Solution: Like in the previous solutions, we do not need all circle pairs to intersect but we assume that the circles form a connected set. Denote by $\mathcal{C}$ and $\mathcal{P}$ the sets of circles and their intersection points, respectively.  The circles divide the plane into several simply connected, bounded regions and one unbounded region. Denote the set of these regions by $\mathcal{R}$. We say that an intersection point or a region is odd or even if it is contained inside an odd or even number of circles, respectively. Let $\mathcal{P}_{\text {odd }}$ and $\mathcal{R}_{\text {odd }}$ be the sets of odd intersection points and odd regions, respectively.  Claim.  $$ \left|\mathcal{R}_{\text {odd }}\right|-\left|\mathcal{P}_{\text {odd }}\right| \equiv n \quad(\bmod 2) . $$  Proof. For each circle $c \in \mathcal{C}$, denote by $R_{c}, P_{c}$, and $X_{c}$ the number of regions inside $c$, the number of intersection points inside $c$, and the number of circles intersecting $c$, respectively. The circles divide each other into several arcs; denote by $A_{c}$ the number of such arcs inside $c$. By double counting the regions and intersection points inside the circles we get  $$ \left|\mathcal{R}_{\mathrm{odd}}\right| \equiv \sum_{c \in \mathcal{C}} R_{c} \quad(\bmod 2) \quad \text { and } \quad\left|\mathcal{P}_{\mathrm{odd}}\right| \equiv \sum_{c \in \mathcal{C}} P_{c} \quad(\bmod 2) . $$  For each circle $c$, apply EULER's polyhedron theorem to the (simply connected) regions in $c$. There are $2 X_{c}$ intersection points on $c$; they divide the circle into $2 X_{c}$ arcs. The polyhedron theorem yields $\left(R_{c}+1\right)+\left(P_{c}+2 X_{c}\right)=\left(A_{c}+2 X_{c}\right)+2$, considering the exterior of $c$ as a single region. Therefore,  $$ R_{c}+P_{c}=A_{c}+1 $$  Moreover, we have four arcs starting from every interior points inside $c$ and a single arc starting into the interior from each intersection point on the circle. By double-counting the end-points of the interior arcs we get $2 A_{c}=4 P_{c}+2 X_{c}$, so  $$ A_{c}=2 P_{c}+X_{c} $$  The relations (2) and (3) together yield  $$ R_{c}-P_{c}=X_{c}+1 $$  By summing up (4) for all circles we obtain  $$ \sum_{c \in \mathcal{C}} R_{c}-\sum_{c \in \mathcal{C}} P_{c}=\sum_{c \in \mathcal{C}} X_{c}+|\mathcal{C}| $$  which yields  $$ \left|\mathcal{R}_{\mathrm{odd}}\right|-\left|\mathcal{P}_{\mathrm{odd}}\right| \equiv \sum_{c \in \mathcal{C}} X_{c}+n \quad(\bmod 2) $$  Notice that in $\sum_{c \in \mathcal{C}} X_{c}$ each intersecting circle pair is counted twice, i.e., for both circles in the pair, so  $$ \sum_{c \in \mathcal{C}} X_{c} \equiv 0 \quad(\bmod 2) $$  which finishes the proof of the Claim. Now insert the same small arcs at the intersections as in the first solution, and suppose that there is a single snail orbit $b$.  First we show that the odd regions are inside the curve $b$, while the even regions are outside. Take a region $r \in \mathcal{R}$ and a point $x$ in its interior, and draw a ray $y$, starting from $x$, that does not pass through any intersection point of the circles and is neither tangent to any of the circles. As is well-known, $x$ is inside the curve $b$ if and only if $y$ intersects $b$ an odd number of times (see Figure 10). Notice that if an arbitrary circle $c$ contains $x$ in its interior, then $c$ intersects $y$ at a single point; otherwise, if $x$ is outside $c$, then $c$ has 2 or 0 intersections with $y$. Therefore, $y$ intersects $b$ an odd number of times if and only if $x$ is contained in an odd number of circles, so if and only if $r$ is odd. ![](https://cdn.mathpix.com/cropped/2024_11_18_62c85fb0b01b41aca29bg-49.jpg?height=321&width=626&top_left_y=2255&top_left_x=715)  Figure 10 Now consider an intersection point $p$ of two circles $c_{1}$ and $c_{2}$ and a small neighbourhood around $p$. Suppose that $p$ is contained inside $k$ circles.  We have four regions that meet at $p$. Let $r_{1}$ be the region that lies outside both $c_{1}$ and $c_{2}$, let $r_{2}$ be the region that lies inside both $c_{1}$ and $c_{2}$, and let $r_{3}$ and $r_{4}$ be the two remaining regions, each lying inside exactly one of $c_{1}$ and $c_{2}$. The region $r_{1}$ is contained inside the same $k$ circles as $p$; the region $r_{2}$ is contained also by $c_{1}$ and $c_{2}$, so by $k+2$ circles in total; each of the regions $r_{3}$ and $r_{4}$ is contained inside $k+1$ circles. After the small arcs have been inserted at $p$, the regions $r_{1}$ and $r_{2}$ get connected, and the regions $r_{3}$ and $r_{4}$ remain separated at $p$ (see Figure 11). If $p$ is an odd point, then $r_{1}$ and $r_{2}$ are odd, so two odd regions are connected at $p$. Otherwise, if $p$ is even, then we have two even regions connected at $p$. ![](https://cdn.mathpix.com/cropped/2024_11_18_62c85fb0b01b41aca29bg-50.jpg?height=261&width=298&top_left_y=658&top_left_x=571)  Figure 11 ![](https://cdn.mathpix.com/cropped/2024_11_18_62c85fb0b01b41aca29bg-50.jpg?height=286&width=481&top_left_y=645&top_left_x=1007)  Figure 12  Consider the system of odd regions and their connections at the odd points as a graph. In this graph the odd regions are the vertices, and each odd point establishes an edge that connects two vertices (see Figure 12). As $b$ is a single closed curve, this graph is connected and contains no cycle, so the graph is a tree. Then the number of vertices must be by one greater than the number of edges, so  $$ \left|\mathcal{R}_{\text {odd }}\right|-\left|\mathcal{P}_{\text {odd }}\right|=1 . $$  The relations (1) and (9) together prove that $n$ must be odd.  Comment. For every odd $n$ there exists at least one configuration of $n$ circles with a single snail orbit. Figure 13 shows a possible configuration with 5 circles. In general, if a circle is rotated by $k \cdot \frac{360^{\circ}}{n}$ $(k=1,2, \ldots, n-1)$ around an interior point other than the centre, the circle and its rotated copies together provide a single snail orbit. ![](https://cdn.mathpix.com/cropped/2024_11_18_62c85fb0b01b41aca29bg-50.jpg?height=584&width=595&top_left_y=1641&top_left_x=736)  Figure 13


The points $P$ and $Q$ are chosen on the side $B C$ of an acute-angled triangle $A B C$ so that $\angle P A B=\angle A C B$ and $\angle Q A C=\angle C B A$. The points $M$ and $N$ are taken on the rays $A P$ and $A Q$, respectively, so that $A P=P M$ and $A Q=Q N$. Prove that the lines $B M$ and $C N$ intersect on the circumcircle of the triangle $A B C$. (Georgia)

Solution: Denote by $S$ the intersection point of the lines $B M$ and $C N$. Let moreover $\beta=\angle Q A C=\angle C B A$ and $\gamma=\angle P A B=\angle A C B$. From these equalities it follows that the triangles $A B P$ and $C A Q$ are similar (see Figure 1). Therefore we obtain  $$ \frac{B P}{P M}=\frac{B P}{P A}=\frac{A Q}{Q C}=\frac{N Q}{Q C} $$  Moreover,  $$ \angle B P M=\beta+\gamma=\angle C Q N $$  Hence the triangles $B P M$ and $N Q C$ are similar. This gives $\angle B M P=\angle N C Q$, so the triangles $B P M$ and $B S C$ are also similar. Thus we get  $$ \angle C S B=\angle B P M=\beta+\gamma=180^{\circ}-\angle B A C, $$  which completes the solution. ![](https://cdn.mathpix.com/cropped/2024_11_18_62c85fb0b01b41aca29bg-51.jpg?height=618&width=595&top_left_y=1299&top_left_x=374)  Figure 1 ![](https://cdn.mathpix.com/cropped/2024_11_18_62c85fb0b01b41aca29bg-51.jpg?height=643&width=624&top_left_y=1272&top_left_x=1090)  Figure 2


Solution: As in the previous solution, denote by $S$ the intersection point of the lines $B M$ and $N C$. Let moreover the circumcircle of the triangle $A B C$ intersect the lines $A P$ and $A Q$ again at $K$ and $L$, respectively (see Figure 2).  Note that $\angle L B C=\angle L A C=\angle C B A$ and similarly $\angle K C B=\angle K A B=\angle B C A$. It implies that the lines $B L$ and $C K$ meet at a point $X$, being symmetric to the point $A$ with respect to the line $B C$. Since $A P=P M$ and $A Q=Q N$, it follows that $X$ lies on the line $M N$. Therefore, using Pascal's theorem for the hexagon $A L B S C K$, we infer that $S$ lies on the circumcircle of the triangle $A B C$, which finishes the proof.  Comment. Both solutions can be modified to obtain a more general result, with the equalities  $$ A P=P M \quad \text { and } \quad A Q=Q N $$  replaced by  $$ \frac{A P}{P M}=\frac{Q N}{A Q} $$


Let $A B C$ be a triangle. The points $K, L$, and $M$ lie on the segments $B C, C A$, and $A B$, respectively, such that the lines $A K, B L$, and $C M$ intersect in a common point. Prove that it is possible to choose two of the triangles $A L M, B M K$, and $C K L$ whose inradii sum up to at least the inradius of the triangle $A B C$. (Estonia)

Solution: Denote  $$ a=\frac{B K}{K C}, \quad b=\frac{C L}{L A}, \quad c=\frac{A M}{M B} . $$  By Ceva's theorem, $a b c=1$, so we may, without loss of generality, assume that $a \geqslant 1$. Then at least one of the numbers $b$ or $c$ is not greater than 1 . Therefore at least one of the pairs $(a, b)$, $(b, c)$ has its first component not less than 1 and the second one not greater than 1 . Without loss of generality, assume that $1 \leqslant a$ and $b \leqslant 1$.  Therefore, we obtain $b c \leqslant 1$ and $1 \leqslant c a$, or equivalently  $$ \frac{A M}{M B} \leqslant \frac{L A}{C L} \quad \text { and } \quad \frac{M B}{A M} \leqslant \frac{B K}{K C} . $$  The first inequality implies that the line passing through $M$ and parallel to $B C$ intersects the segment $A L$ at a point $X$ (see Figure 1). Therefore the inradius of the triangle $A L M$ is not less than the inradius $r_{1}$ of triangle $A M X$.  Similarly, the line passing through $M$ and parallel to $A C$ intersects the segment $B K$ at a point $Y$, so the inradius of the triangle $B M K$ is not less than the inradius $r_{2}$ of the triangle $B M Y$. Thus, to complete our solution, it is enough to show that $r_{1}+r_{2} \geqslant r$, where $r$ is the inradius of the triangle $A B C$. We prove that in fact $r_{1}+r_{2}=r$. ![](https://cdn.mathpix.com/cropped/2024_11_18_62c85fb0b01b41aca29bg-52.jpg?height=597&width=849&top_left_y=1369&top_left_x=621)  Figure 1 Since $M X \| B C$, the dilation with centre $A$ that takes $M$ to $B$ takes the incircle of the triangle $A M X$ to the incircle of the triangle $A B C$. Therefore  $$ \frac{r_{1}}{r}=\frac{A M}{A B}, \quad \text { and similarly } \quad \frac{r_{2}}{r}=\frac{M B}{A B} . $$  Adding these equalities gives $r_{1}+r_{2}=r$, as required. Comment. Alternatively, one can use Desargues' theorem instead of Ceva's theorem, as follows: The lines $A B, B C, C A$ dissect the plane into seven regions. One of them is bounded, and amongst the other six, three are two-sided and three are three-sided. Now define the points $P=B C \cap L M$, $Q=C A \cap M K$, and $R=A B \cap K L$ (in the projective plane). By Desargues' theorem, the points $P$, $Q, R$ lie on a common line $\ell$. This line intersects only unbounded regions. If we now assume (without loss of generality) that $P, Q$ and $R$ lie on $\ell$ in that order, then one of the segments $P Q$ or $Q R$ lies inside a two-sided region. If, for example, this segment is $P Q$, then the triangles $A L M$ and $B M K$ will satisfy the statement of the problem for the same reason.


Let $\Omega$ and $O$ be the circumcircle and the circumcentre of an acute-angled triangle $A B C$ with $A B>B C$. The angle bisector of $\angle A B C$ intersects $\Omega$ at $M \neq B$. Let $\Gamma$ be the circle with diameter $B M$. The angle bisectors of $\angle A O B$ and $\angle B O C$ intersect $\Gamma$ at points $P$ and $Q$, respectively. The point $R$ is chosen on the line $P Q$ so that $B R=M R$. Prove that $B R \| A C$.  (Russia)

Solution: Let $K$ be the midpoint of $B M$, i.e., the centre of $\Gamma$. Notice that $A B \neq B C$ implies $K \neq O$. Clearly, the lines $O M$ and $O K$ are the perpendicular bisectors of $A C$ and $B M$, respectively. Therefore, $R$ is the intersection point of $P Q$ and $O K$.  Let $N$ be the second point of intersection of $\Gamma$ with the line $O M$. Since $B M$ is a diameter of $\Gamma$, the lines $B N$ and $A C$ are both perpendicular to $O M$. Hence $B N \| A C$, and it suffices to prove that $B N$ passes through $R$. Our plan for doing this is to interpret the lines $B N, O K$, and $P Q$ as the radical axes of three appropriate circles.  Let $\omega$ be the circle with diameter $B O$. Since $\angle B N O=\angle B K O=90^{\circ}$, the points $N$ and $K$ lie on $\omega$.  Next we show that the points $O, K, P$, and $Q$ are concyclic. To this end, let $D$ and $E$ be the midpoints of $B C$ and $A B$, respectively. Clearly, $D$ and $E$ lie on the rays $O Q$ and $O P$, respectively. By our assumptions about the triangle $A B C$, the points $B, E, O, K$, and $D$ lie in this order on $\omega$. It follows that $\angle E O R=\angle E B K=\angle K B D=\angle K O D$, so the line $K O$ externally bisects the angle $P O Q$. Since the point $K$ is the centre of $\Gamma$, it also lies on the perpendicular bisector of $P Q$. So $K$ coincides with the midpoint of the $\operatorname{arc} P O Q$ of the circumcircle $\gamma$ of triangle $P O Q$.  Thus the lines $O K, B N$, and $P Q$ are pairwise radical axes of the circles $\omega, \gamma$, and $\Gamma$. Hence they are concurrent at $R$, as required. ![](https://cdn.mathpix.com/cropped/2024_11_18_62c85fb0b01b41aca29bg-53.jpg?height=877&width=1429&top_left_y=1466&top_left_x=316)


Consider a fixed circle $\Gamma$ with three fixed points $A, B$, and $C$ on it. Also, let us fix a real number $\lambda \in(0,1)$. For a variable point $P \notin\{A, B, C\}$ on $\Gamma$, let $M$ be the point on the segment $C P$ such that $C M=\lambda \cdot C P$. Let $Q$ be the second point of intersection of the circumcircles of the triangles $A M P$ and $B M C$. Prove that as $P$ varies, the point $Q$ lies on a fixed circle. (United Kingdom)

Solution: Throughout the solution, we denote by $\Varangle(a, b)$ the directed angle between the lines $a$ and $b$.  Let $D$ be the point on the segment $A B$ such that $B D=\lambda \cdot B A$. We will show that either $Q=D$, or $\Varangle(D Q, Q B)=\Varangle(A B, B C)$; this would mean that the point $Q$ varies over the constant circle through $D$ tangent to $B C$ at $B$, as required.  Denote the circumcircles of the triangles $A M P$ and $B M C$ by $\omega_{A}$ and $\omega_{B}$, respectively. The lines $A P, B C$, and $M Q$ are pairwise radical axes of the circles $\Gamma, \omega_{A}$, and $\omega_{B}$, thus either they are parallel, or they share a common point $X$.  Assume that these lines are parallel (see Figure 1). Then the segments $A P, Q M$, and $B C$ have a common perpendicular bisector; the reflection in this bisector maps the segment $C P$ to $B A$, and maps $M$ to $Q$. Therefore, in this case $Q$ lies on $A B$, and $B Q / A B=C M / C P=$ $B D / A B$; so we have $Q=D$. ![](https://cdn.mathpix.com/cropped/2024_11_18_62c85fb0b01b41aca29bg-54.jpg?height=644&width=549&top_left_y=1300&top_left_x=225)  Figure 1 ![](https://cdn.mathpix.com/cropped/2024_11_18_62c85fb0b01b41aca29bg-54.jpg?height=709&width=1040&top_left_y=1230&top_left_x=802)  Figure 2  Now assume that the lines $A P, Q M$, and $B C$ are concurrent at some point $X$ (see Figure 2). Notice that the points $A, B, Q$, and $X$ lie on a common circle $\Omega$ by Miquel's theorem applied to the triangle $X P C$. Let us denote by $Y$ the symmetric image of $X$ about the perpendicular bisector of $A B$. Clearly, $Y$ lies on $\Omega$, and the triangles $Y A B$ and $\triangle X B A$ are congruent. Moreover, the triangle $X P C$ is similar to the triangle $X B A$, so it is also similar to the triangle $Y A B$.  Next, the points $D$ and $M$ correspond to each other in similar triangles $Y A B$ and $X P C$, since $B D / B A=C M / C P=\lambda$. Moreover, the triangles $Y A B$ and $X P C$ are equi-oriented, so $\Varangle(M X, X P)=\Varangle(D Y, Y A)$. On the other hand, since the points $A, Q, X$, and $Y$ lie on $\Omega$, we have $\Varangle(Q Y, Y A)=\Varangle(M X, X P)$. Therefore, $\Varangle(Q Y, Y A)=\Varangle(D Y, Y A)$, so the points $Y, D$, and $Q$ are collinear.  Finally, we have $\Varangle(D Q, Q B)=\Varangle(Y Q, Q B)=\Varangle(Y A, A B)=\Varangle(A B, B X)=\Varangle(A B, B C)$, as desired.  Comment. In the original proposal, $\lambda$ was supposed to be an arbitrary real number distinct from 0 and 1, and the point $M$ was defined by $\overrightarrow{C M}=\lambda \cdot \overrightarrow{C P}$. The Problem Selection Committee decided to add the restriction $\lambda \in(0,1)$ in order to avoid a large case distinction.


Solution: As in the previous solution, we introduce the radical centre $X=A P \cap B C \cap M Q$ of the circles $\omega_{A}, \omega_{B}$, and $\Gamma$. Next, we also notice that the points $A, Q, B$, and $X$ lie on a common circle $\Omega$.  If the point $P$ lies on the arc $B A C$ of $\Gamma$, then the point $X$ is outside $\Gamma$, thus the point $Q$ belongs to the ray $X M$, and therefore the points $P, A$, and $Q$ lie on the same side of $B C$. Otherwise, if $P$ lies on the arc $B C$ not containing $A$, then $X$ lies inside $\Gamma$, so $M$ and $Q$ lie on different sides of $B C$; thus again $Q$ and $A$ lie on the same side of $B C$. So, in each case the points $Q$ and $A$ lie on the same side of $B C$. ![](https://cdn.mathpix.com/cropped/2024_11_18_62c85fb0b01b41aca29bg-55.jpg?height=715&width=1009&top_left_y=839&top_left_x=523)  Figure 3 Now we prove that the ratio  $$ \frac{Q B}{\sin \angle Q B C}=\frac{Q B}{Q X} \cdot \frac{Q X}{\sin \angle Q B X} $$  is constant. Since the points $A, Q, B$, and $X$ are concyclic, we have  $$ \frac{Q X}{\sin \angle Q B X}=\frac{A X}{\sin \angle A B C} $$  Next, since the points $B, Q, M$, and $C$ are concyclic, the triangles $X B Q$ and $X M C$ are similar, so  $$ \frac{Q B}{Q X}=\frac{C M}{C X}=\lambda \cdot \frac{C P}{C X} $$  Analogously, the triangles $X C P$ and $X A B$ are also similar, so  $$ \frac{C P}{C X}=\frac{A B}{A X} $$  Therefore, we obtain  $$ \frac{Q B}{\sin \angle Q B C}=\lambda \cdot \frac{A B}{A X} \cdot \frac{A X}{\sin \angle A B C}=\lambda \cdot \frac{A B}{\sin \angle A B C} $$  so this ratio is indeed constant. Thus the circle passing through $Q$ and tangent to $B C$ at $B$ is also constant, and $Q$ varies over this fixed circle.  Comment. It is not hard to guess that the desired circle should be tangent to $B C$ at $B$. Indeed, the second paragraph of this solution shows that this circle lies on one side of $B C$; on the other hand, in the limit case $P=B$, the point $Q$ also coincides with $B$.


Solution: Let us perform an inversion centred at $C$. Denote by $X^{\prime}$ the image of a point $X$ under this inversion.  The circle $\Gamma$ maps to the line $\Gamma^{\prime}$ passing through the constant points $A^{\prime}$ and $B^{\prime}$, and containing the variable point $P^{\prime}$. By the problem condition, the point $M$ varies over the circle $\gamma$ which is the homothetic image of $\Gamma$ with centre $C$ and coefficient $\lambda$. Thus $M^{\prime}$ varies over the constant line $\gamma^{\prime} \| A^{\prime} B^{\prime}$ which is the homothetic image of $A^{\prime} B^{\prime}$ with centre $C$ and coefficient $1 / \lambda$, and $M=\gamma^{\prime} \cap C P^{\prime}$. Next, the circumcircles $\omega_{A}$ and $\omega_{B}$ of the triangles $A M P$ and $B M C$ map to the circumcircle $\omega_{A}^{\prime}$ of the triangle $A^{\prime} M^{\prime} P^{\prime}$ and to the line $B^{\prime} M^{\prime}$, respectively; the point $Q$ thus maps to the second point of intersection of $B^{\prime} M^{\prime}$ with $\omega_{A}^{\prime}$ (see Figure 4). ![](https://cdn.mathpix.com/cropped/2024_11_18_62c85fb0b01b41aca29bg-56.jpg?height=675&width=1029&top_left_y=913&top_left_x=519)  Figure 4  Let $J$ be the (constant) common point of the lines $\gamma^{\prime}$ and $C A^{\prime}$, and let $\ell$ be the (constant) line through $J$ parallel to $C B^{\prime}$. Let $V$ be the common point of the lines $\ell$ and $B^{\prime} M^{\prime}$. Applying Pappus' theorem to the triples $\left(C, J, A^{\prime}\right)$ and $\left(V, B^{\prime}, M^{\prime}\right)$ we get that the points $C B^{\prime} \cap J V$, $J M^{\prime} \cap A^{\prime} B^{\prime}$, and $C M^{\prime} \cap A^{\prime} V$ are collinear. The first two of these points are ideal, hence so is the third, which means that $C M^{\prime} \| A^{\prime} V$.  Now we have $\Varangle\left(Q^{\prime} A^{\prime}, A^{\prime} P^{\prime}\right)=\Varangle\left(Q^{\prime} M^{\prime}, M^{\prime} P^{\prime}\right)=\angle\left(V M^{\prime}, A^{\prime} V\right)$, which means that the triangles $B^{\prime} Q^{\prime} A^{\prime}$ and $B^{\prime} A^{\prime} V$ are similar, and $\left(B^{\prime} A^{\prime}\right)^{2}=B^{\prime} Q^{\prime} \cdot B^{\prime} V$. Thus $Q^{\prime}$ is the image of $V$ under the second (fixed) inversion with centre $B^{\prime}$ and radius $B^{\prime} A^{\prime}$. Since $V$ varies over the constant line $\ell, Q^{\prime}$ varies over some constant circle $\Theta$. Thus, applying the first inversion back we get that $Q$ also varies over some fixed circle.  One should notice that this last circle is not a line; otherwise $\Theta$ would contain $C$, and thus $\ell$ would contain the image of $C$ under the second inversion. This is impossible, since $C B^{\prime} \| \ell$.


Let $A B C D$ be a convex quadrilateral with $\angle B=\angle D=90^{\circ}$. Point $H$ is the foot of the perpendicular from $A$ to $B D$. The points $S$ and $T$ are chosen on the sides $A B$ and $A D$, respectively, in such a way that $H$ lies inside triangle $S C T$ and  $$ \angle S H C-\angle B S C=90^{\circ}, \quad \angle T H C-\angle D T C=90^{\circ} . $$  Prove that the circumcircle of triangle $S H T$ is tangent to the line $B D$. (Iran)

Solution: Let the line passing through $C$ and perpendicular to the line $S C$ intersect the line $A B$ at $Q$ (see Figure 1). Then  $$ \angle S Q C=90^{\circ}-\angle B S C=180^{\circ}-\angle S H C, $$  which implies that the points $C, H, S$, and $Q$ lie on a common circle. Moreover, since $S Q$ is a diameter of this circle, we infer that the circumcentre $K$ of triangle $S H C$ lies on the line $A B$. Similarly, we prove that the circumcentre $L$ of triangle $C H T$ lies on the line $A D$. ![](https://cdn.mathpix.com/cropped/2024_11_18_62c85fb0b01b41aca29bg-57.jpg?height=795&width=1052&top_left_y=959&top_left_x=516)  Figure 1 In order to prove that the circumcircle of triangle $S H T$ is tangent to $B D$, it suffices to show that the perpendicular bisectors of $H S$ and $H T$ intersect on the line $A H$. However, these two perpendicular bisectors coincide with the angle bisectors of angles $A K H$ and $A L H$. Therefore, in order to complete the solution, it is enough (by the bisector theorem) to show that  $$ \frac{A K}{K H}=\frac{A L}{L H} $$  We present two proofs of this equality. First proof. Let the lines $K L$ and $H C$ intersect at $M$ (see Figure 2). Since $K H=K C$ and $L H=L C$, the points $H$ and $C$ are symmetric to each other with respect to the line $K L$. Therefore $M$ is the midpoint of $H C$. Denote by $O$ the circumcentre of quadrilateral $A B C D$. Then $O$ is the midpoint of $A C$. Therefore we have $O M \| A H$ and hence $O M \perp B D$. This together with the equality $O B=O D$ implies that $O M$ is the perpendicular bisector of $B D$ and therefore $B M=D M$.  Since $C M \perp K L$, the points $B, C, M$, and $K$ lie on a common circle with diameter $K C$. Similarly, the points $L, C, M$, and $D$ lie on a circle with diameter $L C$. Thus, using the sine law, we obtain  $$ \frac{A K}{A L}=\frac{\sin \angle A L K}{\sin \angle A K L}=\frac{D M}{C L} \cdot \frac{C K}{B M}=\frac{C K}{C L}=\frac{K H}{L H} $$  which finishes the proof of (1). ![](https://cdn.mathpix.com/cropped/2024_11_18_62c85fb0b01b41aca29bg-58.jpg?height=806&width=875&top_left_y=294&top_left_x=202)  Figure 2 ![](https://cdn.mathpix.com/cropped/2024_11_18_62c85fb0b01b41aca29bg-58.jpg?height=744&width=752&top_left_y=359&top_left_x=1126)  Figure 3  Second proof. If the points $A, H$, and $C$ are collinear, then $A K=A L$ and $K H=L H$, so the equality (1) follows. Assume therefore that the points $A, H$, and $C$ do not lie in a line and consider the circle $\omega$ passing through them (see Figure 3). Since the quadrilateral $A B C D$ is cyclic,  $$ \angle B A C=\angle B D C=90^{\circ}-\angle A D H=\angle H A D . $$  Let $N \neq A$ be the intersection point of the circle $\omega$ and the angle bisector of $\angle C A H$. Then $A N$ is also the angle bisector of $\angle B A D$. Since $H$ and $C$ are symmetric to each other with respect to the line $K L$ and $H N=N C$, it follows that both $N$ and the centre of $\omega$ lie on the line $K L$. This means that the circle $\omega$ is an Apollonius circle of the points $K$ and $L$. This immediately yields (1).  Comment. Either proof can be used to obtain the following generalised result: Let $A B C D$ be a convex quadrilateral and let $H$ be a point in its interior with $\angle B A C=\angle D A H$. The points $S$ and $T$ are chosen on the sides $A B$ and $A D$, respectively, in such a way that $H$ lies inside triangle SCT and  $$ \angle S H C-\angle B S C=90^{\circ}, \quad \angle T H C-\angle D T C=90^{\circ} . $$  Then the circumcentre of triangle SHT lies on the line AH (and moreover the circumcentre of triangle SCT lies on $A C$ ).


Let $A B C$ be a fixed acute-angled triangle. Consider some points $E$ and $F$ lying on the sides $A C$ and $A B$, respectively, and let $M$ be the midpoint of $E F$. Let the perpendicular bisector of $E F$ intersect the line $B C$ at $K$, and let the perpendicular bisector of $M K$ intersect the lines $A C$ and $A B$ at $S$ and $T$, respectively. We call the pair $(E, F)$ interesting, if the quadrilateral $K S A T$ is cyclic.  Suppose that the pairs $\left(E_{1}, F_{1}\right)$ and $\left(E_{2}, F_{2}\right)$ are interesting. Prove that  $$ \frac{E_{1} E_{2}}{A B}=\frac{F_{1} F_{2}}{A C} . $$

Solution: For any interesting pair $(E, F)$, we will say that the corresponding triangle $E F K$ is also interesting.  Let $E F K$ be an interesting triangle. Firstly, we prove that $\angle K E F=\angle K F E=\angle A$, which also means that the circumcircle $\omega_{1}$ of the triangle $A E F$ is tangent to the lines $K E$ and $K F$.  Denote by $\omega$ the circle passing through the points $K, S, A$, and $T$. Let the line $A M$ intersect the line $S T$ and the circle $\omega$ (for the second time) at $N$ and $L$, respectively (see Figure 1).  Since $E F \| T S$ and $M$ is the midpoint of $E F, N$ is the midpoint of $S T$. Moreover, since $K$ and $M$ are symmetric to each other with respect to the line $S T$, we have $\angle K N S=\angle M N S=$ $\angle L N T$. Thus the points $K$ and $L$ are symmetric to each other with respect to the perpendicular bisector of $S T$. Therefore $K L \| S T$.  Let $G$ be the point symmetric to $K$ with respect to $N$. Then $G$ lies on the line $E F$, and we may assume that it lies on the ray $M F$. One has  $$ \angle K G E=\angle K N S=\angle S N M=\angle K L A=180^{\circ}-\angle K S A $$  (if $K=L$, then the angle $K L A$ is understood to be the angle between $A L$ and the tangent to $\omega$ at $L$ ). This means that the points $K, G, E$, and $S$ are concyclic. Now, since $K S G T$ is a parallelogram, we obtain $\angle K E F=\angle K S G=180^{\circ}-\angle T K S=\angle A$. Since $K E=K F$, we also have $\angle K F E=\angle K E F=\angle A$.  After having proved this fact, one may finish the solution by different methods. ![](https://cdn.mathpix.com/cropped/2024_11_18_62c85fb0b01b41aca29bg-59.jpg?height=684&width=829&top_left_y=1874&top_left_x=245)  Figure 1 ![](https://cdn.mathpix.com/cropped/2024_11_18_62c85fb0b01b41aca29bg-59.jpg?height=612&width=730&top_left_y=1944&top_left_x=1117)  Figure 2  First method. We have just proved that all interesting triangles are similar to each other. This allows us to use the following lemma.  Lemma. Let $A B C$ be an arbitrary triangle. Choose two points $E_{1}$ and $E_{2}$ on the side $A C$, two points $F_{1}$ and $F_{2}$ on the side $A B$, and two points $K_{1}$ and $K_{2}$ on the side $B C$, in a way that the triangles $E_{1} F_{1} K_{1}$ and $E_{2} F_{2} K_{2}$ are similar. Then the six circumcircles of the triangles $A E_{i} F_{i}$, $B F_{i} K_{i}$, and $C E_{i} K_{i}(i=1,2)$ meet at a common point $Z$. Moreover, $Z$ is the centre of the spiral similarity that takes the triangle $E_{1} F_{1} K_{1}$ to the triangle $E_{2} F_{2} K_{2}$. Proof. Firstly, notice that for each $i=1,2$, the circumcircles of the triangles $A E_{i} F_{i}, B F_{i} K_{i}$, and $C K_{i} E_{i}$ have a common point $Z_{i}$ by Miquel's theorem. Moreover, we have $\Varangle\left(Z_{i} F_{i}, Z_{i} E_{i}\right)=\Varangle(A B, C A), \quad \Varangle\left(Z_{i} K_{i}, Z_{i} F_{i}\right)=\Varangle(B C, A B), \quad \Varangle\left(Z_{i} E_{i}, Z_{i} K_{i}\right)=\Varangle(C A, B C)$. This yields that the points $Z_{1}$ and $Z_{2}$ correspond to each other in similar triangles $E_{1} F_{1} K_{1}$ and $E_{2} F_{2} K_{2}$. Thus, if they coincide, then this common point is indeed the desired centre of a spiral similarity.  Finally, in order to show that $Z_{1}=Z_{2}$, one may notice that $\Varangle\left(A B, A Z_{1}\right)=\Varangle\left(E_{1} F_{1}, E_{1} Z_{1}\right)=$ $\Varangle\left(E_{2} F_{2}, E_{2} Z_{2}\right)=\Varangle\left(A B, A Z_{2}\right)$ (see Figure 2). Similarly, one has $\Varangle\left(B C, B Z_{1}\right)=\Varangle\left(B C, B Z_{2}\right)$ and $\Varangle\left(C A, C Z_{1}\right)=\Varangle\left(C A, C Z_{2}\right)$. This yields $Z_{1}=Z_{2}$.  Now, let $P$ and $Q$ be the feet of the perpendiculars from $B$ and $C$ onto $A C$ and $A B$, respectively, and let $R$ be the midpoint of $B C$ (see Figure 3). Then $R$ is the circumcentre of the cyclic quadrilateral $B C P Q$. Thus we obtain $\angle A P Q=\angle B$ and $\angle R P C=\angle C$, which yields $\angle Q P R=\angle A$. Similarly, we show that $\angle P Q R=\angle A$. Thus, all interesting triangles are similar to the triangle $P Q R$. ![](https://cdn.mathpix.com/cropped/2024_11_18_62c85fb0b01b41aca29bg-60.jpg?height=615&width=715&top_left_y=1326&top_left_x=316)  Figure 3 ![](https://cdn.mathpix.com/cropped/2024_11_18_62c85fb0b01b41aca29bg-60.jpg?height=615&width=692&top_left_y=1326&top_left_x=1090)  Figure 4  Denote now by $Z$ the common point of the circumcircles of $A P Q, B Q R$, and $C P R$. Let $E_{1} F_{1} K_{1}$ and $E_{2} F_{2} K_{2}$ be two interesting triangles. By the lemma, $Z$ is the centre of any spiral similarity taking one of the triangles $E_{1} F_{1} K_{1}, E_{2} F_{2} K_{2}$, and $P Q R$ to some other of them. Therefore the triangles $Z E_{1} E_{2}$ and $Z F_{1} F_{2}$ are similar, as well as the triangles $Z E_{1} F_{1}$ and $Z P Q$. Hence  $$ \frac{E_{1} E_{2}}{F_{1} F_{2}}=\frac{Z E_{1}}{Z F_{1}}=\frac{Z P}{Z Q} $$  Moreover, the equalities $\angle A Z Q=\angle A P Q=\angle A B C=180^{\circ}-\angle Q Z R$ show that the point $Z$ lies on the line $A R$ (see Figure 4). Therefore the triangles $A Z P$ and $A C R$ are similar, as well as the triangles $A Z Q$ and $A B R$. This yields  $$ \frac{Z P}{Z Q}=\frac{Z P}{R C} \cdot \frac{R B}{Z Q}=\frac{A Z}{A C} \cdot \frac{A B}{A Z}=\frac{A B}{A C} $$  which completes the solution.  Second method. Now we will start from the fact that $\omega_{1}$ is tangent to the lines $K E$ and $K F$ (see Figure 5). We prove that if $(E, F)$ is an interesting pair, then  $$ \frac{A E}{A B}+\frac{A F}{A C}=2 \cos \angle A $$  Let $Y$ be the intersection point of the segments $B E$ and $C F$. The points $B, K$, and $C$ are collinear, hence applying PASCAL's theorem to the degenerated hexagon AFFYEE, we infer that $Y$ lies on the circle $\omega_{1}$.  Denote by $Z$ the second intersection point of the circumcircle of the triangle $B F Y$ with the line $B C$ (see Figure 6). By Miquel's theorem, the points $C, Z, Y$, and $E$ are concyclic. Therefore we obtain  $$ B F \cdot A B+C E \cdot A C=B Y \cdot B E+C Y \cdot C F=B Z \cdot B C+C Z \cdot B C=B C^{2} $$  On the other hand, $B C^{2}=A B^{2}+A C^{2}-2 A B \cdot A C \cos \angle A$, by the cosine law. Hence  $$ (A B-A F) \cdot A B+(A C-A E) \cdot A C=A B^{2}+A C^{2}-2 A B \cdot A C \cos \angle A, $$  which simplifies to the desired equality (1). Let now $\left(E_{1}, F_{1}\right)$ and $\left(E_{2}, F_{2}\right)$ be two interesting pairs of points. Then we get  $$ \frac{A E_{1}}{A B}+\frac{A F_{1}}{A C}=\frac{A E_{2}}{A B}+\frac{A F_{2}}{A C} $$  which gives the desired result. ![](https://cdn.mathpix.com/cropped/2024_11_18_62c85fb0b01b41aca29bg-61.jpg?height=709&width=915&top_left_y=1527&top_left_x=248)  Figure 5 ![](https://cdn.mathpix.com/cropped/2024_11_18_62c85fb0b01b41aca29bg-61.jpg?height=569&width=615&top_left_y=1669&top_left_x=1229)  Figure 6  Third method. Again, we make use of the fact that all interesting triangles are similar (and equi-oriented). Let us put the picture onto a complex plane such that $A$ is at the origin, and identify each point with the corresponding complex number.  Let $E F K$ be any interesting triangle. The equalities $\angle K E F=\angle K F E=\angle A$ yield that the ratio $\nu=\frac{K-E}{F-E}$ is the same for all interesting triangles. This in turn means that the numbers $E$, $F$, and $K$ satisfy the linear equation  $$ K=\mu E+\nu F, \quad \text { where } \quad \mu=1-\nu $$  Now let us choose the points $X$ and $Y$ on the rays $A B$ and $A C$, respectively, so that $\angle C X A=\angle A Y B=\angle A=\angle K E F$ (see Figure 7). Then each of the triangles $A X C$ and $Y A B$ is similar to any interesting triangle, which also means that  $$ C=\mu A+\nu X=\nu X \quad \text { and } \quad B=\mu Y+\nu A=\mu Y . $$  Moreover, one has $X / Y=\overline{C / B}$. Since the points $E, F$, and $K$ lie on $A C, A B$, and $B C$, respectively, one gets  $$ E=\rho Y, \quad F=\sigma X, \quad \text { and } \quad K=\lambda B+(1-\lambda) C $$  for some real $\rho, \sigma$, and $\lambda$. In view of (3), the equation (2) now reads $\lambda B+(1-\lambda) C=K=$ $\mu E+\nu F=\rho B+\sigma C$, or  $$ (\lambda-\rho) B=(\sigma+\lambda-1) C $$  Since the nonzero complex numbers $B$ and $C$ have different arguments, the coefficients in the brackets vanish, so $\rho=\lambda$ and $\sigma=1-\lambda$. Therefore,  $$ \frac{E}{Y}+\frac{F}{X}=\rho+\sigma=1 $$  Now, if $\left(E_{1}, F_{1}\right)$ and $\left(E_{2}, F_{2}\right)$ are two distinct interesting pairs, one may apply (4) to both pairs. Subtracting, we get  $$ \frac{E_{1}-E_{2}}{Y}=\frac{F_{2}-F_{1}}{X}, \quad \text { so } \quad \frac{E_{1}-E_{2}}{F_{2}-F_{1}}=\frac{Y}{X}=\frac{\bar{B}}{\bar{C}} $$  Taking absolute values provides the required result. ![](https://cdn.mathpix.com/cropped/2024_11_18_62c85fb0b01b41aca29bg-62.jpg?height=487&width=669&top_left_y=1664&top_left_x=696)  Figure 7  Comment 1. One may notice that the triangle $P Q R$ is also interesting. Comment 2. In order to prove that $\angle K E F=\angle K F E=\angle A$, one may also use the following well-known fact: Let $A E F$ be a triangle with $A E \neq A F$, and let $K$ be the common point of the symmedian taken from $A$ and the perpendicular bisector of $E F$. Then the lines $K E$ and $K F$ are tangent to the circumcircle $\omega_{1}$ of the triangle $A E F$.  In this case, however, one needs to deal with the case $A E=A F$ separately.


Solution: Let $(E, F)$ be an interesting pair. This time we prove that  $$ \frac{A M}{A K}=\cos \angle A $$  As in Solution 1, we introduce the circle $\omega$ passing through the points $K, S$, $A$, and $T$, together with the points $N$ and $L$ at which the line $A M$ intersect the line $S T$ and the circle $\omega$ for the second time, respectively. Let moreover $O$ be the centre of $\omega$ (see Figures 8 and 9). As in Solution 1, we note that $N$ is the midpoint of $S T$ and show that $K L \| S T$, which implies $\angle F A M=\angle E A K$. ![](https://cdn.mathpix.com/cropped/2024_11_18_62c85fb0b01b41aca29bg-63.jpg?height=752&width=1009&top_left_y=749&top_left_x=266)  Figure 8 ![](https://cdn.mathpix.com/cropped/2024_11_18_62c85fb0b01b41aca29bg-63.jpg?height=552&width=549&top_left_y=952&top_left_x=1279)  Figure 9  Suppose now that $K \neq L$ (see Figure 8). Then $K L \| S T$, and consequently the lines $K M$ and $K L$ are perpendicular. It implies that the lines $L O$ and $K M$ meet at a point $X$ lying on the circle $\omega$. Since the lines $O N$ and $X M$ are both perpendicular to the line $S T$, they are parallel to each other, and hence $\angle L O N=\angle L X K=\angle M A K$. On the other hand, $\angle O L N=\angle M K A$, so we infer that triangles $N O L$ and $M A K$ are similar. This yields  $$ \frac{A M}{A K}=\frac{O N}{O L}=\frac{O N}{O T}=\cos \angle T O N=\cos \angle A $$  If, on the other hand, $K=L$, then the points $A, M, N$, and $K$ lie on a common line, and this line is the perpendicular bisector of $S T$ (see Figure 9). This implies that $A K$ is a diameter of $\omega$, which yields $A M=2 O K-2 N K=2 O N$. So also in this case we obtain  $$ \frac{A M}{A K}=\frac{2 O N}{2 O T}=\cos \angle T O N=\cos \angle A $$  Thus (5) is proved. Let $P$ and $Q$ be the feet of the perpendiculars from $B$ and $C$ onto $A C$ and $A B$, respectively (see Figure 10). We claim that the point $M$ lies on the line $P Q$. Consider now the composition of the dilatation with factor $\cos \angle A$ and centre $A$, and the reflection with respect to the angle bisector of $\angle B A C$. This transformation is a similarity that takes $B, C$, and $K$ to $P, Q$, and $M$, respectively. Since $K$ lies on the line $B C$, the point $M$ lies on the line $P Q$. ![](https://cdn.mathpix.com/cropped/2024_11_18_62c85fb0b01b41aca29bg-64.jpg?height=715&width=918&top_left_y=219&top_left_x=586)  Figure 10 Suppose that $E \neq P$. Then also $F \neq Q$, and by Menelaus' theorem, we obtain  $$ \frac{A Q}{F Q} \cdot \frac{F M}{E M} \cdot \frac{E P}{A P}=1 $$  Using the similarity of the triangles $A P Q$ and $A B C$, we infer that  $$ \frac{E P}{F Q}=\frac{A P}{A Q}=\frac{A B}{A C}, \quad \text { and hence } \quad \frac{E P}{A B}=\frac{F Q}{A C} $$  The last equality holds obviously also in case $E=P$, because then $F=Q$. Moreover, since the line $P Q$ intersects the segment $E F$, we infer that the point $E$ lies on the segment $A P$ if and only if the point $F$ lies outside of the segment $A Q$.  Let now $\left(E_{1}, F_{1}\right)$ and $\left(E_{2}, F_{2}\right)$ be two interesting pairs. Then we obtain  $$ \frac{E_{1} P}{A B}=\frac{F_{1} Q}{A C} \quad \text { and } \quad \frac{E_{2} P}{A B}=\frac{F_{2} Q}{A C} . $$  If $P$ lies between the points $E_{1}$ and $E_{2}$, we add the equalities above, otherwise we subtract them. In any case we obtain  $$ \frac{E_{1} E_{2}}{A B}=\frac{F_{1} F_{2}}{A C} $$  which completes the solution.


Let $A B C$ be a triangle with circumcircle $\Omega$ and incentre $I$. Let the line passing through $I$ and perpendicular to $C I$ intersect the segment $B C$ and the $\operatorname{arc} B C($ not containing $A)$ of $\Omega$ at points $U$ and $V$, respectively. Let the line passing through $U$ and parallel to $A I$ intersect $A V$ at $X$, and let the line passing through $V$ and parallel to $A I$ intersect $A B$ at $Y$. Let $W$ and $Z$ be the midpoints of $A X$ and $B C$, respectively. Prove that if the points $I, X$, and $Y$ are collinear, then the points $I, W$, and $Z$ are also collinear.

Solution: We start with some general observations. Set $\alpha=\angle A / 2, \beta=\angle B / 2, \gamma=\angle C / 2$. Then obviously $\alpha+\beta+\gamma=90^{\circ}$. Since $\angle U I C=90^{\circ}$, we obtain $\angle I U C=\alpha+\beta$. Therefore $\angle B I V=\angle I U C-\angle I B C=\alpha=\angle B A I=\angle B Y V$, which implies that the points $B, Y, I$, and $V$ lie on a common circle (see Figure 1).  Assume now that the points $I, X$ and $Y$ are collinear. We prove that $\angle Y I A=90^{\circ}$. Let the line $X U$ intersect $A B$ at $N$. Since the lines $A I, U X$, and $V Y$ are parallel, we get  $$ \frac{N X}{A I}=\frac{Y N}{Y A}=\frac{V U}{V I}=\frac{X U}{A I} $$  implying $N X=X U$. Moreover, $\angle B I U=\alpha=\angle B N U$. This implies that the quadrilateral BUIN is cyclic, and since $B I$ is the angle bisector of $\angle U B N$, we infer that $N I=U I$. Thus in the isosceles triangle $N I U$, the point $X$ is the midpoint of the base $N U$. This gives $\angle I X N=90^{\circ}$, i.e., $\angle Y I A=90^{\circ}$. ![](https://cdn.mathpix.com/cropped/2024_11_18_62c85fb0b01b41aca29bg-65.jpg?height=880&width=966&top_left_y=1256&top_left_x=565)  Figure 1 Let $S$ be the midpoint of the segment $V C$. Let moreover $T$ be the intersection point of the lines $A X$ and $S I$, and set $x=\angle B A V=\angle B C V$. Since $\angle C I A=90^{\circ}+\beta$ and $S I=S C$, we obtain  $$ \angle T I A=180^{\circ}-\angle A I S=90^{\circ}-\beta-\angle C I S=90^{\circ}-\beta-\gamma-x=\alpha-x=\angle T A I, $$  which implies that $T I=T A$. Therefore, since $\angle X I A=90^{\circ}$, the point $T$ is the midpoint of $A X$, i.e., $T=W$.  To complete our solution, it remains to show that the intersection point of the lines $I S$ and $B C$ coincide with the midpoint of the segment $B C$. But since $S$ is the midpoint of the segment $V C$, it suffices to show that the lines $B V$ and $I S$ are parallel.  Since the quadrilateral $B Y I V$ is cyclic, $\angle V B I=\angle V Y I=\angle Y I A=90^{\circ}$. This implies that $B V$ is the external angle bisector of the angle $A B C$, which yields $\angle V A C=\angle V C A$. Therefore $2 \alpha-x=2 \gamma+x$, which gives $\alpha=\gamma+x$. Hence $\angle S C I=\alpha$, so $\angle V S I=2 \alpha$.  On the other hand, $\angle B V C=180^{\circ}-\angle B A C=180^{\circ}-2 \alpha$, which implies that the lines $B V$ and $I S$ are parallel. This completes the solution.


Solution: As in   $$ \angle B A V=\angle C A E $$  Proof. Let $\rho$ be the composition of the inversion with centre $A$ and radius $\sqrt{A B \cdot A C}$, and the symmetry with respect to $A I$. Clearly, $\rho$ interchanges $B$ and $C$.  Let $J$ be the excentre of the triangle $A B C$ opposite to $A$ (see Figure 2). Then we have $\angle J A C=\angle B A I$ and $\angle J C A=90^{\circ}+\gamma=\angle B I A$, so the triangles $A C J$ and $A I B$ are similar, and therefore $A B \cdot A C=A I \cdot A J$. This means that $\rho$ interchanges $I$ and $J$. Moreover, since $Y$ lies on $A B$ and $\angle A I Y=90^{\circ}$, the point $Y^{\prime}=\rho(Y)$ lies on $A C$, and $\angle J Y^{\prime} A=90^{\circ}$. Thus $\rho$ maps the circumcircle $\gamma$ of the triangle $B Y I$ to a circle $\gamma^{\prime}$ with diameter $J C$.  Finally, since $V$ lies on both $\Gamma$ and $\gamma$, the point $V^{\prime}=\rho(V)$ lies on the line $\rho(\Gamma)=A B$ as well as on $\gamma^{\prime}$, which in turn means that $V^{\prime}=E$. This implies the desired result. ![](https://cdn.mathpix.com/cropped/2024_11_18_62c85fb0b01b41aca29bg-66.jpg?height=709&width=572&top_left_y=1713&top_left_x=248)  Figure 2 ![](https://cdn.mathpix.com/cropped/2024_11_18_62c85fb0b01b41aca29bg-66.jpg?height=872&width=989&top_left_y=1549&top_left_x=822)  Figure 3  Now we turn to the solution of the problem. Assume that the incircle $\omega_{1}$ of the triangle $A B C$ is tangent to $B C$ at $D$, and let the excircle $\omega_{2}$ of the triangle $A B C$ opposite to the vertex $A$ touch the side $B C$ at $E$ (see Figure 3). The homothety with centre $A$ that takes $\omega_{2}$ to $\omega_{1}$ takes the point $E$ to some point $F$, and the tangent to $\omega_{1}$ at $F$ is parallel to $B C$. Therefore $D F$ is a diameter of $\omega_{1}$. Moreover, $Z$ is the midpoint of $D E$. This implies that the lines $I Z$ and $F E$ are parallel.  Let $K=Y I \cap A E$. Since $\angle Y I A=90^{\circ}$, the lemma yields that $I$ is the midpoint of $X K$. This implies that the segments $I W$ and $A K$ are parallel. Therefore, the points $W, I$ and $Z$ are collinear.  Comment 1. The properties $\angle Y I A=90^{\circ}$ and $V A=V C$ can be established in various ways. The main difficulty of the problem seems to find out how to use these properties in connection to the points $W$ and $Z$.  In Solution 2 this principal part is more or less covered by the lemma, for which we have presented a direct proof. On the other hand, this lemma appears to be a combination of two well-known facts; let us formulate them in terms of the lemma statement.  Let the line $I Y$ intersect $A C$ at $P$ (see Figure 4). The first fact states that the circumcircle $\omega$ of the triangle $V Y P$ is tangent to the segments $A B$ and $A C$, as well as to the circle $\Gamma$. The second fact states that for such a circle, the angles $B A V$ and $C A E$ are equal.  The awareness of this lemma may help a lot in solving this problem; so the Jury might also consider a variation of the proposed problem, for which the lemma does not seem to be useful; see Comment 3. ![](https://cdn.mathpix.com/cropped/2024_11_18_62c85fb0b01b41aca29bg-67.jpg?height=946&width=1689&top_left_y=1149&top_left_x=189)  Comment 2. The proposed problem stated the equivalence: the point $I$ lies on the line $X Y$ if and only if $I$ lies on the line $W Z$. Here we sketch the proof of the "if" part (see Figure 5). As in Solution 2 , let $B C$ touch the circles $\omega_{1}$ and $\omega_{2}$ at $D$ and $E$, respectively. Since $I Z \| A E$ and $W$ lies on $I Z$, the line $D X$ is also parallel to $A E$. Therefore, the triangles $X U P$ and $A I Q$ are similar. Moreover, the line $D X$ is symmetric to $A E$ with respect to $I$, so $I P=I Q$, where $P=U V \cap X D$ and $Q=U V \cap A E$. Thus we obtain  $$ \frac{U V}{V I}=\frac{U X}{I A}=\frac{U P}{I Q}=\frac{U P}{I P} $$  So the pairs $I U$ and $P V$ are harmonic conjugates, and since $\angle U D I=90^{\circ}$, we get $\angle V D B=\angle B D X=$ $\angle B E A$. Therefore the point $V^{\prime}$ symmetric to $V$ with respect to the perpendicular bisector of $B C$ lies on the line $A E$. So we obtain $\angle B A V=\angle C A E$.  The rest can be obtained by simply reversing the arguments in Solution 2 . The points $B, V, I$, and $Y$ are concyclic. The lemma implies that $\angle Y I A=90^{\circ}$. Moreover, the points $B, U, I$, and $N$, where $N=U X \cap A B$, lie on a common circle, so $I N=I U$. Since $I Y \perp U N$, the point $X^{\prime}=I Y \cap U N$ is the midpoint of $U N$. But in the trapezoid $A Y V I$, the line $X U$ is parallel to the sides $A I$ and $Y V$, so $N X=U X^{\prime}$. This yields $X=X^{\prime}$. The reasoning presented in Solution 1 can also be reversed, but it requires a lot of technicalities. Therefore the Problem Selection Committee proposes to consider only the "only if" part of the original proposal, which is still challenging enough.  Comment 3. The Jury might also consider the following variation of the proposed problem. Let $A B C$ be a triangle with circumcircle $\Omega$ and incentre I. Let the line through I perpendicular to CI intersect the segment $B C$ and the arc $B C$ (not containing $A$ ) of $\Omega$ at $U$ and $V$, respectively. Let the line through $U$ parallel to $A I$ intersect $A V$ at $X$. Prove that if the lines XI and AI are perpendicular, then the midpoint of the segment AC lies on the line XI (see Figure 6). ![](https://cdn.mathpix.com/cropped/2024_11_18_62c85fb0b01b41aca29bg-68.jpg?height=675&width=746&top_left_y=1273&top_left_x=184)  Figure 6 ![](https://cdn.mathpix.com/cropped/2024_11_18_62c85fb0b01b41aca29bg-68.jpg?height=1012&width=980&top_left_y=930&top_left_x=892)  Figure 7  Since the solution contains the arguments used above, we only sketch it. Let $N=X U \cap A B$ (see Figure 7). Then $\angle B N U=\angle B A I=\angle B I U$, so the points $B, U, I$, and $N$ lie on a common circle. Therefore $I U=I N$, and since $I X \perp N U$, it follows that $N X=X U$. Now set $Y=X I \cap A B$. The equality $N X=X U$ implies that  $$ \frac{V X}{V A}=\frac{X U}{A I}=\frac{N X}{A I}=\frac{Y X}{Y I} $$  and therefore $Y V \| A I$. Hence $\angle B Y V=\angle B A I=\angle B I V$, so the points $B, V, I, Y$ are concyclic. Next we have $I Y \perp Y V$, so $\angle I B V=90^{\circ}$. This implies that $B V$ is the external angle bisector of the angle $A B C$, which gives $\angle V A C=\angle V C A$. So in order to show that $M=X I \cap A C$ is the midpoint of $A C$, it suffices to prove that $\angle V M C=90^{\circ}$. But this follows immediately from the observation that the points $V, C, M$, and $I$ are concyclic, as $\angle M I V=\angle Y B V=180^{\circ}-\angle A C V$. The converse statement is also true, but its proof requires some technicalities as well.


Let $n \geqslant 2$ be an integer, and let $A_{n}$ be the set  $$ A_{n}=\left\{2^{n}-2^{k} \mid k \in \mathbb{Z}, 0 \leqslant k<n\right\} . $$  Determine the largest positive integer that cannot be written as the sum of one or more (not necessarily distinct) elements of $A_{n}$. (Serbia)

Solution: Part I. First we show that every integer greater than $(n-2) 2^{n}+1$ can be represented as such a sum. This is achieved by induction on $n$.  For $n=2$, the set $A_{n}$ consists of the two elements 2 and 3 . Every positive integer $m$ except for 1 can be represented as the sum of elements of $A_{n}$ in this case: as $m=2+2+\cdots+2$ if $m$ is even, and as $m=3+2+2+\cdots+2$ if $m$ is odd.  Now consider some $n>2$, and take an integer $m>(n-2) 2^{n}+1$. If $m$ is even, then consider  $$ \frac{m}{2} \geqslant \frac{(n-2) 2^{n}+2}{2}=(n-2) 2^{n-1}+1>(n-3) 2^{n-1}+1 $$  By the induction hypothesis, there is a representation of the form  $$ \frac{m}{2}=\left(2^{n-1}-2^{k_{1}}\right)+\left(2^{n-1}-2^{k_{2}}\right)+\cdots+\left(2^{n-1}-2^{k_{r}}\right) $$  for some $k_{i}$ with $0 \leqslant k_{i}<n-1$. It follows that  $$ m=\left(2^{n}-2^{k_{1}+1}\right)+\left(2^{n}-2^{k_{2}+1}\right)+\cdots+\left(2^{n}-2^{k_{r}+1}\right) $$  giving us the desired representation as a sum of elements of $A_{n}$. If $m$ is odd, we consider  $$ \frac{m-\left(2^{n}-1\right)}{2}>\frac{(n-2) 2^{n}+1-\left(2^{n}-1\right)}{2}=(n-3) 2^{n-1}+1 $$  By the induction hypothesis, there is a representation of the form  $$ \frac{m-\left(2^{n}-1\right)}{2}=\left(2^{n-1}-2^{k_{1}}\right)+\left(2^{n-1}-2^{k_{2}}\right)+\cdots+\left(2^{n-1}-2^{k_{r}}\right) $$  for some $k_{i}$ with $0 \leqslant k_{i}<n-1$. It follows that  $$ m=\left(2^{n}-2^{k_{1}+1}\right)+\left(2^{n}-2^{k_{2}+1}\right)+\cdots+\left(2^{n}-2^{k_{r}+1}\right)+\left(2^{n}-1\right) $$  giving us the desired representation of $m$ once again. Part II. It remains to show that there is no representation for $(n-2) 2^{n}+1$. Let $N$ be the smallest positive integer that satisfies $N \equiv 1\left(\bmod 2^{n}\right)$, and which can be represented as a sum of elements of $A_{n}$. Consider a representation of $N$, i.e.,  $$ N=\left(2^{n}-2^{k_{1}}\right)+\left(2^{n}-2^{k_{2}}\right)+\cdots+\left(2^{n}-2^{k_{r}}\right) $$  where $0 \leqslant k_{1}, k_{2}, \ldots, k_{r}<n$. Suppose first that two of the terms in the sum are the same, i.e., $k_{i}=k_{j}$ for some $i \neq j$. If $k_{i}=k_{j}=n-1$, then we can simply remove these two terms to get a representation for  $$ N-2\left(2^{n}-2^{n-1}\right)=N-2^{n} $$  as a sum of elements of $A_{n}$, which contradicts our choice of $N$. If $k_{i}=k_{j}=k<n-1$, replace the two terms by $2^{n}-2^{k+1}$, which is also an element of $A_{n}$, to get a representation for  $$ N-2\left(2^{n}-2^{k}\right)+2^{n}-2^{k+1}=N-2^{n} . $$  This is a contradiction once again. Therefore, all $k_{i}$ have to be distinct, which means that  $$ 2^{k_{1}}+2^{k_{2}}+\cdots+2^{k_{r}} \leqslant 2^{0}+2^{1}+2^{2}+\cdots+2^{n-1}=2^{n}-1 $$  On the other hand, taking (1) modulo $2^{n}$, we find  $$ 2^{k_{1}}+2^{k_{2}}+\cdots+2^{k_{r}} \equiv-N \equiv-1 \quad\left(\bmod 2^{n}\right) $$  Thus we must have $2^{k_{1}}+2^{k_{2}}+\cdots+2^{k_{r}}=2^{n}-1$, which is only possible if each element of $\{0,1, \ldots, n-1\}$ occurs as one of the $k_{i}$. This gives us  $$ N=n 2^{n}-\left(2^{0}+2^{1}+\cdots+2^{n-1}\right)=(n-1) 2^{n}+1 . $$  In particular, this means that $(n-2) 2^{n}+1$ cannot be represented as a sum of elements of $A_{n}$.


Solution: The fact that $m=(n-2) 2^{n}+1$ cannot be represented as a sum of elements of $A_{n}$ can also be shown in other ways. We prove the following statement by induction on $n$ : Claim. If $a, b$ are integers with $a \geqslant 0, b \geqslant 1$, and $a+b<n$, then $a 2^{n}+b$ cannot be written as a sum of elements of $A_{n}$. Proof. The claim is clearly true for $n=2$ (since $a=0, b=1$ is the only possibility). For $n>2$, assume that there exist integers $a, b$ with $a \geqslant 0, b \geqslant 1$ and $a+b<n$ as well as elements $m_{1}, m_{2}, \ldots, m_{r}$ of $A_{n}$ such that  $$ a 2^{n}+b=m_{1}+m_{2}+\cdots+m_{r} $$  We can suppose, without loss of generality, that $m_{1} \geqslant m_{2} \geqslant \cdots \geqslant m_{r}$. Let $\ell$ be the largest index for which $m_{\ell}=2^{n}-1\left(\ell=0\right.$ if $\left.m_{1} \neq 2^{n}-1\right)$. Clearly, $\ell$ and $b$ must have the same parity. Now  $$ (a-\ell) 2^{n}+(b+\ell)=m_{\ell+1}+m_{\ell+2}+\cdots+m_{r} $$  and thus  $$ (a-\ell) 2^{n-1}+\frac{b+\ell}{2}=\frac{m_{\ell+1}}{2}+\frac{m_{\ell+2}}{2}+\cdots+\frac{m_{r}}{2} . $$  Note that $m_{\ell+1} / 2, m_{\ell+2} / 2, \ldots, m_{r} / 2$ are elements of $A_{n-1}$. Moreover, $a-\ell$ and $(b+\ell) / 2$ are integers, and $(b+\ell) / 2 \geqslant 1$. If $a-\ell$ was negative, then we would have  $$ a 2^{n}+b \geqslant \ell\left(2^{n}-1\right) \geqslant(a+1)\left(2^{n}-1\right)=a 2^{n}+2^{n}-a-1, $$  thus $n \geqslant a+b+1 \geqslant 2^{n}$, which is impossible. So $a-\ell \geqslant 0$. By the induction hypothesis, we must have $a-\ell+\frac{b+\ell}{2} \geqslant n-1$, which gives us a contradiction, since  $$ a-\ell+\frac{b+\ell}{2} \leqslant a-\ell+b+\ell-1=a+b-1<n-1 . $$  Considering the special case $a=n-2, b=1$ now completes the proof.


Solution: Denote by $B_{n}$ the set of all positive integers that can be written as a sum of elements of $A_{n}$. In this solution, we explicitly describe all the numbers in $B_{n}$ by an argument similar to the first solution.  For a positive integer $n$, we denote by $\sigma_{2}(n)$ the sum of its digits in the binary representation. Notice that every positive integer $m$ has a unique representation of the form $m=s 2^{n}-t$ with some positive integer $s$ and $0 \leqslant t \leqslant 2^{n}-1$. Lemma. For any two integers $s \geqslant 1$ and $0 \leqslant t \leqslant 2^{n}-1$, the number $m=s 2^{n}-t$ belongs to $B_{n}$ if and only if $s \geqslant \sigma_{2}(t)$. Proof. For $t=0$, the statement of the Lemma is obvious, since $m=2 s \cdot\left(2^{n}-2^{n-1}\right)$. Now suppose that $t \geqslant 1$, and let  $$ t=2^{k_{1}}+\cdots+2^{k_{\sigma}} \quad\left(0 \leqslant k_{1}<\cdots<k_{\sigma} \leqslant n-1, \quad \sigma=\sigma_{2}(t)\right) $$  be its binary expansion. If $s \geqslant \sigma$, then $m \in B_{n}$ since  $$ m=(s-\sigma) 2^{n}+\left(\sigma 2^{n}-t\right)=2(s-\sigma) \cdot\left(2^{n}-2^{n-1}\right)+\sum_{i=1}^{\sigma}\left(2^{n}-2^{k_{i}}\right) $$  Assume now that there exist integers $s$ and $t$ with $1 \leqslant s<\sigma_{2}(t)$ and $0 \leqslant t \leqslant 2^{n}-1$ such that the number $m=s 2^{n}-t$ belongs to $B_{n}$. Among all such instances, choose the one for which $m$ is smallest, and let  $$ m=\sum_{i=1}^{d}\left(2^{n}-2^{\ell_{i}}\right) \quad\left(0 \leqslant \ell_{i} \leqslant n-1\right) $$  be the corresponding representation. If all the $\ell_{i}$ 's are distinct, then $\sum_{i=1}^{d} 2^{\ell_{i}} \leqslant \sum_{j=0}^{n-1} 2^{j}=2^{n}-1$, so one has $s=d$ and $t=\sum_{i=1}^{d} 2^{\ell_{i}}$, whence $s=d=\sigma_{2}(t)$; this is impossible. Therefore, two of the $\ell_{i}$ 's must be equal, say $\ell_{d-1}=\ell_{d}$. Then $m \geqslant 2\left(2^{n}-2^{\ell_{d}}\right) \geqslant 2^{n}$, so $s \geqslant 2$.  Now we claim that the number $m^{\prime}=m-2^{n}=(s-1) 2^{n}-t$ also belongs to $B_{n}$, which contradicts the minimality assumption. Indeed, one has  $$ \left(2^{n}-2^{\ell_{d-1}}\right)+\left(2^{n}-2^{\ell_{d}}\right)=2\left(2^{n}-2^{\ell_{d}}\right)=2^{n}+\left(2^{n}-2^{\ell_{d}+1}\right) $$  so  $$ m^{\prime}=\sum_{i=1}^{d-2}\left(2^{n}-2^{\ell_{i}}\right)+\left(2^{n}-2^{\ell_{d}+1}\right) $$  is the desired representation of $m^{\prime}$ (if $\ell_{d}=n-1$, then the last summand is simply omitted). This contradiction finishes the proof.  By our lemma, the largest number $M$ which does not belong to $B_{n}$ must have the form  $$ m_{t}=\left(\sigma_{2}(t)-1\right) 2^{n}-t $$  for some $t$ with $1 \leqslant t \leqslant 2^{n}-1$, so $M$ is just the largest of these numbers. For $t_{0}=2^{n}-1$ we have $m_{t_{0}}=(n-1) 2^{n}-\left(2^{n}-1\right)=(n-2) 2^{n}+1$; for every other value of $t$ one has $\sigma_{2}(t) \leqslant n-1$, thus $m_{t} \leqslant(\sigma(t)-1) 2^{n} \leqslant(n-2) 2^{n}<m_{t_{0}}$. This means that $M=m_{t_{0}}=(n-2) 2^{n}+1$.


Determine all pairs $(x, y)$ of positive integers such that  $$ \sqrt[3]{7 x^{2}-13 x y+7 y^{2}}=|x-y|+1 $$

Solution: Let $(x, y)$ be any pair of positive integers solving (1). We shall prove that it appears in the list displayed above. The converse assertion that all these pairs do actually satisfy (1) either may be checked directly by means of a somewhat laborious calculation, or it can be seen by going in reverse order through the displayed equations that follow.  In case $x=y$ the given equation reduces to $x^{2 / 3}=1$, which is equivalent to $x=1$, whereby he have found the first solution.  To find the solutions with $x \neq y$ we may assume $x>y$ due to symmetry. Then the integer $n=x-y$ is positive and (1) may be rewritten as  $$ \sqrt[3]{7(y+n)^{2}-13(y+n) y+7 y^{2}}=n+1 $$  Raising this to the third power and simplifying the result one obtains  $$ y^{2}+y n=n^{3}-4 n^{2}+3 n+1 . $$  To complete the square on the left hand side, we multiply by 4 and add $n^{2}$, thus getting  $$ (2 y+n)^{2}=4 n^{3}-15 n^{2}+12 n+4=(n-2)^{2}(4 n+1) $$  This shows that the cases $n=1$ and $n=2$ are impossible, whence $n>2$, and $4 n+1$ is the square of the rational number $\frac{2 y+n}{n-2}$. Consequently, it has to be a perfect square, and, since it is odd as well, there has to exist some nonnegative integer $m$ such that $4 n+1=(2 m+1)^{2}$, i.e.  $$ n=m^{2}+m $$  Notice that $n>2$ entails $m \geqslant 2$. Substituting the value of $n$ just found into the previous displayed equation we arrive at  $$ \left(2 y+m^{2}+m\right)^{2}=\left(m^{2}+m-2\right)^{2}(2 m+1)^{2}=\left(2 m^{3}+3 m^{2}-3 m-2\right)^{2} . $$  Extracting square roots and taking $2 m^{3}+3 m^{2}-3 m-2=(m-1)\left(2 m^{2}+5 m+2\right)>0$ into account we derive $2 y+m^{2}+m=2 m^{3}+3 m^{2}-3 m-2$, which in turn yields  $$ y=m^{3}+m^{2}-2 m-1 $$  Notice that $m \geqslant 2$ implies that $y=\left(m^{3}-1\right)+(m-2) m$ is indeed positive, as it should be. In view of $x=y+n=y+m^{2}+m$ it also follows that  $$ x=m^{3}+2 m^{2}-m-1, $$  and that this integer is positive as well. Comment. Alternatively one could ask to find all pairs $(x, y)$ of - not necessarily positive - integers solving (1). The answer to that question is a bit nicer than the answer above: the set of solutions are now described by  $$ \{x, y\}=\left\{m^{3}+m^{2}-2 m-1, m^{3}+2 m^{2}-m-1\right\} $$  where $m$ varies through $\mathbb{Z}$. This may be shown using essentially the same arguments as above. We finally observe that the pair $(x, y)=(1,1)$, that appears to be sporadic above, corresponds to $m=-1$.


A coin is called a Cape Town coin if its value is $1 / n$ for some positive integer $n$. Given a collection of Cape Town coins of total value at most $99+\frac{1}{2}$, prove that it is possible to split this collection into at most 100 groups each of total value at most 1. (Luxembourg)

Solution: We will show that for every positive integer $N$ any collection of Cape Town coins of total value at most $N-\frac{1}{2}$ can be split into $N$ groups each of total value at most 1 . The problem statement is a particular case for $N=100$.  We start with some preparations. If several given coins together have a total value also of the form $\frac{1}{k}$ for a positive integer $k$, then we may merge them into one new coin. Clearly, if the resulting collection can be split in the required way then the initial collection can also be split.  After each such merging, the total number of coins decreases, thus at some moment we come to a situation when no more merging is possible. At this moment, for every even $k$ there is at most one coin of value $\frac{1}{k}$ (otherwise two such coins may be merged), and for every odd $k>1$ there are at most $k-1$ coins of value $\frac{1}{k}$ (otherwise $k$ such coins may also be merged).  Now, clearly, each coin of value 1 should form a single group; if there are $d$ such coins then we may remove them from the collection and replace $N$ by $N-d$. So from now on we may assume that there are no coins of value 1.  Finally, we may split all the coins in the following way. For each $k=1,2, \ldots, N$ we put all the coins of values $\frac{1}{2 k-1}$ and $\frac{1}{2 k}$ into a group $G_{k}$; the total value of $G_{k}$ does not exceed  $$ (2 k-2) \cdot \frac{1}{2 k-1}+\frac{1}{2 k}<1 $$  It remains to distribute the "small" coins of values which are less than $\frac{1}{2 N}$; we will add them one by one. In each step, take any remaining small coin. The total value of coins in the groups at this moment is at most $N-\frac{1}{2}$, so there exists a group of total value at most $\frac{1}{N}\left(N-\frac{1}{2}\right)=1-\frac{1}{2 N}$; thus it is possible to put our small coin into this group. Acting so, we will finally distribute all the coins.  Comment 1. The algorithm may be modified, at least the step where one distributes the coins of values $\geqslant \frac{1}{2 N}$. One different way is to put into $G_{k}$ all the coins of values $\frac{1}{(2 k-1) 2^{s}}$ for all integer $s \geqslant 0$. One may easily see that their total value also does not exceed 1.  Comment 2. The original proposal also contained another part, suggesting to show that a required splitting may be impossible if the total value of coins is at most 100 . There are many examples of such a collection, e.g. one may take 98 coins of value 1 , one coin of value $\frac{1}{2}$, two coins of value $\frac{1}{3}$, and four coins of value $\frac{1}{5}$.  The Problem Selection Committee thinks that this part is less suitable for the competition.


Let $n>1$ be a given integer. Prove that infinitely many terms of the sequence $\left(a_{k}\right)_{k \geqslant 1}$, defined by  $$ a_{k}=\left\lfloor\frac{n^{k}}{k}\right\rfloor $$  are odd. (For a real number $x,\lfloor x\rfloor$ denotes the largest integer not exceeding $x$.) (Hong Kong)

Solution: If $n$ is odd, let $k=n^{m}$ for $m=1,2, \ldots$. Then $a_{k}=n^{n^{m}-m}$, which is odd for each $m$.  Henceforth, assume that $n$ is even, say $n=2 t$ for some integer $t \geqslant 1$. Then, for any $m \geqslant 2$, the integer $n^{2^{m}}-2^{m}=2^{m}\left(2^{2^{m}-m} \cdot t^{2^{m}}-1\right)$ has an odd prime divisor $p$, since $2^{m}-m>1$. Then, for $k=p \cdot 2^{m}$, we have  $$ n^{k}=\left(n^{2^{m}}\right)^{p} \equiv\left(2^{m}\right)^{p}=\left(2^{p}\right)^{m} \equiv 2^{m} $$  where the congruences are taken modulo $p\left(\right.$ recall that $2^{p} \equiv 2(\bmod p)$, by Fermat's little theorem). Also, from $n^{k}-2^{m}<n^{k}<n^{k}+2^{m}(p-1)$, we see that the fraction $\frac{n^{k}}{k}$ lies strictly between the consecutive integers $\frac{n^{k}-2^{m}}{p \cdot 2^{m}}$ and $\frac{n^{k}+2^{m}(p-1)}{p \cdot 2^{m}}$, which gives  $$ \left\lfloor\frac{n^{k}}{k}\right\rfloor=\frac{n^{k}-2^{m}}{p \cdot 2^{m}} $$  We finally observe that $\frac{n^{k}-2^{m}}{p \cdot 2^{m}}=\frac{\frac{n^{k}}{2^{m}}-1}{p}$ is an odd integer, since the integer $\frac{n^{k}}{2^{m}}-1$ is odd (recall that $k>m$ ). Note that for different values of $m$, we get different values of $k$, due to the different powers of 2 in the prime factorisation of $k$.


Solution: Treat the (trivial) case when $n$ is odd as in  Now assume that $n$ is even and $n>2$. Let $p$ be a prime divisor of $n-1$. Proceed by induction on $i$ to prove that $p^{i+1}$ is a divisor of $n^{p^{i}}-1$ for every $i \geqslant 0$. The case $i=0$ is true by the way in which $p$ is chosen. Suppose the result is true for some $i \geqslant 0$. The factorisation  $$ n^{p^{i+1}}-1=\left(n^{p^{i}}-1\right)\left[n^{p^{i}(p-1)}+n^{p^{i}(p-2)}+\cdots+n^{p^{i}}+1\right], $$  together with the fact that each of the $p$ terms between the square brackets is congruent to 1 modulo $p$, implies that the result is also true for $i+1$.  Hence $\left\lfloor\frac{n^{p^{i}}}{p^{i}}\right\rfloor=\frac{n^{p^{i}}-1}{p^{i}}$, an odd integer for each $i \geqslant 1$. Finally, we consider the case $n=2$. We observe that $3 \cdot 4^{i}$ is a divisor of $2^{3 \cdot 4^{i}}-4^{i}$ for every $i \geqslant 1$ : Trivially, $4^{i}$ is a divisor of $2^{3 \cdot 4^{i}}-4^{i}$, since $3 \cdot 4^{i}>2 i$. Furthermore, since $2^{3 \cdot 4^{i}}$ and $4^{i}$ are both congruent to 1 modulo 3, we have $3 \mid 2^{3 \cdot 4^{i}}-4^{i}$. Hence, $\left\lfloor\frac{2^{3 \cdot 4^{i}}}{3 \cdot 4^{i}}\right\rfloor=\frac{2^{3 \cdot 4^{i}}-4^{i}}{3 \cdot 4^{i}}=\frac{2^{3 \cdot 4^{i}-2 i}-1}{3}$, which is odd for every $i \geqslant 1$.  Comment. The case $n$ even and $n>2$ can also be solved by recursively defining the sequence $\left(k_{i}\right)_{i \geqslant 1}$ by $k_{1}=1$ and $k_{i+1}=n^{k_{i}}-1$ for $i \geqslant 1$. Then $\left(k_{i}\right)$ is strictly increasing and it follows (by induction on $i$ ) that $k_{i} \mid n^{k_{i}}-1$ for all $i \geqslant 1$, so the $k_{i}$ are as desired.  The case $n=2$ can also be solved as follows: Let $i \geqslant 2$. By Bertrand's postulate, there exists a prime number $p$ such that $2^{2^{i}-1}<p \cdot 2^{i}<2^{2^{i}}$. This gives  $$ p \cdot 2^{i}<2^{2^{i}}<2 p \cdot 2^{i} . $$  Also, we have that $p \cdot 2^{i}$ is a divisor of $2^{p \cdot 2^{i}}-2^{2^{i}}$, hence, using (1), we get that  $$ \left\lfloor\frac{2^{p \cdot 2^{i}}}{p \cdot 2^{i}}\right\rfloor=\frac{2^{p \cdot 2^{i}}-2^{2^{i}}+p \cdot 2^{i}}{p \cdot 2^{i}}=\frac{2^{p \cdot 2^{i}-i}-2^{2^{i}-i}+p}{p} $$  which is an odd integer.


Solution: Treat the (trivial) case when $n$ is odd as in  Let $n$ be even, and let $p$ be a prime divisor of $n+1$. Define the sequence $\left(a_{i}\right)_{i \geqslant 1}$ by  $$ a_{i}=\min \left\{a \in \mathbb{Z}_{>0}: 2^{i} \text { divides } a p+1\right\} $$  Recall that there exists $a$ with $1 \leqslant a<2^{i}$ such that $a p \equiv-1\left(\bmod 2^{i}\right)$, so each $a_{i}$ satisfies $1 \leqslant a_{i}<2^{i}$. This implies that $a_{i} p+1<p \cdot 2^{i}$. Also, $a_{i} \rightarrow \infty$ as $i \rightarrow \infty$, whence there are infinitely many $i$ such that $a_{i}<a_{i+1}$. From now on, we restrict ourselves only to these $i$.  Notice that $p$ is a divisor of $n^{p}+1$, which, in turn, divides $n^{p \cdot 2^{i}}-1$. It follows that $p \cdot 2^{i}$ is a divisor of $n^{p \cdot 2^{i}}-\left(a_{i} p+1\right)$, and we consequently see that the integer $\left\lfloor\frac{n^{p \cdot 2^{i}}}{p \cdot 2^{i}}\right\rfloor=\frac{n^{p \cdot 2^{i}}-\left(a_{i} p+1\right)}{p \cdot 2^{i}}$ is odd, since $2^{i+1}$ divides $n^{p \cdot 2^{i}}$, but not $a_{i} p+1$.


Find all triples $(p, x, y)$ consisting of a prime number $p$ and two positive integers $x$ and $y$ such that $x^{p-1}+y$ and $x+y^{p-1}$ are both powers of $p$. (Belgium)

Solution: For $p=2$, clearly all pairs of two positive integers $x$ and $y$ whose sum is a power of 2 satisfy the condition. Thus we assume in the following that $p>2$, and we let $a$ and $b$ be positive integers such that $x^{p-1}+y=p^{a}$ and $x+y^{p-1}=p^{b}$. Assume further, without loss of generality, that $x \leqslant y$, so that $p^{a}=x^{p-1}+y \leqslant x+y^{p-1}=p^{b}$, which means that $a \leqslant b$ (and thus $\left.p^{a} \mid p^{b}\right)$.  Now we have  $$ p^{b}=y^{p-1}+x=\left(p^{a}-x^{p-1}\right)^{p-1}+x . $$  We take this equation modulo $p^{a}$ and take into account that $p-1$ is even, which gives us  $$ 0 \equiv x^{(p-1)^{2}}+x \quad\left(\bmod p^{a}\right) $$  If $p \mid x$, then $p^{a} \mid x$, since $x^{(p-1)^{2}-1}+1$ is not divisible by $p$ in this case. However, this is impossible, since $x \leqslant x^{p-1}<p^{a}$. Thus we know that $p \nmid x$, which means that  $$ p^{a} \mid x^{(p-1)^{2}-1}+1=x^{p(p-2)}+1 $$  By Fermat's little theorem, $x^{(p-1)^{2}} \equiv 1(\bmod p)$, thus $p$ divides $x+1$. Let $p^{r}$ be the highest power of $p$ that divides $x+1$. By the binomial theorem, we have  $$ x^{p(p-2)}=\sum_{k=0}^{p(p-2)}\binom{p(p-2)}{k}(-1)^{p(p-2)-k}(x+1)^{k} . $$  Except for the terms corresponding to $k=0, k=1$ and $k=2$, all terms in the sum are clearly divisible by $p^{3 r}$ and thus by $p^{r+2}$. The remaining terms are  $$ -\frac{p(p-2)\left(p^{2}-2 p-1\right)}{2}(x+1)^{2} $$  which is divisible by $p^{2 r+1}$ and thus also by $p^{r+2}$,  $$ p(p-2)(x+1) $$  which is divisible by $p^{r+1}$, but not $p^{r+2}$ by our choice of $r$, and the final term -1 corresponding to $k=0$. It follows that the highest power of $p$ that divides $x^{p(p-2)}+1$ is $p^{r+1}$.  On the other hand, we already know that $p^{a}$ divides $x^{p(p-2)}+1$, which means that $a \leqslant r+1$. Moreover,  $$ p^{r} \leqslant x+1 \leqslant x^{p-1}+y=p^{a} . $$  Hence we either have $a=r$ or $a=r+1$. If $a=r$, then $x=y=1$ needs to hold in the inequality above, which is impossible for $p>2$. Thus $a=r+1$. Now since $p^{r} \leqslant x+1$, we get  $$ x=\frac{x^{2}+x}{x+1} \leqslant \frac{x^{p-1}+y}{x+1}=\frac{p^{a}}{x+1} \leqslant \frac{p^{a}}{p^{r}}=p, $$  so we must have $x=p-1$ for $p$ to divide $x+1$. It follows that $r=1$ and $a=2$. If $p \geqslant 5$, we obtain  $$ p^{a}=x^{p-1}+y>(p-1)^{4}=\left(p^{2}-2 p+1\right)^{2}>(3 p)^{2}>p^{2}=p^{a} $$  a contradiction. So the only case that remains is $p=3$, and indeed $x=2$ and $y=p^{a}-x^{p-1}=5$ satisfy the conditions.  Comment 1. In this solution, we are implicitly using a special case of the following lemma known as "lifting the exponent": Lemma. Let $n$ be a positive integer, let $p$ be an odd prime, and let $v_{p}(m)$ denote the exponent of the highest power of $p$ that divides $m$.  If $x$ and $y$ are integers not divisible by $p$ such that $p \mid x-y$, then we have  $$ v_{p}\left(x^{n}-y^{n}\right)=v_{p}(x-y)+v_{p}(n) $$  Likewise, if $x$ and $y$ are integers not divisible by $p$ such that $p \mid x+y$, then we have  $$ v_{p}\left(x^{n}+y^{n}\right)=v_{p}(x+y)+v_{p}(n) . $$  Comment 2. There exist various ways of solving the problem involving the "lifting the exponent" lemma. Let us sketch another one.  The cases $x=y$ and $p \mid x$ are ruled out easily, so we assume that $p>2, x<y$, and $p \nmid x$. In this case we also have $p^{a}<p^{b}$ and $p \mid x+1$.  Now one has  $$ y^{p}-x^{p} \equiv y\left(y^{p-1}+x\right)-x\left(x^{p-1}+y\right) \equiv 0 \quad\left(\bmod p^{a}\right), $$  so by the lemma mentioned above one has $p^{a-1} \mid y-x$ and hence $y=x+t p^{a-1}$ for some positive integer $t$. Thus one gets  $$ x\left(x^{p-2}+1\right)=x^{p-1}+x=\left(x^{p-1}+y\right)-(y-x)=p^{a-1}(p-t) . $$  The factors on the left-hand side are coprime. So if $p \mid x$, then $x^{p-2}+1 \mid p-t$, which is impossible since $x<x^{p-2}+1$. Therefore, $p \nmid x$, and thus $x \mid p-t$. Since $p \mid x+1$, the only remaining case is $x=p-1, t=1$, and $y=p^{a-1}+p-1$. Now the solution can be completed in the same way as before.


Solution: Again, we can focus on the case that $p>2$. If $p \mid x$, then also $p \mid y$. In this case, let $p^{k}$ and $p^{\ell}$ be the highest powers of $p$ that divide $x$ and $y$ respectively, and assume without loss of generality that $k \leqslant \ell$. Then $p^{k}$ divides $x+y^{p-1}$ while $p^{k+1}$ does not, but $p^{k}<x+y^{p-1}$, which yields a contradiction. So $x$ and $y$ are not divisible by $p$. Fermat's little theorem yields $0 \equiv x^{p-1}+y \equiv 1+y(\bmod p)$, so $y \equiv-1(\bmod p)$ and for the same reason $x \equiv-1(\bmod p)$.  In particular, $x, y \geqslant p-1$ and thus $x^{p-1}+y \geqslant 2(p-1)>p$, so $x^{p-1}+y$ and $y^{p-1}+x$ are both at least equal to $p^{2}$. Now we have  $$ x^{p-1} \equiv-y \quad\left(\bmod p^{2}\right) \quad \text { and } \quad y^{p-1} \equiv-x \quad\left(\bmod p^{2}\right) $$  These two congruences, together with the Euler-Fermat theorem, give us  $$ 1 \equiv x^{p(p-1)} \equiv(-y)^{p} \equiv-y^{p} \equiv x y \quad\left(\bmod p^{2}\right) $$  Since $x \equiv y \equiv-1(\bmod p), x-y$ is divisible by $p$, so $(x-y)^{2}$ is divisible by $p^{2}$. This means that  $$ (x+y)^{2}=(x-y)^{2}+4 x y \equiv 4 \quad\left(\bmod p^{2}\right) $$  so $p^{2}$ divides $(x+y-2)(x+y+2)$. We already know that $x+y \equiv-2(\bmod p)$, so $x+y-2 \equiv$ $-4 \not \equiv 0(\bmod p)$. This means that $p^{2}$ divides $x+y+2$.  Using the same notation as in the first solution, we subtract the two original equations to obtain  $$ p^{b}-p^{a}=y^{p-1}-x^{p-1}+x-y=(y-x)\left(y^{p-2}+y^{p-3} x+\cdots+x^{p-2}-1\right) $$  The second factor is symmetric in $x$ and $y$, so it can be written as a polynomial of the elementary symmetric polynomials $x+y$ and $x y$ with integer coefficients. In particular, its value modulo $p^{2}$ is characterised by the two congruences $x y \equiv 1\left(\bmod p^{2}\right)$ and $x+y \equiv-2\left(\bmod p^{2}\right)$. Since both congruences are satisfied when $x=y=-1$, we must have  $$ y^{p-2}+y^{p-3} x+\cdots+x^{p-2}-1 \equiv(-1)^{p-2}+(-1)^{p-3}(-1)+\cdots+(-1)^{p-2}-1 \quad\left(\bmod p^{2}\right), $$  which simplifies to $y^{p-2}+y^{p-3} x+\cdots+x^{p-2}-1 \equiv-p\left(\bmod p^{2}\right)$. Thus the second factor in (1) is divisible by $p$, but not $p^{2}$.  This means that $p^{a-1}$ has to divide the other factor $y-x$. It follows that  $$ 0 \equiv x^{p-1}+y \equiv x^{p-1}+x \equiv x(x+1)\left(x^{p-3}-x^{p-4}+\cdots+1\right) \quad\left(\bmod p^{a-1}\right) . $$  Since $x \equiv-1(\bmod p)$, the last factor is $x^{p-3}-x^{p-4}+\cdots+1 \equiv p-2(\bmod p)$ and in particular not divisible by $p$. We infer that $p^{a-1} \mid x+1$ and continue as in the first solution.  Comment. Instead of reasoning by means of elementary symmetric polynomials, it is possible to provide a more direct argument as well. For odd $r,(x+1)^{2}$ divides $\left(x^{r}+1\right)^{2}$, and since $p$ divides $x+1$, we deduce that $p^{2}$ divides $\left(x^{r}+1\right)^{2}$. Together with the fact that $x y \equiv 1\left(\bmod p^{2}\right)$, we obtain  $$ 0 \equiv y^{r}\left(x^{r}+1\right)^{2} \equiv x^{2 r} y^{r}+2 x^{r} y^{r}+y^{r} \equiv x^{r}+2+y^{r} \quad\left(\bmod p^{2}\right) . $$  We apply this congruence with $r=p-2-2 k$ (where $0 \leqslant k<(p-2) / 2$ ) to find that  $$ x^{k} y^{p-2-k}+x^{p-2-k} y^{k} \equiv(x y)^{k}\left(x^{p-2-2 k}+y^{p-2-2 k}\right) \equiv 1^{k} \cdot(-2) \equiv-2 \quad\left(\bmod p^{2}\right) . $$  Summing over all $k$ yields  $$ y^{p-2}+y^{p-3} x+\cdots+x^{p-2}-1 \equiv \frac{p-1}{2} \cdot(-2)-1 \equiv-p \quad\left(\bmod p^{2}\right) $$  once again.


Let $a_{1}<a_{2}<\cdots<a_{n}$ be pairwise coprime positive integers with $a_{1}$ being prime and $a_{1} \geqslant n+2$. On the segment $I=\left[0, a_{1} a_{2} \cdots a_{n}\right]$ of the real line, mark all integers that are divisible by at least one of the numbers $a_{1}, \ldots, a_{n}$. These points split $I$ into a number of smaller segments. Prove that the sum of the squares of the lengths of these segments is divisible by $a_{1}$. (Serbia)

Solution: Let $A=a_{1} \cdots a_{n}$. Throughout the solution, all intervals will be nonempty and have integer end-points. For any interval $X$, the length of $X$ will be denoted by $|X|$.  Define the following two families of intervals:  $$ \begin{aligned} \mathcal{S} & =\{[x, y]: x<y \text { are consecutive marked points }\} \\ \mathcal{T} & =\{[x, y]: x<y \text { are integers, } 0 \leqslant x \leqslant A-1, \text { and no point is marked in }(x, y)\} \end{aligned} $$  We are interested in computing $\sum_{X \in \mathcal{S}}|X|^{2}$ modulo $a_{1}$. Note that the number $A$ is marked, so in the definition of $\mathcal{T}$ the condition $y \leqslant A$ is enforced without explicitly prescribing it.  Assign weights to the intervals in $\mathcal{T}$, depending only on their lengths. The weight of an arbitrary interval $Y \in \mathcal{T}$ will be $w(|Y|)$, where  $$ w(k)= \begin{cases}1 & \text { if } k=1 \\ 2 & \text { if } k \geqslant 2\end{cases} $$  Consider an arbitrary interval $X \in \mathcal{S}$ and its sub-intervals $Y \in \mathcal{T}$. Clearly, $X$ has one sub-interval of length $|X|$, two sub-intervals of length $|X|-1$ and so on; in general $X$ has $|X|-d+1$ sub-intervals of length $d$ for every $d=1,2, \ldots,|X|$. The sum of the weights of the sub-intervals of $X$ is $\sum_{Y \in \mathcal{T}, Y \subseteq X} w(|Y|)=\sum_{d=1}^{|X|}(|X|-d+1) \cdot w(d)=|X| \cdot 1+((|X|-1)+(|X|-2)+\cdots+1) \cdot 2=|X|^{2}$. Since the intervals in $\mathcal{S}$ are non-overlapping, every interval $Y \in \mathcal{T}$ is a sub-interval of a single interval $X \in \mathcal{S}$. Therefore,  $$ \sum_{X \in \mathcal{S}}|X|^{2}=\sum_{X \in \mathcal{S}}\left(\sum_{Y \in \mathcal{T}, Y \subseteq X} w(|Y|)\right)=\sum_{Y \in \mathcal{T}} w(|Y|) $$  For every $d=1,2, \ldots, a_{1}$, we count how many intervals in $\mathcal{T}$ are of length $d$. Notice that the multiples of $a_{1}$ are all marked, so the lengths of the intervals in $\mathcal{S}$ and $\mathcal{T}$ cannot exceed $a_{1}$. Let $x$ be an arbitrary integer with $0 \leqslant x \leqslant A-1$ and consider the interval $[x, x+d]$. Let $r_{1}$, $\ldots, r_{n}$ be the remainders of $x$ modulo $a_{1}, \ldots, a_{n}$, respectively. Since $a_{1}, \ldots, a_{n}$ are pairwise coprime, the number $x$ is uniquely identified by the sequence $\left(r_{1}, \ldots, r_{n}\right)$, due to the Chinese remainder theorem.  For every $i=1, \ldots, n$, the property that the interval $(x, x+d)$ does not contain any multiple of $a_{i}$ is equivalent with $r_{i}+d \leqslant a_{i}$, i.e. $r_{i} \in\left\{0,1, \ldots, a_{i}-d\right\}$, so there are $a_{i}-d+1$ choices for the number $r_{i}$ for each $i$. Therefore, the number of the remainder sequences $\left(r_{1}, \ldots, r_{n}\right)$ that satisfy $[x, x+d] \in \mathcal{T}$ is precisely $\left(a_{1}+1-d\right) \cdots\left(a_{n}+1-d\right)$. Denote this product by $f(d)$.  Now we can group the last sum in (1) by length of the intervals. As we have seen, for every $d=1, \ldots, a_{1}$ there are $f(d)$ intervals $Y \in \mathcal{T}$ with $|Y|=d$. Therefore, (1) can be continued as  $$ \sum_{X \in \mathcal{S}}|X|^{2}=\sum_{Y \in \mathcal{T}} w(|Y|)=\sum_{d=1}^{a_{1}} f(d) \cdot w(d)=2 \sum_{d=1}^{a_{1}} f(d)-f(1) $$  Having the formula (2), the solution can be finished using the following well-known fact: Lemma. If $p$ is a prime, $F(x)$ is a polynomial with integer coefficients, and $\operatorname{deg} F \leqslant p-2$, then $\sum_{x=1}^{p} F(x)$ is divisible by $p$. Proof. Obviously, it is sufficient to prove the lemma for monomials of the form $x^{k}$ with $k \leqslant p-2$. Apply induction on $k$. If $k=0$ then $F=1$, and the statement is trivial.  Let $1 \leqslant k \leqslant p-2$, and assume that the lemma is proved for all lower degrees. Then  $$ \begin{aligned} 0 & \equiv p^{k+1}=\sum_{x=1}^{p}\left(x^{k+1}-(x-1)^{k+1}\right)=\sum_{x=1}^{p}\left(\sum_{\ell=0}^{k}(-1)^{k-\ell}\binom{k+1}{\ell} x^{\ell}\right) \\ & =(k+1) \sum_{x=1}^{p} x^{k}+\sum_{\ell=0}^{k-1}(-1)^{k-\ell}\binom{k+1}{\ell} \sum_{x=1}^{p} x^{\ell} \equiv(k+1) \sum_{x=1}^{p} x^{k} \quad(\bmod p) . \end{aligned} $$  Since $0<k+1<p$, this proves $\sum_{x=1}^{p} x^{k} \equiv 0(\bmod p)$. In (2), by applying the lemma to the polynomial $f$ and the prime $a_{1}$, we obtain that $\sum_{d=1}^{a_{1}} f(d)$ is divisible by $a_{1}$. The term $f(1)=a_{1} \cdots a_{n}$ is also divisible by $a_{1}$; these two facts together prove that $\sum_{X \in \mathcal{S}}|X|^{2}$ is divisible by $a_{1}$.  Comment 1. With suitable sets of weights, the same method can be used to sum up other expressions on the lengths of the segments. For example, $w(1)=1$ and $w(k)=6(k-1)$ for $k \geqslant 2$ can be used to compute $\sum_{X \in \mathcal{S}}|X|^{3}$ and to prove that this sum is divisible by $a_{1}$ if $a_{1}$ is a prime with $a_{1} \geqslant n+3$. See also Comment 2 after the second solution.


Solution: The conventions from the first paragraph of the first solution are still in force. We shall prove the following more general statement: ( $\boxplus$ ) Let $p$ denote a prime number, let $p=a_{1}<a_{2}<\cdots<a_{n}$ be $n$ pairwise coprime positive integers, and let $d$ be an integer with $1 \leqslant d \leqslant p-n$. Mark all integers that are divisible by at least one of the numbers $a_{1}, \ldots, a_{n}$ on the interval $I=\left[0, a_{1} a_{2} \cdots a_{n}\right]$ of the real line. These points split I into a number of smaller segments, say of lengths $b_{1}, \ldots, b_{k}$. Then the sum $\sum_{i=1}^{k}\binom{b_{i}}{d}$ is divisible by $p$.  Applying ( $\boxplus$ ) to $d=1$ and $d=2$ and using the equation $x^{2}=2\binom{x}{2}+\binom{x}{1}$, one easily gets the statement of the problem.  To prove $(\boxplus)$ itself, we argue by induction on $n$. The base case $n=1$ follows from the known fact that the binomial coefficient $\binom{p}{d}$ is divisible by $p$ whenever $1 \leqslant d \leqslant p-1$.  Let us now assume that $n \geqslant 2$, and that the statement is known whenever $n-1$ rather than $n$ coprime integers are given together with some integer $d \in[1, p-n+1]$. Suppose that the numbers $p=a_{1}<a_{2}<\cdots<a_{n}$ and $d$ are as above. Write $A^{\prime}=\prod_{i=1}^{n-1} a_{i}$ and $A=A^{\prime} a_{n}$. Mark the points on the real axis divisible by one of the numbers $a_{1}, \ldots, a_{n-1}$ green and those divisible by $a_{n}$ red. The green points divide $\left[0, A^{\prime}\right]$ into certain sub-intervals, say $J_{1}, J_{2}, \ldots$, and $J_{\ell}$.  To translate intervals we use the notation $[a, b]+m=[a+m, b+m]$ whenever $a, b, m \in \mathbb{Z}$. For each $i \in\{1,2, \ldots, \ell\}$ let $\mathcal{F}_{i}$ be the family of intervals into which the red points partition the intervals $J_{i}, J_{i}+A^{\prime}, \ldots$, and $J_{i}+\left(a_{n}-1\right) A^{\prime}$. We are to prove that  $$ \sum_{i=1}^{\ell} \sum_{X \in \mathcal{F}_{i}}\binom{|X|}{d} $$  is divisible by $p$. Let us fix any index $i$ with $1 \leqslant i \leqslant \ell$ for a while. Since the numbers $A^{\prime}$ and $a_{n}$ are coprime by hypothesis, the numbers $0, A^{\prime}, \ldots,\left(a_{n}-1\right) A^{\prime}$ form a complete system of residues modulo $a_{n}$. Moreover, we have $\left|J_{i}\right| \leqslant p<a_{n}$, as in particular all multiples of $p$ are green. So each of the intervals $J_{i}, J_{i}+A^{\prime}, \ldots$, and $J_{i}+\left(a_{n}-1\right) A^{\prime}$ contains at most one red point. More precisely, for each $j \in\left\{1, \ldots,\left|J_{i}\right|-1\right\}$ there is exactly one amongst those intervals containing a red point splitting it into an interval of length $j$ followed by an interval of length $\left|J_{i}\right|-j$, while the remaining $a_{n}-\left|J_{i}\right|+1$ such intervals have no red points in their interiors. For these reasons  $$ \begin{aligned} \sum_{X \in \mathcal{F}_{i}}\binom{|X|}{d} & =2\left(\binom{1}{d}+\cdots+\binom{\left|J_{i}\right|-1}{d}\right)+\left(a_{n}-\left|J_{i}\right|+1\right)\binom{\left|J_{i}\right|}{d} \\ & =2\binom{\left|J_{i}\right|}{d+1}+\left(a_{n}-d+1\right)\binom{\left|J_{i}\right|}{d}-(d+1)\binom{\left|J_{i}\right|}{d+1} \\ & =(1-d)\binom{\left|J_{i}\right|}{d+1}+\left(a_{n}-d+1\right)\binom{\left|J_{i}\right|}{d} . \end{aligned} $$  So it remains to prove that  $$ (1-d) \sum_{i=1}^{\ell}\binom{\left|J_{i}\right|}{d+1}+\left(a_{n}-d+1\right) \sum_{i=1}^{\ell}\binom{\left|J_{i}\right|}{d} $$  is divisible by $p$. By the induction hypothesis, however, it is even true that both summands are divisible by $p$, for $1 \leqslant d<d+1 \leqslant p-(n-1)$. This completes the proof of ( $\boxplus$ ) and hence the solution of the problem.  Comment 2. The statement ( $\boxplus$ ) can also be proved by the method of the first solution, using the weights $w(x)=\binom{x-2}{d-2}$.  This page is intentionally left blank


Let $c \geqslant 1$ be an integer. Define a sequence of positive integers by $a_{1}=c$ and  $$ a_{n+1}=a_{n}^{3}-4 c \cdot a_{n}^{2}+5 c^{2} \cdot a_{n}+c $$  for all $n \geqslant 1$. Prove that for each integer $n \geqslant 2$ there exists a prime number $p$ dividing $a_{n}$ but none of the numbers $a_{1}, \ldots, a_{n-1}$. (Austria)

Solution: Let us define $x_{0}=0$ and $x_{n}=a_{n} / c$ for all integers $n \geqslant 1$. It is easy to see that the sequence $\left(x_{n}\right)$ thus obtained obeys the recursive law  $$ x_{n+1}=c^{2}\left(x_{n}^{3}-4 x_{n}^{2}+5 x_{n}\right)+1 $$  for all integers $n \geqslant 0$. In particular, all of its terms are positive integers; notice that $x_{1}=1$ and $x_{2}=2 c^{2}+1$. Since  $$ x_{n+1}=c^{2} x_{n}\left(x_{n}-2\right)^{2}+c^{2} x_{n}+1>x_{n} $$  holds for all integers $n \geqslant 0$, it is also strictly increasing. Since $x_{n+1}$ is by (1) coprime to $c$ for any $n \geqslant 0$, it suffices to prove that for each $n \geqslant 2$ there exists a prime number $p$ dividing $x_{n}$ but none of the numbers $x_{1}, \ldots, x_{n-1}$. Let us begin by establishing three preliminary claims. Claim 1. If $i \equiv j(\bmod m)$ holds for some integers $i, j \geqslant 0$ and $m \geqslant 1$, then $x_{i} \equiv x_{j}\left(\bmod x_{m}\right)$ holds as well. Proof. Evidently, it suffices to show $x_{i+m} \equiv x_{i}\left(\bmod x_{m}\right)$ for all integers $i \geqslant 0$ and $m \geqslant 1$. For this purpose we may argue for fixed $m$ by induction on $i$ using $x_{0}=0$ in the base case $i=0$. Now, if we have $x_{i+m} \equiv x_{i}\left(\bmod x_{m}\right)$ for some integer $i$, then the recursive equation (1) yields  $$ x_{i+m+1} \equiv c^{2}\left(x_{i+m}^{3}-4 x_{i+m}^{2}+5 x_{i+m}\right)+1 \equiv c^{2}\left(x_{i}^{3}-4 x_{i}^{2}+5 x_{i}\right)+1 \equiv x_{i+1} \quad\left(\bmod x_{m}\right), $$  which completes the induction. Claim 2. If the integers $i, j \geqslant 2$ and $m \geqslant 1$ satisfy $i \equiv j(\bmod m)$, then $x_{i} \equiv x_{j}\left(\bmod x_{m}^{2}\right)$ holds as well. Proof. Again it suffices to prove $x_{i+m} \equiv x_{i}\left(\bmod x_{m}^{2}\right)$ for all integers $i \geqslant 2$ and $m \geqslant 1$. As above, we proceed for fixed $m$ by induction on $i$. The induction step is again easy using (1), but this time the base case $i=2$ requires some calculation. Set $L=5 c^{2}$. By (1) we have $x_{m+1} \equiv L x_{m}+1\left(\bmod x_{m}^{2}\right)$, and hence  $$ \begin{aligned} x_{m+1}^{3}-4 x_{m+1}^{2}+5 x_{m+1} & \equiv\left(L x_{m}+1\right)^{3}-4\left(L x_{m}+1\right)^{2}+5\left(L x_{m}+1\right) \\ & \equiv\left(3 L x_{m}+1\right)-4\left(2 L x_{m}+1\right)+5\left(L x_{m}+1\right) \equiv 2 \quad\left(\bmod x_{m}^{2}\right) \end{aligned} $$  which in turn gives indeed $x_{m+2} \equiv 2 c^{2}+1 \equiv x_{2}\left(\bmod x_{m}^{2}\right)$. Claim 3. For each integer $n \geqslant 2$, we have $x_{n}>x_{1} \cdot x_{2} \cdots x_{n-2}$. Proof. The cases $n=2$ and $n=3$ are clear. Arguing inductively, we assume now that the claim holds for some $n \geqslant 3$. Recall that $x_{2} \geqslant 3$, so by monotonicity and (2) we get $x_{n} \geqslant x_{3} \geqslant x_{2}\left(x_{2}-2\right)^{2}+x_{2}+1 \geqslant 7$. It follows that  $$ x_{n+1}>x_{n}^{3}-4 x_{n}^{2}+5 x_{n}>7 x_{n}^{2}-4 x_{n}^{2}>x_{n}^{2}>x_{n} x_{n-1} $$  which by the induction hypothesis yields $x_{n+1}>x_{1} \cdot x_{2} \cdots x_{n-1}$, as desired.  Now we direct our attention to the problem itself: let any integer $n \geqslant 2$ be given. By Claim 3 there exists a prime number $p$ appearing with a higher exponent in the prime factorisation of $x_{n}$ than in the prime factorisation of $x_{1} \cdots x_{n-2}$. In particular, $p \mid x_{n}$, and it suffices to prove that $p$ divides none of $x_{1}, \ldots, x_{n-1}$.  Otherwise let $k \in\{1, \ldots, n-1\}$ be minimal such that $p$ divides $x_{k}$. Since $x_{n-1}$ and $x_{n}$ are coprime by (1) and $x_{1}=1$, we actually have $2 \leqslant k \leqslant n-2$. Write $n=q k+r$ with some integers $q \geqslant 0$ and $0 \leqslant r<k$. By Claim 1 we have $x_{n} \equiv x_{r}\left(\bmod x_{k}\right)$, whence $p \mid x_{r}$. Due to the minimality of $k$ this entails $r=0$, i.e. $k \mid n$.  Thus from Claim 2 we infer  $$ x_{n} \equiv x_{k} \quad\left(\bmod x_{k}^{2}\right) . $$  Now let $\alpha \geqslant 1$ be maximal with the property $p^{\alpha} \mid x_{k}$. Then $x_{k}^{2}$ is divisible by $p^{\alpha+1}$ and by our choice of $p$ so is $x_{n}$. So by the previous congruence $x_{k}$ is a multiple of $p^{\alpha+1}$ as well, contrary to our choice of $\alpha$. This is the final contradiction concluding the solution.


For every real number $x$, let $\|x\|$ denote the distance between $x$ and the nearest integer. Prove that for every pair $(a, b)$ of positive integers there exist an odd prime $p$ and a positive integer $k$ satisfying  $$ \left\|\frac{a}{p^{k}}\right\|+\left\|\frac{b}{p^{k}}\right\|+\left\|\frac{a+b}{p^{k}}\right\|=1 . $$  (Hungary)

Solution: Notice first that $\left\lfloor x+\frac{1}{2}\right\rfloor$ is an integer nearest to $x$, so $\|x\|=\left\lfloor\left.\left\lfloor x+\frac{1}{2}\right\rfloor-x \right\rvert\,\right.$. Thus we have  $$ \left\lfloor x+\frac{1}{2}\right\rfloor=x \pm\|x\| . $$  For every rational number $r$ and every prime number $p$, denote by $v_{p}(r)$ the exponent of $p$ in the prime factorisation of $r$. Recall the notation $(2 n-1)!$ for the product of all odd positive integers not exceeding $2 n-1$, i.e., $(2 n-1)!!=1 \cdot 3 \cdots(2 n-1)$. Lemma. For every positive integer $n$ and every odd prime $p$, we have  $$ v_{p}((2 n-1)!!)=\sum_{k=1}^{\infty}\left\lfloor\frac{n}{p^{k}}+\frac{1}{2}\right\rfloor . $$  Proof. For every positive integer $k$, let us count the multiples of $p^{k}$ among the factors $1,3, \ldots$, $2 n-1$. If $\ell$ is an arbitrary integer, the number $(2 \ell-1) p^{k}$ is listed above if and only if  $$ 0<(2 \ell-1) p^{k} \leqslant 2 n \quad \Longleftrightarrow \quad \frac{1}{2}<\ell \leqslant \frac{n}{p^{k}}+\frac{1}{2} \quad \Longleftrightarrow \quad 1 \leqslant \ell \leqslant\left\lfloor\frac{n}{p^{k}}+\frac{1}{2}\right\rfloor $$  Hence, the number of multiples of $p^{k}$ among the factors is precisely $m_{k}=\left\lfloor\frac{n}{p^{k}}+\frac{1}{2}\right\rfloor$. Thus we obtain  $$ v_{p}((2 n-1)!!)=\sum_{i=1}^{n} v_{p}(2 i-1)=\sum_{i=1}^{n} \sum_{k=1}^{v_{p}(2 i-1)} 1=\sum_{k=1}^{\infty} \sum_{\ell=1}^{m_{k}} 1=\sum_{k=1}^{\infty}\left\lfloor\frac{n}{p^{k}}+\frac{1}{2}\right\rfloor $$  In order to prove the problem statement, consider the rational number  $$ N=\frac{(2 a+2 b-1)!!}{(2 a-1)!!\cdot(2 b-1)!!}=\frac{(2 a+1)(2 a+3) \cdots(2 a+2 b-1)}{1 \cdot 3 \cdots(2 b-1)} $$  Obviously, $N>1$, so there exists a prime $p$ with $v_{p}(N)>0$. Since $N$ is a fraction of two odd numbers, $p$ is odd.  By our lemma,  $$ 0<v_{p}(N)=\sum_{k=1}^{\infty}\left(\left\lfloor\frac{a+b}{p^{k}}+\frac{1}{2}\right\rfloor-\left\lfloor\frac{a}{p^{k}}+\frac{1}{2}\right\rfloor-\left\lfloor\frac{b}{p^{k}}+\frac{1}{2}\right\rfloor\right) . $$  Therefore, there exists some positive integer $k$ such that the integer number  $$ d_{k}=\left\lfloor\frac{a+b}{p^{k}}+\frac{1}{2}\right\rfloor-\left\lfloor\frac{a}{p^{k}}+\frac{1}{2}\right\rfloor-\left\lfloor\frac{b}{p^{k}}+\frac{1}{2}\right\rfloor $$  is positive, so $d_{k} \geqslant 1$. By (2) we have  $$ 1 \leqslant d_{k}=\frac{a+b}{p^{k}}-\frac{a}{p^{k}}-\frac{b}{p^{k}} \pm\left\|\frac{a+b}{p^{k}}\right\| \pm\left\|\frac{a}{p^{k}}\right\| \pm\left\|\frac{b}{p^{k}}\right\|= \pm\left\|\frac{a+b}{p^{k}}\right\| \pm\left\|\frac{a}{p^{k}}\right\| \pm\left\|\frac{b}{p^{k}}\right\| $$  Since $\|x\|<\frac{1}{2}$ for every rational $x$ with odd denominator, the relation (3) can only be satisfied if all three signs on the right-hand side are positive and $d_{k}=1$. Thus we get  $$ \left\|\frac{a}{p^{k}}\right\|+\left\|\frac{b}{p^{k}}\right\|+\left\|\frac{a+b}{p^{k}}\right\|=d_{k}=1, $$  as required. Comment 1. There are various choices for the number $N$ in the solution. Here we sketch such a version.  Let $x$ and $y$ be two rational numbers with odd denominators. It is easy to see that the condition $\|x\|+\|y\|+\|x+y\|=1$ is satisfied if and only if  $$ \text { either } \quad\{x\}<\frac{1}{2}, \quad\{y\}<\frac{1}{2}, \quad\{x+y\}>\frac{1}{2}, \quad \text { or } \quad\{x\}>\frac{1}{2}, \quad\{y\}>\frac{1}{2}, \quad\{x+y\}<\frac{1}{2}, $$  where $\{x\}$ denotes the fractional part of $x$. In the context of our problem, the first condition seems easier to deal with. Also, one may notice that  $$ \{x\}<\frac{1}{2} \Longleftrightarrow \varkappa(x)=0 \quad \text { and } \quad\{x\} \geqslant \frac{1}{2} \Longleftrightarrow \varkappa(x)=1, $$  where  $$ \varkappa(x)=\lfloor 2 x\rfloor-2\lfloor x\rfloor . $$  Now it is natural to consider the number  $$ M=\frac{\binom{2 a+2 b}{a+b}}{\binom{2 a}{a}\binom{2 b}{b}} $$  since  $$ v_{p}(M)=\sum_{k=1}^{\infty}\left(\varkappa\left(\frac{2(a+b)}{p^{k}}\right)-\varkappa\left(\frac{2 a}{p^{k}}\right)-\varkappa\left(\frac{2 b}{p^{k}}\right)\right) . $$  One may see that $M>1$, and that $v_{2}(M) \leqslant 0$. Thus, there exist an odd prime $p$ and a positive integer $k$ with  $$ \varkappa\left(\frac{2(a+b)}{p^{k}}\right)-\varkappa\left(\frac{2 a}{p^{k}}\right)-\varkappa\left(\frac{2 b}{p^{k}}\right)>0 . $$  In view of (4), the last inequality yields  $$ \left\{\frac{a}{p^{k}}\right\}<\frac{1}{2}, \quad\left\{\frac{b}{p^{k}}\right\}<\frac{1}{2}, \quad \text { and } \quad\left\{\frac{a+b}{p^{k}}\right\}>\frac{1}{2} $$  which is what we wanted to obtain. Comment 2. Once one tries to prove the existence of suitable $p$ and $k$ satisfying (5), it seems somehow natural to suppose that $a \leqslant b$ and to add the restriction $p^{k}>a$. In this case the inequalities (5) can be rewritten as  $$ 2 a<p^{k}, \quad 2 m p^{k}<2 b<(2 m+1) p^{k}, \quad \text { and } \quad(2 m+1) p^{k}<2(a+b)<(2 m+2) p^{k} $$  for some positive integer $m$. This means exactly that one of the numbers $2 a+1,2 a+3, \ldots, 2 a+2 b-1$ is divisible by some number of the form $p^{k}$ which is greater than $2 a$.  Using more advanced techniques, one can show that such a number $p^{k}$ exists even with $k=1$. This was shown in 2004 by Laishram and Shorey; the methods used for this proof are elementary but still quite involved. In fact, their result generalises a theorem by SyLvester which states that for every pair of integers $(n, k)$ with $n \geqslant k \geqslant 1$, the product $(n+1)(n+2) \cdots(n+k)$ is divisible by some prime $p>k$. We would like to mention here that Sylvester's theorem itself does not seem to suffice for solving the problem.


Let $z_{0}<z_{1}<z_{2}<\cdots$ be an infinite sequence of positive integers. Prove that there exists a unique integer $n \geqslant 1$ such that  $$ z_{n}<\frac{z_{0}+z_{1}+\cdots+z_{n}}{n} \leqslant z_{n+1} . $$  (Austria)

Solution: For $n=1,2, \ldots$ define  $$ d_{n}=\left(z_{0}+z_{1}+\cdots+z_{n}\right)-n z_{n} $$  The sign of $d_{n}$ indicates whether the first inequality in (1) holds; i.e., it is satisfied if and only if $d_{n}>0$.  Notice that  $$ n z_{n+1}-\left(z_{0}+z_{1}+\cdots+z_{n}\right)=(n+1) z_{n+1}-\left(z_{0}+z_{1}+\cdots+z_{n}+z_{n+1}\right)=-d_{n+1} $$  so the second inequality in (1) is equivalent to $d_{n+1} \leqslant 0$. Therefore, we have to prove that there is a unique index $n \geqslant 1$ that satisfies $d_{n}>0 \geqslant d_{n+1}$.  By its definition the sequence $d_{1}, d_{2}, \ldots$ consists of integers and we have  $$ d_{1}=\left(z_{0}+z_{1}\right)-1 \cdot z_{1}=z_{0}>0 $$  From $d_{n+1}-d_{n}=\left(\left(z_{0}+\cdots+z_{n}+z_{n+1}\right)-(n+1) z_{n+1}\right)-\left(\left(z_{0}+\cdots+z_{n}\right)-n z_{n}\right)=n\left(z_{n}-z_{n+1}\right)<0$ we can see that $d_{n+1}<d_{n}$ and thus the sequence strictly decreases. Hence, we have a decreasing sequence $d_{1}>d_{2}>\ldots$ of integers such that its first element $d_{1}$ is positive. The sequence must drop below 0 at some point, and thus there is a unique index $n$, that is the index of the last positive term, satisfying $d_{n}>0 \geqslant d_{n+1}$.  Comment. Omitting the assumption that $z_{1}, z_{2}, \ldots$ are integers allows the numbers $d_{n}$ to be all positive. In such cases the desired $n$ does not exist. This happens for example if $z_{n}=2-\frac{1}{2^{n}}$ for all integers $n \geqslant 0$.