Solution: Denote by $\omega_{1}$ and $\omega_{2}$ the circumcircles of triangles $A B C$ and $A D C$ respectively. Since $\angle A B C=\angle A D C$, we know that $\omega_{1}$ and $\omega_{2}$ are symmetric with respect to the midpoint $M$ of $A C$.  Firstly, we show that $X$ lies on $\omega_{2}$. Let $X_{1}$ be the second intersection of ray $M B$ and $\omega_{2}$ and $X^{\prime}$ be its symmetric point with respect to $M$. Then $X^{\prime}$ lies on $\omega_{1}$ and $X^{\prime} A X_{1} C$ is a parallelogram. Hence, we have  $$ \begin{aligned} \measuredangle D X_{1} B & =\measuredangle D X_{1} A+\measuredangle A X_{1} B=\measuredangle D C A+\measuredangle A X_{1} X^{\prime}=\measuredangle D C A+\measuredangle C X^{\prime} X_{1} \\ & =\measuredangle D C A+\measuredangle C A B=\measuredangle(C D, A B) . \end{aligned} $$  ![](https://cdn.mathpix.com/cropped/2024_11_18_e5e735288fb13d37ab0cg-65.jpg?height=914&width=1302&top_left_y=1191&top_left_x=449)  Also, we have  $$ \measuredangle D P B=\measuredangle P D C+\angle(C D, A B)+\measuredangle A B P=\angle(C D, A B) . $$  These yield $\measuredangle D X_{1} B=\measuredangle D P B$ and hence $X_{1}$ lies on $\omega$. It follows that $X_{1}=X$ and $X$ lies on $\omega_{2}$. Similarly, $Y$ lies on $\omega_{1}$.  Next, we prove that $Q$ lies on $\omega$. Suppose the perpendicular bisector of $A C$ meet $\omega_{1}$ at $B^{\prime}$ and $M_{1}$ and meet $\omega_{2}$ at $D^{\prime}$ and $M_{2}$, so that $B, M_{1}$ and $D^{\prime}$ lie on the same side of $A C$. Note that $B^{\prime}$ lies on the angle bisector of $\angle A B C$ and similarly $D^{\prime}$ lies on $D P$.  If we denote the area of $W_{1} W_{2} W_{3}$ by $\left[W_{1} W_{2} W_{3}\right]$, then  $$ \frac{B A \cdot X^{\prime} A}{B C \cdot X^{\prime} C}=\frac{\frac{1}{2} B A \cdot X^{\prime} A \sin \angle B A X^{\prime}}{\frac{1}{2} B C \cdot X^{\prime} C \sin \angle B C X^{\prime}}=\frac{\left[B A X^{\prime}\right]}{\left[B C X^{\prime}\right]}=\frac{M A}{M C}=1 $$  As $B E$ is the angle bisector of $\angle A B C$, we have  $$ \frac{E A}{E C}=\frac{B A}{B C}=\frac{X^{\prime} C}{X^{\prime} A}=\frac{X A}{X C} $$  Therefore, $X E$ is the angle bisector of $\angle A X C$, so that $M_{2}$ lies on the line joining $X, E, Q$. Analogously, $M_{1}, F, Q, Y$ are collinear. Thus,  $$ \begin{aligned} \measuredangle X Q Y & =\measuredangle M_{2} Q M_{1}=\measuredangle Q M_{2} M_{1}+\measuredangle M_{2} M_{1} Q=\measuredangle X M_{2} D^{\prime}+\measuredangle B^{\prime} M_{1} Y \\ & =\measuredangle X D D^{\prime}+\measuredangle B^{\prime} B Y=\measuredangle X D P+\measuredangle P B Y=\measuredangle X B P+\measuredangle P B Y=\measuredangle X B Y, \end{aligned} $$  which implies $Q$ lies on $\omega$. Finally, as $M_{1}$ and $M_{2}$ are symmetric with respect to $M$, the quadrilateral $X^{\prime} M_{2} X M_{1}$ is a parallelogram. Consequently,  $$ \measuredangle X Q P=\measuredangle X B P=\measuredangle X^{\prime} B B^{\prime}=\measuredangle X^{\prime} M_{1} B^{\prime}=\measuredangle X M_{2} M_{1} . $$  This shows $Q P / / M_{2} M_{1}$. As $M_{2} M_{1} \perp A C$, we get $Q P \perp A C$.


Let $A B C D$ be a convex quadrilateral with $\angle A B C=\angle A D C<90^{\circ}$. The internal angle bisectors of $\angle A B C$ and $\angle A D C$ meet $A C$ at $E$ and $F$ respectively, and meet each other at point $P$. Let $M$ be the midpoint of $A C$ and let $\omega$ be the circumcircle of triangle $B P D$. Segments $B M$ and $D M$ intersect $\omega$ again at $X$ and $Y$ respectively. Denote by $Q$ the intersection point of lines $X E$ and $Y F$. Prove that $P Q \perp A C$.

Solution: We first state two results which will be needed in our proof.  - Claim 1. In $\triangle X^{\prime} Y^{\prime} Z^{\prime}$ with $X^{\prime} Y^{\prime} \neq X^{\prime} Z^{\prime}$, let $N^{\prime}$ be the midpoint of $Y^{\prime} Z^{\prime}$ and $W^{\prime}$ be the foot of internal angle bisector from $X^{\prime}$. Then $\tan ^{2} \measuredangle W^{\prime} X^{\prime} Z^{\prime}=\tan \measuredangle N^{\prime} X^{\prime} W^{\prime} \tan \measuredangle Z^{\prime} W^{\prime} X^{\prime}$.  Proof. ![](https://cdn.mathpix.com/cropped/2024_11_18_e5e735288fb13d37ab0cg-66.jpg?height=416&width=562&top_left_y=1684&top_left_x=733)  Without loss of generality, assume $X^{\prime} Y^{\prime}>X^{\prime} Z^{\prime}$. Then $W^{\prime}$ lies between $N^{\prime}$ and $Z^{\prime}$. The signs of both sides agree so it suffices to establish the relation for ordinary angles. Let $\angle W^{\prime} X^{\prime} Z^{\prime}=\alpha, \angle N^{\prime} X^{\prime} W^{\prime}=\beta$ and $\angle Z^{\prime} W^{\prime} X^{\prime}=\gamma$. We have  $$ \frac{\sin (\gamma-\alpha)}{\sin (\alpha-\beta)}=\frac{N^{\prime} X^{\prime}}{N^{\prime} Y^{\prime}}=\frac{N^{\prime} X^{\prime}}{N^{\prime} Z^{\prime}}=\frac{\sin (\gamma+\alpha)}{\sin (\alpha+\beta)} $$  This implies  $$ \frac{\tan \gamma-\tan \alpha}{\tan \gamma+\tan \alpha}=\frac{\sin \gamma \cos \alpha-\cos \gamma \sin \alpha}{\sin \gamma \cos \alpha+\cos \gamma \sin \alpha}=\frac{\sin \alpha \cos \beta-\cos \alpha \sin \beta}{\sin \alpha \cos \beta+\cos \alpha \sin \beta}=\frac{\tan \alpha-\tan \beta}{\tan \alpha+\tan \beta} $$  Expanding and simplifying, we get the desired result $\tan ^{2} \alpha=\tan \beta \tan \gamma$.  - Claim 2. Let $A^{\prime} B^{\prime} C^{\prime} D^{\prime}$ be a quadrilateral inscribed in circle $\Gamma$. Let diagonals $A^{\prime} C^{\prime}$ and $B^{\prime} D^{\prime}$ meet at $E^{\prime}$, and $F^{\prime}$ be the intersection of lines $A^{\prime} B^{\prime}$ and $C^{\prime} D^{\prime}$. Let $M^{\prime}$ be the midpoint of $E^{\prime} F^{\prime}$. Then the power of $M^{\prime}$ with respect to $\Gamma$ is equal to $\left(M^{\prime} E^{\prime}\right)^{2}$.  Proof. ![](https://cdn.mathpix.com/cropped/2024_11_18_e5e735288fb13d37ab0cg-67.jpg?height=624&width=904&top_left_y=1000&top_left_x=678)  Let $O^{\prime}$ be the centre of $\Gamma$ and let $\Gamma^{\prime}$ be the circle with centre $M^{\prime}$ passing through $E^{\prime}$. Let $F_{1}$ be the inversion image of $F^{\prime}$ with respect to $\Gamma$. It is well-known that $E^{\prime}$ lies on the polar of $F^{\prime}$ with respect to $\Gamma$. This shows $E^{\prime} F_{1} \perp O^{\prime} F^{\prime}$ and hence $F_{1}$ lies on $\Gamma^{\prime}$. It follows that the inversion image of $\Gamma^{\prime}$ with respect to $\Gamma$ is $\Gamma^{\prime}$ itself. This shows $\Gamma^{\prime}$ is orthogonal to $\Gamma$, and thus the power of $M^{\prime}$ with respect to $\Gamma$ is the square of radius of $\Gamma^{\prime}$, which is $\left(M^{\prime} E^{\prime}\right)^{2}$.  We return to the main problem. Let $Z$ be the intersection of lines $A D$ and $B C$, and $W$ be the intersection of lines $A B$ and $C D$. Since $\measuredangle P D Z=\measuredangle P B C=\measuredangle P B Z$, point $Z$ lies on $\omega$. Similarly, $W$ lies on $\omega$. Applying Claim 2 to the cyclic quadrilateral $Z B D W$, we know that the power of $M$ with respect to $\omega$ is $M A^{2}$. Hence, $M X \cdot M B=M A^{2}$.  Suppose the line through $B$ perpendicular to $B E$ meets line $A C$ at $T$. Then $B E$ and $B T$ are the angle bisectors of $\angle C B A$. This shows $(T, E ; A, C)$ is harmonic. Thus, we have $M E \cdot M T=M A^{2}=M X \cdot M B$. It follows that $E, T, B, X$ are concyclic. ![](https://cdn.mathpix.com/cropped/2024_11_18_e5e735288fb13d37ab0cg-68.jpg?height=808&width=1424&top_left_y=252&top_left_x=285)  The result is trivial for the special case $A D=C D$ since $P, Q$ lie on the perpendicular bisector of $A C$ in that case. Similarly, the case $A B=C B$ is trivial. It remains to consider the general cases where we can apply Claim 1 in the latter part of the proof.  Let the projections from $P$ and $Q$ to $A C$ be $P^{\prime}$ and $Q^{\prime}$ respectively. Then $P Q \perp A C$ if and only if $P^{\prime}=Q^{\prime}$ if and only if $\frac{E P^{\prime}}{F P^{\prime}}=\frac{E Q^{\prime}}{F Q^{\prime}}$ in terms of directed lengths. Note that  $$ \frac{E P^{\prime}}{F P^{\prime}}=\frac{\tan \measuredangle E F P}{\tan \measuredangle F E P}=\frac{\tan \measuredangle A F D}{\tan \measuredangle A E B} $$  Next, we have $\frac{E Q^{\prime}}{F Q^{\prime}}=\frac{\tan \measuredangle E F Q}{\tan \measuredangle F E Q}$ where $\measuredangle F E Q=\measuredangle T E X=\measuredangle T B X=\frac{\pi}{2}+\measuredangle E B M$ and by symmetry $\measuredangle E F Q=\frac{\pi}{2}+\measuredangle F D M$. Combining all these, it suffices to show  $$ \frac{\tan \measuredangle A F D}{\tan \measuredangle A E B}=\frac{\tan \measuredangle M B E}{\tan \measuredangle M D F} $$  We now apply Claim 1 twice to get  $$ \tan \measuredangle A F D \tan \measuredangle M D F=\tan ^{2} \measuredangle F D C=\tan ^{2} \measuredangle E B A=\tan \measuredangle M B E \tan \measuredangle A E B . $$  The result then follows.


Let $I$ be the incentre of a non-equilateral triangle $A B C, I_{A}$ be the $A$-excentre, $I_{A}^{\prime}$ be the reflection of $I_{A}$ in $B C$, and $l_{A}$ be the reflection of line $A I_{A}^{\prime}$ in $A I$. Define points $I_{B}, I_{B}^{\prime}$ and line $l_{B}$ analogously. Let $P$ be the intersection point of $l_{A}$ and $l_{B}$. (a) Prove that $P$ lies on line $O I$ where $O$ is the circumcentre of triangle $A B C$. (b) Let one of the tangents from $P$ to the incircle of triangle $A B C$ meet the circumcircle at points $X$ and $Y$. Show that $\angle X I Y=120^{\circ}$.

Solution: (a) Let $A^{\prime}$ be the reflection of $A$ in $B C$ and let $M$ be the second intersection of line $A I$ and the circumcircle $\Gamma$ of triangle $A B C$. As triangles $A B A^{\prime}$ and $A O C$ are isosceles with $\angle A B A^{\prime}=2 \angle A B C=\angle A O C$, they are similar to each other. Also, triangles $A B I_{A}$ and $A I C$ are similar. Therefore we have  $$ \frac{A A^{\prime}}{A I_{A}}=\frac{A A^{\prime}}{A B} \cdot \frac{A B}{A I_{A}}=\frac{A C}{A O} \cdot \frac{A I}{A C}=\frac{A I}{A O} $$  Together with $\angle A^{\prime} A I_{A}=\angle I A O$, we find that triangles $A A^{\prime} I_{A}$ and $A I O$ are similar. ![](https://cdn.mathpix.com/cropped/2024_11_18_e5e735288fb13d37ab0cg-69.jpg?height=890&width=1025&top_left_y=1192&top_left_x=653)  Denote by $P^{\prime}$ the intersection of line $A P$ and line $O I$. Using directed angles, we have  $$ \begin{aligned} \measuredangle M A P^{\prime} & =\measuredangle I_{A}^{\prime} A I_{A}=\measuredangle I_{A}^{\prime} A A^{\prime}-\measuredangle I_{A} A A^{\prime}=\measuredangle A A^{\prime} I_{A}-\measuredangle(A M, O M) \\ & =\measuredangle A I O-\measuredangle A M O=\measuredangle M O P^{\prime} . \end{aligned} $$  This shows $M, O, A, P^{\prime}$ are concyclic.  Denote by $R$ and $r$ the circumradius and inradius of triangle $A B C$. Then  $$ I P^{\prime}=\frac{I A \cdot I M}{I O}=\frac{I O^{2}-R^{2}}{I O} $$  is independent of $A$. Hence, $B P$ also meets line $O I$ at the same point $P^{\prime}$ so that $P^{\prime}=P$, and $P$ lies on $O I$. (b) By Poncelet's Porism, the other tangents to the incircle of triangle $A B C$ from $X$ and $Y$ meet at a point $Z$ on $\Gamma$. Let $T$ be the touching point of the incircle to $X Y$, and let $D$ be the midpoint of $X Y$. We have  $$ \begin{aligned} O D & =I T \cdot \frac{O P}{I P}=r\left(1+\frac{O I}{I P}\right)=r\left(1+\frac{O I^{2}}{O I \cdot I P}\right)=r\left(1+\frac{R^{2}-2 R r}{R^{2}-I O^{2}}\right) \\ & =r\left(1+\frac{R^{2}-2 R r}{2 R r}\right)=\frac{R}{2}=\frac{O X}{2} \end{aligned} $$  This shows $\angle X Z Y=60^{\circ}$ and hence $\angle X I Y=120^{\circ}$.


Solution: (a) Note that triangles $A I_{B} C$ and $I_{A} B C$ are similar since their corresponding interior angles are equal. Therefore, the four triangles $A I_{B}^{\prime} C, A I_{B} C, I_{A} B C$ and $I_{A}^{\prime} B C$ are all similar. From $\triangle A I_{B}^{\prime} C \sim \triangle I_{A}^{\prime} B C$, we get $\triangle A I_{A}^{\prime} C \sim \triangle I_{B}^{\prime} B C$. From $\measuredangle A B P=\measuredangle I_{B}^{\prime} B C=\measuredangle A I_{A}^{\prime} C$ and $\measuredangle B A P=\measuredangle I_{A}^{\prime} A C$, the triangles $A B P$ and $A I_{A}^{\prime} C$ are directly similar. ![](https://cdn.mathpix.com/cropped/2024_11_18_e5e735288fb13d37ab0cg-70.jpg?height=820&width=1137&top_left_y=1455&top_left_x=494)  Consider the inversion with centre $A$ and radius $\sqrt{A B \cdot A C}$ followed by the reflection in $A I$. Then $B$ and $C$ are mapped to each other, and $I$ and $I_{A}$ are mapped to each other.  From the similar triangles obtained, we have $A P \cdot A I_{A}^{\prime}=A B \cdot A C$ so that $P$ is mapped to $I_{A}^{\prime}$ under the transformation. In addition, line $A O$ is mapped to the altitude from $A$, and hence $O$ is mapped to the reflection of $A$ in $B C$, which we call point $A^{\prime}$. Note that $A A^{\prime} I_{A} I_{A}^{\prime}$ is an isosceles trapezoid, which shows it is inscribed in a circle. The preimage of this circle is a straight line, meaning that $O, I, P$ are collinear. (b) Denote by $R$ and $r$ the circumradius and inradius of triangle $A B C$. Note that by the above transformation, we have $\triangle A P O \sim \triangle A A^{\prime} I_{A}^{\prime}$ and $\triangle A A^{\prime} I_{A} \sim \triangle A I O$. Therefore, we find that  $$ P O=A^{\prime} I_{A}^{\prime} \cdot \frac{A O}{A I_{A}^{\prime}}=A I_{A} \cdot \frac{A O}{A^{\prime} I_{A}}=\frac{A I_{A}}{A^{\prime} I_{A}} \cdot A O=\frac{A O}{I O} \cdot A O $$  This shows $P O \cdot I O=R^{2}$, and it follows that $P$ and $I$ are mapped to each other under the inversion with respect to the circumcircle $\Gamma$ of triangle $A B C$. Then $P X \cdot P Y$, which is the power of $P$ with respect to $\Gamma$, equals $P I \cdot P O$. This yields $X, I, O, Y$ are concyclic.  Let $T$ be the touching point of the incircle to $X Y$, and let $D$ be the midpoint of $X Y$. Then  $$ O D=I T \cdot \frac{P O}{P I}=r \cdot \frac{P O}{P O-I O}=r \cdot \frac{R^{2}}{R^{2}-I O^{2}}=r \cdot \frac{R^{2}}{2 R r}=\frac{R}{2} $$  This shows $\angle D O X=60^{\circ}$ and hence $\angle X I Y=\angle X O Y=120^{\circ}$. Comment. A simplification of this problem is to ask part (a) only. Note that the question in part (b) implicitly requires $P$ to lie on $O I$, or otherwise the angle is not uniquely determined as we can find another tangent from $P$ to the incircle.


Let $A_{1}, B_{1}$ and $C_{1}$ be points on sides $B C, C A$ and $A B$ of an acute triangle $A B C$ respectively, such that $A A_{1}, B B_{1}$ and $C C_{1}$ are the internal angle bisectors of triangle $A B C$. Let $I$ be the incentre of triangle $A B C$, and $H$ be the orthocentre of triangle $A_{1} B_{1} C_{1}$. Show that  $$ A H+B H+C H \geqslant A I+B I+C I $$

Solution: Without loss of generality, assume $\alpha=\angle B A C \leqslant \beta=\angle C B A \leqslant \gamma=\angle A C B$. Denote by $a, b, c$ the lengths of $B C, C A, A B$ respectively. We first show that triangle $A_{1} B_{1} C_{1}$ is acute.  Choose points $D$ and $E$ on side $B C$ such that $B_{1} D / / A B$ and $B_{1} E$ is the internal angle bisector of $\angle B B_{1} C$. As $\angle B_{1} D B=180^{\circ}-\beta$ is obtuse, we have $B B_{1}>B_{1} D$. Thus,  $$ \frac{B E}{E C}=\frac{B B_{1}}{B_{1} C}>\frac{D B_{1}}{B_{1} C}=\frac{B A}{A C}=\frac{B A_{1}}{A_{1} C} $$  Therefore, $B E>B A_{1}$ and $\frac{1}{2} \angle B B_{1} C=\angle B B_{1} E>\angle B B_{1} A_{1}$. Similarly, $\frac{1}{2} \angle B B_{1} A>\angle B B_{1} C_{1}$. It follows that  $$ \angle A_{1} B_{1} C_{1}=\angle B B_{1} A_{1}+\angle B B_{1} C_{1}<\frac{1}{2}\left(\angle B B_{1} C+\angle B B_{1} A\right)=90^{\circ} $$  is acute. By symmetry, triangle $A_{1} B_{1} C_{1}$ is acute. Let $B B_{1}$ meet $A_{1} C_{1}$ at $F$. From $\alpha \leqslant \gamma$, we get $a \leqslant c$, which implies  $$ B A_{1}=\frac{c a}{b+c} \leqslant \frac{a c}{a+b}=B C_{1} $$  and hence $\angle B C_{1} A_{1} \leqslant \angle B A_{1} C_{1}$. As $B F$ is the internal angle bisector of $\angle A_{1} B C_{1}$, this shows $\angle B_{1} F C_{1}=\angle B F A_{1} \leqslant 90^{\circ}$. Hence, $H$ lies on the same side of $B B_{1}$ as $C_{1}$. This shows $H$ lies inside triangle $B B_{1} C_{1}$. Similarly, from $\alpha \leqslant \beta$ and $\beta \leqslant \gamma$, we know that $H$ lies inside triangles $C C_{1} B_{1}$ and $A A_{1} C_{1}$. ![](https://cdn.mathpix.com/cropped/2024_11_18_e5e735288fb13d37ab0cg-72.jpg?height=779&width=1010&top_left_y=1660&top_left_x=506)  As $\alpha \leqslant \beta \leqslant \gamma$, we have $\alpha \leqslant 60^{\circ} \leqslant \gamma$. Then $\angle B I C \leqslant 120^{\circ} \leqslant \angle A I B$. Firstly, suppose $\angle A I C \geqslant 120^{\circ}$.  Rotate points $B, I, H$ through $60^{\circ}$ about $A$ to $B^{\prime}, I^{\prime}, H^{\prime}$ so that $B^{\prime}$ and $C$ lie on different sides of $A B$. Since triangle $A I^{\prime} I$ is equilateral, we have  $$ A I+B I+C I=I^{\prime} I+B^{\prime} I^{\prime}+I C=B^{\prime} I^{\prime}+I^{\prime} I+I C . $$  Similarly,  $$ A H+B H+C H=H^{\prime} H+B^{\prime} H^{\prime}+H C=B^{\prime} H^{\prime}+H^{\prime} H+H C $$  As $\angle A I I^{\prime}=\angle A I^{\prime} I=60^{\circ}, \angle A I^{\prime} B^{\prime}=\angle A I B \geqslant 120^{\circ}$ and $\angle A I C \geqslant 120^{\circ}$, the quadrilateral $B^{\prime} I^{\prime} I C$ is convex and lies on the same side of $B^{\prime} C$ as $A$.  Next, since $H$ lies inside triangle $A C C_{1}, H$ lies outside $B^{\prime} I^{\prime} I C$. Also, $H$ lying inside triangle $A B I$ implies $H^{\prime}$ lies inside triangle $A B^{\prime} I^{\prime}$. This shows $H^{\prime}$ lies outside $B^{\prime} I^{\prime} I C$ and hence the convex quadrilateral $B^{\prime} I^{\prime} I C$ is contained inside the quadrilateral $B^{\prime} H^{\prime} H C$. It follows that the perimeter of $B^{\prime} I^{\prime} I C$ cannot exceed the perimeter of $B^{\prime} H^{\prime} H C$. From (1) and (2), we conclude that  $$ A H+B H+C H \geqslant A I+B I+C I $$  For the case $\angle A I C<120^{\circ}$, we can rotate $B, I, H$ through $60^{\circ}$ about $C$ to $B^{\prime}, I^{\prime}, H^{\prime}$ so that $B^{\prime}$ and $A$ lie on different sides of $B C$. The proof is analogous to the previous case and we still get the desired inequality.


For any positive integer $k$, denote the sum of digits of $k$ in its decimal representation by $S(k)$. Find all polynomials $P(x)$ with integer coefficients such that for any positive integer $n \geqslant 2016$, the integer $P(n)$ is positive and  $$ S(P(n))=P(S(n)) $$

Solution: We consider three cases according to the degree of $P$.  - Case 1. $P(x)$ is a constant polynomial.  Let $P(x)=c$ where $c$ is an integer constant. Then (1) becomes $S(c)=c$. This holds if and only if $1 \leqslant c \leqslant 9$.  - Case 2. $\operatorname{deg} P=1$.  We have the following observation. For any positive integers $m, n$, we have  $$ S(m+n) \leqslant S(m)+S(n) $$  and equality holds if and only if there is no carry in the addition $m+n$. Let $P(x)=a x+b$ for some integers $a, b$ where $a \neq 0$. As $P(n)$ is positive for large $n$, we must have $a \geqslant 1$. The condition (1) becomes $S(a n+b)=a S(n)+b$ for all $n \geqslant 2016$. Setting $n=2025$ and $n=2020$ respectively, we get  $$ S(2025 a+b)-S(2020 a+b)=(a S(2025)+b)-(a S(2020)+b)=5 a $$  On the other hand, (2) implies  $$ S(2025 a+b)=S((2020 a+b)+5 a) \leqslant S(2020 a+b)+S(5 a) $$  These give $5 a \leqslant S(5 a)$. As $a \geqslant 1$, this holds only when $a=1$, in which case (1) reduces to $S(n+b)=S(n)+b$ for all $n \geqslant 2016$. Then we find that  $$ S(n+1+b)-S(n+b)=(S(n+1)+b)-(S(n)+b)=S(n+1)-S(n) $$  If $b>0$, we choose $n$ such that $n+1+b=10^{k}$ for some sufficiently large $k$. Note that all the digits of $n+b$ are 9 's, so that the left-hand side of (3) equals $1-9 k$. As $n$ is a positive integer less than $10^{k}-1$, we have $S(n)<9 k$. Therefore, the right-hand side of (3) is at least $1-(9 k-1)=2-9 k$, which is a contradiction.  The case $b<0$ can be handled similarly by considering $n+1$ to be a large power of 10 . Therefore, we conclude that $P(x)=x$, in which case (1) is trivially satisfied.  - Case 3. $\operatorname{deg} P \geqslant 2$.  Suppose the leading term of $P$ is $a_{d} n^{d}$ where $a_{d} \neq 0$. Clearly, we have $a_{d}>0$. Consider $n=10^{k}-1$ in (1). We get $S(P(n))=P(9 k)$. Note that $P(n)$ grows asymptotically as fast as $n^{d}$, so $S(P(n))$ grows asymptotically as no faster than a constant multiple of $k$. On the other hand, $P(9 k)$ grows asymptotically as fast as $k^{d}$. This shows the two sides of the last equation cannot be equal for sufficiently large $k$ since $d \geqslant 2$.  Therefore, we conclude that $P(x)=c$ where $1 \leqslant c \leqslant 9$ is an integer, or $P(x)=x$.


Solution: Let $P(x)=a_{d} x^{d}+a_{d-1} x^{d-1}+\cdots+a_{0}$. Clearly $a_{d}>0$. There exists an integer $m \geqslant 1$ such that $\left|a_{i}\right|<10^{m}$ for all $0 \leqslant i \leqslant d$. Consider $n=9 \times 10^{k}$ for a sufficiently large integer $k$ in (1). If there exists an index $0 \leqslant i \leqslant d-1$ such that $a_{i}<0$, then all digits of $P(n)$ in positions from $10^{i k+m+1}$ to $10^{(i+1) k-1}$ are all 9 's. Hence, we have $S(P(n)) \geqslant 9(k-m-1)$. On the other hand, $P(S(n))=P(9)$ is a fixed constant. Therefore, (1) cannot hold for large $k$. This shows $a_{i} \geqslant 0$ for all $0 \leqslant i \leqslant d-1$.  Hence, $P(n)$ is an integer formed by the nonnegative integers $a_{d} \times 9^{d}, a_{d-1} \times 9^{d-1}, \ldots, a_{0}$ by inserting some zeros in between. This yields  $$ S(P(n))=S\left(a_{d} \times 9^{d}\right)+S\left(a_{d-1} \times 9^{d-1}\right)+\cdots+S\left(a_{0}\right) . $$  Combining with (1), we have  $$ S\left(a_{d} \times 9^{d}\right)+S\left(a_{d-1} \times 9^{d-1}\right)+\cdots+S\left(a_{0}\right)=P(9)=a_{d} \times 9^{d}+a_{d-1} \times 9^{d-1}+\cdots+a_{0} $$  As $S(m) \leqslant m$ for any positive integer $m$, with equality when $1 \leqslant m \leqslant 9$, this forces each $a_{i} \times 9^{i}$ to be a positive integer between 1 and 9 . In particular, this shows $a_{i}=0$ for $i \geqslant 2$ and hence $d \leqslant 1$. Also, we have $a_{1} \leqslant 1$ and $a_{0} \leqslant 9$. If $a_{1}=1$ and $1 \leqslant a_{0} \leqslant 9$, we take $n=10^{k}+\left(10-a_{0}\right)$ for sufficiently large $k$ in (1). This yields a contradiction since  $$ S(P(n))=S\left(10^{k}+10\right)=2 \neq 11=P\left(11-a_{0}\right)=P(S(n)) $$  The zero polynomial is also rejected since $P(n)$ is positive for large $n$. The remaining candidates are $P(x)=x$ or $P(x)=a_{0}$ where $1 \leqslant a_{0} \leqslant 9$, all of which satisfy (1), and hence are the only solutions.


Let $\tau(n)$ be the number of positive divisors of $n$. Let $\tau_{1}(n)$ be the number of positive divisors of $n$ which have remainders 1 when divided by 3 . Find all possible integral values of the fraction $\frac{\tau(10 n)}{\tau_{1}(10 n)}$.

Solution: In this solution, we always use $p_{i}$ to denote primes congruent to $1 \bmod 3$, and use $q_{j}$ to denote primes congruent to $2 \bmod 3$. When we express a positive integer $m$ using its prime factorization, we also include the special case $m=1$ by allowing the exponents to be zeros. We first compute $\tau_{1}(m)$ for a positive integer $m$.  - Claim. Let $m=3^{x} p_{1}^{a_{1}} p_{2}^{a_{2}} \cdots p_{s}^{a_{s}} q_{1}^{b_{1}} q_{2}^{b_{2}} \cdots q_{t}^{b_{t}}$ be the prime factorization of $m$. Then  $$ \tau_{1}(m)=\prod_{i=1}^{s}\left(a_{i}+1\right)\left\lceil\frac{1}{2} \prod_{j=1}^{t}\left(b_{j}+1\right)\right\rceil . $$  Proof. To choose a divisor of $m$ congruent to $1 \bmod 3$, it cannot have the prime divisor 3, while there is no restriction on choosing prime factors congruent to $1 \bmod 3$. Also, we have to choose an even number of prime factors (counted with multiplicity) congruent to $2 \bmod 3$.  If $\prod_{j=1}^{t}\left(b_{j}+1\right)$ is even, then we may assume without loss of generality $b_{1}+1$ is even. We can choose the prime factors $q_{2}, q_{3}, \ldots, q_{t}$ freely in $\prod_{j=2}^{t}\left(b_{j}+1\right)$ ways. Then the parity of the number of $q_{1}$ is uniquely determined, and hence there are $\frac{1}{2}\left(b_{1}+1\right)$ ways to choose the exponent of $q_{1}$. Hence (1) is verified in this case.  If $\prod_{j=1}^{t}\left(b_{j}+1\right)$ is odd, we use induction on $t$ to count the number of choices. When $t=1$, there are $\left\lceil\frac{b_{1}+1}{2}\right\rceil$ choices for which the exponent is even and $\left\lfloor\frac{b_{1}+1}{2}\right\rfloor$ choices for which the exponent is odd. For the inductive step, we find that there are  $$ \left\lceil\frac{1}{2} \prod_{j=1}^{t-1}\left(b_{j}+1\right)\right\rceil \cdot\left\lceil\frac{b_{t}+1}{2}\right\rceil+\left\lfloor\frac{1}{2} \prod_{j=1}^{t-1}\left(b_{j}+1\right)\right\rfloor \cdot\left\lfloor\frac{b_{t}+1}{2}\right\rfloor=\left\lceil\frac{1}{2} \prod_{j=1}^{t}\left(b_{j}+1\right)\right\rceil $$  choices with an even number of prime factors and hence $\left\lfloor\frac{1}{2} \prod_{j=1}^{t}\left(b_{j}+1\right)\right\rfloor$ choices with an odd number of prime factors. Hence (1) is also true in this case.  Let $n=3^{x} 2^{y} 5^{z} p_{1}^{a_{1}} p_{2}^{a_{2}} \cdots p_{s}^{a_{s}} q_{1}^{b_{1}} q_{2}^{b_{2}} \cdots q_{t}^{b_{t}}$. Using the well-known formula for computing the divisor function, we get  $$ \tau(10 n)=(x+1)(y+2)(z+2) \prod_{i=1}^{s}\left(a_{i}+1\right) \prod_{j=1}^{t}\left(b_{j}+1\right) $$  By the Claim, we have  $$ \tau_{1}(10 n)=\prod_{i=1}^{s}\left(a_{i}+1\right)\left\lceil\frac{1}{2}(y+2)(z+2) \prod_{j=1}^{t}\left(b_{j}+1\right)\right\rceil $$  If $c=(y+2)(z+2) \prod_{j=1}^{t}\left(b_{j}+1\right)$ is even, then (2) and (3) imply  $$ \frac{\tau(10 n)}{\tau_{1}(10 n)}=2(x+1) $$  In this case $\frac{\tau(10 n)}{\tau_{1}(10 n)}$ can be any even positive integer as $x$ runs through all nonnegative integers. If $c$ is odd, which means $y, z$ are odd and each $b_{j}$ is even, then (2) and (3) imply  $$ \frac{\tau(10 n)}{\tau_{1}(10 n)}=\frac{2(x+1) c}{c+1} $$  For this to be an integer, we need $c+1$ divides $2(x+1)$ since $c$ and $c+1$ are relatively prime. Let $2(x+1)=k(c+1)$. Then (4) reduces to  $$ \frac{\tau(10 n)}{\tau_{1}(10 n)}=k c=k(y+2)(z+2) \prod_{j=1}^{t}\left(b_{j}+1\right) $$  Noting that $y, z$ are odd, the integers $y+2$ and $z+2$ are at least 3 . This shows the integer in this case must be composite. On the other hand, for any odd composite number $a b$ with $a, b \geqslant 3$, we may simply take $n=3^{\frac{a b-1}{2}} \cdot 2^{a-2} \cdot 5^{b-2}$ so that $\frac{\tau(10 n)}{\tau_{1}(10 n)}=a b$ from (5).  We conclude that the fraction can be any even integer or any odd composite number. Equivalently, it can be 2 or any composite number.


Define $P(n)=n^{2}+n+1$. For any positive integers $a$ and $b$, the set  $$ \{P(a), P(a+1), P(a+2), \ldots, P(a+b)\} $$  is said to be fragrant if none of its elements is relatively prime to the product of the other elements. Determine the smallest size of a fragrant set.

Solution: We have the following observations. (i) $(P(n), P(n+1))=1$ for any $n$.  We have $(P(n), P(n+1))=\left(n^{2}+n+1, n^{2}+3 n+3\right)=\left(n^{2}+n+1,2 n+2\right)$. Noting that $n^{2}+n+1$ is odd and $\left(n^{2}+n+1, n+1\right)=(1, n+1)=1$, the claim follows. (ii) $(P(n), P(n+2))=1$ for $n \not \equiv 2(\bmod 7)$ and $(P(n), P(n+2))=7$ for $n \equiv 2(\bmod 7)$.  From $(2 n+7) P(n)-(2 n-1) P(n+2)=14$ and the fact that $P(n)$ is odd, $(P(n), P(n+2))$ must be a divisor of 7 . The claim follows by checking $n \equiv 0,1, \ldots, 6(\bmod 7)$ directly. (iii) $(P(n), P(n+3))=1$ for $n \not \equiv 1(\bmod 3)$ and $3 \mid(P(n), P(n+3))$ for $n \equiv 1(\bmod 3)$.  From $(n+5) P(n)-(n-1) P(n+3)=18$ and the fact that $P(n)$ is odd, $(P(n), P(n+3))$ must be a divisor of 9 . The claim follows by checking $n \equiv 0,1,2(\bmod 3)$ directly.  Suppose there exists a fragrant set with at most 5 elements. We may assume it contains exactly 5 elements $P(a), P(a+1), \ldots, P(a+4)$ since the following argument also works with fewer elements. Consider $P(a+2)$. From (i), it is relatively prime to $P(a+1)$ and $P(a+3)$. Without loss of generality, assume $(P(a), P(a+2))>1$. From (ii), we have $a \equiv 2(\bmod 7)$. The same observation implies $(P(a+1), P(a+3))=1$. In order that the set is fragrant, $(P(a), P(a+3))$ and $(P(a+1), P(a+4))$ must both be greater than 1. From (iii), this holds only when both $a$ and $a+1$ are congruent to $1 \bmod 3$, which is a contradiction.  It now suffices to construct a fragrant set of size 6. By the Chinese Remainder Theorem, we can take a positive integer $a$ such that  $$ a \equiv 7 \quad(\bmod 19), \quad a+1 \equiv 2 \quad(\bmod 7), \quad a+2 \equiv 1 \quad(\bmod 3) $$  For example, we may take $a=197$. From (ii), both $P(a+1)$ and $P(a+3)$ are divisible by 7. From (iii), both $P(a+2)$ and $P(a+5)$ are divisible by 3 . One also checks from $19 \mid P(7)=57$ and $19 \mid P(11)=133$ that $P(a)$ and $P(a+4)$ are divisible by 19. Therefore, the set $\{P(a), P(a+1), \ldots, P(a+5)\}$ is fragrant.  Therefore, the smallest size of a fragrant set is 6 . Comment. "Fragrant Harbour" is the English translation of "Hong Kong". A stronger version of this problem is to show that there exists a fragrant set of size $k$ for any $k \geqslant 6$. We present a proof here.  For each even positive integer $m$ which is not divisible by 3 , since $m^{2}+3 \equiv 3(\bmod 4)$, we can find a prime $p_{m} \equiv 3(\bmod 4)$ such that $p_{m} \mid m^{2}+3$. Clearly, $p_{m}>3$.  If $b=2 t \geqslant 6$, we choose $a$ such that $3 \mid 2(a+t)+1$ and $p_{m} \mid 2(a+t)+1$ for each $1 \leqslant m \leqslant b$ with $m \equiv 2,4(\bmod 6)$. For $0 \leqslant r \leqslant t$ and $3 \mid r$, we have $a+t \pm r \equiv 1(\bmod 3)$ so that $3 \mid P(a+t \pm r)$. For $0 \leqslant r \leqslant t$ and $(r, 3)=1$, we have  $$ 4 P(a+t \pm r) \equiv(-1 \pm 2 r)^{2}+2(-1 \pm 2 r)+4=4 r^{2}+3 \equiv 0 \quad\left(\bmod p_{2 r}\right) . $$  Hence, $\{P(a), P(a+1), \ldots, P(a+b)\}$ is fragrant. If $b=2 t+1 \geqslant 7$ (the case $b=5$ has been done in the original problem), we choose $a$ such that $3 \mid 2(a+t)+1$ and $p_{m} \mid 2(a+t)+1$ for $1 \leqslant m \leqslant b$ with $m \equiv 2,4(\bmod 6)$, and that $a+b \equiv 9(\bmod 13)$. Note that $a$ exists by the Chinese Remainder Theorem since $p_{m} \neq 13$ for all $m$. The even case shows that $\{P(a), P(a+1), \ldots, P(a+b-1)\}$ is fragrant. Also, one checks from $13 \mid P(9)=91$ and $13 \mid P(3)=13$ that $P(a+b)$ and $P(a+b-6)$ are divisible by 13. The proof is thus complete.


Let $n, m, k$ and $l$ be positive integers with $n \neq 1$ such that $n^{k}+m n^{l}+1$ divides $n^{k+l}-1$. Prove that  - $m=1$ and $l=2 k$; or - $l \mid k$ and $m=\frac{n^{k-l}-1}{n^{l}-1}$.

Solution: It is given that  $$ n^{k}+m n^{l}+1 \mid n^{k+l}-1 $$  This implies  $$ n^{k}+m n^{l}+1 \mid\left(n^{k+l}-1\right)+\left(n^{k}+m n^{l}+1\right)=n^{k+l}+n^{k}+m n^{l} . $$  We have two cases to discuss.  - Case 1. $l \geqslant k$.  Since $\left(n^{k}+m n^{l}+1, n\right)=1$,(2) yields  $$ n^{k}+m n^{l}+1 \mid n^{l}+m n^{l-k}+1 . $$  In particular, we get $n^{k}+m n^{l}+1 \leqslant n^{l}+m n^{l-k}+1$. As $n \geqslant 2$ and $k \geqslant 1,(m-1) n^{l}$ is at least $2(m-1) n^{l-k}$. It follows that the inequality cannot hold when $m \geqslant 2$. For $m=1$, the above divisibility becomes  $$ n^{k}+n^{l}+1 \mid n^{l}+n^{l-k}+1 . $$  Note that $n^{l}+n^{l-k}+1<n^{l}+n^{l}+1<2\left(n^{k}+n^{l}+1\right)$. Thus we must have $n^{l}+n^{l-k}+1=n^{k}+n^{l}+1$ so that $l=2 k$, which gives the first result.  - Case 2. $l<k$.  This time (2) yields  $$ n^{k}+m n^{l}+1 \mid n^{k}+n^{k-l}+m $$  In particular, we get $n^{k}+m n^{l}+1 \leqslant n^{k}+n^{k-l}+m$, which implies  $$ m \leqslant \frac{n^{k-l}-1}{n^{l}-1} $$  On the other hand, from (1) we may let $n^{k+l}-1=\left(n^{k}+m n^{l}+1\right) t$ for some positive integer $t$. Obviously, $t$ is less than $n^{l}$, which means $t \leqslant n^{l}-1$ as it is an integer. Then we have $n^{k+l}-1 \leqslant\left(n^{k}+m n^{l}+1\right)\left(n^{l}-1\right)$, which is the same as  $$ m \geqslant \frac{n^{k-l}-1}{n^{l}-1} $$  Equations (3) and (4) combine to give $m=\frac{n^{k-l}-1}{n^{l}-1}$. As this is an integer, we have $l \mid k-l$. This means $l \mid k$ and it corresponds to the second result.


Solution: As in   - Case 1. $l \geqslant k$.  Then (2) yields  $$ n^{k}+m n^{l}+1 \mid n^{l}+m n^{l-k}+1 . $$  Since $2\left(n^{k}+m n^{l}+1\right)>2 m n^{l}+1>n^{l}+m n^{l-k}+1$, it follows that $n^{k}+m n^{l}+1=n^{l}+m n^{l-k}+1$, that is,  $$ m\left(n^{l}-n^{l-k}\right)=n^{l}-n^{k} . $$  If $m \geqslant 2$, then $m\left(n^{l}-n^{l-k}\right) \geqslant 2 n^{l}-2 n^{l-k} \geqslant 2 n^{l}-n^{l}>n^{l}-n^{k}$ gives a contradiction. Hence $m=1$ and $l-k=k$, which means $m=1$ and $l=2 k$.  - Case 2. $l<k$.  Then (2) yields  $$ n^{k}+m n^{l}+1 \mid n^{k}+n^{k-l}+m $$  Since $2\left(n^{k}+m n^{l}+1\right)>2 n^{k}+m>n^{k}+n^{k-l}+m$, it follows that $n^{k}+m n^{l}+1=n^{k}+n^{k-l}+m$. This gives $m=\frac{n^{k-l}-1}{n^{l}-1}$. Note that $n^{l}-1 \mid n^{k-l}-1$ implies $l \mid k-l$ and hence $l \mid k$. The proof is thus complete.  Comment. Another version of this problem is as follows: let $n, m, k$ and $l$ be positive integers with $n \neq 1$ such that $k$ and $l$ do not divide each other. Show that $n^{k}+m n^{l}+1$ does not divide $n^{k+l}-1$.


Let $a$ be a positive integer which is not a square number. Denote by $A$ the set of all positive integers $k$ such that  $$ k=\frac{x^{2}-a}{x^{2}-y^{2}} $$  for some integers $x$ and $y$ with $x>\sqrt{a}$. Denote by $B$ the set of all positive integers $k$ such that (1) is satisfied for some integers $x$ and $y$ with $0 \leqslant x<\sqrt{a}$. Prove that $A=B$.

Solution: We first prove the following preliminary result.  - Claim. For fixed $k$, let $x, y$ be integers satisfying (1). Then the numbers $x_{1}, y_{1}$ defined by  $$ x_{1}=\frac{1}{2}\left(x-y+\frac{(x-y)^{2}-4 a}{x+y}\right), \quad y_{1}=\frac{1}{2}\left(x-y-\frac{(x-y)^{2}-4 a}{x+y}\right) $$  are integers and satisfy (1) (with $x, y$ replaced by $x_{1}, y_{1}$ respectively). Proof. Since $x_{1}+y_{1}=x-y$ and  $$ x_{1}=\frac{x^{2}-x y-2 a}{x+y}=-x+\frac{2\left(x^{2}-a\right)}{x+y}=-x+2 k(x-y), $$  both $x_{1}$ and $y_{1}$ are integers. Let $u=x+y$ and $v=x-y$. The relation (1) can be rewritten as  $$ u^{2}-(4 k-2) u v+\left(v^{2}-4 a\right)=0 $$  By Vieta's Theorem, the number $z=\frac{v^{2}-4 a}{u}$ satisfies  $$ v^{2}-(4 k-2) v z+\left(z^{2}-4 a\right)=0 $$  Since $x_{1}$ and $y_{1}$ are defined so that $v=x_{1}+y_{1}$ and $z=x_{1}-y_{1}$, we can reverse the process and verify (1) for $x_{1}, y_{1}$.  We first show that $B \subset A$. Take any $k \in B$ so that (1) is satisfied for some integers $x, y$ with $0 \leqslant x<\sqrt{a}$. Clearly, $y \neq 0$ and we may assume $y$ is positive. Since $a$ is not a square, we have $k>1$. Hence, we get $0 \leqslant x<y<\sqrt{a}$. Define  $$ x_{1}=\frac{1}{2}\left|x-y+\frac{(x-y)^{2}-4 a}{x+y}\right|, \quad y_{1}=\frac{1}{2}\left(x-y-\frac{(x-y)^{2}-4 a}{x+y}\right) . $$  By the Claim, $x_{1}, y_{1}$ are integers satisfying (1). Also, we have  $$ x_{1} \geqslant-\frac{1}{2}\left(x-y+\frac{(x-y)^{2}-4 a}{x+y}\right)=\frac{2 a+x(y-x)}{x+y} \geqslant \frac{2 a}{x+y}>\sqrt{a} $$  This implies $k \in A$ and hence $B \subset A$.  Next, we shall show that $A \subset B$. Take any $k \in A$ so that (1) is satisfied for some integers $x, y$ with $x>\sqrt{a}$. Again, we may assume $y$ is positive. Among all such representations of $k$, we choose the one with smallest $x+y$. Define  $$ x_{1}=\frac{1}{2}\left|x-y+\frac{(x-y)^{2}-4 a}{x+y}\right|, \quad y_{1}=\frac{1}{2}\left(x-y-\frac{(x-y)^{2}-4 a}{x+y}\right) . $$  By the Claim, $x_{1}, y_{1}$ are integers satisfying (1). Since $k>1$, we get $x>y>\sqrt{a}$. Therefore, we have $y_{1}>\frac{4 a}{x+y}>0$ and $\frac{4 a}{x+y}<x+y$. It follows that  $$ x_{1}+y_{1} \leqslant \max \left\{x-y, \frac{4 a-(x-y)^{2}}{x+y}\right\}<x+y $$  If $x_{1}>\sqrt{a}$, we get a contradiction due to the minimality of $x+y$. Therefore, we must have $0 \leqslant x_{1}<\sqrt{a}$, which means $k \in B$ so that $A \subset B$.  The two subset relations combine to give $A=B$.


Solution: The relation (1) is equivalent to  $$ k y^{2}-(k-1) x^{2}=a $$  Motivated by Pell's Equation, we prove the following, which is essentially the same as the Claim in Solution 1.  - Claim. If $\left(x_{0}, y_{0}\right)$ is a solution to $(2)$, then $\left((2 k-1) x_{0} \pm 2 k y_{0},(2 k-1) y_{0} \pm 2(k-1) x_{0}\right)$ is also a solution to (2).  Proof. We check directly that  $$ \begin{aligned} & k\left((2 k-1) y_{0} \pm 2(k-1) x_{0}\right)^{2}-(k-1)\left((2 k-1) x_{0} \pm 2 k y_{0}\right)^{2} \\ = & \left(k(2 k-1)^{2}-(k-1)(2 k)^{2}\right) y_{0}^{2}+\left(k(2(k-1))^{2}-(k-1)(2 k-1)^{2}\right) x_{0}^{2} \\ = & k y_{0}^{2}-(k-1) x_{0}^{2}=a \end{aligned} $$  If (2) is satisfied for some $0 \leqslant x<\sqrt{a}$ and nonnegative integer $y$, then clearly (1) implies $y>x$. Also, we have $k>1$ since $a$ is not a square number. By the Claim, consider another solution to (2) defined by  $$ x_{1}=(2 k-1) x+2 k y, \quad y_{1}=(2 k-1) y+2(k-1) x $$  It satisfies $x_{1} \geqslant(2 k-1) x+2 k(x+1)=(4 k-1) x+2 k>x$. Then we can replace the old solution by a new one which has a larger value in $x$. After a finite number of replacements, we must get a solution with $x>\sqrt{a}$. This shows $B \subset A$.  If (2) is satisfied for some $x>\sqrt{a}$ and nonnegative integer $y$, by the Claim we consider another solution to (2) defined by  $$ x_{1}=|(2 k-1) x-2 k y|, \quad y_{1}=(2 k-1) y-2(k-1) x . $$  From (2), we get $\sqrt{k} y>\sqrt{k-1} x$. This implies $k y>\sqrt{k(k-1)} x>(k-1) x$ and hence $(2 k-1) x-2 k y<x$. On the other hand, the relation (1) implies $x>y$. Then it is clear that $(2 k-1) x-2 k y>-x$. These combine to give $x_{1}<x$, which means we have found a solution to (2) with $x$ having a smaller absolute value. After a finite number of steps, we shall obtain a solution with $0 \leqslant x<\sqrt{a}$. This shows $A \subset B$.  The desired result follows from $B \subset A$ and $A \subset B$.


Solution: It suffices to show $A \cup B$ is a subset of $A \cap B$. We take any $k \in A \cup B$, which means there exist integers $x, y$ satisfying (1). Since $a$ is not a square, it follows that $k \neq 1$. As in   Without loss of generality, assume $x, y \geqslant 0$. Let $u=x+y$ and $v=x-y$. Then $u \geqslant v$ and (1) becomes  $$ k=\frac{(u+v)^{2}-4 a}{4 u v} $$  This is the same as  $$ v^{2}+(2 u-4 k u) v+u^{2}-4 a=0 $$  Let $v_{1}=4 k u-2 u-v$. Then $u+v_{1}=4 k u-u-v \geqslant 8 u-u-v>u+v$. By Vieta's Theorem, $v_{1}$ satisfies  $$ v_{1}^{2}+(2 u-4 k u) v_{1}+u^{2}-4 a=0 $$  This gives $k=\frac{\left(u+v_{1}\right)^{2}-4 a}{4 u v_{1}}$. As $k$ is an integer, $u+v_{1}$ must be even. Therefore, $x_{1}=\frac{u+v_{1}}{2}$ and $y_{1}=\frac{v_{1}-u}{2}$ are integers. By reversing the process, we can see that $\left(x_{1}, y_{1}\right)$ is a solution to (1), with $x_{1}=\frac{u+v_{1}}{2}>\frac{u+v}{2}=x \geqslant 0$. This completes the first half of the proof.  Suppose $x>\sqrt{a}$. Then $u+v>2 \sqrt{a}$ and (3) can be rewritten as  $$ u^{2}+(2 v-4 k v) u+v^{2}-4 a=0 . $$  Let $u_{2}=4 k v-2 v-u$. By Vieta's Theorem, we have $u u_{2}=v^{2}-4 a$ and  $$ u_{2}^{2}+(2 v-4 k v) u_{2}+v^{2}-4 a=0 $$  By $u>0, u+v>2 \sqrt{a}$ and (3), we have $v>0$. If $u_{2} \geqslant 0$, then $v u_{2} \leqslant u u_{2}=v^{2}-4 a<v^{2}$. This shows $u_{2}<v \leqslant u$ and $0<u_{2}+v<u+v$. If $u_{2}<0$, then $\left(u_{2}+v\right)+(u+v)=4 k v>0$ and $u_{2}+v<u+v$ imply $\left|u_{2}+v\right|<u+v$. In any case, since $u_{2}+v$ is even from (4), we can define $x_{2}=\frac{u_{2}+v}{2}$ and $y_{2}=\frac{u_{2}-v}{2}$ so that (1) is satisfied with $\left|x_{2}\right|<x$, as desired. The proof is thus complete.


Denote by $\mathbb{N}$ the set of all positive integers. Find all functions $f: \mathbb{N} \rightarrow \mathbb{N}$ such that for all positive integers $m$ and $n$, the integer $f(m)+f(n)-m n$ is nonzero and divides $m f(m)+n f(n)$.

Solution: It is given that  $$ f(m)+f(n)-m n \mid m f(m)+n f(n) $$  Taking $m=n=1$ in (1), we have $2 f(1)-1 \mid 2 f(1)$. Then $2 f(1)-1 \mid 2 f(1)-(2 f(1)-1)=1$ and hence $f(1)=1$.  Let $p \geqslant 7$ be a prime. Taking $m=p$ and $n=1$ in (1), we have $f(p)-p+1 \mid p f(p)+1$ and hence  $$ f(p)-p+1 \mid p f(p)+1-p(f(p)-p+1)=p^{2}-p+1 $$  If $f(p)-p+1=p^{2}-p+1$, then $f(p)=p^{2}$. If $f(p)-p+1 \neq p^{2}-p+1$, as $p^{2}-p+1$ is an odd positive integer, we have $p^{2}-p+1 \geqslant 3(f(p)-p+1)$, that is,  $$ f(p) \leqslant \frac{1}{3}\left(p^{2}+2 p-2\right) $$  Taking $m=n=p$ in (1), we have $2 f(p)-p^{2} \mid 2 p f(p)$. This implies  $$ 2 f(p)-p^{2} \mid 2 p f(p)-p\left(2 f(p)-p^{2}\right)=p^{3} $$  By (2) and $f(p) \geqslant 1$, we get  $$ -p^{2}<2 f(p)-p^{2} \leqslant \frac{2}{3}\left(p^{2}+2 p-2\right)-p^{2}<-p $$  since $p \geqslant 7$. This contradicts the fact that $2 f(p)-p^{2}$ is a factor of $p^{3}$. Thus we have proved that $f(p)=p^{2}$ for all primes $p \geqslant 7$.  Let $n$ be a fixed positive integer. Choose a sufficiently large prime $p$. Consider $m=p$ in (1). We obtain  $$ f(p)+f(n)-p n \mid p f(p)+n f(n)-n(f(p)+f(n)-p n)=p f(p)-n f(p)+p n^{2} $$  As $f(p)=p^{2}$, this implies $p^{2}-p n+f(n) \mid p\left(p^{2}-p n+n^{2}\right)$. As $p$ is sufficiently large and $n$ is fixed, $p$ cannot divide $f(n)$, and so $\left(p, p^{2}-p n+f(n)\right)=1$. It follows that $p^{2}-p n+f(n) \mid p^{2}-p n+n^{2}$ and hence  $$ p^{2}-p n+f(n) \mid\left(p^{2}-p n+n^{2}\right)-\left(p^{2}-p n+f(n)\right)=n^{2}-f(n) $$  Note that $n^{2}-f(n)$ is fixed while $p^{2}-p n+f(n)$ is chosen to be sufficiently large. Therefore, we must have $n^{2}-f(n)=0$ so that $f(n)=n^{2}$ for any positive integer $n$.  Finally, we check that when $f(n)=n^{2}$ for any positive integer $n$, then  $$ f(m)+f(n)-m n=m^{2}+n^{2}-m n $$  and  $$ m f(m)+n f(n)=m^{3}+n^{3}=(m+n)\left(m^{2}+n^{2}-m n\right) $$  The latter expression is divisible by the former for any positive integers $m, n$. This shows $f(n)=n^{2}$ is the only solution.


Let $n$ be an odd positive integer. In the Cartesian plane, a cyclic polygon $P$ with area $S$ is chosen. All its vertices have integral coordinates, and all squares of its side lengths are divisible by $n$. Prove that $2 S$ is an integer divisible by $n$.

Solution: Let $P=A_{1} A_{2} \ldots A_{k}$ and let $A_{k+i}=A_{i}$ for $i \geqslant 1$. By the Shoelace Formula, the area of any convex polygon with integral coordinates is half an integer. Therefore, $2 S$ is an integer. We shall prove by induction on $k \geqslant 3$ that $2 S$ is divisible by $n$. Clearly, it suffices to consider $n=p^{t}$ where $p$ is an odd prime and $t \geqslant 1$.  For the base case $k=3$, let the side lengths of $P$ be $\sqrt{n a}, \sqrt{n b}, \sqrt{n c}$ where $a, b, c$ are positive integers. By Heron's Formula,  $$ 16 S^{2}=n^{2}\left(2 a b+2 b c+2 c a-a^{2}-b^{2}-c^{2}\right) $$  This shows $16 S^{2}$ is divisible by $n^{2}$. Since $n$ is odd, $2 S$ is divisible by $n$. Assume $k \geqslant 4$. If the square of length of one of the diagonals is divisible by $n$, then that diagonal divides $P$ into two smaller polygons, to which the induction hypothesis applies. Hence we may assume that none of the squares of diagonal lengths is divisible by $n$. As usual, we denote by $\nu_{p}(r)$ the exponent of $p$ in the prime decomposition of $r$. We claim the following.  - Claim. $\nu_{p}\left(A_{1} A_{m}^{2}\right)>\nu_{p}\left(A_{1} A_{m+1}^{2}\right)$ for $2 \leqslant m \leqslant k-1$.  Proof. The case $m=2$ is obvious since $\nu_{p}\left(A_{1} A_{2}^{2}\right) \geqslant p^{t}>\nu_{p}\left(A_{1} A_{3}^{2}\right)$ by the condition and the above assumption.  Suppose $\nu_{p}\left(A_{1} A_{2}^{2}\right)>\nu_{p}\left(A_{1} A_{3}^{2}\right)>\cdots>\nu_{p}\left(A_{1} A_{m}^{2}\right)$ where $3 \leqslant m \leqslant k-1$. For the induction step, we apply Ptolemy's Theorem to the cyclic quadrilateral $A_{1} A_{m-1} A_{m} A_{m+1}$ to get  $$ A_{1} A_{m+1} \times A_{m-1} A_{m}+A_{1} A_{m-1} \times A_{m} A_{m+1}=A_{1} A_{m} \times A_{m-1} A_{m+1} $$  which can be rewritten as  $$ \begin{aligned} A_{1} A_{m+1}^{2} \times A_{m-1} A_{m}^{2}= & A_{1} A_{m-1}^{2} \times A_{m} A_{m+1}^{2}+A_{1} A_{m}^{2} \times A_{m-1} A_{m+1}^{2} \\ & -2 A_{1} A_{m-1} \times A_{m} A_{m+1} \times A_{1} A_{m} \times A_{m-1} A_{m+1} \end{aligned} $$  From this, $2 A_{1} A_{m-1} \times A_{m} A_{m+1} \times A_{1} A_{m} \times A_{m-1} A_{m+1}$ is an integer. We consider the component of $p$ of each term in (1). By the inductive hypothesis, we have $\nu_{p}\left(A_{1} A_{m-1}^{2}\right)>\nu_{p}\left(A_{1} A_{m}^{2}\right)$. Also, we have $\nu_{p}\left(A_{m} A_{m+1}^{2}\right) \geqslant p^{t}>\nu_{p}\left(A_{m-1} A_{m+1}^{2}\right)$. These give  $$ \nu_{p}\left(A_{1} A_{m-1}^{2} \times A_{m} A_{m+1}^{2}\right)>\nu_{p}\left(A_{1} A_{m}^{2} \times A_{m-1} A_{m+1}^{2}\right) $$  Next, we have $\nu_{p}\left(4 A_{1} A_{m-1}^{2} \times A_{m} A_{m+1}^{2} \times A_{1} A_{m}^{2} \times A_{m-1} A_{m+1}^{2}\right)=\nu_{p}\left(A_{1} A_{m-1}^{2} \times A_{m} A_{m+1}^{2}\right)+$ $\nu_{p}\left(A_{1} A_{m}^{2} \times A_{m-1} A_{m+1}^{2}\right)>2 \nu_{p}\left(A_{1} A_{m}^{2} \times A_{m-1} A_{m+1}^{2}\right)$ from (2). This implies  $$ \nu_{p}\left(2 A_{1} A_{m-1} \times A_{m} A_{m+1} \times A_{1} A_{m} \times A_{m-1} A_{m+1}\right)>\nu_{p}\left(A_{1} A_{m}^{2} \times A_{m-1} A_{m+1}^{2}\right) $$  Combining (1), (2) and (3), we conclude that  $$ \nu_{p}\left(A_{1} A_{m+1}^{2} \times A_{m-1} A_{m}^{2}\right)=\nu_{p}\left(A_{1} A_{m}^{2} \times A_{m-1} A_{m+1}^{2}\right) $$  By $\nu_{p}\left(A_{m-1} A_{m}^{2}\right) \geqslant p^{t}>\nu_{p}\left(A_{m-1} A_{m+1}^{2}\right)$, we get $\nu_{p}\left(A_{1} A_{m+1}^{2}\right)<\nu_{p}\left(A_{1} A_{m}^{2}\right)$. The Claim follows by induction.  From the Claim, we get a chain of inequalities  $$ p^{t}>\nu_{p}\left(A_{1} A_{3}^{2}\right)>\nu_{p}\left(A_{1} A_{4}^{2}\right)>\cdots>\nu_{p}\left(A_{1} A_{k}^{2}\right) \geqslant p^{t} $$  which yields a contradiction. Therefore, we can show by induction that $2 S$ is divisible by $n$. Comment. The condition that $P$ is cyclic is crucial. As a counterexample, consider the rhombus with vertices $(0,3),(4,0),(0,-3),(-4,0)$. Each of its squares of side lengths is divisible by 5 , while $2 S=48$ is not.  The proposer also gives a proof for the case $n$ is even. One just needs an extra technical step for the case $p=2$.


Find all polynomials $P(x)$ of odd degree $d$ and with integer coefficients satisfying the following property: for each positive integer $n$, there exist $n$ positive integers $x_{1}, x_{2}, \ldots, x_{n}$ such that $\frac{1}{2}<\frac{P\left(x_{i}\right)}{P\left(x_{j}\right)}<2$ and $\frac{P\left(x_{i}\right)}{P\left(x_{j}\right)}$ is the $d$-th power of a rational number for every pair of indices $i$ and $j$ with $1 \leqslant i, j \leqslant n$.

Solution: Let $P(x)=a_{d} x^{d}+a_{d-1} x^{d-1}+\cdots+a_{0}$. Consider the substitution $y=d a_{d} x+a_{d-1}$. By defining $Q(y)=P(x)$, we find that $Q$ is a polynomial with rational coefficients without the term $y^{d-1}$. Let $Q(y)=b_{d} y^{d}+b_{d-2} y^{d-2}+b_{d-3} y^{d-3}+\cdots+b_{0}$ and $B=\max _{0 \leqslant i \leqslant d}\left\{\left|b_{i}\right|\right\}$ (where $b_{d-1}=0$ ).  The condition shows that for each $n \geqslant 1$, there exist integers $y_{1}, y_{2}, \ldots, y_{n}$ such that $\frac{1}{2}<\frac{Q\left(y_{i}\right)}{Q\left(y_{j}\right)}<2$ and $\frac{Q\left(y_{i}\right)}{Q\left(y_{j}\right)}$ is the $d$-th power of a rational number for $1 \leqslant i, j \leqslant n$. Since $n$ can be arbitrarily large, we may assume all $x_{i}$ 's and hence $y_{i}$ 's are integers larger than some absolute constant in the following.  By Dirichlet's Theorem, since $d$ is odd, we can find a sufficiently large prime $p$ such that $p \equiv 2(\bmod d)$. In particular, we have $(p-1, d)=1$. For this fixed $p$, we choose $n$ to be sufficiently large. Then by the Pigeonhole Principle, there must be $d+1$ of $y_{1}, y_{2}, \ldots, y_{n}$ which are congruent $\bmod p$. Without loss of generality, assume $y_{i} \equiv y_{j}(\bmod p)$ for $1 \leqslant i, j \leqslant d+1$. We shall establish the following.  - Claim. $\frac{Q\left(y_{i}\right)}{Q\left(y_{1}\right)}=\frac{y_{i}^{d}}{y_{1}^{d}}$ for $2 \leqslant i \leqslant d+1$.  Proof. Let $\frac{Q\left(y_{i}\right)}{Q\left(y_{1}\right)}=\frac{l^{d}}{m^{d}}$ where $(l, m)=1$ and $l, m>0$. This can be rewritten in the expanded form  $$ b_{d}\left(m^{d} y_{i}^{d}-l^{d} y_{1}^{d}\right)=-\sum_{j=0}^{d-2} b_{j}\left(m^{d} y_{i}^{j}-l^{d} y_{1}^{j}\right) $$  Let $c$ be the common denominator of $Q$, so that $c Q(k)$ is an integer for any integer $k$. Note that $c$ depends only on $P$ and so we may assume $(p, c)=1$. Then $y_{1} \equiv y_{i}(\bmod p)$ implies $c Q\left(y_{1}\right) \equiv c Q\left(y_{i}\right)(\bmod p)$.  - Case 1. $p \mid c Q\left(y_{1}\right)$.  In this case, there is a cancellation of $p$ in the numerator and denominator of $\frac{c Q\left(y_{i}\right)}{c Q\left(y_{1}\right)}$, so that $m^{d} \leqslant p^{-1}\left|c Q\left(y_{1}\right)\right|$. Noting $\left|Q\left(y_{1}\right)\right|<2 B y_{1}^{d}$ as $y_{1}$ is large, we get  $$ m \leqslant p^{-\frac{1}{d}}(2 c B)^{\frac{1}{d}} y_{1} $$  For large $y_{1}$ and $y_{i}$, the relation $\frac{1}{2}<\frac{Q\left(y_{i}\right)}{Q\left(y_{1}\right)}<2$ implies  $$ \frac{1}{3}<\frac{y_{i}^{d}}{y_{1}^{d}}<3 $$  We also have  $$ \frac{1}{2}<\frac{l^{d}}{m^{d}}<2 $$  Now, the left-hand side of (1) is  $$ b_{d}\left(m y_{i}-l y_{1}\right)\left(m^{d-1} y_{i}^{d-1}+m^{d-2} y_{i}^{d-2} l y_{1}+\cdots+l^{d-1} y_{1}^{d-1}\right) . $$  Suppose on the contrary that $m y_{i}-l y_{1} \neq 0$. Then the absolute value of the above expression is at least $\left|b_{d}\right| m^{d-1} y_{i}^{d-1}$. On the other hand, the absolute value of the right-hand side of (1) is at most  $$ \begin{aligned} \sum_{j=0}^{d-2} B\left(m^{d} y_{i}^{j}+l^{d} y_{1}^{j}\right) & \leqslant(d-1) B\left(m^{d} y_{i}^{d-2}+l^{d} y_{1}^{d-2}\right) \\ & \leqslant(d-1) B\left(7 m^{d} y_{i}^{d-2}\right) \\ & \leqslant 7(d-1) B\left(p^{-\frac{1}{d}}(2 c B)^{\frac{1}{d}} y_{1}\right) m^{d-1} y_{i}^{d-2} \\ & \leqslant 21(d-1) B p^{-\frac{1}{d}}(2 c B)^{\frac{1}{d}} m^{d-1} y_{i}^{d-1} \end{aligned} $$  by using successively (3), (4), (2) and again (3). This shows  $$ \left|b_{d}\right| m^{d-1} y_{i}^{d-1} \leqslant 21(d-1) B p^{-\frac{1}{d}}(2 c B)^{\frac{1}{d}} m^{d-1} y_{i}^{d-1}, $$  which is a contradiction for large $p$ as $b_{d}, B, c, d$ depend only on the polynomial $P$. Therefore, we have $m y_{i}-l y_{1}=0$ in this case.  - Case 2. $\left(p, c Q\left(y_{1}\right)\right)=1$.  From $c Q\left(y_{1}\right) \equiv c Q\left(y_{i}\right)(\bmod p)$, we have $l^{d} \equiv m^{d}(\bmod p)$. Since $(p-1, d)=1$, we use Fermat Little Theorem to conclude $l \equiv m(\bmod p)$. Then $p \mid m y_{i}-l y_{1}$. Suppose on the contrary that $m y_{i}-l y_{1} \neq 0$. Then the left-hand side of (1) has absolute value at least $\left|b_{d}\right| p m^{d-1} y_{i}^{d-1}$. Similar to Case 1, the right-hand side of (1) has absolute value at most  $$ 21(d-1) B(2 c B)^{\frac{1}{d}} m^{d-1} y_{i}^{d-1} $$  which must be smaller than $\left|b_{d}\right| p m^{d-1} y_{i}^{d-1}$ for large $p$. Again this yields a contradiction and hence $m y_{i}-l y_{1}=0$.  In both cases, we find that $\frac{Q\left(y_{i}\right)}{Q\left(y_{1}\right)}=\frac{l^{d}}{m^{d}}=\frac{y_{i}^{d}}{y_{1}^{d}}$. From the Claim, the polynomial $Q\left(y_{1}\right) y^{d}-y_{1}^{d} Q(y)$ has roots $y=y_{1}, y_{2}, \ldots, y_{d+1}$. Since its degree is at most $d$, this must be the zero polynomial. Hence, $Q(y)=b_{d} y^{d}$. This implies $P(x)=a_{d}\left(x+\frac{a_{d-1}}{d a_{d}}\right)^{d}$. Let $\frac{a_{d-1}}{d a_{d}}=\frac{s}{r}$ with integers $r, s$ where $r \geqslant 1$ and $(r, s)=1$. Since $P$ has integer coefficients, we need $r^{d} \mid a_{d}$. Let $a_{d}=r^{d} a$. Then $P(x)=a(r x+s)^{d}$. It is obvious that such a polynomial satisfies the conditions.  Comment. In the proof, the use of prime and Dirichlet's Theorem can be avoided. One can easily show that each $P\left(x_{i}\right)$ can be expressed in the form $u v_{i}^{d}$ where $u, v_{i}$ are integers and $u$ cannot be divisible by the $d$-th power of a prime (note that $u$ depends only on $P$ ). By fixing a large integer $q$ and by choosing a large $n$, we can apply the Pigeonhole Principle and assume $x_{1} \equiv x_{2} \equiv \cdots \equiv x_{d+1}(\bmod q)$ and $v_{1} \equiv v_{2} \equiv \cdots \equiv v_{d+1}(\bmod q)$. Then the remaining proof is similar to Case 2 of the Solution.  Alternatively, we give another modification of the proof as follows. We take a sufficiently large $n$ and consider the corresponding positive integers $y_{1}, y_{2}, \ldots, y_{n}$. For each $2 \leqslant i \leqslant n$, let $\frac{Q\left(y_{i}\right)}{Q\left(y_{1}\right)}=\frac{l_{i}^{d}}{m_{i}^{d}}$.  As in Case 1, if there are $d$ indices $i$ such that the integers $\frac{c\left|Q\left(y_{1}\right)\right|}{m_{i}^{d}}$ are bounded below by a constant depending only on $P$, we can establish the Claim using those $y_{i}$ 's and complete the proof. Similarly, as in Case 2, if there are $d$ indices $i$ such that the integers $\left|m_{i} y_{i}-l_{i} y_{1}\right|$ are bounded below, then the proof goes the same. So it suffices to consider the case where $\frac{c\left|Q\left(y_{1}\right)\right|}{m_{i}^{d}} \leqslant M$ and $\left|m_{i} y_{i}-l_{i} y_{1}\right| \leqslant N$ for all $2 \leqslant i \leqslant n^{\prime}$ where $M, N$ are fixed constants and $n^{\prime}$ is large. Since there are only finitely many choices for $m_{i}$ and $m_{i} y_{i}-l_{i} y_{1}$, by the Pigeonhole Principle, we can assume without loss of generality $m_{i}=m$ and $m_{i} y_{i}-l_{i} y_{1}=t$ for $2 \leqslant i \leqslant d+2$. Then  $$ \frac{Q\left(y_{i}\right)}{Q\left(y_{1}\right)}=\frac{l_{i}^{d}}{m^{d}}=\frac{\left(m y_{i}-t\right)^{d}}{m^{d} y_{1}^{d}} $$  so that $Q\left(y_{1}\right)(m y-t)^{d}-m^{d} y_{1}^{d} Q(y)$ has roots $y=y_{2}, y_{3}, \ldots, y_{d+2}$. Its degree is at most $d$ and hence it is the zero polynomial. Therefore, $Q(y)=\frac{b_{d}}{m^{d}}(m y-t)^{d}$. Indeed, $Q$ does not have the term $y^{d-1}$, which means $t$ should be 0 . This gives the corresponding $P(x)$ of the desired form.  The two modifications of the Solution work equally well when the degree $d$ is even.


Find the smallest real constant $C$ such that for any positive real numbers $a_{1}, a_{2}, a_{3}, a_{4}$ and $a_{5}$ (not necessarily distinct), one can always choose distinct subscripts $i, j, k$ and $l$ such that  $$ \left|\frac{a_{i}}{a_{j}}-\frac{a_{k}}{a_{l}}\right| \leqslant C . $$  Answer. The smallest $C$ is $\frac{1}{2}$.

Solution: We first show that $C \leqslant \frac{1}{2}$. For any positive real numbers $a_{1} \leqslant a_{2} \leqslant a_{3} \leqslant a_{4} \leqslant a_{5}$, consider the five fractions  $$ \frac{a_{1}}{a_{2}}, \frac{a_{3}}{a_{4}}, \frac{a_{1}}{a_{5}}, \frac{a_{2}}{a_{3}}, \frac{a_{4}}{a_{5}} . $$  Each of them lies in the interval $(0,1]$. Therefore, by the Pigeonhole Principle, at least three of them must lie in $\left(0, \frac{1}{2}\right]$ or lie in $\left(\frac{1}{2}, 1\right]$ simultaneously. In particular, there must be two consecutive terms in (2) which belong to an interval of length $\frac{1}{2}$ (here, we regard $\frac{a_{1}}{a_{2}}$ and $\frac{a_{4}}{a_{5}}$ as consecutive). In other words, the difference of these two fractions is less than $\frac{1}{2}$. As the indices involved in these two fractions are distinct, we can choose them to be $i, j, k, l$ and conclude that $C \leqslant \frac{1}{2}$.  Next, we show that $C=\frac{1}{2}$ is best possible. Consider the numbers $1,2,2,2, n$ where $n$ is a large real number. The fractions formed by two of these numbers in ascending order are $\frac{1}{n}, \frac{2}{n}, \frac{1}{2}, \frac{2}{2}, \frac{2}{1}, \frac{n}{2}, \frac{n}{1}$. Since the indices $i, j, k, l$ are distinct, $\frac{1}{n}$ and $\frac{2}{n}$ cannot be chosen simultaneously. Therefore the minimum value of the left-hand side of (1) is $\frac{1}{2}-\frac{2}{n}$. When $n$ tends to infinity, this value approaches $\frac{1}{2}$, and so $C$ cannot be less than $\frac{1}{2}$.  These conclude that $C=\frac{1}{2}$ is the smallest possible choice. Comment. The conclusion still holds if $a_{1}, a_{2}, \ldots, a_{5}$ are pairwise distinct, since in the construction, we may replace the 2's by real numbers sufficiently close to 2 .  There are two possible simplifications for this problem: (i) the answer $C=\frac{1}{2}$ is given to the contestants; or (ii) simply ask the contestants to prove the inequality (1) for $C=\frac{1}{2}$.


Find all integers $n \geqslant 3$ with the following property: for all real numbers $a_{1}, a_{2}, \ldots, a_{n}$ and $b_{1}, b_{2}, \ldots, b_{n}$ satisfying $\left|a_{k}\right|+\left|b_{k}\right|=1$ for $1 \leqslant k \leqslant n$, there exist $x_{1}, x_{2}, \ldots, x_{n}$, each of which is either -1 or 1 , such that  $$ \left|\sum_{k=1}^{n} x_{k} a_{k}\right|+\left|\sum_{k=1}^{n} x_{k} b_{k}\right| \leqslant 1 $$  Answer. $n$ can be any odd integer greater than or equal to 3 .

Solution: For any even integer $n \geqslant 4$, we consider the case  $$ a_{1}=a_{2}=\cdots=a_{n-1}=b_{n}=0 \quad \text { and } \quad b_{1}=b_{2}=\cdots=b_{n-1}=a_{n}=1 $$  The condition $\left|a_{k}\right|+\left|b_{k}\right|=1$ is satisfied for each $1 \leqslant k \leqslant n$. No matter how we choose each $x_{k}$, both sums $\sum_{k=1}^{n} x_{k} a_{k}$ and $\sum_{k=1}^{n} x_{k} b_{k}$ are odd integers. This implies $\left|\sum_{k=1}^{n} x_{k} a_{k}\right| \geqslant 1$ and $\left|\sum_{k=1}^{n} x_{k} b_{k}\right| \geqslant 1$, which shows (1) cannot hold.  For any odd integer $n \geqslant 3$, we may assume without loss of generality $b_{k} \geqslant 0$ for $1 \leqslant k \leqslant n$ (this can be done by flipping the pair $\left(a_{k}, b_{k}\right)$ to $\left(-a_{k},-b_{k}\right)$ and $x_{k}$ to $-x_{k}$ if necessary) and $a_{1} \geqslant a_{2} \geqslant \cdots \geqslant a_{m} \geqslant 0>a_{m+1} \geqslant \cdots \geqslant a_{n}$. We claim that the choice $x_{k}=(-1)^{k+1}$ for $1 \leqslant k \leqslant n$ will work. Define  $$ s=\sum_{k=1}^{m} x_{k} a_{k} \quad \text { and } \quad t=-\sum_{k=m+1}^{n} x_{k} a_{k} . $$  Note that  $$ s=\left(a_{1}-a_{2}\right)+\left(a_{3}-a_{4}\right)+\cdots \geqslant 0 $$  by the assumption $a_{1} \geqslant a_{2} \geqslant \cdots \geqslant a_{m}$ (when $m$ is odd, there is a single term $a_{m}$ at the end, which is also positive). Next, we have  $$ s=a_{1}-\left(a_{2}-a_{3}\right)-\left(a_{4}-a_{5}\right)-\cdots \leqslant a_{1} \leqslant 1 $$  Similarly,  $$ t=\left(-a_{n}+a_{n-1}\right)+\left(-a_{n-2}+a_{n-3}\right)+\cdots \geqslant 0 $$  and  $$ t=-a_{n}+\left(a_{n-1}-a_{n-2}\right)+\left(a_{n-3}-a_{n-4}\right)+\cdots \leqslant-a_{n} \leqslant 1 . $$  From the condition, we have $a_{k}+b_{k}=1$ for $1 \leqslant k \leqslant m$ and $-a_{k}+b_{k}=1$ for $m+1 \leqslant k \leqslant n$. It follows that $\sum_{k=1}^{n} x_{k} a_{k}=s-t$ and $\sum_{k=1}^{n} x_{k} b_{k}=1-s-t$. Hence it remains to prove  $$ |s-t|+|1-s-t| \leqslant 1 $$  under the constraint $0 \leqslant s, t \leqslant 1$. By symmetry, we may assume $s \geqslant t$. If $1-s-t \geqslant 0$, then we have  $$ |s-t|+|1-s-t|=s-t+1-s-t=1-2 t \leqslant 1 $$  If $1-s-t \leqslant 0$, then we have  $$ |s-t|+|1-s-t|=s-t-1+s+t=2 s-1 \leqslant 1 $$  Hence, the inequality is true in both cases. These show $n$ can be any odd integer greater than or equal to 3 .


Solution: The even case can be handled in the same way as   Firstly, for $n=3$, we may assume without loss of generality $a_{1} \geqslant a_{2} \geqslant a_{3} \geqslant 0$ and $b_{1}=a_{1}-1$ (if $b_{1}=1-a_{1}$, we may replace each $b_{k}$ by $-b_{k}$ ).  - Case 1. $b_{2}=a_{2}-1$ and $b_{3}=a_{3}-1$, in which case we take $\left(x_{1}, x_{2}, x_{3}\right)=(1,-1,1)$.  Let $c=a_{1}-a_{2}+a_{3}$ so that $0 \leqslant c \leqslant 1$. Then $\left|b_{1}-b_{2}+b_{3}\right|=\left|a_{1}-a_{2}+a_{3}-1\right|=1-c$ and hence $|c|+\left|b_{1}-b_{2}+b_{3}\right|=1$.  - Case 2. $b_{2}=1-a_{2}$ and $b_{3}=1-a_{3}$, in which case we take $\left(x_{1}, x_{2}, x_{3}\right)=(1,-1,1)$.  Let $c=a_{1}-a_{2}+a_{3}$ so that $0 \leqslant c \leqslant 1$. Since $a_{3} \leqslant a_{2}$ and $a_{1} \leqslant 1$, we have  $$ c-1 \leqslant b_{1}-b_{2}+b_{3}=a_{1}+a_{2}-a_{3}-1 \leqslant 1-c $$  This gives $\left|b_{1}-b_{2}+b_{3}\right| \leqslant 1-c$ and hence $|c|+\left|b_{1}-b_{2}+b_{3}\right| \leqslant 1$.  - Case 3. $b_{2}=a_{2}-1$ and $b_{3}=1-a_{3}$, in which case we take $\left(x_{1}, x_{2}, x_{3}\right)=(-1,1,1)$.  Let $c=-a_{1}+a_{2}+a_{3}$. If $c \geqslant 0$, then $a_{3} \leqslant 1$ and $a_{2} \leqslant a_{1}$ imply  $$ c-1 \leqslant-b_{1}+b_{2}+b_{3}=-a_{1}+a_{2}-a_{3}+1 \leqslant 1-c $$  If $c<0$, then $a_{1} \leqslant a_{2}+1$ and $a_{3} \geqslant 0$ imply  $$ -c-1 \leqslant-b_{1}+b_{2}+b_{3}=-a_{1}+a_{2}-a_{3}+1 \leqslant 1+c $$  In both cases, we get $\left|-b_{1}+b_{2}+b_{3}\right| \leqslant 1-|c|$ and hence $|c|+\left|-b_{1}+b_{2}+b_{3}\right| \leqslant 1$.  - Case 4. $b_{2}=1-a_{2}$ and $b_{3}=a_{3}-1$, in which case we take $\left(x_{1}, x_{2}, x_{3}\right)=(-1,1,1)$.  Let $c=-a_{1}+a_{2}+a_{3}$. If $c \geqslant 0$, then $a_{2} \leqslant 1$ and $a_{3} \leqslant a_{1}$ imply  $$ c-1 \leqslant-b_{1}+b_{2}+b_{3}=-a_{1}-a_{2}+a_{3}+1 \leqslant 1-c $$  If $c<0$, then $a_{1} \leqslant a_{3}+1$ and $a_{2} \geqslant 0$ imply  $$ -c-1 \leqslant-b_{1}+b_{2}+b_{3}=-a_{1}-a_{2}+a_{3}+1 \leqslant 1+c $$  In both cases, we get $\left|-b_{1}+b_{2}+b_{3}\right| \leqslant 1-|c|$ and hence $|c|+\left|-b_{1}+b_{2}+b_{3}\right| \leqslant 1$. We have found $x_{1}, x_{2}, x_{3}$ satisfying (1) in each case for $n=3$. Now, let $n \geqslant 5$ be odd and suppose the result holds for any smaller odd cases. Again we may assume $a_{k} \geqslant 0$ for each $1 \leqslant k \leqslant n$. By the Pigeonhole Principle, there are at least three indices $k$ for which $b_{k}=a_{k}-1$ or $b_{k}=1-a_{k}$. Without loss of generality, suppose $b_{k}=a_{k}-1$ for $k=1,2,3$. Again by the Pigeonhole Principle, as $a_{1}, a_{2}, a_{3}$ lies between 0 and 1 , the difference of two of them is at most $\frac{1}{2}$. By changing indices if necessary, we may assume $0 \leqslant d=a_{1}-a_{2} \leqslant \frac{1}{2}$.  By the inductive hypothesis, we can choose $x_{3}, x_{4}, \ldots, x_{n}$ such that $a^{\prime}=\sum_{k=3}^{n} x_{k} a_{k}$ and $b^{\prime}=\sum_{k=3}^{n} x_{k} b_{k}$ satisfy $\left|a^{\prime}\right|+\left|b^{\prime}\right| \leqslant 1$. We may further assume $a^{\prime} \geqslant 0$.  - Case 1. $b^{\prime} \geqslant 0$, in which case we take $\left(x_{1}, x_{2}\right)=(-1,1)$.  We have $\left|-a_{1}+a_{2}+a^{\prime}\right|+\left|-\left(a_{1}-1\right)+\left(a_{2}-1\right)+b^{\prime}\right|=\left|-d+a^{\prime}\right|+\left|-d+b^{\prime}\right| \leqslant$ $\max \left\{a^{\prime}+b^{\prime}-2 d, a^{\prime}-b^{\prime}, b^{\prime}-a^{\prime}, 2 d-a^{\prime}-b^{\prime}\right\} \leqslant 1$ since $0 \leqslant a^{\prime}, b^{\prime}, a^{\prime}+b^{\prime} \leqslant 1$ and $0 \leqslant d \leqslant \frac{1}{2}$.  - Case 2. $0>b^{\prime} \geqslant-a^{\prime}$, in which case we take $\left(x_{1}, x_{2}\right)=(-1,1)$.  We have $\left|-a_{1}+a_{2}+a^{\prime}\right|+\left|-\left(a_{1}-1\right)+\left(a_{2}-1\right)+b^{\prime}\right|=\left|-d+a^{\prime}\right|+\left|-d+b^{\prime}\right|$. If $-d+a^{\prime} \geqslant 0$, this equals $a^{\prime}-b^{\prime}=\left|a^{\prime}\right|+\left|b^{\prime}\right| \leqslant 1$. If $-d+a^{\prime}<0$, this equals $2 d-a^{\prime}-b^{\prime} \leqslant 2 d \leqslant 1$.  - Case 3. $b^{\prime}<-a^{\prime}$, in which case we take $\left(x_{1}, x_{2}\right)=(1,-1)$.  We have $\left|a_{1}-a_{2}+a^{\prime}\right|+\left|\left(a_{1}-1\right)-\left(a_{2}-1\right)+b^{\prime}\right|=\left|d+a^{\prime}\right|+\left|d+b^{\prime}\right|$. If $d+b^{\prime} \geqslant 0$, this equals $2 d+a^{\prime}+b^{\prime}<2 d \leqslant 1$. If $d+b^{\prime}<0$, this equals $a^{\prime}-b^{\prime}=\left|a^{\prime}\right|+\left|b^{\prime}\right| \leqslant 1$.  Therefore, we have found $x_{1}, x_{2}, \ldots, x_{n}$ satisfying (1) in each case. By induction, the property holds for all odd integers $n \geqslant 3$.


Denote by $\mathbb{R}^{+}$the set of all positive real numbers. Find all functions $f: \mathbb{R}^{+} \rightarrow \mathbb{R}^{+}$such that  $$ x f\left(x^{2}\right) f(f(y))+f(y f(x))=f(x y)\left(f\left(f\left(x^{2}\right)\right)+f\left(f\left(y^{2}\right)\right)\right) $$  for all positive real numbers $x$ and $y$. Answer. $f(x)=\frac{1}{x}$ for any $x \in \mathbb{R}^{+}$. ![](https://cdn.mathpix.com/cropped/2024_11_18_e5e735288fb13d37ab0cg-20.jpg?height=63&width=1663&top_left_y=565&top_left_x=177) hence $f(1)=1$. Swapping $x$ and $y$ in (1) and comparing with (1) again, we find  $$ x f\left(x^{2}\right) f(f(y))+f(y f(x))=y f\left(y^{2}\right) f(f(x))+f(x f(y)) . $$  Taking $y=1$ in (2), we have $x f\left(x^{2}\right)+f(f(x))=f(f(x))+f(x)$, that is,  $$ f\left(x^{2}\right)=\frac{f(x)}{x} $$  Take $y=1$ in (1) and apply (3) to $x f\left(x^{2}\right)$. We get $f(x)+f(f(x))=f(x)\left(f\left(f\left(x^{2}\right)\right)+1\right)$, which implies  $$ f\left(f\left(x^{2}\right)\right)=\frac{f(f(x))}{f(x)} $$  For any $x \in \mathbb{R}^{+}$, we find that  $$ f\left(f(x)^{2}\right) \stackrel{(3)}{=} \frac{f(f(x))}{f(x)} \stackrel{(4)}{=} f\left(f\left(x^{2}\right)\right) \stackrel{(3)}{=} f\left(\frac{f(x)}{x}\right) $$  It remains to show the following key step.  - Claim. The function $f$ is injective.  Proof. Using (3) and (4), we rewrite (1) as  $$ f(x) f(f(y))+f(y f(x))=f(x y)\left(\frac{f(f(x))}{f(x)}+\frac{f(f(y))}{f(y)}\right) . $$  Take $x=y$ in (6) and apply (3). This gives $f(x) f(f(x))+f(x f(x))=2 \frac{f(f(x))}{x}$, which means  $$ f(x f(x))=f(f(x))\left(\frac{2}{x}-f(x)\right) $$  Using (3), equation (2) can be rewritten as  $$ f(x) f(f(y))+f(y f(x))=f(y) f(f(x))+f(x f(y)) $$  Suppose $f(x)=f(y)$ for some $x, y \in \mathbb{R}^{+}$. Then (8) implies  $$ f(y f(y))=f(y f(x))=f(x f(y))=f(x f(x)) $$  Using (7), this gives  $$ f(f(y))\left(\frac{2}{y}-f(y)\right)=f(f(x))\left(\frac{2}{x}-f(x)\right) $$  Noting $f(x)=f(y)$, we find $x=y$. This establishes the injectivity.  By the Claim and (5), we get the only possible solution $f(x)=\frac{1}{x}$. It suffices to check that this is a solution. Indeed, the left-hand side of (1) becomes  $$ x \cdot \frac{1}{x^{2}} \cdot y+\frac{x}{y}=\frac{y}{x}+\frac{x}{y}, $$  while the right-hand side becomes  $$ \frac{1}{x y}\left(x^{2}+y^{2}\right)=\frac{x}{y}+\frac{y}{x} . $$  The two sides agree with each other.

Solution: Taking $x=y=1$ in (1), we get $f(1) f(f(1))+f(f(1))=2 f(1) f(f(1))$ and hence $f(1)=1$. Putting $x=1$ in (1), we have $f(f(y))+f(y)=f(y)\left(1+f\left(f\left(y^{2}\right)\right)\right)$ so that  $$ f(f(y))=f(y) f\left(f\left(y^{2}\right)\right) $$  Putting $y=1$ in (1), we get $x f\left(x^{2}\right)+f(f(x))=f(x)\left(f\left(f\left(x^{2}\right)\right)+1\right)$. Using (9), this gives  $$ x f\left(x^{2}\right)=f(x) $$  Replace $y$ by $\frac{1}{x}$ in (1). Then we have  $$ x f\left(x^{2}\right) f\left(f\left(\frac{1}{x}\right)\right)+f\left(\frac{f(x)}{x}\right)=f\left(f\left(x^{2}\right)\right)+f\left(f\left(\frac{1}{x^{2}}\right)\right) . $$  The relation (10) shows $f\left(\frac{f(x)}{x}\right)=f\left(f\left(x^{2}\right)\right)$. Also, using (9) with $y=\frac{1}{x}$ and using (10) again, the last equation reduces to  $$ f(x) f\left(\frac{1}{x}\right)=1 $$  Replace $x$ by $\frac{1}{x}$ and $y$ by $\frac{1}{y}$ in (1) and apply (11). We get  $$ \frac{1}{x f\left(x^{2}\right) f(f(y))}+\frac{1}{f(y f(x))}=\frac{1}{f(x y)}\left(\frac{1}{f\left(f\left(x^{2}\right)\right)}+\frac{1}{f\left(f\left(y^{2}\right)\right)}\right) . $$  Clearing denominators, we can use (1) to simplify the numerators and obtain  $$ f(x y)^{2} f\left(f\left(x^{2}\right)\right) f\left(f\left(y^{2}\right)\right)=x f\left(x^{2}\right) f(f(y)) f(y f(x)) $$  Using (9) and (10), this is the same as  $$ f(x y)^{2} f(f(x))=f(x)^{2} f(y) f(y f(x)) $$  Substitute $y=f(x)$ in (12) and apply (10) (with $x$ replaced by $f(x)$ ). We have  $$ f(x f(x))^{2}=f(x) f(f(x)) $$  Taking $y=x$ in (12), squaring both sides, and using (10) and (13), we find that  $$ f(f(x))=x^{4} f(x)^{3} $$  Finally, we combine (9), (10) and (14) to get  $$ y^{4} f(y)^{3} \stackrel{(14)}{=} f(f(y)) \stackrel{(9)}{=} f(y) f\left(f\left(y^{2}\right)\right) \stackrel{(14)}{=} f(y) y^{8} f\left(y^{2}\right)^{3} \stackrel{(10)}{=} y^{5} f(y)^{4}, $$  which implies $f(y)=\frac{1}{y}$. This is a solution by the checking in Solution 1.


The equation  $$ (x-1)(x-2) \cdots(x-2016)=(x-1)(x-2) \cdots(x-2016) $$  is written on the board. One tries to erase some linear factors from both sides so that each side still has at least one factor, and the resulting equation has no real roots. Find the least number of linear factors one needs to erase to achieve this. Answer. 2016.

Solution: Since there are 2016 common linear factors on both sides, we need to erase at least 2016 factors. We claim that the equation has no real roots if we erase all factors $(x-k)$ on the left-hand side with $k \equiv 2,3(\bmod 4)$, and all factors $(x-m)$ on the right-hand side with $m \equiv 0,1(\bmod 4)$. Therefore, it suffices to show that no real number $x$ satisfies  $$ \prod_{j=0}^{503}(x-4 j-1)(x-4 j-4)=\prod_{j=0}^{503}(x-4 j-2)(x-4 j-3) $$  - Case 1. $x=1,2, \ldots, 2016$.  In this case, one side of (1) is zero while the other side is not. This shows $x$ cannot satisfy (1).  - Case 2. $4 k+1<x<4 k+2$ or $4 k+3<x<4 k+4$ for some $k=0,1, \ldots, 503$.  For $j=0,1, \ldots, 503$ with $j \neq k$, the product $(x-4 j-1)(x-4 j-4)$ is positive. For $j=k$, the product $(x-4 k-1)(x-4 k-4)$ is negative. This shows the left-hand side of (1) is negative. On the other hand, each product $(x-4 j-2)(x-4 j-3)$ on the right-hand side of (1) is positive. This yields a contradiction.  - Case 3. $x<1$ or $x>2016$ or $4 k<x<4 k+1$ for some $k=1,2, \ldots, 503$.  The equation (1) can be rewritten as  $$ 1=\prod_{j=0}^{503} \frac{(x-4 j-1)(x-4 j-4)}{(x-4 j-2)(x-4 j-3)}=\prod_{j=0}^{503}\left(1-\frac{2}{(x-4 j-2)(x-4 j-3)}\right) $$  Note that $(x-4 j-2)(x-4 j-3)>2$ for $0 \leqslant j \leqslant 503$ in this case. So each term in the product lies strictly between 0 and 1 , and the whole product must be less than 1 , which is impossible.  - Case 4. $4 k+2<x<4 k+3$ for some $k=0,1, \ldots, 503$.  This time we rewrite (1) as  $$ \begin{aligned} 1 & =\frac{x-1}{x-2} \cdot \frac{x-2016}{x-2015} \prod_{j=1}^{503} \frac{(x-4 j)(x-4 j-1)}{(x-4 j+1)(x-4 j-2)} \\ & =\frac{x-1}{x-2} \cdot \frac{x-2016}{x-2015} \prod_{j=1}^{503}\left(1+\frac{2}{(x-4 j+1)(x-4 j-2)}\right) \end{aligned} $$  Clearly, $\frac{x-1}{x-2}$ and $\frac{x-2016}{x-2015}$ are both greater than 1 . For the range of $x$ in this case, each term in the product is also greater than 1 . Then the right-hand side must be greater than 1 and hence a contradiction arises.  From the four cases, we conclude that (1) has no real roots. Hence, the minimum number of linear factors to be erased is 2016.  Comment. We discuss the general case when 2016 is replaced by a positive integer $n$. The above solution works equally well when $n$ is divisible by 4 .  If $n \equiv 2(\bmod 4)$, one may leave $l(x)=(x-1)(x-2) \cdots\left(x-\frac{n}{2}\right)$ on the left-hand side and $r(x)=\left(x-\frac{n}{2}-1\right)\left(x-\frac{n}{2}-2\right) \cdots(x-n)$ on the right-hand side. One checks that for $x<\frac{n+1}{2}$, we have $|l(x)|<|r(x)|$, while for $x>\frac{n+1}{2}$, we have $|l(x)|>|r(x)|$.  If $n \equiv 3(\bmod 4)$, one may leave $l(x)=(x-1)(x-2) \cdots\left(x-\frac{n+1}{2}\right)$ on the left-hand side and $r(x)=\left(x-\frac{n+3}{2}\right)\left(x-\frac{x+5}{2}\right) \cdots(x-n)$ on the right-hand side. For $x<1$ or $\frac{n+1}{2}<x<\frac{n+3}{2}$, we have $l(x)>0>r(x)$. For $1<x<\frac{n+1}{2}$, we have $|l(x)|<|r(x)|$. For $x>\frac{n+3}{2}$, we have $|l(x)|>|r(x)|$.  If $n \equiv 1(\bmod 4)$, as the proposer mentioned, the situation is a bit more out of control. Since the construction for $n-1 \equiv 0(\bmod 4)$ works, the answer can be either $n$ or $n-1$. For $n=5$, we can leave the products $(x-1)(x-2)(x-3)(x-4)$ and $(x-5)$. For $n=9$, the only example that works is $l(x)=(x-1)(x-2)(x-9)$ and $r(x)=(x-3)(x-4) \cdots(x-8)$, while there seems to be no such partition for $n=13$.


Determine the largest real number $a$ such that for all $n \geqslant 1$ and for all real numbers $x_{0}, x_{1}, \ldots, x_{n}$ satisfying $0=x_{0}<x_{1}<x_{2}<\cdots<x_{n}$, we have  $$ \frac{1}{x_{1}-x_{0}}+\frac{1}{x_{2}-x_{1}}+\cdots+\frac{1}{x_{n}-x_{n-1}} \geqslant a\left(\frac{2}{x_{1}}+\frac{3}{x_{2}}+\cdots+\frac{n+1}{x_{n}}\right) . $$  Answer. The largest $a$ is $\frac{4}{9}$.

Solution: We first show that $a=\frac{4}{9}$ is admissible. For each $2 \leqslant k \leqslant n$, by the CauchySchwarz Inequality, we have  $$ \left(x_{k-1}+\left(x_{k}-x_{k-1}\right)\right)\left(\frac{(k-1)^{2}}{x_{k-1}}+\frac{3^{2}}{x_{k}-x_{k-1}}\right) \geqslant(k-1+3)^{2}, $$  which can be rewritten as  $$ \frac{9}{x_{k}-x_{k-1}} \geqslant \frac{(k+2)^{2}}{x_{k}}-\frac{(k-1)^{2}}{x_{k-1}} $$  Summing (2) over $k=2,3, \ldots, n$ and adding $\frac{9}{x_{1}}$ to both sides, we have  $$ 9 \sum_{k=1}^{n} \frac{1}{x_{k}-x_{k-1}} \geqslant 4 \sum_{k=1}^{n} \frac{k+1}{x_{k}}+\frac{n^{2}}{x_{n}}>4 \sum_{k=1}^{n} \frac{k+1}{x_{k}} . $$  This shows (1) holds for $a=\frac{4}{9}$. Next, we show that $a=\frac{4}{9}$ is the optimal choice. Consider the sequence defined by $x_{0}=0$ and $x_{k}=x_{k-1}+k(k+1)$ for $k \geqslant 1$, that is, $x_{k}=\frac{1}{3} k(k+1)(k+2)$. Then the left-hand side of (1) equals  $$ \sum_{k=1}^{n} \frac{1}{k(k+1)}=\sum_{k=1}^{n}\left(\frac{1}{k}-\frac{1}{k+1}\right)=1-\frac{1}{n+1} $$  while the right-hand side equals  $$ a \sum_{k=1}^{n} \frac{k+1}{x_{k}}=3 a \sum_{k=1}^{n} \frac{1}{k(k+2)}=\frac{3}{2} a \sum_{k=1}^{n}\left(\frac{1}{k}-\frac{1}{k+2}\right)=\frac{3}{2}\left(1+\frac{1}{2}-\frac{1}{n+1}-\frac{1}{n+2}\right) a . $$  When $n$ tends to infinity, the left-hand side tends to 1 while the right-hand side tends to $\frac{9}{4} a$. Therefore $a$ has to be at most $\frac{4}{9}$.  Hence the largest value of $a$ is $\frac{4}{9}$.


Solution: We shall give an alternative method to establish (1) with $a=\frac{4}{9}$. We define $y_{k}=x_{k}-x_{k-1}>0$ for $1 \leqslant k \leqslant n$. By the Cauchy-Schwarz Inequality, for $1 \leqslant k \leqslant n$, we have  $$ \left(y_{1}+y_{2}+\cdots+y_{k}\right)\left(\sum_{j=1}^{k} \frac{1}{y_{j}}\binom{j+1}{2}^{2}\right) \geqslant\left(\binom{2}{2}+\binom{3}{2}+\cdots+\binom{k+1}{2}\right)^{2}=\binom{k+2}{3}^{2} $$  This can be rewritten as  $$ \frac{k+1}{y_{1}+y_{2}+\cdots+y_{k}} \leqslant \frac{36}{k^{2}(k+1)(k+2)^{2}}\left(\sum_{j=1}^{k} \frac{1}{y_{j}}\binom{j+1}{2}^{2}\right) . $$  Summing (3) over $k=1,2, \ldots, n$, we get  $$ \frac{2}{y_{1}}+\frac{3}{y_{1}+y_{2}}+\cdots+\frac{n+1}{y_{1}+y_{2}+\cdots+y_{n}} \leqslant \frac{c_{1}}{y_{1}}+\frac{c_{2}}{y_{2}}+\cdots+\frac{c_{n}}{y_{n}} $$  where for $1 \leqslant m \leqslant n$,  $$ \begin{aligned} c_{m} & =36\binom{m+1}{2}^{2} \sum_{k=m}^{n} \frac{1}{k^{2}(k+1)(k+2)^{2}} \\ & =\frac{9 m^{2}(m+1)^{2}}{4} \sum_{k=m}^{n}\left(\frac{1}{k^{2}(k+1)^{2}}-\frac{1}{(k+1)^{2}(k+2)^{2}}\right) \\ & =\frac{9 m^{2}(m+1)^{2}}{4}\left(\frac{1}{m^{2}(m+1)^{2}}-\frac{1}{(n+1)^{2}(n+2)^{2}}\right)<\frac{9}{4} . \end{aligned} $$  From (4), the inequality (1) holds for $a=\frac{4}{9}$. This is also the upper bound as can be verified in the same way as Solution 1.


The leader of an IMO team chooses positive integers $n$ and $k$ with $n>k$, and announces them to the deputy leader and a contestant. The leader then secretly tells the deputy leader an $n$-digit binary string, and the deputy leader writes down all $n$-digit binary strings which differ from the leader's in exactly $k$ positions. (For example, if $n=3$ and $k=1$, and if the leader chooses 101, the deputy leader would write down 001, 111 and 100.) The contestant is allowed to look at the strings written by the deputy leader and guess the leader's string. What is the minimum number of guesses (in terms of $n$ and $k$ ) needed to guarantee the correct answer?  Answer. The minimum number of guesses is 2 if $n=2 k$ and 1 if $n \neq 2 k$.

Solution: Let $X$ be the binary string chosen by the leader and let $X^{\prime}$ be the binary string of length $n$ every digit of which is different from that of $X$. The strings written by the deputy leader are the same as those in the case when the leader's string is $X^{\prime}$ and $k$ is changed to $n-k$. In view of this, we may assume $k \geqslant \frac{n}{2}$. Also, for the particular case $k=\frac{n}{2}$, this argument shows that the strings $X$ and $X^{\prime}$ cannot be distinguished, and hence in that case the contestant has to guess at least twice.  It remains to show that the number of guesses claimed suffices. Consider any string $Y$ which differs from $X$ in $m$ digits where $0<m<2 k$. Without loss of generality, assume the first $m$ digits of $X$ and $Y$ are distinct. Let $Z$ be the binary string obtained from $X$ by changing its first $k$ digits. Then $Z$ is written by the deputy leader. Note that $Z$ differs from $Y$ by $|m-k|$ digits where $|m-k|<k$ since $0<m<2 k$. From this observation, the contestant must know that $Y$ is not the desired string.  As we have assumed $k \geqslant \frac{n}{2}$, when $n<2 k$, every string $Y \neq X$ differs from $X$ in fewer than $2 k$ digits. When $n=2 k$, every string except $X$ and $X^{\prime}$ differs from $X$ in fewer than $2 k$ digits. Hence, the answer is as claimed.


Solution: Firstly, assume $n \neq 2 k$. Without loss of generality suppose the first digit of the leader's string is 1 . Then among the $\binom{n}{k}$ strings written by the deputy leader, $\binom{n-1}{k}$ will begin with 1 and $\binom{n-1}{k-1}$ will begin with 0 . Since $n \neq 2 k$, we have $k+(k-1) \neq n-1$ and so $\binom{n-1}{k} \neq\binom{ n-1}{k-1}$. Thus, by counting the number of strings written by the deputy leader that start with 0 and 1 , the contestant can tell the first digit of the leader's string. The same can be done on the other digits, so 1 guess suffices when $n \neq 2 k$.  Secondly, for the case $n=2$ and $k=1$, the answer is clearly 2 . For the remaining cases where $n=2 k>2$, the deputy leader would write down the same strings if the leader's string $X$ is replaced by $X^{\prime}$ obtained by changing each digit of $X$. This shows at least 2 guesses are needed. We shall show that 2 guesses suffice in this case. Suppose the first two digits of the leader's string are the same. Then among the strings written by the deputy leader, the prefices 01 and 10 will occur $\binom{2 k-2}{k-1}$ times each, while the prefices 00 and 11 will occur $\binom{2 k-2}{k}$ times each. The two numbers are interchanged if the first two digits of the leader's string are different. Since $\binom{2 k-2}{k-1} \neq\binom{ 2 k-2}{k}$, the contestant can tell whether the first two digits of the leader's string are the same or not. He can work out the relation of the first digit and the other digits in the same way and reduce the leader's string to only 2 possibilities. The proof is complete.


Find all positive integers $n$ for which all positive divisors of $n$ can be put into the cells of a rectangular table under the following constraints:  - each cell contains a distinct divisor; - the sums of all rows are equal; and - the sums of all columns are equal.  Answer. 1.

Solution: Suppose all positive divisors of $n$ can be arranged into a rectangular table of size $k \times l$ where the number of rows $k$ does not exceed the number of columns $l$. Let the sum of numbers in each column be $s$. Since $n$ belongs to one of the columns, we have $s \geqslant n$, where equality holds only when $n=1$.  For $j=1,2, \ldots, l$, let $d_{j}$ be the largest number in the $j$-th column. Without loss of generality, assume $d_{1}>d_{2}>\cdots>d_{l}$. Since these are divisors of $n$, we have  $$ d_{l} \leqslant \frac{n}{l} $$  As $d_{l}$ is the maximum entry of the $l$-th column, we must have  $$ d_{l} \geqslant \frac{s}{k} \geqslant \frac{n}{k} . $$  The relations (1) and (2) combine to give $\frac{n}{l} \geqslant \frac{n}{k}$, that is, $k \geqslant l$. Together with $k \leqslant l$, we conclude that $k=l$. Then all inequalities in (1) and (2) are equalities. In particular, $s=n$ and so $n=1$, in which case the conditions are clearly satisfied.


Solution: Clearly $n=1$ works. Then we assume $n>1$ and let its prime factorization be $n=p_{1}^{r_{1}} p_{2}^{r_{2}} \cdots p_{t}^{r_{t}}$. Suppose the table has $k$ rows and $l$ columns with $1<k \leqslant l$. Note that $k l$ is the number of positive divisors of $n$ and the sum of all entries is the sum of positive divisors of $n$, which we denote by $\sigma(n)$. Consider the column containing $n$. Since the column sum is $\frac{\sigma(n)}{l}$, we must have $\frac{\sigma(n)}{l}>n$. Therefore, we have  $$ \begin{aligned} \left(r_{1}+1\right)\left(r_{2}+1\right) \cdots\left(r_{t}+1\right) & =k l \leqslant l^{2}<\left(\frac{\sigma(n)}{n}\right)^{2} \\ & =\left(1+\frac{1}{p_{1}}+\cdots+\frac{1}{p_{1}^{r_{1}}}\right)^{2} \cdots\left(1+\frac{1}{p_{t}}+\cdots+\frac{1}{p_{t}^{r_{t}}}\right)^{2} \end{aligned} $$  This can be rewritten as  $$ f\left(p_{1}, r_{1}\right) f\left(p_{2}, r_{2}\right) \cdots f\left(p_{t}, r_{t}\right)<1 $$  where  $$ f(p, r)=\frac{r+1}{\left(1+\frac{1}{p}+\cdots+\frac{1}{p^{r}}\right)^{2}}=\frac{(r+1)\left(1-\frac{1}{p}\right)^{2}}{\left(1-\frac{1}{p^{r+1}}\right)^{2}} $$  Direct computation yields  $$ f(2,1)=\frac{8}{9}, \quad f(2,2)=\frac{48}{49}, \quad f(3,1)=\frac{9}{8} . $$  Also, we find that  $$ \begin{aligned} & f(2, r) \geqslant\left(1-\frac{1}{2^{r+1}}\right)^{-2}>1 \quad \text { for } r \geqslant 3 \\ & f(3, r) \geqslant \frac{4}{3}\left(1-\frac{1}{3^{r+1}}\right)^{-2}>\frac{4}{3}>\frac{9}{8} \quad \text { for } r \geqslant 2, \text { and } \\ & f(p, r) \geqslant \frac{32}{25}\left(1-\frac{1}{p^{r+1}}\right)^{-2}>\frac{32}{25}>\frac{9}{8} \quad \text { for } p \geqslant 5 \end{aligned} $$  From these values and bounds, it is clear that (3) holds only when $n=2$ or 4 . In both cases, it is easy to see that the conditions are not satisfied. Hence, the only possible $n$ is 1 .


Let $n$ be a positive integer relatively prime to 6 . We paint the vertices of a regular $n$-gon with three colours so that there is an odd number of vertices of each colour. Show that there exists an isosceles triangle whose three vertices are of different colours.

Solution: For $k=1,2,3$, let $a_{k}$ be the number of isosceles triangles whose vertices contain exactly $k$ colours. Suppose on the contrary that $a_{3}=0$. Let $b, c, d$ be the number of vertices of the three different colours respectively. We now count the number of pairs $(\triangle, E)$ where $\triangle$ is an isosceles triangle and $E$ is a side of $\triangle$ whose endpoints are of different colours.  On the one hand, since we have assumed $a_{3}=0$, each triangle in the pair must contain exactly two colours, and hence each triangle contributes twice. Thus the number of pairs is $2 a_{2}$ 。  On the other hand, if we pick any two vertices $A, B$ of distinct colours, then there are three isosceles triangles having these as vertices, two when $A B$ is not the base and one when $A B$ is the base since $n$ is odd. Note that the three triangles are all distinct as $(n, 3)=1$. In this way, we count the number of pairs to be $3(b c+c d+d b)$. However, note that $2 a_{2}$ is even while $3(b c+c d+d b)$ is odd, as each of $b, c, d$ is. This yields a contradiction and hence $a_{3} \geqslant 1$.  Comment. A slightly stronger version of this problem is to replace the condition $(n, 6)=1$ by $n$ being odd (where equilateral triangles are regarded as isosceles triangles). In that case, the only difference in the proof is that by fixing any two vertices $A, B$, one can find exactly one or three isosceles triangles having these as vertices. But since only parity is concerned in the solution, the proof goes the same way.  The condition that there is an odd number of vertices of each colour is necessary, as can be seen from the following example. Consider $n=25$ and we label the vertices $A_{0}, A_{1}, \ldots, A_{24}$. Suppose colour 1 is used for $A_{0}$, colour 2 is used for $A_{5}, A_{10}, A_{15}, A_{20}$, while colour 3 is used for the remaining vertices. Then any isosceles triangle having colours 1 and 2 must contain $A_{0}$ and one of $A_{5}, A_{10}, A_{15}, A_{20}$. Clearly, the third vertex must have index which is a multiple of 5 so it is not of colour 3 .


Find all positive integers $n$ for which we can fill in the entries of an $n \times n$ table with the following properties:  - each entry can be one of $I, M$ and $O$; - in each row and each column, the letters $I, M$ and $O$ occur the same number of times; and - in any diagonal whose number of entries is a multiple of three, the letters $I, M$ and $O$ occur the same number of times.  Answer. $n$ can be any multiple of 9 .

Solution: We first show that such a table exists when $n$ is a multiple of 9 . Consider the following $9 \times 9$ table.  $$ \left(\begin{array}{ccccccccc} I & I & I & M & M & M & O & O & O \\ M & M & M & O & O & O & I & I & I \\ O & O & O & I & I & I & M & M & M \\ I & I & I & M & M & M & O & O & O \\ M & M & M & O & O & O & I & I & I \\ O & O & O & I & I & I & M & M & M \\ I & I & I & M & M & M & O & O & O \\ M & M & M & O & O & O & I & I & I \\ O & O & O & I & I & I & M & M & M \end{array}\right) $$  It is a direct checking that the table (1) satisfies the requirements. For $n=9 k$ where $k$ is a positive integer, we form an $n \times n$ table using $k \times k$ copies of (1). For each row and each column of the table of size $n$, since there are three $I$ 's, three $M$ 's and three $O$ 's for any nine consecutive entries, the numbers of $I, M$ and $O$ are equal. In addition, every diagonal of the large table whose number of entries is divisible by 3 intersects each copy of (1) at a diagonal with number of entries divisible by 3 (possibly zero). Therefore, every such diagonal also contains the same number of $I, M$ and $O$.  Next, consider any $n \times n$ table for which the requirements can be met. As the number of entries of each row should be a multiple of 3 , we let $n=3 k$ where $k$ is a positive integer. We divide the whole table into $k \times k$ copies of $3 \times 3$ blocks. We call the entry at the centre of such a $3 \times 3$ square a vital entry. We also call any row, column or diagonal that contains at least one vital entry a vital line. We compute the number of pairs $(l, c)$ where $l$ is a vital line and $c$ is an entry belonging to $l$ that contains the letter $M$. We let this number be $N$.  On the one hand, since each vital line contains the same number of $I, M$ and $O$, it is obvious that each vital row and each vital column contain $k$ occurrences of $M$. For vital diagonals in either direction, we count there are exactly  $$ 1+2+\cdots+(k-1)+k+(k-1)+\cdots+2+1=k^{2} $$  occurrences of $M$. Therefore, we have $N=4 k^{2}$.  On the other hand, there are $3 k^{2}$ occurrences of $M$ in the whole table. Note that each entry belongs to exactly 1 or 4 vital lines. Therefore, $N$ must be congruent to $3 k^{2} \bmod 3$.  From the double counting, we get $4 k^{2} \equiv 3 k^{2}(\bmod 3)$, which forces $k$ to be a multiple of 3. Therefore, $n$ has to be a multiple of 9 and the proof is complete.


Let $n \geqslant 3$ be a positive integer. Find the maximum number of diagonals of a regular $n$-gon one can select, so that any two of them do not intersect in the interior or they are perpendicular to each other.  Answer. $n-2$ if $n$ is even and $n-3$ if $n$ is odd.

Solution: We consider two cases according to the parity of $n$.  - Case 1. $n$ is odd.  We first claim that no pair of diagonals is perpendicular. Suppose $A, B, C, D$ are vertices where $A B$ and $C D$ are perpendicular, and let $E$ be the vertex lying on the perpendicular bisector of $A B$. Let $E^{\prime}$ be the opposite point of $E$ on the circumcircle of the regular polygon. Since $E C=E^{\prime} D$ and $C, D, E$ are vertices of the regular polygon, $E^{\prime}$ should also belong to the polygon. This contradicts the fact that a regular polygon with an odd number of vertices does not contain opposite points on the circumcircle. ![](https://cdn.mathpix.com/cropped/2024_11_18_e5e735288fb13d37ab0cg-39.jpg?height=513&width=1250&top_left_y=1061&top_left_x=493)  Therefore in the odd case we can only select diagonals which do not intersect. In the maximal case these diagonals should divide the regular $n$-gon into $n-2$ triangles, so we can select at most $n-3$ diagonals. This can be done, for example, by selecting all diagonals emanated from a particular vertex.  - Case 2. $n$ is even.  If there is no intersection, then the proof in the odd case works. Suppose there are two perpendicular diagonals selected. We consider the set $S$ of all selected diagonals parallel to one of them which intersect with some selected diagonals. Suppose $S$ contains $k$ diagonals and the number of distinct endpoints of the $k$ diagonals is $l$.  Firstly, consider the longest diagonal in one of the two directions in $S$. No other diagonal in $S$ can start from either endpoint of that diagonal, since otherwise it has to meet another longer diagonal in $S$. The same holds true for the other direction. Ignoring these two longest diagonals and their four endpoints, the remaining $k-2$ diagonals share $l-4$ endpoints where each endpoint can belong to at most two diagonals. This gives $2(l-4) \geqslant 2(k-2)$, so that $k \leqslant l-2$. ![](https://cdn.mathpix.com/cropped/2024_11_18_e5e735288fb13d37ab0cg-40.jpg?height=430&width=1442&top_left_y=289&top_left_x=294)  Consider a group of consecutive vertices of the regular $n$-gon so that each of the two outermost vertices is an endpoint of a diagonal in $S$, while the interior points are not. There are $l$ such groups. We label these groups $P_{1}, P_{2}, \ldots, P_{l}$ in this order. We claim that each selected diagonal outside $S$ must connect vertices of the same group $P_{i}$. Consider any diagonal $d$ joining vertices from distinct groups $P_{i}$ and $P_{j}$. Let $d_{1}$ and $d_{2}$ be two diagonals in $S$ each having one of the outermost points of $P_{i}$ as endpoint. Then $d$ must meet either $d_{1}, d_{2}$ or a diagonal in $S$ which is perpendicular to both $d_{1}$ and $d_{2}$. In any case $d$ should belong to $S$ by definition, which is a contradiction.  Within the same group $P_{i}$, there are no perpendicular diagonals since the vertices belong to the same side of a diameter of the circumcircle. Hence there can be at most $\left|P_{i}\right|-2$ selected diagonals within $P_{i}$, including the one joining the two outermost points of $P_{i}$ when $\left|P_{i}\right|>2$. Therefore, the maximum number of diagonals selected is  $$ \sum_{i=1}^{l}\left(\left|P_{i}\right|-2\right)+k=\sum_{i=1}^{l}\left|P_{i}\right|-2 l+k=(n+l)-2 l+k=n-l+k \leqslant n-2 $$  This upper bound can be attained as follows. We take any vertex $A$ and let $A^{\prime}$ be the vertex for which $A A^{\prime}$ is a diameter of the circumcircle. If we select all diagonals emanated from $A$ together with the diagonal $d^{\prime}$ joining the two neighbouring vertices of $A^{\prime}$, then the only pair of diagonals that meet each other is $A A^{\prime}$ and $d^{\prime}$, which are perpendicular to each other. In total we can take $n-2$ diagonals. ![](https://cdn.mathpix.com/cropped/2024_11_18_e5e735288fb13d37ab0cg-40.jpg?height=459&width=421&top_left_y=1798&top_left_x=803)


Solution: The constructions and the odd case are the same as   The base case $n=3$ is trivial since there is no diagonal at all. Suppose the upper bound holds for any cyclic polygon with fewer than $n$ sides. For a cyclic $n$-gon, if there is a selected diagonal which does not intersect any other selected diagonal, then this diagonal divides the $n$-gon into an $m$-gon and an $l$-gon (with $m+l=n+2$ ) so that each selected diagonal belongs to one of them. Without loss of generality, we may assume the $m$-gon lies on the same side of a diameter of $\Gamma$. Then no two selected diagonals of the $m$-gon can intersect, and hence we can select at most $m-3$ diagonals. Also, we can apply the inductive hypothesis to the $l$-gon. This shows the maximum number of selected diagonals is $(m-3)+(l-2)+1=n-2$.  It remains to consider the case when all selected diagonals meet at least one other selected diagonal. Consider a pair of selected perpendicular diagonals $d_{1}, d_{2}$. They divide the circumference of $\Gamma$ into four arcs, each of which lies on the same side of a diameter of $\Gamma$. If there are two selected diagonals intersecting each other and neither is parallel to $d_{1}$ or $d_{2}$, then their endpoints must belong to the same arc determined by $d_{1}, d_{2}$, and hence they cannot be perpendicular. This violates the condition, and hence all selected diagonals must have the same direction as one of $d_{1}, d_{2}$. ![](https://cdn.mathpix.com/cropped/2024_11_18_e5e735288fb13d37ab0cg-41.jpg?height=435&width=1234&top_left_y=1036&top_left_x=494)  Take the longest selected diagonal in one of the two directions. We argue as in Solution 1 that its endpoints do not belong to any other selected diagonal. The same holds true for the longest diagonal in the other direction. Apart from these four endpoints, each of the remaining $n-4$ vertices can belong to at most two selected diagonals. Thus we can select at most $\frac{1}{2}(2(n-4)+4)=n-2$ diagonals. Then the proof follows by induction.


Let $n$ be a positive integer. Determine the smallest positive integer $k$ with the following property: it is possible to mark $k$ cells on a $2 n \times 2 n$ board so that there exists a unique partition of the board into $1 \times 2$ and $2 \times 1$ dominoes, none of which contains two marked cells.  Answer. $2 n$.

Solution: We first construct an example of marking $2 n$ cells satisfying the requirement. Label the rows and columns $1,2, \ldots, 2 n$ and label the cell in the $i$-th row and the $j$-th column $(i, j)$.  For $i=1,2, \ldots, n$, we mark the cells $(i, i)$ and $(i, i+1)$. We claim that the required partition exists and is unique. The two diagonals of the board divide the board into four regions. Note that the domino covering cell $(1,1)$ must be vertical. This in turn shows that each domino covering $(2,2),(3,3), \ldots,(n, n)$ is vertical. By induction, the dominoes in the left region are all vertical. By rotational symmetry, the dominoes in the bottom region are horizontal, and so on. This shows that the partition exists and is unique. ![](https://cdn.mathpix.com/cropped/2024_11_18_e5e735288fb13d37ab0cg-45.jpg?height=323&width=820&top_left_y=1080&top_left_x=696)  It remains to show that this value of $k$ is the smallest possible. Assume that only $k<2 n$ cells are marked, and there exists a partition $P$ satisfying the requirement. It suffices to show there exists another desirable partition distinct from $P$. Let $d$ be the main diagonal of the board.  Construct the following graph with edges of two colours. Its vertices are the cells of the board. Connect two vertices with a red edge if they belong to the same domino of $P$. Connect two vertices with a blue edge if their reflections in $d$ are connected by a red edge. It is possible that two vertices are connected by edges of both colours. Clearly, each vertex has both red and blue degrees equal to 1 . Thus the graph splits into cycles where the colours of edges in each cycle alternate (a cycle may have length 2).  Consider any cell $c$ lying on the diagonal $d$. Its two edges are symmetrical with respect to $d$. Thus they connect $c$ to different cells. This shows $c$ belongs to a cycle $C(c)$ of length at least 4. Consider a part of this cycle $c_{0}, c_{1}, \ldots, c_{m}$ where $c_{0}=c$ and $m$ is the least positive integer such that $c_{m}$ lies on $d$. Clearly, $c_{m}$ is distinct from $c$. From the construction, the path symmetrical to this with respect to $d$ also lies in the graph, and so these paths together form $C(c)$. Hence, $C(c)$ contains exactly two cells from $d$. Then all $2 n$ cells in $d$ belong to $n$ cycles $C_{1}, C_{2}, \ldots, C_{n}$, each has length at least 4.  By the Pigeonhole Principle, there exists a cycle $C_{i}$ containing at most one of the $k$ marked cells. We modify $P$ as follows. We remove all dominoes containing the vertices of $C_{i}$, which correspond to the red edges of $C_{i}$. Then we put the dominoes corresponding to the blue edges of $C_{i}$. Since $C_{i}$ has at least 4 vertices, the resultant partition $P^{\prime}$ is different from $P$. Clearly, no domino in $P^{\prime}$ contains two marked cells as $C_{i}$ contains at most one marked cell. This shows the partition is not unique and hence $k$ cannot be less than $2 n$.