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$14^{\text {th }}$ Middle European

Mathematical Olympiad 2020

# Contest Problems with Solutions 

The Problem Selection Committee

The Problem Selection Committee

|  | Algebra | Combinatorics | Geometry | Number Theory |
| :--- | :---: | :---: | :---: | :---: |
| Dominika Regiec (Poland) | Secondary |  | Primary |  |
| Dominik Burek (Poland) |  |  | Problem Captain | Secondary |
| Gerhard Kirchner (Austria) |  | Secondary |  | Primary |
| Ivan Krijan (Croatia) |  | Problem Captain |  | Secondary |
| Jakob Jurij Snoj (Slovenia) | Primary | Secondary |  | Problem Captain |
| Jörg Jahnel (Germany) | Secondary |  | Secondary |  |
| Josef Tkadlec (Czechia) |  | Primary | Secondary |  |
| Tomasz Cieśla (Poland) | Problem Captain | Chair of the Jury |  |  |
| Birgit Vera Schmidt (Austria) |  |  |  |  |

gratefully received

## 63 problem proposals submitted by 5 countries:

Austria - Croatia - Czech Republic - Poland - Slovakia

## Authors of the selected problems:

- I-1: Ivan Novak (Croatia)
- I-2: Stephan Wagner (Austria)
- I-3: Dominik Burek (Poland)
- I-4: Daniel Paleka (Croatia)

The Problem Selection Committee would also like to thank Roger Labahn for providing the $\mathrm{I}_{\mathrm{A}} \mathrm{TEX}_{\mathrm{E}}$ templates.

## Contents

Algebra ..... 4
I-1 ..... 4
Combinatorics ..... 7
I-2 ..... 7
Geometry ..... 10
I-3 ..... 10
Number Theory ..... 15
I-4 ..... 15

## $\mathrm{I}-1$

Let $\mathbb{N}$ be the set of positive integers. Determine all positive integers $k$ for which there exist functions $f: \mathbb{N} \rightarrow \mathbb{N}$ and $g: \mathbb{N} \rightarrow \mathbb{N}$ such that $g$ assumes infinitely many values and such that

$$
f^{g(n)}(n)=f(n)+k
$$

holds for every positive integer $n$.

(Remark. Here, $f^{i}$ denotes the function $f$ applied $i$ times, i.e., $f^{i}(j)=\underbrace{f(f(\ldots f(f}_{i \text { times }}(j)) \ldots))$.)

Answer. Such functions exist if and only if $k \geq 2$.

Solution 1. Suppose that $k=1$ and that $f$ and $g$ satisfy the desired conditions.

Claim. There exist no positive integers $m$ and $n$ with $f^{m}(n)=n$.

Proof. Suppose that $f^{m}(n)=n$ for some positive integers $m$ and $n$. Consider the orbit of $n$, i.e. the set $A=\left\{f(n), f^{2}(n), \ldots, f^{m-1}(n), f^{m}(n)\right\}$. Clearly, $f^{r}(n) \in A$ for all integers $r \geq 0$. Let $t=f^{s}(n)=\max A$ and denote $u=f^{s+m-1}(n)$ so that $f(u)=t$. Then $t+1=f(u)+1=$ $f^{g(u)}(u)=f^{g(u)+s+m-1}(n) \in A$, contradicting $t=\max A$.

Claim. $f(n) \geq n+1$ for all positive integers $n$.

Proof. Suppose there exists a positive integer $n$ with $f(n) \leq n$.

We show inductively that for each integers $r \geq 0$ there exists an integer $s \geq 1$ with $f^{s}(n)=$ $f(n)+r$. For $r=0$ take $s=1$. Induction step: Suppose that $f^{s}(n)=f(n)+r$ for some integer $s \geq 1$. Denote $t=f^{s-1}(n)$ and note that $f(n)+r+1=f^{s}(n)+1=f(t)+1=f^{g(t)}(t)=$ $f^{g(t)+s-1}(n)$, hence $s^{\prime}=g(t)+s-1 \geq 1$ works for $r+1$.

In particular, setting $r=n-f(n) \geq 0$ we see that $f^{s}(n)=n$ for some $s \geq 1$. This contradicts the previous claim.

Now, for all positive integers $m$ and $n$ we obtain

$$
f^{m}(n) \geq f^{m-1}(n)+1 \geq \ldots \geq f(n)+m-1
$$

Set $m=g(n)$. We obtain $f(n)+1=f^{g(n)}(n) \geq f(n)+g(n)-1$, hence $g(n) \leq 2$ for all $n$. This contradicts the assumption that $g$ is unbounded.

Now, let $k \geq 2$. We construct $f$ and $g$ satisfying desired conditions.

Let $n_{1}<n_{2}<n_{3}<\ldots$ be the sequence consisting of all positive integers not divisible by $k$ (i.e. $n_{i(k-1)+j}=i k+j$ for any $i \geq 0, j \in\{1,2, \ldots, k-1\}$ ). Consider the sequence

$$
k, n_{1}, 2 k, n_{2}, n_{3}, n_{4}, 3 k, n_{5}, \ldots, n_{9}, 4 k, n_{10}, \ldots, n_{16}, 5 k, \ldots, i k, n_{(i-1)^{2}+1}, \ldots, n_{i^{2}},(i+1) k, \ldots
$$

Note that every positive integer occurs in this sequence exactly once and for every $n$ the number $n+k$ appears after $n$. For every $n$ let $f(n)$ be the successor of $n$ in this sequence and let $g(n)$ be the number of terms in this sequence between $f(n)$ and $f(n)+k$ (inclusive - we count $f(n)$ and $f(n)+k$ as well). By previous remarks, $f$ and $g$ are well defined and satisfy $f^{g(n)}(n)=f(n)+k$. Moreover, $g\left(n_{i^{2}}\right)=2 i+3$ for any $i$, hence $g$ is unbounded.

Solution 2. We will prove that if $k=1$ then $g$ is necessarily bounded.

The given equation implies that if $m=f(n)$ is in the image of $f$ then $m+1=f(n)+1=$ $f\left(f^{g(n)-1}(n)\right)$ is in the image of $f$ as well. Let $f(a)$ be the minimum of the image of $f$. Then the image of $f$ is equal to $\{f(a), f(a)+1, \ldots\}$.

However, an easy inductive argument shows that for every $m$ the number $f(a)+m$ is of the form $f^{n}(a)$. Hence the set $\left\{f(a), f^{2}(a), \ldots\right\}$ is also equal to the image of $f$.

If $f^{x}(a)=f^{y}(a)$ for some $x>y$, then the sequence $a_{n}=f^{n}(a)$ is eventually periodic with a period $x-y$, but then the set $\left\{f(a), f^{2}(a), \ldots\right\}$ is finite, which is a contradiction.

Therefore, for every $n \in \mathbb{N}_{0}$, there exists a unique positive integer $x_{n}$ such that $f^{x_{n}}(a)=n+f(a)$, and conversely, for all $x \geq 1$ the number $f^{x}(a)$ is of the form $m+f(a)$ for some $m \geq 0$. In other words, the map $\mathbb{N}_{0} \ni n \mapsto x_{n} \in \mathbb{N}$ is bijective.

Furthermore, $f^{x_{n+1}}(a)=(f(a)+n)+1=f^{x_{n}}(a)+1=f\left(f^{x_{n}-1}(a)\right)+1=f^{g\left(f^{x_{n}-1}(a)\right)}\left(f^{x_{n}-1}(a)\right)=$ $f^{x_{n}-1+g\left(f^{x_{n}-1}(a)\right)}(a)$, which implies $x_{n+1}=x_{n}-1+g\left(f^{x_{n}-1}(a)\right)>x_{n}$, since $g(t)>1$ for all $t$.

Therefore, the map $n \mapsto x_{n}$ is a strictly increasing bijection from $\mathbb{N}_{0}$ to $\mathbb{N}$ which gives $x_{n}=n+1$ for all $n$. Thus $f^{n+1}(a)=f(a)+n$, which implies $f(f(n))=f(n)+1$ for all $n \in \mathbb{N}$, hence $g(n)=2$ for all $n \in \mathbb{N}$. Obviously, this means $g$ is bounded.

Suppose now that $k \geq 2$. We will give an explicit example of functions $f$ and $g$ satisfying required properties.

- For each positive integer $n$, let $f\left(k^{n}\right)=n k+1$ and let $f(n k+1)=k^{n}+2$.
- For each positive integer $a$ which is not a power of $k$, let $f(a k)=a k+2$.
- For any other positive integer $x$, let $f(x)=x+1$.

In other words: We take the sequence of all positive integers $1,2,3, \ldots$, remove from it all numbers congruent to 1 modulo $k$ except for 1 itself, then insert them again a bit later, with each $n k+1$ occurring directly after $k^{n}$. That is, we get the sequence (shown here for a $k \geq 4$ )

$1,2, \ldots, k, k+1, k+2, \ldots, 2 k, 2 k+2, \ldots, 3 k, 3 k+2, \ldots, k^{2}, 2 k+1, k^{2}+2, \ldots, k^{3}, 3 k+1, k^{3}+2, \ldots$ and for every $n$ we let $f(n)$ be the successor of $n$ in this sequence.

For example in the case $k=2$, we get the sequence

$$
1, \mathbf{2}, 3, \mathbf{4}, 5,6, \mathbf{8}, 7,10,12,14, \mathbf{1 6}, 9,18,20, \ldots, \mathbf{3 2}, 11,34 \ldots
$$

and in the case $k=5$, we get the sequence

$$
1,2,3,4,5,6,7,8,9,10,12,13,14,15,17,18,19,20,22,23,24, \mathbf{2 5}, 11,27,28,29,30,32,33, \ldots
$$

with the powers of $k$ highlighted in each case.

Like in the first solution we note that for every $n$ the number $n+k$ appears after $n$, so we again define $g(n)$ as the number of terms in the sequence between $f(n)$ and $f(n)+k$ (inclusively), or equivalently, define $g(n)$ to be the smallest $m$ such that $f^{m}(n)=f(n)+k$.

Now, $g\left(k^{n}\right)$ (denoting the distance between $f\left(k^{n}\right)=n k+1$ and $f\left(k^{n}\right)+k=n k+1+k=$ $(n+1) k+1=f\left(k^{n+1}\right)$ which occur after $k^{n}$ and $k^{n+1}$, respectively) is equal to two plus the number of integers $i$ such that $k^{n}+2 \leq i \leq k^{n+1}$ and $i$ is not congruent to 1 modulo $k$. For easier calculation we note that this is equivalent to the amount of integers $i$ with $k^{n}<i \leq k^{n+1}$ and $i$ not congruent to 1 modulo $k$, and observe that there exist exactly

$$
\left(k^{n+1}-k^{n}\right) \cdot \frac{k-1}{k}=\left(k^{n}-k^{n-1}\right) \cdot(k-1)=k^{n-1} \cdot(k-1)^{2}
$$

such numbers, which grows arbitrarily large for sufficiently large $n$. Therefore, $g$ is also unbounded.

A note on possible constructions. The two shown constructions share a common idea that can be generalized further: Taking the sequence of all positive integers, removing any one residue class modulo $k$, and sprinkling it back in at increasingly large distances will always result in functions $f$ and $g$ with the desired properties.

## $\mathrm{I}-2$

We call a positive integer $N$ contagious if there exist 1000 consecutive non-negative integers such that the sum of all their digits is $N$. Find all contagious positive integers.

Answer. All $N \geq 13500$.

Any solution naturally splits into two parts:

Part 1: Showing that no $N<13500$ is contagious.

Part 2: Showing that all $N \geq 13500$ are contagious.

We present one approach to Part 1 and three approaches to Part 2 (by direct construction, by induction, and by discrete continuity).

Part 1. We make the following observation:

(T) Consider a block of 1000 consecutive non-negative integers. Then the last three digits of those numbers (prepended by zeros if needed) form a set $\{000,001, \ldots, 999\}$.

Thus, given any such block, the sum of the last three digits alone equals $3 \cdot 100 \cdot(0+1+\cdots+9)=$ 13500 (since each of the digits $0,1, \ldots, 9$ occurs 100 times in each of the 3 positions). Therefore no integer less than 13500 is contagious.

Part 2, by direct construction. Fix $N \geq 13500$ and write the "remaining" digit sum as $N-13500=d \cdot 1000+r$, where $d \geq 0$ and $r \in[0,999]$ are non-negative integers. Write $r=\overline{r_{2} r_{1} r_{0}}$ as a 3 -digit number (prepended by zeros if needed). Consider a number

$$
X=\underbrace{\overline{11 \ldots 1} r_{2} r_{1} r_{0}}_{d \text { times }}
$$

formed by concatenating $d$ copies of the digit 1 and the digits $r_{2}, r_{1}, r_{0}$. (If $d=0$ set $X=r$.) We claim that the total digit sum of the 1000 consecutive non-negative integers $X, X+1, \ldots, X+999$ equals $N$. Note that:

(a) Ignoring the last three digits, the $1000-r$ numbers $X, \ldots, X+(999-r)$ have digit sum $d \cdot 1=d$ each and the next $r$ numbers $X+(1000-r), \ldots, X+999$ have digit sum $(d-1) \cdot 1+2=d+1$ each.

(b) As in Part 1, the last three digits of all the 1000 numbers add up to 13500 .

Therefore, all in all, we obtain that the total digit sum of $X, X+1, \ldots, X+999$ equals

$$
(1000-r) \cdot d+r \cdot(d+1)+13500=1000 d+r+13500=N
$$

as required.

Part 2, by induction. Given a non-negative integer $n$, denote by $s_{n}$ the digit sum of $n$ and by $S(n)$ the total digit sum of $n, n+1, \ldots, n+999$, that is,

$$
S(n)=s_{n}+s_{n+1}+\cdots+s_{n+999}
$$

We proceed by induction. As a first step, we show that the 1000 numbers $N \in\{13500, \ldots, 14499\}$ are all contagious. As a second step, we show that if $N$ is contagious, then $N+1000$ is contagious. Combined, this implies that all $N \geq 13500$ are contagious.

For the first step, note that for any integer $n \geq 0$ we have

$$
S(n+1)-S(n)=\left(s_{n+1}+\cdots+s_{n+1000}\right)-\left(s_{n}+\cdots+s_{n+999}\right)=s_{n+1000}-s_{n}
$$

Thus, for $0 \leq X \leq 999$, we have $S(X+1)=S(X)+1$, since the number $X+1000$ has an extra digit 1 in front of the (up to three-digit) number $X$. Since $S(0)=13500$ by Part 1, we get $S(X)=13500+X$ for $0 \leq X \leq 999$. Therefore all $N \in\{13500, \ldots, 14499\}$ are indeed contagious.

For the second step, suppose that $N$ is contagious, that is, there exist 1000 consecutive integers $X, X+1, \ldots, X+999$ with total digit sum $N$. Take any integer $i$ such that $10^{i}>X+999$. Then the 1000 consecutive integers

$$
10^{i}+X, 10^{i}+X+1, \ldots, 10^{i}+X+999
$$

have a total digit sum equal to $N+1000$ (since each number got an extra digit 1 and, possibly, several zeroes).

Part 2, by discrete continuity. We make three observations:

(A) For any integer $n \geq 0$, we have $S(n+1)-S(n) \leq 1$.

- Indeed, as before, we have

$$
S(n+1)-S(n)=\left(s_{n+1}+\cdots+s_{n+1000}\right)-\left(s_{n}+\cdots+s_{n+999}\right)=s_{n+1000}-s_{n}
$$

Note that the numbers $n+1000$ and $n$ have the same last three digits. We distinguish two cases:
a) If the fourth digit of $n$ from the right is less than 9 , then the digits of $n+1000$ and $n$ differ only in that position and we have $s_{n+1000}-s_{n}=1$. (If $n$ is 3 -digit, this is true too.)

b) Otherwise, suppose that there are $d \geq 1$ consecutive digits 9 just in front of the last three digits of $n$. Then $s_{n+1000}-s_{n}=1-9 d<1$, because the resulting number will have $d$ zeroes in place of the nines, and the digit to the left of the nines increased by one.

(B) We have $S(0)=13500$.

- By the same argument as in Part 1 we get $S(0)=3 \cdot 100 \cdot(0+1+\cdots+9)=13500$.

(C) The sequence $S(n)$ is unbounded as $n \rightarrow \infty$.

- For instance, setting $n=10^{k}-1$ we get $S(n) \geq s_{n}=9 k$.

It remains to put the observations together. By (B), the number $n=13500$ is contagious. Now fix $N \geq 13501$. Since the sequence $(S(n))_{n=0}^{\infty}$ is unbounded, there exists an integer $k \geq 1$ such that $S(k) \geq N$. Take the smallest such $k$. By minimality of $k$ we have $S(k-1) \leq N-1$. Combining this with (A) we now deduce

$$
N \leq S(k) \leq 1+S(k-1) \leq 1+(N-1)=N
$$

hence $S(k)=N$ implying that $N$ is contagious.

## $\mathrm{I}-3$

Let $A B C$ be an acute scalene triangle with circumcircle $\omega$ and incenter $I$. Suppose the orthocenter $H$ of $B I C$ lies inside $\omega$. Let $M$ be the midpoint of the longer arc $B C$ of $\omega$. Let $N$ be the midpoint of the shorter arc $A M$ of $\omega$.

Prove that there exists a circle tangent to $\omega$ at $N$ and tangent to the circumcircles of $B H I$ and CHI.

Solution 1. Denote the circumcircles of $B H I$ and $C H I$ by $\omega_{1}$ and $\omega_{2}$ and their centers by $O_{1}$ and $O_{2}$, respectively. Let $O$ be the center of $\omega$. Let $R$ be the radius of $\omega$.

Since $H$ is the orthocenter of triangle $B I C$ it follows that $I$ is the orthocenter of triangle $B H C$. Therefore

$\angle H I B=180^{\circ}-(\angle B H I+\angle I B H)=180^{\circ}-\left(90^{\circ}-\angle C B H+90^{\circ}-\angle B H C\right)=180^{\circ}-\angle H C B$,

and analogously we get $\angle C I H=180^{\circ}-\angle C B H$ and $\angle B I C=180^{\circ}-\angle B H C$.

Denote by $r$ the radius of circle $\omega_{1}$, then from sine law we get

$$
\begin{aligned}
2 r & =\frac{H B}{\sin \angle H I B}=\frac{H B}{\sin \left(180^{\circ}-\angle H I B\right)}=\frac{H B}{\sin \angle H C B}= \\
& =\text { diameter of circumcircle of the triangle } B H C
\end{aligned}
$$

Using the same argument for triangles $C I H$ and $B I C$ we see that $r$ is equal to radii of $\omega_{1}, \omega_{2}$, circumcircles of $B I C$ and $B H C$.

From the following angle chase it follows that

$$
\begin{aligned}
\angle B H C & =180^{\circ}-\angle B I C=180^{\circ}-\left(180^{\circ}-\frac{1}{2} \angle C B A-\frac{1}{2} \angle B C A\right)= \\
& =\frac{1}{2}(\angle C B A+\angle B C A)=90^{\circ}-\frac{1}{2} \angle B A C
\end{aligned}
$$

Since $H$ lies inside $\omega$ and $\angle B A C$ is acute we conclude that

$$
\angle B A C<\angle B H C=90^{\circ}-\frac{1}{2} \angle B A C<90^{\circ}
$$

so

$$
2 r=\text { diameter of circumcircle of } B H C=\frac{B C}{\sin \angle B H C}<\frac{B C}{\sin \angle B A C}=2 R
$$

thus $r<R$.

Let $\angle B A C=\alpha, \angle C B A=\beta, \angle A C B=\gamma$. Then

$$
\angle B O_{1} I=2 \angle B H I=2\left(90^{\circ}-\angle C B H\right)=2 \angle I C B=\gamma
$$

so

$$
\angle I B O_{1}=90^{\circ}-\frac{1}{2} \angle B O_{1} I=90^{\circ}-\frac{\gamma}{2}=\frac{\alpha+\beta}{2}
$$

and finally

$$
\angle A B O_{1}=\angle I B O_{1}-\angle I B A=\frac{\alpha+\beta}{2}-\frac{\beta}{2}=\frac{\alpha}{2}=\angle B A I .
$$

This shows that $B O_{1} \| A I$, and moreover, rays $B O_{1}^{\rightarrow}, A I \rightarrow$ determine opposite directions. Similarly, rays $\mathrm{CO}_{2}, A I^{\rightarrow}$ are parallel and determine opposite directions. Therefore these rays are parallel and $B O_{1}, C O_{2} \rightarrow$ determine the same direction. Since $B O_{1}=r=C O_{2}$, it follows that vectors $\overrightarrow{B O_{1}}, \overrightarrow{C O_{2}}$ are equal. Denote this vector by $\vec{v}$.

![](https://cdn.mathpix.com/cropped/2024_06_05_6a77b54051eac76d613fg-11.jpg?height=1242&width=1288&top_left_y=1001&top_left_x=384)

Note that $O N \perp A M$. Moreover

$$
\begin{aligned}
\angle I A M & =\angle I A C+\angle C A M=\angle I A C+\angle C B M= \\
& =\angle I A C+\frac{1}{2}\left(180^{\circ}-\angle B M C\right)=\angle I A C+\frac{1}{2}\left(180^{\circ}-\angle B A C\right)=90^{\circ}
\end{aligned}
$$

so $A M \perp A I$, hence $O N\|A I\| B O_{1} \| \vec{v}$. Let $X$ be a point such that $\overrightarrow{O X}=\vec{v}$. Since $O N \| \vec{v}, X$ lies on line $O N$. It actually lies on ray $O N \rightarrow$ since rays $O N \rightarrow, A I^{\rightarrow}$ determine opposite directions.

Note that translation by $\vec{v}$ maps triangle $B C O$ to triangle $O_{1} O_{2} X$. Therefore $O_{1} X=B O=R$ and $O_{2} X=C O=R$.

Let $\omega^{\prime}$ be the circle centered at $X$ with radius $R-r>0$.

Observe that $O_{1} X=R=r+(R-r)$, so $\omega^{\prime}$ is tangent externally to $\omega_{1}$. For similar reason it is tangent externally to $\omega_{2}$. Moreover $O X=r=R-(R-r)=O N-X N$, so $\omega^{\prime}$ is tangent to $\omega$ internally at point $N$.

Solution 2. Let $\beta:=\angle A B I=\angle I B C$ and $\gamma:=\angle B C I=\angle I C A$. Without loss of generality, we assume that $\beta>\gamma$. Furthermore, we define $P$ to be the intersection of $N B$ and the circumcircle of $B H I$ and $Q$ to be the intersection of $N C$ and the circumcircle of $C H I$.

![](https://cdn.mathpix.com/cropped/2024_06_05_6a77b54051eac76d613fg-12.jpg?height=1002&width=945&top_left_y=1172&top_left_x=561)

Our goal is to prove that the circumcircle of $N P Q$ satisfies the desired conditions. To this end, define the tangent $t_{N}$ to $\omega$ through $N$, tangent $t_{P}$ to the circumcircle of $B H I$ through $P$ and tangent $t_{Q}$ to the circumcircle of $C H I$ through $Q$. Let $X$ be the intersection of $t_{P}$ and $t_{Q}, Y$ the intersection of $t_{N}$ and $t_{Q}$ and $Z$ the intersection of $t_{N}$ and $t_{P}$. If we can prove that $P X=Q X$, $N Y=Q Y$ and $N Z=P Z$, it follows that the circumcircle of $N P Q$ is the incircle of $X Y Z$ and $N, P$ and $Q$ are the contact points. By definition of $t_{N}, t_{P}$ and $t_{Q}$, this would imply that the circumcircle of $N P Q$ satisfies the desired properties.

Using that the arcs $M N$ and $N A$ have the same lengths and the triangle $B M C$ is isosceles, we calculate some angles:

$$
\begin{aligned}
\angle A C N=\angle A B N & =\frac{1}{2} \angle A B M \\
& =\frac{1}{2}(\angle A B C-\angle M B C) \\
& =\frac{1}{2}\left(2 \beta-\frac{180^{\circ}-\angle C M B}{2}\right) \\
& =\frac{1}{2}\left(2 \beta-\frac{180^{\circ}-\left(180^{\circ}-2 \beta-2 \gamma\right)}{2}\right) \\
& =\frac{1}{2}(\beta-\gamma)
\end{aligned}
$$

Furthermore, using that $H$ is the orthocenter of $B I C$ and has to lie outside of it but inside $\omega$, we compute

$$
\begin{aligned}
\angle I C Q & =\angle I C A+\angle A C N=\gamma+\frac{1}{2}(\beta-\gamma)=\frac{1}{2}(\beta+\gamma) \\
\angle P B I & =\angle A B I-\angle A B N=\beta-\frac{1}{2}(\beta-\gamma)=\frac{1}{2}(\beta+\gamma) \\
\angle B C H & =90^{\circ}-\angle I B C=90^{\circ}-\beta \\
\angle H I Q & =\angle H C Q \\
& =\angle B C I+\angle I C Q-\angle B C H \\
& =\gamma+\frac{1}{2}(\beta+\gamma)-\left(90^{\circ}-\beta\right) \\
& =\frac{1}{2}\left(3 \beta+3 \gamma-180^{\circ}\right)
\end{aligned}
$$

By an analogous computation, we find that

$$
\angle P I H=\frac{1}{2}\left(3 \beta+3 \gamma-180^{\circ}\right)
$$

If we now define $X^{\prime}$ as the intersection of $t_{P}$ and $H I$ as well as $X^{\prime \prime}$ the intersection of $t_{Q}$ and $H I$, we obtain by tangency:

$$
\angle X^{\prime} P I=\frac{1}{2} \angle P B I=\frac{1}{2}(\beta+\gamma)=\frac{1}{2} \angle I C Q=\angle I Q X^{\prime \prime}
$$

Since also

$$
\angle P I X^{\prime}=\angle P I H=\frac{1}{2}\left(3 \beta+3 \gamma-180^{\circ}\right)=\angle H I Q=\angle X^{\prime \prime} I Q
$$

we get two similar triangles $X^{\prime} P I$ and $X^{\prime \prime} Q I$. Also, since $H I$ is the power line of the circumcircles of $B H I$ and $C H I$, we have

$$
X^{\prime} P^{2}=X^{\prime} H \cdot X^{\prime} I, \quad X^{\prime \prime} Q^{2}=X^{\prime \prime} H \cdot X^{\prime \prime} I
$$

and hence

$$
\frac{X^{\prime} H \cdot X^{\prime} I}{X^{\prime \prime} H \cdot X^{\prime \prime} I}=\left(\frac{X^{\prime} P}{X^{\prime \prime} Q}\right)^{2}=\left(\frac{X^{\prime} I}{X^{\prime \prime} I}\right)^{2}
$$

using similarity. By simplifying those terms, we get

$$
\frac{X^{\prime} H}{X^{\prime \prime} H}=\frac{X^{\prime} I}{X^{\prime \prime} I}=\frac{X^{\prime} H+H I}{X^{\prime \prime} H+H I}
$$

hence $X^{\prime} H \cdot H I=X^{\prime \prime} H \cdot H I$ and therefore $X^{\prime} H=X^{\prime \prime} H$, which implies that $X^{\prime}=X^{\prime \prime}=X$. By the power of $X$ to $B H I$ and $C H I$, we finally get $X P^{2}=X Q^{2}$, so $X P=X Q$ as desired.

To see that $N Y=Q Y$, define $D$ as the second intersection of the circumcircles of $C H I$ and $A B C$ and let $Y^{\prime}$ be the intersection of $C D$ and $t_{N}$. We want to show that $Y^{\prime}=Y$. Using that $\angle C Q I=\angle C H I=90^{\circ}-\angle B C H=90^{\circ}-\left(90^{\circ}-\angle I B C\right)=\beta$, we compute

$$
\begin{aligned}
\angle Q D Y^{\prime} & =180^{\circ}-\angle C D Q \\
& =\angle Q I C \\
& =180^{\circ}-\angle C Q I-\angle I C Q \\
& =180^{\circ}-\beta-\frac{1}{2}(\beta+\gamma)
\end{aligned}
$$

On the other hand, the tangency to $\omega$ at $N$ yields

$$
\angle Y^{\prime} N Q=\angle Y^{\prime} N C=\angle N B C=\angle I B C+\angle N B I=\beta+\frac{1}{2}(\beta+\gamma)
$$

Now, since $\angle Q D Y^{\prime}+\angle Y^{\prime} N Q=180^{\circ}$, we conclude that $N Q D Y^{\prime}$ is a cyclic quadrilateral. The same is true for $N Q D Y$ because of

$$
\angle Y N D=\angle N C D=\angle Q C D=\angle Y Q D
$$

where we used tangency of $t_{N}$ to $\omega$ and of $t_{Q}$ to the circumcircle of $C H I$. Since both $Y$ and $Y^{\prime}$ lie on $t_{N}$, they have to be the same point. Since $C D$ is the power line of circle $\omega$ and the circumcircle of $C H I$, we obtain $Y Q^{2}=Y N^{2}$, so $Q Y=N Y$. We can find a completely analogous argument for $P Z=N Z$ to conclude.

## $\mathrm{I}-4$

Find all positive integers $n$ for which there exist positive integers $x_{1}, x_{2}, \ldots, x_{n}$ such that

$$
\frac{1}{x_{1}^{2}}+\frac{2}{x_{2}^{2}}+\frac{4}{x_{3}^{2}}+\cdots+\frac{2^{n-1}}{x_{n}^{2}}=1
$$

Answer. Solutions exist for all positive integers $n$ except for $n=2$.

## Solution 1.

- $n=1$ :

Here, $x_{1}:=1$ provides a solution, since

$$
\frac{1}{1^{2}}=1
$$

- $n=2$ :

Here, no solution exists. Indeed, $x_{1}=1$ or $x_{2}=1$ yields $\frac{1}{x_{1}^{2}}+\frac{2}{x_{2}^{2}}>1$, while $x_{1}, x_{2} \geq 2$ leads to

$$
\frac{1}{x_{1}^{2}}+\frac{2}{x_{2}^{2}} \leq \frac{1}{4}+\frac{2}{4}=\frac{3}{4}<1
$$

- $n=4$ :

Here, $\left(x_{1}, x_{2}, x_{3}, x_{4}\right):=(3,3,3,6)$ provides a solution, since

$$
\frac{1}{3^{2}}+\frac{2}{3^{2}}+\frac{4}{3^{2}}+\frac{8}{6^{2}}=\frac{7}{9}+\frac{8}{36}=\frac{7}{9}+\frac{2}{9}=1
$$

- Induction step from $n$ to $(n+2)$ :

Let $\left(y_{1}, y_{2}, \ldots, y_{n}\right)$ be a solution for $n$, i.e.,

$$
\frac{1}{y_{1}^{2}}+\frac{2}{y_{2}^{2}}+\frac{4}{y_{3}^{2}}+\cdots+\frac{2^{n-1}}{y_{n}^{2}}=1
$$

Then

$$
\left(x_{1}, x_{2}, \ldots, x_{n+2}\right):=\left(2,2,4 y_{1}, 4 y_{2}, \ldots, 4 y_{n}\right)
$$

is a solution for $(n+2)$, since

$$
\begin{aligned}
\frac{1}{x_{1}^{2}}+\frac{2}{x_{2}^{2}}+\frac{4}{x_{3}^{2}}+\cdots+\frac{2^{n+1}}{x_{n+2}^{2}} & =\frac{1}{2^{2}}+\frac{2}{2^{2}}+\frac{4}{\left(4 y_{1}\right)^{2}}+\cdots+\frac{2^{n+1}}{\left(4 y_{n}\right)^{2}} \\
& =\frac{1}{4}+\frac{2}{4}+\frac{4}{16}\left[\frac{1}{\left(y_{1}\right)^{2}}+\cdots+\frac{2^{n-1}}{\left(y_{n}\right)^{2}}\right] \\
& =\frac{3}{4}+\frac{1}{4} \cdot 1 \\
& =1
\end{aligned}
$$

Using this induction step and the solutions for $n=1$ and $n=4$, we can construct solutions for all $n \geq 3$.

Solution 1a. There are other induction steps possible. For example from $n$ to $(n+3)$ :

Let $\left(y_{1}, y_{2}, \ldots, y_{n}\right)$ be a solution for $n$, i.e.,

$$
\frac{1}{y_{1}^{2}}+\frac{2}{y_{2}^{2}}+\frac{4}{y_{3}^{2}}+\cdots+\frac{2^{n-1}}{y_{n}^{2}}=1
$$

Then

$$
\left(x_{1}, x_{2}, \ldots, x_{n+3}\right):=\left(3,3,3,6 y_{1}, 6 y_{2}, \ldots, 6 y_{n}\right)
$$

is a solution for $(n+3)$, since

$$
\begin{aligned}
\frac{1}{x_{1}^{2}}+\frac{2}{x_{2}^{2}}+\frac{4}{x_{3}^{2}}+\cdots+\frac{2^{n+2}}{x_{n+3}^{2}} & =\frac{1}{3^{2}}+\frac{2}{3^{2}}+\frac{4}{3^{2}}+\frac{8}{\left(6 y_{1}\right)^{2}}+\cdots+\frac{2^{n+2}}{\left(6 y_{n}\right)^{2}} \\
& =\frac{1}{9}+\frac{2}{9}+\frac{4}{9}+\frac{8}{36}\left[\frac{1}{\left(y_{1}\right)^{2}}+\cdots+\frac{2^{n-1}}{\left(y_{n}\right)^{2}}\right] \\
& =\frac{7}{9}+\frac{2}{9} \cdot 1 \\
& =1
\end{aligned}
$$

In order to complete this approach, of course, solutions have to be provided for $n=1,3$, and 5 .

In fact, every solution $\left(z_{1}, \ldots, z_{k}\right)$ for $k$ yields an induction step from $n$ to $(n+k-1)$. Indeed, if $\left(y_{1}, \ldots, y_{n}\right)$ is a solution then

$$
\left(z_{1}, \ldots, z_{k-1}, z_{k} \cdot y_{1}, \ldots, z_{k} \cdot y_{n}\right)
$$

is a solution, too. The two constructions presented above belong to $(2,2,4)$ and $(3,3,3,6)$.

So it is conceivable that someone finds an induction from $n$ to, say, $(n+6)$. In this case, solutions for six suitable small values of $n$ would be necessary in order to complete the approach.

Solution 2. As in the previous solution, we can show that there does not exist a solution for $n=2$ and find explicit solutions for $n=1$ and $n=3$.

We construct further solutions by induction. Let $\left(y_{1}, y_{2}, \ldots, y_{n}\right)$ be a solution for $n \geq 3$.

Then setting $x_{n-2}=x_{n+1}:=3 y_{n-2}$ and $x_{i}:=y_{i}$ for all other $i$ is a solution for $n+1$, since for the sum of the terms corresponding to $x_{n-2}$ and $x_{n+1}$ (which are the only ones that differ between the solutions for $n$ and $n+1$ ) we get

$$
\frac{2^{n-3}}{x_{n-2}^{2}}+\frac{2^{n}}{x_{n+1}^{2}}=\frac{2^{n-3}}{9 y_{n-2}^{2}}+\frac{2^{n}}{9 y_{n-2}^{2}}=\frac{(1+8) \cdot 2^{n-3}}{9 y_{n-2}^{2}}=\frac{2^{n-3}}{y_{n-2}^{2}}
$$

thereby keeping the sum of all terms equal.

Solution 3. As in the other solutions, we can show that there does not exist a solution for $n=2$. Also, we can find explicit solutions for $n=4, n=6$ and $n=8$. Now we prove that we can find suitable integers for all other $n$ :

For $n=2 k+1$ (where $k$ is a suitable non-negative integer), we can choose

$$
x_{1}=\cdots=x_{n-1}=2^{k}, \quad x_{n}=2^{2 k}
$$

to obtain

$$
\sum_{i=1}^{n} \frac{2^{i-1}}{x_{i}^{2}}=\sum_{i=1}^{2 k} \frac{2^{i-1}}{2^{2 k}}+\frac{2^{2 k}}{2^{4 k}}=\frac{\frac{2^{2 k}-1}{2-1}}{2^{2 k}}+\frac{1}{2^{2 k}}=1
$$

For $n=2 k$ with an integer $k \geq 5$, observe that, by setting $x_{i}=2^{\frac{i-1}{2}}$ for odd $i$ and $x_{i}=2^{\frac{i}{2}-1}$ for even $i$, we have

$$
\sum_{i=1}^{2 k} \frac{2^{i-1}}{x_{i}^{2}}=\sum_{j=1}^{k}\left(\frac{2^{2 j-2}}{2^{2 j-2}}+\frac{2^{2 j-1}}{2^{2 j-2}}\right)=\sum_{j=1}^{k}(1+2)=3 k
$$

If we can modify the $x_{i}$ in order to obtain some square number $m^{2}$ on the right hand side, we can in a second step multiply each $x_{i}$ by $m$ to obtain a sum of 1 .

Observe that all $x_{i}$ for $i>2$ are even. We show that there is a square divisible by 3 that can be obtained by replacing some of the $x_{i}$ (with $i>2$ ) by $x_{i}^{\prime}:=x_{i} / 2$ :

Note that for odd $i$, this increases the sum by $4-1=3$ and for even $i$, this increases the sum by $8-2=6$. Since there are $k-1$ odd and $k-1$ even indexes to choose from, we can increase the sum by every number $l$ that is divisible by three and satisfies $0 \leq l \leq 3(k-1)+6(k-1)=9 k-9$.

Because of $k \geq 5$ and therefore $\sqrt{3 k}<k$, the smallest square $m^{2} \geq 3 k$ which is divisible by 3 certainly satisfies

$$
(\sqrt{3 k})^{2}=3 k \leq m^{2} \leq(\sqrt{3 k}+3)^{2}=3 k+6 \sqrt{3 k}+9<9 k+9
$$

In particular, $m^{2} \leq 9 k+6$ because $m$ is divisible by 3 . This means that in order to increase $3 k$ to $m^{2}$, we have to add a number between 0 and $6 k+6$ to the sum above. However, $6 k+6 \leq 9 k-9$ because $k \geq 5$ and by the above argument, we can always do that.

To summarize, we first set the $x_{i}$ as above, then select up to $2 k-2$ of them to be divided by 2 , then multiply all of them by $m$, yielding a right hand side of 1 .