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


Question: Let $\mathcal{R}$ denote the circular region bounded by $x^{2}+y^{2}=36$. The lines $x=4$ and $y=3$ partition $\mathcal{R}$ into four regions $\mathcal{R}_{1}, \mathcal{R}_{2}, \mathcal{R}_{3}$, and $\mathcal{R}_{4}$. $\left[\mathcal{R}_{i}\right]$ denotes the area of region $\mathcal{R}_{i}$. If $\left[\mathcal{R}_{1}\right]>\left[\mathcal{R}_{2}\right]>\left[\mathcal{R}_{3}\right]>\left[\mathcal{R}_{4}\right]$, compute $\left[\mathcal{R}_{1}\right]-\left[\mathcal{R}_{2}\right]-\left[\mathcal{R}_{3}\right]+\left[\mathcal{R}_{4}\right]$.

Solution:
Draw the lines $x=-4$ and $y=-3$, creating regions $\mathcal{R}_{21}, \mathcal{R}_{22}, \mathcal{R}_{11}, \mathcal{R}_{12}, \mathcal{R}_{13}, \mathcal{R}_{14}$ as shown below.

<img_3593>



Then $\left[\mathcal{R}_{21}\right]=\left[\mathcal{R}_{4}\right]=\left[\mathcal{R}_{13}\right],\left[\mathcal{R}_{22}\right]=\left[\mathcal{R}_{14}\right]$, and $\left[\mathcal{R}_{3}\right]=\left[\mathcal{R}_{12}\right]+\left[\mathcal{R}_{13}\right]$. Therefore

$$
\begin{aligned}
{\left[\mathcal{R}_{1}\right]-\left[\mathcal{R}_{2}\right]-\left[\mathcal{R}_{3}\right]+\left[\mathcal{R}_{4}\right] } & =\left(\left[\mathcal{R}_{1}\right]-\left[\mathcal{R}_{2}\right]\right)-\left(\left[\mathcal{R}_{3}\right]-\left[\mathcal{R}_{4}\right]\right) \\
& =\left(\left[\mathcal{R}_{1}\right]-\left[\mathcal{R}_{13}\right]-\left[\mathcal{R}_{14}\right]\right)-\left(\left[\mathcal{R}_{12}\right]+\left[\mathcal{R}_{13}\right]-\left[\mathcal{R}_{21}\right]\right) \\
& =\left(\left[\mathcal{R}_{11}\right]+\left[\mathcal{R}_{12}\right]\right)-\left[\mathcal{R}_{12}\right] \\
& =\left[\mathcal{R}_{11}\right] .
\end{aligned}
$$

This last region is simply a rectangle of height 6 and width 8 , so its area is 48 .

Final answer: 48
Subfield: Algebra


Question: Let $a_{1}, a_{2}, a_{3}, \ldots$ be an arithmetic sequence, and let $b_{1}, b_{2}, b_{3}, \ldots$ be a geometric sequence. The sequence $c_{1}, c_{2}, c_{3}, \ldots$ has $c_{n}=a_{n}+b_{n}$ for each positive integer $n$. If $c_{1}=1, c_{2}=4, c_{3}=15$, and $c_{4}=2$, compute $c_{5}$.

Solution:
Let $a_{2}-a_{1}=d$ and $\frac{b_{2}}{b_{1}}=r$. Using $a=a_{1}$ and $b=b_{1}$, write the system of equations:

$$
\begin{aligned}
a+b & =1 \\
(a+d)+b r & =4 \\
(a+2 d)+b r^{2} & =15 \\
(a+3 d)+b r^{3} & =2 .
\end{aligned}
$$

Subtract the first equation from the second, the second from the third, and the third from the fourth to obtain three equations:

$$
\begin{aligned}
d+b(r-1) & =3 \\
d+b\left(r^{2}-r\right) & =11 \\
d+b\left(r^{3}-r^{2}\right) & =-13
\end{aligned}
$$

Notice that the $a$ terms have canceled. Repeat to find the second differences:

$$
\begin{aligned}
b\left(r^{2}-2 r+1\right) & =8 \\
b\left(r^{3}-2 r^{2}+r\right) & =-24
\end{aligned}
$$

Now divide the second equation by the first to obtain $r=-3$. Substituting back into either of these two last equations yields $b=\frac{1}{2}$. Continuing in the same vein yields $d=5$ and $a=\frac{1}{2}$. Then $a_{5}=\frac{41}{2}$ and $b_{5}=\frac{81}{2}$, so $c_{5}=\mathbf{6 1}$.

Final answer: 61
Subfield: Geometry


Question: In square $A B C D$ with diagonal $1, E$ is on $\overline{A B}$ and $F$ is on $\overline{B C}$ with $\mathrm{m} \angle B C E=\mathrm{m} \angle B A F=$ $30^{\circ}$. If $\overline{C E}$ and $\overline{A F}$ intersect at $G$, compute the distance between the incenters of triangles $A G E$ and $C G F$.

Solution:
Let $M$ be the midpoint of $\overline{A G}$, and $I$ the incenter of $\triangle A G E$ as shown below.

<img_3715>

Because $\frac{A B}{A C}=\sin 45^{\circ}$ and $\frac{E B}{A B}=\frac{E B}{B C}=\tan 30^{\circ}$,

$$
\begin{aligned}
A E & =A B-E B=A B\left(1-\tan 30^{\circ}\right) \\
& =\sin 45^{\circ}\left(1-\tan 30^{\circ}\right) \\
& =\frac{\sin 45^{\circ} \cos 30^{\circ}-\cos 45^{\circ} \sin 30^{\circ}}{\cos 30^{\circ}} \\
& =\frac{\sin \left(45^{\circ}-30^{\circ}\right)}{\cos 30^{\circ}} \\
& =\frac{\sin 15^{\circ}}{\cos 30^{\circ}} .
\end{aligned}
$$



Note that $\frac{A M}{A E}=\cos 30^{\circ}$ and $\frac{A M}{A I}=\cos 15^{\circ}$. Therefore

$$
\begin{aligned}
\frac{A I}{A E} & =\frac{\cos 30^{\circ}}{\cos 15^{\circ}} \\
& =\frac{\sin 60^{\circ}}{\cos 15^{\circ}} \\
& =\frac{2 \sin 30^{\circ} \cos 30^{\circ}}{\cos 15^{\circ}} \\
& =\frac{2\left(2 \sin 15^{\circ} \cos 15^{\circ}\right) \cos 30^{\circ}}{\cos 15^{\circ}} \\
& =4 \sin 15^{\circ} \cos 30^{\circ} .
\end{aligned}
$$

Thus $A I=\left(4 \sin 15^{\circ} \cos 30^{\circ}\right)\left(\frac{\sin 15^{\circ}}{\cos 30^{\circ}}\right)=4 \sin ^{2} 15^{\circ}=4 \cdot\left(\frac{1-\cos 30^{\circ}}{2}\right)=2-\sqrt{3}$. Finally, the desired distance is $2 I G=2 A I=4-2 \sqrt{3}$.

Final answer: $4-2 \sqrt{3}$
Subfield: Geometry


Question: Let $a, b, m, n$ be positive integers with $a m=b n=120$ and $a \neq b$. In the coordinate plane, let $A=(a, m), B=(b, n)$, and $O=(0,0)$. If $X$ is a point in the plane such that $A O B X$ is a parallelogram, compute the minimum area of $A O B X$.

Solution:
The area of parallelogram $A O B X$ is given by the absolute value of the cross product $|\langle a, m\rangle \times\langle b, n\rangle|=|a n-m b|$. Because $m=\frac{120}{a}$ and $n=\frac{120}{b}$, the desired area of $A O B X$ equals $120\left|\frac{a}{b}-\frac{b}{a}\right|$. Note that the function $f(x)=x-\frac{1}{x}$ is monotone increasing for $x>1$. (Proof: if $x_{1}>x_{2}>0$, then $f\left(x_{1}\right)-f\left(x_{2}\right)=\left(x_{1}-x_{2}\right)+\frac{x_{1}-x_{2}}{x_{1} x_{2}}$, where both terms are positive because $x_{1} x_{2}>0$.) So the minimum value of $[A O B X]$ is attained when $\frac{a}{b}$ is as close as possible to 1 , that is, when $a$ and $b$ are consecutive divisors of 120. By symmetry, consider only $a<b$; notice too that because $\frac{120 / a}{120 / b}=\frac{b}{a}$, only values with $b \leq \sqrt{120}$ need be considered. These observations can be used to generate the table below:

| $a, m$ | 1,120 | 2,60 | 3,40 | 4,30 | 5,24 | 6,20 | 8,15 | 10,12 |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| $b, n$ | 2,60 | 3,40 | 4,30 | 5,24 | 6,20 | 8,15 | 10,12 | 12,10 |
| $[A O B X]$ | 180 | 100 | 70 | 54 | 44 | 70 | 54 | 44 |

The smallest value is $\mathbf{4 4}$, achieved using $(5,24)$ and $(6,20)$, or using $(10,12)$ and $(12,10)$.

Note: The fact that $a$ and $b$ must be consecutive divisors of 120 can also be established by the following geometric argument. Notice that $[A O B X]=2[A O B]$. Suppose $C$ is a point on the hyperbola $y=120 / x$ between $A$ and $B$, as shown in the diagram below.



<img_3796>

Because the hyperbola is concave up, $[O A C]+[O C B]<[O A B]$, so in particular, $[O A C]<$ $[O A B]$. Thus, if $[O A B]$ is minimal, there can be no point $C$ with integer coordinates between $A$ and $B$ on the hyperbola.

Final answer: 44
Subfield: Combinatorics


Question: Let $\mathcal{S}$ be the set of integers from 0 to 9999 inclusive whose base- 2 and base- 5 representations end in the same four digits. (Leading zeros are allowed, so $1=0001_{2}=0001_{5}$ is one such number.) Compute the remainder when the sum of the elements of $\mathcal{S}$ is divided by 10,000.

Solution:
The remainders of an integer $N$ modulo $2^{4}=16$ and $5^{4}=625$ uniquely determine its remainder modulo 10000. There are only 16 strings of four 0's and 1's. In addition, because 16 and 625 are relatively prime, it will be shown below that for each such string $s$, there exists exactly one integer $x_{s}$ in the range $0 \leq x_{s}<10000$ such that the base- 2 and base- 5 representations of $x_{s}$ end in the digits of $s$ (e.g., $x_{1001}$ is the unique positive integer less than 10000 such that $x$ 's base- 5 representation and base- 2 representation both end in 1001).

Here is a proof of the preceding claim: Let $p(s)$ be the number whose digits in base 5 are the string $s$, and $b(s)$ be the number whose digits in base 2 are the string $s$. Then the system $x \equiv$ $p(s) \bmod 625$ and $x \equiv b(s) \bmod 16$ can be rewritten as $x=p(s)+625 m$ and $x=b(s)+16 n$ for integers $m$ and $n$. These reduce to the Diophantine equation $16 n-625 m=p(s)-b(s)$, which has solutions $m, n$ in $\mathbb{Z}$, with at least one of $m, n \geq 0$. Assuming without loss of generality that $m>0$ yields $x=p(s)+625 m \geq 0$. To show that there exists an $x_{s}<10000$ and that it is unique, observe that the general form of the solution is $m^{\prime}=m-16 t, n^{\prime}=n+625 t$. Thus if $p(s)+625 m>10000$, an appropriate $t$ can be found by writing $0 \leq p(s)+625(m-16 t)<10000$, which yields $p(s)+625 m-10000<10000 t \leq p(s)+625 m$. Because there are exactly 10000 integers in that interval, exactly one of them is divisible by 10000 , so there is exactly one value of $t$ satisfying $0 \leq p(s)+625(m-16 t)<10000$, and set $x_{s}=625(m-16 t)$.

Therefore there will be 16 integers whose base- 2 and base- 5 representations end in the same four digits, possibly with leading 0 's as in the example. Let $X=x_{0000}+\cdots+x_{1111}$. Then $X$ is congruent modulo 16 to $0000_{2}+\cdots+1111_{2}=8 \cdot\left(1111_{2}\right)=8 \cdot 15 \equiv 8$. Similarly, $X$ is congruent modulo 625 to $0000_{5}+\cdots+1111_{5}=8 \cdot 1111_{5}=2 \cdot 4444_{5} \equiv 2 \cdot(-1)=-2$.

So $X$ must be $8(\bmod 16)$ and $-2(\bmod 625)$. Noticing that $625 \equiv 1(\bmod 16)$, conclude that the answer is $-2+10 \cdot 625=\mathbf{6 2 4 8}$.

Final answer: 6248
Subfield: Number Theory


Question: If $A, R, M$, and $L$ are positive integers such that $A^{2}+R^{2}=20$ and $M^{2}+L^{2}=10$, compute the product $A \cdot R \cdot M \cdot L$.

Solution:
The only positive integers whose squares sum to 20 are 2 and 4 . The only positive integers whose squares sum to 10 are 1 and 3 . Thus $A \cdot R=8$ and $M \cdot L=3$, so $A \cdot R \cdot M \cdot L=\mathbf{2 4}$.

Final answer: 24
Subfield: Geometry


Question: Let $T=24$. A regular $n$-gon is inscribed in a circle; $P$ and $Q$ are consecutive vertices of the polygon, and $A$ is another vertex of the polygon as shown. If $\mathrm{m} \angle A P Q=\mathrm{m} \angle A Q P=T \cdot \mathrm{m} \angle Q A P$, compute the value of $n$.

![](https://cdn.mathpix.com/cropped/2023_12_21_daa62eb0f647b311d7dag-1.jpg?height=352&width=290&top_left_y=732&top_left_x=1357)

Solution:
Let $\mathrm{m} \angle A=x$. Then $\mathrm{m} \angle P=\mathrm{m} \angle Q=T x$, and $(2 T+1) x=180^{\circ}$, so $x=\frac{180^{\circ}}{2 T+1}$. Let $O$ be the center of the circle, as shown below.

![](https://cdn.mathpix.com/cropped/2023_12_21_44186b4d618c78311efdg-1.jpg?height=439&width=393&top_left_y=710&top_left_x=909)

Then $\mathrm{m} \angle P O Q=2 \mathrm{~m} \angle P A Q=2\left(\frac{180^{\circ}}{2 T+1}\right)=\frac{360^{\circ}}{2 T+1}$. Because $\mathrm{m} \angle P O Q=\frac{360^{\circ}}{n}$, the denominators must be equal: $n=2 T+1$. Substitute $T=24$ to find $n=\mathbf{4 9}$.

Final answer: 49
Subfield: Geometry


Question: Let $T=24$. A regular $n$-gon is inscribed in a circle; $P$ and $Q$ are consecutive vertices of the polygon, and $A$ is another vertex of the polygon as shown. If $\mathrm{m} \angle A P Q=\mathrm{m} \angle A Q P=T \cdot \mathrm{m} \angle Q A P$, compute the value of $n$.

<img_3413>

Solution:
Let $\mathrm{m} \angle A=x$. Then $\mathrm{m} \angle P=\mathrm{m} \angle Q=T x$, and $(2 T+1) x=180^{\circ}$, so $x=\frac{180^{\circ}}{2 T+1}$. Let $O$ be the center of the circle, as shown below.

<img_3423>

Then $\mathrm{m} \angle P O Q=2 \mathrm{~m} \angle P A Q=2\left(\frac{180^{\circ}}{2 T+1}\right)=\frac{360^{\circ}}{2 T+1}$. Because $\mathrm{m} \angle P O Q=\frac{360^{\circ}}{n}$, the denominators must be equal: $n=2 T+1$. Substitute $T=24$ to find $n=\mathbf{4 9}$.

Final answer: 49
Subfield: Number Theory


Question: Let $T=49$. Compute the last digit, in base 10, of the sum

$$
T^{2}+(2 T)^{2}+(3 T)^{2}+\ldots+\left(T^{2}\right)^{2}
$$

Solution:
Let $S$ be the required sum. Factoring $T^{2}$ from the sum yields

$$
\begin{aligned}
S & =T^{2}\left(1+4+9+\ldots+T^{2}\right) \\
& =T^{2}\left(\frac{T(T+1)(2 T+1)}{6}\right) \\
& =\frac{T^{3}(T+1)(2 T+1)}{6} .
\end{aligned}
$$

Further analysis makes the final computation simpler. If $T \equiv 0,2$, or $3 \bmod 4$, then $S$ is even. Otherwise, $S$ is odd. And if $T \equiv 0,2$, or $4 \bmod 5$, then $S \equiv 0 \bmod 5$; otherwise, $S \equiv 1 \bmod 5$. These observations yield the following table:

| $T \bmod 4$ | $T \bmod 5$ | $S \bmod 10$ |
| :---: | :---: | :---: |
| $0,2,3$ | $0,2,4$ | 0 |
| $0,2,3$ | 1,3 | 6 |
| 1 | $0,2,4$ | 5 |
| 1 | 1,3 | 1 |

Because $T=49$, the value corresponds to the third case above; the last digit is $\mathbf{5}$.

Final answer: 5
Subfield: Combinatorics


Question: A fair coin is flipped $n$ times. Compute the smallest positive integer $n$ for which the probability that the coin has the same result every time is less than $10 \%$.

Solution:
After the first throw, the probability that the succeeding $n-1$ throws have the same result is $\frac{1}{2^{n-1}}$. Thus $\frac{1}{2^{n-1}}<\frac{1}{10} \Rightarrow 2^{n-1}>10 \Rightarrow n-1 \geq 4$, so $n=5$ is the smallest possible value.

Final answer: 5
Subfield: Algebra


Question: Let $T=5$. Compute the smallest positive integer $n$ such that there are at least $T$ positive integers in the domain of $f(x)=\sqrt{-x^{2}-2 x+n}$.

Solution:
Completing the square under the radical yields $\sqrt{n+1-(x+1)^{2}}$. The larger zero of the radicand is $-1+\sqrt{n+1}$, and the smaller zero is negative because $-1-\sqrt{n+1}<0$, so the $T$ positive integers in the domain of $f$ must be $1,2,3, \ldots, T$. Therefore $-1+\sqrt{n+1} \geq T$. Hence $\sqrt{n+1} \geq T+1$, and $n+1 \geq(T+1)^{2}$. Therefore $n \geq T^{2}+2 T$, and substituting $T=5$ yields $n \geq 35$. So $n=\mathbf{3 5}$ is the smallest such value.

Final answer: 35
Subfield: Algebra


Question: Let $T=35$. Compute the smallest positive real number $x$ such that $\frac{\lfloor x\rfloor}{x-\lfloor x\rfloor}=T$.

Solution:
If $\frac{\lfloor x\rfloor}{x-\lfloor x\rfloor}=T$, the equation can be rewritten as follows:

$$
\begin{aligned}
\frac{x-\lfloor x\rfloor}{\lfloor x\rfloor} & =\frac{1}{T} \\
\frac{x}{\lfloor x\rfloor}-1 & =\frac{1}{T} \\
\frac{x}{\lfloor x\rfloor} & =\frac{T+1}{T} .
\end{aligned}
$$

Now $0<x<1$ is impossible because it makes the numerator of the original expression 0 . To make $x$ as small as possible, place it in the interval $1<x<2$, so that $\lfloor x\rfloor=1$. Then $x=\frac{T+1}{T}$. When $T=35, x=\frac{36}{35}$.

Final answer: $\frac{36}{35}$
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 $d$.
Without using Descartes' Circle Formula, Show that the circles marked off and sold on day 1 are centered at $\left(0, \pm \frac{2}{3}\right)$ with radius $\frac{1}{3}$.

Solution:
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 $r+\frac{1}{2}$. Therefore

$$
\begin{aligned}
\left(\frac{1}{2}\right)^{2}+(1-r)^{2} & =\left(r+\frac{1}{2}\right)^{2} \\
1-2 r & =r \\
r & =\frac{1}{3} \\
y=1-r & =\frac{2}{3} .
\end{aligned}
$$

Thus the plots are centered at $\left(0, \pm \frac{2}{3}\right)$ and have radius $\frac{1}{3}$.

Final answer: 证明题，略
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, Show that the circles marked off and sold on day 1 are centered at $\left(0, \pm \frac{2}{3}\right)$ with radius $\frac{1}{3}$.

Solution:
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 $r+\frac{1}{2}$. Therefore

$$
\begin{aligned}
\left(\frac{1}{2}\right)^{2}+(1-r)^{2} & =\left(r+\frac{1}{2}\right)^{2} \\
1-2 r & =r \\
r & =\frac{1}{3} \\
y=1-r & =\frac{2}{3} .
\end{aligned}
$$

Thus the plots are centered at $\left(0, \pm \frac{2}{3}\right)$ and have radius $\frac{1}{3}$.
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 $d$.
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: $\frac{5 \pi}{18}$