In the plane, there are $n \geqslant 6$ pairwise disjoint disks $D_{1}, D_{2}, \ldots, D_{n}$ with radii $R_{1} \geqslant R_{2} \geqslant \ldots \geqslant R_{n}$. For every $i=1,2, \ldots, n$, a point $P_{i}$ is chosen in disk $D_{i}$. Let $O$ be an arbitrary point in the plane. Prove that  $$ O P_{1}+O P_{2}+\ldots+O P_{n} \geqslant R_{6}+R_{7}+\ldots+R_{n} $$  (A disk is assumed to contain its boundary.) (Iran)

Solution: We will make use of the following lemma. Lemma. Let $D_{1}, \ldots, D_{6}$ be disjoint disks in the plane with radii $R_{1}, \ldots, R_{6}$. Let $P_{i}$ be a point in $D_{i}$, and let $O$ be an arbitrary point. Then there exist indices $i$ and $j$ such that $O P_{i} \geqslant R_{j}$. Proof. Let $O_{i}$ be the center of $D_{i}$. Consider six rays $O O_{1}, \ldots, O O_{6}$ (if $O=O_{i}$, then the ray $O O_{i}$ may be assumed to have an arbitrary direction). These rays partition the plane into six angles (one of which may be non-convex) whose measures sum up to $360^{\circ}$; hence one of the angles, say $\angle O_{i} O O_{j}$, has measure at most $60^{\circ}$. Then $O_{i} O_{j}$ cannot be the unique largest side in (possibly degenerate) triangle $O O_{i} O_{j}$, so, without loss of generality, $O O_{i} \geqslant O_{i} O_{j} \geqslant R_{i}+R_{j}$. Therefore, $O P_{i} \geqslant O O_{i}-R_{i} \geqslant\left(R_{i}+R_{j}\right)-R_{i}=R_{j}$, as desired.  Now we prove the required inequality by induction on $n \geqslant 5$. The base case $n=5$ is trivial. For the inductive step, apply the Lemma to the six largest disks, in order to find indices $i$ and $j$ such that $1 \leqslant i, j \leqslant 6$ and $O P_{i} \geqslant R_{j} \geqslant R_{6}$. Removing $D_{i}$ from the configuration and applying the inductive hypothesis, we get  $$ \sum_{k \neq i} O P_{k} \geqslant \sum_{\ell \geqslant 7} R_{\ell} . $$  Adding up this inequality with $O P_{i} \geqslant R_{6}$ we establish the inductive step. Comment 1. It is irrelevant to the problem whether the disks contain their boundaries or not. This condition is included for clarity reasons only. The problem statement remains true, and the solution works verbatim, if the disks are assumed to have disjoint interiors.  Comment 2. There are several variations of the above solution. In particular, while performing the inductive step, one may remove the disk with the largest value of $O P_{i}$ and apply the inductive hypothesis to the remaining disks (the Lemma should still be applied to the six largest disks).  Comment 3. While proving the Lemma, one may reduce it to a particular case when the disks are congruent, as follows: Choose the smallest radius $r$ of the disks in the Lemma statement, and then replace, for each $i$, the $i^{\text {th }}$ disk with its homothetic copy, using the homothety centered at $P_{i}$ with ratio $r / R_{i}$.  This argument shows that the Lemma is tightly connected to a circle packing problem, see, e.g., https://en.wikipedia.org/wiki/Circle_packing_in_a_circle. The known results on that problem provide versions of the Lemma for different numbers of disks, which lead to different inequalities of the same kind. E.g., for 4 disks the best possible estimate in the Lemma is $O P_{i} \geqslant(\sqrt{2}-1) R_{j}$, while for 13 disks it has the form $O P_{i} \geqslant \sqrt{5} R_{j}$. Arguing as in the above solution, one obtains the inequalities  $$ \sum_{i=1}^{n} O P_{i} \geqslant(\sqrt{2}-1) \sum_{j=4}^{n} R_{j} \quad \text { and } \quad \sum_{i=1}^{n} O P_{i} \geqslant \sqrt{5} \sum_{j=13}^{n} R_{j} . $$  However, there are some harder arguments which allow to improve these inequalities, meaning that the $R_{j}$ with large indices may be taken with much greater factors.


Let $A B C D$ be a cyclic quadrilateral with no two sides parallel. Let $K, L, M$, and $N$ be points lying on sides $A B, B C, C D$, and $D A$, respectively, such that $K L M N$ is a rhombus with $K L \| A C$ and $L M \| B D$. Let $\omega_{1}, \omega_{2}, \omega_{3}$, and $\omega_{4}$ be the incircles of triangles $A N K$, $B K L, C L M$, and $D M N$, respectively. Prove that the internal common tangents to $\omega_{1}$ and $\omega_{3}$ and the internal common tangents to $\omega_{2}$ and $\omega_{4}$ are concurrent. (Poland)

Solution: Let $I_{i}$ be the center of $\omega_{i}$, and let $r_{i}$ be its radius for $i=1,2,3,4$. Denote by $T_{1}$ and $T_{3}$ the points of tangency of $\omega_{1}$ and $\omega_{3}$ with $N K$ and $L M$, respectively. Suppose that the internal common tangents to $\omega_{1}$ and $\omega_{3}$ meet at point $S$, which is the center of homothety $h$ with negative ratio (namely, with ratio $-\frac{r_{3}}{r_{1}}$ ) mapping $\omega_{1}$ to $\omega_{3}$. This homothety takes $T_{1}$ to $T_{3}$ (since the tangents to $\omega_{1}$ and $\omega_{3}$ at $T_{1}$ to $T_{3}$ are parallel), hence $S$ is a point on the segment $T_{1} T_{3}$ with $T_{1} S: S T_{3}=r_{1}: r_{3}$.  Construct segments $S_{1} S_{3} \| K L$ and $S_{2} S_{4} \| L M$ through $S$ with $S_{1} \in N K, S_{2} \in K L$, $S_{3} \in L M$, and $S_{4} \in M N$. Note that $h$ takes $S_{1}$ to $S_{3}$, hence $I_{1} S_{1} \| I_{3} S_{3}$, and $S_{1} S: S S_{3}=r_{1}: r_{3}$. We will prove that $S_{2} S: S S_{4}=r_{2}: r_{4}$ or, equivalently, $K S_{1}: S_{1} N=r_{2}: r_{4}$. This will yield the problem statement; indeed, applying similar arguments to the intersection point $S^{\prime}$ of the internal common tangents to $\omega_{2}$ and $\omega_{4}$, we see that $S^{\prime}$ satisfies similar relations, and there is a unique point inside $K L M N$ satisfying them. Therefore, $S^{\prime}=S$. ![](https://cdn.mathpix.com/cropped/2024_11_18_bcdd154f2030efbe2fa4g-56.jpg?height=670&width=1409&top_left_y=1207&top_left_x=326)  Further, denote by $I_{A}, I_{B}, I_{C}, I_{D}$ and $r_{A}, r_{B}, r_{C}, r_{D}$ the incenters and inradii of triangles $D A B, A B C, B C D$, and $C D A$, respectively. One can shift triangle $C L M$ by $\overrightarrow{L K}$ to glue it with triangle $A K N$ into a quadrilateral $A K C^{\prime} N$ similar to $A B C D$. In particular, this shows that $r_{1}: r_{3}=r_{A}: r_{C}$; similarly, $r_{2}: r_{4}=r_{B}: r_{D}$. Moreover, the same shift takes $S_{3}$ to $S_{1}$, and it also takes $I_{3}$ to the incenter $I_{3}^{\prime}$ of triangle $K C^{\prime} N$. Since $I_{1} S_{1} \| I_{3} S_{3}$, the points $I_{1}, S_{1}, I_{3}^{\prime}$ are collinear. Thus, in order to complete the solution, it suffices to apply the following Lemma to quadrilateral $A K C^{\prime} N$. Lemma 1. Let $A B C D$ be a cyclic quadrilateral, and define $I_{A}, I_{C}, r_{B}$, and $r_{D}$ as above. Let $I_{A} I_{C}$ meet $B D$ at $X$; then $B X: X D=r_{B}: r_{D}$. Proof. Consider an inversion centered at $X$; the images under that inversion will be denoted by primes, e.g., $A^{\prime}$ is the image of $A$.  By properties of inversion, we have  $$ \angle I_{C}^{\prime} I_{A}^{\prime} D^{\prime}=\angle X I_{A}^{\prime} D^{\prime}=\angle X D I_{A}=\angle B D A / 2=\angle B C A / 2=\angle A C I_{B} $$  We obtain $\angle I_{A}^{\prime} I_{C}^{\prime} D^{\prime}=\angle C A I_{B}$ likewise; therefore, $\triangle I_{C}^{\prime} I_{A}^{\prime} D^{\prime} \sim \triangle A C I_{B}$. In the same manner, we get $\triangle I_{C}^{\prime} I_{A}^{\prime} B^{\prime} \sim \triangle A C I_{D}$, hence the quadrilaterals $I_{C}^{\prime} B^{\prime} I_{A}^{\prime} D^{\prime}$ and $A I_{D} C I_{B}$ are also similar. But the diagonals $A C$ and $I_{B} I_{D}$ of quadrilateral $A I_{D} C I_{B}$ meet at a point $Y$ such that $I_{B} Y$ : $Y I_{D}=r_{B}: r_{D}$. By similarity, we get $D^{\prime} X: B^{\prime} X=r_{B}: r_{D}$ and hence $B X: X D=D^{\prime} X:$ $B^{\prime} X=r_{B}: r_{D}$.  Comment 1. The solution above shows that the problem statement holds also for any parallelogram $K L M N$ whose sides are parallel to the diagonals of $A B C D$, as no property specific for a rhombus has been used. This solution works equally well when two sides of quadrilateral $A B C D$ are parallel.  Comment 2. The problem may be reduced to Lemma 1 by using different tools, e.g., by using mass point geometry, linear motion of $K, L, M$, and $N$, etc.  Lemma 1 itself also can be proved in different ways. We present below one alternative proof. Proof. In the circumcircle of $A B C D$, let $K^{\prime}, L^{\prime} . M^{\prime}$, and $N^{\prime}$ be the midpoints of arcs $A B, B C$, $C D$, and $D A$ containing no other vertices of $A B C D$, respectively. Thus, $K^{\prime}=C I_{B} \cap D I_{A}$, etc. In the computations below, we denote by $[P]$ the area of a polygon $P$. We use similarities $\triangle I_{A} B K^{\prime} \sim$ $\triangle I_{A} D N^{\prime}, \triangle I_{B} K^{\prime} L^{\prime} \sim \triangle I_{B} A C$, etc., as well as congruences $\triangle I_{B} K^{\prime} L^{\prime}=\triangle B K^{\prime} L^{\prime}$ and $\triangle I_{D} M^{\prime} N^{\prime}=$ $\triangle D M^{\prime} N^{\prime}$ (e.g., the first congruence holds because $K^{\prime} L^{\prime}$ is a common internal bisector of angles $B K^{\prime} I_{B}$ and $B L^{\prime} I_{B}$ ).  We have  $$ \begin{aligned} & \frac{B X}{D X}=\frac{\left[I_{A} B I_{C}\right]}{\left[I_{A} D I_{C}\right]}=\frac{B I_{A} \cdot B I_{C} \cdot \sin I_{A} B I_{C}}{D I_{A} \cdot D I_{C} \cdot \sin I_{A} D I_{C}}=\frac{B I_{A}}{D I_{A}} \cdot \frac{B I_{C}}{D I_{C}} \cdot \frac{\sin N^{\prime} B M^{\prime}}{\sin K^{\prime} D L^{\prime}} \\ & =\frac{B K^{\prime}}{D N^{\prime}} \cdot \frac{B L^{\prime}}{D M^{\prime}} \cdot \frac{\sin N^{\prime} D M^{\prime}}{\sin K^{\prime} B L^{\prime}}=\frac{B K^{\prime} \cdot B L^{\prime} \cdot \sin K^{\prime} B L^{\prime}}{D N^{\prime} \cdot D M^{\prime} \cdot \sin N^{\prime} D M^{\prime}} \cdot \frac{\sin ^{2} N^{\prime} D M^{\prime}}{\sin ^{2} K^{\prime} B L^{\prime}} \\ & \quad=\frac{\left[K^{\prime} B L^{\prime}\right]}{\left[M^{\prime} D N^{\prime}\right]} \cdot \frac{N^{\prime} M^{\prime 2}}{K^{\prime} L^{\prime 2}}=\frac{\left[K^{\prime} I_{B} L^{\prime}\right] \cdot \frac{A^{\prime} C^{\prime 2}}{K^{\prime} L^{\prime 2}}}{\left[M^{\prime} I_{D} N^{\prime}\right] \cdot \frac{A^{\prime} C^{\prime 2}}{N^{\prime} M^{\prime 2}}}=\frac{\left[A I_{B} C\right]}{\left[A I_{D} C\right]}=\frac{r_{B}}{r_{D}}, \end{aligned} $$  as required.


Solution: This solution is based on the following general Lemma. Lemma 2. Let $E$ and $F$ be distinct points, and let $\omega_{i}, i=1,2,3,4$, be circles lying in the same halfplane with respect to $E F$. For distinct indices $i, j \in\{1,2,3,4\}$, denote by $O_{i j}^{+}$ (respectively, $O_{i j}^{-}$) the center of homothety with positive (respectively, negative) ratio taking $\omega_{i}$ to $\omega_{j}$. Suppose that $E=O_{12}^{+}=O_{34}^{+}$and $F=O_{23}^{+}=O_{41}^{+}$. Then $O_{13}^{-}=O_{24}^{-}$. Proof. Applying Monge's theorem to triples of circles $\omega_{1}, \omega_{2}, \omega_{4}$ and $\omega_{1}, \omega_{3}, \omega_{4}$, we get that both points $O_{24}^{-}$and $O_{13}^{-}$lie on line $E O_{14}^{-}$. Notice that this line is distinct from $E F$. Similarly we obtain that both points $O_{24}^{-}$and $O_{13}^{-}$lie on $F O_{34}^{-}$. Since the lines $E O_{14}^{-}$and $F O_{34}^{-}$are distinct, both points coincide with the meeting point of those lines. ![](https://cdn.mathpix.com/cropped/2024_11_18_bcdd154f2030efbe2fa4g-57.jpg?height=716&width=1837&top_left_y=2073&top_left_x=108)  Turning back to the problem, let $A B$ intersect $C D$ at $E$ and let $B C$ intersect $D A$ at $F$. Assume, without loss of generality, that $B$ lies on segments $A E$ and $C F$. We will show that the points $E$ and $F$, and the circles $\omega_{i}$ satisfy the conditions of Lemma 2 , so the problem statement follows. In the sequel, we use the notation of $O_{i j}^{ \pm}$from the statement of Lemma 2, applied to circles $\omega_{1}, \omega_{2}, \omega_{3}$, and $\omega_{4}$.  Using the relations $\triangle E C A \sim \triangle E B D, K N \| B D$, and $M N \| A C$. we get  $$ \frac{A N}{N D}=\frac{A N}{A D} \cdot \frac{A D}{N D}=\frac{K N}{B D} \cdot \frac{A C}{N M}=\frac{A C}{B D}=\frac{A E}{E D} $$  Therefore, by the angle bisector theorem, point $N$ lies on the internal angle bisector of $\angle A E D$. We prove similarly that $L$ also lies on that bisector, and that the points $K$ and $M$ lie on the internal angle bisector of $\angle A F B$.  Since $K L M N$ is a rhombus, points $K$ and $M$ are symmetric in line $E L N$. Hence, the convex quadrilateral determined by the lines $E K, E M, K L$, and $M L$ is a kite, and therefore it has an incircle $\omega_{0}$. Applying Monge's theorem to $\omega_{0}, \omega_{2}$, and $\omega_{3}$, we get that $O_{23}^{+}$lies on $K M$. On the other hand, $O_{23}^{+}$lies on $B C$, as $B C$ is an external common tangent to $\omega_{2}$ and $\omega_{3}$. It follows that $F=O_{23}^{+}$. Similarly, $E=O_{12}^{+}=O_{34}^{+}$, and $F=O_{41}^{+}$.  Comment 3. The reduction to Lemma 2 and the proof of Lemma 2 can be performed with the use of different tools, e.g., by means of Menelaus theorem, by projecting harmonic quadruples, by applying Monge's theorem in some other ways, etc.  This page is intentionally left blank


Let $I$ and $I_{A}$ be the incenter and the $A$-excenter of an acute-angled triangle $A B C$ with $A B<A C$. Let the incircle meet $B C$ at $D$. The line $A D$ meets $B I_{A}$ and $C I_{A}$ at $E$ and $F$, respectively. Prove that the circumcircles of triangles $A I D$ and $I_{A} E F$ are tangent to each other.

Solution: Let $\Varangle(p, q)$ denote the directed angle between lines $p$ and $q$. The points $B, C, I$, and $I_{A}$ lie on the circle $\Gamma$ with diameter $I I_{A}$. Let $\omega$ and $\Omega$ denote the circles $\left(I_{A} E F\right)$ and $(A I D)$, respectively. Let $T$ be the second intersection point of $\omega$ and $\Gamma$. Then $T$ is the Miquel point of the complete quadrilateral formed by the lines $B C, B I_{A}, C I_{A}$, and $D E F$, so $T$ also lies on circle $(B D E)$ (as well as on circle $(C D F)$ ). We claim that $T$ is a desired tangency point of $\omega$ and $\Omega$.  In order to show that $T$ lies on $\Omega$, use cyclic quadrilaterals $B D E T$ and $B I I_{A} T$ to write  $$ \Varangle(D T, D A)=\Varangle(D T, D E)=\Varangle(B T, B E)=\Varangle\left(B T, B I_{A}\right)=\Varangle\left(I T, I I_{A}\right)=\Varangle(I T, I A) . $$  ![](https://cdn.mathpix.com/cropped/2024_11_18_bcdd154f2030efbe2fa4g-60.jpg?height=1121&width=794&top_left_y=1047&top_left_x=651)  To show that $\omega$ and $\Omega$ are tangent at $T$, let $\ell$ be the tangent to $\omega$ at $T$, so that $\Varangle\left(T I_{A}, \ell\right)=$ $\Varangle\left(E I_{A}, E T\right)$. Using circles $(B D E T)$ and $\left(B I C I_{A}\right)$, we get  $$ \Varangle\left(E I_{A}, E T\right)=\Varangle(E B, E T)=\Varangle(D B, D T) . $$  Therefore,  $$ \Varangle(T I, \ell)=90^{\circ}+\Varangle\left(T I_{A}, \ell\right)=90^{\circ}+\Varangle(D B, D T)=\Varangle(D I, D T), $$  which shows that $\ell$ is tangent to $\Omega$ at $T$.


Solution: We use the notation of circles $\Gamma, \omega$, and $\Omega$ as in the previous solution. Let $L$ be the point opposite to $I$ in circle $\Omega$. Then $\angle I A L=\angle I D L=90^{\circ}$, which means that $L$ is the foot of the external bisector of $\angle A$ in triangle $A B C$. Let $L I$ cross $\Gamma$ again at $M$.  Let $T$ be the foot of the perpendicular from $I$ onto $I_{A} L$. Then $T$ is the second intersection point of $\Gamma$ and $\Gamma$. We will show that $T$ is the desired tangency point.  First, we show that $T$ lies on circle $\omega$. Notice that  $$ \Varangle(L T, L M)=\Varangle(A T, A I) \quad \text { and } \quad \Varangle(M T, M L)=\Varangle(M T, M I)=\Varangle\left(I_{A} T, I_{A} I\right), $$  which shows that triangles $T M L$ and $T I_{A} A$ are similar and equioriented. So there exists a rotational homothety $\tau$ mapping $T M L$ to $T I_{A} A$.  Since $\Varangle(M L, L D)=\Varangle(A I, A D)$, we get $\tau(B C)=A D$. Next, since  $$ \Varangle(M B, M L)=\Varangle(M B, M I)=\Varangle\left(I_{A} B, I_{A} I\right)=\Varangle\left(I_{A} E, I_{A} A\right), $$  we get $\tau(B)=E$. Similarly, $\tau(C)=F$. Since the points $M, B, C$, and $T$ are concyclic, so are their $\tau$-images, which means that $T$ lies on $\omega=\tau(\Gamma)$. ![](https://cdn.mathpix.com/cropped/2024_11_18_bcdd154f2030efbe2fa4g-61.jpg?height=1055&width=1400&top_left_y=909&top_left_x=334)  Finally, since $\tau(L)=A$ and $\tau(B)=E$, triangles $A T L$ and $E T B$ are similar so that  $$ \Varangle(A T, A L)=\Varangle(E T, E B)=\Varangle\left(E I_{A}, E T\right) . $$  This means that the tangents to $\Omega$ and $\omega$ at $T$ make the same angle with the line $I_{A} T L$, so the circles are indeed tangent at $T$.  Comment. In both solutions above, a crucial step is a guess that the desired tangency point lies on $\Gamma$. There are several ways to recognize this helpful property. E.g. one may perform some angle chasing to see that the tangents to $\Omega$ at $L$ and to $\omega$ at $I_{A}$ are parallel (and the circles lie on different sides of the tangents). This yields that, under the assumption that the circles are tangent externally, the tangency point must lie on $I_{A} L$. Since $I L$ is a diameter in $\Omega$, this, in turn, implies that $T$ is the projection of $I$ onto $I_{A} L$.  Another way to see the same fact is to perform a homothety centered at $A$ and mapping $I$ to $I_{A}$ (and $D$ to some point $D^{\prime}$ ). The image $\Omega^{\prime}$ of $\Omega$ is tangent to $\omega$ at $I_{A}$, because $\angle B I_{A} A+\angle C I_{A} D^{\prime}=180^{\circ}$. This yields that the tangents to $\Omega$ at $I$ and to $\omega$ at $I_{A}$ are parallel.  There are other ways to describe the tangency point. The next solution presents one of them.


Solution: We also use the notation of circles $\omega$, and $\Omega$ from the previous solutions. Perform an inversion centered at $D$. The images of the points will be denoted by primes, e.g., $A^{\prime}$ is the image of $A$.  For convenience, we use the notation $\angle B I D=\beta, \angle C I D=\gamma$, and $\alpha=180^{\circ}-\beta-\gamma=$ $90^{\circ}-\angle B A I$. We start with computing angles appearing after inversion. We get  $$ \begin{gathered} \angle D B^{\prime} I^{\prime}=\beta, \quad \angle D C^{\prime} I^{\prime}=\gamma, \quad \text { and hence } \angle B^{\prime} I^{\prime} C^{\prime}=\alpha ; \\ \angle E^{\prime} I_{A}^{\prime} F^{\prime}=\angle E^{\prime} I_{A}^{\prime} D-\angle F^{\prime} I_{A}^{\prime} D=\angle I_{A} E D-\angle I_{A} F D=\angle E I_{A} F=180^{\circ}-\alpha . \end{gathered} $$  Next, we have  $$ \angle A^{\prime} E^{\prime} B^{\prime}=\angle D E^{\prime} B^{\prime}=\angle D B E=\beta=90^{\circ}-\frac{\angle D B A}{2}=90^{\circ}-\frac{\angle E^{\prime} A^{\prime} B^{\prime}}{2} $$  which yields that triangle $A^{\prime} B^{\prime} E^{\prime}$ is isosceles with $A^{\prime} B^{\prime}=A^{\prime} E^{\prime}$. Similarly, $A^{\prime} F^{\prime}=A^{\prime} C^{\prime}$. Finally, we get  $$ \begin{aligned} \angle A^{\prime} B^{\prime} I^{\prime}=\angle I^{\prime} B^{\prime} D-\angle A^{\prime} B^{\prime} D=\beta- & \angle B A D=\beta-\left(90^{\circ}-\alpha\right)+\angle I A D \\ & =\angle I C D+\angle I A D=\angle C^{\prime} I^{\prime} D+\angle A^{\prime} I^{\prime} D=\angle C^{\prime} I^{\prime} A^{\prime} \end{aligned} $$  similarly, $\angle A^{\prime} C^{\prime} I^{\prime}=\angle A^{\prime} I^{\prime} B^{\prime}$, so that triangles $A^{\prime} B^{\prime} I^{\prime}$ and $A^{\prime} I^{\prime} C^{\prime}$ are similar. Therefore, $A^{\prime} I^{\prime 2}=A^{\prime} B^{\prime} \cdot A^{\prime} C^{\prime}$.  Recall that we need to prove the tangency of line $A^{\prime} I^{\prime}=\Omega^{\prime}$ with circle $\left(E^{\prime} F^{\prime} I_{A}^{\prime}\right)=\omega^{\prime}$. A desired tangency point $T^{\prime}$ must satisfy $A^{\prime} T^{\prime 2}=A^{\prime} E^{\prime} \cdot A^{\prime} F^{\prime}$; the relations obtained above yield  $$ A^{\prime} E^{\prime} \cdot A^{\prime} F^{\prime}=A^{\prime} B^{\prime} \cdot A^{\prime} C^{\prime}=A^{\prime} I^{\prime 2} $$  so that $T^{\prime}$ should be symmetric to $I^{\prime}$ with respect to $A^{\prime}$. Thus, let us define a point $T^{\prime}$ as the reflection of $I^{\prime}$ in $A^{\prime}$, and show that $T^{\prime}$ lies on circle $\Omega^{\prime}$; the equalities above will then imply that $A^{\prime} T^{\prime}$ is tangent to $\Omega^{\prime}$, as desired. ![](https://cdn.mathpix.com/cropped/2024_11_18_bcdd154f2030efbe2fa4g-62.jpg?height=734&width=1156&top_left_y=1569&top_left_x=451)  The property that triangles $B^{\prime} A^{\prime} I^{\prime}$ and $I^{\prime} A^{\prime} C^{\prime}$ are similar means that quadrilateral $B^{\prime} I^{\prime} C^{\prime} T^{\prime}$ is harmonic. Indeed, let $C^{*}$ be the reflection of $C^{\prime}$ in the perpendicular bisector of $I^{\prime} T^{\prime}$; then $C^{*}$ lies on $B^{\prime} A^{\prime}$ by $\angle B^{\prime} A^{\prime} I^{\prime}=\angle A^{\prime} I^{\prime} C^{\prime}=\angle T^{\prime} I^{\prime} C^{*}$, and then $C^{*}$ lies on circle ( $I^{\prime} B^{\prime} T^{\prime}$ ) since $A^{\prime} B^{\prime} \cdot A^{\prime} C^{*}=A^{\prime} B^{\prime} \cdot A^{\prime} C^{\prime}=A^{\prime} I^{\prime 2}=A^{\prime} I^{\prime} \cdot A^{\prime} T^{\prime}$. Therefore, $C^{\prime}$ also lies on that circle (and the circle is $\left.\left(B^{\prime} I^{\prime} C^{\prime}\right)=\Gamma^{\prime}\right)$. Moreover, $B^{\prime} C^{*}$ is a median in triangle $B^{\prime} I^{\prime} T^{\prime}$, so $B^{\prime} C^{\prime}$ is its symmedian, which establishes harmonicity.  Now we have $\angle A^{\prime} B^{\prime} T^{\prime}=\angle I^{\prime} B^{\prime} C^{\prime}=\beta=\angle A^{\prime} B^{\prime} E^{\prime}$; which shows that $E^{\prime}$ lies on $B^{\prime} T^{\prime}$. Similarly, $F^{\prime}$ lies on $C^{\prime} T^{\prime}$. Hence, $\angle E^{\prime} T^{\prime} F^{\prime}=\angle B^{\prime} I^{\prime} C^{\prime}=180^{\circ}-\angle E^{\prime} I_{A}^{\prime} F^{\prime}$, which establishes $T^{\prime} \in \omega^{\prime}$.  Comment 2. The solution above could be finished without use of harmonicity. E.g., one may notice that both triangles $A^{\prime} T^{\prime} F^{\prime}$ and $A^{\prime} E^{\prime} T^{\prime}$ are similar to triangle $B^{\prime} I^{\prime} J$, where $J$ is the point symmetric to $I^{\prime}$ in the perpendicular bisector of $B^{\prime} C^{\prime}$; indeed, we have $\angle T^{\prime} A^{\prime} E^{\prime}=\gamma-\beta=\angle I^{\prime} B^{\prime} J^{\prime}$ and $\frac{B^{\prime} I^{\prime}}{B^{\prime} J}=\frac{B^{\prime} I^{\prime}}{C^{\prime} I}=$ $\frac{B^{\prime} A^{\prime}}{A^{\prime} I^{\prime}}=\frac{A^{\prime} E^{\prime}}{A^{\prime} T^{\prime}}$. This also allows to compute $\angle E^{\prime} T^{\prime} F^{\prime}=\angle E^{\prime} T^{\prime} A^{\prime}-\angle F^{\prime} T^{\prime} A^{\prime}=\angle I^{\prime} J B^{\prime}-\angle J I^{\prime} B^{\prime}=\alpha$.  Comment 3. Here we list several properties of the configuration in the problem, which can be derived from the solutions above.  The quadrilateral $I B T C$ (as well as $I^{\prime} B^{\prime} T^{\prime} C^{\prime}$ ) is harmonic. Hence, line $I T$ contains the meeting point of tangents to $\Gamma$ at $B$ and $C$, i.e., the midpoint $N$ of $\operatorname{arc} B A C$ in the circumcircle of $\triangle A B C$.


Let $P$ be a point on the circumcircle of an acute-angled triangle $A B C$. Let $D$, $E$, and $F$ be the reflections of $P$ in the midlines of triangle $A B C$ parallel to $B C, C A$, and $A B$, respectively. Denote by $\omega_{A}, \omega_{B}$, and $\omega_{C}$ the circumcircles of triangles $A D P, B E P$, and $C F P$, respectively. Denote by $\omega$ the circumcircle of the triangle formed by the perpendicular bisectors of segments $A D, B E$ and $C F$.  Show that $\omega_{A}, \omega_{B}, \omega_{C}$, and $\omega$ have a common point. (Denmark)

Solution: Let $A A_{1}, B B_{1}$, and $C C_{1}$ be the altitudes in triangle $A B C$, and let $m_{A}, m_{B}$, and $m_{C}$ be the midlines parallel to $B C, C A$, and $A B$, respectively. We always denote by $\Varangle(p, q)$ the directed angle from a line $p$ to a line $q$, taken modulo $180^{\circ}$.  Step 1: Circles $\omega_{A}, \omega_{B}$, and $\omega_{C}$ share a common point $Q$ different from $P$. Notice that $m_{A}$ is the perpendicular bisector of $P D$, so $\omega_{A}$ is symmetric with respect to $m_{A}$. Since $A$ and $A_{1}$ are also symmetric to each other in $m_{A}$, the point $A_{1}$ lies on $\omega_{A}$. Similarly, $B_{1}$ and $C_{1}$ lie on $\omega_{B}$ and $\omega_{C}$, respectively.  Let $H$ be the orthocenter of $\triangle A B C$. Quadrilaterals $A B A_{1} B_{1}$ and $B C B_{1} C_{1}$ are cyclic, so $A H \cdot H A_{1}=B H \cdot H B_{1}=C H \cdot H C_{1}$. This means that $H$ lies on pairwise radical axes of $\omega_{A}$, $\omega_{B}$, and $\omega_{C}$. Point $P$ also lies on those radical axes; hence the three circles have a common radical axis $\ell=P H$, and the second meeting point $Q$ of $\ell$ with $\omega_{A}$ is the second common point of the three circles. Notice here that $H$ lies inside all three circles, hence $Q \neq P$. ![](https://cdn.mathpix.com/cropped/2024_11_18_bcdd154f2030efbe2fa4g-64.jpg?height=931&width=1634&top_left_y=1282&top_left_x=215)  Step 2: Point $Q$ lies on $\omega$. Let $p_{A}, p_{B}$, and $p_{C}$ denote the perpendicular bisectors of $A D, B E$, and $C F$, respectively; denote by $\Delta$ the triangle formed by those perpendicular bisectors. By Simson's theorem, in order to show that $Q$ lies on the circumcircle $\omega$ of $\Delta$, it suffices to prove that the projections of $Q$ onto the sidelines $p_{A}, p_{B}$, and $p_{C}$ are collinear. Alternatively, but equivalently, it suffices to prove that the reflections $Q_{A}, Q_{B}$, and $Q_{C}$ of $Q$ in those lines, respectively, are collinear. In fact, we will show that four points $P, Q_{A}, Q_{B}$, and $Q_{C}$ are collinear.  Since $p_{A}$ is the common perpendicular bisector of $A D$ and $Q Q_{A}$, the point $Q_{A}$ lies on $\omega_{A}$, and, moreover, $\Varangle\left(D A, D Q_{A}\right)=\Varangle(A Q, A D)$. Therefore,  $$ \Varangle\left(P A, P Q_{A}\right)=\Varangle\left(D A, D Q_{A}\right)=\Varangle(A Q, A D)=\Varangle(P Q, P D)=\Varangle(P Q, B C)+90^{\circ} . $$  Similarly, we get $\Varangle\left(P B, P Q_{B}\right)=\Varangle(P Q, C A)+90^{\circ}$. Therefore,  $$ \begin{aligned} \Varangle\left(P Q_{A}, P Q_{B}\right)=\Varangle\left(P Q_{A}, P A\right) & +\Varangle(P A, P B)+\Varangle\left(P B, P Q_{B}\right) \\ & =\Varangle(B C, P Q)+90^{\circ}+\Varangle(C A, C B)+\Varangle(P Q, C A)+90^{\circ}=0, \end{aligned} $$  which shows that $P, Q_{A}$, and $Q_{B}$ are collinear. Similarly, $Q_{C}$ also lies on $P Q_{A}$. Comment 1. There are several variations of Step 2. In particular, let $O_{A}, O_{B}$, and $O_{C}$ denote the centers of $\omega_{A}, \omega_{B}$, and $\omega_{C}$, respectively; they lie on $p_{A}, p_{B}$, and $p_{C}$, respectively. Moreover, all those centers lie on the perpendicular bisector $p$ of $P Q$. Now one can show that $\Varangle\left(Q O_{A}, p_{A}\right)=$ $\Varangle\left(Q O_{B}, p_{B}\right)=\Varangle\left(Q O_{C}, p_{C}\right)$, and then finish by applying generalized Simson's theorem, Alternatively, but equivalently, those relations show that $Q$ is the Miquel point of the lines $p_{A}, p_{B}, p_{C}$, and $p$.  To establish $\Varangle\left(Q O_{A}, p_{A}\right)=\Varangle\left(Q O_{C}, p_{C}\right)$, notice that it is equivalent to $\Varangle\left(Q O_{A}, Q O_{C}\right)=\Varangle\left(p_{A}, p_{C}\right)$ which may be obtained, e.g., as follows:  $$ \begin{aligned} & \Varangle\left(Q O_{A}, Q O_{C}\right)=\Varangle\left(Q O_{A}, p\right)+\Varangle\left(p, Q O_{C}\right)=\Varangle(A Q, A P)+\Varangle(C P, C Q) \\ & =\Varangle(A Q, C Q)+\Varangle(C P, A P)=\Varangle(A Q, P Q)+\Varangle(P Q, C Q)+\Varangle(C B, A B) \\ & \quad=\Varangle\left(A D, A A_{1}\right)+\Varangle\left(C C_{1}, C F\right)+\Varangle\left(A A_{1}, C C_{1}\right)=\Varangle(A D, C F)=\Varangle\left(p_{A}, p_{C}\right) . \end{aligned} $$  ![](https://cdn.mathpix.com/cropped/2024_11_18_bcdd154f2030efbe2fa4g-65.jpg?height=949&width=843&top_left_y=1145&top_left_x=612)  Comment 2. The inversion at $H$ with (negative) power $-A H \cdot H A_{1}$ maps $P$ to $Q$, and the circumcircle of $\triangle A B C$ to its Euler circle. Therefore, $Q$ lies on that Euler circle.


Let $\Gamma$ and $I$ be the circumcircle and the incenter of an acute-angled triangle $A B C$. Two circles $\omega_{B}$ and $\omega_{C}$ passing through $B$ and $C$, respectively, are tangent at $I$. Let $\omega_{B}$ meet the shorter arc $A B$ of $\Gamma$ and segment $A B$ again at $P$ and $M$, respectively. Similarly, let $\omega_{C}$ meet the shorter arc $A C$ of $\Gamma$ and segment $A C$ again at $Q$ and $N$, respectively. The rays $P M$ and $Q N$ meet at $X$, and the tangents to $\omega_{B}$ and $\omega_{C}$ at $B$ and $C$, respectively, meet at $Y$.  Prove that the points $A, X$, and $Y$ are collinear. (Netherlands)

Solution: Let $A I, B I$, and $C I$ meet $\Gamma$ again at $D, E$, and $F$, respectively. Let $\ell$ be the common tangent to $\omega_{B}$ and $\omega_{C}$ at $I$. We always denote by $\Varangle(p, q)$ the directed angle from a line $p$ to a line $q$, taken modulo $180^{\circ}$.  Step 1: We show that $Y$ lies on $\Gamma$. Recall that any chord of a circle makes complementary directed angles with the tangents to the circle at its endpoints. Hence,  $$ \Varangle(B Y, B I)+\Varangle(C I, C Y)=\Varangle(I B, \ell)+\Varangle(\ell, I C)=\Varangle(I B, I C) . $$  Therefore,  $$ \begin{aligned} \Varangle(B Y, B A)+\Varangle(C A, C Y)=\Varangle(B I, B A)+ & \Varangle(B Y, B I)+\Varangle(C I, C Y)+\Varangle(C A, C I) \\ & =\Varangle(B C, B I)+\Varangle(I B, I C)+\Varangle(C I, C B)=0, \end{aligned} $$  which yields $Y \in \Gamma$. ![](https://cdn.mathpix.com/cropped/2024_11_18_bcdd154f2030efbe2fa4g-66.jpg?height=932&width=1029&top_left_y=1481&top_left_x=519)  Step 2: We show that $X=\ell \cap E F$. Let $X_{*}=\ell \cap E F$. To prove our claim, it suffices to show that $X_{*}$ lies on both $P M$ and $Q N$; this will yield $X_{*}=X$. Due to symmetry, it suffices to show $X_{*} \in Q N$.  Notice that  $$ \Varangle\left(I X_{*}, I Q\right)=\Varangle(C I, C Q)=\Varangle(C F, C Q)=\Varangle(E F, E Q)=\Varangle\left(E X_{*}, E Q\right) ; $$  therefore, the points $X_{*}, I, Q$, and $E$ are concyclic (if $Q=E$, then the direction of $E Q$ is supposed to be the direction of a tangent to $\Gamma$ at $Q$; in this case, the equality means that the circle $\left(X_{*} I Q\right)$ is tangent to $\Gamma$ at $\left.Q\right)$. Then we have  $$ \Varangle\left(Q X_{*}, Q I\right)=\Varangle\left(E X_{*}, E I\right)=\Varangle(E F, E B)=\Varangle(C A, C F)=\Varangle(C N, C I)=\Varangle(Q N, Q I), $$  which shows that $X_{*} \in Q N$. Step 3: We finally show that $A, X$, and $Y$ are collinear. Recall that $I$ is the orthocenter of triangle $D E F$, and $A$ is symmetric to $I$ with respect to $E F$. Therefore,  $$ \Varangle(A X, A E)=\Varangle(I E, I X)=\Varangle(B I, \ell)=\Varangle(B Y, B I)=\Varangle(B Y, B E)=\Varangle(A Y, A E), $$  which yields the desired collinearity. Comment 1. Step 2 in the above solution seems to be crucial. After it has been performed (even without Step 1), there are different ways of finishing the solution. E.g., one may involve the notion of isogonal conjugacy. Let $X_{1}$ and $Y_{1}$ be isogonal conjugates of $X$ and $Y$, respectively, with respect to triangle $A B C$. Since $X A=X I$, triangle $A I X$ is isosceles, and hence the lines $A X$ and $X I$ form equal angles with the internal bisector $A I$ of $\angle B A C$. This means that $A X_{1} \| X I$, or $A X_{1} \| \ell$.  On the other hand, the lines $B Y$ and $\ell$ form equal angles with $B I$, so that $B Y_{1} \| \ell$. Similarly, $C Y_{1} \| \ell$. This means that $Y_{1}$ is an ideal point, and $A Y_{1} \| \ell$ as well. Therefore, points $A, X_{1}$, and $Y_{1}$ are collinear, and hence $A, X$, and $Y$ are such.


Solution: Perform an inversion centered at $I$; the images of the points are denoted by primes, e.g., $A^{\prime}$ is the image of $A$.  On the inverted figure, $I$ and $\Gamma^{\prime}$ are the orthocenter and the circumcircle of triangle $A^{\prime} B^{\prime} C^{\prime}$, respectively. The points $P^{\prime}$ and $Q^{\prime}$ lie on $\Gamma^{\prime}$ such that $B^{\prime} P^{\prime} \| C^{\prime} Q^{\prime}$ (since $B^{\prime} P^{\prime}=\omega_{B}^{\prime}$ and $C^{\prime} Q^{\prime}=\omega_{C}^{\prime}$ ). The points $M^{\prime}$ and $N^{\prime}$ are the second intersections of lines $B^{\prime} P^{\prime}$ and $C^{\prime} Q^{\prime}$ with the circumcircles $\gamma_{B}$ and $\gamma_{C}$ of triangles $A^{\prime} I B^{\prime}$ and $A^{\prime} I C^{\prime}$, respectively. Notice here that $\gamma_{C}$ is obtained from $\gamma_{B}$ by the translation at $\overrightarrow{B^{\prime} C^{\prime}}$; the same translation maps line $B^{\prime} P^{\prime}$ to $C^{\prime} Q^{\prime}$, and hence $M^{\prime}$ to $N^{\prime}$. In other words, $B^{\prime} M^{\prime} N^{\prime} C^{\prime}$ is a parallelogram, and $P^{\prime} Q^{\prime}$ partitions it into two isosceles trapezoids.  Point $X^{\prime}$ is the second intersection point of circles $\left(I P^{\prime} M^{\prime}\right)$ and $\left(I Q^{\prime} N^{\prime}\right)$ that is - the reflection of $I$ in their line of centers. But the centers lie on the common perpendicular bisector $p$ of $P^{\prime} M^{\prime}$ and $Q^{\prime} N^{\prime}$, so $p$ is that line of centers. Hence, $I X^{\prime} \| B^{\prime} P^{\prime}$, as both lines are perpendicular to $p$.  Finally, the point $Y$ satisfies $\Varangle(B Y, B I)=\Varangle(P B, P I)$ and $\Varangle(C Y, C I)=\Varangle(Q C, Q I)$, which yields $\Varangle\left(Y^{\prime} B^{\prime}, Y^{\prime} I\right)=\Varangle\left(B^{\prime} P^{\prime}, B^{\prime} I\right)$ and $\Varangle\left(Y^{\prime} C^{\prime}, Y^{\prime} I\right)=\Varangle\left(C^{\prime} Q^{\prime}, C^{\prime} I\right)$. Therefore,  $$ \Varangle\left(Y^{\prime} B^{\prime}, Y^{\prime} C^{\prime}\right)=\Varangle\left(B^{\prime} P^{\prime}, B^{\prime} I\right)+\Varangle\left(C^{\prime} I, C^{\prime} Q^{\prime}\right)=\Varangle\left(C^{\prime} I, B^{\prime} I\right)=\Varangle\left(A^{\prime} B^{\prime}, A^{\prime} C^{\prime}\right), $$  which shows that $Y^{\prime} \in \Gamma^{\prime}$. In congruent circles $\Gamma^{\prime}$ and $\gamma_{B}$, the chords $A^{\prime} P^{\prime}$ and $A^{\prime} M^{\prime}$ subtend the same angle $\angle A^{\prime} B^{\prime} P^{\prime}$; therefore, $A^{\prime} P^{\prime}=A^{\prime} M^{\prime}$, and hence $A^{\prime} \in p$. This yields $A^{\prime} X^{\prime}=A^{\prime} I$, and hence $\Varangle\left(I A^{\prime}, I X^{\prime}\right)=$ $\Varangle\left(X^{\prime} I, X^{\prime} A^{\prime}\right)$ 。  Finally, we have  $$ \begin{aligned} \Varangle\left(Y^{\prime} I, Y^{\prime} A^{\prime}\right) & =\Varangle\left(Y^{\prime} I, Y^{\prime} B^{\prime}\right)+\Varangle\left(Y^{\prime} B^{\prime}, Y^{\prime} A^{\prime}\right) \\ & =\Varangle\left(B^{\prime} I, B^{\prime} P^{\prime}\right)+\Varangle\left(I A^{\prime}, I B^{\prime}\right)=\Varangle\left(I A^{\prime}, B^{\prime} P^{\prime}\right)=\Varangle\left(I A^{\prime}, I X^{\prime}\right)=\Varangle\left(X^{\prime} I, X^{\prime} A^{\prime}\right), \end{aligned} $$  which yields that the points $A^{\prime}, X^{\prime}, Y^{\prime}$, and $I$ are concyclic. This means exactly that $A, X$, and $Y$ are collinear. ![](https://cdn.mathpix.com/cropped/2024_11_18_bcdd154f2030efbe2fa4g-68.jpg?height=967&width=1560&top_left_y=199&top_left_x=248)  Comment 2. An inversion at $I$ may also help in establishing Step 2 in Solution 1. Indeed, relation $A^{\prime} X^{\prime}=A^{\prime} I$ yields $X A=X I$, so that $X \in E F$. On the other hand, $I X^{\prime} \| B^{\prime} P^{\prime}$ yields $I X \| \ell$, i.e., $X \in \ell$.


Prove that there exists a positive constant $c$ such that the following statement is true:  Assume that $n$ is an integer with $n \geqslant 2$, and let $\mathcal{S}$ be a set of $n$ points in the plane such that the distance between any two distinct points in $\mathcal{S}$ is at least 1 . Then there is a line $\ell$ separating $\mathcal{S}$ such that the distance from any point of $\mathcal{S}$ to $\ell$ is at least $c n^{-1 / 3}$. (A line $\ell$ separates a point set $\mathcal{S}$ if some segment joining two points in $\mathcal{S}$ crosses $\ell$.)

Solution: We prove that the desired statement is true with $c=\frac{1}{8}$. Set $\delta=\frac{1}{8} n^{-1 / 3}$. For any line $\ell$ and any point $X$, let $X_{\ell}$ denote the projection of $X$ to $\ell$; a similar notation applies to sets of points.  Suppose that, for some line $\ell$, the set $\mathcal{S}_{\ell}$ contains two adjacent points $X$ and $Y$ with $X Y=2 d$. Then the line perpendicular to $\ell$ and passing through the midpoint of segment $X Y$ separates $\mathcal{S}$, and all points in $\mathcal{S}$ are at least $d$ apart from $\ell$. Thus, if $d \geqslant \delta$, then a desired line has been found. For the sake of contradiction, we assume that no such points exist, in any projection.  Choose two points $A$ and $B$ in $\mathcal{S}$ with the maximal distance $M=A B$ (i.e., $A B$ is a diameter of $\mathcal{S}$ ); by the problem condition, $M \geqslant 1$. Denote by $\ell$ the line $A B$. The set $\mathcal{S}$ is contained in the intersection of two disks $D_{A}$ and $D_{B}$ of radius $M$ centered at $A$ and $B$, respectively. Hence, the projection $\mathcal{S}_{\ell}$ is contained in the segment $A B$. Moreover, the points in $\mathcal{S}_{\ell}$ divide that segment into at most $n-1$ parts, each of length less than $2 \delta$. Therefore,  $$ M<n \cdot 2 \delta $$  ![](https://cdn.mathpix.com/cropped/2024_11_18_bcdd154f2030efbe2fa4g-69.jpg?height=418&width=924&top_left_y=1404&top_left_x=566)  Choose a point $H$ on segment $A B$ with $A H=\frac{1}{2}$. Let $P$ be a strip between the lines $a$ and $h$ perpendicular to $A B$ and passing through $A$ and $H$, respectively; we assume that $P$ contains its boundary, which consists of lines $a$ and $h$. Set $\mathcal{T}=P \cap \mathcal{S}$ and let $t=|\mathcal{T}|$. By our assumption, segment $A H$ contains at least $\left\lceil\frac{1}{2}:(2 \delta)\right\rceil$ points of $S_{\ell}$, which yields  $$ t \geqslant \frac{1}{4 \delta} $$  Notice that $\mathcal{T}$ is contained in $Q=P \cap D_{B}$. The set $Q$ is a circular segment, and its projection $Q_{a}$ is a line segment of length  $$ 2 \sqrt{M^{2}-\left(M-\frac{1}{2}\right)^{2}}<2 \sqrt{M} $$  On the other hand, for any two points $X, Y \in \mathcal{T}$, we have $X Y \geqslant 1$ and $X_{\ell} Y_{\ell} \leqslant \frac{1}{2}$, so $X_{a} Y_{a}=$ $\sqrt{X Y^{2}-X_{\ell} Y_{\ell}^{2}} \geqslant \frac{\sqrt{3}}{2}$. To summarize, $t$ points constituting $\mathcal{T}_{a}$ lie on the segment of length less than $2 \sqrt{M}$, and are at least $\frac{\sqrt{3}}{2}$ apart from each other. This yields $2 \sqrt{M}>(t-1) \frac{\sqrt{3}}{2}$, or  $$ t<1+\frac{4 \sqrt{M}}{\sqrt{3}}<4 \sqrt{M} $$  as $M \geqslant 1$. Combining the estimates (1), (2), and (3), we finally obtain  $$ \frac{1}{4 \delta} \leqslant t<4 \sqrt{M}<4 \sqrt{2 n \delta}, \quad \text { or } \quad 512 n \delta^{3}>1 $$  which does not hold for the chosen value of $\delta$. Comment 1. As the proposer mentions, the exponent $-1 / 3$ in the problem statement is optimal. In fact, for any $n \geqslant 2$, there is a configuration $\mathcal{S}$ of $n$ points in the plane such that any two points in $\mathcal{S}$ are at least 1 apart, but every line $\ell$ separating $\mathcal{S}$ is at most $c^{\prime} n^{-1 / 3} \log n$ apart from some point in $\mathcal{S}$; here $c^{\prime}$ is some absolute constant.  The original proposal suggested to prove the estimate of the form $\mathrm{cn}^{-1 / 2}$. That version admits much easier solutions. E.g., setting $\delta=\frac{1}{16} n^{-1 / 2}$ and applying (1), we see that $\mathcal{S}$ is contained in a disk $D$ of radius $\frac{1}{8} n^{1 / 2}$. On the other hand, for each point $X$ of $\mathcal{S}$, let $D_{X}$ be the disk of radius $\frac{1}{2}$ centered at $X$; all these disks have disjoint interiors and lie within the disk concentric to $D$, of radius $\frac{1}{16} n^{1 / 2}+\frac{1}{2}<\frac{1}{2} n^{1 / 2}$. Comparing the areas, we get  $$ n \cdot \frac{\pi}{4} \leqslant \pi\left(\frac{n^{1 / 2}}{16}+\frac{1}{2}\right)^{2}<\frac{\pi n}{4} $$  which is a contradiction. The Problem Selection Committee decided to choose a harder version for the Shortlist. Comment 2. In this comment, we discuss some versions of the solution above, which avoid concentrating on the diameter of $\mathcal{S}$. We start with introducing some terminology suitable for those versions.  Put $\delta=c n^{-1 / 3}$ for a certain sufficiently small positive constant $c$. For the sake of contradiction, suppose that, for some set $\mathcal{S}$ satisfying the conditions in the problem statement, there is no separating line which is at least $\delta$ apart from each point of $\mathcal{S}$.  Let $C$ be the convex hull of $\mathcal{S}$. A line is separating if and only if it meets $C$ (we assume that a line passing through a point of $\mathcal{S}$ is always separating). Consider a strip between two parallel separating lines $a$ and $a^{\prime}$ which are, say, $\frac{1}{4}$ apart from each other. Define a slice determined by the strip as the intersection of $\mathcal{S}$ with the strip. The length of the slice is the diameter of the projection of the slice to $a$.  In this terminology, the arguments used in the proofs of (2) and (3) show that for any slice $\mathcal{T}$ of length $L$, we have  $$ \frac{1}{8 \delta} \leqslant|\mathcal{T}| \leqslant 1+\frac{4}{\sqrt{15}} L $$  The key idea of the solution is to apply these estimates to a peel slice, where line $a$ does not cross the interior of $C$. In the above solution, this idea was applied to one carefully chosen peel slice. Here, we outline some different approach involving many of them. We always assume that $n$ is sufficiently large.  Consider a peel slice determined by lines $a$ and $a^{\prime}$, where $a$ contains no interior points of $C$. We orient $a$ so that $C$ lies to the left of $a$. Line $a$ is called a supporting line of the slice, and the obtained direction is the direction of the slice; notice that the direction determines uniquely the supporting line and hence the slice. Fix some direction $\mathbf{v}_{0}$, and for each $\alpha \in[0,2 \pi)$ denote by $\mathcal{T}_{\alpha}$ the peel slice whose direction is $\mathbf{v}_{0}$ rotated by $\alpha$ counterclockwise.  When speaking about the slice, we always assume that the figure is rotated so that its direction is vertical from the bottom to the top; then the points in $\mathcal{T}$ get a natural order from the bottom to the top. In particular, we may speak about the top half $\mathrm{T}(\mathcal{T})$ consisting of $\lfloor|\mathcal{T}| / 2\rfloor$ topmost points in $\mathcal{T}$, and similarly about its bottom half $\mathrm{B}(\mathcal{T})$. By (4), each half contains at least 10 points when $n$ is large. Claim. Consider two angles $\alpha, \beta \in[0, \pi / 2]$ with $\beta-\alpha \geqslant 40 \delta=: \phi$. Then all common points of $\mathcal{T}_{\alpha}$ and $\mathcal{T}_{\beta}$ lie in $\mathrm{T}\left(\mathcal{T}_{\alpha}\right) \cap \mathrm{B}\left(\mathcal{T}_{\beta}\right)$. ![](https://cdn.mathpix.com/cropped/2024_11_18_bcdd154f2030efbe2fa4g-71.jpg?height=738&width=1338&top_left_y=201&top_left_x=366)  Proof. By symmetry, it suffices to show that all those points lie in $\mathrm{T}\left(\mathcal{T}_{\alpha}\right)$. Let $a$ be the supporting line of $\mathcal{T}_{\alpha}$, and let $\ell$ be a line perpendicular to the direction of $\mathcal{T}_{\beta}$. Let $P_{1}, \ldots, P_{k}$ list all points in $\mathcal{T}_{\alpha}$, numbered from the bottom to the top; by (4), we have $k \geqslant \frac{1}{8} \delta^{-1}$.  Introduce the Cartesian coordinates so that the (oriented) line $a$ is the $y$-axis. Let $P_{i}$ be any point in $\mathrm{B}\left(\mathcal{T}_{\alpha}\right)$. The difference of ordinates of $P_{k}$ and $P_{i}$ is at least $\frac{\sqrt{15}}{4}(k-i)>\frac{1}{3} k$, while their abscissas differ by at most $\frac{1}{4}$. This easily yields that the projections of those points to $\ell$ are at least  $$ \frac{k}{3} \sin \phi-\frac{1}{4} \geqslant \frac{1}{24 \delta} \cdot 20 \delta-\frac{1}{4}>\frac{1}{4} $$  apart from each other, and $P_{k}$ is closer to the supporting line of $\mathcal{T}_{\beta}$ than $P_{i}$, so that $P_{i}$ does not belong to $\mathcal{T}_{\beta}$.  Now, put $\alpha_{i}=40 \delta i$, for $i=0,1, \ldots,\left\lfloor\frac{1}{40} \delta^{-1} \cdot \frac{\pi}{2}\right\rfloor$, and consider the slices $\mathcal{T}_{\alpha_{i}}$. The Claim yields that each point in $\mathcal{S}$ is contained in at most two such slices. Hence, the union $\mathcal{U}$ of those slices contains at least  $$ \frac{1}{2} \cdot \frac{1}{8 \delta} \cdot \frac{1}{40 \delta} \cdot \frac{\pi}{2}=\frac{\lambda}{\delta^{2}} $$  points (for some constant $\lambda$ ), and each point in $\mathcal{U}$ is at most $\frac{1}{4}$ apart from the boundary of $C$. It is not hard now to reach a contradiction with (1). E.g., for each point $X \in \mathcal{U}$, consider a closest point $f(X)$ on the boundary of $C$. Obviously, $f(X) f(Y) \geqslant X Y-\frac{1}{2} \geqslant \frac{1}{2} X Y$ for all $X, Y \in \mathcal{U}$. This yields that the perimeter of $C$ is at least $\mu \delta^{-2}$, for some constant $\mu$, and hence the diameter of $\mathcal{S}$ is of the same order.  Alternatively, one may show that the projection of $\mathcal{U}$ to the line at the angle of $\pi / 4$ with $\mathbf{v}_{0}$ has diameter at least $\mu \delta^{-2}$ for some constant $\mu$.


Given a positive integer $k$, show that there exists a prime $p$ such that one can choose distinct integers $a_{1}, a_{2}, \ldots, a_{k+3} \in\{1,2, \ldots, p-1\}$ such that $p$ divides $a_{i} a_{i+1} a_{i+2} a_{i+3}-i$ for all $i=1,2, \ldots, k$. (South Africa)

Solution: First we choose distinct positive rational numbers $r_{1}, \ldots, r_{k+3}$ such that  $$ r_{i} r_{i+1} r_{i+2} r_{i+3}=i \quad \text { for } 1 \leqslant i \leqslant k $$  Let $r_{1}=x, r_{2}=y, r_{3}=z$ be some distinct primes greater than $k$; the remaining terms satisfy $r_{4}=\frac{1}{r_{1} r_{2} r_{3}}$ and $r_{i+4}=\frac{i+1}{i} r_{i}$. It follows that if $r_{i}$ are represented as irreducible fractions, the numerators are divisible by $x$ for $i \equiv 1(\bmod 4)$, by $y$ for $i \equiv 2(\bmod 4)$, by $z$ for $i \equiv 3(\bmod 4)$ and by none for $i \equiv 0(\bmod 4)$. Notice that $r_{i}<r_{i+4}$; thus the sequences $r_{1}<r_{5}<r_{9}<\ldots$, $r_{2}<r_{6}<r_{10}<\ldots, r_{3}<r_{7}<r_{11}<\ldots, r_{4}<r_{8}<r_{12}<\ldots$ are increasing and have no common terms, that is, all $r_{i}$ are distinct.  If each $r_{i}$ is represented by an irreducible fraction $\frac{u_{i}}{v_{i}}$, choose a prime $p$ which divides neither $v_{i}, 1 \leqslant i \leqslant k+1$, nor $v_{i} v_{j}\left(r_{i}-r_{j}\right)=v_{j} u_{i}-v_{i} u_{j}$ for $i<j$, and define $a_{i}$ by the congruence $a_{i} v_{i} \equiv u_{i}(\bmod p)$. Since $r_{i} r_{i+1} r_{i+2} r_{i+3}=i$, we have  $$ \begin{aligned} i v_{i} v_{i+1} v_{i+2} v_{i+3}=r_{i} v_{i} r_{i+1} v_{i+1} r_{i+2} & v_{i+2} r_{i+3} v_{i+3} \\ & =u_{i} u_{i+1} u_{i+2} u_{i+3} \equiv a_{i} v_{i} a_{i+1} v_{i+1} a_{i+2} v_{i+2} a_{i+3} v_{i+3} \quad(\bmod p) \end{aligned} $$  and therefore $a_{i} a_{i+1} a_{i+2} a_{i+3} \equiv i(\bmod p)$ for $1 \leqslant i \leqslant k$. If $a_{i} \equiv a_{j}(\bmod p)$, then $u_{i} v_{j} \equiv a_{i} v_{i} v_{j} \equiv u_{j} v_{i}(\bmod p)$, a contradiction. Comment. One can explicitly express residues $b_{i} \equiv a_{1} a_{2} \cdot \ldots \cdot a_{i}(\bmod p)$ in terms of $b_{1}, b_{2}, b_{3}$ and $b_{0}=1$ :  $$ b_{i+3}=i(i-4)(i-8) \cdot \ldots \cdot(i-4 k+4) b_{r} $$  where $i+3=4 k+r, 0 \leqslant r<4$. Then the numbers $a_{i}$ are found from the congruences $b_{i-1} a_{i} \equiv b_{i}$ $(\bmod p)$, and choosing $p$ so that $a_{i}$ are not congruent modulo $p$ is done in a way very similar to the above solution.  This page is intentionally left blank


For each prime $p$, there is a kingdom of $p$-Landia consisting of $p$ islands numbered $1,2, \ldots, p$. Two distinct islands numbered $n$ and $m$ are connected by a bridge if and only if $p$ divides $\left(n^{2}-m+1\right)\left(m^{2}-n+1\right)$. The bridges may pass over each other, but cannot cross. Prove that for infinitely many $p$ there are two islands in $p$-Landia not connected by a chain of bridges. (Denmark)

Solution: We prove that for each prime $p>3$ dividing a number of the form $x^{2}-x+1$ with integer $x$ there are two unconnected islands in $p$-Landia.  For brevity's sake, when a bridge connects the islands numbered $m$ and $n$, we shall speak simply that it connects $m$ and $n$.  A bridge connects $m$ and $n$ if $n \equiv m^{2}+1(\bmod p)$ or $m \equiv n^{2}+1(\bmod p)$. If $m^{2}+1 \equiv n$ $(\bmod p)$, we draw an arrow starting at $m$ on the bridge connecting $m$ and $n$. Clearly only one arrow starts at $m$ if $m^{2}+1 \not \equiv m(\bmod p)$, and no arrows otherwise. The total number of bridges does not exceed the total number of arrows.  Suppose $x^{2}-x+1 \equiv 0(\bmod p)$. We may assume that $1 \leqslant x \leqslant p$; then there is no arrow starting at $x$. Since $(1-x)^{2}-(1-x)+1=x^{2}-x+1,(p+1-x)^{2}+1 \equiv(p+1-x)(\bmod p)$, and there is also no arrow starting at $p+1-x$. If $x=p+1-x$, that is, $x=\frac{p+1}{2}$, then $4\left(x^{2}-x+1\right)=p^{2}+3$ and therefore $x^{2}-x+1$ is not divisible by $p$. Thus the islands $x$ and $p+1-x$ are different, and no arrows start at either of them. It follows that the total number of bridges in $p$-Landia does not exceed $p-2$.  Let $1,2, \ldots, p$ be the vertices of a graph $G_{p}$, where an edge connects $m$ and $n$ if and only if there is a bridge between $m$ and $n$. The number of vertices of $G_{p}$ is $p$ and the number of edges is less than $p-1$. This means that the graph is not connected, which means that there are two islands not connected by a chain of bridges.  It remains to prove that there are infinitely many primes $p$ dividing $x^{2}-x+1$ for some integer $x$. Let $p_{1}, p_{2}, \ldots, p_{k}$ be any finite set of such primes. The number $\left(p_{1} p_{2} \cdot \ldots \cdot p_{k}\right)^{2}-p_{1} p_{2} \cdot \ldots \cdot p_{k}+1$ is greater than 1 and not divisible by any $p_{i}$; therefore it has another prime divisor with the required property.


Solution: One can show, by using only arithmetical methods, that for infinitely many $p$, the kingdom of $p$-Ladia contains two islands connected to no other island, except for each other.  Let arrows between islands have the same meaning as in the previous solution. Suppose that positive $a<p$ satisfies the congruence $x^{2}-x+1 \equiv 0(\bmod p)$. We have seen in the first solution that $b=p+1-a$ satisfies it too, and $b \neq a$ when $p>3$. It follows that $a b \equiv a(1-a) \equiv 1$ $(\bmod p)$. If an arrow goes from $t$ to $a$, then $t$ must satisfy the congruence $t^{2}+1 \equiv a \equiv a^{2}+1$ $(\bmod p)$; the only such $t \neq a$ is $p-a$. Similarly, the only arrow going to $b$ goes from $p-b$. If one of the numbers $p-a$ and $p-b$, say, $p-a$, is not at the end of any arrow, the pair $a, p-a$ is not connected with the rest of the islands. This is true if at least one of the congruences $x^{2}+1 \equiv-a, x^{2}+1 \equiv-b$ has no solutions, that is, either $-a-1$ or $-b-1$ is a quadratic non-residue modulo $p$.  Note that $x^{2}-x+1 \equiv x^{2}-(a+b) x+a b \equiv(x-a)(x-b)(\bmod p)$. Substituting $x=-1$ we get $(-1-a)(-1-b) \equiv 3(\bmod p)$. If 3 is a quadratic non-residue modulo $p$, so is one of the numbers $-1-a$ and $-1-b$.  Thus it is enough to find infinitely many primes $p>3$ dividing $x^{2}-x+1$ for some integer $x$ and such that 3 is a quadratic non-residue modulo $p$.  If $x^{2}-x+1 \equiv 0(\bmod p)$ then $(2 x-1)^{2} \equiv-3(\bmod p)$, that is, -3 is a quadratic residue modulo $p$, so 3 is a quadratic non-residue if and only if -1 is also a non-residue, in other words, $p \equiv-1(\bmod 4)$.  Similarly to the first solution, let $p_{1}, \ldots, p_{k}$ be primes congruent to -1 modulo 4 and dividing numbers of the form $x^{2}-x+1$. The number $\left(2 p_{1} \cdot \ldots \cdot p_{k}\right)^{2}-2 p_{1} \cdot \ldots \cdot p_{k}+1$ is not divisible by any $p_{i}$ and is congruent to -1 modulo 4 , therefore, it has some prime divisor $p \equiv-1(\bmod 4)$ which has the required properties.


Let $n$ be an integer with $n \geqslant 2$. Does there exist a sequence $\left(a_{1}, \ldots, a_{n}\right)$ of positive integers with not all terms being equal such that the arithmetic mean of every two terms is equal to the geometric mean of some (one or more) terms in this sequence? (Estonia)

Solution: Suppose that $a_{1}, \ldots, a_{n}$ satisfy the required properties. Let $d=\operatorname{gcd}\left(a_{1} \ldots, a_{n}\right)$. If $d>1$ then replace the numbers $a_{1}, \ldots, a_{n}$ by $\frac{a_{1}}{d}, \ldots, \frac{a_{n}}{d}$; all arithmetic and all geometric means will be divided by $d$, so we obtain another sequence satisfying the condition. Hence, without loss of generality, we can assume that $\operatorname{gcd}\left(a_{1} \ldots, a_{n}\right)=1$.  We show two numbers, $a_{m}$ and $a_{k}$ such that their arithmetic mean, $\frac{a_{m}+a_{k}}{2}$ is different from the geometric mean of any (nonempty) subsequence of $a_{1} \ldots, a_{n}$. That proves that there cannot exist such a sequence.  Choose the index $m \in\{1, \ldots, n\}$ such that $a_{m}=\max \left(a_{1}, \ldots, a_{n}\right)$. Note that $a_{m} \geqslant 2$, because $a_{1}, \ldots, a_{n}$ are not all equal. Let $p$ be a prime divisor of $a_{m}$.  Let $k \in\{1, \ldots, n\}$ be an index such that $a_{k}=\max \left\{a_{i}: p \nmid a_{i}\right\}$. Due to $\operatorname{gcd}\left(a_{1} \ldots, a_{n}\right)=1$, not all $a_{i}$ are divisible by $p$, so such a $k$ exists. Note that $a_{m}>a_{k}$ because $a_{m} \geqslant a_{k}, p \mid a_{m}$ and $p \nmid a_{k}$.  Let $b=\frac{a_{m}+a_{k}}{2}$; we will show that $b$ cannot be the geometric mean of any subsequence of $a_{1}, \ldots, a_{n}$.  Consider the geometric mean, $g=\sqrt[t]{a_{i_{1}} \cdot \ldots \cdot a_{i_{t}}}$ of an arbitrary subsequence of $a_{1}, \ldots, a_{n}$. If none of $a_{i_{1}}, \ldots, a_{i_{t}}$ is divisible by $p$, then they are not greater than $a_{k}$, so  $$ g=\sqrt[t]{a_{i_{1}} \cdot \ldots \cdot a_{i_{t}}} \leqslant a_{k}<\frac{a_{m}+a_{k}}{2}=b $$  and therefore $g \neq b$. Otherwise, if at least one of $a_{i_{1}}, \ldots, a_{i_{t}}$ is divisible by $p$, then $2 g=2 \sqrt{t} \sqrt{a_{i_{1}} \cdot \ldots \cdot a_{i_{t}}}$ is either not an integer or is divisible by $p$, while $2 b=a_{m}+a_{k}$ is an integer not divisible by $p$, so $g \neq b$ again.


Solution: Like in the previous solution, we assume that the numbers $a_{1}, \ldots, a_{n}$ have no common divisor greater than 1 . The arithmetic mean of any two numbers in the sequence is half of an integer; on the other hand, it is a (some integer order) root of an integer. This means each pair's mean is an integer, so all terms in the sequence must be of the same parity; hence they all are odd. Let $d=\min \left\{\operatorname{gcd}\left(a_{i}, a_{j}\right): a_{i} \neq a_{j}\right\}$. By reordering the sequence we can assume that $\operatorname{gcd}\left(a_{1}, a_{2}\right)=d$, the sum $a_{1}+a_{2}$ is maximal among such pairs, and $a_{1}>a_{2}$.  We will show that $\frac{a_{1}+a_{2}}{2}$ cannot be the geometric mean of any subsequence of $a_{1} \ldots, a_{n}$. Let $a_{1}=x d$ and $a_{2}=y d$ where $x, y$ are coprime, and suppose that there exist some $b_{1}, \ldots, b_{t} \in\left\{a_{1}, \ldots, a_{n}\right\}$ whose geometric mean is $\frac{a_{1}+a_{2}}{2}$. Let $d_{i}=\operatorname{gcd}\left(a_{1}, b_{i}\right)$ for $i=1,2, \ldots, t$ and let $D=d_{1} d_{2} \cdot \ldots \cdot d_{t}$. Then  $$ D=d_{1} d_{2} \cdot \ldots \cdot d_{t} \left\lvert\, b_{1} b_{2} \cdot \ldots \cdot b_{t}=\left(\frac{a_{1}+a_{2}}{2}\right)^{t}=\left(\frac{x+y}{2}\right)^{t} d^{t}\right. $$  We claim that $D \mid d^{t}$. Consider an arbitrary prime divisor $p$ of $D$. Let $\nu_{p}(x)$ denote the exponent of $p$ in the prime factorization of $x$. If $p \left\lvert\, \frac{x+y}{2}\right.$, then $p \nmid x, y$, so $p$ is coprime with $x$; hence, $\nu_{p}\left(d_{i}\right) \leqslant \nu_{p}\left(a_{1}\right)=\nu_{p}(x d)=\nu_{p}(d)$ for every $1 \leqslant i \leqslant t$, therefore $\nu_{p}(D)=\sum_{i} \nu_{p}\left(d_{i}\right) \leqslant$ $t \nu_{p}(d)=\nu_{p}\left(d^{t}\right)$. Otherwise, if $p$ is coprime to $\frac{x+y}{2}$, we have $\nu_{p}(D) \leqslant \nu_{p}\left(d^{t}\right)$ trivially. The claim has been proved.  Notice that $d_{i}=\operatorname{gcd}\left(b_{i}, a_{1}\right) \geqslant d$ for $1 \leqslant i \leqslant t$ : if $b_{i} \neq a_{1}$ then this follows from the definition of $d$; otherwise we have $b_{i}=a_{1}$, so $d_{i}=a_{1} \geqslant d$. Hence, $D=d_{1} \cdot \ldots \cdot d_{t} \geqslant d^{t}$, and the claim forces $d_{1}=\ldots=d_{t}=d$.  Finally, by $\frac{a_{1}+a_{2}}{2}>a_{2}$ there must be some $b_{k}$ which is greater than $a_{2}$. From $a_{1}>a_{2} \geqslant$ $d=\operatorname{gcd}\left(a_{1}, b_{k}\right)$ it follows that $a_{1} \neq b_{k}$. Now the have a pair $a_{1}, b_{k}$ such that $\operatorname{gcd}\left(a_{1}, b_{k}\right)=d$ but $a_{1}+b_{k}>a_{1}+a_{2}$; that contradicts the choice of $a_{1}$ and $a_{2}$.  Comment. The original problem proposal contained a second question asking if there exists a nonconstant sequence $\left(a_{1}, \ldots, a_{n}\right)$ of positive integers such that the geometric mean of every two terms is equal the arithmetic mean of some terms.  For $n \geqslant 3$ such a sequence is $(4,1,1, \ldots, 1)$. The case $n=2$ can be done by the trivial estimates  $$ \min \left(a_{1}, a_{2}\right)<\sqrt{a_{1} a_{2}}<\frac{a_{1}+a_{2}}{2}<\max \left(a_{1}, a_{2}\right) $$  The Problem Selection Committee found this variant less interesting and suggests using only the first question.


For any odd prime $p$ and any integer $n$, let $d_{p}(n) \in\{0,1, \ldots, p-1\}$ denote the remainder when $n$ is divided by $p$. We say that $\left(a_{0}, a_{1}, a_{2}, \ldots\right)$ is a $p$-sequence, if $a_{0}$ is a positive integer coprime to $p$, and $a_{n+1}=a_{n}+d_{p}\left(a_{n}\right)$ for $n \geqslant 0$. (a) Do there exist infinitely many primes $p$ for which there exist $p$-sequences $\left(a_{0}, a_{1}, a_{2}, \ldots\right)$ and $\left(b_{0}, b_{1}, b_{2}, \ldots\right)$ such that $a_{n}>b_{n}$ for infinitely many $n$, and $b_{n}>a_{n}$ for infinitely many $n$ ? (b) Do there exist infinitely many primes $p$ for which there exist $p$-sequences $\left(a_{0}, a_{1}, a_{2}, \ldots\right)$ and $\left(b_{0}, b_{1}, b_{2}, \ldots\right)$ such that $a_{0}<b_{0}$, but $a_{n}>b_{n}$ for all $n \geqslant 1$ ? (United Kingdom)

Solution: Fix some odd prime $p$, and let $T$ be the smallest positive integer such that $p \mid 2^{T}-1$; in other words, $T$ is the multiplicative order of 2 modulo $p$.  Consider any $p$-sequence $\left(x_{n}\right)=\left(x_{0}, x_{1}, x_{2}, \ldots\right)$. Obviously, $x_{n+1} \equiv 2 x_{n}(\bmod p)$ and therefore $x_{n} \equiv 2^{n} x_{0}(\bmod p)$. This yields $x_{n+T} \equiv x_{n}(\bmod p)$ and therefore $d\left(x_{n+T}\right)=d\left(x_{n}\right)$ for all $n \geqslant 0$. It follows that the sum $d\left(x_{n}\right)+d\left(x_{n+1}\right)+\ldots+d\left(x_{n+T-1}\right)$ does not depend on $n$ and is thus a function of $x_{0}$ (and $p$ ) only; we shall denote this sum by $S_{p}\left(x_{0}\right)$, and extend the function $S_{p}(\cdot)$ to all (not necessarily positive) integers. Therefore, we have $x_{n+k T}=x_{n}+k S_{p}\left(x_{0}\right)$ for all positive integers $n$ and $k$. Clearly, $S_{p}\left(x_{0}\right)=S_{p}\left(2^{t} x_{0}\right)$ for every integer $t \geqslant 0$.  In both parts, we use the notation  $$ S_{p}^{+}=S_{p}(1)=\sum_{i=0}^{T-1} d_{p}\left(2^{i}\right) \quad \text { and } \quad S_{p}^{-}=S_{p}(-1)=\sum_{i=0}^{T-1} d_{p}\left(p-2^{i}\right) $$  (a) Let $q>3$ be a prime and $p$ a prime divisor of $2^{q}+1$ that is greater than 3 . We will show that $p$ is suitable for part (a). Notice that $9 \nmid 2^{q}+1$, so that such a $p$ exists. Moreover, for any two odd primes $q<r$ we have $\operatorname{gcd}\left(2^{q}+1,2^{r}+1\right)=2^{\operatorname{gcd}(q, r)}+1=3$, thus there exist infinitely many such primes $p$.  For the chosen $p$, we have $T=2 q$. Since $2^{q} \equiv-1(\bmod p)$, we have $S_{p}^{+}=S_{p}^{-}$. Now consider the $p$-sequences $\left(a_{n}\right)$ and $\left(b_{n}\right)$ with $a_{0}=p+1$ and $b_{0}=p-1$; we claim that these sequences satisfy the required conditions. We have $a_{0}>b_{0}$ and $a_{1}=p+2<b_{1}=2 p-2$. It follows then that  $$ a_{k \cdot 2 q}=a_{0}+k S_{p}^{+}>b_{0}+k S_{p}^{+}=b_{k \cdot 2 q} \quad \text { and } \quad a_{k \cdot 2 q+1}=a_{1}+k S_{p}^{+}<b_{1}+k S_{p}^{+}=b_{k \cdot 2 q+1} $$  for all $k=0,1, \ldots$, as desired. (b) Let $q$ be an odd prime and $p$ a prime divisor of $2^{q}-1$; thus we have $T=q$. We will show that $p$ is suitable for part (b). Notice that the numbers of the form $2^{q}-1$ are pairwise coprime (since $\operatorname{gcd}\left(2^{q}-1,2^{r}-1\right)=2^{\operatorname{gcd}(q, r)}-1=1$ for any two distinct primes $q$ and $r$ ), thus there exist infinitely many such primes $p$. Notice that $d_{p}(x)+d_{p}(p-x)=p$ for all $x$ with $p \nmid x$, so that the sum $S_{p}^{+}+S_{p}^{-}=p q$ is odd, which yields $S_{p}^{+}=S_{p}(1) \neq S_{p}(-1)=S_{p}^{-}$.  Assume that $\left(x_{n}\right)$ and $\left(y_{n}\right)$ are two $p$-sequences with $S_{p}\left(x_{0}\right)>S_{p}\left(y_{0}\right)$ but $x_{0}<y_{0}$. The first condition yields that  $$ x_{M q+r}-y_{M q+r}=\left(x_{r}-y_{r}\right)+M\left(S_{p}\left(x_{0}\right)-S_{p}\left(y_{0}\right)\right) \geqslant\left(x_{r}-y_{r}\right)+M $$  for all nonnegative integers $M$ and every $r=0,1, \ldots, q-1$. Thus, we have $x_{n}>y_{n}$ for every $n \geqslant q+q \cdot \max \left\{y_{r}-x_{r}: r=0,1, \ldots, q-1\right\}$. Now, since $x_{0}<y_{0}$, there exists the largest $n_{0}$ with $x_{n_{0}}<y_{n_{0}}$. In this case the $p$-sequences $a_{n}=x_{n-n_{0}}$ and $b_{n}=y_{n-n_{0}}$ possess the desired property (notice here that $x_{n} \neq y_{n}$ for all $n \geqslant 0$, as otherwise we would have $\left.S_{p}\left(x_{0}\right)=S_{p}\left(x_{n}\right)=S_{p}\left(y_{n}\right)=S_{p}\left(y_{0}\right)\right)$.  It remains to find $p$-sequences $\left(x_{n}\right)$ and $\left(y_{n}\right)$ satisfying the two conditions. Recall that $S_{p}^{+} \neq S_{p}^{-}$. Now, if $S_{p}^{+}>S_{p}^{-}$, then we can put $x_{0}=1$ and $y_{0}=p-1$. Otherwise, if $S_{p}^{+}<S_{p}^{-}$, then we put $x_{0}=p-1$ and $y_{0}=p+1$.  This page is intentionally left blank


Determine all functions $f$ defined on the set of all positive integers and taking non-negative integer values, satisfying the three conditions: (i) $f(n) \neq 0$ for at least one $n$; (ii) $f(x y)=f(x)+f(y)$ for every positive integers $x$ and $y$; (iii) there are infinitely many positive integers $n$ such that $f(k)=f(n-k)$ for all $k<n$. (Croatia)

Solution: If a number $n$ is a product of primes, $n=p_{1} p_{2} \cdot \ldots \cdot p_{k}$, then  $$ f(n)=f\left(p_{1}\right)+\ldots+f\left(p_{k}\right) $$  in particular, $f(1)=0($ since $f(1)=f(1)+f(1))$. It is also clear that $f(n)=0$ implies $f(p)=0$ for all primes $p$ dividing $n$. Let us call positive integer $n \operatorname{good}$ if $f(k)=f(n-k)$ for $0<k<n$. If $n$ is good then each its divisor $d$ is also good; indeed, if $n=d m$ then  $$ f(k)=f(m k)-f(m)=f(n-m k)-f(m)=f(m(d-k))-f(m)=f(d-k) $$  for $0<k<d$. Thus, good numbers are products of good primes. It follows immediately from (i) that there exists a prime $p$ such that $f(p) \neq 0$; let $p$ be the smallest such prime. Then $f(r)=0$ for all $r<p$ (since all prime divisors of $r<p$ are less than $p$ ). Now every good number $n>p$ must be divisible by $p$. Indeed, if $n=p k+r$ is a good number, $k>0,0<r<p$, then $f(p) \leqslant f(p k)=f(n-p k)=f(r)=0$, a contradiction. Since any divisor of a good number is also good, this means that if a divisor $r$ of a good number is not divisible by $p$, it is less than $p$. Thus all good numbers have the form $r \cdot p^{k}$ with $r<p$. The condition (iii) implies that $k$ can be arbitrarily large, consequently all powers of $p$ are good.  If $q \neq p$ is a prime, $p^{q-1}-1$ is divisible by $q$ and $p^{q-1}$ is good. Then $f(q) \leqslant f\left(p^{q-1}-1\right)=$ $f(1)=0$, that is, $f(q)=0$.  Now we see that $f(n)=\nu_{p}(n) \cdot c$, where $c=f(p)$. The conditions (i) and (ii) for all such functions with $c \neq 0$ are obvious; the condition (iii) holds for all $n=p^{m}$, since $\nu_{p}\left(p^{m}-k\right)=\nu_{p}(k)$ when $0<k<p^{m}$.


Solution: We use the notion of a good number from the previous solution. As above, we also denote by $\nu_{p}(n)$ the exponent of a prime $p$ in the prime decomposition of $n$.  Say that a positive integer $k$ is big if $f(k)>0$. Let $\mathcal{B}$ be the set of big primes, and let $p_{1}<p_{2}<\ldots$ list the elements of $\mathcal{B}$ (this set might be either finite or infinite). By the problem conditions, we have  $$ f(n)=\sum_{i} \nu_{p_{i}}(n) f\left(p_{i}\right) $$  thus, the big numbers are those divisible by at least one big prime. For a positive integer $k$, define its essence $e(k)$ to be the largest product $e$ of (not necessarily different) big primes such that $e \mid k$. In other words,  $$ e(n)=\prod_{p_{i} \in \mathcal{B}} p_{i}^{\nu_{p_{i}}(n)} $$  This yields that $k / e(k)$ is not big, so $f(k)=f(e(k))+f(k / e(k))=f(e(k))$. Lemma. Assume that $n$ is a good number. Then $e(k)=e(n-k)$ for all $k<n$.  Proof. Arguing indirectly, choose a minimal $k$ for which the claim of the lemma is violated. Clearly, $k$ is big, as otherwise $f(k)=f(n-k)=0$ and hence $e(k)=e(n-k)=1$.  There are $t=k / e(k)$ multiples of $e(k)$ in each of the segments $[1, k]$ and $[n-k, n-1]$. On the other hand, there are $t-1$ such multiples on $[1, k-1]$ - and, by minimality of $k$, on $[n-k+1, n-1]$ as well. This yields that $n-k$ is a multiple of $e(k)$. Therefore,  $$ f(e(k))=f(k)=f(n-k)=f(e(k))+f\left(\frac{n-k}{e(k)}\right), $$  so the last summand vanishes, hence $\frac{n-k}{e(k)}$ has no big prime divisors, that is, $e(n-k)=e(k)$. This contradicts our choice.  Back to the problem, assume that $|\mathcal{B}| \geqslant 2$. Take any good number $n>p_{1} p_{2}$, and let $p_{1}^{\alpha}$ be the largest power of $p_{1}$ smaller than $n$, so that $n \leqslant p_{1}^{\alpha+1}<p_{1}^{\alpha} p_{2}$. By the lemma, $e\left(n-p_{1}^{\alpha}\right)=e\left(p_{1}^{\alpha}\right)=p_{1}^{\alpha}$, which yields $p_{1}^{\alpha} \mid n$. Similarly, $p_{2} \mid n$, so that $n \geqslant p_{1}^{\alpha} p_{2}$. This contradiction shows that $|\mathcal{B}| \leqslant 1$, which by (1) yields that $f$ is listed in the answer.


Solution: We have $f\left(\prod p_{i}^{\alpha_{i}}\right)=\sum \alpha_{i} f\left(p_{i}\right)$. Note that  $$ f(n-1)+f(n-2)+\ldots+\ldots f(n-k) \geqslant f(1)+\ldots+f(k) $$  for all $k=1,2, \ldots, n-1$, since the difference LHS-RHS is just $\left.f\binom{n-1}{k}\right)$. Assume that $f(p)>0$. If $f(k)=f(n-k)$ for all $k$, it implies that $\binom{n-1}{k}$ is not divisible by $p$ for all $k=1,2, \ldots, n-2$. It is well known that it implies $n=a \cdot p^{s}, a<p$. If there are two primes $p, q$ such that $f(p)>0, f(q)>0$, there exist only finitely many $n$ which are equal both to $a \cdot p^{s}, a<p$, and $b \cdot q^{t}, b<q$. So there exists at most one such $p$, and therefore $f(n)=C \cdot \nu_{p}(n)$ for some constant $C$.


Solution: We call a function $f: \mathbb{N} \rightarrow \mathbb{N}_{0}$ satisfying (ii) additive. We call a pair $(f, n)$, where $f$ is an additive function and $n \in \mathbb{N}$, good, if for all $k<n$ it holds $f(k)=f(n-k)$. For an additive function $f$ and a prime number $p$ the number $\frac{f(p)}{\ln p}$ is denoted by $g(f, p)$.  Let $(f, n)$ be a good pair such that $f(p)>0$ for at least two primes less than $n$. Let $p_{0}$ be the prime with maximal $g(f, p)$ among all primes $p<n$. Let $a_{0}$ be the maximal exponent such that $p_{0}^{a_{0}}<n$. Then $f(k)<f\left(p_{0}^{a_{0}}\right)$ for all $k<p_{0}^{a_{0}}$. Indeed, if $k=p_{1}^{a_{1}} \ldots p_{m}^{a_{m}}<p_{0}^{a_{0}}$, then  $$ \begin{aligned} f(k) & =a_{1} f\left(p_{1}\right)+\ldots+a_{m} f\left(p_{m}\right)=g\left(f, p_{1}\right) a_{1} \ln p_{1}+\ldots+g\left(f, p_{m}\right) a_{m} \ln a_{m} \\ & <g\left(f, p_{0}\right) a_{0} \ln p_{0}=f\left(p_{0}^{a_{0}}\right) . \end{aligned} $$  Let $n=b p_{0}^{a_{0}}+r$, where $0<r<p_{0}^{a_{0}}$. Then $f(r)=f\left(b p_{0}^{a_{0}}\right) \geqslant f\left(p_{0}^{a_{0}}\right)$. This contradiction shows that $p_{0}^{a_{0}} \mid n$. Then $n=p_{0}^{\nu_{p_{0}}(n)} n^{\prime}$, where $n^{\prime} \leqslant p_{0}$.  The functions $f_{1}(m):=f\left(p_{0}\right) \nu_{p_{0}}(m)$ and $f_{2}:=f-f_{1}$ are additive (obviously $f(m) \geqslant$ $f\left(p_{0}^{\nu_{p_{0}}(m)}\right)=f_{1}(m)$, since $p_{0}^{\nu_{p_{0}}(m)}$ divides $\left.m\right)$. For $k<n, \nu_{p}(k)=\nu_{p}(n-k)$. Hence the pair $\left(f_{2}, n\right)$ is also good. Note that $f_{2}\left(p_{0}\right)=0$.  Choose among all primes $p<n$ the prime $q_{0}$ with maximal $g\left(f_{2}, p\right)$. As above we can prove that $n=q_{0}^{\nu_{0}(n)} n^{\prime \prime}$ with $n^{\prime \prime}<q_{0}$. Since $p_{0} \neq q_{0}$, we get a contradiction. Thus $f(n)=f(p) \cdot \nu_{p}(n)$.


For a positive integer $n$, let $d(n)$ be the number of positive divisors of $n$, and let $\varphi(n)$ be the number of positive integers not exceeding $n$ which are coprime to $n$. Does there exist a constant $C$ such that  $$ \frac{\varphi(d(n))}{d(\varphi(n))} \leqslant C $$  for all $n \geqslant 1$ ? (Cyprus)

Solution: Fix $N>1$, let $p_{1}, \ldots, p_{k}$ be all primes between 1 and $N$ and $p_{k+1}, \ldots, p_{k+s}$ be all primes between $N+1$ and $2 N$. Since for $j \leqslant k+s$ all prime divisors of $p_{j}-1$ do not exceed $N$, we have  $$ \prod_{j=1}^{k+s}\left(p_{j}-1\right)=\prod_{i=1}^{k} p_{i}^{c_{i}} $$  with some fixed exponents $c_{1}, \ldots, c_{k}$. Choose a huge prime number $q$ and consider a number  $$ n=\left(p_{1} \cdot \ldots \cdot p_{k}\right)^{q-1} \cdot\left(p_{k+1} \cdot \ldots \cdot p_{k+s}\right) $$  Then  $$ \varphi(d(n))=\varphi\left(q^{k} \cdot 2^{s}\right)=q^{k-1}(q-1) 2^{s-1} $$  and  $$ d(\varphi(n))=d\left(\left(p_{1} \cdot \ldots \cdot p_{k}\right)^{q-2} \prod_{i=1}^{k+s}\left(p_{i}-1\right)\right)=d\left(\prod_{i=1}^{k} p_{i}^{q-2+c_{i}}\right)=\prod_{i=1}^{k}\left(q-1+c_{i}\right) $$  so  $$ \frac{\varphi(d(n))}{d(\varphi(n))}=\frac{q^{k-1}(q-1) 2^{s-1}}{\prod_{i=1}^{k}\left(q-1+c_{i}\right)}=2^{s-1} \cdot \frac{q-1}{q} \cdot \prod_{i=1}^{k} \frac{q}{q-1+c_{i}} $$  which can be made arbitrarily close to $2^{s-1}$ by choosing $q$ large enough. It remains to show that $s$ can be arbitrarily large, i.e. that there can be arbitrarily many primes between $N$ and $2 N$.  This follows, for instance, from the well-known fact that $\sum \frac{1}{p}=\infty$, where the sum is taken over the set $\mathbb{P}$ of prime numbers. Indeed, if, for some constant $\stackrel{p}{C}$, there were always at most $C$ primes between $2^{\ell}$ and $2^{\ell+1}$, we would have  $$ \sum_{p \in \mathbb{P}} \frac{1}{p}=\sum_{\ell=0}^{\infty} \sum_{\substack{p \in \mathbb{P} \\ p \in\left[2^{\ell}, 2^{\ell+1}\right)}} \frac{1}{p} \leqslant \sum_{\ell=0}^{\infty} \frac{C}{2^{\ell}}<\infty, $$  which is a contradiction. Comment 1. Here we sketch several alternative elementary self-contained ways to perform the last step of the solution above. In particular, they avoid using divergence of $\sum \frac{1}{p}$.  Suppose that for some constant $C$ and for every $k=1,2, \ldots$ there exist at most $C$ prime numbers between $2^{k}$ and $2^{k+1}$. Consider the prime factorization of the factorial $\left(2^{n}\right)!=\prod p^{\alpha_{p}}$. We have $\alpha_{p}=\left\lfloor 2^{n} / p\right\rfloor+\left\lfloor 2^{n} / p^{2}\right\rfloor+\ldots$. Thus, for $p \in\left[2^{k}, 2^{k+1}\right)$, we get $\alpha_{p} \leqslant 2^{n} / 2^{k}+2^{n} / 2^{k+1}+\ldots=2^{n-k+1}$, therefore $p^{\alpha_{p}} \leqslant 2^{(k+1) 2^{n-k+1}}$. Combining this with the bound $(2 m)!\geqslant m(m+1) \cdot \ldots \cdot(2 m-1) \geqslant m^{m}$ for $m=2^{n-1}$ we get  $$ 2^{(n-1) \cdot 2^{n-1}} \leqslant\left(2^{n}\right)!\leqslant \prod_{k=1}^{n-1} 2^{C(k+1) 2^{n-k+1}} $$  or  $$ \sum_{k=1}^{n-1} C(k+1) 2^{1-k} \geqslant \frac{n-1}{2} $$  that fails for large $n$ since $C(k+1) 2^{1-k}<1 / 3$ for all but finitely many $k$. In fact, a much stronger inequality can be obtained in an elementary way: Note that the formula for $\nu_{p}(n!)$ implies that if $p^{\alpha}$ is the largest power of $p$ dividing $\binom{n}{n / 2}$, then $p^{\alpha} \leqslant n$. By looking at prime factorization of $\binom{n}{n / 2}$ we instantaneously infer that  $$ \pi(n) \geqslant \log _{n}\binom{n}{n / 2} \geqslant \frac{\log \left(2^{n} / n\right)}{\log n} \geqslant \frac{n}{2 \log n} . $$  This, in particular, implies that for infinitely many $n$ there are at least $\frac{n}{3 \log n}$ primes between $n$ and $2 n$.


Solution: In this solution we will use the Prime Number Theorem which states that  $$ \pi(m)=\frac{m}{\log m} \cdot(1+o(1)) $$  as $m$ tends to infinity. Here and below $\pi(m)$ denotes the number of primes not exceeding $m$, and $\log$ the natural logarithm.  Let $m>5$ be a large positive integer and let $n:=p_{1} p_{2} \cdot \ldots \cdot p_{\pi(m)}$ be the product of all primes not exceeding $m$. Then $\varphi(d(n))=\varphi\left(2^{\pi(m)}\right)=2^{\pi(m)-1}$. Consider the number  $$ \varphi(n)=\prod_{k=1}^{\pi(m)}\left(p_{k}-1\right)=\prod_{s=1}^{\pi(m / 2)} q_{s}^{\alpha_{s}} $$  where $q_{1}, \ldots, q_{\pi(m / 2)}$ are primes not exceeding $m / 2$. Note that every term $p_{k}-1$ contributes at most one prime $q_{s}>\sqrt{m}$ into the product $\prod_{s} q_{s}^{\alpha_{s}}$, so we have  $$ \sum_{s: q_{s}>\sqrt{m}} \alpha_{s} \leqslant \pi(m) \Longrightarrow \sum_{s: q_{s}>\sqrt{m}}\left(1+\alpha_{s}\right) \leqslant \pi(m)+\pi(m / 2) . $$  Hence, applying the AM-GM inequality and the inequality $(A / x)^{x} \leqslant e^{A / e}$, we obtain  $$ \prod_{s: q_{s}>\sqrt{m}}\left(\alpha_{s}+1\right) \leqslant\left(\frac{\pi(m)+\pi(m / 2)}{\ell}\right)^{\ell} \leqslant \exp \left(\frac{\pi(m)+\pi(m / 2)}{e}\right) $$  where $\ell$ is the number of primes in the interval $(\sqrt{m}, m]$. We then use a trivial bound $\alpha_{i} \leqslant \log _{2}(\varphi(n)) \leqslant \log _{2} n<\log _{2}\left(m^{m}\right)<m^{2}$ for each $i$ with $q_{i}<\sqrt{m}$ to obtain  $$ \prod_{s=1}^{\pi(\sqrt{m})}\left(\alpha_{s}+1\right) \leqslant\left(m^{2}\right)^{\sqrt{m}}=m^{2 \sqrt{m}} $$  Putting this together we obtain  $$ d(\varphi(n))=\prod_{s=1}^{\pi(m / 2)}\left(\alpha_{s}+1\right) \leqslant \exp \left(2 \sqrt{m} \cdot \log m+\frac{\pi(m)+\pi(m / 2)}{e}\right) $$  The prime number theorem then implies that  $$ \limsup _{m \rightarrow \infty} \frac{\log (d(\varphi(n)))}{m / \log m} \leqslant \limsup _{m \rightarrow \infty} \frac{2 \sqrt{m} \cdot \log m}{m / \log m}+\limsup _{m \rightarrow \infty} \frac{\pi(m)+\pi(m / 2)}{e \cdot m / \log m}=\frac{3}{2 e} . $$  Whereas, again by prime number theorem, we have  $$ \liminf _{m \rightarrow \infty} \frac{\log (\varphi(d(n)))}{m / \log m}=\liminf _{m \rightarrow \infty} \frac{\log \left(2^{\pi(m)-1}\right)}{m / \log m}=\log 2 $$  Since $\frac{3}{2 e}<\frac{3}{5}<\log 2$, this implies that $\varphi(d(n)) / d(\varphi(n))$ can be arbitrarily large.  Comment 2. The original formulation of the problem was asking whether $d(\varphi(n)) \geqslant \varphi(d(n))$ for all but finitely many values of $n$. The Problem Selection Committee decided that the presented version is better suited for the Shortlist.  This page is intentionally left blank


Let $\mathcal{S}$ be a set consisting of $n \geqslant 3$ positive integers, none of which is a sum of two other distinct members of $\mathcal{S}$. Prove that the elements of $\mathcal{S}$ may be ordered as $a_{1}, a_{2}, \ldots, a_{n}$ so that $a_{i}$ does not divide $a_{i-1}+a_{i+1}$ for all $i=2,3, \ldots, n-1$. (Ukraine)

Solution: We prove the following stronger statement. Claim. Let $\mathcal{S}$ be a good set consisting of $n \geqslant 2$ positive integers. Then the elements of $\mathcal{S}$ may be ordered as $a_{1}, a_{2}, \ldots, a_{n}$ so that $a_{i} \nmid a_{i-1}+a_{i+1}$ and $a_{i} \nmid a_{i-1}-a_{i+1}$, for all $i=2,3, \ldots, n-1$. Proof. Say that the ordering $a_{1}, \ldots, a_{n}$ of $\mathcal{S}$ is nice if it satisfies the required property.  We proceed by induction on $n$. The base case $n=2$ is trivial, as there are no restrictions on the ordering.  To perform the step of induction, suppose that $n \geqslant 3$. Let $a=\max \mathcal{S}$, and set $\mathcal{T}=\mathcal{S} \backslash\{a\}$. Use the inductive hypothesis to find a nice ordering $b_{1}, \ldots, b_{n-1}$ of $\mathcal{T}$. We will show that $a$ may be inserted into this sequence so as to reach a nice ordering of $\mathcal{S}$. In other words, we will show that there exists a $j \in\{1,2, \ldots, n\}$ such that the ordering  $$ N_{j}=\left(b_{1}, \ldots, b_{j-1}, a, b_{j}, b_{j+1}, \ldots, b_{n-1}\right) $$  is nice. Assume that, for some $j$, the ordering $N_{j}$ is not nice, so that some element $x$ in it divides either the sum or the difference of two adjacent ones. This did not happen in the ordering of $\mathcal{T}$, hence $x \in\left\{b_{j-1}, a, b_{j}\right\}$ (if, say, $b_{j-1}$ does not exist, then $x \in\left\{a, b_{j}\right\}$; a similar agreement is applied hereafter). But the case $x=a$ is impossible: $a$ cannot divide $b_{j-1}-b_{j}$, since $0<\left|b_{j-1}-b_{j}\right|<a$, while $a \nmid b_{j-1}+b_{j}$ by Observation A. Therefore $x \in\left\{b_{j-1}, b_{j}\right\}$. In this case, assign the number $x$ to the index $j$.  Suppose now that none of the $N_{j}$ is nice. Since there are $n$ possible indices $j$, and only $n-1$ elements in $\mathcal{T}$, one of those elements (say, $b_{k}$ ) is assigned to two different indices, which then should equal $k$ and $k+1$. This means that $b_{k}$ divides the numbers $b_{k-1}+\varepsilon_{1} a$ and $a+\varepsilon_{2} b_{k+1}$, for some signs $\varepsilon_{1}, \varepsilon_{2} \in\{-1,1\}$. But then  $$ b_{k-1} \equiv-\varepsilon_{1} a \equiv \varepsilon_{1} \varepsilon_{2} b_{k+1} \quad\left(\bmod b_{k}\right), $$  and therefore $b_{k} \mid b_{k-1}-\varepsilon_{1} \varepsilon_{2} b_{k+1}$, which means that the ordering of $\mathcal{T}$ was not nice. This contradiction proves the step of induction.


Solution: We again prove a stronger statement. Claim. Let $\mathcal{S}$ be an arbitrary set of $n \geqslant 3$ positive integers. Then its elements can be ordered as $a_{1}, \ldots, a_{n}$ so that, if $a_{i} \mid a_{i-1}+a_{i+1}$, then $a_{i}=\max \mathcal{S}$.  The claim easily implies what we need to prove, due to Observation A. To prove the Claim, introduce the function $f$ which assigns to any two elements $a, b \in \mathcal{S}$ with $a<b$ the unique integer $f(a, b) \in\{1,2, \ldots, a\}$ such that $a \mid b+f(a, b)$. Hence, if $b \mid a+c$ for some $a, b, c \in \mathcal{S}$ with $a<b<c$, then $a=f(b, c)$. Therefore, the Claim is a consequence of the following combinatorial lemma.  Lemma. Let $\mathcal{S}$ be a set of $n \geqslant 3$ positive integers, and let $f$ be a function which assigns to any $a, b \in \mathcal{S}$ with $a<b$ some integer from the range $\{1, \ldots, a\}$. Then the elements of $\mathcal{S}$ may be ordered as $a_{1}, a_{2}, \ldots, a_{n}$ so as to satisfy the following two conditions simultaneously: (i) Unimodality: There exists a $j \in\{1,2, \ldots, n\}$ such that $a_{1}<a_{2}<\ldots<a_{j}>a_{j+1}>\ldots>$ $a_{n}$; and (ii) $f$-avoidance: If $a<b$ are two elements of $\mathcal{S}$, which are adjacent in the ordering, then $f(a, b)$ is not adjacent to $a$. Proof. We call an ordering of $\mathcal{S}$ satisfying (i) and (ii) $f$-nice. We agree that $f(x, y)=x$ for $x \geqslant y$; this agreement puts no extra restriction.  We proceed by induction; for the base case $n=3$, it suffices to put the maximal element in $\mathcal{S}$ onto the middle position.  To perform the step of induction, let $p<q$ be the two minimal elements of $\mathcal{S}$, and set $\mathcal{T}=\mathcal{S} \backslash\{p\}$. Define a function $g$ by assigning to any elements $a<b$ of $\mathcal{T}$ the value  $$ g(a, b)= \begin{cases}q, & \text { if } f(a, b)=p \\ f(a, b), & \text { otherwise }\end{cases} $$  Notice that $g(a, b) \leqslant a$ for all $a, b \in \mathcal{T}$. Use the inductive hypothesis to get a $g$-nice ordering $b_{1}, b_{2}, \ldots, b_{n-1}$ of $\mathcal{T}$. By unimodality, either $b_{1}$ or $b_{n-1}$ equals $q$; these cases differ only by reverting the order, so we assume $b_{1}=q$.  Notice that, according to (1), the number $f\left(b_{2}, b_{3}\right)$ differs from both $p$ and $q$. On the other hand, the number $f\left(b_{n-1}, b_{n-2}\right)$ differs from at least one of them - say, from $r$; set $s=p+q-r$, so that $\{r, s\}=\{p, q\}$. Now, order $\mathcal{S}$ as  $$ s, b_{2}, b_{3}, \ldots, b_{n-1}, r $$  By the induction hypothesis and the above choice, this ordering is nice. Comment. In the original proposal, the numbers in the set were assumed to be odd (which implies that none is a sum of two others); moreover, the proposal requested to arrange in a row all numbers but one.  On the other hand, Solution 2 shows that the condition of $\mathcal{S}$ being good may be relaxed to the condition that the maximal element of $\mathcal{S}$ is not a sum of two other elements in $\mathcal{S}$. On the other hand, the set $\{1,2,3\}$ shows that the condition cannot be merely omitted.  The Problem Selection Committee considered several versions of the problem and chose the best version in their opinion for the Shortlist.  61st International Mathematical Olympiad Saint Petersburg Russia


Version 1. Let $n$ be a positive integer, and set $N=2^{n}$. Determine the smallest real number $a_{n}$ such that, for all real $x$,  $$ \sqrt[N]{\frac{x^{2 N}+1}{2}} \leqslant a_{n}(x-1)^{2}+x $$  Version 2. For every positive integer $N$, determine the smallest real number $b_{N}$ such that, for all real $x$,  $$ \sqrt[N]{\frac{x^{2 N}+1}{2}} \leqslant b_{N}(x-1)^{2}+x $$  (Ireland) Answer for both versions : $a_{n}=b_{N}=N / 2$. Solution 1 (for Version 1). First of all, assume that $a_{n}<N / 2$ satisfies the condition. Take $x=1+t$ for $t>0$, we should have  $$ \frac{(1+t)^{2 N}+1}{2} \leqslant\left(1+t+a_{n} t^{2}\right)^{N} . $$  Expanding the brackets we get  $$ \left(1+t+a_{n} t^{2}\right)^{N}-\frac{(1+t)^{2 N}+1}{2}=\left(N a_{n}-\frac{N^{2}}{2}\right) t^{2}+c_{3} t^{3}+\ldots+c_{2 N} t^{2 N} $$  with some coefficients $c_{3}, \ldots, c_{2 N}$. Since $a_{n}<N / 2$, the right hand side of (1) is negative for sufficiently small $t$. A contradiction.  It remains to prove the following inequality  $$ \sqrt[N]{\frac{1+x^{2 N}}{2}} \leqslant x+\frac{N}{2}(x-1)^{2} $$  where $N=2^{n}$. Use induction in $n$. The base case $n=0$ is trivial: $N=1$ and both sides of $\mathcal{I}(N, x)$ are equal to $\left(1+x^{2}\right) / 2$. For completing the induction we prove $\mathcal{I}(2 N, x)$ assuming that $\mathcal{I}(N, y)$ is established for all real $y$. We have  $$ \begin{aligned} \left(x+N(x-1)^{2}\right)^{2} & =x^{2}+N^{2}(x-1)^{4}+N(x-1)^{2} \frac{(x+1)^{2}-(x-1)^{2}}{2} \\ & =x^{2}+\frac{N}{2}\left(x^{2}-1\right)^{2}+\left(N^{2}-\frac{N}{2}\right)(x-1)^{4} \geqslant x^{2}+\frac{N}{2}\left(x^{2}-1\right)^{2} \geqslant \sqrt[N]{\frac{1+x^{4 N}}{2}} \end{aligned} $$  where the last inequality is $\mathcal{I}\left(N, x^{2}\right)$. Since  $$ x+N(x-1)^{2} \geqslant x+\frac{(x-1)^{2}}{2}=\frac{x^{2}+1}{2} \geqslant 0 $$  taking square root we get $\mathcal{I}(2 N, x)$. The inductive step is complete.

Solution: 1 (for Version 2). Like in   First of all, $\mathcal{I}(N, 0)$ is obvious. Further, if $x>0$, then the left hand sides of $\mathcal{I}(N,-x)$ and $\mathcal{I}(N, x)$ coincide, while the right hand side of $\mathcal{I}(N,-x)$ is larger than that of $\mathcal{I}(N,-x)$ (their difference equals $2(N-1) x \geqslant 0)$. Therefore, $\mathcal{I}(N,-x)$ follows from $\mathcal{I}(N, x)$. So, hereafter we suppose that $x>0$.  Divide $\mathcal{I}(N, x)$ by $x$ and let $t=(x-1)^{2} / x=x-2+1 / x$; then $\mathcal{I}(n, x)$ reads as  $$ f_{N}:=\frac{x^{N}+x^{-N}}{2} \leqslant\left(1+\frac{N}{2} t\right)^{N} $$  The key identity is the expansion of $f_{N}$ as a polynomial in $t$ :


Solution: 2 (for Version 2). Here we present another proof of the inequality (2) for $x>0$, or, equivalently, for $t=(x-1)^{2} / x \geqslant 0$. Instead of finding the coefficients of the polynomial $f_{N}=f_{N}(t)$ we may find its roots, which is in a sense more straightforward. Note that the recurrence (4) and the initial conditions $f_{0}=1, f_{1}=1+t / 2$ imply that $f_{N}$ is a polynomial in $t$ of degree $N$. It also follows by induction that $f_{N}(0)=1, f_{N}^{\prime}(0)=N^{2} / 2$ : the recurrence relations read as $f_{N+1}(0)+f_{N-1}(0)=2 f_{N}(0)$ and $f_{N+1}^{\prime}(0)+f_{N-1}^{\prime}(0)=2 f_{N}^{\prime}(0)+f_{N}(0)$, respectively.  Next, if $x_{k}=\exp \left(\frac{i \pi(2 k-1)}{2 N}\right)$ for $k \in\{1,2, \ldots, N\}$, then  $$ -t_{k}:=2-x_{k}-\frac{1}{x_{k}}=2-2 \cos \frac{\pi(2 k-1)}{2 N}=4 \sin ^{2} \frac{\pi(2 k-1)}{4 N}>0 $$  and  $$ f_{N}\left(t_{k}\right)=\frac{x_{k}^{N}+x_{k}^{-N}}{2}=\frac{\exp \left(\frac{i \pi(2 k-1)}{2}\right)+\exp \left(-\frac{i \pi(2 k-1)}{2}\right)}{2}=0 . $$  So the roots of $f_{N}$ are $t_{1}, \ldots, t_{N}$ and by the AM-GM inequality we have  $$ \begin{aligned} f_{N}(t)=\left(1-\frac{t}{t_{1}}\right)\left(1-\frac{t}{t_{2}}\right) \ldots\left(1-\frac{t}{t_{N}}\right) & \leqslant\left(1-\frac{t}{N}\left(\frac{1}{t_{1}}+\ldots+\frac{1}{t_{n}}\right)\right)^{N}= \\ \left(1+\frac{t f_{N}^{\prime}(0)}{N}\right)^{N} & =\left(1+\frac{N}{2} t\right)^{N} \end{aligned} $$  Comment. The polynomial $f_{N}(t)$ equals to $\frac{1}{2} T_{N}(t+2)$, where $T_{n}$ is the $n^{\text {th }}$ Chebyshev polynomial of the first kind: $T_{n}(2 \cos s)=2 \cos n s, T_{n}(x+1 / x)=x^{n}+1 / x^{n}$.


Solution: 3 (for Version 2). Here we solve the problem when $N \geqslant 1$ is an arbitrary real number. For a real number $a$ let  $$ f(x)=\left(\frac{x^{2 N}+1}{2}\right)^{\frac{1}{N}}-a(x-1)^{2}-x $$  Then $f(1)=0$,  $$ f^{\prime}(x)=\left(\frac{x^{2 N}+1}{2}\right)^{\frac{1}{N}-1} x^{2 N-1}-2 a(x-1)-1 \quad \text { and } \quad f^{\prime}(1)=0 $$  $f^{\prime \prime}(x)=(1-N)\left(\frac{x^{2 N}+1}{2}\right)^{\frac{1}{N}-2} x^{4 N-2}+(2 N-1)\left(\frac{x^{2 N}+1}{2}\right)^{\frac{1}{N}-1} x^{2 N-2}-2 a \quad$ and $\quad f^{\prime \prime}(1)=N-2 a$. So if $a<\frac{N}{2}$, the function $f$ has a strict local minimum at point 1 , and the inequality $f(x) \leqslant$ $0=f(1)$ does not hold. This proves $b_{N} \geqslant N / 2$.  For $a=\frac{N}{2}$ we have $f^{\prime \prime}(1)=0$ and  $$ f^{\prime \prime \prime}(x)=\frac{1}{2}(1-N)(1-2 N)\left(\frac{x^{2 N}+1}{2}\right)^{\frac{1}{N}-3} x^{2 N-3}\left(1-x^{2 N}\right) \quad \begin{cases}>0 & \text { if } 0<x<1 \text { and } \\ <0 & \text { if } x>1\end{cases} $$  Hence, $f^{\prime \prime}(x)<0$ for $x \neq 1 ; f^{\prime}(x)>0$ for $x<1$ and $f^{\prime}(x)<0$ for $x>1$, finally $f(x)<0$ for $x \neq 1$.  Comment. Version 2 is much more difficult, of rather A5 or A6 difficulty. The induction in Version 1 is rather straightforward, while all three above solutions of Version 2 require some creativity.  This page is intentionally left blank


Let $\mathcal{A}$ denote the set of all polynomials in three variables $x, y, z$ with integer coefficients. Let $\mathcal{B}$ denote the subset of $\mathcal{A}$ formed by all polynomials which can be expressed as  $$ (x+y+z) P(x, y, z)+(x y+y z+z x) Q(x, y, z)+x y z R(x, y, z) $$  with $P, Q, R \in \mathcal{A}$. Find the smallest non-negative integer $n$ such that $x^{i} y^{j} z^{k} \in \mathcal{B}$ for all nonnegative integers $i, j, k$ satisfying $i+j+k \geqslant n$. (Venezuela) Answer: $n=4$.

Solution: We start by showing that $n \leqslant 4$, i.e., any monomial $f=x^{i} y^{j} z^{k}$ with $i+j+k \geqslant 4$ belongs to $\mathcal{B}$. Assume that $i \geqslant j \geqslant k$, the other cases are analogous.  Let $x+y+z=p, x y+y z+z x=q$ and $x y z=r$. Then  $$ 0=(x-x)(x-y)(x-z)=x^{3}-p x^{2}+q x-r $$  therefore $x^{3} \in \mathcal{B}$. Next, $x^{2} y^{2}=x y q-(x+y) r \in \mathcal{B}$. If $k \geqslant 1$, then $r$ divides $f$, thus $f \in \mathcal{B}$. If $k=0$ and $j \geqslant 2$, then $x^{2} y^{2}$ divides $f$, thus we have $f \in \mathcal{B}$ again. Finally, if $k=0, j \leqslant 1$, then $x^{3}$ divides $f$ and $f \in \mathcal{B}$ in this case also.  In order to prove that $n \geqslant 4$, we show that the monomial $x^{2} y$ does not belong to $\mathcal{B}$. Assume the contrary:  $$ x^{2} y=p P+q Q+r R $$  for some polynomials $P, Q, R$. If polynomial $P$ contains the monomial $x^{2}$ (with nonzero coefficient), then $p P+q Q+r R$ contains the monomial $x^{3}$ with the same nonzero coefficient. So $P$ does not contain $x^{2}, y^{2}, z^{2}$ and we may write  $$ x^{2} y=(x+y+z)(a x y+b y z+c z x)+(x y+y z+z x)(d x+e y+f z)+g x y z $$  where $a, b, c ; d, e, f ; g$ are the coefficients of $x y, y z, z x ; x, y, z ; x y z$ in the polynomials $P$; $Q ; R$, respectively (the remaining coefficients do not affect the monomials of degree 3 in $p P+q Q+r R)$. By considering the coefficients of $x y^{2}$ we get $e=-a$, analogously $e=-b$, $f=-b, f=-c, d=-c$, thus $a=b=c$ and $f=e=d=-a$, but then the coefficient of $x^{2} y$ in the right hand side equals $a+d=0 \neq 1$.  Comment 1. The general question is the following. Call a polynomial $f\left(x_{1}, \ldots, x_{n}\right)$ with integer coefficients nice, if $f(0,0, \ldots, 0)=0$ and $f\left(x_{\pi_{1}}, \ldots, x_{\pi_{n}}\right)=f\left(x_{1}, \ldots, x_{n}\right)$ for any permutation $\pi$ of $1, \ldots, n$ (in other words, $f$ is symmetric and its constant term is zero.) Denote by $\mathcal{I}$ the set of polynomials of the form  $$ p_{1} q_{1}+p_{2} q_{2}+\ldots+p_{m} q_{m} $$  where $m$ is an integer, $q_{1}, \ldots, q_{m}$ are polynomials with integer coefficients, and $p_{1}, \ldots, p_{m}$ are nice polynomials. Find the least $N$ for which any monomial of degree at least $N$ belongs to $\mathcal{I}$.  The answer is $n(n-1) / 2+1$. The lower bound follows from the following claim: the polynomial  $$ F\left(x_{1}, \ldots, x_{n}\right)=x_{2} x_{3}^{2} x_{4}^{3} \cdot \ldots \cdot x_{n}^{n-1} $$  does not belong to $\mathcal{I}$. Assume that $F=\sum p_{i} q_{i}$, according to (2). By taking only the monomials of degree $n(n-1) / 2$, we can additionally assume that every $p_{i}$ and every $q_{i}$ is homogeneous, $\operatorname{deg} p_{i}>0$, and $\operatorname{deg} p_{i}+\operatorname{deg} q_{i}=$ $\operatorname{deg} F=n(n-1) / 2$ for all $i$.  Consider the alternating sum  $$ \sum_{\pi} \operatorname{sign}(\pi) F\left(x_{\pi_{1}}, \ldots, x_{\pi_{n}}\right)=\sum_{i=1}^{m} p_{i} \sum_{\pi} \operatorname{sign}(\pi) q_{i}\left(x_{\pi_{1}}, \ldots, x_{\pi_{n}}\right):=S $$  where the summation is done over all permutations $\pi$ of $1, \ldots n$, and $\operatorname{sign}(\pi)$ denotes the sign of the permutation $\pi$. Since $\operatorname{deg} q_{i}=n(n-1) / 2-\operatorname{deg} p_{i}<n(n-1) / 2$, in any monomial $Q$ of $q_{i}$, there are at least two variables, say $x_{\alpha}$ and $x_{\beta}$, with equal exponents. Therefore $\sum_{\pi} \operatorname{sign}(\pi) Q\left(x_{\pi_{1}}, \ldots, x_{\pi_{n}}\right)=0$, because each pair of terms that corresponds to permutations which differ by the transposition of $\alpha$ and $\beta$, cancels out. This holds for any $i=1, \ldots, m$ and any monomial of $q_{i}$, so $S=0$. But the left hand side of (3) is a non-zero polynomial. This is a contradiction.  Let us now prove, using induction on $n$, that any monomial $h=x_{1}^{c_{1}} \ldots x_{n}^{c_{n}}$ of degree $n(n-1) / 2+1$ belongs to $\mathcal{I}$, and additionally all $p_{i}, q_{i}$ in the representation (2) can be chosen homogeneous with sum of degrees equal to $n(n-1) / 2+1$. (Obviously, any monomial of degree at least $n(n-1) / 2+1$ is divisible by a monomial of degree exactly $n(n-1) / 2+1$, thus this suffices.) The proposition is true for $n=1$, so assume that $n>1$ and that the proposition is proved for smaller values of $n$.  We proceed by an internal induction on $S:=\left|\left\{i: c_{i}=0\right\}\right|$. In the base case $S=0$ the monomial $h$ is divisible by the nice polynomial $x_{1} \cdot \ldots x_{n}$, therefore $h \in \mathcal{I}$. Now assume that $S>0$ and that the claim holds for smaller values of $S$. Let $T=n-S$. We may assume that $c_{T+1}=\ldots=c_{n}=0$ and $h=x_{1} \cdot \ldots \cdot x_{T} g\left(x_{1}, \ldots, x_{n-1}\right)$, where $\operatorname{deg} g=n(n-1) / 2-T+1 \geqslant(n-1)(n-2) / 2+1$. Using the outer induction hypothesis we represent $g$ as $p_{1} q_{1}+\ldots+p_{m} q_{m}$, where $p_{i}\left(x_{1}, \ldots, x_{n-1}\right)$ are nice polynomials in $n-1$ variables. There exist nice homogeneous polynomials $P_{i}\left(x_{1}, \ldots, x_{n}\right)$ such that $P_{i}\left(x_{1}, \ldots, x_{n-1}, 0\right)=p_{i}\left(x_{1}, \ldots, x_{n-1}\right)$. In other words, $\Delta_{i}:=p_{i}\left(x_{1}, \ldots, x_{n-1}\right)-P_{i}\left(x_{1}, \ldots, x_{n-1}, x_{n}\right)$ is divisible by $x_{n}$, let $\Delta_{i}=x_{n} g_{i}$. We get  $$ h=x_{1} \cdot \ldots \cdot x_{T} \sum p_{i} q_{i}=x_{1} \cdot \ldots \cdot x_{T} \sum\left(P_{i}+x_{n} g_{i}\right) q_{i}=\left(x_{1} \cdot \ldots \cdot x_{T} x_{n}\right) \sum g_{i} q_{i}+\sum P_{i} q_{i} \in \mathcal{I} $$  The first term belongs to $\mathcal{I}$ by the inner induction hypothesis. This completes both inductions. Comment 2. The solutions above work smoothly for the versions of the original problem and its extensions to the case of $n$ variables, where all polynomials are assumed to have real coefficients. In the version with integer coefficients, the argument showing that $x^{2} y \notin \mathcal{B}$ can be simplified: it is not hard to show that in every polynomial $f \in \mathcal{B}$, the sum of the coefficients of $x^{2} y, x^{2} z, y^{2} x, y^{2} z, z^{2} x$ and $z^{2} y$ is even. A similar fact holds for any number of variables and also implies that $N \geqslant n(n-1) / 2+1$ in terms of the previous comment.


Suppose that $a, b, c, d$ are positive real numbers satisfying $(a+c)(b+d)=a c+b d$. Find the smallest possible value of  $$ S=\frac{a}{b}+\frac{b}{c}+\frac{c}{d}+\frac{d}{a} $$  (Israel) Answer: The smallest possible value is 8.

Solution: To show that $S \geqslant 8$, apply the AM-GM inequality twice as follows:  $$ \left(\frac{a}{b}+\frac{c}{d}\right)+\left(\frac{b}{c}+\frac{d}{a}\right) \geqslant 2 \sqrt{\frac{a c}{b d}}+2 \sqrt{\frac{b d}{a c}}=\frac{2(a c+b d)}{\sqrt{a b c d}}=\frac{2(a+c)(b+d)}{\sqrt{a b c d}} \geqslant 2 \cdot \frac{2 \sqrt{a c} \cdot 2 \sqrt{b d}}{\sqrt{a b c d}}=8 . $$  The above inequalities turn into equalities when $a=c$ and $b=d$. Then the condition $(a+c)(b+d)=a c+b d$ can be rewritten as $4 a b=a^{2}+b^{2}$. So it is satisfied when $a / b=2 \pm \sqrt{3}$. Hence, $S$ attains value 8 , e.g., when $a=c=1$ and $b=d=2+\sqrt{3}$.


Solution: By homogeneity we may suppose that $a b c d=1$. Let $a b=C, b c=A$ and $c a=B$. Then $a, b, c$ can be reconstructed from $A, B$ and $C$ as $a=\sqrt{B C / A}, b=\sqrt{A C / B}$ and $c=\sqrt{A B / C}$. Moreover, the condition $(a+c)(b+d)=a c+b d$ can be written in terms of $A, B, C$ as  $$ A+\frac{1}{A}+C+\frac{1}{C}=b c+a d+a b+c d=(a+c)(b+d)=a c+b d=B+\frac{1}{B} . $$  We then need to minimize the expression  $$ \begin{aligned} S & :=\frac{a d+b c}{b d}+\frac{a b+c d}{a c}=\left(A+\frac{1}{A}\right) B+\left(C+\frac{1}{C}\right) \frac{1}{B} \\ & =\left(A+\frac{1}{A}\right)\left(B-\frac{1}{B}\right)+\left(A+\frac{1}{A}+C+\frac{1}{C}\right) \frac{1}{B} \\ & =\left(A+\frac{1}{A}\right)\left(B-\frac{1}{B}\right)+\left(B+\frac{1}{B}\right) \frac{1}{B} . \end{aligned} $$  Without loss of generality assume that $B \geqslant 1$ (otherwise, we may replace $B$ by $1 / B$ and swap $A$ and $C$, this changes neither the relation nor the function to be maximized). Therefore, we can write  $$ S \geqslant 2\left(B-\frac{1}{B}\right)+\left(B+\frac{1}{B}\right) \frac{1}{B}=2 B+\left(1-\frac{1}{B}\right)^{2}=: f(B) $$  Clearly, $f$ increases on $[1, \infty)$. Since  $$ B+\frac{1}{B}=A+\frac{1}{A}+C+\frac{1}{C} \geqslant 4, $$  we have $B \geqslant B^{\prime}$, where $B^{\prime}=2+\sqrt{3}$ is the unique root greater than 1 of the equation $B^{\prime}+1 / B^{\prime}=4$. Hence,  $$ S \geqslant f(B) \geqslant f\left(B^{\prime}\right)=2\left(B^{\prime}-\frac{1}{B^{\prime}}\right)+\left(B^{\prime}+\frac{1}{B^{\prime}}\right) \frac{1}{B^{\prime}}=2 B^{\prime}-\frac{2}{B^{\prime}}+\frac{4}{B^{\prime}}=8 $$  It remains to note that when $A=C=1$ and $B=B^{\prime}$ we have the equality $S=8$.


Solution: We present another proof of the inequality $S \geqslant 8$. We start with the estimate  $$ \left(\frac{a}{b}+\frac{c}{d}\right)+\left(\frac{b}{c}+\frac{d}{a}\right) \geqslant 2 \sqrt{\frac{a c}{b d}}+2 \sqrt{\frac{b d}{a c}} $$  Let $y=\sqrt{a c}$ and $z=\sqrt{b d}$, and assume, without loss of generality, that $a c \geqslant b d$. By the AM-GM inequality, we have  $$ y^{2}+z^{2}=a c+b d=(a+c)(b+d) \geqslant 2 \sqrt{a c} \cdot 2 \sqrt{b d}=4 y z . $$  Substituting $x=y / z$, we get $4 x \leqslant x^{2}+1$. For $x \geqslant 1$, this holds if and only if $x \geqslant 2+\sqrt{3}$. Now we have  $$ 2 \sqrt{\frac{a c}{b d}}+2 \sqrt{\frac{b d}{a c}}=2\left(x+\frac{1}{x}\right) . $$  Clearly, this is minimized by setting $x(\geqslant 1)$ as close to 1 as possible, i.e., by taking $x=2+\sqrt{3}$. Then $2(x+1 / x)=2((2+\sqrt{3})+(2-\sqrt{3}))=8$, as required.