di-zhang-fdu/OlympiadExtended huggingface dataset:
Subfield: Number Theory


Question: Compute the integer $n$ such that $2009<n<3009$ and the sum of the odd positive divisors of $n$ is 1024 .

Solution:
Suppose that $n=2^{k} p_{1}^{a_{1}} \cdots p_{r}^{a_{r}}$, where the $p_{i}$ are distinct odd primes, $k$ is a nonnegative integer, and $a_{1}, \ldots, a_{r}$ are positive integers. Then the sum of the odd positive divisors of $n$ is equal to

$$
\prod_{i=1}^{r}\left(1+p_{i}+\cdots+p_{i}^{a_{i}}\right)=\prod_{i=1}^{r} \frac{p_{i}^{a_{i}+1}-1}{p_{i}-1}=1024=2^{10}
$$

Note that $1+p_{i}+\cdots+p_{i}^{a_{i}}$ is the sum of $a_{i}+1$ odd numbers. Because the product of those sums is a power of two, each sum must be even (in fact, a power of 2). Thus, each $a_{i}$ must be odd.

Because $1+11+11^{2}+11^{3}>1024$, if $p_{i} \geq 11$, then $a_{i}=1$ and $1+p_{i}$ must be a power of 2 that is no greater than 1024. The possible values of $p_{i}$, with $p_{i} \geq 11$, are 31 and 127 (as 5 divides 255, 7 divides 511, and 3 divides 1023).

If $p_{1}<11$, then $p_{i}$ can be $3,5,7$. It is routine to check that $a_{i}=1$ and $p_{i}=3$ or 7 .

Thus $a_{i}=1$ for all $i$, and the possible values of $p_{i}$ are $3,7,31,127$. The only combinations of these primes that yield 1024 are $(1+3) \cdot(1+7) \cdot(1+31)\left(\right.$ with $\left.n=2^{k} \cdot 3 \cdot 7 \cdot 31=651 \cdot 2^{k}\right)$ and $(1+7) \cdot(1+127)$ (with $n=7 \cdot 127=889 \cdot 2^{k}$ ). Thus $n=651 \cdot 2^{2}=\mathbf{2 6 0 4}$ is the unique value of $n$ satisfying the conditions of the problem.

Final answer: 2604
Subfield: Geometry


Question: Points $A, R, M$, and $L$ are consecutively the midpoints of the sides of a square whose area is 650. The coordinates of point $A$ are $(11,5)$. If points $R, M$, and $L$ are all lattice points, and $R$ is in Quadrant I, compute the number of possible ordered pairs $(x, y)$ of coordinates for point $R$.

Solution:
Write $x=11+c$ and $y=5+d$. Then $A R^{2}=c^{2}+d^{2}=\frac{1}{2} \cdot 650=325$. Note that $325=18^{2}+1^{2}=17^{2}+6^{2}=15^{2}+10^{2}$. Temporarily restricting ourselves to the case where $c$ and $d$ are both positive, there are three classes of solutions: $\{c, d\}=\{18,1\},\{c, d\}=\{17,6\}$, or $\{c, d\}=\{15,10\}$. In fact, $c$ and $d$ can be negative, so long as those values do not cause $x$ or $y$ to be negative. So there are 10 solutions:

| $(c, d)$ | $(x, y)$ |
| :---: | :---: |
| $(18,1)$ | $(29,6)$ |
| $(18,-1)$ | $(29,4)$ |
| $(1,18)$ | $(12,23)$ |
| $(-1,18)$ | $(10,23)$ |
| $(17,6)$ | $(28,11)$ |
| $(6,17)$ | $(17,22)$ |
| $(-6,17)$ | $(5,22)$ |
| $(15,10)$ | $(26,15)$ |
| $(10,15)$ | $(21,20)$ |
| $(-10,15)$ | $(1,20)$ |

Final answer: 10
Subfield: Geometry


Question: The taxicab distance between points $\left(x_{1}, y_{1}, z_{1}\right)$ and $\left(x_{2}, y_{2}, z_{2}\right)$ is given by

$$
d\left(\left(x_{1}, y_{1}, z_{1}\right),\left(x_{2}, y_{2}, z_{2}\right)\right)=\left|x_{1}-x_{2}\right|+\left|y_{1}-y_{2}\right|+\left|z_{1}-z_{2}\right| .
$$

The region $\mathcal{R}$ is obtained by taking the cube $\{(x, y, z): 0 \leq x, y, z \leq 1\}$ and removing every point whose taxicab distance to any vertex of the cube is less than $\frac{3}{5}$. Compute the volume of $\mathcal{R}$.

Solution:
For a fixed vertex $V$ on the cube, the locus of points on or inside the cube that are at most $\frac{3}{5}$ away from $V$ form a corner at $V$ (that is, the right pyramid $V W_{1} W_{2} W_{3}$ in the figure shown at left below, with equilateral triangular base $W_{1} W_{2} W_{3}$ and three isosceles right triangular lateral faces $V W_{1} W_{2}, V W_{2} W_{3}, V W_{3} W_{1}$ ). Thus $\mathcal{R}$ is formed by removing eight such congruent corners from the cube. However, each two neighboring corners share a common region along their shared edge. This common region is the union of two smaller right pyramids, each similar to the original corners. (See the figure shown at right below.)
<img_4047>

We compute the volume of $\mathcal{R}$ as

$$
1-8 \cdot \frac{1}{6}\left(\frac{3}{5}\right)^{3}+12 \cdot 2 \cdot \frac{1}{6}\left(\frac{1}{10}\right)^{3}=\frac{\mathbf{1 7 9}}{\mathbf{2 5 0}}
$$

Final answer: $\frac{179}{250}$
Subfield: Algebra


Question: $\quad$ Let $a$ and $b$ be real numbers such that

$$
a^{3}-15 a^{2}+20 a-50=0 \quad \text { and } \quad 8 b^{3}-60 b^{2}-290 b+2575=0
$$

Compute $a+b$.

Solution:
Each cubic expression can be depressed - that is, the quadratic term can be eliminated-by substituting as follows. Because $(a-p)^{3}=a^{3}-3 a^{2} p+3 a p^{2}-p^{3}$, setting $p=-\frac{(-15)}{3}=5$ and substituting $c+p=a$ transforms the expression $a^{3}-15 a^{2}+20 a-50$ into the equivalent expression $(c+5)^{3}-15(c+5)^{2}+20(c+5)-50$, which simplifies to $c^{3}-55 c-200$. Similarly, the substitution $d=b-\frac{5}{2}$ yields the equation $d^{3}-55 d=-200$. [This procedure, which is analogous to completing the square, is an essential step in the algebraic solution to the general cubic equation.]

Consider the function $f(x)=x^{3}-55 x$. It has three zeros, namely, 0 and $\pm \sqrt{55}$. Therefore, it has a relative maximum and a relative minimum in the interval $[-\sqrt{55}, \sqrt{55}]$. Note that for $0 \leq x \leq 5.5,|f(x)|<\left|x^{3}\right|<5.5^{3}=166.375$, and for $5.5<x \leq \sqrt{55}<8$, we have

$$
|f(x)|=\left|x^{3}-55 x\right|<x\left|x^{2}-55\right|<8\left(55-5.5^{2}\right)=198
$$

Because $f(x)$ is an odd function of $x$ (its graph is symmetric about the origin), we conclude that for $-\sqrt{55} \leq x \leq \sqrt{55},|f(x)|<198$. Therefore, for constant $m$ with $|m|>198$, there is a unique real number $x_{0}$ such that $f\left(x_{0}\right)=m$.

In particular, since $200>198$, the values of $c$ and $d$ are uniquely determined. Because $f(x)$ is odd, we conclude that $c=-d$, or $a+b=\frac{\mathbf{1 5}}{\mathbf{2}}$.
Set $a=x-b$ and substitute into the first equation. We get

$$
\begin{aligned}
(x-b)^{3}-15(x-b)^{2}+20(x-b)-50 & =0 \\
-b^{3}+b^{2}(3 x-15)+b\left(-3 x^{2}+30 x-20\right)+\left(x^{3}-15 x^{2}+20 x-50\right) & =0 \\
8 b^{3}+b^{2}(-24 x+120)+b\left(24 x^{2}-240 x+160\right)-8\left(x^{3}-15 x^{2}+20 x-50\right) & =0 .
\end{aligned}
$$

If we equate coefficients, we see that

$$
\begin{aligned}
-24 x+120 & =-60 \\
24 x^{2}-240 x+160 & =-290 \\
-8\left(x^{3}-15 x^{2}+20 x-50\right) & =2575
\end{aligned}
$$

are all satisfied by $x=\frac{15}{2}$. This means that any real solution $b$ to the second equation yields a real solution of $\frac{15}{2}-b$ to the first equation. We can follow the reasoning of the previous solution to establish the existence of exactly one real solution to the second cubic equation. Thus $a$ and $b$ are unique, and their sum is $\left(\frac{15}{2}-b\right)+b=\frac{\mathbf{1 5}}{\mathbf{2}}$.

Final answer: $\frac{15}{2}$
Subfield: Number Theory


Question: For a positive integer $n$, define $s(n)$ to be the sum of $n$ and its digits. For example, $s(2009)=2009+2+0+0+9=2020$. Compute the number of elements in the set $\{s(0), s(1), s(2), \ldots, s(9999)\}$.

Solution:
If $s(10 x)=a$, then the values of $s$ over $\{10 x+0,10 x+1, \ldots, 10 x+9\}$ are $a, a+2, a+4, \ldots, a+18$. Furthermore, if $x$ is not a multiple of 10 , then $s(10(x+1))=a+11$. This indicates that the values of $s$ "interweave" somewhat from one group of 10 to the next: the sets alternate between even and odd. Because the $s$-values for starting blocks of ten differ by 11, consecutive blocks of the same parity differ by 22 , so the values of $s$ do not overlap. That is, $s$ takes on 100 distinct values over any range of the form $\{100 y+0,100 y+1, \ldots, 100 y+99\}$.

First determine how many values are repeated between consecutive hundreds. Let $y$ be an integer that is not a multiple of 10 . Then the largest value for $s(100 y+k)(0 \leq k \leq 99)$ is $100 y+(s(y)-y)+99+s(99)=100 y+s(y)-y+117$, whereas the smallest value in the next group of 100 is for

$$
\begin{aligned}
s(100(y+1)) & =100(y+1)+(s(y+1)-(y+1))=100 y+(s(y)+2)-(y+1)+100 \\
& =100 y+s(y)-y+101
\end{aligned}
$$

This result implies that the values for $s(100 y+91)$ through $s(100 y+99)$ match the values of $s(100 y+100)$ through $s(100 y+108)$. So there are 9 repeated values.

Now determine how many values are repeated between consecutive thousands. Let $z$ be a digit, and consider $s(1000 z+999)$ versus $s(1000(z+1))$. The first value equals

$$
1000 z+(s(z)-z)+999+s(999)=1000 z+z+1026=1001 z+1026
$$

The latter value equals $1000(z+1)+(s(z+1)-(z+1))=1001(z+1)=1001 z+1001$. These values differ by an odd number. We have overlap between the $982,983, \ldots, 989$ terms and the $000,001, \ldots, 007$ terms. We also have overlap between the $992,993, \ldots, 999$ terms and the $010,011, \ldots, 017$ terms, for a total of 16 repeated values in all.

There are 90 instances in which we have 9 repeated terms, and 9 instances in which we have 16 repeated terms, so there are a total of $10000-90 \cdot 9-9 \cdot 16=\mathbf{9 0 4 6}$ unique values.

Final answer: 9046
Subfield: Geometry


Question: Quadrilateral $A R M L$ is a kite with $A R=R M=5, A M=8$, and $R L=11$. Compute $A L$.

Solution:
Let $K$ be the midpoint of $\overline{A M}$. Then $A K=K M=8 / 2=4, R K=\sqrt{5^{2}-4^{2}}=3$, and $K L=11-3=8$. Thus $A L=\sqrt{A K^{2}+K L^{2}}=\sqrt{4^{2}+8^{2}}=4 \sqrt{5}$.

Final answer: $4 \sqrt{5}$
Subfield: Algebra


Question: Let $T=4 \sqrt{5}$. If $x y=\sqrt{5}, y z=5$, and $x z=T$, compute the positive value of $x$.

Solution:
Multiply the three given equations to obtain $x^{2} y^{2} z^{2}=5 T \sqrt{5}$. Thus $x y z= \pm \sqrt[4]{125 T^{2}}$, and the positive value of $x$ is $x=x y z / y z=\sqrt[4]{125 T^{2}} / 5=\sqrt[4]{T^{2} / 5}$. With $T=4 \sqrt{5}$, we have $x=\mathbf{2}$.

Final answer: 2
Subfield: Combinatorics


Question: $\quad$ Let $T=2$. In how many ways can $T$ boys and $T+1$ girls be arranged in a row if all the girls must be standing next to each other?

Solution:
First choose the position of the first girl, starting from the left. There are $T+1$ possible positions, and then the positions for the girls are all determined. There are $(T+1)$ ! ways to arrange the girls, and there are $T$ ! ways to arrange the boys, for a total of $(T+1) \cdot(T+1) ! \cdot T !=$ $((T+1) !)^{2}$ arrangements. With $T=2$, the answer is $\mathbf{3 6}$.

Final answer: 36
Subfield: Geometry


Question: $\triangle A B C$ is on a coordinate plane such that $A=(3,6)$, $B=(T, 0)$, and $C=(2 T-1,1-T)$. Let $\ell$ be the line containing the altitude to $\overline{B C}$. Compute the $y$-intercept of $\ell$.

Solution:
The slope of $\overleftrightarrow{B C}$ is $\frac{(1-T)-0}{(2 T-1)-T}=-1$, and since $\ell$ is perpendicular to $\overleftrightarrow{B C}$, the slope of $\ell$ is 1. Because $\ell$ passes through $A=(3,6)$, the equation of $\ell$ is $y=x+3$, and its $y$-intercept is 3 (independent of $T$ ).

Final answer: 3
Subfield: Geometry


Question: Let $T=3$. In triangle $A B C, A B=A C-2=T$, and $\mathrm{m} \angle A=60^{\circ}$. Compute $B C^{2}$.

Solution:
By the Law of Cosines, $B C^{2}=A B^{2}+A C^{2}-2 \cdot A B \cdot A C \cdot \cos A=T^{2}+(T+2)^{2}-2 \cdot T \cdot(T+2) \cdot \frac{1}{2}=$ $T^{2}+2 T+4$. With $T=3$, the answer is 19 .

Final answer: 19
Subfield: Algebra


Question: Let $T=19$. Let $\mathcal{S}_{1}$ denote the arithmetic sequence $0, \frac{1}{4}, \frac{1}{2}, \ldots$, and let $\mathcal{S}_{2}$ denote the arithmetic sequence $0, \frac{1}{6}, \frac{1}{3}, \ldots$ Compute the $T^{\text {th }}$ smallest number that occurs in both sequences $\mathcal{S}_{1}$ and $\mathcal{S}_{2}$.

Solution:
$\mathcal{S}_{1}$ consists of all numbers of the form $\frac{n}{4}$, and $\mathcal{S}_{2}$ consists of all numbers of the form $\frac{n}{6}$, where $n$ is a nonnegative integer. Since $\operatorname{gcd}(4,6)=2$, the numbers that are in both sequences are of the form $\frac{n}{2}$, and the $T^{\text {th }}$ smallest such number is $\frac{T-1}{2}$. With $T=19$, the answer is 9 .

Final answer: 9
Subfield: Combinatorics


Question: $\quad$ Let $T=9$. An integer $n$ is randomly selected from the set $\{1,2,3, \ldots, 2 T\}$. Compute the probability that the integer $\left|n^{3}-7 n^{2}+13 n-6\right|$ is a prime number.

Solution:
Let $P(n)=n^{3}-7 n^{2}+13 n-6$, and note that $P(n)=(n-2)\left(n^{2}-5 n+3\right)$. Thus $|P(n)|$ is prime if either $|n-2|=1$ and $\left|n^{2}-5 n+3\right|$ is prime or if $\left|n^{2}-5 n+3\right|=1$ and $|n-2|$ is prime. Solving $|n-2|=1$ gives $n=1$ or 3 , and solving $\left|n^{2}-5 n+3\right|=1$ gives $n=1$ or 4 or $\frac{5 \pm \sqrt{17}}{2}$. Note that $P(1)=1, P(3)=-3$, and $P(4)=-2$. Thus $|P(n)|$ is prime only when $n$ is 3 or 4 , and if $T \geq 2$, then the desired probability is $\frac{2}{2 T}=\frac{1}{T}$. With $T=9$, the answer is $\frac{\mathbf{1}}{\mathbf{9}}$.

Final answer: $\frac{1}{9}$
Subfield: Algebra


Question: Let $A=\frac{1}{9}$, and let $B=\frac{1}{25}$. In $\frac{1}{A}$ minutes, 20 frogs can eat 1800 flies. At this rate, in $\frac{1}{B}$ minutes, how many flies will 15 frogs be able to eat?

Solution:
In $\frac{1}{A}$ minutes, 1 frog can eat $1800 / 20=90$ flies; thus in $\frac{1}{B}$ minutes, 1 frog can eat $\frac{A}{B} \cdot 90$ flies. Thus in $\frac{1}{B}$ minutes, 15 frogs can eat $15 \cdot 90 \cdot \frac{A}{B}$ flies. With $A=\frac{1}{9}$ and $B=\frac{1}{25}$, this simplifies to $15 \cdot 250=\mathbf{3 7 5 0}$.

Final answer: 3750
Subfield: Algebra


Question: Let $T=5$. If $|T|-1+3 i=\frac{1}{z}$, compute the sum of the real and imaginary parts of $z$.

Solution:
Let $t=|T|$. Note that $z=\frac{1}{t-1+3 i}=\frac{1}{t-1+3 i} \cdot \frac{t-1-3 i}{t-1-3 i}=\frac{t-1-3 i}{t^{2}-2 t+10}$. Thus the sum of the real and imaginary parts of $z$ is $\frac{t-1}{t^{2}-2 t+10}+\frac{-3}{t^{2}-2 t+10}=\frac{|T|-4}{|T|^{2}-2|T|+10}$. With $T=5$, the answer is $\frac{1}{25}$.

Final answer: $\frac{1}{25}$
Subfield: Geometry


Question: Let $T=80$. In circle $O$, diagrammed at right, minor arc $\widehat{A B}$ measures $\frac{T}{4}$ degrees. If $\mathrm{m} \angle O A C=10^{\circ}$ and $\mathrm{m} \angle O B D=5^{\circ}$, compute the degree measure of $\angle A E B$. Just pass the number without the units.

![](https://cdn.mathpix.com/cropped/2023_12_21_8fe750e0fde7a21c6855g-1.jpg?height=293&width=477&top_left_y=2111&top_left_x=1472)

Solution:
Note that $\mathrm{m} \angle A E B=\frac{1}{2}(\mathrm{~m} \widehat{A B}-m \widehat{C D})=\frac{1}{2}(\mathrm{~m} \widehat{A B}-\mathrm{m} \angle C O D)$. Also note that $\mathrm{m} \angle C O D=$ $360^{\circ}-(\mathrm{m} \angle A O C+\mathrm{m} \angle B O D+\mathrm{m} \angle A O B)=360^{\circ}-\left(180^{\circ}-2 \mathrm{~m} \angle O A C\right)-\left(180^{\circ}-2 \mathrm{~m} \angle O B D\right)-$ $\mathrm{m} \widehat{A B}=2(\mathrm{~m} \angle O A C+\mathrm{m} \angle O B D)-\mathrm{m} \widehat{A B}$. Thus $\mathrm{m} \angle A E B=\mathrm{m} \widehat{A B}-\mathrm{m} \angle O A C-\mathrm{m} \angle O B D=$ $\frac{T}{4}-10^{\circ}-5^{\circ}$, and with $T=80$, the answer is 5 .

Final answer: 5
Subfield: Geometry


Question: Let $T=80$. In circle $O$, diagrammed at right, minor arc $\widehat{A B}$ measures $\frac{T}{4}$ degrees. If $\mathrm{m} \angle O A C=10^{\circ}$ and $\mathrm{m} \angle O B D=5^{\circ}$, compute the degree measure of $\angle A E B$. Just pass the number without the units.

<img_3347>

Solution:
Note that $\mathrm{m} \angle A E B=\frac{1}{2}(\mathrm{~m} \widehat{A B}-m \widehat{C D})=\frac{1}{2}(\mathrm{~m} \widehat{A B}-\mathrm{m} \angle C O D)$. Also note that $\mathrm{m} \angle C O D=$ $360^{\circ}-(\mathrm{m} \angle A O C+\mathrm{m} \angle B O D+\mathrm{m} \angle A O B)=360^{\circ}-\left(180^{\circ}-2 \mathrm{~m} \angle O A C\right)-\left(180^{\circ}-2 \mathrm{~m} \angle O B D\right)-$ $\mathrm{m} \widehat{A B}=2(\mathrm{~m} \angle O A C+\mathrm{m} \angle O B D)-\mathrm{m} \widehat{A B}$. Thus $\mathrm{m} \angle A E B=\mathrm{m} \widehat{A B}-\mathrm{m} \angle O A C-\mathrm{m} \angle O B D=$ $\frac{T}{4}-10^{\circ}-5^{\circ}$, and with $T=80$, the answer is 5 .

Final answer: 5
Subfield: Algebra


Question: Let $T=10$. Ann spends 80 seconds climbing up a $T$ meter rope at a constant speed, and she spends 70 seconds climbing down the same rope at a constant speed (different from her upward speed). Ann begins climbing up and down the rope repeatedly, and she does not pause after climbing the length of the rope. After $T$ minutes, how many meters will Ann have climbed in either direction?

Solution:
In 150 seconds (or 2.5 minutes), Ann climbs up and down the entire rope. Thus in $T$ minutes, she makes $\left\lfloor\frac{T}{2.5}\right\rfloor$ round trips, and therefore climbs $2 T\left\lfloor\frac{T}{2.5}\right\rfloor$ meters. After making all her round trips, there are $t=60\left(T-2.5\left\lfloor\frac{T}{2.5}\right\rfloor\right)$ seconds remaining. If $t \leq 80$, then the remaining distance climbed is $T \cdot \frac{t}{80}$ meters, and if $t>80$, then the distance climbed is $T+T \cdot\left(\frac{t-80}{70}\right)$ meters. In general, the total distance in meters that Ann climbs is

$$
2 T\left\lfloor\frac{T}{2.5}\right\rfloor+T \cdot \min \left(1, \frac{60\left(T-2.5\left\lfloor\frac{T}{2.5}\right\rfloor\right)}{80}\right)+T \cdot \max \left(0, \frac{60\left(T-2.5\left\lfloor\frac{T}{2.5}\right\rfloor\right)-80}{70}\right) .
$$

With $T=10$, Ann makes exactly 4 round trips, and therefore climbs a total of $4 \cdot 2 \cdot 10=\mathbf{8 0}$ meters.

Final answer: 80
Subfield: Algebra


Question: Let $T=800$. Simplify $2^{\log _{4} T} / 2^{\log _{16} 64}$.

Solution:
Note that $2^{\log _{4} T}=4^{\left(\frac{1}{2} \log _{4} T\right)}=4^{\log _{4} T^{\frac{1}{2}}}=\sqrt{T}$. Letting $\log _{16} 64=x$, we see that $2^{4 x}=2^{6}$, thus $x=\frac{3}{2}$, and $2^{x}=\sqrt{8}$. Thus the given expression equals $\sqrt{\frac{T}{8}}$, and with $T=800$, this is equal to 10 .

Final answer: 10
Subfield: Algebra


Question: Let $P(x)=x^{2}+T x+800$, and let $r_{1}$ and $r_{2}$ be the roots of $P(x)$. The polynomial $Q(x)$ is quadratic, it has leading coefficient 1, and it has roots $r_{1}+1$ and $r_{2}+1$. Find the sum of the coefficients of $Q(x)$.

Solution:
Let $Q(x)=x^{2}+A x+B$. Then $A=-\left(r_{1}+1+r_{2}+1\right)$ and $B=\left(r_{1}+1\right)\left(r_{2}+1\right)$. Thus the sum of the coefficients of $Q(x)$ is $1+\left(-r_{1}-r_{2}-2\right)+\left(r_{1} r_{2}+r_{1}+r_{2}+1\right)=r_{1} r_{2}$. Note that $T=-\left(r_{1}+r_{2}\right)$ and $800=r_{1} r_{2}$, so the answer is $\mathbf{8 0 0}$ (independent of $T$ ). [Note: With $T=108,\left\{r_{1}, r_{2}\right\}=\{-8,-100\}$.

Final answer: 800
Subfield: Geometry


Question: Let $T=12$. Equilateral triangle $A B C$ is given with side length $T$. Points $D$ and $E$ are the midpoints of $\overline{A B}$ and $\overline{A C}$, respectively. Point $F$ lies in space such that $\triangle D E F$ is equilateral and $\triangle D E F$ lies in a plane perpendicular to the plane containing $\triangle A B C$. Compute the volume of tetrahedron $A B C F$.

Solution:
The volume of tetrahedron $A B C F$ is one-third the area of $\triangle A B C$ times the distance from $F$ to $\triangle A B C$. Since $D$ and $E$ are midpoints, $D E=\frac{B C}{2}=\frac{T}{2}$, and the distance from $F$ to $\triangle A B C$ is $\frac{T \sqrt{3}}{4}$. Thus the volume of $A B C F$ is $\frac{1}{3} \cdot \frac{T^{2} \sqrt{3}}{4} \cdot \frac{T \sqrt{3}}{4}=\frac{T^{3}}{16}$. With $T=12$, the answer is $\mathbf{1 0 8}$.

Final answer: 108
Subfield: Geometry


Question: In triangle $A B C, A B=5, A C=6$, and $\tan \angle B A C=-\frac{4}{3}$. Compute the area of $\triangle A B C$.

Solution:
Let $s=\sin \angle B A C$. Then $s>0$ and $\frac{s}{-\sqrt{1-s^{2}}}=-\frac{4}{3}$, which gives $s=\frac{4}{5}$. The area of triangle $A B C$ is therefore $\frac{1}{2} \cdot A B \cdot A C \cdot \sin \angle B A C=\frac{1}{2} \cdot 5 \cdot 6 \cdot \frac{4}{5}=\mathbf{1 2}$.

Final answer: 12
Subfield: Number Theory


Question: Compute the number of positive integers less than 25 that cannot be written as the difference of two squares of integers.

Solution:
Suppose $n=a^{2}-b^{2}=(a+b)(a-b)$, where $a$ and $b$ are integers. Because $a+b$ and $a-b$ differ by an even number, they have the same parity. Thus $n$ must be expressible as the product of two even integers or two odd integers. This condition is sufficient for $n$ to be a difference of squares, because if $n$ is odd, then $n=(k+1)^{2}-k^{2}=(2 k+1) \cdot 1$ for some integer $k$, and if $n$ is a multiple of 4 , then $n=(k+1)^{2}-(k-1)^{2}=2 k \cdot 2$ for some integer $k$. Therefore any integer of the form $4 k+2$ for integral $k$ cannot be expressed as the difference of two squares of integers, hence the desired integers in the given range are $2,6,10,14,18$, and 22 , for a total of 6 values.
Suppose that an integer $n$ can be expressed as the difference of squares of two integers, and let the squares be $a^{2}$ and $(a+b)^{2}$, with $a, b \geq 0$. Then

$$
\begin{aligned}
& n=(a+b)^{2}-a^{2}=2 a b+b^{2} \\
& =2 a+1 \quad(b=1) \\
& =4 a+4 \quad(b=2) \\
& =6 a+9 \quad(b=3) \\
& =8 a+16 \quad(b=4) \\
& =10 a+25 \quad(b=5) .
\end{aligned}
$$

Setting $b=1$ generates all odd integers. If $b=3$ or $b=5$, then the values of $n$ are still odd, hence are already accounted for. If $b=2$, then the values of $4 a+4=4(a+1)$ yield all multiples of $4 ; b=8$ yields multiples of 8 (hence are already accounted for). The remaining integers are even numbers that are not multiples of $4: 2,6,10,14,18,22$, for a total of 6 such numbers.

Final answer: 6
Subfield: Algebra


Question: For digits $A, B$, and $C,(\underline{A} \underline{B})^{2}+(\underline{A} \underline{C})^{2}=1313$. Compute $A+B+C$.

Solution:
Because $10 A \leq \underline{A} \underline{B}<10(A+1), 200 A^{2}<(\underline{A} \underline{B})^{2}+(\underline{A} \underline{C})^{2}<200(A+1)^{2}$. So $200 A^{2}<$ $1313<200(A+1)^{2}$, and $A=2$. Note that $B$ and $C$ must have opposite parity, so without loss of generality, assume that $B$ is even. Consider the numbers modulo 10: for any integer $n, n^{2} \equiv 0,1,4,5,6$, or $9 \bmod 10$. The only combination whose sum is congruent to $3 \bmod 10$ is $4+9$. So $B=2$ or 8 and $C=3$ or 7 . Checking cases shows that $28^{2}+23^{2}=1313$, so $B=8, C=3$, and $A+B+C=\mathbf{1 3}$.
Rewrite $1313=13 \cdot 101=\left(3^{2}+2^{2}\right)\left(10^{2}+1^{2}\right)$. The two-square identity states:

$$
\begin{aligned}
\left(a^{2}+b^{2}\right)\left(x^{2}+y^{2}\right) & =(a x+b y)^{2}+(a y-b x)^{2} \\
& =(a y+b x)^{2}+(a x-b y)^{2}
\end{aligned}
$$

Therefore

$$
\begin{aligned}
1313=(30+2)^{2}+(3-20)^{2} & =32^{2}+17^{2} \\
& =(3+20)^{2}+(30-2)^{2}=23^{2}+28^{2}
\end{aligned}
$$



Hence $A=2, B=3, C=8$, and $A+B+C=\mathbf{1 3}$.

Final answer: 13
Subfield: Geometry


Question: Points $P, Q, R$, and $S$ lie in the interior of square $A B C D$ such that triangles $A B P, B C Q$, $C D R$, and $D A S$ are equilateral. If $A B=1$, compute the area of quadrilateral $P Q R S$.

Solution:
$P Q R S$ is a square with diagonal $\overline{R P}$. Extend $\overline{R P}$ to intersect $\overline{A B}$ and $\overline{C D}$ at $M$ and $N$ respectively, as shown in the diagram below.

<img_3457>

Then $\overline{M P}$ is an altitude of $\triangle A B P$ and $\overline{R N}$ is an altitude of $\triangle C D R$. Adding lengths, $M P+R N=M R+2 R P+P N=1+R P$, so $R P=\sqrt{3}-1$. Therefore $[P Q R S]=\frac{1}{2}(R P)^{2}=$ $2-\sqrt{3}$.

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


Question: For real numbers $\alpha, B$, and $C$, the zeros of $T(x)=x^{3}+x^{2}+B x+C \operatorname{are~}^{2} \alpha$, $\cos ^{2} \alpha$, and $-\csc ^{2} \alpha$. Compute $T(5)$.

Solution:
Use the sum of the roots formula to obtain $\sin ^{2} \alpha+\cos ^{2} \alpha+-\csc ^{2} \alpha=-1$, so $\csc ^{2} \alpha=2$, and $\sin ^{2} \alpha=\frac{1}{2}$. Therefore $\cos ^{2} \alpha=\frac{1}{2}$. T(x) has leading coefficient 1 , so by the factor theorem, $T(x)=\left(x-\frac{1}{2}\right)\left(x-\frac{1}{2}\right)(x+2)$. Then $T(5)=\left(5-\frac{1}{2}\right)\left(5-\frac{1}{2}\right)(5+2)=\frac{567}{4}$.

Final answer: $\frac{567}{4}$