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


Question: In right triangle $A B C$ with right angle $C$, line $\ell$ is drawn through $C$ and is parallel to $\overline{A B}$. Points $P$ and $Q$ lie on $\overline{A B}$ with $P$ between $A$ and $Q$, and points $R$ and $S$ lie on $\ell$ with $C$ between $R$ and $S$ such that $P Q R S$ is a square. Let $\overline{P S}$ intersect $\overline{A C}$ in $X$, and let $\overline{Q R}$ intersect $\overline{B C}$ in $Y$. The inradius of triangle $A B C$ is 10 , and the area of square $P Q R S$ is 576 . Compute the sum of the inradii of triangles $A X P, C X S, C Y R$, and $B Y Q$.

Solution:
Note that in right triangle $A B C$ with right angle $C$, the inradius $r$ is equal to $\frac{a+b-c}{2}$, where $a=B C, b=A C$, and $c=A B$, because the inradius equals the distance from the vertex of the right angle $C$ to (either) point of tangency along $\overline{A C}$ or $\overline{B C}$. Thus the sum of the inradii of triangles $A X P, C X S, C Y R$, and $B Y Q$ is equal to one-half the difference between the sum of the lengths of the legs of these triangles and the sum of the lengths of the hypotenuses of these triangles. Let $t$ be the side length of square $P Q R S$. Then the sum of the lengths of the legs of triangles $A X P, C X S, C Y R$, and $B Y Q$ is

$$
\begin{aligned}
& A P+P X+X S+S C+C R+R Y+Y Q+Q B \\
= & A P+P S+S R+R Q+Q B \\
= & A P+t+t+t+Q B \\
= & A B-P Q+3 t \\
= & c-t+3 t \\
= & c+2 t .
\end{aligned}
$$



The sum of the lengths of the hypotenuses of triangles $A X P, C X S, C Y R$, and $B Y Q$ is $A X+X C+C Y+Y B=A C+C B=b+a$. Hence the sum of the inradii of triangles $A X P, C X S, C Y R$, and $B Y Q$ is $\frac{c+2 t-(a+b)}{2}=t-r$. Thus the desired sum equals $\sqrt{576}-10=24-10=\mathbf{1 4}$.

Final answer: 14
Subfield: Number Theory


Question: Compute the sum of all real numbers $x$ such that

$$
\left\lfloor\frac{x}{2}\right\rfloor-\left\lfloor\frac{x}{3}\right\rfloor=\frac{x}{7}
$$

Solution:
Because the quantity on the left side is the difference of two integers, $x / 7$ must be an integer, hence $x$ is an integer (in fact a multiple of 7). Because the denominators on the left side are 2 and 3 , it is convenient to write $x=6 q+r$, where $0 \leq r \leq 5$, so that $\lfloor x / 2\rfloor=3 q+\lfloor r / 2\rfloor$ and $\lfloor x / 3\rfloor=2 q+\lfloor r / 3\rfloor$. Then for $r=0,1, \ldots, 5$ these expressions can be simplified as shown in the table below.

| $r$ | 0 | 1 | 2 | 3 | 4 | 5 |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| $\left\lfloor\frac{x}{2}\right\rfloor$ | $3 q$ | $3 q$ | $3 q+1$ | $3 q+1$ | $3 q+2$ | $3 q+2$ |
| $\left\lfloor\frac{x}{3}\right\rfloor$ | $2 q$ | $2 q$ | $2 q$ | $2 q+1$ | $2 q+1$ | $2 q+1$ |
| $\left\lfloor\frac{x}{2}\right\rfloor-\left\lfloor\frac{x}{3}\right\rfloor$ | $q$ | $q$ | $q+1$ | $q$ | $q+1$ | $q+1$ |

Now proceed by cases:

$r=0:$ Then $q=x / 6$. But from the statement of the problem, $q=x / 7$, so $x=0$.

$r=1: \quad$ Then $q=(x-1) / 6=x / 7 \Rightarrow x=7$.

$r=2: \quad$ Then $q=(x-2) / 6$ and $q+1=x / 7$, so $(x+4) / 6=x / 7$, and $x=-28$.

$r=3$ : Then $q=(x-3) / 6$ and $q=x / 7$, so $x=21$.

$r=4: \quad$ Then $q=(x-4) / 6$ and $q+1=x / 7$, so $(x+2) / 6=x / 7$, and $x=-14$.

$r=5$ : Then $q=(x-5) / 6$ and $q+1=x / 7$, so $(x+1) / 6=x / 7$, and $x=-7$.

The sum of these values is $0+7+-28+21+-14+-7=\mathbf{- 2 1}$.

Final answer: -21
Subfield: Geometry


Question: Two equilateral triangles of side length 1 and six isosceles triangles with legs of length $x$ and base of length 1 are joined as shown below; the net is folded to make a solid. If the volume of the solid is 6 , compute $x$.

![](https://cdn.mathpix.com/cropped/2023_12_21_71dab4a47f4ff807d6edg-1.jpg?height=393&width=330&top_left_y=497&top_left_x=949)

Solution:
First consider a regular octahedron of side length 1. To compute its volume, divide it into two square-based pyramids with edges of length 1 . Such a pyramid has slant height $\frac{\sqrt{3}}{2}$ and height $\sqrt{\left(\frac{\sqrt{3}}{2}\right)^{2}-\left(\frac{1}{2}\right)^{2}}=\sqrt{\frac{1}{2}}=\frac{\sqrt{2}}{2}$, so its volume is $\frac{1}{3} \cdot 1^{2} \cdot \frac{\sqrt{2}}{2}=\frac{\sqrt{2}}{6}$. Thus the octahedron has volume twice that, or $\frac{\sqrt{2}}{3}$. The result of folding the net shown is actually the image of a regular octahedron after being stretched along an axis perpendicular to one face by a factor of $r$. Because the octahedron is only being stretched in one dimension, the volume changes by the same factor $r$. So the problem reduces to computing the factor $r$ and the edge length of the resulting octahedron.

For convenience, imagine that one face of the octahedron rests on a plane. Seen from above the plane, the octahedron appears as shown below.



![](https://cdn.mathpix.com/cropped/2023_12_21_6337ac12aa6d82067ee9g-1.jpg?height=407&width=395&top_left_y=249&top_left_x=903)

Let $P$ be the projection of $A$ onto the plane on which the octahedron rests, and let $Q$ be the foot of the perpendicular from $P$ to $\overline{B C}$. Then $P Q=R_{C}-R_{I}$, where $R_{C}$ is the circumradius and $R_{I}$ the inradius of the equilateral triangle. Thus $P Q=\frac{2}{3}\left(\frac{\sqrt{3}}{2}\right)-\frac{1}{3}\left(\frac{\sqrt{3}}{2}\right)=\frac{\sqrt{3}}{6}$. Then $B P^{2}=P Q^{2}+B Q^{2}=\frac{3}{36}+\frac{1}{4}=\frac{1}{3}$, so $A P^{2}=A B^{2}-B P^{2}=\frac{2}{3}$, and $A P=\frac{\sqrt{6}}{3}$.

Now let a vertical stretch take place along an axis parallel to $\overleftrightarrow{A P}$. If the scale factor is $r$, then $A P=\frac{r \sqrt{6}}{3}$, and because the stretch occurs on an axis perpendicular to $\overline{B P}$, the length $B P$ is unchanged, as can be seen below.

![](https://cdn.mathpix.com/cropped/2023_12_21_6337ac12aa6d82067ee9g-1.jpg?height=563&width=401&top_left_y=1239&top_left_x=903)

Thus $A B^{2}=\frac{6 r^{2}}{9}+\frac{1}{3}=\frac{6 r^{2}+3}{9}$. It remains to compute $r$. But $r$ is simply the ratio of the new volume to the old volume:

$$
r=\frac{6}{\frac{\sqrt{2}}{3}}=\frac{18}{\sqrt{2}}=9 \sqrt{2}
$$

Thus $A B^{2}=\frac{6(9 \sqrt{2})^{2}+3}{9}=\frac{975}{9}=\frac{325}{3}$, and $A B=\frac{5 \sqrt{39}}{3}$.

Final answer: $\frac{5 \sqrt{39}}{3}$
Subfield: Geometry


Question: Two equilateral triangles of side length 1 and six isosceles triangles with legs of length $x$ and base of length 1 are joined as shown below; the net is folded to make a solid. If the volume of the solid is 6 , compute $x$.

<img_3327>

Solution:
First consider a regular octahedron of side length 1. To compute its volume, divide it into two square-based pyramids with edges of length 1 . Such a pyramid has slant height $\frac{\sqrt{3}}{2}$ and height $\sqrt{\left(\frac{\sqrt{3}}{2}\right)^{2}-\left(\frac{1}{2}\right)^{2}}=\sqrt{\frac{1}{2}}=\frac{\sqrt{2}}{2}$, so its volume is $\frac{1}{3} \cdot 1^{2} \cdot \frac{\sqrt{2}}{2}=\frac{\sqrt{2}}{6}$. Thus the octahedron has volume twice that, or $\frac{\sqrt{2}}{3}$. The result of folding the net shown is actually the image of a regular octahedron after being stretched along an axis perpendicular to one face by a factor of $r$. Because the octahedron is only being stretched in one dimension, the volume changes by the same factor $r$. So the problem reduces to computing the factor $r$ and the edge length of the resulting octahedron.

For convenience, imagine that one face of the octahedron rests on a plane. Seen from above the plane, the octahedron appears as shown below.



<img_3368>

Let $P$ be the projection of $A$ onto the plane on which the octahedron rests, and let $Q$ be the foot of the perpendicular from $P$ to $\overline{B C}$. Then $P Q=R_{C}-R_{I}$, where $R_{C}$ is the circumradius and $R_{I}$ the inradius of the equilateral triangle. Thus $P Q=\frac{2}{3}\left(\frac{\sqrt{3}}{2}\right)-\frac{1}{3}\left(\frac{\sqrt{3}}{2}\right)=\frac{\sqrt{3}}{6}$. Then $B P^{2}=P Q^{2}+B Q^{2}=\frac{3}{36}+\frac{1}{4}=\frac{1}{3}$, so $A P^{2}=A B^{2}-B P^{2}=\frac{2}{3}$, and $A P=\frac{\sqrt{6}}{3}$.

Now let a vertical stretch take place along an axis parallel to $\overleftrightarrow{A P}$. If the scale factor is $r$, then $A P=\frac{r \sqrt{6}}{3}$, and because the stretch occurs on an axis perpendicular to $\overline{B P}$, the length $B P$ is unchanged, as can be seen below.

<img_3791>

Thus $A B^{2}=\frac{6 r^{2}}{9}+\frac{1}{3}=\frac{6 r^{2}+3}{9}$. It remains to compute $r$. But $r$ is simply the ratio of the new volume to the old volume:

$$
r=\frac{6}{\frac{\sqrt{2}}{3}}=\frac{18}{\sqrt{2}}=9 \sqrt{2}
$$

Thus $A B^{2}=\frac{6(9 \sqrt{2})^{2}+3}{9}=\frac{975}{9}=\frac{325}{3}$, and $A B=\frac{5 \sqrt{39}}{3}$.

Final answer: $\frac{5 \sqrt{39}}{3}$
Subfield: Number Theory


Question: Compute the smallest positive integer $n$ such that $n^{2}+n^{0}+n^{1}+n^{3}$ is a multiple of 13 .

Solution:
Note that $n^{2}+n^{0}+n^{1}+n^{3}=n^{2}+1+n+n^{3}=\left(n^{2}+1\right)(1+n)$. Because 13 is prime, 13 must be a divisor of one of these factors. The smallest positive integer $n$ such that $13 \mid 1+n$ is $n=12$, whereas the smallest positive integer $n$ such that $13 \mid n^{2}+1$ is $n=\mathbf{5}$.

Final answer: 5
Subfield: Geometry


Question: Let $T=T N Y W R$. The diagram at right consists of $T$ congruent circles, each of radius 1 , whose centers are collinear, and each pair of adjacent circles are externally tangent to each other. Compute the length of the tangent segment $\overline{A B}$.

![](https://cdn.mathpix.com/cropped/2023_12_21_7104b4f1e75a0066b281g-1.jpg?height=160&width=433&top_left_y=516&top_left_x=1472)

Solution:
For each point of tangency of consecutive circles, drop a perpendicular from that point to $\overline{A B}$. For each of the $T-2$ circles between the first and last circles, the distance between consecutive perpendiculars is $2 \cdot 1=2$. Furthermore, the distance from $A$ to the first perpendicular equals 1 (i.e., the common radius of the circles), which also equals the distance from the last perpendicular to $B$. Thus $A B=1+(T-2) \cdot 2+1=2(T-1)$. With $T=5$, it follows that $A B=2 \cdot 4=8$.

Final answer: 8
Subfield: Geometry


Question: Let $T=T N Y W R$. The diagram at right consists of $T$ congruent circles, each of radius 1 , whose centers are collinear, and each pair of adjacent circles are externally tangent to each other. Compute the length of the tangent segment $\overline{A B}$.

<img_3453>

Solution:
For each point of tangency of consecutive circles, drop a perpendicular from that point to $\overline{A B}$. For each of the $T-2$ circles between the first and last circles, the distance between consecutive perpendiculars is $2 \cdot 1=2$. Furthermore, the distance from $A$ to the first perpendicular equals 1 (i.e., the common radius of the circles), which also equals the distance from the last perpendicular to $B$. Thus $A B=1+(T-2) \cdot 2+1=2(T-1)$. With $T=5$, it follows that $A B=2 \cdot 4=8$.

Final answer: 8
Subfield: Algebra


Question: Let $T=T N Y W R$. Compute $2^{\log _{T} 8}-8^{\log _{T} 2}$.

Solution:
Let $\log _{T} 8=x$. Then $T^{x}=8$. Thus the given expression equals $2^{x}-\left(T^{x}\right)^{\log _{T} 2}=2^{x}-T^{x \log _{T} 2}=$ $2^{x}-T^{\log _{T} 2^{x}}=2^{x}-2^{x}=\mathbf{0}$ (independent of $T$ ).

Final answer: 0
Subfield: Combinatorics


Question: Let $T=T N Y W R$. At some point during a given week, a law enforcement officer had issued $T+2$ traffic warnings, 20 tickets, and had made $T+5$ arrests. How many more tickets must the officer issue in order for the combined number of tickets and arrests to be 20 times the number of warnings issued that week?

Solution:
The problem requests the value of $k$ such that $20+k+T+5=20(T+2)$, thus $k=19 T+15$. With $T=0$, it follows that $k=\mathbf{1 5}$.

Final answer: 15
Subfield: Geometry


Question: $\quad$ Let $T=T N Y W R$. In parallelogram $A R M L$, points $P$ and $Q$ trisect $\overline{A R}$ and points $W, X, Y, Z$ divide $\overline{M L}$ into fifths (where $W$ is closest to $M$, and points $X$ and $Y$ are both between $W$ and $Z$ ). If $[A R M L]=T$, compute $[P Q W Z]$.

Solution:
Let $h$ be the distance between $\overline{A R}$ and $\overline{M L}$, and for simplicity, let $A R=M L=15 n$. Then $[A R M L]=15 n h$, and $[P Q W Z]=(1 / 2)(P Q+W Z) h$. Note that $P Q=15 n / 3=5 n$ and $W Z=15 n-3 n-3 n=9 n$. Thus $[P Q W Z]=7 n h=(7 / 15) \cdot[A R M L]=7 T / 15$. With $T=15$, the answer is 7 .

Final answer: 7
Subfield: Number Theory


Question: Let $T=T N Y W R$. Compute the number of positive perfect cubes that are divisors of $(T+10) !$.

Solution:
Let $N=T+10$. In order for $k^{3}(k \in \mathbb{N})$ to be a divisor of $N$ !, the largest odd prime factor of $k$ (call it $p$ ) must be less than or equal to $N / 3$ so that there are at least three multiples of $p$ among the product of the first $N$ positive integers. If $p=3$, then the smallest possible value of $N$ is 9 , and the largest perfect cube factor of 9 ! is $2^{6} \cdot 3^{3}$. Similarly, if $p=5$, then the smallest possible value of $N$ is 15 , and the largest perfect cube factor of 15 ! is $2^{9} \cdot 3^{6} \cdot 5^{3}$. With $T=7, N=17$, and the largest perfect cube factor of 17 ! is $2^{15} \cdot 3^{6} \cdot 5^{3}$. Thus $k^{3} \mid 17$ ! if and only if $k \mid 2^{5} \cdot 3^{2} \cdot 5^{1}$. Therefore $k=2^{x} 3^{y} 5^{z}$, where $x, y, z$ are nonnegative integers with $x \leq 5, y \leq 2, z \leq 1$, yielding $6 \cdot 3 \cdot 2=\mathbf{3 6}$ possible values of $k$.

Final answer: 36
Subfield: Geometry


Question: Let $T=T N Y W R$. The graph of $y=x^{2}+2 x-T$ intersects the $x$-axis at points $A$ and $M$, which are diagonally opposite vertices of square $A R M L$. Compute $[A R M L]$.

Solution:
Note that the $x$-coordinates of $A$ and $M$ correspond to the two roots $r_{1}, r_{2}$ of $x^{2}+2 x-T$. If $s$ is the side length of square $A R M L$, then $A M=s \sqrt{2}=\left|r_{1}-r_{2}\right|=\sqrt{\left(r_{1}-r_{2}\right)^{2}}=$ $\sqrt{\left(r_{1}+r_{2}\right)^{2}-4 r_{1} r_{2}}=\sqrt{(-2)^{2}-4(-T)}=2 \sqrt{1+T}$. Thus $[A R M L]=s^{2}=2(1+T)$. With $T=36,[A R M L]=\mathbf{7 4}$.

Final answer: 74
Subfield: Number Theory


Question: Let $S$ be the set of prime factors of the numbers you receive from positions 7 and 9 , and let $p$ and $q$ be the two least distinct elements of $S$, with $p<q$. Hexagon HEXAGO is inscribed in circle $\omega$, and every angle of $H E X A G O$ is $120^{\circ}$. If $H E=X A=G O=p$ and $E X=A G=O H=q$, compute the area of circle $\omega$.

Solution:
The given information implies that triangles $H E X, X A G$, and $G O H$ are congruent, hence triangle $H X G$ is equilateral. If $H X=s$, then the radius of the circle circumscribing $\triangle H X G$ is $s / \sqrt{3}$ so that the circle's area is $\pi s^{2} / 3$. It remains to compute $s$. With $\mathrm{m} \angle H E X=120^{\circ}$, use the Law of Cosines to find

$$
\begin{aligned}
H X^{2} & =H E^{2}+E X^{2}-2 H E \cdot E X \cdot \cos 120^{\circ} \\
& =p^{2}+q^{2}-2 p q(-1 / 2) \\
& =p^{2}+q^{2}+p q .
\end{aligned}
$$

Using the answers 74 and 7 from positions 7 and 9 , respectively, conclude that $S=\{2,7,37\}$ and that $(p, q)=(2,7)$. Hence the foregoing yields $H X^{2}=4+49+14=67$. Thus the area of circle $\omega$ is $\frac{\mathbf{6 7 \pi}}{\mathbf{3}}$.

Final answer: $\frac{67 \pi}{3}$
Subfield: Combinatorics


Question: Let $T=T N Y W R$. A group of $n$ friends goes camping; two of them are selected to set up the campsite when they arrive and two others are selected to take down the campsite the next day. Compute the smallest possible value of $n$ such that there are at least $T$ ways of selecting the four helpers.

Solution:
There are $\left(\begin{array}{l}n \\ 2\end{array}\right)$ ways of choosing the two people to set up and $\left(\begin{array}{c}n-2 \\ 2\end{array}\right)$ ways of choosing the two people to take down the campsite, so there are $\frac{n(n-1)}{2} \cdot \frac{(n-2)(n-3)}{2}$ ways of choosing the four people, or $\frac{n(n-1)(n-2)(n-3)}{4}$ ways total; call this function $C(n)$. For the least $n$ such that $\frac{n(n-1)(n-2)(n-3)}{4} \geq T$, as a rough approximation, note that $n-3<\sqrt[4]{4 T}<n$. With $T=184$, the approximation becomes $n-3<\sqrt[4]{736}<n$. Now $5^{4}=625$ while $6^{4}=1296$, so $5<n<9$. Try values starting from $n=6$ :

$$
\begin{aligned}
& C(6)=\frac{6 \cdot 5 \cdot 4 \cdot 3}{4}=90 \\
& C(7)=\frac{7 \cdot 6 \cdot 5 \cdot 4}{4}=210 .
\end{aligned}
$$

Thus $n=7$.

Final answer: 7
Subfield: Geometry


Question: Let $T=T N Y W R$. The parabola $y=x^{2}+T x$ is tangent to the parabola $y=-(x-2 T)^{2}+b$. Compute $b$.

Solution:
In this case, the two parabolas are tangent exactly when the system of equations has a unique solution. (Query: Is this the case for every pair of equations representing parabolas?) So set the right sides equal to each other: $x^{2}+T x=-(x-2 T)^{2}+b$. Then $x^{2}+T x=$ $-x^{2}+4 T x-4 T^{2}+b$, or equivalently, $2 x^{2}-3 T x+4 T^{2}-b=0$. The equation has a double root when the discriminant is 0 , so set $(-3 T)^{2}-4\left(4 T^{2}-b\right)(2)=0$ and solve: $9 T^{2}-32 T^{2}+8 b=0$ implies $-23 T^{2}+8 b=0$, or $b=23 T^{2} / 8$. Using $T=8$ yields $b=\mathbf{1 8 4}$.

Final answer: 184
Subfield: Algebra


Question: Let $T=T N Y W R$. The first two terms of a sequence are $a_{1}=3 / 5$ and $a_{2}=4 / 5$. For $n>2$, if $n$ is odd, then $a_{n}=a_{n-1}^{2}-a_{n-2}^{2}$, while if $n$ is even, then $a_{n}=2 a_{n-2} a_{n-3}$. Compute the sum of the squares of the first $T-3$ terms of the sequence.

Solution:
Using the identity $\left(x^{2}-y^{2}\right)^{2}+(2 x y)^{2}=\left(x^{2}+y^{2}\right)^{2}$, notice that $a_{2 n+1}^{2}+a_{2 n+2}^{2}=\left(a_{2 n}^{2}-a_{2 n-1}^{2}\right)^{2}+$ $\left(2 a_{2 n} a_{2 n-1}\right)^{2}=\left(a_{2 n}^{2}+a_{2 n-1}^{2}\right)^{2}$. So surprisingly, for all $n \in \mathbb{N}, a_{2 n+1}^{2}+a_{2 n+2}^{2}=1$. Thus if $n$ is even, the sum of the squares of the first $n$ terms is $n / 2$. With $T=19, T-3=16$, and the sum is 8 .

Final answer: 8
Subfield: Geometry


Question: Let $T=T N Y W R$. A regular $n$-gon has exactly $T$ more diagonals than a regular $(n-1)$-gon. Compute the value of $n$.

Solution:
Using the formula $D(n)=\frac{n(n-3)}{2}$ twice yields $D(n)-D(n-1)=\frac{n^{2}-3 n}{2}-\frac{n^{2}-5 n+4}{2}=\frac{2 n-4}{2}=n-2$. So $T=n-2$, thus $n=T+2$, and with $T=17, n=19$.

Final answer: 19
Subfield: Algebra


Question: Let $T=T N Y W R$. The sequence $a_{1}, a_{2}, a_{3}, \ldots$, is arithmetic with $a_{16}=13$ and $a_{30}=20$. Compute the value of $k$ for which $a_{k}=T$.

Solution:
If $d$ is the common difference of the sequence, then the $n^{\text {th }}$ term of the sequence is $a_{n}=$ $a_{16}+d(n-16)$. The values $a_{16}=13$ and $a_{30}=20$ yield $d=(20-13) /(30-16)=1 / 2$, hence $a_{n}=13+(1 / 2)(n-16)$. If $a_{n}=T$, then $n=2(T-13)+16=2 T-10$. With $T=27 / 2$, it follows that $n=\mathbf{1 7}$.

Final answer: 17
Subfield: Geometry


Question: Let $T=T N Y W R$. A rectangular prism has a length of 1 , a width of 3 , a height of $h$, and has a total surface area of $T$. Compute the value of $h$.

Solution:
The surface area is given by the expression $2 \cdot 1 \cdot 3+2 \cdot 1 \cdot h+2 \cdot 3 \cdot h=6+8 h$. Because $6+8 h=T, h=\frac{T-6}{8}$. With $T=114, h=108 / 8=\mathbf{2 7} / \mathbf{2}$.

Final answer: $\frac{27}{2}$
Subfield: Algebra


Question: The zeros of $x^{2}+b x+93$ are $r$ and $s$. If the zeros of $x^{2}-22 x+c$ are $r+1$ and $s+1$, compute $c$.

Solution:
Use sums and products of roots formulas: the desired quantity $c=(r+1)(s+1)=r s+r+s+1$. From the first equation, $r s=93$, while from the second equation, $(r+1)+(s+1)=r+s+2=$ 22. So $r s+r+s+1=93+22-1=\mathbf{1 1 4}$.

Final answer: 114
Subfield: Number Theory


Question: Let $N=888,888 \times 9,999,999$. Compute the sum of the digits of $N$.

Solution:
Write $N$ as

$$
\begin{aligned}
& (10,000,000-1) \cdot 888,888 \\
= & 8,888,880,000,000-888,888 \\
= & 8,888,879,111,112 .
\end{aligned}
$$

The sum of the digits of $N$ is 63 .

Final answer: 63
Subfield: Combinatorics


Question: Five equilateral triangles are drawn in the plane so that no two sides of any of the triangles are parallel. Compute the maximum number of points of intersection among all five triangles.

Solution:
Any two of the triangles intersect in at most six points, because each side of one triangle can intersect the other triangle in at most two points. To count the total number of intersections among the five triangles, note that there are $\left(\begin{array}{l}5 \\ 2\end{array}\right)=10$ ways to select a pair of triangles, and each pair may result in 6 intersections. Thus $10 \times 6=60$ is an upper bound.

This can be achieved, for example, by taking six equilateral triangles of equal size, centered at a single point, and rotating them different amounts so that no three sides intersect at a single point. Thus the answer is 60.

Final answer: 60
Subfield: Combinatorics


Question: $\quad$ Let $S$ be the set of four-digit positive integers for which the sum of the squares of their digits is 17 . For example, $2023 \in S$ because $2^{2}+0^{2}+2^{2}+3^{2}=17$. Compute the median of $S$.

Solution:
In order for the sums of the squares of four digits to be 17 , the digits must be either $0,2,2$, and 3 , or $0,0,1$, and 4 , in some order. If the leading digit is 2 , there are $3 !=6$ possible four-digit numbers. If the leading digit is 1,3 , or 4 , there are $\frac{3 !}{2 !}=3$ possible four-digit numbers. In total, there are $6+3 \cdot 3=15$ four-digit integers in $S$, and the median will be the eighth least. The least eight integers in $S$, from least to greatest, are: 1004, 1040, 1400, 2023, 2032, 2203, 2230, 2302. Thus the median of $S$ is 2302.

Final answer: 2302
Subfield: Geometry


Question: Let $E U C L I D$ be a hexagon inscribed in a circle of radius 5 . Given that $E U=U C=L I=I D=6$, and $C L=D E$, compute $C L$.

Solution:
Let $C L=x$. Because the quadrilaterals $E U C L$ and $L I D E$ are congruent, $\overline{E L}$ is a diameter of the circle in which the hexagon is inscribed, so $E L=10$. Furthermore, because $\overline{E L}$ is a diameter of the circle, it follows that the inscribed $\angle E U L$ is a right angle, hence $U L=8$.



<img_3485>

Using Ptolemy's Theorem for cyclic quadrilaterals and the fact that $\triangle E C L$ is also a right triangle,

$$
\begin{aligned}
& U C \cdot E L+E U \cdot C L=E C \cdot U L \\
\Longrightarrow & 6(10+x)=8 \sqrt{100-x^{2}} \\
\Longrightarrow & 36(10+x)^{2}=64(10+x)(10-x) \\
\Longrightarrow & 6 \sqrt{10+x}=8 \sqrt{10-x} \\
\Longrightarrow & 36(10+x)=64(10-x) \\
\Longrightarrow & 360+36 x=640-64 x \\
\Longrightarrow & 100 x=280 \\
\Longrightarrow & x=\frac{\mathbf{1 4}}{\mathbf{5}} .
\end{aligned}
$$

Final answer: $\frac{14}{5}$