Points $A_{1}, B_{1}, C_{1}$ are chosen on the sides $B C, C A, A B$ of a triangle $A B C$, respectively. The circumcircles of triangles $A B_{1} C_{1}, B C_{1} A_{1}, C A_{1} B_{1}$ intersect the circumcircle of triangle $A B C$ again at points $A_{2}, B_{2}, C_{2}$, respectively $\left(A_{2} \neq A, B_{2} \neq B, C_{2} \neq C\right)$. Points $A_{3}, B_{3}, C_{3}$ are symmetric to $A_{1}, B_{1}, C_{1}$ with respect to the midpoints of the sides $B C, C A, A B$ respectively. Prove that the triangles $A_{2} B_{2} C_{2}$ and $A_{3} B_{3} C_{3}$ are similar.  (Russia)

Solution: We will work with oriented angles between lines. For two straight lines $\ell, m$ in the plane, $\angle(\ell, m)$ denotes the angle of counterclockwise rotation which transforms line $\ell$ into a line parallel to $m$ (the choice of the rotation centre is irrelevant). This is a signed quantity; values differing by a multiple of $\pi$ are identified, so that  $$ \angle(\ell, m)=-\angle(m, \ell), \quad \angle(\ell, m)+\angle(m, n)=\angle(\ell, n) . $$  If $\ell$ is the line through points $K, L$ and $m$ is the line through $M, N$, one writes $\angle(K L, M N)$ for $\angle(\ell, m)$; the characters $K, L$ are freely interchangeable; and so are $M, N$.  The counterpart of the classical theorem about cyclic quadrilaterals is the following: If $K, L, M, N$ are four noncollinear points in the plane then  $$ K, L, M, N \text { are concyclic if and only if } \angle(K M, L M)=\angle(K N, L N) \text {. } $$  Passing to the solution proper, we first show that the three circles $\left(A B_{1} C_{1}\right),\left(B C_{1} A_{1}\right)$, $\left(C A_{1} B_{1}\right)$ have a common point. So, let $\left(A B_{1} C_{1}\right)$ and $\left(B C_{1} A_{1}\right)$ intersect at the points $C_{1}$ and $P$. Then by (1)  $$ \begin{gathered} \angle\left(P A_{1}, C A_{1}\right)=\angle\left(P A_{1}, B A_{1}\right)=\angle\left(P C_{1}, B C_{1}\right) \\ =\angle\left(P C_{1}, A C_{1}\right)=\angle\left(P B_{1}, A B_{1}\right)=\angle\left(P B_{1}, C B_{1}\right) \end{gathered} $$  Denote this angle by $\varphi$.  The equality between the outer terms shows, again by (1), that the points $A_{1}, B_{1}, P, C$ are concyclic. Thus $P$ is the common point of the three mentioned circles.  From now on the basic property (1) will be used without explicit reference. We have  $$ \varphi=\angle\left(P A_{1}, B C\right)=\angle\left(P B_{1}, C A\right)=\angle\left(P C_{1}, A B\right) . $$  ![](https://cdn.mathpix.com/cropped/2024_04_17_e55ccc62f43b046023cag-49.jpg?height=778&width=1620&top_left_y=1938&top_left_x=226)  Let lines $A_{2} P, B_{2} P, C_{2} P$ meet the circle $(A B C)$ again at $A_{4}, B_{4}, C_{4}$, respectively. As  $$ \angle\left(A_{4} A_{2}, A A_{2}\right)=\angle\left(P A_{2}, A A_{2}\right)=\angle\left(P C_{1}, A C_{1}\right)=\angle\left(P C_{1}, A B\right)=\varphi \text {, } $$  we see that line $A_{2} A$ is the image of line $A_{2} A_{4}$ under rotation about $A_{2}$ by the angle $\varphi$. Hence the point $A$ is the image of $A_{4}$ under rotation by $2 \varphi$ about $O$, the centre of $(A B C)$. The same rotation sends $B_{4}$ to $B$ and $C_{4}$ to $C$. Triangle $A B C$ is the image of $A_{4} B_{4} C_{4}$ in this map. Thus  $$ \angle\left(A_{4} B_{4}, A B\right)=\angle\left(B_{4} C_{4}, B C\right)=\angle\left(C_{4} A_{4}, C A\right)=2 \varphi \text {. } $$  Since the rotation by $2 \varphi$ about $O$ takes $B_{4}$ to $B$, we have $\angle\left(A B_{4}, A B\right)=\varphi$. Hence by (2)  $$ \angle\left(A B_{4}, P C_{1}\right)=\angle\left(A B_{4}, A B\right)+\angle\left(A B, P C_{1}\right)=\varphi+(-\varphi)=0 $$  which means that $A B_{4} \| P C_{1}$. ![](https://cdn.mathpix.com/cropped/2024_04_17_e55ccc62f43b046023cag-50.jpg?height=784&width=1682&top_left_y=978&top_left_x=197)  Let $C_{5}$ be the intersection of lines $P C_{1}$ and $A_{4} B_{4}$; define $A_{5}, B_{5}$ analogously. So $A B_{4} \| C_{1} C_{5}$ and, by (3) and (2),  $$ \angle\left(A_{4} B_{4}, P C_{1}\right)=\angle\left(A_{4} B_{4}, A B\right)+\angle\left(A B, P C_{1}\right)=2 \varphi+(-\varphi)=\varphi $$  i.e., $\angle\left(B_{4} C_{5}, C_{5} C_{1}\right)=\varphi$. This combined with $\angle\left(C_{5} C_{1}, C_{1} A\right)=\angle\left(P C_{1}, A B\right)=\varphi$ (see (2)) proves that the quadrilateral $A B_{4} C_{5} C_{1}$ is an isosceles trapezoid with $A C_{1}=B_{4} C_{5}$.  Interchanging the roles of $A$ and $B$ we infer that also $B C_{1}=A_{4} C_{5}$. And since $A C_{1}+B C_{1}=$ $A B=A_{4} B_{4}$, it follows that the point $C_{5}$ lies on the line segment $A_{4} B_{4}$ and partitions it into segments $A_{4} C_{5}, B_{4} C_{5}$ of lengths $B C_{1}\left(=A C_{3}\right)$ and $A C_{1}\left(=B C_{3}\right)$. In other words, the rotation which maps triangle $A_{4} B_{4} C_{4}$ onto $A B C$ carries $C_{5}$ onto $C_{3}$. Likewise, it sends $A_{5}$ to $A_{3}$ and $B_{5}$ to $B_{3}$. So the triangles $A_{3} B_{3} C_{3}$ and $A_{5} B_{5} C_{5}$ are congruent. It now suffices to show that the latter is similar to $A_{2} B_{2} C_{2}$.  Lines $B_{4} C_{5}$ and $P C_{5}$ coincide respectively with $A_{4} B_{4}$ and $P C_{1}$. Thus by (4)  $$ \angle\left(B_{4} C_{5}, P C_{5}\right)=\varphi $$  Analogously (by cyclic shift) $\varphi=\angle\left(C_{4} A_{5}, P A_{5}\right)$, which rewrites as  $$ \varphi=\angle\left(B_{4} A_{5}, P A_{5}\right) $$  These relations imply that the points $P, B_{4}, C_{5}, A_{5}$ are concyclic. Analogously, $P, C_{4}, A_{5}, B_{5}$ and $P, A_{4}, B_{5}, C_{5}$ are concyclic quadruples. Therefore  $$ \angle\left(A_{5} B_{5}, C_{5} B_{5}\right)=\angle\left(A_{5} B_{5}, P B_{5}\right)+\angle\left(P B_{5}, C_{5} B_{5}\right)=\angle\left(A_{5} C_{4}, P C_{4}\right)+\angle\left(P A_{4}, C_{5} A_{4}\right) . $$  On the other hand, since the points $A_{2}, B_{2}, C_{2}, A_{4}, B_{4}, C_{4}$ all lie on the circle $(A B C)$, we have  $$ \angle\left(A_{2} B_{2}, C_{2} B_{2}\right)=\angle\left(A_{2} B_{2}, B_{4} B_{2}\right)+\angle\left(B_{4} B_{2}, C_{2} B_{2}\right)=\angle\left(A_{2} A_{4}, B_{4} A_{4}\right)+\angle\left(B_{4} C_{4}, C_{2} C_{4}\right) . $$  But the lines $A_{2} A_{4}, B_{4} A_{4}, B_{4} C_{4}, C_{2} C_{4}$ coincide respectively with $P A_{4}, C_{5} A_{4}, A_{5} C_{4}, P C_{4}$. So the sums on the right-hand sides of (5) and (6) are equal, leading to equality between their left-hand sides: $\angle\left(A_{5} B_{5}, C_{5} B_{5}\right)=\angle\left(A_{2} B_{2}, C_{2} B_{2}\right)$. Hence (by cyclic shift, once more) also $\angle\left(B_{5} C_{5}, A_{5} C_{5}\right)=\angle\left(B_{2} C_{2}, A_{2} C_{2}\right)$ and $\angle\left(C_{5} A_{5}, B_{5} A_{5}\right)=\angle\left(C_{2} A_{2}, B_{2} A_{2}\right)$. This means that the triangles $A_{5} B_{5} C_{5}$ and $A_{2} B_{2} C_{2}$ have their corresponding angles equal, and consequently they are similar.  Comment 1. This is the way in which the proof has been presented by the proposer. Trying to work it out in the language of classical geometry, so as to avoid oriented angles, one is led to difficulties due to the fact that the reasoning becomes heavily case-dependent. Disposition of relevant points can vary in many respects. Angles which are equal in one case become supplementary in another. Although it seems not hard to translate all formulas from the shapes they have in one situation to the one they have in another, the real trouble is to identify all cases possible and rigorously verify that the key conclusions retain validity in each case.  The use of oriented angles is a very efficient method to omit this trouble. It seems to be the most appropriate environment in which the solution can be elaborated.  Comment 2. Actually, the fact that the circles $\left(A B_{1} C_{1}\right),\left(B C_{1} A_{1}\right)$ and $\left(C A_{1} B_{1}\right)$ have a common point does not require a proof; it is known as Miquel's theorem.


To each side $a$ of a convex polygon we assign the maximum area of a triangle contained in the polygon and having $a$ as one of its sides. Show that the sum of the areas assigned to all sides of the polygon is not less than twice the area of the polygon.  (Serbia)

Solution: Lemma. Every convex $(2 n)$-gon, of area $S$, has a side and a vertex that jointly span a triangle of area not less than $S / n$.  Proof. By main diagonals of the $(2 n)$-gon we shall mean those which partition the $(2 n)$-gon into two polygons with equally many sides. For any side $b$ of the $(2 n)$-gon denote by $\Delta_{b}$ the triangle $A B P$ where $A, B$ are the endpoints of $b$ and $P$ is the intersection point of the main diagonals $A A^{\prime}, B B^{\prime}$. We claim that the union of triangles $\Delta_{b}$, taken over all sides, covers the whole polygon.  To show this, choose any side $A B$ and consider the main diagonal $A A^{\prime}$ as a directed segment. Let $X$ be any point in the polygon, not on any main diagonal. For definiteness, let $X$ lie on the left side of the ray $A A^{\prime}$. Consider the sequence of main diagonals $A A^{\prime}, B B^{\prime}, C C^{\prime}, \ldots$, where $A, B, C, \ldots$ are consecutive vertices, situated right to $A A^{\prime}$.  The $n$-th item in this sequence is the diagonal $A^{\prime} A$ (i.e. $A A^{\prime}$ reversed), having $X$ on its right side. So there are two successive vertices $K, L$ in the sequence $A, B, C, \ldots$ before $A^{\prime}$ such that $X$ still lies to the left of $K K^{\prime}$ but to the right of $L L^{\prime}$. And this means that $X$ is in the triangle $\Delta_{\ell^{\prime}}, \ell^{\prime}=K^{\prime} L^{\prime}$. Analogous reasoning applies to points $X$ on the right of $A A^{\prime}$ (points lying on main diagonals can be safely ignored). Thus indeed the triangles $\Delta_{b}$ jointly cover the whole polygon.  The sum of their areas is no less than $S$. So we can find two opposite sides, say $b=A B$ and $b^{\prime}=A^{\prime} B^{\prime}$ (with $A A^{\prime}, B B^{\prime}$ main diagonals) such that $\left[\Delta_{b}\right]+\left[\Delta_{b^{\prime}}\right] \geq S / n$, where $[\cdots]$ stands for the area of a region. Let $A A^{\prime}, B B^{\prime}$ intersect at $P$; assume without loss of generality that $P B \geq P B^{\prime}$. Then  $$ \left[A B A^{\prime}\right]=[A B P]+\left[P B A^{\prime}\right] \geq[A B P]+\left[P A^{\prime} B^{\prime}\right]=\left[\Delta_{b}\right]+\left[\Delta_{b^{\prime}}\right] \geq S / n $$  proving the lemma.  Now, let $\mathcal{P}$ be any convex polygon, of area $S$, with $m$ sides $a_{1}, \ldots, a_{m}$. Let $S_{i}$ be the area of the greatest triangle in $\mathcal{P}$ with side $a_{i}$. Suppose, contrary to the assertion, that  $$ \sum_{i=1}^{m} \frac{S_{i}}{S}<2 $$  Then there exist rational numbers $q_{1}, \ldots, q_{m}$ such that $\sum q_{i}=2$ and $q_{i}>S_{i} / S$ for each $i$.  Let $n$ be a common denominator of the $m$ fractions $q_{1}, \ldots, q_{m}$. Write $q_{i}=k_{i} / n$; so $\sum k_{i}=2 n$. Partition each side $a_{i}$ of $\mathcal{P}$ into $k_{i}$ equal segments, creating a convex $(2 n)$-gon of area $S$ (with some angles of size $180^{\circ}$ ), to which we apply the lemma. Accordingly, this refined polygon has a side $b$ and a vertex $H$ spanning a triangle $T$ of area $[T] \geq S / n$. If $b$ is a piece of a side $a_{i}$ of $\mathcal{P}$, then the triangle $W$ with base $a_{i}$ and summit $H$ has area  $$ [W]=k_{i} \cdot[T] \geq k_{i} \cdot S / n=q_{i} \cdot S>S_{i}, $$  in contradiction with the definition of $S_{i}$. This ends the proof.


Solution: As in the first solution, we allow again angles of size $180^{\circ}$ at some vertices of the convex polygons considered.  To each convex $n$-gon $\mathcal{P}=A_{1} A_{2} \ldots A_{n}$ we assign a centrally symmetric convex $(2 n)$-gon $\mathcal{Q}$ with side vectors $\pm \overrightarrow{A_{i} A_{i+1}}, 1 \leq i \leq n$. The construction is as follows. Attach the $2 n$ vectors $\pm \overrightarrow{A_{i} A_{i+1}}$ at a common origin and label them $\overrightarrow{\mathbf{b}_{1}}, \overrightarrow{\mathbf{b}_{2}}, \ldots, \overrightarrow{\mathbf{b}_{2 n}}$ in counterclockwise direction; the choice of the first vector $\overrightarrow{\mathbf{b}_{1}}$ is irrelevant. The order of labelling is well-defined if $\mathcal{P}$ has neither parallel sides nor angles equal to $180^{\circ}$. Otherwise several collinear vectors with the same direction are labelled consecutively $\overrightarrow{\mathbf{b}_{j}}, \overrightarrow{\mathbf{b}_{j+1}}, \ldots, \overrightarrow{\mathbf{b}_{j+r}}$. One can assume that in such cases the respective opposite vectors occur in the order $-\overrightarrow{\mathbf{b}_{j}},-\overrightarrow{\mathbf{b}_{j+1}}, \ldots,-\overrightarrow{\mathbf{b}_{j+r}}$, ensuring that $\overrightarrow{\mathbf{b}_{j+n}}=-\overrightarrow{\mathbf{b}_{j}}$ for $j=1, \ldots, 2 n$. Indices are taken cyclically here and in similar situations below.  Choose points $B_{1}, B_{2}, \ldots, B_{2 n}$ satisfying $\overrightarrow{B_{j} B_{j+1}}=\overrightarrow{\mathbf{b}_{j}}$ for $j=1, \ldots, 2 n$. The polygonal line $\mathcal{Q}=B_{1} B_{2} \ldots B_{2 n}$ is closed, since $\sum_{j=1}^{2 n} \overrightarrow{\mathbf{b}_{j}}=\overrightarrow{0}$. Moreover, $\mathcal{Q}$ is a convex $(2 n)$-gon due to the arrangement of the vectors $\overrightarrow{\mathbf{b}_{j}}$, possibly with $180^{\circ}$-angles. The side vectors of $\mathcal{Q}$ are $\pm \overrightarrow{A_{i} A_{i+1}}$, $1 \leq i \leq n$. So in particular $\mathcal{Q}$ is centrally symmetric, because it contains as side vectors $\overrightarrow{A_{i} A_{i+1}}$ and $-\overrightarrow{A_{i} A_{i+1}}$ for each $i=1, \ldots, n$. Note that $B_{j} B_{j+1}$ and $B_{j+n} B_{j+n+1}$ are opposite sides of $\mathcal{Q}$, $1 \leq j \leq n$. We call $\mathcal{Q}$ the associate of $\mathcal{P}$.  Let $S_{i}$ be the maximum area of a triangle with side $A_{i} A_{i+1}$ in $\mathcal{P}, 1 \leq i \leq n$. We prove that  $$ \left[B_{1} B_{2} \ldots B_{2 n}\right]=2 \sum_{i=1}^{n} S_{i} $$  and  $$ \left[B_{1} B_{2} \ldots B_{2 n}\right] \geq 4\left[A_{1} A_{2} \ldots A_{n}\right] $$  It is clear that (1) and (2) imply the conclusion of the original problem.  Lemma. For a side $A_{i} A_{i+1}$ of $\mathcal{P}$, let $h_{i}$ be the maximum distance from a point of $\mathcal{P}$ to line $A_{i} A_{i+1}$, $i=1, \ldots, n$. Denote by $B_{j} B_{j+1}$ the side of $\mathcal{Q}$ such that $\overrightarrow{A_{i} A_{i+1}}=\overrightarrow{B_{j} B_{j+1}}$. Then the distance between $B_{j} B_{j+1}$ and its opposite side in $\mathcal{Q}$ is equal to $2 h_{i}$.  Proof. Choose a vertex $A_{k}$ of $\mathcal{P}$ at distance $h_{i}$ from line $A_{i} A_{i+1}$. Let $\mathbf{u}$ be the unit vector perpendicular to $A_{i} A_{i+1}$ and pointing inside $\mathcal{P}$. Denoting by $\mathbf{x} \cdot \mathbf{y}$ the dot product of vectors $\mathbf{x}$ and $\mathbf{y}$, we have  $$ h=\mathbf{u} \cdot \overrightarrow{A_{i} A_{k}}=\mathbf{u} \cdot\left(\overrightarrow{A_{i} A_{i+1}}+\cdots+\overrightarrow{A_{k-1} A_{k}}\right)=\mathbf{u} \cdot\left(\overrightarrow{A_{i} A_{i-1}}+\cdots+\overrightarrow{A_{k+1} A_{k}}\right) . $$  In $\mathcal{Q}$, the distance $H_{i}$ between the opposite sides $B_{j} B_{j+1}$ and $B_{j+n} B_{j+n+1}$ is given by  $$ H_{i}=\mathbf{u} \cdot\left(\overrightarrow{B_{j} B_{j+1}}+\cdots+\overrightarrow{B_{j+n-1} B_{j+n}}\right)=\mathbf{u} \cdot\left(\overrightarrow{\mathbf{b}_{j}}+\overrightarrow{\mathbf{b}_{j+1}}+\cdots+\overrightarrow{\mathbf{b}_{j+n-1}}\right) . $$  The choice of vertex $A_{k}$ implies that the $n$ consecutive vectors $\overrightarrow{\mathbf{b}_{j}}, \overrightarrow{\mathbf{b}_{j+1}}, \ldots, \overrightarrow{\mathbf{b}_{j+n-1}}$ are precisely $\overrightarrow{A_{i} A_{i+1}}, \ldots, \overrightarrow{A_{k-1} A_{k}}$ and $\overrightarrow{A_{i} A_{i-1}}, \ldots, \overrightarrow{A_{k+1} A_{k}}$, taken in some order. This implies $H_{i}=2 h_{i}$.  For a proof of (1), apply the lemma to each side of $\mathcal{P}$. If $O$ the centre of $\mathcal{Q}$ then, using the notation of the lemma,  $$ \left[B_{j} B_{j+1} O\right]=\left[B_{j+n} B_{j+n+1} O\right]=\left[A_{i} A_{i+1} A_{k}\right]=S_{i} $$  Summation over all sides of $\mathcal{P}$ yields (1).  Set $d(\mathcal{P})=[\mathcal{Q}]-4[\mathcal{P}]$ for a convex polygon $\mathcal{P}$ with associate $\mathcal{Q}$. Inequality $(2)$ means that $d(\mathcal{P}) \geq 0$ for each convex polygon $\mathcal{P}$. The last inequality will be proved by induction on the number $\ell$ of side directions of $\mathcal{P}$, i. e. the number of pairwise nonparallel lines each containing a side of $\mathcal{P}$.  We choose to start the induction with $\ell=1$ as a base case, meaning that certain degenerate polygons are allowed. More exactly, we regard as degenerate convex polygons all closed polygonal lines of the form $X_{1} X_{2} \ldots X_{k} Y_{1} Y_{2} \ldots Y_{m} X_{1}$, where $X_{1}, X_{2}, \ldots, X_{k}$ are points in this order on a line segment $X_{1} Y_{1}$, and so are $Y_{m}, Y_{m-1}, \ldots, Y_{1}$. The initial construction applies to degenerate polygons; their associates are also degenerate, and the value of $d$ is zero. For the inductive step, consider a convex polygon $\mathcal{P}$ which determines $\ell$ side directions, assuming that $d(\mathcal{P}) \geq 0$ for polygons with smaller values of $\ell$.  Suppose first that $\mathcal{P}$ has a pair of parallel sides, i. e. sides on distinct parallel lines. Let $A_{i} A_{i+1}$ and $A_{j} A_{j+1}$ be such a pair, and let $A_{i} A_{i+1} \leq A_{j} A_{j+1}$. Remove from $\mathcal{P}$ the parallelogram $R$ determined by vectors $\overrightarrow{A_{i} A_{i+1}}$ and $\overrightarrow{A_{i} A_{j+1}}$. Two polygons are obtained in this way. Translating one of them by vector $\overrightarrow{A_{i} A_{i+1}}$ yields a new convex polygon $\mathcal{P}^{\prime}$, of area $[\mathcal{P}]-[R]$ and with value of $\ell$ not exceeding the one of $\mathcal{P}$. The construction just described will be called operation A. ![](https://cdn.mathpix.com/cropped/2024_04_17_e55ccc62f43b046023cag-54.jpg?height=464&width=1440&top_left_y=1098&top_left_x=316)  The associate of $\mathcal{P}^{\prime}$ is obtained from $\mathcal{Q}$ upon decreasing the lengths of two opposite sides by an amount of $2 A_{i} A_{i+1}$. By the lemma, the distance between these opposite sides is twice the distance between $A_{i} A_{i+1}$ and $A_{j} A_{j+1}$. Thus operation $\mathbf{A}$ decreases $[\mathcal{Q}]$ by the area of a parallelogram with base and respective altitude twice the ones of $R$, i. e. by $4[R]$. Hence $\mathbf{A}$ leaves the difference $d(\mathcal{P})=[\mathcal{Q}]-4[\mathcal{P}]$ unchanged.  Now, if $\mathcal{P}^{\prime}$ also has a pair of parallel sides, apply operation $\mathbf{A}$ to it. Keep doing so with the subsequent polygons obtained for as long as possible. Now, A decreases the number $p$ of pairs of parallel sides in $\mathcal{P}$. Hence its repeated applications gradually reduce $p$ to 0 , and further applications of $\mathbf{A}$ will be impossible after several steps. For clarity, let us denote by $\mathcal{P}$ again the polygon obtained at that stage.  The inductive step is complete if $\mathcal{P}$ is degenerate. Otherwise $\ell>1$ and $p=0$, i. e. there are no parallel sides in $\mathcal{P}$. Observe that then $\ell \geq 3$. Indeed, $\ell=2$ means that the vertices of $\mathcal{P}$ all lie on the boundary of a parallelogram, implying $p>0$.  Furthermore, since $\mathcal{P}$ has no parallel sides, consecutive collinear vectors in the sequence $\left(\overrightarrow{\mathbf{b}_{k}}\right)$ (if any) correspond to consecutive $180^{\circ}$-angles in $\mathcal{P}$. Removing the vertices of such angles, we obtain a convex polygon with the same value of $d(\mathcal{P})$.  In summary, if operation $\mathbf{A}$ is impossible for a nondegenerate polygon $\mathcal{P}$, then $\ell \geq 3$. In addition, one may assume that $\mathcal{P}$ has no angles of size $180^{\circ}$.  The last two conditions then also hold for the associate $\mathcal{Q}$ of $\mathcal{P}$, and we perform the following construction. Since $\ell \geq 3$, there is a side $B_{j} B_{j+1}$ of $\mathcal{Q}$ such that the sum of the angles at $B_{j}$ and $B_{j+1}$ is greater than $180^{\circ}$. (Such a side exists in each convex $k$-gon for $k>4$.) Naturally, $B_{j+n} B_{j+n+1}$ is a side with the same property. Extend the pairs of sides $B_{j-1} B_{j}, B_{j+1} B_{j+2}$ and $B_{j+n-1} B_{j+n}, B_{j+n+1} B_{j+n+2}$ to meet at $U$ and $V$, respectively. Let $\mathcal{Q}^{\prime}$ be the centrally symmetric convex $2(n+1)$-gon obtained from $\mathcal{Q}$ by inserting $U$ and $V$ into the sequence $B_{1}, \ldots, B_{2 n}$ as new vertices between $B_{j}, B_{j+1}$ and $B_{j+n}, B_{j+n+1}$, respectively. Informally, we adjoin to $\mathcal{Q}$ the congruent triangles $B_{j} B_{j+1} U$ and $B_{j+n} B_{j+n+1} V$. Note that $B_{j}, B_{j+1}, B_{j+n}$ and $B_{j+n+1}$ are kept as vertices of $\mathcal{Q}^{\prime}$, although $B_{j} B_{j+1}$ and $B_{j+n} B_{j+n+1}$ are no longer its sides.  Let $A_{i} A_{i+1}$ be the side of $\mathcal{P}$ such that $\overrightarrow{A_{i} A_{i+1}}=\overrightarrow{B_{j} B_{j+1}}=\overrightarrow{\mathbf{b}_{j}}$. Consider the point $W$ such that triangle $A_{i} A_{i+1} W$ is congruent to triangle $B_{j} B_{j+1} U$ and exterior to $\mathcal{P}$. Insert $W$ into the sequence $A_{1}, A_{2}, \ldots, A_{n}$ as a new vertex between $A_{i}$ and $A_{i+1}$ to obtain an $(n+1)$-gon $\mathcal{P}^{\prime}$. We claim that $\mathcal{P}^{\prime}$ is convex and its associate is $\mathcal{Q}^{\prime}$. ![](https://cdn.mathpix.com/cropped/2024_04_17_e55ccc62f43b046023cag-55.jpg?height=532&width=1388&top_left_y=796&top_left_x=344)  Vectors $\overrightarrow{A_{i} W}$ and $\overrightarrow{\mathbf{b}_{j-1}}$ are collinear and have the same direction, as well as vectors $\overrightarrow{W A_{i+1}}$ and $\overrightarrow{\mathbf{b}_{j+1}}$. Since $\overrightarrow{\mathbf{b}_{j-1}}, \overrightarrow{\mathbf{b}_{j}}, \overrightarrow{\mathbf{b}_{j+1}}$ are consecutive terms in the sequence $\left(\overrightarrow{\mathbf{b}_{k}}\right)$, the angle inequalities $\angle\left(\overrightarrow{\mathbf{b}_{j-1}}, \overrightarrow{\mathbf{b}_{j}}\right) \leq \angle\left(\overrightarrow{A_{i-1} A_{i}}, \overrightarrow{\mathbf{b}_{j}}\right)$ and $\angle\left(\overrightarrow{\mathbf{b}_{j}}, \overrightarrow{\mathbf{b}_{j+1}}\right) \leq \angle\left(\overrightarrow{\mathbf{b}_{j}}, \overrightarrow{A_{i+1} A_{i+2}}\right)$ hold true. They show that $\mathcal{P}^{\prime}$ is a convex polygon. To construct its associate, vectors $\pm \overrightarrow{A_{i} A_{i+1}}= \pm \overrightarrow{\mathbf{b}_{j}}$ must be deleted from the defining sequence $\left(\overrightarrow{\mathbf{b}_{k}}\right.$ ) of $\mathcal{Q}$, and the vectors $\pm \overrightarrow{A_{i} W}, \pm \overrightarrow{W A_{i+1}}$ must be inserted appropriately into it. The latter can be done as follows:  $$ \ldots, \overrightarrow{\mathbf{b}_{j-1}}, \overrightarrow{A_{i} W}, \overrightarrow{W A_{i+1}}, \overrightarrow{\mathbf{b}_{j+1}}, \ldots,-\overrightarrow{\mathbf{b}_{j-1}},-\overrightarrow{A_{i} W},-\overrightarrow{W A_{i+1}},-\overrightarrow{\mathbf{b}_{j+1}}, \ldots $$  This updated sequence produces $\mathcal{Q}^{\prime}$ as the associate of $\mathcal{P}^{\prime}$.  It follows from the construction that $\left[\mathcal{P}^{\prime}\right]=[\mathcal{P}]+\left[A_{i} A_{i+1} W\right]$ and $\left[\mathcal{Q}^{\prime}\right]=[\mathcal{Q}]+2\left[A_{i} A_{i+1} W\right]$. Therefore $d\left(\mathcal{P}^{\prime}\right)=d(\mathcal{P})-2\left[A_{i} A_{i+1} W\right]<d(\mathcal{P})$.  To finish the induction, it remains to notice that the value of $\ell$ for $\mathcal{P}^{\prime}$ is less than the one for $\mathcal{P}$. This is because side $A_{i} A_{i+1}$ was removed. The newly added sides $A_{i} W$ and $W A_{i+1}$ do not introduce new side directions. Each one of them is either parallel to a side of $\mathcal{P}$ or lies on the line determined by such a side. The proof is complete.


Determine all pairs $(x, y)$ of integers satisfying the equation  $$ 1+2^{x}+2^{2 x+1}=y^{2} . $$

Solution: If $(x, y)$ is a solution then obviously $x \geq 0$ and $(x,-y)$ is a solution too. For $x=0$ we get the two solutions $(0,2)$ and $(0,-2)$.  Now let $(x, y)$ be a solution with $x>0$; without loss of generality confine attention to $y>0$. The equation rewritten as  $$ 2^{x}\left(1+2^{x+1}\right)=(y-1)(y+1) $$  shows that the factors $y-1$ and $y+1$ are even, exactly one of them divisible by 4 . Hence $x \geq 3$ and one of these factors is divisible by $2^{x-1}$ but not by $2^{x}$. So  $$ y=2^{x-1} m+\epsilon, \quad m \text { odd }, \quad \epsilon= \pm 1 $$  Plugging this into the original equation we obtain  $$ 2^{x}\left(1+2^{x+1}\right)=\left(2^{x-1} m+\epsilon\right)^{2}-1=2^{2 x-2} m^{2}+2^{x} m \epsilon, $$  or, equivalently  $$ 1+2^{x+1}=2^{x-2} m^{2}+m \epsilon . $$  Therefore  $$ 1-\epsilon m=2^{x-2}\left(m^{2}-8\right) . $$  For $\epsilon=1$ this yields $m^{2}-8 \leq 0$, i.e., $m=1$, which fails to satisfy (2).  For $\epsilon=-1$ equation (2) gives us  $$ 1+m=2^{x-2}\left(m^{2}-8\right) \geq 2\left(m^{2}-8\right), $$  implying $2 m^{2}-m-17 \leq 0$. Hence $m \leq 3$; on the other hand $m$ cannot be 1 by (2). Because $m$ is odd, we obtain $m=3$, leading to $x=4$. From (1) we get $y=23$. These values indeed satisfy the given equation. Recall that then $y=-23$ is also good. Thus we have the complete list of solutions $(x, y):(0,2),(0,-2),(4,23),(4,-23)$.


For $x \in(0,1)$ let $y \in(0,1)$ be the number whose $n$th digit after the decimal point is the $\left(2^{n}\right)$ th digit after the decimal point of $x$. Show that if $x$ is rational then so is $y$.  (Canada)

Solution: Since $x$ is rational, its digits repeat periodically starting at some point. We wish to show that this is also true for the digits of $y$, implying that $y$ is rational.  Let $d$ be the length of the period of $x$ and let $d=2^{u} \cdot v$, where $v$ is odd. There is a positive integer $w$ such that  $$ 2^{w} \equiv 1 \quad(\bmod v) $$  (For instance, one can choose $w$ to be $\varphi(v)$, the value of Euler's function at $v$.) Therefore  $$ 2^{n+w}=2^{n} \cdot 2^{w} \equiv 2^{n} \quad(\bmod v) $$  for each $n$. Also, for $n \geq u$ we have  $$ 2^{n+w} \equiv 2^{n} \equiv 0 \quad\left(\bmod 2^{u}\right) $$  It follows that, for all $n \geq u$, the relation  $$ 2^{n+w} \equiv 2^{n} \quad(\bmod d) $$  holds. Thus, for $n$ sufficiently large, the $2^{n+w}$ th digit of $x$ is in the same spot in the cycle of $x$ as its $2^{n}$ th digit, and so these digits are equal. Hence the $(n+w)$ th digit of $y$ is equal to its $n$th digit. This means that the digits of $y$ repeat periodically with period $w$ from some point on, as required.


The sequence $f(1), f(2), f(3), \ldots$ is defined by  $$ f(n)=\frac{1}{n}\left(\left\lfloor\frac{n}{1}\right\rfloor+\left\lfloor\frac{n}{2}\right\rfloor+\cdots+\left\lfloor\frac{n}{n}\right\rfloor\right), $$  where $\lfloor x\rfloor$ denotes the integer part of $x$.  (a) Prove that $f(n+1)>f(n)$ infinitely often.  (b) Prove that $f(n+1)<f(n)$ infinitely often.  (South Africa)

Solution: Let $g(n)=n f(n)$ for $n \geq 1$ and $g(0)=0$. We note that, for $k=1, \ldots, n$,  $$ \left\lfloor\frac{n}{k}\right\rfloor-\left\lfloor\frac{n-1}{k}\right\rfloor=0 $$  if $k$ is not a divisor of $n$ and  $$ \left\lfloor\frac{n}{k}\right\rfloor-\left\lfloor\frac{n-1}{k}\right\rfloor=1 $$  if $k$ divides $n$. It therefore follows that if $d(n)$ is the number of positive divisors of $n \geq 1$ then  $$ \begin{aligned} g(n) & =\left\lfloor\frac{n}{1}\right\rfloor+\left\lfloor\frac{n}{2}\right\rfloor+\cdots+\left\lfloor\frac{n}{n-1}\right\rfloor+\left\lfloor\frac{n}{n}\right\rfloor \\ & =\left\lfloor\frac{n-1}{1}\right\rfloor+\left\lfloor\frac{n-1}{2}\right\rfloor+\cdots+\left\lfloor\frac{n-1}{n-1}\right\rfloor+\left\lfloor\frac{n-1}{n}\right\rfloor+d(n) \\ & =g(n-1)+d(n) . \end{aligned} $$  Hence  $$ g(n)=g(n-1)+d(n)=g(n-2)+d(n-1)+d(n)=\cdots=d(1)+d(2)+\cdots+d(n), $$  meaning that  $$ f(n)=\frac{d(1)+d(2)+\cdots+d(n)}{n} . $$  In other words, $f(n)$ is equal to the arithmetic mean of $d(1), d(2), \ldots, d(n)$. In order to prove the claims, it is therefore sufficient to show that $d(n+1)>f(n)$ and $d(n+1)<f(n)$ both hold infinitely often.  We note that $d(1)=1$. For $n>1, d(n) \geq 2$ holds, with equality if and only if $n$ is prime. Since $f(6)=7 / 3>2$, it follows that $f(n)>2$ holds for all $n \geq 6$.  Since there are infinitely many primes, $d(n+1)=2$ holds for infinitely many values of $n$, and for each such $n \geq 6$ we have $d(n+1)=2<f(n)$. This proves claim (b).  To prove (a), notice that the sequence $d(1), d(2), d(3), \ldots$ is unbounded (e. g. $d\left(2^{k}\right)=k+1$ for all $k$ ). Hence $d(n+1)>\max \{d(1), d(2), \ldots, d(n)\}$ for infinitely many $n$. For all such $n$, we have $d(n+1)>f(n)$. This completes the solution.


Let $P$ be a polynomial of degree $n>1$ with integer coefficients and let $k$ be any positive integer. Consider the polynomial $Q(x)=P(P(\ldots P(P(x)) \ldots))$, with $k$ pairs of parentheses. Prove that $Q$ has no more than $n$ integer fixed points, i.e. integers satisfying the equation $Q(x)=x$.  (Romania)

Solution: The claim is obvious if every integer fixed point of $Q$ is a fixed point of $P$ itself. For the sequel assume that this is not the case. Take any integer $x_{0}$ such that $Q\left(x_{0}\right)=x_{0}$, $P\left(x_{0}\right) \neq x_{0}$ and define inductively $x_{i+1}=P\left(x_{i}\right)$ for $i=0,1,2, \ldots$; then $x_{k}=x_{0}$.  It is evident that  $$ P(u)-P(v) \text { is divisible by } u-v \text { for distinct integers } u, v \text {. } $$  (Indeed, if $P(x)=\sum a_{i} x^{i}$ then each $a_{i}\left(u^{i}-v^{i}\right)$ is divisible by $u-v$.) Therefore each term in the chain of (nonzero) differences  $$ x_{0}-x_{1}, \quad x_{1}-x_{2}, \quad \ldots, \quad x_{k-1}-x_{k}, \quad x_{k}-x_{k+1} $$  is a divisor of the next one; and since $x_{k}-x_{k+1}=x_{0}-x_{1}$, all these differences have equal absolute values. For $x_{m}=\min \left(x_{1}, \ldots, x_{k}\right)$ this means that $x_{m-1}-x_{m}=-\left(x_{m}-x_{m+1}\right)$. Thus $x_{m-1}=x_{m+1}\left(\neq x_{m}\right)$. It follows that consecutive differences in the sequence (2) have opposite signs. Consequently, $x_{0}, x_{1}, x_{2}, \ldots$ is an alternating sequence of two distinct values. In other words, every integer fixed point of $Q$ is a fixed point of the polynomial $P(P(x))$. Our task is to prove that there are at most $n$ such points.  Let $a$ be one of them so that $b=P(a) \neq a$ (we have assumed that such an $a$ exists); then $a=P(b)$. Take any other integer fixed point $\alpha$ of $P(P(x))$ and let $P(\alpha)=\beta$, so that $P(\beta)=\alpha$; the numbers $\alpha$ and $\beta$ need not be distinct ( $\alpha$ can be a fixed point of $P$ ), but each of $\alpha, \beta$ is different from each of $a, b$. Applying property (1) to the four pairs of integers $(\alpha, a),(\beta, b)$, $(\alpha, b),(\beta, a)$ we get that the numbers $\alpha-a$ and $\beta-b$ divide each other, and also $\alpha-b$ and $\beta-a$ divide each other. Consequently  $$ \alpha-b= \pm(\beta-a), \quad \alpha-a= \pm(\beta-b) . $$  Suppose we have a plus in both instances: $\alpha-b=\beta-a$ and $\alpha-a=\beta-b$. Subtraction yields $a-b=b-a$, a contradiction, as $a \neq b$. Therefore at least one equality in (3) holds with a minus sign. For each of them this means that $\alpha+\beta=a+b$; equivalently $a+b-\alpha-P(\alpha)=0$.  Denote $a+b$ by $C$. We have shown that every integer fixed point of $Q$ other that $a$ and $b$ is a root of the polynomial $F(x)=C-x-P(x)$. This is of course true for $a$ and $b$ as well. And since $P$ has degree $n>1$, the polynomial $F$ has the same degree, so it cannot have more than $n$ roots. Hence the result.  Comment. The first part of the solution, showing that integer fixed points of any iterate of $P$ are in fact fixed points of the second iterate $P \circ P$ is standard; moreover, this fact has already appeared in contests. We however do not consider this as a major drawback to the problem because the only tricky moment comes up only at the next stage of the reasoning - to apply the divisibility property (1) to points from distinct 2-orbits of $P$. Yet maybe it would be more appropriate to state the problem in a version involving $k=2$ only.


Find all integer solutions of the equation  $$ \frac{x^{7}-1}{x-1}=y^{5}-1 $$  (Russia)

Solution: The equation has no integer solutions. To show this, we first prove a lemma.  Lemma. If $x$ is an integer and $p$ is a prime divisor of $\frac{x^{7}-1}{x-1}$ then either $p \equiv 1(\bmod 7)$ or $p=7$.  Proof. Both $x^{7}-1$ and $x^{p-1}-1$ are divisible by $p$, by hypothesis and by Fermat's little theorem, respectively. Suppose that 7 does not divide $p-1$. Then $\operatorname{gcd}(p-1,7)=1$, so there exist integers $k$ and $m$ such that $7 k+(p-1) m=1$. We therefore have  $$ x \equiv x^{7 k+(p-1) m} \equiv\left(x^{7}\right)^{k} \cdot\left(x^{p-1}\right)^{m} \equiv 1 \quad(\bmod p) $$  and so  $$ \frac{x^{7}-1}{x-1}=1+x+\cdots+x^{6} \equiv 7 \quad(\bmod p) $$  It follows that $p$ divides 7 , hence $p=7$ must hold if $p \equiv 1(\bmod 7)$ does not, as stated.  The lemma shows that each positive divisor $d$ of $\frac{x^{7}-1}{x-1}$ satisfies either $d \equiv 0(\bmod 7)$ or $d \equiv 1(\bmod 7)$.  Now assume that $(x, y)$ is an integer solution of the original equation. Notice that $y-1>0$, because $\frac{x^{7}-1}{x-1}>0$ for all $x \neq 1$. Since $y-1$ divides $\frac{x^{7}-1}{x-1}=y^{5}-1$, we have $y \equiv 1(\bmod 7)$ or $y \equiv 2(\bmod 7)$ by the previous paragraph. In the first case, $1+y+y^{2}+y^{3}+y^{4} \equiv 5(\bmod 7)$, and in the second $1+y+y^{2}+y^{3}+y^{4} \equiv 3(\bmod 7)$. Both possibilities contradict the fact that the positive divisor $1+y+y^{2}+y^{3}+y^{4}$ of $\frac{x^{7}-1}{x-1}$ is congruent to 0 or 1 modulo 7 . So the given equation has no integer solutions.


Let $a>b>1$ be relatively prime positive integers. Define the weight of an integer $c$, denoted by $w(c)$, to be the minimal possible value of $|x|+|y|$ taken over all pairs of integers $x$ and $y$ such that  $$ a x+b y=c . $$  An integer $c$ is called a local champion if $w(c) \geq w(c \pm a)$ and $w(c) \geq w(c \pm b)$.  Find all local champions and determine their number.

Solution: Call the pair of integers $(x, y)$ a representation of $c$ if $a x+b y=c$ and $|x|+|y|$ has the smallest possible value, i.e. $|x|+|y|=w(c)$.  We characterise the local champions by the following three observations.  Lemma 1. If $(x, y)$ a representation of a local champion $c$ then $x y<0$.  Proof. Suppose indirectly that $x \geq 0$ and $y \geq 0$ and consider the values $w(c)$ and $w(c+a)$. All representations of the numbers $c$ and $c+a$ in the form $a u+b v$ can be written as  $$ c=a(x-k b)+b(y+k a), \quad c+a=a(x+1-k b)+b(y+k a) $$  where $k$ is an arbitrary integer.  Since $|x|+|y|$ is minimal, we have  $$ x+y=|x|+|y| \leq|x-k b|+|y+k a| $$  for all $k$. On the other hand, $w(c+a) \leq w(c)$, so there exists a $k$ for which  $$ |x+1-k b|+|y+k a| \leq|x|+|y|=x+y . $$  Then  $$ (x+1-k b)+(y+k a) \leq|x+1-k b|+|y+k a| \leq x+y \leq|x-k b|+|y+k a| . $$  Comparing the first and the third expressions, we find $k(a-b)+1 \leq 0$ implying $k<0$. Comparing the second and fourth expressions, we get $|x+1-k b| \leq|x-k b|$, therefore $k b>x$; this is a contradiction.  If $x, y \leq 0$ then we can switch to $-c,-x$ and $-y$.  From this point, write $c=a x-b y$ instead of $c=a x+b y$ and consider only those cases where $x$ and $y$ are nonzero and have the same sign. By Lemma 1, there is no loss of generality in doing so.  Lemma 2. Let $c=a x-b y$ where $|x|+|y|$ is minimal and $x, y$ have the same sign. The number $c$ is a local champion if and only if $|x|<b$ and $|x|+|y|=\left\lfloor\frac{a+b}{2}\right\rfloor$.  Proof. Without loss of generality we may assume $x, y>0$.  The numbers $c-a$ and $c+b$ can be written as  $$ c-a=a(x-1)-b y \quad \text { and } \quad c+b=a x-b(y-1) $$  and trivially $w(c-a) \leq(x-1)+y<w(c)$ and $w(c+b) \leq x+(y-1)<w(c)$ in all cases.  Now assume that $c$ is a local champion and consider $w(c+a)$. Since $w(c+a) \leq w(c)$, there exists an integer $k$ such that  $$ c+a=a(x+1-k b)-b(y-k a) \text { and } \quad|x+1-k b|+|y-k a| \leq x+y \text {. } $$  This inequality cannot hold if $k \leq 0$, therefore $k>0$. We prove that we can choose $k=1$.  Consider the function $f(t)=|x+1-b t|+|y-a t|-(x+y)$. This is a convex function and we have $f(0)=1$ and $f(k) \leq 0$. By Jensen's inequality, $f(1) \leq\left(1-\frac{1}{k}\right) f(0)+\frac{1}{k} f(k)<1$. But $f(1)$ is an integer. Therefore $f(1) \leq 0$ and  $$ |x+1-b|+|y-a| \leq x+y . $$  Knowing $c=a(x-b)-b(y-a)$, we also have  $$ x+y \leq|x-b|+|y-a| . $$  Combining the two inequalities yields $|x+1-b| \leq|x-b|$ which is equivalent to $x<b$.  Considering $w(c-b)$, we obtain similarly that $y<a$.  Now $|x-b|=b-x,|x+1-b|=b-x-1$ and $|y-a|=a-y$, therefore we have  $$ \begin{gathered} (b-x-1)+(a-y) \leq x+y \leq(b-x)+(a-y), \\ \frac{a+b-1}{2} \leq x+y \leq \frac{a+b}{2} . \end{gathered} $$  Hence $x+y=\left\lfloor\frac{a+b}{2}\right\rfloor$.  To prove the opposite direction, assume $0<x<b$ and $x+y=\left\lfloor\frac{a+b}{2}\right\rfloor$. Since $a>b$, we also have $0<y<a$. Then  $$ w(c+a) \leq|x+1-b|+|y-a|=a+b-1-(x+y) \leq x+y=w(c) $$  and  $$ w(c-b) \leq|x-b|+|y+1-a|=a+b-1-(x+y) \leq x+y=w(c) $$  therefore $c$ is a local champion indeed.  Lemma 3. Let $c=a x-b y$ and assume that $x$ and $y$ have the same sign, $|x|<b,|y|<a$ and $|x|+|y|=\left\lfloor\frac{a+b}{2}\right\rfloor$. Then $w(c)=x+y$.  Proof. By definition $w(c)=\min \{|x-k b|+|y-k a|: k \in \mathbb{Z}\}$. If $k \leq 0$ then obviously $|x-k b|+|y-k a| \geq x+y$. If $k \geq 1$ then  $$ |x-k b|+|y-k a|=(k b-x)+(k a-y)=k(a+b)-(x+y) \geq(2 k-1)(x+y) \geq x+y . $$  Therefore $w(c)=x+y$ indeed.  Lemmas 1, 2 and 3 together yield that the set of local champions is  $$ C=\left\{ \pm(a x-b y): 0<x<b, x+y=\left\lfloor\frac{a+b}{2}\right\rfloor\right\} \text {. } $$  Denote by $C^{+}$and $C^{-}$the two sets generated by the expressions $+(a x-b y)$ and $-(a x-b y)$, respectively. It is easy to see that both sets are arithmetic progressions of length $b-1$, with difference $a+b$.  If $a$ and $b$ are odd, then $C^{+}=C^{-}$, because $a(-x)-b(-y)=a(b-x)-b(a-y)$ and $x+y=\frac{a+b}{2}$ is equivalent to $(b-x)+(a-y)=\frac{a+b}{2}$. In this case there exist $b-1$ local champions.  If $a$ and $b$ have opposite parities then the answer is different. For any $c_{1} \in C^{+}$and $c_{2} \in C^{-}$,  $$ 2 c_{1} \equiv-2 c_{2} \equiv 2\left(a \frac{a+b-1}{2}-b \cdot 0\right) \equiv-a \quad(\bmod a+b) $$  and  $$ 2 c_{1}-2 c_{2} \equiv-2 a \quad(\bmod a+b) $$  The number $a+b$ is odd and relatively prime to $a$, therefore the elements of $C^{+}$and $C^{-}$belong to two different residue classes modulo $a+b$. Hence, the set $C$ is the union of two disjoint arithmetic progressions and the number of all local champions is $2(b-1)$.  So the number of local champions is $b-1$ if both $a$ and $b$ are odd and $2(b-1)$ otherwise.  Comment. The original question, as stated by the proposer, was:  (a) Show that there exists only finitely many local champions;  (b) Show that there exists at least one local champion.


Prove that, for every positive integer $n$, there exists an integer $m$ such that $2^{m}+m$ is divisible by $n$.  (Estonia)

Solution: We will prove by induction on $d$ that, for every positive integer $N$, there exist positive integers $b_{0}, b_{1}, \ldots, b_{d-1}$ such that, for each $i=0,1,2, \ldots, d-1$, we have $b_{i}>N$ and  $$ 2^{b_{i}}+b_{i} \equiv i \quad(\bmod d) . $$  This yields the claim for $m=b_{0}$.  The base case $d=1$ is trivial. Take an $a>1$ and assume that the statement holds for all $d<a$. Note that the remainders of $2^{i}$ modulo $a$ repeat periodically starting with some exponent $M$. Let $k$ be the length of the period; this means that $2^{M+k^{\prime}} \equiv 2^{M}(\bmod a)$ holds only for those $k^{\prime}$ which are multiples of $k$. Note further that the period cannot contain all the $a$ remainders, since 0 either is missing or is the only number in the period. Thus $k<a$.  Let $d=\operatorname{gcd}(a, k)$ and let $a^{\prime}=a / d, k^{\prime}=k / d$. Since $0<k<a$, we also have $0<d<a$. By the induction hypothesis, there exist positive integers $b_{0}, b_{1}, \ldots, b_{d-1}$ such that $b_{i}>\max \left(2^{M}, N\right)$ and  $$ 2^{b_{i}}+b_{i} \equiv i \quad(\bmod d) \quad \text { for } \quad i=0,1,2, \ldots, d-1 $$  For each $i=0,1, \ldots, d-1$ consider the sequence  $$ 2^{b_{i}}+b_{i}, \quad 2^{b_{i}+k}+\left(b_{i}+k\right), \ldots, \quad 2^{b_{i}+\left(a^{\prime}-1\right) k}+\left(b_{i}+\left(a^{\prime}-1\right) k\right) . $$  Modulo $a$, these numbers are congruent to  $$ 2^{b_{i}}+b_{i}, \quad 2^{b_{i}}+\left(b_{i}+k\right), \ldots, \quad 2^{b_{i}}+\left(b_{i}+\left(a^{\prime}-1\right) k\right), $$  respectively. The $d$ sequences contain $a^{\prime} d=a$ numbers altogether. We shall now prove that no two of these numbers are congruent modulo $a$.  Suppose that  $$ 2^{b_{i}}+\left(b_{i}+m k\right) \equiv 2^{b_{j}}+\left(b_{j}+n k\right) \quad(\bmod a) $$  for some values of $i, j \in\{0,1, \ldots, d-1\}$ and $m, n \in\left\{0,1, \ldots, a^{\prime}-1\right\}$. Since $d$ is a divisor of $a$, we also have  $$ 2^{b_{i}}+\left(b_{i}+m k\right) \equiv 2^{b_{j}}+\left(b_{j}+n k\right) \quad(\bmod d) . $$  Because $d$ is a divisor of $k$ and in view of $(1)$, we obtain $i \equiv j(\bmod d)$. As $i, j \in\{0,1, \ldots, d-1\}$, this just means that $i=j$. Substituting this into (3) yields $m k \equiv n k(\bmod a)$. Therefore $m k^{\prime} \equiv n k^{\prime}\left(\bmod a^{\prime}\right)$; and since $a^{\prime}$ and $k^{\prime}$ are coprime, we get $m \equiv n\left(\bmod a^{\prime}\right)$. Hence also $m=n$.  It follows that the $a$ numbers that make up the $d$ sequences (2) satisfy all the requirements; they are certainly all greater than $N$ because we chose each $b_{i}>\max \left(2^{M}, N\right)$. So the statement holds for $a$, completing the induction.