exam: IZho

problem: Each point with integral coordinates in the plane is coloured white or blue. Prove that one can choose a colour so that for every positive integer $n$ there exists a triangle of area $n$ having its vertices of the chosen colour.

solution:
If there exists some $c$-monochromatic horizontal row $y=k$, then if both rows $y=k-1$ and $y=k+1$ are $\bar{c}$-monochromatic we can find $\bar{c}$-monochromatic triangles of any positive integer area, otherwise they must contain at least a $c$-point, and we can find $c$-monochromatic triangles of any positive integer area. So assume there exists no monochromatic horizontal row. Consider the horizontal row $y=0$. If it contains two $c$-points at distance 1 , then for any positive integer $n$, together with a $c$-point on row $y=2 n$ they will make a $c$-monochromatic triangle of area $n$. Otherwise it will contain two $\bar{c}$-points at distance 2 , which for any positive integer $n$, together with a $\bar{c}$-point on row $y=n$ will make a $\bar{c}$-monochromatic triangle of area $n$.

Comentarii. Extrem de puţin din întreaga diversitate de colorări este utilizat, ceea ce face problema oarecum trivială ... De remarcat că nu este adevărat că se pot întotdeauna obţine triunghiuri monocromatice de orice arie $n / 2$; de exemplu pentru colorarea "în tablă de şah" nu se vor obţine triunghiuri monocromatice de arie $1 / 2$. Iar dacă mărim numărul de culori la trei, există o (simplă) colorare fără triunghiuri monocromatice de arie 1. Presupun că întrebări mult mai interesante şi dificile pot fi imaginate ...

problem_match: Problem 1.

solution_match: Solution.


problem: Inside the triangle $A B C$ a point $M$ is given. The line $B M$ meets the side $A C$ at $N$. The point $K$ is symmetrical to $M$ with respect to $A C$. The line $B K$ meets $A C$ at $P$. If $\angle A M P=\angle C M N$, prove that $\angle A B P=\angle C B N$.

solution:
(AoPS - user TelvCohl) Let $M^{*}$ be the isogonal conjugate of $M$ with respect to $\triangle A B C$.

LEMMA. Let $\ell$ be the isogonal conjugate of $A M$ with respect to $\angle B M C$, and let $\ell^{*}$ be the isogonal conjugate of $A M^{*}$ with respect to $\angle B M^{*} C$. The lines $\ell, \ell^{*}$ meet on $B C$ and are symmetrical with respect to $B C$.

Proof. Let $X$ be the meeting point of $B M^{*}$ and $C M$. Let $Y \in B C$ be a point such that $X A, X Y$ are isogonal conjugate with respect to $\angle B X C$.

Since $C A, C Y$ are isogonal conjugate with respect to $\angle M^{*} C M$, so $A$ and $Y$ are isogonal conjugate with respect to $\triangle C X M^{*}$, it follows that $M^{*} Y$, $M^{*} A$ are isogonal conjugate with respect to $\angle B M^{*} C$. Similarly, we can prove $M Y, M A$ are isogonal conjugate with respect to $\angle B M C$.

By easy angle chasing we get $\angle M Y B=\angle C Y M^{*}$, thus $M Y \equiv \ell$ and $M^{*} Y \equiv \ell^{*}$ are symmetrical with respect to $B C$.

From the LemmA we now get that $P K$ is the isogonal conjugate of $B M^{*}$ with respect to $\angle A M^{*} C$, thus from $B \in P K$ we get $P K$ to be the angle bisector of $\angle A M^{*} C$ and $M^{*} \in B K$, therefore $\angle A B P=\angle C B N$.

problem_match: Problem 2.


solution:
(Ştefan Tudose on AoPS) Notice that the thesis is equivalent (by Steiner's theorem) with $\frac{B C}{B A}=\frac{M C}{M A}$; stated in other words, that $B$ lies on the $M$-Apollonius circle of $\triangle M C A$.

Let $F$ be the foot of the angle bisector of $\angle C M A$. By dint of the above observation, it is enough to prove that $B \in \odot(M F K)$, which is trivial by simple angle chasing; suppose wlog that $M A>M C$, it then follows $\angle M F K=\pi+\angle M A C-\angle M C A=\pi-\angle K B M$.

solution_match: Alternative Solution.


problem:
Determine all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ such that

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

for all $x, y \in \mathbb{R}$.

solution:
(Ştefan Tudose on AoPS) Let $P(x, y)$ be the assertion that $f\left(x^{3}+y^{3}+x y\right)=x^{2} f(x)+y^{2} f(y)+f(x y)$. From $P(1,0)$ we get that $f(0)=0$, hence from $P(x, 0)$ we get $f\left(x^{3}\right)=x^{2} f(x) . P(x,-x)$ yields $f(x)=-f(-x)$.

From $P(x, y)-P(x,-y)$ we get

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

Substituting back into $P(x, y)$ we get that

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

Let $a, b$ be real numbers, and let $x=\sqrt[3]{a}$. The polynomial $P_{b}(\alpha)=\alpha^{3}+x \alpha-b$ has odd degree, hence at least one real root. Let $y$ be one of the real roots. Plugging $x$ and $y$ in $(*)$, we get that $f(a+b)=2 f(a)+f(b-a)$ for all $a, b \in \mathbb{R}$. Taking $a=b$, we get $f(2 a)=2 f(a)$, so

$$
f(x+y)=f(x)+f(y) \text { for all } x, y \in \mathbb{R}
$$

So $f(x+y)=f(x)+f(y)$ and

$$
f\left(x^{3}\right)=x^{2} f(x) \quad(* *)
$$

Taking $x+1$ and $x-1$ in $(* *)$ and summing up the two relations we get that $2 x^{2} f(x)+6 f(x)=f(x)\left(2 x^{2}+2\right)+4 x f(1)$, i.e. $f(x)=x f(1)$.

problem_match: Problem 3.


solution:
(AoPS - user pco) Under the notations of above we similarly get $f(0)=0, f\left(x^{3}\right)=x^{2} f(x)$, and $f(x)$ odd.

Let then $u, v$ be such that $u+v \leq 0$. It's easy to see that the system $x^{3}+y^{3}+x y=u,-x^{3}-y^{3}+x y=v$ always has a solution, and then

$$
\begin{aligned}
& P(x, y) \text { yields } f(u)=x^{2} f(x)+y^{2} f(y)+f\left(\frac{u+v}{2}\right) \\
& P(-x,-y) \text { yields } f(v)=-x^{2} f(x)-y^{2} f(y)+f\left(\frac{u+v}{2}\right)
\end{aligned}
$$

Just adding the two lines above gives $f\left(\frac{u+v}{2}\right)=\frac{f(u)+f(v)}{2}$ for all $u, v$ such that $u+v \leq 0$. If $u+v \geq 0$, we then have $-u-v \leq 0$ and so $f\left(\frac{-u-v}{2}\right)=\frac{f(-u)+f(-v)}{2}$ and, since the function $f(x)$ is odd, we also have $f\left(\frac{u+v}{2}\right)=\frac{f(u)+f(v)}{2}$ for all $u, v$ such that $u+v \geq 0$.

Then $f\left(\frac{2 u+0}{2}\right)=\frac{f(2 u)+f(0)}{2}$, and so $f(2 u)=2 f(u)$. Therefore $f\left(\frac{2 u+2 v}{2}\right)=\frac{f(2 u)+f(2 v)}{2}=f(u)+f(v)$, i.e. $f(x)$ is additive.

Let then be $n \in \mathbb{N} ; P(x+n, 0)$ yields $f\left((x+n)^{3}\right)=(x+n)^{2} f(x+n)$, whence $n(2 f(x)-2 x f(1))=-3 f\left(x^{2}\right)+2 x f(x)+x^{2} f(1)$ for all $n \in \mathbb{N}$. So $f(x)=x f(1)$ for all $x$, and so $f(x)=a x$ for all $x$, which indeed is a solution, whatever $a \in \mathbb{R}$.

## Second Day - Solutions


problem: Determine the maximum integer $n$ with the property that for each positive integer $k \leq \frac{n}{2}$ there exist two positive divisors of $n$ with difference $k$.

solution:
If there exists a positive integer $p \leq\lfloor n / 6\rfloor$ such that $p \nmid n$, then we have $\lfloor n / 2\rfloor>\lfloor n / 6\rfloor$, and taking $k=\lfloor n / 2\rfloor-p \geq 2$ and two positive divisors $d, d+k$ of $n$, we need $d+(\lfloor n / 2\rfloor-p)$ to divide $n$. But $d+(\lfloor n / 2\rfloor-p) \geq d+\lfloor n / 2\rfloor-\lfloor n / 6\rfloor>d+(n / 2-1)-n / 6 \geq n / 3$, so $d+(\lfloor n / 2\rfloor-p) \in\{n / 2, n\}$, the only possible divisors of $n$ larger than $n / 3$. However, $d+(\lfloor n / 2\rfloor-p)=n / 2$ yields $d=p$, absurd (since $d \mid n$ but $p \nmid n$ ), while $d+(\lfloor n / 2\rfloor-p)=n$ yields $d>n / 2$, thus $d=n$ (since $d \mid n$ ), forcing $p=\lfloor n / 2\rfloor>\lfloor n / 6\rfloor$, again absurd. Therefore all positive integers not larger than $\lfloor n / 6\rfloor$ must divide $n$.

Denote $u=\lfloor n / 6\rfloor$. Since $\operatorname{gcd}(u, u-1)=1$, it follows $u(u-1) \mid n$, so $u(u-1) \leq n=6(n / 6)<6(u+1)$, forcing $u \leq 7$. For $u \geq 4$ we need $\operatorname{lcm}[1,2,3,4]=12 \mid n$, and we can see that $n=24$ satisfies, and moreover is an acceptable value. For $n=36$ we get $u=6$, but $\operatorname{lcm}[1,2,3,4,5,6]=60 \nmid n$. And for $n \geq 48$ we have $u \geq 8$, not acceptable. Thus the answer is $n=24$.

We may in fact quite easily exhibit the full set $\{1,2,4,6,8,12,18,24\}$ of such positive integers $n$ (for $n=1$ the condition is vacuously fulfilled). The related question of which are the positive integers satisfying the above property for all $1 \leq k \leq n-1$ can also easily be answered; the full set is $\{1,2,4,6\}$.

Comentarii. O problemă extrem de drăguţă, şi nu tocmai simplă dacă ne străduim să evităm discutarea a prea multe cazuri. Analiza numerelor $n$ mici ne sugerează imediat că $n>1$ nu poate să fie impar (ceea ce este trivial), şi prin faptul că singurul $k$ defect pentru $n=36$ este $k=13$, ideea pentru soluţia dată mai sus. Oricum, o idee proaspătă, şi care se implementează elegant şi cu calcule minime.

problem_match: Problem 4.


problem:
Let $A_{n}$ be the set of partitions of the sequence $(1,2, \ldots, n)$ into several subsequences such that every two neighbouring terms of each subsequence have different parity, and let $B_{n}$ be the set of partitions of the sequence $(1,2, \ldots, n)$ into several subsequences such that all the terms of each subsequence have the same parity. 2

Prove that for every positive integer $n$ the sets $A_{n}$ and $B_{n+1}$ contain the same number of elements.

solution:
For each partition $\pi$ of $\{1,2, \ldots, n\}$, with the elements within each block written in ascending order, denote by $k(\pi)$ the number of blocks of $\pi$ ending in an even number and by $\ell(\pi)$ the number of blocks of $\pi$ ending in an odd number.

Also denote $f_{n}(x, y)=\sum_{\pi \in A_{n}} x^{k(\pi)} y^{\ell(\pi)}$ and $g_{n}(x, y)=\sum_{\pi \in B_{n}} x^{k(\pi)} y^{\ell(\pi)}$.

We will have $\left|A_{n}\right|=f_{n}(1,1)$.

For example

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

A moment of reflection will show us that

$$
\begin{aligned}
& f_{n+1}(x, y)=y\left(f_{n}(x, y)+\frac{\mathrm{d}}{\mathrm{d} x} f_{n}(x, y)\right) \text { for even } n \\
& f_{n+1}(x, y)=x\left(f_{n}(x, y)+\frac{\mathrm{d}}{\mathrm{d} y} f_{n}(x, y)\right) \text { for odd } n
\end{aligned}
$$

This comes from considering where the element $n+1$ may go, and how this affects the number of blocks.

We will also have $\left|B_{n}\right|=g_{n}(1,1) 3^{3}$[^1]

For example

$$
\begin{aligned}
& g_{1}(x, y)=y \\
& g_{2}(x, y)=x y \\
& g_{3}(x, y)=y x(y+1) \\
& g_{4}(x, y)=x y(x y+y+x+1) \\
& g_{5}(x, y)=y x\left(x y^{2}+y^{2}+3 x y+3 y+x+1\right)
\end{aligned}
$$

Another moment of reflection will show us that

$$
\begin{aligned}
& g_{n+1}(x, y)=y\left(g_{n}(x, y)+\frac{\mathrm{d}}{\mathrm{d} y} g_{n}(x, y)\right) \text { for even } n \\
& g_{n+1}(x, y)=x\left(g_{n}(x, y)+\frac{\mathrm{d}}{\mathrm{d} x} g_{n}(x, y)\right) \text { for odd } n
\end{aligned}
$$

This also comes from considering where the element $n+1$ may go, and how this affects the number of blocks.

It is not hard to check that $g_{n+1}(x, y)=y f_{n}(x, y)$ for even $n$ and that $g_{n+1}(x, y)=x f_{n}(x, y)$ for odd $n$. This is seen to hold true for small values of $n$. Henceforth, for even $n, g_{n+1}(x, y)=y\left(g_{n}(x, y)+\frac{\mathrm{d}}{\mathrm{d} y} g_{n}(x, y)\right)=$ $y x f_{n-1}(x, y)+y x \frac{\mathrm{d}}{\mathrm{d} y} f_{n-1}(x, y)$ by induction step on $g_{n}(x, y)=x f_{n-1}(x, y)$, while $y f_{n}(x, y)=y x\left(f_{n-1}(x, y)+\frac{\mathrm{d}}{\mathrm{d} y} f_{n-1}(x, y)\right)$. Alike computation holds for odd $n$. So $g_{n+1}(1,1)=f_{n}(1,1)$ in all cases, and so $\left|B_{n+1}\right|=\left|A_{n}\right|$.

Comentarii. De fapt o partiţie de unul din cele două tipuri poate fi privită şi ca scrierea în cicluri disjuncte a unei permutări convenabile din $\mathcal{S}_{n}$. Odată ce ideea (folosirea unui fel de funcţii generatoare) se iveşte, problema devine aproape trivială. Consideraţii asupra acestor partiţii din $A_{n}$, numite parity-alternating, sunt numeroase, relativ şi la numerele Stirling de a doua speţă $\left\{\begin{array}{l}n \\ k\end{array}\right\}$, legate de numerele Bell $\beta_{n}$ prin relaţia $\beta_{n}=\sum_{k=1}^{n}\left\{\begin{array}{l}n \\ k\end{array}\right\}$; de exemplu http://www.sciencedirect.com/science/article/pii/S0024379513004758

problem_match: Problem 5.