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


Question: The arithmetic sequences $a_{1}, a_{2}, a_{3}, \ldots, a_{20}$ and $b_{1}, b_{2}, b_{3}, \ldots, b_{20}$ consist of 40 distinct positive integers, and $a_{20}+b_{14}=1000$. Compute the least possible value for $b_{20}+a_{14}$.

Solution:
Write $a_{n}=a_{1}+r(n-1)$ and $b_{n}=b_{1}+s(n-1)$. Then $a_{20}+b_{14}=a_{1}+b_{1}+19 r+13 s$, while $b_{20}+a_{14}=a_{1}+b_{1}+13 r+19 s=a_{20}+b_{14}+6(s-r)$. Because both sequences consist only of integers, $r$ and $s$ must be integers, so $b_{20}+a_{14} \equiv a_{20}+b_{14} \bmod 6$. Thus the least possible value of $b_{20}+a_{14}$ is 4 . If $b_{20}=3$ and $a_{14}=1$, then $\left\{a_{n}\right\}$ must be a decreasing sequence (else $a_{13}$ would not be positive) and $a_{20} \leq-5$, which is impossible. The case $b_{20}=a_{14}=2$ violates the requirement that the terms be distinct, and by reasoning analogous to the first case, $b_{20}=1, a_{14}=3$ is also impossible. Hence the sum $b_{20}+a_{14}$ is at least 10 . To show that 10 is attainable, make $\left\{a_{n}\right\}$ decreasing and $b_{20}$ as small as possible: set $b_{20}=1, a_{14}=9$, and $a_{n}=23-n$. Then $a_{20}=3$, yielding $b_{14}=997$. Hence $s=\frac{997-1}{14-20}=\frac{996}{-6}=-166$ and



$b_{1}=997-(13)(-166)=3155$, yielding $b_{n}=3155-166(n-1)$. Because $b_{20}=1 \leq a_{20}$ and $b_{19}=167 \geq a_{1}$, the sequences $\left\{b_{n}\right\}$ and $\left\{a_{n}\right\}$ are distinct for $1 \leq n \leq 20$, completing the proof. Hence the minimum possible value of $b_{20}+a_{14}$ is 10.

Final answer: 10
Subfield: Algebra


Question: Compute the ordered triple $(x, y, z)$ representing the farthest lattice point from the origin that satisfies $x y-z^{2}=y^{2} z-x=14$.

Solution:
First, eliminate $x: y\left(y^{2} z-x\right)+\left(x y-z^{2}\right)=14(y+1) \Rightarrow z^{2}-y^{3} z+14(y+1)=0$. Viewed as a quadratic in $z$, this equation implies $z=\frac{y^{3} \pm \sqrt{y^{6}-56(y+1)}}{2}$. In order for $z$ to be an integer, the discriminant must be a perfect square. Because $y^{6}=\left(y^{3}\right)^{2}$ and $\left(y^{3}-1\right)^{2}=y^{6}-2 y^{3}+1$, it follows that $|56(y+1)| \geq 2\left|y^{3}\right|-1$. This inequality only holds for $|y| \leq 5$. Within that range, the only values of $y$ for which $y^{6}-56 y-56$ is a perfect square are -1 and -3 . If $y=-1$, then $z=-1$ or $z=0$. If $y=-3$, then $z=1$ or $z=-28$. After solving for the respective values of $x$ in the various cases, the four lattice points satisfying the system are $(-15,-1,-1),(-14,-1,0),(-5,-3,1)$, and $(-266,-3,-28)$. The farthest solution point from the origin is therefore $(-\mathbf{2 6 6}, \mathbf{- 3 , - 2 8})$.

Final answer: $(-266,-3,-28)$
Subfield: Algebra


Question: The sequence $a_{1}, a_{2}, a_{3}, \ldots$ is a geometric sequence with $a_{20}=8$ and $a_{14}=2^{21}$. Compute $a_{21}$.

Solution:
Let $r$ be the common ratio of the sequence. Then $a_{20}=r^{20-14} \cdot a_{14}$, hence $8=r^{6} \cdot 2^{21} \Rightarrow r^{6}=$ $\frac{2^{3}}{2^{21}}=2^{-18}$, so $r=2^{-3}=\frac{1}{8}$. Thus $a_{21}=r \cdot a_{20}=\frac{1}{8} \cdot 8=\mathbf{1}$.

Final answer: 1
Subfield: Geometry


Question: Let $T=1$. Circles $L$ and $O$ are internally tangent and have radii $T$ and $4 T$, respectively. Point $E$ lies on circle $L$ such that $\overline{O E}$ is tangent to circle $L$. Compute $O E$.

Solution:
Because $\overline{O E}$ is tangent to circle $L, \overline{L E} \perp \overline{O E}$. Also note that $L O=4 T-T=3 T$. Hence, by the Pythagorean Theorem, $O E=\sqrt{(3 T)^{2}-T^{2}}=2 T \sqrt{2}$ (this also follows from the TangentSecant Theorem). With $T=1, O E=\mathbf{2} \sqrt{\mathbf{2}}$.

Final answer: $2 \sqrt{2}$
Subfield: Geometry


Question: Let $T=2 \sqrt{2}$. In a right triangle, one leg has length $T^{2}$ and the other leg is 2 less than the hypotenuse. Compute the triangle's perimeter.

Solution:
Let $c$ be the length of the hypotenuse. Then, by the Pythagorean Theorem, $\left(T^{2}\right)^{2}+(c-2)^{2}=$ $c^{2} \Rightarrow c=\frac{T^{4}}{4}+1$. With $T=2 \sqrt{2}, T^{4}=64$, and $c=17$. So the triangle is a $8-15-17$ triangle with perimeter 40 .

Final answer: 40
Subfield: Algebra


Question: $\quad$ Let $T=40$. If $x+9 y=17$ and $T x+(T+1) y=T+2$, compute $20 x+14 y$.

Solution:
Multiply each side of the first equation by $T$ to obtain $T x+9 T y=17 T$. Subtract the second equation to yield $9 T y-T y-y=16 T-2 \Rightarrow y(8 T-1)=2(8 T-1)$. Hence either $T=\frac{1}{8}$ (in which case, the value of $y$ is not uniquely determined) or $y=2$. Plug $y=2$ into the first equation to obtain $x=-1$. Hence $20 x+14 y=-20+28=\mathbf{8}$.

Final answer: 8
Subfield: Algebra


Question: Let $T=8$. Let $f(x)=a x^{2}+b x+c$. The product of the roots of $f$ is $T$. If $(-2,20)$ and $(1,14)$ lie on the graph of $f$, compute $a$.

Solution:
Using Vièta's Formula, write $f(x)=a x^{2}+b x+T a$. Substituting the coordinates of the given points yields the system of equations: $4 a-2 b+T a=20$ and $a+b+T a=14$. Multiply each side of the latter equation by 2 and add the resulting equation to the former equation to eliminate $b$. Simplifying yields $a=\frac{16}{T+2}$. With $T=8, a=8 / 5$.

Final answer: $\frac{8}{5}$
Subfield: Number Theory


Question: Let $T=\frac{8}{5}$. Let $z_{1}=15+5 i$ and $z_{2}=1+K i$. Compute the smallest positive integral value of $K$ such that $\left|z_{1}-z_{2}\right| \geq 15 T$.

Solution:
Note that $z_{1}-z_{2}=14+(5-K) i$, hence $\left|z_{1}-z_{2}\right|=\sqrt{14^{2}+(5-K)^{2}}$. With $T=8 / 5,15 T=24$, hence $14^{2}+(5-K)^{2} \geq 24^{2}$. Thus $|5-K| \geq \sqrt{24^{2}-14^{2}}=\sqrt{380}$. Because $K$ is a positive integer, it follows that $K-5 \geq 20$, hence the desired value of $K$ is $\mathbf{2 5}$.

Final answer: 25
Subfield: Combinatorics


Question: Let $T=25$. Suppose that $T$ people are standing in a line, including three people named Charlie, Chris, and Abby. If the people are assigned their positions in line at random, compute the probability that Charlie is standing next to at least one of Chris or Abby.

Solution:
First count the number of arrangements in which Chris stands next to Charlie. This is $(T-1) \cdot 2 ! \cdot(T-2) !=2 \cdot(T-1)$ ! because there are $T-1$ possible leftmost positions for the pair $\{$ Charlie, Chris $\}$, there are 2 ! orderings of this pair, and there are $(T-2)$ ! ways to arrange the remaining people. There are equally many arrangements in which Abby stands next to Charlie. However, adding these overcounts the arrangements in which Abby, Charlie, and Chris are standing next to each other, with Charlie in the middle. Using similar reasoning as above, there are $(T-2) \cdot 2 ! \cdot(T-3) !=2 \cdot(T-2)$ ! such arrangements. Hence the desired probability is $\frac{2 \cdot 2 \cdot(T-1) !-2 \cdot(T-2) !}{T !}=\frac{2 \cdot(T-2) !(2 T-2-1)}{T !}=\frac{2(2 T-3)}{T(T-1)}$. With $T=25$, the fraction simplifies to $\frac{\mathbf{4 7}}{\mathbf{3 0 0}}$.

Final answer: $\frac{47}{300}$
Subfield: Geometry


Question: Let $A$ be the number you will receive from position 7 and let $B$ be the number you will receive from position 9. Let $\alpha=\sin ^{-1} A$ and let $\beta=\cos ^{-1} B$. Compute $\sin (\alpha+\beta)+\sin (\alpha-\beta)$.

Solution:
The given conditions are equivalent to $\sin \alpha=A$ and $\cos \beta=B$. Using either the sumto-product or the sine of a sum/difference identities, the desired expression is equivalent to $2(\sin \alpha)(\cos \beta)=2 \cdot A \cdot B$. With $A=\frac{47}{300}$ and $B=\frac{12}{169}, 2 \cdot A \cdot B=\frac{2 \cdot 47}{25 \cdot 169}=\frac{\mathbf{9 4}}{\mathbf{4 2 2 5}}$.

Final answer: $\frac{94}{4225}$
Subfield: Geometry


Question: Let $T=13$. If $r$ is the radius of a right circular cone and the cone's height is $T-r^{2}$, let $V$ be the maximum possible volume of the cone. Compute $\pi / V$.

Solution:
The cone's volume is $\frac{1}{3} \pi r^{2}\left(T-r^{2}\right)$. Maximizing this is equivalent to maximizing $x(T-x)$, where $x=r^{2}$. Using the formula for the vertex of a parabola (or the AM-GM inequality), the maximum value occurs when $x=\frac{T}{2}$. Hence $V=\frac{1}{3} \pi \cdot \frac{T}{2} \cdot \frac{T}{2}=\frac{\pi T^{2}}{12}$, and $\pi / V=12 / T^{2}$. With $T=13, V=\frac{\mathbf{1 2}}{\mathbf{1 6 9}}$.

Final answer: $\frac{12}{169}$
Subfield: Algebra


Question: Let $T=650$. If $\log T=2-\log 2+\log k$, compute the value of $k$.

Solution:
Write $2=\log 100$ and use the well-known properties for the sum/difference of two logs to obtain $\log T=\log \left(\frac{100 k}{2}\right)$, hence $k=\frac{T}{50}$. With $T=650, k=13$.

Final answer: 13
Subfield: Algebra


Question: Let $T=100$. Nellie has a flight from Rome to Athens that is scheduled to last for $T+30$ minutes. However, owing to a tailwind, her flight only lasts for $T$ minutes. The plane's speed is 1.5 miles per minute faster than what it would have been for the originally scheduled flight. Compute the distance (in miles) that the plane travels.

Solution:
Let $D$ be the distance in miles traveled by the plane. The given conditions imply that $\frac{D}{T}-\frac{D}{T+30}=1.5 \Rightarrow \frac{30 D}{T(T+30)}=1.5 \Rightarrow D=\frac{T(T+30)}{20}$. With $T=100, D=5 \cdot 130=\mathbf{6 5 0}$.

Final answer: 650
Subfield: Algebra


Question: Let $T=9$. Compute $\sqrt{\sqrt{\sqrt[T]{10^{T^{2}-T}}}}$.

Solution:
The given radical equals $\left(\left(\left(10^{T^{2}-T}\right)^{\frac{1}{T}}\right)^{\frac{1}{2}}\right)^{\frac{1}{2}}=10^{(T-1) / 4}$. With $T=9$, this simplifies to $10^{2}=100$

Final answer: 100
Subfield: Geometry


Question: Let $T=3$. Regular hexagon $S U P E R B$ has side length $\sqrt{T}$. Compute the value of $B E \cdot S U \cdot R E$.

Solution:
Because $\overline{S U}$ and $\overline{R E}$ are sides of the hexagon, $S U=R E=\sqrt{T}$. Let $H$ be the foot of the altitude from $R$ to $\overline{B E}$ in $\triangle B R E$ and note that each interior angle of a regular hexagon is $120^{\circ}$. Thus $B E=B H+H E=2\left(\frac{\sqrt{3}}{2}\right)(\sqrt{T})=\sqrt{3 T}$. Thus $B E \cdot S U \cdot R E=\sqrt{3 T} \cdot \sqrt{T} \cdot \sqrt{T}=T \sqrt{3 T}$. With $T=3$, the answer is $\mathbf{9}$.

Final answer: 9
Subfield: Combinatorics


Question: Let $T=70$. Chef Selma is preparing a burrito menu. A burrito consists of: (1) a choice of chicken, beef, turkey, or no meat, (2) exactly one of three types of beans, (3) exactly one of two types of rice, and (4) exactly one of $K$ types of cheese. Compute the smallest value of $K$ such that Chef Selma can make at least $T$ different burrito varieties.

Solution:
Using the Multiplication Principle, Chef Selma can make $4 \cdot 3 \cdot 2 \cdot K=24 K$ different burrito varieties. With $T=70$, the smallest integral value of $K$ such that $24 K \geq 70$ is $\left\lceil\frac{70}{24}\right\rceil=3$.

Final answer: 3
Subfield: Number Theory


Question: Compute the smallest positive integer $N$ such that $20 N$ is a multiple of 14 and $14 N$ is a multiple of 20 .

Solution:
Because $\operatorname{gcd}(14,20)=2$, the problem is equivalent to computing the smallest positive integer $N$ such that $7 \mid 10 N$ and $10 \mid 7 N$. Thus $7 \mid N$ and $10 \mid N$, and the desired value of $N$ is $\operatorname{lcm}(7,10)=\mathbf{7 0}$.

Final answer: 70
Subfield: Number Theory


Question: Call a positive integer fibbish if each digit, after the leftmost two, is at least the sum of the previous two digits. Compute the greatest fibbish number.

Solution:
The largest fibbish number is 10112369. First, if $\underline{A_{1}} \underline{A_{2}} \cdots \underline{A_{n}}$ is an $n$-digit fibbish number with $A_{1}$ and $A_{2} \neq 0$, the number created by prepending the ${\text { digits }} A_{1}$ and 0 to the number is larger and still fibbish: $\underline{A_{1}} \underline{0} \underline{A_{1}} \underline{A_{2}} \cdots \underline{A_{n}}>\underline{A_{1}} \underline{A_{2}} \cdots \underline{A_{n}}$. Suppose that $A_{2}=0$ and $A_{3}=A_{1}$, so that the number begins $\underline{A_{1}} \underline{0} \underline{A_{1}} \underline{A_{4}}$. If the number is to be fibbish, $A_{4} \geq A_{1}>0$. Then if $A_{1} \geq 2$ and $A_{4} \geq 2$, because the number is fibbish, $A_{5} \geq 4$, and $A_{6} \geq 6$. In this case there can be no more digits, because $A_{5}+A_{6} \geq 10$. So the largest possible fibbish number beginning with 20 is 202246. If $A_{1}=2$ and $A_{2}=1$, then $A_{3}$ must be at least 3 , and the largest possible number is 21459; changing $A_{3}$ to 3 does not increase the length. Now consider $A_{1}=1$. If $A_{2}=1$, then $A_{3} \geq 2, A_{4} \geq 3, A_{5} \geq 5$, and $A_{6} \geq 8$. There can be no seventh digit because that digit would have to be at least 13 . Increasing $A_{3}$ to 3 yields only two additional digits, because $A_{4} \geq 4, A_{5} \geq 7$. So $A_{3}=2$ yields a longer (and thus larger) number. Increasing $A_{4}$ to 4 yields only one additional digit, $A_{5} \geq 6$, because $A_{4}+A_{5} \geq 10$. But if $A_{4}=3$, increasing $A_{5}$ to 6 still allows $A_{6}=9$, yielding the largest possible number of digits (8) and the largest fibbish number with that many digits.

Final answer: 10112369
Subfield: Combinatorics


Question: An ARMLbar is a $7 \times 7$ grid of unit squares with the center unit square removed. A portion of an ARMLbar is a square section of the bar, cut along the gridlines of the original bar. Compute the number of different ways there are to cut a single portion from an ARMLbar.

Solution:
Note that any portion of side length $m \geq 4$ will overlap the center square, so consider only portions of side length 3 or less. If there were no hole in the candy bar, the number of portions could be counted by conditioning on the possible location of the upper-left corner of the portion. If the portion is of size $1 \times 1$, then the corner can occupy any of the $7^{2}$ squares of the bar. If the portion is of size $2 \times 2$, then the corner can occupy any of the top 6 rows and any of the left 6 columns, for $6^{2}=36$ possible $2 \times 2$ portions. In general, the upper-left corner of an $m \times m$ portion can occupy any of the top $8-m$ rows and any of the left $8-m$ columns. So the total number of portions from an intact bar would be $7^{2}+6^{2}+5^{2}$. Now when $m \leq 3$, the number of $m \times m$ portions that include the missing square is simply $m^{2}$, because the missing square could be any square of the portion. So the net number of portions is

$$
\begin{aligned}
7^{2}+6^{2}+5^{2}-3^{2}-2^{2}-1^{2} & =(49+36+25)-(9+4+1) \\
& =110-14 \\
& =\mathbf{9 6}
\end{aligned}
$$
First ignore the missing square. As in the previous solution, the number of $m \times m$ portions that can fit in the bar is $(8-m)^{2}$. So the total number of portions of all sizes is simply

$$
7^{2}+6^{2}+\cdots+1^{2}=\frac{7(7+1)(2 \cdot 7+1)}{6}=140
$$

To exclude portions that overlap the missing center square, it is useful to consider the location of the missing square within the portion. If an $m \times m$ portion includes the missing center



square, and $m \leq 4$, then the missing square could be any one of the $m^{2}$ squares in the portion. If $m=5$, then the missing square cannot be in the leftmost or rightmost columns of the portion, because then the entire bar would have to extend at least four squares past the hole, and it only extends three. By similar logic, the square cannot be in the top or bottom rows of the portion. So for $m=5$, there are $3 \cdot 3=9$ possible positions. For $m=6$, the two left and two right columns are excluded, as are the two top and the two bottom rows, for $2 \cdot 2=4$ possible positions for the portion. And in a $7 \times 7$ square, the only possible location of the hole is in the center. So the total number of portions overlapping the missing square is

$$
1^{2}+2^{2}+3^{2}+4^{2}+3^{2}+2^{2}+1^{2}=44 .
$$

The difference is thus $140-44=\mathbf{9 6}$

Final answer: 96
Subfield: Geometry


Question: Regular hexagon $A B C D E F$ and regular hexagon $G H I J K L$ both have side length 24 . The hexagons overlap, so that $G$ is on $\overline{A B}, B$ is on $\overline{G H}, K$ is on $\overline{D E}$, and $D$ is on $\overline{J K}$. If $[G B C D K L]=\frac{1}{2}[A B C D E F]$, compute $L F$.

Solution:
The diagram below shows the hexagons.

<img_3234>

The area of hexagon $G B C D K L$ can be computed as $[G B C D K L]=[A B C D E F]-[A G L K E F]$, and $[A G L K E F]$ can be computed by dividing concave hexagon $A G L K E F$ into two parallelograms sharing $\overline{F L}$. If $A B=s$, then the height $A E$ is $s \sqrt{3}$, so the height of parallelogram $A G L F$ is $\frac{s \sqrt{3}}{2}$. Thus $[A G L F]=L F \cdot \frac{s \sqrt{3}}{2}$ and $[A G L K E F]=L F \cdot s \sqrt{3}$. On the other hand, the area of a regular hexagon of side length $s$ is $\frac{3 s^{2} \sqrt{3}}{2}$. Because $[G B C D K L]=\frac{1}{2}[A B C D E F]$, it follows that $[A G L K E F]=\frac{1}{2}[A B C D E F]$, and

$$
L F \cdot s \sqrt{3}=\frac{1}{2}\left(\frac{3 s^{2} \sqrt{3}}{2}\right)=\frac{3 s^{2} \sqrt{3}}{4}
$$

whence $L F=\frac{3}{4} s$. With $s=24$, the answer is $\mathbf{1 8}$.
Compute $[B C D K L G]$ as twice the area of trapezoid $B C L G$. If $A B=s$, then $B G=s-L F$ and $C L=2 s-L F$, while the height of the trapezoid is $\frac{s \sqrt{3}}{2}$.[^0]


[^0]:    ${ }^{1}$ The answer 115 was also accepted for this problem because of an alternate (and unintended) reasonable interpretation of the problem statement. Some students also counted portions that contained the "hole", with the hole being strictly inside the portion, and not along its edges.



Thus the area of the trapezoid is:

$$
\frac{1}{2}\left(\frac{s \sqrt{3}}{2}\right)((s-L F)+(2 s-L F))=\frac{s \sqrt{3}(3 s-2 L F)}{4}
$$

Double that area to obtain

$$
[B C D K L G]=\frac{s \sqrt{3}(3 s-2 L F)}{2}
$$

On the other hand, $[A B C D E F]=\frac{3 s^{2} \sqrt{3}}{2}$, so

$$
\begin{aligned}
\frac{s \sqrt{3}(3 s-2 L F)}{2} & =\frac{3 s^{2} \sqrt{3}}{4} \\
3 s-2 L F & =\frac{3 s}{2} \\
L F & =\frac{3}{4} s .
\end{aligned}
$$

Substituting $s=24$ yields $L F=\mathbf{1 8}$.

Final answer: 18
Subfield: Number Theory


Question: Compute the largest base-10 integer $\underline{A} \underline{B} \underline{C} \underline{D}$, with $A>0$, such that $\underline{A} \underline{B} \underline{C} \underline{D}=B !+C !+D !$.

Solution:
Let $\underline{A} \underline{B} \underline{C} \underline{D}=N$. Because $7 !=5040$ and $8 !=40,320, N$ must be no greater than $7 !+6 !+6 !=6480$. This value of $N$ does not work, so work through the list of possible sums in decreasing order: $7 !+6 !+5 !, 7 !+6 !+4$ !, etc. The first value that works is $N=5762=7 !+6 !+2 !$.
Let $\underline{A} \underline{B} \underline{C} \underline{D}=N$. Because $7 !=5040$ and $8 !=40,320$, to find the maximal value, first consider values of $N$ that include 7 as a digit. Suppose then that $N=5040+X !+Y$ !. To force a 7 to appear in this sum with maximal $N$, let $X=6$, which yields $N=5040+720+Y !=5760+Y$ !. This value of $N$ has a 7 (and a 6 ), so search for values of $Y$ to find ones that satisfy the conditions of the problem. Only $Y=1$ and $Y=2$ will do, giving 5761 and 5762 . Hence $\mathbf{5 7 6 2}$ is the maximum possible value of $N$.

Final answer: 5762
Subfield: Number Theory


Question: Let $X$ be the number of digits in the decimal expansion of $100^{1000^{10,000}}$, and let $Y$ be the number of digits in the decimal expansion of $1000^{10,000^{100,000}}$. Compute $\left\lfloor\log _{X} Y\right\rfloor$.

Solution:
The number of digits of $n$ is $\lfloor\log n\rfloor+1$. Because $100^{1000^{10,000}}=\left(10^{2}\right)^{1000^{10,000}}, X=2$. $1000^{10,000}+1$. Similarly, $Y=3 \cdot 10,000^{100,000}+1$. Using the change-of-base formula,

$$
\begin{aligned}
\log _{X} Y=\frac{\log Y}{\log X} & \approx \frac{\log 3+\log 10,000^{100,000}}{\log 2+\log 1000^{10,000}} \\
& =\frac{\log 3+100,000 \log 10,000}{\log 2+10,000 \log 1000} \\
& =\frac{\log 3+100,000 \cdot 4}{\log 2+10,000 \cdot 3} \\
& =\frac{400,000+\log 3}{30,000+\log 2}
\end{aligned}
$$



Both $\log 3$ and $\log 2$ are tiny compared to the integers to which they are being added. If the quotient 400,000/30,000 were an integer (or extremely close to an integer), the values of these logarithms might matter, but $400,000 / 30,000=40 / 3=13 . \overline{3}$, so in this case, they are irrelevant. Hence

$$
\left\lfloor\log _{X} Y\right\rfloor=\left\lfloor\frac{400,000}{30,000}\right\rfloor=\left\lfloor\frac{40}{3}\right\rfloor=13
$$

Final answer: 13
Subfield: Geometry


Question: Compute the smallest possible value of $n$ such that two diagonals of a regular $n$-gon intersect at an angle of 159 degrees.

Solution:
Let the vertices of the polygon be $A_{0}, A_{1}, \ldots, A_{n-1}$. Considering the polygon as inscribed in a circle, the angle between diagonals $\overline{A_{0} A_{i}}$ and $\overline{A_{0} A_{j}}$ is $\frac{1}{2} \cdot\left(\frac{360^{\circ}}{n}\right) \cdot|j-i|=\left(\frac{180|j-i|}{n}\right)^{\circ}$. The diagonal $\overline{A_{k} A_{k+j}}$ can be considered as the rotation of $\overline{A_{0} A_{j}}$ through $k / n$ of a circle, or $\left(\frac{360 k}{n}\right)^{\circ}$. So the diagonals $A_{0} A_{i}$ and $A_{k} A_{k+j}$ intersect at a combined angle of $\left(\frac{180|j-i|}{n}\right)^{\circ}+\left(\frac{360 k}{n}\right)^{\circ}$. Without loss of generality, assume $i<j$ (otherwise relabel vertices in the opposite order, with $A_{k}$ becoming $A_{0}$ ). Then the desired number is the least $n$ such that

$$
\left(\frac{180(j-i)}{n}\right)+\frac{360 k}{n}=159
$$

Multiply both sides of the equation by $n$ and factor to obtain $180(j-i+2 k)=159 n$, thus $60(j-i+2 k)=53 n$. Because 53 and 60 are relatively prime and $(j-i+2 k)$ is an integer, it follows that $60 \mid n$. So the smallest possible value is $n=\mathbf{6 0}$; one set of values that satisfies the equation is $i=1, j=54, k=0$.

Final answer: 60
Subfield: Algebra


Question: Compute the number of quadratic functions $f(x)=a x^{2}+b x+c$ with integer roots and integer coefficients whose graphs pass through the points $(0,0)$ and $(15,225)$.

Solution:
Because the graph passes through $(0,0)$, conclude that $c=0$. Then

$$
f(15)=225 \Rightarrow a(15)^{2}+b(15)=225 a+15 b=225
$$

from which $b=15-15 a$. On the other hand, $f$ can be factored as $f(x)=a x(x+b / a)$, so if the roots are integers, $b / a$ must be an integer. Divide both sides of the equation $b=15-15 a$ by $a$ to obtain $b / a=15 / a-15$. Thus $15 / a$ must be an integer, and $a \in\{ \pm 1, \pm 3, \pm 5, \pm 15\}$. Because $b=15-15 a$ is linear, each of these values for $a$ yields a unique integer value for $b$, so there are 8 such ordered pairs. The values of $a, b$, and the nonnegative root are given in the table below.

| $a$ | $b$ | Root |
| ---: | ---: | ---: |
| 1 | 0 | 0 |
| 3 | -30 | 10 |
| 5 | -60 | 12 |
| 15 | -210 | 14 |
| -1 | 30 | 30 |
| -3 | 60 | 20 |
| -5 | 90 | 18 |
| -15 | 240 | 16 |

Final answer: 8