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 |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
2,941 | 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 |
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