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


Question: Given a regular 16-gon, extend three of its sides to form a triangle none of whose vertices lie on the 16-gon itself. Compute the number of noncongruent triangles that can be formed in this manner.

Solution:
Label the sides of the polygon, in order, $s_{0}, s_{1}, \ldots, s_{15}$. First note that two sides of the polygon intersect at a vertex if and only if the sides are adjacent. So the sides chosen must be nonconsecutive. Second, if nonparallel sides $s_{i}$ and $s_{j}$ are extended, the angle of intersection is determined by $|i-j|$, as are the lengths of the extended portions of the segments. In other words, the spacing of the extended sides completely determines the shape of the triangle. So the problem reduces to selecting appropriate spacings, that is, finding integers $a, b, c \geq 2$ whose sum is 16 . However, diametrically opposite sides are parallel, so (for example) the sides $s_{3}$ and $s_{11}$ cannot both be used. Thus none of $a, b, c$ may equal 8 . Taking $s_{0}$ as the first side, the second side would be $s_{0+a}=s_{a}$, and the third side would be $s_{a+b}$, with $c$ sides between $s_{a+b}$ and $s_{0}$. To eliminate reflections and rotations, specify additionally that $a \geq b \geq c$. The allowable partitions are in the table below.

| $a$ | $b$ | $c$ | triangle |
| :---: | :---: | :---: | :---: |
| 12 | 2 | 2 | $s_{0} s_{12} s_{14}$ |
| 11 | 3 | 2 | $s_{0} s_{11} s_{14}$ |
| 10 | 4 | 2 | $s_{0} s_{10} s_{14}$ |
| 10 | 3 | 3 | $s_{0} s_{10} s_{13}$ |
| 9 | 5 | 2 | $s_{0} s_{9} s_{14}$ |
| 9 | 4 | 3 | $s_{0} s_{9} s_{13}$ |
| 7 | 7 | 2 | $s_{0} s_{7} s_{14}$ |
| 7 | 6 | 3 | $s_{0} s_{7} s_{13}$ |
| 7 | 5 | 4 | $s_{0} s_{7} s_{12}$ |
| 6 | 6 | 4 | $s_{0} s_{6} s_{12}$ |
| 6 | 5 | 5 | $s_{0} s_{6} s_{11}$ |

Thus there are $\mathbf{1 1}$ distinct such triangles.

Final answer: 11
Subfield: Geometry


Question: Two square tiles of area 9 are placed with one directly on top of the other. The top tile is then rotated about its center by an acute angle $\theta$. If the area of the overlapping region is 8 , compute $\sin \theta+\cos \theta$.

Solution:
In the diagram below, $O$ is the center of both squares $A_{1} A_{2} A_{3} A_{4}$ and $B_{1} B_{2} B_{3} B_{4}$. Let $P_{1}, P_{2}, P_{3}, P_{4}$ and $Q_{1}, Q_{2}, Q_{3}, Q_{4}$ be the intersections of the sides of the squares as shown. Let $H_{A}$ be on $\overline{A_{3} A_{4}}$ so that $\angle A_{3} H_{A} O$ is right. Similarly, let $H_{B}$ be on $\overline{B_{3} B_{4}}$ such that $\angle B_{3} H_{B} O$ is right. Then the angle by which $B_{1} B_{2} B_{3} B_{4}$ was rotated is $\angle H_{A} O H_{B}$. Extend $\overline{O H_{B}}$ to meet $\overline{A_{3} A_{4}}$ at $M$.

<img_3949>

Both $\triangle H_{A} O M$ and $\triangle H_{B} P_{3} M$ are right triangles sharing acute $\angle M$, so $\triangle H_{A} O M \sim \triangle H_{B} P_{3} M$. By an analogous argument, both triangles are similar to $\triangle B_{3} P_{3} Q_{3}$. Thus $\mathrm{m} \angle Q_{3} P_{3} B_{3}=\theta$. Now let $B_{3} P_{3}=x, B_{3} Q_{3}=y$, and $P_{3} Q_{3}=z$. By symmetry, notice that $B_{3} P_{3}=B_{2} P_{2}$ and that $P_{3} Q_{3}=P_{2} Q_{3}$. Thus

$$
x+y+z=B_{3} Q_{3}+Q_{3} P_{2}+P_{2} B_{2}=B_{2} B_{3}=3 .
$$

By the Pythagorean Theorem, $x^{2}+y^{2}=z^{2}$. Therefore

$$
\begin{aligned}
x+y & =3-z \\
x^{2}+y^{2}+2 x y & =9-6 z+z^{2} \\
2 x y & =9-6 z .
\end{aligned}
$$

The value of $x y$ can be determined from the areas of the four triangles $\triangle B_{i} P_{i} Q_{i}$. By symmetry, these four triangles are congruent to each other. Their total area is the area not in both squares, i.e., $9-8=1$. Thus $\frac{x y}{2}=\frac{1}{4}$, so $2 x y=1$. Applying this result to the above equation,

$$
\begin{aligned}
1 & =9-6 z \\
z & =\frac{4}{3}
\end{aligned}
$$



The desired quantity is $\sin \theta+\cos \theta=\frac{x}{z}+\frac{y}{z}$, and

$$
\begin{aligned}
\frac{x}{z}+\frac{y}{z} & =\frac{x+y+z}{z}-\frac{z}{z} \\
& =\frac{3}{z}-1 \\
& =\frac{\mathbf{5}}{\mathbf{4}}
\end{aligned}
$$

Final answer: $\frac{5}{4}$
Subfield: Number Theory


Question: Suppose that neither of the three-digit numbers $M=\underline{4} \underline{A} \underline{6}$ and $N=\underline{1} \underline{B} \underline{7}$ is divisible by 9 , but the product $M \cdot N$ is divisible by 9 . Compute the largest possible value of $A+B$.

Solution:
In order for the conditions of the problem to be satisfied, $M$ and $N$ must both be divisible by 3 , but not by 9 . Thus the largest possible value of $A$ is 5 , and the largest possible value of $B$ is 7 , so $A+B=\mathbf{1 2}$.

Final answer: 12
Subfield: Geometry


Question: Let $T=12$. Each interior angle of a regular $T$-gon has measure $d^{\circ}$. Compute $d$.

Solution:
From the angle sum formula, $d^{\circ}=\frac{180^{\circ} \cdot(T-2)}{T}$. With $T=12, d=\mathbf{1 5 0}$.

Final answer: 150
Subfield: Algebra


Question: Suppose that $r$ and $s$ are the two roots of the equation $F_{k} x^{2}+F_{k+1} x+F_{k+2}=0$, where $F_{n}$ denotes the $n^{\text {th }}$ Fibonacci number. Compute the value of $(r+1)(s+1)$.

Solution:
$\quad$ Distributing, $(r+1)(s+1)=r s+(r+s)+1=\frac{F_{k+2}}{F_{k}}+\left(-\frac{F_{k+1}}{F_{k}}\right)+1=\frac{F_{k+2}-F_{k+1}}{F_{k}}+1=\frac{F_{k}}{F_{k}}+1=\mathbf{2}$.

Final answer: 2
Subfield: Algebra


Question: Let $T=2$. Compute the product of $-T-i$ and $i-T$, where $i=\sqrt{-1}$.

Solution:
Multiplying, $(-T-i)(i-T)=-(i+T)(i-T)=-\left(i^{2}-T^{2}\right)=1+T^{2}$. With $T=2,1+T^{2}=\mathbf{5}$.

Final answer: 5
Subfield: Combinatorics


Question: Let $T=5$. Compute the number of positive divisors of the number $20^{4} \cdot 11^{T}$ that are perfect cubes.

Solution:
Let $N=20^{4} \cdot 11^{T}=2^{8} \cdot 5^{4} \cdot 11^{T}$. If $m \mid N$, then $m=2^{a} \cdot 5^{b} \cdot 11^{c}$ where $a, b$, and $c$ are nonnegative integers such that $a \leq 8, b \leq 4$, and $c \leq T$. If $m$ is a perfect cube, then $a, b$, and $c$ must be divisible by 3 . So $a=0,3$, or $6 ; b=0$ or 3 , and $c \in\{0,3, \ldots, 3 \cdot\lfloor T / 3\rfloor\}$. There are a total of $3 \cdot 2 \cdot(\lfloor T / 3\rfloor+1)$ possible values of $m$. For $T=5,\lfloor T / 3\rfloor+1=2$, so the number of possible values of $m$ is $\mathbf{1 2}$.

Final answer: 12
Subfield: Geometry


Question: $\quad$ Let $T=12$. As shown, three circles are mutually externally tangent. The large circle has a radius of $T$, and the smaller two circles each have radius $\frac{T}{2}$. Compute the area of the triangle whose vertices are the centers of the three circles.

<img_3787>

Solution:
The desired triangle is an isosceles triangle whose base vertices are the centers of the two smaller circles. The congruent sides of the triangle have length $T+\frac{T}{2}$. Thus the altitude to the base has length $\sqrt{\left(\frac{3 T}{2}\right)^{2}-\left(\frac{T}{2}\right)^{2}}=T \sqrt{2}$. Thus the area of the triangle is $\frac{1}{2} \cdot\left(\frac{T}{2}+\frac{T}{2}\right) \cdot T \sqrt{2}=\frac{T^{2} \sqrt{2}}{2}$. With $T=12$, the area is $\mathbf{7 2} \sqrt{\mathbf{2}}$.

Final answer: $72 \sqrt{2}$
Subfield: Geometry


Question: $\quad$ Let $T=12$. As shown, three circles are mutually externally tangent. The large circle has a radius of $T$, and the smaller two circles each have radius $\frac{T}{2}$. Compute the area of the triangle whose vertices are the centers of the three circles.

![](https://cdn.mathpix.com/cropped/2023_12_21_9a82243db0768b0d2f1cg-1.jpg?height=436&width=632&top_left_y=1273&top_left_x=1343)

Solution:
The desired triangle is an isosceles triangle whose base vertices are the centers of the two smaller circles. The congruent sides of the triangle have length $T+\frac{T}{2}$. Thus the altitude to the base has length $\sqrt{\left(\frac{3 T}{2}\right)^{2}-\left(\frac{T}{2}\right)^{2}}=T \sqrt{2}$. Thus the area of the triangle is $\frac{1}{2} \cdot\left(\frac{T}{2}+\frac{T}{2}\right) \cdot T \sqrt{2}=\frac{T^{2} \sqrt{2}}{2}$. With $T=12$, the area is $\mathbf{7 2} \sqrt{\mathbf{2}}$.

Final answer: $72 \sqrt{2}$
Subfield: Algebra


Question: Let $T=72 \sqrt{2}$, and let $K=\left(\frac{T}{12}\right)^{2}$. In the sequence $0.5,1,-1.5,2,2.5,-3, \ldots$, every third term is negative, and the absolute values of the terms form an arithmetic sequence. Compute the sum of the first $K$ terms of this sequence.

Solution:
The general sequence looks like $x, x+d,-(x+2 d), x+3 d, x+4 d,-(x+5 d), \ldots$ The sum of the first three terms is $x-d$; the sum of the second three terms is $x+2 d$; the sum of the third three terms is $x+5 d$, and so on. Thus the sequence of sums of terms $3 k-2,3 k-1$, and $3 k$ is an arithmetic sequence. Notice that $x=d=0.5$ and so $x-d=0$. If there are $n$ triads of terms of the original sequence, then their common difference is 1.5 and their sum is $n \cdot\left(\frac{0+0+(n-1) \cdot 1.5}{2}\right) \cdot T=72 \sqrt{2}$, so $K=72$, and $n=24$. Thus the desired sum is 414.

Final answer: 414
Subfield: Algebra


Question: Let $T=9.5$. If $\log _{2} x^{T}-\log _{4} x=\log _{8} x^{k}$ is an identity for all $x>0$, compute the value of $k$.

Solution:
Note that in general, $\log _{b} c=\log _{b^{n}} c^{n}$. Using this identity yields $\log _{2} x^{T}=\log _{2^{2}}\left(x^{T}\right)^{2}=$ $\log _{4} x^{2 T}$. Thus the left hand side of the given equation simplifies to $\log _{4} x^{2 T-1}$. Express each side in base 64: $\log _{4} x^{2 T-1}=\log _{64} x^{6 T-3}=\log _{64} x^{2 k}=\log _{8} x^{k}$. Thus $k=3 T-\frac{3}{2}$. With $T=9.5, k=\mathbf{2 7}$.

Final answer: 27
Subfield: Geometry


Question: Let $T=16$. An isosceles trapezoid has an area of $T+1$, a height of 2 , and the shorter base is 3 units shorter than the longer base. Compute the sum of the length of the shorter base and the length of one of the congruent sides.

Solution:
Let $x$ be the length of the shorter base of the trapezoid. The area of the trapezoid is $\frac{1}{2} \cdot 2$. $(x+x+3)=T+1$, so $x=\frac{T}{2}-1$. Drop perpendiculars from each vertex of the shorter base to the longer base, and note that by symmetry, the feet of these perpendiculars lie $\frac{3}{2}=1.5$ units away from their nearest vertices of the trapezoid. Hence the congruent sides have length $\sqrt{1.5^{2}+2^{2}}=2.5$. With $T=16, x=7$, and the desired sum of the lengths is $\mathbf{9 . 5}$.

Final answer: 9.5
Subfield: Combinatorics


Question: Let $T=10$. Susan flips a fair coin $T$ times. Leo has an unfair coin such that the probability of flipping heads is $\frac{1}{3}$. Leo gets to flip his coin the least number of times so that Leo's expected number of heads will exceed Susan's expected number of heads. Compute the number of times Leo gets to flip his coin.

Solution:
The expected number of heads for Susan is $\frac{T}{2}$. If Leo flips his coin $N$ times, the expected number of heads for Leo is $\frac{N}{3}$. Thus $\frac{N}{3}>\frac{T}{2}$, so $N>\frac{3 T}{2}$. With $T=10$, the smallest possible value of $N$ is $\mathbf{1 6}$.

Final answer: 16
Subfield: Algebra


Question: Let $T=1$. Dennis and Edward each take 48 minutes to mow a lawn, and Shawn takes 24 minutes to mow a lawn. Working together, how many lawns can Dennis, Edward, and Shawn mow in $2 \cdot T$ hours? (For the purposes of this problem, you may assume that after they complete mowing a lawn, they immediately start mowing the next lawn.)

Solution:
Working together, Dennis and Edward take $\frac{48}{2}=24$ minutes to mow a lawn. When the three of them work together, it takes them $\frac{24}{2}=12$ minutes to mow a lawn. Thus they can mow 5 lawns per hour. With $T=1$, they can mow $5 \cdot 2=\mathbf{1 0}$ lawns in 2 hours.

Final answer: 10
Subfield: Algebra


Question: Let T be a rational number. Compute $\sin ^{2} \frac{T \pi}{2}+\sin ^{2} \frac{(5-T) \pi}{2}$.

Solution:
Note that $\sin \frac{(5-T) \pi}{2}=\cos \left(\frac{\pi}{2}-\frac{(5-T) \pi}{2}\right)=\cos \left(\frac{T \pi}{2}-2 \pi\right)=\cos \frac{T \pi}{2}$. Thus the desired quantity is $\sin ^{2} \frac{T \pi}{2}+\cos ^{2} \frac{T \pi}{2}=\mathbf{1}$ (independent of $T$ ).

Final answer: 1
Subfield: Algebra


Question: Let $T=11$. Compute the value of $x$ that satisfies $\sqrt{20+\sqrt{T+x}}=5$.

Solution:
Squaring each side gives $20+\sqrt{T+x}=25$, thus $\sqrt{T+x}=5$, and $x=25-T$. With $T=11$, $x=14$.

Final answer: 14
Subfield: Geometry


Question: The sum of the interior angles of an $n$-gon equals the sum of the interior angles of a pentagon plus the sum of the interior angles of an octagon. Compute $n$.

Solution:
Using the angle sum formula, $180^{\circ} \cdot(n-2)=180^{\circ} \cdot 3+180^{\circ} \cdot 6=180^{\circ} \cdot 9$. Thus $n-2=9$, and $n=11$.

Final answer: 11