Solution: We will use Lemma 1 from   $$ f_{X}=f_{Y} \Longleftrightarrow f_{X}\left(\mathbb{Z}_{>0}\right)=f_{Y}\left(\mathbb{Z}_{>0}\right) \Longleftrightarrow\left(\mathbb{Z}_{>0} \backslash X\right)=\left(\mathbb{Z}_{>0} \backslash Y\right) \Longleftrightarrow X=Y, $$  where the first equivalence is because $f_{X}$ and $f_{Y}$ are strictly increasing functions, and the second equivalence is because $f_{X}\left(\mathbb{Z}_{>0}\right)=\mathbb{Z}_{>0} \backslash X$ and $f_{Y}\left(\mathbb{Z}_{>0}\right)=\mathbb{Z}_{>0} \backslash Y$.  Denote $g=f_{A}$ and $h=f_{B}$. The given relation $A * B=B * A$ is equivalent to $f_{A * B}=f_{B * A}$ because of (3), and by Lemma 1 of the first solution, this is equivalent to $g \circ h=h \circ g$. Similarly, the required relation $A^{* b}=B^{* a}$ is equivalent to $g^{b}=h^{a}$. We will show that  $$ g^{b}(n)=h^{a}(n) $$  for all $n \in \mathbb{Z}_{>0}$, which suffices to solve the problem. To start, we claim that (4) holds for all sufficiently large $n$. Indeed, let $p$ and $q$ be the maximal elements of $A$ and $B$, respectively; we may assume that $p \geqslant q$. Then, for every $n \geqslant p$ we have $g(n)=n+a$ and $h(n)=n+b$, whence $g^{b}(n)=n+a b=h^{a}(n)$, as was claimed.  In view of this claim, if (4) is not identically true, then there exists a maximal $s$ with $g^{b}(s) \neq$ $h^{a}(s)$. Without loss of generality, we may assume that $g(s) \neq s$, for if we had $g(s)=h(s)=s$, then $s$ would satisfy (4). As $g$ is increasing, we then have $g(s)>s$, so (4) holds for $n=g(s)$. But then we have  $$ g\left(g^{b}(s)\right)=g^{b+1}(s)=g^{b}(n)=h^{a}(n)=h^{a}(g(s))=g\left(h^{a}(s)\right), $$  where the last equality holds in view of $g \circ h=h \circ g$. By the injectivity of $g$, the above equality yields $g^{b}(s)=h^{a}(s)$, which contradicts the choice of $s$. Thus, we have proved that (4) is identically true on $\mathbb{Z}_{>0}$, as desired.  Comment 2. We present another proof of Lemma 2 of the first solution. Let $x=|X|=|Y|$. Say that $u$ is the smallest number in $X$ and $v$ is the smallest number in $Y$; assume without loss of generality that $u \leqslant v$.  Let $T$ be any finite set of positive integers, and define $t=|T|$. Enumerate the elements of $X$ as $x_{1}<x_{2}<\cdots<x_{n}$. Define $S_{m}=f_{\left(T * X^{*(m-1)}\right)}(X)$, and enumerate its elements $s_{m, 1}<s_{m, 2}<\cdots<$ $s_{m, n}$. Note that the $S_{m}$ are pairwise disjoint; indeed, if we have $m<m^{\prime}$, then  $$ S_{m} \subset T * X^{* m} \subset T * X^{*\left(m^{\prime}-1\right)} \quad \text { and } \quad S_{m^{\prime}}=\left(T * X^{* m^{\prime}}\right) \backslash\left(T * X^{*\left(m^{\prime}-1\right)}\right) $$  We claim the following statement, which essentially says that the $S_{m}$ are eventually linear translates of each other:  Claim. For every $i$, there exists some $m_{i}$ and $c_{i}$ such that for all $m>m_{i}$, we have that $s_{m, i}=t+m n-c_{i}$. Furthermore, the $c_{i}$ do not depend on the choice of $T$.  First, we show that this claim implies Lemma 2. We may choose $T=X$ and $T=Y$. Then, there is some $m^{\prime}$ such that for all $m \geqslant m^{\prime}$, we have  $$ f_{X * m}(X)=f_{(Y * X *(m-1))}(X) $$  Because $u$ is the minimum element of $X, v$ is the minimum element of $Y$, and $u \leqslant v$, we have that  $$ \left(\bigcup_{m=m^{\prime}}^{\infty} f_{X * m}(X)\right) \cup X^{* m^{\prime}}=\left(\bigcup_{m=m^{\prime}}^{\infty} f_{\left(Y * X^{*(m-1)}\right)}(X)\right) \cup\left(Y * X^{*\left(m^{\prime}-1\right)}\right)=\{u, u+1, \ldots\}, $$  and in both the first and second expressions, the unions are of pairwise distinct sets. By (5), we obtain $X^{* m^{\prime}}=Y * X^{*\left(m^{\prime}-1\right)}$. Now, because $X$ and $Y$ commute, we get $X^{* m^{\prime}}=X^{*\left(m^{\prime}-1\right)} * Y$, and so $X=Y$.  We now prove the claim. Proof of the claim. We induct downwards on $i$, first proving the statement for $i=n$, and so on. Assume that $m$ is chosen so that all elements of $S_{m}$ are greater than all elements of $T$ (which is possible because $T$ is finite). For $i=n$, we have that $s_{m, n}>s_{k, n}$ for every $k<m$. Thus, all ( $m-1$ ) $n$ numbers of the form $s_{k, u}$ for $k<m$ and $1 \leqslant u \leqslant n$ are less than $s_{m, n}$. We then have that $s_{m, n}$ is the $\left((m-1) n+x_{n}\right)^{t h}$ number not in $T$, which is equal to $t+(m-1) n+x_{n}$. So we may choose $c_{n}=x_{n}-n$, which does not depend on $T$, which proves the base case for the induction.  For $i<n$, we have again that all elements $s_{m, j}$ for $j<i$ and $s_{p, i}$ for $p<m$ are less than $s_{m, i}$, so $s_{m, i}$ is the $\left((m-1) i+x_{i}\right)^{t h}$ element not in $T$ or of the form $s_{p, j}$ for $j>i$ and $p<m$. But by the inductive hypothesis, each of the sequences $s_{p, j}$ is eventually periodic with period $n$, and thus the sequence $s_{m, i}$ such must be as well. Since each of the sequences $s_{p, j}-t$ with $j>i$ eventually do not depend on $T$, the sequence $s_{m, i}-t$ eventually does not depend on $T$ either, so the inductive step is complete. This proves the claim and thus Lemma 2.


Let $n$ be a given positive integer. In the Cartesian plane, each lattice point with nonnegative coordinates initially contains a butterfly, and there are no other butterflies. The neighborhood of a lattice point $c$ consists of all lattice points within the axis-aligned $(2 n+1) \times(2 n+1)$ square centered at $c$, apart from $c$ itself. We call a butterfly lonely, crowded, or comfortable, depending on whether the number of butterflies in its neighborhood $N$ is respectively less than, greater than, or equal to half of the number of lattice points in $N$.  Every minute, all lonely butterflies fly away simultaneously. This process goes on for as long as there are any lonely butterflies. Assuming that the process eventually stops, determine the number of comfortable butterflies at the final state.

Solution: We always identify a butterfly with the lattice point it is situated at. For two points $p$ and $q$, we write $p \geqslant q$ if each coordinate of $p$ is at least the corresponding coordinate of $q$. Let $O$ be the origin, and let $\mathcal{Q}$ be the set of initially occupied points, i.e., of all lattice points with nonnegative coordinates. Let $\mathcal{R}_{\mathrm{H}}=\{(x, 0): x \geqslant 0\}$ and $\mathcal{R}_{\mathrm{V}}=\{(0, y): y \geqslant 0\}$ be the sets of the lattice points lying on the horizontal and vertical boundary rays of $\mathcal{Q}$. Denote by $N(a)$ the neighborhood of a lattice point $a$.  1. Initial observations. We call a set of lattice points up-right closed if its points stay in the set after being shifted by any lattice vector $(i, j)$ with $i, j \geqslant 0$. Whenever the butterflies form a up-right closed set $\mathcal{S}$, we have $|N(p) \cap \mathcal{S}| \geqslant|N(q) \cap \mathcal{S}|$ for any two points $p, q \in \mathcal{S}$ with $p \geqslant q$. So, since $\mathcal{Q}$ is up-right closed, the set of butterflies at any moment also preserves this property. We assume all forthcoming sets of lattice points to be up-right closed.  When speaking of some set $\mathcal{S}$ of lattice points, we call its points lonely, comfortable, or crowded with respect to this set (i.e., as if the butterflies were exactly at all points of $\mathcal{S}$ ). We call a set $\mathcal{S} \subset \mathcal{Q}$ stable if it contains no lonely points. In what follows, we are interested only in those stable sets whose complements in $\mathcal{Q}$ are finite, because one can easily see that only a finite number of butterflies can fly away on each minute.  If the initial set $\mathcal{Q}$ of butterflies contains some stable set $\mathcal{S}$, then, clearly no butterfly of this set will fly away. On the other hand, the set $\mathcal{F}$ of all butterflies in the end of the process is stable. This means that $\mathcal{F}$ is the largest (with respect to inclusion) stable set within $\mathcal{Q}$, and we are about to describe this set. 2. A description of a final set. The following notion will be useful. Let $\mathcal{U}=\left\{\vec{u}_{1}, \vec{u}_{2}, \ldots, \vec{u}_{d}\right\}$ be a set of $d$ pairwise non-parallel lattice vectors, each having a positive $x$ - and a negative $y$-coordinate. Assume that they are numbered in increasing order according to slope. We now define a $\mathcal{U}$-curve to be the broken line $p_{0} p_{1} \ldots p_{d}$ such that $p_{0} \in \mathcal{R}_{\mathrm{V}}, p_{d} \in \mathcal{R}_{\mathrm{H}}$, and $\overrightarrow{p_{i-1} p_{i}}=\vec{u}_{i}$ for all $i=1,2, \ldots, m$ (see the Figure below to the left). ![](https://cdn.mathpix.com/cropped/2024_11_18_ff08e3c077fa2f8c16b2g-53.jpg?height=546&width=766&top_left_y=2194&top_left_x=191)  Construction of $\mathcal{U}$-curve ![](https://cdn.mathpix.com/cropped/2024_11_18_ff08e3c077fa2f8c16b2g-53.jpg?height=415&width=803&top_left_y=2322&top_left_x=1066)  Construction of $\mathcal{D}$  Now, let $\mathcal{K}_{n}=\{(i, j): 1 \leqslant i \leqslant n,-n \leqslant j \leqslant-1\}$. Consider all the rays emerging at $O$ and passing through a point from $\mathcal{K}_{n}$; number them as $r_{1}, \ldots, r_{m}$ in increasing order according to slope. Let $A_{i}$ be the farthest from $O$ lattice point in $r_{i} \cap \mathcal{K}_{n}$, set $k_{i}=\left|r_{i} \cap \mathcal{K}_{n}\right|$, let $\vec{v}_{i}=\overrightarrow{O A_{i}}$, and finally denote $\mathcal{V}=\left\{\vec{v}_{i}: 1 \leqslant i \leqslant m\right\}$; see the Figure above to the right. We will concentrate on the $\mathcal{V}$-curve $d_{0} d_{1} \ldots d_{m}$; let $\mathcal{D}$ be the set of all lattice points $p$ such that $p \geqslant p^{\prime}$ for some (not necessarily lattice) point $p^{\prime}$ on the $\mathcal{V}$-curve. In fact, we will show that $\mathcal{D}=\mathcal{F}$.  Clearly, the $\mathcal{V}$-curve is symmetric in the line $y=x$. Denote by $D$ the convex hull of $\mathcal{D}$. 3. We prove that the set $\mathcal{D}$ contains all stable sets. Let $\mathcal{S} \subset \mathcal{Q}$ be a stable set (recall that it is assumed to be up-right closed and to have a finite complement in $\mathcal{Q}$ ). Denote by $S$ its convex hull; clearly, the vertices of $S$ are lattice points. The boundary of $S$ consists of two rays (horizontal and vertical ones) along with some $\mathcal{V}_{*}$-curve for some set of lattice vectors $\mathcal{V}_{*}$. Claim 1. For every $\vec{v}_{i} \in \mathcal{V}$, there is a $\vec{v}_{i}^{*} \in \mathcal{V}_{*}$ co-directed with $\vec{v}$ with $\left|\vec{v}_{i}^{*}\right| \geqslant|\vec{v}|$. Proof. Let $\ell$ be the supporting line of $S$ parallel to $\vec{v}_{i}$ (i.e., $\ell$ contains some point of $S$, and the set $S$ lies on one side of $\ell$ ). Take any point $b \in \ell \cap \mathcal{S}$ and consider $N(b)$. The line $\ell$ splits the set $N(b) \backslash \ell$ into two congruent parts, one having an empty intersection with $\mathcal{S}$. Hence, in order for $b$ not to be lonely, at least half of the set $\ell \cap N(b)$ (which contains $2 k_{i}$ points) should lie in $S$. Thus, the boundary of $S$ contains a segment $\ell \cap S$ with at least $k_{i}+1$ lattice points (including $b$ ) on it; this segment corresponds to the required vector $\vec{v}_{i}^{*} \in \mathcal{V}_{*}$. ![](https://cdn.mathpix.com/cropped/2024_11_18_ff08e3c077fa2f8c16b2g-54.jpg?height=600&width=1626&top_left_y=1199&top_left_x=221)  Claim 2. Each stable set $\mathcal{S} \subseteq \mathcal{Q}$ lies in $\mathcal{D}$. Proof. To show this, it suffices to prove that the $\mathcal{V}_{*}$-curve lies in $D$, i.e., that all its vertices do so. Let $p^{\prime}$ be an arbitrary vertex of the $\mathcal{V}_{*^{-}}$-curve; $p^{\prime}$ partitions this curve into two parts, $\mathcal{X}$ (being down-right of $p$ ) and $\mathcal{Y}$ (being up-left of $p$ ). The set $\mathcal{V}$ is split now into two parts: $\mathcal{V}_{\mathcal{X}}$ consisting of those $\vec{v}_{i} \in \mathcal{V}$ for which $\vec{v}_{i}^{*}$ corresponds to a segment in $\mathcal{X}$, and a similar part $\mathcal{V}_{\mathcal{Y}}$. Notice that the $\mathcal{V}$-curve consists of several segments corresponding to $\mathcal{V}_{\mathcal{X}}$, followed by those corresponding to $\mathcal{V}_{\mathcal{Y}}$. Hence there is a vertex $p$ of the $\mathcal{V}$-curve separating $\mathcal{V}_{\mathcal{X}}$ from $\mathcal{V}_{\mathcal{Y}}$. Claim 1 now yields that $p^{\prime} \geqslant p$, so $p^{\prime} \in \mathcal{D}$, as required.  Claim 2 implies that the final set $\mathcal{F}$ is contained in $\mathcal{D}$. 4. $\mathcal{D}$ is stable, and its comfortable points are known. Recall the definitions of $r_{i}$; let $r_{i}^{\prime}$ be the ray complementary to $r_{i}$. By our definitions, the set $N(O)$ contains no points between the rays $r_{i}$ and $r_{i+1}$, as well as between $r_{i}^{\prime}$ and $r_{i+1}^{\prime}$. Claim 3. In the set $\mathcal{D}$, all lattice points of the $\mathcal{V}$-curve are comfortable. Proof. Let $p$ be any lattice point of the $\mathcal{V}$-curve, belonging to some segment $d_{i} d_{i+1}$. Draw the line $\ell$ containing this segment. Then $\ell \cap \mathcal{D}$ contains exactly $k_{i}+1$ lattice points, all of which lie in $N(p)$ except for $p$. Thus, exactly half of the points in $N(p) \cap \ell$ lie in $\mathcal{D}$. It remains to show that all points of $N(p)$ above $\ell$ lie in $\mathcal{D}$ (recall that all the points below $\ell$ lack this property).  Notice that each vector in $\mathcal{V}$ has one coordinate greater than $n / 2$; thus the neighborhood of $p$ contains parts of at most two segments of the $\mathcal{V}$-curve succeeding $d_{i} d_{i+1}$, as well as at most two of those preceding it.  The angles formed by these consecutive segments are obtained from those formed by $r_{j}$ and $r_{j-1}^{\prime}$ (with $i-1 \leqslant j \leqslant i+2$ ) by shifts; see the Figure below. All the points in $N(p)$ above $\ell$ which could lie outside $\mathcal{D}$ lie in shifted angles between $r_{j}, r_{j+1}$ or $r_{j}^{\prime}, r_{j-1}^{\prime}$. But those angles, restricted to $N(p)$, have no lattice points due to the above remark. The claim is proved. ![](https://cdn.mathpix.com/cropped/2024_11_18_ff08e3c077fa2f8c16b2g-55.jpg?height=404&width=1052&top_left_y=592&top_left_x=503)  Proof of Claim 3 Claim 4. All the points of $\mathcal{D}$ which are not on the boundary of $D$ are crowded. Proof. Let $p \in \mathcal{D}$ be such a point. If it is to the up-right of some point $p^{\prime}$ on the curve, then the claim is easy: the shift of $N\left(p^{\prime}\right) \cap \mathcal{D}$ by $\overrightarrow{p^{\prime} p}$ is still in $\mathcal{D}$, and $N(p)$ contains at least one more point of $\mathcal{D}$ - either below or to the left of $p$. So, we may assume that $p$ lies in a right triangle constructed on some hypothenuse $d_{i} d_{i+1}$. Notice here that $d_{i}, d_{i+1} \in N(p)$.  Draw a line $\ell \| d_{i} d_{i+1}$ through $p$, and draw a vertical line $h$ through $d_{i}$; see Figure below. Let $\mathcal{D}_{\mathrm{L}}$ and $\mathcal{D}_{\mathrm{R}}$ be the parts of $\mathcal{D}$ lying to the left and to the right of $h$, respectively (points of $\mathcal{D} \cap h$ lie in both parts). ![](https://cdn.mathpix.com/cropped/2024_11_18_ff08e3c077fa2f8c16b2g-55.jpg?height=538&width=1120&top_left_y=1556&top_left_x=468)  Notice that the vectors $\overrightarrow{d_{i} p}, \overrightarrow{d_{i+1} d_{i+2}}, \overrightarrow{d_{i} d_{i+1}}, \overrightarrow{d_{i-1} d_{i}}$, and $\overrightarrow{p d_{i+1}}$ are arranged in non-increasing order by slope. This means that $\mathcal{D}_{\mathrm{L}}$ shifted by $\overrightarrow{d_{i} p}$ still lies in $\mathcal{D}$, as well as $\mathcal{D}_{\mathrm{R}}$ shifted by $\overrightarrow{d_{i+1} p}$. As we have seen in the proof of Claim 3, these two shifts cover all points of $N(p)$ above $\ell$, along with those on $\ell$ to the left of $p$. Since $N(p)$ contains also $d_{i}$ and $d_{i+1}$, the point $p$ is crowded.  Thus, we have proved that $\mathcal{D}=\mathcal{F}$, and have shown that the lattice points on the $\mathcal{V}$-curve are exactly the comfortable points of $\mathcal{D}$. It remains to find their number.  Recall the definition of $\mathcal{K}_{n}$ (see Figure on the first page of the solution). Each segment $d_{i} d_{i+1}$ contains $k_{i}$ lattice points different from $d_{i}$. Taken over all $i$, these points exhaust all the lattice points in the $\mathcal{V}$-curve, except for $d_{1}$, and thus the number of lattice points on the $\mathcal{V}$-curve is $1+\sum_{i=1}^{m} k_{i}$. On the other hand, $\sum_{i=1}^{m} k_{i}$ is just the number of points in $\mathcal{K}_{n}$, so it equals $n^{2}$. Hence the answer to the problem is $n^{2}+1$.  Comment 1. The assumption that the process eventually stops is unnecessary for the problem, as one can see that, in fact, the process stops for every $n \geqslant 1$. Indeed, the proof of Claims 3 and 4 do not rely essentially on this assumption, and they together yield that the set $\mathcal{D}$ is stable. So, only butterflies that are not in $\mathcal{D}$ may fly away, and this takes only a finite time.  This assumption has been inserted into the problem statement in order to avoid several technical details regarding finiteness issues. It may also simplify several other arguments.  Comment 2. The description of the final set $\mathcal{F}(=\mathcal{D})$ seems to be crucial for the solution; the Problem Selection Committee is not aware of any solution that completely avoids such a description.  On the other hand, after the set $\mathcal{D}$ has been defined, the further steps may be performed in several ways. For example, in order to prove that all butterflies outside $\mathcal{D}$ will fly away, one may argue as follows. (Here we will also make use of the assumption that the process eventually stops.)  First of all, notice that the process can be modified in the following manner: Each minute, exactly one of the lonely butterflies flies away, until there are no more lonely butterflies. The modified process necessarily stops at the same state as the initial one. Indeed, one may observe, as in solution above, that the (unique) largest stable set is still the final set for the modified process.  Thus, in order to prove our claim, it suffices to indicate an order in which the butterflies should fly away in the new process; if we are able to exhaust the whole set $\mathcal{Q} \backslash \mathcal{D}$, we are done.  Let $\mathcal{C}_{0}=d_{0} d_{1} \ldots d_{m}$ be the $\mathcal{V}$-curve. Take its copy $\mathcal{C}$ and shift it downwards so that $d_{0}$ comes to some point below the origin $O$. Now we start moving $\mathcal{C}$ upwards continuously, until it comes back to its initial position $\mathcal{C}_{0}$. At each moment when $\mathcal{C}$ meets some lattice points, we convince all the butterflies at those points to fly away in a certain order. We will now show that we always have enough arguments for butterflies to do so, which will finish our argument for the claim..  Let $\mathcal{C}^{\prime}=d_{0}^{\prime} d_{1}^{\prime} \ldots d_{m}^{\prime}$ be a position of $\mathcal{C}$ when it meets some butterflies. We assume that all butterflies under this current position of $\mathcal{C}$ were already convinced enough and flied away. Consider the lowest butterfly $b$ on $\mathcal{C}^{\prime}$. Let $d_{i}^{\prime} d_{i+1}^{\prime}$ be the segment it lies on; we choose $i$ so that $b \neq d_{i+1}^{\prime}$ (this is possible because $\mathcal{C}$ as not yet reached $\mathcal{C}_{0}$ ).  Draw a line $\ell$ containing the segment $d_{i}^{\prime} d_{i+1}^{\prime}$. Then all the butterflies in $N(b)$ are situated on or above $\ell$; moreover, those on $\ell$ all lie on the segment $d_{i} d_{i+1}$. But this segment now contains at most $k_{i}$ butterflies (including $b$ ), since otherwise some butterfly had to occupy $d_{i+1}^{\prime}$ which is impossible by the choice of $b$. Thus, $b$ is lonely and hence may be convinced to fly away.  After $b$ has flied away, we switch to the lowest of the remaining butterflies on $\mathcal{C}^{\prime}$, and so on. Claims 3 and 4 also allow some different proofs which are not presented here.  This page is intentionally left blank


Let $A B C D E$ be a convex pentagon such that $A B=B C=C D, \angle E A B=\angle B C D$, and $\angle E D C=\angle C B A$. Prove that the perpendicular line from $E$ to $B C$ and the line segments $A C$ and $B D$ are concurrent. (Italy)

Solution: Throughout the solution, we refer to $\angle A, \angle B, \angle C, \angle D$, and $\angle E$ as internal angles of the pentagon $A B C D E$. Let the perpendicular bisectors of $A C$ and $B D$, which pass respectively through $B$ and $C$, meet at point $I$. Then $B D \perp C I$ and, similarly, $A C \perp B I$. Hence $A C$ and $B D$ meet at the orthocenter $H$ of the triangle $B I C$, and $I H \perp B C$. It remains to prove that $E$ lies on the line $I H$ or, equivalently, $E I \perp B C$.  Lines $I B$ and $I C$ bisect $\angle B$ and $\angle C$, respectively. Since $I A=I C, I B=I D$, and $A B=$ $B C=C D$, the triangles $I A B, I C B$ and $I C D$ are congruent. Hence $\angle I A B=\angle I C B=$ $\angle C / 2=\angle A / 2$, so the line $I A$ bisects $\angle A$. Similarly, the line $I D$ bisects $\angle D$. Finally, the line $I E$ bisects $\angle E$ because $I$ lies on all the other four internal bisectors of the angles of the pentagon.  The sum of the internal angles in a pentagon is $540^{\circ}$, so  $$ \angle E=540^{\circ}-2 \angle A+2 \angle B . $$  In quadrilateral $A B I E$,  $$ \begin{aligned} \angle B I E & =360^{\circ}-\angle E A B-\angle A B I-\angle A E I=360^{\circ}-\angle A-\frac{1}{2} \angle B-\frac{1}{2} \angle E \\ & =360^{\circ}-\angle A-\frac{1}{2} \angle B-\left(270^{\circ}-\angle A-\angle B\right) \\ & =90^{\circ}+\frac{1}{2} \angle B=90^{\circ}+\angle I B C, \end{aligned} $$  which means that $E I \perp B C$, completing the proof. ![](https://cdn.mathpix.com/cropped/2024_11_18_ff08e3c077fa2f8c16b2g-58.jpg?height=618&width=730&top_left_y=1801&top_left_x=663)


Solution: We present another proof of the fact that $E$ lies on line $I H$. Since all five internal bisectors of $A B C D E$ meet at $I$, this pentagon has an inscribed circle with center $I$. Let this circle touch side $B C$ at $T$.  Applying Brianchon's theorem to the (degenerate) hexagon $A B T C D E$ we conclude that $A C, B D$ and $E T$ are concurrent, so point $E$ also lies on line $I H T$, completing the proof.


Solution: We present yet another proof that $E I \perp B C$. In pentagon $A B C D E, \angle E<$ $180^{\circ} \Longleftrightarrow \angle A+\angle B+\angle C+\angle D>360^{\circ}$. Then $\angle A+\angle B=\angle C+\angle D>180^{\circ}$, so rays $E A$ and $C B$ meet at a point $P$, and rays $B C$ and $E D$ meet at a point $Q$. Now,  $$ \angle P B A=180^{\circ}-\angle B=180^{\circ}-\angle D=\angle Q D C $$  and, similarly, $\angle P A B=\angle Q C D$. Since $A B=C D$, the triangles $P A B$ and $Q C D$ are congruent with the same orientation. Moreover, $P Q E$ is isosceles with $E P=E Q$. ![](https://cdn.mathpix.com/cropped/2024_11_18_ff08e3c077fa2f8c16b2g-59.jpg?height=524&width=1546&top_left_y=612&top_left_x=255)  In Solution 1 we have proved that triangles $I A B$ and $I C D$ are also congruent with the same orientation. Then we conclude that quadrilaterals $P B I A$ and $Q D I C$ are congruent, which implies $I P=I Q$. Then $E I$ is the perpendicular bisector of $P Q$ and, therefore, $E I \perp$ $P Q \Longleftrightarrow E I \perp B C$.  Comment. Even though all three solutions used the point $I$, there are solutions that do not need it. We present an outline of such a solution: if $J$ is the incenter of $\triangle Q C D$ (with $P$ and $Q$ as defined in Solution 3), then a simple angle chasing shows that triangles $C J D$ and $B H C$ are congruent. Then if $S$ is the projection of $J$ onto side $C D$ and $T$ is the orthogonal projection of $H$ onto side $B C$, one can verify that  $$ Q T=Q C+C T=Q C+D S=Q C+\frac{C D+D Q-Q C}{2}=\frac{P B+B C+Q C}{2}=\frac{P Q}{2}, $$  so $T$ is the midpoint of $P Q$, and $E, H$ and $T$ all lie on the perpendicular bisector of $P Q$.


Let $R$ and $S$ be distinct points on circle $\Omega$, and let $t$ denote the tangent line to $\Omega$ at $R$. Point $R^{\prime}$ is the reflection of $R$ with respect to $S$. A point $I$ is chosen on the smaller arc $R S$ of $\Omega$ so that the circumcircle $\Gamma$ of triangle $I S R^{\prime}$ intersects $t$ at two different points. Denote by $A$ the common point of $\Gamma$ and $t$ that is closest to $R$. Line $A I$ meets $\Omega$ again at $J$. Show that $J R^{\prime}$ is tangent to $\Gamma$. (Luxembourg)

Solution: In the circles $\Omega$ and $\Gamma$ we have $\angle J R S=\angle J I S=\angle A R^{\prime} S$. On the other hand, since $R A$ is tangent to $\Omega$, we get $\angle S J R=\angle S R A$. So the triangles $A R R^{\prime}$ and $S J R$ are similar, and  $$ \frac{R^{\prime} R}{R J}=\frac{A R^{\prime}}{S R}=\frac{A R^{\prime}}{S R^{\prime}} $$  The last relation, together with $\angle A R^{\prime} S=\angle J R R^{\prime}$, yields $\triangle A S R^{\prime} \sim \triangle R^{\prime} J R$, hence $\angle S A R^{\prime}=\angle R R^{\prime} J$. It follows that $J R^{\prime}$ is tangent to $\Gamma$ at $R^{\prime}$. ![](https://cdn.mathpix.com/cropped/2024_11_18_ff08e3c077fa2f8c16b2g-60.jpg?height=880&width=757&top_left_y=1082&top_left_x=244)  Solution 1 ![](https://cdn.mathpix.com/cropped/2024_11_18_ff08e3c077fa2f8c16b2g-60.jpg?height=1029&width=758&top_left_y=933&top_left_x=1066)  Solution 2


Solution: As in  Let $A^{\prime}$ be the reflection of $A$ about $S$; then $A R A^{\prime} R^{\prime}$ is a parallelogram with center $S$, and hence the point $J$ lies on the line $R A^{\prime}$.  From $\angle S R^{\prime} A^{\prime}=\angle S R A=\angle S J R$ we get that the points $S, J, A^{\prime}, R^{\prime}$ are concyclic. This proves that $\angle S R^{\prime} J=\angle S A^{\prime} J=\angle S A^{\prime} R=\angle S A R^{\prime}$, so $J R^{\prime}$ is tangent to $\Gamma$ at $R^{\prime}$.


Let $O$ be the circumcenter of an acute scalene triangle $A B C$. Line $O A$ intersects the altitudes of $A B C$ through $B$ and $C$ at $P$ and $Q$, respectively. The altitudes meet at $H$. Prove that the circumcenter of triangle $P Q H$ lies on a median of triangle $A B C$. (Ukraine)

Solution: Suppose, without loss of generality, that $A B<A C$. We have $\angle P Q H=90^{\circ}-$ $\angle Q A B=90^{\circ}-\angle O A B=\frac{1}{2} \angle A O B=\angle A C B$, and similarly $\angle Q P H=\angle A B C$. Thus triangles $A B C$ and $H P Q$ are similar. Let $\Omega$ and $\omega$ be the circumcircles of $A B C$ and $H P Q$, respectively. Since $\angle A H P=90^{\circ}-\angle H A C=\angle A C B=\angle H Q P$, line $A H$ is tangent to $\omega$. ![](https://cdn.mathpix.com/cropped/2024_11_18_ff08e3c077fa2f8c16b2g-61.jpg?height=894&width=1355&top_left_y=678&top_left_x=356)  Let $T$ be the center of $\omega$ and let lines $A T$ and $B C$ meet at $M$. We will take advantage of the similarity between $A B C$ and $H P Q$ and the fact that $A H$ is tangent to $\omega$ at $H$, with $A$ on line $P Q$. Consider the corresponding tangent $A S$ to $\Omega$, with $S \in B C$. Then $S$ and $A$ correspond to each other in $\triangle A B C \sim \triangle H P Q$, and therefore $\angle O S M=\angle O A T=\angle O A M$. Hence quadrilateral $S A O M$ is cyclic, and since the tangent line $A S$ is perpendicular to $A O$, $\angle O M S=180^{\circ}-\angle O A S=90^{\circ}$. This means that $M$ is the orthogonal projection of $O$ onto $B C$, which is its midpoint. So $T$ lies on median $A M$ of triangle $A B C$.


In triangle $A B C$, let $\omega$ be the excircle opposite $A$. Let $D, E$, and $F$ be the points where $\omega$ is tangent to lines $B C, C A$, and $A B$, respectively. The circle $A E F$ intersects line $B C$ at $P$ and $Q$. Let $M$ be the midpoint of $A D$. Prove that the circle $M P Q$ is tangent to $\omega$. (Denmark)

Solution: Denote by $\Omega$ the circle $A E F P Q$, and denote by $\gamma$ the circle $P Q M$. Let the line $A D$ meet $\omega$ again at $T \neq D$. We will show that $\gamma$ is tangent to $\omega$ at $T$.  We first prove that points $P, Q, M, T$ are concyclic. Let $A^{\prime}$ be the center of $\omega$. Since $A^{\prime} E \perp A E$ and $A^{\prime} F \perp A F, A A^{\prime}$ is a diameter in $\Omega$. Let $N$ be the midpoint of $D T$; from $A^{\prime} D=A^{\prime} T$ we can see that $\angle A^{\prime} N A=90^{\circ}$ and therefore $N$ also lies on the circle $\Omega$. Now, from the power of $D$ with respect to the circles $\gamma$ and $\Omega$ we get  $$ D P \cdot D Q=D A \cdot D N=2 D M \cdot \frac{D T}{2}=D M \cdot D T $$  so $P, Q, M, T$ are concyclic. If $E F \| B C$, then $A B C$ is isosceles and the problem is now immediate by symmetry. Otherwise, let the tangent line to $\omega$ at $T$ meet line $B C$ at point $R$. The tangent line segments $R D$ and $R T$ have the same length, so $A^{\prime} R$ is the perpendicular bisector of $D T$; since $N D=N T$, $N$ lies on this perpendicular bisector.  In right triangle $A^{\prime} R D, R D^{2}=R N \cdot R A^{\prime}=R P \cdot R Q$, in which the last equality was obtained from the power of $R$ with respect to $\Omega$. Hence $R T^{2}=R P \cdot R Q$, which implies that $R T$ is also tangent to $\gamma$. Because $R T$ is a common tangent to $\omega$ and $\gamma$, these two circles are tangent at $T$. ![](https://cdn.mathpix.com/cropped/2024_11_18_ff08e3c077fa2f8c16b2g-62.jpg?height=841&width=883&top_left_y=1330&top_left_x=592)


Solution: After proving that $P, Q, M, T$ are concyclic, we finish the problem in a different fashion. We only consider the case in which $E F$ and $B C$ are not parallel. Let lines $P Q$ and $E F$ meet at point $R$. Since $P Q$ and $E F$ are radical axes of $\Omega, \gamma$ and $\omega, \gamma$, respectively, $R$ is the radical center of these three circles.  With respect to the circle $\omega$, the line $D R$ is the polar of $D$, and the line $E F$ is the polar of $A$. So the pole of line $A D T$ is $D R \cap E F=R$, and therefore $R T$ is tangent to $\omega$.  Finally, since $T$ belongs to $\gamma$ and $\omega$ and $R$ is the radical center of $\gamma, \omega$ and $\Omega$, line $R T$ is the radical axis of $\gamma$ and $\omega$, and since it is tangent to $\omega$, it is also tangent to $\gamma$. Because $R T$ is a common tangent to $\omega$ and $\gamma$, these two circles are tangent at $T$.  Comment. In Solution 2 we defined the point $R$ from Solution 1 in a different way.


Solution: We give an alternative proof that the circles are tangent at the common point $T$. Again, we start from the fact that $P, Q, M, T$ are concyclic. Let point $O$ be the midpoint of diameter $A A^{\prime}$. Then $M O$ is the midline of triangle $A D A^{\prime}$, so $M O \| A^{\prime} D$. Since $A^{\prime} D \perp P Q$, $M O$ is perpendicular to $P Q$ as well.  Looking at circle $\Omega$, which has center $O, M O \perp P Q$ implies that $M O$ is the perpendicular bisector of the chord $P Q$. Thus $M$ is the midpoint of arc $\widehat{P Q}$ from $\gamma$, and the tangent line $m$ to $\gamma$ at $M$ is parallel to $P Q$. ![](https://cdn.mathpix.com/cropped/2024_11_18_ff08e3c077fa2f8c16b2g-63.jpg?height=804&width=644&top_left_y=569&top_left_x=706)  Consider the homothety with center $T$ and ratio $\frac{T D}{T M}$. It takes $D$ to $M$, and the line $P Q$ to the line $m$. Since the circle that is tangent to a line at a given point and that goes through another given point is unique, this homothety also takes $\omega$ (tangent to $P Q$ and going through $T$ ) to $\gamma$ (tangent to $m$ and going through $T$ ). We conclude that $\omega$ and $\gamma$ are tangent at $T$.


Let $A B C C_{1} B_{1} A_{1}$ be a convex hexagon such that $A B=B C$, and suppose that the line segments $A A_{1}, B B_{1}$, and $C C_{1}$ have the same perpendicular bisector. Let the diagonals $A C_{1}$ and $A_{1} C$ meet at $D$, and denote by $\omega$ the circle $A B C$. Let $\omega$ intersect the circle $A_{1} B C_{1}$ again at $E \neq B$. Prove that the lines $B B_{1}$ and $D E$ intersect on $\omega$. (Ukraine)

Solution: If $A A_{1}=C C_{1}$, then the hexagon is symmetric about the line $B B_{1}$; in particular the circles $A B C$ and $A_{1} B C_{1}$ are tangent to each other. So $A A_{1}$ and $C C_{1}$ must be different. Since the points $A$ and $A_{1}$ can be interchanged with $C$ and $C_{1}$, respectively, we may assume $A A_{1}<C C_{1}$.  Let $R$ be the radical center of the circles $A E B C$ and $A_{1} E B C_{1}$, and the circumcircle of the symmetric trapezoid $A C C_{1} A_{1}$; that is the common point of the pairwise radical axes $A C, A_{1} C_{1}$, and $B E$. By the symmetry of $A C$ and $A_{1} C_{1}$, the point $R$ lies on the common perpendicular bisector of $A A_{1}$ and $C C_{1}$, which is the external bisector of $\angle A D C$.  Let $F$ be the second intersection of the line $D R$ and the circle $A C D$. From the power of $R$ with respect to the circles $\omega$ and $A C F D$ we have $R B \cdot R E=R A \cdot R C=R D \cdot D F$, so the points $B, E, D$ and $F$ are concyclic.  The line $R D F$ is the external bisector of $\angle A D C$, so the point $F$ bisects the arc $\widehat{C D A}$. By $A B=B C$, on circle $\omega$, the point $B$ is the midpoint of arc $\widehat{A E C}$; let $M$ be the point diametrically opposite to $B$, that is the midpoint of the opposite $\operatorname{arc} \overline{C A}$ of $\omega$. Notice that the points $B, F$ and $M$ lie on the perpendicular bisector of $A C$, so they are collinear. ![](https://cdn.mathpix.com/cropped/2024_11_18_ff08e3c077fa2f8c16b2g-64.jpg?height=1183&width=1355&top_left_y=1273&top_left_x=356)  Finally, let $X$ be the second intersection point of $\omega$ and the line $D E$. Since $B M$ is a diameter in $\omega$, we have $\angle B X M=90^{\circ}$. Moreover,  $$ \angle E X M=180^{\circ}-\angle M B E=180^{\circ}-\angle F B E=\angle E D F, $$  so $M X$ and $F D$ are parallel. Since $B X$ is perpendicular to $M X$ and $B B_{1}$ is perpendicular to $F D$, this shows that $X$ lies on line $B B_{1}$.


Solution: Define point $M$ as the point opposite to $B$ on circle $\omega$, and point $R$ as the intersection of lines $A C, A_{1} C_{1}$ and $B E$, and show that $R$ lies on the external bisector of $\angle A D C$, like in the first solution.  Since $B$ is the midpoint of the arc $\widehat{A E C}$, the line $B E R$ is the external bisector of $\angle C E A$. Now we show that the internal angle bisectors of $\angle A D C$ and $\angle C E A$ meet on the segment $A C$. Let the angle bisector of $\angle A D C$ meet $A C$ at $S$, and let the angle bisector of $\angle C E A$, which is line $E M$, meet $A C$ at $S^{\prime}$. By applying the angle bisector theorem to both internal and external bisectors of $\angle A D C$ and $\angle C E A$,  $$ A S: C S=A D: C D=A R: C R=A E: C E=A S^{\prime}: C S^{\prime} $$  so indeed $S=S^{\prime}$. By $\angle R D S=\angle S E R=90^{\circ}$ the points $R, S, D$ and $E$ are concyclic. ![](https://cdn.mathpix.com/cropped/2024_11_18_ff08e3c077fa2f8c16b2g-65.jpg?height=1035&width=1084&top_left_y=839&top_left_x=492)  Now let the lines $B B_{1}$ and $D E$ meet at point $X$. Notice that $\angle E X B=\angle E D S$ because both $B B_{1}$ and $D S$ are perpendicular to the line $D R$, we have that $\angle E D S=\angle E R S$ in circle $S R D E$, and $\angle E R S=\angle E M B$ because $S R \perp B M$ and $E R \perp M E$. Therefore, $\angle E X B=\angle E M B$, so indeed, the point $X$ lies on $\omega$.


Let $n \geqslant 3$ be an integer. Two regular $n$-gons $\mathcal{A}$ and $\mathcal{B}$ are given in the plane. Prove that the vertices of $\mathcal{A}$ that lie inside $\mathcal{B}$ or on its boundary are consecutive. (That is, prove that there exists a line separating those vertices of $\mathcal{A}$ that lie inside $\mathcal{B}$ or on its boundary from the other vertices of $\mathcal{A}$.) (Czech Republic)

Solution: In both solutions, by a polygon we always mean its interior together with its boundary.  We start with finding a regular $n$-gon $\mathcal{C}$ which $(i)$ is inscribed into $\mathcal{B}$ (that is, all vertices of $\mathcal{C}$ lie on the perimeter of $\mathcal{B}$ ); and (ii) is either a translation of $\mathcal{A}$, or a homothetic image of $\mathcal{A}$ with a positive factor.  Such a polygon may be constructed as follows. Let $O_{A}$ and $O_{B}$ be the centers of $\mathcal{A}$ and $\mathcal{B}$, respectively, and let $A$ be an arbitrary vertex of $\mathcal{A}$. Let $\overrightarrow{O_{B} C}$ be the vector co-directional to $\overrightarrow{O_{A} A}$, with $C$ lying on the perimeter of $\mathcal{B}$. The rotations of $C$ around $O_{B}$ by multiples of $2 \pi / n$ form the required polygon. Indeed, it is regular, inscribed into $\mathcal{B}$ (due to the rotational symmetry of $\mathcal{B}$ ), and finally the translation/homothety mapping $\overrightarrow{O_{A} A}$ to $\overrightarrow{O_{B} C}$ maps $\mathcal{A}$ to $\mathcal{C}$.  Now we separate two cases. ![](https://cdn.mathpix.com/cropped/2024_11_18_ff08e3c077fa2f8c16b2g-66.jpg?height=507&width=549&top_left_y=1157&top_left_x=365)  Construction of $\mathcal{C}$ ![](https://cdn.mathpix.com/cropped/2024_11_18_ff08e3c077fa2f8c16b2g-66.jpg?height=589&width=700&top_left_y=1076&top_left_x=1009)  Case 1: Translation  Case 1: $\mathcal{C}$ is a translation of $\mathcal{A}$ by a vector $\vec{v}$. Denote by $t$ the translation transform by vector $\vec{v}$. We need to prove that the vertices of $\mathcal{C}$ which stay in $\mathcal{B}$ under $t$ are consecutive. To visualize the argument, we refer the plane to Cartesian coordinates so that the $x$-axis is co-directional with $\vec{v}$. This way, the notions of right/left and top/bottom are also introduced, according to the $x$ - and $y$-coordinates, respectively.  Let $B_{\mathrm{T}}$ and $B_{\mathrm{B}}$ be the top and the bottom vertices of $\mathcal{B}$ (if several vertices are extremal, we take the rightmost of them). They split the perimeter of $\mathcal{B}$ into the right part $\mathcal{B}_{\mathrm{R}}$ and the left part $\mathcal{B}_{\mathrm{L}}$ (the vertices $B_{\mathrm{T}}$ and $B_{\mathrm{B}}$ are assumed to lie in both parts); each part forms a connected subset of the perimeter of $\mathcal{B}$. So the vertices of $\mathcal{C}$ are also split into two parts $\mathcal{C}_{\mathrm{L}} \subset \mathcal{B}_{\mathrm{L}}$ and $\mathcal{C}_{\mathrm{R}} \subset \mathcal{B}_{\mathrm{R}}$, each of which consists of consecutive vertices.  Now, all the points in $\mathcal{B}_{\mathrm{R}}$ (and hence in $\mathcal{C}_{\mathrm{R}}$ ) move out from $\mathcal{B}$ under $t$, since they are the rightmost points of $\mathcal{B}$ on the corresponding horizontal lines. It remains to prove that the vertices of $\mathcal{C}_{\mathrm{L}}$ which stay in $\mathcal{B}$ under $t$ are consecutive.  For this purpose, let $C_{1}, C_{2}$, and $C_{3}$ be three vertices in $\mathcal{C}_{\mathrm{L}}$ such that $C_{2}$ is between $C_{1}$ and $C_{3}$, and $t\left(C_{1}\right)$ and $t\left(C_{3}\right)$ lie in $\mathcal{B}$; we need to prove that $t\left(C_{2}\right) \in \mathcal{B}$ as well. Let $A_{i}=t\left(C_{i}\right)$. The line through $C_{2}$ parallel to $\vec{v}$ crosses the segment $C_{1} C_{3}$ to the right of $C_{2}$; this means that this line crosses $A_{1} A_{3}$ to the right of $A_{2}$, so $A_{2}$ lies inside the triangle $A_{1} C_{2} A_{3}$ which is contained in $\mathcal{B}$. This yields the desired result.  Case 2: $\mathcal{C}$ is a homothetic image of $\mathcal{A}$ centered at $X$ with factor $k>0$.  Denote by $h$ the homothety mapping $\mathcal{C}$ to $\mathcal{A}$. We need now to prove that the vertices of $\mathcal{C}$ which stay in $\mathcal{B}$ after applying $h$ are consecutive. If $X \in \mathcal{B}$, the claim is easy. Indeed, if $k<1$, then the vertices of $\mathcal{A}$ lie on the segments of the form $X C$ ( $C$ being a vertex of $\mathcal{C}$ ) which lie in $\mathcal{B}$. If $k>1$, then the vertices of $\mathcal{A}$ lie on the extensions of such segments $X C$ beyond $C$, and almost all these extensions lie outside $\mathcal{B}$. The exceptions may occur only in case when $X$ lies on the boundary of $\mathcal{B}$, and they may cause one or two vertices of $\mathcal{A}$ stay on the boundary of $\mathcal{B}$. But even in this case those vertices are still consecutive.  So, from now on we assume that $X \notin \mathcal{B}$. Now, there are two vertices $B_{\mathrm{T}}$ and $\mathcal{B}_{\mathrm{B}}$ of $\mathcal{B}$ such that $\mathcal{B}$ is contained in the angle $\angle B_{\mathrm{T}} X B_{\mathrm{B}}$; if there are several options, say, for $B_{\mathrm{T}}$, then we choose the farthest one from $X$ if $k>1$, and the nearest one if $k<1$. For the visualization purposes, we refer the plane to Cartesian coordinates so that the $y$-axis is co-directional with $\overrightarrow{B_{\mathrm{B}} B_{\mathrm{T}}}$, and $X$ lies to the left of the line $B_{\mathrm{T}} B_{\mathrm{B}}$. Again, the perimeter of $\mathcal{B}$ is split by $B_{\mathrm{T}}$ and $B_{\mathrm{B}}$ into the right part $\mathcal{B}_{\mathrm{R}}$ and the left part $\mathcal{B}_{\mathrm{L}}$, and the set of vertices of $\mathcal{C}$ is split into two subsets $\mathcal{C}_{\mathrm{R}} \subset \mathcal{B}_{\mathrm{R}}$ and $\mathcal{C}_{\mathrm{L}} \subset \mathcal{B}_{\mathrm{L}}$. ![](https://cdn.mathpix.com/cropped/2024_11_18_ff08e3c077fa2f8c16b2g-67.jpg?height=572&width=546&top_left_y=959&top_left_x=281)  Case 2, $X$ inside $\mathcal{B}$ ![](https://cdn.mathpix.com/cropped/2024_11_18_ff08e3c077fa2f8c16b2g-67.jpg?height=623&width=906&top_left_y=908&top_left_x=866)  Subcase 2.1: $k>1$  Subcase 2.1: $k>1$. In this subcase, all points from $\mathcal{B}_{\mathrm{R}}$ (and hence from $\mathcal{C}_{\mathrm{R}}$ ) move out from $\mathcal{B}$ under $h$, because they are the farthest points of $\mathcal{B}$ on the corresponding rays emanated from $X$. It remains to prove that the vertices of $\mathcal{C}_{\mathrm{L}}$ which stay in $\mathcal{B}$ under $h$ are consecutive.  Again, let $C_{1}, C_{2}, C_{3}$ be three vertices in $\mathcal{C}_{\mathrm{L}}$ such that $C_{2}$ is between $C_{1}$ and $C_{3}$, and $h\left(C_{1}\right)$ and $h\left(C_{3}\right)$ lie in $\mathcal{B}$. Let $A_{i}=h\left(C_{i}\right)$. Then the ray $X C_{2}$ crosses the segment $C_{1} C_{3}$ beyond $C_{2}$, so this ray crosses $A_{1} A_{3}$ beyond $A_{2}$; this implies that $A_{2}$ lies in the triangle $A_{1} C_{2} A_{3}$, which is contained in $\mathcal{B}$. ![](https://cdn.mathpix.com/cropped/2024_11_18_ff08e3c077fa2f8c16b2g-67.jpg?height=515&width=909&top_left_y=2021&top_left_x=576)  Subcase 2.2: $k<1$ Subcase 2.2: $k<1$. This case is completely similar to the previous one. All points from $\mathcal{B}_{\mathrm{L}}$ (and hence from $\mathcal{C}_{\mathrm{L}}$ move out from $\mathcal{B}$ under $h$, because they are the nearest points of $\mathcal{B}$ on the corresponding rays emanated from $X$. Assume that $C_{1}, C_{2}$, and $C_{3}$ are three vertices in $\mathcal{C}_{\mathrm{R}}$ such that $C_{2}$ lies between $C_{1}$ and $C_{3}$, and $h\left(C_{1}\right)$ and $h\left(C_{3}\right)$ lie in $\mathcal{B}$; let $A_{i}=h\left(C_{i}\right)$. Then $A_{2}$ lies on the segment $X C_{2}$, and the segments $X A_{2}$ and $A_{1} A_{3}$ cross each other. Thus $A_{2}$ lies in the triangle $A_{1} C_{2} A_{3}$, which is contained in $\mathcal{B}$.  Comment 1. In fact, Case 1 can be reduced to Case 2 via the following argument. Assume that $\mathcal{A}$ and $\mathcal{C}$ are congruent. Apply to $\mathcal{A}$ a homothety centered at $O_{B}$ with a factor slightly smaller than 1 to obtain a polygon $\mathcal{A}^{\prime}$. With appropriately chosen factor, the vertices of $\mathcal{A}$ which were outside $/$ inside $\mathcal{B}$ stay outside/inside it, so it suffices to prove our claim for $\mathcal{A}^{\prime}$ instead of $\mathcal{A}$. And now, the polygon $\mathcal{A}^{\prime}$ is a homothetic image of $\mathcal{C}$, so the arguments from Case 2 apply.  Comment 2. After the polygon $\mathcal{C}$ has been found, the rest of the solution uses only the convexity of the polygons, instead of regularity. Thus, it proves a more general statement:  Assume that $\mathcal{A}, \mathcal{B}$, and $\mathcal{C}$ are three convex polygons in the plane such that $\mathcal{C}$ is inscribed into $\mathcal{B}$, and $\mathcal{A}$ can be obtained from it via either translation or positive homothety. Then the vertices of $\mathcal{A}$ that lie inside $\mathcal{B}$ or on its boundary are consecutive.


Solution: Let $O_{A}$ and $O_{B}$ be the centers of $\mathcal{A}$ and $\mathcal{B}$, respectively. Denote $[n]=\{1,2, \ldots, n\}$. We start with introducing appropriate enumerations and notations. Enumerate the sidelines of $\mathcal{B}$ clockwise as $\ell_{1}, \ell_{2}, \ldots, \ell_{n}$. Denote by $\mathcal{H}_{i}$ the half-plane of $\ell_{i}$ that contains $\mathcal{B}$ ( $\mathcal{H}_{i}$ is assumed to contain $\left.\ell_{i}\right)$; by $B_{i}$ the midpoint of the side belonging to $\ell_{i}$; and finally denote $\overrightarrow{b_{i}}=\overrightarrow{B_{i} O_{B}}$. (As usual, the numbering is cyclic modulo $n$, so $\ell_{n+i}=\ell_{i}$ etc.)  Now, choose a vertex $A_{1}$ of $\mathcal{A}$ such that the vector $\overrightarrow{O_{A} A_{1}}$ points "mostly outside $\mathcal{H}_{1}$ "; strictly speaking, this means that the scalar product $\left\langle\overrightarrow{O_{A} A_{1}}, \overrightarrow{b_{1}}\right\rangle$ is minimal. Starting from $A_{1}$, enumerate the vertices of $\mathcal{A}$ clockwise as $A_{1}, A_{2}, \ldots, A_{n}$; by the rotational symmetry, the choice of $A_{1}$ yields that the vector $\overrightarrow{O_{A} A_{i}}$ points "mostly outside $\mathcal{H}_{i}$ ", i.e.,  $$ \left\langle\overrightarrow{O_{A} A_{i}}, \overrightarrow{b_{i}}\right\rangle=\min _{j \in[n]}\left\langle\overrightarrow{O_{A} A_{j}}, \overrightarrow{b_{i}}\right\rangle $$  ![](https://cdn.mathpix.com/cropped/2024_11_18_ff08e3c077fa2f8c16b2g-68.jpg?height=560&width=898&top_left_y=1516&top_left_x=579)  Enumerations and notations We intend to reformulate the problem in more combinatorial terms, for which purpose we introduce the following notion. Say that a subset $I \subseteq[n]$ is connected if the elements of this set are consecutive in the cyclic order (in other words, if we join each $i$ with $i+1 \bmod n$ by an edge, this subset is connected in the usual graph sense). Clearly, the union of two connected subsets sharing at least one element is connected too. Next, for any half-plane $\mathcal{H}$ the indices of vertices of, say, $\mathcal{A}$ that lie in $\mathcal{H}$ form a connected set.  To access the problem, we denote  $$ M=\left\{j \in[n]: A_{j} \notin \mathcal{B}\right\}, \quad M_{i}=\left\{j \in[n]: A_{j} \notin \mathcal{H}_{i}\right\} \quad \text { for } i \in[n] $$  We need to prove that $[n] \backslash M$ is connected, which is equivalent to $M$ being connected. On the other hand, since $\mathcal{B}=\bigcap_{i \in[n]} \mathcal{H}_{i}$, we have $M=\bigcup_{i \in[n]} M_{i}$, where the sets $M_{i}$ are easier to investigate. We will utilize the following properties of these sets; the first one holds by the definition of $M_{i}$, along with the above remark. ![](https://cdn.mathpix.com/cropped/2024_11_18_ff08e3c077fa2f8c16b2g-69.jpg?height=529&width=1183&top_left_y=175&top_left_x=439)  The sets $M_{i}$  Property 1: Each set $M_{i}$ is connected. Property 2: If $M_{i}$ is nonempty, then $i \in M_{i}$. Proof. Indeed, we have  $$ j \in M_{i} \Longleftrightarrow A_{j} \notin \mathcal{H}_{i} \Longleftrightarrow\left\langle\overrightarrow{B_{i} A_{j}}, \overrightarrow{b_{i}}\right\rangle<0 \Longleftrightarrow\left\langle\overrightarrow{O_{A} A_{j}}, \overrightarrow{b_{i}}\right\rangle<\left\langle\overrightarrow{O_{A} B_{i}}, \overrightarrow{b_{i}}\right\rangle $$  The right-hand part of the last inequality does not depend on $j$. Therefore, if some $j$ lies in $M_{i}$, then by (1) so does $i$.  In view of Property 2, it is useful to define the set  $$ M^{\prime}=\left\{i \in[n]: i \in M_{i}\right\}=\left\{i \in[n]: M_{i} \neq \varnothing\right\} . $$  Property 3: The set $M^{\prime}$ is connected. Proof. To prove this property, we proceed on with the investigation started in (2) to write  $$ i \in M^{\prime} \Longleftrightarrow A_{i} \in M_{i} \Longleftrightarrow\left\langle\overrightarrow{B_{i}} \overrightarrow{A_{i}}, \overrightarrow{b_{i}}\right\rangle<0 \Longleftrightarrow\left\langle\overrightarrow{O_{B} O_{A}}, \overrightarrow{b_{i}}\right\rangle<\left\langle\overrightarrow{O_{B} B_{i}}, \overrightarrow{b_{i}}\right\rangle+\left\langle\overrightarrow{A_{i} O_{A}}, \overrightarrow{b_{i}}\right\rangle $$  The right-hand part of the obtained inequality does not depend on $i$, due to the rotational symmetry; denote its constant value by $\mu$. Thus, $i \in M^{\prime}$ if and only if $\left\langle\overrightarrow{O_{B} O_{A}}, \overrightarrow{b_{i}}\right\rangle<\mu$. This condition is in turn equivalent to the fact that $B_{i}$ lies in a certain (open) half-plane whose boundary line is orthogonal to $O_{B} O_{A}$; thus, it defines a connected set.  Now we can finish the solution. Since $M^{\prime} \subseteq M$, we have  $$ M=\bigcup_{i \in[n]} M_{i}=M^{\prime} \cup \bigcup_{i \in[n]} M_{i}, $$  so $M$ can be obtained from $M^{\prime}$ by adding all the sets $M_{i}$ one by one. All these sets are connected, and each nonempty $M_{i}$ contains an element of $M^{\prime}$ (namely, $i$ ). Thus their union is also connected.  Comment 3. Here we present a way in which one can come up with a solution like the one above. Assume, for sake of simplicity, that $O_{A}$ lies inside $\mathcal{B}$. Let us first put onto the plane a very small regular $n$-gon $\mathcal{A}^{\prime}$ centered at $O_{A}$ and aligned with $\mathcal{A}$; all its vertices lie inside $\mathcal{B}$. Now we start blowing it up, looking at the order in which the vertices leave $\mathcal{B}$. To go out of $\mathcal{B}$, a vertex should cross a certain side of $\mathcal{B}$ (which is hard to describe), or, equivalently, to cross at least one sideline of $\mathcal{B}$ - and this event is easier to describe. Indeed, the first vertex of $\mathcal{A}^{\prime}$ to cross $\ell_{i}$ is the vertex $A_{i}^{\prime}$ (corresponding to $A_{i}$ in $\mathcal{A}$ ); more generally, the vertices $A_{j}^{\prime}$ cross $\ell_{i}$ in such an order that the scalar product $\left\langle\overrightarrow{O_{A}} \overrightarrow{A_{j}}, \overrightarrow{b_{i}}\right\rangle$ does not increase. For different indices $i$, these orders are just cyclic shifts of each other; and this provides some intuition for the notions and claims from Solution 2.


A convex quadrilateral $A B C D$ has an inscribed circle with center $I$. Let $I_{a}, I_{b}, I_{c}$, and $I_{d}$ be the incenters of the triangles $D A B, A B C, B C D$, and $C D A$, respectively. Suppose that the common external tangents of the circles $A I_{b} I_{d}$ and $C I_{b} I_{d}$ meet at $X$, and the common external tangents of the circles $B I_{a} I_{c}$ and $D I_{a} I_{c}$ meet at $Y$. Prove that $\angle X I Y=90^{\circ}$. (Kazakhstan)

Solution: Denote by $\omega_{a}, \omega_{b}, \omega_{c}$ and $\omega_{d}$ the circles $A I_{b} I_{d}, B I_{a} I_{c}, C I_{b} I_{d}$, and $D I_{a} I_{c}$, let their centers be $O_{a}, O_{b}, O_{c}$ and $O_{d}$, and let their radii be $r_{a}, r_{b}, r_{c}$ and $r_{d}$, respectively. Claim 1. $I_{b} I_{d} \perp A C$ and $I_{a} I_{c} \perp B D$. Proof. Let the incircles of triangles $A B C$ and $A C D$ be tangent to the line $A C$ at $T$ and $T^{\prime}$, respectively. (See the figure to the left.) We have $A T=\frac{A B+A C-B C}{2}$ in triangle $A B C, A T^{\prime}=$ $\frac{A D+A C-C D}{2}$ in triangle $A C D$, and $A B-B C=A D-C D$ in quadrilateral $A B C D$, so  $$ A T=\frac{A C+A B-B C}{2}=\frac{A C+A D-C D}{2}=A T^{\prime} $$  This shows $T=T^{\prime}$. As an immediate consequence, $I_{b} I_{d} \perp A C$. The second statement can be shown analogously. ![](https://cdn.mathpix.com/cropped/2024_11_18_ff08e3c077fa2f8c16b2g-70.jpg?height=726&width=1729&top_left_y=1076&top_left_x=168)  Claim 2. The points $O_{a}, O_{b}, O_{c}$ and $O_{d}$ lie on the lines $A I, B I, C I$ and $D I$, respectively. Proof. By symmetry it suffices to prove the claim for $O_{a}$. (See the figure to the right above.)  Notice first that the incircles of triangles $A B C$ and $A C D$ can be obtained from the incircle of the quadrilateral $A B C D$ with homothety centers $B$ and $D$, respectively, and homothety factors less than 1 , therefore the points $I_{b}$ and $I_{d}$ lie on the line segments $B I$ and $D I$, respectively.  As is well-known, in every triangle the altitude and the diameter of the circumcircle starting from the same vertex are symmetric about the angle bisector. By Claim 1, in triangle $A I_{d} I_{b}$, the segment $A T$ is the altitude starting from $A$. Since the foot $T$ lies inside the segment $I_{b} I_{d}$, the circumcenter $O_{a}$ of triangle $A I_{d} I_{b}$ lies in the angle domain $I_{b} A I_{d}$ in such a way that $\angle I_{b} A T=\angle O_{a} A I_{d}$. The points $I_{b}$ and $I_{d}$ are the incenters of triangles $A B C$ and $A C D$, so the lines $A I_{b}$ and $A I_{d}$ bisect the angles $\angle B A C$ and $\angle C A D$, respectively. Then  $$ \angle O_{a} A D=\angle O_{a} A I_{d}+\angle I_{d} A D=\angle I_{b} A T+\angle I_{d} A D=\frac{1}{2} \angle B A C+\frac{1}{2} \angle C A D=\frac{1}{2} \angle B A D, $$  so $O_{a}$ lies on the angle bisector of $\angle B A D$, that is, on line $A I$. The point $X$ is the external similitude center of $\omega_{a}$ and $\omega_{c}$; let $U$ be their internal similitude center. The points $O_{a}$ and $O_{c}$ lie on the perpendicular bisector of the common chord $I_{b} I_{d}$ of $\omega_{a}$ and $\omega_{c}$, and the two similitude centers $X$ and $U$ lie on the same line; by Claim 2, that line is parallel to $A C$. ![](https://cdn.mathpix.com/cropped/2024_11_18_ff08e3c077fa2f8c16b2g-71.jpg?height=1149&width=1529&top_left_y=179&top_left_x=269)  From the similarity of the circles $\omega_{a}$ and $\omega_{c}$, from $O_{a} I_{b}=O_{a} I_{d}=O_{a} A=r_{a}$ and $O_{c} I_{b}=$ $O_{c} I_{d}=O_{c} C=r_{c}$, and from $A C \| O_{a} O_{c}$ we can see that  $$ \frac{O_{a} X}{O_{c} X}=\frac{O_{a} U}{O_{c} U}=\frac{r_{a}}{r_{c}}=\frac{O_{a} I_{b}}{O_{c} I_{b}}=\frac{O_{a} I_{d}}{O_{c} I_{d}}=\frac{O_{a} A}{O_{c} C}=\frac{O_{a} I}{O_{c} I} $$  So the points $X, U, I_{b}, I_{d}, I$ lie on the Apollonius circle of the points $O_{a}, O_{c}$ with ratio $r_{a}: r_{c}$. In this Apollonius circle $X U$ is a diameter, and the lines $I U$ and $I X$ are respectively the internal and external bisectors of $\angle O_{a} I O_{c}=\angle A I C$, according to the angle bisector theorem. Moreover, in the Apollonius circle the diameter $U X$ is the perpendicular bisector of $I_{b} I_{d}$, so the lines $I X$ and $I U$ are the internal and external bisectors of $\angle I_{b} I I_{d}=\angle B I D$, respectively.  Repeating the same argument for the points $B, D$ instead of $A, C$, we get that the line $I Y$ is the internal bisector of $\angle A I C$ and the external bisector of $\angle B I D$. Therefore, the lines $I X$ and $I Y$ respectively are the internal and external bisectors of $\angle B I D$, so they are perpendicular.  Comment. In fact the points $O_{a}, O_{b}, O_{c}$ and $O_{d}$ lie on the line segments $A I, B I, C I$ and $D I$, respectively. For the point $O_{a}$ this can be shown for example by $\angle I_{d} O_{a} A+\angle A O_{a} I_{b}=\left(180^{\circ}-\right.$ $\left.2 \angle O_{a} A I_{d}\right)+\left(180^{\circ}-2 \angle I_{b} A O_{a}\right)=360^{\circ}-\angle B A D=\angle A D I+\angle D I A+\angle A I B+\angle I B A>\angle I_{d} I A+\angle A I I_{b}$.  The solution also shows that the line $I Y$ passes through the point $U$, and analogously, $I X$ passes through the internal similitude center of $\omega_{b}$ and $\omega_{d}$.


There are 2017 mutually external circles drawn on a blackboard, such that no two are tangent and no three share a common tangent. A tangent segment is a line segment that is a common tangent to two circles, starting at one tangent point and ending at the other one. Luciano is drawing tangent segments on the blackboard, one at a time, so that no tangent segment intersects any other circles or previously drawn tangent segments. Luciano keeps drawing tangent segments until no more can be drawn. Find all possible numbers of tangent segments when he stops drawing. (Australia)

Solution: First, consider a particular arrangement of circles $C_{1}, C_{2}, \ldots, C_{n}$ where all the centers are aligned and each $C_{i}$ is eclipsed from the other circles by its neighbors - for example, taking $C_{i}$ with center $\left(i^{2}, 0\right)$ and radius $i / 2$ works. Then the only tangent segments that can be drawn are between adjacent circles $C_{i}$ and $C_{i+1}$, and exactly three segments can be drawn for each pair. So Luciano will draw exactly $3(n-1)$ segments in this case. ![](https://cdn.mathpix.com/cropped/2024_11_18_ff08e3c077fa2f8c16b2g-72.jpg?height=258&width=1335&top_left_y=1059&top_left_x=355)  For the general case, start from a final configuration (that is, an arrangement of circles and segments in which no further segments can be drawn). The idea of the solution is to continuously resize and move the circles around the plane, one by one (in particular, making sure we never have 4 circles with a common tangent line), and show that the number of segments drawn remains constant as the picture changes. This way, we can reduce any circle/segment configuration to the particular one mentioned above, and the final number of segments must remain at $3 n-3$.  Some preliminary considerations: look at all possible tangent segments joining any two circles. A segment that is tangent to a circle $A$ can do so in two possible orientations - it may come out of $A$ in clockwise or counterclockwise orientation. Two segments touching the same circle with the same orientation will never intersect each other. Each pair $(A, B)$ of circles has 4 choices of tangent segments, which can be identified by their orientations - for example, $(A+, B-)$ would be the segment which comes out of $A$ in clockwise orientation and comes out of $B$ in counterclockwise orientation. In total, we have $2 n(n-1)$ possible segments, disregarding intersections.  Now we pick a circle $C$ and start to continuously move and resize it, maintaining all existing tangent segments according to their identifications, including those involving $C$. We can keep our choice of tangent segments until the configuration reaches a transition. We lose nothing if we assume that $C$ is kept at least $\varepsilon$ units away from any other circle, where $\varepsilon$ is a positive, fixed constant; therefore at a transition either: (1) a currently drawn tangent segment $t$ suddenly becomes obstructed; or (2) a currently absent tangent segment $t$ suddenly becomes unobstructed and available. Claim. A transition can only occur when three circles $C_{1}, C_{2}, C_{3}$ are tangent to a common line $\ell$ containing $t$, in a way such that the three tangent segments lying on $\ell$ (joining the three circles pairwise) are not obstructed by any other circles or tangent segments (other than $C_{1}, C_{2}, C_{3}$ ). Proof. Since (2) is effectively the reverse of (1), it suffices to prove the claim for (1). Suppose $t$ has suddenly become obstructed, and let us consider two cases.  Case 1: $t$ becomes obstructed by a circle ![](https://cdn.mathpix.com/cropped/2024_11_18_ff08e3c077fa2f8c16b2g-73.jpg?height=318&width=1437&top_left_y=292&top_left_x=315)  Then the new circle becomes the third circle tangent to $\ell$, and no other circles or tangent segments are obstructing $t$.


Solution: First note that all tangent segments lying on the boundary of the convex hull of the circles are always drawn since they do not intersect anything else. Now in the final picture, aside from the $n$ circles, the blackboard is divided into regions. We can consider the picture as a plane (multi-)graph $G$ in which the circles are the vertices and the tangent segments are the edges. The idea of this solution is to find a relation between the number of edges and the number of regions in $G$; then, once we prove that $G$ is connected, we can use Euler's formula to finish the problem.  The boundary of each region consists of 1 or more (for now) simple closed curves, each made of arcs and tangent segments. The segment and the arc might meet smoothly (as in $S_{i}$, $i=1,2, \ldots, 6$ in the figure below) or not (as in $P_{1}, P_{2}, P_{3}, P_{4}$; call such points sharp corners of the boundary). In other words, if a person walks along the border, her direction would suddenly turn an angle of $\pi$ at a sharp corner. ![](https://cdn.mathpix.com/cropped/2024_11_18_ff08e3c077fa2f8c16b2g-75.jpg?height=626&width=577&top_left_y=178&top_left_x=748)  Claim 1. The outer boundary $B_{1}$ of any internal region has at least 3 sharp corners. Proof. Let a person walk one lap along $B_{1}$ in the counterclockwise orientation. As she does so, she will turn clockwise as she moves along the circle arcs, and not turn at all when moving along the lines. On the other hand, her total rotation after one lap is $2 \pi$ in the counterclockwise direction! Where could she be turning counterclockwise? She can only do so at sharp corners, and, even then, she turns only an angle of $\pi$ there. But two sharp corners are not enough, since at least one arc must be present-so she must have gone through at least 3 sharp corners.  Claim 2. Each internal region is simply connected, that is, has only one boundary curve. Proof. Suppose, by contradiction, that some region has an outer boundary $B_{1}$ and inner boundaries $B_{2}, B_{3}, \ldots, B_{m}(m \geqslant 2)$. Let $P_{1}$ be one of the sharp corners of $B_{1}$.  Now consider a car starting at $P_{1}$ and traveling counterclockwise along $B_{1}$. It starts in reverse, i.e., it is initially facing the corner $P_{1}$. Due to the tangent conditions, the car may travel in a way so that its orientation only changes when it is moving along an arc. In particular, this means the car will sometimes travel forward. For example, if the car approaches a sharp corner when driving in reverse, it would continue travel forward after the corner, instead of making an immediate half-turn. This way, the orientation of the car only changes in a clockwise direction since the car always travels clockwise around each arc.  Now imagine there is a laser pointer at the front of the car, pointing directly ahead. Initially, the laser endpoint hits $P_{1}$, but, as soon as the car hits an arc, the endpoint moves clockwise around $B_{1}$. In fact, the laser endpoint must move continuously along $B_{1}$ ! Indeed, if the endpoint ever jumped (within $B_{1}$, or from $B_{1}$ to one of the inner boundaries), at the moment of the jump the interrupted laser would be a drawable tangent segment that Luciano missed (see figure below for an example). ![](https://cdn.mathpix.com/cropped/2024_11_18_ff08e3c077fa2f8c16b2g-75.jpg?height=637&width=587&top_left_y=2140&top_left_x=740)  Now, let $P_{2}$ and $P_{3}$ be the next two sharp corners the car goes through, after $P_{1}$ (the previous lemma assures their existence). At $P_{2}$ the car starts moving forward, and at $P_{3}$ it will start to move in reverse again. So, at $P_{3}$, the laser endpoint is at $P_{3}$ itself. So while the car moved counterclockwise between $P_{1}$ and $P_{3}$, the laser endpoint moved clockwise between $P_{1}$ and $P_{3}$. That means the laser beam itself scanned the whole region within $B_{1}$, and it should have crossed some of the inner boundaries.  Claim 3. Each region has exactly 3 sharp corners. Proof. Consider again the car of the previous claim, with its laser still firmly attached to its front, traveling the same way as before and going through the same consecutive sharp corners $P_{1}, P_{2}$ and $P_{3}$. As we have seen, as the car goes counterclockwise from $P_{1}$ to $P_{3}$, the laser endpoint goes clockwise from $P_{1}$ to $P_{3}$, so together they cover the whole boundary. If there were a fourth sharp corner $P_{4}$, at some moment the laser endpoint would pass through it. But, since $P_{4}$ is a sharp corner, this means the car must be on the extension of a tangent segment going through $P_{4}$. Since the car is not on that segment itself (the car never goes through $P_{4}$ ), we would have 3 circles with a common tangent line, which is not allowed. ![](https://cdn.mathpix.com/cropped/2024_11_18_ff08e3c077fa2f8c16b2g-76.jpg?height=538&width=792&top_left_y=1007&top_left_x=643)  We are now ready to finish the solution. Let $r$ be the number of internal regions, and $s$ be the number of tangent segments. Since each tangent segment contributes exactly 2 sharp corners to the diagram, and each region has exactly 3 sharp corners, we must have $2 s=3 r$. Since the graph corresponding to the diagram is connected, we can use Euler's formula $n-s+r=1$ and find $s=3 n-3$ and $r=2 n-2$.


The sequence $a_{0}, a_{1}, a_{2}, \ldots$ of positive integers satisfies  $$ a_{n+1}=\left\{\begin{array}{ll} \sqrt{a_{n}}, & \text { if } \sqrt{a_{n}} \text { is an integer } \\ a_{n}+3, & \text { otherwise } \end{array} \quad \text { for every } n \geqslant 0\right. $$  Determine all values of $a_{0}>1$ for which there is at least one number $a$ such that $a_{n}=a$ for infinitely many values of $n$. (South Africa)

Solution: Since the value of $a_{n+1}$ only depends on the value of $a_{n}$, if $a_{n}=a_{m}$ for two different indices $n$ and $m$, then the sequence is eventually periodic. So we look for the values of $a_{0}$ for which the sequence is eventually periodic. Claim 1. If $a_{n} \equiv-1(\bmod 3)$, then, for all $m>n, a_{m}$ is not a perfect square. It follows that the sequence is eventually strictly increasing, so it is not eventually periodic. Proof. A square cannot be congruent to -1 modulo 3 , so $a_{n} \equiv-1(\bmod 3)$ implies that $a_{n}$ is not a square, therefore $a_{n+1}=a_{n}+3>a_{n}$. As a consequence, $a_{n+1} \equiv a_{n} \equiv-1(\bmod 3)$, so $a_{n+1}$ is not a square either. By repeating the argument, we prove that, from $a_{n}$ on, all terms of the sequence are not perfect squares and are greater than their predecessors, which completes the proof. Claim 2. If $a_{n} \not \equiv-1(\bmod 3)$ and $a_{n}>9$ then there is an index $m>n$ such that $a_{m}<a_{n}$. Proof. Let $t^{2}$ be the largest perfect square which is less than $a_{n}$. Since $a_{n}>9, t$ is at least 3. The first square in the sequence $a_{n}, a_{n}+3, a_{n}+6, \ldots$ will be $(t+1)^{2},(t+2)^{2}$ or $(t+3)^{2}$, therefore there is an index $m>n$ such that $a_{m} \leqslant t+3<t^{2}<a_{n}$, as claimed. Claim 3. If $a_{n} \equiv 0(\bmod 3)$, then there is an index $m>n$ such that $a_{m}=3$. Proof. First we notice that, by the definition of the sequence, a multiple of 3 is always followed by another multiple of 3 . If $a_{n} \in\{3,6,9\}$ the sequence will eventually follow the periodic pattern $3,6,9,3,6,9, \ldots$ If $a_{n}>9$, let $j$ be an index such that $a_{j}$ is equal to the minimum value of the set $\left\{a_{n+1}, a_{n+2}, \ldots\right\}$. We must have $a_{j} \leqslant 9$, otherwise we could apply Claim 2 to $a_{j}$ and get a contradiction on the minimality hypothesis. It follows that $a_{j} \in\{3,6,9\}$, and the proof is complete. Claim 4. If $a_{n} \equiv 1(\bmod 3)$, then there is an index $m>n$ such that $a_{m} \equiv-1(\bmod 3)$. Proof. In the sequence, 4 is always followed by $2 \equiv-1(\bmod 3)$, so the claim is true for $a_{n}=4$. If $a_{n}=7$, the next terms will be $10,13,16,4,2, \ldots$ and the claim is also true. For $a_{n} \geqslant 10$, we again take an index $j>n$ such that $a_{j}$ is equal to the minimum value of the set $\left\{a_{n+1}, a_{n+2}, \ldots\right\}$, which by the definition of the sequence consists of non-multiples of 3 . Suppose $a_{j} \equiv 1(\bmod 3)$. Then we must have $a_{j} \leqslant 9$ by Claim 2 and the minimality of $a_{j}$. It follows that $a_{j} \in\{4,7\}$, so $a_{m}=2<a_{j}$ for some $m>j$, contradicting the minimality of $a_{j}$. Therefore, we must have $a_{j} \equiv-1(\bmod 3)$.  It follows from the previous claims that if $a_{0}$ is a multiple of 3 the sequence will eventually reach the periodic pattern $3,6,9,3,6,9, \ldots$; if $a_{0} \equiv-1(\bmod 3)$ the sequence will be strictly increasing; and if $a_{0} \equiv 1(\bmod 3)$ the sequence will be eventually strictly increasing.  So the sequence will be eventually periodic if, and only if, $a_{0}$ is a multiple of 3 .


Let $p \geqslant 2$ be a prime number. Eduardo and Fernando play the following game making moves alternately: in each move, the current player chooses an index $i$ in the set $\{0,1, \ldots, p-1\}$ that was not chosen before by either of the two players and then chooses an element $a_{i}$ of the set $\{0,1,2,3,4,5,6,7,8,9\}$. Eduardo has the first move. The game ends after all the indices $i \in\{0,1, \ldots, p-1\}$ have been chosen. Then the following number is computed:  $$ M=a_{0}+10 \cdot a_{1}+\cdots+10^{p-1} \cdot a_{p-1}=\sum_{j=0}^{p-1} a_{j} \cdot 10^{j} $$  The goal of Eduardo is to make the number $M$ divisible by $p$, and the goal of Fernando is to prevent this.  Prove that Eduardo has a winning strategy. (Morocco)

Solution: We say that a player makes the move $\left(i, a_{i}\right)$ if he chooses the index $i$ and then the element $a_{i}$ of the set $\{0,1,2,3,4,5,6,7,8,9\}$ in this move.  If $p=2$ or $p=5$ then Eduardo chooses $i=0$ and $a_{0}=0$ in the first move, and wins, since, independently of the next moves, $M$ will be a multiple of 10 .  Now assume that the prime number $p$ does not belong to $\{2,5\}$. Eduardo chooses $i=p-1$ and $a_{p-1}=0$ in the first move. By Fermat's Little Theorem, $\left(10^{(p-1) / 2}\right)^{2}=10^{p-1} \equiv 1(\bmod p)$, so $p \mid\left(10^{(p-1) / 2}\right)^{2}-1=\left(10^{(p-1) / 2}+1\right)\left(10^{(p-1) / 2}-1\right)$. Since $p$ is prime, either $p \mid 10^{(p-1) / 2}+1$ or $p \mid 10^{(p-1) / 2}-1$. Thus we have two cases: Case a: $10^{(p-1) / 2} \equiv-1(\bmod p)$ In this case, for each move $\left(i, a_{i}\right)$ of Fernando, Eduardo immediately makes the move $\left(j, a_{j}\right)=$ $\left(i+\frac{p-1}{2}, a_{i}\right)$, if $0 \leqslant i \leqslant \frac{p-3}{2}$, or $\left(j, a_{j}\right)=\left(i-\frac{p-1}{2}, a_{i}\right)$, if $\frac{p-1}{2} \leqslant i \leqslant p-2$. We will have $10^{j} \equiv-10^{i}$ $(\bmod p)$, and so $a_{j} \cdot 10^{j}=a_{i} \cdot 10^{j} \equiv-a_{i} \cdot 10^{i}(\bmod p)$. Notice that this move by Eduardo is always possible. Indeed, immediately before a move by Fernando, for any set of the type $\{r, r+(p-1) / 2\}$ with $0 \leqslant r \leqslant(p-3) / 2$, either no element of this set was chosen as an index by the players in the previous moves or else both elements of this set were chosen as indices by the players in the previous moves. Therefore, after each of his moves, Eduardo always makes the sum of the numbers $a_{k} \cdot 10^{k}$ corresponding to the already chosen pairs $\left(k, a_{k}\right)$ divisible by $p$, and thus wins the game. Case b: $10^{(p-1) / 2} \equiv 1(\bmod p)$ In this case, for each move $\left(i, a_{i}\right)$ of Fernando, Eduardo immediately makes the move $\left(j, a_{j}\right)=$ $\left(i+\frac{p-1}{2}, 9-a_{i}\right)$, if $0 \leqslant i \leqslant \frac{p-3}{2}$, or $\left(j, a_{j}\right)=\left(i-\frac{p-1}{2}, 9-a_{i}\right)$, if $\frac{p-1}{2} \leqslant i \leqslant p-2$. The same argument as above shows that Eduardo can always make such move. We will have $10^{j} \equiv 10^{i}$ $(\bmod p)$, and so $a_{j} \cdot 10^{j}+a_{i} \cdot 10^{i} \equiv\left(a_{i}+a_{j}\right) \cdot 10^{i}=9 \cdot 10^{i}(\bmod p)$. Therefore, at the end of the game, the sum of all terms $a_{k} \cdot 10^{k}$ will be congruent to  $$ \sum_{i=0}^{\frac{p-3}{2}} 9 \cdot 10^{i}=10^{(p-1) / 2}-1 \equiv 0 \quad(\bmod p) $$  and Eduardo wins the game.


Determine all integers $n \geqslant 2$ with the following property: for any integers $a_{1}, a_{2}, \ldots, a_{n}$ whose sum is not divisible by $n$, there exists an index $1 \leqslant i \leqslant n$ such that none of the numbers  $$ a_{i}, a_{i}+a_{i+1}, \ldots, a_{i}+a_{i+1}+\cdots+a_{i+n-1} $$  is divisible by $n$. (We let $a_{i}=a_{i-n}$ when $i>n$.) (Thailand)

Solution: Let us first show that, if $n=a b$, with $a, b \geqslant 2$ integers, then the property in the statement of the problem does not hold. Indeed, in this case, let $a_{k}=a$ for $1 \leqslant k \leqslant n-1$ and $a_{n}=0$. The sum $a_{1}+a_{2}+\cdots+a_{n}=a \cdot(n-1)$ is not divisible by $n$. Let $i$ with $1 \leqslant i \leqslant n$ be an arbitrary index. Taking $j=b$ if $1 \leqslant i \leqslant n-b$, and $j=b+1$ if $n-b<i \leqslant n$, we have  $$ a_{i}+a_{i+1}+\cdots+a_{i+j-1}=a \cdot b=n \equiv 0 \quad(\bmod n) $$  It follows that the given example is indeed a counterexample to the property of the statement. Now let $n$ be a prime number. Suppose by contradiction that the property in the statement of the problem does not hold. Then there are integers $a_{1}, a_{2}, \ldots, a_{n}$ whose sum is not divisible by $n$ such that for each $i, 1 \leqslant i \leqslant n$, there is $j, 1 \leqslant j \leqslant n$, for which the number $a_{i}+a_{i+1}+$ $\cdots+a_{i+j-1}$ is divisible by $n$. Notice that, in any such case, we should have $1 \leqslant j \leqslant n-1$, since $a_{1}+a_{2}+\cdots+a_{n}$ is not divisible by $n$. So we may construct recursively a finite sequence of integers $0=i_{0}<i_{1}<i_{2}<\cdots<i_{n}$ with $i_{s+1}-i_{s} \leqslant n-1$ for $0 \leqslant s \leqslant n-1$ such that, for $0 \leqslant s \leqslant n-1$,  $$ a_{i_{s}+1}+a_{i_{s}+2}+\cdots+a_{i_{s+1}} \equiv 0 \quad(\bmod n) $$  (where we take indices modulo $n$ ). Indeed, for $0 \leqslant s<n$, we apply the previous observation to $i=i_{s}+1$ in order to define $i_{s+1}=i_{s}+j$.  In the sequence of $n+1$ indices $i_{0}, i_{1}, i_{2}, \ldots, i_{n}$, by the pigeonhole principle, we have two distinct elements which are congruent modulo $n$. So there are indices $r, s$ with $0 \leqslant r<s \leqslant n$ such that $i_{s} \equiv i_{r}(\bmod n)$ and  $$ a_{i_{r}+1}+a_{i_{r}+2}+\cdots+a_{i_{s}}=\sum_{j=r}^{s-1}\left(a_{i_{j}+1}+a_{i_{j}+2}+\cdots+a_{i_{j+1}}\right) \equiv 0 \quad(\bmod n) $$  Since $i_{s} \equiv i_{r}(\bmod n)$, we have $i_{s}-i_{r}=k \cdot n$ for some positive integer $k$, and, since $i_{j+1}-i_{j} \leqslant$ $n-1$ for $0 \leqslant j \leqslant n-1$, we have $i_{s}-i_{r} \leqslant(n-1) \cdot n$, so $k \leqslant n-1$. But in this case  $$ a_{i_{r}+1}+a_{i_{r}+2}+\cdots+a_{i_{s}}=k \cdot\left(a_{1}+a_{2}+\cdots+a_{n}\right) $$  cannot be a multiple of $n$, since $n$ is prime and neither $k$ nor $a_{1}+a_{2}+\cdots+a_{n}$ is a multiple of $n$. A contradiction.


Call a rational number short if it has finitely many digits in its decimal expansion. For a positive integer $m$, we say that a positive integer $t$ is $m$-tastic if there exists a number $c \in\{1,2,3, \ldots, 2017\}$ such that $\frac{10^{t}-1}{c \cdot m}$ is short, and such that $\frac{10^{k}-1}{c \cdot m}$ is not short for any $1 \leqslant k<t$. Let $S(m)$ be the set of $m$-tastic numbers. Consider $S(m)$ for $m=1,2, \ldots$. What is the maximum number of elements in $S(m)$ ? (Turkey)

Solution: First notice that $x \in \mathbb{Q}$ is short if and only if there are exponents $a, b \geqslant 0$ such that $2^{a} \cdot 5^{b} \cdot x \in \mathbb{Z}$. In fact, if $x$ is short, then $x=\frac{n}{10^{k}}$ for some $k$ and we can take $a=b=k$; on the other hand, if $2^{a} \cdot 5^{b} \cdot x=q \in \mathbb{Z}$ then $x=\frac{2^{b} \cdot 5^{a} q}{10^{a+b}}$, so $x$ is short.  If $m=2^{a} \cdot 5^{b} \cdot s$, with $\operatorname{gcd}(s, 10)=1$, then $\frac{10^{t}-1}{m}$ is short if and only if $s$ divides $10^{t}-1$. So we may (and will) suppose without loss of generality that $\operatorname{gcd}(m, 10)=1$. Define  $$ C=\{1 \leqslant c \leqslant 2017: \operatorname{gcd}(c, 10)=1\} . $$  The $m$-tastic numbers are then precisely the smallest exponents $t>0$ such that $10^{t} \equiv 1$ $(\bmod \mathrm{cm})$ for some integer $c \in C$, that is, the set of orders of 10 modulo cm . In other words,  $$ S(m)=\left\{\operatorname{ord}_{c m}(10): c \in C\right\} $$  Since there are $4 \cdot 201+3=807$ numbers $c$ with $1 \leqslant c \leqslant 2017$ and $\operatorname{gcd}(c, 10)=1$, namely those such that $c \equiv 1,3,7,9(\bmod 10)$,  $$ |S(m)| \leqslant|C|=807 $$  Now we find $m$ such that $|S(m)|=807$. Let  $$ P=\{1<p \leqslant 2017: p \text { is prime, } p \neq 2,5\} $$  and choose a positive integer $\alpha$ such that every $p \in P$ divides $10^{\alpha}-1$ (e.g. $\alpha=\varphi(T), T$ being the product of all primes in $P$ ), and let $m=10^{\alpha}-1$. Claim. For every $c \in C$, we have  $$ \operatorname{ord}_{c m}(10)=c \alpha $$  As an immediate consequence, this implies $|S(m)|=|C|=807$, finishing the problem. Proof. Obviously $\operatorname{ord}_{m}(10)=\alpha$. Let $t=\operatorname{ord}_{c m}(10)$. Then  $$ c m\left|10^{t}-1 \quad \Longrightarrow \quad m\right| 10^{t}-1 \quad \Longrightarrow \quad \alpha \mid t $$  Hence $t=k \alpha$ for some $k \in \mathbb{Z}_{>0}$. We will show that $k=c$. Denote by $\nu_{p}(n)$ the number of prime factors $p$ in $n$, that is, the maximum exponent $\beta$ for which $p^{\beta} \mid n$. For every $\ell \geqslant 1$ and $p \in P$, the Lifting the Exponent Lemma provides  $$ \nu_{p}\left(10^{\ell \alpha}-1\right)=\nu_{p}\left(\left(10^{\alpha}\right)^{\ell}-1\right)=\nu_{p}\left(10^{\alpha}-1\right)+\nu_{p}(\ell)=\nu_{p}(m)+\nu_{p}(\ell) $$  so  $$ \begin{aligned} c m \mid 10^{k \alpha}-1 & \Longleftrightarrow \forall p \in P ; \nu_{p}(c m) \leqslant \nu_{p}\left(10^{k \alpha}-1\right) \\ & \Longleftrightarrow \forall p \in P ; \nu_{p}(m)+\nu_{p}(c) \leqslant \nu_{p}(m)+\nu_{p}(k) \\ & \Longleftrightarrow \forall p \in P ; \nu_{p}(c) \leqslant \nu_{p}(k) \\ & \Longleftrightarrow c \mid k . \end{aligned} $$  The first such $k$ is $k=c$, so $\operatorname{ord}_{c m}(10)=c \alpha$.  Comment. The Lifting the Exponent Lemma states that, for any odd prime $p$, any integers $a, b$ coprime with $p$ such that $p \mid a-b$, and any positive integer exponent $n$,  $$ \nu_{p}\left(a^{n}-b^{n}\right)=\nu_{p}(a-b)+\nu_{p}(n), $$  and, for $p=2$,  $$ \nu_{2}\left(a^{n}-b^{n}\right)=\nu_{2}\left(a^{2}-b^{2}\right)+\nu_{p}(n)-1 . $$  Both claims can be proved by induction on $n$.


Find all pairs $(p, q)$ of prime numbers with $p>q$ for which the number  $$ \frac{(p+q)^{p+q}(p-q)^{p-q}-1}{(p+q)^{p-q}(p-q)^{p+q}-1} $$  is an integer.

Solution: Let $M=(p+q)^{p-q}(p-q)^{p+q}-1$, which is relatively prime with both $p+q$ and $p-q$. Denote by $(p-q)^{-1}$ the multiplicative inverse of $(p-q)$ modulo $M$.  By eliminating the term -1 in the numerator,  $$ \begin{aligned} (p+q)^{p+q}(p-q)^{p-q}-1 & \equiv(p+q)^{p-q}(p-q)^{p+q}-1 \quad(\bmod M) \\ (p+q)^{2 q} & \equiv(p-q)^{2 q} \quad(\bmod M) \\ \left((p+q) \cdot(p-q)^{-1}\right)^{2 q} & \equiv 1 \quad(\bmod M) \end{aligned} $$  Case 1: $q \geqslant 5$. Consider an arbitrary prime divisor $r$ of $M$. Notice that $M$ is odd, so $r \geqslant 3$. By (2), the multiplicative order of $\left((p+q) \cdot(p-q)^{-1}\right)$ modulo $r$ is a divisor of the exponent $2 q$ in (2), so it can be $1,2, q$ or $2 q$.  By Fermat's theorem, the order divides $r-1$. So, if the order is $q$ or $2 q$ then $r \equiv 1(\bmod q)$. If the order is 1 or 2 then $r \mid(p+q)^{2}-(p-q)^{2}=4 p q$, so $r=p$ or $r=q$. The case $r=p$ is not possible, because, by applying Fermat's theorem, $M=(p+q)^{p-q}(p-q)^{p+q}-1 \equiv q^{p-q}(-q)^{p+q}-1=\left(q^{2}\right)^{p}-1 \equiv q^{2}-1=(q+1)(q-1) \quad(\bmod p)$ and the last factors $q-1$ and $q+1$ are less than $p$ and thus $p \nmid M$. Hence, all prime divisors of $M$ are either $q$ or of the form $k q+1$; it follows that all positive divisors of $M$ are congruent to 0 or 1 modulo $q$.  Now notice that  $$ M=\left((p+q)^{\frac{p-q}{2}}(p-q)^{\frac{p+q}{2}}-1\right)\left((p+q)^{\frac{p-q}{2}}(p-q)^{\frac{p+q}{2}}+1\right) $$  is the product of two consecutive positive odd numbers; both should be congruent to 0 or 1 modulo $q$. But this is impossible by the assumption $q \geqslant 5$. So, there is no solution in Case 1 . Case 2: $q=2$. By (1), we have $M \mid(p+q)^{2 q}-(p-q)^{2 q}=(p+2)^{4}-(p-2)^{4}$, so  $$ \begin{gathered} (p+2)^{p-2}(p-2)^{p+2}-1=M \leqslant(p+2)^{4}-(p-2)^{4} \leqslant(p+2)^{4}-1, \\ (p+2)^{p-6}(p-2)^{p+2} \leqslant 1 . \end{gathered} $$  If $p \geqslant 7$ then the left-hand side is obviously greater than 1 . For $p=5$ we have $(p+2)^{p-6}(p-2)^{p+2}=7^{-1} \cdot 3^{7}$ which is also too large.  There remains only one candidate, $p=3$, which provides a solution:  $$ \frac{(p+q)^{p+q}(p-q)^{p-q}-1}{(p+q)^{p-q}(p-q)^{p+q}-1}=\frac{5^{5} \cdot 1^{1}-1}{5^{1} \cdot 1^{5}-1}=\frac{3124}{4}=781 . $$  So in Case 2 the only solution is $(p, q)=(3,2)$.  Case 3: $q=3$. Similarly to Case 2, we have  $$ M \left\lvert\,(p+q)^{2 q}-(p-q)^{2 q}=64 \cdot\left(\left(\frac{p+3}{2}\right)^{6}-\left(\frac{p-3}{2}\right)^{6}\right)\right. $$  Since $M$ is odd, we conclude that  $$ M \left\lvert\,\left(\frac{p+3}{2}\right)^{6}-\left(\frac{p-3}{2}\right)^{6}\right. $$  and  $$ \begin{gathered} (p+3)^{p-3}(p-3)^{p+3}-1=M \leqslant\left(\frac{p+3}{2}\right)^{6}-\left(\frac{p-3}{2}\right)^{6} \leqslant\left(\frac{p+3}{2}\right)^{6}-1 \\ 64(p+3)^{p-9}(p-3)^{p+3} \leqslant 1 \end{gathered} $$  If $p \geqslant 11$ then the left-hand side is obviously greater than 1 . If $p=7$ then the left-hand side is $64 \cdot 10^{-2} \cdot 4^{10}>1$. If $p=5$ then the left-hand side is $64 \cdot 8^{-4} \cdot 2^{8}=2^{2}>1$. Therefore, there is no solution in Case 3.


Find the smallest positive integer $n$, or show that no such $n$ exists, with the following property: there are infinitely many distinct $n$-tuples of positive rational numbers ( $a_{1}, a_{2}, \ldots, a_{n}$ ) such that both  $$ a_{1}+a_{2}+\cdots+a_{n} \quad \text { and } \quad \frac{1}{a_{1}}+\frac{1}{a_{2}}+\cdots+\frac{1}{a_{n}} $$  are integers. (Singapore)

Solution: For $n=1, a_{1} \in \mathbb{Z}_{>0}$ and $\frac{1}{a_{1}} \in \mathbb{Z}_{>0}$ if and only if $a_{1}=1$. Next we show that (i) There are finitely many $(x, y) \in \mathbb{Q}_{>0}^{2}$ satisfying $x+y \in \mathbb{Z}$ and $\frac{1}{x}+\frac{1}{y} \in \mathbb{Z}$  Write $x=\frac{a}{b}$ and $y=\frac{c}{d}$ with $a, b, c, d \in \mathbb{Z}_{>0}$ and $\operatorname{gcd}(a, b)=\operatorname{gcd}(c, d)=1$. Then $x+y \in \mathbb{Z}$ and $\frac{1}{x}+\frac{1}{y} \in \mathbb{Z}$ is equivalent to the two divisibility conditions  $$ b d \mid a d+b c \quad(1) \quad \text { and } \quad a c \mid a d+b c $$  Condition (1) implies that $d|a d+b c \Longleftrightarrow d| b c \Longleftrightarrow d \mid b$ since $\operatorname{gcd}(c, d)=1$. Still from (1) we get $b|a d+b c \Longleftrightarrow b| a d \Longleftrightarrow b \mid d$ since $\operatorname{gcd}(a, b)=1$. From $b \mid d$ and $d \mid b$ we have $b=d$. An analogous reasoning with condition (2) shows that $a=c$. Hence $x=\frac{a}{b}=\frac{c}{d}=y$, i.e., the problem amounts to finding all $x \in \mathbb{Q}_{>0}$ such that $2 x \in \mathbb{Z}_{>0}$ and $\frac{2}{x} \in \mathbb{Z}_{>0}$. Letting $n=2 x \in \mathbb{Z}_{>0}$, we have that $\frac{2}{x} \in \mathbb{Z}_{>0} \Longleftrightarrow \frac{4}{n} \in \mathbb{Z}_{>0} \Longleftrightarrow n=1,2$ or 4 , and there are finitely many solutions, namely $(x, y)=\left(\frac{1}{2}, \frac{1}{2}\right),(1,1)$ or $(2,2)$. (ii) There are infinitely many triples $(x, y, z) \in \mathbb{Q}_{>0}^{2}$ such that $x+y+z \in \mathbb{Z}$ and $\frac{1}{x}+\frac{1}{y}+\frac{1}{z} \in \mathbb{Z}$. We will look for triples such that $x+y+z=1$, so we may write them in the form  $$ (x, y, z)=\left(\frac{a}{a+b+c}, \frac{b}{a+b+c}, \frac{c}{a+b+c}\right) \quad \text { with } a, b, c \in \mathbb{Z}_{>0} $$  We want these to satisfy  $$ \frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{a+b+c}{a}+\frac{a+b+c}{b}+\frac{a+b+c}{c} \in \mathbb{Z} \Longleftrightarrow \frac{b+c}{a}+\frac{a+c}{b}+\frac{a+b}{c} \in \mathbb{Z} $$  Fixing $a=1$, it suffices to find infinitely many pairs $(b, c) \in \mathbb{Z}_{>0}^{2}$ such that  $$ \frac{1}{b}+\frac{1}{c}+\frac{c}{b}+\frac{b}{c}=3 \Longleftrightarrow b^{2}+c^{2}-3 b c+b+c=0 $$  To show that equation (*) has infinitely many solutions, we use Vieta jumping (also known as root flipping): starting with $b=2, c=3$, the following algorithm generates infinitely many solutions. Let $c \geqslant b$, and view (*) as a quadratic equation in $b$ for $c$ fixed:  $$ b^{2}-(3 c-1) \cdot b+\left(c^{2}+c\right)=0 $$  Then there exists another root $b_{0} \in \mathbb{Z}$ of ( $\left.* *\right)$ which satisfies $b+b_{0}=3 c-1$ and $b \cdot b_{0}=c^{2}+c$. Since $c \geqslant b$ by assumption,  $$ b_{0}=\frac{c^{2}+c}{b} \geqslant \frac{c^{2}+c}{c}>c $$  Hence from the solution $(b, c)$ we obtain another one $\left(c, b_{0}\right)$ with $b_{0}>c$, and we can then "jump" again, this time with $c$ as the "variable" in the quadratic (*). This algorithm will generate an infinite sequence of distinct solutions, whose first terms are $(2,3),(3,6),(6,14),(14,35),(35,90),(90,234),(234,611),(611,1598),(1598,4182), \ldots$  Comment. Although not needed for solving this problem, we may also explicitly solve the recursion given by the Vieta jumping. Define the sequence $\left(x_{n}\right)$ as follows:  $$ x_{0}=2, \quad x_{1}=3 \quad \text { and } \quad x_{n+2}=3 x_{n+1}-x_{n}-1 \text { for } n \geqslant 0 $$  Then the triple  $$ (x, y, z)=\left(\frac{1}{1+x_{n}+x_{n+1}}, \frac{x_{n}}{1+x_{n}+x_{n+1}}, \frac{x_{n+1}}{1+x_{n}+x_{n+1}}\right) $$  satisfies the problem conditions for all $n \in \mathbb{N}$. It is easy to show that $x_{n}=F_{2 n+1}+1$, where $F_{n}$ denotes the $n$-th term of the Fibonacci sequence ( $F_{0}=0, F_{1}=1$, and $F_{n+2}=F_{n+1}+F_{n}$ for $n \geqslant 0$ ).


Solution: Call the $n$-tuples $\left(a_{1}, a_{2}, \ldots, a_{n}\right) \in \mathbb{Q}_{>0}^{n}$ satisfying the conditions of the problem statement good, and those for which  $$ f\left(a_{1}, \ldots, a_{n}\right) \stackrel{\text { def }}{=}\left(a_{1}+a_{2}+\cdots+a_{n}\right)\left(\frac{1}{a_{1}}+\frac{1}{a_{2}}+\cdots+\frac{1}{a_{n}}\right) $$  is an integer pretty. Then good $n$-tuples are pretty, and if $\left(b_{1}, \ldots, b_{n}\right)$ is pretty then  $$ \left(\frac{b_{1}}{b_{1}+b_{2}+\cdots+b_{n}}, \frac{b_{2}}{b_{1}+b_{2}+\cdots+b_{n}}, \ldots, \frac{b_{n}}{b_{1}+b_{2}+\cdots+b_{n}}\right) $$  is good since the sum of its components is 1 , and the sum of the reciprocals of its components equals $f\left(b_{1}, \ldots, b_{n}\right)$. We declare pretty $n$-tuples proportional to each other equivalent since they are precisely those which give rise to the same good $n$-tuple. Clearly, each such equivalence class contains exactly one $n$-tuple of positive integers having no common prime divisors. Call such $n$-tuple a primitive pretty tuple. Our task is to find infinitely many primitive pretty $n$-tuples.  For $n=1$, there is clearly a single primitive 1 -tuple. For $n=2$, we have $f(a, b)=\frac{(a+b)^{2}}{a b}$, which can be integral (for coprime $a, b \in \mathbb{Z}_{>0}$ ) only if $a=b=1$ (see for instance (i) in the first solution).  Now we construct infinitely many primitive pretty triples for $n=3$. Fix $b, c, k \in \mathbb{Z}_{>0}$; we will try to find sufficient conditions for the existence of an $a \in \mathbb{Q}_{>0}$ such that $f(a, b, c)=k$. Write $\sigma=b+c, \tau=b c$. From $f(a, b, c)=k$, we have that $a$ should satisfy the quadratic equation  $$ a^{2} \cdot \sigma+a \cdot\left(\sigma^{2}-(k-1) \tau\right)+\sigma \tau=0 $$  whose discriminant is  $$ \Delta=\left(\sigma^{2}-(k-1) \tau\right)^{2}-4 \sigma^{2} \tau=\left((k+1) \tau-\sigma^{2}\right)^{2}-4 k \tau^{2} $$  We need it to be a square of an integer, say, $\Delta=M^{2}$ for some $M \in \mathbb{Z}$, i.e., we want  $$ \left((k+1) \tau-\sigma^{2}\right)^{2}-M^{2}=2 k \cdot 2 \tau^{2} $$  so that it suffices to set  $$ (k+1) \tau-\sigma^{2}=\tau^{2}+k, \quad M=\tau^{2}-k . $$  The first relation reads $\sigma^{2}=(\tau-1)(k-\tau)$, so if $b$ and $c$ satisfy  $$ \tau-1 \mid \sigma^{2} \quad \text { i.e. } \quad b c-1 \mid(b+c)^{2} $$  then $k=\frac{\sigma^{2}}{\tau-1}+\tau$ will be integral, and we find rational solutions to (1), namely  $$ a=\frac{\sigma}{\tau-1}=\frac{b+c}{b c-1} \quad \text { or } \quad a=\frac{\tau^{2}-\tau}{\sigma}=\frac{b c \cdot(b c-1)}{b+c} $$  We can now find infinitely many pairs ( $b, c$ ) satisfying (2) by Vieta jumping. For example, if we impose  $$ (b+c)^{2}=5 \cdot(b c-1) $$  then all pairs $(b, c)=\left(v_{i}, v_{i+1}\right)$ satisfy the above condition, where  $$ v_{1}=2, v_{2}=3, \quad v_{i+2}=3 v_{i+1}-v_{i} \quad \text { for } i \geqslant 0 $$  For $(b, c)=\left(v_{i}, v_{i+1}\right)$, one of the solutions to (1) will be $a=(b+c) /(b c-1)=5 /(b+c)=$ $5 /\left(v_{i}+v_{i+1}\right)$. Then the pretty triple $(a, b, c)$ will be equivalent to the integral pretty triple  $$ \left(5, v_{i}\left(v_{i}+v_{i+1}\right), v_{i+1}\left(v_{i}+v_{i+1}\right)\right) $$  After possibly dividing by 5 , we obtain infinitely many primitive pretty triples, as required. Comment. There are many other infinite series of $(b, c)=\left(v_{i}, v_{i+1}\right)$ with $b c-1 \mid(b+c)^{2}$. Some of them are:  $$ \begin{array}{llll} v_{1}=1, & v_{2}=3, & v_{i+1}=6 v_{i}-v_{i-1}, & \left(v_{i}+v_{i+1}\right)^{2}=8 \cdot\left(v_{i} v_{i+1}-1\right) ; \\ v_{1}=1, & v_{2}=2, & v_{i+1}=7 v_{i}-v_{i-1}, & \left(v_{i}+v_{i+1}\right)^{2}=9 \cdot\left(v_{i} v_{i+1}-1\right) ; \\ v_{1}=1, & v_{2}=5, & v_{i+1}=7 v_{i}-v_{i-1}, & \left(v_{i}+v_{i+1}\right)^{2}=9 \cdot\left(v_{i} v_{i+1}-1\right) \end{array} $$  (the last two are in fact one sequence prolonged in two possible directions).


Say that an ordered pair $(x, y)$ of integers is an irreducible lattice point if $x$ and $y$ are relatively prime. For any finite set $S$ of irreducible lattice points, show that there is a homogenous polynomial in two variables, $f(x, y)$, with integer coefficients, of degree at least 1 , such that $f(x, y)=1$ for each $(x, y)$ in the set $S$.  Note: A homogenous polynomial of degree $n$ is any nonzero polynomial of the form  $$ f(x, y)=a_{0} x^{n}+a_{1} x^{n-1} y+a_{2} x^{n-2} y^{2}+\cdots+a_{n-1} x y^{n-1}+a_{n} y^{n} . $$

Solution: First of all, we note that finding a homogenous polynomial $f(x, y)$ such that $f(x, y)= \pm 1$ is enough, because we then have $f^{2}(x, y)=1$. Label the irreducible lattice points $\left(x_{1}, y_{1}\right)$ through $\left(x_{n}, y_{n}\right)$. If any two of these lattice points $\left(x_{i}, y_{i}\right)$ and $\left(x_{j}, y_{j}\right)$ lie on the same line through the origin, then $\left(x_{j}, y_{j}\right)=\left(-x_{i},-y_{i}\right)$ because both of the points are irreducible. We then have $f\left(x_{j}, y_{j}\right)= \pm f\left(x_{i}, y_{i}\right)$ whenever $f$ is homogenous, so we can assume that no two of the lattice points are collinear with the origin by ignoring the extra lattice points.  Consider the homogenous polynomials $\ell_{i}(x, y)=y_{i} x-x_{i} y$ and define  $$ g_{i}(x, y)=\prod_{j \neq i} \ell_{j}(x, y) $$  Then $\ell_{i}\left(x_{j}, y_{j}\right)=0$ if and only if $j=i$, because there is only one lattice point on each line through the origin. Thus, $g_{i}\left(x_{j}, y_{j}\right)=0$ for all $j \neq i$. Define $a_{i}=g_{i}\left(x_{i}, y_{i}\right)$, and note that $a_{i} \neq 0$.  Note that $g_{i}(x, y)$ is a degree $n-1$ polynomial with the following two properties:  1. $g_{i}\left(x_{j}, y_{j}\right)=0$ if $j \neq i$. 2. $g_{i}\left(x_{i}, y_{i}\right)=a_{i}$.  For any $N \geqslant n-1$, there also exists a polynomial of degree $N$ with the same two properties. Specifically, let $I_{i}(x, y)$ be a degree 1 homogenous polynomial such that $I_{i}\left(x_{i}, y_{i}\right)=1$, which exists since $\left(x_{i}, y_{i}\right)$ is irreducible. Then $I_{i}(x, y)^{N-(n-1)} g_{i}(x, y)$ satisfies both of the above properties and has degree $N$.  We may now reduce the problem to the following claim: Claim: For each positive integer $a$, there is a homogenous polynomial $f_{a}(x, y)$, with integer coefficients, of degree at least 1 , such that $f_{a}(x, y) \equiv 1(\bmod a)$ for all relatively prime $(x, y)$.  To see that this claim solves the problem, take $a$ to be the least common multiple of the numbers $a_{i}(1 \leqslant i \leqslant n)$. Take $f_{a}$ given by the claim, choose some power $f_{a}(x, y)^{k}$ that has degree at least $n-1$, and subtract appropriate multiples of the $g_{i}$ constructed above to obtain the desired polynomial.  We prove the claim by factoring $a$. First, if $a$ is a power of a prime ( $a=p^{k}$ ), then we may choose either:  - $f_{a}(x, y)=\left(x^{p-1}+y^{p-1}\right)^{\phi(a)}$ if $p$ is odd; - $f_{a}(x, y)=\left(x^{2}+x y+y^{2}\right)^{\phi(a)}$ if $p=2$.  Now suppose $a$ is any positive integer, and let $a=q_{1} q_{2} \cdots q_{k}$, where the $q_{i}$ are prime powers, pairwise relatively prime. Let $f_{q_{i}}$ be the polynomials just constructed, and let $F_{q_{i}}$ be powers of these that all have the same degree. Note that  $$ \frac{a}{q_{i}} F_{q_{i}}(x, y) \equiv \frac{a}{q_{i}} \quad(\bmod a) $$  for any relatively prime $x, y$. By Bézout's lemma, there is an integer linear combination of the $\frac{a}{q_{i}}$ that equals 1 . Thus, there is a linear combination of the $F_{q_{i}}$ such that $F_{q_{i}}(x, y) \equiv 1$ $(\bmod a)$ for any relatively prime $(x, y)$; and this polynomial is homogenous because all the $F_{q_{i}}$ have the same degree.


Solution: As in the previous solution, label the irreducible lattice points $\left(x_{1}, y_{1}\right), \ldots,\left(x_{n}, y_{n}\right)$ and assume without loss of generality that no two of the points are collinear with the origin. We induct on $n$ to construct a homogenous polynomial $f(x, y)$ such that $f\left(x_{i}, y_{i}\right)=1$ for all $1 \leqslant i \leqslant n$.  If $n=1$ : Since $x_{1}$ and $y_{1}$ are relatively prime, there exist some integers $c, d$ such that $c x_{1}+d y_{1}=1$. Then $f(x, y)=c x+d y$ is suitable.  If $n \geqslant 2$ : By the induction hypothesis we already have a homogeneous polynomial $g(x, y)$ with $g\left(x_{1}, y_{1}\right)=\ldots=g\left(x_{n-1}, y_{n-1}\right)=1$. Let $j=\operatorname{deg} g$,  $$ g_{n}(x, y)=\prod_{k=1}^{n-1}\left(y_{k} x-x_{k} y\right) $$  and $a_{n}=g_{n}\left(x_{n}, y_{n}\right)$. By assumption, $a_{n} \neq 0$. Take some integers $c, d$ such that $c x_{n}+d y_{n}=1$. We will construct $f(x, y)$ in the form  $$ f(x, y)=g(x, y)^{K}-C \cdot g_{n}(x, y) \cdot(c x+d y)^{L} $$  where $K$ and $L$ are some positive integers and $C$ is some integer. We assume that $L=K j-n+1$ so that $f$ is homogenous.  Due to $g\left(x_{1}, y_{1}\right)=\ldots=g\left(x_{n-1}, y_{n-1}\right)=1$ and $g_{n}\left(x_{1}, y_{1}\right)=\ldots=g_{n}\left(x_{n-1}, y_{n-1}\right)=0$, the property $f\left(x_{1}, y_{1}\right)=\ldots=f\left(x_{n-1}, y_{n-1}\right)=1$ is automatically satisfied with any choice of $K, L$, and $C$.  Furthermore,  $$ f\left(x_{n}, y_{n}\right)=g\left(x_{n}, y_{n}\right)^{K}-C \cdot g_{n}\left(x_{n}, y_{n}\right) \cdot\left(c x_{n}+d y_{n}\right)^{L}=g\left(x_{n}, y_{n}\right)^{K}-C a_{n} $$  If we have an exponent $K$ such that $g\left(x_{n}, y_{n}\right)^{K} \equiv 1\left(\bmod a_{n}\right)$, then we may choose $C$ such that $f\left(x_{n}, y_{n}\right)=1$. We now choose such a $K$.  Consider an arbitrary prime divisor $p$ of $a_{n}$. By  $$ p \mid a_{n}=g_{n}\left(x_{n}, y_{n}\right)=\prod_{k=1}^{n-1}\left(y_{k} x_{n}-x_{k} y_{n}\right) $$  there is some $1 \leqslant k<n$ such that $x_{k} y_{n} \equiv x_{n} y_{k}(\bmod p)$. We first show that $x_{k} x_{n}$ or $y_{k} y_{n}$ is relatively prime with $p$. This is trivial in the case $x_{k} y_{n} \equiv x_{n} y_{k} \not \equiv 0(\bmod p)$. In the other case, we have $x_{k} y_{n} \equiv x_{n} y_{k} \equiv 0(\bmod p)$, If, say $p \mid x_{k}$, then $p \nmid y_{k}$ because $\left(x_{k}, y_{k}\right)$ is irreducible, so $p \mid x_{n}$; then $p \nmid y_{n}$ because $\left(x_{k}, y_{k}\right)$ is irreducible. In summary, $p \mid x_{k}$ implies $p \nmid y_{k} y_{n}$. Similarly, $p \mid y_{n}$ implies $p \nmid x_{k} x_{n}$.  By the homogeneity of $g$ we have the congruences  $$ x_{k}^{d} \cdot g\left(x_{n}, y_{n}\right)=g\left(x_{k} x_{n}, x_{k} y_{n}\right) \equiv g\left(x_{k} x_{n}, y_{k} x_{n}\right)=x_{n}^{d} \cdot g\left(x_{k}, y_{k}\right)=x_{n}^{d} \quad(\bmod p) $$  and  $$ y_{k}^{d} \cdot g\left(x_{n}, y_{n}\right)=g\left(y_{k} x_{n}, y_{k} y_{n}\right) \equiv g\left(x_{k} y_{n}, y_{k} y_{n}\right)=y_{n}^{d} \cdot g\left(x_{k}, y_{k}\right)=y_{n}^{d} \quad(\bmod p) . $$  If $p \nmid x_{k} x_{n}$, then take the $(p-1)^{s t}$ power of (1.1); otherwise take the $(p-1)^{s t}$ power of (1.2); by Fermat's theorem, in both cases we get  $$ g\left(x_{n}, y_{n}\right)^{p-1} \equiv 1 \quad(\bmod p) $$  If $p^{\alpha} \mid m$, then we have  $$ g\left(x_{n}, y_{n}\right)^{p^{\alpha-1}(p-1)} \equiv 1 \quad\left(\bmod p^{\alpha}\right) $$  which implies that the exponent $K=n \cdot \varphi\left(a_{n}\right)$, which is a multiple of all $p^{\alpha-1}(p-1)$, is a suitable choice. (The factor $n$ is added only so that $K \geqslant n$ and so $L>0$.)  Comment. It is possible to show that there is no constant $C$ for which, given any two irreducible lattice points, there is some homogenous polynomial $f$ of degree at most $C$ with integer coefficients that takes the value 1 on the two points. Indeed, if one of the points is $(1,0)$ and the other is $(a, b)$, the polynomial $f(x, y)=a_{0} x^{n}+a_{1} x^{n-1} y+\cdots+a_{n} y^{n}$ should satisfy $a_{0}=1$, and so $a^{n} \equiv 1(\bmod b)$. If $a=3$ and $b=2^{k}$ with $k \geqslant 3$, then $n \geqslant 2^{k-2}$. If we choose $2^{k-2}>C$, this gives a contradiction.