di-zhang-fdu/OlympiadExtended huggingface dataset:
Subfield: Geometry


Question: 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.

<img_4001>

Circle $A$ has curvature 2; circle $B$ has curvature 1 .

<img_3649>

Circle $A$ has curvature 2; circle $B$ has curvature -1 .

Using these conventions allows us to express a beautiful theorem of Descartes: when four circles $A, B, C, D$ are pairwise tangent, with respective curvatures $a, b, c, d$, then

$$
(a+b+c+d)^{2}=2\left(a^{2}+b^{2}+c^{2}+d^{2}\right),
$$

where (as before) $a$ is taken to be negative if $B, C, D$ are internally tangent to $A$, and correspondingly for $b, c$, or $d$.
Without using Descartes' Circle Formula, Find the combined area of the six remaining curvilinear territories after day 1.

Solution:
The four "removed" circles have radii $\frac{1}{2}, \frac{1}{2}, \frac{1}{3}, \frac{1}{3}$ so the combined area of the six remaining curvilinear territories is:

$$
\pi\left(1^{2}-\left(\frac{1}{2}\right)^{2}-\left(\frac{1}{2}\right)^{2}-\left(\frac{1}{3}\right)^{2}-\left(\frac{1}{3}\right)^{2}\right)=\frac{5 \pi}{18}
$$

Final answer: $\frac{5 \pi}{18}$
Subfield: Geometry


Question: 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.

<img_4001>

Circle $A$ has curvature 2; circle $B$ has curvature 1 .

<img_3649>

Circle $A$ has curvature 2; circle $B$ has curvature -1 .

Using these conventions allows us to express a beautiful theorem of Descartes: when four circles $A, B, C, D$ are pairwise tangent, with respective curvatures $a, b, c, d$, then

$$
(a+b+c+d)^{2}=2\left(a^{2}+b^{2}+c^{2}+d^{2}\right),
$$

where (as before) $a$ is taken to be negative if $B, C, D$ are internally tangent to $A$, and correspondingly for $b, c$, or $d$.
Determine the number of curvilinear territories remaining at the end of day 3.

Solution:
At the beginning of day 2, there are six c-triangles, so six incircles are sold, dividing each of the six territories into three smaller curvilinear triangles. So a total of 18 curvilinear triangles exist at the start of day 3, each of which is itself divided into three pieces that day (by the sale of a total of 18 regions bounded by the territories' incircles). Therefore there are 54 regions at the end of day 3.

Final answer: 54
Subfield: Geometry


Question: 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.

![](https://cdn.mathpix.com/cropped/2023_12_21_de9643dc5cb5c1da0c52g-1.jpg?height=304&width=455&top_left_y=1886&top_left_x=347)

Circle $A$ has curvature 2; circle $B$ has curvature 1 .

![](https://cdn.mathpix.com/cropped/2023_12_21_de9643dc5cb5c1da0c52g-1.jpg?height=301&width=307&top_left_y=1888&top_left_x=1386)

Circle $A$ has curvature 2; circle $B$ has curvature -1 .

Using these conventions allows us to express a beautiful theorem of Descartes: when four circles $A, B, C, D$ are pairwise tangent, with respective curvatures $a, b, c, d$, then

$$
(a+b+c+d)^{2}=2\left(a^{2}+b^{2}+c^{2}+d^{2}\right),
$$

where (as before) $a$ is taken to be negative if $B, C, D$ are internally tangent to $A$, and correspondingly for $b, c$, or $d$.
Determine the number of curvilinear territories remaining at the end of day 3.

Solution:
At the beginning of day 2, there are six c-triangles, so six incircles are sold, dividing each of the six territories into three smaller curvilinear triangles. So a total of 18 curvilinear triangles exist at the start of day 3, each of which is itself divided into three pieces that day (by the sale of a total of 18 regions bounded by the territories' incircles). Therefore there are 54 regions at the end of day 3.

Final answer: 54
Subfield: Geometry


Question: 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.

![](https://cdn.mathpix.com/cropped/2023_12_21_de9643dc5cb5c1da0c52g-1.jpg?height=304&width=455&top_left_y=1886&top_left_x=347)

Circle $A$ has curvature 2; circle $B$ has curvature 1 .

![](https://cdn.mathpix.com/cropped/2023_12_21_de9643dc5cb5c1da0c52g-1.jpg?height=301&width=307&top_left_y=1888&top_left_x=1386)

Circle $A$ has curvature 2; circle $B$ has curvature -1 .

Using these conventions allows us to express a beautiful theorem of Descartes: when four circles $A, B, C, D$ are pairwise tangent, with respective curvatures $a, b, c, d$, then

$$
(a+b+c+d)^{2}=2\left(a^{2}+b^{2}+c^{2}+d^{2}\right),
$$

where (as before) $a$ is taken to be negative if $B, C, D$ are internally tangent to $A$, and correspondingly for $b, c$, or $d$.
Let $X_{n}$ be the number of plots sold on day $n$. Find a formula for $X_{n}$ in terms of $n$.

Solution:
Each day, every curvilinear territory is divided into three smaller curvilinear territories. Let $R_{n}$ be the number of regions at the end of day $n$. Then $R_{0}=2$, and $R_{n+1}=3 \cdot R_{n}$. Thus $R_{n}$ is a geometric sequence, so $R_{n}=2 \cdot 3^{n}$. For $n>0$, the number of plots sold on day $n$ equals the number of territories existing at the end of day $n-1$, i.e., $X_{n}=R_{n-1}$, so $X_{0}=X_{1}=2$, and for $n>1, X_{n}=2 \cdot 3^{n-1}$.

Final answer: $X_{0}=X_{1}=2$, and for $n>1, X_{n}=2 \cdot 3^{n-1}$
Subfield: Geometry


Question: 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.

![](https://cdn.mathpix.com/cropped/2023_12_21_de9643dc5cb5c1da0c52g-1.jpg?height=304&width=455&top_left_y=1886&top_left_x=347)

Circle $A$ has curvature 2; circle $B$ has curvature 1 .

![](https://cdn.mathpix.com/cropped/2023_12_21_de9643dc5cb5c1da0c52g-1.jpg?height=301&width=307&top_left_y=1888&top_left_x=1386)

Circle $A$ has curvature 2; circle $B$ has curvature -1 .

Using these conventions allows us to express a beautiful theorem of Descartes: when four circles $A, B, C, D$ are pairwise tangent, with respective curvatures $a, b, c, d$, then

$$
(a+b+c+d)^{2}=2\left(a^{2}+b^{2}+c^{2}+d^{2}\right),
$$

where (as before) $a$ is taken to be negative if $B, C, D$ are internally tangent to $A$, and correspondingly for $b, c$, or $d$.
Determine the total number of plots sold up to and including day $n$.

Solution:
The total number of plots sold up to and including day $n$ is

$$
\begin{aligned}
2+\sum_{k=1}^{n} X_{k} & =2+2 \sum_{k=1}^{n} 3^{k-1} \\
& =2+2 \cdot\left(1+3+3^{2}+\ldots+3^{n-1}\right) \\
& =3^{n}+1
\end{aligned}
$$

Alternatively, proceed by induction: on day 0 , there are $2=3^{0}+1$ plots sold, and for $n \geq 0$,

$$
\begin{aligned}
\left(3^{n}+1\right)+X_{n+1} & =\left(3^{n}+1\right)+2 \cdot 3^{n} \\
& =3 \cdot 3^{n}+1 \\
& =3^{n+1}+1 .
\end{aligned}
$$

Final answer: $3^{n}+1$
Subfield: Geometry


Question: 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.

<img_4001>

Circle $A$ has curvature 2; circle $B$ has curvature 1 .

<img_3649>

Circle $A$ has curvature 2; circle $B$ has curvature -1 .

Using these conventions allows us to express a beautiful theorem of Descartes: when four circles $A, B, C, D$ are pairwise tangent, with respective curvatures $a, b, c, d$, then

$$
(a+b+c+d)^{2}=2\left(a^{2}+b^{2}+c^{2}+d^{2}\right),
$$

where (as before) $a$ is taken to be negative if $B, C, D$ are internally tangent to $A$, and correspondingly for $b, c$, or $d$.
Determine the total number of plots sold up to and including day $n$.

Solution:
The total number of plots sold up to and including day $n$ is

$$
\begin{aligned}
2+\sum_{k=1}^{n} X_{k} & =2+2 \sum_{k=1}^{n} 3^{k-1} \\
& =2+2 \cdot\left(1+3+3^{2}+\ldots+3^{n-1}\right) \\
& =3^{n}+1
\end{aligned}
$$

Alternatively, proceed by induction: on day 0 , there are $2=3^{0}+1$ plots sold, and for $n \geq 0$,

$$
\begin{aligned}
\left(3^{n}+1\right)+X_{n+1} & =\left(3^{n}+1\right)+2 \cdot 3^{n} \\
& =3 \cdot 3^{n}+1 \\
& =3^{n+1}+1 .
\end{aligned}
$$

Final answer: $3^{n}+1$
Subfield: Geometry


Question: 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.

<img_4001>

Circle $A$ has curvature 2; circle $B$ has curvature 1 .

<img_3649>

Circle $A$ has curvature 2; circle $B$ has curvature -1 .

Using these conventions allows us to express a beautiful theorem of Descartes: when four circles $A, B, C, D$ are pairwise tangent, with respective curvatures $a, b, c, d$, then

$$
(a+b+c+d)^{2}=2\left(a^{2}+b^{2}+c^{2}+d^{2}\right),
$$

where (as before) $a$ is taken to be negative if $B, C, D$ are internally tangent to $A$, and correspondingly for $b, c$, or $d$.
Algebraically, it is possible for a quadruple $(a, b, c, 0)$ to satisfy Descartes' Circle Formula, as occurs when $a=b=1$ and $c=4$. Find a geometric interpretation for this situation.

Solution:
In this case, the fourth "circle" is actually a line tangent to all three circles, as shown in the diagram below.

<img_3396>
Subfield: Geometry


Question: 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.

![](https://cdn.mathpix.com/cropped/2023_12_21_de9643dc5cb5c1da0c52g-1.jpg?height=304&width=455&top_left_y=1886&top_left_x=347)

Circle $A$ has curvature 2; circle $B$ has curvature 1 .

![](https://cdn.mathpix.com/cropped/2023_12_21_de9643dc5cb5c1da0c52g-1.jpg?height=301&width=307&top_left_y=1888&top_left_x=1386)

Circle $A$ has curvature 2; circle $B$ has curvature -1 .

Using these conventions allows us to express a beautiful theorem of Descartes: when four circles $A, B, C, D$ are pairwise tangent, with respective curvatures $a, b, c, d$, then

$$
(a+b+c+d)^{2}=2\left(a^{2}+b^{2}+c^{2}+d^{2}\right),
$$

where (as before) $a$ is taken to be negative if $B, C, D$ are internally tangent to $A$, and correspondingly for $b, c$, or $d$.
Algebraically, it is possible for a quadruple $(a, b, c, 0)$ to satisfy Descartes' Circle Formula, as occurs when $a=b=1$ and $c=4$. Find a geometric interpretation for this situation.

Solution:
In this case, the fourth "circle" is actually a line tangent to all three circles, as shown in the diagram below.

![](https://cdn.mathpix.com/cropped/2023_12_21_a1e4d46b9ec7c4ff99e7g-1.jpg?height=417&width=952&top_left_y=1667&top_left_x=627)

Final answer: 证明题，略