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sequence | final_answer
sequence | 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
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string | question_type
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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 circles; in what follows, we will call such regions curvilinear triangles, or $c$-triangles ( $\mathrm{c} \triangle$ ) for short.
This sad day marks day 0 of a new fiscal era. Unfortunately, these drastic measures are not enough, and so each day thereafter, court geometers mark off the largest possible circle contained in each c-triangle in the remaining property. This circle is tangent to all three arcs of the c-triangle, and will be referred to as the incircle of the c-triangle. At the end of the day, all incircles demarcated that day are sold off, and the following day, the remaining c-triangles are partitioned in the same manner.
Some notation: when discussing mutually tangent circles (or arcs), it is convenient to refer to the curvature of a circle rather than its radius. We define curvature as follows. Suppose that circle $A$ of radius $r_{a}$ is externally tangent to circle $B$ of radius $r_{b}$. Then the curvatures of the circles are simply the reciprocals of their radii, $\frac{1}{r_{a}}$ and $\frac{1}{r_{b}}$. If circle $A$ is internally tangent to circle $B$, however, as in the right diagram below, the curvature of circle $A$ is still $\frac{1}{r_{a}}$, while the curvature of circle $B$ is $-\frac{1}{r_{b}}$, the opposite of the reciprocal of its radius.
<image_1>
Circle $A$ has curvature 2; circle $B$ has curvature 1 .
<image_2>
Circle $A$ has curvature 2; circle $B$ has curvature -1 .
Using these conventions allows us to express a beautiful theorem of Descartes: when four circles $A, B, C, D$ are pairwise tangent, with respective curvatures $a, b, c, d$, then
$$
(a+b+c+d)^{2}=2\left(a^{2}+b^{2}+c^{2}+d^{2}\right),
$$
where (as before) $a$ is taken to be negative if $B, C, D$ are internally tangent to $A$, and correspondingly for $b, c$, or $d$.
Show that the plots sold on day 3 have mean curvature of 23. | [
"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 circles; in what follows, we will call such regions curvilinear triangles, or $c$-triangles ( $\mathrm{c} \triangle$ ) for short.
This sad day marks day 0 of a new fiscal era. Unfortunately, these drastic measures are not enough, and so each day thereafter, court geometers mark off the largest possible circle contained in each c-triangle in the remaining property. This circle is tangent to all three arcs of the c-triangle, and will be referred to as the incircle of the c-triangle. At the end of the day, all incircles demarcated that day are sold off, and the following day, the remaining c-triangles are partitioned in the same manner.
Some notation: when discussing mutually tangent circles (or arcs), it is convenient to refer to the curvature of a circle rather than its radius. We define curvature as follows. Suppose that circle $A$ of radius $r_{a}$ is externally tangent to circle $B$ of radius $r_{b}$. Then the curvatures of the circles are simply the reciprocals of their radii, $\frac{1}{r_{a}}$ and $\frac{1}{r_{b}}$. If circle $A$ is internally tangent to circle $B$, however, as in the right diagram below, the curvature of circle $A$ is still $\frac{1}{r_{a}}$, while the curvature of circle $B$ is $-\frac{1}{r_{b}}$, the opposite of the reciprocal of its radius.
<image_1>
Circle $A$ has curvature 2; circle $B$ has curvature 1 .
<image_2>
Circle $A$ has curvature 2; circle $B$ has curvature -1 .
Using these conventions allows us to express a beautiful theorem of Descartes: when four circles $A, B, C, D$ are pairwise tangent, with respective curvatures $a, b, c, d$, then
$$
(a+b+c+d)^{2}=2\left(a^{2}+b^{2}+c^{2}+d^{2}\right),
$$
where (as before) $a$ is taken to be negative if $B, C, D$ are internally tangent to $A$, and correspondingly for $b, c$, or $d$.
Prove that the curvature of each circular plot is an integer. | [
"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 curvatures are integers."
] | 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 circles; in what follows, we will call such regions curvilinear triangles, or $c$-triangles ( $\mathrm{c} \triangle$ ) for short.
This sad day marks day 0 of a new fiscal era. Unfortunately, these drastic measures are not enough, and so each day thereafter, court geometers mark off the largest possible circle contained in each c-triangle in the remaining property. This circle is tangent to all three arcs of the c-triangle, and will be referred to as the incircle of the c-triangle. At the end of the day, all incircles demarcated that day are sold off, and the following day, the remaining c-triangles are partitioned in the same manner.
Some notation: when discussing mutually tangent circles (or arcs), it is convenient to refer to the curvature of a circle rather than its radius. We define curvature as follows. Suppose that circle $A$ of radius $r_{a}$ is externally tangent to circle $B$ of radius $r_{b}$. Then the curvatures of the circles are simply the reciprocals of their radii, $\frac{1}{r_{a}}$ and $\frac{1}{r_{b}}$. If circle $A$ is internally tangent to circle $B$, however, as in the right diagram below, the curvature of circle $A$ is still $\frac{1}{r_{a}}$, while the curvature of circle $B$ is $-\frac{1}{r_{b}}$, the opposite of the reciprocal of its radius.
<image_1>
Circle $A$ has curvature 2; circle $B$ has curvature 1 .
<image_2>
Circle $A$ has curvature 2; circle $B$ has curvature -1 .
Using these conventions allows us to express a beautiful theorem of Descartes: when four circles $A, B, C, D$ are pairwise tangent, with respective curvatures $a, b, c, d$, then
$$
(a+b+c+d)^{2}=2\left(a^{2}+b^{2}+c^{2}+d^{2}\right),
$$
where (as before) $a$ is taken to be negative if $B, C, D$ are internally tangent to $A$, and correspondingly for $b, c$, or $d$.
Descartes' Circle Formula can be extended by interpreting the coordinates of points on the plane as complex numbers in the usual way: the point $(x, y)$ represents the complex number $x+y i$. On the complex plane, let $z_{A}, z_{B}, z_{C}, z_{D}$ be the centers of circles $A, B, C, D$ respectively; as before, $a, b, c, d$ are the curvatures of their respective circles. Then Descartes' Extended Circle Formula states
$$
\left(a \cdot z_{A}+b \cdot z_{B}+c \cdot z_{C}+d \cdot z_{D}\right)^{2}=2\left(a^{2} z_{A}^{2}+b^{2} z_{B}^{2}+c^{2} z_{C}^{2}+d^{2} z_{D}^{2}\right) .
$$
Prove that the center of each circular plot has coordinates $\left(\frac{u}{c}, \frac{v}{c}\right)$ where $u$ and $v$ are integers, and $c$ is the curvature of the plot. | [
"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 case, check the original kingdom and the first four plots: $U, C, C^{\\prime}, P_{1}$, and $P_{2}$. Circle $U$ is centered at $(0,0)$, yielding $-1 \\cdot z_{U}=0+0 i$. Circles $C$ and $C^{\\prime}$ are symmetric about the $y$-axis, so it suffices to check just one of them. Circle $C$ has radius $\\frac{1}{2}$ and therefore curvature 2 . It is centered at $\\left(\\frac{1}{2}, \\frac{0}{2}\\right)$, yielding $2 z_{C}=2\\left(\\frac{1}{2}+0 i\\right)=1$. Circles $P_{1}$ and $P_{2}$ are symmetric about the $x$-axis, so it suffices to check just one of them. Circle $P_{1}$ has radius $\\frac{1}{3}$ and therefore curvature 3 . It is centered at $\\left(\\frac{0}{3}, \\frac{2}{3}\\right)$, yielding $3 z_{P_{3}}=0+2 i$.\n\nFor the inductive step, suppose that $a z_{A}, b z_{B}, c z_{C}, d z_{D}$ have integer real and imaginary parts. Then by closure of addition and multiplication in the integers, $a^{\\prime} z_{A^{\\prime}}=2 \\hat{s}-3 a z_{A}$ also has integer real and imaginary parts, and similarly for $b^{\\prime} z_{B^{\\prime}}, c^{\\prime} z_{C^{\\prime}}, d^{\\prime} z_{D^{\\prime}}$.\n\nSo for all plots $A$ sold, $a z_{A}$ has integer real and imaginary parts, so each is centered at $\\left(\\frac{u}{c}, \\frac{v}{c}\\right)$ where $u$ and $v$ are integers, and $c$ is the curvature."
] | 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 circles; in what follows, we will call such regions curvilinear triangles, or $c$-triangles ( $\mathrm{c} \triangle$ ) for short.
This sad day marks day 0 of a new fiscal era. Unfortunately, these drastic measures are not enough, and so each day thereafter, court geometers mark off the largest possible circle contained in each c-triangle in the remaining property. This circle is tangent to all three arcs of the c-triangle, and will be referred to as the incircle of the c-triangle. At the end of the day, all incircles demarcated that day are sold off, and the following day, the remaining c-triangles are partitioned in the same manner.
Some notation: when discussing mutually tangent circles (or arcs), it is convenient to refer to the curvature of a circle rather than its radius. We define curvature as follows. Suppose that circle $A$ of radius $r_{a}$ is externally tangent to circle $B$ of radius $r_{b}$. Then the curvatures of the circles are simply the reciprocals of their radii, $\frac{1}{r_{a}}$ and $\frac{1}{r_{b}}$. If circle $A$ is internally tangent to circle $B$, however, as in the right diagram below, the curvature of circle $A$ is still $\frac{1}{r_{a}}$, while the curvature of circle $B$ is $-\frac{1}{r_{b}}$, the opposite of the reciprocal of its radius.
<image_1>
Circle $A$ has curvature 2; circle $B$ has curvature 1 .
<image_2>
Circle $A$ has curvature 2; circle $B$ has curvature -1 .
Using these conventions allows us to express a beautiful theorem of Descartes: when four circles $A, B, C, D$ are pairwise tangent, with respective curvatures $a, b, c, d$, then
$$
(a+b+c+d)^{2}=2\left(a^{2}+b^{2}+c^{2}+d^{2}\right),
$$
where (as before) $a$ is taken to be negative if $B, C, D$ are internally tangent to $A$, and correspondingly for $b, c$, or $d$.
Descartes' Circle Formula can be extended by interpreting the coordinates of points on the plane as complex numbers in the usual way: the point $(x, y)$ represents the complex number $x+y i$. On the complex plane, let $z_{A}, z_{B}, z_{C}, z_{D}$ be the centers of circles $A, B, C, D$ respectively; as before, $a, b, c, d$ are the curvatures of their respective circles. Then Descartes' Extended Circle Formula states
$$
\left(a \cdot z_{A}+b \cdot z_{B}+c \cdot z_{C}+d \cdot z_{D}\right)^{2}=2\left(a^{2} z_{A}^{2}+b^{2} z_{B}^{2}+c^{2} z_{C}^{2}+d^{2} z_{D}^{2}\right) .
$$
Given a c-triangle $T$, let $a, b$, and $c$ be the curvatures of the three $\operatorname{arcs}$ bounding $T$, with $a \leq b \leq c$, and let $d$ be the curvature of the incircle of $T$. Define the circle configuration associated with $T$ to be $\mathcal{C}(T)=(a, b, c, d)$. Define the c-triangle $T$ to be proper if $c \leq d$. For example, circles of curvatures $-1,2$, and 3 determine two c-triangles. The incircle of one has curvature 6 , so it is proper; the incircle of the other has curvature 2 , so it is not proper.
Let $P$ and $Q$ be two c-triangles, with associated configurations $\mathcal{C}(P)=(a, b, c, d)$ and $\mathcal{C}(Q)=$ $(w, x, y, z)$. We say that $P$ dominates $Q$ if $a \leq w, b \leq x, c \leq y$, and $d \leq z$. (The term "dominates" refers to the fact that the radii of the arcs defining $Q$ cannot be larger than the radii of the arcs defining $P$.)
Removing the incircle from $T$ gives three c-triangles, $T^{(1)}, T^{(2)}, T^{(3)}$, each bounded by the incircle of $T$ and two of the arcs that bound $T$. These triangles have associated configurations
$$
\begin{aligned}
\mathcal{C}\left(T^{(1)}\right) & =\left(b, c, d, a^{\prime}\right), \\
\mathcal{C}\left(T^{(2)}\right) & =\left(a, c, d, b^{\prime}\right), \\
\mathcal{C}\left(T^{(3)}\right) & =\left(a, b, d, c^{\prime}\right),
\end{aligned}
$$
Let $P$ and $Q$ be two proper c-triangles such that $P$ dominates $Q$. Let $\mathcal{C}(P)=(a, b, c, d)$ and $\mathcal{C}(Q)=(w, x, y, z)$.
Show that $P^{(3)}$ dominates $P^{(2)}$ and that $P^{(2)}$ dominates $P^{(1)}$. | [
"The equation $(x+b+c+d)^{2}=2\\left(x^{2}+b^{2}+c^{2}+d^{2}\\right)$ is quadratic with two solutions. Call them $a$ and $a^{\\prime}$. These are the curvatures of the two circles which are tangent to circles with curvatures $b, c$, and $d$. Rewrite the equation in standard form to obtain $x^{2}-2(b+c+d) x+$ $\\ldots=0$. Using the sum of the roots formula, $a+a^{\\prime}=2(b+c+d)=2(s-a)$. So $a^{\\prime}=2 s-3 a$, and therefore\n\n$$\n\\begin{aligned}\ns^{\\prime} & =a^{\\prime}+b+c+d \\\\\n& =2 s-3 a+s-a \\\\\n& =3 s-4 a .\n\\end{aligned}\n$$\n\n\nLet $s=a+b+c+d$. From above induction, it follows that\n\n$$\n\\begin{aligned}\n\\mathcal{C}\\left(P^{(1)}\\right) & =\\left(b, c, d, a^{\\prime}\\right) \\\\\n\\mathcal{C}\\left(P^{(2)}\\right) & =\\left(a, c, d, b^{\\prime}\\right) \\\\\n\\mathcal{C}\\left(P^{(3)}\\right) & =\\left(a, b, d, c^{\\prime}\\right)\n\\end{aligned}\n$$\n\nwhere $a^{\\prime}=2 s-3 a, b^{\\prime}=2 s-3 b$, and $c^{\\prime}=2 s-3 c$. Because $a \\leq b \\leq c$, it follows that $c^{\\prime} \\leq b^{\\prime} \\leq a^{\\prime}$. Therefore $P^{(3)}$ dominates $P^{(2)}$ and $P^{(2)}$ dominates $P^{(1)}$."
] | 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 |
||
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 circles; in what follows, we will call such regions curvilinear triangles, or $c$-triangles ( $\mathrm{c} \triangle$ ) for short.
This sad day marks day 0 of a new fiscal era. Unfortunately, these drastic measures are not enough, and so each day thereafter, court geometers mark off the largest possible circle contained in each c-triangle in the remaining property. This circle is tangent to all three arcs of the c-triangle, and will be referred to as the incircle of the c-triangle. At the end of the day, all incircles demarcated that day are sold off, and the following day, the remaining c-triangles are partitioned in the same manner.
Some notation: when discussing mutually tangent circles (or arcs), it is convenient to refer to the curvature of a circle rather than its radius. We define curvature as follows. Suppose that circle $A$ of radius $r_{a}$ is externally tangent to circle $B$ of radius $r_{b}$. Then the curvatures of the circles are simply the reciprocals of their radii, $\frac{1}{r_{a}}$ and $\frac{1}{r_{b}}$. If circle $A$ is internally tangent to circle $B$, however, as in the right diagram below, the curvature of circle $A$ is still $\frac{1}{r_{a}}$, while the curvature of circle $B$ is $-\frac{1}{r_{b}}$, the opposite of the reciprocal of its radius.
<image_1>
Circle $A$ has curvature 2; circle $B$ has curvature 1 .
<image_2>
Circle $A$ has curvature 2; circle $B$ has curvature -1 .
Using these conventions allows us to express a beautiful theorem of Descartes: when four circles $A, B, C, D$ are pairwise tangent, with respective curvatures $a, b, c, d$, then
$$
(a+b+c+d)^{2}=2\left(a^{2}+b^{2}+c^{2}+d^{2}\right),
$$
where (as before) $a$ is taken to be negative if $B, C, D$ are internally tangent to $A$, and correspondingly for $b, c$, or $d$.
Descartes' Circle Formula can be extended by interpreting the coordinates of points on the plane as complex numbers in the usual way: the point $(x, y)$ represents the complex number $x+y i$. On the complex plane, let $z_{A}, z_{B}, z_{C}, z_{D}$ be the centers of circles $A, B, C, D$ respectively; as before, $a, b, c, d$ are the curvatures of their respective circles. Then Descartes' Extended Circle Formula states
$$
\left(a \cdot z_{A}+b \cdot z_{B}+c \cdot z_{C}+d \cdot z_{D}\right)^{2}=2\left(a^{2} z_{A}^{2}+b^{2} z_{B}^{2}+c^{2} z_{C}^{2}+d^{2} z_{D}^{2}\right) .
$$
Given a c-triangle $T$, let $a, b$, and $c$ be the curvatures of the three $\operatorname{arcs}$ bounding $T$, with $a \leq b \leq c$, and let $d$ be the curvature of the incircle of $T$. Define the circle configuration associated with $T$ to be $\mathcal{C}(T)=(a, b, c, d)$. Define the c-triangle $T$ to be proper if $c \leq d$. For example, circles of curvatures $-1,2$, and 3 determine two c-triangles. The incircle of one has curvature 6 , so it is proper; the incircle of the other has curvature 2 , so it is not proper.
Let $P$ and $Q$ be two c-triangles, with associated configurations $\mathcal{C}(P)=(a, b, c, d)$ and $\mathcal{C}(Q)=$ $(w, x, y, z)$. We say that $P$ dominates $Q$ if $a \leq w, b \leq x, c \leq y$, and $d \leq z$. (The term "dominates" refers to the fact that the radii of the arcs defining $Q$ cannot be larger than the radii of the arcs defining $P$.)
Removing the incircle from $T$ gives three c-triangles, $T^{(1)}, T^{(2)}, T^{(3)}$, each bounded by the incircle of $T$ and two of the arcs that bound $T$. These triangles have associated configurations
$$
\begin{aligned}
\mathcal{C}\left(T^{(1)}\right) & =\left(b, c, d, a^{\prime}\right), \\
\mathcal{C}\left(T^{(2)}\right) & =\left(a, c, d, b^{\prime}\right), \\
\mathcal{C}\left(T^{(3)}\right) & =\left(a, b, d, c^{\prime}\right),
\end{aligned}
$$
Let $P$ and $Q$ be two proper c-triangles such that $P$ dominates $Q$. Let $\mathcal{C}(P)=(a, b, c, d)$ and $\mathcal{C}(Q)=(w, x, y, z)$.
Prove that $P^{(1)}$ dominates $Q^{(1)}$. | [
"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$$\n(X+b+c+d)^{2}=2\\left(X^{2}+b^{2}+c^{2}+d^{2}\\right)\n$$\n\nSolve by completing the square:\n\n$$\n\\begin{aligned}\nX^{2}-2(b+c+d) X+2\\left(b^{2}+c^{2}+d^{2}\\right) & =(b+c+d)^{2} \\\\\n(X-(b+c+d))^{2} & =2(b+c+d)^{2}-2\\left(b^{2}+c^{2}+d^{2}\\right) \\\\\n& =4(b c+b d+c d) .\n\\end{aligned}\n$$\n\nThus $a, a^{\\prime}=b+c+d \\pm 2 \\sqrt{b c+b d+c d}$.\n\nBecause $a \\leq b \\leq c \\leq d$, and only $a$ can be less than zero, $a$ must get the minus sign, and $a^{\\prime}$ gets the plus sign:\n\n$$\na^{\\prime}=b+c+d+2 \\sqrt{b c+b d+c d} \\text {. }\n$$\n\nSimilarly,\n\n$$\nw^{\\prime}=x+y+z+2 \\sqrt{x y+x z+y z} .\n$$\n\nBecause $P$ dominates $Q$, each term in the expression for $a^{\\prime}$ is less than or equal to the corresponding term in the expression for $w^{\\prime}$, thus $a^{\\prime} \\leq w^{\\prime}$."
] | 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 |
||
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 circles; in what follows, we will call such regions curvilinear triangles, or $c$-triangles ( $\mathrm{c} \triangle$ ) for short.
This sad day marks day 0 of a new fiscal era. Unfortunately, these drastic measures are not enough, and so each day thereafter, court geometers mark off the largest possible circle contained in each c-triangle in the remaining property. This circle is tangent to all three arcs of the c-triangle, and will be referred to as the incircle of the c-triangle. At the end of the day, all incircles demarcated that day are sold off, and the following day, the remaining c-triangles are partitioned in the same manner.
Some notation: when discussing mutually tangent circles (or arcs), it is convenient to refer to the curvature of a circle rather than its radius. We define curvature as follows. Suppose that circle $A$ of radius $r_{a}$ is externally tangent to circle $B$ of radius $r_{b}$. Then the curvatures of the circles are simply the reciprocals of their radii, $\frac{1}{r_{a}}$ and $\frac{1}{r_{b}}$. If circle $A$ is internally tangent to circle $B$, however, as in the right diagram below, the curvature of circle $A$ is still $\frac{1}{r_{a}}$, while the curvature of circle $B$ is $-\frac{1}{r_{b}}$, the opposite of the reciprocal of its radius.
<image_1>
Circle $A$ has curvature 2; circle $B$ has curvature 1 .
<image_2>
Circle $A$ has curvature 2; circle $B$ has curvature -1 .
Using these conventions allows us to express a beautiful theorem of Descartes: when four circles $A, B, C, D$ are pairwise tangent, with respective curvatures $a, b, c, d$, then
$$
(a+b+c+d)^{2}=2\left(a^{2}+b^{2}+c^{2}+d^{2}\right),
$$
where (as before) $a$ is taken to be negative if $B, C, D$ are internally tangent to $A$, and correspondingly for $b, c$, or $d$.
Descartes' Circle Formula can be extended by interpreting the coordinates of points on the plane as complex numbers in the usual way: the point $(x, y)$ represents the complex number $x+y i$. On the complex plane, let $z_{A}, z_{B}, z_{C}, z_{D}$ be the centers of circles $A, B, C, D$ respectively; as before, $a, b, c, d$ are the curvatures of their respective circles. Then Descartes' Extended Circle Formula states
$$
\left(a \cdot z_{A}+b \cdot z_{B}+c \cdot z_{C}+d \cdot z_{D}\right)^{2}=2\left(a^{2} z_{A}^{2}+b^{2} z_{B}^{2}+c^{2} z_{C}^{2}+d^{2} z_{D}^{2}\right) .
$$
Given a c-triangle $T$, let $a, b$, and $c$ be the curvatures of the three $\operatorname{arcs}$ bounding $T$, with $a \leq b \leq c$, and let $d$ be the curvature of the incircle of $T$. Define the circle configuration associated with $T$ to be $\mathcal{C}(T)=(a, b, c, d)$. Define the c-triangle $T$ to be proper if $c \leq d$. For example, circles of curvatures $-1,2$, and 3 determine two c-triangles. The incircle of one has curvature 6 , so it is proper; the incircle of the other has curvature 2 , so it is not proper.
Let $P$ and $Q$ be two c-triangles, with associated configurations $\mathcal{C}(P)=(a, b, c, d)$ and $\mathcal{C}(Q)=$ $(w, x, y, z)$. We say that $P$ dominates $Q$ if $a \leq w, b \leq x, c \leq y$, and $d \leq z$. (The term "dominates" refers to the fact that the radii of the arcs defining $Q$ cannot be larger than the radii of the arcs defining $P$.)
Removing the incircle from $T$ gives three c-triangles, $T^{(1)}, T^{(2)}, T^{(3)}$, each bounded by the incircle of $T$ and two of the arcs that bound $T$. These triangles have associated configurations
$$
\begin{aligned}
\mathcal{C}\left(T^{(1)}\right) & =\left(b, c, d, a^{\prime}\right), \\
\mathcal{C}\left(T^{(2)}\right) & =\left(a, c, d, b^{\prime}\right), \\
\mathcal{C}\left(T^{(3)}\right) & =\left(a, b, d, c^{\prime}\right),
\end{aligned}
$$
Let $P$ and $Q$ be two proper c-triangles such that $P$ dominates $Q$. Let $\mathcal{C}(P)=(a, b, c, d)$ and $\mathcal{C}(Q)=(w, x, y, z)$.
Prove that $P^{(3)}$ dominates $Q^{(3)}$. | [
"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, $c, c^{\\prime}=a+b+d \\pm 2 \\sqrt{a b+a d+b d}$. If $a<0$, then the other three circles are internally tangent to the circle of curvature $a$, so this circle has the largest radius. In particular, $\\frac{1}{|a|}>\\frac{1}{b}$. Thus $b>|a|=-a$, which shows that $a+b>0$. Therefore $c$ must get the minus sign, and $c^{\\prime}$ gets the plus sign. The same argument applies to $y$ and $y^{\\prime}$.\n\nWhen $a<0$, it is also worth considering whether the square roots are defined (and real). In fact, they are. Consider the diameters of the circles with curvatures $b$ and $d$ along the line through the centers of these circles. These two diameters form a single segment inside the circle with curvature $a$, so the sum of the diameters is at most the diameter of that circle: $\\frac{2}{b}+\\frac{2}{d} \\leq \\frac{2}{|a|}$. It follows that $-a d-a b=|a| d+|a| b \\leq b d$, or $a b+a d+b d \\geq 0$. This is the argument of the square root in the expressions for $c$ and $c^{\\prime}$. An analogous argument shows that the radicands are nonnegative in the expressions for $b$ and $b^{\\prime}$.\n\nThe foregoing shows that\n\n$$\nc^{\\prime}=a+b+d+2 \\sqrt{a b+a d+b d}\n$$\n\nand, by an analogous argument for $w<0$,\n\n$$\ny^{\\prime}=w+x+z+2 \\sqrt{w x+w z+x z} .\n$$\n\nIt remains to prove that $c^{\\prime} \\leq y^{\\prime}$. Note that only $a$ and $w$ may be negative; $b, c, d, x, y$, and $z$ are all positive. There are three cases.\n\n\n\n(i) If $0 \\leq a \\leq w$, then $a b \\leq w x, a d \\leq w z$, and $b d \\leq x z$, so $c^{\\prime} \\leq y^{\\prime}$.\n\n(ii) If $a<0 \\leq w$, then $a b+a d+b d \\leq b d$, and $b d \\leq x z \\leq w x+w z+x z$, so $c^{\\prime} \\leq y^{\\prime}$. (As noted above, both radicands are nonnegative.)\n\n(iii) If $a \\leq w<0$, then it has already been established that $a+b$ is positive. Analogously, $a+d, w+x$, and $w+z$ are positive. Furthermore, $a^{2} \\geq w^{2}$. Thus $(a+b)(a+d)-a^{2} \\leq$ $(w+x)(w+z)-w^{2}$, which establishes that $a b+a d+b d \\leq w x+w z+x z$, so $c^{\\prime} \\leq y^{\\prime}$."
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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 the Meruprastāra, or "Mountain of Gems". In this Power Question, we'll explore some properties of Pingala's/Pascal's Triangle ("PT") and its variants.
Unless otherwise specified, the only definition, notation, and formulas you may use for PT are the definition, notation, and formulas given below.
PT consists of an infinite number of rows, numbered from 0 onwards. The $n^{\text {th }}$ row contains $n+1$ numbers, identified as $\mathrm{Pa}(n, k)$, where $0 \leq k \leq n$. For all $n$, define $\mathrm{Pa}(n, 0)=\operatorname{Pa}(n, n)=1$. Then for $n>1$ and $1 \leq k \leq n-1$, define $\mathrm{Pa}(n, k)=\mathrm{Pa}(n-1, k-1)+\mathrm{Pa}(n-1, k)$. It is convenient to define $\mathrm{Pa}(n, k)=0$ when $k<0$ or $k>n$. We write the nonzero values of $\mathrm{PT}$ in the familiar pyramid shown below.
<image_1>
As is well known, $\mathrm{Pa}(n, k)$ gives the number of ways of choosing a committee of $k$ people from a set of $n$ people, so a simple formula for $\mathrm{Pa}(n, k)$ is $\mathrm{Pa}(n, k)=\frac{n !}{k !(n-k) !}$. You may use this formula or the recursive definition above throughout this Power Question.
Consider the parity of each entry: define
$$
\operatorname{PaP}(n, k)= \begin{cases}1 & \text { if } \mathrm{Pa}(n, k) \text { is odd } \\ 0 & \text { if } \mathrm{Pa}(n, k) \text { is even }\end{cases}
$$
Prove that $\operatorname{PaP}(n, 0)=\operatorname{PaP}(n, n)=1$ for all nonnegative integers $n$. | [
"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."
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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 the Meruprastāra, or "Mountain of Gems". In this Power Question, we'll explore some properties of Pingala's/Pascal's Triangle ("PT") and its variants.
Unless otherwise specified, the only definition, notation, and formulas you may use for PT are the definition, notation, and formulas given below.
PT consists of an infinite number of rows, numbered from 0 onwards. The $n^{\text {th }}$ row contains $n+1$ numbers, identified as $\mathrm{Pa}(n, k)$, where $0 \leq k \leq n$. For all $n$, define $\mathrm{Pa}(n, 0)=\operatorname{Pa}(n, n)=1$. Then for $n>1$ and $1 \leq k \leq n-1$, define $\mathrm{Pa}(n, k)=\mathrm{Pa}(n-1, k-1)+\mathrm{Pa}(n-1, k)$. It is convenient to define $\mathrm{Pa}(n, k)=0$ when $k<0$ or $k>n$. We write the nonzero values of $\mathrm{PT}$ in the familiar pyramid shown below.
<image_1>
As is well known, $\mathrm{Pa}(n, k)$ gives the number of ways of choosing a committee of $k$ people from a set of $n$ people, so a simple formula for $\mathrm{Pa}(n, k)$ is $\mathrm{Pa}(n, k)=\frac{n !}{k !(n-k) !}$. You may use this formula or the recursive definition above throughout this Power Question.
Consider the parity of each entry: define
$$
\operatorname{PaP}(n, k)= \begin{cases}1 & \text { if } \mathrm{Pa}(n, k) \text { is odd } \\ 0 & \text { if } \mathrm{Pa}(n, k) \text { is even }\end{cases}
$$
If $n=2^{j}$ for some nonnegative integer $j$, and $0<k<n$, show that $\operatorname{PaP}(n, k)=0$. | [
"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 factor, $\\mathrm{Pa}(n-1, k-1)$ is an integer, and the first factor has an even numerator, so $\\mathrm{Pa}(n, k)$ is even. Therefore $\\operatorname{PaP}(n, k)=0$."
] | 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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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 the Meruprastāra, or "Mountain of Gems". In this Power Question, we'll explore some properties of Pingala's/Pascal's Triangle ("PT") and its variants.
Unless otherwise specified, the only definition, notation, and formulas you may use for PT are the definition, notation, and formulas given below.
PT consists of an infinite number of rows, numbered from 0 onwards. The $n^{\text {th }}$ row contains $n+1$ numbers, identified as $\mathrm{Pa}(n, k)$, where $0 \leq k \leq n$. For all $n$, define $\mathrm{Pa}(n, 0)=\operatorname{Pa}(n, n)=1$. Then for $n>1$ and $1 \leq k \leq n-1$, define $\mathrm{Pa}(n, k)=\mathrm{Pa}(n-1, k-1)+\mathrm{Pa}(n-1, k)$. It is convenient to define $\mathrm{Pa}(n, k)=0$ when $k<0$ or $k>n$. We write the nonzero values of $\mathrm{PT}$ in the familiar pyramid shown below.
<image_1>
As is well known, $\mathrm{Pa}(n, k)$ gives the number of ways of choosing a committee of $k$ people from a set of $n$ people, so a simple formula for $\mathrm{Pa}(n, k)$ is $\mathrm{Pa}(n, k)=\frac{n !}{k !(n-k) !}$. You may use this formula or the recursive definition above throughout this Power Question.
Consider the parity of each entry: define
$$
\operatorname{PaP}(n, k)= \begin{cases}1 & \text { if } \mathrm{Pa}(n, k) \text { is odd } \\ 0 & \text { if } \mathrm{Pa}(n, k) \text { is even }\end{cases}
$$
Let $j \geq 0$, and suppose $n \geq 2^{j}$. Prove that $\mathrm{Pa}(n, k)$ has the same parity as the sum $\operatorname{Pa}\left(n-2^{j}, k-2^{j}\right)+\operatorname{Pa}\left(n-2^{j}, k\right)$, i.e., either both $\operatorname{Pa}(n, k)$ and the given sum are even, or both are odd. | [
"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 $k=n$, the left-hand side is 1 and the right-hand side is either $0+1$ or $1+0$. For other values of $k$, all three terms are zero.\n\nNow, assume $\\mathrm{Pa}(n, k) \\equiv \\operatorname{Pa}\\left(n-2^{j}, k-2^{j}\\right)+\\mathrm{Pa}\\left(n-2^{j}, k\\right) \\bmod 2$ for some $j \\geq 0$, and let $n \\geq 2^{j+1}$. For any $k$, apply the inductive hypothesis three times:\n\n$$\n\\begin{aligned}\n\\operatorname{Pa}(n, k) \\equiv & \\operatorname{Pa}\\left(n-2^{j}, k-2^{j}\\right)+\\operatorname{Pa}\\left(n-2^{j}, k\\right) \\bmod 2 \\\\\n\\equiv & \\operatorname{Pa}\\left(n-2^{j}-2^{j}, k-2^{j}-2^{j}\\right)+\\operatorname{Pa}\\left(n-2^{j}-2^{j}, k-2^{j}\\right) \\\\\n& \\quad+\\operatorname{Pa}\\left(n-2^{j}-2^{j}, k-2^{j}\\right)+\\operatorname{Pa}\\left(n-2^{j}-2^{j}, k\\right)\n\\end{aligned}\n$$\n\nBut $u+u \\equiv 0 \\bmod 2$ for any integer $u$. Thus the two occurrences of $\\mathrm{Pa}\\left(n-2^{j}-2^{j}, k-2^{j}\\right)$ cancel each other out, and\n\n$$\n\\begin{aligned}\n\\operatorname{Pa}(n, k) & \\equiv \\operatorname{Pa}\\left(n-2^{j}-2^{j}, k-2^{j}-2^{j}\\right)+\\operatorname{Pa}\\left(n-2^{j}-2^{j}, k\\right) \\\\\n& \\equiv \\operatorname{Pa}\\left(n-2^{j+1}, k-2^{j+1}\\right)+\\operatorname{Pa}\\left(n-2^{j+1}, k\\right) \\bmod 2 .\n\\end{aligned}\n$$"
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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 the Meruprastāra, or "Mountain of Gems". In this Power Question, we'll explore some properties of Pingala's/Pascal's Triangle ("PT") and its variants.
Unless otherwise specified, the only definition, notation, and formulas you may use for PT are the definition, notation, and formulas given below.
PT consists of an infinite number of rows, numbered from 0 onwards. The $n^{\text {th }}$ row contains $n+1$ numbers, identified as $\mathrm{Pa}(n, k)$, where $0 \leq k \leq n$. For all $n$, define $\mathrm{Pa}(n, 0)=\operatorname{Pa}(n, n)=1$. Then for $n>1$ and $1 \leq k \leq n-1$, define $\mathrm{Pa}(n, k)=\mathrm{Pa}(n-1, k-1)+\mathrm{Pa}(n-1, k)$. It is convenient to define $\mathrm{Pa}(n, k)=0$ when $k<0$ or $k>n$. We write the nonzero values of $\mathrm{PT}$ in the familiar pyramid shown below.
<image_1>
As is well known, $\mathrm{Pa}(n, k)$ gives the number of ways of choosing a committee of $k$ people from a set of $n$ people, so a simple formula for $\mathrm{Pa}(n, k)$ is $\mathrm{Pa}(n, k)=\frac{n !}{k !(n-k) !}$. You may use this formula or the recursive definition above throughout this Power Question.
Consider the parity of each entry: define
$$
\operatorname{PaP}(n, k)= \begin{cases}1 & \text { if } \mathrm{Pa}(n, k) \text { is odd } \\ 0 & \text { if } \mathrm{Pa}(n, k) \text { is even }\end{cases}
$$
If $j$ is an integer such that $2^{j} \leq n<2^{j+1}$, and $k<2^{j}$, prove that
$$
\operatorname{PaP}(n, k)=\operatorname{PaP}\left(n-2^{j}, k\right) .
$$ | [
"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 $k=n$, the left-hand side is 1 and the right-hand side is either $0+1$ or $1+0$. For other values of $k$, all three terms are zero.\n\nNow, assume $\\mathrm{Pa}(n, k) \\equiv \\operatorname{Pa}\\left(n-2^{j}, k-2^{j}\\right)+\\mathrm{Pa}\\left(n-2^{j}, k\\right) \\bmod 2$ for some $j \\geq 0$, and let $n \\geq 2^{j+1}$. For any $k$, apply the inductive hypothesis three times:\n\n$$\n\\begin{aligned}\n\\operatorname{Pa}(n, k) \\equiv & \\operatorname{Pa}\\left(n-2^{j}, k-2^{j}\\right)+\\operatorname{Pa}\\left(n-2^{j}, k\\right) \\bmod 2 \\\\\n\\equiv & \\operatorname{Pa}\\left(n-2^{j}-2^{j}, k-2^{j}-2^{j}\\right)+\\operatorname{Pa}\\left(n-2^{j}-2^{j}, k-2^{j}\\right) \\\\\n& \\quad+\\operatorname{Pa}\\left(n-2^{j}-2^{j}, k-2^{j}\\right)+\\operatorname{Pa}\\left(n-2^{j}-2^{j}, k\\right)\n\\end{aligned}\n$$\n\nBut $u+u \\equiv 0 \\bmod 2$ for any integer $u$. Thus the two occurrences of $\\mathrm{Pa}\\left(n-2^{j}-2^{j}, k-2^{j}\\right)$ cancel each other out, and\n\n$$\n\\begin{aligned}\n\\operatorname{Pa}(n, k) & \\equiv \\operatorname{Pa}\\left(n-2^{j}-2^{j}, k-2^{j}-2^{j}\\right)+\\operatorname{Pa}\\left(n-2^{j}-2^{j}, k\\right) \\\\\n& \\equiv \\operatorname{Pa}\\left(n-2^{j+1}, k-2^{j+1}\\right)+\\operatorname{Pa}\\left(n-2^{j+1}, k\\right) \\bmod 2 .\n\\end{aligned}\n$$\n\n$\\operatorname{Pa}(n, k) \\equiv \\operatorname{Pa}\\left(n-2^{j}, k-2^{j}\\right)+\\operatorname{Pa}\\left(n-2^{j}, k\\right) \\bmod 2$ when $j \\geq 0$ and $n \\geq 2^{j}$. If $2^{j} \\leq n<2^{j+1}$ and $0 \\leq k<2^{j}$, then $\\mathrm{Pa}\\left(n-2^{j}, k-2^{j}\\right)=0$, so\n\n$$\n\\mathrm{Pa}(n, k) \\equiv \\operatorname{Pa}\\left(n-2^{j}, k-2^{j}\\right)+\\mathrm{Pa}\\left(n-2^{j}, k\\right) \\equiv \\operatorname{Pa}\\left(n-2^{j}, k\\right) \\bmod 2 .\n$$\n\nThus $\\operatorname{PaP}(n, k)=\\operatorname{PaP}\\left(n-2^{j}, k\\right)$.",
"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 factor, $\\mathrm{Pa}(n-1, k-1)$ is an integer, and the first factor has an even numerator, so $\\mathrm{Pa}(n, k)$ is even. Therefore $\\operatorname{PaP}(n, k)=0$.\n\nNow when $n=2^{j}$. For $2^{j}<n<2^{j+1}$, proceed by induction on $n$. Then $\\operatorname{PaP}(n, k) \\equiv \\operatorname{PaP}(n-1, k-1)+\\operatorname{PaP}(n-1, k) \\bmod 2$, while $\\operatorname{PaP}(n-1, k-1)=\\operatorname{PaP}\\left(n-1-2^{j}, k-1\\right)$ and $\\operatorname{PaP}(n-1, k)=\\operatorname{PaP}\\left(n-1-2^{j}, k\\right)$. But $\\operatorname{PaP}\\left(n-1-2^{j}, k-1\\right)+\\mathrm{PaP}\\left(n-1-2^{j}, k\\right) \\equiv \\operatorname{PaP}\\left(n-2^{j}, k\\right) \\bmod 2$, establishing the statement."
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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 the Meruprastāra, or "Mountain of Gems". In this Power Question, we'll explore some properties of Pingala's/Pascal's Triangle ("PT") and its variants.
Unless otherwise specified, the only definition, notation, and formulas you may use for PT are the definition, notation, and formulas given below.
PT consists of an infinite number of rows, numbered from 0 onwards. The $n^{\text {th }}$ row contains $n+1$ numbers, identified as $\mathrm{Pa}(n, k)$, where $0 \leq k \leq n$. For all $n$, define $\mathrm{Pa}(n, 0)=\operatorname{Pa}(n, n)=1$. Then for $n>1$ and $1 \leq k \leq n-1$, define $\mathrm{Pa}(n, k)=\mathrm{Pa}(n-1, k-1)+\mathrm{Pa}(n-1, k)$. It is convenient to define $\mathrm{Pa}(n, k)=0$ when $k<0$ or $k>n$. We write the nonzero values of $\mathrm{PT}$ in the familiar pyramid shown below.
<image_1>
As is well known, $\mathrm{Pa}(n, k)$ gives the number of ways of choosing a committee of $k$ people from a set of $n$ people, so a simple formula for $\mathrm{Pa}(n, k)$ is $\mathrm{Pa}(n, k)=\frac{n !}{k !(n-k) !}$. You may use this formula or the recursive definition above throughout this Power Question.
Clark's Triangle: If the left side of PT is replaced with consecutive multiples of 6 , starting with 0 , but the right entries (except the first) and the generating rule are left unchanged, the result is called Clark's Triangle. If the $k^{\text {th }}$ entry of the $n^{\text {th }}$ row is denoted by $\mathrm{Cl}(n, k)$, then the formal rule is:
$$
\begin{cases}\mathrm{Cl}(n, 0)=6 n & \text { for all } n \\ \mathrm{Cl}(n, n)=1 & \text { for } n \geq 1 \\ \mathrm{Cl}(n, k)=\mathrm{Cl}(n-1, k-1)+\mathrm{Cl}(n-1, k) & \text { for } n \geq 1 \text { and } 1 \leq k \leq n-1\end{cases}
$$
The first four rows of Clark's Triangle are given below.
<image_2>
Prove the formula $\mathrm{Cl}(n, 1)=3 n^{2}-3 n+1$. | [
"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 $\\mathrm{Cl}(n+1,1)=\\mathrm{Cl}(n, 0)+\\mathrm{Cl}(n, 1)=6 n+\\left(3 n^{2}-3 n+1\\right)=\\left(3 n^{2}+6 n+3\\right)-$ $(3 n+3)+1=3(n+1)^{2}-3(n+1)+1$."
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3,072 | Leibniz's Harmonic Triangle: Consider the triangle formed by the rule
$$
\begin{cases}\operatorname{Le}(n, 0)=\frac{1}{n+1} & \text { for all } n \\ \operatorname{Le}(n, n)=\frac{1}{n+1} & \text { for all } n \\ \operatorname{Le}(n, k)=\operatorname{Le}(n+1, k)+\operatorname{Le}(n+1, k+1) & \text { for all } n \text { and } 0 \leq k \leq n\end{cases}
$$
This triangle, discovered first by Leibniz, consists of reciprocals of integers as shown below.
<image_1>
For this contest, you may assume that $\operatorname{Le}(n, k)>0$ whenever $0 \leq k \leq n$, and that $\operatorname{Le}(n, k)$ is undefined if $k<0$ or $k>n$.
If $\sum_{i=m}^{\infty} \operatorname{Le}(i, m)=\operatorname{Le}(n, k)$, prove that $n=k=m-1$. | [
"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$$\n\\begin{aligned}\n\\sum_{i=m}^{n} \\operatorname{Le}(i, m)= & \\sum_{i=m}^{n}(\\operatorname{Le}(i-1, m-1)-\\operatorname{Le}(i, m-1)) \\\\\n= & (\\operatorname{Le}(m-1, m-1)-\\operatorname{Le}(m, m-1))+(\\operatorname{Le}(m, m-1)-\\operatorname{Le}(m+1, m-1))+ \\\\\n& \\cdots+(\\operatorname{Le}(n-1, m-1)-\\operatorname{Le}(n, m-1)) \\\\\n= & \\operatorname{Le}(m-1, m-1)-\\operatorname{Le}(n, m-1) .\n\\end{aligned}\n$$\n\nBecause the values of $\\operatorname{Le}(n, m-1)$ get arbitrarily small as $n$ increases (proof: $\\operatorname{Le}(i, j)<$ $\\operatorname{Le}(i-1, j-1)$ by construction, so $\\left.\\operatorname{Le}(n, m-1)<\\operatorname{Le}(n-m+1,0)=\\frac{1}{n-m+1}\\right)$, the limit of these partial sums is $\\operatorname{Le}(m-1, m-1)$. So $n=k=m-1$.\n\n\n\nNote: This result can be extended even further. In fact, for every value of $k<n$,\n\n$$\n\\operatorname{Le}(n, k)=\\sum_{i=n+1}^{\\infty} \\operatorname{Le}(i, k+1)\n$$\n\nIn other words, each entry in Leibniz's triangle is an infinite sum of the entries in the diagonal directly to its right, beginning with the entry below and to the right of the given one."
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