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


Question: Leibniz's Harmonic Triangle: Consider the triangle formed by the rule

$$
\begin{cases}\operatorname{Le}(n, 0)=\frac{1}{n+1} & \text { for all } n \\ \operatorname{Le}(n, n)=\frac{1}{n+1} & \text { for all } n \\ \operatorname{Le}(n, k)=\operatorname{Le}(n+1, k)+\operatorname{Le}(n+1, k+1) & \text { for all } n \text { and } 0 \leq k \leq n\end{cases}
$$

This triangle, discovered first by Leibniz, consists of reciprocals of integers as shown below.

![](https://cdn.mathpix.com/cropped/2023_12_21_cadedb21e813fea89a84g-1.jpg?height=414&width=1174&top_left_y=1056&top_left_x=470)

For this contest, you may assume that $\operatorname{Le}(n, k)>0$ whenever $0 \leq k \leq n$, and that $\operatorname{Le}(n, k)$ is undefined if $k<0$ or $k>n$.
If $\sum_{i=m}^{\infty} \operatorname{Le}(i, m)=\operatorname{Le}(n, k)$, prove that $n=k=m-1$.

Solution:
Because in general $\operatorname{Le}(i, m)=\operatorname{Le}(i-1, m-1)-\operatorname{Le}(i, m-1)$, a partial sum can be rewritten as follows:

$$
\begin{aligned}
\sum_{i=m}^{n} \operatorname{Le}(i, m)= & \sum_{i=m}^{n}(\operatorname{Le}(i-1, m-1)-\operatorname{Le}(i, m-1)) \\
= & (\operatorname{Le}(m-1, m-1)-\operatorname{Le}(m, m-1))+(\operatorname{Le}(m, m-1)-\operatorname{Le}(m+1, m-1))+ \\
& \cdots+(\operatorname{Le}(n-1, m-1)-\operatorname{Le}(n, m-1)) \\
= & \operatorname{Le}(m-1, m-1)-\operatorname{Le}(n, m-1) .
\end{aligned}
$$

Because the values of $\operatorname{Le}(n, m-1)$ get arbitrarily small as $n$ increases (proof: $\operatorname{Le}(i, j)<$ $\operatorname{Le}(i-1, j-1)$ by construction, so $\left.\operatorname{Le}(n, m-1)<\operatorname{Le}(n-m+1,0)=\frac{1}{n-m+1}\right)$, the limit of these partial sums is $\operatorname{Le}(m-1, m-1)$. So $n=k=m-1$.



Note: This result can be extended even further. In fact, for every value of $k<n$,

$$
\operatorname{Le}(n, k)=\sum_{i=n+1}^{\infty} \operatorname{Le}(i, k+1)
$$

In other words, each entry in Leibniz's triangle is an infinite sum of the entries in the diagonal directly to its right, beginning with the entry below and to the right of the given one.

Final answer: 证明题，略
Subfield: Combinatorics


Question: Leibniz's Harmonic Triangle: Consider the triangle formed by the rule

$$
\begin{cases}\operatorname{Le}(n, 0)=\frac{1}{n+1} & \text { for all } n \\ \operatorname{Le}(n, n)=\frac{1}{n+1} & \text { for all } n \\ \operatorname{Le}(n, k)=\operatorname{Le}(n+1, k)+\operatorname{Le}(n+1, k+1) & \text { for all } n \text { and } 0 \leq k \leq n\end{cases}
$$

This triangle, discovered first by Leibniz, consists of reciprocals of integers as shown below.

<img_3684>

For this contest, you may assume that $\operatorname{Le}(n, k)>0$ whenever $0 \leq k \leq n$, and that $\operatorname{Le}(n, k)$ is undefined if $k<0$ or $k>n$.
If $\sum_{i=m}^{\infty} \operatorname{Le}(i, m)=\operatorname{Le}(n, k)$, prove that $n=k=m-1$.

Solution:
Because in general $\operatorname{Le}(i, m)=\operatorname{Le}(i-1, m-1)-\operatorname{Le}(i, m-1)$, a partial sum can be rewritten as follows:

$$
\begin{aligned}
\sum_{i=m}^{n} \operatorname{Le}(i, m)= & \sum_{i=m}^{n}(\operatorname{Le}(i-1, m-1)-\operatorname{Le}(i, m-1)) \\
= & (\operatorname{Le}(m-1, m-1)-\operatorname{Le}(m, m-1))+(\operatorname{Le}(m, m-1)-\operatorname{Le}(m+1, m-1))+ \\
& \cdots+(\operatorname{Le}(n-1, m-1)-\operatorname{Le}(n, m-1)) \\
= & \operatorname{Le}(m-1, m-1)-\operatorname{Le}(n, m-1) .
\end{aligned}
$$

Because the values of $\operatorname{Le}(n, m-1)$ get arbitrarily small as $n$ increases (proof: $\operatorname{Le}(i, j)<$ $\operatorname{Le}(i-1, j-1)$ by construction, so $\left.\operatorname{Le}(n, m-1)<\operatorname{Le}(n-m+1,0)=\frac{1}{n-m+1}\right)$, the limit of these partial sums is $\operatorname{Le}(m-1, m-1)$. So $n=k=m-1$.



Note: This result can be extended even further. In fact, for every value of $k<n$,

$$
\operatorname{Le}(n, k)=\sum_{i=n+1}^{\infty} \operatorname{Le}(i, k+1)
$$

In other words, each entry in Leibniz's triangle is an infinite sum of the entries in the diagonal directly to its right, beginning with the entry below and to the right of the given one.
Subfield: Geometry


Question: Spheres centered at points $P, Q, R$ are externally tangent to each other, and are tangent to plane $\mathcal{M}$ at points $P^{\prime}, Q^{\prime}, R^{\prime}$, respectively. All three spheres are on the same side of the plane. If $P^{\prime} Q^{\prime}=Q^{\prime} R^{\prime}=12$ and $P^{\prime} R^{\prime}=6$, compute the area of $\triangle P Q R$.

Solution:
Let the radii be $p, q, r$ respectively. Looking at a cross-section of the spheres through $\overline{P Q}$ perpendicular to the plane, the points $P^{\prime}, P, Q, Q^{\prime}$ form a right trapezoid with $\overline{P^{\prime} P} \perp \overline{P^{\prime} Q^{\prime}}$ and $\overline{Q^{\prime} Q} \perp \overline{P^{\prime} Q^{\prime}}$. Draw $\overline{P M}$ perpendicular to $\overline{Q Q^{\prime}}$ as shown.

<img_3907>

Then $P P^{\prime}=M Q^{\prime}=p$ and $Q M=q-p$, while $P Q=p+q$ and $P M=P^{\prime} Q^{\prime}$. By the Pythagorean Theorem, $(q-p)^{2}+P^{\prime} Q^{\prime 2}=(p+q)^{2}$, so $q=\frac{\left(P^{\prime} Q^{\prime}\right)^{2}}{4 p}$. Thus $4 p q=P^{\prime} Q^{\prime 2}=12^{2}$. Similarly, $4 p r=P^{\prime} R^{\prime 2}=6^{2}$ and $4 q r=Q^{\prime} R^{\prime 2}=12^{2}$. Dividing the first equation by the third shows that $p=r$ (which can also be inferred from the symmetry of $\triangle P^{\prime} Q^{\prime} R^{\prime}$ ) and the equation $p r=9$ yields 3 as their common value; substitute in either of the other two equations to obtain $q=12$. Therefore the sides of $\triangle P Q R$ are $P Q=Q R=12+3=15$ and $P R=6$. The altitude to $\overline{P R}$ has length $\sqrt{15^{2}-3^{2}}=6 \sqrt{6}$, so the triangle's area is $\frac{1}{2}(6)(6 \sqrt{6})=\mathbf{1 8} \sqrt{\mathbf{6}}$.

Final answer: $18 \sqrt{6}$
Subfield: Algebra


Question: Let $f(x)=x^{1}+x^{2}+x^{4}+x^{8}+x^{16}+x^{32}+\cdots$. Compute the coefficient of $x^{10}$ in $f(f(x))$.

Solution:
By the definition of $f$,

$$
f(f(x))=f(x)+(f(x))^{2}+(f(x))^{4}+(f(x))^{8}+\cdots
$$

Consider this series term by term. The first term, $f(x)$, contains no $x^{10}$ terms, so its contribution is 0 . The second term, $(f(x))^{2}$, can produce terms of $x^{10}$ in two ways: as $x^{2} \cdot x^{8}$ or as $x^{8} \cdot x^{2}$. So its contribution is 2 .

Now consider the third term:

$$
\begin{aligned}
(f(x))^{4}= & f(x) \cdot f(x) \cdot f(x) \cdot f(x) \\
= & \left(x^{1}+x^{2}+x^{4}+x^{8}+x^{16}+x^{32}+\cdots\right) \cdot\left(x^{1}+x^{2}+x^{4}+x^{8}+x^{16}+x^{32}+\cdots\right) \cdot \\
& \left(x^{1}+x^{2}+x^{4}+x^{8}+x^{16}+x^{32}+\cdots\right) \cdot\left(x^{1}+x^{2}+x^{4}+x^{8}+x^{16}+x^{32}+\cdots\right) .
\end{aligned}
$$

Each $x^{10}$ term in the product is the result of multiplying four terms whose exponents sum to 10 , one from each factor of $f(x)$. Thus this product contains a term of $x^{10}$ for each quadruple



of nonnegative integers $(i, j, k, l)$ such that $2^{i}+2^{j}+2^{k}+2^{l}=10$; the order of the quadruple is relevant because rearrangements of the integers correspond to choosing terms from different factors. Note that none of the exponents can exceed 2 because $2^{3}+2^{0}+2^{0}+2^{0}>10$. Therefore $i, j, k, l \leq 2$. Considering cases from largest values to smallest yields two basic cases. First, $10=4+4+1+1=2^{2}+2^{2}+2^{0}+2^{0}$, which yields $\frac{4 !}{2 ! \cdot 2 !}=6$ ordered quadruples. Second, $10=4+2+2+2=2^{2}+2^{1}+2^{1}+2^{1}$, which yields 4 ordered quadruples. Thus the contribution of the $(f(x))^{4}$ term is $6+4=10$.

The last term to consider is $f(x)^{8}$, because $(f(x))^{n}$ contains no terms of degree less than $n$. An analogous analysis to the case of $(f(x))^{4}$ suggests that the expansion of $(f(x))^{8}$ has an $x^{10}$ term for every ordered partition of 10 into a sum of eight powers of two. Up to order, there is only one such partition: $2^{1}+2^{1}+2^{0}+2^{0}+2^{0}+2^{0}+2^{0}+2^{0}$, which yields $\frac{8 !}{6 ! \cdot 2 !}=28$ ordered quadruples.

Therefore the coefficient of $x^{10}$ is $2+10+28=\mathbf{4 0}$.

Final answer: 40
Subfield: Algebra


Question: Compute $\left\lfloor 100000(1.002)^{10}\right\rfloor$.

Solution:
Consider the expansion of $(1.002)^{10}$ as $(1+0.002)^{10}$. Using the Binomial Theorem yields the following:

$$
(1+0.002)^{10}=1+\left(\begin{array}{c}
10 \\
1
\end{array}\right)(0.002)+\left(\begin{array}{c}
10 \\
2
\end{array}\right)(0.002)^{2}+\left(\begin{array}{c}
10 \\
3
\end{array}\right)(0.002)^{3}+\cdots+(0.002)^{10} .
$$

However, when $k>3$, the terms $\left(\begin{array}{c}10 \\ k\end{array}\right)(0.002)^{k}$ do not affect the final answer, because $0.002^{4}=$ $0.000000000016=\frac{16}{10^{12}}$, and the maximum binomial coefficient is $\left(\begin{array}{c}10 \\ 5\end{array}\right)=252$, so

$$
\left(\begin{array}{c}
10 \\
4
\end{array}\right)(0.002)^{4}+\left(\begin{array}{c}
10 \\
5
\end{array}\right)(0.002)^{5}+\cdots+(0.002)^{10}<\frac{252 \cdot 16}{10^{12}}+\frac{252 \cdot 16}{10^{12}}+\cdots+\frac{252 \cdot 16}{10^{12}},
$$

where the right side of the inequality contains seven terms, giving an upper bound of $\frac{7 \cdot 252 \cdot 16}{10^{12}}$. The numerator is approximately 28000 , but $\frac{28000}{10^{12}}=2.8 \times 10^{-8}$. So even when multiplied by $100000=10^{5}$, these terms contribute at most $3 \times 10^{-3}$ to the value of the expression before rounding.

The result of adding the first four terms $(k=0$ through $k=3)$ and multiplying by 100,000 is given by the following sum:

$$
100000+10(200)+45(0.4)+120(0.0008)=100000+2000+18+0.096=102018.096 .
$$

Then the desired quantity is $\lfloor 102018.096\rfloor=\mathbf{1 0 2 , 0 1 8}$.

Final answer: 102018
Subfield: Algebra


Question: If $1, x, y$ is a geometric sequence and $x, y, 3$ is an arithmetic sequence, compute the maximum value of $x+y$.

Solution:
The common ratio in the geometric sequence $1, x, y$ is $\frac{x}{1}=x$, so $y=x^{2}$. The arithmetic sequence $x, y, 3$ has a common difference, so $y-x=3-y$. Substituting $y=x^{2}$ in the equation yields

$$
\begin{aligned}
x^{2}-x & =3-x^{2} \\
2 x^{2}-x-3 & =0
\end{aligned}
$$

from which $x=\frac{3}{2}$ or -1 . The respective values of $y$ are $y=x^{2}=\frac{9}{4}$ or 1 . Thus the possible values of $x+y$ are $\frac{15}{4}$ and 0 , so the answer is $\frac{\mathbf{1 5}}{\mathbf{4}}$.

Final answer: $\frac{15}{4}$
Subfield: Algebra


Question: Define the sequence of positive integers $\left\{a_{n}\right\}$ as follows:

$$
\left\{\begin{array}{l}
a_{1}=1 \\
\text { for } n \geq 2, a_{n} \text { is the smallest possible positive value of } n-a_{k}^{2}, \text { for } 1 \leq k<n .
\end{array}\right.
$$

For example, $a_{2}=2-1^{2}=1$, and $a_{3}=3-1^{2}=2$. Compute $a_{1}+a_{2}+\cdots+a_{50}$.

Solution:
The requirement that $a_{n}$ be the smallest positive value of $n-a_{k}^{2}$ for $k<n$ is equivalent to determining the largest value of $a_{k}$ such that $a_{k}^{2}<n$. For $n=3$, use either $a_{1}=a_{2}=1$ to find $a_{3}=3-1^{2}=2$. For $n=4$, the strict inequality eliminates $a_{3}$, so $a_{4}=4-1^{2}=3$, but $a_{3}$ can be used to compute $a_{5}=5-2^{2}=1$. In fact, until $n=10$, the largest allowable prior value of $a_{k}$ is $a_{3}=2$, yielding the values $a_{6}=2, a_{7}=3, a_{8}=4, a_{9}=5$. In general, this pattern continues: from $n=m^{2}+1$ until $n=(m+1)^{2}$, the values of $a_{n}$ increase from 1 to $2 m+1$. Let $S_{m}=1+2+\cdots+(2 m+1)$. Then the problem reduces to computing $S_{0}+S_{1}+\cdots+S_{6}+1$, because $a_{49}=49-6^{2}$ while $a_{50}=50-7^{2}=1 . S_{m}=\frac{(2 m+1)(2 m+2)}{2}=2 m^{2}+3 m+1$, so

$$
\begin{aligned}
S_{0}+S_{1}+S_{2}+S_{3}+S_{4}+S_{5}+S_{6} & =1+6+15+28+45+66+91 \\
& =252
\end{aligned}
$$

Therefore the desired sum is $252+1=\mathbf{2 5 3}$.

Final answer: 253
Subfield: Number Theory


Question: Compute the base $b$ for which $253_{b} \cdot 341_{b}=\underline{7} \underline{4} \underline{X} \underline{Y} \underline{Z}_{b}$, for some base- $b$ digits $X, Y, Z$.

Solution:
Write $253_{b} \cdot 341_{b}=\left(2 b^{2}+5 b+3\right)\left(3 b^{2}+4 b+1\right)=6 b^{4}+23 b^{3}+31 b^{2}+17 b+3$. Compare the coefficients in this polynomial to the digits in the numeral $\underline{7} \underline{4} \underline{X} \underline{Y} \underline{Z}$. In the polynomial, the coefficient of $b^{4}$ is 6 , so there must be a carry from the $b^{3}$ place to get the $7 b^{4}$ in the numeral. After the carry, there should be no more than 4 left for the coefficient of $b^{3}$ as only one $b$ is carried. Therefore $23-b \leq 4$ or $b \geq 19$. By comparing digits, note that $Z=3$. Then

$$
\begin{aligned}
6 b^{4}+23 b^{3}+31 b^{2}+17 b & =\underline{7} \underline{4} \underline{X} \underline{Y} \underline{0} \\
& =7 b^{4}+4 b^{3}+X \cdot b^{2}+Y \cdot b
\end{aligned}
$$

Because $b>0$, this equation can be simplified to

$$
b^{3}+X \cdot b+Y=19 b^{2}+31 b+17
$$

Thus $Y=17$ and $b^{2}+X=19 b+31$, from which $b(b-19)=31-X$. The expression on the left side is positive (because $b>19$ ) and the expression on the right side is at most 31 (because $X>0$ ), so the only possible solution is $b=20, X=11$. The answer is 20 .

Final answer: 20
Subfield: Geometry


Question: Some portions of the line $y=4 x$ lie below the curve $y=10 \pi \sin ^{2} x$, and other portions lie above the curve. Compute the sum of the lengths of all the segments of the graph of $y=4 x$ that lie in the first quadrant, below the graph of $y=10 \pi \sin ^{2} x$.

Solution:
Notice first that all intersections of the two graphs occur in the interval $0 \leq x \leq \frac{5 \pi}{2}$, because the maximum value of $10 \pi \sin ^{2} x$ is $10 \pi$ (at odd multiples of $\frac{\pi}{2}$ ), and $4 x>10 \pi$ when $x>\frac{5 \pi}{2}$. The graphs are shown below.

<img_3576>

Within that interval, both graphs are symmetric about the point $A=\left(\frac{5 \pi}{4}, 5 \pi\right)$. For the case of $y=10 \pi \sin ^{2} x$, this symmetry can be seen by using the power-reducing identity $\sin ^{2} x=$ $\frac{1-\cos 2 x}{2}$. Then the equation becomes $y=5 \pi-5 \pi \cos 2 x$, which has amplitude $5 \pi$ about the line $y=5 \pi$, and which crosses the line $y=5 \pi$ for $x=\frac{\pi}{4}, \frac{3 \pi}{4}, \frac{5 \pi}{4}, \ldots$ Label the points of intersection $A, B, C, D, E, F$, and $O$ as shown. Then $\overline{A B} \cong \overline{A C}, \overline{B D} \cong \overline{C E}$, and $\overline{O D} \cong \overline{E F}$. Thus

$$
\begin{aligned}
B D+A C+E F & =O D+D B+B A \\
& =O A .
\end{aligned}
$$

By the Pythagorean Theorem,

$$
\begin{aligned}
O A & =\sqrt{\left(\frac{5 \pi}{4}\right)^{2}+(5 \pi)^{2}} \\
& =\frac{5 \pi}{4} \sqrt{1^{2}+4^{2}} \\
& =\frac{5 \pi}{\mathbf{4}} \sqrt{\mathbf{1 7}}
\end{aligned}
$$

Final answer: $\frac{5 \pi}{4} \sqrt{17}$
Subfield: Geometry


Question: In equilateral hexagon $A B C D E F, \mathrm{~m} \angle A=2 \mathrm{~m} \angle C=2 \mathrm{~m} \angle E=5 \mathrm{~m} \angle D=10 \mathrm{~m} \angle B=10 \mathrm{~m} \angle F$, and diagonal $B E=3$. Compute $[A B C D E F]$, that is, the area of $A B C D E F$.

Solution:
Let $\mathrm{m} \angle B=\alpha$. Then the sum of the measures of the angles in the hexagon is:

$$
\begin{aligned}
720^{\circ} & =\mathrm{m} \angle A+\mathrm{m} \angle C+\mathrm{m} \angle E+\mathrm{m} \angle D+\mathrm{m} \angle B+\mathrm{m} \angle F \\
& =10 \alpha+5 \alpha+5 \alpha+2 \alpha+\alpha+\alpha=24 \alpha .
\end{aligned}
$$



Thus $30^{\circ}=\alpha$ and $\mathrm{m} \angle A=300^{\circ}$, so the exterior angle at $A$ has measure $60^{\circ}=\mathrm{m} \angle D$. Further, because $A B=C D$ and $D E=A F$, it follows that $\triangle C D E \cong \triangle B A F$. Thus

$$
[A B C D E F]=[A B C E F]+[C D E]=[A B C E F]+[A B F]=[B C E F] .
$$

<img_3798>

To compute $[B C E F]$, notice that because $\mathrm{m} \angle D=60^{\circ}, \triangle C D E$ is equilateral. In addition,

$$
\begin{aligned}
150^{\circ} & =\mathrm{m} \angle B C D \\
& =\mathrm{m} \angle B C E+\mathrm{m} \angle D C E=\mathrm{m} \angle B C E+60^{\circ} .
\end{aligned}
$$

Therefore $\mathrm{m} \angle B C E=90^{\circ}$. Similarly, because the hexagon is symmetric, $\mathrm{m} \angle C E F=90^{\circ}$, so quadrilateral $B C E F$ is actually a square with side length 3 . Thus $C E=\frac{B E}{\sqrt{2}}=\frac{3}{\sqrt{2}}$, and $[A B C D E F]=[B C E F]=\frac{9}{2}$.

Alternate Solution: Calculate the angles of the hexagon as in the first solution. Then proceed as follows.

First, $A B C D E F$ can be partitioned into four congruent triangles. Because the hexagon is equilateral and $\mathrm{m} \angle A B C=\mathrm{m} \angle A F E=30^{\circ}$, it follows that $\triangle A B C$ and $\triangle A F E$ are congruent isosceles triangles whose base angles measure $75^{\circ}$. Next, $\mathrm{m} \angle A B C+\mathrm{m} \angle B C D=30^{\circ}+150^{\circ}=$ $180^{\circ}$, so $\overline{A B} \| \overline{C D}$. Because these two segments are also congruent, quadrilateral $A B C D$ is a parallelogram. In particular, $\triangle C D A \cong \triangle A B C$. Similarly, $\triangle E D A \cong \triangle A F E$.

Now let $a=A C=A E$ be the length of the base of these isosceles triangles, and let $b=A B$ be the length of the other sides (or of the equilateral hexagon). Because the four triangles are congruent, $[A B C D E F]=[A B C]+[A C D]+[A D E]+[A E F]=4[A B C]=4 \cdot \frac{1}{2} b^{2} \sin 30^{\circ}=b^{2}$. Applying the Law of Cosines to $\triangle A B C$ gives $a^{2}=b^{2}+b^{2}-2 b^{2} \cos 30^{\circ}=(2-\sqrt{3}) b^{2}$. Because $4-2 \sqrt{3}=(\sqrt{3}-1)^{2}$, this gives $a=\left(\frac{\sqrt{3}-1}{\sqrt{2}}\right) b$. Using the given length $B E=3$ and applying the Law of Cosines to $\triangle A B E$ gives

$$
\begin{aligned}
9 & =a^{2}+b^{2}-2 a b \cos 135^{\circ} \\
& =a^{2}+b^{2}+\sqrt{2} a b \\
& =(2-\sqrt{3}) b^{2}+b^{2}+(\sqrt{3}-1) b^{2} \\
& =2 b^{2} .
\end{aligned}
$$

Thus $[A B C D E F]=b^{2}=\frac{9}{2}$.

Final answer: $\frac{9}{2}$
Subfield: Geometry


Question: The taxicab distance between points $A=\left(x_{A}, y_{A}\right)$ and $B=\left(x_{B}, y_{B}\right)$ is defined as $d(A, B)=$ $\left|x_{A}-x_{B}\right|+\left|y_{A}-y_{B}\right|$. Given some $s>0$ and points $A=\left(x_{A}, y_{A}\right)$ and $B=\left(x_{B}, y_{B}\right)$, define the taxicab ellipse with foci $A=\left(x_{A}, y_{A}\right)$ and $B=\left(x_{B}, y_{B}\right)$ to be the set of points $\{Q \mid d(A, Q)+d(B, Q)=s\}$. Compute the area enclosed by the taxicab ellipse with foci $(0,5)$ and $(12,0)$, passing through $(1,-1)$.

Solution:
Let $A=(0,5)$ and $B=(12,0)$, and let $C=(1,-1)$. First compute the distance sum: $d(A, C)+d(B, C)=19$. Notice that if $P=(x, y)$ is on the segment from $(0,-1)$ to $(12,-1)$, then $d(A, P)+d(B, P)$ is constant. This is because if $0<x<12$,

$$
\begin{aligned}
d(A, P)+d(B, P) & =|0-x|+|5-(-1)|+|12-x|+|0-(-1)| \\
& =x+6+(12-x)+1 \\
& =19
\end{aligned}
$$

Similarly, $d(A, P)+d(P, B)=19$ whenever $P$ is on the segment from $(0,6)$ to $(12,6)$. If $P$ is on the segment from $(13,0)$ to $(13,5)$, then $P$ 's coordinates are $(13, y)$, with $0 \leq y \leq 5$, and thus

$$
\begin{aligned}
d(A, P)+d(B, P) & =|0-13|+|5-y|+|12-13|+|0-y| \\
& =13+(5-y)+1+y \\
& =19
\end{aligned}
$$

Similarly, $d(A, P)+d(P, B)=19$ whenever $P$ is on the segment from $(-1,0)$ to $(-1,5)$.

Finally, if $P$ is on the segment from $(12,-1)$ to $(13,0)$, then $d(A, P)+d(B, P)$ is constant:

$$
\begin{aligned}
d(A, P)+d(B, P) & =|0-x|+|5-y|+|12-x|+|0-y| \\
& =x+(5-y)+(x-12)+(-y) \\
& =2 x-2 y-7
\end{aligned}
$$

and because the line segment has equation $x-y=13$, this expression reduces to

$$
\begin{aligned}
d(A, P)+d(B, P) & =2(x-y)-7 \\
& =2(13)-7 \\
& =19
\end{aligned}
$$

Similarly, $d(A, P)+d(B, P)=19$ on the segments joining $(13,5)$ and $(12,6),(0,6)$ and $(-1,5)$, and $(-1,0)$ to $(0,-1)$. The shape of the "ellipse" is given below.

<img_3562>



The simplest way to compute the polygon's area is to subtract the areas of the four corner triangles from that of the enclosing rectangle. The enclosing rectangle's area is $14 \cdot 7=98$, while each triangle has area $\frac{1}{2} \cdot 1 \cdot 1=\frac{1}{2}$. Thus the area is $98-4 \cdot \frac{1}{2}=\mathbf{9 6}$.

Final answer: 96
Subfield: Algebra


Question: The function $f$ satisfies the relation $f(n)=f(n-1) f(n-2)$ for all integers $n$, and $f(n)>0$ for all positive integers $n$. If $f(1)=\frac{f(2)}{512}$ and $\frac{1}{f(1)}=2 f(2)$, compute $f(f(4))$.

Solution:
Substituting yields $\frac{512}{f(2)}=2 f(2) \Rightarrow(f(2))^{2}=256 \Rightarrow f(2)=16$. Therefore $f(1)=\frac{1}{32}$. Using the recursion, $f(3)=\frac{1}{2}$ and $f(4)=8$. So $f(f(4))=f(8)$. Continue to apply the recursion:

$$
f(5)=4, \quad f(6)=32, \quad f(7)=128, \quad f(8)=\mathbf{4 0 9 6} .
$$

Alternate Solution: Let $g(n)=\log _{2} f(n)$. Then $g(n)=g(n-1)+g(n-2)$, with initial conditions $g(1)=g(2)-9$ and $-g(1)=1+g(2)$. From this, $g(1)=-5$ and $g(2)=4$, and from the recursion,

$$
g(3)=-1, \quad g(4)=3
$$

so $f(4)=2^{g(4)}=8$. Continue to apply the recursion:

$$
g(5)=2, \quad g(6)=5, \quad g(7)=7, \quad g(8)=12
$$

Because $g(f(4))=12$, it follows that $f(f(4))=2^{12}=\mathbf{4 0 9 6}$.

Final answer: 4096
Subfield: Algebra


Question: Frank Narf accidentally read a degree $n$ polynomial with integer coefficients backwards. That is, he read $a_{n} x^{n}+\ldots+a_{1} x+a_{0}$ as $a_{0} x^{n}+\ldots+a_{n-1} x+a_{n}$. Luckily, the reversed polynomial had the same zeros as the original polynomial. All the reversed polynomial's zeros were real, and also integers. If $1 \leq n \leq 7$, compute the number of such polynomials such that $\operatorname{GCD}\left(a_{0}, a_{1}, \ldots, a_{n}\right)=1$.

Solution:
When the coefficients of a polynomial $f$ are reversed to form a new polynomial $g$, the zeros of $g$ are the reciprocals of the zeros of $f: r$ is a zero of $f$ if and only if $r^{-1}$ is a zero of $g$. In this case, the two polynomials have the same zeros; that is, whenever $r$ is a zero of either, so must be $r^{-1}$. Furthermore, both $r$ and $r^{-1}$ must be real as well as integers, so $r= \pm 1$. As the only zeros are \pm 1 , and the greatest common divisor of all the coefficients is 1 , the polynomial must have leading coefficient 1 or -1 . Thus

$$
\begin{aligned}
f(x) & = \pm(x \pm 1)(x \pm 1) \cdots(x \pm 1) \\
& = \pm(x+1)^{k}(x-1)^{n-k}
\end{aligned}
$$

If $A_{n}$ is the number of such degree $n$ polynomials, then there are $n+1$ choices for $k, 0 \leq k \leq n$. Thus $A_{n}=2(n+1)$. The number of such degree $n$ polynomials for $1 \leq n \leq 7$ is the sum:

$$
A_{1}+A_{2}+\ldots+A_{7}=2(2+3+\ldots+8)=2 \cdot 35=\mathbf{7 0}
$$

Final answer: 70