exam: IZho

problem: Points $M, N, K$ lie on the sides $B C, C A, A B$ respectively, of a triangle $A B C$, and are different from its vertices. The triangle $M N K$ is called beautiful if the triangles $M N K$ and $A B C$ are similar (with the vertices respectively in this order). Show that if in the triangle $A B C$ there are two beautiful triangles with a common vertex, then $\triangle A B C$ is right-angled.

solution:
(L. Ploscaru) Say $M N_{1} K_{1}$ and $M N_{2} K_{2}$ are such two beautiful triangles. Let $T=N_{1} K_{1} \cap N_{2} K_{2} ; T$ exists, and belongs to the interior of $\triangle A B C$, since angles at $M$ are equal to $\angle A$. Then $\angle N_{1} M N_{2}=\angle K_{1} M K_{2}$. We then have $\angle M N_{1} T=\angle M N_{2} T=\angle B$ and $\angle M K_{1} T=\angle M K_{2} T=\angle C$. It follows that $M N_{2} N_{1} T$ and $M K_{1} K_{2} T$ are cyclic quadrilaterals.

So $\angle K_{2} K_{1} T=\angle K_{2} M T$ and $\angle A N_{1} T=\pi-\angle N_{2} N_{1} T=\angle N_{2} M T$, whence $\pi-\angle A=\angle A K_{1} N_{1}+\angle A N_{1} K_{1}=\angle K_{2} M N_{2}=\angle A$, yielding $\pi-\angle A=\angle A$, so $\angle A=\pi / 2$. Moreover, notice it forces $M$ to be the midpoint $M_{0}$ of $B C$; conversely then, all triangles $M_{0} N K$ with $\angle K M_{0} N=\pi / 2$ are beautiful.

Comentarii. Ceea ce se numeste, în "English mathematical parlance", "simple angle chasing". Reciproca merita şi ea a fi remarcată.

problem_match: Problem 1.

solution_match: Solution.


problem:
Do there exist? functions $f: \mathbb{R} \rightarrow \mathbb{R}$ such that
a) $f$ is a surjective function; and
b) $f(f(x))=(x-1) f(x)+2$ for all $x$ real.

solution:
We have $f(f(f(x)))=f(f(f(x)))=f((x-1) f(x)+2)$ and also $f(f(f(x)))=(f(x)-1) f(f(x))+2=(f(x)-1)((x-1) f(x)+2)+2=$ $(f(x)-1)(x-1) f(x)+2(f(x)-1)+2=f(x)((f(x)-1)(x-1)+2)$, so

$$
f((x-1) f(x)+2)=f(x)((x-1) f(x)+2-(x-1))
$$

Let $f(a)=0$; then for $x=a$ we get $f(0)=2$, and then for $x=0$ we get $f(2)=0$. Now for $x=1$ we get $f(1)=0$. Taking $f(b)=1$ leads us to $b=-1$, and then to $f(-1)=1$. Finally, taking $f(c)=-1$ leads to $c=2$, but that means $0=f(2)=-1$, absurd. Thus the answer is No.

Comentarii. Jonglerii cu valori particulare (mici), până la obţinerea unei contradicţii ... Important este că undeva pe parcurs trebuie exprimat $f(f(f(z)))$ în două feluri diferite, ceea ce este tipic pentru ecuaţii funcţionale conţinând iterata functुiei.

Ar fi interesant de văzut în ce măsură putem relaxa condiţile, sau ce fenomen se ascunde aici!?! Fără condiţia a) de surjectivitate, o soluţie banală este $f(0)=2, f(x)=0$ pentru $x \neq 0$. Putem oare găsi toate soluţiile?

problem_match: Problem 2.


problem: There are given 100 distinct positive integers. We call a pair of integers among them good if the ratio of its elements is either 2 or 3 . What is the maximum number $g$ of good pairs that these 100 numbers can form? (A same number can be used in several pairs.)

solution:
Like so often in Russian problems, numbers are used instead of generic symbols. Let us therefore denote $10=n>1,2=k>1,3=\ell>1$, with the extra condition both $k$ and $\ell$ aren't powers of a same number. Consider the digraph $G$ whose set of vertices $V(G)$ is made of $v=n^{2}$ distinct positive integers, and whose set of edges $E(G)$ is made by the pairs $(a, b) \in V(G) \times V(G)$ with $a \mid b$. For each positive integer $m$ consider now the (not induced) spanning subdigraph $G_{m}$ of $G$ (with $V\left(G_{m}\right)=V(G)$ and so $v_{m}=v=n^{2}$ vertices), and whose edges are the pairs $(a, b) \in G \times G$ with $b=m a$. Moreover, it is clear that $E\left(G_{m^{\prime}}\right) \cap E\left(G_{m^{\prime \prime}}\right)=\emptyset$ for $m^{\prime} \neq m^{\prime \prime}$ (since if $(a, b) \in E(G)$ then $(a, b) \in E\left(G_{b / a}\right)$ only), and also $\bigcup_{m \geq 1} E\left(G_{m}\right)=E(G)$ (but that is irrelevant). Since the good pairs are precisely the edges of $G_{k}$ and $G_{\ell}$ together, we need to maximize their number $g$.

A digraph $G_{m}$ is clearly a union of some $n_{m}$ disjoint (directed) paths $P_{m, i}$, with lengths $\lambda\left(P_{m, i}\right)=\lambda_{m, i}, 0 \leq \lambda_{m, i} \leq n^{2}-1$, such that $\sum_{i=1}^{n_{m}}\left(\lambda_{m, i}+1\right)=n^{2}$, and containing $e_{m}=\sum_{i=1}^{n_{m}} \lambda_{m, i}$ edges (zero-length paths, i.e. isolated vertices, are possible, allowed, and duly considered). The defect of the graph $G_{m}$ is taken to be $v-e_{m}=n_{m}$. We therefore need to maximize $g=e_{k}+e_{\ell}$, or equivalently, to minimize the defect $\delta=n_{k}+n_{\ell}$.

Using the model $V(G)=V_{x}=\left\{k^{i-1} \ell^{j-1} x \mid 1 \leq i, j \leq n\right\}$, we have $n_{k}=n_{\ell}=n$, therefore $\delta=2 n$, so $g=2 n(n-1)$. To prove value $2 n$ is a minimum for $\delta$ is almost obvious. We have $\lambda_{k, i} \leq n_{\ell}-1$ for all $1 \leq i \leq n_{k}$ (by the condition on $k$ and $\ell$, we have $\left|P_{k, i} \cap P_{\ell, j}\right| \leq 1$ for all $1 \leq i \leq n_{k}$ and $\left.1 \leq j \leq n_{\ell}\right),{ }^{2}$ so $n^{2}-n_{k}=e_{k}=\sum_{i=1}^{n_{k}} \lambda_{k, i} \leq \sum_{i=1}^{n_{k}}\left(n_{\ell}-1\right)=n_{k} n_{\ell}-n_{k}$, therefore $n^{2} \leq n_{k} n_{\ell}$, and so $\delta=n_{k}+n_{\ell} \geq 2 \sqrt{n_{k} n_{\ell}}=2 n$. Moreover, we see equality occurs if and only if $n_{k}=n_{\ell}=n$ and $\lambda_{k, i}=\lambda_{\ell, i}=n-1$ for all $1 \leq i \leq n$, thus only for the sets $V_{x}$ described above. Răspunsul este deci $g=180$.

Comentarii. Odată ce ideea vine, problema este aproape trivială, cu detaliile tehnice fiind aproape "forţate". Valorile particulare folosite aruncă doar un văl de umbră asupra situaţiei de fapt (mai ales ocultul $100=10^{2}$ )! Laticea de divizibilitate a celor $n^{2}$ numere este considerată în mod natural, şi conduce la facila numărătoare de mai sus.[^1]

## Second Day - Solutions

problem_match: Problem 3.


problem: Does it exist? a polynomial $P(x)$ with integer coefficients, such that $P(1+\sqrt{3})=2+\sqrt{3}$ and $P(3+\sqrt{5})=3+\sqrt{5}$.

solution:
The answer is No. The polynomial $P(x)-x$ has root $3+\sqrt{5}$, and since it has integer coefficients, it also has root $3-\sqrt{5}$. The quadratic having these two roots is $x^{2}-6 x+4$. Therefore $P(x)-x=\left(x^{2}-6 x+4\right) Q(x)$ for some polynomial $Q(x)$ with integer coefficients. Plugging in $x=1+\sqrt{3}$ we find $1=(2-4 \sqrt{3}) Q(1+\sqrt{3})$. From known conjugates properties we also do have $1=(2+4 \sqrt{3}) Q(1-\sqrt{3})$, so (multiply) $1=-44 Q(1+\sqrt{3}) Q(1-\sqrt{3})$, impossible, since again by the properties of conjugates

$$
Q(1+\sqrt{3}) Q(1-\sqrt{3})=Q(1+\sqrt{3}) Q(\overline{1+\sqrt{3}})=Q(1+\sqrt{3}) \overline{Q(1+\sqrt{3})}
$$

must be some integer $k$, and we cannot have $1=-44 k$.

Comentarii. Simple consideraţii legate de rădăcinile conjugate ale unor iraţionale pătratice pentru un polinom cu coeficienţi întregi. Multiple soluţii, bazate toate pe aceste proprietăţi, sunt posibile, iar problema este de o tehnicalitate elementară şi artificială.

problem_match: Problem 4.


problem:
Let $U=\{1,2,3, \ldots, 2014\}$. For all $a, b, c \in \mathbb{N}$ let $f(a, b, c)$ be the number of ordered sextuplets $\left(X_{1}, X_{2}, X_{3}, Y_{1}, Y_{2}, Y_{3}\right)$ of subsets of $U$, satisfying the following conditions

(i) $Y_{1} \subseteq X_{1} \subseteq U$ and $\left|X_{1}\right|=a$;

(ii) $Y_{2} \subseteq X_{2} \subseteq U \backslash Y_{1}$ and $\left|X_{2}\right|=b$;

(iii) $Y_{3} \subseteq X_{3} \subseteq U \backslash\left(Y_{1} \cup Y_{2}\right)$ and $\left|X_{3}\right|=c$.

Prove $f(\sigma(a), \sigma(b), \sigma(c))$ does not change, for permutations $\sigma$ of $a, b, c$.

solution:
In order to avoid any confusion between the letters $a, b, c$ and their numerical values (as cardinalities of sets), the most convenient way will be to denote by $|\ell|$ the cardinality symbolized by any such letter $\ell$. We can now consider the true 3 -element set $\{a, b, c\}$, and the canonical bijection $\phi:\{1,2,3\} \rightarrow\{a, b, c\}$ given by $\phi(1)=a, \phi(2)=b, \phi(3)=c$.

Let us now consider any permutation $\sigma$ of $\{a, b, c\}$. We will denote by $\mathcal{F}_{\sigma}$ the family of ordered sextuplets $\left(X_{1}, X_{2}, X_{3}, Y_{1}, Y_{2}, Y_{3}\right)$ of subsets of $U$ satisfying the conditions of the statement but, under our notations, having $\left|X_{1}\right|=|\sigma(a)|,\left|X_{2}\right|=|\sigma(b)|,\left|X_{3}\right|=|\sigma(c)|$. We will also denote by $\mathcal{F}$ the family of doubletons $\left\{\left(X_{a}, X_{b}, X_{c}\right), Y\right\}$, with $X_{a}, X_{b}, X_{c}$ subsets of $U$ having $\left|X_{a}\right|=|a|,\left|X_{b}\right|=|b|,\left|X_{c}\right|=|c|$, and $\emptyset \subseteq Y \subseteq X=X_{a} \cup X_{b} \cup X_{c}$. The set $Y$ uniquely partitions into 7 classes (some of them maybe empty), indexed by the non-empty subsets $S$ of $\{a, b, c\}$ via $Y_{S}=Y \cap\left(\bigcap_{\ell \in S} X_{\ell}\right) \cap\left(\bigcap_{\ell \notin S}\left(U \backslash X_{\ell}\right)\right)$.

We will now establish a bijection between $\mathcal{F}$ and $\mathcal{F}_{\sigma}$. This will show that $f(\sigma(a), \sigma(b), \sigma(c))=|\mathcal{F}|$ is constant over all permutations $\sigma$. We send an element $\left\{\left(X_{a}, X_{b}, X_{c}\right), Y\right\} \in \mathcal{F}$ into the sextuplet $\left(X_{1}, X_{2}, X_{3}, Y_{1}, Y_{2}, Y_{3}\right)$ given by, and easily verified it actually belongs to $\mathcal{F}_{\sigma}$,

- $X_{1}=X_{\sigma(a)}, X_{2}=X_{\sigma(b)}, X_{3}=X_{\sigma(c)}$ (i.e. $X_{i}=X_{\sigma(\phi(i))}$ for $\left.i \in\{1,2,3\}\right)$,
- $Y_{1}=Y_{\{\sigma(a)\}}$,
- $Y_{2}=Y_{\{\sigma(b)\}} \cup Y_{\{\sigma(a), \sigma(b)\}}$,
- $Y_{3}=Y_{\{\sigma(c)\}} \cup Y_{\{\sigma(a), \sigma(c)\}} \cup Y_{\{\sigma(b), \sigma(c)\}} \cup Y_{\{\sigma(a), \sigma(b), \sigma(c)\}}$.

We also send an element $\left(X_{1}, X_{2}, X_{3}, Y_{1}, Y_{2}, Y_{3}\right) \in \mathcal{F}_{\sigma}$ into the doubleton $\left\{\left(X_{a}, X_{b}, X_{c}\right), Y\right\}$ given by, and easily verified it actually belongs to $\mathcal{F}$, with $\tau$ the permutation of $\{1,2,3\}$ induced by $\sigma$ via $\tau(i)=\phi^{-1}(\sigma(\phi(i)))$ for all $i \in\{1,2,3\}$,

- $X_{a}=X_{\tau(1)}, X_{b}=X_{\tau(2)}, X_{c}=X_{\tau(3)}$ (i.e. $X_{\phi(i)}=X_{\tau(i)}$ for $\left.i \in\{1,2,3\}\right)$, - $Y=Y_{1} \cup Y_{2} \cup Y_{3}$.

It is immediate to see this mapping is a bijection, due to the unicity of the partitioning described in the above. Visualizing the Venn diagrams should tremendously help in understanding our considerations. The only difficulty resides in providing a luminous write-up of the argumentation, the underlying phenomenon being in fact almost trivial. The key element of this solution is to consider the unique partitioning of the set $Y=Y_{1} \cup Y_{2} \cup Y_{3}$ induced by the three sets of cardinalities $a, b, c$.

Comentarii. Pouah ... ce urâţenie de enunţ! Iar cerinţa este aproape evidentă, doar că soluţia este cam lung de scris, formalizat, şi explicat. Evident, valoarea 2014 nu joacă niciun rol.

problem_match: Problem 5.


problem:
Let $A B C D$ be a convex quadrilateral, partitioned by four lines as in the picture below, such that the meeting points of these lines all lie on the diagonals of $A B C D$. Prove that if the corner quadrilaterals 1, 2,3 and the center quadrilateral 4 are all tangential, then the corner quadrilateral 5 is also tangential.

![](https://cdn.mathpix.com/cropped/2024_06_04_b42e6a69c29e8caed2bcg-06.jpg?height=411&width=508&top_left_y=1825&top_left_x=817)

solution: It seems the corner quadrilateral 5 is tangential if and only if quadrilateral $A B C D$ is itself tangential; but no progress is thus yet made. Not being a geometer, I will wait for a decent solution to present itself (and it did! - see the presentation on the last page).

problem_match: Problem 6.