# Solutions of EGMO 2020 

Problem 1. The positive integers $a_{0}, a_{1}, a_{2}, \ldots, a_{3030}$ satisfy

$$
2 a_{n+2}=a_{n+1}+4 a_{n} \text { for } n=0,1,2, \ldots, 3028
$$

Prove that at least one of the numbers $a_{0}, a_{1}, a_{2}, \ldots, a_{3030}$ is divisible by $2^{2020}$.

Problem 2. Find all lists $\left(x_{1}, x_{2}, \ldots, x_{2020}\right)$ of non-negative real numbers such that the following three conditions are all satisfied:
(i) $x_{1} \leq x_{2} \leq \ldots \leq x_{2020}$ ；
(ii) $x_{2020} \leq x_{1}+1$;
(iii) there is a permutation $\left(y_{1}, y_{2}, \ldots, y_{2020}\right)$ of $\left(x_{1}, x_{2}, \ldots, x_{2020}\right)$ such that

$$
\sum_{i=1}^{2020}\left(\left(x_{i}+1\right)\left(y_{i}+1\right)\right)^{2}=8 \sum_{i=1}^{2020} x_{i}^{3}
$$

A permutation of a list is a list of the same length, with the same entries, but the entries are allowed to be in any order. For example, $(2,1,2)$ is a permutation of $(1,2,2)$, and they are both permutations of $(2,2,1)$. Note that any list is a permutation of itself.

Problem 3. Let $A B C D E F$ be a convex hexagon such that $\angle A=\angle C=\angle E$ and $\angle B=\angle D=$ $\angle F$ and the (interior) angle bisectors of $\angle A, \angle C$, and $\angle E$ are concurrent.

Prove that the (interior) angle bisectors of $\angle B, \angle D$, and $\angle F$ must also be concurrent.

Note that $\angle A=\angle F A B$. The other interior angles of the hexagon are similarly described.

Problem 4. A permutation of the integers $1,2, \ldots, m$ is called fresh if there exists no positive integer $k<m$ such that the first $k$ numbers in the permutation are $1,2, \ldots, k$ in some order. Let $f_{m}$ be the number of fresh permutations of the integers $1,2, \ldots, m$.
Prove that $f_{n} \geq n \cdot f_{n-1}$ for all $n \geq 3$.
For example, if $m=4$, then the permutation $(3,1,4,2)$ is fresh, whereas the permutation $(2,3,1,4)$ is not.

Problem 5. Consider the triangle $A B C$ with $\angle B C A>90^{\circ}$. The circumcircle $\Gamma$ of $A B C$ has radius $R$. There is a point $P$ in the interior of the line segment $A B$ such that $P B=P C$ and the length of $P A$ is $R$. The perpendicular bisector of $P B$ intersects $\Gamma$ at the points $D$ and $E$.
Prove that $P$ is the incentre of triangle $C D E$.
Problem 6. Let $m>1$ be an integer. A sequence $a_{1}, a_{2}, a_{3}, \ldots$ is defined by $a_{1}=a_{2}=1$, $a_{3}=4$, and for all $n \geq 4$,

$$
a_{n}=m\left(a_{n-1}+a_{n-2}\right)-a_{n-3} .
$$

Determine all integers $m$ such that every term of the sequence is a square.

## Solutions to Problem 1

There are different ways of solving the problem. All of these use some induction argument. Most of these proofs use one of the following two lemmas. In many places, they can be used interchangeably.

Lemma. If $a, b, c, d$ are integers with $2 c=b+4 a$ and $2 d=c+4 b$, then $4 \mid b$.
Proof: From $2 d=c+4 b$ we have that $c$ is even, and then from $2 c=b+4 a$ it follows that $b$ is divisible by 4 .

Lemma：For $0 \leq n \leq 3030$, denote by $v_{n}$ the largest integer such that $2^{v_{n}}$ divides $a_{n}$. We claim the following:

$$
(*) \quad v_{n+1} \geq \min \left(v_{n}+2, v_{n+2}+1\right) \quad \text { for } n=0,1, \ldots, 3028
$$

Proof: Let $0 \leq n \leq 3028$ and let $s=\min \left(v_{n}+2, v_{n+2}+1\right)$. Then $s \leq v_{n}+2$ implies $2^{s} \mid 4 a_{n}$ and $s \leq v_{n+2}+1$ implies $2^{s} \mid 2 a_{n+2}$. It follows that $a_{n+1}=2 a_{n+2}-4 a_{n}$ is also divisible by $2^{s}$, hence $s \leq v_{n+1}$, which proves $(*)$.

Here are different ways of working out the induction argument that is crucial in the proofs.

## Induction part, alternative A.

Statement: For $k=0,1, \ldots, 1010$, the terms $a_{k}, a_{k+1}, \ldots, a_{3030-2 k}$ are all divisible by $2^{2 k}$.
Reformulation of the statement using notation $v_{n}$ is the largest integer such that $2^{v_{n}}$ divides $a_{n}$ : we have $v_{n} \geq k$ for any $n$ satisfying $\left\lceil\frac{1}{2} k\right\rceil \leq n \leq 3030-k$. Here $\lceil x\rceil$ denotes the smallest integer not smaller than $x$.

Proof 1: We proceed by induction on $k$. For $k=0$ the statement is obvious, so, for the inductive step, suppose that $a_{k}, a_{k+1}, \ldots, a_{3030-2 k}$ are all divisible by $2^{2 k}$. Apply the Lemma with

$$
(a, b, c, d)=\left(\frac{a_{i-1}}{2^{2 k}}, \frac{a_{i}}{2^{2 k}}, \frac{a_{i+1}}{2^{2 k}}, \frac{a_{i+2}}{2^{2 k}}\right)
$$

for $i=k+1, k+2, \ldots, 3030-2 k-2$. We obtain that $\frac{a_{i}}{2^{2 k}}$ is divisible by 4 (and hence $a_{i}$ is divisible by $2^{2 k+2}$ ) for $i=k+1, k+2, \ldots, i=3030-2 k-2$. This completes the induction. For $k=1010$ we obtain that $a_{1010}$ is divisible by $2^{2020}$, and the solution is complete.

Remark. Remark, notice that by replacing 1010 with $n$ (and hence, $2 n$ with 2020 and $3 n$ with 3030, this argument works, too. Then the claim is the following: if $a_{0}, a_{1}, \ldots, a_{3 n}$ are integers that satisfy the recursion in the problem, then $a_{n}$ is divisible by $2^{2 n}$.

Proof 2: We will show by two step induction to $k \geq 0$ : we have $v_{n} \geq k$ for any $n$ satisfying $\left\lceil\frac{1}{2} k\right\rceil \leq n \leq 3030-k$. Here $\lceil x\rceil$ denotes the smallest integer not smaller than $x$. Plugging in $k=2020$ and $n=1010$ will give the desired result.
The case $k=0$ is trivial, since the $v_{n}$ are non-negative. For the case $k=1$, let $1 \leq n \leq 3029$, then $v_{n} \geq \min \left(v_{n-1}+2, v_{n+1}+1\right) \geq 1$.
Suppose we have it proven for some $k \geq 1$ and for $k-1$. Let $\left\lceil\frac{1}{2}(k+1)\right\rceil \leq n \leq 3030-(k+1)$. Then $\left\lceil\frac{1}{2}(k-1)\right\rceil \leq n-1 \leq 3030-(k-1)$ and also $\left\lceil\frac{1}{2} k\right\rceil \leq n+1 \leq 3030-k$. By induction hypothesis we have $v_{n-1} \geq k-1$ and $v_{n+1} \geq k$. Then $v_{n} \geq \min \left(v_{n-1}+2, v_{n+1}+1\right) \geq k$, finishing the induction.

Proof 3: The notation is the same as in alternative A1, but the induction step has two steps. We use the lemma stated in the beginning as the first step. Assume now that $2^{2 k} \mid a_{k}, \ldots, a_{3030-2 k}$ for some $k \geq 1$. We claim that then $2^{2 k+2} \mid a_{k+1}, \ldots, a_{3030-2(k+1)}$. Notice first that any $i$ on the interval $[k+1,3030-2 k-1]$ satisfies the equation $2 a_{i+1}=a_{i}+4 a_{i-2}$, we have $2^{2 k+1} \mid a_{i}$. Furthermore, if $i \in[k+1,3030-2 k-2]$, we have $2^{2 k+2} \mid a_{i}$ since $2^{2 k+1} \mid a_{i+1}$ and hence
$2^{2 k+2} \mid 2 a_{i+1}$ and $2^{2 k+2} \mid 4 a_{i-1}$.

## Induction part, alternative B.

Statement: if $a_{0}, a_{1}, \ldots, a_{3 n}$ are integers that satisfy the recursion in the problem, then $a_{n}$ is divisible by $2^{2 n}$. The problem statement follows for $n=1010$.

Proof: The base case of the induction is exactly the Lemma we just proved. Now, for the inductive step, suppose that the statement holds for $n=k-1$, and consider integers $a_{0}, a_{1}$, $\ldots, a_{3 k}$ that satisfy the recursion. By applying the induction hypothesis to the four sequences $\left(a_{0}, a_{1}, \ldots, a_{3 k-3}\right),\left(a_{1}, a_{2}, \ldots, a_{3 k-2}\right),\left(a_{2}, a_{3}, \ldots, a_{3 k-1}\right)$ and $\left(a_{3}, a_{4}, \ldots, a_{3 k}\right)$, we find that $a_{k-1}$, $a_{k}, a_{k+1}$ and $a_{k+2}$ are all divisible by $2^{2 k-2}$. If we now apply the lemma to $a_{k-1} / 2^{2 k-2}, a_{k} / 2^{2 k-2}$, $a_{k+1} / 2^{2 k-2}$ and $a_{k+2} / 2^{2 k-2}$, we find that $a_{k} / 2^{2 k-2}$ is divisible by 4 , so $a_{k}$ is divisible by $2^{2 k}$, as desired.

## Induction part, alternative C.

Statement: Given positive integers $a_{0}, a_{1}, a_{2}, \ldots, a_{3 k}$ such that

$$
2 a_{n+2}=a_{n+1}+4 a_{n} \text { for } n=0,1,2, \ldots, 3 k-2
$$

then $2^{2 k}$ divides at least one of the numbers $a_{0}, a_{1}, a_{2}, \ldots, a_{3 k}$.
Proof: The case $k=1$ is obtained from Lemma.
Suppose that for some $k \geq 1$, our claim is true for any sequence of $3 k+1$ positive integers that satisfy similar defining relations. Then consider a sequence of $3(k+1)+1=3 k+4$ positive integers $a_{0}, a_{1}, a_{2}, \ldots, a_{3 k+3}$ such that

$$
2 a_{n+2}=a_{n+1}+4 a_{n} \text { for } n=0,1,2, \ldots, 3 k+1
$$

Then, we see that all numbers $a_{1}, a_{2}, \ldots, a_{3 k+2}$ are even, and so we may set $a_{i}=2 b_{i}$ for some positive integers $b_{i}$ for $i=1,2, \ldots, 3 k+2$. Then $2 a_{2}=a_{1}+4 a_{0}$ becomes

$$
2 b_{2}=b_{1}+2 a_{0}
$$

Note that

$$
2 b_{n+2}=b_{n+1}+4 b_{n} \text { for } n=1,2, \ldots, 3 k
$$

Hence the positive integers $b_{1}, b_{2}, \ldots, b_{3 k+1}$ are even, and so we may write $b_{i}=2 c_{i}$ for some positive integers $c_{i}$ for $i=1,2, \ldots, 3 k+1$. Then $2 b_{2}=b_{1}+2 a_{0}$ becomes $2 c_{2}=c_{1}+a_{0}$, which does not give us anything. However, we have

$$
2 c_{n+2}=c_{n+1}+4 c_{n} \text { for } n=1,2, \ldots, 3 k-1
$$

So the sequence of positive integers $c_{1}, c_{2}, \ldots, c_{3 k+1}$ is a sequence of $3 k+1$ positive integers that satisfy the defining relations. By the inductive hypothesis, $2^{2 k}$ divides at least one of the numbers $c_{1}, c_{2}, \ldots, c_{3 k+1}$. Since $a_{i}=4 c_{i}$ for all $i=1,2, \ldots, 3 k+1$, it follows that $4 \cdot 2^{2 k}=2^{2(k+1)}$ divides at least one of the numbers $a_{1}, \ldots, a_{3 k+1}$. This completes the induction, and hence the proof.

## Induction part, alternative D.

Statement: If $v_{0}, v_{1}, \ldots, v_{3030}$ is a sequence of non-negative integers satisfying $(*)$, there must be a $k$ such that $v_{k} \geq 2020$. In fact, we will show that $v_{1010} \geq 2020$.

These proofs use Lemma'.

Proof 1: We will show by induction on $k$ that $v_{n} \geq 2 k$ for $k \leq n \leq 3030-2 k$ and $v_{n} \geq 2 k+1$ for $k+1 \leq n \leq 3030-2 k-1$. For $k=1010$, the first statement implies that $v_{1010} \geq 2020$.
For $k=0$ the first statement $v_{n} \geq 0$ is obvious, and the second statement follows using ( $*$ ): we have $v_{n} \geq v_{n+1}+1 \geq 1$ for $1 \leq n \leq 3029$.
Now suppose that the inductive hypothesis holds for $k=\ell$ : we have $v_{n} \geq 2 \ell$ for $\ell \leq n \leq 3030-2 \ell$ and $v_{n} \geq 2 \ell+1$ for $\ell+1 \leq n \leq 3030-2 \ell-1$. For the first statement, consider an $n$ with $\ell+1 \leq n \leq 3030-2 \ell-2$. Then using $(*)$ and the inductive hypothesis, we obtain

$$
v_{n} \geq \min \left(v_{n-1}+2, v_{n+1}+1\right) \geq \min (2 \ell+2,2 \ell+2)=2 \ell+2
$$

because $v_{n-1} \geq 2 \ell$ (as $\left.\ell \leq n-1 \leq 3030-2 \ell\right)$ and $v_{n+1} \geq 2 \ell+1($ as $\ell+1 \leq n+1 \leq 3030-2 \ell-1)$. Similarly, for $\ell+2 \leq n \leq 3030-2 \ell-3$ we find

$$
v_{n} \geq \min \left(v_{n-1}+2, v_{n+1}+1\right) \geq \min (2 \ell+3,2 \ell+3)=2 \ell+3
$$

because $v_{n-1} \geq 2 \ell+1$ (as $\left.\ell+1 \leq n-1 \leq 3030-2 \ell-1\right)$ and $v_{n+1} \geq 2 \ell+2($ as $\ell+1 \leq n \leq$ $3030-2 \ell-2$ ). This completes the induction.

Proof 2: The inequality $v_{1010} \geq \min \left(v_{1009}+2, v_{1011}+1\right)$ gives us two cases to consider: either $v_{1010} \geq v_{1009}+2$ or $v_{1010} \geq v_{1011}+1$.
Suppose first that $v_{1010} \geq v_{1009}+2$. In this case we will show by induction on $k$ that $v_{1010-k} \geq$ $v_{1009-k}+2$ for $0 \leq k \leq 1009$. The case $k=0$ is assumed, so suppose that $v_{1010-\ell} \geq v_{1009-\ell}+2$ holds for some $0 \leq \ell<1009$. We obtain

$$
v_{1009-\ell} \geq \min \left(v_{1010-\ell}+1, v_{1008-\ell}+2\right) \geq \min \left(v_{1009-\ell}+3, v_{1008-\ell}+2\right)
$$

Because $v_{1009-\ell}<v_{1009-\ell}+3$ we must have $v_{1009-\ell} \geq v_{1008-\ell}+2$, completing the induction. We conclude that

$$
v_{1010} \geq v_{1009}+2 \geq v_{1008}+4 \geq \cdots \geq v_{0}+2020 \geq 2020
$$

In the second case, we show by induction on $k$ that $v_{1010+k} \geq v_{1011+k}+1$ for $0 \leq k \leq 2019$. Again, the base case $k=0$ is assumed, so suppose that $v_{1010+\ell} \geq v_{1011+\ell}+1$ holds for some $0 \leq \ell<2019$. We obtain

$$
v_{1011+\ell} \geq \min \left(v_{1010+\ell}+2, v_{1012+\ell}+1\right) \geq \min \left(v_{1011+\ell}+3, v_{1012+\ell}+1\right)
$$

Because $v_{1011+\ell}<v_{1011+\ell}+3$, we must have $v_{1011+\ell} \geq v_{1012+\ell}+1$, completing the induction. We conclude that

$$
v_{1010} \geq v_{1011}+1 \geq v_{1012}+2 \geq \cdots \geq v_{3030}+2020 \geq 2020
$$

as desired.

## Alternative E.

This solution is different from the other solutions. Shift the sequence so that it starts at $a_{-1010}$ and ends at $a_{2020}$. We will show that $a_{0}$ is divisible by $2^{2020}$. Consider $a_{0}$. It either has at least as many factors 2 as $2 a_{1}$, or at least as many factors 2 as $4 a_{-1}$ (this follows from $2 a_{1}=a_{0}+4 a_{-1}$ ). Consider the first case, so $e_{2}\left(a_{0}\right) \geq e_{2}\left(2 a_{1}\right)$. By multiplying the original recursion by $2^{n-1}$, we note that $b_{n}=2^{n} a_{n}$ satisfies the recursion $b_{n}=b_{n-1}+8 b_{n-2}$. Furthermore, we assumed that $e_{2}\left(b_{0}\right) \geq e_{2}\left(b_{1}\right)$. This implies that $e_{2}\left(b_{n}\right)$ is constant for $n \geq 1$. Furthermore, clearly $2^{2020} \mid b_{2020}$, so $2^{2020} \mid b_{1}$ and hence $2^{2020} \mid a_{0}$. The case where $e_{2}\left(a_{0}\right) \geq e_{2}\left(4 a_{-1}\right)$ is similar; we then look at $b_{n}=4^{n} a_{-n}$ which satisfies $b_{n}=8 b_{n-2}-b_{n-1}$ and use that $2^{2020} \mid b_{1010}$. The rest of the argument is the same.

## Solutions to Problem 2

Answer. There are two solutions: $(\underbrace{0,0, \ldots, 0}_{1010}, \underbrace{1,1, \ldots, 1}_{1010})$ and $(\underbrace{1,1, \ldots, 1}_{1010}, \underbrace{2,2, \ldots, 2}_{1010})$.
Solution A. We first prove the inequality

$$
((x+1)(y+1))^{2} \geq 4\left(x^{3}+y^{3}\right)
$$

for real numbers $x, y \geq 0$ satisfying $|x-y| \leq 1$, with equality if and only if $\{x, y\}=\{0,1\}$ or $\{x, y\}=\{1,2\}$.
Indeed,

$$
\begin{aligned}
4\left(x^{3}+y^{3}\right) & =4(x+y)\left(x^{2}-x y+y^{2}\right) \\
& \leq\left((x+y)+\left(x^{2}-x y+y^{2}\right)\right)^{2} \\
& =\left(x y+x+y+(x-y)^{2}\right)^{2} \\
& \leq(x y+x+y+1)^{2} \\
& =((x+1)(y+1))^{2},
\end{aligned}
$$

where the first inequality follows by applying the AM-GM inequality on $x+y$ and $x^{2}-x y+y^{2}$ (which are clearly nonnegative). Equality holds in the first inequality precisely if $x+y=$ $x^{2}-x y+y^{2}$ and in the second one if and only if $|x-y|=1$. Combining these equalities we have $x+y=(x-y)^{2}+x y=1+x y$ or $(x-1)(y-1)=0$, which yields the solutions $\{x, y\}=\{0,1\}$ or $\{x, y\}=\{1,2\}$. Now, let $\left(x_{1}, x_{2}, \ldots, x_{2020}\right)$ be any sequence satisfying conditions (i) and (ii) and let $\left(y_{1}, y_{2}, \ldots, y_{2020}\right)$ be any permutation of $\left(x_{1}, x_{2}, \ldots, x_{2020}\right)$. As $0 \leq \min \left(x_{i}, y_{i}\right) \leq$ $\max \left(x_{i}, y_{i}\right) \leq \min \left(x_{i}, y_{i}\right)+1$, we can applying inequality (1) to the pair $\left(x_{i}, y_{i}\right)$ and sum over all $1 \leq i \leq 2020$ to conclude that

$$
\sum_{i=1}^{2020}\left(\left(x_{i}+1\right)\left(y_{i}+1\right)\right)^{2} \geq 4 \sum_{i=1}^{2020}\left(x_{i}^{3}+y_{i}^{3}\right)=8 \sum_{i=1}^{2020} x_{i}^{3}
$$

Therefore, in order to satisfy condition (iii), every inequality must be an equality. Hence, for every $1 \leq i \leq 2020$ we must have $\left\{x_{i}, y_{i}\right\}=\{0,1\}$ or $\left\{x_{i}, y_{i}\right\}=\{1,2\}$. By condition (ii), we see that either $\left\{x_{i}, y_{i}\right\}=\{0,1\}$ for all $i$ or $\left\{x_{i}, y_{i}\right\}=\{1,2\}$ for all $i$.
If $\left\{x_{i}, y_{i}\right\}=\{0,1\}$ for every $1 \leq i \leq 2020$, this implies that the sequences $\left(x_{1}, x_{2}, \ldots, x_{2020}\right)$ and $\left(y_{1}, y_{2}, \ldots, y_{2020}\right)$ together have 2020 zeroes and 2020 ones. As $\left(y_{1}, y_{2}, \ldots, y_{2020}\right)$ is a permutation of $\left(x_{1}, x_{2}, \ldots, x_{2020}\right)$ this implies that $\left(x_{1}, x_{2}, \ldots, x_{2020}\right)=(0,0, \ldots, 0,1,1, \ldots, 1)$ with 1010 zeroes and 1010 ones. Conversely, note that this sequence satisfies conditions (i), (ii), and (iii) (in (iii), we take $\left.\left(y_{1}, y_{2}, \ldots, y_{2020}\right)=\left(x_{2020}, x_{2019}, \ldots, x_{1}\right)\right)$, showing that this sequence indeed works. The same reasoning holds for the case that $\left\{x_{i}, y_{i}\right\}=\{1,2\}$ for all $i$.

Comment. There are multiple ways to show the main inequality (1):

- Write

$$
\begin{aligned}
((x+1)(y+1))^{2} & \geq((x+1)(y+1))^{2}-((x-1)(y-1))^{2} \\
& =4(x+y)(x y+1) \\
& \geq 4(x+y)\left(x y+(x-y)^{2}\right) \\
& =4\left(x^{3}+y^{3}\right),
\end{aligned}
$$

where equality holds precisely if $|x-y|=1$ and $(x-1)(y-1)=0$.

- Write

$$
(x+1)^{2}(y+1)^{2}-4 x^{3}-4 y^{3}=(x-1)^{2}(y-1)^{2}+4(x+y)\left(1-(x-y)^{2}\right) \geq 0
$$

where equality holds precisely if $(x-1)(y-1)=0$ and $(x-y)^{2}=1$.

- One can rewrite the difference between the two sides as a sum of nonnegative expressions. One such way is to assume that $y \geq x$ and then to rewrite the difference as

$$
x^{2}(y-2)^{2}+(x+1-y)\left(4 y^{2}-4 x^{2}+2 x y+x+3 y+1\right)
$$

where $x^{2}(y-2)^{2} \geq 0, x+1-y \geq 0$ and $4 y^{2}-4 x^{2}+2 x y+x+3 y+1>0$, so in the equality case we must have $x+1-y=0$ and $x(y-2)=0$.

- Again assume $y \geq x$; substitute $y=x+u$ with $0 \leq u \leq 1$ and rewrite the difference as

$$
x^{2}(x+u-2)^{2}+x^{2}(2-2 u)+\left(4+6 u-10 u^{2}\right) x+\left(1+2 u+u^{2}-4 u^{3}\right),
$$

of which each summand is nonnegative, with equality case $u=1$ and $x \in\{0,1\}$.

- Fix $x \geq 0$. We aim to show that the function

$$
f(y)=((x+1)(y+1))^{2}-4\left(x^{3}+y^{3}\right)
$$

viewed as polynomial in $y$, is nonnegative on the interval $[x, x+1]$. First note that for $y=x$ and $y=x+1$ the function equals

$$
\left(x^{2}-2 x\right)^{2}+2 x^{2}+4 x+1>0 \quad \text { and } \quad\left(x^{2}-x\right)^{2} \geq 0
$$

respectively. The derivative of $f$ with respect to $y$ equals

$$
2(x+1)^{2}+2(x+1)^{2} y-12 y^{2}
$$

which is a quadratic with negative leading coefficient that evaluates as $2(x+1)^{2}>0$ for $y=0$. Therefore, this quadratic has one positive and one negative root. Therefore, on $[0, \infty)$, the function $f$ will be initially increasing and eventually decreasing, hence the minimum on the interval $[x, x+1]$ will be achieved on one of the endpoints. To have equality, we must have $x^{2}-x=0$, hence $x \in\{0,1\}$.

- Observe that $((x+1)(y+1))^{2}-4\left(x^{3}+y^{3}\right)$ is the discriminant of

$$
p(z)=\left(x^{2}-x y+y^{2}\right) z^{2}-(x+1)(y+1) z+(x+y)
$$

Note that the leading coefficient $x^{2}-x y+y^{2}=(x-y)^{2}+x y$ is positive unless $x=y=0$, in which case $\left((0+1)^{2}(0+1)^{2}\right)>4 \cdot 0^{3}+4 \cdot 0^{3}$. Substituting $z=1$, we get

$$
p(1)=\left(x^{2}-x y+y^{2}\right)-(x+1)(y+1)+(x+y)=(x-y)^{2}-1 \leq 0 .
$$

It follows that the discriminant is non-negative. It equals zero if and only if $|x-y|=1$ and $p(z)$ attains its minimum at $z=1$. Without loss of generality $y=x+1$. The minimum is attained at

$$
1=\frac{(x+1)(y+1)}{2\left(x^{2}-x y+y^{2}\right)}=\frac{x^{2}+3 x+2}{2 x^{2}+2 x+2}
$$

which reduces to $x^{2}=x$. Therefore, the only critical points are $(x, y)=(0,1)$ and $(x, y)=$ $(1,2)$.

- Without loss of generality $y \geq x$. Let $z=\frac{x+y}{2}$ and $a=y-z$. Note that $z \geq 0$ and $0 \leq a \leq \min \left\{\frac{1}{2}, z\right\}$. We can rewrite the inequality in terms of $z$ and $a$.

$$
\begin{aligned}
((x+1)(y+1))^{2}-4 x^{3}-4 y^{3} & =((z+1+a)(z+1-a))^{2}-4(z-a)^{3}-4(z+a)^{3} \\
& =\left((z+1)^{2}-a^{2}\right)^{2}-8 z^{3}-24 z a^{2} \\
& =a^{4}-\left(24 z+2(z+1)^{2}\right) a^{2}+\left((z+1)^{4}-8 z^{3}\right)
\end{aligned}
$$

This is a quadratic in $a^{2}$. It attains its minimum at $a^{2}=12 z+(z+1)^{2} \geq 1$. Therefore it is strictly decreasing in $a$ on the interval $\left[0, \min \left\{\frac{1}{2}, z\right\}\right]$. If $a=\frac{1}{2}$, then $y=x+1$. It follows that

$$
((x+1)(y+1))^{2}-4 x^{3}-4 y^{3}=((x+1)(x+2))^{2}-4 x^{3}-4(x+1)^{3}=x^{2}(x-1)^{2} \geq 0
$$

with equality if and only if $x=0$ or $x=1$. If $a=z$, then $x=0$. It follows that

$$
((x+1)(y+1))^{2}-4 x^{3}-4 y^{3}=(y+1)^{2}-4 y^{3}=y^{2}(1-y)+2 y\left(1-y^{2}\right)+\left(1-y^{3}\right) \geq 0
$$

with equality if and only if $y=1$. Therefore $(0,1)$ and $(1,2)$ are the only critical points.

## Solutions to Problem 3

Solution A. Denote the angle bisector of $A$ by $a$ and similarly for the other bisectors. Thus, given that $a, c, e$ have a common point $M$, we need to prove that $b, d, f$ are concurrent. We write $\angle(x, y)$ for the value of the directed angle between the lines $x$ and $y$, i.e. the angle of the counterclockwise rotation from $x$ to $y\left(\right.$ defined $\left.\left(\bmod 180^{\circ}\right)\right)$.
Since the sum of the angles of a convex hexagon is $720^{\circ}$, from the angle conditions we get that the sum of any two consecutive angles is equal to $240^{\circ}$. In particular, it now follows that $\angle(b, a)=\angle(c, b)=\angle(d, c)=\angle(e, d)=\angle(f, e)=\angle(a, f)=60^{\circ}$ (assuming the hexagon is clockwise oriented).
Let $X=A B \cap C D, Y=C D \cap E F$ and $Z=E F \cap A B$. Similarly, let $P=B C \cap D E, Q=D E \cap F A$ and $R=F A \cap B C$. From $\angle B+\angle C=240^{\circ}$ it follows that $\angle(Z X, X Y)=\angle(B X, X C)=60^{\circ}$. Similarly we have $\angle(X Y, Y Z)=\angle(Y Z, Z X)=60^{\circ}$, so triangle $X Y Z$ (and similarly triangle $P Q R$ ) is equilateral. We see that the hexagon $A B C D E F$ is obtained by intersecting the two equilateral triangles $X Y Z$ and $P Q R$.
We have $\angle(A M, M C)=\angle(a, c)=\angle(a, b)+\angle(b, c)=60^{\circ}$, and since

$$
\angle(A M, M C)=\angle(A X, X C)=\angle(A R, R C)=60^{\circ}
$$

$A, C, M, X, R$ are concyclic. Because $M$ lies on the bisector of angle $\angle X A R$, we must have $M R=M X$, so triangle $M R X$ is isosceles. Moreover, we have $\angle(M R, M X)=\angle(C R, C X)=$ $\angle(B C, C D)$, which is angle $C$ of the hexagon. We now see that the triangles $M R X, M P Y$ and $M Q Z$ are isosceles and similar. This implies that there is a rotation centered at $M$ that sends $X, Y$ and $Z$ to $R, P$ and $Q$ respectively. In particular, the equilateral triangles $X Y Z$ and $P Q R$ are congruent.
It follows that there also exists a rotation sending $X, Y, Z$ to $P, Q, R$ respectively. Define $N$ as the center of this rotation. Triangles $N X Z$ and $N P R$ are congruent and equally oriented, hence $N$ is equidistant from $X Z$ and $P R$ and lies on the inner bisector $b$ of $\angle B$ (we know $N$ lies on the inner, not the outer bisector because the rotation centered at $N$ is clockwise). In the same way we can show that $N$ is on $d$ and on $f$, so $b, d, f$ are concurrent at $N$.

Remark. The key observation (a rotation centered at $M$ sends $\triangle X Y Z$ to $\triangle R Q P$ ) can be established in slightly different ways. E.g., since $A, M, R, X$ are concyclic and $A, M, Q, Z$ are concyclic, $M$ is the Miquel point of the lines $X Z, R Q, X R, Z Q$, hence it is the center of similitude $s$ sending $\overrightarrow{X Z}$ to $\overrightarrow{R Q}$. Repeating the same argument for the other pairs of vectors, we obtain that $s$ sends $\triangle X Y Z$ to $\triangle R Q P$. Moreover, $s$ is a rotation, since $M$ is equidistant from $X Z$ and $R Q$.

Remark. The reverse argument can be derived in a different way, e.g., defining $N$ as the common point of the circles $B X P D, D Y Q F, F Z R B$, and showing that $\triangle N X Z=\triangle N P R$, etc.
![](https://cdn.mathpix.com/cropped/2024_11_22_b34699d446f3f5711934g-08.jpg?height=952&width=1021&top_left_y=341&top_left_x=523)

Solution B. As in Solution A, we prove that the hexagon $A B C D E F$ is the intersection of the equilateral triangles $P Q R$ and $X Y Z$.
Let $d(S, A B)$ denote the signed distance from the point $S$ to the line $A B$, where the negative sign is taken if $A B$ separates $S$ and the hexagon. We define similarly the other distances $(d(S, B C)$, etc). Since $M \in a$, we have $d(M, Z X)=d(M, Q R)$. In the same way, we have $d(M, X Y)=d(M, R P)$ and $d(M, Y Z)=d(M, P Q)$. Therefore $d(M, Z X)+d(M, X Y)+$ $d(M, Y Z)=d(M, Q R)+d(M, R P)+d(M, P Q)$.
We now use of the following well-known lemma (which can be easily proved using areas) to deduce that triangles $P Q R$ and $X Y Z$ are congruent.

Lemma. The sum of the signed distances from any point to the sidelines of an equilateral triangle (where the signs are taken such that all distances are positive inside the triangle) is constant and equals the length of the altitude.

For $N=b \cap d$ we now find $d(N, Z X)=d(N, R P)$ and $d(N, X Y)=d(N, P Q)$. Using again the lemma for the point $N$, we get $d(N, Z X)+d(N, X Y)+d(N, Y Z)=d(N, Q R)+d(N, R P)+$ $d(N, P Q)$. Therefore $d(N, Y Z)=d(N, Q R)$, thus $N \in f$.

Remark. Instead of using the lemma, it is possible to use some equivalent observation in terms of signed areas.

Solution C. We use the same notations as in Solution A. We will show that $a, c$ and $e$ are concurrent if and only if

$$
A B+C D+E F=B C+D E+F A,
$$

which clearly implies the problem statement by symmetry.
Let $\vec{a}$ be the vector of unit length parallel to $a$ directed from $A$ towards the interior of the hexagon. We define analogously $\vec{b}$, etc. The angle conditions imply that opposite bisectors of the hexagon are parallel, so we have $\vec{a}\|\vec{d}, \vec{b}\| \vec{e}$ and $\vec{c} \| \vec{f}$. Moreover, as in the previous solutions, we know that $\vec{a}, \vec{c}$ and $\vec{e}$ make angles of $120^{\circ}$ with each other. Let $M_{A}=c \cap e, M_{C}=e \cap a$ and $M_{E}=a \cap c$. Then $M_{A}, M_{C}, M_{E}$ form an equilateral triangle with side length denoted by $s$. Note that the case $s=0$ is equivalent to $a, c$ and $e$ being concurrent.
Projecting $\overrightarrow{M_{E}} A+\overrightarrow{A B}=\overrightarrow{M_{E}} B=\overrightarrow{M_{E}} C+\overrightarrow{C B}$ onto $\vec{e}=-\vec{b}$, we obtain

$$
\overrightarrow{A B} \cdot \vec{b}-\overrightarrow{C B} \cdot \vec{b}=\overrightarrow{M_{E} C} \cdot \vec{b}-\overrightarrow{M_{E} A} \cdot \vec{b}=\overrightarrow{M_{E}} A \cdot \vec{e}-\overrightarrow{M_{E}} C \cdot \vec{e}
$$

Writing $\varphi=\frac{1}{2} \angle B=\frac{1}{2} \angle D=\frac{1}{2} \angle F$, we know that $\overrightarrow{A B} \cdot \vec{b}=-A B \cdot \cos (\varphi)$, and similarly $\overrightarrow{C B} \cdot \vec{b}=$ $-C B \cdot \cos (\varphi)$. Because $M_{E} A$ and $M_{E} C$ intersect $e$ at $120^{\circ}$ angles, we have $\overrightarrow{M_{E}} A \cdot \vec{e}=\frac{1}{2} M_{E} A$ and $\overrightarrow{M_{E}} C \cdot \vec{e}=\frac{1}{2} M_{E} C$. We conclude that

$$
2 \cos (\varphi)(A B-C B)=M_{E} C-M_{E} A
$$

Adding the analogous equalities $2 \cos (\varphi)(C D-E D)=M_{A} E-M_{A} C$ and $2 \cos (\varphi)(E F-A F)=$ $M_{C} A-M_{C} E$, we obtain
$2 \cos (\varphi)(A B+C D+E F-C B-E D-A F)=M_{E} C-M_{E} A+M_{A} E-M_{A} C+M_{C} A-M_{C} E$.
Because $M_{A}, M_{C}$ and $M_{E}$ form an equilateral triangle with side length $s$, we have $M_{E} C-$ $M_{A} C= \pm s, M_{C} A-M_{E} A= \pm s$, and $M_{A} E-M_{C} E= \pm s$. Therefore, the right hand side $M_{E} C-M_{E} A+M_{A} E-M_{A} C+M_{C} A-M_{C} E$ equals $\pm s \pm s \pm s$, which (irrespective of the choices of the $\pm$-signs) is 0 if and only if $s=0$. Because $\cos (\varphi) \neq 0$, we conclude that

$$
A B+C D+E F=C B+E D+A F \Longleftrightarrow s=0 \Longleftrightarrow a, c, e \text { concurrent, }
$$

as desired.
Remark. Equalities used in the solution could appear in different forms, in particular, in terms of signed lengths.

Remark. Similar solutions could be obtained by projecting onto the line perpendicular to $b$ instead of $b$.

Solution D. We use the the same notations as in previous solutions and the fact that $a \| d$, $b \| e$ and $c \| f$ make angles of $120^{\circ}$. Also, we may assume that $E$ and $C$ are not symmetric in $a$ (if they are, the entire figure is symmetric and the conclusion is immediate).
We consider two mappings: the first one $s: a \rightarrow B C \rightarrow d$ sending $A^{\prime} \mapsto B^{\prime} \mapsto S$ is defined such that $A^{\prime} B^{\prime} \| A B$ and $B^{\prime} S \| b$, and the second one $t: a \rightarrow E F \rightarrow d$ sending $A^{\prime} \mapsto F^{\prime} \mapsto T$ is defined such that $A^{\prime} F^{\prime} \| A F$ and $F^{\prime} T \| f$. Both maps are affine linear since they are compositions of affine tranformations. We will prove that they coincide by finding two distinct points $A^{\prime}, A^{\prime \prime} \in a$ for which $s\left(A^{\prime}\right)=t\left(A^{\prime}\right)$ and $s\left(A^{\prime \prime}\right)=t\left(A^{\prime \prime}\right)$. Then we will obtain that $s(A)=t(A)$, which by construction implies that the bisectors of $\angle B, \angle D$ and $\angle F$ are concurrent.
We will choose $A^{\prime}$ to be the reflection of $C$ in $e$ and $A^{\prime \prime}$ to be the reflection of $E$ in $c$. They are distinct since otherwise $C$ and $E$ would be symmetric in $a$. Applying the above maps $a \rightarrow B C$ and $a \rightarrow E F$ to $A^{\prime}$, we get points $B^{\prime}$ and $F^{\prime}$ such that $A^{\prime} B^{\prime} C D E F^{\prime}$ satisfies the problem statement. However, this hexagon is symmetric in $e$, hence the bisectors of $\angle B^{\prime}, \angle D, \angle F^{\prime}$ are concurrent and $s\left(A^{\prime}\right)=t\left(A^{\prime}\right)$. The same reasoning yields $s\left(A^{\prime \prime}\right)=t\left(A^{\prime \prime}\right)$, which finishes the solution.

Remark. This solution is based on the fact that that two specific affine linear maps coincide. Here it was proved by exhibiting two points where they coincide. One could prove it in another way, exhibiting one such point and proving that the 'slopes' are equal.

Remark. There are similar solutions where claims and proofs could be presented in more 'elementary' terms. For example, an elementary reformulation of the 'slopes' being equal is: if $b^{\prime}$ passes through $B^{\prime}$ parallel to $b$, and $f^{\prime}$ passes through $F^{\prime}$ parallel to $f$, then the line through $b \cap f$ and $b^{\prime} \cap f^{\prime}$ is parallel to $a$ (which is parallel to $d$ ).

Solution E. We use the same notations as in previous solutions.
Since the sum of the angles of a convex hexagon is $720^{\circ}$, from the angle conditions we get $\angle B+\angle C=720^{\circ} / 3=240^{\circ}$. From $\angle B+\angle C=240^{\circ}$ it follows that the angle between $c$ and $b$ equals $60^{\circ}$. The same is analogously true for other pairs of bisectors of neighboring angles.
Consider the points $O_{a} \in a, O_{c} \in c, O_{e} \in e$, each at the same distance $d^{\prime}$ from $M$, where $d^{\prime}>\max \{M A, M C, M E\}$, and such that the rays $A O_{a}, C O_{c}, E O_{e}$ point out of the hexagon. By
construcion, $O_{a}$ and $O_{c}$ are symmetrical in $e$, hence $O_{a} O_{c} \perp b$. Similarly, $O_{c} O_{e} \perp d, O_{e} O_{a} \perp f$. Thus it suffices to prove that perpendiculars from $B, D, F$ to the sidelines of $\triangle O_{a} O_{c} O_{e}$ are concurrent. By a well-known criteria, this condition is equivalent to equality

$$
O_{a} B^{2}-O_{c} B^{2}+O_{c} D^{2}-O_{e} D^{2}+O_{e} F^{2}-O_{a} F^{2}=0
$$

To prove ( $*$ ) consider a circle $\omega_{a}$ centered at $O_{a}$ and tangent to $A B$ and $A F$ and define circles $\omega_{c}$ and $\omega_{e}$ in the same way. Rewrite $O_{a} B^{2}$ as $r_{a}^{2}+B_{a} B^{2}$, where $r_{a}$ is the radius of $\omega_{a}$, and $B_{a}$ is the touch point of $\omega_{a}$ with $A B$. Using similar notation for the other tangent points, transform (*) into

$$
B_{a} B^{2}-B_{c} B^{2}+D_{c} D^{2}-D_{e} D^{2}+F_{e} F^{2}-F_{a} F^{2}=0 .
$$

Furthermore, $\angle O_{c} O_{a} B_{a}=\angle M O_{a} B_{a}+\angle O_{c} O_{a} M=\left(90^{\circ}-\varphi\right)+30^{\circ}=120^{\circ}-\varphi$, where $\varphi=$ $\frac{1}{2} \angle A$. (Note that $\varphi>30^{\circ}$, since $A B C D E F$ is convex.) By analogous arguments, $\angle O_{a} O_{c} B_{c}=$ $\angle O_{e} O_{c} D_{c}=\angle O_{c} O_{e} D_{e}=\angle O_{a} O_{e} F_{e}=\angle O_{e} O_{a} F_{a}=120^{\circ}-\varphi$. It follows that rays $O_{a} B_{a}$ and $O_{c} B_{c}$ (being symmetrical in $e$ ) intersect at $U_{e} \in e$ forming an isosceles triangle $\triangle O_{a} U_{e} O_{c}$. Similarly define $\triangle O_{c} U_{a} O_{e}$ and $\triangle O_{e} U_{c} O_{a}$. These triangles are congruent (equal bases and corresponding angles). Therefore we have $O_{a} U_{c}=U_{c} O_{e}=O_{e} U_{a}=U_{a} O_{c}=O_{c} U_{e}=U_{e} O_{a}$. Moreover, we also have $B_{a} U_{e}=O_{a} U_{e}-r_{a}=O_{a} U_{c}-r_{a}=F_{a} U_{c}=x$, and thus similarly $D_{c} U_{a}=B_{c} U_{e}=y$, $F_{e} U_{c}=D_{e} U_{a}=z$.
Now from quadrilateral $B B_{a} U_{e} B_{c}$ with two opposite right angles $B_{a} B^{2}-B_{c} B^{2}=B_{c} U_{e}^{2}-B_{a} U_{e}^{2}=$ $y^{2}-x^{2}$. Similarly $D_{c} D^{2}-D_{e} D^{2}=D_{e} U_{a}^{2}-D_{c} U_{a}^{2}=z^{2}-y^{2}$ and $F_{e} F^{2}-F_{a} F^{2}=F_{a} U_{c}^{2}-F_{e} U_{c}^{2}=$ $x^{2}-z^{2}$. Finally, we substitute this into $(* *)$, and the claim is proved.

Remark. Circles $\omega_{a}, \omega_{c}$ and $\omega_{e}$ could be helpful in some other solutions. In particular, the movement of $A$ along $a$ in Solution D is equivalent to varying $r_{a}$.

## Solutions to Problem 4

Solution A. Let $\sigma=\left(\sigma_{1}, \ldots, \sigma_{n-1}\right)$ be a fresh permutation of the integers $1,2, \ldots, n-1$. We claim that for any $1 \leq i \leq n-1$ the permutation

$$
\sigma^{(i)}=\left(\sigma_{1}, \ldots, \sigma_{i-1}, n, \sigma_{i}, \ldots, \sigma_{n-1}\right)
$$

is a fresh permutation of the integers $1,2, \ldots, n$. Indeed, let $1 \leq k \leq n-1$. If $k \geq i$ then we have $n \in\left\{\sigma_{1}^{(i)}, \ldots, \sigma_{k}^{(i)}\right\}$, but $n \notin\{1,2, \ldots, k\}$. And if $k<i$ we have $k<n-1$, and $\left\{\sigma_{1}^{(i)}, \ldots, \sigma_{k}^{(i)}\right\}=\left\{\sigma_{1}, \ldots, \sigma_{k}\right\} \neq\{1,2, \ldots, k\}$, since $\sigma$ is fresh. Moreover, it is easy to see that when we apply this construction to all fresh permutations of $1,2, \ldots, n-1$, we obtain $(n-1) \cdot f_{n-1}$ distinct fresh permutations of $1,2, \ldots, n$.
Note that a fresh permutation of $1,2, \ldots, n-1$ cannot end in $n-1$, and hence none of the previously constructed fresh permutations of $1,2, \ldots, n$ will end in $n-1$ either. Therefore, we will complete the proof by finding $f_{n-1}$ fresh permutations of $1,2, \ldots, n$ that end in $n-1$. To do this, let $\sigma=\left(\sigma_{1}, \ldots, \sigma_{n-1}\right)$ be a fresh permutation of $1,2, \ldots, n-1$ and let $j$ be such that $\sigma_{j}=n-1$. Define

$$
\sigma^{\prime}=\left(\sigma_{1}, \ldots, \sigma_{j-1}, n, \sigma_{j+1}, \ldots, \sigma_{n-1}, n-1\right)
$$

then clearly $\sigma^{\prime}$ is a permutation of $1,2, \ldots, n$ that ends in $n-1$. We show that $\sigma^{\prime}$ is fresh, so let $1 \leq k \leq n-1$. If $k \geq j$ then $n \in\left\{\sigma_{1}^{\prime}, \ldots, \sigma_{k}^{\prime}\right\}$ but $n \notin\{1,2, \ldots, k\}$; if $k<j$, then $k<n-1$ and $\left\{\sigma_{1}^{\prime}, \ldots, \sigma_{k}^{\prime}\right\}=\left\{\sigma_{1}, \ldots, \sigma_{k}\right\} \neq\{1,2, \ldots, k\}$, since $\sigma$ is fresh. So we have constructed $f_{n-1}$ additional fresh permutations of $1,2, \ldots, n$ (which again are all different), and the total number $f_{n}$ of fresh permutations of $1,2, \ldots, n$ must at least be $(n-1) f_{n-1}+f_{n-1}=n f_{n-1}$, as required.

Comment. A similar way to construct $n$ fresh permutations of $1,2, \ldots, n$ for each fresh permutation $\sigma$ of $1,2, \ldots, n-1$ is as follows: increase all entries of $\sigma$ by 1 , and then add the number 1 anywhere; all these permutations are fresh, except the ones where 1 is added at the front, which can be made fresh by swapping the 1 and the 2 . It is again straightforward to check that we obtain $n f_{n-1}$ permutations that are fresh and distinct, although a little extra care is needed to account for the fact that we increased the entries of our original permutation.

Solution B. Assuming $n \geq 3$, we construct $f_{n-1} \cdot n$ different fresh permutations.
Consider a fresh permutation of the $n-1$ numbers $1,3,4, \ldots, n$ (the number 2 has been removed), by which we mean a permutation $\left(x_{1}, \ldots, x_{n-1}\right)$ such that $x_{1} \neq 1$ and $\left\{x_{1}, \ldots, x_{k}\right\} \neq$ $\{1,3, \ldots, k+1\}$ for all $k$ with $2 \leq k \leq n-2$. There are exactly $f_{n-1}$ such permutations.
By inserting 2 anywhere in such a permutation, i.e. before or after all entries, or between two entries $x_{i}, x_{i+1}$, we generate the following list of $n$ distinct permutations of $1,2, \ldots, n$, which we claim are all fresh:

$$
\left(2, x_{1}, \ldots, x_{n-1}\right), \quad \ldots, \quad\left(x_{1}, \ldots, x_{i-1}, 2, x_{i}, \ldots, x_{n-1}\right), \quad \ldots, \quad\left(x_{1}, \ldots, x_{n-1}, 2\right)
$$

In order to verify freshness, suppose that some permutation above is not fresh, that is, for some $1 \leq k \leq n-1$, the first $k$ elements are $1, \ldots, k$. If $k=1$ then the first element is 1 ; but the first element is either 2 or $x_{1}$ and $x_{1} \neq 2$, so this is not possible. If $k \geq 2$ then the first $k$ elements must contain 2 and $x_{1}, \ldots, x_{k-1}$, so $\left\{x_{1}, \ldots, x_{k-1}\right\}=\{1,3, \ldots, k\}$, in contradiction to the fact that $\left(x_{1}, \ldots, x_{n-1}\right)$ is fresh.
We have thus constructed $f_{n-1} \cdot n$ different fresh permutations, so $f_{n} \geq f_{n-1} \cdot n$.
Comment. Note that the construction from solution B can also be performed by placing $n-1$ instead of 2 , or indeed any $k$ between 2 and $n-1$ (in which case it will work for $n \geq k+1$ ).
Moreover, similar arguments can be made while choosing to work with non-fresh permutations rather than fresh permutations. Indeed, in order to show that $n!-f_{n} \leq n\left((n-1)!-f_{n-1}\right)$ for all $n \geq 3$, which is equivalent to the problem statement, one can argue as follows. Given a non-fresh permutation $\left(\sigma_{1}, \ldots, \sigma_{n}\right)$ of $1,2, \ldots, n$, erasing $n-1$ from it and changing $n$ to $n-1$ yields a permutation which cannot be fresh. Moreover, each non-fresh permutation of $1,2, \ldots, n-1$ can be obtained in this way from at most $n$ different non-fresh permutations of $1,2, \ldots, n$. This type of argument would essentially mirror the one carried out in Solution B.

Solution C. By considering all permutations of $1,2, \ldots, n$ and considering the smallest $k$ for which the first $k$ numbers are $1,2, \ldots, k$ in some order, one can deduce the recursion

$$
\sum_{k=1}^{n}(n-k)!\cdot f_{k}=n!
$$

which holds for any $n \geq 1$.
Therefore, if $n \geq 3$ we have

$$
\begin{aligned}
0 & =n!-(n+1)(n-1)!+(n-1)(n-2)! \\
& =\sum_{k=1}^{n}(n-k)!\cdot f_{k}-(n+1) \sum_{k=1}^{n-1}(n-1-k)!\cdot f_{k}+(n-1) \sum_{k=1}^{n-2}(n-2-k)!\cdot f_{k} \\
& =f_{n}-n f_{n-1}+\sum_{k=1}^{n-2}(n-2-k)!\cdot((n-k)(n-k-1)-(n+1)(n-k-1)+(n-1)) \cdot f_{k} \\
& =f_{n}-n f_{n-1}+\sum_{k=1}^{n-2}(n-2-k)!\cdot k(k+2-n) \cdot f_{k} \leq f_{n}-n f_{n-1}
\end{aligned}
$$

which rewrites as $f_{n} \geq n f_{n-1}$.
Solution D. We first show that for any $n \geq 2$ we have

$$
f_{n}=\sum_{k=1}^{n-1} k(n-1-k)!\cdot f_{k}
$$

To deduce this relation, we imagine obtaining permutations of $1,2, \ldots, n$ by inserting $n$ into a permutation of the numbers $1,2, \ldots, n-1$. Specifically, let $\left(\sigma_{1}, \ldots, \sigma_{n-1}\right)$ be any permutation of $1,2, \ldots, n-1$, and let $1 \leq k \leq n-1$ be minimal with $\left\{\sigma_{1}, \ldots, \sigma_{k}\right\}=\{1,2, \ldots, k\}$. Note that there are $(n-1-k)!\cdot f_{k}$ such permutations. Furthermore, inserting $n$ in this permutation will give a fresh permutation if and only if $n$ is inserted before $\sigma_{k}$, so there are $k$ options to do so. Consequently, if $n \geq 3$,

$$
\begin{aligned}
f_{n} & =\sum_{k=1}^{n-1} k(n-1-k)!\cdot f_{k} \\
& =(n-1) \cdot f_{n-1}+\sum_{k=1}^{n-2} k(n-1-k)!\cdot f_{k} \\
& \geq(n-1) \cdot f_{n-1}+\sum_{k=1}^{n-2} k(n-2-k)!\cdot f_{k} \\
& =(n-1) \cdot f_{n-1}+f_{n-1}=n \cdot f_{n-1}
\end{aligned}
$$

completing the proof.
Comment. Fresh permutations are known as indecomposable permutations or irreducible permutations in the literature. The problem asks to prove that the probability that a randomly chosen permutation of $1,2, \ldots, n$ is indecomposable is a non-decreasing function of $n$. In fact, it turns out that this probability goes to 1 as $n \rightarrow \infty$ : for large $n$, almost all permutations of 1 , $2, \ldots, n$ are indecomposable. More can be found in:
Y. Koh \& S. Ree. Connected permutation graphs. Discrete Mathematics 307 (21):2628-2635, 2007.

## Solutions to Problem 5

Solution A. The angle bisector of $\angle E C D$ intersects the circumcircle of $C D E$ (which is $\Gamma$ ) at the midpoint $M$ of arc $D B E$. It is well-known that the incentre is the intersection of the angle bisector segment $C M$ and the circle with centre at $M$ and passing through $D, E$. We will verify this property for $P$.
![](https://cdn.mathpix.com/cropped/2024_11_22_b34699d446f3f5711934g-13.jpg?height=729&width=1023&top_left_y=635&top_left_x=519)

By the conditions we have $A P=O A=O B=O C=O D=O E=R$. Both lines $O M$ and $A P B$ are perpendicular to $E D$, therefore $A P \| O M$; in the quadrilateral $A O M P$ we have $A P=O A=A M=R$ and $A P \| O M$, so $A O M P$ is a rhombus and its fourth side is $P M=R$.
In the convex quadrilateral $O M B P$ we have $O M \| P B$, so $O M B P$ is a symmetric trapezoid; the perpendicular bisector of its bases $A O$ and $P B$ coincide. From this symmetry we obtain $M D=O D=R$ and $M E=O E=R$. (Note that the triangles $O E M$ ad $O M D$ are equilateral.) We already have $M P=M D=M E=R$, so $P$ indeed lies on the circle with center $M$ and passing through $D, E$. (Notice that this circle is the reflection of $\Gamma$ about $D E$.)
From $P B=P C$ and $O B=O C$ we know that $B$ and $C$ are symmetrical about $O P$; from the rhombus $A O M P$ we find that $A$ and $M$ are also symmetrical about $O P$. By reflecting the collinear points $B, P, A$ (with $P$ lying in the middle) we get that $C, P, M$ are collinear (and $P$ is in the middle). Hence, $P$ lies on the line segments $C M$.

Solution B. Let $X$ be the second intersection of $C P$ with $\Gamma$. Using the power of the point $P$ in the circle $\Gamma$ and the fact that $P B=P C$, we find that $P X=P A=R$. The quadrilateral $A O X P$ has four sides of equal length, so it is a rhombus and in particular $O X$ is parallel to $A P$. This proves that $O X B P$ is a trapezoid, and because the diagonals $P X$ and $O B$ have equal length, this is even an isosceles trapezoid. Because of that, $D E$ is not only the perpendicular bisector of $P B$, but also of $O X$.
![](https://cdn.mathpix.com/cropped/2024_11_22_b34699d446f3f5711934g-14.jpg?height=635&width=918&top_left_y=748&top_left_x=672)

In particular we have $X D=X P=X O=X E=R$, which proves that $X$ is the middle of the arc $D E$ and $P$ belongs to the circle with center $X$ going through $D$ and $E$. These properties, together with the fact that $C, P, X$ are collinear, determine uniquely the incenter of $C D E$.

Solution C. Let $Y$ be the circumcenter of triangle $B P C$. Then from $Y B=Y P$ it follows that $Y$ lies on $D E$ (we assume $D$ lies in between $Y$ and $E$ ), and from $Y B=Y C$ it follows that $Y$ lies on $O P$, where $O$ is the center of $\Gamma$.
From $\angle A O C=2 \angle A B C=\angle A P C$ (because $\angle P B C=\angle P C B$ ) we deduce that $A O P C$ is a cyclic quadrilateral, and from $A P=R$ it follows that $A O P C$ is an isosceles trapezoid. We now find that $\angle Y C P=\angle Y P C=180^{\circ}-\angle O P C=180^{\circ}-\angle A C P$, so $Y$ lies on $A C$.
![](https://cdn.mathpix.com/cropped/2024_11_22_b34699d446f3f5711934g-15.jpg?height=940&width=601&top_left_y=632&top_left_x=736)

Power of a point gives $Y O \cdot Y P=Y C \cdot Y A=Y D \cdot Y E$, so $D, P, O$ and $E$ are concyclic. It follows that $2 \angle D A E=\angle D O E=\angle D P E=\angle D B E=180^{\circ}-\angle D A E$, so $\angle D A E=60^{\circ}$. We can now finish the proof by angle chasing.
From $A B \perp D E$ we have $\angle A O D+\angle B O E=180^{\circ}$ and from $\angle D O E=2 \angle D A E=120^{\circ}$ it follows that $\angle B O D+\angle B O E=120^{\circ}$. It follows that $\angle A O D-\angle B O D=180^{\circ}-120^{\circ}=60^{\circ}$. Let $\angle O A B=\angle O B A=2 \beta$; then $\angle A O D+\angle B O D=\angle A O B=180^{\circ}-4 \beta$. Together with $\angle A O D-\angle B O D=60^{\circ}$, this yields $\angle A O D=120^{\circ}-2 \beta$ and $\angle B O D=60^{\circ}-2 \beta$. We now find $\angle A E D=\frac{1}{2} \angle A O D=60^{\circ}-\beta$, which together with $\angle D A E=60^{\circ}$ yields $\angle A D E=60^{\circ}+\beta$. From the isosceles trapezoid $A O P C$ we have $\angle C D A=\angle C B A=\frac{1}{2} \angle C P A=\frac{1}{2} \angle P A O=\beta$, so $\angle C D E=\angle C D A+\angle A D E=\beta+60^{\circ}+\beta=60^{\circ}+2 \beta$.
From $\angle B O D=60^{\circ}-2 \beta$ we deduce that $\angle B E D=30^{\circ}-\beta$; together with $\angle D B E=120^{\circ}$ this yields $\angle E D B=30^{\circ}+\beta$. We now see that $\angle P D E=\angle B D E=30^{\circ}+\beta=\frac{1}{2} \angle C D E$, so $P$ is on the angle bisector of $\angle C D E$. Similarly, $P$ lies on the angle bisector of $\angle C E D$, so $P$ is the incenter of $C D E$.

Solution D. We draw the lines $D P$ and $E P$ and let $D^{\prime}$ resp. $E^{\prime}$ be the second intersection point with $\Gamma$.
![](https://cdn.mathpix.com/cropped/2024_11_22_b34699d446f3f5711934g-16.jpg?height=806&width=775&top_left_y=599&top_left_x=732)

The triangles $A P D^{\prime}$ and $D P B$ are similar, and the triangles $A P E^{\prime}$ and $E P B$ are also similar, hence they are all isosceles and it follows that $E^{\prime}, O, P, D^{\prime}$ lie on a circle with center $A$. In particular $A O D^{\prime}$ and $A O E^{\prime}$ are equilateral triangles. Angle chasing gives

$$
\begin{gathered}
\angle C D P=\angle C D D^{\prime}=\frac{1}{2} \angle C O D^{\prime}=\frac{1}{2}\left(60^{\circ}+\angle C O A\right) \\
\angle E D P=\angle E D D^{\prime}=\angle E E^{\prime} D^{\prime}=\angle P E^{\prime} D^{\prime}=\frac{1}{2} \angle P A D^{\prime}=\frac{1}{2}\left(60^{\circ}+\angle P A O\right)
\end{gathered}
$$

Similarly we prove $\angle C E P=\frac{1}{2}\left(60^{\circ}-\angle C O A\right)$ and $\angle D E P=\frac{1}{2}\left(60^{\circ}-\angle P A O\right)$ so if we can prove that $\angle C O A=\angle P A O$, we will have proven that $P$ belongs to the angle bisector of $\angle C E D$ and to the angle bisector of $\angle C D E$, which is enough to prove that $P$ is the incenter of the triangle $C D E$.
Let $\beta=\angle A B C=\angle P C B$. We have $\angle A P C=2 \beta$ and $\angle A O C=2 \beta$, so $A O P C$ is an inscribed quadrilateral. Moreover, since the diagonals $A P$ and $C O$ have equal length, this is actually an isosceles trapezoid, and hence $\angle P A O=\angle C P A=\angle C O A$ which concludes the proof.

Solution E. Without loss of generality we assume that $D$ and $C$ are in the same half-plane regarding line $A B$.
Since $P C=B P$ and $A B C D$ is inscribed quadrilateral we have $\angle P C B=\angle C B P=\angle C E A=\alpha$. As in the other solutions, $A O P C$ is an isosceles trapezoid and $2 \alpha=\angle C P A=\angle P C O$
![](https://cdn.mathpix.com/cropped/2024_11_22_b34699d446f3f5711934g-17.jpg?height=675&width=692&top_left_y=608&top_left_x=671)

Let $K$ be intersection of $E O$ and $\Gamma$. Then $\angle K D E=90^{\circ}, A B \| D K$ and $K D B A$ is isosceles trapezoid. We obtain $D P=B D=A K$, which implies that $D P A K$ is a parallelogram and hence $D K=P A=R=O D=O K$. We see that $D O K$ is an equilateral triangle. Then $\angle E C D=\angle E K D=60^{\circ}$.
Further we prove that $P C$ bisects $\angle E C D$ using $\angle D C B=\angle D K B=\angle K D A$ (from isosceles trapeziod $D K A B$ ) and that $\angle K E C=\angle O E C=\angle O C E$ (from isosceles triangle $O C E$ ):

$$
\angle D C P=\angle D C B+\angle B C P=\angle K D A+\alpha=\angle K E C+\angle C E A+\alpha=\angle O C E+2 \alpha=\angle P C E
$$

Further by $\angle D C P=\angle P C E=\frac{1}{2} \angle E C D=30^{\circ}$ distances between $P$ and sides $\triangle C D E$ are $\frac{1}{2} P C=\frac{1}{2} P B$ (as ratio between cathetus and hypotenuse in right triangle with angles $60^{\circ}$ and $30^{\circ}$ ). So, we have found incentre.

Solution F. Let $F, G, H$ be the projections of $P$ on the sides $D E, D C$ resp. $C E$. If $P$ is indeed the incenter, then the three line segments $P F, P G, P H$ have the same length. This means that the problem is equivalent to proving that $P G=P H=P F=\frac{1}{2} P B=\frac{1}{2} P C$ and thus trigonometry in the right-angled triangles $C P G$ and $C P H$ tells us that it is enough to prove that $\angle D C P=\angle E C P=30^{\circ}$.
![](https://cdn.mathpix.com/cropped/2024_11_22_b34699d446f3f5711934g-18.jpg?height=618&width=643&top_left_y=705&top_left_x=678)

We introduce the point $X$ as the second intersection of the line $C P$ with $\Gamma$. Because $O$ is the center of $\Gamma$ we can reduce the problem to proving that $\angle X O D=\angle X O E=60^{\circ}$, or equivalently that $X O D$ and $X O E$ are equilateral triangles. This last condition is equivalent to $X$ being the reflection of $O$ on the line $D E$. Following the chain of equivalences, we see therefore that in order to solve the problem it is enough to prove that $X$ is the reflection of $O$ on $D E$. We prove this property as in Solution B, using the fact that $O X B P$ is an isosceles trapezoid.

Solution G. Assume $A B$ is parallel to the horizontal axis, and that $\Gamma$ is the unit circle. Write $f(\theta)$ for the point $(\cos (\theta), \sin (\theta))$ on $\Gamma$. As in Solution C, assume that $\angle A O B=180^{\circ}-4 \beta$; then we can take $B=f(2 \beta)$ and $A=f\left(180^{\circ}-2 \beta\right)$. As in Solution C, we observe that $A O P C$ is an isosceles trapezoid, which we use to deduce that $\angle A B C=\frac{1}{2} \angle A P C=\frac{1}{2} \angle O A B=\beta$. We now know that $C=f\left(180^{\circ}-4 \beta\right)$.
The point $P$ lies on $A B$ with $A P=R=1$, so $P=\left(\cos \left(180^{\circ}-2 \beta\right)+1, \sin (2 \beta)\right)=(1-$ $\cos (2 \beta), \sin (2 \beta))$. The midpoint of $B P$ therefore has coordinates $\left(\frac{1}{2}, \sin (2 \beta)\right)$, so $D$ and $E$ have $x$-coordinate $\frac{1}{2}$. Without loss of generality, we take $D=f\left(60^{\circ}\right)$ and $E=f\left(-60^{\circ}\right)$.
We have now obtained coordinates for all points in the problem, with one free parameter $(\beta)$. To show that $P$ is the incenter of $C D E$, we will show that $P$ lies on the bisector of $\angle C D E$; analogously, one can show that $P$ lies on the bisector of angle $C E D$. The bisector of angle $C D E$ passes through the midpoint $M$ of the arc $C E$ not containing $D E$; because $C=f\left(180^{\circ}-4 \beta\right)$ and $E=f\left(-60^{\circ}\right)$, we have $M=f\left(240^{\circ}-2 \beta\right)$.
It remains to show that $P=(1-\cos (2 \beta), \sin (2 \beta))$ lies on the line connecting the points $D=$ $\left(\cos \left(60^{\circ}\right), \sin \left(60^{\circ}\right)\right)$ and $M=\left(\cos \left(240^{\circ}-2 \beta\right), \sin \left(240^{\circ}-2 \beta\right)\right)=\left(-\cos \left(60^{\circ}-2 \beta\right),-\sin \left(60^{\circ}-2 \beta\right)\right)$. The equation for the line $D M$ is
$\left(Y+\sin \left(60^{\circ}-2 \beta\right)\right)\left(\cos \left(60^{\circ}\right)+\cos \left(60^{\circ}-2 \beta\right)\right)=\left(\sin \left(60^{\circ}\right)+\sin \left(60^{\circ}-2 \beta\right)\right)\left(X+\cos \left(60^{\circ}-2 \beta\right)\right)$,
which, using the fact that $\cos \left(60^{\circ}\right)+\cos \left(60^{\circ}-2 \beta\right)=2 \cos (\beta) \cos \left(60^{\circ}-\beta\right)$ and $\sin \left(60^{\circ}\right)+\sin \left(60^{\circ}-\right.$ $2 \beta)=2 \cos (\beta) \sin \left(60^{\circ}-\beta\right)$, simplifies to

$$
\left(Y+\sin \left(60^{\circ}-2 \beta\right)\right) \cos \left(60^{\circ}-\beta\right)=\left(X+\cos \left(60^{\circ}-2 \beta\right)\right) \sin \left(60^{\circ}-\beta\right)
$$

Because $\cos \left(60^{\circ}-2 \beta\right) \sin \left(60^{\circ}-\beta\right)-\sin \left(60^{\circ}-2 \beta\right) \cos \left(60^{\circ}-\beta\right)=\sin (\beta)$, this equation further simplifies to

$$
Y \cos \left(60^{\circ}-\beta\right)-X \sin \left(60^{\circ}-\beta\right)=\sin (\beta) .
$$

Plugging in the coordinates of $P$, i.e., $X=1-\cos (2 \beta)$ and $Y=\sin (2 \beta)$, shows that $P$ is on this line: for this choice of $X$ and $Y$, the left hand side equals $\sin \left(60^{\circ}+\beta\right)-\sin \left(60^{\circ}-\beta\right)=$ $2 \cos \left(60^{\circ}\right) \sin (\beta)$, which is indeed equal to $\sin (\beta)$. So $P$ lies on the bisector $D M$ of $\angle C D E$, as desired.

Solution H. Let $\Gamma$ be the complex unit circle and let $A B$ be parallel with the real line and $0<\varphi=\arg b<\frac{\pi}{2}$. Then

$$
|b|=1, \quad a=-\bar{b}, \quad p=a+1=1-\bar{b}
$$

From $\operatorname{Re} d=\operatorname{Re} e=\operatorname{Re} \frac{p+b}{2}=\frac{1}{2}$ we get that $d=\frac{1}{2}+\frac{\sqrt{3}}{2} i$ and $e=\frac{1}{2}-\frac{\sqrt{3}}{2} i$ are conjugate 6 th roots of unity; $d^{3}=e^{3}=-1, d+e=1, d^{2}=-e, e^{2}=-d$ etc.
Point $C$ is the reflection of $B$ in line $O P$. From $\arg p=\arg (1-\bar{b})=\frac{1}{2}(\pi-\varphi)$, we can get $\arg c=2 \arg p-\arg b=\pi-2 \varphi$, so $c=-\bar{b}^{2}$.
Now we can verify that $E P$ bisects $\angle C E D$. This happens if and only if $(p-e)^{2}(\bar{c}-\bar{e})(\bar{d}-\bar{e})$ is real. Since $\bar{d}-\bar{e}=-\sqrt{3} i$, this is equivalent with $\operatorname{Re}\left[(p-e)^{2}(\bar{c}-\bar{e})\right]=0$. Here

$$
\begin{aligned}
(p-e)^{2}(\bar{c}-\bar{e}) & =(1-\bar{b}-e)^{2}\left(-b^{2}-d\right)=(d-\bar{b})^{2}\left(-b^{2}-d\right) \\
& =-|b|^{4}+2 d|b|^{2} b-d^{2} b^{2}-d \bar{b}^{2}+2 d^{2} \bar{b}-d^{3} \\
& =-1-2 d b+\bar{d} b^{2}-d \bar{b}^{2}-2 \overline{d b}+1 \\
& =-2(d b-\overline{d b})+\left(\bar{d} b^{2}-d \bar{b}^{2}\right),
\end{aligned}
$$

whose real part is zero. It can be proved similarly that $D P$ bisects $\angle E D C$.

## Solutions to Problem 6

Answer. The only such $m$ are $m=2$ and $m=10$.
Solution A. Consider an integer $m>1$ for which the sequence defined in the problem statement contains only perfect squares. We shall first show that $m-1$ is a power of 3 .
Suppose that $m-1$ is even. Then $a_{4}=5 m-1$ should be divisible by 4 and hence $m \equiv 1(\bmod 4)$. But then $a_{5}=5 m^{2}+3 m-1 \equiv 3(\bmod 4)$ cannot be a square, a contradiction. Therefore $m-1$ is odd.
Suppose that an odd prime $p \neq 3$ divides $m-1$. Note that $a_{n}-a_{n-1} \equiv a_{n-2}-a_{n-3}(\bmod p)$. It follows that modulo $p$ the sequence takes the form $1,1,4,4,7,7,10,10, \ldots$; indeed, a simple induction shows that $a_{2 k} \equiv a_{2 k-1} \equiv 3 k-2(\bmod p)$ for $k \geq 1$. Since $\operatorname{gcd}(p, 3)=1$ we get that the sequence $a_{n}(\bmod p)$ contains all the residues modulo $p$, a contradiction since only $(p+1) / 2$ residues modulo $p$ are squares. This shows that $m-1$ is a power of 3 .
Let $h, k$ be integers such that $m=3^{k}+1$ and $a_{4}=h^{2}$. We then have $5 \cdot 3^{k}=(h-2)(h+2)$. Since $\operatorname{gcd}(h-2, h+2)=1$, it follows that $h-2$ equals either $1,3^{k}$ or 5 , and $h+2$ equals either $5 \cdot 3^{k}, 5$ or $3^{k}$, respectively. In the first two cases we get $k=0$ and in the last case we get $k=2$. This implies that either $m=2$ or $m=10$.
We now show the converse. Suppose that $m=2$ or $m=10$. Let $t=1$ or $t=3$ so that $m=t^{2}+1$. Let $b_{1}, b_{2}, b_{3}, \ldots$ be a sequence of integers defined by $b_{1}=1, b_{2}=1, b_{3}=2$, and

$$
b_{n}=t b_{n-1}+b_{n-2}, \quad \text { for all } n \geq 4
$$

Clearly, $a_{n}=b_{n}^{2}$ for $n=1,2,3$. Note that if $m=2$ then $a_{4}=9$ and $b_{4}=3$, and if $m=10$ then $a_{4}=49$ and $b_{4}=7$. In both the cases we have $a_{4}=b_{4}^{2}$.
If $n \geq 5$ then we have

$$
b_{n}^{2}+b_{n-3}^{2}=\left(t b_{n-1}+b_{n-2}\right)^{2}+\left(b_{n-1}-t b_{n-2}\right)^{2}=\left(t^{2}+1\right)\left(b_{n-1}^{2}+b_{n-2}^{2}\right)=m\left(b_{n-1}^{2}+b_{n-2}^{2}\right) .
$$

Therefore, it follows by induction that $a_{n}=b_{n}^{2}$ for all $n \geq 1$. This completes the solution.
Solution B. We present an alternate proof that $m=2$ and $m=10$ are the only possible values of $m$ with the required property.
Note that

$$
\begin{aligned}
& a_{4}=5 m-1 \\
& a_{5}=5 m^{2}+3 m-1 \\
& a_{6}=5 m^{3}+8 m^{2}-2 m-4
\end{aligned}
$$

Since $a_{4}$ and $a_{6}$ are squares, so is $a_{4} a_{6}$. We have

$$
4 a_{4} a_{6}=100 m^{4}+140 m^{3}-72 m^{2}-72 m+16
$$

Notice that

$$
\begin{aligned}
& \left(10 m^{2}+7 m-7\right)^{2}=100 m^{4}+140 m^{3}-91 m^{2}-98 m+49<4 a_{4} a_{6} \\
& \left(10 m^{2}+7 m-5\right)^{2}=100 m^{4}+140 m^{3}-51 m^{2}-70 m+25>4 a_{4} a_{6}
\end{aligned}
$$

so we must have

$$
4 a_{4} a_{6}=\left(10 m^{2}+7 m-6\right)^{2}=100 m^{4}+140 m^{3}-71 m^{2}-84 m+36
$$

This implies that $m^{2}-12 m+20=0$, so $m=2$ or $m=10$.