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


Question: David builds a circular table; he then carves one or more positive integers into the table at points equally spaced around its circumference. He considers two tables to be the same if one can be rotated so that it has the same numbers in the same positions as the other. For example, a table with the numbers $8,4,5$ (in clockwise order) is considered the same as a table with the numbers 4, 5,8 (in clockwise order), but both tables are different from a table with the numbers 8, 5, 4 (in clockwise order). Given that the numbers he carves sum to 17 , compute the number of different tables he can make.

Solution:
The problem calls for the number of ordered partitions of 17 , where two partitions are considered the same if they are cyclic permutations of each other. Because 17 is prime, each ordered partition of 17 into $n$ parts will be a cyclic permutation of exactly $n$ such partitions (including itself), unless $n=17$. (If $n=17$, then all the numbers are 1s, and there is exactly one table David can make.) By the sticks and stones method, the number of ordered partitions of 17 into $n$ nonzero parts is $\left(\begin{array}{c}16 \\ n-1\end{array}\right)$, and this overcounts the number of tables by a factor of $n$, except when $n=17$. Thus the number of possible tables is

$$
1+\sum_{n=1}^{16}\left(\begin{array}{c}
16 \\
n-1
\end{array}\right) \cdot \frac{1}{n}=1+\sum_{n=1}^{16}\left(\begin{array}{c}
17 \\
n
\end{array}\right) \cdot \frac{1}{17}=1+\frac{2^{17}-2}{17}=\mathbf{7 7 1 1}
$$

Final answer: 7711
Subfield: Geometry


Question: In quadrilateral $A B C D, \mathrm{~m} \angle B+\mathrm{m} \angle D=270^{\circ}$. The circumcircle of $\triangle A B D$ intersects $\overline{C D}$ at point $E$, distinct from $D$. Given that $B C=4, C E=5$, and $D E=7$, compute the diameter of the circumcircle of $\triangle A B D$.

Solution:
Note that $\mathrm{m} \angle A+\mathrm{m} \angle C=90^{\circ}$ in quadrilateral $A B C D$. Because quadrilateral $A B E D$ is cyclic, it follows that $\mathrm{m} \angle A D E+\mathrm{m} \angle A B E=180^{\circ}$. Moreover, because $\mathrm{m} \angle A B E+\mathrm{m} \angle E B C+\mathrm{m} \angle A D E=270^{\circ}$, it follows that $\angle E B C$ is a right angle. Thus $B E=\sqrt{C E^{2}-B C^{2}}=\sqrt{5^{2}-4^{2}}=3$. Let $\mathrm{m} \angle B E C=\theta$; then $\cos \theta=\frac{3}{5}$ and $\sin \theta=\frac{4}{5}$.

<img_3458>

Applying the Law of Cosines to $\triangle B E D$ yields

$$
B D^{2}=3^{2}+7^{2}-2 \cdot 3 \cdot 7 \cos \left(180^{\circ}-\theta\right)=3^{2}+7^{2}+2 \cdot 3 \cdot 7 \cos \theta=\frac{416}{5}
$$

Thus $B D=\frac{4 \sqrt{26}}{\sqrt{5}}$. Let $R$ be the circumradius of $\triangle A B D$ and $\triangle B E D$. Then the requested diameter is $2 R$, and



applying the Law of Sines to $\triangle B E D$ yields

$$
2 R=\frac{B D}{\sin \left(180^{\circ}-\theta\right)}=\frac{B D}{\sin \theta}=\frac{4 \sqrt{26}}{\sqrt{5}} \cdot \frac{5}{4}=\sqrt{\mathbf{1 3 0}}
$$

Final answer: $\sqrt{130}$
Subfield: Geometry


Question: Suppose that Xena traces a path along the segments in the figure shown, starting and ending at point $A$. The path passes through each of the eleven vertices besides $A$ exactly once, and only visits $A$ at the beginning and end of the path. Compute the number of possible paths Xena could trace.

![](https://cdn.mathpix.com/cropped/2023_12_21_5b86f26b7ea18a50f6b5g-1.jpg?height=330&width=307&top_left_y=1727&top_left_x=947)

Solution:
Count the number of complete paths that pass through all vertices exactly once (such a path is called a Hamiltonian path). The set of vertices can be split into two rings:

$$
\mathcal{I}=\left\{A_{1}, A_{2}, \ldots, A_{6}\right\} \text { (i.e., the inner ring), } \quad \mathcal{O}=\left\{B_{1}, B_{2}, \ldots, B_{6}\right\} \text { (i.e., the outer ring). }
$$

where $A_{1}=A$. The two rings are connected by the edges $E=\left\{A_{1} B_{1}, A_{2} B_{2}, \ldots, A_{6} B_{6}\right\}$. Each vertex in the figure has exactly three edges joining it with the neighboring vertices. Also note that any closed loop must use exactly two edges (out of three) for each vertex.

Further note that a loop must use at least one edge from $E$ to move from one ring to the other. Consider two cases: the loop uses all six edges from $E$, or it uses some but not all of them.

If all edges from $E$ are used, there are two possible undirected loops. It is not possible to use both edges $A_{1} A_{2}$ and $B_{1} B_{2}$, so either $A_{1} A_{2}$ or $B_{1} B_{2}$ will be used. This choice determines how the entire loop is constructed.

If not all edges from $E$ are used, then there must be some $i$ for which the loop uses the edge $A_{i} B_{i}$ and does not use $A_{i+1} B_{i+1}$ (where $A_{j} B_{j}$ represents $A_{j-6} B_{j-6}$ if $7 \leq j \leq 12$ ). Because $A_{i+1}$ is only connected to three other vertices, the loop must use $A_{i} A_{i+1}$ and $A_{i+1} A_{i+2}$, and similarly must use $B_{i} B_{i+1}$ and $B_{i+1} B_{i+2}$. This also precludes using $A_{i+2} B_{i+2}$, because doing so would close the loop before it visits all 12 vertices. Therefore the loop must also use $A_{i+2} A_{i+3}$ and $B_{i+2} B_{i+3}$, which now precludes using $A_{i+3} B_{i+3}$. This continues to force the structure of the loop until it closes by using $A_{i+5} B_{i+5}=A_{i-1} B_{i-1}$. Hence the loop must use exactly two edges from $E$, and they must be consecutive: $A_{i-1} B_{i-1}$ and $A_{i} B_{i}$. There are 6 ways to choose those two consecutive edges, so there are 6 possible undirected loops in this case.

The forced path is a loop, and the only way the given conditions are satisfied if $A_{i+k}=A_{i-1}$ and $B_{i+k}=B_{i-1}$. Hence the loop must use (precisely) two consecutive edges from $E: A_{i-1} B_{i-1}$ and $A_{i} B_{i}$. There are 6 ways to choose two consecutive edges, so there are 6 possible undirected loops in this case.

Each undirected loop can be traced in two ways, and thus the number of ways for Xena to trace the path is $(6+2) \cdot 2=16$.

Final answer: 16
Subfield: Geometry


Question: Suppose that Xena traces a path along the segments in the figure shown, starting and ending at point $A$. The path passes through each of the eleven vertices besides $A$ exactly once, and only visits $A$ at the beginning and end of the path. Compute the number of possible paths Xena could trace.

<img_3947>

Solution:
Count the number of complete paths that pass through all vertices exactly once (such a path is called a Hamiltonian path). The set of vertices can be split into two rings:

$$
\mathcal{I}=\left\{A_{1}, A_{2}, \ldots, A_{6}\right\} \text { (i.e., the inner ring), } \quad \mathcal{O}=\left\{B_{1}, B_{2}, \ldots, B_{6}\right\} \text { (i.e., the outer ring). }
$$

where $A_{1}=A$. The two rings are connected by the edges $E=\left\{A_{1} B_{1}, A_{2} B_{2}, \ldots, A_{6} B_{6}\right\}$. Each vertex in the figure has exactly three edges joining it with the neighboring vertices. Also note that any closed loop must use exactly two edges (out of three) for each vertex.

Further note that a loop must use at least one edge from $E$ to move from one ring to the other. Consider two cases: the loop uses all six edges from $E$, or it uses some but not all of them.

If all edges from $E$ are used, there are two possible undirected loops. It is not possible to use both edges $A_{1} A_{2}$ and $B_{1} B_{2}$, so either $A_{1} A_{2}$ or $B_{1} B_{2}$ will be used. This choice determines how the entire loop is constructed.

If not all edges from $E$ are used, then there must be some $i$ for which the loop uses the edge $A_{i} B_{i}$ and does not use $A_{i+1} B_{i+1}$ (where $A_{j} B_{j}$ represents $A_{j-6} B_{j-6}$ if $7 \leq j \leq 12$ ). Because $A_{i+1}$ is only connected to three other vertices, the loop must use $A_{i} A_{i+1}$ and $A_{i+1} A_{i+2}$, and similarly must use $B_{i} B_{i+1}$ and $B_{i+1} B_{i+2}$. This also precludes using $A_{i+2} B_{i+2}$, because doing so would close the loop before it visits all 12 vertices. Therefore the loop must also use $A_{i+2} A_{i+3}$ and $B_{i+2} B_{i+3}$, which now precludes using $A_{i+3} B_{i+3}$. This continues to force the structure of the loop until it closes by using $A_{i+5} B_{i+5}=A_{i-1} B_{i-1}$. Hence the loop must use exactly two edges from $E$, and they must be consecutive: $A_{i-1} B_{i-1}$ and $A_{i} B_{i}$. There are 6 ways to choose those two consecutive edges, so there are 6 possible undirected loops in this case.

The forced path is a loop, and the only way the given conditions are satisfied if $A_{i+k}=A_{i-1}$ and $B_{i+k}=B_{i-1}$. Hence the loop must use (precisely) two consecutive edges from $E: A_{i-1} B_{i-1}$ and $A_{i} B_{i}$. There are 6 ways to choose two consecutive edges, so there are 6 possible undirected loops in this case.

Each undirected loop can be traced in two ways, and thus the number of ways for Xena to trace the path is $(6+2) \cdot 2=16$.

Final answer: 16
Subfield: Algebra


Question: Let $i=\sqrt{-1}$. The complex number $z=-142+333 \sqrt{5} i$ can be expressed as a product of two complex numbers in multiple different ways, two of which are $(57-8 \sqrt{5} i)(-6+5 \sqrt{5} i)$ and $(24+\sqrt{5} i)(-3+14 \sqrt{5} i)$. Given that $z=-142+333 \sqrt{5} i$ can be written as $(a+b \sqrt{5} i)(c+d \sqrt{5} i)$, where $a, b, c$, and $d$ are positive integers, compute the lesser of $a+b$ and $c+d$.

Solution:
Multiply each of the given parenthesized expressions by its complex conjugate to obtain

$$
\begin{aligned}
142^{2}+5 \cdot 333^{2} & =\left(57^{2}+5 \cdot 8^{2}\right)\left(6^{2}+5 \cdot 5^{2}\right) \\
& =\left(24^{2}+5 \cdot 1^{2}\right)\left(3^{2}+5 \cdot 14^{2}\right) \\
& =\left(a^{2}+5 b^{2}\right)\left(c^{2}+5 d^{2}\right) .
\end{aligned}
$$

The expression on the second line is equal to $581 \cdot 989=7 \cdot 83 \cdot 23 \cdot 43$ (one can perhaps factor 989 a little faster by noting that 23 divides $6^{2}+5 \cdot 5^{2}=7 \cdot 23$ but not 581 , so it must divide 989 ). Thus $a^{2}+5 b^{2}$ and $c^{2}+5 d^{2}$ must be a factor pair of this number. It is not possible to express $1,7,23,43$, or 83 in the form $x^{2}+5 y^{2}$ for integers $x, y$.

Let $N=a^{2}+5 b^{2}$, and without loss of generality, assume that 7 divides $N$. From the above analysis, $N$ must be $7 \cdot 23,7 \cdot 43$, or $7 \cdot 83$. By direct computation of checking all positive integers $b$ less than $\sqrt{\frac{N}{5}}$, the only possibilities for $(a, b)$ are:

- when $N=7 \cdot 23$, either $(9,4)$ or $(6,5)$;
- when $N=7 \cdot 43$, either $(16,3)$ or $(11,6)$; and
- when $N=7 \cdot 83$, either $(24,1)$ or $(9,10)$.

Next, observe that

$$
\frac{-142+333 \sqrt{5} i}{a+b \sqrt{5} i}=\frac{(-142 a+1665 b)+(333 a+142 b) \sqrt{5} i}{N}
$$

must equal $c+d \sqrt{5} i$, so $N$ must divide $-142 a+1665 b$ and $333 a+142 b$. But

- 7 does not divide $333 \cdot 9+142 \cdot 4$ or $333 \cdot 6+142 \cdot 5$;
- 43 does not divide $333 \cdot 16+142 \cdot 3$; and
- 83 does not divide $333 \cdot 9+142 \cdot 10$.

Thus the only candidates are $(a, b)=(11,6)$ and $(a, b)=(24,1)$. Note that $(24,1)$ yields the second factorization given in the problem statement, which has a negative real part in one of its factors. Thus the only remaining candidate for $(a, b)$ is $(11,6)$, which yields $(c, d)=(28,15)$, thus the answer is $11+6=\mathbf{1 7}$.

Final answer: 17
Subfield: Number Theory


Question: Compute the least integer greater than 2023 , the sum of whose digits is 17 .

Solution:
A candidate for desired number is $\underline{2} \underline{0} \underline{X} \underline{Y}$, where $X$ and $Y$ are digits and $X+Y=15$. To minimize this number, take $Y=9$. Then $X=6$, and the desired number is 2069 .

Final answer: 2069
Subfield: Algebra


Question: Let $T$ = 2069, and let $K$ be the sum of the digits of $T$. Let $r$ and $s$ be the two roots of the polynomial $x^{2}-18 x+K$. Compute $|r-s|$.

Solution:
Note that $|r-s|=\sqrt{r^{2}-2 r s+s^{2}}=\sqrt{(r+s)^{2}-4 r s}$. By Vieta's Formulas, $r+s=-(-18)$ and $r s=K$, so $|r-s|=\sqrt{18^{2}-4 K}$. With $T=2069, K=17$, and the answer is $\sqrt{324-68}=\sqrt{256}=16$.

Final answer: 16
Subfield: Geometry


Question: Let $T$ be a rational number. Two coplanar squares $\mathcal{S}_{1}$ and $\mathcal{S}_{2}$ each have area $T$ and are arranged as shown to form a nonconvex octagon. The center of $\mathcal{S}_{1}$ is a vertex of $\mathcal{S}_{2}$, and the center of $\mathcal{S}_{2}$ is a vertex of $\mathcal{S}_{1}$. Compute $\frac{\text { area of the union of } \mathcal{S}_{1} \text { and } \mathcal{S}_{2}}{\text { area of the intersection of } \mathcal{S}_{1} \text { and } \mathcal{S}_{2}}$.

<img_3476>

Solution:
Let $2 x$ be the side length of the squares. Then the intersection of $\mathcal{S}_{1}$ and $\mathcal{S}_{2}$ is a square of side length $x$, so its area is $x^{2}$. The area of the union of $\mathcal{S}_{1}$ and $\mathcal{S}_{2}$ is $(2 x)^{2}+(2 x)^{2}-x^{2}=7 x^{2}$. Thus the desired ratio of areas is $\frac{7 x^{2}}{x^{2}}=7$ (independent of $T$ ).

Final answer: 7
Subfield: Geometry


Question: Let $T$ be a rational number. Two coplanar squares $\mathcal{S}_{1}$ and $\mathcal{S}_{2}$ each have area $T$ and are arranged as shown to form a nonconvex octagon. The center of $\mathcal{S}_{1}$ is a vertex of $\mathcal{S}_{2}$, and the center of $\mathcal{S}_{2}$ is a vertex of $\mathcal{S}_{1}$. Compute $\frac{\text { area of the union of } \mathcal{S}_{1} \text { and } \mathcal{S}_{2}}{\text { area of the intersection of } \mathcal{S}_{1} \text { and } \mathcal{S}_{2}}$.

![](https://cdn.mathpix.com/cropped/2023_12_21_a3766b499ee5b1dfa2f0g-1.jpg?height=241&width=290&top_left_y=411&top_left_x=1487)

Solution:
Let $2 x$ be the side length of the squares. Then the intersection of $\mathcal{S}_{1}$ and $\mathcal{S}_{2}$ is a square of side length $x$, so its area is $x^{2}$. The area of the union of $\mathcal{S}_{1}$ and $\mathcal{S}_{2}$ is $(2 x)^{2}+(2 x)^{2}-x^{2}=7 x^{2}$. Thus the desired ratio of areas is $\frac{7 x^{2}}{x^{2}}=7$ (independent of $T$ ).

Final answer: 7
Subfield: Algebra


Question: Let $T=$ 7, and let $K=9 T$. Let $A_{1}=2$, and for $n \geq 2$, let

$$
A_{n}= \begin{cases}A_{n-1}+1 & \text { if } n \text { is not a perfect square } \\ \sqrt{n} & \text { if } n \text { is a perfect square. }\end{cases}
$$

Compute $A_{K}$.

Solution:
Let $\lfloor\sqrt{n}\rfloor=x$. Then $n$ can be written as $x^{2}+y$, where $y$ is an integer such that $0 \leq y<2 x+1$. Let $m$ be the greatest perfect square less than or equal to $9 T$. Then the definition of the sequence and the previous observation imply that $A_{K}=A_{9 T}=\sqrt{m}+(9 T-m)=\lfloor\sqrt{9 T}\rfloor+\left(9 T-\lfloor\sqrt{9 T}\rfloor^{2}\right)$. With $T=7, K=9 T=63$, $\lfloor\sqrt{9 T}\rfloor=7$, and the answer is therefore $7+\left(63-7^{2}\right)=\mathbf{2 1}$.

Final answer: 21
Subfield: Number Theory


Question: Let $T=$ 21. The number $20^{T} \cdot 23^{T}$ has $K$ positive divisors. Compute the greatest prime factor of $K$.

Solution:
Write $20^{T} \cdot 23^{T}$ as $2^{2 T} \cdot 5^{T} \cdot 23^{T}$. This number has $K=(2 T+1)(T+1)^{2}$ positive divisors. With $T=21, K=43 \cdot 22^{2}$. The greatest prime factor of $K$ is $\mathbf{4 3}$.

Final answer: 43
Subfield: Number Theory


Question: Let $T=43$. Compute the positive integer $n \neq 17$ for which $\left(\begin{array}{c}T-3 \\ 17\end{array}\right)=\left(\begin{array}{c}T-3 \\ n\end{array}\right)$.

Solution:
Using the symmetry property of binomial coefficients, the desired value of $n$ is $T-3-17=T-20$. With $T=43$, the answer is $\mathbf{2 3}$.

Final answer: 23
Subfield: Number Theory


Question: Let $T=23$ . Compute the units digit of $T^{2023}+T^{20}-T^{23}$.

Solution:
Assuming that $T$ is a positive integer, because units digits of powers of $T$ cycle in groups of at most 4, the numbers $T^{2023}$ and $T^{23}$ have the same units digit, hence the number $T^{2023}-T^{23}$ has a units digit of 0 , and the answer is thus the units digit of $T^{20}$. With $T=23$, the units digit of $23^{20}$ is the same as the units digit of $3^{20}$, which is the same as the units digit of $3^{4}=81$, so the answer is $\mathbf{1}$.

Final answer: 1
Subfield: Geometry


Question: In acute triangle $I L K$, shown in the figure, point $G$ lies on $\overline{L K}$ so that $\overline{I G} \perp \overline{L K}$. Given that $I L=\sqrt{41}$ and $L G=I K=5$, compute $G K$.

<img_3922>

Solution:
Using the Pythagorean Theorem, $I G=\sqrt{(I L)^{2}-(L G)^{2}}=\sqrt{41-25}=4$, and $G K=\sqrt{(I K)^{2}-(I G)^{2}}=$ $\sqrt{25-16}=3$.

Final answer: 3
Subfield: Geometry


Question: In acute triangle $I L K$, shown in the figure, point $G$ lies on $\overline{L K}$ so that $\overline{I G} \perp \overline{L K}$. Given that $I L=\sqrt{41}$ and $L G=I K=5$, compute $G K$.

![](https://cdn.mathpix.com/cropped/2023_12_21_a3766b499ee5b1dfa2f0g-1.jpg?height=225&width=330&top_left_y=1210&top_left_x=1277)

Solution:
Using the Pythagorean Theorem, $I G=\sqrt{(I L)^{2}-(L G)^{2}}=\sqrt{41-25}=4$, and $G K=\sqrt{(I K)^{2}-(I G)^{2}}=$ $\sqrt{25-16}=3$.

Final answer: 3
Subfield: Combinatorics


Question: Let $T=$ 3. Suppose that $T$ fair coins are flipped. Compute the probability that at least one tails is flipped.

Solution:
The probability of flipping all heads is $\left(\frac{1}{2}\right)^{T}$, so the probability of flipping at least one tails is $1-\frac{1}{2^{T}}$. With $T=3$, the desired probability is $1-\frac{1}{8}=\frac{7}{8}$.

Final answer: $\frac{7}{8}$
Subfield: Algebra


Question: Let $T=$ $\frac{7}{8}$. The number $T$ can be expressed as a reduced fraction $\frac{m}{n}$, where $m$ and $n$ are positive integers whose greatest common divisor is 1 . The equation $x^{2}+(m+n) x+m n=0$ has two distinct real solutions. Compute the lesser of these two solutions.

Solution:
The left-hand side of the given equation can be factored as $(x+m)(x+n)$. The two solutions are therefore $-m$ and $-n$, so the answer is $\min \{-m,-n\}$. With $T=\frac{7}{8}, m=7, n=8$, and $\min \{-7,-8\}$ is $\mathbf{- 8}$.

Final answer: -8
Subfield: Algebra


Question: Let $T=$ -8, and let $i=\sqrt{-1}$. Compute the positive integer $k$ for which $(-1+i)^{k}=\frac{1}{2^{T}}$.

Solution:
Note that $(-1+i)^{2}=1+2 i-1=2 i$. Thus $(-1+i)^{4}=(2 i)^{2}=-4$, and $(-1+i)^{8}=(-4)^{2}=16$. The expression $\frac{1}{2^{T}}$ is a power of 16 if $T$ is a negative multiple of 4 . With $T=-8, \frac{1}{2^{-8}}=2^{8}=16^{2}=\left((-1+i)^{8}\right)^{2}=$ $(-1+i)^{16}$, so the desired value of $k$ is $\mathbf{1 6}$.

Final answer: 16
Subfield: Algebra


Question: Let $T=$ 16. Compute the value of $x$ that satisfies $\log _{4} T=\log _{2} x$.

Solution:
By the change of base rule and a property of $\operatorname{logs}, \log _{4} T=\frac{\log _{2} T}{\log _{2} 4}=\frac{\log _{2} T}{2}=\log _{2} \sqrt{T}$. Thus $x=\sqrt{T}$, and with $T=16, x=4$.

Final answer: 4
Subfield: Geometry


Question: Let $T=$ 4. Pyramid $L E O J S$ is a right square pyramid with base $E O J S$, whose area is $T$. Given that $L E=5 \sqrt{2}$, compute $[L E O]$.

Solution:
Let the side length of square base $E O J S$ be $2 x$, and let $M$ be the midpoint of $\overline{E O}$. Then $\overline{L M} \perp \overline{E O}$, and $L M=\sqrt{(5 \sqrt{2})^{2}-x^{2}}$ by the Pythagorean Theorem. Thus $[L E O]=\frac{1}{2} \cdot 2 x \sqrt{(5 \sqrt{2})^{2}-x^{2}}=$



$x \sqrt{(5 \sqrt{2})^{2}-x^{2}}$. With $T=4, x=1$, and the answer is $1 \cdot \sqrt{50-1}=\mathbf{7}$.

Final answer: 7
Subfield: Number Theory


Question: Let $T=$ 7. Compute the units digit of $T^{2023}+(T-2)^{20}-(T+10)^{23}$.

Solution:
Note that $T$ and $T+10$ have the same units digit. Because units digits of powers of $T$ cycle in groups of at most 4 , the numbers $T^{2023}$ and $(T+10)^{23}$ have the same units digit, hence the number $T^{2023}-(T+10)^{23}$ has a units digit of 0 , and the answer is thus the units digit of $(T-2)^{20}$. With $T=7$, the units digit of $5^{20}$ is 5 .

Final answer: 5
Subfield: Geometry


Question: Let $r=1$ and $R=5$. A circle with radius $r$ is centered at $A$, and a circle with radius $R$ is centered at $B$. The two circles are internally tangent. Point $P$ lies on the smaller circle so that $\overline{B P}$ is tangent to the smaller circle. Compute $B P$.

Solution:
Draw radius $A P$ and note that $A P B$ is a right triangle with $\mathrm{m} \angle A P B=90^{\circ}$. Note that $A B=R-r$ and $A P=r$, so by the Pythagorean Theorem, $B P=\sqrt{(R-r)^{2}-r^{2}}=\sqrt{R^{2}-2 R r}$. With $r=1$ and $R=5$, it follows that $B P=\sqrt{\mathbf{1 5}}$.

Final answer: $\sqrt{15}$
Subfield: Number Theory


Question: Compute the largest prime divisor of $15 !-13$ !.

Solution:
Factor 15 ! -13 ! to obtain $13 !(15 \cdot 14-1)=13$ ! $\cdot 209$. The largest prime divisor of 13 ! is 13 , so continue by factoring $209=11 \cdot 19$. Thus the largest prime divisor of 15 ! - 13 ! is 19 .

Final answer: 19
Subfield: Geometry


Question: A circle with center $O$ and radius 1 contains chord $\overline{A B}$ of length 1 , and point $M$ is the midpoint of $\overline{A B}$. If the perpendicular to $\overline{A O}$ through $M$ intersects $\overline{A O}$ at $P$, compute $[M A P]$.

Solution:
Draw auxiliary segment $\overline{O B}$, as shown in the diagram below.

<img_4031>

Triangle $O A B$ is equilateral, so $\mathrm{m} \angle O A B=60^{\circ}$. Then $\triangle M A P$ is a $30^{\circ}-60^{\circ}-90^{\circ}$ triangle with hypotenuse $A M=1 / 2$. Thus $A P=1 / 4$ and $M P=\sqrt{3} / 4$, so

$$
\begin{aligned}
{[M A P] } & =\frac{1}{2}\left(\frac{1}{4}\right)\left(\frac{\sqrt{3}}{4}\right) \\
& =\frac{\sqrt{3}}{\mathbf{3 2}} .
\end{aligned}
$$

Final answer: $\frac{\sqrt{3}}{32}$
Subfield: Number Theory


Question: $\quad$ Suppose that $p$ and $q$ are two-digit prime numbers such that $p^{2}-q^{2}=2 p+6 q+8$. Compute the largest possible value of $p+q$.

Solution:
Subtract from both sides and regroup to obtain $p^{2}-2 p-\left(q^{2}+6 q\right)=8$. Completing both squares yields $(p-1)^{2}-(q+3)^{2}=0$. The left side is a difference of two squares; factor to obtain $((p-1)+(q+3))((p-1)-(q+3))=0$, whence $(p+q+2)(p-q-4)=0$. For positive primes $p$ and $q$, the first factor $p+q+2$ must also be positive. Therefore the second factor $p-q-4$ must be zero, hence $p-4=q$. Now look for primes starting with 97 and working downward. If $p=97$, then $q=93$, which is not prime; if $p=89$, then $q=85$, which is also not prime. But if $p=83$, then $q=79$, which is prime. Thus the largest possible value of $p+q$ is $83+79=\mathbf{1 6 2}$.

Final answer: 162
Subfield: Algebra


Question: The four zeros of the polynomial $x^{4}+j x^{2}+k x+225$ are distinct real numbers in arithmetic progression. Compute the value of $j$.

Solution:
Let the four zeros be $p \leq q \leq r \leq s$. The coefficient of $x^{3}$ is 0 , so $p+q+r+s=0$. The mean of four numbers in arithmetic progression is the mean of the middle two numbers, so $q=-r$. Then the common difference is $r-q=r-(-r)=2 r$, so $s=r+2 r=3 r$ and $p=q-2 r=-3 r$. Therefore the four zeros are $-3 r,-r, r, 3 r$. The product of



the zeros is $9 r^{4}$; referring to the original polynomial and using the product of roots formula gives $9 r^{4}=225$. Thus $r=\sqrt{5}$, the zeros are $-3 \sqrt{5},-\sqrt{5}, \sqrt{5}, 3 \sqrt{5}$, and the polynomial can be factored as $(x-\sqrt{5})(x+\sqrt{5})(x-3 \sqrt{5})(x+3 \sqrt{5})$. Expanding this product yields $\left(x^{2}-5\right)\left(x^{2}-45\right)=x^{4}-50 x^{2}+225$, so $j=-50$.
Proceed as in the original solution, finding the values $-3 \sqrt{5},-\sqrt{5}, \sqrt{5}$, and $3 \sqrt{5}$ for the zeros. By the sums and products of roots formulas, the coefficient of $x^{2}$ is the sum of all six possible products of pairs of roots:

$$
(-3 \sqrt{5})(-\sqrt{5})+(-3 \sqrt{5})(\sqrt{5})+(-3 \sqrt{5})(3 \sqrt{5})+(-\sqrt{5})(\sqrt{5})+(-\sqrt{5})(3 \sqrt{5})+(\sqrt{5})(3 \sqrt{5})
$$

Observing that some of these terms will cancel yields the simpler expression

$$
(-3 \sqrt{5})(3 \sqrt{5})+(-\sqrt{5})(\sqrt{5})=-45+-5=-50
$$

Final answer: -50