# Problem 1. 

Solution. As $a^{3} b-1=b\left(a^{3}+1\right)-(b+1)$ and $a+1 \mid a^{3}+1$, we have $a+1 \mid b+1$.

As $b^{3} a+1=a\left(b^{3}-1\right)+(a+1)$ and $b-1 \mid b^{3}-1$, we have $b-1 \mid a+1$.

So $b-1 \mid b+1$ and hence $b-1 \mid 2$.

- If $b=2$, then $a+1 \mid b+1=3$ gives $a=2$. Hence $(a, b)=(2,2)$ is the only solution in this case.
- If $b=3$, then $a+1 \mid b+1=4$ gives $a=1$ or $a=3$. Hence $(a, b)=(1,3)$ and $(3,3)$ are the only solutions in this case.

To summarize, $(a, b)=(1,3),(2,2)$ and $(3,3)$ are the only solutions.

![](https://cdn.mathpix.com/cropped/2024_06_05_325ea7a5a575cb0d7ed2g-1.jpg?height=460&width=460&top_left_y=2300&top_left_x=1415)

## Problem 2.

Solution. We will show that $M O P D$ is a parallelogram. From this it follows that $M, N, P$ are collinear.

Since $\angle B A D=\angle C A O=90^{\circ}-\angle A B C, D$ is the foot of the perpendicular from $A$ to side $B C$. Since $M$ is the midpoint of the line segment $B E$, we have $B M=M E=M D$ and hence $\angle M D E=\angle M E D=\angle A C B$.

Let the line $M D$ intersect the line $A C$ at $D_{1}$. Since $\angle A D D_{1}=\angle M D E=\angle A C D, M D$ is perpendicular to $A C$. On the other hand, since $O$ is the center of the circumcircle of triangle $A B C$ and $P$ is the midpoint of the side $A C, O P$ is perpendicular to $A C$. Therefore $M D$ and $O P$ are parallel.

Similarly, since $P$ is the midpoint of the side $A C$, we have $A P=P C=D P$ and hence $\angle P D C=\angle A C B$. Let the line $P D$ intersect the line $B E$ at $D_{2}$. Since $\angle B D D_{2}=\angle P D C=$ $\angle A C B=\angle B E D$, we conclude that $P D$ is perpendicular to $B E$. Since $M$ is the midpoint of the line segment $B E, O M$ is perpendicular to $B E$ and hence $O M$ and $P D$ are parallel.

![](https://cdn.mathpix.com/cropped/2024_06_05_325ea7a5a575cb0d7ed2g-2.jpg?height=1648&width=1534&top_left_y=1112&top_left_x=341)

## Problem 3.

Solution 1. By the AM-GM Inequality we have:

$$
\frac{a+1}{2}+\frac{2}{a+1} \geq 2
$$

Therefore

$$
a+2 b+\frac{2}{a+1} \geq \frac{a+3}{2}+2 b
$$

and, similarly,

$$
b+2 a+\frac{2}{b+1} \geq 2 a+\frac{b+3}{2}
$$

On the other hand,

$$
(a+4 b+3)(b+4 a+3) \geq(\sqrt{a b}+4 \sqrt{a b}+3)^{2} \geq 64
$$

by the Cauchy-Schwarz Inequality as $a b \geq 1$, and we are done.

Solution 2. Since $a b \geq 1$, we have $a+b \geq a+1 / a \geq 2 \sqrt{a \cdot(1 / a)}=2$.

Then

$$
\begin{aligned}
a+2 b+\frac{2}{a+1} & =b+(a+b)+\frac{2}{a+1} \\
& \geq b+2+\frac{2}{a+1} \\
& =\frac{b+1}{2}+\frac{b+1}{2}+1+\frac{2}{a+1} \\
& \geq 4 \sqrt[4]{\frac{(b+1)^{2}}{2(a+1)}}
\end{aligned}
$$

by the AM-GM Inequality. Similarly,

$$
b+2 a+\frac{2}{b+1} \geq 4 \sqrt[4]{\frac{(a+1)^{2}}{2(b+1)}}
$$

![](https://cdn.mathpix.com/cropped/2024_06_05_325ea7a5a575cb0d7ed2g-3.jpg?height=454&width=460&top_left_y=2303&top_left_x=1412)

Now using these and applying the AM-GM Inequality another time we obtain:

$$
\begin{aligned}
\left(a+2 b+\frac{2}{a+1}\right)\left(b+2 a+\frac{2}{b+1}\right) & \geq 16 \sqrt[4]{\frac{(a+1)(b+1)}{4}} \\
& \geq 16 \sqrt[4]{\frac{(2 \sqrt{a})(2 \sqrt{b})}{4}} \\
& =16 \sqrt[8]{a b} \\
& \geq 16
\end{aligned}
$$

Solution 3. We have

$$
\begin{aligned}
\left(a+2 b+\frac{2}{a+1}\right)\left(b+2 a+\frac{2}{b+1}\right) & =\left((a+b)+b+\frac{2}{a+1}\right)\left((a+b)+a+\frac{2}{b+1}\right) \\
& \geq\left(a+b+\sqrt{a b}+\frac{2}{\sqrt{(a+1)(b+1)}}\right)^{2}
\end{aligned}
$$

by the Cauchy-Schwarz Inequality.

On the other hand,

$$
\frac{2}{\sqrt{(a+1)(b+1)}} \geq \frac{4}{a+b+2}
$$

by the AM-GM Inequality and

$$
a+b+\sqrt{a b}+\frac{2}{\sqrt{(a+1)(b+1)}} \geq a+b+1+\frac{4}{a+b+2}=\frac{(a+b+1)(a+b-2)}{a+b+2}+4 \geq 4
$$

as $a+b \geq 2 \sqrt{a b} \geq 2$, finishing the proof.

![](https://cdn.mathpix.com/cropped/2024_06_05_325ea7a5a575cb0d7ed2g-4.jpg?height=454&width=460&top_left_y=2303&top_left_x=1415)

## Problem 4.

Solution. a. Yes. Let $a \leq b \leq c \leq d \leq e$ be the numbers chosen by Alice. As each number appears in a pairwise sum 4 times, by adding all 10 pairwise sums and dividing the result by 4, Bob obtains $a+b+c+d+e$. Subtracting the smallest and the largest pairwise sums $a+b$ and $d+e$ from this he obtains $c$. Subtracting $c$ from the second largest pairwise sum $c+e$ he obtains $e$. Subtracting $e$ from the largest pairwise sum $d+e$ he obtains $d$. He can similarly determine $a$ and $b$.

b. Yes. Let $a \leq b \leq c \leq d \leq e \leq f$ be the numbers chosen by Alice. As each number appears in a pairwise sum 5 times, by adding all 15 pairwise sums and dividing the result by 5 , Bob obtains $a+b+c+d+e+f$. Subtracting the smallest and the largest pairwise sums $a+b$ and $e+f$ from this he obtains $c+d$.

Subtracting the smallest and the second largest pairwise sums $a+b$ and $d+f$ from $a+b+c+$ $d+e+f$ he obtains $c+e$. Similarly he can obtain $b+d$. He uses these to obtain $a+f$ and $b+e$.

Now $a+d, a+e, b+c$ are the three smallest among the remaining six pairwise sums. If Bob adds these up, subtracts the known sums $c+d$ and $b+e$ from the result and divides the difference by 2 , he obtains $a$. Then he can determine the remaining numbers.

c. No. If Alice chooses the eight numbers $1,5,7,9,12,14,16,20$, then Bob cannot be sure to guess these numbers correctly as the eight numbers $2,4,6,10,11,15,17,19$ also give exactly the same 28 pairwise sums as these numbers.

![](https://cdn.mathpix.com/cropped/2024_06_05_325ea7a5a575cb0d7ed2g-5.jpg?height=452&width=460&top_left_y=2304&top_left_x=1415)