{63}$, 每组中的各数之和均为 32 , 称这种组为 $B$ 类组.\n\n显然,每个项 $x_{k}$ 恰属于一个 $A$ 类组和一个 $B$ 类组，即同类组之间没有公共项; 若两个组 $A_{i} 、 B_{j}$ 中有两个公共项 $x_{r} 、 x_{t}$, 则可将这两个数合并为一项 $x_{r}+x_{t}$, 使 $n$ 值减少,故不妨设每对 $A_{i} 、 B_{j}$ 至多有一... | [
"$94$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Combinatorics | Math | Chinese |
751 | 对于平面 $\alpha$, 若多面体的各个顶点到平面 $\alpha$ 的距离均相等, 则称平面 $\alpha$ 为多面体的“中位面”.
四面体有多少个互不相同的中位面? | [
"将所考虑的四面体记作 $A B C D$.\n\n若四个顶点均在平面 $\\alpha$ 的一侧,则这四个顶点必位于一个与平面 $\\alpha$ 平行的平面内,不符合条件.\n\n只考虑以下两种情形.\n\n(i) 平面 $\\alpha$ 的一侧有三个顶点、另一侧有一个顶点.\n\n不妨设点 $A 、 B 、 C$ 在平面 $\\alpha$ 的一侧、点 $D$在另一侧. 则 $A 、 B 、 C$ 三点所确定的平面必平行于平面 $\\alpha$.\n\n由点 $D$ 作平面 $A B C$ 的垂线 $D D_{1}, D_{1}$ 为垂足. 则中位面 $\\alpha$ 必为经过 $D D_{1}$ 的中点且与 $... | [
"$7$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Geometry | Math | Chinese |
752 | 对于平面 $\alpha$, 若多面体的各个顶点到平面 $\alpha$ 的距离均相等, 则称平面 $\alpha$ 为多面体的“中位面”.
平行六面体有多少个互不相同的中位面? | [
"由于平行六面体任意五个顶点不共面,于是, 平行六面体的八个顶点分布在平面 $\\alpha$两侧的情形: 8-0、7-1、6-2、5-3 均不符合条件.\n\n只考虑八个顶点在平面 $\\alpha$ 两侧各四点的情形. 此时,平面 $\\alpha$ 一侧的 4 个顶点必位于一个与平面 $\\alpha$ 平行的平面上.\n\n将所考虑的平行六面体记作 $A B C D-$ $A_{1} B_{1} C_{1} D_{1}$.\n\n\n由于面 $A B C D / /$ 面 $A_{1} B_{1} C_{1} D_{1}$, 则经过这两个面的任意一条公垂线中点且与这两个面平行的平面, 到这两个面的距离相等. 从而,该... | [
"$3$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Geometry | Math | Chinese |
753 | 对于平面 $\alpha$, 若多面体的各个顶点到平面 $\alpha$ 的距离均相等, 则称平面 $\alpha$ 为多面体的“中位面”.
给定三维空间内不共面的四个点, 以这四点作为平行六面体的顶点 (中的四个),共可得到多少个互不相同的平行六面体? | [
"由于平行六面体任意五个顶点不共面,于是, 平行六面体的八个顶点分布在平面 $\\alpha$两侧的情形: 8-0、7-1、6-2、5-3 均不符合条件.\n\n只考虑八个顶点在平面 $\\alpha$ 两侧各四点的情形. 此时,平面 $\\alpha$ 一侧的 4 个顶点必位于一个与平面 $\\alpha$ 平行的平面上.\n\n将所考虑的平行六面体记作 $A B C D-$ $A_{1} B_{1} C_{1} D_{1}$.\n\n由于面 $A B C D / /$ 面 $A_{1} B_{1} C_{1} D_{1}$, 则经过这两个面的任意一条公垂线中点且与这两个面平行的平面, 到这两个面的距离相等. 从而,该平面... | [
"$29$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Geometry | Math | Chinese |
755 | 求所有的正整数 $x, y$, 使得 $\left(x^{2}+y\right)\left(x+y^{2}\right)$ 为素数的 5 次幂. | [
"易知 $x \\neq y$, 不妨设 $x<y$, 则 $x^{2}+y<x+y^{2}$.\n\n设 $p$ 为素数, $s<t \\in \\mathbb{N}^{*}, s+t=5$, 使得 $x^{2}+y=p^{s}, x+y^{2}=p^{t}$.\n\n若 $s=1(t=4)$, 则 $p^{2}=\\left(x^{2}+y\\right)^{2}>x+y^{2}=p^{4}$, 矛盾. 故 $s=2, t=3$. 即\n\n$$\nx^{2}+y=p^{2}, x+y^{2}=p^{3}, x<y \\text {. }\n$$\n\n注意到 $x<p$, 且\n\n$$\n\\left(x^{2}+y... | [
"$(2,5)，(5,2)$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | true | null | Tuple | null | Open-ended | Number Theory | Math | Chinese |
757 | 已知过点 $P(3,0)$ 斜率为 $k$ 的直线 $l$ 交双曲线 $C: x^{2}-\frac{y^{2}}{3}=1$ 右支于 $A 、 B$ 两点, $F$ 为双曲线 $C$ 的右焦点, 且 $|A F|+|B F|=16$, 求 $k$ 的值. | [
"设 $A\\left(x_{1}, y_{1}\\right), B\\left(x_{2}, y_{2}\\right)$, 则直线 $l$ : $y=k(x-3)$. 离心率 $e=2$\n\n联立 $\\left\\{\\begin{array}{c}y=k(x-3) \\\\ x^{2}-\\frac{y^{2}}{3}=1\\end{array}\\right.$ 得 $\\left(k^{2}-3\\right) x^{2}-6 k^{2} x+9 k^{2}+3=0$\n\n$\\therefore x_{1}+x_{2}=\\frac{6 k^{2}}{k^{2}-3} $.\n\n$\\therefore... | [
"$3，-3$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | true | null | Numerical | null | Open-ended | Conic Sections | Math | Chinese |
761 | 设点 $A$ 的坐标为 $(0,3)$, 点 $B, C$ 为圆 $O: x^{2}+y^{2}=25$ 上的两动点, 满足 $\angle B A C=90^{\circ}$, 求 $\triangle A B C$ 面积的最大值. | [
"<img_4208>\n\n如上图, 设 $B\\left(x_{1}, y_{1}\\right), C\\left(x_{2}, y_{2}\\right), P(x, y)$ 为线段 $B C$ 的中点.\n\n则\n\n$$\n\\begin{aligned}\n& x_{1}^{2}+y_{1}^{2}=25 \\\\\n& x_{2}^{2}+y_{2}^{2}=25 \\\\\n& x_{1} x_{2}+\\left(y_{1}-3\\right)\\left(y_{2}-3\\right)=0 \\\\\n& x_{1}+x_{2}=2 x, y_{1}+y_{2}=2 y\n\\end{aligned}... | [
"$\\frac{25+3 \\sqrt{41}}{2}$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Plane Geometry | Math | Chinese |
762 | 设 $a, b, c \in(0,1], \lambda$ 为实数, 使得
$$
\frac{\sqrt{3}}{\sqrt{a+b+c}} \geq 1+\lambda(1-a)(1-b)(1-c)
$$
恒成立, 求 $\lambda$ 的最大值. | [
"取 $a=b=c=\\frac{1}{4}$ 时, $\\lambda \\leq \\frac{64}{27}$.\n\n下证: $\\lambda=\\frac{64}{27}$ 满足条件, 即证 $\\frac{\\sqrt{3}}{\\sqrt{a+b+c}} \\geq 1+\\frac{64}{27}(1-a)(1-b)(1-c)$\n\n注意到: $(1-a)(1-b)(1-c) \\leq\\left(1-\\frac{a+b+c}{3}\\right)^{3}$\n\n令 $a+b+c=3 x^{2}$, 其中 $x>0$, 则 $0<x \\leq 1$.\n\n只须证 $\\frac{1}{x} \\... | [
"$\\frac{64}{27}$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Inequality | Math | Chinese |
764 | 已知函数 $f(x)=x \ln x-a x^{2}, a \in \mathbf{R}$.
设函数 $F(x)=|f(x)|(x \in[1, \mathrm{e}])$ 有极小值, 求 $a$ 的取值范围. | [
"$F(x)=|f(x)|=x^{2}\\left|\\frac{\\ln x}{x}-a\\right|, \\quad x \\in[1, \\mathrm{e}]$\n\n令 $t(x)=\\frac{\\ln x}{x}-a, x \\in[1, \\mathrm{e}]$, 则 $t^{\\prime}(x)=\\frac{1-\\ln x}{x^{2}}$,\n\n当 $x \\in[1, \\mathrm{e}]$ 时, $t^{\\prime}(x) \\geq 0$,\n\n故 $t(x)$ 在 $[1, \\mathrm{e}]$ 上单调递增.\n\n于是 $t(1) \\leq t(x) \\leq t... | [
"$(0, \\frac{1}{e}) \\cup (\\frac{1}{e}, \\frac{1}{2})$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Interval | null | Open-ended | Elementary Functions | Math | Chinese |
765 | 设 $F$ 是椭圆 $E: \frac{x^{2}}{3}+y^{2}=1$ 的左焦点, 过点 $F$ 斜率为正的直线 $l$ 与 $E$ 相交于 $A, B$ 两点, 过点 $A, B$ 分别作直线 $A M$ 和 $B N$ 满足 $A M \perp l, B N \perp l$, 且直线 $A M, B N$ 分别与 $x$ 轴相交于 $M$ 和 $N$. 试求 $|M N|$ 的最小值. | [
"设过椭圆 $E$ 的左焦点 $F$ 的直线 $l$ 的倾斜角为 $\\alpha$, 依题意知 $\\frac{\\pi}{2}>\\alpha>0$.\n\n<img_4159>\n\n如上图, 设 $F^{\\prime}$ 为椭圆 $E$ 的右焦点, 在 $R t \\triangle M A F$ 中 $\\cos \\angle M F A=\\frac{|A F|}{|M F|}$, 所以有 $|M F|=\\frac{|A F|}{\\cos \\angle M F A}$. 在 $R t \\triangle N B F$ 中, 同理有 $|N F|=\\frac{|B F|}{\\cos \\angle ... | [
"$\\sqrt{6}$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Plane Geometry | Math | Chinese |
766 | 设 $a_{0}, a_{1}, a_{2}, \ldots, a_{n}$ 是正整数, 且 $a_{0}>a_{1}>a_{2}>\cdots>a_{n}>1$并满足
$$
\left(1-\frac{1}{a_{1}}\right)+\left(1-\frac{1}{a_{2}}\right)+\cdots+\left(1-\frac{1}{a_{n}}\right)=2\left(1-\frac{1}{a_{0}}\right) .
$$
试求出 $\left(a_{0}, a_{1}, a_{2}, \ldots, a_{n}\right)$ 所有可能的解. | [
"由题意知, 对 $0 \\leq i \\leq n$ 均有 $a_{i} \\geq 2$. 于是有\n\n$$\n2>2\\left(1-\\frac{1}{a_{0}}\\right)=\\left(1-\\frac{1}{a_{1}}\\right)+\\left(1-\\frac{1}{a_{2}}\\right)+\\cdots+\\left(1-\\frac{1}{a_{n}}\\right)>\\frac{n}{2}\n$$\n\n可得 $n<4$. 由于 $n$ 是正整数, 故 $n \\in\\{1,2,3\\}$.\n\n(i) 当 $n=1$ 时, 我们有 $\\frac{2}{a_{0}}=1+\... | [
"$(24,4,3,2),(60,5,3,2)$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | true | null | Tuple | null | Open-ended | Algebra | Math | Chinese |
767 | 给定正实数 $0<a<b$, 设 $x_{1}, x_{2}, x_{3}, x_{4} \in[a, b]$. 试求$\frac{\frac{x_{1}^{2}}{x_{2}}+\frac{x_{2}^{2}}{x_{3}}+\frac{x_{3}^{2}}{x_{4}}+\frac{x_{4}^{2}}{x_{1}}}{x_{1}+x_{2}+x_{3}+x_{4}} $的最小值与最大值. | [
"(i) 因为 $x_{1}, x_{2}, x_{3}, x_{4} \\in[a, b]$ 且 $0<a<b$, 所以有\n\n$$\n\\left\\{\\begin{array}{l}\n\\frac{x_{1}^{2}}{x_{2}}+x_{2} \\geq 2 \\sqrt{\\frac{x_{1}^{2}}{x_{2}} \\cdot x_{2}}=2 x_{1} \\\\\n\\frac{x_{2}^{2}}{x_{3}}+x_{3} \\geq 2 \\sqrt{\\frac{x_{2}^{2}}{x_{3}} \\cdot x_{3}}=2 x_{2} \\\\\n\\frac{x_{3}^{2}}{x_... | [
"$1 , \\frac{b}{a}+\\frac{a}{b}-1$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | true | null | Numerical,Expression | null | Open-ended | Inequality | Math | Chinese |
770 | 试求所有由互异正奇数构成的三元集 $\{a, b, c\}$, 使其满足: $a^{2}+b^{2}+c^{2}=2019$. | [
"据对称性, 不妨设 $a<b<c$, 由于奇平方数的末位数字只具有 $1,5,9$ 形式, 于是 $a^{2}, b^{2}, c^{2}$ 的末位数字, 要么是 $5,5,9$ 形式, 要么是 $1,9,9$ 形式; 又知, 如果正整数 $n$ 是 3 的倍数, 那么 $n^{2}$ 必是 9 的倍数; 如果 $n$ 不是 3 的倍数, 那么 $n^{2}$ 被 3 除余 1 . 由于 2019 是 3 的倍数, 但不是 9 的倍数, 因此奇数 $a, b, c$ 皆不是 3 的倍数. 注意 $c \\leq[\\sqrt{2019}]=44$, 即奇数 $c \\leq 43$, 而 $3 c^{2}>a^{2}+b^... | [
"$(1,13,43),(7,11,43),(13,25,35),(5,25,37),(11,23,37),(17,19,37),(7,17,41)$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | true | null | Tuple | null | Open-ended | Number Theory | Math | Chinese |
772 | 已知函数 $f(x)=\left(a+\frac{1}{a}\right) \ln x+\frac{1}{x}-x$.
设 $a>1$, 则求$f(x)$ 在区间 $(0,1)$ 上的单调递增区间; | [
"$f^{\\prime}(x)=\\left(a+\\frac{1}{a}\\right) \\frac{1}{x}-\\frac{1}{x^{2}}-1=-\\frac{(x-a)\\left(x-\\frac{1}{a}\\right)}{x^{2}}$.\n\n由 $a>1$ 可知 $0<\\frac{1}{a}<1<a$, 所以 $f(x)$ 的减区间是 $\\left(0, \\frac{1}{a}\\right]$, 增区间是 $\\left[\\frac{1}{a}, 1\\right)$.\n\n"
] | [
"$\\left[\\frac{1}{a}, 1\\right)$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Interval | null | Open-ended | Elementary Functions | Math | Chinese |
773 | 已知函数 $f(x)=\left(a+\frac{1}{a}\right) \ln x+\frac{1}{x}-x$.
设 $a>0$, 求 $f(x)$ 的极值. | [
"$f^{\\prime}(x)=-\\frac{(x-a)\\left(x-\\frac{1}{a}\\right)}{x^{2}}, x>0$.\n\n当 $a>1$ 时, $f(x)$ 的减区间是 $\\left(0, \\frac{1}{a}\\right]$ 和 $[a,+\\infty)$, 增区间是 $\\left[\\frac{1}{a}, a\\right]$.\n\n$f(x)$ 的极小值为 $f\\left(\\frac{1}{a}\\right)=-\\left(a+\\frac{1}{a}\\right) \\ln a+a-\\frac{1}{a}$, 极大值为 $f(a)=\\left(a+\\f... | [
"$\\left(a+\\frac{1}{a}\\right) \\ln a+\\frac{1}{a}-a , -\\left(a+\\frac{1}{a}\\right) \\ln a+a-\\frac{1}{a}$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | true | null | Expression | null | Open-ended | Elementary Functions | Math | Chinese |
779 | 在平面直角坐标系 $x O y$ 中, 设椭圆 $C: \frac{x^{2}}{a^{2}}+\frac{y^{2}}{a^{2}-9}=1(a>3)$.
过椭圆 $C$ 的左焦点, 且垂直于 $x$ 轴的直线与椭圆 $C$ 交于 $M, N$ 两点, 若 $M N=9$, 求实数 $a$ 的值. | [
"记椭圆 $C: \\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{a^{2}-9}=1$ 的左焦点为 $F$, 则点 $F$ 的横坐标为 -3 .\n\n因为过点 $F$ 且垂直于 $x$ 轴的直线与椭圆 $C$ 交于 $M, N$ 两点, $M N=9$,\n\n所以 $M\\left(-3, \\frac{9}{2}\\right), N\\left(-3,-\\frac{9}{2}\\right)$,\n\n从而 $\\frac{9}{a^{2}}+\\frac{81}{4\\left(a^{2}-9\\right)}=1$.\n\n解得 $a^{2}=36$ 或 $a^{2}=\\frac{9}... | [
"$6$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Plane Geometry | Math | Chinese |
780 | 在平面直角坐标系 $x O y$ 中, 设椭圆 $C: \frac{x^{2}}{a^{2}}+\frac{y^{2}}{a^{2}-9}=1(a>3)$.
若直线 $l: \frac{x}{a}+\frac{y}{a-3}=1$ 与椭圆 $C$ 交于 $A, B$ 两点, 其中对任意大于 3 的实数 $a$, 以 $A B$ 为直径的圆过定点, 求定点坐标. | [
"由 $\\left\\{\\begin{array}{l}\\frac{x}{a}+\\frac{y}{a-3}=1, \\\\ \\frac{x^{2}}{a^{2}}+\\frac{y^{2}}{a^{2}-9}=1\\end{array}\\right.$\n\n得 $A(a, 0), B\\left(-3, \\frac{a^{2}-9}{a}\\right)$.\n\n从而以 $A B$ 为直径的圆的方程为 $(x-a)(x+3)+y\\left(y-\\frac{a^{2}-9}{a}\\right)=0$,\n\n即 $(x+y+3) a^{2}-\\left(x^{2}+3 x+y^{2}\\right) ... | [
"$(-3,0)$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Tuple | null | Open-ended | Plane Geometry | Math | Chinese |
781 | 在数列 $\left\{a_{n}\right\}$ 中, 已知 $a_{1}=\frac{1}{p}, a_{n+1}=\frac{a_{n}}{n a_{n}+1}, p>0, n \in \mathbf{N}^{*}$.
若 $p=1$, 求数列 $\left\{a_{n}\right\}$ 的通项公式. | [
"因为 $a_{n+1}=\\frac{a_{n}}{n a_{n}+1}, a_{1}=\\frac{1}{p}>0$, 故 $a_{n}>0, \\frac{1}{a_{n+1}}=\\frac{1}{a_{n}}+n$,\n\n因此 $\\frac{1}{a_{n}}=(n-1)+(n-2)+\\ldots+1+p=\\frac{n^{2}-n+2 p}{2}$, 即 $a_{n}=\\frac{2}{n^{2}-n+2 p}$.\n\n因为 $p=1$, 所以 $a_{n}=\\frac{2}{n^{2}-n+2}$."
] | [
"$a_{n}=\\frac{2}{n^{2}-n+2}$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Expression | null | Open-ended | Sequence | Math | Chinese |
782 | 在数列 $\left\{a_{n}\right\}$ 中, 已知 $a_{1}=\frac{1}{p}, a_{n+1}=\frac{a_{n}}{n a_{n}+1}, p>0, n \in \mathbf{N}^{*}$.
记 $b_{n}=n a_{n}$. 若在数列 $\left\{b_{n}\right\}$ 中, $b_{n} \leq b_{8}\left(n \in \mathbf{N}^{*}\right)$, 求实数 $p$ 的取值范围. | [
"$b_{n}=n a_{n}=\\frac{2 n}{n^{2}-n+2 p}$.\n\n在数列 $\\left\\{b_{n}\\right\\}$ 中,\n\n因为 $b_{n} \\leq b_{8}\\left(n \\in \\mathbf{N}^{*}\\right)$, 所以 $b_{7} \\leq b_{8}, b_{8} \\leq b_{8}$, 即 $\\left\\{\\begin{array}{c}\\frac{7}{21+p} \\leq \\frac{8}{28+p} \\\\ \\frac{9}{36+p} \\leq \\frac{8}{28+p}\\end{array}\\right.... | [
"$[28,36]$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Interval | null | Open-ended | Sequence | Math | Chinese |
785 | 已知数列 $\left\{a_{n}\right\}$ 满足 $a_{1}=\frac{9}{4}, 2 a_{n+1} a_{n}-7 a_{n+1}-3 a_{n}+12=0 \quad\left(n \in N^{*}\right)$ .
记 $c_{n}=a_{n}-2$, 求数列 $\left\{c_{n}\right\}$ 的通项公式; | [
"由 $c_{n}=a_{n}-2$, 得 $a_{n}=c_{n}+2$, 代入条件递推式, 得\n\n$2\\left(c_{n+1}+2\\right)\\left(c_{n}+2\\right)-7\\left(c_{n+1}+2\\right)-3\\left(c_{n}+2\\right)+12=0$ .\n\n整理, 得 $2 c_{n+1} c_{n}-3 c_{n+1}+c_{n}=0$, 即 $\\frac{1}{c_{n+1}}=\\frac{3}{c_{n}}-2$ .\n\n$\\therefore \\frac{1}{c_{n+1}}-1=3\\left(\\frac{1}{c_{n}}-1\\r... | [
"$c_{n}=\\frac{1}{3^{n}+1}$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Expression | null | Open-ended | Sequence | Math | Chinese |
786 | 已知数列 $\left\{a_{n}\right\}$ 满足 $a_{1}=\frac{9}{4}, 2 a_{n+1} a_{n}-7 a_{n+1}-3 a_{n}+12=0 \quad\left(n \in N^{*}\right)$ .
记 $b_{n}=\frac{n^{2}}{n+1} a_{n}$, 求使 $\left[b_{1}\right]+\left[b_{2}\right]+\left[b_{3}\right]+\cdots+\left[b_{n}\right] \leq 2019$ 成立的最大正整数 $n$ 的值.
（其中, 符号 $[x]$ 表示不超过 $x$ 的最大整数) | [
"记 $c_{n}=a_{n}-2$, 则 $a_{n}=c_{n}+2$, 代入条件递推式, 得\n\n$2\\left(c_{n+1}+2\\right)\\left(c_{n}+2\\right)-7\\left(c_{n+1}+2\\right)-3\\left(c_{n}+2\\right)+12=0$ .\n\n整理, 得 $2 c_{n+1} c_{n}-3 c_{n+1}+c_{n}=0$, 即 $\\frac{1}{c_{n+1}}=\\frac{3}{c_{n}}-2$ .\n\n$\\therefore \\frac{1}{c_{n+1}}-1=3\\left(\\frac{1}{c_{n}}-1\\r... | [
"$45$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Sequence | Math | Chinese |
791 | 已知 $f(x)=e^{x}$ .
若 $x \geq 0$ 时, 不等式 $(x-1) f(x) \geq m x^{2}-1$ 恒成立, 求 $m$ 的取值范围; | [
"依题意, 当 $x \\geq 0$ 时, $(x-1) e^{x} \\geq m x^{2}-1$ 恒成立.\n\n设 $k(x)=(x-1) e^{x}-m x^{2}+1$, 则 $x \\geq 0$ 时, $k(x) \\geq 0$ 恒成立.\n\n$k^{\\prime}(x)=e^{x}+(x-1) e^{x}-2 m x=x\\left(e^{x}-2 m\\right) .$\n\n若 $m \\leq \\frac{1}{2}$, 则 $x>0$ 时, $k^{\\prime}(x)=x\\left(e^{x}-2 m\\right)>0, k(x)$ 在 $[0,+\\infty)$ 上为增函数.... | [
"$\\left(-\\infty, \\frac{1}{2}\\right]$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Interval | null | Open-ended | Elementary Functions | Math | Chinese |
794 | 已知 $a, b, c, d \in \mathbb{R}, f(x)=a x^{3}+b x^{2}+c x+d$, 且 $f(2019-i)=i$ $(i=1,2,3,4)$, 求 $8 a+4 b+2 c+d$ 的值. | [
"令 $g(x)=f(x)-(2019-x)$, 由已知可得\n\n$$\ng(2019-i)=f(2019-i)-2019+(2019-i)=0\n$$\n\n即 $2019-i(i=1,2,3,4)$ 为方程 $g(x)=0$ 的根, 令\n\n$$\ng(x)=f(x)-(2019-x)=c(x)(x-2018)(x-2017)(x-2016)(x-2015) \\cdots (1)\n$$\n\n(1) 式左边是三次, 右边高于三次, 所以 $c(x)=0$, 因此 $f(x)=2019-x$ .\n\n所以 $8 a+4 b+2 c+d=f(2)=2017$ ."
] | [
"$2017$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Elementary Functions | Math | Chinese |
796 | 已知椭圆 $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$, 过两个焦点分别作两条相互垂直的直线 $l_{1}, l_{2}$ .假设两条直线 $l_{1}, l_{2}$ 相交于 $P, l_{1}, l_{2}$ 在椭圆内所截的线段长度为 $m_{1}, m_{2}$ .
求 $m_{1} m_{2}$ 的最小值和最大值. | [
"设 $l_{1}$ 的斜率为 $k$, 则 $l_{1}$ 的直线方程为 $y=k(x-c)$, 联立椭圆方程, 求得 $m_{1}=\\frac{2 a b^{2}\\left(1+k^{2}\\right)}{a^{2} k^{2}+b^{2}}$\n\n同理求得 $m_{2}=\\frac{2 a b^{2}\\left(1+k^{2}\\right)}{a^{2}+b^{2} k^{2}}$ $m_{1} m_{2}=\\frac{4 a^{2} b^{4}\\left(1+k^{2}\\right)^{2}}{\\left(a^{2} k^{2}+b^{2}\\right)\\left(a^{2}+b^{2} k... | [
"$\\frac{16 a^{2} b^{4}}{\\left(a^{2}+b^{2}\\right)^{2}}, 4 b^{2}$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | true | null | Expression | null | Open-ended | Plane Geometry | Math | Chinese |
798 | 设 $A$ 为 20 ! 的所有因数构成的集合, $x_{i} \in A(1 \leq i \leq n)$ .若对任意的正整数 $y_{i}(1 \leq i \leq n)$, 存在正整数 $z$, 使得 $\frac{y_{i}+z}{x_{i}}$ 为正整数.
求 $n$ 的最大值. | [
"$20 !=2^{18} \\times 3^{8} \\times 5^{4} \\times 7^{2} \\times 11 \\times 13 \\times 17 \\times 19, A=\\left\\{x|x|20!,x \\in \\mathbb{N}_{+}\\right\\}$.\n\n令 $\\left(x_{i}, x_{j}\\right)=d\\left(x_{i}, x_{j} \\in A\\right), x_{i}=a d, x_{j}=b d$, 且 $(a, b)=1$ .\n\n取 $y_{1}=1, y_{2}=2$, 得 $\\frac{1+z}{a d}, \\frac... | [
"$9$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Number Theory | Math | Chinese |
799 | 设 $A$ 为 20 ! 的所有因数构成的集合, $x_{i} \in A(1 \leq i \leq n)$ .若对任意的正整数 $y_{i}(1 \leq i \leq n)$, 存在正整数 $z$, 使得 $\frac{y_{i}+z}{x_{i}}$ 为正整数.
$n$ 取最大值时，求所有可能的 $n$ 元数组 $\left(x_{1}, x_{2}, \cdots, x_{n}\right)$ 的组数. | [
"$20 !=2^{18} \\times 3^{8} \\times 5^{4} \\times 7^{2} \\times 11 \\times 13 \\times 17 \\times 19, A=\\left\\{x|x| 20！\\right.$ , $\\left.x \\in \\mathbb{N}_{+}\\right\\}$.\n\n令 $\\left(x_{i}, x_{j}\\right)=d\\left(x_{i}, x_{j} \\in A\\right), x_{i}=a d, x_{j}=b d$, 且 $(a, b)=1$ .\n\n取 $y_{1}=1, y_{2}=2$, 得 $\\fr... | [
"$418037760$"
] | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | false | null | Numerical | null | Open-ended | Number Theory | Math | Chinese |
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 | ||
2,941 | 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... | [
"Because 18 plots were sold on day 3 , the mean curvature is\n\n$$\n\\frac{2(38+38+35)+4(23+14+11)}{18}=23 .\n$$"
] | 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 | Geometry | Math | English | ||
2,942 | 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... | [
"Proceed by induction. The base case, that all curvatures prior to day 2 are integers. Using the formula $a^{\\prime}=2 s-3 a$, if $a, b, c, d$, and $s$ are integers on day $n$, then $a^{\\prime}, b^{\\prime}, c^{\\prime}$, and $d^{\\prime}$ are integer curvatures on day $n+1$, proving inductively that all curvatur... | 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 | ||
2,943 | 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... | [
"It suffices to show that for each circle $C$ with curvature $c$ and center $z_{C}$ (in the complex plane), $c z_{C}$ is of the form $u+i v$ where $u$ and $v$ are integers. If this is the case, then each center is of the form $\\left(\\frac{u}{c}, \\frac{v}{c}\\right)$.\n\nProceed by induction. To check the base ca... | 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 | ||
2,944 | 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 | Combinatorics | Math | English | ||
2,945 | 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... | [
"Because $\\mathcal{C}\\left(P^{(1)}\\right)=\\left(b, c, d, a^{\\prime}\\right)$ and $\\mathcal{C}\\left(Q^{(1)}\\right)=\\left(x, y, z, w^{\\prime}\\right)$, it is enough to show that $a^{\\prime} \\leq w^{\\prime}$. $a$ and $a^{\\prime}$ are the two roots of the quadratic given by Descartes' Circle Formula:\n\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 | ||
2,946 | 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... | [
"Because $\\mathcal{C}\\left(P^{(3)}\\right)=\\left(a, b, d, c^{\\prime}\\right)$ and $\\mathcal{C}\\left(Q^{(3)}\\right)=\\left(w, x, z, y^{\\prime}\\right)$, it suffices to show that $c^{\\prime} \\leq y^{\\prime}$. If $a \\geq 0$, but if $a<0$, then there is more to be done.\n\nArguing as in 9b, $c, c^{\\prime}=... | 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 | ||
3,052 | 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... | [
"By definition of $\\mathrm{Pa}, \\operatorname{Pa}(n, 0)=\\operatorname{Pa}(n, n)=1$ for all nonnegative integers $n$, and this value is odd, so $\\operatorname{PaP}(n, 0)=\\operatorname{PaP}(n, n)=1$ by definition."
] | 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 | |
3,054 | 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\n\n$$\n\\begin{aligned}\n\\operatorname{Pa}(n, k) & =\\frac{n !}{k !(n-k) !} \\\\\n& =\\frac{n}{k} \\frac{(n-1) !}{(k-1) !((n-1)-(k-1)) !} \\\\\n& =\\frac{n}{k} \\cdot \\operatorname{Pa}(n-1, k-1) .\n\\end{aligned}\n$$\n\nExamining the right hand side in the case where $n=2^{j}$ and $0<k<n$, the second... | 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 | |
3,055 | 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... | [
"Proceed by induction on $j$. The claim is that $\\mathrm{Pa}(n, k) \\equiv \\mathrm{Pa}\\left(n-2^{j}, k-2^{j}\\right)+\\mathrm{Pa}\\left(n-2^{j}, k\\right) \\bmod 2$. If $j=0$, so that\n\n\n\n$2^{j}=1$, then this congruence is exactly the recursive definition of $\\mathrm{Pa}(n, k)$ when $0<k<n$. For $k=0$ and $k... | 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 | |
3,056 | 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... | [
"Proceed by induction on $j$. The claim is that $\\mathrm{Pa}(n, k) \\equiv \\mathrm{Pa}\\left(n-2^{j}, k-2^{j}\\right)+\\mathrm{Pa}\\left(n-2^{j}, k\\right) \\bmod 2$. If $j=0$, so that\n\n\n\n$2^{j}=1$, then this congruence is exactly the recursive definition of $\\mathrm{Pa}(n, k)$ when $0<k<n$. For $k=0$ and $k... | 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 | |
3,059 | 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... | [
"Use induction on $n$. For $n=1,2,3$, the values above demonstrate the theorem. If $\\mathrm{Cl}(n, 1)=$ $3 n^{2}-3 n+1$, then $\\mathrm{Cl}(n+1,1)=\\mathrm{Cl}(n, 0)+\\mathrm{Cl}(n, 1)=6 n+\\left(3 n^{2}-3 n+1\\right)=\\left(3 n^{2}+6 n+3\\right)-$ $(3 n+3)+1=3(n+1)^{2}-3(n+1)+1$."
] | 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 | ||
3,072 | Leibniz's Harmonic Triangle: Consider the triangle formed by the rule
$$
\begin{cases}\operatorname{Le}(n, 0)=\frac{1}{n+1} & \text { for all } n \\ \operatorname{Le}(n, n)=\frac{1}{n+1} & \text { for all } n \\ \operatorname{Le}(n, k)=\operatorname{Le}(n+1, k)+\operatorname{Le}(n+1, k+1) & \text { for all } n \text {... | [
"Because in general $\\operatorname{Le}(i, m)=\\operatorname{Le}(i-1, m-1)-\\operatorname{Le}(i, m-1)$, a partial sum can be rewritten as follows:\n\n$$\n\\begin{aligned}\n\\sum_{i=m}^{n} \\operatorname{Le}(i, m)= & \\sum_{i=m}^{n}(\\operatorname{Le}(i-1, m-1)-\\operatorname{Le}(i, m-1)) \\\\\n= & (\\operatorname{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 | Combinatorics | Math | English | |
1,602 | Let $P$ and $P^{\prime}$ be two convex quadrilateral regions in the plane (regions contain their boundary). Let them intersect, with $O$ a point in the intersection. Suppose that for every line $\ell$ through $O$ the segment $\ell \cap P$ is strictly longer than the segment $\ell \cap P^{\prime}$. Is it possible that t... | [
"The answer is in the affirmative: Given a positive $\\epsilon<2$, the ratio in question may indeed be greater than $2-\\epsilon$.\n\n\n\nTo show this, consider a square $A B C D$ centred at $O$, and let $A^{\\prime}, B^{\\prime}$, and $C^{\\prime}$ be the reflections of $O$ in $A, B$, and $C$, respectively. Notice... | null | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | true | null | null | null | Theorem proof | Geometry | Math | English |
1,603 | Given an integer $k \geq 2$, set $a_{1}=1$ and, for every integer $n \geq 2$, let $a_{n}$ be the smallest $x>a_{n-1}$ such that:
$$
x=1+\sum_{i=1}^{n-1}\left\lfloor\sqrt[k]{\frac{x}{a_{i}}}\right\rfloor
$$
Prove that every prime occurs in the sequence $a_{1}, a_{2}, \ldots$ | [
"We prove that the $a_{n}$ are precisely the $k$ th-power-free positive integers, that is, those divisible by the $k$ th power of no prime. The conclusion then follows.\n\n\n\nLet $B$ denote the set of all $k$ th-power-free positive integers. We first show that, given a positive integer $c$,\n\n\n\n$$\n\n\\sum_{b \... | null | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | true | null | null | null | Theorem proof | Number Theory | Math | English |
1,604 | $2 n$ distinct tokens are placed at the vertices of a regular $2 n$-gon, with one token placed at each vertex. A move consists of choosing an edge of the $2 n$-gon and interchanging the two tokens at the endpoints of that edge. Suppose that after a finite number of moves, every pair of tokens have been interchanged exa... | [
"Step 1. Enumerate all the tokens in the initial arrangement in clockwise circular order; also enumerate the vertices of the $2 n$-gon accordingly. Consider any three tokens $i<j<k$. At each moment, their cyclic order may be either $i, j, k$ or $i, k, j$, counted clockwise. This order changes exactly when two of th... | null | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | true | null | null | null | Theorem proof | Combinatorics | Math | English |
1,605 | Let $T_{1}, T_{2}, T_{3}, T_{4}$ be pairwise distinct collinear points such that $T_{2}$ lies between $T_{1}$ and $T_{3}$, and $T_{3}$ lies between $T_{2}$ and $T_{4}$. Let $\omega_{1}$ be a circle through $T_{1}$ and $T_{4}$; let $\omega_{2}$ be the circle through $T_{2}$ and internally tangent to $\omega_{1}$ at $T_{... | [
"Let $O_{i}$ be the centre of $\\omega_{i}, i=1,2,3,4$. Notice that the isosceles triangles $O_{i} T_{i} T_{i-1}$ are similar (indices are reduced modulo 4), to infer that $\\omega_{4}$ is internally tangent to $\\omega_{1}$ at $T_{4}$, and $\\mathrm{O}_{1} \\mathrm{O}_{2} \\mathrm{O}_{3} \\mathrm{O}_{4}$ is a (pos... | null | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | true | null | null | null | Theorem proof | Geometry | Math | English |
1,607 | A number of 17 workers stand in a row. Every contiguous group of at least 2 workers is a brigade. The chief wants to assign each brigade a leader (which is a member of the brigade) so that each worker's number of assignments is divisible by 4 . Prove that the number of such ways to assign the leaders is divisible by 17... | [
"Assume that every single worker also forms a brigade (with a unique possible leader). In this modified setting, we are interested in the number $N$ of ways to assign leadership so that each worker's number of assignments is congruent to 1 modulo 4.\n\n\n\nConsider the variables $x_{1}, x_{2}, \\ldots, x_{17}$ corr... | null | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | true | null | null | null | Theorem proof | Number Theory | Math | English |
1,608 | Let $A B C$ be a triangle, let $D$ be the touchpoint of the side $B C$ and the incircle of the triangle $A B C$, and let $J_{b}$ and $J_{c}$ be the incentres of the triangles $A B D$ and $A C D$, respectively. Prove that the circumcentre of the triangle $A J_{b} J_{c}$ lies on the bisectrix of the angle $B A C$. | [
"Let the incircle of the triangle $A B C$ meet $C A$ and $A B$ at points $E$ and $F$, respectively. Let the incircles of the triangles $A B D$ and $A C D$ meet $A D$ at points $X$ and $Y$, respectively. Then $2 D X=D A+D B-A B=D A+D B-B F-A F=D A-A F$; similarly, $2 D Y=D A-A E=2 D X$. Hence the points $X$ and $Y$ ... | null | null | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Not supported with pagination yet | Text-only | Competition | true | null | null | null | Theorem proof | Geometry | Math | English |
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