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


Question: Let $T=6$. In the square $D E F G$ diagrammed at right, points $M$ and $N$ trisect $\overline{F G}$, points $A$ and $B$ are the midpoints of $\overline{E F}$ and $\overline{D G}$, respectively, and $\overline{E M} \cap \overline{A B}=S$ and $\overline{D N} \cap \overline{A B}=H$. If the side length of square $D E F G$ is $T$, compute $[D E S H]$.

![](https://cdn.mathpix.com/cropped/2023_12_21_da8b9ad9706d1f3f1497g-1.jpg?height=372&width=390&top_left_y=909&top_left_x=1540)

Solution:
Note that $D E S H$ is a trapezoid with height $\frac{T}{2}$. Because $\overline{A S}$ and $\overline{B H}$ are midlines of triangles $E F M$ and $D G N$ respectively, it follows that $A S=B H=\frac{T}{6}$. Thus $S H=T-2 \cdot \frac{T}{6}=\frac{2 T}{3}$. Thus $[D E S H]=\frac{1}{2}\left(T+\frac{2 T}{3}\right) \cdot \frac{T}{2}=\frac{5 T^{2}}{12}$. With $T=6$, the desired area is 15 .

Final answer: 15
Subfield: Algebra


Question: Let $T=15$. For complex $z$, define the function $f_{1}(z)=z$, and for $n>1, f_{n}(z)=$ $f_{n-1}(\bar{z})$. If $f_{1}(z)+2 f_{2}(z)+3 f_{3}(z)+4 f_{4}(z)+5 f_{5}(z)=T+T i$, compute $|z|$.

Solution:
Because $\overline{\bar{z}}=z$, it follows that $f_{n}(z)=z$ when $n$ is odd, and $f_{n}(z)=\bar{z}$ when $n$ is even. Taking $z=a+b i$, where $a$ and $b$ are real, it follows that $\sum_{k=1}^{5} k f_{k}(z)=15 a+3 b i$. Thus $a=\frac{T}{15}, b=\frac{T}{3}$, and $|z|=\sqrt{a^{2}+b^{2}}=\frac{|T| \sqrt{26}}{15}$. With $T=15$, the answer is $\sqrt{\mathbf{2 6}}$.

Final answer: $\sqrt{26}$
Subfield: Number Theory


Question: Let $T=\sqrt{26}$. Compute the number of ordered pairs of positive integers $(a, b)$ with the property that $a b=T^{20} \cdot 210^{12}$, and the greatest common divisor of $a$ and $b$ is 1 .

Solution:
If the prime factorization of $a b$ is $p_{1}^{e_{1}} p_{2}^{e_{2}} \ldots p_{k}^{e_{k}}$, where the $p_{i}$ 's are distinct primes and the $e_{i}$ 's are positive integers, then in order for $\operatorname{gcd}(a, b)$ to equal 1 , each $p_{i}$ must be a divisor of exactly one of $a$ or $b$. Thus the desired number of ordered pairs is $2^{k}$ because there are 2 choices for each prime divisor (i.e., $p_{i} \mid a$ or $p_{i} \mid b$ ). With $T=\sqrt{26}$, it follows that $(\sqrt{26})^{20} \cdot 210^{12}=\left(2^{10} \cdot 13^{10}\right) \cdot 210^{12}=2^{22} \cdot 3^{12} \cdot 5^{12} \cdot 7^{12} \cdot 13^{10}$. Thus there are five distinct prime divisors, and the answer is $2^{5}=\mathbf{3 2}$.

Final answer: 32
Subfield: Algebra


Question: Let $T=32$. Given that $\sin \theta=\frac{\sqrt{T^{2}-64}}{T}$, compute the largest possible value of the infinite series $\cos \theta+\cos ^{2} \theta+\cos ^{3} \theta+\ldots$.

Solution:
Using $\sin ^{2} \theta+\cos ^{2} \theta=1$ gives $\cos ^{2} \theta=\frac{64}{T^{2}}$, so to maximize the sum, take $\cos \theta=\frac{8}{|T|}$. Using the formula for the sum of an infinite geometric series gives $\frac{8 /|T|}{1-8 /|T|}=\frac{8}{|T|-8}$. With $T=32$, the answer is $\frac{8}{24}=\frac{1}{3}$.

Final answer: $\frac{1}{3}$
Subfield: Geometry


Question: Let $R$ be the larger number you will receive, and let $r$ be the smaller number you will receive. In the diagram at right (not drawn to scale), circle $D$ has radius $R$, circle $K$ has radius $r$, and circles $D$ and $K$ are tangent at $C$. Line $\overleftrightarrow{Y P}$ is tangent to circles $D$ and $K$. Compute $Y P$.

![](https://cdn.mathpix.com/cropped/2023_12_21_da8b9ad9706d1f3f1497g-1.jpg?height=304&width=531&top_left_y=1943&top_left_x=1472)

Solution:
Note that $\overline{D Y}$ and $\overline{K P}$ are both perpendicular to line $\overleftrightarrow{Y P}$. Let $J$ be the foot of the perpendicular from $K$ to $\overline{D Y}$. Then $P K J Y$ is a rectangle and $Y P=J K=\sqrt{D K^{2}-D J^{2}}=$ $\sqrt{(R+r)^{2}-(R-r)^{2}}=2 \sqrt{R r}$. With $R=450$ and $r=\frac{1}{3}$, the answer is $2 \sqrt{150}=\mathbf{1 0} \sqrt{\mathbf{6}}$.

Final answer: $10 \sqrt{6}$
Subfield: Geometry


Question: Let $R$ be the larger number you will receive, and let $r$ be the smaller number you will receive. In the diagram at right (not drawn to scale), circle $D$ has radius $R$, circle $K$ has radius $r$, and circles $D$ and $K$ are tangent at $C$. Line $\overleftrightarrow{Y P}$ is tangent to circles $D$ and $K$. Compute $Y P$.

<img_3573>

Solution:
Note that $\overline{D Y}$ and $\overline{K P}$ are both perpendicular to line $\overleftrightarrow{Y P}$. Let $J$ be the foot of the perpendicular from $K$ to $\overline{D Y}$. Then $P K J Y$ is a rectangle and $Y P=J K=\sqrt{D K^{2}-D J^{2}}=$ $\sqrt{(R+r)^{2}-(R-r)^{2}}=2 \sqrt{R r}$. With $R=450$ and $r=\frac{1}{3}$, the answer is $2 \sqrt{150}=\mathbf{1 0} \sqrt{\mathbf{6}}$.

Final answer: $10 \sqrt{6}$
Subfield: Geometry


Question: Let $T=\frac{9}{17}$. When $T$ is expressed as a reduced fraction, let $m$ and $n$ be the numerator and denominator, respectively. A square pyramid has base $A B C D$, the distance from vertex $P$ to the base is $n-m$, and $P A=P B=P C=P D=n$. Compute the area of square $A B C D$.

Solution:
By the Pythagorean Theorem, half the diagonal of the square is $\sqrt{n^{2}-(n-m)^{2}}=\sqrt{2 m n-m^{2}}$. Thus the diagonal of the square is $2 \sqrt{2 m n-m^{2}}$, and the square's area is $4 m n-2 m^{2}$. With $T=\frac{9}{17}, m=9, n=17$, and the answer is 450 .

Final answer: 450
Subfield: Combinatorics


Question: Let $T=-14$, and let $d=|T|$. A person whose birthday falls between July 23 and August 22 inclusive is called a Leo. A person born in July is randomly selected, and it is given that her birthday is before the $d^{\text {th }}$ day of July. Another person who was also born in July is randomly selected, and it is given that his birthday is after the $d^{\text {th }}$ day of July. Compute the probability that exactly one of these people is a Leo.

Solution:
Note that there are 9 days in July in which a person could be a Leo (July 23-31). Let the woman (born before the $d^{\text {th }}$ day of July) be called Carol, and let the man (born after the $d^{\text {th }}$ day of July) be called John, and consider the possible values of $d$. If $d \leq 21$, then Carol will not be a Leo, and the probability that John is a Leo is $\frac{9}{31-d}$. If $d=22$ or 23 , then the probability is 1 . If $d \geq 24$, then John will be a Leo, and Carol will not be a Leo with probability $1-\frac{d-23}{d-1}$. With $T=-14$, the first case applies, and the desired probability is $\frac{\mathbf{9}}{\mathbf{1 7}}$.

Final answer: $\frac{9}{17}$
Subfield: Algebra


Question: Let $T=-10$. Given that $\log _{2} 4^{8 !}+\log _{4} 2^{8 !}=6 ! \cdot T \cdot x$, compute $x$.

Solution:
Note that $4^{8 !}=2^{2 \cdot 8 !}$, thus $\log _{2} 4^{8 !}=2 \cdot 8$ !. Similarly, $\log _{4} 2^{8 !}=\frac{8 !}{2}$. Thus $2 \cdot 8 !+\frac{8 !}{2}=$ $6 !\left(2 \cdot 7 \cdot 8+7 \cdot \frac{8}{2}\right)=6 ! \cdot 140$. Thus $140=T x$, and with $T=-10, x=\mathbf{- 1 4}$.

Final answer: -14
Subfield: Algebra


Question: Let $T=20$. For some real constants $a$ and $b$, the solution sets of the equations $x^{2}+(5 b-T-a) x=T+1$ and $2 x^{2}+(T+8 a-2) x=-10 b$ are the same. Compute $a$.

Solution:
Divide each side of the second equation by 2 and equate coefficients to obtain $5 b-T-a=$ $\frac{T}{2}+4 a-1$ and $T+1=-5 b$. Thus $b=\frac{T+1}{-5}$, and plugging this value into the first equation yields $a=-\frac{T}{2}$. With $T=20$, the answer is $\mathbf{- 1 0}$.

Final answer: -10
Subfield: Algebra


Question: Let T be a rational number, and let $K=T-2$. If $K$ workers can produce 9 widgets in 1 hour, compute the number of workers needed to produce $\frac{720}{K}$ widgets in 4 hours.

Solution:
Because $T$ workers produce 9 widgets in 1 hour, 1 worker will produce $\frac{9}{T}$ widgets in 1 hour. Thus 1 worker will produce $\frac{36}{T}$ widgets in 4 hours. In order to produce $\frac{720}{T}$ widgets in 4 hours, it will require $\frac{720 / T}{36 / T}=\mathbf{2 0}$ workers (independent of $T$ ).

Final answer: 20
Subfield: Algebra


Question: Let $T=2018$, and append the digits of $T$ to $\underline{A} \underline{A} \underline{B}$ (for example, if $T=17$, then the result would be $\underline{1} \underline{\underline{A}} \underline{A} \underline{B}$ ). If the resulting number is divisible by 11 , compute the largest possible value of $A+B$.

Solution:
Let $R$ be the remainder when $T$ is divided by 11 . Note that the alternating sum of the digits of the number must be divisible by 11 . This sum will be congruent $\bmod 11$ to $B-A+A-R=$ $B-R$, thus $B=R$. Because $A$ 's value is irrelevant, to maximize $A+B$, set $A=9$ to yield $A+B=9+R$. For $T=2018, R=5$, and the answer is $9+5=\mathbf{1 4}$.

Final answer: 14
Subfield: Algebra


Question: Given that April $1^{\text {st }}, 2012$ fell on a Sunday, what is the next year in which April $1^{\text {st }}$ will fall on a Sunday?

Solution:
Note that $365=7 \cdot 52+1$. Thus over the next few years after 2012 , the day of the week for April $1^{\text {st }}$ will advance by one day in a non-leap year, and it will advance by two days in a leap year. Thus in six years, the day of the week will have rotated a complete cycle, and the answer is 2018 .

Final answer: 2018
Subfield: Number Theory


Question: Let $p$ be a prime number. If $p$ years ago, the ages of three children formed a geometric sequence with a sum of $p$ and a common ratio of 2 , compute the sum of the children's current ages.

Solution:
Let $x, 2 x$, and $4 x$ be the ages of the children $p$ years ago. Then $x+2 x+4 x=p$, so $7 x=p$. Since $p$ is prime, $x=1$. Thus the sum of the children's current ages is $(1+7)+(2+7)+(4+7)=\mathbf{2 8}$.

Final answer: 28
Subfield: Number Theory


Question: Define a reverse prime to be a positive integer $N$ such that when the digits of $N$ are read in reverse order, the resulting number is a prime. For example, the numbers 5, 16, and 110 are all reverse primes. Compute the largest two-digit integer $N$ such that the numbers $N, 4 \cdot N$, and $5 \cdot N$ are all reverse primes.

Solution:
Because $N<100,5 \cdot N<500$. Since no primes end in 4, it follows that $5 \cdot N<400$, hence $N \leq 79$. The reverses of $5 \cdot 79=395,4 \cdot 79=316$, and 79 are 593,613 , and 97 , respectively. All three of these numbers are prime, thus 79 is the largest two-digit integer $N$ for which $N$, $4 \cdot N$, and $5 \cdot N$ are all reverse primes.

Final answer: 79
Subfield: Combinatorics


Question: Some students in a gym class are wearing blue jerseys, and the rest are wearing red jerseys. There are exactly 25 ways to pick a team of three players that includes at least one player wearing each color. Compute the number of students in the class.

Solution:
Let $r$ and $b$ be the number of students wearing red and blue jerseys, respectively. Then either we choose two blues and one red or one blue and two reds. Thus

$$
\begin{aligned}
& \left(\begin{array}{l}
b \\
2
\end{array}\right)\left(\begin{array}{l}
r \\
1
\end{array}\right)+\left(\begin{array}{l}
b \\
1
\end{array}\right)\left(\begin{array}{l}
r \\
2
\end{array}\right)=25 \\
\Rightarrow & \frac{r b(b-1)}{2}+\frac{b r(r-1)}{2}=25 \\
\Rightarrow & r b((r-1)+(b-1))=50 \\
\Rightarrow & r b(r+b-2)=50 .
\end{aligned}
$$

Now because $r, b$, and $r+b-2$ are positive integer divisors of 50 , and $r, b \geq 2$, we have only a few possibilities to check. If $r=2$, then $b^{2}=25$, so $b=5$; the case $r=5$ is symmetric. If $r=10$, then $b(b+8)=5$, which is impossible. If $r=25$, then $b(b+23)=2$, which is also impossible. So $\{r, b\}=\{2,5\}$, and $r+b=7$.

Final answer: 7
Subfield: Geometry


Question: Point $P$ is on the hypotenuse $\overline{E N}$ of right triangle $B E N$ such that $\overline{B P}$ bisects $\angle E B N$. Perpendiculars $\overline{P R}$ and $\overline{P S}$ are drawn to sides $\overline{B E}$ and $\overline{B N}$, respectively. If $E N=221$ and $P R=60$, compute $\frac{1}{B E}+\frac{1}{B N}$.

Solution:
We observe that $\frac{1}{B E}+\frac{1}{B N}=\frac{B E+B N}{B E \cdot B N}$. The product in the denominator suggests that we compare areas. Let $[B E N]$ denote the area of $\triangle B E N$. Then $[B E N]=\frac{1}{2} B E \cdot B N$, but because $P R=P S=60$, we can also write $[B E N]=[B E P]+[B N P]=\frac{1}{2} \cdot 60 \cdot B E+\frac{1}{2} \cdot 60 \cdot B N$. Therefore $B E \cdot B N=60(B E+B N)$, so $\frac{1}{B E}+\frac{1}{B N}=\frac{B E+B N}{B E \cdot B N}=\frac{1}{\mathbf{6 0}}$. Note that this value does not depend on the length of the hypotenuse $\overline{E N}$; for a given location of point $P, \frac{1}{B E}+\frac{1}{B N}$ is invariant.
Using similar triangles, we have $\frac{E R}{P R}=\frac{P S}{S N}=\frac{B E}{B N}$, so $\frac{B E-60}{60}=$ $\frac{60}{B N-60}=\frac{B E}{B N}$ and $B E^{2}+B N^{2}=221^{2}$. Using algebra, we find that $B E=204, B N=85$, and $\frac{1}{204}+\frac{1}{85}=\frac{1}{60}$.

Final answer: $\frac{1}{60}$
Subfield: Algebra


Question: $\quad$ Compute all real values of $x$ such that $\log _{2}\left(\log _{2} x\right)=\log _{4}\left(\log _{4} x\right)$.

Solution:
If $y=\log _{a}\left(\log _{a} x\right)$, then $a^{a^{y}}=x$. Let $y=\log _{2}\left(\log _{2} x\right)=\log _{4}\left(\log _{4} x\right)$. Then $2^{2^{y}}=4^{4^{y}}=$ $\left(2^{2}\right)^{\left(2^{2}\right)^{y}}=2^{2^{2 y+1}}$, so $2 y+1=y, y=-1$, and $x=\sqrt{\mathbf{2}}$. (This problem is based on one submitted by ARML alum James Albrecht, 1986-2007.)
Raise 4 (or $2^{2}$ ) to the power of both sides to get $\left(\log _{2} x\right)^{2}=\log _{4} x$. By the change of base formula, $\frac{(\log x)^{2}}{(\log 2)^{2}}=\frac{\log x}{2 \log 2}$, so $\log x=\frac{\log 2}{2}$, thus $x=2^{1 / 2}=\sqrt{\mathbf{2}}$.
Let $x=4^{a}$. The equation then becomes $\log _{2}(2 a)=\log _{4} a$. Raising 4 to the power of each side, we get $4 a^{2}=a$. Since $a \neq 0$, we get $4 a=1$, thus $a=\frac{1}{4}$ and $x=\sqrt{2}$.

Final answer: $\sqrt{2}$
Subfield: Number Theory


Question: Let $k$ be the least common multiple of the numbers in the set $\mathcal{S}=\{1,2, \ldots, 30\}$. Determine the number of positive integer divisors of $k$ that are divisible by exactly 28 of the numbers in the set $\mathcal{S}$.

Solution:
We know that $k=2^{4} \cdot 3^{3} \cdot 5^{2} \cdot 7 \cdot 11 \cdot 13 \cdot 17 \cdot 19 \cdot 23 \cdot 29$. It is not difficult to see that the set $\mathcal{T}_{1}=\left\{\frac{k}{2}, \frac{k}{3}, \frac{k}{5}, \frac{k}{17}, \frac{k}{19}, \frac{k}{23}, \frac{k}{29}\right\}$ comprises all divisors of $k$ that are divisible by exactly 29 of the numbers in the set $\mathcal{S}$. Let $\mathcal{P}=\{2,3,5,17,19,23,29\}$. Then

$$
\mathcal{T}_{2}=\left\{\frac{k}{p_{1} p_{2}}, \text { where } p_{1} \text { and } p_{2} \text { are distinct elements of } \mathcal{P}\right\}
$$

consists of divisors of $k$ that are divisible by exactly 28 of the numbers in the set $\mathcal{S}$. There are $\left(\begin{array}{l}7 \\ 2\end{array}\right)=21$ elements in $\mathcal{T}_{2}$.

Furthermore, note that $\frac{k}{7}$ is only divisible by 26 of the numbers in $\mathcal{S}$ (since it is not divisible by $7,14,21$, or 28 ) while $\frac{k}{11}$ and $\frac{k}{13}$ are each divisible by 28 of the numbers in $\mathcal{S}$. We can also rule out $\frac{k}{4}$ (27 divisors: all but 8,16 , and 24 ), $\frac{k}{9}$ (27 divisors), $\frac{k}{25}$ (24 divisors), and all other numbers, thus the answer is $21+2=\mathbf{2 3}$.

Final answer: 23
Subfield: Number Theory


Question: Let $A$ and $B$ be digits from the set $\{0,1,2, \ldots, 9\}$. Let $r$ be the two-digit integer $\underline{A} \underline{B}$ and let $s$ be the two-digit integer $\underline{B} \underline{A}$, so that $r$ and $s$ are members of the set $\{00,01, \ldots, 99\}$. Compute the number of ordered pairs $(A, B)$ such that $|r-s|=k^{2}$ for some integer $k$.

Solution:
Because $|(10 A+B)-(10 B+A)|=9|A-B|=k^{2}$, it follows that $|A-B|$ is a perfect square. $|A-B|=0$ yields 10 pairs of integers: $(A, B)=(0,0),(1,1), \ldots,(9,9)$.

$|A-B|=1$ yields 18 pairs: the nine $(A, B)=(0,1),(1,2), \ldots,(8,9)$, and their reverses.

$|A-B|=4$ yields 12 pairs: the six $(A, B)=(0,4),(1,5), \ldots,(5,9)$, and their reverses.

$|A-B|=9$ yields 2 pairs: $(A, B)=(0,9)$ and its reverse.

Thus the total number of possible ordered pairs $(A, B)$ is $10+18+12+2=\mathbf{4 2}$.

Final answer: 42
Subfield: Geometry


Question: A cylinder with radius $r$ and height $h$ has volume 1 and total surface area 12. Compute $\frac{1}{r}+\frac{1}{h}$.

Solution:
Since $\pi r^{2} h=1$, we have $h=\frac{1}{\pi r^{2}}$ and $\pi r^{2}=\frac{1}{h}$. Consequently,

$$
2 \pi r h+2 \pi r^{2}=12 \Rightarrow(2 \pi r)\left(\frac{1}{\pi r^{2}}\right)+2\left(\frac{1}{h}\right)=12 \Rightarrow \frac{2}{r}+\frac{2}{h}=12 \Rightarrow \frac{1}{r}+\frac{1}{h}=\mathbf{6}
$$
The total surface area is $2 \pi r h+2 \pi r^{2}=12$ and the volume is $\pi r^{2} h=1$. Dividing, we obtain $\frac{12}{1}=\frac{2 \pi r h+2 \pi r^{2}}{\pi r^{2} h}=\frac{2}{r}+\frac{2}{h}$, thus $\frac{1}{r}+\frac{1}{h}=\frac{12}{2}=\mathbf{6}$.

Final answer: 6
Subfield: Geometry


Question: A rectangular box has dimensions $8 \times 10 \times 12$. Compute the fraction of the box's volume that is not within 1 unit of any of the box's faces.

Solution:
Let the box be defined by the product of the intervals on the $x, y$, and $z$ axes as $[0,8] \times$ $[0,10] \times[0,12]$ with volume $8 \times 10 \times 12$. The set of points inside the box that are not within 1 unit of any face is defined by the product of the intervals $[1,7] \times[1,9] \times[1,11]$ with volume $6 \times 8 \times 10$. This volume is $\frac{6 \times 8 \times 10}{8 \times 10 \times 12}=\frac{1}{2}$ of the whole box.

Final answer: $\frac{1}{2}$
Subfield: Algebra


Question: Let $T=T N Y W R$. Compute the largest real solution $x$ to $(\log x)^{2}-\log \sqrt{x}=T$.

Solution:
Let $u=\log x$. Then the given equation can be rewritten as $u^{2}-\frac{1}{2} u-T=0 \rightarrow 2 u^{2}-u-2 T=0$. This quadratic has solutions $u=\frac{1 \pm \sqrt{1+16 T}}{4}$. As we are looking for the largest real solution for $x$ (and therefore, for $u$ ), we want $u=\frac{1+\sqrt{1+16 T}}{4}=1$ when $T=\frac{1}{2}$. Therefore, $x=10^{1}=\mathbf{1 0}$.

Final answer: 10
Subfield: Combinatorics


Question: Let $T=T N Y W R$. Kay has $T+1$ different colors of fingernail polish. Compute the number of ways that Kay can paint the five fingernails on her left hand by using at least three colors and such that no two consecutive fingernails have the same color.

Solution:
There are $T+1$ possible colors for the first nail. Each remaining nail may be any color except that of the preceding nail, that is, there are $T$ possible colors. Thus, using at least two colors, there are $(T+1) T^{4}$ possible colorings. The problem requires that at least three colors be used, so we must subtract the number of colorings that use only two colors. As before, there are $T+1$ possible colors for the first nail and $T$ colors for the second. With only two colors, there are no remaining choices; the colors simply alternate. The answer is therefore $(T+1) T^{4}-(T+1) T$, and with $T=10$, this expression is equal to $110000-110=\mathbf{1 0 9 8 9 0}$.

Final answer: 109890
Subfield: Number Theory


Question: Compute the number of ordered pairs $(x, y)$ of positive integers satisfying $x^{2}-8 x+y^{2}+4 y=5$.

Solution:
Completing the square twice in $x$ and $y$, we obtain the equivalent equation $(x-4)^{2}+(y+2)^{2}=$ 25 , which describes a circle centered at $(4,-2)$ with radius 5 . The lattice points on this circle are points 5 units up, down, left, or right of the center, or points 3 units away on one axis and 4 units away on the other. Because the center is below the $x$-axis, we know that $y$ must increase by at least 2 units; $x$ cannot decrease by 4 or more units if it is to remain positive. Thus, we have:

$$
\begin{aligned}
& (x, y)=(4,-2)+(-3,4)=(1,2) \\
& (x, y)=(4,-2)+(0,5)=(4,3) \\
& (x, y)=(4,-2)+(3,4)=(7,2) \\
& (x, y)=(4,-2)+(4,3)=(8,1) .
\end{aligned}
$$

There are $\mathbf{4}$ such ordered pairs.

Final answer: 4
Subfield: Number Theory


Question: Let $T=T N Y W R$ and let $k=21+2 T$. Compute the largest integer $n$ such that $2 n^{2}-k n+77$ is a positive prime number.

Solution:
If $k$ is positive, there are only four possible factorizations of $2 n^{2}-k n+77$ over the integers, namely

$$
\begin{aligned}
& (2 n-77)(n-1)=2 n^{2}-79 n+77 \\
& (2 n-1)(n-77)=2 n^{2}-145 n+77 \\
& (2 n-11)(n-7)=2 n^{2}-25 n+77 \\
& (2 n-7)(n-11)=2 n^{2}-29 n+77
\end{aligned}
$$



Because $T=4, k=29$, and so the last factorization is the correct one. Because $2 n-7$ and $n-11$ are both integers, in order for their product to be prime, one factor must equal 1 or -1 , so $n=3,4,10$, or 12 . Checking these possibilities from the greatest downward, $n=12$ produces $17 \cdot 1=17$, which is prime. So the answer is $\mathbf{1 2}$.

Final answer: 12
Subfield: Geometry


Question: Let $T=T N Y W R$. In triangle $A B C, B C=T$ and $\mathrm{m} \angle B=30^{\circ}$. Compute the number of integer values of $A C$ for which there are two possible values for side length $A B$.

Solution:
By the Law of Cosines, $(A C)^{2}=T^{2}+(A B)^{2}-2 T(A B) \cos 30^{\circ} \rightarrow(A B)^{2}-2 T \cos 30^{\circ}(A B)+$ $\left(T^{2}-(A C)^{2}\right)=0$. This quadratic in $A B$ has two positive solutions when the discriminant and product of the roots are both positive. Thus $\left(2 T \cos 30^{\circ}\right)^{2}-4\left(T^{2}-(A C)^{2}\right)>0$, and $\left(T^{2}-(A C)^{2}\right)>0$. The second inequality implies that $A C<T$. The first inequality simplifies to $4(A C)^{2}-T^{2}>0$, so $T / 2<A C$. Since $T=12$, we have that $6<A C<12$, giving 5 integral values for $A C$.

Final answer: 5
Subfield: Combinatorics


Question: An $\boldsymbol{n}$-label is a permutation of the numbers 1 through $n$. For example, $J=35214$ is a 5 -label and $K=132$ is a 3 -label. For a fixed positive integer $p$, where $p \leq n$, consider consecutive blocks of $p$ numbers in an $n$-label. For example, when $p=3$ and $L=263415$, the blocks are 263,634,341, and 415. We can associate to each of these blocks a $p$-label that corresponds to the relative order of the numbers in that block. For $L=263415$, we get the following:

$$
\underline{263} 415 \rightarrow 132 ; \quad 2 \underline{63415} \rightarrow 312 ; \quad 26 \underline{341} 5 \rightarrow 231 ; \quad 263 \underline{415} \rightarrow 213
$$

Moving from left to right in the $n$-label, there are $n-p+1$ such blocks, which means we obtain an $(n-p+1)$-tuple of $p$-labels. For $L=263415$, we get the 4 -tuple $(132,312,231,213)$. We will call this $(n-p+1)$-tuple the $\boldsymbol{p}$-signature of $L$ (or signature, if $p$ is clear from the context) and denote it by $S_{p}[L]$; the $p$-labels in the signature are called windows. For $L=263415$, the windows are $132,312,231$, and 213 , and we write

$$
S_{3}[263415]=(132,312,231,213)
$$

More generally, we will call any $(n-p+1)$-tuple of $p$-labels a $p$-signature, even if we do not know of an $n$-label to which it corresponds (and even if no such label exists). A signature that occurs for exactly one $n$-label is called unique, and a signature that doesn't occur for any $n$-labels is called impossible. A possible signature is one that occurs for at least one $n$-label.

In this power question, you will be asked to analyze some of the properties of labels and signatures.
Compute the 3 -signature for 52341.

Solution:
$(312,123,231)$

Final answer: $(312,123,231)$