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Let $A B C$ be a triangle with $C A \neq C B$. Let $D, F$, and $G$ be the midpoints of the sides $A B, A C$, and $B C$, respectively. A circle $\Gamma$ passing through $C$ and tangent to $A B$ at $D$ meets the segments $A F$ and $B G$ at $H$ and $I$, respectively. The points $H^{\prime}$ and $I^{\prime}$ are symmetric to $H$ and $I$ about $F$ and $G$, respectively. The line $H^{\prime} I^{\prime}$ meets $C D$ and $F G$ at $Q$ and $M$, respectively. The line $C M$ meets $\Gamma$ again at $P$. Prove that $C Q=Q P$. (El Salvador)
|
We may assume that $C A>C B$. Observe that $H^{\prime}$ and $I^{\prime}$ lie inside the segments $C F$ and $C G$, respectively. Therefore, $M$ lies outside $\triangle A B C$ (see Figure 1). Due to the powers of points $A$ and $B$ with respect to the circle $\Gamma$, we have $$ C H^{\prime} \cdot C A=A H \cdot A C=A D^{2}=B D^{2}=B I \cdot B C=C I^{\prime} \cdot C B $$ Therefore, $C H^{\prime} \cdot C F=C I^{\prime} \cdot C G$. Hence, the quadrilateral $H^{\prime} I^{\prime} G F$ is cyclic, and so $\angle I^{\prime} H^{\prime} C=$ $\angle C G F$. Let $D F$ and $D G$ meet $\Gamma$ again at $R$ and $S$, respectively. We claim that the points $R$ and $S$ lie on the line $H^{\prime} I^{\prime}$. Observe that $F H^{\prime} \cdot F A=F H \cdot F C=F R \cdot F D$. Thus, the quadrilateral $A D H^{\prime} R$ is cyclic, and hence $\angle R H^{\prime} F=\angle F D A=\angle C G F=\angle I^{\prime} H^{\prime} C$. Therefore, the points $R, H^{\prime}$, and $I^{\prime}$ are collinear. Similarly, the points $S, H^{\prime}$, and $I^{\prime}$ are also collinear, and so all the points $R, H^{\prime}, Q, I^{\prime}, S$, and $M$ are all collinear.  Figure 1  Figure 2 Then, $\angle R S D=\angle R D A=\angle D F G$. Hence, the quadrilateral $R S G F$ is cyclic (see Figure 2). Therefore, $M H^{\prime} \cdot M I^{\prime}=M F \cdot M G=M R \cdot M S=M P \cdot M C$. Thus, the quadrilateral $C P I^{\prime} H^{\prime}$ is also cyclic. Let $\omega$ be its circumcircle. Notice that $\angle H^{\prime} C Q=\angle S D C=\angle S R C$ and $\angle Q C I^{\prime}=\angle C D R=\angle C S R$. Hence, $\triangle C H^{\prime} Q \sim \triangle R C Q$ and $\triangle C I^{\prime} Q \sim \triangle S C Q$, and therefore $Q H^{\prime} \cdot Q R=Q C^{2}=Q I^{\prime} \cdot Q S$. We apply the inversion with center $Q$ and radius $Q C$. Observe that the points $R, C$, and $S$ are mapped to $H^{\prime}, C$, and $I^{\prime}$, respectively. Therefore, the circumcircle $\Gamma$ of $\triangle R C S$ is mapped to the circumcircle $\omega$ of $\triangle H^{\prime} C I^{\prime}$. Since $P$ and $C$ belong to both circles and the point $C$ is preserved by the inversion, we have that $P$ is also mapped to itself. We then get $Q P^{2}=Q C^{2}$. Hence, $Q P=Q C$. Comment 1. The problem statement still holds when $\Gamma$ intersects the sides $C A$ and $C B$ outside segments $A F$ and $B G$, respectively.
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
Let $A B C$ be a triangle with $C A \neq C B$. Let $D, F$, and $G$ be the midpoints of the sides $A B, A C$, and $B C$, respectively. A circle $\Gamma$ passing through $C$ and tangent to $A B$ at $D$ meets the segments $A F$ and $B G$ at $H$ and $I$, respectively. The points $H^{\prime}$ and $I^{\prime}$ are symmetric to $H$ and $I$ about $F$ and $G$, respectively. The line $H^{\prime} I^{\prime}$ meets $C D$ and $F G$ at $Q$ and $M$, respectively. The line $C M$ meets $\Gamma$ again at $P$. Prove that $C Q=Q P$. (El Salvador)
|
We may assume that $C A>C B$. Observe that $H^{\prime}$ and $I^{\prime}$ lie inside the segments $C F$ and $C G$, respectively. Therefore, $M$ lies outside $\triangle A B C$ (see Figure 1). Due to the powers of points $A$ and $B$ with respect to the circle $\Gamma$, we have $$ C H^{\prime} \cdot C A=A H \cdot A C=A D^{2}=B D^{2}=B I \cdot B C=C I^{\prime} \cdot C B $$ Therefore, $C H^{\prime} \cdot C F=C I^{\prime} \cdot C G$. Hence, the quadrilateral $H^{\prime} I^{\prime} G F$ is cyclic, and so $\angle I^{\prime} H^{\prime} C=$ $\angle C G F$. Let $D F$ and $D G$ meet $\Gamma$ again at $R$ and $S$, respectively. We claim that the points $R$ and $S$ lie on the line $H^{\prime} I^{\prime}$. Observe that $F H^{\prime} \cdot F A=F H \cdot F C=F R \cdot F D$. Thus, the quadrilateral $A D H^{\prime} R$ is cyclic, and hence $\angle R H^{\prime} F=\angle F D A=\angle C G F=\angle I^{\prime} H^{\prime} C$. Therefore, the points $R, H^{\prime}$, and $I^{\prime}$ are collinear. Similarly, the points $S, H^{\prime}$, and $I^{\prime}$ are also collinear, and so all the points $R, H^{\prime}, Q, I^{\prime}, S$, and $M$ are all collinear.  Figure 1  Figure 2 Then, $\angle R S D=\angle R D A=\angle D F G$. Hence, the quadrilateral $R S G F$ is cyclic (see Figure 2). Therefore, $M H^{\prime} \cdot M I^{\prime}=M F \cdot M G=M R \cdot M S=M P \cdot M C$. Thus, the quadrilateral $C P I^{\prime} H^{\prime}$ is also cyclic. Let $\omega$ be its circumcircle. Notice that $\angle H^{\prime} C Q=\angle S D C=\angle S R C$ and $\angle Q C I^{\prime}=\angle C D R=\angle C S R$. Hence, $\triangle C H^{\prime} Q \sim \triangle R C Q$ and $\triangle C I^{\prime} Q \sim \triangle S C Q$, and therefore $Q H^{\prime} \cdot Q R=Q C^{2}=Q I^{\prime} \cdot Q S$. We apply the inversion with center $Q$ and radius $Q C$. Observe that the points $R, C$, and $S$ are mapped to $H^{\prime}, C$, and $I^{\prime}$, respectively. Therefore, the circumcircle $\Gamma$ of $\triangle R C S$ is mapped to the circumcircle $\omega$ of $\triangle H^{\prime} C I^{\prime}$. Since $P$ and $C$ belong to both circles and the point $C$ is preserved by the inversion, we have that $P$ is also mapped to itself. We then get $Q P^{2}=Q C^{2}$. Hence, $Q P=Q C$. Comment 1. The problem statement still holds when $\Gamma$ intersects the sides $C A$ and $C B$ outside segments $A F$ and $B G$, respectively.
|
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324b8872-1804-57c3-8ef3-3e944754ae99
| 24,620
|
Let $A B C$ be a triangle with $C A \neq C B$. Let $D, F$, and $G$ be the midpoints of the sides $A B, A C$, and $B C$, respectively. A circle $\Gamma$ passing through $C$ and tangent to $A B$ at $D$ meets the segments $A F$ and $B G$ at $H$ and $I$, respectively. The points $H^{\prime}$ and $I^{\prime}$ are symmetric to $H$ and $I$ about $F$ and $G$, respectively. The line $H^{\prime} I^{\prime}$ meets $C D$ and $F G$ at $Q$ and $M$, respectively. The line $C M$ meets $\Gamma$ again at $P$. Prove that $C Q=Q P$. (El Salvador)
|
Let $X=H I \cap A B$, and let the tangent to $\Gamma$ at $C$ meet $A B$ at $Y$. Let $X C$ meet $\Gamma$ again at $X^{\prime}$ (see Figure 3). Projecting from $C, X$, and $C$ again, we have $(X, A ; D, B)=$ $\left(X^{\prime}, H ; D, I\right)=(C, I ; D, H)=(Y, B ; D, A)$. Since $A$ and $B$ are symmetric about $D$, it follows that $X$ and $Y$ are also symmetric about $D$. Now, Menelaus' theorem applied to $\triangle A B C$ with the line $H I X$ yields $$ 1=\frac{C H}{H A} \cdot \frac{B I}{I C} \cdot \frac{A X}{X B}=\frac{A H^{\prime}}{H^{\prime} C} \cdot \frac{C I^{\prime}}{I^{\prime} B} \cdot \frac{B Y}{Y A} . $$ By the converse of Menelaus' theorem applied to $\triangle A B C$ with points $H^{\prime}, I^{\prime}, Y$, we get that the points $H^{\prime}, I^{\prime}, Y$ are collinear.  Figure 3 Let $T$ be the midpoint of $C D$, and let $O$ be the center of $\Gamma$. Let $C M$ meet $T Y$ at $N$. To avoid confusion, we clean some superfluous details out of the picture (see Figure 4). Let $V=M T \cap C Y$. Since $M T \| Y D$ and $D T=T C$, we get $C V=V Y$. Then CevA's theorem applied to $\triangle C T Y$ with the point $M$ yields $$ 1=\frac{T Q}{Q C} \cdot \frac{C V}{V Y} \cdot \frac{Y N}{N T}=\frac{T Q}{Q C} \cdot \frac{Y N}{N T} $$ Therefore, $\frac{T Q}{Q C}=\frac{T N}{N Y}$. So, $N Q \| C Y$, and thus $N Q \perp O C$. Note that the points $O, N, T$, and $Y$ are collinear. Therefore, $C Q \perp O N$. So, $Q$ is the orthocenter of $\triangle O C N$, and hence $O Q \perp C P$. Thus, $Q$ lies on the perpendicular bisector of $C P$, and therefore $C Q=Q P$, as required.  Figure 4 Comment 2. The second part of Solution 2 provides a proof of the following more general statement, which does not involve a specific choice of $Q$ on $C D$. Let $Y C$ and $Y D$ be two tangents to a circle $\Gamma$ with center $O$ (see Figure 4). Let $\ell$ be the midline of $\triangle Y C D$ parallel to $Y D$. Let $Q$ and $M$ be two points on $C D$ and $\ell$, respectively, such that the line $Q M$ passes through $Y$. Then $O Q \perp C M$.
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
Let $A B C$ be a triangle with $C A \neq C B$. Let $D, F$, and $G$ be the midpoints of the sides $A B, A C$, and $B C$, respectively. A circle $\Gamma$ passing through $C$ and tangent to $A B$ at $D$ meets the segments $A F$ and $B G$ at $H$ and $I$, respectively. The points $H^{\prime}$ and $I^{\prime}$ are symmetric to $H$ and $I$ about $F$ and $G$, respectively. The line $H^{\prime} I^{\prime}$ meets $C D$ and $F G$ at $Q$ and $M$, respectively. The line $C M$ meets $\Gamma$ again at $P$. Prove that $C Q=Q P$. (El Salvador)
|
Let $X=H I \cap A B$, and let the tangent to $\Gamma$ at $C$ meet $A B$ at $Y$. Let $X C$ meet $\Gamma$ again at $X^{\prime}$ (see Figure 3). Projecting from $C, X$, and $C$ again, we have $(X, A ; D, B)=$ $\left(X^{\prime}, H ; D, I\right)=(C, I ; D, H)=(Y, B ; D, A)$. Since $A$ and $B$ are symmetric about $D$, it follows that $X$ and $Y$ are also symmetric about $D$. Now, Menelaus' theorem applied to $\triangle A B C$ with the line $H I X$ yields $$ 1=\frac{C H}{H A} \cdot \frac{B I}{I C} \cdot \frac{A X}{X B}=\frac{A H^{\prime}}{H^{\prime} C} \cdot \frac{C I^{\prime}}{I^{\prime} B} \cdot \frac{B Y}{Y A} . $$ By the converse of Menelaus' theorem applied to $\triangle A B C$ with points $H^{\prime}, I^{\prime}, Y$, we get that the points $H^{\prime}, I^{\prime}, Y$ are collinear.  Figure 3 Let $T$ be the midpoint of $C D$, and let $O$ be the center of $\Gamma$. Let $C M$ meet $T Y$ at $N$. To avoid confusion, we clean some superfluous details out of the picture (see Figure 4). Let $V=M T \cap C Y$. Since $M T \| Y D$ and $D T=T C$, we get $C V=V Y$. Then CevA's theorem applied to $\triangle C T Y$ with the point $M$ yields $$ 1=\frac{T Q}{Q C} \cdot \frac{C V}{V Y} \cdot \frac{Y N}{N T}=\frac{T Q}{Q C} \cdot \frac{Y N}{N T} $$ Therefore, $\frac{T Q}{Q C}=\frac{T N}{N Y}$. So, $N Q \| C Y$, and thus $N Q \perp O C$. Note that the points $O, N, T$, and $Y$ are collinear. Therefore, $C Q \perp O N$. So, $Q$ is the orthocenter of $\triangle O C N$, and hence $O Q \perp C P$. Thus, $Q$ lies on the perpendicular bisector of $C P$, and therefore $C Q=Q P$, as required.  Figure 4 Comment 2. The second part of Solution 2 provides a proof of the following more general statement, which does not involve a specific choice of $Q$ on $C D$. Let $Y C$ and $Y D$ be two tangents to a circle $\Gamma$ with center $O$ (see Figure 4). Let $\ell$ be the midline of $\triangle Y C D$ parallel to $Y D$. Let $Q$ and $M$ be two points on $C D$ and $\ell$, respectively, such that the line $Q M$ passes through $Y$. Then $O Q \perp C M$.
|
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324b8872-1804-57c3-8ef3-3e944754ae99
| 24,620
|
Let $A B C$ be an acute triangle with $A B>A C$, and let $\Gamma$ be its circumcircle. Let $H$, $M$, and $F$ be the orthocenter of the triangle, the midpoint of $B C$, and the foot of the altitude from $A$, respectively. Let $Q$ and $K$ be the two points on $\Gamma$ that satisfy $\angle A Q H=90^{\circ}$ and $\angle Q K H=90^{\circ}$. Prove that the circumcircles of the triangles $K Q H$ and $K F M$ are tangent to each other. (Ukraine)
|
Let $A^{\prime}$ be the point diametrically opposite to $A$ on $\Gamma$. Since $\angle A Q A^{\prime}=90^{\circ}$ and $\angle A Q H=90^{\circ}$, the points $Q, H$, and $A^{\prime}$ are collinear. Similarly, if $Q^{\prime}$ denotes the point on $\Gamma$ diametrically opposite to $Q$, then $K, H$, and $Q^{\prime}$ are collinear. Let the line $A H F$ intersect $\Gamma$ again at $E$; it is known that $M$ is the midpoint of the segment $H A^{\prime}$ and that $F$ is the midpoint of $H E$. Let $J$ be the midpoint of $H Q^{\prime}$. Consider any point $T$ such that $T K$ is tangent to the circle $K Q H$ at $K$ with $Q$ and $T$ lying on different sides of $K H$ (see Figure 1). Then $\angle H K T=\angle H Q K$ and we are to prove that $\angle M K T=\angle C F K$. Thus it remains to show that $\angle H Q K=\angle C F K+\angle H K M$. Due to $\angle H Q K=90^{\circ}-\angle Q^{\prime} H A^{\prime}$ and $\angle C F K=90^{\circ}-\angle K F A$, this means the same as $\angle Q^{\prime} H A^{\prime}=$ $\angle K F A-\angle H K M$. Now, since the triangles $K H E$ and $A H Q^{\prime}$ are similar with $F$ and $J$ being the midpoints of corresponding sides, we have $\angle K F A=\angle H J A$, and analogously one may obtain $\angle H K M=\angle J Q H$. Thereby our task is reduced to verifying $$ \angle Q^{\prime} H A^{\prime}=\angle H J A-\angle J Q H $$  Figure 1  Figure 2 To avoid confusion, let us draw a new picture at this moment (see Figure 2). Owing to $\angle Q^{\prime} H A^{\prime}=\angle J Q H+\angle H J Q$ and $\angle H J A=\angle Q J A+\angle H J Q$, we just have to show that $2 \angle J Q H=\angle Q J A$. To this end, it suffices to remark that $A Q A^{\prime} Q^{\prime}$ is a rectangle and that $J$, being defined to be the midpoint of $H Q^{\prime}$, has to lie on the mid parallel of $Q A^{\prime}$ and $Q^{\prime} A$.
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
Let $A B C$ be an acute triangle with $A B>A C$, and let $\Gamma$ be its circumcircle. Let $H$, $M$, and $F$ be the orthocenter of the triangle, the midpoint of $B C$, and the foot of the altitude from $A$, respectively. Let $Q$ and $K$ be the two points on $\Gamma$ that satisfy $\angle A Q H=90^{\circ}$ and $\angle Q K H=90^{\circ}$. Prove that the circumcircles of the triangles $K Q H$ and $K F M$ are tangent to each other. (Ukraine)
|
Let $A^{\prime}$ be the point diametrically opposite to $A$ on $\Gamma$. Since $\angle A Q A^{\prime}=90^{\circ}$ and $\angle A Q H=90^{\circ}$, the points $Q, H$, and $A^{\prime}$ are collinear. Similarly, if $Q^{\prime}$ denotes the point on $\Gamma$ diametrically opposite to $Q$, then $K, H$, and $Q^{\prime}$ are collinear. Let the line $A H F$ intersect $\Gamma$ again at $E$; it is known that $M$ is the midpoint of the segment $H A^{\prime}$ and that $F$ is the midpoint of $H E$. Let $J$ be the midpoint of $H Q^{\prime}$. Consider any point $T$ such that $T K$ is tangent to the circle $K Q H$ at $K$ with $Q$ and $T$ lying on different sides of $K H$ (see Figure 1). Then $\angle H K T=\angle H Q K$ and we are to prove that $\angle M K T=\angle C F K$. Thus it remains to show that $\angle H Q K=\angle C F K+\angle H K M$. Due to $\angle H Q K=90^{\circ}-\angle Q^{\prime} H A^{\prime}$ and $\angle C F K=90^{\circ}-\angle K F A$, this means the same as $\angle Q^{\prime} H A^{\prime}=$ $\angle K F A-\angle H K M$. Now, since the triangles $K H E$ and $A H Q^{\prime}$ are similar with $F$ and $J$ being the midpoints of corresponding sides, we have $\angle K F A=\angle H J A$, and analogously one may obtain $\angle H K M=\angle J Q H$. Thereby our task is reduced to verifying $$ \angle Q^{\prime} H A^{\prime}=\angle H J A-\angle J Q H $$  Figure 1  Figure 2 To avoid confusion, let us draw a new picture at this moment (see Figure 2). Owing to $\angle Q^{\prime} H A^{\prime}=\angle J Q H+\angle H J Q$ and $\angle H J A=\angle Q J A+\angle H J Q$, we just have to show that $2 \angle J Q H=\angle Q J A$. To this end, it suffices to remark that $A Q A^{\prime} Q^{\prime}$ is a rectangle and that $J$, being defined to be the midpoint of $H Q^{\prime}$, has to lie on the mid parallel of $Q A^{\prime}$ and $Q^{\prime} A$.
|
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1b8f5b16-fcbb-58a4-b2f3-13d3e9010c5f
| 24,624
|
Let $A B C$ be an acute triangle with $A B>A C$, and let $\Gamma$ be its circumcircle. Let $H$, $M$, and $F$ be the orthocenter of the triangle, the midpoint of $B C$, and the foot of the altitude from $A$, respectively. Let $Q$ and $K$ be the two points on $\Gamma$ that satisfy $\angle A Q H=90^{\circ}$ and $\angle Q K H=90^{\circ}$. Prove that the circumcircles of the triangles $K Q H$ and $K F M$ are tangent to each other. (Ukraine)
|
We define the points $A^{\prime}$ and $E$ and prove that the ray $M H$ passes through $Q$ in the same way as in the first solution. Notice that the points $A^{\prime}$ and $E$ can play analogous roles to the points $Q$ and $K$, respectively: point $A^{\prime}$ is the second intersection of the line $M H$ with $\Gamma$, and $E$ is the point on $\Gamma$ with the property $\angle H E A^{\prime}=90^{\circ}$ (see Figure 3). In the circles $K Q H$ and $E A^{\prime} H$, the line segments $H Q$ and $H A^{\prime}$ are diameters, respectively; so, these circles have a common tangent $t$ at $H$, perpendicular to $M H$. Let $R$ be the radical center of the circles $A B C, K Q H$ and $E A^{\prime} H$. Their pairwise radical axes are the lines $Q K$, $A^{\prime} E$ and the line $t$; they all pass through $R$. Let $S$ be the midpoint of $H R$; by $\angle Q K H=$  Figure 3 $\angle H E A^{\prime}=90^{\circ}$, the quadrilateral $H E R K$ is cyclic and its circumcenter is $S$; hence we have $S K=S E=S H$. The line $B C$, being the perpendicular bisector of $H E$, passes through $S$. The circle $H M F$ also is tangent to $t$ at $H$; from the power of $S$ with respect to the circle $H M F$ we have $$ S M \cdot S F=S H^{2}=S K^{2} $$ So, the power of $S$ with respect to the circles $K Q H$ and $K F M$ is $S K^{2}$. Therefore, the line segment $S K$ is tangent to both circles at $K$.
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
Let $A B C$ be an acute triangle with $A B>A C$, and let $\Gamma$ be its circumcircle. Let $H$, $M$, and $F$ be the orthocenter of the triangle, the midpoint of $B C$, and the foot of the altitude from $A$, respectively. Let $Q$ and $K$ be the two points on $\Gamma$ that satisfy $\angle A Q H=90^{\circ}$ and $\angle Q K H=90^{\circ}$. Prove that the circumcircles of the triangles $K Q H$ and $K F M$ are tangent to each other. (Ukraine)
|
We define the points $A^{\prime}$ and $E$ and prove that the ray $M H$ passes through $Q$ in the same way as in the first solution. Notice that the points $A^{\prime}$ and $E$ can play analogous roles to the points $Q$ and $K$, respectively: point $A^{\prime}$ is the second intersection of the line $M H$ with $\Gamma$, and $E$ is the point on $\Gamma$ with the property $\angle H E A^{\prime}=90^{\circ}$ (see Figure 3). In the circles $K Q H$ and $E A^{\prime} H$, the line segments $H Q$ and $H A^{\prime}$ are diameters, respectively; so, these circles have a common tangent $t$ at $H$, perpendicular to $M H$. Let $R$ be the radical center of the circles $A B C, K Q H$ and $E A^{\prime} H$. Their pairwise radical axes are the lines $Q K$, $A^{\prime} E$ and the line $t$; they all pass through $R$. Let $S$ be the midpoint of $H R$; by $\angle Q K H=$  Figure 3 $\angle H E A^{\prime}=90^{\circ}$, the quadrilateral $H E R K$ is cyclic and its circumcenter is $S$; hence we have $S K=S E=S H$. The line $B C$, being the perpendicular bisector of $H E$, passes through $S$. The circle $H M F$ also is tangent to $t$ at $H$; from the power of $S$ with respect to the circle $H M F$ we have $$ S M \cdot S F=S H^{2}=S K^{2} $$ So, the power of $S$ with respect to the circles $K Q H$ and $K F M$ is $S K^{2}$. Therefore, the line segment $S K$ is tangent to both circles at $K$.
|
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1b8f5b16-fcbb-58a4-b2f3-13d3e9010c5f
| 24,624
|
Let $A B C D$ be a convex quadrilateral, and let $P, Q, R$, and $S$ be points on the sides $A B, B C, C D$, and $D A$, respectively. Let the line segments $P R$ and $Q S$ meet at $O$. Suppose that each of the quadrilaterals $A P O S, B Q O P, C R O Q$, and $D S O R$ has an incircle. Prove that the lines $A C, P Q$, and $R S$ are either concurrent or parallel to each other. (Bulgaria)
|
Denote by $\gamma_{A}, \gamma_{B}, \gamma_{C}$, and $\gamma_{D}$ the incircles of the quadrilaterals $A P O S, B Q O P$, $C R O Q$, and $D S O R$, respectively. We start with proving that the quadrilateral $A B C D$ also has an incircle which will be referred to as $\Omega$. Denote the points of tangency as in Figure 1. It is well-known that $Q Q_{1}=O O_{1}$ (if $B C \| P R$, this is obvious; otherwise, one may regard the two circles involved as the incircle and an excircle of the triangle formed by the lines $O Q, P R$, and $B C$ ). Similarly, $O O_{1}=P P_{1}$. Hence we have $Q Q_{1}=P P_{1}$. The other equalities of segment lengths marked in Figure 1 can be proved analogously. These equalities, together with $A P_{1}=A S_{1}$ and similar ones, yield $A B+C D=A D+B C$, as required.  Figure 1 Next, let us draw the lines parallel to $Q S$ through $P$ and $R$, and also draw the lines parallel to $P R$ through $Q$ and $S$. These lines form a parallelogram; denote its vertices by $A^{\prime}, B^{\prime}, C^{\prime}$, and $D^{\prime}$ as shown in Figure 2. Since the quadrilateral $A P O S$ has an incircle, we have $A P-A S=O P-O S=A^{\prime} S-A^{\prime} P$. It is well-known that in this case there also exists a circle $\omega_{A}$ tangent to the four rays $A P$, $A S, A^{\prime} P$, and $A^{\prime} S$. It is worth mentioning here that in case when, say, the lines $A B$ and $A^{\prime} B^{\prime}$ coincide, the circle $\omega_{A}$ is just tangent to $A B$ at $P$. We introduce the circles $\omega_{B}, \omega_{C}$, and $\omega_{D}$ in a similar manner. Assume that the radii of the circles $\omega_{A}$ and $\omega_{C}$ are different. Let $X$ be the center of the homothety having a positive scale factor and mapping $\omega_{A}$ to $\omega_{C}$. Now, Monge's theorem applied to the circles $\omega_{A}, \Omega$, and $\omega_{C}$ shows that the points $A, C$, and $X$ are collinear. Applying the same theorem to the circles $\omega_{A}, \omega_{B}$, and $\omega_{C}$, we see that the points $P, Q$, and $X$ are also collinear. Similarly, the points $R, S$, and $X$ are collinear, as required. If the radii of $\omega_{A}$ and $\omega_{C}$ are equal but these circles do not coincide, then the degenerate version of the same theorem yields that the three lines $A C, P Q$, and $R S$ are parallel to the line of centers of $\omega_{A}$ and $\omega_{C}$. Finally, we need to say a few words about the case when $\omega_{A}$ and $\omega_{C}$ coincide (and thus they also coincide with $\Omega, \omega_{B}$, and $\omega_{D}$ ). It may be regarded as the limit case in the following manner.  Figure 2 Let us fix the positions of $A, P, O$, and $S$ (thus we also fix the circles $\omega_{A}, \gamma_{A}, \gamma_{B}$, and $\gamma_{D}$ ). Now we vary the circle $\gamma_{C}$ inscribed into $\angle Q O R$; for each of its positions, one may reconstruct the lines $B C$ and $C D$ as the external common tangents to $\gamma_{B}, \gamma_{C}$ and $\gamma_{C}, \gamma_{D}$ different from $P R$ and $Q S$, respectively. After such variation, the circle $\Omega$ changes, so the result obtained above may be applied.
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
Let $A B C D$ be a convex quadrilateral, and let $P, Q, R$, and $S$ be points on the sides $A B, B C, C D$, and $D A$, respectively. Let the line segments $P R$ and $Q S$ meet at $O$. Suppose that each of the quadrilaterals $A P O S, B Q O P, C R O Q$, and $D S O R$ has an incircle. Prove that the lines $A C, P Q$, and $R S$ are either concurrent or parallel to each other. (Bulgaria)
|
Denote by $\gamma_{A}, \gamma_{B}, \gamma_{C}$, and $\gamma_{D}$ the incircles of the quadrilaterals $A P O S, B Q O P$, $C R O Q$, and $D S O R$, respectively. We start with proving that the quadrilateral $A B C D$ also has an incircle which will be referred to as $\Omega$. Denote the points of tangency as in Figure 1. It is well-known that $Q Q_{1}=O O_{1}$ (if $B C \| P R$, this is obvious; otherwise, one may regard the two circles involved as the incircle and an excircle of the triangle formed by the lines $O Q, P R$, and $B C$ ). Similarly, $O O_{1}=P P_{1}$. Hence we have $Q Q_{1}=P P_{1}$. The other equalities of segment lengths marked in Figure 1 can be proved analogously. These equalities, together with $A P_{1}=A S_{1}$ and similar ones, yield $A B+C D=A D+B C$, as required.  Figure 1 Next, let us draw the lines parallel to $Q S$ through $P$ and $R$, and also draw the lines parallel to $P R$ through $Q$ and $S$. These lines form a parallelogram; denote its vertices by $A^{\prime}, B^{\prime}, C^{\prime}$, and $D^{\prime}$ as shown in Figure 2. Since the quadrilateral $A P O S$ has an incircle, we have $A P-A S=O P-O S=A^{\prime} S-A^{\prime} P$. It is well-known that in this case there also exists a circle $\omega_{A}$ tangent to the four rays $A P$, $A S, A^{\prime} P$, and $A^{\prime} S$. It is worth mentioning here that in case when, say, the lines $A B$ and $A^{\prime} B^{\prime}$ coincide, the circle $\omega_{A}$ is just tangent to $A B$ at $P$. We introduce the circles $\omega_{B}, \omega_{C}$, and $\omega_{D}$ in a similar manner. Assume that the radii of the circles $\omega_{A}$ and $\omega_{C}$ are different. Let $X$ be the center of the homothety having a positive scale factor and mapping $\omega_{A}$ to $\omega_{C}$. Now, Monge's theorem applied to the circles $\omega_{A}, \Omega$, and $\omega_{C}$ shows that the points $A, C$, and $X$ are collinear. Applying the same theorem to the circles $\omega_{A}, \omega_{B}$, and $\omega_{C}$, we see that the points $P, Q$, and $X$ are also collinear. Similarly, the points $R, S$, and $X$ are collinear, as required. If the radii of $\omega_{A}$ and $\omega_{C}$ are equal but these circles do not coincide, then the degenerate version of the same theorem yields that the three lines $A C, P Q$, and $R S$ are parallel to the line of centers of $\omega_{A}$ and $\omega_{C}$. Finally, we need to say a few words about the case when $\omega_{A}$ and $\omega_{C}$ coincide (and thus they also coincide with $\Omega, \omega_{B}$, and $\omega_{D}$ ). It may be regarded as the limit case in the following manner.  Figure 2 Let us fix the positions of $A, P, O$, and $S$ (thus we also fix the circles $\omega_{A}, \gamma_{A}, \gamma_{B}$, and $\gamma_{D}$ ). Now we vary the circle $\gamma_{C}$ inscribed into $\angle Q O R$; for each of its positions, one may reconstruct the lines $B C$ and $C D$ as the external common tangents to $\gamma_{B}, \gamma_{C}$ and $\gamma_{C}, \gamma_{D}$ different from $P R$ and $Q S$, respectively. After such variation, the circle $\Omega$ changes, so the result obtained above may be applied.
|
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c9f439a3-5788-536c-94de-63db725b78bc
| 24,628
|
Let $A B C D$ be a convex quadrilateral, and let $P, Q, R$, and $S$ be points on the sides $A B, B C, C D$, and $D A$, respectively. Let the line segments $P R$ and $Q S$ meet at $O$. Suppose that each of the quadrilaterals $A P O S, B Q O P, C R O Q$, and $D S O R$ has an incircle. Prove that the lines $A C, P Q$, and $R S$ are either concurrent or parallel to each other. (Bulgaria)
|
Applying Menelaus' theorem to $\triangle A B C$ with the line $P Q$ and to $\triangle A C D$ with the line $R S$, we see that the line $A C$ meets $P Q$ and $R S$ at the same point (possibly at infinity) if and only if $$ \frac{A P}{P B} \cdot \frac{B Q}{Q C} \cdot \frac{C R}{R D} \cdot \frac{D S}{S A}=1 $$ So, it suffices to prove (1). We start with the following result. Lemma 1. Let $E F G H$ be a circumscribed quadrilateral, and let $M$ be its incenter. Then $$ \frac{E F \cdot F G}{G H \cdot H E}=\frac{F M^{2}}{H M^{2}} $$ Proof. Notice that $\angle E M H+\angle G M F=\angle F M E+\angle H M G=180^{\circ}, \angle F G M=\angle M G H$, and $\angle H E M=\angle M E F$ (see Figure 3). By the law of sines, we get $$ \frac{E F}{F M} \cdot \frac{F G}{F M}=\frac{\sin \angle F M E \cdot \sin \angle G M F}{\sin \angle M E F \cdot \sin \angle F G M}=\frac{\sin \angle H M G \cdot \sin \angle E M H}{\sin \angle M G H \cdot \sin \angle H E M}=\frac{G H}{H M} \cdot \frac{H E}{H M} . $$  Figure 3  Figure 4 We denote by $I, J, K$, and $L$ the incenters of the quadrilaterals $A P O S, B Q O P, C R O Q$, and $D S O R$, respectively. Applying Lemma 1 to these four quadrilaterals we get $$ \frac{A P \cdot P O}{O S \cdot S A} \cdot \frac{B Q \cdot Q O}{O P \cdot P B} \cdot \frac{C R \cdot R O}{O Q \cdot Q C} \cdot \frac{D S \cdot S O}{O R \cdot R D}=\frac{P I^{2}}{S I^{2}} \cdot \frac{Q J^{2}}{P J^{2}} \cdot \frac{R K^{2}}{Q K^{2}} \cdot \frac{S L^{2}}{R L^{2}} $$ which reduces to $$ \frac{A P}{P B} \cdot \frac{B Q}{Q C} \cdot \frac{C R}{R D} \cdot \frac{D S}{S A}=\frac{P I^{2}}{P J^{2}} \cdot \frac{Q J^{2}}{Q K^{2}} \cdot \frac{R K^{2}}{R L^{2}} \cdot \frac{S L^{2}}{S I^{2}} $$ Next, we have $\angle I P J=\angle J O I=90^{\circ}$, and the line $O P$ separates $I$ and $J$ (see Figure 4). This means that the quadrilateral $I P J O$ is cyclic. Similarly, we get that the quadrilateral $J Q K O$ is cyclic with $\angle J Q K=90^{\circ}$. Thus, $\angle Q K J=\angle Q O J=\angle J O P=\angle J I P$. Hence, the right triangles $I P J$ and $K Q J$ are similar. Therefore, $\frac{P I}{P J}=\frac{Q K}{Q J}$. Likewise, we obtain $\frac{R K}{R L}=\frac{S I}{S L}$. These two equations together with (2) yield (1). Comment. Instead of using the sine law, one may prove Lemma 1 by the following approach.  Figure 5 Let $N$ be the point such that $\triangle N H G \sim \triangle M E F$ and such that $N$ and $M$ lie on different sides of the line $G H$, as shown in Figure 5. Then $\angle G N H+\angle H M G=\angle F M E+\angle H M G=180^{\circ}$. So, the quadrilateral $G N H M$ is cyclic. Thus, $\angle M N H=\angle M G H=\angle F G M$ and $\angle H M N=\angle H G N=$ $\angle E F M=\angle M F G$. Hence, $\triangle H M N \sim \triangle M F G$. Therefore, $\frac{H M}{H G}=\frac{H M}{H N} \cdot \frac{H N}{H G}=\frac{M F}{M G} \cdot \frac{E M}{E F}$. Similarly, we obtain $\frac{H M}{H E}=\frac{M F}{M E} \cdot \frac{G M}{G F}$. By multiplying these two equations, we complete the proof.
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
Let $A B C D$ be a convex quadrilateral, and let $P, Q, R$, and $S$ be points on the sides $A B, B C, C D$, and $D A$, respectively. Let the line segments $P R$ and $Q S$ meet at $O$. Suppose that each of the quadrilaterals $A P O S, B Q O P, C R O Q$, and $D S O R$ has an incircle. Prove that the lines $A C, P Q$, and $R S$ are either concurrent or parallel to each other. (Bulgaria)
|
Applying Menelaus' theorem to $\triangle A B C$ with the line $P Q$ and to $\triangle A C D$ with the line $R S$, we see that the line $A C$ meets $P Q$ and $R S$ at the same point (possibly at infinity) if and only if $$ \frac{A P}{P B} \cdot \frac{B Q}{Q C} \cdot \frac{C R}{R D} \cdot \frac{D S}{S A}=1 $$ So, it suffices to prove (1). We start with the following result. Lemma 1. Let $E F G H$ be a circumscribed quadrilateral, and let $M$ be its incenter. Then $$ \frac{E F \cdot F G}{G H \cdot H E}=\frac{F M^{2}}{H M^{2}} $$ Proof. Notice that $\angle E M H+\angle G M F=\angle F M E+\angle H M G=180^{\circ}, \angle F G M=\angle M G H$, and $\angle H E M=\angle M E F$ (see Figure 3). By the law of sines, we get $$ \frac{E F}{F M} \cdot \frac{F G}{F M}=\frac{\sin \angle F M E \cdot \sin \angle G M F}{\sin \angle M E F \cdot \sin \angle F G M}=\frac{\sin \angle H M G \cdot \sin \angle E M H}{\sin \angle M G H \cdot \sin \angle H E M}=\frac{G H}{H M} \cdot \frac{H E}{H M} . $$  Figure 3  Figure 4 We denote by $I, J, K$, and $L$ the incenters of the quadrilaterals $A P O S, B Q O P, C R O Q$, and $D S O R$, respectively. Applying Lemma 1 to these four quadrilaterals we get $$ \frac{A P \cdot P O}{O S \cdot S A} \cdot \frac{B Q \cdot Q O}{O P \cdot P B} \cdot \frac{C R \cdot R O}{O Q \cdot Q C} \cdot \frac{D S \cdot S O}{O R \cdot R D}=\frac{P I^{2}}{S I^{2}} \cdot \frac{Q J^{2}}{P J^{2}} \cdot \frac{R K^{2}}{Q K^{2}} \cdot \frac{S L^{2}}{R L^{2}} $$ which reduces to $$ \frac{A P}{P B} \cdot \frac{B Q}{Q C} \cdot \frac{C R}{R D} \cdot \frac{D S}{S A}=\frac{P I^{2}}{P J^{2}} \cdot \frac{Q J^{2}}{Q K^{2}} \cdot \frac{R K^{2}}{R L^{2}} \cdot \frac{S L^{2}}{S I^{2}} $$ Next, we have $\angle I P J=\angle J O I=90^{\circ}$, and the line $O P$ separates $I$ and $J$ (see Figure 4). This means that the quadrilateral $I P J O$ is cyclic. Similarly, we get that the quadrilateral $J Q K O$ is cyclic with $\angle J Q K=90^{\circ}$. Thus, $\angle Q K J=\angle Q O J=\angle J O P=\angle J I P$. Hence, the right triangles $I P J$ and $K Q J$ are similar. Therefore, $\frac{P I}{P J}=\frac{Q K}{Q J}$. Likewise, we obtain $\frac{R K}{R L}=\frac{S I}{S L}$. These two equations together with (2) yield (1). Comment. Instead of using the sine law, one may prove Lemma 1 by the following approach.  Figure 5 Let $N$ be the point such that $\triangle N H G \sim \triangle M E F$ and such that $N$ and $M$ lie on different sides of the line $G H$, as shown in Figure 5. Then $\angle G N H+\angle H M G=\angle F M E+\angle H M G=180^{\circ}$. So, the quadrilateral $G N H M$ is cyclic. Thus, $\angle M N H=\angle M G H=\angle F G M$ and $\angle H M N=\angle H G N=$ $\angle E F M=\angle M F G$. Hence, $\triangle H M N \sim \triangle M F G$. Therefore, $\frac{H M}{H G}=\frac{H M}{H N} \cdot \frac{H N}{H G}=\frac{M F}{M G} \cdot \frac{E M}{E F}$. Similarly, we obtain $\frac{H M}{H E}=\frac{M F}{M E} \cdot \frac{G M}{G F}$. By multiplying these two equations, we complete the proof.
|
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c9f439a3-5788-536c-94de-63db725b78bc
| 24,628
|
Let $a$ and $b$ be positive integers such that $a!b$ ! is a multiple of $a!+b!$. Prove that $3 a \geqslant 2 b+2$. (United Kingdom)
|
If $a>b$, we immediately get $3 a \geqslant 2 b+2$. In the case $a=b$, the required inequality is equivalent to $a \geqslant 2$, which can be checked easily since $(a, b)=(1,1)$ does not satisfy $a!+b!\mid a!b!$. We now assume $a<b$ and denote $c=b-a$. The required inequality becomes $a \geqslant 2 c+2$. Suppose, to the contrary, that $a \leqslant 2 c+1$. Define $M=\frac{b!}{a!}=(a+1)(a+2) \cdots(a+c)$. Since $a!+b!\mid a!b!$ implies $1+M \mid a!M$, we obtain $1+M \mid a!$. Note that we must have $c<a$; otherwise $1+M>a$ !, which is impossible. We observe that $c!\mid M$ since $M$ is a product of $c$ consecutive integers. Thus $\operatorname{gcd}(1+M, c!)=1$, which implies $$ 1+M \left\lvert\, \frac{a!}{c!}=(c+1)(c+2) \cdots a\right. $$ If $a \leqslant 2 c$, then $\frac{a!}{c!}$ is a product of $a-c \leqslant c$ integers not exceeding $a$ whereas $M$ is a product of $c$ integers exceeding $a$. Therefore, $1+M>\frac{a!}{c!}$, which is a contradiction. It remains to exclude the case $a=2 c+1$. Since $a+1=2(c+1)$, we have $c+1 \mid M$. Hence, we can deduce from (1) that $1+M \mid(c+2)(c+3) \cdots a$. Now $(c+2)(c+3) \cdots a$ is a product of $a-c-1=c$ integers not exceeding $a$; thus it is smaller than $1+M$. Again, we arrive at a contradiction. Comment 1. One may derive a weaker version of (1) and finish the problem as follows. After assuming $a \leqslant 2 c+1$, we have $\left\lfloor\frac{a}{2}\right\rfloor \leqslant c$, so $\left.\left\lfloor\frac{a}{2}\right\rfloor!\right\rvert\, M$. Therefore, $$ 1+M \left\lvert\,\left(\left\lfloor\frac{a}{2}\right\rfloor+1\right)\left(\left\lfloor\frac{a}{2}\right\rfloor+2\right) \cdots a\right. $$ Observe that $\left(\left\lfloor\frac{a}{2}\right\rfloor+1\right)\left(\left\lfloor\frac{a}{2}\right\rfloor+2\right) \cdots a$ is a product of $\left\lceil\frac{a}{2}\right\rceil$ integers not exceeding $a$. This leads to a contradiction when $a$ is even since $\left\lceil\frac{a}{2}\right\rceil=\frac{a}{2} \leqslant c$ and $M$ is a product of $c$ integers exceeding $a$. When $a$ is odd, we can further deduce that $1+M \left\lvert\,\left(\frac{a+3}{2}\right)\left(\frac{a+5}{2}\right) \cdots a\right.$ since $\left.\left\lfloor\frac{a}{2}\right\rfloor+1=\frac{a+1}{2} \right\rvert\, a+1$. Now $\left(\frac{a+3}{2}\right)\left(\frac{a+5}{2}\right) \cdots a$ is a product of $\frac{a-1}{2} \leqslant c$ numbers not exceeding $a$, and we get a contradiction.
|
proof
|
Yes
|
Yes
|
proof
|
Number Theory
|
Let $a$ and $b$ be positive integers such that $a!b$ ! is a multiple of $a!+b!$. Prove that $3 a \geqslant 2 b+2$. (United Kingdom)
|
If $a>b$, we immediately get $3 a \geqslant 2 b+2$. In the case $a=b$, the required inequality is equivalent to $a \geqslant 2$, which can be checked easily since $(a, b)=(1,1)$ does not satisfy $a!+b!\mid a!b!$. We now assume $a<b$ and denote $c=b-a$. The required inequality becomes $a \geqslant 2 c+2$. Suppose, to the contrary, that $a \leqslant 2 c+1$. Define $M=\frac{b!}{a!}=(a+1)(a+2) \cdots(a+c)$. Since $a!+b!\mid a!b!$ implies $1+M \mid a!M$, we obtain $1+M \mid a!$. Note that we must have $c<a$; otherwise $1+M>a$ !, which is impossible. We observe that $c!\mid M$ since $M$ is a product of $c$ consecutive integers. Thus $\operatorname{gcd}(1+M, c!)=1$, which implies $$ 1+M \left\lvert\, \frac{a!}{c!}=(c+1)(c+2) \cdots a\right. $$ If $a \leqslant 2 c$, then $\frac{a!}{c!}$ is a product of $a-c \leqslant c$ integers not exceeding $a$ whereas $M$ is a product of $c$ integers exceeding $a$. Therefore, $1+M>\frac{a!}{c!}$, which is a contradiction. It remains to exclude the case $a=2 c+1$. Since $a+1=2(c+1)$, we have $c+1 \mid M$. Hence, we can deduce from (1) that $1+M \mid(c+2)(c+3) \cdots a$. Now $(c+2)(c+3) \cdots a$ is a product of $a-c-1=c$ integers not exceeding $a$; thus it is smaller than $1+M$. Again, we arrive at a contradiction. Comment 1. One may derive a weaker version of (1) and finish the problem as follows. After assuming $a \leqslant 2 c+1$, we have $\left\lfloor\frac{a}{2}\right\rfloor \leqslant c$, so $\left.\left\lfloor\frac{a}{2}\right\rfloor!\right\rvert\, M$. Therefore, $$ 1+M \left\lvert\,\left(\left\lfloor\frac{a}{2}\right\rfloor+1\right)\left(\left\lfloor\frac{a}{2}\right\rfloor+2\right) \cdots a\right. $$ Observe that $\left(\left\lfloor\frac{a}{2}\right\rfloor+1\right)\left(\left\lfloor\frac{a}{2}\right\rfloor+2\right) \cdots a$ is a product of $\left\lceil\frac{a}{2}\right\rceil$ integers not exceeding $a$. This leads to a contradiction when $a$ is even since $\left\lceil\frac{a}{2}\right\rceil=\frac{a}{2} \leqslant c$ and $M$ is a product of $c$ integers exceeding $a$. When $a$ is odd, we can further deduce that $1+M \left\lvert\,\left(\frac{a+3}{2}\right)\left(\frac{a+5}{2}\right) \cdots a\right.$ since $\left.\left\lfloor\frac{a}{2}\right\rfloor+1=\frac{a+1}{2} \right\rvert\, a+1$. Now $\left(\frac{a+3}{2}\right)\left(\frac{a+5}{2}\right) \cdots a$ is a product of $\frac{a-1}{2} \leqslant c$ numbers not exceeding $a$, and we get a contradiction.
|
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|
9afe038a-41d4-5e8c-8020-d97fb48a7a90
| 24,642
|
Let $m$ and $n$ be positive integers such that $m>n$. Define $x_{k}=(m+k) /(n+k)$ for $k=$ $1,2, \ldots, n+1$. Prove that if all the numbers $x_{1}, x_{2}, \ldots, x_{n+1}$ are integers, then $x_{1} x_{2} \cdots x_{n+1}-1$ is divisible by an odd prime. (Austria)
|
Assume that $x_{1}, x_{2}, \ldots, x_{n+1}$ are integers. Define the integers $$ a_{k}=x_{k}-1=\frac{m+k}{n+k}-1=\frac{m-n}{n+k}>0 $$ for $k=1,2, \ldots, n+1$. Let $P=x_{1} x_{2} \cdots x_{n+1}-1$. We need to prove that $P$ is divisible by an odd prime, or in other words, that $P$ is not a power of 2 . To this end, we investigate the powers of 2 dividing the numbers $a_{k}$. Let $2^{d}$ be the largest power of 2 dividing $m-n$, and let $2^{c}$ be the largest power of 2 not exceeding $2 n+1$. Then $2 n+1 \leqslant 2^{c+1}-1$, and so $n+1 \leqslant 2^{c}$. We conclude that $2^{c}$ is one of the numbers $n+1, n+2, \ldots, 2 n+1$, and that it is the only multiple of $2^{c}$ appearing among these numbers. Let $\ell$ be such that $n+\ell=2^{c}$. Since $\frac{m-n}{n+\ell}$ is an integer, we have $d \geqslant c$. Therefore, $2^{d-c+1} \nmid a_{\ell}=\frac{m-n}{n+\ell}$, while $2^{d-c+1} \mid a_{k}$ for all $k \in\{1, \ldots, n+1\} \backslash\{\ell\}$. Computing modulo $2^{d-c+1}$, we get $$ P=\left(a_{1}+1\right)\left(a_{2}+1\right) \cdots\left(a_{n+1}+1\right)-1 \equiv\left(a_{\ell}+1\right) \cdot 1^{n}-1 \equiv a_{\ell} \not \equiv 0 \quad\left(\bmod 2^{d-c+1}\right) $$ Therefore, $2^{d-c+1} \nmid P$. On the other hand, for any $k \in\{1, \ldots, n+1\} \backslash\{\ell\}$, we have $2^{d-c+1} \mid a_{k}$. So $P \geqslant a_{k} \geqslant 2^{d-c+1}$, and it follows that $P$ is not a power of 2 . Comment. Instead of attempting to show that $P$ is not a power of 2 , one may try to find an odd factor of $P$ (greater than 1) as follows: From $a_{k}=\frac{m-n}{n+k} \in \mathbb{Z}_{>0}$, we get that $m-n$ is divisible by $n+1, n+2, \ldots, 2 n+1$, and thus it is also divisible by their least common multiple $L$. So $m-n=q L$ for some positive integer $q$; hence $x_{k}=q \cdot \frac{L}{n+k}+1$. Then, since $n+1 \leqslant 2^{c}=n+\ell \leqslant 2 n+1 \leqslant 2^{c+1}-1$, we have $2^{c} \mid L$, but $2^{c+1} \nmid L$. So $\frac{L}{n+\ell}$ is odd, while $\frac{L}{n+k}$ is even for $k \neq \ell$. Computing modulo $2 q$ yields $$ x_{1} x_{2} \cdots x_{n+1}-1 \equiv(q+1) \cdot 1^{n}-1 \equiv q \quad(\bmod 2 q) $$ Thus, $x_{1} x_{2} \cdots x_{n+1}-1=2 q r+q=q(2 r+1)$ for some integer $r$. Since $x_{1} x_{2} \cdots x_{n+1}-1 \geqslant x_{1} x_{2}-1 \geqslant(q+1)^{2}-1>q$, we have $r \geqslant 1$. This implies that $x_{1} x_{2} \cdots x_{n+1}-1$ is divisible by an odd prime.
|
proof
|
Yes
|
Yes
|
proof
|
Number Theory
|
Let $m$ and $n$ be positive integers such that $m>n$. Define $x_{k}=(m+k) /(n+k)$ for $k=$ $1,2, \ldots, n+1$. Prove that if all the numbers $x_{1}, x_{2}, \ldots, x_{n+1}$ are integers, then $x_{1} x_{2} \cdots x_{n+1}-1$ is divisible by an odd prime. (Austria)
|
Assume that $x_{1}, x_{2}, \ldots, x_{n+1}$ are integers. Define the integers $$ a_{k}=x_{k}-1=\frac{m+k}{n+k}-1=\frac{m-n}{n+k}>0 $$ for $k=1,2, \ldots, n+1$. Let $P=x_{1} x_{2} \cdots x_{n+1}-1$. We need to prove that $P$ is divisible by an odd prime, or in other words, that $P$ is not a power of 2 . To this end, we investigate the powers of 2 dividing the numbers $a_{k}$. Let $2^{d}$ be the largest power of 2 dividing $m-n$, and let $2^{c}$ be the largest power of 2 not exceeding $2 n+1$. Then $2 n+1 \leqslant 2^{c+1}-1$, and so $n+1 \leqslant 2^{c}$. We conclude that $2^{c}$ is one of the numbers $n+1, n+2, \ldots, 2 n+1$, and that it is the only multiple of $2^{c}$ appearing among these numbers. Let $\ell$ be such that $n+\ell=2^{c}$. Since $\frac{m-n}{n+\ell}$ is an integer, we have $d \geqslant c$. Therefore, $2^{d-c+1} \nmid a_{\ell}=\frac{m-n}{n+\ell}$, while $2^{d-c+1} \mid a_{k}$ for all $k \in\{1, \ldots, n+1\} \backslash\{\ell\}$. Computing modulo $2^{d-c+1}$, we get $$ P=\left(a_{1}+1\right)\left(a_{2}+1\right) \cdots\left(a_{n+1}+1\right)-1 \equiv\left(a_{\ell}+1\right) \cdot 1^{n}-1 \equiv a_{\ell} \not \equiv 0 \quad\left(\bmod 2^{d-c+1}\right) $$ Therefore, $2^{d-c+1} \nmid P$. On the other hand, for any $k \in\{1, \ldots, n+1\} \backslash\{\ell\}$, we have $2^{d-c+1} \mid a_{k}$. So $P \geqslant a_{k} \geqslant 2^{d-c+1}$, and it follows that $P$ is not a power of 2 . Comment. Instead of attempting to show that $P$ is not a power of 2 , one may try to find an odd factor of $P$ (greater than 1) as follows: From $a_{k}=\frac{m-n}{n+k} \in \mathbb{Z}_{>0}$, we get that $m-n$ is divisible by $n+1, n+2, \ldots, 2 n+1$, and thus it is also divisible by their least common multiple $L$. So $m-n=q L$ for some positive integer $q$; hence $x_{k}=q \cdot \frac{L}{n+k}+1$. Then, since $n+1 \leqslant 2^{c}=n+\ell \leqslant 2 n+1 \leqslant 2^{c+1}-1$, we have $2^{c} \mid L$, but $2^{c+1} \nmid L$. So $\frac{L}{n+\ell}$ is odd, while $\frac{L}{n+k}$ is even for $k \neq \ell$. Computing modulo $2 q$ yields $$ x_{1} x_{2} \cdots x_{n+1}-1 \equiv(q+1) \cdot 1^{n}-1 \equiv q \quad(\bmod 2 q) $$ Thus, $x_{1} x_{2} \cdots x_{n+1}-1=2 q r+q=q(2 r+1)$ for some integer $r$. Since $x_{1} x_{2} \cdots x_{n+1}-1 \geqslant x_{1} x_{2}-1 \geqslant(q+1)^{2}-1>q$, we have $r \geqslant 1$. This implies that $x_{1} x_{2} \cdots x_{n+1}-1$ is divisible by an odd prime.
|
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|
513ae0a6-9080-5ad3-8bf4-a00fe8aebd0e
| 24,646
|
Suppose that $a_{0}, a_{1}, \ldots$ and $b_{0}, b_{1}, \ldots$ are two sequences of positive integers satisfying $a_{0}, b_{0} \geqslant 2$ and $$ a_{n+1}=\operatorname{gcd}\left(a_{n}, b_{n}\right)+1, \quad b_{n+1}=\operatorname{lcm}\left(a_{n}, b_{n}\right)-1 $$ for all $n \geqslant 0$. Prove that the sequence $\left(a_{n}\right)$ is eventually periodic; in other words, there exist integers $N \geqslant 0$ and $t>0$ such that $a_{n+t}=a_{n}$ for all $n \geqslant N$. (France)
|
Let $s_{n}=a_{n}+b_{n}$. Notice that if $a_{n} \mid b_{n}$, then $a_{n+1}=a_{n}+1, b_{n+1}=b_{n}-1$ and $s_{n+1}=s_{n}$. So, $a_{n}$ increases by 1 and $s_{n}$ does not change until the first index is reached with $a_{n} \nmid s_{n}$. Define $$ W_{n}=\left\{m \in \mathbb{Z}_{>0}: m \geqslant a_{n} \text { and } m \nmid s_{n}\right\} \quad \text { and } \quad w_{n}=\min W_{n} $$ Claim 1. The sequence $\left(w_{n}\right)$ is non-increasing. Proof. If $a_{n} \mid b_{n}$ then $a_{n+1}=a_{n}+1$. Due to $a_{n} \mid s_{n}$, we have $a_{n} \notin W_{n}$. Moreover $s_{n+1}=s_{n}$; therefore, $W_{n+1}=W_{n}$ and $w_{n+1}=w_{n}$. Otherwise, if $a_{n} \nmid b_{n}$, then $a_{n} \nmid s_{n}$, so $a_{n} \in W_{n}$ and thus $w_{n}=a_{n}$. We show that $a_{n} \in W_{n+1}$; this implies $w_{n+1} \leqslant a_{n}=w_{n}$. By the definition of $W_{n+1}$, we need that $a_{n} \geqslant a_{n+1}$ and $a_{n} \nmid s_{n+1}$. The first relation holds because of $\operatorname{gcd}\left(a_{n}, b_{n}\right)<a_{n}$. For the second relation, observe that in $s_{n+1}=\operatorname{gcd}\left(a_{n}, b_{n}\right)+\operatorname{lcm}\left(a_{n}, b_{n}\right)$, the second term is divisible by $a_{n}$, but the first term is not. So $a_{n} \nmid s_{n+1}$; that completes the proof of the claim. Let $w=\min _{n} w_{n}$ and let $N$ be an index with $w=w_{N}$. Due to Claim 1, we have $w_{n}=w$ for all $n \geqslant N$. Let $g_{n}=\operatorname{gcd}\left(w, s_{n}\right)$. As we have seen, starting from an arbitrary index $n \geqslant N$, the sequence $a_{n}, a_{n+1}, \ldots$ increases by 1 until it reaches $w$, which is the first value not dividing $s_{n}$; then it drops to $\operatorname{gcd}\left(w, s_{n}\right)+1=g_{n}+1$. Claim 2. The sequence $\left(g_{n}\right)$ is constant for $n \geqslant N$. Proof. If $a_{n} \mid b_{n}$, then $s_{n+1}=s_{n}$ and hence $g_{n+1}=g_{n}$. Otherwise we have $a_{n}=w$, $$ \begin{aligned} \operatorname{gcd}\left(a_{n}, b_{n}\right) & =\operatorname{gcd}\left(a_{n}, s_{n}\right)=\operatorname{gcd}\left(w, s_{n}\right)=g_{n} \\ s_{n+1} & =\operatorname{gcd}\left(a_{n}, b_{n}\right)+\operatorname{lcm}\left(a_{n}, b_{n}\right)=g_{n}+\frac{a_{n} b_{n}}{g_{n}}=g_{n}+\frac{w\left(s_{n}-w\right)}{g_{n}} \end{aligned} $$ and $\quad g_{n+1}=\operatorname{gcd}\left(w, s_{n+1}\right)=\operatorname{gcd}\left(w, g_{n}+\frac{s_{n}-w}{g_{n}} w\right)=\operatorname{gcd}\left(w, g_{n}\right)=g_{n}$. Let $g=g_{N}$. We have proved that the sequence $\left(a_{n}\right)$ eventually repeats the following cycle: $$ g+1 \mapsto g+2 \mapsto \ldots \mapsto w \mapsto g+1 $$
|
proof
|
Yes
|
Yes
|
proof
|
Number Theory
|
Suppose that $a_{0}, a_{1}, \ldots$ and $b_{0}, b_{1}, \ldots$ are two sequences of positive integers satisfying $a_{0}, b_{0} \geqslant 2$ and $$ a_{n+1}=\operatorname{gcd}\left(a_{n}, b_{n}\right)+1, \quad b_{n+1}=\operatorname{lcm}\left(a_{n}, b_{n}\right)-1 $$ for all $n \geqslant 0$. Prove that the sequence $\left(a_{n}\right)$ is eventually periodic; in other words, there exist integers $N \geqslant 0$ and $t>0$ such that $a_{n+t}=a_{n}$ for all $n \geqslant N$. (France)
|
Let $s_{n}=a_{n}+b_{n}$. Notice that if $a_{n} \mid b_{n}$, then $a_{n+1}=a_{n}+1, b_{n+1}=b_{n}-1$ and $s_{n+1}=s_{n}$. So, $a_{n}$ increases by 1 and $s_{n}$ does not change until the first index is reached with $a_{n} \nmid s_{n}$. Define $$ W_{n}=\left\{m \in \mathbb{Z}_{>0}: m \geqslant a_{n} \text { and } m \nmid s_{n}\right\} \quad \text { and } \quad w_{n}=\min W_{n} $$ Claim 1. The sequence $\left(w_{n}\right)$ is non-increasing. Proof. If $a_{n} \mid b_{n}$ then $a_{n+1}=a_{n}+1$. Due to $a_{n} \mid s_{n}$, we have $a_{n} \notin W_{n}$. Moreover $s_{n+1}=s_{n}$; therefore, $W_{n+1}=W_{n}$ and $w_{n+1}=w_{n}$. Otherwise, if $a_{n} \nmid b_{n}$, then $a_{n} \nmid s_{n}$, so $a_{n} \in W_{n}$ and thus $w_{n}=a_{n}$. We show that $a_{n} \in W_{n+1}$; this implies $w_{n+1} \leqslant a_{n}=w_{n}$. By the definition of $W_{n+1}$, we need that $a_{n} \geqslant a_{n+1}$ and $a_{n} \nmid s_{n+1}$. The first relation holds because of $\operatorname{gcd}\left(a_{n}, b_{n}\right)<a_{n}$. For the second relation, observe that in $s_{n+1}=\operatorname{gcd}\left(a_{n}, b_{n}\right)+\operatorname{lcm}\left(a_{n}, b_{n}\right)$, the second term is divisible by $a_{n}$, but the first term is not. So $a_{n} \nmid s_{n+1}$; that completes the proof of the claim. Let $w=\min _{n} w_{n}$ and let $N$ be an index with $w=w_{N}$. Due to Claim 1, we have $w_{n}=w$ for all $n \geqslant N$. Let $g_{n}=\operatorname{gcd}\left(w, s_{n}\right)$. As we have seen, starting from an arbitrary index $n \geqslant N$, the sequence $a_{n}, a_{n+1}, \ldots$ increases by 1 until it reaches $w$, which is the first value not dividing $s_{n}$; then it drops to $\operatorname{gcd}\left(w, s_{n}\right)+1=g_{n}+1$. Claim 2. The sequence $\left(g_{n}\right)$ is constant for $n \geqslant N$. Proof. If $a_{n} \mid b_{n}$, then $s_{n+1}=s_{n}$ and hence $g_{n+1}=g_{n}$. Otherwise we have $a_{n}=w$, $$ \begin{aligned} \operatorname{gcd}\left(a_{n}, b_{n}\right) & =\operatorname{gcd}\left(a_{n}, s_{n}\right)=\operatorname{gcd}\left(w, s_{n}\right)=g_{n} \\ s_{n+1} & =\operatorname{gcd}\left(a_{n}, b_{n}\right)+\operatorname{lcm}\left(a_{n}, b_{n}\right)=g_{n}+\frac{a_{n} b_{n}}{g_{n}}=g_{n}+\frac{w\left(s_{n}-w\right)}{g_{n}} \end{aligned} $$ and $\quad g_{n+1}=\operatorname{gcd}\left(w, s_{n+1}\right)=\operatorname{gcd}\left(w, g_{n}+\frac{s_{n}-w}{g_{n}} w\right)=\operatorname{gcd}\left(w, g_{n}\right)=g_{n}$. Let $g=g_{N}$. We have proved that the sequence $\left(a_{n}\right)$ eventually repeats the following cycle: $$ g+1 \mapsto g+2 \mapsto \ldots \mapsto w \mapsto g+1 $$
|
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bd0732e1-6728-597c-8d9a-38d395637399
| 24,649
|
Suppose that $a_{0}, a_{1}, \ldots$ and $b_{0}, b_{1}, \ldots$ are two sequences of positive integers satisfying $a_{0}, b_{0} \geqslant 2$ and $$ a_{n+1}=\operatorname{gcd}\left(a_{n}, b_{n}\right)+1, \quad b_{n+1}=\operatorname{lcm}\left(a_{n}, b_{n}\right)-1 $$ for all $n \geqslant 0$. Prove that the sequence $\left(a_{n}\right)$ is eventually periodic; in other words, there exist integers $N \geqslant 0$ and $t>0$ such that $a_{n+t}=a_{n}$ for all $n \geqslant N$. (France)
|
By Claim 1 in the first solution, we have $a_{n} \leqslant w_{n} \leqslant w_{0}$, so the sequence $\left(a_{n}\right)$ is bounded, and hence it has only finitely many values. Let $M=\operatorname{lcm}\left(a_{1}, a_{2}, \ldots\right)$, and consider the sequence $b_{n}$ modulo $M$. Let $r_{n}$ be the remainder of $b_{n}$, divided by $M$. For every index $n$, since $a_{n}|M| b_{n}-r_{n}$, we have $\operatorname{gcd}\left(a_{n}, b_{n}\right)=\operatorname{gcd}\left(a_{n}, r_{n}\right)$, and therefore $$ a_{n+1}=\operatorname{gcd}\left(a_{n}, r_{n}\right)+1 $$ Moreover, $$ \begin{aligned} r_{n+1} & \equiv b_{n+1}=\operatorname{lcm}\left(a_{n}, b_{n}\right)-1=\frac{a_{n}}{\operatorname{gcd}\left(a_{n}, b_{n}\right)} b_{n}-1 \\ & =\frac{a_{n}}{\operatorname{gcd}\left(a_{n}, r_{n}\right)} b_{n}-1 \equiv \frac{a_{n}}{\operatorname{gcd}\left(a_{n}, r_{n}\right)} r_{n}-1 \quad(\bmod M) \end{aligned} $$ Hence, the pair $\left(a_{n}, r_{n}\right)$ uniquely determines the pair $\left(a_{n+1}, r_{n+1}\right)$. Since there are finitely many possible pairs, the sequence of pairs $\left(a_{n}, r_{n}\right)$ is eventually periodic; in particular, the sequence $\left(a_{n}\right)$ is eventually periodic. Comment. We show that there are only four possibilities for $g$ and $w$ (as defined in Solution 1), namely $$ (w, g) \in\{(2,1),(3,1),(4,2),(5,1)\} $$ This means that the sequence $\left(a_{n}\right)$ eventually repeats one of the following cycles: $$ (2), \quad(2,3), \quad(3,4), \quad \text { or } \quad(2,3,4,5) $$ Using the notation of Solution 1 , for $n \geqslant N$ the sequence $\left(a_{n}\right)$ has a cycle $(g+1, g+2, \ldots, w)$ such that $g=\operatorname{gcd}\left(w, s_{n}\right)$. By the observations in the proof of Claim 2, the numbers $g+1, \ldots, w-1$ all divide $s_{n}$; so the number $L=\operatorname{lcm}(g+1, g+2, \ldots, w-1)$ also divides $s_{n}$. Moreover, $g$ also divides $w$. Now choose any $n \geqslant N$ such that $a_{n}=w$. By (1), we have $$ s_{n+1}=g+\frac{w\left(s_{n}-w\right)}{g}=s_{n} \cdot \frac{w}{g}-\frac{w^{2}-g^{2}}{g} . $$ Since $L$ divides both $s_{n}$ and $s_{n+1}$, it also divides the number $T=\frac{w^{2}-g^{2}}{g}$. Suppose first that $w \geqslant 6$, which yields $g+1 \leqslant \frac{w}{2}+1 \leqslant w-2$. Then $(w-2)(w-1)|L| T$, so we have either $w^{2}-g^{2} \geqslant 2(w-1)(w-2)$, or $g=1$ and $w^{2}-g^{2}=(w-1)(w-2)$. In the former case we get $(w-1)(w-5)+\left(g^{2}-1\right) \leqslant 0$ which is false by our assumption. The latter equation rewrites as $3 w=3$, so $w=1$, which is also impossible. Now we are left with the cases when $w \leqslant 5$ and $g \mid w$. The case $(w, g)=(4,1)$ violates the condition $L \left\lvert\, \frac{w^{2}-g^{2}}{g}\right.$; all other such pairs are listed in (2). In the table below, for each pair $(w, g)$, we provide possible sequences $\left(a_{n}\right)$ and $\left(b_{n}\right)$. That shows that the cycles shown in (3) are indeed possible. $$ \begin{array}{llll} w=2 & g=1 & a_{n}=2 & b_{n}=2 \cdot 2^{n}+1 \\ w=3 & g=1 & \left(a_{2 k}, a_{2 k+1}\right)=(2,3) & \left(b_{2 k}, b_{2 k+1}\right)=\left(6 \cdot 3^{k}+2,6 \cdot 3^{k}+1\right) \\ w=4 & g=2 & \left(a_{2 k}, a_{2 k+1}\right)=(3,4) & \left(b_{2 k}, b_{2 k+1}\right)=\left(12 \cdot 2^{k}+3,12 \cdot 2^{k}+2\right) \\ w=5 & g=1 & \left(a_{4 k}, \ldots, a_{4 k+3}\right)=(2,3,4,5) & \left(b_{4 k}, \ldots, b_{4 k+3}\right)=\left(6 \cdot 5^{k}+4, \ldots, 6 \cdot 5^{k}+1\right) \end{array} $$ $$ b_{n}=2 \cdot 2^{n}+1 $$
|
proof
|
Yes
|
Yes
|
proof
|
Number Theory
|
Suppose that $a_{0}, a_{1}, \ldots$ and $b_{0}, b_{1}, \ldots$ are two sequences of positive integers satisfying $a_{0}, b_{0} \geqslant 2$ and $$ a_{n+1}=\operatorname{gcd}\left(a_{n}, b_{n}\right)+1, \quad b_{n+1}=\operatorname{lcm}\left(a_{n}, b_{n}\right)-1 $$ for all $n \geqslant 0$. Prove that the sequence $\left(a_{n}\right)$ is eventually periodic; in other words, there exist integers $N \geqslant 0$ and $t>0$ such that $a_{n+t}=a_{n}$ for all $n \geqslant N$. (France)
|
By Claim 1 in the first solution, we have $a_{n} \leqslant w_{n} \leqslant w_{0}$, so the sequence $\left(a_{n}\right)$ is bounded, and hence it has only finitely many values. Let $M=\operatorname{lcm}\left(a_{1}, a_{2}, \ldots\right)$, and consider the sequence $b_{n}$ modulo $M$. Let $r_{n}$ be the remainder of $b_{n}$, divided by $M$. For every index $n$, since $a_{n}|M| b_{n}-r_{n}$, we have $\operatorname{gcd}\left(a_{n}, b_{n}\right)=\operatorname{gcd}\left(a_{n}, r_{n}\right)$, and therefore $$ a_{n+1}=\operatorname{gcd}\left(a_{n}, r_{n}\right)+1 $$ Moreover, $$ \begin{aligned} r_{n+1} & \equiv b_{n+1}=\operatorname{lcm}\left(a_{n}, b_{n}\right)-1=\frac{a_{n}}{\operatorname{gcd}\left(a_{n}, b_{n}\right)} b_{n}-1 \\ & =\frac{a_{n}}{\operatorname{gcd}\left(a_{n}, r_{n}\right)} b_{n}-1 \equiv \frac{a_{n}}{\operatorname{gcd}\left(a_{n}, r_{n}\right)} r_{n}-1 \quad(\bmod M) \end{aligned} $$ Hence, the pair $\left(a_{n}, r_{n}\right)$ uniquely determines the pair $\left(a_{n+1}, r_{n+1}\right)$. Since there are finitely many possible pairs, the sequence of pairs $\left(a_{n}, r_{n}\right)$ is eventually periodic; in particular, the sequence $\left(a_{n}\right)$ is eventually periodic. Comment. We show that there are only four possibilities for $g$ and $w$ (as defined in Solution 1), namely $$ (w, g) \in\{(2,1),(3,1),(4,2),(5,1)\} $$ This means that the sequence $\left(a_{n}\right)$ eventually repeats one of the following cycles: $$ (2), \quad(2,3), \quad(3,4), \quad \text { or } \quad(2,3,4,5) $$ Using the notation of Solution 1 , for $n \geqslant N$ the sequence $\left(a_{n}\right)$ has a cycle $(g+1, g+2, \ldots, w)$ such that $g=\operatorname{gcd}\left(w, s_{n}\right)$. By the observations in the proof of Claim 2, the numbers $g+1, \ldots, w-1$ all divide $s_{n}$; so the number $L=\operatorname{lcm}(g+1, g+2, \ldots, w-1)$ also divides $s_{n}$. Moreover, $g$ also divides $w$. Now choose any $n \geqslant N$ such that $a_{n}=w$. By (1), we have $$ s_{n+1}=g+\frac{w\left(s_{n}-w\right)}{g}=s_{n} \cdot \frac{w}{g}-\frac{w^{2}-g^{2}}{g} . $$ Since $L$ divides both $s_{n}$ and $s_{n+1}$, it also divides the number $T=\frac{w^{2}-g^{2}}{g}$. Suppose first that $w \geqslant 6$, which yields $g+1 \leqslant \frac{w}{2}+1 \leqslant w-2$. Then $(w-2)(w-1)|L| T$, so we have either $w^{2}-g^{2} \geqslant 2(w-1)(w-2)$, or $g=1$ and $w^{2}-g^{2}=(w-1)(w-2)$. In the former case we get $(w-1)(w-5)+\left(g^{2}-1\right) \leqslant 0$ which is false by our assumption. The latter equation rewrites as $3 w=3$, so $w=1$, which is also impossible. Now we are left with the cases when $w \leqslant 5$ and $g \mid w$. The case $(w, g)=(4,1)$ violates the condition $L \left\lvert\, \frac{w^{2}-g^{2}}{g}\right.$; all other such pairs are listed in (2). In the table below, for each pair $(w, g)$, we provide possible sequences $\left(a_{n}\right)$ and $\left(b_{n}\right)$. That shows that the cycles shown in (3) are indeed possible. $$ \begin{array}{llll} w=2 & g=1 & a_{n}=2 & b_{n}=2 \cdot 2^{n}+1 \\ w=3 & g=1 & \left(a_{2 k}, a_{2 k+1}\right)=(2,3) & \left(b_{2 k}, b_{2 k+1}\right)=\left(6 \cdot 3^{k}+2,6 \cdot 3^{k}+1\right) \\ w=4 & g=2 & \left(a_{2 k}, a_{2 k+1}\right)=(3,4) & \left(b_{2 k}, b_{2 k+1}\right)=\left(12 \cdot 2^{k}+3,12 \cdot 2^{k}+2\right) \\ w=5 & g=1 & \left(a_{4 k}, \ldots, a_{4 k+3}\right)=(2,3,4,5) & \left(b_{4 k}, \ldots, b_{4 k+3}\right)=\left(6 \cdot 5^{k}+4, \ldots, 6 \cdot 5^{k}+1\right) \end{array} $$ $$ b_{n}=2 \cdot 2^{n}+1 $$
|
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bd0732e1-6728-597c-8d9a-38d395637399
| 24,649
|
Let $n$ be a fixed integer with $n \geqslant 2$. We say that two polynomials $P$ and $Q$ with real coefficients are block-similar if for each $i \in\{1,2, \ldots, n\}$ the sequences $$ \begin{aligned} & P(2015 i), P(2015 i-1), \ldots, P(2015 i-2014) \quad \text { and } \\ & Q(2015 i), Q(2015 i-1), \ldots, Q(2015 i-2014) \end{aligned} $$ are permutations of each other. (a) Prove that there exist distinct block-similar polynomials of degree $n+1$. (b) Prove that there do not exist distinct block-similar polynomials of degree $n$. (Canada)
|
We provide an alternative argument for part (b). Assume again that there exist two distinct block-similar polynomials $P(x)$ and $Q(x)$ of degree $n$. Let $R(x)=P(x)-Q(x)$ and $S(x)=P(x)+Q(x)$. For brevity, we also denote the segment $[(i-1) k+1, i k]$ by $I_{i}$, and the set $\{(i-1) k+1,(i-1) k+2, \ldots, i k\}$ of all integer points in $I_{i}$ by $Z_{i}$. Step 1. We prove that $R(x)$ has exactly one root in each segment $I_{i}, i=1,2, \ldots, n$, and all these roots are simple. Indeed, take any $i \in\{1,2, \ldots, n\}$ and choose some points $p^{-}, p^{+} \in Z_{i}$ so that $$ P\left(p^{-}\right)=\min _{x \in Z_{i}} P(x) \quad \text { and } \quad P\left(p^{+}\right)=\max _{x \in Z_{i}} P(x) $$ Since the sequences of values of $P$ and $Q$ in $Z_{i}$ are permutations of each other, we have $R\left(p^{-}\right)=P\left(p^{-}\right)-Q\left(p^{-}\right) \leqslant 0$ and $R\left(p^{+}\right)=P\left(p^{+}\right)-Q\left(p^{+}\right) \geqslant 0$. Since $R(x)$ is continuous, there exists at least one root of $R(x)$ between $p^{-}$and $p^{+}$- thus in $I_{i}$. So, $R(x)$ has at least one root in each of the $n$ disjoint segments $I_{i}$ with $i=1,2, \ldots, n$. Since $R(x)$ is nonzero and its degree does not exceed $n$, it should have exactly one root in each of these segments, and all these roots are simple, as required.
|
proof
|
Yes
|
Yes
|
proof
|
Algebra
|
Let $n$ be a fixed integer with $n \geqslant 2$. We say that two polynomials $P$ and $Q$ with real coefficients are block-similar if for each $i \in\{1,2, \ldots, n\}$ the sequences $$ \begin{aligned} & P(2015 i), P(2015 i-1), \ldots, P(2015 i-2014) \quad \text { and } \\ & Q(2015 i), Q(2015 i-1), \ldots, Q(2015 i-2014) \end{aligned} $$ are permutations of each other. (a) Prove that there exist distinct block-similar polynomials of degree $n+1$. (b) Prove that there do not exist distinct block-similar polynomials of degree $n$. (Canada)
|
We provide an alternative argument for part (b). Assume again that there exist two distinct block-similar polynomials $P(x)$ and $Q(x)$ of degree $n$. Let $R(x)=P(x)-Q(x)$ and $S(x)=P(x)+Q(x)$. For brevity, we also denote the segment $[(i-1) k+1, i k]$ by $I_{i}$, and the set $\{(i-1) k+1,(i-1) k+2, \ldots, i k\}$ of all integer points in $I_{i}$ by $Z_{i}$. Step 1. We prove that $R(x)$ has exactly one root in each segment $I_{i}, i=1,2, \ldots, n$, and all these roots are simple. Indeed, take any $i \in\{1,2, \ldots, n\}$ and choose some points $p^{-}, p^{+} \in Z_{i}$ so that $$ P\left(p^{-}\right)=\min _{x \in Z_{i}} P(x) \quad \text { and } \quad P\left(p^{+}\right)=\max _{x \in Z_{i}} P(x) $$ Since the sequences of values of $P$ and $Q$ in $Z_{i}$ are permutations of each other, we have $R\left(p^{-}\right)=P\left(p^{-}\right)-Q\left(p^{-}\right) \leqslant 0$ and $R\left(p^{+}\right)=P\left(p^{+}\right)-Q\left(p^{+}\right) \geqslant 0$. Since $R(x)$ is continuous, there exists at least one root of $R(x)$ between $p^{-}$and $p^{+}$- thus in $I_{i}$. So, $R(x)$ has at least one root in each of the $n$ disjoint segments $I_{i}$ with $i=1,2, \ldots, n$. Since $R(x)$ is nonzero and its degree does not exceed $n$, it should have exactly one root in each of these segments, and all these roots are simple, as required.
|
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|
142121a5-aa92-50df-8e47-a6fa7885a2d5
| 24,683
|
Consider an infinite sequence $a_{1}, a_{2}, \ldots$ of positive integers with $a_{i} \leqslant 2015$ for all $i \geqslant 1$. Suppose that for any two distinct indices $i$ and $j$ we have $i+a_{i} \neq j+a_{j}$. Prove that there exist two positive integers $b$ and $N$ such that $$ \left|\sum_{i=m+1}^{n}\left(a_{i}-b\right)\right| \leqslant 1007^{2} $$ whenever $n>m \geqslant N$. (Australia)
|
We visualize the set of positive integers as a sequence of points. For each $n$ we draw an arrow emerging from $n$ that points to $n+a_{n}$; so the length of this arrow is $a_{n}$. Due to the condition that $m+a_{m} \neq n+a_{n}$ for $m \neq n$, each positive integer receives at most one arrow. There are some positive integers, such as 1 , that receive no arrows; these will be referred to as starting points in the sequel. When one starts at any of the starting points and keeps following the arrows, one is led to an infinite path, called its ray, that visits a strictly increasing sequence of positive integers. Since the length of any arrow is at most 2015, such a ray, say with starting point $s$, meets every interval of the form $[n, n+2014]$ with $n \geqslant s$ at least once. Suppose for the sake of contradiction that there would be at least 2016 starting points. Then we could take an integer $n$ that is larger than the first 2016 starting points. But now the interval $[n, n+2014]$ must be met by at least 2016 rays in distinct points, which is absurd. We have thereby shown that the number $b$ of starting points satisfies $1 \leqslant b \leqslant 2015$. Let $N$ denote any integer that is larger than all starting points. We contend that $b$ and $N$ are as required. To see this, let any two integers $m$ and $n$ with $n>m \geqslant N$ be given. The sum $\sum_{i=m+1}^{n} a_{i}$ gives the total length of the arrows emerging from $m+1, \ldots, n$. Taken together, these arrows form $b$ subpaths of our rays, some of which may be empty. Now on each ray we look at the first number that is larger than $m$; let $x_{1}, \ldots, x_{b}$ denote these numbers, and let $y_{1}, \ldots, y_{b}$ enumerate in corresponding order the numbers defined similarly with respect to $n$. Then the list of differences $y_{1}-x_{1}, \ldots, y_{b}-x_{b}$ consists of the lengths of these paths and possibly some zeros corresponding to empty paths. Consequently, we obtain $$ \sum_{i=m+1}^{n} a_{i}=\sum_{j=1}^{b}\left(y_{j}-x_{j}\right) $$ whence $$ \sum_{i=m+1}^{n}\left(a_{i}-b\right)=\sum_{j=1}^{b}\left(y_{j}-n\right)-\sum_{j=1}^{b}\left(x_{j}-m\right) . $$ Now each of the $b$ rays meets the interval $[m+1, m+2015]$ at some point and thus $x_{1}-$ $m, \ldots, x_{b}-m$ are $b$ distinct members of the set $\{1,2, \ldots, 2015\}$. Moreover, since $m+1$ is not a starting point, it must belong to some ray; so 1 has to appear among these numbers, wherefore $$ 1+\sum_{j=1}^{b-1}(j+1) \leqslant \sum_{j=1}^{b}\left(x_{j}-m\right) \leqslant 1+\sum_{j=1}^{b-1}(2016-b+j) $$ The same argument applied to $n$ and $y_{1}, \ldots, y_{b}$ yields $$ 1+\sum_{j=1}^{b-1}(j+1) \leqslant \sum_{j=1}^{b}\left(y_{j}-n\right) \leqslant 1+\sum_{j=1}^{b-1}(2016-b+j) $$ So altogether we get $$ \begin{aligned} \left|\sum_{i=m+1}^{n}\left(a_{i}-b\right)\right| & \leqslant \sum_{j=1}^{b-1}((2016-b+j)-(j+1))=(b-1)(2015-b) \\ & \leqslant\left(\frac{(b-1)+(2015-b)}{2}\right)^{2}=1007^{2}, \end{aligned} $$ as desired.
|
proof
|
Yes
|
Yes
|
proof
|
Combinatorics
|
Consider an infinite sequence $a_{1}, a_{2}, \ldots$ of positive integers with $a_{i} \leqslant 2015$ for all $i \geqslant 1$. Suppose that for any two distinct indices $i$ and $j$ we have $i+a_{i} \neq j+a_{j}$. Prove that there exist two positive integers $b$ and $N$ such that $$ \left|\sum_{i=m+1}^{n}\left(a_{i}-b\right)\right| \leqslant 1007^{2} $$ whenever $n>m \geqslant N$. (Australia)
|
We visualize the set of positive integers as a sequence of points. For each $n$ we draw an arrow emerging from $n$ that points to $n+a_{n}$; so the length of this arrow is $a_{n}$. Due to the condition that $m+a_{m} \neq n+a_{n}$ for $m \neq n$, each positive integer receives at most one arrow. There are some positive integers, such as 1 , that receive no arrows; these will be referred to as starting points in the sequel. When one starts at any of the starting points and keeps following the arrows, one is led to an infinite path, called its ray, that visits a strictly increasing sequence of positive integers. Since the length of any arrow is at most 2015, such a ray, say with starting point $s$, meets every interval of the form $[n, n+2014]$ with $n \geqslant s$ at least once. Suppose for the sake of contradiction that there would be at least 2016 starting points. Then we could take an integer $n$ that is larger than the first 2016 starting points. But now the interval $[n, n+2014]$ must be met by at least 2016 rays in distinct points, which is absurd. We have thereby shown that the number $b$ of starting points satisfies $1 \leqslant b \leqslant 2015$. Let $N$ denote any integer that is larger than all starting points. We contend that $b$ and $N$ are as required. To see this, let any two integers $m$ and $n$ with $n>m \geqslant N$ be given. The sum $\sum_{i=m+1}^{n} a_{i}$ gives the total length of the arrows emerging from $m+1, \ldots, n$. Taken together, these arrows form $b$ subpaths of our rays, some of which may be empty. Now on each ray we look at the first number that is larger than $m$; let $x_{1}, \ldots, x_{b}$ denote these numbers, and let $y_{1}, \ldots, y_{b}$ enumerate in corresponding order the numbers defined similarly with respect to $n$. Then the list of differences $y_{1}-x_{1}, \ldots, y_{b}-x_{b}$ consists of the lengths of these paths and possibly some zeros corresponding to empty paths. Consequently, we obtain $$ \sum_{i=m+1}^{n} a_{i}=\sum_{j=1}^{b}\left(y_{j}-x_{j}\right) $$ whence $$ \sum_{i=m+1}^{n}\left(a_{i}-b\right)=\sum_{j=1}^{b}\left(y_{j}-n\right)-\sum_{j=1}^{b}\left(x_{j}-m\right) . $$ Now each of the $b$ rays meets the interval $[m+1, m+2015]$ at some point and thus $x_{1}-$ $m, \ldots, x_{b}-m$ are $b$ distinct members of the set $\{1,2, \ldots, 2015\}$. Moreover, since $m+1$ is not a starting point, it must belong to some ray; so 1 has to appear among these numbers, wherefore $$ 1+\sum_{j=1}^{b-1}(j+1) \leqslant \sum_{j=1}^{b}\left(x_{j}-m\right) \leqslant 1+\sum_{j=1}^{b-1}(2016-b+j) $$ The same argument applied to $n$ and $y_{1}, \ldots, y_{b}$ yields $$ 1+\sum_{j=1}^{b-1}(j+1) \leqslant \sum_{j=1}^{b}\left(y_{j}-n\right) \leqslant 1+\sum_{j=1}^{b-1}(2016-b+j) $$ So altogether we get $$ \begin{aligned} \left|\sum_{i=m+1}^{n}\left(a_{i}-b\right)\right| & \leqslant \sum_{j=1}^{b-1}((2016-b+j)-(j+1))=(b-1)(2015-b) \\ & \leqslant\left(\frac{(b-1)+(2015-b)}{2}\right)^{2}=1007^{2}, \end{aligned} $$ as desired.
|
{
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|
985c561b-8d4e-5b28-90e9-4ccf43ee5d1d
| 24,697
|
Consider an infinite sequence $a_{1}, a_{2}, \ldots$ of positive integers with $a_{i} \leqslant 2015$ for all $i \geqslant 1$. Suppose that for any two distinct indices $i$ and $j$ we have $i+a_{i} \neq j+a_{j}$. Prove that there exist two positive integers $b$ and $N$ such that $$ \left|\sum_{i=m+1}^{n}\left(a_{i}-b\right)\right| \leqslant 1007^{2} $$ whenever $n>m \geqslant N$. (Australia)
|
Set $s_{n}=n+a_{n}$ for all positive integers $n$. By our assumptions, we have $$ n+1 \leqslant s_{n} \leqslant n+2015 $$ for all $n \in \mathbb{Z}_{>0}$. The members of the sequence $s_{1}, s_{2}, \ldots$ are distinct. We shall investigate the set $$ M=\mathbb{Z}_{>0} \backslash\left\{s_{1}, s_{2}, \ldots\right\} $$ Claim. At most 2015 numbers belong to $M$. Proof. Otherwise let $m_{1}<m_{2}<\cdots<m_{2016}$ be any 2016 distinct elements from $M$. For $n=m_{2016}$ we have $$ \left\{s_{1}, \ldots, s_{n}\right\} \cup\left\{m_{1}, \ldots, m_{2016}\right\} \subseteq\{1,2, \ldots, n+2015\} $$ where on the left-hand side we have a disjoint union containing altogether $n+2016$ elements. But the set on the right-hand side has only $n+2015$ elements. This contradiction proves our claim. Now we work towards proving that the positive integers $b=|M|$ and $N=\max (M)$ are as required. Recall that we have just shown $b \leqslant 2015$. Let us consider any integer $r \geqslant N$. As in the proof of the above claim, we see that $$ B_{r}=M \cup\left\{s_{1}, \ldots, s_{r}\right\} $$ is a subset of $[1, r+2015] \cap \mathbb{Z}$ with precisely $b+r$ elements. Due to the definitions of $M$ and $N$, we also know $[1, r+1] \cap \mathbb{Z} \subseteq B_{r}$. It follows that there is a set $C_{r} \subseteq\{1,2, \ldots, 2014\}$ with $\left|C_{r}\right|=b-1$ and $$ B_{r}=([1, r+1] \cap \mathbb{Z}) \cup\left\{r+1+x \mid x \in C_{r}\right\} $$ For any finite set of integers $J$ we denote the sum of its elements by $\sum J$. Now the equations (1) and (2) give rise to two ways of computing $\sum B_{r}$ and the comparison of both methods leads to $$ \sum M+\sum_{i=1}^{r} s_{i}=\sum_{i=1}^{r} i+b(r+1)+\sum C_{r} $$ or in other words to $$ \sum M+\sum_{i=1}^{r}\left(a_{i}-b\right)=b+\sum C_{r} $$ After this preparation, we consider any two integers $m$ and $n$ with $n>m \geqslant N$. Plugging $r=n$ and $r=m$ into (3) and subtracting the estimates that result, we deduce $$ \sum_{i=m+1}^{n}\left(a_{i}-b\right)=\sum C_{n}-\sum C_{m} $$ Since $C_{n}$ and $C_{m}$ are subsets of $\{1,2, \ldots, 2014\}$ with $\left|C_{n}\right|=\left|C_{m}\right|=b-1$, it is clear that the absolute value of the right-hand side of the above inequality attains its largest possible value if either $C_{m}=\{1,2, \ldots, b-1\}$ and $C_{n}=\{2016-b, \ldots, 2014\}$, or the other way around. In these two cases we have $$ \left|\sum C_{n}-\sum C_{m}\right|=(b-1)(2015-b) $$ so in the general case we find $$ \left|\sum_{i=m+1}^{n}\left(a_{i}-b\right)\right| \leqslant(b-1)(2015-b) \leqslant\left(\frac{(b-1)+(2015-b)}{2}\right)^{2}=1007^{2} $$ as desired. Comment. The sets $C_{n}$ may be visualized by means of the following process: Start with an empty blackboard. For $n \geqslant 1$, the following happens during the $n^{\text {th }}$ step. The number $a_{n}$ gets written on the blackboard, then all numbers currently on the blackboard are decreased by 1 , and finally all zeros that have arisen get swept away. It is not hard to see that the numbers present on the blackboard after $n$ steps are distinct and form the set $C_{n}$. Moreover, it is possible to complete a solution based on this idea.
|
proof
|
Yes
|
Yes
|
proof
|
Combinatorics
|
Consider an infinite sequence $a_{1}, a_{2}, \ldots$ of positive integers with $a_{i} \leqslant 2015$ for all $i \geqslant 1$. Suppose that for any two distinct indices $i$ and $j$ we have $i+a_{i} \neq j+a_{j}$. Prove that there exist two positive integers $b$ and $N$ such that $$ \left|\sum_{i=m+1}^{n}\left(a_{i}-b\right)\right| \leqslant 1007^{2} $$ whenever $n>m \geqslant N$. (Australia)
|
Set $s_{n}=n+a_{n}$ for all positive integers $n$. By our assumptions, we have $$ n+1 \leqslant s_{n} \leqslant n+2015 $$ for all $n \in \mathbb{Z}_{>0}$. The members of the sequence $s_{1}, s_{2}, \ldots$ are distinct. We shall investigate the set $$ M=\mathbb{Z}_{>0} \backslash\left\{s_{1}, s_{2}, \ldots\right\} $$ Claim. At most 2015 numbers belong to $M$. Proof. Otherwise let $m_{1}<m_{2}<\cdots<m_{2016}$ be any 2016 distinct elements from $M$. For $n=m_{2016}$ we have $$ \left\{s_{1}, \ldots, s_{n}\right\} \cup\left\{m_{1}, \ldots, m_{2016}\right\} \subseteq\{1,2, \ldots, n+2015\} $$ where on the left-hand side we have a disjoint union containing altogether $n+2016$ elements. But the set on the right-hand side has only $n+2015$ elements. This contradiction proves our claim. Now we work towards proving that the positive integers $b=|M|$ and $N=\max (M)$ are as required. Recall that we have just shown $b \leqslant 2015$. Let us consider any integer $r \geqslant N$. As in the proof of the above claim, we see that $$ B_{r}=M \cup\left\{s_{1}, \ldots, s_{r}\right\} $$ is a subset of $[1, r+2015] \cap \mathbb{Z}$ with precisely $b+r$ elements. Due to the definitions of $M$ and $N$, we also know $[1, r+1] \cap \mathbb{Z} \subseteq B_{r}$. It follows that there is a set $C_{r} \subseteq\{1,2, \ldots, 2014\}$ with $\left|C_{r}\right|=b-1$ and $$ B_{r}=([1, r+1] \cap \mathbb{Z}) \cup\left\{r+1+x \mid x \in C_{r}\right\} $$ For any finite set of integers $J$ we denote the sum of its elements by $\sum J$. Now the equations (1) and (2) give rise to two ways of computing $\sum B_{r}$ and the comparison of both methods leads to $$ \sum M+\sum_{i=1}^{r} s_{i}=\sum_{i=1}^{r} i+b(r+1)+\sum C_{r} $$ or in other words to $$ \sum M+\sum_{i=1}^{r}\left(a_{i}-b\right)=b+\sum C_{r} $$ After this preparation, we consider any two integers $m$ and $n$ with $n>m \geqslant N$. Plugging $r=n$ and $r=m$ into (3) and subtracting the estimates that result, we deduce $$ \sum_{i=m+1}^{n}\left(a_{i}-b\right)=\sum C_{n}-\sum C_{m} $$ Since $C_{n}$ and $C_{m}$ are subsets of $\{1,2, \ldots, 2014\}$ with $\left|C_{n}\right|=\left|C_{m}\right|=b-1$, it is clear that the absolute value of the right-hand side of the above inequality attains its largest possible value if either $C_{m}=\{1,2, \ldots, b-1\}$ and $C_{n}=\{2016-b, \ldots, 2014\}$, or the other way around. In these two cases we have $$ \left|\sum C_{n}-\sum C_{m}\right|=(b-1)(2015-b) $$ so in the general case we find $$ \left|\sum_{i=m+1}^{n}\left(a_{i}-b\right)\right| \leqslant(b-1)(2015-b) \leqslant\left(\frac{(b-1)+(2015-b)}{2}\right)^{2}=1007^{2} $$ as desired. Comment. The sets $C_{n}$ may be visualized by means of the following process: Start with an empty blackboard. For $n \geqslant 1$, the following happens during the $n^{\text {th }}$ step. The number $a_{n}$ gets written on the blackboard, then all numbers currently on the blackboard are decreased by 1 , and finally all zeros that have arisen get swept away. It is not hard to see that the numbers present on the blackboard after $n$ steps are distinct and form the set $C_{n}$. Moreover, it is possible to complete a solution based on this idea.
|
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|
985c561b-8d4e-5b28-90e9-4ccf43ee5d1d
| 24,697
|
Let $a, b$ and $c$ be positive real numbers such that $\min \{a b, b c, c a\} \geqslant 1$. Prove that $$ \sqrt[3]{\left(a^{2}+1\right)\left(b^{2}+1\right)\left(c^{2}+1\right)} \leqslant\left(\frac{a+b+c}{3}\right)^{2}+1 $$
|
We first show the following. - Claim. For any positive real numbers $x, y$ with $x y \geqslant 1$, we have $$ \left(x^{2}+1\right)\left(y^{2}+1\right) \leqslant\left(\left(\frac{x+y}{2}\right)^{2}+1\right)^{2} $$ Proof. Note that $x y \geqslant 1$ implies $\left(\frac{x+y}{2}\right)^{2}-1 \geqslant x y-1 \geqslant 0$. We find that $\left(x^{2}+1\right)\left(y^{2}+1\right)=(x y-1)^{2}+(x+y)^{2} \leqslant\left(\left(\frac{x+y}{2}\right)^{2}-1\right)^{2}+(x+y)^{2}=\left(\left(\frac{x+y}{2}\right)^{2}+1\right)^{2}$. Without loss of generality, assume $a \geqslant b \geqslant c$. This implies $a \geqslant 1$. Let $d=\frac{a+b+c}{3}$. Note that $$ a d=\frac{a(a+b+c)}{3} \geqslant \frac{1+1+1}{3}=1 . $$ Then we can apply (2) to the pair $(a, d)$ and the pair $(b, c)$. We get $$ \left(a^{2}+1\right)\left(d^{2}+1\right)\left(b^{2}+1\right)\left(c^{2}+1\right) \leqslant\left(\left(\frac{a+d}{2}\right)^{2}+1\right)^{2}\left(\left(\frac{b+c}{2}\right)^{2}+1\right)^{2} $$ Next, from $$ \frac{a+d}{2} \cdot \frac{b+c}{2} \geqslant \sqrt{a d} \cdot \sqrt{b c} \geqslant 1 $$ we can apply (2) again to the pair $\left(\frac{a+d}{2}, \frac{b+c}{2}\right)$. Together with (3), we have $$ \left(a^{2}+1\right)\left(d^{2}+1\right)\left(b^{2}+1\right)\left(c^{2}+1\right) \leqslant\left(\left(\frac{a+b+c+d}{4}\right)^{2}+1\right)^{4}=\left(d^{2}+1\right)^{4} $$ Therefore, $\left(a^{2}+1\right)\left(b^{2}+1\right)\left(c^{2}+1\right) \leqslant\left(d^{2}+1\right)^{3}$, and (1) follows by taking cube root of both sides. Comment. After justifying the Claim, one may also obtain (1) by mixing variables. Indeed, the function involved is clearly continuous, and hence it suffices to check that the condition $x y \geqslant 1$ is preserved under each mixing step. This is true since whenever $a b, b c, c a \geqslant 1$, we have $$ \frac{a+b}{2} \cdot \frac{a+b}{2} \geqslant a b \geqslant 1 \quad \text { and } \quad \frac{a+b}{2} \cdot c \geqslant \frac{1+1}{2}=1 . $$
|
proof
|
Yes
|
Yes
|
proof
|
Inequalities
|
Let $a, b$ and $c$ be positive real numbers such that $\min \{a b, b c, c a\} \geqslant 1$. Prove that $$ \sqrt[3]{\left(a^{2}+1\right)\left(b^{2}+1\right)\left(c^{2}+1\right)} \leqslant\left(\frac{a+b+c}{3}\right)^{2}+1 $$
|
We first show the following. - Claim. For any positive real numbers $x, y$ with $x y \geqslant 1$, we have $$ \left(x^{2}+1\right)\left(y^{2}+1\right) \leqslant\left(\left(\frac{x+y}{2}\right)^{2}+1\right)^{2} $$ Proof. Note that $x y \geqslant 1$ implies $\left(\frac{x+y}{2}\right)^{2}-1 \geqslant x y-1 \geqslant 0$. We find that $\left(x^{2}+1\right)\left(y^{2}+1\right)=(x y-1)^{2}+(x+y)^{2} \leqslant\left(\left(\frac{x+y}{2}\right)^{2}-1\right)^{2}+(x+y)^{2}=\left(\left(\frac{x+y}{2}\right)^{2}+1\right)^{2}$. Without loss of generality, assume $a \geqslant b \geqslant c$. This implies $a \geqslant 1$. Let $d=\frac{a+b+c}{3}$. Note that $$ a d=\frac{a(a+b+c)}{3} \geqslant \frac{1+1+1}{3}=1 . $$ Then we can apply (2) to the pair $(a, d)$ and the pair $(b, c)$. We get $$ \left(a^{2}+1\right)\left(d^{2}+1\right)\left(b^{2}+1\right)\left(c^{2}+1\right) \leqslant\left(\left(\frac{a+d}{2}\right)^{2}+1\right)^{2}\left(\left(\frac{b+c}{2}\right)^{2}+1\right)^{2} $$ Next, from $$ \frac{a+d}{2} \cdot \frac{b+c}{2} \geqslant \sqrt{a d} \cdot \sqrt{b c} \geqslant 1 $$ we can apply (2) again to the pair $\left(\frac{a+d}{2}, \frac{b+c}{2}\right)$. Together with (3), we have $$ \left(a^{2}+1\right)\left(d^{2}+1\right)\left(b^{2}+1\right)\left(c^{2}+1\right) \leqslant\left(\left(\frac{a+b+c+d}{4}\right)^{2}+1\right)^{4}=\left(d^{2}+1\right)^{4} $$ Therefore, $\left(a^{2}+1\right)\left(b^{2}+1\right)\left(c^{2}+1\right) \leqslant\left(d^{2}+1\right)^{3}$, and (1) follows by taking cube root of both sides. Comment. After justifying the Claim, one may also obtain (1) by mixing variables. Indeed, the function involved is clearly continuous, and hence it suffices to check that the condition $x y \geqslant 1$ is preserved under each mixing step. This is true since whenever $a b, b c, c a \geqslant 1$, we have $$ \frac{a+b}{2} \cdot \frac{a+b}{2} \geqslant a b \geqslant 1 \quad \text { and } \quad \frac{a+b}{2} \cdot c \geqslant \frac{1+1}{2}=1 . $$
|
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|
2ab5d9ed-56d8-549c-ba29-6d86d99a5866
| 24,728
|
Let $a, b$ and $c$ be positive real numbers such that $\min \{a b, b c, c a\} \geqslant 1$. Prove that $$ \sqrt[3]{\left(a^{2}+1\right)\left(b^{2}+1\right)\left(c^{2}+1\right)} \leqslant\left(\frac{a+b+c}{3}\right)^{2}+1 $$
|
Let $f(x)=\ln \left(1+x^{2}\right)$. Then the inequality (1) to be shown is equivalent to $$ \frac{f(a)+f(b)+f(c)}{3} \leqslant f\left(\frac{a+b+c}{3}\right), $$ while (2) becomes $$ \frac{f(x)+f(y)}{2} \leqslant f\left(\frac{x+y}{2}\right) $$ for $x y \geqslant 1$. Without loss of generality, assume $a \geqslant b \geqslant c$. From the Claim in Solution 1, we have $$ \frac{f(a)+f(b)+f(c)}{3} \leqslant \frac{f(a)+2 f\left(\frac{b+c}{2}\right)}{3} . $$ Note that $a \geqslant 1$ and $\frac{b+c}{2} \geqslant \sqrt{b c} \geqslant 1$. Since $$ f^{\prime \prime}(x)=\frac{2\left(1-x^{2}\right)}{\left(1+x^{2}\right)^{2}} $$ we know that $f$ is concave on $[1, \infty)$. Then we can apply Jensen's Theorem to get $$ \frac{f(a)+2 f\left(\frac{b+c}{2}\right)}{3} \leqslant f\left(\frac{a+2 \cdot \frac{b+c}{2}}{3}\right)=f\left(\frac{a+b+c}{3}\right) . $$ This completes the proof.
|
proof
|
Yes
|
Yes
|
proof
|
Inequalities
|
Let $a, b$ and $c$ be positive real numbers such that $\min \{a b, b c, c a\} \geqslant 1$. Prove that $$ \sqrt[3]{\left(a^{2}+1\right)\left(b^{2}+1\right)\left(c^{2}+1\right)} \leqslant\left(\frac{a+b+c}{3}\right)^{2}+1 $$
|
Let $f(x)=\ln \left(1+x^{2}\right)$. Then the inequality (1) to be shown is equivalent to $$ \frac{f(a)+f(b)+f(c)}{3} \leqslant f\left(\frac{a+b+c}{3}\right), $$ while (2) becomes $$ \frac{f(x)+f(y)}{2} \leqslant f\left(\frac{x+y}{2}\right) $$ for $x y \geqslant 1$. Without loss of generality, assume $a \geqslant b \geqslant c$. From the Claim in Solution 1, we have $$ \frac{f(a)+f(b)+f(c)}{3} \leqslant \frac{f(a)+2 f\left(\frac{b+c}{2}\right)}{3} . $$ Note that $a \geqslant 1$ and $\frac{b+c}{2} \geqslant \sqrt{b c} \geqslant 1$. Since $$ f^{\prime \prime}(x)=\frac{2\left(1-x^{2}\right)}{\left(1+x^{2}\right)^{2}} $$ we know that $f$ is concave on $[1, \infty)$. Then we can apply Jensen's Theorem to get $$ \frac{f(a)+2 f\left(\frac{b+c}{2}\right)}{3} \leqslant f\left(\frac{a+2 \cdot \frac{b+c}{2}}{3}\right)=f\left(\frac{a+b+c}{3}\right) . $$ This completes the proof.
|
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|
2ab5d9ed-56d8-549c-ba29-6d86d99a5866
| 24,728
|
(a) Prove that for every positive integer $n$, there exists a fraction $\frac{a}{b}$ where $a$ and $b$ are integers satisfying $0<b \leqslant \sqrt{n}+1$ and $\sqrt{n} \leqslant \frac{a}{b} \leqslant \sqrt{n+1}$. (b) Prove that there are infinitely many positive integers $n$ such that there is no fraction $\frac{a}{b}$ where $a$ and $b$ are integers satisfying $0<b \leqslant \sqrt{n}$ and $\sqrt{n} \leqslant \frac{a}{b} \leqslant \sqrt{n+1}$.
|
(a) Let $r$ be the unique positive integer for which $r^{2} \leqslant n<(r+1)^{2}$. Write $n=r^{2}+s$. Then we have $0 \leqslant s \leqslant 2 r$. We discuss in two cases according to the parity of $s$. - Case 1. $s$ is even. Consider the number $\left(r+\frac{s}{2 r}\right)^{2}=r^{2}+s+\left(\frac{s}{2 r}\right)^{2}$. We find that $$ n=r^{2}+s \leqslant r^{2}+s+\left(\frac{s}{2 r}\right)^{2} \leqslant r^{2}+s+1=n+1 $$ It follows that $$ \sqrt{n} \leqslant r+\frac{s}{2 r} \leqslant \sqrt{n+1} $$ Since $s$ is even, we can choose the fraction $r+\frac{s}{2 r}=\frac{r^{2}+(s / 2)}{r}$ since $r \leqslant \sqrt{n}$. - Case 2. $s$ is odd. Consider the number $\left(r+1-\frac{2 r+1-s}{2(r+1)}\right)^{2}=(r+1)^{2}-(2 r+1-s)+\left(\frac{2 r+1-s}{2(r+1)}\right)^{2}$. We find that $$ \begin{aligned} n=r^{2}+s=(r+1)^{2}-(2 r+1-s) & \leqslant(r+1)^{2}-(2 r+1-s)+\left(\frac{2 r+1-s}{2(r+1)}\right)^{2} \\ & \leqslant(r+1)^{2}-(2 r+1-s)+1=n+1 \end{aligned} $$ It follows that $$ \sqrt{n} \leqslant r+1-\frac{2 r+1-s}{2(r+1)} \leqslant \sqrt{n+1} $$ Since $s$ is odd, we can choose the fraction $(r+1)-\frac{2 r+1-s}{2(r+1)}=\frac{(r+1)^{2}-r+((s-1) / 2)}{r+1}$ since $r+1 \leqslant \sqrt{n}+1$. (b) We show that for every positive integer $r$, there is no fraction $\frac{a}{b}$ with $b \leqslant \sqrt{r^{2}+1}$ such that $\sqrt{r^{2}+1} \leqslant \frac{a}{b} \leqslant \sqrt{r^{2}+2}$. Suppose on the contrary that such a fraction exists. Since $b \leqslant \sqrt{r^{2}+1}<r+1$ and $b$ is an integer, we have $b \leqslant r$. Hence, $$ (b r)^{2}<b^{2}\left(r^{2}+1\right) \leqslant a^{2} \leqslant b^{2}\left(r^{2}+2\right) \leqslant b^{2} r^{2}+2 b r<(b r+1)^{2} $$ This shows the square number $a^{2}$ is strictly bounded between the two consecutive squares $(b r)^{2}$ and $(b r+1)^{2}$, which is impossible. Hence, we have found infinitely many $n=r^{2}+1$ for which there is no fraction of the desired form.
|
proof
|
Yes
|
Yes
|
proof
|
Number Theory
|
(a) Prove that for every positive integer $n$, there exists a fraction $\frac{a}{b}$ where $a$ and $b$ are integers satisfying $0<b \leqslant \sqrt{n}+1$ and $\sqrt{n} \leqslant \frac{a}{b} \leqslant \sqrt{n+1}$. (b) Prove that there are infinitely many positive integers $n$ such that there is no fraction $\frac{a}{b}$ where $a$ and $b$ are integers satisfying $0<b \leqslant \sqrt{n}$ and $\sqrt{n} \leqslant \frac{a}{b} \leqslant \sqrt{n+1}$.
|
(a) Let $r$ be the unique positive integer for which $r^{2} \leqslant n<(r+1)^{2}$. Write $n=r^{2}+s$. Then we have $0 \leqslant s \leqslant 2 r$. We discuss in two cases according to the parity of $s$. - Case 1. $s$ is even. Consider the number $\left(r+\frac{s}{2 r}\right)^{2}=r^{2}+s+\left(\frac{s}{2 r}\right)^{2}$. We find that $$ n=r^{2}+s \leqslant r^{2}+s+\left(\frac{s}{2 r}\right)^{2} \leqslant r^{2}+s+1=n+1 $$ It follows that $$ \sqrt{n} \leqslant r+\frac{s}{2 r} \leqslant \sqrt{n+1} $$ Since $s$ is even, we can choose the fraction $r+\frac{s}{2 r}=\frac{r^{2}+(s / 2)}{r}$ since $r \leqslant \sqrt{n}$. - Case 2. $s$ is odd. Consider the number $\left(r+1-\frac{2 r+1-s}{2(r+1)}\right)^{2}=(r+1)^{2}-(2 r+1-s)+\left(\frac{2 r+1-s}{2(r+1)}\right)^{2}$. We find that $$ \begin{aligned} n=r^{2}+s=(r+1)^{2}-(2 r+1-s) & \leqslant(r+1)^{2}-(2 r+1-s)+\left(\frac{2 r+1-s}{2(r+1)}\right)^{2} \\ & \leqslant(r+1)^{2}-(2 r+1-s)+1=n+1 \end{aligned} $$ It follows that $$ \sqrt{n} \leqslant r+1-\frac{2 r+1-s}{2(r+1)} \leqslant \sqrt{n+1} $$ Since $s$ is odd, we can choose the fraction $(r+1)-\frac{2 r+1-s}{2(r+1)}=\frac{(r+1)^{2}-r+((s-1) / 2)}{r+1}$ since $r+1 \leqslant \sqrt{n}+1$. (b) We show that for every positive integer $r$, there is no fraction $\frac{a}{b}$ with $b \leqslant \sqrt{r^{2}+1}$ such that $\sqrt{r^{2}+1} \leqslant \frac{a}{b} \leqslant \sqrt{r^{2}+2}$. Suppose on the contrary that such a fraction exists. Since $b \leqslant \sqrt{r^{2}+1}<r+1$ and $b$ is an integer, we have $b \leqslant r$. Hence, $$ (b r)^{2}<b^{2}\left(r^{2}+1\right) \leqslant a^{2} \leqslant b^{2}\left(r^{2}+2\right) \leqslant b^{2} r^{2}+2 b r<(b r+1)^{2} $$ This shows the square number $a^{2}$ is strictly bounded between the two consecutive squares $(b r)^{2}$ and $(b r+1)^{2}$, which is impossible. Hence, we have found infinitely many $n=r^{2}+1$ for which there is no fraction of the desired form.
|
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6e65626a-1606-5247-b7ba-7f896c969107
| 24,742
|
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. $\mathbf{C 4}$. 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.
|
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 .
|
proof
|
Yes
|
Yes
|
proof
|
Combinatorics
|
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. $\mathbf{C 4}$. 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.
|
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 .
|
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1e8a0027-511b-58ea-a42b-9161c6ecbec8
| 24,766
|
There are $n \geqslant 3$ islands in a city. Initially, the ferry company offers some routes between some pairs of islands so that it is impossible to divide the islands into two groups such that no two islands in different groups are connected by a ferry route. After each year, the ferry company will close a ferry route between some two islands $X$ and $Y$. At the same time, in order to maintain its service, the company will open new routes according to the following rule: for any island which is connected by a ferry route to exactly one of $X$ and $Y$, a new route between this island and the other of $X$ and $Y$ is added. Suppose at any moment, if we partition all islands into two nonempty groups in any way, then it is known that the ferry company will close a certain route connecting two islands from the two groups after some years. Prove that after some years there will be an island which is connected to all other islands by ferry routes.
|
Initially, we pick any pair of islands $A$ and $B$ which are connected by a ferry route and put $A$ in set $\mathcal{A}$ and $B$ in set $\mathcal{B}$. From the condition, without loss of generality there must be another island which is connected to $A$. We put such an island $C$ in set $\mathcal{B}$. We say that two sets of islands form a network if each island in one set is connected to each island in the other set. Next, we shall included all islands to $\mathcal{A} \cup \mathcal{B}$ one by one. Suppose we have two sets $\mathcal{A}$ and $\mathcal{B}$ which form a network where $3 \leqslant|\mathcal{A} \cup \mathcal{B}|<n$. This relation no longer holds only when a ferry route between islands $A \in \mathcal{A}$ and $B \in \mathcal{B}$ is closed. In that case, we define $\mathcal{A}^{\prime}=\{A, B\}$, and $\mathcal{B}^{\prime}=(\mathcal{A} \cup \mathcal{B})-\{A, B\}$. Note that $\mathcal{B}^{\prime}$ is nonempty. Consider any island $C \in \mathcal{A}-\{A\}$. From the relation of $\mathcal{A}$ and $\mathcal{B}$, we know that $C$ is connected to $B$. If $C$ was not connected to $A$ before the route between $A$ and $B$ closes, then there will be a route added between $C$ and $A$ afterwards. Hence, $C$ must now be connected to both $A$ and $B$. The same holds true for any island in $\mathcal{B}-\{B\}$. Therefore, $\mathcal{A}^{\prime}$ and $\mathcal{B}^{\prime}$ form a network, and $\mathcal{A}^{\prime} \cup \mathcal{B}^{\prime}=\mathcal{A} \cup \mathcal{B}$. Hence these islands can always be partitioned into sets $\mathcal{A}$ and $\mathcal{B}$ which form a network. As $|\mathcal{A} \cup \mathcal{B}|<n$, there are some islands which are not included in $\mathcal{A} \cup \mathcal{B}$. From the condition, after some years there must be a ferry route between an island $A$ in $\mathcal{A} \cup \mathcal{B}$ and an island $D$ outside $\mathcal{A} \cup \mathcal{B}$ which closes. Without loss of generality assume $A \in \mathcal{A}$. Then each island in $\mathcal{B}$ must then be connected to $D$, no matter it was or not before. Hence, we can put $D$ in set $\mathcal{A}$ so that the new sets $\mathcal{A}$ and $\mathcal{B}$ still form a network and the size of $\mathcal{A} \cup \mathcal{B}$ is increased by 1 . The same process can be done to increase the size of $\mathcal{A} \cup \mathcal{B}$. Eventually, all islands are included in this way so we may now assume $|\mathcal{A} \cup \mathcal{B}|=n$. Suppose a ferry route between $A \in \mathcal{A}$ and $B \in \mathcal{B}$ is closed after some years. We put $A$ and $B$ in set $\mathcal{A}^{\prime}$ and all remaining islands in set $\mathcal{B}^{\prime}$. Then $\mathcal{A}^{\prime}$ and $\mathcal{B}^{\prime}$ form a network. This relation no longer holds only when a route between $A$, without loss of generality, and $C \in \mathcal{B}^{\prime}$ is closed. Since this must eventually occur, at that time island $B$ will be connected to all other islands and the result follows.
|
proof
|
Yes
|
Yes
|
proof
|
Combinatorics
|
There are $n \geqslant 3$ islands in a city. Initially, the ferry company offers some routes between some pairs of islands so that it is impossible to divide the islands into two groups such that no two islands in different groups are connected by a ferry route. After each year, the ferry company will close a ferry route between some two islands $X$ and $Y$. At the same time, in order to maintain its service, the company will open new routes according to the following rule: for any island which is connected by a ferry route to exactly one of $X$ and $Y$, a new route between this island and the other of $X$ and $Y$ is added. Suppose at any moment, if we partition all islands into two nonempty groups in any way, then it is known that the ferry company will close a certain route connecting two islands from the two groups after some years. Prove that after some years there will be an island which is connected to all other islands by ferry routes.
|
Initially, we pick any pair of islands $A$ and $B$ which are connected by a ferry route and put $A$ in set $\mathcal{A}$ and $B$ in set $\mathcal{B}$. From the condition, without loss of generality there must be another island which is connected to $A$. We put such an island $C$ in set $\mathcal{B}$. We say that two sets of islands form a network if each island in one set is connected to each island in the other set. Next, we shall included all islands to $\mathcal{A} \cup \mathcal{B}$ one by one. Suppose we have two sets $\mathcal{A}$ and $\mathcal{B}$ which form a network where $3 \leqslant|\mathcal{A} \cup \mathcal{B}|<n$. This relation no longer holds only when a ferry route between islands $A \in \mathcal{A}$ and $B \in \mathcal{B}$ is closed. In that case, we define $\mathcal{A}^{\prime}=\{A, B\}$, and $\mathcal{B}^{\prime}=(\mathcal{A} \cup \mathcal{B})-\{A, B\}$. Note that $\mathcal{B}^{\prime}$ is nonempty. Consider any island $C \in \mathcal{A}-\{A\}$. From the relation of $\mathcal{A}$ and $\mathcal{B}$, we know that $C$ is connected to $B$. If $C$ was not connected to $A$ before the route between $A$ and $B$ closes, then there will be a route added between $C$ and $A$ afterwards. Hence, $C$ must now be connected to both $A$ and $B$. The same holds true for any island in $\mathcal{B}-\{B\}$. Therefore, $\mathcal{A}^{\prime}$ and $\mathcal{B}^{\prime}$ form a network, and $\mathcal{A}^{\prime} \cup \mathcal{B}^{\prime}=\mathcal{A} \cup \mathcal{B}$. Hence these islands can always be partitioned into sets $\mathcal{A}$ and $\mathcal{B}$ which form a network. As $|\mathcal{A} \cup \mathcal{B}|<n$, there are some islands which are not included in $\mathcal{A} \cup \mathcal{B}$. From the condition, after some years there must be a ferry route between an island $A$ in $\mathcal{A} \cup \mathcal{B}$ and an island $D$ outside $\mathcal{A} \cup \mathcal{B}$ which closes. Without loss of generality assume $A \in \mathcal{A}$. Then each island in $\mathcal{B}$ must then be connected to $D$, no matter it was or not before. Hence, we can put $D$ in set $\mathcal{A}$ so that the new sets $\mathcal{A}$ and $\mathcal{B}$ still form a network and the size of $\mathcal{A} \cup \mathcal{B}$ is increased by 1 . The same process can be done to increase the size of $\mathcal{A} \cup \mathcal{B}$. Eventually, all islands are included in this way so we may now assume $|\mathcal{A} \cup \mathcal{B}|=n$. Suppose a ferry route between $A \in \mathcal{A}$ and $B \in \mathcal{B}$ is closed after some years. We put $A$ and $B$ in set $\mathcal{A}^{\prime}$ and all remaining islands in set $\mathcal{B}^{\prime}$. Then $\mathcal{A}^{\prime}$ and $\mathcal{B}^{\prime}$ form a network. This relation no longer holds only when a route between $A$, without loss of generality, and $C \in \mathcal{B}^{\prime}$ is closed. Since this must eventually occur, at that time island $B$ will be connected to all other islands and the result follows.
|
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c7e5ec23-cd36-5489-a9ac-e7544a2997f7
| 24,773
|
Let $n \geqslant 2$ be an integer. In the plane, there are $n$ segments given in such a way that any two segments have an intersection point in the interior, and no three segments intersect at a single point. Jeff places a snail at one of the endpoints of each of the segments and claps his hands $n-1$ times. Each time when he claps his hands, all the snails move along their own segments and stay at the next intersection points until the next clap. Since there are $n-1$ intersection points on each segment, all snails will reach the furthest intersection points from their starting points after $n-1$ claps. (a) Prove that if $n$ is odd then Jeff can always place the snails so that no two of them ever occupy the same intersection point. (b) Prove that if $n$ is even then there must be a moment when some two snails occupy the same intersection point no matter how Jeff places the snails.
|
We consider a big disk which contains all the segments. We extend each segment to a line $l_{i}$ so that each of them cuts the disk at two distinct points $A_{i}, B_{i}$. (a) For odd $n$, we travel along the circumference of the disk and mark each of the points $A_{i}$ or $B_{i}$ 'in' and 'out' alternately. Since every pair of lines intersect in the disk, there are exactly $n-1$ points between $A_{i}$ and $B_{i}$ for any fixed $1 \leqslant i \leqslant n$. As $n$ is odd, this means one of $A_{i}$ and $B_{i}$ is marked 'in' and the other is marked 'out'. Then Jeff can put a snail on the endpoint of each segment which is closer to the 'in' side of the corresponding line. We claim that the snails on $l_{i}$ and $l_{j}$ do not meet for any pairs $i, j$, hence proving part (a).  Without loss of generality, we may assume the snails start at $A_{i}$ and $A_{j}$ respectively. Let $l_{i}$ intersect $l_{j}$ at $P$. Note that there is an odd number of points between arc $A_{i} A_{j}$. Each of these points belongs to a line $l_{k}$. Such a line $l_{k}$ must intersect exactly one of the segments $A_{i} P$ and $A_{j} P$, making an odd number of intersections. For the other lines, they may intersect both segments $A_{i} P$ and $A_{j} P$, or meet none of them. Therefore, the total number of intersection points on segments $A_{i} P$ and $A_{j} P$ (not counting $P$ ) is odd. However, if the snails arrive at $P$ at the same time, then there should be the same number of intersections on $A_{i} P$ and $A_{j} P$, which gives an even number of intersections. This is a contradiction so the snails do not meet each other. (b) For even $n$, we consider any way that Jeff places the snails and mark each of the points $A_{i}$ or $B_{i}$ 'in' and 'out' according to the directions travelled by the snails. In this case there must be two neighbouring points $A_{i}$ and $A_{j}$ both of which are marked 'in'. Let $P$ be the intersection of the segments $A_{i} B_{i}$ and $A_{j} B_{j}$. Then any other segment meeting one of the segments $A_{i} P$ and $A_{j} P$ must also meet the other one, and so the number of intersections on $A_{i} P$ and $A_{j} P$ are the same. This shows the snails starting from $A_{i}$ and $A_{j}$ will meet at $P$. Comment. The conclusions do not hold for pseudosegments, as can be seen from the following examples. 
|
proof
|
Yes
|
Yes
|
proof
|
Combinatorics
|
Let $n \geqslant 2$ be an integer. In the plane, there are $n$ segments given in such a way that any two segments have an intersection point in the interior, and no three segments intersect at a single point. Jeff places a snail at one of the endpoints of each of the segments and claps his hands $n-1$ times. Each time when he claps his hands, all the snails move along their own segments and stay at the next intersection points until the next clap. Since there are $n-1$ intersection points on each segment, all snails will reach the furthest intersection points from their starting points after $n-1$ claps. (a) Prove that if $n$ is odd then Jeff can always place the snails so that no two of them ever occupy the same intersection point. (b) Prove that if $n$ is even then there must be a moment when some two snails occupy the same intersection point no matter how Jeff places the snails.
|
We consider a big disk which contains all the segments. We extend each segment to a line $l_{i}$ so that each of them cuts the disk at two distinct points $A_{i}, B_{i}$. (a) For odd $n$, we travel along the circumference of the disk and mark each of the points $A_{i}$ or $B_{i}$ 'in' and 'out' alternately. Since every pair of lines intersect in the disk, there are exactly $n-1$ points between $A_{i}$ and $B_{i}$ for any fixed $1 \leqslant i \leqslant n$. As $n$ is odd, this means one of $A_{i}$ and $B_{i}$ is marked 'in' and the other is marked 'out'. Then Jeff can put a snail on the endpoint of each segment which is closer to the 'in' side of the corresponding line. We claim that the snails on $l_{i}$ and $l_{j}$ do not meet for any pairs $i, j$, hence proving part (a).  Without loss of generality, we may assume the snails start at $A_{i}$ and $A_{j}$ respectively. Let $l_{i}$ intersect $l_{j}$ at $P$. Note that there is an odd number of points between arc $A_{i} A_{j}$. Each of these points belongs to a line $l_{k}$. Such a line $l_{k}$ must intersect exactly one of the segments $A_{i} P$ and $A_{j} P$, making an odd number of intersections. For the other lines, they may intersect both segments $A_{i} P$ and $A_{j} P$, or meet none of them. Therefore, the total number of intersection points on segments $A_{i} P$ and $A_{j} P$ (not counting $P$ ) is odd. However, if the snails arrive at $P$ at the same time, then there should be the same number of intersections on $A_{i} P$ and $A_{j} P$, which gives an even number of intersections. This is a contradiction so the snails do not meet each other. (b) For even $n$, we consider any way that Jeff places the snails and mark each of the points $A_{i}$ or $B_{i}$ 'in' and 'out' according to the directions travelled by the snails. In this case there must be two neighbouring points $A_{i}$ and $A_{j}$ both of which are marked 'in'. Let $P$ be the intersection of the segments $A_{i} B_{i}$ and $A_{j} B_{j}$. Then any other segment meeting one of the segments $A_{i} P$ and $A_{j} P$ must also meet the other one, and so the number of intersections on $A_{i} P$ and $A_{j} P$ are the same. This shows the snails starting from $A_{i}$ and $A_{j}$ will meet at $P$. Comment. The conclusions do not hold for pseudosegments, as can be seen from the following examples. 
|
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252affe6-a3cd-5c1b-8918-efa8b67d8edc
| 24,776
|
In a convex pentagon $A B C D E$, let $F$ be a point on $A C$ such that $\angle F B C=90^{\circ}$. Suppose triangles $A B F, A C D$ and $A D E$ are similar isosceles triangles with $$ \angle F A B=\angle F B A=\angle D A C=\angle D C A=\angle E A D=\angle E D A $$ Let $M$ be the midpoint of $C F$. Point $X$ is chosen such that $A M X E$ is a parallelogram. Show that $B D, E M$ and $F X$ are concurrent.
|
Denote the common angle in (1) by $\theta$. As $\triangle A B F \sim \triangle A C D$, we have $\frac{A B}{A C}=\frac{A F}{A D}$ so that $\triangle A B C \sim \triangle A F D$. From $E A=E D$, we get $$ \angle A F D=\angle A B C=90^{\circ}+\theta=180^{\circ}-\frac{1}{2} \angle A E D . $$ Hence, $F$ lies on the circle with centre $E$ and radius $E A$. In particular, $E F=E A=E D$. As $\angle E F A=\angle E A F=2 \theta=\angle B F C$, points $B, F, E$ are collinear. As $\angle E D A=\angle M A D$, we have $E D / / A M$ and hence $E, D, X$ are collinear. As $M$ is the midpoint of $C F$ and $\angle C B F=90^{\circ}$, we get $M F=M B$. In the isosceles triangles $E F A$ and $M F B$, we have $\angle E F A=\angle M F B$ and $A F=B F$. Therefore, they are congruent to each other. Then we have $B M=A E=X M$ and $B E=B F+F E=A F+F M=A M=E X$. This shows $\triangle E M B \cong \triangle E M X$. As $F$ and $D$ lie on $E B$ and $E X$ respectively and $E F=E D$, we know that lines $B D$ and $X F$ are symmetric with respect to $E M$. It follows that the three lines are concurrent. 
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
In a convex pentagon $A B C D E$, let $F$ be a point on $A C$ such that $\angle F B C=90^{\circ}$. Suppose triangles $A B F, A C D$ and $A D E$ are similar isosceles triangles with $$ \angle F A B=\angle F B A=\angle D A C=\angle D C A=\angle E A D=\angle E D A $$ Let $M$ be the midpoint of $C F$. Point $X$ is chosen such that $A M X E$ is a parallelogram. Show that $B D, E M$ and $F X$ are concurrent.
|
Denote the common angle in (1) by $\theta$. As $\triangle A B F \sim \triangle A C D$, we have $\frac{A B}{A C}=\frac{A F}{A D}$ so that $\triangle A B C \sim \triangle A F D$. From $E A=E D$, we get $$ \angle A F D=\angle A B C=90^{\circ}+\theta=180^{\circ}-\frac{1}{2} \angle A E D . $$ Hence, $F$ lies on the circle with centre $E$ and radius $E A$. In particular, $E F=E A=E D$. As $\angle E F A=\angle E A F=2 \theta=\angle B F C$, points $B, F, E$ are collinear. As $\angle E D A=\angle M A D$, we have $E D / / A M$ and hence $E, D, X$ are collinear. As $M$ is the midpoint of $C F$ and $\angle C B F=90^{\circ}$, we get $M F=M B$. In the isosceles triangles $E F A$ and $M F B$, we have $\angle E F A=\angle M F B$ and $A F=B F$. Therefore, they are congruent to each other. Then we have $B M=A E=X M$ and $B E=B F+F E=A F+F M=A M=E X$. This shows $\triangle E M B \cong \triangle E M X$. As $F$ and $D$ lie on $E B$ and $E X$ respectively and $E F=E D$, we know that lines $B D$ and $X F$ are symmetric with respect to $E M$. It follows that the three lines are concurrent. 
|
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|
07e6f558-2934-5040-989d-58b6a169e993
| 24,782
|
In a convex pentagon $A B C D E$, let $F$ be a point on $A C$ such that $\angle F B C=90^{\circ}$. Suppose triangles $A B F, A C D$ and $A D E$ are similar isosceles triangles with $$ \angle F A B=\angle F B A=\angle D A C=\angle D C A=\angle E A D=\angle E D A $$ Let $M$ be the midpoint of $C F$. Point $X$ is chosen such that $A M X E$ is a parallelogram. Show that $B D, E M$ and $F X$ are concurrent.
|
From $\angle C A D=\angle E D A$, we have $A C / / E D$. Together with $A C / / E X$, we know that $E, D, X$ are collinear. Denote the common angle in (1) by $\theta$. From $\triangle A B F \sim \triangle A C D$, we get $\frac{A B}{A C}=\frac{A F}{A D}$ so that $\triangle A B C \sim \triangle A F D$. This yields $\angle A F D=\angle A B C=90^{\circ}+\theta$ and hence $\angle F D C=90^{\circ}$, implying that $B C D F$ is cyclic. Let $\Gamma_{1}$ be its circumcircle. Next, from $\triangle A B F \sim \triangle A D E$, we have $\frac{A B}{A D}=\frac{A F}{A E}$ so that $\triangle A B D \sim \triangle A F E$. Therefore, $$ \angle A F E=\angle A B D=\theta+\angle F B D=\theta+\angle F C D=2 \theta=180^{\circ}-\angle B F A . $$ This implies $B, F, E$ are collinear. Note that $F$ is the incentre of triangle $D A B$. Point $E$ lies on the internal angle bisector of $\angle D B A$ and lies on the perpendicular bisector of $A D$. It follows that $E$ lies on the circumcircle $\Gamma_{2}$ of triangle $A B D$, and $E A=E F=E D$. Also, since $C F$ is a diameter of $\Gamma_{1}$ and $M$ is the midpoint of $C F, M$ is the centre of $\Gamma_{1}$ and hence $\angle A M D=2 \theta=\angle A B D$. This shows $M$ lies on $\Gamma_{2}$. Next, $\angle M D X=\angle M A E=\angle D X M$ since $A M X E$ is a parallelogram. Hence $M D=M X$ and $X$ lies on $\Gamma_{1}$.  We now have two ways to complete the solution. - Method 1. From $E F=E A=X M$ and $E X / / F M, E F M X$ is an isosceles trapezoid and is cyclic. Denote its circumcircle by $\Gamma_{3}$. Since $B D, E M, F X$ are the three radical axes of $\Gamma_{1}, \Gamma_{2}, \Gamma_{3}$, they must be concurrent. - Method 2. As $\angle D M F=2 \theta=\angle B F M$, we have $D M / / E B$. Also, $$ \angle B F D+\angle X B F=\angle B F C+\angle C F D+90^{\circ}-\angle C B X=2 \theta+\left(90^{\circ}-\theta\right)+90^{\circ}-\theta=180^{\circ} $$ implies $D F / / X B$. These show the corresponding sides of triangles $D M F$ and $B E X$ are parallel. By Desargues' Theorem, the two triangles are perspective and hence $D B, M E, F X$ meet at a point. Comment. In Solution 2, both the Radical Axis Theorem and Desargues' Theorem could imply $D B, M E, F X$ are parallel. However, this is impossible as can be seen from the configuration. For example, it is obvious that $D B$ and $M E$ meet each other.
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
In a convex pentagon $A B C D E$, let $F$ be a point on $A C$ such that $\angle F B C=90^{\circ}$. Suppose triangles $A B F, A C D$ and $A D E$ are similar isosceles triangles with $$ \angle F A B=\angle F B A=\angle D A C=\angle D C A=\angle E A D=\angle E D A $$ Let $M$ be the midpoint of $C F$. Point $X$ is chosen such that $A M X E$ is a parallelogram. Show that $B D, E M$ and $F X$ are concurrent.
|
From $\angle C A D=\angle E D A$, we have $A C / / E D$. Together with $A C / / E X$, we know that $E, D, X$ are collinear. Denote the common angle in (1) by $\theta$. From $\triangle A B F \sim \triangle A C D$, we get $\frac{A B}{A C}=\frac{A F}{A D}$ so that $\triangle A B C \sim \triangle A F D$. This yields $\angle A F D=\angle A B C=90^{\circ}+\theta$ and hence $\angle F D C=90^{\circ}$, implying that $B C D F$ is cyclic. Let $\Gamma_{1}$ be its circumcircle. Next, from $\triangle A B F \sim \triangle A D E$, we have $\frac{A B}{A D}=\frac{A F}{A E}$ so that $\triangle A B D \sim \triangle A F E$. Therefore, $$ \angle A F E=\angle A B D=\theta+\angle F B D=\theta+\angle F C D=2 \theta=180^{\circ}-\angle B F A . $$ This implies $B, F, E$ are collinear. Note that $F$ is the incentre of triangle $D A B$. Point $E$ lies on the internal angle bisector of $\angle D B A$ and lies on the perpendicular bisector of $A D$. It follows that $E$ lies on the circumcircle $\Gamma_{2}$ of triangle $A B D$, and $E A=E F=E D$. Also, since $C F$ is a diameter of $\Gamma_{1}$ and $M$ is the midpoint of $C F, M$ is the centre of $\Gamma_{1}$ and hence $\angle A M D=2 \theta=\angle A B D$. This shows $M$ lies on $\Gamma_{2}$. Next, $\angle M D X=\angle M A E=\angle D X M$ since $A M X E$ is a parallelogram. Hence $M D=M X$ and $X$ lies on $\Gamma_{1}$.  We now have two ways to complete the solution. - Method 1. From $E F=E A=X M$ and $E X / / F M, E F M X$ is an isosceles trapezoid and is cyclic. Denote its circumcircle by $\Gamma_{3}$. Since $B D, E M, F X$ are the three radical axes of $\Gamma_{1}, \Gamma_{2}, \Gamma_{3}$, they must be concurrent. - Method 2. As $\angle D M F=2 \theta=\angle B F M$, we have $D M / / E B$. Also, $$ \angle B F D+\angle X B F=\angle B F C+\angle C F D+90^{\circ}-\angle C B X=2 \theta+\left(90^{\circ}-\theta\right)+90^{\circ}-\theta=180^{\circ} $$ implies $D F / / X B$. These show the corresponding sides of triangles $D M F$ and $B E X$ are parallel. By Desargues' Theorem, the two triangles are perspective and hence $D B, M E, F X$ meet at a point. Comment. In Solution 2, both the Radical Axis Theorem and Desargues' Theorem could imply $D B, M E, F X$ are parallel. However, this is impossible as can be seen from the configuration. For example, it is obvious that $D B$ and $M E$ meet each other.
|
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07e6f558-2934-5040-989d-58b6a169e993
| 24,782
|
In a convex pentagon $A B C D E$, let $F$ be a point on $A C$ such that $\angle F B C=90^{\circ}$. Suppose triangles $A B F, A C D$ and $A D E$ are similar isosceles triangles with $$ \angle F A B=\angle F B A=\angle D A C=\angle D C A=\angle E A D=\angle E D A $$ Let $M$ be the midpoint of $C F$. Point $X$ is chosen such that $A M X E$ is a parallelogram. Show that $B D, E M$ and $F X$ are concurrent.
|
Let the common angle in (1) be $\theta$. From $\triangle A B F \sim \triangle A C D$, we have $\frac{A B}{A C}=\frac{A F}{A D}$ so that $\triangle A B C \sim \triangle A F D$. Then $\angle A D F=\angle A C B=90^{\circ}-2 \theta=90^{\circ}-\angle B A D$ and hence $D F \perp A B$. As $F A=F B$, this implies $\triangle D A B$ is isosceles with $D A=D B$. Then $F$ is the incentre of $\triangle D A B$. Next, from $\angle A E D=180^{\circ}-2 \theta=180^{\circ}-\angle D B A$, points $A, B, D, E$ are concyclic. Since we also have $E A=E D$, this shows $E, F, B$ are collinear and $E A=E F=E D$.  Note that $C$ lies on the internal angle bisector of $\angle B A D$ and lies on the external angle bisector of $\angle D B A$. It follows that it is the $A$-excentre of triangle $D A B$. As $M$ is the midpoint of $C F, M$ lies on the circumcircle of triangle $D A B$ and it is the centre of the circle passing through $D, F, B, C$. By symmetry, $D E F M$ is a rhombus. Then the midpoints of $A X, E M$ and $D F$ coincide, and it follows that $D A F X$ is a parallelogram. Let $P$ be the intersection of $B D$ and $E M$, and $Q$ be the intersection of $A D$ and $B E$. From $\angle B A C=\angle D C A$, we know that $D C, A B, E M$ are parallel. Thus we have $\frac{D P}{P B}=\frac{C M}{M A}$. This is further equal to $\frac{A E}{B E}$ since $C M=D M=D E=A E$ and $M A=B E$. From $\triangle A E Q \sim \triangle B E A$, we find that $$ \frac{D P}{P B}=\frac{A E}{B E}=\frac{A Q}{B A}=\frac{Q F}{F B} $$ by the Angle Bisector Theorem. This implies $Q D / / F P$ and hence $F, P, X$ are collinear, as desired.
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
In a convex pentagon $A B C D E$, let $F$ be a point on $A C$ such that $\angle F B C=90^{\circ}$. Suppose triangles $A B F, A C D$ and $A D E$ are similar isosceles triangles with $$ \angle F A B=\angle F B A=\angle D A C=\angle D C A=\angle E A D=\angle E D A $$ Let $M$ be the midpoint of $C F$. Point $X$ is chosen such that $A M X E$ is a parallelogram. Show that $B D, E M$ and $F X$ are concurrent.
|
Let the common angle in (1) be $\theta$. From $\triangle A B F \sim \triangle A C D$, we have $\frac{A B}{A C}=\frac{A F}{A D}$ so that $\triangle A B C \sim \triangle A F D$. Then $\angle A D F=\angle A C B=90^{\circ}-2 \theta=90^{\circ}-\angle B A D$ and hence $D F \perp A B$. As $F A=F B$, this implies $\triangle D A B$ is isosceles with $D A=D B$. Then $F$ is the incentre of $\triangle D A B$. Next, from $\angle A E D=180^{\circ}-2 \theta=180^{\circ}-\angle D B A$, points $A, B, D, E$ are concyclic. Since we also have $E A=E D$, this shows $E, F, B$ are collinear and $E A=E F=E D$.  Note that $C$ lies on the internal angle bisector of $\angle B A D$ and lies on the external angle bisector of $\angle D B A$. It follows that it is the $A$-excentre of triangle $D A B$. As $M$ is the midpoint of $C F, M$ lies on the circumcircle of triangle $D A B$ and it is the centre of the circle passing through $D, F, B, C$. By symmetry, $D E F M$ is a rhombus. Then the midpoints of $A X, E M$ and $D F$ coincide, and it follows that $D A F X$ is a parallelogram. Let $P$ be the intersection of $B D$ and $E M$, and $Q$ be the intersection of $A D$ and $B E$. From $\angle B A C=\angle D C A$, we know that $D C, A B, E M$ are parallel. Thus we have $\frac{D P}{P B}=\frac{C M}{M A}$. This is further equal to $\frac{A E}{B E}$ since $C M=D M=D E=A E$ and $M A=B E$. From $\triangle A E Q \sim \triangle B E A$, we find that $$ \frac{D P}{P B}=\frac{A E}{B E}=\frac{A Q}{B A}=\frac{Q F}{F B} $$ by the Angle Bisector Theorem. This implies $Q D / / F P$ and hence $F, P, X$ are collinear, as desired.
|
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|
07e6f558-2934-5040-989d-58b6a169e993
| 24,782
|
Let $A B C$ be a triangle with circumcircle $\Gamma$ and incentre $I$. Let $M$ be the midpoint of side $B C$. Denote by $D$ the foot of perpendicular from $I$ to side $B C$. The line through $I$ perpendicular to $A I$ meets sides $A B$ and $A C$ at $F$ and $E$ respectively. Suppose the circumcircle of triangle $A E F$ intersects $\Gamma$ at a point $X$ other than $A$. Prove that lines $X D$ and $A M$ meet on $\Gamma$.
|
Let $A M$ meet $\Gamma$ again at $Y$ and $X Y$ meet $B C$ at $D^{\prime}$. It suffices to show $D^{\prime}=D$. We shall apply the following fact. - Claim. For any cyclic quadrilateral $P Q R S$ whose diagonals meet at $T$, we have $$ \frac{Q T}{T S}=\frac{P Q \cdot Q R}{P S \cdot S R} $$ Proof. We use $\left[W_{1} W_{2} W_{3}\right]$ to denote the area of $W_{1} W_{2} W_{3}$. Then $$ \frac{Q T}{T S}=\frac{[P Q R]}{[P S R]}=\frac{\frac{1}{2} P Q \cdot Q R \sin \angle P Q R}{\frac{1}{2} P S \cdot S R \sin \angle P S R}=\frac{P Q \cdot Q R}{P S \cdot S R} $$ Applying the Claim to $A B Y C$ and $X B Y C$ respectively, we have $1=\frac{B M}{M C}=\frac{A B \cdot B Y}{A C \cdot C Y}$ and $\frac{B D^{\prime}}{D^{\prime} C}=\frac{X B \cdot B Y}{X C \cdot C Y}$. These combine to give $$ \frac{B D^{\prime}}{C D^{\prime}}=\frac{X B}{X C} \cdot \frac{B Y}{C Y}=\frac{X B}{X C} \cdot \frac{A C}{A B} $$ Next, we use directed angles to find that $\measuredangle X B F=\measuredangle X B A=\measuredangle X C A=\measuredangle X C E$ and $\measuredangle X F B=\measuredangle X F A=\measuredangle X E A=\measuredangle X E C$. This shows triangles $X B F$ and $X C E$ are directly similar. In particular, we have $$ \frac{X B}{X C}=\frac{B F}{C E} $$ In the following, we give two ways to continue the proof. - Method 1. Here is a geometrical method. As $\angle F I B=\angle A I B-90^{\circ}=\frac{1}{2} \angle A C B=\angle I C B$ and $\angle F B I=\angle I B C$, the triangles $F B I$ and $I B C$ are similar. Analogously, triangles $E I C$ and $I B C$ are also similar. Hence, we get $$ \frac{F B}{I B}=\frac{B I}{B C} \quad \text { and } \quad \frac{E C}{I C}=\frac{I C}{B C} $$  Next, construct a line parallel to $B C$ and tangent to the incircle. Suppose it meets sides $A B$ and $A C$ at $B_{1}$ and $C_{1}$ respectively. Let the incircle touch $A B$ and $A C$ at $B_{2}$ and $C_{2}$ respectively. By homothety, the line $B_{1} I$ is parallel to the external angle bisector of $\angle A B C$, and hence $\angle B_{1} I B=90^{\circ}$. Since $\angle B B_{2} I=90^{\circ}$, we get $B B_{2} \cdot B B_{1}=B I^{2}$, and similarly $C C_{2} \cdot C C_{1}=C I^{2}$. Hence, $$ \frac{B I^{2}}{C I^{2}}=\frac{B B_{2} \cdot B B_{1}}{C C_{2} \cdot C C_{1}}=\frac{B B_{1}}{C C_{1}} \cdot \frac{B D}{C D}=\frac{A B}{A C} \cdot \frac{B D}{C D} $$ Combining (1), (2), (3) and (4), we conclude $$ \frac{B D^{\prime}}{C D^{\prime}}=\frac{X B}{X C} \cdot \frac{A C}{A B}=\frac{B F}{C E} \cdot \frac{A C}{A B}=\frac{B I^{2}}{C I^{2}} \cdot \frac{A C}{A B}=\frac{B D}{C D} $$ so that $D^{\prime}=D$. The result then follows. - Method 2. We continue the proof of Solution 1 using trigonometry. Let $\beta=\frac{1}{2} \angle A B C$ and $\gamma=\frac{1}{2} \angle A C B$. Observe that $\angle F I B=\angle A I B-90^{\circ}=\gamma$. Hence, $\frac{B F}{F I}=\frac{\sin \angle F I B}{\sin \angle I B F}=\frac{\sin \gamma}{\sin \beta}$. Similarly, $\frac{C E}{E I}=\frac{\sin \beta}{\sin \gamma}$. As $F I=E I$, we get $$ \frac{B F}{C E}=\frac{B F}{F I} \cdot \frac{E I}{C E}=\left(\frac{\sin \gamma}{\sin \beta}\right)^{2} $$ Together with (1) and (2), we find that $$ \frac{B D^{\prime}}{C D^{\prime}}=\frac{A C}{A B} \cdot\left(\frac{\sin \gamma}{\sin \beta}\right)^{2}=\frac{\sin 2 \beta}{\sin 2 \gamma} \cdot\left(\frac{\sin \gamma}{\sin \beta}\right)^{2}=\frac{\tan \gamma}{\tan \beta}=\frac{I D / C D}{I D / B D}=\frac{B D}{C D} $$ This shows $D^{\prime}=D$ and the result follows.
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
Let $A B C$ be a triangle with circumcircle $\Gamma$ and incentre $I$. Let $M$ be the midpoint of side $B C$. Denote by $D$ the foot of perpendicular from $I$ to side $B C$. The line through $I$ perpendicular to $A I$ meets sides $A B$ and $A C$ at $F$ and $E$ respectively. Suppose the circumcircle of triangle $A E F$ intersects $\Gamma$ at a point $X$ other than $A$. Prove that lines $X D$ and $A M$ meet on $\Gamma$.
|
Let $A M$ meet $\Gamma$ again at $Y$ and $X Y$ meet $B C$ at $D^{\prime}$. It suffices to show $D^{\prime}=D$. We shall apply the following fact. - Claim. For any cyclic quadrilateral $P Q R S$ whose diagonals meet at $T$, we have $$ \frac{Q T}{T S}=\frac{P Q \cdot Q R}{P S \cdot S R} $$ Proof. We use $\left[W_{1} W_{2} W_{3}\right]$ to denote the area of $W_{1} W_{2} W_{3}$. Then $$ \frac{Q T}{T S}=\frac{[P Q R]}{[P S R]}=\frac{\frac{1}{2} P Q \cdot Q R \sin \angle P Q R}{\frac{1}{2} P S \cdot S R \sin \angle P S R}=\frac{P Q \cdot Q R}{P S \cdot S R} $$ Applying the Claim to $A B Y C$ and $X B Y C$ respectively, we have $1=\frac{B M}{M C}=\frac{A B \cdot B Y}{A C \cdot C Y}$ and $\frac{B D^{\prime}}{D^{\prime} C}=\frac{X B \cdot B Y}{X C \cdot C Y}$. These combine to give $$ \frac{B D^{\prime}}{C D^{\prime}}=\frac{X B}{X C} \cdot \frac{B Y}{C Y}=\frac{X B}{X C} \cdot \frac{A C}{A B} $$ Next, we use directed angles to find that $\measuredangle X B F=\measuredangle X B A=\measuredangle X C A=\measuredangle X C E$ and $\measuredangle X F B=\measuredangle X F A=\measuredangle X E A=\measuredangle X E C$. This shows triangles $X B F$ and $X C E$ are directly similar. In particular, we have $$ \frac{X B}{X C}=\frac{B F}{C E} $$ In the following, we give two ways to continue the proof. - Method 1. Here is a geometrical method. As $\angle F I B=\angle A I B-90^{\circ}=\frac{1}{2} \angle A C B=\angle I C B$ and $\angle F B I=\angle I B C$, the triangles $F B I$ and $I B C$ are similar. Analogously, triangles $E I C$ and $I B C$ are also similar. Hence, we get $$ \frac{F B}{I B}=\frac{B I}{B C} \quad \text { and } \quad \frac{E C}{I C}=\frac{I C}{B C} $$  Next, construct a line parallel to $B C$ and tangent to the incircle. Suppose it meets sides $A B$ and $A C$ at $B_{1}$ and $C_{1}$ respectively. Let the incircle touch $A B$ and $A C$ at $B_{2}$ and $C_{2}$ respectively. By homothety, the line $B_{1} I$ is parallel to the external angle bisector of $\angle A B C$, and hence $\angle B_{1} I B=90^{\circ}$. Since $\angle B B_{2} I=90^{\circ}$, we get $B B_{2} \cdot B B_{1}=B I^{2}$, and similarly $C C_{2} \cdot C C_{1}=C I^{2}$. Hence, $$ \frac{B I^{2}}{C I^{2}}=\frac{B B_{2} \cdot B B_{1}}{C C_{2} \cdot C C_{1}}=\frac{B B_{1}}{C C_{1}} \cdot \frac{B D}{C D}=\frac{A B}{A C} \cdot \frac{B D}{C D} $$ Combining (1), (2), (3) and (4), we conclude $$ \frac{B D^{\prime}}{C D^{\prime}}=\frac{X B}{X C} \cdot \frac{A C}{A B}=\frac{B F}{C E} \cdot \frac{A C}{A B}=\frac{B I^{2}}{C I^{2}} \cdot \frac{A C}{A B}=\frac{B D}{C D} $$ so that $D^{\prime}=D$. The result then follows. - Method 2. We continue the proof of Solution 1 using trigonometry. Let $\beta=\frac{1}{2} \angle A B C$ and $\gamma=\frac{1}{2} \angle A C B$. Observe that $\angle F I B=\angle A I B-90^{\circ}=\gamma$. Hence, $\frac{B F}{F I}=\frac{\sin \angle F I B}{\sin \angle I B F}=\frac{\sin \gamma}{\sin \beta}$. Similarly, $\frac{C E}{E I}=\frac{\sin \beta}{\sin \gamma}$. As $F I=E I$, we get $$ \frac{B F}{C E}=\frac{B F}{F I} \cdot \frac{E I}{C E}=\left(\frac{\sin \gamma}{\sin \beta}\right)^{2} $$ Together with (1) and (2), we find that $$ \frac{B D^{\prime}}{C D^{\prime}}=\frac{A C}{A B} \cdot\left(\frac{\sin \gamma}{\sin \beta}\right)^{2}=\frac{\sin 2 \beta}{\sin 2 \gamma} \cdot\left(\frac{\sin \gamma}{\sin \beta}\right)^{2}=\frac{\tan \gamma}{\tan \beta}=\frac{I D / C D}{I D / B D}=\frac{B D}{C D} $$ This shows $D^{\prime}=D$ and the result follows.
|
{
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|
7a6402af-b080-5cf9-8fbc-389b3f16a721
| 24,787
|
Let $A B C$ be a triangle with circumcircle $\Gamma$ and incentre $I$. Let $M$ be the midpoint of side $B C$. Denote by $D$ the foot of perpendicular from $I$ to side $B C$. The line through $I$ perpendicular to $A I$ meets sides $A B$ and $A C$ at $F$ and $E$ respectively. Suppose the circumcircle of triangle $A E F$ intersects $\Gamma$ at a point $X$ other than $A$. Prove that lines $X D$ and $A M$ meet on $\Gamma$.
|
Let $\omega_{A}$ be the $A$-mixtilinear incircle of triangle $A B C$. From the properties of mixtilinear incircles, $\omega_{A}$ touches sides $A B$ and $A C$ at $F$ and $E$ respectively. Suppose $\omega_{A}$ is tangent to $\Gamma$ at $T$. Let $A M$ meet $\Gamma$ again at $Y$, and let $D_{1}, T_{1}$ be the reflections of $D$ and $T$ with respect to the perpendicular bisector of $B C$ respectively. It is well-known that $\angle B A T=\angle D_{1} A C$ so that $A, D_{1}, T_{1}$ are collinear.  We then show that $X, M, T_{1}$ are collinear. Let $R$ be the radical centre of $\omega_{A}, \Gamma$ and the circumcircle of triangle $A E F$. Then $R$ lies on $A X, E F$ and the tangent at $T$ to $\Gamma$. Let $A T$ meet $\omega_{A}$ again at $S$ and meet $E F$ at $P$. Obviously, $S F T E$ is a harmonic quadrilateral. Projecting from $T$, the pencil $(R, P ; F, E)$ is harmonic. We further project the pencil onto $\Gamma$ from $A$, so that $X B T C$ is a harmonic quadrilateral. As $T T_{1} / / B C$, the projection from $T_{1}$ onto $B C$ maps $T$ to a point at infinity, and hence maps $X$ to the midpoint of $B C$, which is $M$. This shows $X, M, T_{1}$ are collinear. We have two ways to finish the proof. - Method 1. Note that both $A Y$ and $X T_{1}$ are chords of $\Gamma$ passing through the midpoint $M$ of the chord $B C$. By the Butterfly Theorem, $X Y$ and $A T_{1}$ cut $B C$ at a pair of symmetric points with respect to $M$, and hence $X, D, Y$ are collinear. The proof is thus complete. - Method 2. Here, we finish the proof without using the Butterfly Theorem. As $D T T_{1} D_{1}$ is an isosceles trapezoid, we have $$ \measuredangle Y T D=\measuredangle Y T T_{1}+\measuredangle T_{1} T D=\measuredangle Y A T_{1}+\measuredangle A D_{1} D=\measuredangle Y M D $$ so that $D, T, Y, M$ are concyclic. As $X, M, T_{1}$ are collinear, we have $$ \measuredangle A Y D=\measuredangle M T D=\measuredangle D_{1} T_{1} M=\measuredangle A T_{1} X=\measuredangle A Y X $$ This shows $X, D, Y$ are collinear.
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
Let $A B C$ be a triangle with circumcircle $\Gamma$ and incentre $I$. Let $M$ be the midpoint of side $B C$. Denote by $D$ the foot of perpendicular from $I$ to side $B C$. The line through $I$ perpendicular to $A I$ meets sides $A B$ and $A C$ at $F$ and $E$ respectively. Suppose the circumcircle of triangle $A E F$ intersects $\Gamma$ at a point $X$ other than $A$. Prove that lines $X D$ and $A M$ meet on $\Gamma$.
|
Let $\omega_{A}$ be the $A$-mixtilinear incircle of triangle $A B C$. From the properties of mixtilinear incircles, $\omega_{A}$ touches sides $A B$ and $A C$ at $F$ and $E$ respectively. Suppose $\omega_{A}$ is tangent to $\Gamma$ at $T$. Let $A M$ meet $\Gamma$ again at $Y$, and let $D_{1}, T_{1}$ be the reflections of $D$ and $T$ with respect to the perpendicular bisector of $B C$ respectively. It is well-known that $\angle B A T=\angle D_{1} A C$ so that $A, D_{1}, T_{1}$ are collinear.  We then show that $X, M, T_{1}$ are collinear. Let $R$ be the radical centre of $\omega_{A}, \Gamma$ and the circumcircle of triangle $A E F$. Then $R$ lies on $A X, E F$ and the tangent at $T$ to $\Gamma$. Let $A T$ meet $\omega_{A}$ again at $S$ and meet $E F$ at $P$. Obviously, $S F T E$ is a harmonic quadrilateral. Projecting from $T$, the pencil $(R, P ; F, E)$ is harmonic. We further project the pencil onto $\Gamma$ from $A$, so that $X B T C$ is a harmonic quadrilateral. As $T T_{1} / / B C$, the projection from $T_{1}$ onto $B C$ maps $T$ to a point at infinity, and hence maps $X$ to the midpoint of $B C$, which is $M$. This shows $X, M, T_{1}$ are collinear. We have two ways to finish the proof. - Method 1. Note that both $A Y$ and $X T_{1}$ are chords of $\Gamma$ passing through the midpoint $M$ of the chord $B C$. By the Butterfly Theorem, $X Y$ and $A T_{1}$ cut $B C$ at a pair of symmetric points with respect to $M$, and hence $X, D, Y$ are collinear. The proof is thus complete. - Method 2. Here, we finish the proof without using the Butterfly Theorem. As $D T T_{1} D_{1}$ is an isosceles trapezoid, we have $$ \measuredangle Y T D=\measuredangle Y T T_{1}+\measuredangle T_{1} T D=\measuredangle Y A T_{1}+\measuredangle A D_{1} D=\measuredangle Y M D $$ so that $D, T, Y, M$ are concyclic. As $X, M, T_{1}$ are collinear, we have $$ \measuredangle A Y D=\measuredangle M T D=\measuredangle D_{1} T_{1} M=\measuredangle A T_{1} X=\measuredangle A Y X $$ This shows $X, D, Y$ are collinear.
|
{
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"problem_match": null,
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}
|
7a6402af-b080-5cf9-8fbc-389b3f16a721
| 24,787
|
Let $A B C$ be a triangle with $A B=A C \neq B C$ and let $I$ be its incentre. The line $B I$ meets $A C$ at $D$, and the line through $D$ perpendicular to $A C$ meets $A I$ at $E$. Prove that the reflection of $I$ in $A C$ lies on the circumcircle of triangle $B D E$.
|
 Let $\Gamma$ be the circle with centre $E$ passing through $B$ and $C$. Since $E D \perp A C$, the point $F$ symmetric to $C$ with respect to $D$ lies on $\Gamma$. From $\angle D C I=\angle I C B=\angle C B I$, the line $D C$ is a tangent to the circumcircle of triangle $I B C$. Let $J$ be the symmetric point of $I$ with respect to $D$. Using directed lengths, from $$ D C \cdot D F=-D C^{2}=-D I \cdot D B=D J \cdot D B $$ the point $J$ also lies on $\Gamma$. Let $I^{\prime}$ be the reflection of $I$ in $A C$. Since $I J$ and $C F$ bisect each other, $C J F I$ is a parallelogram. From $\angle F I^{\prime} C=\angle C I F=\angle F J C$, we find that $I^{\prime}$ lies on $\Gamma$. This gives $E I^{\prime}=E B$. Note that $A C$ is the internal angle bisector of $\angle B D I^{\prime}$. This shows $D E$ is the external angle bisector of $\angle B D I^{\prime}$ as $D E \perp A C$. Together with $E I^{\prime}=E B$, it is well-known that $E$ lies on the circumcircle of triangle $B D I^{\prime}$.
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
Let $A B C$ be a triangle with $A B=A C \neq B C$ and let $I$ be its incentre. The line $B I$ meets $A C$ at $D$, and the line through $D$ perpendicular to $A C$ meets $A I$ at $E$. Prove that the reflection of $I$ in $A C$ lies on the circumcircle of triangle $B D E$.
|
 Let $\Gamma$ be the circle with centre $E$ passing through $B$ and $C$. Since $E D \perp A C$, the point $F$ symmetric to $C$ with respect to $D$ lies on $\Gamma$. From $\angle D C I=\angle I C B=\angle C B I$, the line $D C$ is a tangent to the circumcircle of triangle $I B C$. Let $J$ be the symmetric point of $I$ with respect to $D$. Using directed lengths, from $$ D C \cdot D F=-D C^{2}=-D I \cdot D B=D J \cdot D B $$ the point $J$ also lies on $\Gamma$. Let $I^{\prime}$ be the reflection of $I$ in $A C$. Since $I J$ and $C F$ bisect each other, $C J F I$ is a parallelogram. From $\angle F I^{\prime} C=\angle C I F=\angle F J C$, we find that $I^{\prime}$ lies on $\Gamma$. This gives $E I^{\prime}=E B$. Note that $A C$ is the internal angle bisector of $\angle B D I^{\prime}$. This shows $D E$ is the external angle bisector of $\angle B D I^{\prime}$ as $D E \perp A C$. Together with $E I^{\prime}=E B$, it is well-known that $E$ lies on the circumcircle of triangle $B D I^{\prime}$.
|
{
"resource_path": "IMO/segmented/en-IMO2016SL.jsonl",
"problem_match": null,
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}
|
264af218-b34b-5426-9880-59764679f217
| 24,794
|
Let $A B C$ be a triangle with $A B=A C \neq B C$ and let $I$ be its incentre. The line $B I$ meets $A C$ at $D$, and the line through $D$ perpendicular to $A C$ meets $A I$ at $E$. Prove that the reflection of $I$ in $A C$ lies on the circumcircle of triangle $B D E$.
|
Let $I^{\prime}$ be the reflection of $I$ in $A C$ and let $S$ be the intersection of $I^{\prime} C$ and $A I$. Using directed angles, we let $\theta=\measuredangle A C I=\measuredangle I C B=\measuredangle C B I$. We have $$ \measuredangle I^{\prime} S E=\measuredangle I^{\prime} C A+\measuredangle C A I=\theta+\left(\frac{\pi}{2}+2 \theta\right)=3 \theta+\frac{\pi}{2} $$ and $$ \measuredangle I^{\prime} D E=\measuredangle I^{\prime} D C+\frac{\pi}{2}=\measuredangle C D I+\frac{\pi}{2}=\measuredangle D C B+\measuredangle C B D+\frac{\pi}{2}=3 \theta+\frac{\pi}{2} $$ This shows $I^{\prime}, D, E, S$ are concyclic. Next, we find $\measuredangle I^{\prime} S B=2 \measuredangle I^{\prime} S E=6 \theta$ and $\measuredangle I^{\prime} D B=2 \measuredangle C D I=6 \theta$. Therefore, $I^{\prime}, D, B, S$ are concyclic so that $I^{\prime}, D, E, B, S$ lie on the same circle. The result then follows.  Comment. The point $S$ constructed in Solution 2 may lie on the same side as $A$ of $B C$. Also, since $S$ lies on the circumcircle of the non-degenerate triangle $B D E$, we implicitly know that $S$ is not an ideal point. Indeed, one can verify that $I^{\prime} C$ and $A I$ are parallel if and only if triangle $A B C$ is equilateral.
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
Let $A B C$ be a triangle with $A B=A C \neq B C$ and let $I$ be its incentre. The line $B I$ meets $A C$ at $D$, and the line through $D$ perpendicular to $A C$ meets $A I$ at $E$. Prove that the reflection of $I$ in $A C$ lies on the circumcircle of triangle $B D E$.
|
Let $I^{\prime}$ be the reflection of $I$ in $A C$ and let $S$ be the intersection of $I^{\prime} C$ and $A I$. Using directed angles, we let $\theta=\measuredangle A C I=\measuredangle I C B=\measuredangle C B I$. We have $$ \measuredangle I^{\prime} S E=\measuredangle I^{\prime} C A+\measuredangle C A I=\theta+\left(\frac{\pi}{2}+2 \theta\right)=3 \theta+\frac{\pi}{2} $$ and $$ \measuredangle I^{\prime} D E=\measuredangle I^{\prime} D C+\frac{\pi}{2}=\measuredangle C D I+\frac{\pi}{2}=\measuredangle D C B+\measuredangle C B D+\frac{\pi}{2}=3 \theta+\frac{\pi}{2} $$ This shows $I^{\prime}, D, E, S$ are concyclic. Next, we find $\measuredangle I^{\prime} S B=2 \measuredangle I^{\prime} S E=6 \theta$ and $\measuredangle I^{\prime} D B=2 \measuredangle C D I=6 \theta$. Therefore, $I^{\prime}, D, B, S$ are concyclic so that $I^{\prime}, D, E, B, S$ lie on the same circle. The result then follows.  Comment. The point $S$ constructed in Solution 2 may lie on the same side as $A$ of $B C$. Also, since $S$ lies on the circumcircle of the non-degenerate triangle $B D E$, we implicitly know that $S$ is not an ideal point. Indeed, one can verify that $I^{\prime} C$ and $A I$ are parallel if and only if triangle $A B C$ is equilateral.
|
{
"resource_path": "IMO/segmented/en-IMO2016SL.jsonl",
"problem_match": null,
"solution_match": null
}
|
264af218-b34b-5426-9880-59764679f217
| 24,794
|
Let $A B C$ be a triangle with $A B=A C \neq B C$ and let $I$ be its incentre. The line $B I$ meets $A C$ at $D$, and the line through $D$ perpendicular to $A C$ meets $A I$ at $E$. Prove that the reflection of $I$ in $A C$ lies on the circumcircle of triangle $B D E$.
|
Let $I^{\prime}$ be the reflection of $I$ in $A C$, and let $D^{\prime}$ be the second intersection of $A I$ and the circumcircle of triangle $A B D$. Since $A D^{\prime}$ bisects $\angle B A D$, point $D^{\prime}$ is the midpoint of the arc $B D$ and $D D^{\prime}=B D^{\prime}=C D^{\prime}$. Obviously, $A, E, D^{\prime}$ lie on $A I$ in this order.  We find that $\angle E D^{\prime} D=\angle A D^{\prime} D=\angle A B D=\angle I B C=\angle I C B$. Next, since $D^{\prime}$ is the circumcentre of triangle $B C D$, we have $\angle E D D^{\prime}=90^{\circ}-\angle D^{\prime} D C=\angle C B D=\angle I B C$. The two relations show that triangles $E D^{\prime} D$ and $I C B$ are similar. Therefore, we have $$ \frac{B C}{C I^{\prime}}=\frac{B C}{C I}=\frac{D D^{\prime}}{D^{\prime} E}=\frac{B D^{\prime}}{D^{\prime} E} $$ Also, we get $$ \angle B C I^{\prime}=\angle B C A+\angle A C I^{\prime}=\angle B C A+\angle I C A=\angle B C A+\angle D B C=\angle B D A=\angle B D^{\prime} E . $$ These show triangles $B C I^{\prime}$ and $B D^{\prime} E$ are similar, and hence triangles $B C D^{\prime}$ and $B I^{\prime} E$ are similar. As $B C D^{\prime}$ is isosceles, we obtain $B E=I^{\prime} E$. As $D E$ is the external angle bisector of $\angle B D I^{\prime}$ and $E I^{\prime}=E B$, we know that $E$ lies on the circumcircle of triangle $B D I^{\prime}$.
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
Let $A B C$ be a triangle with $A B=A C \neq B C$ and let $I$ be its incentre. The line $B I$ meets $A C$ at $D$, and the line through $D$ perpendicular to $A C$ meets $A I$ at $E$. Prove that the reflection of $I$ in $A C$ lies on the circumcircle of triangle $B D E$.
|
Let $I^{\prime}$ be the reflection of $I$ in $A C$, and let $D^{\prime}$ be the second intersection of $A I$ and the circumcircle of triangle $A B D$. Since $A D^{\prime}$ bisects $\angle B A D$, point $D^{\prime}$ is the midpoint of the arc $B D$ and $D D^{\prime}=B D^{\prime}=C D^{\prime}$. Obviously, $A, E, D^{\prime}$ lie on $A I$ in this order.  We find that $\angle E D^{\prime} D=\angle A D^{\prime} D=\angle A B D=\angle I B C=\angle I C B$. Next, since $D^{\prime}$ is the circumcentre of triangle $B C D$, we have $\angle E D D^{\prime}=90^{\circ}-\angle D^{\prime} D C=\angle C B D=\angle I B C$. The two relations show that triangles $E D^{\prime} D$ and $I C B$ are similar. Therefore, we have $$ \frac{B C}{C I^{\prime}}=\frac{B C}{C I}=\frac{D D^{\prime}}{D^{\prime} E}=\frac{B D^{\prime}}{D^{\prime} E} $$ Also, we get $$ \angle B C I^{\prime}=\angle B C A+\angle A C I^{\prime}=\angle B C A+\angle I C A=\angle B C A+\angle D B C=\angle B D A=\angle B D^{\prime} E . $$ These show triangles $B C I^{\prime}$ and $B D^{\prime} E$ are similar, and hence triangles $B C D^{\prime}$ and $B I^{\prime} E$ are similar. As $B C D^{\prime}$ is isosceles, we obtain $B E=I^{\prime} E$. As $D E$ is the external angle bisector of $\angle B D I^{\prime}$ and $E I^{\prime}=E B$, we know that $E$ lies on the circumcircle of triangle $B D I^{\prime}$.
|
{
"resource_path": "IMO/segmented/en-IMO2016SL.jsonl",
"problem_match": null,
"solution_match": null
}
|
264af218-b34b-5426-9880-59764679f217
| 24,794
|
Let $A B C$ be a triangle with $A B=A C \neq B C$ and let $I$ be its incentre. The line $B I$ meets $A C$ at $D$, and the line through $D$ perpendicular to $A C$ meets $A I$ at $E$. Prove that the reflection of $I$ in $A C$ lies on the circumcircle of triangle $B D E$.
|
Let $A I$ and $B I$ meet the circumcircle of triangle $A B C$ again at $A^{\prime}$ and $B^{\prime}$ respectively, and let $E^{\prime}$ be the reflection of $E$ in $A C$. From $$ \begin{aligned} \angle B^{\prime} A E^{\prime} & =\angle B^{\prime} A D-\angle E^{\prime} A D=\frac{\angle A B C}{2}-\frac{\angle B A C}{2}=90^{\circ}-\angle B A C-\frac{\angle A B C}{2} \\ & =90^{\circ}-\angle B^{\prime} D A=\angle B^{\prime} D E^{\prime} \end{aligned} $$ points $B^{\prime}, A, D, E^{\prime}$ are concyclic. Then $$ \angle D B^{\prime} E^{\prime}=\angle D A E^{\prime}=\frac{\angle B A C}{2}=\angle B A A^{\prime}=\angle D B^{\prime} A^{\prime} $$ and hence $B^{\prime}, E^{\prime}, A^{\prime}$ are collinear. It is well-known that $A^{\prime} B^{\prime}$ is the perpendicular bisector of $C I$, so that $C E^{\prime}=I E^{\prime}$. Let $I^{\prime}$ be the reflection of $I$ in $A C$. This implies $B E=C E=I^{\prime} E$. As $D E$ is the external angle bisector of $\angle B D I^{\prime}$ and $E I^{\prime}=E B$, we know that $E$ lies on the circumcircle of triangle $B D I^{\prime}$. 
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
Let $A B C$ be a triangle with $A B=A C \neq B C$ and let $I$ be its incentre. The line $B I$ meets $A C$ at $D$, and the line through $D$ perpendicular to $A C$ meets $A I$ at $E$. Prove that the reflection of $I$ in $A C$ lies on the circumcircle of triangle $B D E$.
|
Let $A I$ and $B I$ meet the circumcircle of triangle $A B C$ again at $A^{\prime}$ and $B^{\prime}$ respectively, and let $E^{\prime}$ be the reflection of $E$ in $A C$. From $$ \begin{aligned} \angle B^{\prime} A E^{\prime} & =\angle B^{\prime} A D-\angle E^{\prime} A D=\frac{\angle A B C}{2}-\frac{\angle B A C}{2}=90^{\circ}-\angle B A C-\frac{\angle A B C}{2} \\ & =90^{\circ}-\angle B^{\prime} D A=\angle B^{\prime} D E^{\prime} \end{aligned} $$ points $B^{\prime}, A, D, E^{\prime}$ are concyclic. Then $$ \angle D B^{\prime} E^{\prime}=\angle D A E^{\prime}=\frac{\angle B A C}{2}=\angle B A A^{\prime}=\angle D B^{\prime} A^{\prime} $$ and hence $B^{\prime}, E^{\prime}, A^{\prime}$ are collinear. It is well-known that $A^{\prime} B^{\prime}$ is the perpendicular bisector of $C I$, so that $C E^{\prime}=I E^{\prime}$. Let $I^{\prime}$ be the reflection of $I$ in $A C$. This implies $B E=C E=I^{\prime} E$. As $D E$ is the external angle bisector of $\angle B D I^{\prime}$ and $E I^{\prime}=E B$, we know that $E$ lies on the circumcircle of triangle $B D I^{\prime}$. 
|
{
"resource_path": "IMO/segmented/en-IMO2016SL.jsonl",
"problem_match": null,
"solution_match": null
}
|
264af218-b34b-5426-9880-59764679f217
| 24,794
|
Let $A B C$ be a triangle with $A B=A C \neq B C$ and let $I$ be its incentre. The line $B I$ meets $A C$ at $D$, and the line through $D$ perpendicular to $A C$ meets $A I$ at $E$. Prove that the reflection of $I$ in $A C$ lies on the circumcircle of triangle $B D E$.
|
Let $F$ be the intersection of $C I$ and $A B$. Clearly, $F$ and $D$ are symmetric with respect to $A I$. Let $O$ be the circumcentre of triangle $B I F$, and let $I^{\prime}$ be the reflection of $I$ in $A C$.  From $\angle B F O=90^{\circ}-\angle F I B=\frac{1}{2} \angle B A C=\angle B A I$, we get $E I / / F O$. Also, from the relation $\angle O I B=90^{\circ}-\angle B F I=90^{\circ}-\angle C D I=\angle I^{\prime} I D$, we know that $O, I, I^{\prime}$ are collinear. Note that $E D / / O I$ since both are perpendicular to $A C$. Then $\angle F E I=\angle D E I=\angle O I E$. Together with $E I / / F O$, the quadrilateral $E F O I$ is an isosceles trapezoid. Therefore, we find that $\angle D I E=\angle F I E=\angle O E I$ so $O E / / I D$. Then $D E O I$ is a parallelogram. Hence, we have $D I^{\prime}=D I=E O$, which shows $D E O I^{\prime}$ is an isosceles trapezoid. In addition, $E D=O I=O B$ and $O E / / B D$ imply $E O B D$ is another isosceles trapezoid. In particular, both $D E O I^{\prime}$ and $E O B D$ are cyclic. This shows $B, D, E, I^{\prime}$ are concyclic.
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
Let $A B C$ be a triangle with $A B=A C \neq B C$ and let $I$ be its incentre. The line $B I$ meets $A C$ at $D$, and the line through $D$ perpendicular to $A C$ meets $A I$ at $E$. Prove that the reflection of $I$ in $A C$ lies on the circumcircle of triangle $B D E$.
|
Let $F$ be the intersection of $C I$ and $A B$. Clearly, $F$ and $D$ are symmetric with respect to $A I$. Let $O$ be the circumcentre of triangle $B I F$, and let $I^{\prime}$ be the reflection of $I$ in $A C$.  From $\angle B F O=90^{\circ}-\angle F I B=\frac{1}{2} \angle B A C=\angle B A I$, we get $E I / / F O$. Also, from the relation $\angle O I B=90^{\circ}-\angle B F I=90^{\circ}-\angle C D I=\angle I^{\prime} I D$, we know that $O, I, I^{\prime}$ are collinear. Note that $E D / / O I$ since both are perpendicular to $A C$. Then $\angle F E I=\angle D E I=\angle O I E$. Together with $E I / / F O$, the quadrilateral $E F O I$ is an isosceles trapezoid. Therefore, we find that $\angle D I E=\angle F I E=\angle O E I$ so $O E / / I D$. Then $D E O I$ is a parallelogram. Hence, we have $D I^{\prime}=D I=E O$, which shows $D E O I^{\prime}$ is an isosceles trapezoid. In addition, $E D=O I=O B$ and $O E / / B D$ imply $E O B D$ is another isosceles trapezoid. In particular, both $D E O I^{\prime}$ and $E O B D$ are cyclic. This shows $B, D, E, I^{\prime}$ are concyclic.
|
{
"resource_path": "IMO/segmented/en-IMO2016SL.jsonl",
"problem_match": null,
"solution_match": null
}
|
264af218-b34b-5426-9880-59764679f217
| 24,794
|
Let $A B C$ be a triangle with $A B=A C \neq B C$ and let $I$ be its incentre. The line $B I$ meets $A C$ at $D$, and the line through $D$ perpendicular to $A C$ meets $A I$ at $E$. Prove that the reflection of $I$ in $A C$ lies on the circumcircle of triangle $B D E$.
|
Let $I^{\prime}$ be the reflection of $I$ in $A C$. Denote by $T$ and $M$ the projections from $I$ to sides $A B$ and $B C$ respectively. Since $B I$ is the perpendicular bisector of $T M$, we have $$ D T=D M $$ Since $\angle A D E=\angle A T I=90^{\circ}$ and $\angle D A E=\angle T A I$, we have $\triangle A D E \sim \triangle A T I$. This shows $\frac{A D}{A E}=\frac{A T}{A I}=\frac{A T}{A I^{\prime}}$. Together with $\angle D A T=2 \angle D A E=\angle E A I^{\prime}$, this yields $\triangle D A T \sim \triangle E A I^{\prime}$. In particular, we have $$ \frac{D T}{E I^{\prime}}=\frac{A T}{A I^{\prime}}=\frac{A T}{A I} $$ Obviously, the right-angled triangles $A M B$ and $A T I$ are similar. Then we get $$ \frac{A M}{A B}=\frac{A T}{A I} $$ Next, from $\triangle A M B \sim \triangle A T I \sim \triangle A D E$, we have $\frac{A M}{A B}=\frac{A D}{A E}$ so that $\triangle A D M \sim \triangle A E B$. It follows that $$ \frac{D M}{E B}=\frac{A M}{A B} . $$ Combining (1), (2), (3) and (4), we get $E B=E I^{\prime}$. As $D E$ is the external angle bisector of $\angle B D I^{\prime}$, we know that $E$ lies on the circumcircle of triangle $B D I^{\prime}$.  Comment. A stronger version of this problem is to ask the contestants to prove the reflection of $I$ in $A C$ lies on the circumcircle of triangle $B D E$ if and only if $A B=A C$. Some of the above solutions can be modified to prove the converse statement to the original problem. For example, we borrow some ideas from Solution 2 to establish the converse as follows.  Let $I^{\prime}$ be the reflection of $I$ in $A C$ and suppose $B, E, D, I^{\prime}$ lie on a circle $\Gamma$. Let $A I$ meet $\Gamma$ again at $S$. As $D E$ is the external angle bisector of $\angle B D I^{\prime}$, we have $E B=E I^{\prime}$. Using directed angles, we get $$ \measuredangle C I^{\prime} S=\measuredangle C I^{\prime} D+\measuredangle D I^{\prime} S=\measuredangle D I C+\measuredangle D E S=\measuredangle D I C+\measuredangle D E A=\measuredangle D I C+\measuredangle D C B=0 . $$ This means $I^{\prime}, C, S$ are collinear. From this we get $\measuredangle B S E=\measuredangle E S I^{\prime}=\measuredangle E S C$ and hence $A S$ bisects both $\angle B A C$ and $\angle B S C$. Clearly, $S$ and $A$ are distinct points. It follows that $\triangle B A S \cong \triangle C A S$ and thus $A B=A C$. As in some of the above solutions, an obvious way to prove the stronger version is to establish the following equivalence: $B E=E I^{\prime}$ if and only if $A B=A C$. In addition to the ideas used in those solutions, one can use trigonometry to express the lengths of $B E$ and $E I^{\prime}$ in terms of the side lengths of triangle $A B C$ and to establish the equivalence.
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
Let $A B C$ be a triangle with $A B=A C \neq B C$ and let $I$ be its incentre. The line $B I$ meets $A C$ at $D$, and the line through $D$ perpendicular to $A C$ meets $A I$ at $E$. Prove that the reflection of $I$ in $A C$ lies on the circumcircle of triangle $B D E$.
|
Let $I^{\prime}$ be the reflection of $I$ in $A C$. Denote by $T$ and $M$ the projections from $I$ to sides $A B$ and $B C$ respectively. Since $B I$ is the perpendicular bisector of $T M$, we have $$ D T=D M $$ Since $\angle A D E=\angle A T I=90^{\circ}$ and $\angle D A E=\angle T A I$, we have $\triangle A D E \sim \triangle A T I$. This shows $\frac{A D}{A E}=\frac{A T}{A I}=\frac{A T}{A I^{\prime}}$. Together with $\angle D A T=2 \angle D A E=\angle E A I^{\prime}$, this yields $\triangle D A T \sim \triangle E A I^{\prime}$. In particular, we have $$ \frac{D T}{E I^{\prime}}=\frac{A T}{A I^{\prime}}=\frac{A T}{A I} $$ Obviously, the right-angled triangles $A M B$ and $A T I$ are similar. Then we get $$ \frac{A M}{A B}=\frac{A T}{A I} $$ Next, from $\triangle A M B \sim \triangle A T I \sim \triangle A D E$, we have $\frac{A M}{A B}=\frac{A D}{A E}$ so that $\triangle A D M \sim \triangle A E B$. It follows that $$ \frac{D M}{E B}=\frac{A M}{A B} . $$ Combining (1), (2), (3) and (4), we get $E B=E I^{\prime}$. As $D E$ is the external angle bisector of $\angle B D I^{\prime}$, we know that $E$ lies on the circumcircle of triangle $B D I^{\prime}$.  Comment. A stronger version of this problem is to ask the contestants to prove the reflection of $I$ in $A C$ lies on the circumcircle of triangle $B D E$ if and only if $A B=A C$. Some of the above solutions can be modified to prove the converse statement to the original problem. For example, we borrow some ideas from Solution 2 to establish the converse as follows.  Let $I^{\prime}$ be the reflection of $I$ in $A C$ and suppose $B, E, D, I^{\prime}$ lie on a circle $\Gamma$. Let $A I$ meet $\Gamma$ again at $S$. As $D E$ is the external angle bisector of $\angle B D I^{\prime}$, we have $E B=E I^{\prime}$. Using directed angles, we get $$ \measuredangle C I^{\prime} S=\measuredangle C I^{\prime} D+\measuredangle D I^{\prime} S=\measuredangle D I C+\measuredangle D E S=\measuredangle D I C+\measuredangle D E A=\measuredangle D I C+\measuredangle D C B=0 . $$ This means $I^{\prime}, C, S$ are collinear. From this we get $\measuredangle B S E=\measuredangle E S I^{\prime}=\measuredangle E S C$ and hence $A S$ bisects both $\angle B A C$ and $\angle B S C$. Clearly, $S$ and $A$ are distinct points. It follows that $\triangle B A S \cong \triangle C A S$ and thus $A B=A C$. As in some of the above solutions, an obvious way to prove the stronger version is to establish the following equivalence: $B E=E I^{\prime}$ if and only if $A B=A C$. In addition to the ideas used in those solutions, one can use trigonometry to express the lengths of $B E$ and $E I^{\prime}$ in terms of the side lengths of triangle $A B C$ and to establish the equivalence.
|
{
"resource_path": "IMO/segmented/en-IMO2016SL.jsonl",
"problem_match": null,
"solution_match": null
}
|
264af218-b34b-5426-9880-59764679f217
| 24,794
|
Let $D$ be the foot of perpendicular from $A$ to the Euler line (the line passing through the circumcentre and the orthocentre) of an acute scalene triangle $A B C$. A circle $\omega$ with centre $S$ passes through $A$ and $D$, and it intersects sides $A B$ and $A C$ at $X$ and $Y$ respectively. Let $P$ be the foot of altitude from $A$ to $B C$, and let $M$ be the midpoint of $B C$. Prove that the circumcentre of triangle $X S Y$ is equidistant from $P$ and $M$.
|
Denote the orthocentre and circumcentre of triangle $A B C$ by $H$ and $O$ respectively. Let $Q$ be the midpoint of $A H$ and $N$ be the nine-point centre of triangle $A B C$. It is known that $Q$ lies on the nine-point circle of triangle $A B C, N$ is the midpoint of $Q M$ and that $Q M$ is parallel to $A O$. Let the perpendicular from $S$ to $X Y$ meet line $Q M$ at $S^{\prime}$. Let $E$ be the foot of altitude from $B$ to side $A C$. Since $Q$ and $S$ lie on the perpendicular bisector of $A D$, using directed angles, we have $$ \begin{aligned} \measuredangle S D Q & =\measuredangle Q A S=\measuredangle X A S-\measuredangle X A Q=\left(\frac{\pi}{2}-\measuredangle A Y X\right)-\measuredangle B A P=\measuredangle C B A-\measuredangle A Y X \\ & =(\measuredangle C B A-\measuredangle A C B)-\measuredangle B C A-\measuredangle A Y X=\measuredangle P E M-(\measuredangle B C A+\measuredangle A Y X) \\ & =\measuredangle P Q M-\measuredangle(B C, X Y)=\frac{\pi}{2}-\measuredangle\left(S^{\prime} Q, B C\right)-\angle(B C, X Y)=\measuredangle S S^{\prime} Q \end{aligned} $$ This shows $D, S^{\prime}, S, Q$ are concyclic.  Let the perpendicular from $N$ to $B C$ intersect line $S S^{\prime}$ at $O_{1}$. (Note that the two lines coincide when $S$ is the midpoint of $A O$, in which case the result is true since the circumcentre of triangle $X S Y$ must lie on this line.) It suffices to show that $O_{1}$ is the circumcentre of triangle $X S Y$ since $N$ lies on the perpendicular bisector of $P M$. From $$ \measuredangle D S^{\prime} O_{1}=\measuredangle D Q S=\measuredangle S Q A=\angle(S Q, Q A)=\angle\left(O D, O_{1} N\right)=\measuredangle D N O_{1} $$ since $S Q / / O D$ and $Q A / / O_{1} N$, we know that $D, O_{1}, S^{\prime}, N$ are concyclic. Therefore, we get $$ \measuredangle S D S^{\prime}=\measuredangle S Q S^{\prime}=\angle\left(S Q, Q S^{\prime}\right)=\angle\left(N D, N S^{\prime}\right)=\measuredangle D N S^{\prime} $$ so that $S D$ is a tangent to the circle through $D, O_{1}, S^{\prime}, N$. Then we have $$ S S^{\prime} \cdot S O_{1}=S D^{2}=S X^{2} $$ Next, we show that $S$ and $S^{\prime}$ are symmetric with respect to $X Y$. By the Sine Law, we have $$ \frac{S S^{\prime}}{\sin \angle S Q S^{\prime}}=\frac{S Q}{\sin \angle S S^{\prime} Q}=\frac{S Q}{\sin \angle S D Q}=\frac{S Q}{\sin \angle S A Q}=\frac{S A}{\sin \angle S Q A} $$ It follows that $$ S S^{\prime}=S A \cdot \frac{\sin \angle S Q S^{\prime}}{\sin \angle S Q A}=S A \cdot \frac{\sin \angle H O A}{\sin \angle O H A}=S A \cdot \frac{A H}{A O}=S A \cdot 2 \cos A, $$ which is twice the distance from $S$ to $X Y$. Note that $S$ and $C$ lie on the same side of the perpendicular bisector of $P M$ if and only if $\angle S A C<\angle O A C$ if and only if $\angle Y X A>\angle C B A$. This shows $S$ and $O_{1}$ lie on different sides of $X Y$. As $S^{\prime}$ lies on ray $S O_{1}$, it follows that $S$ and $S^{\prime}$ cannot lie on the same side of $X Y$. Therefore, $S$ and $S^{\prime}$ are symmetric with respect to $X Y$. Let $d$ be the diameter of the circumcircle of triangle $X S Y$. As $S S^{\prime}$ is twice the distance from $S$ to $X Y$ and $S X=S Y$, we have $S S^{\prime}=2 \frac{S X^{2}}{d}$. It follows from (1) that $d=2 S O_{1}$. As $S O_{1}$ is the perpendicular bisector of $X Y$, point $O_{1}^{d}$ is the circumcentre of triangle $X S Y$.
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
Let $D$ be the foot of perpendicular from $A$ to the Euler line (the line passing through the circumcentre and the orthocentre) of an acute scalene triangle $A B C$. A circle $\omega$ with centre $S$ passes through $A$ and $D$, and it intersects sides $A B$ and $A C$ at $X$ and $Y$ respectively. Let $P$ be the foot of altitude from $A$ to $B C$, and let $M$ be the midpoint of $B C$. Prove that the circumcentre of triangle $X S Y$ is equidistant from $P$ and $M$.
|
Denote the orthocentre and circumcentre of triangle $A B C$ by $H$ and $O$ respectively. Let $Q$ be the midpoint of $A H$ and $N$ be the nine-point centre of triangle $A B C$. It is known that $Q$ lies on the nine-point circle of triangle $A B C, N$ is the midpoint of $Q M$ and that $Q M$ is parallel to $A O$. Let the perpendicular from $S$ to $X Y$ meet line $Q M$ at $S^{\prime}$. Let $E$ be the foot of altitude from $B$ to side $A C$. Since $Q$ and $S$ lie on the perpendicular bisector of $A D$, using directed angles, we have $$ \begin{aligned} \measuredangle S D Q & =\measuredangle Q A S=\measuredangle X A S-\measuredangle X A Q=\left(\frac{\pi}{2}-\measuredangle A Y X\right)-\measuredangle B A P=\measuredangle C B A-\measuredangle A Y X \\ & =(\measuredangle C B A-\measuredangle A C B)-\measuredangle B C A-\measuredangle A Y X=\measuredangle P E M-(\measuredangle B C A+\measuredangle A Y X) \\ & =\measuredangle P Q M-\measuredangle(B C, X Y)=\frac{\pi}{2}-\measuredangle\left(S^{\prime} Q, B C\right)-\angle(B C, X Y)=\measuredangle S S^{\prime} Q \end{aligned} $$ This shows $D, S^{\prime}, S, Q$ are concyclic.  Let the perpendicular from $N$ to $B C$ intersect line $S S^{\prime}$ at $O_{1}$. (Note that the two lines coincide when $S$ is the midpoint of $A O$, in which case the result is true since the circumcentre of triangle $X S Y$ must lie on this line.) It suffices to show that $O_{1}$ is the circumcentre of triangle $X S Y$ since $N$ lies on the perpendicular bisector of $P M$. From $$ \measuredangle D S^{\prime} O_{1}=\measuredangle D Q S=\measuredangle S Q A=\angle(S Q, Q A)=\angle\left(O D, O_{1} N\right)=\measuredangle D N O_{1} $$ since $S Q / / O D$ and $Q A / / O_{1} N$, we know that $D, O_{1}, S^{\prime}, N$ are concyclic. Therefore, we get $$ \measuredangle S D S^{\prime}=\measuredangle S Q S^{\prime}=\angle\left(S Q, Q S^{\prime}\right)=\angle\left(N D, N S^{\prime}\right)=\measuredangle D N S^{\prime} $$ so that $S D$ is a tangent to the circle through $D, O_{1}, S^{\prime}, N$. Then we have $$ S S^{\prime} \cdot S O_{1}=S D^{2}=S X^{2} $$ Next, we show that $S$ and $S^{\prime}$ are symmetric with respect to $X Y$. By the Sine Law, we have $$ \frac{S S^{\prime}}{\sin \angle S Q S^{\prime}}=\frac{S Q}{\sin \angle S S^{\prime} Q}=\frac{S Q}{\sin \angle S D Q}=\frac{S Q}{\sin \angle S A Q}=\frac{S A}{\sin \angle S Q A} $$ It follows that $$ S S^{\prime}=S A \cdot \frac{\sin \angle S Q S^{\prime}}{\sin \angle S Q A}=S A \cdot \frac{\sin \angle H O A}{\sin \angle O H A}=S A \cdot \frac{A H}{A O}=S A \cdot 2 \cos A, $$ which is twice the distance from $S$ to $X Y$. Note that $S$ and $C$ lie on the same side of the perpendicular bisector of $P M$ if and only if $\angle S A C<\angle O A C$ if and only if $\angle Y X A>\angle C B A$. This shows $S$ and $O_{1}$ lie on different sides of $X Y$. As $S^{\prime}$ lies on ray $S O_{1}$, it follows that $S$ and $S^{\prime}$ cannot lie on the same side of $X Y$. Therefore, $S$ and $S^{\prime}$ are symmetric with respect to $X Y$. Let $d$ be the diameter of the circumcircle of triangle $X S Y$. As $S S^{\prime}$ is twice the distance from $S$ to $X Y$ and $S X=S Y$, we have $S S^{\prime}=2 \frac{S X^{2}}{d}$. It follows from (1) that $d=2 S O_{1}$. As $S O_{1}$ is the perpendicular bisector of $X Y$, point $O_{1}^{d}$ is the circumcentre of triangle $X S Y$.
|
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81e0df74-86e0-5973-a601-e8358ad3ded5
| 24,804
|
Let $D$ be the foot of perpendicular from $A$ to the Euler line (the line passing through the circumcentre and the orthocentre) of an acute scalene triangle $A B C$. A circle $\omega$ with centre $S$ passes through $A$ and $D$, and it intersects sides $A B$ and $A C$ at $X$ and $Y$ respectively. Let $P$ be the foot of altitude from $A$ to $B C$, and let $M$ be the midpoint of $B C$. Prove that the circumcentre of triangle $X S Y$ is equidistant from $P$ and $M$.
|
Denote the orthocentre and circumcentre of triangle $A B C$ by $H$ and $O$ respectively. Let $O_{1}$ be the circumcentre of triangle $X S Y$. Consider two other possible positions of $S$. We name them $S^{\prime}$ and $S^{\prime \prime}$ and define the analogous points $X^{\prime}, Y^{\prime}, O_{1}^{\prime}, X^{\prime \prime}, Y^{\prime \prime} O_{1}^{\prime \prime}$ accordingly. Note that $S, S^{\prime}, S^{\prime \prime}$ lie on the perpendicular bisector of $A D$. As $X X^{\prime}$ and $Y Y^{\prime}$ meet at $A$ and the circumcircles of triangles $A X Y$ and $A X^{\prime} Y^{\prime}$ meet at $D$, there is a spiral similarity with centre $D$ mapping $X Y$ to $X^{\prime} Y^{\prime}$. We find that $$ \measuredangle S X Y=\frac{\pi}{2}-\measuredangle Y A X=\frac{\pi}{2}-\measuredangle Y^{\prime} A X^{\prime}=\measuredangle S^{\prime} X^{\prime} Y^{\prime} $$ and similarly $\measuredangle S Y X=\measuredangle S^{\prime} Y^{\prime} X^{\prime}$. This shows triangles $S X Y$ and $S^{\prime} X^{\prime} Y^{\prime}$ are directly similar. Then the spiral similarity with centre $D$ takes points $S, X, Y, O_{1}$ to $S^{\prime}, X^{\prime}, Y^{\prime}, O_{1}^{\prime}$. Similarly, there is a spiral similarity with centre $D$ mapping $S, X, Y, O_{1}$ to $S^{\prime \prime}, X^{\prime \prime}, Y^{\prime \prime}, O_{1}^{\prime \prime}$. From these, we see that there is a spiral similarity taking the corresponding points $S, S^{\prime}, S^{\prime \prime}$ to points $O_{1}, O_{1}^{\prime}, O_{1}^{\prime \prime}$. In particular, $O_{1}, O_{1}^{\prime}, O_{1}^{\prime \prime}$ are collinear.  It now suffices to show that $O_{1}$ lies on the perpendicular bisector of $P M$ for two special cases. Firstly, we take $S$ to be the midpoint of $A H$. Then $X$ and $Y$ are the feet of altitudes from $C$ and $B$ respectively. It is well-known that the circumcircle of triangle $X S Y$ is the nine-point circle of triangle $A B C$. Then $O_{1}$ is the nine-point centre and $O_{1} P=O_{1} M$. Indeed, $P$ and $M$ also lies on the nine-point circle. Secondly, we take $S^{\prime}$ to be the midpoint of $A O$. Then $X^{\prime}$ and $Y^{\prime}$ are the midpoints of $A B$ and $A C$ respectively. Then $X^{\prime} Y^{\prime} / / B C$. Clearly, $S^{\prime}$ lies on the perpendicular bisector of $P M$. This shows the perpendicular bisectors of $X^{\prime} Y^{\prime}$ and $P M$ coincide. Hence, we must have $O_{1}^{\prime} P=O_{1}^{\prime} M$. 
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
Let $D$ be the foot of perpendicular from $A$ to the Euler line (the line passing through the circumcentre and the orthocentre) of an acute scalene triangle $A B C$. A circle $\omega$ with centre $S$ passes through $A$ and $D$, and it intersects sides $A B$ and $A C$ at $X$ and $Y$ respectively. Let $P$ be the foot of altitude from $A$ to $B C$, and let $M$ be the midpoint of $B C$. Prove that the circumcentre of triangle $X S Y$ is equidistant from $P$ and $M$.
|
Denote the orthocentre and circumcentre of triangle $A B C$ by $H$ and $O$ respectively. Let $O_{1}$ be the circumcentre of triangle $X S Y$. Consider two other possible positions of $S$. We name them $S^{\prime}$ and $S^{\prime \prime}$ and define the analogous points $X^{\prime}, Y^{\prime}, O_{1}^{\prime}, X^{\prime \prime}, Y^{\prime \prime} O_{1}^{\prime \prime}$ accordingly. Note that $S, S^{\prime}, S^{\prime \prime}$ lie on the perpendicular bisector of $A D$. As $X X^{\prime}$ and $Y Y^{\prime}$ meet at $A$ and the circumcircles of triangles $A X Y$ and $A X^{\prime} Y^{\prime}$ meet at $D$, there is a spiral similarity with centre $D$ mapping $X Y$ to $X^{\prime} Y^{\prime}$. We find that $$ \measuredangle S X Y=\frac{\pi}{2}-\measuredangle Y A X=\frac{\pi}{2}-\measuredangle Y^{\prime} A X^{\prime}=\measuredangle S^{\prime} X^{\prime} Y^{\prime} $$ and similarly $\measuredangle S Y X=\measuredangle S^{\prime} Y^{\prime} X^{\prime}$. This shows triangles $S X Y$ and $S^{\prime} X^{\prime} Y^{\prime}$ are directly similar. Then the spiral similarity with centre $D$ takes points $S, X, Y, O_{1}$ to $S^{\prime}, X^{\prime}, Y^{\prime}, O_{1}^{\prime}$. Similarly, there is a spiral similarity with centre $D$ mapping $S, X, Y, O_{1}$ to $S^{\prime \prime}, X^{\prime \prime}, Y^{\prime \prime}, O_{1}^{\prime \prime}$. From these, we see that there is a spiral similarity taking the corresponding points $S, S^{\prime}, S^{\prime \prime}$ to points $O_{1}, O_{1}^{\prime}, O_{1}^{\prime \prime}$. In particular, $O_{1}, O_{1}^{\prime}, O_{1}^{\prime \prime}$ are collinear.  It now suffices to show that $O_{1}$ lies on the perpendicular bisector of $P M$ for two special cases. Firstly, we take $S$ to be the midpoint of $A H$. Then $X$ and $Y$ are the feet of altitudes from $C$ and $B$ respectively. It is well-known that the circumcircle of triangle $X S Y$ is the nine-point circle of triangle $A B C$. Then $O_{1}$ is the nine-point centre and $O_{1} P=O_{1} M$. Indeed, $P$ and $M$ also lies on the nine-point circle. Secondly, we take $S^{\prime}$ to be the midpoint of $A O$. Then $X^{\prime}$ and $Y^{\prime}$ are the midpoints of $A B$ and $A C$ respectively. Then $X^{\prime} Y^{\prime} / / B C$. Clearly, $S^{\prime}$ lies on the perpendicular bisector of $P M$. This shows the perpendicular bisectors of $X^{\prime} Y^{\prime}$ and $P M$ coincide. Hence, we must have $O_{1}^{\prime} P=O_{1}^{\prime} M$. 
|
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81e0df74-86e0-5973-a601-e8358ad3ded5
| 24,804
|
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 $P Q \perp A C$.
|
 Let $\omega_{1}$ be the circumcircle of triangle $A B C$. We first prove that $Y$ lies on $\omega_{1}$. Let $Y^{\prime}$ be the point on ray $M D$ such that $M Y^{\prime} \cdot M D=M A^{2}$. Then triangles $M A Y^{\prime}$ and $M D A$ are oppositely similar. Since $M C^{2}=M A^{2}=M Y^{\prime} \cdot M D$, triangles $M C Y^{\prime}$ and $M D C$ are also oppositely similar. Therefore, using directed angles, we have $$ \measuredangle A Y^{\prime} C=\measuredangle A Y^{\prime} M+\measuredangle M Y^{\prime} C=\measuredangle M A D+\measuredangle D C M=\measuredangle C D A=\measuredangle A B C $$ so that $Y^{\prime}$ lies on $\omega_{1}$. Let $Z$ be the intersection point of lines $B C$ and $A D$. Since $\measuredangle P D Z=\measuredangle P B C=\measuredangle P B Z$, point $Z$ lies on $\omega$. In addition, from $\measuredangle Y^{\prime} B Z=\measuredangle Y^{\prime} B C=\measuredangle Y^{\prime} A C=\measuredangle Y^{\prime} A M=\measuredangle Y^{\prime} D Z$, we also know that $Y^{\prime}$ lies on $\omega$. Note that $\angle A D C$ is acute implies $M A \neq M D$ so $M Y^{\prime} \neq M D$. Therefore, $Y^{\prime}$ is the second intersection of $D M$ and $\omega$. Then $Y^{\prime}=Y$ and hence $Y$ lies on $\omega_{1}$. Next, by the Angle Bisector Theorem and the similar triangles, we have $$ \frac{F A}{F C}=\frac{A D}{C D}=\frac{A D}{A M} \cdot \frac{C M}{C D}=\frac{Y A}{Y M} \cdot \frac{Y M}{Y C}=\frac{Y A}{Y C} $$ Hence, $F Y$ is the internal angle bisector of $\angle A Y C$. Let $B^{\prime}$ be the second intersection of the internal angle bisector of $\angle C B A$ and $\omega_{1}$. Then $B^{\prime}$ is the midpoint of arc $A C$ not containing $B$. Therefore, $Y B^{\prime}$ is the external angle bisector of $\angle A Y C$, so that $B^{\prime} Y \perp F Y$. Denote by $l$ the line through $P$ parallel to $A C$. Suppose $l$ meets line $B^{\prime} Y$ at $S$. From $$ \begin{aligned} \measuredangle P S Y & =\measuredangle\left(A C, B^{\prime} Y\right)=\measuredangle A C Y+\measuredangle C Y B^{\prime}=\measuredangle A C Y+\measuredangle C A B^{\prime}=\measuredangle A C Y+\measuredangle B^{\prime} C A \\ & =\measuredangle B^{\prime} C Y=\measuredangle B^{\prime} B Y=\measuredangle P B Y \end{aligned} $$ the point $S$ lies on $\omega$. Similarly, the line through $X$ perpendicular to $X E$ also passes through the second intersection of $l$ and $\omega$, which is the point $S$. From $Q Y \perp Y S$ and $Q X \perp X S$, point $Q$ lies on $\omega$ and $Q S$ is a diameter of $\omega$. Therefore, $P Q \perp P S$ so that $P Q \perp A C$.
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
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 $P Q \perp A C$.
|
 Let $\omega_{1}$ be the circumcircle of triangle $A B C$. We first prove that $Y$ lies on $\omega_{1}$. Let $Y^{\prime}$ be the point on ray $M D$ such that $M Y^{\prime} \cdot M D=M A^{2}$. Then triangles $M A Y^{\prime}$ and $M D A$ are oppositely similar. Since $M C^{2}=M A^{2}=M Y^{\prime} \cdot M D$, triangles $M C Y^{\prime}$ and $M D C$ are also oppositely similar. Therefore, using directed angles, we have $$ \measuredangle A Y^{\prime} C=\measuredangle A Y^{\prime} M+\measuredangle M Y^{\prime} C=\measuredangle M A D+\measuredangle D C M=\measuredangle C D A=\measuredangle A B C $$ so that $Y^{\prime}$ lies on $\omega_{1}$. Let $Z$ be the intersection point of lines $B C$ and $A D$. Since $\measuredangle P D Z=\measuredangle P B C=\measuredangle P B Z$, point $Z$ lies on $\omega$. In addition, from $\measuredangle Y^{\prime} B Z=\measuredangle Y^{\prime} B C=\measuredangle Y^{\prime} A C=\measuredangle Y^{\prime} A M=\measuredangle Y^{\prime} D Z$, we also know that $Y^{\prime}$ lies on $\omega$. Note that $\angle A D C$ is acute implies $M A \neq M D$ so $M Y^{\prime} \neq M D$. Therefore, $Y^{\prime}$ is the second intersection of $D M$ and $\omega$. Then $Y^{\prime}=Y$ and hence $Y$ lies on $\omega_{1}$. Next, by the Angle Bisector Theorem and the similar triangles, we have $$ \frac{F A}{F C}=\frac{A D}{C D}=\frac{A D}{A M} \cdot \frac{C M}{C D}=\frac{Y A}{Y M} \cdot \frac{Y M}{Y C}=\frac{Y A}{Y C} $$ Hence, $F Y$ is the internal angle bisector of $\angle A Y C$. Let $B^{\prime}$ be the second intersection of the internal angle bisector of $\angle C B A$ and $\omega_{1}$. Then $B^{\prime}$ is the midpoint of arc $A C$ not containing $B$. Therefore, $Y B^{\prime}$ is the external angle bisector of $\angle A Y C$, so that $B^{\prime} Y \perp F Y$. Denote by $l$ the line through $P$ parallel to $A C$. Suppose $l$ meets line $B^{\prime} Y$ at $S$. From $$ \begin{aligned} \measuredangle P S Y & =\measuredangle\left(A C, B^{\prime} Y\right)=\measuredangle A C Y+\measuredangle C Y B^{\prime}=\measuredangle A C Y+\measuredangle C A B^{\prime}=\measuredangle A C Y+\measuredangle B^{\prime} C A \\ & =\measuredangle B^{\prime} C Y=\measuredangle B^{\prime} B Y=\measuredangle P B Y \end{aligned} $$ the point $S$ lies on $\omega$. Similarly, the line through $X$ perpendicular to $X E$ also passes through the second intersection of $l$ and $\omega$, which is the point $S$. From $Q Y \perp Y S$ and $Q X \perp X S$, point $Q$ lies on $\omega$ and $Q S$ is a diameter of $\omega$. Therefore, $P Q \perp P S$ so that $P Q \perp A C$.
|
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4a0f0e68-06f1-5b11-84ac-db452d90625c
| 24,808
|
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 $P Q \perp A C$.
|
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 $$ \begin{aligned} \measuredangle D X_{1} B & =\measuredangle D X_{1} A+\measuredangle A X_{1} B=\measuredangle D C A+\measuredangle A X_{1} X^{\prime}=\measuredangle D C A+\measuredangle C X^{\prime} X_{1} \\ & =\measuredangle D C A+\measuredangle C A B=\measuredangle(C D, A B) . \end{aligned} $$  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 $Q P \perp A C$.
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
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 $P Q \perp A C$.
|
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 $$ \begin{aligned} \measuredangle D X_{1} B & =\measuredangle D X_{1} A+\measuredangle A X_{1} B=\measuredangle D C A+\measuredangle A X_{1} X^{\prime}=\measuredangle D C A+\measuredangle C X^{\prime} X_{1} \\ & =\measuredangle D C A+\measuredangle C A B=\measuredangle(C D, A B) . \end{aligned} $$  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 $Q P \perp A C$.
|
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4a0f0e68-06f1-5b11-84ac-db452d90625c
| 24,808
|
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 $\angle X I Y=120^{\circ}$.
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(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.  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 $\angle X I Y=120^{\circ}$.
|
proof
|
Yes
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Yes
|
proof
|
Geometry
|
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 $\angle X I Y=120^{\circ}$.
|
(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.  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 $\angle X I Y=120^{\circ}$.
|
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7aa54f19-7001-58f3-bebb-c8525675f018
| 24,813
|
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 $\angle X I Y=120^{\circ}$.
|
(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.  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.
|
proof
|
Yes
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Yes
|
proof
|
Geometry
|
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 $\angle X I Y=120^{\circ}$.
|
(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.  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.
|
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7aa54f19-7001-58f3-bebb-c8525675f018
| 24,813
|
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 $$ A H+B H+C H \geqslant A I+B I+C I $$
|
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 $A A_{1} C_{1}$.  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.
|
proof
|
Yes
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Yes
|
proof
|
Geometry
|
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 $$ A H+B H+C H \geqslant A I+B I+C I $$
|
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 $A A_{1} C_{1}$.  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.
|
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0bb9d9f5-4950-5764-afed-d572528ad73b
| 24,818
|
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 $m=\frac{n^{k-l}-1}{n^{l}-1}$.
|
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.
|
proof
|
Yes
|
Yes
|
proof
|
Number Theory
|
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 $m=\frac{n^{k-l}-1}{n^{l}-1}$.
|
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.
|
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|
6034ae4f-da26-5804-aff8-f1ab3adf235b
| 24,831
|
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 $m=\frac{n^{k-l}-1}{n^{l}-1}$.
|
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 $n^{k+l}-1$.
|
proof
|
Yes
|
Yes
|
proof
|
Number Theory
|
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 $m=\frac{n^{k-l}-1}{n^{l}-1}$.
|
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 $n^{k+l}-1$.
|
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6034ae4f-da26-5804-aff8-f1ab3adf235b
| 24,831
|
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 $A=B$.
|
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 $A=B$.
|
proof
|
Yes
|
Yes
|
proof
|
Number Theory
|
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 $A=B$.
|
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 $A=B$.
|
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95fa4c0d-c215-5bf0-8ae4-8fed0df8eac4
| 24,836
|
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 $A=B$.
|
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 $A \subset B$.
|
proof
|
Yes
|
Yes
|
proof
|
Number Theory
|
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 $A=B$.
|
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 $A \subset B$.
|
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95fa4c0d-c215-5bf0-8ae4-8fed0df8eac4
| 24,836
|
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 $A=B$.
|
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.
|
proof
|
Yes
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Yes
|
proof
|
Number Theory
|
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 $A=B$.
|
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.
|
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95fa4c0d-c215-5bf0-8ae4-8fed0df8eac4
| 24,836
|
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.
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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 .
|
proof
|
Yes
|
Yes
|
proof
|
Combinatorics
|
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.
|
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 .
|
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23be2ebb-5023-5c4d-a7a0-e473aabde18a
| 24,876
|
Let $a_{1}, a_{2}, \ldots, a_{n}, k$, and $M$ be positive integers such that $$ \frac{1}{a_{1}}+\frac{1}{a_{2}}+\cdots+\frac{1}{a_{n}}=k \quad \text { and } \quad a_{1} a_{2} \ldots a_{n}=M $$ If $M>1$, prove that the polynomial $$ P(x)=M(x+1)^{k}-\left(x+a_{1}\right)\left(x+a_{2}\right) \cdots\left(x+a_{n}\right) $$ has no positive roots. (Trinidad and Tobago)
|
We first prove that, for $x>0$, $$ a_{i}(x+1)^{1 / a_{i}} \leqslant x+a_{i}, $$ with equality if and only if $a_{i}=1$. It is clear that equality occurs if $a_{i}=1$. If $a_{i}>1$, the AM-GM inequality applied to a single copy of $x+1$ and $a_{i}-1$ copies of 1 yields $$ \frac{(x+1)+\overbrace{1+1+\cdots+1}^{a_{i}-1 \text { ones }}}{a_{i}} \geqslant \sqrt[a_{i}]{(x+1) \cdot 1^{a_{i}-1}} \Longrightarrow a_{i}(x+1)^{1 / a_{i}} \leqslant x+a_{i} $$ Since $x+1>1$, the inequality is strict for $a_{i}>1$. Multiplying the inequalities (1) for $i=1,2, \ldots, n$ yields $$ \prod_{i=1}^{n} a_{i}(x+1)^{1 / a_{i}} \leqslant \prod_{i=1}^{n}\left(x+a_{i}\right) \Longleftrightarrow M(x+1)^{\sum_{i=1}^{n} 1 / a_{i}}-\prod_{i=1}^{n}\left(x+a_{i}\right) \leqslant 0 \Longleftrightarrow P(x) \leqslant 0 $$ with equality iff $a_{i}=1$ for all $i \in\{1,2, \ldots, n\}$. But this implies $M=1$, which is not possible. Hence $P(x)<0$ for all $x \in \mathbb{R}^{+}$, and $P$ has no positive roots. Comment 1. Inequality (1) can be obtained in several ways. For instance, we may also use the binomial theorem: since $a_{i} \geqslant 1$, $$ \left(1+\frac{x}{a_{i}}\right)^{a_{i}}=\sum_{j=0}^{a_{i}}\binom{a_{i}}{j}\left(\frac{x}{a_{i}}\right)^{j} \geqslant\binom{ a_{i}}{0}+\binom{a_{i}}{1} \cdot \frac{x}{a_{i}}=1+x $$ Both proofs of (1) mimic proofs to Bernoulli's inequality for a positive integer exponent $a_{i}$; we can use this inequality directly: $$ \left(1+\frac{x}{a_{i}}\right)^{a_{i}} \geqslant 1+a_{i} \cdot \frac{x}{a_{i}}=1+x $$ and so $$ x+a_{i}=a_{i}\left(1+\frac{x}{a_{i}}\right) \geqslant a_{i}(1+x)^{1 / a_{i}} $$ or its (reversed) formulation, with exponent $1 / a_{i} \leqslant 1$ : $$ (1+x)^{1 / a_{i}} \leqslant 1+\frac{1}{a_{i}} \cdot x=\frac{x+a_{i}}{a_{i}} \Longrightarrow a_{i}(1+x)^{1 / a_{i}} \leqslant x+a_{i} . $$
|
proof
|
Yes
|
Yes
|
proof
|
Algebra
|
Let $a_{1}, a_{2}, \ldots, a_{n}, k$, and $M$ be positive integers such that $$ \frac{1}{a_{1}}+\frac{1}{a_{2}}+\cdots+\frac{1}{a_{n}}=k \quad \text { and } \quad a_{1} a_{2} \ldots a_{n}=M $$ If $M>1$, prove that the polynomial $$ P(x)=M(x+1)^{k}-\left(x+a_{1}\right)\left(x+a_{2}\right) \cdots\left(x+a_{n}\right) $$ has no positive roots. (Trinidad and Tobago)
|
We first prove that, for $x>0$, $$ a_{i}(x+1)^{1 / a_{i}} \leqslant x+a_{i}, $$ with equality if and only if $a_{i}=1$. It is clear that equality occurs if $a_{i}=1$. If $a_{i}>1$, the AM-GM inequality applied to a single copy of $x+1$ and $a_{i}-1$ copies of 1 yields $$ \frac{(x+1)+\overbrace{1+1+\cdots+1}^{a_{i}-1 \text { ones }}}{a_{i}} \geqslant \sqrt[a_{i}]{(x+1) \cdot 1^{a_{i}-1}} \Longrightarrow a_{i}(x+1)^{1 / a_{i}} \leqslant x+a_{i} $$ Since $x+1>1$, the inequality is strict for $a_{i}>1$. Multiplying the inequalities (1) for $i=1,2, \ldots, n$ yields $$ \prod_{i=1}^{n} a_{i}(x+1)^{1 / a_{i}} \leqslant \prod_{i=1}^{n}\left(x+a_{i}\right) \Longleftrightarrow M(x+1)^{\sum_{i=1}^{n} 1 / a_{i}}-\prod_{i=1}^{n}\left(x+a_{i}\right) \leqslant 0 \Longleftrightarrow P(x) \leqslant 0 $$ with equality iff $a_{i}=1$ for all $i \in\{1,2, \ldots, n\}$. But this implies $M=1$, which is not possible. Hence $P(x)<0$ for all $x \in \mathbb{R}^{+}$, and $P$ has no positive roots. Comment 1. Inequality (1) can be obtained in several ways. For instance, we may also use the binomial theorem: since $a_{i} \geqslant 1$, $$ \left(1+\frac{x}{a_{i}}\right)^{a_{i}}=\sum_{j=0}^{a_{i}}\binom{a_{i}}{j}\left(\frac{x}{a_{i}}\right)^{j} \geqslant\binom{ a_{i}}{0}+\binom{a_{i}}{1} \cdot \frac{x}{a_{i}}=1+x $$ Both proofs of (1) mimic proofs to Bernoulli's inequality for a positive integer exponent $a_{i}$; we can use this inequality directly: $$ \left(1+\frac{x}{a_{i}}\right)^{a_{i}} \geqslant 1+a_{i} \cdot \frac{x}{a_{i}}=1+x $$ and so $$ x+a_{i}=a_{i}\left(1+\frac{x}{a_{i}}\right) \geqslant a_{i}(1+x)^{1 / a_{i}} $$ or its (reversed) formulation, with exponent $1 / a_{i} \leqslant 1$ : $$ (1+x)^{1 / a_{i}} \leqslant 1+\frac{1}{a_{i}} \cdot x=\frac{x+a_{i}}{a_{i}} \Longrightarrow a_{i}(1+x)^{1 / a_{i}} \leqslant x+a_{i} . $$
|
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013585db-1fec-52eb-8d43-03e53435c361
| 24,909
|
Let $a_{1}, a_{2}, \ldots, a_{n}, k$, and $M$ be positive integers such that $$ \frac{1}{a_{1}}+\frac{1}{a_{2}}+\cdots+\frac{1}{a_{n}}=k \quad \text { and } \quad a_{1} a_{2} \ldots a_{n}=M $$ If $M>1$, prove that the polynomial $$ P(x)=M(x+1)^{k}-\left(x+a_{1}\right)\left(x+a_{2}\right) \cdots\left(x+a_{n}\right) $$ has no positive roots. (Trinidad and Tobago)
|
We will prove that, in fact, all coefficients of the polynomial $P(x)$ are non-positive, and at least one of them is negative, which implies that $P(x)<0$ for $x>0$. Indeed, since $a_{j} \geqslant 1$ for all $j$ and $a_{j}>1$ for some $j$ (since $a_{1} a_{2} \ldots a_{n}=M>1$ ), we have $k=\frac{1}{a_{1}}+\frac{1}{a_{2}}+\cdots+\frac{1}{a_{n}}<n$, so the coefficient of $x^{n}$ in $P(x)$ is $-1<0$. Moreover, the coefficient of $x^{r}$ in $P(x)$ is negative for $k<r \leqslant n=\operatorname{deg}(P)$. For $0 \leqslant r \leqslant k$, the coefficient of $x^{r}$ in $P(x)$ is $$ M \cdot\binom{k}{r}-\sum_{1 \leqslant i_{1}<i_{2}<\cdots<i_{n-r} \leqslant n} a_{i_{1}} a_{i_{2}} \cdots a_{i_{n-r}}=a_{1} a_{2} \cdots a_{n} \cdot\binom{k}{r}-\sum_{1 \leqslant i_{1}<i_{2}<\cdots<i_{n-r} \leqslant n} a_{i_{1}} a_{i_{2}} \cdots a_{i_{n-r}} $$ which is non-positive iff $$ \binom{k}{r} \leqslant \sum_{1 \leqslant j_{1}<j_{2}<\cdots<j_{r} \leqslant n} \frac{1}{a_{j_{1}} a_{j_{2}} \cdots a_{j_{r}}} $$ We will prove (2) by induction on $r$. For $r=0$ it is an equality because the constant term of $P(x)$ is $P(0)=0$, and if $r=1$, (2) becomes $k=\sum_{i=1}^{n} \frac{1}{a_{i}}$. For $r>1$, if (2) is true for a given $r<k$, we have $$ \binom{k}{r+1}=\frac{k-r}{r+1} \cdot\binom{k}{r} \leqslant \frac{k-r}{r+1} \cdot \sum_{1 \leqslant j_{1}<j_{2}<\cdots<j_{r} \leqslant n} \frac{1}{a_{j_{1}} a_{j_{2}} \cdots a_{j_{r}}} $$ and it suffices to prove that $$ \frac{k-r}{r+1} \cdot \sum_{1 \leqslant j_{1}<j_{2}<\cdots<j_{r} \leqslant n} \frac{1}{a_{j_{1}} a_{j_{2}} \cdots a_{j_{r}}} \leqslant \sum_{1 \leqslant j_{1}<\cdots<j_{r}<j_{r+1} \leqslant n} \frac{1}{a_{j_{1}} a_{j_{2}} \cdots a_{j_{r}} a_{j_{r+1}}} $$ which is equivalent to $$ \left(\frac{1}{a_{1}}+\frac{1}{a_{2}}+\cdots+\frac{1}{a_{n}}-r\right) \sum_{1 \leqslant j_{1}<j_{2}<\cdots<j_{r} \leqslant n} \frac{1}{a_{j_{1}} a_{j_{2}} \cdots a_{j_{r}}} \leqslant(r+1) \sum_{1 \leqslant j_{1}<\cdots<j_{r}<j_{r+1} \leqslant n} \frac{1}{a_{j_{1}} a_{j_{2}} \cdots a_{j_{r}} a_{j_{r+1}}} $$ Since there are $r+1$ ways to choose a fraction $\frac{1}{a_{j_{i}}}$ from $\frac{1}{a_{j_{1}} a_{j_{2}} \cdots a_{j_{r}} a_{j_{r}+1}}$ to factor out, every term $\frac{1}{a_{j_{1}} a_{j_{2}} \cdots a_{j_{r}} a_{j_{r+1}}}$ in the right hand side appears exactly $r+1$ times in the product $$ \left(\frac{1}{a_{1}}+\frac{1}{a_{2}}+\cdots+\frac{1}{a_{n}}\right)_{1 \leqslant j_{1}<j_{2}<\cdots<j_{r} \leqslant n} \sum_{a_{j_{1}} a_{j_{2}} \cdots a_{j_{r}}} $$ Hence all terms in the right hand side cancel out. The remaining terms in the left hand side can be grouped in sums of the type $$ \begin{aligned} \frac{1}{a_{j_{1}}^{2} a_{j_{2}} \cdots a_{j_{r}}}+\frac{1}{a_{j_{1}} a_{j_{2}}^{2} \cdots a_{j_{r}}}+\cdots+\frac{1}{a_{j_{1}} a_{j_{2}} \cdots a_{j_{r}}^{2}} & -\frac{r}{a_{j_{1}} a_{j_{2}} \cdots a_{j_{r}}} \\ & =\frac{1}{a_{j_{1}} a_{j_{2}} \cdots a_{j_{r}}}\left(\frac{1}{a_{j_{1}}}+\frac{1}{a_{j_{2}}}+\cdots+\frac{1}{a_{j_{r}}}-r\right) \end{aligned} $$ which are all non-positive because $a_{i} \geqslant 1 \Longrightarrow \frac{1}{a_{i}} \leqslant 1, i=1,2, \ldots, n$. Comment 2. The result is valid for any real numbers $a_{i}, i=1,2, \ldots, n$ with $a_{i} \geqslant 1$ and product $M$ greater than 1. A variation of Solution 1, namely using weighted AM-GM (or the Bernoulli inequality for real exponents), actually proves that $P(x)<0$ for $x>-1$ and $x \neq 0$.
|
proof
|
Yes
|
Yes
|
proof
|
Algebra
|
Let $a_{1}, a_{2}, \ldots, a_{n}, k$, and $M$ be positive integers such that $$ \frac{1}{a_{1}}+\frac{1}{a_{2}}+\cdots+\frac{1}{a_{n}}=k \quad \text { and } \quad a_{1} a_{2} \ldots a_{n}=M $$ If $M>1$, prove that the polynomial $$ P(x)=M(x+1)^{k}-\left(x+a_{1}\right)\left(x+a_{2}\right) \cdots\left(x+a_{n}\right) $$ has no positive roots. (Trinidad and Tobago)
|
We will prove that, in fact, all coefficients of the polynomial $P(x)$ are non-positive, and at least one of them is negative, which implies that $P(x)<0$ for $x>0$. Indeed, since $a_{j} \geqslant 1$ for all $j$ and $a_{j}>1$ for some $j$ (since $a_{1} a_{2} \ldots a_{n}=M>1$ ), we have $k=\frac{1}{a_{1}}+\frac{1}{a_{2}}+\cdots+\frac{1}{a_{n}}<n$, so the coefficient of $x^{n}$ in $P(x)$ is $-1<0$. Moreover, the coefficient of $x^{r}$ in $P(x)$ is negative for $k<r \leqslant n=\operatorname{deg}(P)$. For $0 \leqslant r \leqslant k$, the coefficient of $x^{r}$ in $P(x)$ is $$ M \cdot\binom{k}{r}-\sum_{1 \leqslant i_{1}<i_{2}<\cdots<i_{n-r} \leqslant n} a_{i_{1}} a_{i_{2}} \cdots a_{i_{n-r}}=a_{1} a_{2} \cdots a_{n} \cdot\binom{k}{r}-\sum_{1 \leqslant i_{1}<i_{2}<\cdots<i_{n-r} \leqslant n} a_{i_{1}} a_{i_{2}} \cdots a_{i_{n-r}} $$ which is non-positive iff $$ \binom{k}{r} \leqslant \sum_{1 \leqslant j_{1}<j_{2}<\cdots<j_{r} \leqslant n} \frac{1}{a_{j_{1}} a_{j_{2}} \cdots a_{j_{r}}} $$ We will prove (2) by induction on $r$. For $r=0$ it is an equality because the constant term of $P(x)$ is $P(0)=0$, and if $r=1$, (2) becomes $k=\sum_{i=1}^{n} \frac{1}{a_{i}}$. For $r>1$, if (2) is true for a given $r<k$, we have $$ \binom{k}{r+1}=\frac{k-r}{r+1} \cdot\binom{k}{r} \leqslant \frac{k-r}{r+1} \cdot \sum_{1 \leqslant j_{1}<j_{2}<\cdots<j_{r} \leqslant n} \frac{1}{a_{j_{1}} a_{j_{2}} \cdots a_{j_{r}}} $$ and it suffices to prove that $$ \frac{k-r}{r+1} \cdot \sum_{1 \leqslant j_{1}<j_{2}<\cdots<j_{r} \leqslant n} \frac{1}{a_{j_{1}} a_{j_{2}} \cdots a_{j_{r}}} \leqslant \sum_{1 \leqslant j_{1}<\cdots<j_{r}<j_{r+1} \leqslant n} \frac{1}{a_{j_{1}} a_{j_{2}} \cdots a_{j_{r}} a_{j_{r+1}}} $$ which is equivalent to $$ \left(\frac{1}{a_{1}}+\frac{1}{a_{2}}+\cdots+\frac{1}{a_{n}}-r\right) \sum_{1 \leqslant j_{1}<j_{2}<\cdots<j_{r} \leqslant n} \frac{1}{a_{j_{1}} a_{j_{2}} \cdots a_{j_{r}}} \leqslant(r+1) \sum_{1 \leqslant j_{1}<\cdots<j_{r}<j_{r+1} \leqslant n} \frac{1}{a_{j_{1}} a_{j_{2}} \cdots a_{j_{r}} a_{j_{r+1}}} $$ Since there are $r+1$ ways to choose a fraction $\frac{1}{a_{j_{i}}}$ from $\frac{1}{a_{j_{1}} a_{j_{2}} \cdots a_{j_{r}} a_{j_{r}+1}}$ to factor out, every term $\frac{1}{a_{j_{1}} a_{j_{2}} \cdots a_{j_{r}} a_{j_{r+1}}}$ in the right hand side appears exactly $r+1$ times in the product $$ \left(\frac{1}{a_{1}}+\frac{1}{a_{2}}+\cdots+\frac{1}{a_{n}}\right)_{1 \leqslant j_{1}<j_{2}<\cdots<j_{r} \leqslant n} \sum_{a_{j_{1}} a_{j_{2}} \cdots a_{j_{r}}} $$ Hence all terms in the right hand side cancel out. The remaining terms in the left hand side can be grouped in sums of the type $$ \begin{aligned} \frac{1}{a_{j_{1}}^{2} a_{j_{2}} \cdots a_{j_{r}}}+\frac{1}{a_{j_{1}} a_{j_{2}}^{2} \cdots a_{j_{r}}}+\cdots+\frac{1}{a_{j_{1}} a_{j_{2}} \cdots a_{j_{r}}^{2}} & -\frac{r}{a_{j_{1}} a_{j_{2}} \cdots a_{j_{r}}} \\ & =\frac{1}{a_{j_{1}} a_{j_{2}} \cdots a_{j_{r}}}\left(\frac{1}{a_{j_{1}}}+\frac{1}{a_{j_{2}}}+\cdots+\frac{1}{a_{j_{r}}}-r\right) \end{aligned} $$ which are all non-positive because $a_{i} \geqslant 1 \Longrightarrow \frac{1}{a_{i}} \leqslant 1, i=1,2, \ldots, n$. Comment 2. The result is valid for any real numbers $a_{i}, i=1,2, \ldots, n$ with $a_{i} \geqslant 1$ and product $M$ greater than 1. A variation of Solution 1, namely using weighted AM-GM (or the Bernoulli inequality for real exponents), actually proves that $P(x)<0$ for $x>-1$ and $x \neq 0$.
|
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013585db-1fec-52eb-8d43-03e53435c361
| 24,909
|
Let $S$ be a finite set, and let $\mathcal{A}$ be the set of all functions from $S$ to $S$. Let $f$ be an element of $\mathcal{A}$, and let $T=f(S)$ be the image of $S$ under $f$. Suppose that $f \circ g \circ f \neq g \circ f \circ g$ for every $g$ in $\mathcal{A}$ with $g \neq f$. Show that $f(T)=T$. (India)
|
For $n \geqslant 1$, denote the $n$-th composition of $f$ with itself by $$ f^{n} \stackrel{\text { def }}{=} \underbrace{f \circ f \circ \cdots \circ f}_{n \text { times }} $$ By hypothesis, if $g \in \mathcal{A}$ satisfies $f \circ g \circ f=g \circ f \circ g$, then $g=f$. A natural idea is to try to plug in $g=f^{n}$ for some $n$ in the expression $f \circ g \circ f=g \circ f \circ g$ in order to get $f^{n}=f$, which solves the problem: Claim. If there exists $n \geqslant 3$ such that $f^{n+2}=f^{2 n+1}$, then the restriction $f: T \rightarrow T$ of $f$ to $T$ is a bijection. Proof. Indeed, by hypothesis, $f^{n+2}=f^{2 n+1} \Longleftrightarrow f \circ f^{n} \circ f=f^{n} \circ f \circ f^{n} \Longrightarrow f^{n}=f$. Since $n-2 \geqslant 1$, the image of $f^{n-2}$ is contained in $T=f(S)$, hence $f^{n-2}$ restricts to a function $f^{n-2}: T \rightarrow T$. This is the inverse of $f: T \rightarrow T$. In fact, given $t \in T$, say $t=f(s)$ with $s \in S$, we have $$ t=f(s)=f^{n}(s)=f^{n-2}(f(t))=f\left(f^{n-2}(t)\right), \quad \text { i.e., } \quad f^{n-2} \circ f=f \circ f^{n-2}=\operatorname{id} \text { on } T $$ (here id stands for the identity function). Hence, the restriction $f: T \rightarrow T$ of $f$ to $T$ is bijective with inverse given by $f^{n-2}: T \rightarrow T$. It remains to show that $n$ as in the claim exists. For that, define $$ S_{m} \stackrel{\text { def }}{=} f^{m}(S) \quad\left(S_{m} \text { is image of } f^{m}\right) $$ Clearly the image of $f^{m+1}$ is contained in the image of $f^{m}$, i.e., there is a descending chain of subsets of $S$ $$ S \supseteq S_{1} \supseteq S_{2} \supseteq S_{3} \supseteq S_{4} \supseteq \cdots, $$ which must eventually stabilise since $S$ is finite, i.e., there is a $k \geqslant 1$ such that $$ S_{k}=S_{k+1}=S_{k+2}=S_{k+3}=\cdots \stackrel{\text { def }}{=} S_{\infty} $$ Hence $f$ restricts to a surjective function $f: S_{\infty} \rightarrow S_{\infty}$, which is also bijective since $S_{\infty} \subseteq S$ is finite. To sum up, $f: S_{\infty} \rightarrow S_{\infty}$ is a permutation of the elements of the finite set $S_{\infty}$, hence there exists an integer $r \geqslant 1$ such that $f^{r}=\mathrm{id}$ on $S_{\infty}$ (for example, we may choose $r=\left|S_{\infty}\right|!$ ). In other words, $$ f^{m+r}=f^{m} \text { on } S \text { for all } m \geqslant k $$ Clearly, (*) also implies that $f^{m+t r}=f^{m}$ for all integers $t \geqslant 1$ and $m \geqslant k$. So, to find $n$ as in the claim and finish the problem, it is enough to choose $m$ and $t$ in order to ensure that there exists $n \geqslant 3$ satisfying $$ \left\{\begin{array} { l } { 2 n + 1 = m + t r } \\ { n + 2 = m } \end{array} \Longleftrightarrow \left\{\begin{array}{l} m=3+t r \\ n=m-2 \end{array}\right.\right. $$ This can be clearly done by choosing $m$ large enough with $m \equiv 3(\bmod r)$. For instance, we may take $n=2 k r+1$, so that $$ f^{n+2}=f^{2 k r+3}=f^{4 k r+3}=f^{2 n+1} $$ where the middle equality follows by (*) since $2 k r+3 \geqslant k$.
|
proof
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Yes
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Yes
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proof
|
Algebra
|
Let $S$ be a finite set, and let $\mathcal{A}$ be the set of all functions from $S$ to $S$. Let $f$ be an element of $\mathcal{A}$, and let $T=f(S)$ be the image of $S$ under $f$. Suppose that $f \circ g \circ f \neq g \circ f \circ g$ for every $g$ in $\mathcal{A}$ with $g \neq f$. Show that $f(T)=T$. (India)
|
For $n \geqslant 1$, denote the $n$-th composition of $f$ with itself by $$ f^{n} \stackrel{\text { def }}{=} \underbrace{f \circ f \circ \cdots \circ f}_{n \text { times }} $$ By hypothesis, if $g \in \mathcal{A}$ satisfies $f \circ g \circ f=g \circ f \circ g$, then $g=f$. A natural idea is to try to plug in $g=f^{n}$ for some $n$ in the expression $f \circ g \circ f=g \circ f \circ g$ in order to get $f^{n}=f$, which solves the problem: Claim. If there exists $n \geqslant 3$ such that $f^{n+2}=f^{2 n+1}$, then the restriction $f: T \rightarrow T$ of $f$ to $T$ is a bijection. Proof. Indeed, by hypothesis, $f^{n+2}=f^{2 n+1} \Longleftrightarrow f \circ f^{n} \circ f=f^{n} \circ f \circ f^{n} \Longrightarrow f^{n}=f$. Since $n-2 \geqslant 1$, the image of $f^{n-2}$ is contained in $T=f(S)$, hence $f^{n-2}$ restricts to a function $f^{n-2}: T \rightarrow T$. This is the inverse of $f: T \rightarrow T$. In fact, given $t \in T$, say $t=f(s)$ with $s \in S$, we have $$ t=f(s)=f^{n}(s)=f^{n-2}(f(t))=f\left(f^{n-2}(t)\right), \quad \text { i.e., } \quad f^{n-2} \circ f=f \circ f^{n-2}=\operatorname{id} \text { on } T $$ (here id stands for the identity function). Hence, the restriction $f: T \rightarrow T$ of $f$ to $T$ is bijective with inverse given by $f^{n-2}: T \rightarrow T$. It remains to show that $n$ as in the claim exists. For that, define $$ S_{m} \stackrel{\text { def }}{=} f^{m}(S) \quad\left(S_{m} \text { is image of } f^{m}\right) $$ Clearly the image of $f^{m+1}$ is contained in the image of $f^{m}$, i.e., there is a descending chain of subsets of $S$ $$ S \supseteq S_{1} \supseteq S_{2} \supseteq S_{3} \supseteq S_{4} \supseteq \cdots, $$ which must eventually stabilise since $S$ is finite, i.e., there is a $k \geqslant 1$ such that $$ S_{k}=S_{k+1}=S_{k+2}=S_{k+3}=\cdots \stackrel{\text { def }}{=} S_{\infty} $$ Hence $f$ restricts to a surjective function $f: S_{\infty} \rightarrow S_{\infty}$, which is also bijective since $S_{\infty} \subseteq S$ is finite. To sum up, $f: S_{\infty} \rightarrow S_{\infty}$ is a permutation of the elements of the finite set $S_{\infty}$, hence there exists an integer $r \geqslant 1$ such that $f^{r}=\mathrm{id}$ on $S_{\infty}$ (for example, we may choose $r=\left|S_{\infty}\right|!$ ). In other words, $$ f^{m+r}=f^{m} \text { on } S \text { for all } m \geqslant k $$ Clearly, (*) also implies that $f^{m+t r}=f^{m}$ for all integers $t \geqslant 1$ and $m \geqslant k$. So, to find $n$ as in the claim and finish the problem, it is enough to choose $m$ and $t$ in order to ensure that there exists $n \geqslant 3$ satisfying $$ \left\{\begin{array} { l } { 2 n + 1 = m + t r } \\ { n + 2 = m } \end{array} \Longleftrightarrow \left\{\begin{array}{l} m=3+t r \\ n=m-2 \end{array}\right.\right. $$ This can be clearly done by choosing $m$ large enough with $m \equiv 3(\bmod r)$. For instance, we may take $n=2 k r+1$, so that $$ f^{n+2}=f^{2 k r+3}=f^{4 k r+3}=f^{2 n+1} $$ where the middle equality follows by (*) since $2 k r+3 \geqslant k$.
|
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0b5ea90d-5192-58c4-b0d2-7032c3ef1ff3
| 24,917
|
A sequence of real numbers $a_{1}, a_{2}, \ldots$ satisfies the relation $$ a_{n}=-\max _{i+j=n}\left(a_{i}+a_{j}\right) \quad \text { for all } n>2017 $$ Prove that this sequence is bounded, i.e., there is a constant $M$ such that $\left|a_{n}\right| \leqslant M$ for all positive integers $n$.
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Set $D=2017$. Denote $$ M_{n}=\max _{k<n} a_{k} \quad \text { and } \quad m_{n}=-\min _{k<n} a_{k}=\max _{k<n}\left(-a_{k}\right) . $$ Clearly, the sequences $\left(m_{n}\right)$ and $\left(M_{n}\right)$ are nondecreasing. We need to prove that both are bounded. Consider an arbitrary $n>D$; our first aim is to bound $a_{n}$ in terms of $m_{n}$ and $M_{n}$. (i) There exist indices $p$ and $q$ such that $a_{n}=-\left(a_{p}+a_{q}\right)$ and $p+q=n$. Since $a_{p}, a_{q} \leqslant M_{n}$, we have $a_{n} \geqslant-2 M_{n}$. (ii) On the other hand, choose an index $k<n$ such that $a_{k}=M_{n}$. Then, we have $$ a_{n}=-\max _{\ell<n}\left(a_{n-\ell}+a_{\ell}\right) \leqslant-\left(a_{n-k}+a_{k}\right)=-a_{n-k}-M_{n} \leqslant m_{n}-M_{n} $$ Summarizing (i) and (ii), we get $$ -2 M_{n} \leqslant a_{n} \leqslant m_{n}-M_{n}, $$ whence $$ m_{n} \leqslant m_{n+1} \leqslant \max \left\{m_{n}, 2 M_{n}\right\} \quad \text { and } \quad M_{n} \leqslant M_{n+1} \leqslant \max \left\{M_{n}, m_{n}-M_{n}\right\} $$ Now, say that an index $n>D$ is lucky if $m_{n} \leqslant 2 M_{n}$. Two cases are possible. Case 1. Assume that there exists a lucky index $n$. In this case, (1) yields $m_{n+1} \leqslant 2 M_{n}$ and $M_{n} \leqslant M_{n+1} \leqslant M_{n}$. Therefore, $M_{n+1}=M_{n}$ and $m_{n+1} \leqslant 2 M_{n}=2 M_{n+1}$. So, the index $n+1$ is also lucky, and $M_{n+1}=M_{n}$. Applying the same arguments repeatedly, we obtain that all indices $k>n$ are lucky (i.e., $m_{k} \leqslant 2 M_{k}$ for all these indices), and $M_{k}=M_{n}$ for all such indices. Thus, all of the $m_{k}$ and $M_{k}$ are bounded by $2 M_{n}$. Case 2. Assume now that there is no lucky index, i.e., $2 M_{n}<m_{n}$ for all $n>D$. Then (1) shows that for all $n>D$ we have $m_{n} \leqslant m_{n+1} \leqslant m_{n}$, so $m_{n}=m_{D+1}$ for all $n>D$. Since $M_{n}<m_{n} / 2$ for all such indices, all of the $m_{n}$ and $M_{n}$ are bounded by $m_{D+1}$. Thus, in both cases the sequences $\left(m_{n}\right)$ and $\left(M_{n}\right)$ are bounded, as desired.
|
proof
|
Yes
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Yes
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proof
|
Algebra
|
A sequence of real numbers $a_{1}, a_{2}, \ldots$ satisfies the relation $$ a_{n}=-\max _{i+j=n}\left(a_{i}+a_{j}\right) \quad \text { for all } n>2017 $$ Prove that this sequence is bounded, i.e., there is a constant $M$ such that $\left|a_{n}\right| \leqslant M$ for all positive integers $n$.
|
Set $D=2017$. Denote $$ M_{n}=\max _{k<n} a_{k} \quad \text { and } \quad m_{n}=-\min _{k<n} a_{k}=\max _{k<n}\left(-a_{k}\right) . $$ Clearly, the sequences $\left(m_{n}\right)$ and $\left(M_{n}\right)$ are nondecreasing. We need to prove that both are bounded. Consider an arbitrary $n>D$; our first aim is to bound $a_{n}$ in terms of $m_{n}$ and $M_{n}$. (i) There exist indices $p$ and $q$ such that $a_{n}=-\left(a_{p}+a_{q}\right)$ and $p+q=n$. Since $a_{p}, a_{q} \leqslant M_{n}$, we have $a_{n} \geqslant-2 M_{n}$. (ii) On the other hand, choose an index $k<n$ such that $a_{k}=M_{n}$. Then, we have $$ a_{n}=-\max _{\ell<n}\left(a_{n-\ell}+a_{\ell}\right) \leqslant-\left(a_{n-k}+a_{k}\right)=-a_{n-k}-M_{n} \leqslant m_{n}-M_{n} $$ Summarizing (i) and (ii), we get $$ -2 M_{n} \leqslant a_{n} \leqslant m_{n}-M_{n}, $$ whence $$ m_{n} \leqslant m_{n+1} \leqslant \max \left\{m_{n}, 2 M_{n}\right\} \quad \text { and } \quad M_{n} \leqslant M_{n+1} \leqslant \max \left\{M_{n}, m_{n}-M_{n}\right\} $$ Now, say that an index $n>D$ is lucky if $m_{n} \leqslant 2 M_{n}$. Two cases are possible. Case 1. Assume that there exists a lucky index $n$. In this case, (1) yields $m_{n+1} \leqslant 2 M_{n}$ and $M_{n} \leqslant M_{n+1} \leqslant M_{n}$. Therefore, $M_{n+1}=M_{n}$ and $m_{n+1} \leqslant 2 M_{n}=2 M_{n+1}$. So, the index $n+1$ is also lucky, and $M_{n+1}=M_{n}$. Applying the same arguments repeatedly, we obtain that all indices $k>n$ are lucky (i.e., $m_{k} \leqslant 2 M_{k}$ for all these indices), and $M_{k}=M_{n}$ for all such indices. Thus, all of the $m_{k}$ and $M_{k}$ are bounded by $2 M_{n}$. Case 2. Assume now that there is no lucky index, i.e., $2 M_{n}<m_{n}$ for all $n>D$. Then (1) shows that for all $n>D$ we have $m_{n} \leqslant m_{n+1} \leqslant m_{n}$, so $m_{n}=m_{D+1}$ for all $n>D$. Since $M_{n}<m_{n} / 2$ for all such indices, all of the $m_{n}$ and $M_{n}$ are bounded by $m_{D+1}$. Thus, in both cases the sequences $\left(m_{n}\right)$ and $\left(M_{n}\right)$ are bounded, as desired.
|
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|
0741b8a0-b65b-53b3-acd9-eb06285def7d
| 24,920
|
Let $a_{0}, a_{1}, a_{2}, \ldots$ be a sequence of integers and $b_{0}, b_{1}, b_{2}, \ldots$ be a sequence of positive integers such that $a_{0}=0, a_{1}=1$, and $$ a_{n+1}=\left\{\begin{array}{ll} a_{n} b_{n}+a_{n-1}, & \text { if } b_{n-1}=1 \\ a_{n} b_{n}-a_{n-1}, & \text { if } b_{n-1}>1 \end{array} \quad \text { for } n=1,2, \ldots\right. $$ Prove that at least one of the two numbers $a_{2017}$ and $a_{2018}$ must be greater than or equal to 2017 . (Australia)
|
The value of $b_{0}$ is irrelevant since $a_{0}=0$, so we may assume that $b_{0}=1$. Lemma. We have $a_{n} \geqslant 1$ for all $n \geqslant 1$. Proof. Let us suppose otherwise in order to obtain a contradiction. Let $$ n \geqslant 1 \text { be the smallest integer with } a_{n} \leqslant 0 \text {. } $$ Note that $n \geqslant 2$. It follows that $a_{n-1} \geqslant 1$ and $a_{n-2} \geqslant 0$. Thus we cannot have $a_{n}=$ $a_{n-1} b_{n-1}+a_{n-2}$, so we must have $a_{n}=a_{n-1} b_{n-1}-a_{n-2}$. Since $a_{n} \leqslant 0$, we have $a_{n-1} \leqslant a_{n-2}$. Thus we have $a_{n-2} \geqslant a_{n-1} \geqslant a_{n}$. Let $$ r \text { be the smallest index with } a_{r} \geqslant a_{r+1} \geqslant a_{r+2} \text {. } $$ Then $r \leqslant n-2$ by the above, but also $r \geqslant 2$ : if $b_{1}=1$, then $a_{2}=a_{1}=1$ and $a_{3}=a_{2} b_{2}+a_{1}>a_{2}$; if $b_{1}>1$, then $a_{2}=b_{1}>1=a_{1}$. By the minimal choice (2) of $r$, it follows that $a_{r-1}<a_{r}$. And since $2 \leqslant r \leqslant n-2$, by the minimal choice (1) of $n$ we have $a_{r-1}, a_{r}, a_{r+1}>0$. In order to have $a_{r+1} \geqslant a_{r+2}$, we must have $a_{r+2}=a_{r+1} b_{r+1}-a_{r}$ so that $b_{r} \geqslant 2$. Putting everything together, we conclude that $$ a_{r+1}=a_{r} b_{r} \pm a_{r-1} \geqslant 2 a_{r}-a_{r-1}=a_{r}+\left(a_{r}-a_{r-1}\right)>a_{r} $$ which contradicts (2). To complete the problem, we prove that $\max \left\{a_{n}, a_{n+1}\right\} \geqslant n$ by induction. The cases $n=0,1$ are given. Assume it is true for all non-negative integers strictly less than $n$, where $n \geqslant 2$. There are two cases: Case 1: $b_{n-1}=1$. Then $a_{n+1}=a_{n} b_{n}+a_{n-1}$. By the inductive assumption one of $a_{n-1}, a_{n}$ is at least $n-1$ and the other, by the lemma, is at least 1 . Hence $$ a_{n+1}=a_{n} b_{n}+a_{n-1} \geqslant a_{n}+a_{n-1} \geqslant(n-1)+1=n . $$ Thus $\max \left\{a_{n}, a_{n+1}\right\} \geqslant n$, as desired. Case 2: $b_{n-1}>1$. Since we defined $b_{0}=1$ there is an index $r$ with $1 \leqslant r \leqslant n-1$ such that $$ b_{n-1}, b_{n-2}, \ldots, b_{r} \geqslant 2 \quad \text { and } \quad b_{r-1}=1 $$ We have $a_{r+1}=a_{r} b_{r}+a_{r-1} \geqslant 2 a_{r}+a_{r-1}$. Thus $a_{r+1}-a_{r} \geqslant a_{r}+a_{r-1}$. Now we claim that $a_{r}+a_{r-1} \geqslant r$. Indeed, this holds by inspection for $r=1$; for $r \geqslant 2$, one of $a_{r}, a_{r-1}$ is at least $r-1$ by the inductive assumption, while the other, by the lemma, is at least 1 . Hence $a_{r}+a_{r-1} \geqslant r$, as claimed, and therefore $a_{r+1}-a_{r} \geqslant r$ by the last inequality in the previous paragraph. Since $r \geqslant 1$ and, by the lemma, $a_{r} \geqslant 1$, from $a_{r+1}-a_{r} \geqslant r$ we get the following two inequalities: $$ a_{r+1} \geqslant r+1 \quad \text { and } \quad a_{r+1}>a_{r} . $$ Now observe that $$ a_{m}>a_{m-1} \Longrightarrow a_{m+1}>a_{m} \text { for } m=r+1, r+2, \ldots, n-1 \text {, } $$ since $a_{m+1}=a_{m} b_{m}-a_{m-1} \geqslant 2 a_{m}-a_{m-1}=a_{m}+\left(a_{m}-a_{m-1}\right)>a_{m}$. Thus $$ a_{n}>a_{n-1}>\cdots>a_{r+1} \geqslant r+1 \Longrightarrow a_{n} \geqslant n $$ So $\max \left\{a_{n}, a_{n+1}\right\} \geqslant n$, as desired.
|
proof
|
Yes
|
Yes
|
proof
|
Number Theory
|
Let $a_{0}, a_{1}, a_{2}, \ldots$ be a sequence of integers and $b_{0}, b_{1}, b_{2}, \ldots$ be a sequence of positive integers such that $a_{0}=0, a_{1}=1$, and $$ a_{n+1}=\left\{\begin{array}{ll} a_{n} b_{n}+a_{n-1}, & \text { if } b_{n-1}=1 \\ a_{n} b_{n}-a_{n-1}, & \text { if } b_{n-1}>1 \end{array} \quad \text { for } n=1,2, \ldots\right. $$ Prove that at least one of the two numbers $a_{2017}$ and $a_{2018}$ must be greater than or equal to 2017 . (Australia)
|
The value of $b_{0}$ is irrelevant since $a_{0}=0$, so we may assume that $b_{0}=1$. Lemma. We have $a_{n} \geqslant 1$ for all $n \geqslant 1$. Proof. Let us suppose otherwise in order to obtain a contradiction. Let $$ n \geqslant 1 \text { be the smallest integer with } a_{n} \leqslant 0 \text {. } $$ Note that $n \geqslant 2$. It follows that $a_{n-1} \geqslant 1$ and $a_{n-2} \geqslant 0$. Thus we cannot have $a_{n}=$ $a_{n-1} b_{n-1}+a_{n-2}$, so we must have $a_{n}=a_{n-1} b_{n-1}-a_{n-2}$. Since $a_{n} \leqslant 0$, we have $a_{n-1} \leqslant a_{n-2}$. Thus we have $a_{n-2} \geqslant a_{n-1} \geqslant a_{n}$. Let $$ r \text { be the smallest index with } a_{r} \geqslant a_{r+1} \geqslant a_{r+2} \text {. } $$ Then $r \leqslant n-2$ by the above, but also $r \geqslant 2$ : if $b_{1}=1$, then $a_{2}=a_{1}=1$ and $a_{3}=a_{2} b_{2}+a_{1}>a_{2}$; if $b_{1}>1$, then $a_{2}=b_{1}>1=a_{1}$. By the minimal choice (2) of $r$, it follows that $a_{r-1}<a_{r}$. And since $2 \leqslant r \leqslant n-2$, by the minimal choice (1) of $n$ we have $a_{r-1}, a_{r}, a_{r+1}>0$. In order to have $a_{r+1} \geqslant a_{r+2}$, we must have $a_{r+2}=a_{r+1} b_{r+1}-a_{r}$ so that $b_{r} \geqslant 2$. Putting everything together, we conclude that $$ a_{r+1}=a_{r} b_{r} \pm a_{r-1} \geqslant 2 a_{r}-a_{r-1}=a_{r}+\left(a_{r}-a_{r-1}\right)>a_{r} $$ which contradicts (2). To complete the problem, we prove that $\max \left\{a_{n}, a_{n+1}\right\} \geqslant n$ by induction. The cases $n=0,1$ are given. Assume it is true for all non-negative integers strictly less than $n$, where $n \geqslant 2$. There are two cases: Case 1: $b_{n-1}=1$. Then $a_{n+1}=a_{n} b_{n}+a_{n-1}$. By the inductive assumption one of $a_{n-1}, a_{n}$ is at least $n-1$ and the other, by the lemma, is at least 1 . Hence $$ a_{n+1}=a_{n} b_{n}+a_{n-1} \geqslant a_{n}+a_{n-1} \geqslant(n-1)+1=n . $$ Thus $\max \left\{a_{n}, a_{n+1}\right\} \geqslant n$, as desired. Case 2: $b_{n-1}>1$. Since we defined $b_{0}=1$ there is an index $r$ with $1 \leqslant r \leqslant n-1$ such that $$ b_{n-1}, b_{n-2}, \ldots, b_{r} \geqslant 2 \quad \text { and } \quad b_{r-1}=1 $$ We have $a_{r+1}=a_{r} b_{r}+a_{r-1} \geqslant 2 a_{r}+a_{r-1}$. Thus $a_{r+1}-a_{r} \geqslant a_{r}+a_{r-1}$. Now we claim that $a_{r}+a_{r-1} \geqslant r$. Indeed, this holds by inspection for $r=1$; for $r \geqslant 2$, one of $a_{r}, a_{r-1}$ is at least $r-1$ by the inductive assumption, while the other, by the lemma, is at least 1 . Hence $a_{r}+a_{r-1} \geqslant r$, as claimed, and therefore $a_{r+1}-a_{r} \geqslant r$ by the last inequality in the previous paragraph. Since $r \geqslant 1$ and, by the lemma, $a_{r} \geqslant 1$, from $a_{r+1}-a_{r} \geqslant r$ we get the following two inequalities: $$ a_{r+1} \geqslant r+1 \quad \text { and } \quad a_{r+1}>a_{r} . $$ Now observe that $$ a_{m}>a_{m-1} \Longrightarrow a_{m+1}>a_{m} \text { for } m=r+1, r+2, \ldots, n-1 \text {, } $$ since $a_{m+1}=a_{m} b_{m}-a_{m-1} \geqslant 2 a_{m}-a_{m-1}=a_{m}+\left(a_{m}-a_{m-1}\right)>a_{m}$. Thus $$ a_{n}>a_{n-1}>\cdots>a_{r+1} \geqslant r+1 \Longrightarrow a_{n} \geqslant n $$ So $\max \left\{a_{n}, a_{n+1}\right\} \geqslant n$, as desired.
|
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9649a8b9-8009-5133-9407-80bb94b8e8f8
| 24,932
|
Let $a_{0}, a_{1}, a_{2}, \ldots$ be a sequence of integers and $b_{0}, b_{1}, b_{2}, \ldots$ be a sequence of positive integers such that $a_{0}=0, a_{1}=1$, and $$ a_{n+1}=\left\{\begin{array}{ll} a_{n} b_{n}+a_{n-1}, & \text { if } b_{n-1}=1 \\ a_{n} b_{n}-a_{n-1}, & \text { if } b_{n-1}>1 \end{array} \quad \text { for } n=1,2, \ldots\right. $$ Prove that at least one of the two numbers $a_{2017}$ and $a_{2018}$ must be greater than or equal to 2017 . (Australia)
|
We say that an index $n>1$ is bad if $b_{n-1}=1$ and $b_{n-2}>1$; otherwise $n$ is good. The value of $b_{0}$ is irrelevant to the definition of $\left(a_{n}\right)$ since $a_{0}=0$; so we assume that $b_{0}>1$. Lemma 1. (a) $a_{n} \geqslant 1$ for all $n>0$. (b) If $n>1$ is good, then $a_{n}>a_{n-1}$. Proof. Induction on $n$. In the base cases $n=1,2$ we have $a_{1}=1 \geqslant 1, a_{2}=b_{1} a_{1} \geqslant 1$, and finally $a_{2}>a_{1}$ if 2 is good, since in this case $b_{1}>1$. Now we assume that the lemma statement is proved for $n=1,2, \ldots, k$ with $k \geqslant 2$, and prove it for $n=k+1$. Recall that $a_{k}$ and $a_{k-1}$ are positive by the induction hypothesis. Case 1: $k$ is bad. We have $b_{k-1}=1$, so $a_{k+1}=b_{k} a_{k}+a_{k-1} \geqslant a_{k}+a_{k-1}>a_{k} \geqslant 1$, as required. Case 2: $k$ is good. We already have $a_{k}>a_{k-1} \geqslant 1$ by the induction hypothesis. We consider three easy subcases. Subcase 2.1: $b_{k}>1$. Then $a_{k+1} \geqslant b_{k} a_{k}-a_{k-1} \geqslant a_{k}+\left(a_{k}-a_{k-1}\right)>a_{k} \geqslant 1$. Subcase 2.2: $b_{k}=b_{k-1}=1$. Then $a_{k+1}=a_{k}+a_{k-1}>a_{k} \geqslant 1$. Subcase 2.3: $b_{k}=1$ but $b_{k-1}>1$. Then $k+1$ is bad, and we need to prove only (a), which is trivial: $a_{k+1}=a_{k}-a_{k-1} \geqslant 1$. So, in all three subcases we have verified the required relations. Lemma 2. Assume that $n>1$ is bad. Then there exists a $j \in\{1,2,3\}$ such that $a_{n+j} \geqslant$ $a_{n-1}+j+1$, and $a_{n+i} \geqslant a_{n-1}+i$ for all $1 \leqslant i<j$. Proof. Recall that $b_{n-1}=1$. Set $$ m=\inf \left\{i>0: b_{n+i-1}>1\right\} $$ (possibly $m=+\infty$ ). We claim that $j=\min \{m, 3\}$ works. Again, we distinguish several cases, according to the value of $m$; in each of them we use Lemma 1 without reference. Case 1: $m=1$, so $b_{n}>1$. Then $a_{n+1} \geqslant 2 a_{n}+a_{n-1} \geqslant a_{n-1}+2$, as required. Case 2: $m=2$, so $b_{n}=1$ and $b_{n+1}>1$. Then we successively get $$ \begin{gathered} a_{n+1}=a_{n}+a_{n-1} \geqslant a_{n-1}+1 \\ a_{n+2} \geqslant 2 a_{n+1}+a_{n} \geqslant 2\left(a_{n-1}+1\right)+a_{n}=a_{n-1}+\left(a_{n-1}+a_{n}+2\right) \geqslant a_{n-1}+4 \end{gathered} $$ which is even better than we need. Case 3: $m>2$, so $b_{n}=b_{n+1}=1$. Then we successively get $$ \begin{gathered} a_{n+1}=a_{n}+a_{n-1} \geqslant a_{n-1}+1, \quad a_{n+2}=a_{n+1}+a_{n} \geqslant a_{n-1}+1+a_{n} \geqslant a_{n-1}+2, \\ a_{n+3} \geqslant a_{n+2}+a_{n+1} \geqslant\left(a_{n-1}+1\right)+\left(a_{n-1}+2\right) \geqslant a_{n-1}+4 \end{gathered} $$ as required. Lemmas 1 (b) and 2 provide enough information to prove that $\max \left\{a_{n}, a_{n+1}\right\} \geqslant n$ for all $n$ and, moreover, that $a_{n} \geqslant n$ often enough. Indeed, assume that we have found some $n$ with $a_{n-1} \geqslant n-1$. If $n$ is good, then by Lemma 1 (b) we have $a_{n} \geqslant n$ as well. If $n$ is bad, then Lemma 2 yields $\max \left\{a_{n+i}, a_{n+i+1}\right\} \geqslant a_{n-1}+i+1 \geqslant n+i$ for all $0 \leqslant i<j$ and $a_{n+j} \geqslant a_{n-1}+j+1 \geqslant n+j$; so $n+j$ is the next index to start with.
|
proof
|
Yes
|
Yes
|
proof
|
Number Theory
|
Let $a_{0}, a_{1}, a_{2}, \ldots$ be a sequence of integers and $b_{0}, b_{1}, b_{2}, \ldots$ be a sequence of positive integers such that $a_{0}=0, a_{1}=1$, and $$ a_{n+1}=\left\{\begin{array}{ll} a_{n} b_{n}+a_{n-1}, & \text { if } b_{n-1}=1 \\ a_{n} b_{n}-a_{n-1}, & \text { if } b_{n-1}>1 \end{array} \quad \text { for } n=1,2, \ldots\right. $$ Prove that at least one of the two numbers $a_{2017}$ and $a_{2018}$ must be greater than or equal to 2017 . (Australia)
|
We say that an index $n>1$ is bad if $b_{n-1}=1$ and $b_{n-2}>1$; otherwise $n$ is good. The value of $b_{0}$ is irrelevant to the definition of $\left(a_{n}\right)$ since $a_{0}=0$; so we assume that $b_{0}>1$. Lemma 1. (a) $a_{n} \geqslant 1$ for all $n>0$. (b) If $n>1$ is good, then $a_{n}>a_{n-1}$. Proof. Induction on $n$. In the base cases $n=1,2$ we have $a_{1}=1 \geqslant 1, a_{2}=b_{1} a_{1} \geqslant 1$, and finally $a_{2}>a_{1}$ if 2 is good, since in this case $b_{1}>1$. Now we assume that the lemma statement is proved for $n=1,2, \ldots, k$ with $k \geqslant 2$, and prove it for $n=k+1$. Recall that $a_{k}$ and $a_{k-1}$ are positive by the induction hypothesis. Case 1: $k$ is bad. We have $b_{k-1}=1$, so $a_{k+1}=b_{k} a_{k}+a_{k-1} \geqslant a_{k}+a_{k-1}>a_{k} \geqslant 1$, as required. Case 2: $k$ is good. We already have $a_{k}>a_{k-1} \geqslant 1$ by the induction hypothesis. We consider three easy subcases. Subcase 2.1: $b_{k}>1$. Then $a_{k+1} \geqslant b_{k} a_{k}-a_{k-1} \geqslant a_{k}+\left(a_{k}-a_{k-1}\right)>a_{k} \geqslant 1$. Subcase 2.2: $b_{k}=b_{k-1}=1$. Then $a_{k+1}=a_{k}+a_{k-1}>a_{k} \geqslant 1$. Subcase 2.3: $b_{k}=1$ but $b_{k-1}>1$. Then $k+1$ is bad, and we need to prove only (a), which is trivial: $a_{k+1}=a_{k}-a_{k-1} \geqslant 1$. So, in all three subcases we have verified the required relations. Lemma 2. Assume that $n>1$ is bad. Then there exists a $j \in\{1,2,3\}$ such that $a_{n+j} \geqslant$ $a_{n-1}+j+1$, and $a_{n+i} \geqslant a_{n-1}+i$ for all $1 \leqslant i<j$. Proof. Recall that $b_{n-1}=1$. Set $$ m=\inf \left\{i>0: b_{n+i-1}>1\right\} $$ (possibly $m=+\infty$ ). We claim that $j=\min \{m, 3\}$ works. Again, we distinguish several cases, according to the value of $m$; in each of them we use Lemma 1 without reference. Case 1: $m=1$, so $b_{n}>1$. Then $a_{n+1} \geqslant 2 a_{n}+a_{n-1} \geqslant a_{n-1}+2$, as required. Case 2: $m=2$, so $b_{n}=1$ and $b_{n+1}>1$. Then we successively get $$ \begin{gathered} a_{n+1}=a_{n}+a_{n-1} \geqslant a_{n-1}+1 \\ a_{n+2} \geqslant 2 a_{n+1}+a_{n} \geqslant 2\left(a_{n-1}+1\right)+a_{n}=a_{n-1}+\left(a_{n-1}+a_{n}+2\right) \geqslant a_{n-1}+4 \end{gathered} $$ which is even better than we need. Case 3: $m>2$, so $b_{n}=b_{n+1}=1$. Then we successively get $$ \begin{gathered} a_{n+1}=a_{n}+a_{n-1} \geqslant a_{n-1}+1, \quad a_{n+2}=a_{n+1}+a_{n} \geqslant a_{n-1}+1+a_{n} \geqslant a_{n-1}+2, \\ a_{n+3} \geqslant a_{n+2}+a_{n+1} \geqslant\left(a_{n-1}+1\right)+\left(a_{n-1}+2\right) \geqslant a_{n-1}+4 \end{gathered} $$ as required. Lemmas 1 (b) and 2 provide enough information to prove that $\max \left\{a_{n}, a_{n+1}\right\} \geqslant n$ for all $n$ and, moreover, that $a_{n} \geqslant n$ often enough. Indeed, assume that we have found some $n$ with $a_{n-1} \geqslant n-1$. If $n$ is good, then by Lemma 1 (b) we have $a_{n} \geqslant n$ as well. If $n$ is bad, then Lemma 2 yields $\max \left\{a_{n+i}, a_{n+i+1}\right\} \geqslant a_{n-1}+i+1 \geqslant n+i$ for all $0 \leqslant i<j$ and $a_{n+j} \geqslant a_{n-1}+j+1 \geqslant n+j$; so $n+j$ is the next index to start with.
|
{
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9649a8b9-8009-5133-9407-80bb94b8e8f8
| 24,932
|
Assume that a function $f: \mathbb{R} \rightarrow \mathbb{R}$ satisfies the following condition: For every $x, y \in \mathbb{R}$ such that $(f(x)+y)(f(y)+x)>0$, we have $f(x)+y=f(y)+x$. Prove that $f(x)+y \leqslant f(y)+x$ whenever $x>y$. (Netherlands)
|
Define $g(x)=x-f(x)$. The condition on $f$ then rewrites as follows: For every $x, y \in \mathbb{R}$ such that $((x+y)-g(x))((x+y)-g(y))>0$, we have $g(x)=g(y)$. This condition may in turn be rewritten in the following form: If $g(x) \neq g(y)$, then the number $x+y$ lies (non-strictly) between $g(x)$ and $g(y)$. Notice here that the function $g_{1}(x)=-g(-x)$ also satisfies $(*)$, since $$ \begin{aligned} g_{1}(x) \neq g_{1}(y) \Longrightarrow & g(-x) \neq g(-y) \Longrightarrow \quad-(x+y) \text { lies between } g(-x) \text { and } g(-y) \\ & \Longrightarrow \quad x+y \text { lies between } g_{1}(x) \text { and } g_{1}(y) \end{aligned} $$ On the other hand, the relation we need to prove reads now as $$ g(x) \leqslant g(y) \quad \text { whenever } x<y $$ Again, this condition is equivalent to the same one with $g$ replaced by $g_{1}$. If $g(x)=2 x$ for all $x \in \mathbb{R}$, then $(*)$ is obvious; so in what follows we consider the other case. We split the solution into a sequence of lemmas, strengthening one another. We always consider some value of $x$ with $g(x) \neq 2 x$ and denote $X=g(x)$. Lemma 1. Assume that $X<2 x$. Then on the interval $(X-x ; x]$ the function $g$ attains at most two values - namely, $X$ and, possibly, some $Y>X$. Similarly, if $X>2 x$, then $g$ attains at most two values on $[x ; X-x)$ - namely, $X$ and, possibly, some $Y<X$. Proof. We start with the first claim of the lemma. Notice that $X-x<x$, so the considered interval is nonempty. Take any $a \in(X-x ; x)$ with $g(a) \neq X$ (if it exists). If $g(a)<X$, then (*) yields $g(a) \leqslant$ $a+x \leqslant g(x)=X$, so $a \leqslant X-x$ which is impossible. Thus, $g(a)>X$ and hence by (*) we get $X \leqslant a+x \leqslant g(a)$. Now, for any $b \in(X-x ; x)$ with $g(b) \neq X$ we similarly get $b+x \leqslant g(b)$. Therefore, the number $a+b$ (which is smaller than each of $a+x$ and $b+x$ ) cannot lie between $g(a)$ and $g(b)$, which by (*) implies that $g(a)=g(b)$. Hence $g$ may attain only two values on $(X-x ; x]$, namely $X$ and $g(a)>X$. To prove the second claim, notice that $g_{1}(-x)=-X<2 \cdot(-x)$, so $g_{1}$ attains at most two values on $(-X+x,-x]$, i.e., $-X$ and, possibly, some $-Y>-X$. Passing back to $g$, we get what we need. Lemma 2. If $X<2 x$, then $g$ is constant on $(X-x ; x)$. Similarly, if $X>2 x$, then $g$ is constant on $(x ; X-x)$. Proof. Again, it suffices to prove the first claim only. Assume, for the sake of contradiction, that there exist $a, b \in(X-x ; x)$ with $g(a) \neq g(b)$; by Lemma 1, we may assume that $g(a)=X$ and $Y=g(b)>X$. Notice that $\min \{X-a, X-b\}>X-x$, so there exists a $u \in(X-x ; x)$ such that $u<\min \{X-a, X-b\}$. By Lemma 1, we have either $g(u)=X$ or $g(u)=Y$. In the former case, by (*) we have $X \leqslant u+b \leqslant Y$ which contradicts $u<X-b$. In the second case, by (*) we have $X \leqslant u+a \leqslant Y$ which contradicts $u<X-a$. Thus the lemma is proved. Lemma 3. If $X<2 x$, then $g(a)=X$ for all $a \in(X-x ; x)$. Similarly, if $X>2 x$, then $g(a)=X$ for all $a \in(x ; X-x)$. Proof. Again, we only prove the first claim. By Lemmas 1 and 2, this claim may be violated only if $g$ takes on a constant value $Y>X$ on $(X-x, x)$. Choose any $a, b \in(X-x ; x)$ with $a<b$. By (*), we have $$ Y \geqslant b+x \geqslant X $$ In particular, we have $Y \geqslant b+x>2 a$. Applying Lemma 2 to $a$ in place of $x$, we obtain that $g$ is constant on $(a, Y-a)$. By (2) again, we have $x \leqslant Y-b<Y-a$; so $x, b \in(a ; Y-a)$. But $X=g(x) \neq g(b)=Y$, which is a contradiction. Now we are able to finish the solution. Assume that $g(x)>g(y)$ for some $x<y$. Denote $X=g(x)$ and $Y=g(y)$; by (*), we have $X \geqslant x+y \geqslant Y$, so $Y-y \leqslant x<y \leqslant X-x$, and hence $(Y-y ; y) \cap(x ; X-x)=(x, y) \neq \varnothing$. On the other hand, since $Y-y<y$ and $x<X-x$, Lemma 3 shows that $g$ should attain a constant value $X$ on $(x ; X-x)$ and a constant value $Y \neq X$ on $(Y-y ; y)$. Since these intervals overlap, we get the final contradiction.
|
proof
|
Yes
|
Yes
|
proof
|
Algebra
|
Assume that a function $f: \mathbb{R} \rightarrow \mathbb{R}$ satisfies the following condition: For every $x, y \in \mathbb{R}$ such that $(f(x)+y)(f(y)+x)>0$, we have $f(x)+y=f(y)+x$. Prove that $f(x)+y \leqslant f(y)+x$ whenever $x>y$. (Netherlands)
|
Define $g(x)=x-f(x)$. The condition on $f$ then rewrites as follows: For every $x, y \in \mathbb{R}$ such that $((x+y)-g(x))((x+y)-g(y))>0$, we have $g(x)=g(y)$. This condition may in turn be rewritten in the following form: If $g(x) \neq g(y)$, then the number $x+y$ lies (non-strictly) between $g(x)$ and $g(y)$. Notice here that the function $g_{1}(x)=-g(-x)$ also satisfies $(*)$, since $$ \begin{aligned} g_{1}(x) \neq g_{1}(y) \Longrightarrow & g(-x) \neq g(-y) \Longrightarrow \quad-(x+y) \text { lies between } g(-x) \text { and } g(-y) \\ & \Longrightarrow \quad x+y \text { lies between } g_{1}(x) \text { and } g_{1}(y) \end{aligned} $$ On the other hand, the relation we need to prove reads now as $$ g(x) \leqslant g(y) \quad \text { whenever } x<y $$ Again, this condition is equivalent to the same one with $g$ replaced by $g_{1}$. If $g(x)=2 x$ for all $x \in \mathbb{R}$, then $(*)$ is obvious; so in what follows we consider the other case. We split the solution into a sequence of lemmas, strengthening one another. We always consider some value of $x$ with $g(x) \neq 2 x$ and denote $X=g(x)$. Lemma 1. Assume that $X<2 x$. Then on the interval $(X-x ; x]$ the function $g$ attains at most two values - namely, $X$ and, possibly, some $Y>X$. Similarly, if $X>2 x$, then $g$ attains at most two values on $[x ; X-x)$ - namely, $X$ and, possibly, some $Y<X$. Proof. We start with the first claim of the lemma. Notice that $X-x<x$, so the considered interval is nonempty. Take any $a \in(X-x ; x)$ with $g(a) \neq X$ (if it exists). If $g(a)<X$, then (*) yields $g(a) \leqslant$ $a+x \leqslant g(x)=X$, so $a \leqslant X-x$ which is impossible. Thus, $g(a)>X$ and hence by (*) we get $X \leqslant a+x \leqslant g(a)$. Now, for any $b \in(X-x ; x)$ with $g(b) \neq X$ we similarly get $b+x \leqslant g(b)$. Therefore, the number $a+b$ (which is smaller than each of $a+x$ and $b+x$ ) cannot lie between $g(a)$ and $g(b)$, which by (*) implies that $g(a)=g(b)$. Hence $g$ may attain only two values on $(X-x ; x]$, namely $X$ and $g(a)>X$. To prove the second claim, notice that $g_{1}(-x)=-X<2 \cdot(-x)$, so $g_{1}$ attains at most two values on $(-X+x,-x]$, i.e., $-X$ and, possibly, some $-Y>-X$. Passing back to $g$, we get what we need. Lemma 2. If $X<2 x$, then $g$ is constant on $(X-x ; x)$. Similarly, if $X>2 x$, then $g$ is constant on $(x ; X-x)$. Proof. Again, it suffices to prove the first claim only. Assume, for the sake of contradiction, that there exist $a, b \in(X-x ; x)$ with $g(a) \neq g(b)$; by Lemma 1, we may assume that $g(a)=X$ and $Y=g(b)>X$. Notice that $\min \{X-a, X-b\}>X-x$, so there exists a $u \in(X-x ; x)$ such that $u<\min \{X-a, X-b\}$. By Lemma 1, we have either $g(u)=X$ or $g(u)=Y$. In the former case, by (*) we have $X \leqslant u+b \leqslant Y$ which contradicts $u<X-b$. In the second case, by (*) we have $X \leqslant u+a \leqslant Y$ which contradicts $u<X-a$. Thus the lemma is proved. Lemma 3. If $X<2 x$, then $g(a)=X$ for all $a \in(X-x ; x)$. Similarly, if $X>2 x$, then $g(a)=X$ for all $a \in(x ; X-x)$. Proof. Again, we only prove the first claim. By Lemmas 1 and 2, this claim may be violated only if $g$ takes on a constant value $Y>X$ on $(X-x, x)$. Choose any $a, b \in(X-x ; x)$ with $a<b$. By (*), we have $$ Y \geqslant b+x \geqslant X $$ In particular, we have $Y \geqslant b+x>2 a$. Applying Lemma 2 to $a$ in place of $x$, we obtain that $g$ is constant on $(a, Y-a)$. By (2) again, we have $x \leqslant Y-b<Y-a$; so $x, b \in(a ; Y-a)$. But $X=g(x) \neq g(b)=Y$, which is a contradiction. Now we are able to finish the solution. Assume that $g(x)>g(y)$ for some $x<y$. Denote $X=g(x)$ and $Y=g(y)$; by (*), we have $X \geqslant x+y \geqslant Y$, so $Y-y \leqslant x<y \leqslant X-x$, and hence $(Y-y ; y) \cap(x ; X-x)=(x, y) \neq \varnothing$. On the other hand, since $Y-y<y$ and $x<X-x$, Lemma 3 shows that $g$ should attain a constant value $X$ on $(x ; X-x)$ and a constant value $Y \neq X$ on $(Y-y ; y)$. Since these intervals overlap, we get the final contradiction.
|
{
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|
981702bc-8d17-50f1-89a8-3ee38a1387e1
| 24,935
|
A rectangle $\mathcal{R}$ with odd integer side lengths is divided into small rectangles with integer side lengths. Prove that there is at least one among the small rectangles whose distances from the four sides of $\mathcal{R}$ are either all odd or all even. (Singapore)
|
Let the width and height of $\mathcal{R}$ be odd numbers $a$ and $b$. Divide $\mathcal{R}$ into $a b$ unit squares and color them green and yellow in a checkered pattern. Since the side lengths of $a$ and $b$ are odd, the corner squares of $\mathcal{R}$ will all have the same color, say green. Call a rectangle (either $\mathcal{R}$ or a small rectangle) green if its corners are all green; call it yellow if the corners are all yellow, and call it mixed if it has both green and yellow corners. In particular, $\mathcal{R}$ is a green rectangle. We will use the following trivial observations. - Every mixed rectangle contains the same number of green and yellow squares; - Every green rectangle contains one more green square than yellow square; - Every yellow rectangle contains one more yellow square than green square. The rectangle $\mathcal{R}$ is green, so it contains more green unit squares than yellow unit squares. Therefore, among the small rectangles, at least one is green. Let $\mathcal{S}$ be such a small green rectangle, and let its distances from the sides of $\mathcal{R}$ be $x, y, u$ and $v$, as shown in the picture. The top-left corner of $\mathcal{R}$ and the top-left corner of $\mathcal{S}$ have the same color, which happen if and only if $x$ and $u$ have the same parity. Similarly, the other three green corners of $\mathcal{S}$ indicate that $x$ and $v$ have the same parity, $y$ and $u$ have the same parity, i.e. $x, y, u$ and $v$ are all odd or all even. 
|
proof
|
Yes
|
Yes
|
proof
|
Combinatorics
|
A rectangle $\mathcal{R}$ with odd integer side lengths is divided into small rectangles with integer side lengths. Prove that there is at least one among the small rectangles whose distances from the four sides of $\mathcal{R}$ are either all odd or all even. (Singapore)
|
Let the width and height of $\mathcal{R}$ be odd numbers $a$ and $b$. Divide $\mathcal{R}$ into $a b$ unit squares and color them green and yellow in a checkered pattern. Since the side lengths of $a$ and $b$ are odd, the corner squares of $\mathcal{R}$ will all have the same color, say green. Call a rectangle (either $\mathcal{R}$ or a small rectangle) green if its corners are all green; call it yellow if the corners are all yellow, and call it mixed if it has both green and yellow corners. In particular, $\mathcal{R}$ is a green rectangle. We will use the following trivial observations. - Every mixed rectangle contains the same number of green and yellow squares; - Every green rectangle contains one more green square than yellow square; - Every yellow rectangle contains one more yellow square than green square. The rectangle $\mathcal{R}$ is green, so it contains more green unit squares than yellow unit squares. Therefore, among the small rectangles, at least one is green. Let $\mathcal{S}$ be such a small green rectangle, and let its distances from the sides of $\mathcal{R}$ be $x, y, u$ and $v$, as shown in the picture. The top-left corner of $\mathcal{R}$ and the top-left corner of $\mathcal{S}$ have the same color, which happen if and only if $x$ and $u$ have the same parity. Similarly, the other three green corners of $\mathcal{S}$ indicate that $x$ and $v$ have the same parity, $y$ and $u$ have the same parity, i.e. $x, y, u$ and $v$ are all odd or all even. 
|
{
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|
2c1065a9-506d-5133-8db9-3d161c71e61d
| 24,940
|
Let $n$ be a positive integer. Define a chameleon to be any sequence of $3 n$ letters, with exactly $n$ occurrences of each of the letters $a, b$, and $c$. Define a swap to be the transposition of two adjacent letters in a chameleon. Prove that for any chameleon $X$, there exists a chameleon $Y$ such that $X$ cannot be changed to $Y$ using fewer than $3 n^{2} / 2$ swaps. (Australia)
|
To start, notice that the swap of two identical letters does not change a chameleon, so we may assume there are no such swaps. For any two chameleons $X$ and $Y$, define their distance $d(X, Y)$ to be the minimal number of swaps needed to transform $X$ into $Y$ (or vice versa). Clearly, $d(X, Y)+d(Y, Z) \geqslant d(X, Z)$ for any three chameleons $X, Y$, and $Z$. Lemma. Consider two chameleons $$ P=\underbrace{a a \ldots a}_{n} \underbrace{b b \ldots b}_{n} \underbrace{c c \ldots c}_{n} \text { and } Q=\underbrace{c c \ldots c}_{n} \underbrace{b b \ldots b}_{n} \underbrace{a a \ldots a}_{n} . $$ Then $d(P, Q) \geqslant 3 n^{2}$. Proof. For any chameleon $X$ and any pair of distinct letters $u, v \in\{a, b, c\}$, we define $f_{u, v}(X)$ to be the number of pairs of positions in $X$ such that the left one is occupied by $u$, and the right one is occupied by $v$. Define $f(X)=f_{a, b}(X)+f_{a, c}(X)+f_{b, c}(X)$. Notice that $f_{a, b}(P)=f_{a, c}(P)=f_{b, c}(P)=n^{2}$ and $f_{a, b}(Q)=f_{a, c}(Q)=f_{b, c}(Q)=0$, so $f(P)=3 n^{2}$ and $f(Q)=0$. Now consider some swap changing a chameleon $X$ to $X^{\prime}$; say, the letters $a$ and $b$ are swapped. Then $f_{a, b}(X)$ and $f_{a, b}\left(X^{\prime}\right)$ differ by exactly 1 , while $f_{a, c}(X)=f_{a, c}\left(X^{\prime}\right)$ and $f_{b, c}(X)=f_{b, c}\left(X^{\prime}\right)$. This yields $\left|f(X)-f\left(X^{\prime}\right)\right|=1$, i.e., on any swap the value of $f$ changes by 1 . Hence $d(X, Y) \geqslant$ $|f(X)-f(Y)|$ for any two chameleons $X$ and $Y$. In particular, $d(P, Q) \geqslant|f(P)-f(Q)|=3 n^{2}$, as desired. Back to the problem, take any chameleon $X$ and notice that $d(X, P)+d(X, Q) \geqslant d(P, Q) \geqslant$ $3 n^{2}$ by the lemma. Consequently, $\max \{d(X, P), d(X, Q)\} \geqslant \frac{3 n^{2}}{2}$, which establishes the problem statement. Comment 1. The problem may be reformulated in a graph language. Construct a graph $G$ with the chameleons as vertices, two vertices being connected with an edge if and only if these chameleons differ by a single swap. Then $d(X, Y)$ is the usual distance between the vertices $X$ and $Y$ in this graph. Recall that the radius of a connected graph $G$ is defined as $$ r(G)=\min _{v \in V} \max _{u \in V} d(u, v) . $$ So we need to prove that the radius of the constructed graph is at least $3 n^{2} / 2$. It is well-known that the radius of any connected graph is at least the half of its diameter (which is simply $\max _{u, v \in V} d(u, v)$ ). Exactly this fact has been used above in order to finish the solution.
|
proof
|
Yes
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Yes
|
proof
|
Combinatorics
|
Let $n$ be a positive integer. Define a chameleon to be any sequence of $3 n$ letters, with exactly $n$ occurrences of each of the letters $a, b$, and $c$. Define a swap to be the transposition of two adjacent letters in a chameleon. Prove that for any chameleon $X$, there exists a chameleon $Y$ such that $X$ cannot be changed to $Y$ using fewer than $3 n^{2} / 2$ swaps. (Australia)
|
To start, notice that the swap of two identical letters does not change a chameleon, so we may assume there are no such swaps. For any two chameleons $X$ and $Y$, define their distance $d(X, Y)$ to be the minimal number of swaps needed to transform $X$ into $Y$ (or vice versa). Clearly, $d(X, Y)+d(Y, Z) \geqslant d(X, Z)$ for any three chameleons $X, Y$, and $Z$. Lemma. Consider two chameleons $$ P=\underbrace{a a \ldots a}_{n} \underbrace{b b \ldots b}_{n} \underbrace{c c \ldots c}_{n} \text { and } Q=\underbrace{c c \ldots c}_{n} \underbrace{b b \ldots b}_{n} \underbrace{a a \ldots a}_{n} . $$ Then $d(P, Q) \geqslant 3 n^{2}$. Proof. For any chameleon $X$ and any pair of distinct letters $u, v \in\{a, b, c\}$, we define $f_{u, v}(X)$ to be the number of pairs of positions in $X$ such that the left one is occupied by $u$, and the right one is occupied by $v$. Define $f(X)=f_{a, b}(X)+f_{a, c}(X)+f_{b, c}(X)$. Notice that $f_{a, b}(P)=f_{a, c}(P)=f_{b, c}(P)=n^{2}$ and $f_{a, b}(Q)=f_{a, c}(Q)=f_{b, c}(Q)=0$, so $f(P)=3 n^{2}$ and $f(Q)=0$. Now consider some swap changing a chameleon $X$ to $X^{\prime}$; say, the letters $a$ and $b$ are swapped. Then $f_{a, b}(X)$ and $f_{a, b}\left(X^{\prime}\right)$ differ by exactly 1 , while $f_{a, c}(X)=f_{a, c}\left(X^{\prime}\right)$ and $f_{b, c}(X)=f_{b, c}\left(X^{\prime}\right)$. This yields $\left|f(X)-f\left(X^{\prime}\right)\right|=1$, i.e., on any swap the value of $f$ changes by 1 . Hence $d(X, Y) \geqslant$ $|f(X)-f(Y)|$ for any two chameleons $X$ and $Y$. In particular, $d(P, Q) \geqslant|f(P)-f(Q)|=3 n^{2}$, as desired. Back to the problem, take any chameleon $X$ and notice that $d(X, P)+d(X, Q) \geqslant d(P, Q) \geqslant$ $3 n^{2}$ by the lemma. Consequently, $\max \{d(X, P), d(X, Q)\} \geqslant \frac{3 n^{2}}{2}$, which establishes the problem statement. Comment 1. The problem may be reformulated in a graph language. Construct a graph $G$ with the chameleons as vertices, two vertices being connected with an edge if and only if these chameleons differ by a single swap. Then $d(X, Y)$ is the usual distance between the vertices $X$ and $Y$ in this graph. Recall that the radius of a connected graph $G$ is defined as $$ r(G)=\min _{v \in V} \max _{u \in V} d(u, v) . $$ So we need to prove that the radius of the constructed graph is at least $3 n^{2} / 2$. It is well-known that the radius of any connected graph is at least the half of its diameter (which is simply $\max _{u, v \in V} d(u, v)$ ). Exactly this fact has been used above in order to finish the solution.
|
{
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|
5b23ea01-9d92-52b4-ad5a-c6655cdc6994
| 24,943
|
Let $N \geqslant 2$ be an integer. $N(N+1)$ soccer players, no two of the same height, stand in a row in some order. Coach Ralph wants to remove $N(N-1)$ people from this row so that in the remaining row of $2 N$ players, no one stands between the two tallest ones, no one stands between the third and the fourth tallest ones, ..., and finally no one stands between the two shortest ones. Show that this is always possible. (Russia)
|
Split the row into $N$ blocks with $N+1$ consecutive people each. We will show how to remove $N-1$ people from each block in order to satisfy the coach's wish. First, construct a $(N+1) \times N$ matrix where $x_{i, j}$ is the height of the $i^{\text {th }}$ tallest person of the $j^{\text {th }}$ block-in other words, each column lists the heights within a single block, sorted in decreasing order from top to bottom. We will reorder this matrix by repeatedly swapping whole columns. First, by column permutation, make sure that $x_{2,1}=\max \left\{x_{2, i}: i=1,2, \ldots, N\right\}$ (the first column contains the largest height of the second row). With the first column fixed, permute the other ones so that $x_{3,2}=\max \left\{x_{3, i}: i=2, \ldots, N\right\}$ (the second column contains the tallest person of the third row, first column excluded). In short, at step $k(k=1,2, \ldots, N-1)$, we permute the columns from $k$ to $N$ so that $x_{k+1, k}=\max \left\{x_{i, k}: i=k, k+1, \ldots, N\right\}$, and end up with an array like this: | $\boldsymbol{x}_{\mathbf{1 , 1}}$ | $x_{1,2}$ | $x_{1,3}$ | $\cdots$ | $x_{1, N-1}$ | $x_{1, N}$ | | :---: | :---: | :---: | :---: | :---: | :---: | | $\vee$ | $\vee$ | $\vee$ | | $\vee$ | $\vee$ | | $\boldsymbol{x}_{\mathbf{2 , 1}}$ | $\boldsymbol{x}_{\mathbf{2 , 2}}$ | $x_{2,3}$ | $\cdots$ | $x_{2, N-1}$ | $x_{2, N}$ | | $\vee$ | $\vee$ | $\vee$ | | $\vee$ | $\vee$ | | $x_{3,1}$ | $\boldsymbol{x}_{\mathbf{3 , 2}}$ | $\boldsymbol{x}_{\mathbf{3 , 3}}$ | $\cdots$ | $x_{3, N-1}$ | $x_{3, N}$ | | $\vee$ | $\vee$ | $\vee$ | | $\vee$ | $\vee$ | | $\vdots$ | $\vdots$ | $\vdots$ | $\ddots$ | $\vdots$ | $\vdots$ | | $\vee$ | $\vee$ | $\vee$ | | $\vee$ | $\vee$ | | $x_{N, 1}$ | $x_{N, 2}$ | $x_{N, 3}$ | $\cdots$ | $\boldsymbol{x}_{\boldsymbol{N}, \boldsymbol{N}-\mathbf{1}}>$ | $\boldsymbol{x}_{\boldsymbol{N}, \boldsymbol{N}}$ | | $\vee$ | $\vee$ | $\vee$ | | $\vee$ | $\vee$ | | $x_{N+1,1}$ | $x_{N+1,2}$ | $x_{N+1,3}$ | $\cdots$ | $x_{N+1, N-1}$ | $\boldsymbol{x}_{\boldsymbol{N + 1 , N}}$ | Now we make the bold choice: from the original row of people, remove everyone but those with heights $$ x_{1,1}>x_{2,1}>x_{2,2}>x_{3,2}>\cdots>x_{N, N-1}>x_{N, N}>x_{N+1, N} $$ Of course this height order (*) is not necessarily their spatial order in the new row. We now need to convince ourselves that each pair $\left(x_{k, k} ; x_{k+1, k}\right)$ remains spatially together in this new row. But $x_{k, k}$ and $x_{k+1, k}$ belong to the same column/block of consecutive $N+1$ people; the only people that could possibly stand between them were also in this block, and they are all gone.
|
proof
|
Yes
|
Yes
|
proof
|
Combinatorics
|
Let $N \geqslant 2$ be an integer. $N(N+1)$ soccer players, no two of the same height, stand in a row in some order. Coach Ralph wants to remove $N(N-1)$ people from this row so that in the remaining row of $2 N$ players, no one stands between the two tallest ones, no one stands between the third and the fourth tallest ones, ..., and finally no one stands between the two shortest ones. Show that this is always possible. (Russia)
|
Split the row into $N$ blocks with $N+1$ consecutive people each. We will show how to remove $N-1$ people from each block in order to satisfy the coach's wish. First, construct a $(N+1) \times N$ matrix where $x_{i, j}$ is the height of the $i^{\text {th }}$ tallest person of the $j^{\text {th }}$ block-in other words, each column lists the heights within a single block, sorted in decreasing order from top to bottom. We will reorder this matrix by repeatedly swapping whole columns. First, by column permutation, make sure that $x_{2,1}=\max \left\{x_{2, i}: i=1,2, \ldots, N\right\}$ (the first column contains the largest height of the second row). With the first column fixed, permute the other ones so that $x_{3,2}=\max \left\{x_{3, i}: i=2, \ldots, N\right\}$ (the second column contains the tallest person of the third row, first column excluded). In short, at step $k(k=1,2, \ldots, N-1)$, we permute the columns from $k$ to $N$ so that $x_{k+1, k}=\max \left\{x_{i, k}: i=k, k+1, \ldots, N\right\}$, and end up with an array like this: | $\boldsymbol{x}_{\mathbf{1 , 1}}$ | $x_{1,2}$ | $x_{1,3}$ | $\cdots$ | $x_{1, N-1}$ | $x_{1, N}$ | | :---: | :---: | :---: | :---: | :---: | :---: | | $\vee$ | $\vee$ | $\vee$ | | $\vee$ | $\vee$ | | $\boldsymbol{x}_{\mathbf{2 , 1}}$ | $\boldsymbol{x}_{\mathbf{2 , 2}}$ | $x_{2,3}$ | $\cdots$ | $x_{2, N-1}$ | $x_{2, N}$ | | $\vee$ | $\vee$ | $\vee$ | | $\vee$ | $\vee$ | | $x_{3,1}$ | $\boldsymbol{x}_{\mathbf{3 , 2}}$ | $\boldsymbol{x}_{\mathbf{3 , 3}}$ | $\cdots$ | $x_{3, N-1}$ | $x_{3, N}$ | | $\vee$ | $\vee$ | $\vee$ | | $\vee$ | $\vee$ | | $\vdots$ | $\vdots$ | $\vdots$ | $\ddots$ | $\vdots$ | $\vdots$ | | $\vee$ | $\vee$ | $\vee$ | | $\vee$ | $\vee$ | | $x_{N, 1}$ | $x_{N, 2}$ | $x_{N, 3}$ | $\cdots$ | $\boldsymbol{x}_{\boldsymbol{N}, \boldsymbol{N}-\mathbf{1}}>$ | $\boldsymbol{x}_{\boldsymbol{N}, \boldsymbol{N}}$ | | $\vee$ | $\vee$ | $\vee$ | | $\vee$ | $\vee$ | | $x_{N+1,1}$ | $x_{N+1,2}$ | $x_{N+1,3}$ | $\cdots$ | $x_{N+1, N-1}$ | $\boldsymbol{x}_{\boldsymbol{N + 1 , N}}$ | Now we make the bold choice: from the original row of people, remove everyone but those with heights $$ x_{1,1}>x_{2,1}>x_{2,2}>x_{3,2}>\cdots>x_{N, N-1}>x_{N, N}>x_{N+1, N} $$ Of course this height order (*) is not necessarily their spatial order in the new row. We now need to convince ourselves that each pair $\left(x_{k, k} ; x_{k+1, k}\right)$ remains spatially together in this new row. But $x_{k, k}$ and $x_{k+1, k}$ belong to the same column/block of consecutive $N+1$ people; the only people that could possibly stand between them were also in this block, and they are all gone.
|
{
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|
1746fbd3-dcc2-5230-a3c7-c84e28a0aef2
| 24,951
|
Let $N \geqslant 2$ be an integer. $N(N+1)$ soccer players, no two of the same height, stand in a row in some order. Coach Ralph wants to remove $N(N-1)$ people from this row so that in the remaining row of $2 N$ players, no one stands between the two tallest ones, no one stands between the third and the fourth tallest ones, ..., and finally no one stands between the two shortest ones. Show that this is always possible. (Russia)
|
Split the people into $N$ groups by height: group $G_{1}$ has the $N+1$ tallest ones, group $G_{2}$ has the next $N+1$ tallest, and so on, up to group $G_{N}$ with the $N+1$ shortest people. Now scan the original row from left to right, stopping as soon as you have scanned two people (consecutively or not) from the same group, say, $G_{i}$. Since we have $N$ groups, this must happen before or at the $(N+1)^{\text {th }}$ person of the row. Choose this pair of people, removing all the other people from the same group $G_{i}$ and also all people that have been scanned so far. The only people that could separate this pair's heights were in group $G_{i}$ (and they are gone); the only people that could separate this pair's positions were already scanned (and they are gone too). We are now left with $N-1$ groups (all except $G_{i}$ ). Since each of them lost at most one person, each one has at least $N$ unscanned people left in the row. Repeat the scanning process from left to right, choosing the next two people from the same group, removing this group and everyone scanned up to that point. Once again we end up with two people who are next to each other in the remaining row and whose heights cannot be separated by anyone else who remains (since the rest of their group is gone). After picking these 2 pairs, we still have $N-2$ groups with at least $N-1$ people each. If we repeat the scanning process a total of $N$ times, it is easy to check that we will end up with 2 people from each group, for a total of $2 N$ people remaining. The height order is guaranteed by the grouping, and the scanning construction from left to right guarantees that each pair from a group stand next to each other in the final row. We are done.
|
proof
|
Yes
|
Yes
|
proof
|
Combinatorics
|
Let $N \geqslant 2$ be an integer. $N(N+1)$ soccer players, no two of the same height, stand in a row in some order. Coach Ralph wants to remove $N(N-1)$ people from this row so that in the remaining row of $2 N$ players, no one stands between the two tallest ones, no one stands between the third and the fourth tallest ones, ..., and finally no one stands between the two shortest ones. Show that this is always possible. (Russia)
|
Split the people into $N$ groups by height: group $G_{1}$ has the $N+1$ tallest ones, group $G_{2}$ has the next $N+1$ tallest, and so on, up to group $G_{N}$ with the $N+1$ shortest people. Now scan the original row from left to right, stopping as soon as you have scanned two people (consecutively or not) from the same group, say, $G_{i}$. Since we have $N$ groups, this must happen before or at the $(N+1)^{\text {th }}$ person of the row. Choose this pair of people, removing all the other people from the same group $G_{i}$ and also all people that have been scanned so far. The only people that could separate this pair's heights were in group $G_{i}$ (and they are gone); the only people that could separate this pair's positions were already scanned (and they are gone too). We are now left with $N-1$ groups (all except $G_{i}$ ). Since each of them lost at most one person, each one has at least $N$ unscanned people left in the row. Repeat the scanning process from left to right, choosing the next two people from the same group, removing this group and everyone scanned up to that point. Once again we end up with two people who are next to each other in the remaining row and whose heights cannot be separated by anyone else who remains (since the rest of their group is gone). After picking these 2 pairs, we still have $N-2$ groups with at least $N-1$ people each. If we repeat the scanning process a total of $N$ times, it is easy to check that we will end up with 2 people from each group, for a total of $2 N$ people remaining. The height order is guaranteed by the grouping, and the scanning construction from left to right guarantees that each pair from a group stand next to each other in the final row. We are done.
|
{
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|
1746fbd3-dcc2-5230-a3c7-c84e28a0aef2
| 24,951
|
Let $N \geqslant 2$ be an integer. $N(N+1)$ soccer players, no two of the same height, stand in a row in some order. Coach Ralph wants to remove $N(N-1)$ people from this row so that in the remaining row of $2 N$ players, no one stands between the two tallest ones, no one stands between the third and the fourth tallest ones, ..., and finally no one stands between the two shortest ones. Show that this is always possible. (Russia)
|
This is essentially the same as solution 1, but presented inductively. The essence of the argument is the following lemma. Lemma. Assume that we have $N$ disjoint groups of at least $N+1$ people in each, all people have distinct heights. Then one can choose two people from each group so that among the chosen people, the two tallest ones are in one group, the third and the fourth tallest ones are in one group, ..., and the two shortest ones are in one group. Proof. Induction on $N \geqslant 1$; for $N=1$, the statement is trivial. Consider now $N$ groups $G_{1}, \ldots, G_{N}$ with at least $N+1$ people in each for $N \geqslant 2$. Enumerate the people by $1,2, \ldots, N(N+1)$ according to their height, say, from tallest to shortest. Find the least $s$ such that two people among $1,2, \ldots, s$ are in one group (without loss of generality, say this group is $G_{N}$ ). By the minimality of $s$, the two mentioned people in $G_{N}$ are $s$ and some $i<s$. Now we choose people $i$ and $s$ in $G_{N}$, forget about this group, and remove the people $1,2, \ldots, s$ from $G_{1}, \ldots, G_{N-1}$. Due to minimality of $s$ again, each of the obtained groups $G_{1}^{\prime}, \ldots, G_{N-1}^{\prime}$ contains at least $N$ people. By the induction hypothesis, one can choose a pair of people from each of $G_{1}^{\prime}, \ldots, G_{N-1}^{\prime}$ so as to satisfy the required conditions. Since all these people have numbers greater than $s$, addition of the pair $(s, i)$ from $G_{N}$ does not violate these requirements. To solve the problem, it suffices now to split the row into $N$ contiguous groups with $N+1$ people in each and apply the Lemma to those groups. Comment 1. One can identify each person with a pair of indices $(p, h)(p, h \in\{1,2, \ldots, N(N+1)\})$ so that the $p^{\text {th }}$ person in the row (say, from left to right) is the $h^{\text {th }}$ tallest person in the group. Say that $(a, b)$ separates $\left(x_{1}, y_{1}\right)$ and $\left(x_{2}, y_{2}\right)$ whenever $a$ is strictly between $x_{1}$ and $y_{1}$, or $b$ is strictly between $x_{2}$ and $y_{2}$. So the coach wants to pick $2 N$ people $\left(p_{i}, h_{i}\right)(i=1,2, \ldots, 2 N)$ such that no chosen person separates ( $p_{1}, h_{1}$ ) from ( $p_{2}, h_{2}$ ), no chosen person separates ( $p_{3}, h_{3}$ ) and ( $p_{4}, h_{4}$ ), and so on. This formulation reveals a duality between positions and heights. In that sense, solutions 1 and 2 are dual of each other. Comment 2. The number $N(N+1)$ is sharp for $N=2$ and $N=3$, due to arrangements $1,5,3,4,2$ and $1,10,6,4,3,9,5,8,7,2,11$.
|
proof
|
Yes
|
Yes
|
proof
|
Combinatorics
|
Let $N \geqslant 2$ be an integer. $N(N+1)$ soccer players, no two of the same height, stand in a row in some order. Coach Ralph wants to remove $N(N-1)$ people from this row so that in the remaining row of $2 N$ players, no one stands between the two tallest ones, no one stands between the third and the fourth tallest ones, ..., and finally no one stands between the two shortest ones. Show that this is always possible. (Russia)
|
This is essentially the same as solution 1, but presented inductively. The essence of the argument is the following lemma. Lemma. Assume that we have $N$ disjoint groups of at least $N+1$ people in each, all people have distinct heights. Then one can choose two people from each group so that among the chosen people, the two tallest ones are in one group, the third and the fourth tallest ones are in one group, ..., and the two shortest ones are in one group. Proof. Induction on $N \geqslant 1$; for $N=1$, the statement is trivial. Consider now $N$ groups $G_{1}, \ldots, G_{N}$ with at least $N+1$ people in each for $N \geqslant 2$. Enumerate the people by $1,2, \ldots, N(N+1)$ according to their height, say, from tallest to shortest. Find the least $s$ such that two people among $1,2, \ldots, s$ are in one group (without loss of generality, say this group is $G_{N}$ ). By the minimality of $s$, the two mentioned people in $G_{N}$ are $s$ and some $i<s$. Now we choose people $i$ and $s$ in $G_{N}$, forget about this group, and remove the people $1,2, \ldots, s$ from $G_{1}, \ldots, G_{N-1}$. Due to minimality of $s$ again, each of the obtained groups $G_{1}^{\prime}, \ldots, G_{N-1}^{\prime}$ contains at least $N$ people. By the induction hypothesis, one can choose a pair of people from each of $G_{1}^{\prime}, \ldots, G_{N-1}^{\prime}$ so as to satisfy the required conditions. Since all these people have numbers greater than $s$, addition of the pair $(s, i)$ from $G_{N}$ does not violate these requirements. To solve the problem, it suffices now to split the row into $N$ contiguous groups with $N+1$ people in each and apply the Lemma to those groups. Comment 1. One can identify each person with a pair of indices $(p, h)(p, h \in\{1,2, \ldots, N(N+1)\})$ so that the $p^{\text {th }}$ person in the row (say, from left to right) is the $h^{\text {th }}$ tallest person in the group. Say that $(a, b)$ separates $\left(x_{1}, y_{1}\right)$ and $\left(x_{2}, y_{2}\right)$ whenever $a$ is strictly between $x_{1}$ and $y_{1}$, or $b$ is strictly between $x_{2}$ and $y_{2}$. So the coach wants to pick $2 N$ people $\left(p_{i}, h_{i}\right)(i=1,2, \ldots, 2 N)$ such that no chosen person separates ( $p_{1}, h_{1}$ ) from ( $p_{2}, h_{2}$ ), no chosen person separates ( $p_{3}, h_{3}$ ) and ( $p_{4}, h_{4}$ ), and so on. This formulation reveals a duality between positions and heights. In that sense, solutions 1 and 2 are dual of each other. Comment 2. The number $N(N+1)$ is sharp for $N=2$ and $N=3$, due to arrangements $1,5,3,4,2$ and $1,10,6,4,3,9,5,8,7,2,11$.
|
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1746fbd3-dcc2-5230-a3c7-c84e28a0aef2
| 24,951
|
For any finite sets $X$ and $Y$ of positive integers, denote by $f_{X}(k)$ the $k^{\text {th }}$ smallest positive integer not in $X$, and let $$ X * Y=X \cup\left\{f_{X}(y): y \in Y\right\} $$ Let $A$ be a set of $a>0$ positive integers, and let $B$ be a set of $b>0$ positive integers. Prove that if $A * B=B * A$, then $$ \underbrace{A *(A * \cdots *(A *(A * A)) \ldots)}_{A \text { appears } b \text { times }}=\underbrace{B *(B * \cdots *(B *(B * B)) \ldots)}_{B \text { appears } a \text { times }} . $$
|
For any function $g: \mathbb{Z}_{>0} \rightarrow \mathbb{Z}_{>0}$ and any subset $X \subset \mathbb{Z}_{>0}$, we define $g(X)=$ $\{g(x): x \in X\}$. We have that the image of $f_{X}$ is $f_{X}\left(\mathbb{Z}_{>0}\right)=\mathbb{Z}_{>0} \backslash X$. We now show a general lemma about the operation *, with the goal of showing that $*$ is associative. Lemma 1. Let $X$ and $Y$ be finite sets of positive integers. The functions $f_{X * Y}$ and $f_{X} \circ f_{Y}$ are equal. Proof. We have $f_{X * Y}\left(\mathbb{Z}_{>0}\right)=\mathbb{Z}_{>0} \backslash(X * Y)=\left(\mathbb{Z}_{>0} \backslash X\right) \backslash f_{X}(Y)=f_{X}\left(\mathbb{Z}_{>0}\right) \backslash f_{X}(Y)=f_{X}\left(\mathbb{Z}_{>0} \backslash Y\right)=f_{X}\left(f_{Y}\left(\mathbb{Z}_{>0}\right)\right)$. Thus, the functions $f_{X * Y}$ and $f_{X} \circ f_{Y}$ are strictly increasing functions with the same range. Because a strictly function is uniquely defined by its range, we have $f_{X * Y}=f_{X} \circ f_{Y}$. Lemma 1 implies that * is associative, in the sense that $(A * B) * C=A *(B * C)$ for any finite sets $A, B$, and $C$ of positive integers. We prove the associativity by noting $$ \begin{gathered} \mathbb{Z}_{>0} \backslash((A * B) * C)=f_{(A * B) * C}\left(\mathbb{Z}_{>0}\right)=f_{A * B}\left(f_{C}\left(\mathbb{Z}_{>0}\right)\right)=f_{A}\left(f_{B}\left(f_{C}\left(\mathbb{Z}_{>0}\right)\right)\right) \\ =f_{A}\left(f_{B * C}\left(\mathbb{Z}_{>0}\right)=f_{A *(B * C)}\left(\mathbb{Z}_{>0}\right)=\mathbb{Z}_{>0} \backslash(A *(B * C))\right. \end{gathered} $$ In light of the associativity of *, we may drop the parentheses when we write expressions like $A *(B * C)$. We also introduce the notation $$ X^{* k}=\underbrace{X *(X * \cdots *(X *(X * X)) \ldots)}_{X \text { appears } k \text { times }} $$ Our goal is then to show that $A * B=B * A$ implies $A^{* b}=B^{* a}$. We will do so via the following general lemma. Lemma 2. Suppose that $X$ and $Y$ are finite sets of positive integers satisfying $X * Y=Y * X$ and $|X|=|Y|$. Then, we must have $X=Y$. Proof. Assume that $X$ and $Y$ are not equal. Let $s$ be the largest number in exactly one of $X$ and $Y$. Without loss of generality, say that $s \in X \backslash Y$. The number $f_{X}(s)$ counts the $s^{t h}$ number not in $X$, which implies that $$ f_{X}(s)=s+\left|X \cap\left\{1,2, \ldots, f_{X}(s)\right\}\right| $$ Since $f_{X}(s) \geqslant s$, we have that $$ \left\{f_{X}(s)+1, f_{X}(s)+2, \ldots\right\} \cap X=\left\{f_{X}(s)+1, f_{X}(s)+2, \ldots\right\} \cap Y $$ which, together with the assumption that $|X|=|Y|$, gives $$ \left|X \cap\left\{1,2, \ldots, f_{X}(s)\right\}\right|=\left|Y \cap\left\{1,2, \ldots, f_{X}(s)\right\}\right| $$ Now consider the equation $$ t-|Y \cap\{1,2, \ldots, t\}|=s $$ This equation is satisfied only when $t \in\left[f_{Y}(s), f_{Y}(s+1)\right)$, because the left hand side counts the number of elements up to $t$ that are not in $Y$. We have that the value $t=f_{X}(s)$ satisfies the above equation because of (1) and (2). Furthermore, since $f_{X}(s) \notin X$ and $f_{X}(s) \geqslant s$, we have that $f_{X}(s) \notin Y$ due to the maximality of $s$. Thus, by the above discussion, we must have $f_{X}(s)=f_{Y}(s)$. Finally, we arrive at a contradiction. The value $f_{X}(s)$ is neither in $X$ nor in $f_{X}(Y)$, because $s$ is not in $Y$ by assumption. Thus, $f_{X}(s) \notin X * Y$. However, since $s \in X$, we have $f_{Y}(s) \in Y * X$, a contradiction. We are now ready to finish the proof. Note first of all that $\left|A^{* b}\right|=a b=\left|B^{* a}\right|$. Moreover, since $A * B=B * A$, and $*$ is associative, it follows that $A^{* b} * B^{* a}=B^{* a} * A^{* b}$. Thus, by Lemma 2, we have $A^{* b}=B^{* a}$, as desired. Comment 1. Taking $A=X^{* k}$ and $B=X^{* l}$ generates many non-trivial examples where $A * B=B * A$. There are also other examples not of this form. For example, if $A=\{1,2,4\}$ and $B=\{1,3\}$, then $A * B=\{1,2,3,4,6\}=B * A$.
|
proof
|
Yes
|
Yes
|
proof
|
Combinatorics
|
For any finite sets $X$ and $Y$ of positive integers, denote by $f_{X}(k)$ the $k^{\text {th }}$ smallest positive integer not in $X$, and let $$ X * Y=X \cup\left\{f_{X}(y): y \in Y\right\} $$ Let $A$ be a set of $a>0$ positive integers, and let $B$ be a set of $b>0$ positive integers. Prove that if $A * B=B * A$, then $$ \underbrace{A *(A * \cdots *(A *(A * A)) \ldots)}_{A \text { appears } b \text { times }}=\underbrace{B *(B * \cdots *(B *(B * B)) \ldots)}_{B \text { appears } a \text { times }} . $$
|
For any function $g: \mathbb{Z}_{>0} \rightarrow \mathbb{Z}_{>0}$ and any subset $X \subset \mathbb{Z}_{>0}$, we define $g(X)=$ $\{g(x): x \in X\}$. We have that the image of $f_{X}$ is $f_{X}\left(\mathbb{Z}_{>0}\right)=\mathbb{Z}_{>0} \backslash X$. We now show a general lemma about the operation *, with the goal of showing that $*$ is associative. Lemma 1. Let $X$ and $Y$ be finite sets of positive integers. The functions $f_{X * Y}$ and $f_{X} \circ f_{Y}$ are equal. Proof. We have $f_{X * Y}\left(\mathbb{Z}_{>0}\right)=\mathbb{Z}_{>0} \backslash(X * Y)=\left(\mathbb{Z}_{>0} \backslash X\right) \backslash f_{X}(Y)=f_{X}\left(\mathbb{Z}_{>0}\right) \backslash f_{X}(Y)=f_{X}\left(\mathbb{Z}_{>0} \backslash Y\right)=f_{X}\left(f_{Y}\left(\mathbb{Z}_{>0}\right)\right)$. Thus, the functions $f_{X * Y}$ and $f_{X} \circ f_{Y}$ are strictly increasing functions with the same range. Because a strictly function is uniquely defined by its range, we have $f_{X * Y}=f_{X} \circ f_{Y}$. Lemma 1 implies that * is associative, in the sense that $(A * B) * C=A *(B * C)$ for any finite sets $A, B$, and $C$ of positive integers. We prove the associativity by noting $$ \begin{gathered} \mathbb{Z}_{>0} \backslash((A * B) * C)=f_{(A * B) * C}\left(\mathbb{Z}_{>0}\right)=f_{A * B}\left(f_{C}\left(\mathbb{Z}_{>0}\right)\right)=f_{A}\left(f_{B}\left(f_{C}\left(\mathbb{Z}_{>0}\right)\right)\right) \\ =f_{A}\left(f_{B * C}\left(\mathbb{Z}_{>0}\right)=f_{A *(B * C)}\left(\mathbb{Z}_{>0}\right)=\mathbb{Z}_{>0} \backslash(A *(B * C))\right. \end{gathered} $$ In light of the associativity of *, we may drop the parentheses when we write expressions like $A *(B * C)$. We also introduce the notation $$ X^{* k}=\underbrace{X *(X * \cdots *(X *(X * X)) \ldots)}_{X \text { appears } k \text { times }} $$ Our goal is then to show that $A * B=B * A$ implies $A^{* b}=B^{* a}$. We will do so via the following general lemma. Lemma 2. Suppose that $X$ and $Y$ are finite sets of positive integers satisfying $X * Y=Y * X$ and $|X|=|Y|$. Then, we must have $X=Y$. Proof. Assume that $X$ and $Y$ are not equal. Let $s$ be the largest number in exactly one of $X$ and $Y$. Without loss of generality, say that $s \in X \backslash Y$. The number $f_{X}(s)$ counts the $s^{t h}$ number not in $X$, which implies that $$ f_{X}(s)=s+\left|X \cap\left\{1,2, \ldots, f_{X}(s)\right\}\right| $$ Since $f_{X}(s) \geqslant s$, we have that $$ \left\{f_{X}(s)+1, f_{X}(s)+2, \ldots\right\} \cap X=\left\{f_{X}(s)+1, f_{X}(s)+2, \ldots\right\} \cap Y $$ which, together with the assumption that $|X|=|Y|$, gives $$ \left|X \cap\left\{1,2, \ldots, f_{X}(s)\right\}\right|=\left|Y \cap\left\{1,2, \ldots, f_{X}(s)\right\}\right| $$ Now consider the equation $$ t-|Y \cap\{1,2, \ldots, t\}|=s $$ This equation is satisfied only when $t \in\left[f_{Y}(s), f_{Y}(s+1)\right)$, because the left hand side counts the number of elements up to $t$ that are not in $Y$. We have that the value $t=f_{X}(s)$ satisfies the above equation because of (1) and (2). Furthermore, since $f_{X}(s) \notin X$ and $f_{X}(s) \geqslant s$, we have that $f_{X}(s) \notin Y$ due to the maximality of $s$. Thus, by the above discussion, we must have $f_{X}(s)=f_{Y}(s)$. Finally, we arrive at a contradiction. The value $f_{X}(s)$ is neither in $X$ nor in $f_{X}(Y)$, because $s$ is not in $Y$ by assumption. Thus, $f_{X}(s) \notin X * Y$. However, since $s \in X$, we have $f_{Y}(s) \in Y * X$, a contradiction. We are now ready to finish the proof. Note first of all that $\left|A^{* b}\right|=a b=\left|B^{* a}\right|$. Moreover, since $A * B=B * A$, and $*$ is associative, it follows that $A^{* b} * B^{* a}=B^{* a} * A^{* b}$. Thus, by Lemma 2, we have $A^{* b}=B^{* a}$, as desired. Comment 1. Taking $A=X^{* k}$ and $B=X^{* l}$ generates many non-trivial examples where $A * B=B * A$. There are also other examples not of this form. For example, if $A=\{1,2,4\}$ and $B=\{1,3\}$, then $A * B=\{1,2,3,4,6\}=B * A$.
|
{
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|
1ab131b9-2f83-5e16-87bb-7119d20ca25f
| 24,965
|
Let $A B C D E$ be a convex pentagon such that $A B=B C=C D, \angle E A B=\angle B C D$, and $\angle E D C=\angle C B A$. Prove that the perpendicular line from $E$ to $B C$ and the line segments $A C$ and $B D$ are concurrent. (Italy)
|
Throughout the solution, we refer to $\angle A, \angle B, \angle C, \angle D$, and $\angle E$ as internal angles of the pentagon $A B C D E$. Let the perpendicular bisectors of $A C$ and $B D$, which pass respectively through $B$ and $C$, meet at point $I$. Then $B D \perp C I$ and, similarly, $A C \perp B I$. Hence $A C$ and $B D$ meet at the orthocenter $H$ of the triangle $B I C$, and $I H \perp B C$. It remains to prove that $E$ lies on the line $I H$ or, equivalently, $E I \perp B C$. Lines $I B$ and $I C$ bisect $\angle B$ and $\angle C$, respectively. Since $I A=I C, I B=I D$, and $A B=$ $B C=C D$, the triangles $I A B, I C B$ and $I C D$ are congruent. Hence $\angle I A B=\angle I C B=$ $\angle C / 2=\angle A / 2$, so the line $I A$ bisects $\angle A$. Similarly, the line $I D$ bisects $\angle D$. Finally, the line $I E$ bisects $\angle E$ because $I$ lies on all the other four internal bisectors of the angles of the pentagon. The sum of the internal angles in a pentagon is $540^{\circ}$, so $$ \angle E=540^{\circ}-2 \angle A+2 \angle B . $$ In quadrilateral $A B I E$, $$ \begin{aligned} \angle B I E & =360^{\circ}-\angle E A B-\angle A B I-\angle A E I=360^{\circ}-\angle A-\frac{1}{2} \angle B-\frac{1}{2} \angle E \\ & =360^{\circ}-\angle A-\frac{1}{2} \angle B-\left(270^{\circ}-\angle A-\angle B\right) \\ & =90^{\circ}+\frac{1}{2} \angle B=90^{\circ}+\angle I B C, \end{aligned} $$ which means that $E I \perp B C$, completing the proof. 
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
Let $A B C D E$ be a convex pentagon such that $A B=B C=C D, \angle E A B=\angle B C D$, and $\angle E D C=\angle C B A$. Prove that the perpendicular line from $E$ to $B C$ and the line segments $A C$ and $B D$ are concurrent. (Italy)
|
Throughout the solution, we refer to $\angle A, \angle B, \angle C, \angle D$, and $\angle E$ as internal angles of the pentagon $A B C D E$. Let the perpendicular bisectors of $A C$ and $B D$, which pass respectively through $B$ and $C$, meet at point $I$. Then $B D \perp C I$ and, similarly, $A C \perp B I$. Hence $A C$ and $B D$ meet at the orthocenter $H$ of the triangle $B I C$, and $I H \perp B C$. It remains to prove that $E$ lies on the line $I H$ or, equivalently, $E I \perp B C$. Lines $I B$ and $I C$ bisect $\angle B$ and $\angle C$, respectively. Since $I A=I C, I B=I D$, and $A B=$ $B C=C D$, the triangles $I A B, I C B$ and $I C D$ are congruent. Hence $\angle I A B=\angle I C B=$ $\angle C / 2=\angle A / 2$, so the line $I A$ bisects $\angle A$. Similarly, the line $I D$ bisects $\angle D$. Finally, the line $I E$ bisects $\angle E$ because $I$ lies on all the other four internal bisectors of the angles of the pentagon. The sum of the internal angles in a pentagon is $540^{\circ}$, so $$ \angle E=540^{\circ}-2 \angle A+2 \angle B . $$ In quadrilateral $A B I E$, $$ \begin{aligned} \angle B I E & =360^{\circ}-\angle E A B-\angle A B I-\angle A E I=360^{\circ}-\angle A-\frac{1}{2} \angle B-\frac{1}{2} \angle E \\ & =360^{\circ}-\angle A-\frac{1}{2} \angle B-\left(270^{\circ}-\angle A-\angle B\right) \\ & =90^{\circ}+\frac{1}{2} \angle B=90^{\circ}+\angle I B C, \end{aligned} $$ which means that $E I \perp B C$, completing the proof. 
|
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|
bcef7c5b-7311-5932-a53f-1c6cd6823269
| 24,972
|
Let $A B C D E$ be a convex pentagon such that $A B=B C=C D, \angle E A B=\angle B C D$, and $\angle E D C=\angle C B A$. Prove that the perpendicular line from $E$ to $B C$ and the line segments $A C$ and $B D$ are concurrent. (Italy)
|
We present another proof of the fact that $E$ lies on line $I H$. Since all five internal bisectors of $A B C D E$ meet at $I$, this pentagon has an inscribed circle with center $I$. Let this circle touch side $B C$ at $T$. Applying Brianchon's theorem to the (degenerate) hexagon $A B T C D E$ we conclude that $A C, B D$ and $E T$ are concurrent, so point $E$ also lies on line $I H T$, completing the proof.
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
Let $A B C D E$ be a convex pentagon such that $A B=B C=C D, \angle E A B=\angle B C D$, and $\angle E D C=\angle C B A$. Prove that the perpendicular line from $E$ to $B C$ and the line segments $A C$ and $B D$ are concurrent. (Italy)
|
We present another proof of the fact that $E$ lies on line $I H$. Since all five internal bisectors of $A B C D E$ meet at $I$, this pentagon has an inscribed circle with center $I$. Let this circle touch side $B C$ at $T$. Applying Brianchon's theorem to the (degenerate) hexagon $A B T C D E$ we conclude that $A C, B D$ and $E T$ are concurrent, so point $E$ also lies on line $I H T$, completing the proof.
|
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|
bcef7c5b-7311-5932-a53f-1c6cd6823269
| 24,972
|
Let $A B C D E$ be a convex pentagon such that $A B=B C=C D, \angle E A B=\angle B C D$, and $\angle E D C=\angle C B A$. Prove that the perpendicular line from $E$ to $B C$ and the line segments $A C$ and $B D$ are concurrent. (Italy)
|
We present yet another proof that $E I \perp B C$. In pentagon $A B C D E, \angle E<$ $180^{\circ} \Longleftrightarrow \angle A+\angle B+\angle C+\angle D>360^{\circ}$. Then $\angle A+\angle B=\angle C+\angle D>180^{\circ}$, so rays $E A$ and $C B$ meet at a point $P$, and rays $B C$ and $E D$ meet at a point $Q$. Now, $$ \angle P B A=180^{\circ}-\angle B=180^{\circ}-\angle D=\angle Q D C $$ and, similarly, $\angle P A B=\angle Q C D$. Since $A B=C D$, the triangles $P A B$ and $Q C D$ are congruent with the same orientation. Moreover, $P Q E$ is isosceles with $E P=E Q$.  In Solution 1 we have proved that triangles $I A B$ and $I C D$ are also congruent with the same orientation. Then we conclude that quadrilaterals $P B I A$ and $Q D I C$ are congruent, which implies $I P=I Q$. Then $E I$ is the perpendicular bisector of $P Q$ and, therefore, $E I \perp$ $P Q \Longleftrightarrow E I \perp B C$. Comment. Even though all three solutions used the point $I$, there are solutions that do not need it. We present an outline of such a solution: if $J$ is the incenter of $\triangle Q C D$ (with $P$ and $Q$ as defined in Solution 3), then a simple angle chasing shows that triangles $C J D$ and $B H C$ are congruent. Then if $S$ is the projection of $J$ onto side $C D$ and $T$ is the orthogonal projection of $H$ onto side $B C$, one can verify that $$ Q T=Q C+C T=Q C+D S=Q C+\frac{C D+D Q-Q C}{2}=\frac{P B+B C+Q C}{2}=\frac{P Q}{2}, $$ so $T$ is the midpoint of $P Q$, and $E, H$ and $T$ all lie on the perpendicular bisector of $P Q$.
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
Let $A B C D E$ be a convex pentagon such that $A B=B C=C D, \angle E A B=\angle B C D$, and $\angle E D C=\angle C B A$. Prove that the perpendicular line from $E$ to $B C$ and the line segments $A C$ and $B D$ are concurrent. (Italy)
|
We present yet another proof that $E I \perp B C$. In pentagon $A B C D E, \angle E<$ $180^{\circ} \Longleftrightarrow \angle A+\angle B+\angle C+\angle D>360^{\circ}$. Then $\angle A+\angle B=\angle C+\angle D>180^{\circ}$, so rays $E A$ and $C B$ meet at a point $P$, and rays $B C$ and $E D$ meet at a point $Q$. Now, $$ \angle P B A=180^{\circ}-\angle B=180^{\circ}-\angle D=\angle Q D C $$ and, similarly, $\angle P A B=\angle Q C D$. Since $A B=C D$, the triangles $P A B$ and $Q C D$ are congruent with the same orientation. Moreover, $P Q E$ is isosceles with $E P=E Q$.  In Solution 1 we have proved that triangles $I A B$ and $I C D$ are also congruent with the same orientation. Then we conclude that quadrilaterals $P B I A$ and $Q D I C$ are congruent, which implies $I P=I Q$. Then $E I$ is the perpendicular bisector of $P Q$ and, therefore, $E I \perp$ $P Q \Longleftrightarrow E I \perp B C$. Comment. Even though all three solutions used the point $I$, there are solutions that do not need it. We present an outline of such a solution: if $J$ is the incenter of $\triangle Q C D$ (with $P$ and $Q$ as defined in Solution 3), then a simple angle chasing shows that triangles $C J D$ and $B H C$ are congruent. Then if $S$ is the projection of $J$ onto side $C D$ and $T$ is the orthogonal projection of $H$ onto side $B C$, one can verify that $$ Q T=Q C+C T=Q C+D S=Q C+\frac{C D+D Q-Q C}{2}=\frac{P B+B C+Q C}{2}=\frac{P Q}{2}, $$ so $T$ is the midpoint of $P Q$, and $E, H$ and $T$ all lie on the perpendicular bisector of $P Q$.
|
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|
bcef7c5b-7311-5932-a53f-1c6cd6823269
| 24,972
|
Let $R$ and $S$ be distinct points on circle $\Omega$, and let $t$ denote the tangent line to $\Omega$ at $R$. Point $R^{\prime}$ is the reflection of $R$ with respect to $S$. A point $I$ is chosen on the smaller arc $R S$ of $\Omega$ so that the circumcircle $\Gamma$ of triangle $I S R^{\prime}$ intersects $t$ at two different points. Denote by $A$ the common point of $\Gamma$ and $t$ that is closest to $R$. Line $A I$ meets $\Omega$ again at $J$. Show that $J R^{\prime}$ is tangent to $\Gamma$. (Luxembourg)
|
In the circles $\Omega$ and $\Gamma$ we have $\angle J R S=\angle J I S=\angle A R^{\prime} S$. On the other hand, since $R A$ is tangent to $\Omega$, we get $\angle S J R=\angle S R A$. So the triangles $A R R^{\prime}$ and $S J R$ are similar, and $$ \frac{R^{\prime} R}{R J}=\frac{A R^{\prime}}{S R}=\frac{A R^{\prime}}{S R^{\prime}} $$ The last relation, together with $\angle A R^{\prime} S=\angle J R R^{\prime}$, yields $\triangle A S R^{\prime} \sim \triangle R^{\prime} J R$, hence $\angle S A R^{\prime}=\angle R R^{\prime} J$. It follows that $J R^{\prime}$ is tangent to $\Gamma$ at $R^{\prime}$.  Solution 1  Solution 2
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
Let $R$ and $S$ be distinct points on circle $\Omega$, and let $t$ denote the tangent line to $\Omega$ at $R$. Point $R^{\prime}$ is the reflection of $R$ with respect to $S$. A point $I$ is chosen on the smaller arc $R S$ of $\Omega$ so that the circumcircle $\Gamma$ of triangle $I S R^{\prime}$ intersects $t$ at two different points. Denote by $A$ the common point of $\Gamma$ and $t$ that is closest to $R$. Line $A I$ meets $\Omega$ again at $J$. Show that $J R^{\prime}$ is tangent to $\Gamma$. (Luxembourg)
|
In the circles $\Omega$ and $\Gamma$ we have $\angle J R S=\angle J I S=\angle A R^{\prime} S$. On the other hand, since $R A$ is tangent to $\Omega$, we get $\angle S J R=\angle S R A$. So the triangles $A R R^{\prime}$ and $S J R$ are similar, and $$ \frac{R^{\prime} R}{R J}=\frac{A R^{\prime}}{S R}=\frac{A R^{\prime}}{S R^{\prime}} $$ The last relation, together with $\angle A R^{\prime} S=\angle J R R^{\prime}$, yields $\triangle A S R^{\prime} \sim \triangle R^{\prime} J R$, hence $\angle S A R^{\prime}=\angle R R^{\prime} J$. It follows that $J R^{\prime}$ is tangent to $\Gamma$ at $R^{\prime}$.  Solution 1  Solution 2
|
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|
42e0e1bf-e980-5894-98bf-1d66d8510012
| 24,977
|
Let $R$ and $S$ be distinct points on circle $\Omega$, and let $t$ denote the tangent line to $\Omega$ at $R$. Point $R^{\prime}$ is the reflection of $R$ with respect to $S$. A point $I$ is chosen on the smaller arc $R S$ of $\Omega$ so that the circumcircle $\Gamma$ of triangle $I S R^{\prime}$ intersects $t$ at two different points. Denote by $A$ the common point of $\Gamma$ and $t$ that is closest to $R$. Line $A I$ meets $\Omega$ again at $J$. Show that $J R^{\prime}$ is tangent to $\Gamma$. (Luxembourg)
|
As in Let $A^{\prime}$ be the reflection of $A$ about $S$; then $A R A^{\prime} R^{\prime}$ is a parallelogram with center $S$, and hence the point $J$ lies on the line $R A^{\prime}$. From $\angle S R^{\prime} A^{\prime}=\angle S R A=\angle S J R$ we get that the points $S, J, A^{\prime}, R^{\prime}$ are concyclic. This proves that $\angle S R^{\prime} J=\angle S A^{\prime} J=\angle S A^{\prime} R=\angle S A R^{\prime}$, so $J R^{\prime}$ is tangent to $\Gamma$ at $R^{\prime}$.
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
Let $R$ and $S$ be distinct points on circle $\Omega$, and let $t$ denote the tangent line to $\Omega$ at $R$. Point $R^{\prime}$ is the reflection of $R$ with respect to $S$. A point $I$ is chosen on the smaller arc $R S$ of $\Omega$ so that the circumcircle $\Gamma$ of triangle $I S R^{\prime}$ intersects $t$ at two different points. Denote by $A$ the common point of $\Gamma$ and $t$ that is closest to $R$. Line $A I$ meets $\Omega$ again at $J$. Show that $J R^{\prime}$ is tangent to $\Gamma$. (Luxembourg)
|
As in Let $A^{\prime}$ be the reflection of $A$ about $S$; then $A R A^{\prime} R^{\prime}$ is a parallelogram with center $S$, and hence the point $J$ lies on the line $R A^{\prime}$. From $\angle S R^{\prime} A^{\prime}=\angle S R A=\angle S J R$ we get that the points $S, J, A^{\prime}, R^{\prime}$ are concyclic. This proves that $\angle S R^{\prime} J=\angle S A^{\prime} J=\angle S A^{\prime} R=\angle S A R^{\prime}$, so $J R^{\prime}$ is tangent to $\Gamma$ at $R^{\prime}$.
|
{
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|
42e0e1bf-e980-5894-98bf-1d66d8510012
| 24,977
|
Let $O$ be the circumcenter of an acute scalene triangle $A B C$. Line $O A$ intersects the altitudes of $A B C$ through $B$ and $C$ at $P$ and $Q$, respectively. The altitudes meet at $H$. Prove that the circumcenter of triangle $P Q H$ lies on a median of triangle $A B C$. (Ukraine)
|
Suppose, without loss of generality, that $A B<A C$. We have $\angle P Q H=90^{\circ}-$ $\angle Q A B=90^{\circ}-\angle O A B=\frac{1}{2} \angle A O B=\angle A C B$, and similarly $\angle Q P H=\angle A B C$. Thus triangles $A B C$ and $H P Q$ are similar. Let $\Omega$ and $\omega$ be the circumcircles of $A B C$ and $H P Q$, respectively. Since $\angle A H P=90^{\circ}-\angle H A C=\angle A C B=\angle H Q P$, line $A H$ is tangent to $\omega$.  Let $T$ be the center of $\omega$ and let lines $A T$ and $B C$ meet at $M$. We will take advantage of the similarity between $A B C$ and $H P Q$ and the fact that $A H$ is tangent to $\omega$ at $H$, with $A$ on line $P Q$. Consider the corresponding tangent $A S$ to $\Omega$, with $S \in B C$. Then $S$ and $A$ correspond to each other in $\triangle A B C \sim \triangle H P Q$, and therefore $\angle O S M=\angle O A T=\angle O A M$. Hence quadrilateral $S A O M$ is cyclic, and since the tangent line $A S$ is perpendicular to $A O$, $\angle O M S=180^{\circ}-\angle O A S=90^{\circ}$. This means that $M$ is the orthogonal projection of $O$ onto $B C$, which is its midpoint. So $T$ lies on median $A M$ of triangle $A B C$.
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
Let $O$ be the circumcenter of an acute scalene triangle $A B C$. Line $O A$ intersects the altitudes of $A B C$ through $B$ and $C$ at $P$ and $Q$, respectively. The altitudes meet at $H$. Prove that the circumcenter of triangle $P Q H$ lies on a median of triangle $A B C$. (Ukraine)
|
Suppose, without loss of generality, that $A B<A C$. We have $\angle P Q H=90^{\circ}-$ $\angle Q A B=90^{\circ}-\angle O A B=\frac{1}{2} \angle A O B=\angle A C B$, and similarly $\angle Q P H=\angle A B C$. Thus triangles $A B C$ and $H P Q$ are similar. Let $\Omega$ and $\omega$ be the circumcircles of $A B C$ and $H P Q$, respectively. Since $\angle A H P=90^{\circ}-\angle H A C=\angle A C B=\angle H Q P$, line $A H$ is tangent to $\omega$.  Let $T$ be the center of $\omega$ and let lines $A T$ and $B C$ meet at $M$. We will take advantage of the similarity between $A B C$ and $H P Q$ and the fact that $A H$ is tangent to $\omega$ at $H$, with $A$ on line $P Q$. Consider the corresponding tangent $A S$ to $\Omega$, with $S \in B C$. Then $S$ and $A$ correspond to each other in $\triangle A B C \sim \triangle H P Q$, and therefore $\angle O S M=\angle O A T=\angle O A M$. Hence quadrilateral $S A O M$ is cyclic, and since the tangent line $A S$ is perpendicular to $A O$, $\angle O M S=180^{\circ}-\angle O A S=90^{\circ}$. This means that $M$ is the orthogonal projection of $O$ onto $B C$, which is its midpoint. So $T$ lies on median $A M$ of triangle $A B C$.
|
{
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|
221e94a2-9a1b-5b2c-8a64-8d419caeefa6
| 24,982
|
In triangle $A B C$, let $\omega$ be the excircle opposite $A$. Let $D, E$, and $F$ be the points where $\omega$ is tangent to lines $B C, C A$, and $A B$, respectively. The circle $A E F$ intersects line $B C$ at $P$ and $Q$. Let $M$ be the midpoint of $A D$. Prove that the circle $M P Q$ is tangent to $\omega$. (Denmark)
|
Denote by $\Omega$ the circle $A E F P Q$, and denote by $\gamma$ the circle $P Q M$. Let the line $A D$ meet $\omega$ again at $T \neq D$. We will show that $\gamma$ is tangent to $\omega$ at $T$. We first prove that points $P, Q, M, T$ are concyclic. Let $A^{\prime}$ be the center of $\omega$. Since $A^{\prime} E \perp A E$ and $A^{\prime} F \perp A F, A A^{\prime}$ is a diameter in $\Omega$. Let $N$ be the midpoint of $D T$; from $A^{\prime} D=A^{\prime} T$ we can see that $\angle A^{\prime} N A=90^{\circ}$ and therefore $N$ also lies on the circle $\Omega$. Now, from the power of $D$ with respect to the circles $\gamma$ and $\Omega$ we get $$ D P \cdot D Q=D A \cdot D N=2 D M \cdot \frac{D T}{2}=D M \cdot D T $$ so $P, Q, M, T$ are concyclic. If $E F \| B C$, then $A B C$ is isosceles and the problem is now immediate by symmetry. Otherwise, let the tangent line to $\omega$ at $T$ meet line $B C$ at point $R$. The tangent line segments $R D$ and $R T$ have the same length, so $A^{\prime} R$ is the perpendicular bisector of $D T$; since $N D=N T$, $N$ lies on this perpendicular bisector. In right triangle $A^{\prime} R D, R D^{2}=R N \cdot R A^{\prime}=R P \cdot R Q$, in which the last equality was obtained from the power of $R$ with respect to $\Omega$. Hence $R T^{2}=R P \cdot R Q$, which implies that $R T$ is also tangent to $\gamma$. Because $R T$ is a common tangent to $\omega$ and $\gamma$, these two circles are tangent at $T$. 
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
In triangle $A B C$, let $\omega$ be the excircle opposite $A$. Let $D, E$, and $F$ be the points where $\omega$ is tangent to lines $B C, C A$, and $A B$, respectively. The circle $A E F$ intersects line $B C$ at $P$ and $Q$. Let $M$ be the midpoint of $A D$. Prove that the circle $M P Q$ is tangent to $\omega$. (Denmark)
|
Denote by $\Omega$ the circle $A E F P Q$, and denote by $\gamma$ the circle $P Q M$. Let the line $A D$ meet $\omega$ again at $T \neq D$. We will show that $\gamma$ is tangent to $\omega$ at $T$. We first prove that points $P, Q, M, T$ are concyclic. Let $A^{\prime}$ be the center of $\omega$. Since $A^{\prime} E \perp A E$ and $A^{\prime} F \perp A F, A A^{\prime}$ is a diameter in $\Omega$. Let $N$ be the midpoint of $D T$; from $A^{\prime} D=A^{\prime} T$ we can see that $\angle A^{\prime} N A=90^{\circ}$ and therefore $N$ also lies on the circle $\Omega$. Now, from the power of $D$ with respect to the circles $\gamma$ and $\Omega$ we get $$ D P \cdot D Q=D A \cdot D N=2 D M \cdot \frac{D T}{2}=D M \cdot D T $$ so $P, Q, M, T$ are concyclic. If $E F \| B C$, then $A B C$ is isosceles and the problem is now immediate by symmetry. Otherwise, let the tangent line to $\omega$ at $T$ meet line $B C$ at point $R$. The tangent line segments $R D$ and $R T$ have the same length, so $A^{\prime} R$ is the perpendicular bisector of $D T$; since $N D=N T$, $N$ lies on this perpendicular bisector. In right triangle $A^{\prime} R D, R D^{2}=R N \cdot R A^{\prime}=R P \cdot R Q$, in which the last equality was obtained from the power of $R$ with respect to $\Omega$. Hence $R T^{2}=R P \cdot R Q$, which implies that $R T$ is also tangent to $\gamma$. Because $R T$ is a common tangent to $\omega$ and $\gamma$, these two circles are tangent at $T$. 
|
{
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|
98d1f6bf-dac0-5c9c-8b91-60f13e7112c7
| 24,985
|
In triangle $A B C$, let $\omega$ be the excircle opposite $A$. Let $D, E$, and $F$ be the points where $\omega$ is tangent to lines $B C, C A$, and $A B$, respectively. The circle $A E F$ intersects line $B C$ at $P$ and $Q$. Let $M$ be the midpoint of $A D$. Prove that the circle $M P Q$ is tangent to $\omega$. (Denmark)
|
After proving that $P, Q, M, T$ are concyclic, we finish the problem in a different fashion. We only consider the case in which $E F$ and $B C$ are not parallel. Let lines $P Q$ and $E F$ meet at point $R$. Since $P Q$ and $E F$ are radical axes of $\Omega, \gamma$ and $\omega, \gamma$, respectively, $R$ is the radical center of these three circles. With respect to the circle $\omega$, the line $D R$ is the polar of $D$, and the line $E F$ is the polar of $A$. So the pole of line $A D T$ is $D R \cap E F=R$, and therefore $R T$ is tangent to $\omega$. Finally, since $T$ belongs to $\gamma$ and $\omega$ and $R$ is the radical center of $\gamma, \omega$ and $\Omega$, line $R T$ is the radical axis of $\gamma$ and $\omega$, and since it is tangent to $\omega$, it is also tangent to $\gamma$. Because $R T$ is a common tangent to $\omega$ and $\gamma$, these two circles are tangent at $T$. Comment. In Solution 2 we defined the point $R$ from Solution 1 in a different way.
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
In triangle $A B C$, let $\omega$ be the excircle opposite $A$. Let $D, E$, and $F$ be the points where $\omega$ is tangent to lines $B C, C A$, and $A B$, respectively. The circle $A E F$ intersects line $B C$ at $P$ and $Q$. Let $M$ be the midpoint of $A D$. Prove that the circle $M P Q$ is tangent to $\omega$. (Denmark)
|
After proving that $P, Q, M, T$ are concyclic, we finish the problem in a different fashion. We only consider the case in which $E F$ and $B C$ are not parallel. Let lines $P Q$ and $E F$ meet at point $R$. Since $P Q$ and $E F$ are radical axes of $\Omega, \gamma$ and $\omega, \gamma$, respectively, $R$ is the radical center of these three circles. With respect to the circle $\omega$, the line $D R$ is the polar of $D$, and the line $E F$ is the polar of $A$. So the pole of line $A D T$ is $D R \cap E F=R$, and therefore $R T$ is tangent to $\omega$. Finally, since $T$ belongs to $\gamma$ and $\omega$ and $R$ is the radical center of $\gamma, \omega$ and $\Omega$, line $R T$ is the radical axis of $\gamma$ and $\omega$, and since it is tangent to $\omega$, it is also tangent to $\gamma$. Because $R T$ is a common tangent to $\omega$ and $\gamma$, these two circles are tangent at $T$. Comment. In Solution 2 we defined the point $R$ from Solution 1 in a different way.
|
{
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|
98d1f6bf-dac0-5c9c-8b91-60f13e7112c7
| 24,985
|
In triangle $A B C$, let $\omega$ be the excircle opposite $A$. Let $D, E$, and $F$ be the points where $\omega$ is tangent to lines $B C, C A$, and $A B$, respectively. The circle $A E F$ intersects line $B C$ at $P$ and $Q$. Let $M$ be the midpoint of $A D$. Prove that the circle $M P Q$ is tangent to $\omega$. (Denmark)
|
We give an alternative proof that the circles are tangent at the common point $T$. Again, we start from the fact that $P, Q, M, T$ are concyclic. Let point $O$ be the midpoint of diameter $A A^{\prime}$. Then $M O$ is the midline of triangle $A D A^{\prime}$, so $M O \| A^{\prime} D$. Since $A^{\prime} D \perp P Q$, $M O$ is perpendicular to $P Q$ as well. Looking at circle $\Omega$, which has center $O, M O \perp P Q$ implies that $M O$ is the perpendicular bisector of the chord $P Q$. Thus $M$ is the midpoint of arc $\widehat{P Q}$ from $\gamma$, and the tangent line $m$ to $\gamma$ at $M$ is parallel to $P Q$.  Consider the homothety with center $T$ and ratio $\frac{T D}{T M}$. It takes $D$ to $M$, and the line $P Q$ to the line $m$. Since the circle that is tangent to a line at a given point and that goes through another given point is unique, this homothety also takes $\omega$ (tangent to $P Q$ and going through $T$ ) to $\gamma$ (tangent to $m$ and going through $T$ ). We conclude that $\omega$ and $\gamma$ are tangent at $T$.
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
In triangle $A B C$, let $\omega$ be the excircle opposite $A$. Let $D, E$, and $F$ be the points where $\omega$ is tangent to lines $B C, C A$, and $A B$, respectively. The circle $A E F$ intersects line $B C$ at $P$ and $Q$. Let $M$ be the midpoint of $A D$. Prove that the circle $M P Q$ is tangent to $\omega$. (Denmark)
|
We give an alternative proof that the circles are tangent at the common point $T$. Again, we start from the fact that $P, Q, M, T$ are concyclic. Let point $O$ be the midpoint of diameter $A A^{\prime}$. Then $M O$ is the midline of triangle $A D A^{\prime}$, so $M O \| A^{\prime} D$. Since $A^{\prime} D \perp P Q$, $M O$ is perpendicular to $P Q$ as well. Looking at circle $\Omega$, which has center $O, M O \perp P Q$ implies that $M O$ is the perpendicular bisector of the chord $P Q$. Thus $M$ is the midpoint of arc $\widehat{P Q}$ from $\gamma$, and the tangent line $m$ to $\gamma$ at $M$ is parallel to $P Q$.  Consider the homothety with center $T$ and ratio $\frac{T D}{T M}$. It takes $D$ to $M$, and the line $P Q$ to the line $m$. Since the circle that is tangent to a line at a given point and that goes through another given point is unique, this homothety also takes $\omega$ (tangent to $P Q$ and going through $T$ ) to $\gamma$ (tangent to $m$ and going through $T$ ). We conclude that $\omega$ and $\gamma$ are tangent at $T$.
|
{
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|
98d1f6bf-dac0-5c9c-8b91-60f13e7112c7
| 24,985
|
Let $A B C C_{1} B_{1} A_{1}$ be a convex hexagon such that $A B=B C$, and suppose that the line segments $A A_{1}, B B_{1}$, and $C C_{1}$ have the same perpendicular bisector. Let the diagonals $A C_{1}$ and $A_{1} C$ meet at $D$, and denote by $\omega$ the circle $A B C$. Let $\omega$ intersect the circle $A_{1} B C_{1}$ again at $E \neq B$. Prove that the lines $B B_{1}$ and $D E$ intersect on $\omega$. (Ukraine)
|
If $A A_{1}=C C_{1}$, then the hexagon is symmetric about the line $B B_{1}$; in particular the circles $A B C$ and $A_{1} B C_{1}$ are tangent to each other. So $A A_{1}$ and $C C_{1}$ must be different. Since the points $A$ and $A_{1}$ can be interchanged with $C$ and $C_{1}$, respectively, we may assume $A A_{1}<C C_{1}$. Let $R$ be the radical center of the circles $A E B C$ and $A_{1} E B C_{1}$, and the circumcircle of the symmetric trapezoid $A C C_{1} A_{1}$; that is the common point of the pairwise radical axes $A C, A_{1} C_{1}$, and $B E$. By the symmetry of $A C$ and $A_{1} C_{1}$, the point $R$ lies on the common perpendicular bisector of $A A_{1}$ and $C C_{1}$, which is the external bisector of $\angle A D C$. Let $F$ be the second intersection of the line $D R$ and the circle $A C D$. From the power of $R$ with respect to the circles $\omega$ and $A C F D$ we have $R B \cdot R E=R A \cdot R C=R D \cdot D F$, so the points $B, E, D$ and $F$ are concyclic. The line $R D F$ is the external bisector of $\angle A D C$, so the point $F$ bisects the arc $\widehat{C D A}$. By $A B=B C$, on circle $\omega$, the point $B$ is the midpoint of arc $\widehat{A E C}$; let $M$ be the point diametrically opposite to $B$, that is the midpoint of the opposite $\operatorname{arc} \overline{C A}$ of $\omega$. Notice that the points $B, F$ and $M$ lie on the perpendicular bisector of $A C$, so they are collinear.  Finally, let $X$ be the second intersection point of $\omega$ and the line $D E$. Since $B M$ is a diameter in $\omega$, we have $\angle B X M=90^{\circ}$. Moreover, $$ \angle E X M=180^{\circ}-\angle M B E=180^{\circ}-\angle F B E=\angle E D F, $$ so $M X$ and $F D$ are parallel. Since $B X$ is perpendicular to $M X$ and $B B_{1}$ is perpendicular to $F D$, this shows that $X$ lies on line $B B_{1}$.
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
Let $A B C C_{1} B_{1} A_{1}$ be a convex hexagon such that $A B=B C$, and suppose that the line segments $A A_{1}, B B_{1}$, and $C C_{1}$ have the same perpendicular bisector. Let the diagonals $A C_{1}$ and $A_{1} C$ meet at $D$, and denote by $\omega$ the circle $A B C$. Let $\omega$ intersect the circle $A_{1} B C_{1}$ again at $E \neq B$. Prove that the lines $B B_{1}$ and $D E$ intersect on $\omega$. (Ukraine)
|
If $A A_{1}=C C_{1}$, then the hexagon is symmetric about the line $B B_{1}$; in particular the circles $A B C$ and $A_{1} B C_{1}$ are tangent to each other. So $A A_{1}$ and $C C_{1}$ must be different. Since the points $A$ and $A_{1}$ can be interchanged with $C$ and $C_{1}$, respectively, we may assume $A A_{1}<C C_{1}$. Let $R$ be the radical center of the circles $A E B C$ and $A_{1} E B C_{1}$, and the circumcircle of the symmetric trapezoid $A C C_{1} A_{1}$; that is the common point of the pairwise radical axes $A C, A_{1} C_{1}$, and $B E$. By the symmetry of $A C$ and $A_{1} C_{1}$, the point $R$ lies on the common perpendicular bisector of $A A_{1}$ and $C C_{1}$, which is the external bisector of $\angle A D C$. Let $F$ be the second intersection of the line $D R$ and the circle $A C D$. From the power of $R$ with respect to the circles $\omega$ and $A C F D$ we have $R B \cdot R E=R A \cdot R C=R D \cdot D F$, so the points $B, E, D$ and $F$ are concyclic. The line $R D F$ is the external bisector of $\angle A D C$, so the point $F$ bisects the arc $\widehat{C D A}$. By $A B=B C$, on circle $\omega$, the point $B$ is the midpoint of arc $\widehat{A E C}$; let $M$ be the point diametrically opposite to $B$, that is the midpoint of the opposite $\operatorname{arc} \overline{C A}$ of $\omega$. Notice that the points $B, F$ and $M$ lie on the perpendicular bisector of $A C$, so they are collinear.  Finally, let $X$ be the second intersection point of $\omega$ and the line $D E$. Since $B M$ is a diameter in $\omega$, we have $\angle B X M=90^{\circ}$. Moreover, $$ \angle E X M=180^{\circ}-\angle M B E=180^{\circ}-\angle F B E=\angle E D F, $$ so $M X$ and $F D$ are parallel. Since $B X$ is perpendicular to $M X$ and $B B_{1}$ is perpendicular to $F D$, this shows that $X$ lies on line $B B_{1}$.
|
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|
f0dcd808-d335-5ed8-a711-87eb0f73ca77
| 24,990
|
Let $A B C C_{1} B_{1} A_{1}$ be a convex hexagon such that $A B=B C$, and suppose that the line segments $A A_{1}, B B_{1}$, and $C C_{1}$ have the same perpendicular bisector. Let the diagonals $A C_{1}$ and $A_{1} C$ meet at $D$, and denote by $\omega$ the circle $A B C$. Let $\omega$ intersect the circle $A_{1} B C_{1}$ again at $E \neq B$. Prove that the lines $B B_{1}$ and $D E$ intersect on $\omega$. (Ukraine)
|
Define point $M$ as the point opposite to $B$ on circle $\omega$, and point $R$ as the intersection of lines $A C, A_{1} C_{1}$ and $B E$, and show that $R$ lies on the external bisector of $\angle A D C$, like in the first solution. Since $B$ is the midpoint of the arc $\widehat{A E C}$, the line $B E R$ is the external bisector of $\angle C E A$. Now we show that the internal angle bisectors of $\angle A D C$ and $\angle C E A$ meet on the segment $A C$. Let the angle bisector of $\angle A D C$ meet $A C$ at $S$, and let the angle bisector of $\angle C E A$, which is line $E M$, meet $A C$ at $S^{\prime}$. By applying the angle bisector theorem to both internal and external bisectors of $\angle A D C$ and $\angle C E A$, $$ A S: C S=A D: C D=A R: C R=A E: C E=A S^{\prime}: C S^{\prime} $$ so indeed $S=S^{\prime}$. By $\angle R D S=\angle S E R=90^{\circ}$ the points $R, S, D$ and $E$ are concyclic.  Now let the lines $B B_{1}$ and $D E$ meet at point $X$. Notice that $\angle E X B=\angle E D S$ because both $B B_{1}$ and $D S$ are perpendicular to the line $D R$, we have that $\angle E D S=\angle E R S$ in circle $S R D E$, and $\angle E R S=\angle E M B$ because $S R \perp B M$ and $E R \perp M E$. Therefore, $\angle E X B=\angle E M B$, so indeed, the point $X$ lies on $\omega$.
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
Let $A B C C_{1} B_{1} A_{1}$ be a convex hexagon such that $A B=B C$, and suppose that the line segments $A A_{1}, B B_{1}$, and $C C_{1}$ have the same perpendicular bisector. Let the diagonals $A C_{1}$ and $A_{1} C$ meet at $D$, and denote by $\omega$ the circle $A B C$. Let $\omega$ intersect the circle $A_{1} B C_{1}$ again at $E \neq B$. Prove that the lines $B B_{1}$ and $D E$ intersect on $\omega$. (Ukraine)
|
Define point $M$ as the point opposite to $B$ on circle $\omega$, and point $R$ as the intersection of lines $A C, A_{1} C_{1}$ and $B E$, and show that $R$ lies on the external bisector of $\angle A D C$, like in the first solution. Since $B$ is the midpoint of the arc $\widehat{A E C}$, the line $B E R$ is the external bisector of $\angle C E A$. Now we show that the internal angle bisectors of $\angle A D C$ and $\angle C E A$ meet on the segment $A C$. Let the angle bisector of $\angle A D C$ meet $A C$ at $S$, and let the angle bisector of $\angle C E A$, which is line $E M$, meet $A C$ at $S^{\prime}$. By applying the angle bisector theorem to both internal and external bisectors of $\angle A D C$ and $\angle C E A$, $$ A S: C S=A D: C D=A R: C R=A E: C E=A S^{\prime}: C S^{\prime} $$ so indeed $S=S^{\prime}$. By $\angle R D S=\angle S E R=90^{\circ}$ the points $R, S, D$ and $E$ are concyclic.  Now let the lines $B B_{1}$ and $D E$ meet at point $X$. Notice that $\angle E X B=\angle E D S$ because both $B B_{1}$ and $D S$ are perpendicular to the line $D R$, we have that $\angle E D S=\angle E R S$ in circle $S R D E$, and $\angle E R S=\angle E M B$ because $S R \perp B M$ and $E R \perp M E$. Therefore, $\angle E X B=\angle E M B$, so indeed, the point $X$ lies on $\omega$.
|
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f0dcd808-d335-5ed8-a711-87eb0f73ca77
| 24,990
|
A convex quadrilateral $A B C D$ has an inscribed circle with center $I$. Let $I_{a}, I_{b}, I_{c}$, and $I_{d}$ be the incenters of the triangles $D A B, A B C, B C D$, and $C D A$, respectively. Suppose that the common external tangents of the circles $A I_{b} I_{d}$ and $C I_{b} I_{d}$ meet at $X$, and the common external tangents of the circles $B I_{a} I_{c}$ and $D I_{a} I_{c}$ meet at $Y$. Prove that $\angle X I Y=90^{\circ}$. (Kazakhstan)
|
Denote by $\omega_{a}, \omega_{b}, \omega_{c}$ and $\omega_{d}$ the circles $A I_{b} I_{d}, B I_{a} I_{c}, C I_{b} I_{d}$, and $D I_{a} I_{c}$, let their centers be $O_{a}, O_{b}, O_{c}$ and $O_{d}$, and let their radii be $r_{a}, r_{b}, r_{c}$ and $r_{d}$, respectively. Claim 1. $I_{b} I_{d} \perp A C$ and $I_{a} I_{c} \perp B D$. Proof. Let the incircles of triangles $A B C$ and $A C D$ be tangent to the line $A C$ at $T$ and $T^{\prime}$, respectively. (See the figure to the left.) We have $A T=\frac{A B+A C-B C}{2}$ in triangle $A B C, A T^{\prime}=$ $\frac{A D+A C-C D}{2}$ in triangle $A C D$, and $A B-B C=A D-C D$ in quadrilateral $A B C D$, so $$ A T=\frac{A C+A B-B C}{2}=\frac{A C+A D-C D}{2}=A T^{\prime} $$ This shows $T=T^{\prime}$. As an immediate consequence, $I_{b} I_{d} \perp A C$. The second statement can be shown analogously.  Claim 2. The points $O_{a}, O_{b}, O_{c}$ and $O_{d}$ lie on the lines $A I, B I, C I$ and $D I$, respectively. Proof. By symmetry it suffices to prove the claim for $O_{a}$. (See the figure to the right above.) Notice first that the incircles of triangles $A B C$ and $A C D$ can be obtained from the incircle of the quadrilateral $A B C D$ with homothety centers $B$ and $D$, respectively, and homothety factors less than 1 , therefore the points $I_{b}$ and $I_{d}$ lie on the line segments $B I$ and $D I$, respectively. As is well-known, in every triangle the altitude and the diameter of the circumcircle starting from the same vertex are symmetric about the angle bisector. By Claim 1, in triangle $A I_{d} I_{b}$, the segment $A T$ is the altitude starting from $A$. Since the foot $T$ lies inside the segment $I_{b} I_{d}$, the circumcenter $O_{a}$ of triangle $A I_{d} I_{b}$ lies in the angle domain $I_{b} A I_{d}$ in such a way that $\angle I_{b} A T=\angle O_{a} A I_{d}$. The points $I_{b}$ and $I_{d}$ are the incenters of triangles $A B C$ and $A C D$, so the lines $A I_{b}$ and $A I_{d}$ bisect the angles $\angle B A C$ and $\angle C A D$, respectively. Then $$ \angle O_{a} A D=\angle O_{a} A I_{d}+\angle I_{d} A D=\angle I_{b} A T+\angle I_{d} A D=\frac{1}{2} \angle B A C+\frac{1}{2} \angle C A D=\frac{1}{2} \angle B A D, $$ so $O_{a}$ lies on the angle bisector of $\angle B A D$, that is, on line $A I$. The point $X$ is the external similitude center of $\omega_{a}$ and $\omega_{c}$; let $U$ be their internal similitude center. The points $O_{a}$ and $O_{c}$ lie on the perpendicular bisector of the common chord $I_{b} I_{d}$ of $\omega_{a}$ and $\omega_{c}$, and the two similitude centers $X$ and $U$ lie on the same line; by Claim 2, that line is parallel to $A C$.  From the similarity of the circles $\omega_{a}$ and $\omega_{c}$, from $O_{a} I_{b}=O_{a} I_{d}=O_{a} A=r_{a}$ and $O_{c} I_{b}=$ $O_{c} I_{d}=O_{c} C=r_{c}$, and from $A C \| O_{a} O_{c}$ we can see that $$ \frac{O_{a} X}{O_{c} X}=\frac{O_{a} U}{O_{c} U}=\frac{r_{a}}{r_{c}}=\frac{O_{a} I_{b}}{O_{c} I_{b}}=\frac{O_{a} I_{d}}{O_{c} I_{d}}=\frac{O_{a} A}{O_{c} C}=\frac{O_{a} I}{O_{c} I} $$ So the points $X, U, I_{b}, I_{d}, I$ lie on the Apollonius circle of the points $O_{a}, O_{c}$ with ratio $r_{a}: r_{c}$. In this Apollonius circle $X U$ is a diameter, and the lines $I U$ and $I X$ are respectively the internal and external bisectors of $\angle O_{a} I O_{c}=\angle A I C$, according to the angle bisector theorem. Moreover, in the Apollonius circle the diameter $U X$ is the perpendicular bisector of $I_{b} I_{d}$, so the lines $I X$ and $I U$ are the internal and external bisectors of $\angle I_{b} I I_{d}=\angle B I D$, respectively. Repeating the same argument for the points $B, D$ instead of $A, C$, we get that the line $I Y$ is the internal bisector of $\angle A I C$ and the external bisector of $\angle B I D$. Therefore, the lines $I X$ and $I Y$ respectively are the internal and external bisectors of $\angle B I D$, so they are perpendicular. Comment. In fact the points $O_{a}, O_{b}, O_{c}$ and $O_{d}$ lie on the line segments $A I, B I, C I$ and $D I$, respectively. For the point $O_{a}$ this can be shown for example by $\angle I_{d} O_{a} A+\angle A O_{a} I_{b}=\left(180^{\circ}-\right.$ $\left.2 \angle O_{a} A I_{d}\right)+\left(180^{\circ}-2 \angle I_{b} A O_{a}\right)=360^{\circ}-\angle B A D=\angle A D I+\angle D I A+\angle A I B+\angle I B A>\angle I_{d} I A+\angle A I I_{b}$. The solution also shows that the line $I Y$ passes through the point $U$, and analogously, $I X$ passes through the internal similitude center of $\omega_{b}$ and $\omega_{d}$.
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
A convex quadrilateral $A B C D$ has an inscribed circle with center $I$. Let $I_{a}, I_{b}, I_{c}$, and $I_{d}$ be the incenters of the triangles $D A B, A B C, B C D$, and $C D A$, respectively. Suppose that the common external tangents of the circles $A I_{b} I_{d}$ and $C I_{b} I_{d}$ meet at $X$, and the common external tangents of the circles $B I_{a} I_{c}$ and $D I_{a} I_{c}$ meet at $Y$. Prove that $\angle X I Y=90^{\circ}$. (Kazakhstan)
|
Denote by $\omega_{a}, \omega_{b}, \omega_{c}$ and $\omega_{d}$ the circles $A I_{b} I_{d}, B I_{a} I_{c}, C I_{b} I_{d}$, and $D I_{a} I_{c}$, let their centers be $O_{a}, O_{b}, O_{c}$ and $O_{d}$, and let their radii be $r_{a}, r_{b}, r_{c}$ and $r_{d}$, respectively. Claim 1. $I_{b} I_{d} \perp A C$ and $I_{a} I_{c} \perp B D$. Proof. Let the incircles of triangles $A B C$ and $A C D$ be tangent to the line $A C$ at $T$ and $T^{\prime}$, respectively. (See the figure to the left.) We have $A T=\frac{A B+A C-B C}{2}$ in triangle $A B C, A T^{\prime}=$ $\frac{A D+A C-C D}{2}$ in triangle $A C D$, and $A B-B C=A D-C D$ in quadrilateral $A B C D$, so $$ A T=\frac{A C+A B-B C}{2}=\frac{A C+A D-C D}{2}=A T^{\prime} $$ This shows $T=T^{\prime}$. As an immediate consequence, $I_{b} I_{d} \perp A C$. The second statement can be shown analogously.  Claim 2. The points $O_{a}, O_{b}, O_{c}$ and $O_{d}$ lie on the lines $A I, B I, C I$ and $D I$, respectively. Proof. By symmetry it suffices to prove the claim for $O_{a}$. (See the figure to the right above.) Notice first that the incircles of triangles $A B C$ and $A C D$ can be obtained from the incircle of the quadrilateral $A B C D$ with homothety centers $B$ and $D$, respectively, and homothety factors less than 1 , therefore the points $I_{b}$ and $I_{d}$ lie on the line segments $B I$ and $D I$, respectively. As is well-known, in every triangle the altitude and the diameter of the circumcircle starting from the same vertex are symmetric about the angle bisector. By Claim 1, in triangle $A I_{d} I_{b}$, the segment $A T$ is the altitude starting from $A$. Since the foot $T$ lies inside the segment $I_{b} I_{d}$, the circumcenter $O_{a}$ of triangle $A I_{d} I_{b}$ lies in the angle domain $I_{b} A I_{d}$ in such a way that $\angle I_{b} A T=\angle O_{a} A I_{d}$. The points $I_{b}$ and $I_{d}$ are the incenters of triangles $A B C$ and $A C D$, so the lines $A I_{b}$ and $A I_{d}$ bisect the angles $\angle B A C$ and $\angle C A D$, respectively. Then $$ \angle O_{a} A D=\angle O_{a} A I_{d}+\angle I_{d} A D=\angle I_{b} A T+\angle I_{d} A D=\frac{1}{2} \angle B A C+\frac{1}{2} \angle C A D=\frac{1}{2} \angle B A D, $$ so $O_{a}$ lies on the angle bisector of $\angle B A D$, that is, on line $A I$. The point $X$ is the external similitude center of $\omega_{a}$ and $\omega_{c}$; let $U$ be their internal similitude center. The points $O_{a}$ and $O_{c}$ lie on the perpendicular bisector of the common chord $I_{b} I_{d}$ of $\omega_{a}$ and $\omega_{c}$, and the two similitude centers $X$ and $U$ lie on the same line; by Claim 2, that line is parallel to $A C$.  From the similarity of the circles $\omega_{a}$ and $\omega_{c}$, from $O_{a} I_{b}=O_{a} I_{d}=O_{a} A=r_{a}$ and $O_{c} I_{b}=$ $O_{c} I_{d}=O_{c} C=r_{c}$, and from $A C \| O_{a} O_{c}$ we can see that $$ \frac{O_{a} X}{O_{c} X}=\frac{O_{a} U}{O_{c} U}=\frac{r_{a}}{r_{c}}=\frac{O_{a} I_{b}}{O_{c} I_{b}}=\frac{O_{a} I_{d}}{O_{c} I_{d}}=\frac{O_{a} A}{O_{c} C}=\frac{O_{a} I}{O_{c} I} $$ So the points $X, U, I_{b}, I_{d}, I$ lie on the Apollonius circle of the points $O_{a}, O_{c}$ with ratio $r_{a}: r_{c}$. In this Apollonius circle $X U$ is a diameter, and the lines $I U$ and $I X$ are respectively the internal and external bisectors of $\angle O_{a} I O_{c}=\angle A I C$, according to the angle bisector theorem. Moreover, in the Apollonius circle the diameter $U X$ is the perpendicular bisector of $I_{b} I_{d}$, so the lines $I X$ and $I U$ are the internal and external bisectors of $\angle I_{b} I I_{d}=\angle B I D$, respectively. Repeating the same argument for the points $B, D$ instead of $A, C$, we get that the line $I Y$ is the internal bisector of $\angle A I C$ and the external bisector of $\angle B I D$. Therefore, the lines $I X$ and $I Y$ respectively are the internal and external bisectors of $\angle B I D$, so they are perpendicular. Comment. In fact the points $O_{a}, O_{b}, O_{c}$ and $O_{d}$ lie on the line segments $A I, B I, C I$ and $D I$, respectively. For the point $O_{a}$ this can be shown for example by $\angle I_{d} O_{a} A+\angle A O_{a} I_{b}=\left(180^{\circ}-\right.$ $\left.2 \angle O_{a} A I_{d}\right)+\left(180^{\circ}-2 \angle I_{b} A O_{a}\right)=360^{\circ}-\angle B A D=\angle A D I+\angle D I A+\angle A I B+\angle I B A>\angle I_{d} I A+\angle A I I_{b}$. The solution also shows that the line $I Y$ passes through the point $U$, and analogously, $I X$ passes through the internal similitude center of $\omega_{b}$ and $\omega_{d}$.
|
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701c0cbe-2398-5597-b9f4-2a76ae35aec0
| 24,998
|
Let $p \geqslant 2$ be a prime number. Eduardo and Fernando play the following game making moves alternately: in each move, the current player chooses an index $i$ in the set $\{0,1, \ldots, p-1\}$ that was not chosen before by either of the two players and then chooses an element $a_{i}$ of the set $\{0,1,2,3,4,5,6,7,8,9\}$. Eduardo has the first move. The game ends after all the indices $i \in\{0,1, \ldots, p-1\}$ have been chosen. Then the following number is computed: $$ M=a_{0}+10 \cdot a_{1}+\cdots+10^{p-1} \cdot a_{p-1}=\sum_{j=0}^{p-1} a_{j} \cdot 10^{j} $$ The goal of Eduardo is to make the number $M$ divisible by $p$, and the goal of Fernando is to prevent this. Prove that Eduardo has a winning strategy. (Morocco)
|
We say that a player makes the move $\left(i, a_{i}\right)$ if he chooses the index $i$ and then the element $a_{i}$ of the set $\{0,1,2,3,4,5,6,7,8,9\}$ in this move. If $p=2$ or $p=5$ then Eduardo chooses $i=0$ and $a_{0}=0$ in the first move, and wins, since, independently of the next moves, $M$ will be a multiple of 10 . Now assume that the prime number $p$ does not belong to $\{2,5\}$. Eduardo chooses $i=p-1$ and $a_{p-1}=0$ in the first move. By Fermat's Little Theorem, $\left(10^{(p-1) / 2}\right)^{2}=10^{p-1} \equiv 1(\bmod p)$, so $p \mid\left(10^{(p-1) / 2}\right)^{2}-1=\left(10^{(p-1) / 2}+1\right)\left(10^{(p-1) / 2}-1\right)$. Since $p$ is prime, either $p \mid 10^{(p-1) / 2}+1$ or $p \mid 10^{(p-1) / 2}-1$. Thus we have two cases: Case a: $10^{(p-1) / 2} \equiv-1(\bmod p)$ In this case, for each move $\left(i, a_{i}\right)$ of Fernando, Eduardo immediately makes the move $\left(j, a_{j}\right)=$ $\left(i+\frac{p-1}{2}, a_{i}\right)$, if $0 \leqslant i \leqslant \frac{p-3}{2}$, or $\left(j, a_{j}\right)=\left(i-\frac{p-1}{2}, a_{i}\right)$, if $\frac{p-1}{2} \leqslant i \leqslant p-2$. We will have $10^{j} \equiv-10^{i}$ $(\bmod p)$, and so $a_{j} \cdot 10^{j}=a_{i} \cdot 10^{j} \equiv-a_{i} \cdot 10^{i}(\bmod p)$. Notice that this move by Eduardo is always possible. Indeed, immediately before a move by Fernando, for any set of the type $\{r, r+(p-1) / 2\}$ with $0 \leqslant r \leqslant(p-3) / 2$, either no element of this set was chosen as an index by the players in the previous moves or else both elements of this set were chosen as indices by the players in the previous moves. Therefore, after each of his moves, Eduardo always makes the sum of the numbers $a_{k} \cdot 10^{k}$ corresponding to the already chosen pairs $\left(k, a_{k}\right)$ divisible by $p$, and thus wins the game. Case b: $10^{(p-1) / 2} \equiv 1(\bmod p)$ In this case, for each move $\left(i, a_{i}\right)$ of Fernando, Eduardo immediately makes the move $\left(j, a_{j}\right)=$ $\left(i+\frac{p-1}{2}, 9-a_{i}\right)$, if $0 \leqslant i \leqslant \frac{p-3}{2}$, or $\left(j, a_{j}\right)=\left(i-\frac{p-1}{2}, 9-a_{i}\right)$, if $\frac{p-1}{2} \leqslant i \leqslant p-2$. The same argument as above shows that Eduardo can always make such move. We will have $10^{j} \equiv 10^{i}$ $(\bmod p)$, and so $a_{j} \cdot 10^{j}+a_{i} \cdot 10^{i} \equiv\left(a_{i}+a_{j}\right) \cdot 10^{i}=9 \cdot 10^{i}(\bmod p)$. Therefore, at the end of the game, the sum of all terms $a_{k} \cdot 10^{k}$ will be congruent to $$ \sum_{i=0}^{\frac{p-3}{2}} 9 \cdot 10^{i}=10^{(p-1) / 2}-1 \equiv 0 \quad(\bmod p) $$ and Eduardo wins the game.
|
proof
|
Yes
|
Yes
|
proof
|
Number Theory
|
Let $p \geqslant 2$ be a prime number. Eduardo and Fernando play the following game making moves alternately: in each move, the current player chooses an index $i$ in the set $\{0,1, \ldots, p-1\}$ that was not chosen before by either of the two players and then chooses an element $a_{i}$ of the set $\{0,1,2,3,4,5,6,7,8,9\}$. Eduardo has the first move. The game ends after all the indices $i \in\{0,1, \ldots, p-1\}$ have been chosen. Then the following number is computed: $$ M=a_{0}+10 \cdot a_{1}+\cdots+10^{p-1} \cdot a_{p-1}=\sum_{j=0}^{p-1} a_{j} \cdot 10^{j} $$ The goal of Eduardo is to make the number $M$ divisible by $p$, and the goal of Fernando is to prevent this. Prove that Eduardo has a winning strategy. (Morocco)
|
We say that a player makes the move $\left(i, a_{i}\right)$ if he chooses the index $i$ and then the element $a_{i}$ of the set $\{0,1,2,3,4,5,6,7,8,9\}$ in this move. If $p=2$ or $p=5$ then Eduardo chooses $i=0$ and $a_{0}=0$ in the first move, and wins, since, independently of the next moves, $M$ will be a multiple of 10 . Now assume that the prime number $p$ does not belong to $\{2,5\}$. Eduardo chooses $i=p-1$ and $a_{p-1}=0$ in the first move. By Fermat's Little Theorem, $\left(10^{(p-1) / 2}\right)^{2}=10^{p-1} \equiv 1(\bmod p)$, so $p \mid\left(10^{(p-1) / 2}\right)^{2}-1=\left(10^{(p-1) / 2}+1\right)\left(10^{(p-1) / 2}-1\right)$. Since $p$ is prime, either $p \mid 10^{(p-1) / 2}+1$ or $p \mid 10^{(p-1) / 2}-1$. Thus we have two cases: Case a: $10^{(p-1) / 2} \equiv-1(\bmod p)$ In this case, for each move $\left(i, a_{i}\right)$ of Fernando, Eduardo immediately makes the move $\left(j, a_{j}\right)=$ $\left(i+\frac{p-1}{2}, a_{i}\right)$, if $0 \leqslant i \leqslant \frac{p-3}{2}$, or $\left(j, a_{j}\right)=\left(i-\frac{p-1}{2}, a_{i}\right)$, if $\frac{p-1}{2} \leqslant i \leqslant p-2$. We will have $10^{j} \equiv-10^{i}$ $(\bmod p)$, and so $a_{j} \cdot 10^{j}=a_{i} \cdot 10^{j} \equiv-a_{i} \cdot 10^{i}(\bmod p)$. Notice that this move by Eduardo is always possible. Indeed, immediately before a move by Fernando, for any set of the type $\{r, r+(p-1) / 2\}$ with $0 \leqslant r \leqslant(p-3) / 2$, either no element of this set was chosen as an index by the players in the previous moves or else both elements of this set were chosen as indices by the players in the previous moves. Therefore, after each of his moves, Eduardo always makes the sum of the numbers $a_{k} \cdot 10^{k}$ corresponding to the already chosen pairs $\left(k, a_{k}\right)$ divisible by $p$, and thus wins the game. Case b: $10^{(p-1) / 2} \equiv 1(\bmod p)$ In this case, for each move $\left(i, a_{i}\right)$ of Fernando, Eduardo immediately makes the move $\left(j, a_{j}\right)=$ $\left(i+\frac{p-1}{2}, 9-a_{i}\right)$, if $0 \leqslant i \leqslant \frac{p-3}{2}$, or $\left(j, a_{j}\right)=\left(i-\frac{p-1}{2}, 9-a_{i}\right)$, if $\frac{p-1}{2} \leqslant i \leqslant p-2$. The same argument as above shows that Eduardo can always make such move. We will have $10^{j} \equiv 10^{i}$ $(\bmod p)$, and so $a_{j} \cdot 10^{j}+a_{i} \cdot 10^{i} \equiv\left(a_{i}+a_{j}\right) \cdot 10^{i}=9 \cdot 10^{i}(\bmod p)$. Therefore, at the end of the game, the sum of all terms $a_{k} \cdot 10^{k}$ will be congruent to $$ \sum_{i=0}^{\frac{p-3}{2}} 9 \cdot 10^{i}=10^{(p-1) / 2}-1 \equiv 0 \quad(\bmod p) $$ and Eduardo wins the game.
|
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8fda95e0-f10c-535f-afe9-d6af7e5aef3f
| 25,008
|
Determine all integers $n \geqslant 2$ with the following property: for any integers $a_{1}, a_{2}, \ldots, a_{n}$ whose sum is not divisible by $n$, there exists an index $1 \leqslant i \leqslant n$ such that none of the numbers $$ a_{i}, a_{i}+a_{i+1}, \ldots, a_{i}+a_{i+1}+\cdots+a_{i+n-1} $$ is divisible by $n$. (We let $a_{i}=a_{i-n}$ when $i>n$.) (Thailand)
|
Let us first show that, if $n=a b$, with $a, b \geqslant 2$ integers, then the property in the statement of the problem does not hold. Indeed, in this case, let $a_{k}=a$ for $1 \leqslant k \leqslant n-1$ and $a_{n}=0$. The sum $a_{1}+a_{2}+\cdots+a_{n}=a \cdot(n-1)$ is not divisible by $n$. Let $i$ with $1 \leqslant i \leqslant n$ be an arbitrary index. Taking $j=b$ if $1 \leqslant i \leqslant n-b$, and $j=b+1$ if $n-b<i \leqslant n$, we have $$ a_{i}+a_{i+1}+\cdots+a_{i+j-1}=a \cdot b=n \equiv 0 \quad(\bmod n) $$ It follows that the given example is indeed a counterexample to the property of the statement. Now let $n$ be a prime number. Suppose by contradiction that the property in the statement of the problem does not hold. Then there are integers $a_{1}, a_{2}, \ldots, a_{n}$ whose sum is not divisible by $n$ such that for each $i, 1 \leqslant i \leqslant n$, there is $j, 1 \leqslant j \leqslant n$, for which the number $a_{i}+a_{i+1}+$ $\cdots+a_{i+j-1}$ is divisible by $n$. Notice that, in any such case, we should have $1 \leqslant j \leqslant n-1$, since $a_{1}+a_{2}+\cdots+a_{n}$ is not divisible by $n$. So we may construct recursively a finite sequence of integers $0=i_{0}<i_{1}<i_{2}<\cdots<i_{n}$ with $i_{s+1}-i_{s} \leqslant n-1$ for $0 \leqslant s \leqslant n-1$ such that, for $0 \leqslant s \leqslant n-1$, $$ a_{i_{s}+1}+a_{i_{s}+2}+\cdots+a_{i_{s+1}} \equiv 0 \quad(\bmod n) $$ (where we take indices modulo $n$ ). Indeed, for $0 \leqslant s<n$, we apply the previous observation to $i=i_{s}+1$ in order to define $i_{s+1}=i_{s}+j$. In the sequence of $n+1$ indices $i_{0}, i_{1}, i_{2}, \ldots, i_{n}$, by the pigeonhole principle, we have two distinct elements which are congruent modulo $n$. So there are indices $r, s$ with $0 \leqslant r<s \leqslant n$ such that $i_{s} \equiv i_{r}(\bmod n)$ and $$ a_{i_{r}+1}+a_{i_{r}+2}+\cdots+a_{i_{s}}=\sum_{j=r}^{s-1}\left(a_{i_{j}+1}+a_{i_{j}+2}+\cdots+a_{i_{j+1}}\right) \equiv 0 \quad(\bmod n) $$ Since $i_{s} \equiv i_{r}(\bmod n)$, we have $i_{s}-i_{r}=k \cdot n$ for some positive integer $k$, and, since $i_{j+1}-i_{j} \leqslant$ $n-1$ for $0 \leqslant j \leqslant n-1$, we have $i_{s}-i_{r} \leqslant(n-1) \cdot n$, so $k \leqslant n-1$. But in this case $$ a_{i_{r}+1}+a_{i_{r}+2}+\cdots+a_{i_{s}}=k \cdot\left(a_{1}+a_{2}+\cdots+a_{n}\right) $$ cannot be a multiple of $n$, since $n$ is prime and neither $k$ nor $a_{1}+a_{2}+\cdots+a_{n}$ is a multiple of $n$. A contradiction.
|
proof
|
Yes
|
Yes
|
proof
|
Number Theory
|
Determine all integers $n \geqslant 2$ with the following property: for any integers $a_{1}, a_{2}, \ldots, a_{n}$ whose sum is not divisible by $n$, there exists an index $1 \leqslant i \leqslant n$ such that none of the numbers $$ a_{i}, a_{i}+a_{i+1}, \ldots, a_{i}+a_{i+1}+\cdots+a_{i+n-1} $$ is divisible by $n$. (We let $a_{i}=a_{i-n}$ when $i>n$.) (Thailand)
|
Let us first show that, if $n=a b$, with $a, b \geqslant 2$ integers, then the property in the statement of the problem does not hold. Indeed, in this case, let $a_{k}=a$ for $1 \leqslant k \leqslant n-1$ and $a_{n}=0$. The sum $a_{1}+a_{2}+\cdots+a_{n}=a \cdot(n-1)$ is not divisible by $n$. Let $i$ with $1 \leqslant i \leqslant n$ be an arbitrary index. Taking $j=b$ if $1 \leqslant i \leqslant n-b$, and $j=b+1$ if $n-b<i \leqslant n$, we have $$ a_{i}+a_{i+1}+\cdots+a_{i+j-1}=a \cdot b=n \equiv 0 \quad(\bmod n) $$ It follows that the given example is indeed a counterexample to the property of the statement. Now let $n$ be a prime number. Suppose by contradiction that the property in the statement of the problem does not hold. Then there are integers $a_{1}, a_{2}, \ldots, a_{n}$ whose sum is not divisible by $n$ such that for each $i, 1 \leqslant i \leqslant n$, there is $j, 1 \leqslant j \leqslant n$, for which the number $a_{i}+a_{i+1}+$ $\cdots+a_{i+j-1}$ is divisible by $n$. Notice that, in any such case, we should have $1 \leqslant j \leqslant n-1$, since $a_{1}+a_{2}+\cdots+a_{n}$ is not divisible by $n$. So we may construct recursively a finite sequence of integers $0=i_{0}<i_{1}<i_{2}<\cdots<i_{n}$ with $i_{s+1}-i_{s} \leqslant n-1$ for $0 \leqslant s \leqslant n-1$ such that, for $0 \leqslant s \leqslant n-1$, $$ a_{i_{s}+1}+a_{i_{s}+2}+\cdots+a_{i_{s+1}} \equiv 0 \quad(\bmod n) $$ (where we take indices modulo $n$ ). Indeed, for $0 \leqslant s<n$, we apply the previous observation to $i=i_{s}+1$ in order to define $i_{s+1}=i_{s}+j$. In the sequence of $n+1$ indices $i_{0}, i_{1}, i_{2}, \ldots, i_{n}$, by the pigeonhole principle, we have two distinct elements which are congruent modulo $n$. So there are indices $r, s$ with $0 \leqslant r<s \leqslant n$ such that $i_{s} \equiv i_{r}(\bmod n)$ and $$ a_{i_{r}+1}+a_{i_{r}+2}+\cdots+a_{i_{s}}=\sum_{j=r}^{s-1}\left(a_{i_{j}+1}+a_{i_{j}+2}+\cdots+a_{i_{j+1}}\right) \equiv 0 \quad(\bmod n) $$ Since $i_{s} \equiv i_{r}(\bmod n)$, we have $i_{s}-i_{r}=k \cdot n$ for some positive integer $k$, and, since $i_{j+1}-i_{j} \leqslant$ $n-1$ for $0 \leqslant j \leqslant n-1$, we have $i_{s}-i_{r} \leqslant(n-1) \cdot n$, so $k \leqslant n-1$. But in this case $$ a_{i_{r}+1}+a_{i_{r}+2}+\cdots+a_{i_{s}}=k \cdot\left(a_{1}+a_{2}+\cdots+a_{n}\right) $$ cannot be a multiple of $n$, since $n$ is prime and neither $k$ nor $a_{1}+a_{2}+\cdots+a_{n}$ is a multiple of $n$. A contradiction.
|
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5832c527-6e6d-5e6f-8b5e-0da3138c3e9c
| 25,011
|
Say that an ordered pair $(x, y)$ of integers is an irreducible lattice point if $x$ and $y$ are relatively prime. For any finite set $S$ of irreducible lattice points, show that there is a homogenous polynomial in two variables, $f(x, y)$, with integer coefficients, of degree at least 1 , such that $f(x, y)=1$ for each $(x, y)$ in the set $S$. Note: A homogenous polynomial of degree $n$ is any nonzero polynomial of the form $$ f(x, y)=a_{0} x^{n}+a_{1} x^{n-1} y+a_{2} x^{n-2} y^{2}+\cdots+a_{n-1} x y^{n-1}+a_{n} y^{n} . $$
|
First of all, we note that finding a homogenous polynomial $f(x, y)$ such that $f(x, y)= \pm 1$ is enough, because we then have $f^{2}(x, y)=1$. Label the irreducible lattice points $\left(x_{1}, y_{1}\right)$ through $\left(x_{n}, y_{n}\right)$. If any two of these lattice points $\left(x_{i}, y_{i}\right)$ and $\left(x_{j}, y_{j}\right)$ lie on the same line through the origin, then $\left(x_{j}, y_{j}\right)=\left(-x_{i},-y_{i}\right)$ because both of the points are irreducible. We then have $f\left(x_{j}, y_{j}\right)= \pm f\left(x_{i}, y_{i}\right)$ whenever $f$ is homogenous, so we can assume that no two of the lattice points are collinear with the origin by ignoring the extra lattice points. Consider the homogenous polynomials $\ell_{i}(x, y)=y_{i} x-x_{i} y$ and define $$ g_{i}(x, y)=\prod_{j \neq i} \ell_{j}(x, y) $$ Then $\ell_{i}\left(x_{j}, y_{j}\right)=0$ if and only if $j=i$, because there is only one lattice point on each line through the origin. Thus, $g_{i}\left(x_{j}, y_{j}\right)=0$ for all $j \neq i$. Define $a_{i}=g_{i}\left(x_{i}, y_{i}\right)$, and note that $a_{i} \neq 0$. Note that $g_{i}(x, y)$ is a degree $n-1$ polynomial with the following two properties: 1. $g_{i}\left(x_{j}, y_{j}\right)=0$ if $j \neq i$. 2. $g_{i}\left(x_{i}, y_{i}\right)=a_{i}$. For any $N \geqslant n-1$, there also exists a polynomial of degree $N$ with the same two properties. Specifically, let $I_{i}(x, y)$ be a degree 1 homogenous polynomial such that $I_{i}\left(x_{i}, y_{i}\right)=1$, which exists since $\left(x_{i}, y_{i}\right)$ is irreducible. Then $I_{i}(x, y)^{N-(n-1)} g_{i}(x, y)$ satisfies both of the above properties and has degree $N$. We may now reduce the problem to the following claim: Claim: For each positive integer $a$, there is a homogenous polynomial $f_{a}(x, y)$, with integer coefficients, of degree at least 1 , such that $f_{a}(x, y) \equiv 1(\bmod a)$ for all relatively prime $(x, y)$. To see that this claim solves the problem, take $a$ to be the least common multiple of the numbers $a_{i}(1 \leqslant i \leqslant n)$. Take $f_{a}$ given by the claim, choose some power $f_{a}(x, y)^{k}$ that has degree at least $n-1$, and subtract appropriate multiples of the $g_{i}$ constructed above to obtain the desired polynomial. We prove the claim by factoring $a$. First, if $a$ is a power of a prime ( $a=p^{k}$ ), then we may choose either: - $f_{a}(x, y)=\left(x^{p-1}+y^{p-1}\right)^{\phi(a)}$ if $p$ is odd; - $f_{a}(x, y)=\left(x^{2}+x y+y^{2}\right)^{\phi(a)}$ if $p=2$. Now suppose $a$ is any positive integer, and let $a=q_{1} q_{2} \cdots q_{k}$, where the $q_{i}$ are prime powers, pairwise relatively prime. Let $f_{q_{i}}$ be the polynomials just constructed, and let $F_{q_{i}}$ be powers of these that all have the same degree. Note that $$ \frac{a}{q_{i}} F_{q_{i}}(x, y) \equiv \frac{a}{q_{i}} \quad(\bmod a) $$ for any relatively prime $x, y$. By Bézout's lemma, there is an integer linear combination of the $\frac{a}{q_{i}}$ that equals 1 . Thus, there is a linear combination of the $F_{q_{i}}$ such that $F_{q_{i}}(x, y) \equiv 1$ $(\bmod a)$ for any relatively prime $(x, y)$; and this polynomial is homogenous because all the $F_{q_{i}}$ have the same degree.
|
proof
|
Yes
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Yes
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proof
|
Number Theory
|
Say that an ordered pair $(x, y)$ of integers is an irreducible lattice point if $x$ and $y$ are relatively prime. For any finite set $S$ of irreducible lattice points, show that there is a homogenous polynomial in two variables, $f(x, y)$, with integer coefficients, of degree at least 1 , such that $f(x, y)=1$ for each $(x, y)$ in the set $S$. Note: A homogenous polynomial of degree $n$ is any nonzero polynomial of the form $$ f(x, y)=a_{0} x^{n}+a_{1} x^{n-1} y+a_{2} x^{n-2} y^{2}+\cdots+a_{n-1} x y^{n-1}+a_{n} y^{n} . $$
|
First of all, we note that finding a homogenous polynomial $f(x, y)$ such that $f(x, y)= \pm 1$ is enough, because we then have $f^{2}(x, y)=1$. Label the irreducible lattice points $\left(x_{1}, y_{1}\right)$ through $\left(x_{n}, y_{n}\right)$. If any two of these lattice points $\left(x_{i}, y_{i}\right)$ and $\left(x_{j}, y_{j}\right)$ lie on the same line through the origin, then $\left(x_{j}, y_{j}\right)=\left(-x_{i},-y_{i}\right)$ because both of the points are irreducible. We then have $f\left(x_{j}, y_{j}\right)= \pm f\left(x_{i}, y_{i}\right)$ whenever $f$ is homogenous, so we can assume that no two of the lattice points are collinear with the origin by ignoring the extra lattice points. Consider the homogenous polynomials $\ell_{i}(x, y)=y_{i} x-x_{i} y$ and define $$ g_{i}(x, y)=\prod_{j \neq i} \ell_{j}(x, y) $$ Then $\ell_{i}\left(x_{j}, y_{j}\right)=0$ if and only if $j=i$, because there is only one lattice point on each line through the origin. Thus, $g_{i}\left(x_{j}, y_{j}\right)=0$ for all $j \neq i$. Define $a_{i}=g_{i}\left(x_{i}, y_{i}\right)$, and note that $a_{i} \neq 0$. Note that $g_{i}(x, y)$ is a degree $n-1$ polynomial with the following two properties: 1. $g_{i}\left(x_{j}, y_{j}\right)=0$ if $j \neq i$. 2. $g_{i}\left(x_{i}, y_{i}\right)=a_{i}$. For any $N \geqslant n-1$, there also exists a polynomial of degree $N$ with the same two properties. Specifically, let $I_{i}(x, y)$ be a degree 1 homogenous polynomial such that $I_{i}\left(x_{i}, y_{i}\right)=1$, which exists since $\left(x_{i}, y_{i}\right)$ is irreducible. Then $I_{i}(x, y)^{N-(n-1)} g_{i}(x, y)$ satisfies both of the above properties and has degree $N$. We may now reduce the problem to the following claim: Claim: For each positive integer $a$, there is a homogenous polynomial $f_{a}(x, y)$, with integer coefficients, of degree at least 1 , such that $f_{a}(x, y) \equiv 1(\bmod a)$ for all relatively prime $(x, y)$. To see that this claim solves the problem, take $a$ to be the least common multiple of the numbers $a_{i}(1 \leqslant i \leqslant n)$. Take $f_{a}$ given by the claim, choose some power $f_{a}(x, y)^{k}$ that has degree at least $n-1$, and subtract appropriate multiples of the $g_{i}$ constructed above to obtain the desired polynomial. We prove the claim by factoring $a$. First, if $a$ is a power of a prime ( $a=p^{k}$ ), then we may choose either: - $f_{a}(x, y)=\left(x^{p-1}+y^{p-1}\right)^{\phi(a)}$ if $p$ is odd; - $f_{a}(x, y)=\left(x^{2}+x y+y^{2}\right)^{\phi(a)}$ if $p=2$. Now suppose $a$ is any positive integer, and let $a=q_{1} q_{2} \cdots q_{k}$, where the $q_{i}$ are prime powers, pairwise relatively prime. Let $f_{q_{i}}$ be the polynomials just constructed, and let $F_{q_{i}}$ be powers of these that all have the same degree. Note that $$ \frac{a}{q_{i}} F_{q_{i}}(x, y) \equiv \frac{a}{q_{i}} \quad(\bmod a) $$ for any relatively prime $x, y$. By Bézout's lemma, there is an integer linear combination of the $\frac{a}{q_{i}}$ that equals 1 . Thus, there is a linear combination of the $F_{q_{i}}$ such that $F_{q_{i}}(x, y) \equiv 1$ $(\bmod a)$ for any relatively prime $(x, y)$; and this polynomial is homogenous because all the $F_{q_{i}}$ have the same degree.
|
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7a1f5a3e-0408-59d4-86e8-7979b55d4ec1
| 25,024
|
A sequence of real numbers $a_{1}, a_{2}, \ldots$ satisfies the relation $$ a_{n}=-\max _{i+j=n}\left(a_{i}+a_{j}\right) \quad \text { for all } n>2017 $$ Prove that this sequence is bounded, i.e., there is a constant $M$ such that $\left|a_{n}\right| \leqslant M$ for all positive integers $n$. (Russia)
|
Set $D=2017$. Denote $$ M_{n}=\max _{k<n} a_{k} \quad \text { and } \quad m_{n}=-\min _{k<n} a_{k}=\max _{k<n}\left(-a_{k}\right) . $$ Clearly, the sequences $\left(m_{n}\right)$ and $\left(M_{n}\right)$ are nondecreasing. We need to prove that both are bounded. Consider an arbitrary $n>D$; our first aim is to bound $a_{n}$ in terms of $m_{n}$ and $M_{n}$. (i) There exist indices $p$ and $q$ such that $a_{n}=-\left(a_{p}+a_{q}\right)$ and $p+q=n$. Since $a_{p}, a_{q} \leqslant M_{n}$, we have $a_{n} \geqslant-2 M_{n}$. (ii) On the other hand, choose an index $k<n$ such that $a_{k}=M_{n}$. Then, we have $$ a_{n}=-\max _{\ell<n}\left(a_{n-\ell}+a_{\ell}\right) \leqslant-\left(a_{n-k}+a_{k}\right)=-a_{n-k}-M_{n} \leqslant m_{n}-M_{n} $$ Summarizing (i) and (ii), we get $$ -2 M_{n} \leqslant a_{n} \leqslant m_{n}-M_{n}, $$ whence $$ m_{n} \leqslant m_{n+1} \leqslant \max \left\{m_{n}, 2 M_{n}\right\} \quad \text { and } \quad M_{n} \leqslant M_{n+1} \leqslant \max \left\{M_{n}, m_{n}-M_{n}\right\} $$ Now, say that an index $n>D$ is lucky if $m_{n} \leqslant 2 M_{n}$. Two cases are possible. Case 1. Assume that there exists a lucky index $n$. In this case, (1) yields $m_{n+1} \leqslant 2 M_{n}$ and $M_{n} \leqslant M_{n+1} \leqslant M_{n}$. Therefore, $M_{n+1}=M_{n}$ and $m_{n+1} \leqslant 2 M_{n}=2 M_{n+1}$. So, the index $n+1$ is also lucky, and $M_{n+1}=M_{n}$. Applying the same arguments repeatedly, we obtain that all indices $k>n$ are lucky (i.e., $m_{k} \leqslant 2 M_{k}$ for all these indices), and $M_{k}=M_{n}$ for all such indices. Thus, all of the $m_{k}$ and $M_{k}$ are bounded by $2 M_{n}$. Case 2. Assume now that there is no lucky index, i.e., $2 M_{n}<m_{n}$ for all $n>D$. Then (1) shows that for all $n>D$ we have $m_{n} \leqslant m_{n+1} \leqslant m_{n}$, so $m_{n}=m_{D+1}$ for all $n>D$. Since $M_{n}<m_{n} / 2$ for all such indices, all of the $m_{n}$ and $M_{n}$ are bounded by $m_{D+1}$. Thus, in both cases the sequences $\left(m_{n}\right)$ and $\left(M_{n}\right)$ are bounded, as desired.
|
proof
|
Yes
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Yes
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proof
|
Algebra
|
A sequence of real numbers $a_{1}, a_{2}, \ldots$ satisfies the relation $$ a_{n}=-\max _{i+j=n}\left(a_{i}+a_{j}\right) \quad \text { for all } n>2017 $$ Prove that this sequence is bounded, i.e., there is a constant $M$ such that $\left|a_{n}\right| \leqslant M$ for all positive integers $n$. (Russia)
|
Set $D=2017$. Denote $$ M_{n}=\max _{k<n} a_{k} \quad \text { and } \quad m_{n}=-\min _{k<n} a_{k}=\max _{k<n}\left(-a_{k}\right) . $$ Clearly, the sequences $\left(m_{n}\right)$ and $\left(M_{n}\right)$ are nondecreasing. We need to prove that both are bounded. Consider an arbitrary $n>D$; our first aim is to bound $a_{n}$ in terms of $m_{n}$ and $M_{n}$. (i) There exist indices $p$ and $q$ such that $a_{n}=-\left(a_{p}+a_{q}\right)$ and $p+q=n$. Since $a_{p}, a_{q} \leqslant M_{n}$, we have $a_{n} \geqslant-2 M_{n}$. (ii) On the other hand, choose an index $k<n$ such that $a_{k}=M_{n}$. Then, we have $$ a_{n}=-\max _{\ell<n}\left(a_{n-\ell}+a_{\ell}\right) \leqslant-\left(a_{n-k}+a_{k}\right)=-a_{n-k}-M_{n} \leqslant m_{n}-M_{n} $$ Summarizing (i) and (ii), we get $$ -2 M_{n} \leqslant a_{n} \leqslant m_{n}-M_{n}, $$ whence $$ m_{n} \leqslant m_{n+1} \leqslant \max \left\{m_{n}, 2 M_{n}\right\} \quad \text { and } \quad M_{n} \leqslant M_{n+1} \leqslant \max \left\{M_{n}, m_{n}-M_{n}\right\} $$ Now, say that an index $n>D$ is lucky if $m_{n} \leqslant 2 M_{n}$. Two cases are possible. Case 1. Assume that there exists a lucky index $n$. In this case, (1) yields $m_{n+1} \leqslant 2 M_{n}$ and $M_{n} \leqslant M_{n+1} \leqslant M_{n}$. Therefore, $M_{n+1}=M_{n}$ and $m_{n+1} \leqslant 2 M_{n}=2 M_{n+1}$. So, the index $n+1$ is also lucky, and $M_{n+1}=M_{n}$. Applying the same arguments repeatedly, we obtain that all indices $k>n$ are lucky (i.e., $m_{k} \leqslant 2 M_{k}$ for all these indices), and $M_{k}=M_{n}$ for all such indices. Thus, all of the $m_{k}$ and $M_{k}$ are bounded by $2 M_{n}$. Case 2. Assume now that there is no lucky index, i.e., $2 M_{n}<m_{n}$ for all $n>D$. Then (1) shows that for all $n>D$ we have $m_{n} \leqslant m_{n+1} \leqslant m_{n}$, so $m_{n}=m_{D+1}$ for all $n>D$. Since $M_{n}<m_{n} / 2$ for all such indices, all of the $m_{n}$ and $M_{n}$ are bounded by $m_{D+1}$. Thus, in both cases the sequences $\left(m_{n}\right)$ and $\left(M_{n}\right)$ are bounded, as desired.
|
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619e7e11-2bc1-59fc-a84b-cbc92d092671
| 25,036
|
For any finite sets $X$ and $Y$ of positive integers, denote by $f_{X}(k)$ the $k^{\mathrm{th}}$ smallest positive integer not in $X$, and let $$ X * Y=X \cup\left\{f_{X}(y): y \in Y\right\} $$ Let $A$ be a set of $a>0$ positive integers, and let $B$ be a set of $b>0$ positive integers. Prove that if $A * B=B * A$, then $$ \underbrace{A *(A * \cdots *(A *(A * A)) \ldots)}_{A \text { appears } b \text { times }}=\underbrace{B *(B * \cdots *(B *(B * B)) \ldots)}_{B \text { appears } a \text { times }} . $$
|
For any function $g: \mathbb{Z}_{>0} \rightarrow \mathbb{Z}_{>0}$ and any subset $X \subset \mathbb{Z}_{>0}$, we define $g(X)=$ $\{g(x): x \in X\}$. We have that the image of $f_{X}$ is $f_{X}\left(\mathbb{Z}_{>0}\right)=\mathbb{Z}_{>0} \backslash X$. We now show a general lemma about the operation *, with the goal of showing that $*$ is associative. Lemma 1. Let $X$ and $Y$ be finite sets of positive integers. The functions $f_{X * Y}$ and $f_{X} \circ f_{Y}$ are equal. Proof. We have $f_{X * Y}\left(\mathbb{Z}_{>0}\right)=\mathbb{Z}_{>0} \backslash(X * Y)=\left(\mathbb{Z}_{>0} \backslash X\right) \backslash f_{X}(Y)=f_{X}\left(\mathbb{Z}_{>0}\right) \backslash f_{X}(Y)=f_{X}\left(\mathbb{Z}_{>0} \backslash Y\right)=f_{X}\left(f_{Y}\left(\mathbb{Z}_{>0}\right)\right)$. Thus, the functions $f_{X * Y}$ and $f_{X} \circ f_{Y}$ are strictly increasing functions with the same range. Because a strictly function is uniquely defined by its range, we have $f_{X * Y}=f_{X} \circ f_{Y}$. Lemma 1 implies that * is associative, in the sense that $(A * B) * C=A *(B * C)$ for any finite sets $A, B$, and $C$ of positive integers. We prove the associativity by noting $$ \begin{gathered} \mathbb{Z}_{>0} \backslash((A * B) * C)=f_{(A * B) * C}\left(\mathbb{Z}_{>0}\right)=f_{A * B}\left(f_{C}\left(\mathbb{Z}_{>0}\right)\right)=f_{A}\left(f_{B}\left(f_{C}\left(\mathbb{Z}_{>0}\right)\right)\right) \\ =f_{A}\left(f_{B * C}\left(\mathbb{Z}_{>0}\right)=f_{A *(B * C)}\left(\mathbb{Z}_{>0}\right)=\mathbb{Z}_{>0} \backslash(A *(B * C))\right. \end{gathered} $$ In light of the associativity of *, we may drop the parentheses when we write expressions like $A *(B * C)$. We also introduce the notation $$ X^{* k}=\underbrace{X *(X * \cdots *(X *(X * X)) \ldots)}_{X \text { appears } k \text { times }} $$ Our goal is then to show that $A * B=B * A$ implies $A^{* b}=B^{* a}$. We will do so via the following general lemma. Lemma 2. Suppose that $X$ and $Y$ are finite sets of positive integers satisfying $X * Y=Y * X$ and $|X|=|Y|$. Then, we must have $X=Y$. Proof. Assume that $X$ and $Y$ are not equal. Let $s$ be the largest number in exactly one of $X$ and $Y$. Without loss of generality, say that $s \in X \backslash Y$. The number $f_{X}(s)$ counts the $s^{t h}$ number not in $X$, which implies that $$ f_{X}(s)=s+\left|X \cap\left\{1,2, \ldots, f_{X}(s)\right\}\right| $$ Since $f_{X}(s) \geqslant s$, we have that $$ \left\{f_{X}(s)+1, f_{X}(s)+2, \ldots\right\} \cap X=\left\{f_{X}(s)+1, f_{X}(s)+2, \ldots\right\} \cap Y $$ which, together with the assumption that $|X|=|Y|$, gives $$ \left|X \cap\left\{1,2, \ldots, f_{X}(s)\right\}\right|=\left|Y \cap\left\{1,2, \ldots, f_{X}(s)\right\}\right| $$ Now consider the equation $$ t-|Y \cap\{1,2, \ldots, t\}|=s $$ This equation is satisfied only when $t \in\left[f_{Y}(s), f_{Y}(s+1)\right)$, because the left hand side counts the number of elements up to $t$ that are not in $Y$. We have that the value $t=f_{X}(s)$ satisfies the above equation because of (1) and (2). Furthermore, since $f_{X}(s) \notin X$ and $f_{X}(s) \geqslant s$, we have that $f_{X}(s) \notin Y$ due to the maximality of $s$. Thus, by the above discussion, we must have $f_{X}(s)=f_{Y}(s)$. Finally, we arrive at a contradiction. The value $f_{X}(s)$ is neither in $X$ nor in $f_{X}(Y)$, because $s$ is not in $Y$ by assumption. Thus, $f_{X}(s) \notin X * Y$. However, since $s \in X$, we have $f_{Y}(s) \in Y * X$, a contradiction. We are now ready to finish the proof. Note first of all that $\left|A^{* b}\right|=a b=\left|B^{* a}\right|$. Moreover, since $A * B=B * A$, and $*$ is associative, it follows that $A^{* b} * B^{* a}=B^{* a} * A^{* b}$. Thus, by Lemma 2, we have $A^{* b}=B^{* a}$, as desired. Comment 1. Taking $A=X^{* k}$ and $B=X^{* l}$ generates many non-trivial examples where $A * B=B * A$. There are also other examples not of this form. For example, if $A=\{1,2,4\}$ and $B=\{1,3\}$, then $A * B=\{1,2,3,4,6\}=B * A$.
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proof
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Yes
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Yes
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proof
|
Combinatorics
|
For any finite sets $X$ and $Y$ of positive integers, denote by $f_{X}(k)$ the $k^{\mathrm{th}}$ smallest positive integer not in $X$, and let $$ X * Y=X \cup\left\{f_{X}(y): y \in Y\right\} $$ Let $A$ be a set of $a>0$ positive integers, and let $B$ be a set of $b>0$ positive integers. Prove that if $A * B=B * A$, then $$ \underbrace{A *(A * \cdots *(A *(A * A)) \ldots)}_{A \text { appears } b \text { times }}=\underbrace{B *(B * \cdots *(B *(B * B)) \ldots)}_{B \text { appears } a \text { times }} . $$
|
For any function $g: \mathbb{Z}_{>0} \rightarrow \mathbb{Z}_{>0}$ and any subset $X \subset \mathbb{Z}_{>0}$, we define $g(X)=$ $\{g(x): x \in X\}$. We have that the image of $f_{X}$ is $f_{X}\left(\mathbb{Z}_{>0}\right)=\mathbb{Z}_{>0} \backslash X$. We now show a general lemma about the operation *, with the goal of showing that $*$ is associative. Lemma 1. Let $X$ and $Y$ be finite sets of positive integers. The functions $f_{X * Y}$ and $f_{X} \circ f_{Y}$ are equal. Proof. We have $f_{X * Y}\left(\mathbb{Z}_{>0}\right)=\mathbb{Z}_{>0} \backslash(X * Y)=\left(\mathbb{Z}_{>0} \backslash X\right) \backslash f_{X}(Y)=f_{X}\left(\mathbb{Z}_{>0}\right) \backslash f_{X}(Y)=f_{X}\left(\mathbb{Z}_{>0} \backslash Y\right)=f_{X}\left(f_{Y}\left(\mathbb{Z}_{>0}\right)\right)$. Thus, the functions $f_{X * Y}$ and $f_{X} \circ f_{Y}$ are strictly increasing functions with the same range. Because a strictly function is uniquely defined by its range, we have $f_{X * Y}=f_{X} \circ f_{Y}$. Lemma 1 implies that * is associative, in the sense that $(A * B) * C=A *(B * C)$ for any finite sets $A, B$, and $C$ of positive integers. We prove the associativity by noting $$ \begin{gathered} \mathbb{Z}_{>0} \backslash((A * B) * C)=f_{(A * B) * C}\left(\mathbb{Z}_{>0}\right)=f_{A * B}\left(f_{C}\left(\mathbb{Z}_{>0}\right)\right)=f_{A}\left(f_{B}\left(f_{C}\left(\mathbb{Z}_{>0}\right)\right)\right) \\ =f_{A}\left(f_{B * C}\left(\mathbb{Z}_{>0}\right)=f_{A *(B * C)}\left(\mathbb{Z}_{>0}\right)=\mathbb{Z}_{>0} \backslash(A *(B * C))\right. \end{gathered} $$ In light of the associativity of *, we may drop the parentheses when we write expressions like $A *(B * C)$. We also introduce the notation $$ X^{* k}=\underbrace{X *(X * \cdots *(X *(X * X)) \ldots)}_{X \text { appears } k \text { times }} $$ Our goal is then to show that $A * B=B * A$ implies $A^{* b}=B^{* a}$. We will do so via the following general lemma. Lemma 2. Suppose that $X$ and $Y$ are finite sets of positive integers satisfying $X * Y=Y * X$ and $|X|=|Y|$. Then, we must have $X=Y$. Proof. Assume that $X$ and $Y$ are not equal. Let $s$ be the largest number in exactly one of $X$ and $Y$. Without loss of generality, say that $s \in X \backslash Y$. The number $f_{X}(s)$ counts the $s^{t h}$ number not in $X$, which implies that $$ f_{X}(s)=s+\left|X \cap\left\{1,2, \ldots, f_{X}(s)\right\}\right| $$ Since $f_{X}(s) \geqslant s$, we have that $$ \left\{f_{X}(s)+1, f_{X}(s)+2, \ldots\right\} \cap X=\left\{f_{X}(s)+1, f_{X}(s)+2, \ldots\right\} \cap Y $$ which, together with the assumption that $|X|=|Y|$, gives $$ \left|X \cap\left\{1,2, \ldots, f_{X}(s)\right\}\right|=\left|Y \cap\left\{1,2, \ldots, f_{X}(s)\right\}\right| $$ Now consider the equation $$ t-|Y \cap\{1,2, \ldots, t\}|=s $$ This equation is satisfied only when $t \in\left[f_{Y}(s), f_{Y}(s+1)\right)$, because the left hand side counts the number of elements up to $t$ that are not in $Y$. We have that the value $t=f_{X}(s)$ satisfies the above equation because of (1) and (2). Furthermore, since $f_{X}(s) \notin X$ and $f_{X}(s) \geqslant s$, we have that $f_{X}(s) \notin Y$ due to the maximality of $s$. Thus, by the above discussion, we must have $f_{X}(s)=f_{Y}(s)$. Finally, we arrive at a contradiction. The value $f_{X}(s)$ is neither in $X$ nor in $f_{X}(Y)$, because $s$ is not in $Y$ by assumption. Thus, $f_{X}(s) \notin X * Y$. However, since $s \in X$, we have $f_{Y}(s) \in Y * X$, a contradiction. We are now ready to finish the proof. Note first of all that $\left|A^{* b}\right|=a b=\left|B^{* a}\right|$. Moreover, since $A * B=B * A$, and $*$ is associative, it follows that $A^{* b} * B^{* a}=B^{* a} * A^{* b}$. Thus, by Lemma 2, we have $A^{* b}=B^{* a}$, as desired. Comment 1. Taking $A=X^{* k}$ and $B=X^{* l}$ generates many non-trivial examples where $A * B=B * A$. There are also other examples not of this form. For example, if $A=\{1,2,4\}$ and $B=\{1,3\}$, then $A * B=\{1,2,3,4,6\}=B * A$.
|
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6926ad78-026c-5ed1-bdc1-210f4bc42a8b
| 25,060
|
Determine all integers $n \geqslant 2$ with the following property: for any integers $a_{1}, a_{2}, \ldots, a_{n}$ whose sum is not divisible by $n$, there exists an index $1 \leqslant i \leqslant n$ such that none of the numbers $$ a_{i}, a_{i}+a_{i+1}, \ldots, a_{i}+a_{i+1}+\cdots+a_{i+n-1} $$ is divisible by $n$. (We let $a_{i}=a_{i-n}$ when $i>n$.) (Thailand) Answer: These integers are exactly the prime numbers.
|
Let us first show that, if $n=a b$, with $a, b \geqslant 2$ integers, then the property in the statement of the problem does not hold. Indeed, in this case, let $a_{k}=a$ for $1 \leqslant k \leqslant n-1$ and $a_{n}=0$. The sum $a_{1}+a_{2}+\cdots+a_{n}=a \cdot(n-1)$ is not divisible by $n$. Let $i$ with $1 \leqslant i \leqslant n$ be an arbitrary index. Taking $j=b$ if $1 \leqslant i \leqslant n-b$, and $j=b+1$ if $n-b<i \leqslant n$, we have $$ a_{i}+a_{i+1}+\cdots+a_{i+j-1}=a \cdot b=n \equiv 0 \quad(\bmod n) $$ It follows that the given example is indeed a counterexample to the property of the statement. Now let $n$ be a prime number. Suppose by contradiction that the property in the statement of the problem does not hold. Then there are integers $a_{1}, a_{2}, \ldots, a_{n}$ whose sum is not divisible by $n$ such that for each $i, 1 \leqslant i \leqslant n$, there is $j, 1 \leqslant j \leqslant n$, for which the number $a_{i}+a_{i+1}+$ $\cdots+a_{i+j-1}$ is divisible by $n$. Notice that, in any such case, we should have $1 \leqslant j \leqslant n-1$, since $a_{1}+a_{2}+\cdots+a_{n}$ is not divisible by $n$. So we may construct recursively a finite sequence of integers $0=i_{0}<i_{1}<i_{2}<\cdots<i_{n}$ with $i_{s+1}-i_{s} \leqslant n-1$ for $0 \leqslant s \leqslant n-1$ such that, for $0 \leqslant s \leqslant n-1$, $$ a_{i_{s}+1}+a_{i_{s}+2}+\cdots+a_{i_{s+1}} \equiv 0 \quad(\bmod n) $$ (where we take indices modulo $n$ ). Indeed, for $0 \leqslant s<n$, we apply the previous observation to $i=i_{s}+1$ in order to define $i_{s+1}=i_{s}+j$. In the sequence of $n+1$ indices $i_{0}, i_{1}, i_{2}, \ldots, i_{n}$, by the pigeonhole principle, we have two distinct elements which are congruent modulo $n$. So there are indices $r, s$ with $0 \leqslant r<s \leqslant n$ such that $i_{s} \equiv i_{r}(\bmod n)$ and $$ a_{i_{r}+1}+a_{i_{r}+2}+\cdots+a_{i_{s}}=\sum_{j=r}^{s-1}\left(a_{i_{j}+1}+a_{i_{j}+2}+\cdots+a_{i_{j+1}}\right) \equiv 0 \quad(\bmod n) $$ Since $i_{s} \equiv i_{r}(\bmod n)$, we have $i_{s}-i_{r}=k \cdot n$ for some positive integer $k$, and, since $i_{j+1}-i_{j} \leqslant$ $n-1$ for $0 \leqslant j \leqslant n-1$, we have $i_{s}-i_{r} \leqslant(n-1) \cdot n$, so $k \leqslant n-1$. But in this case $$ a_{i_{r}+1}+a_{i_{r}+2}+\cdots+a_{i_{s}}=k \cdot\left(a_{1}+a_{2}+\cdots+a_{n}\right) $$ cannot be a multiple of $n$, since $n$ is prime and neither $k$ nor $a_{1}+a_{2}+\cdots+a_{n}$ is a multiple of $n$. A contradiction.
|
proof
|
Yes
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Yes
|
proof
|
Number Theory
|
Determine all integers $n \geqslant 2$ with the following property: for any integers $a_{1}, a_{2}, \ldots, a_{n}$ whose sum is not divisible by $n$, there exists an index $1 \leqslant i \leqslant n$ such that none of the numbers $$ a_{i}, a_{i}+a_{i+1}, \ldots, a_{i}+a_{i+1}+\cdots+a_{i+n-1} $$ is divisible by $n$. (We let $a_{i}=a_{i-n}$ when $i>n$.) (Thailand) Answer: These integers are exactly the prime numbers.
|
Let us first show that, if $n=a b$, with $a, b \geqslant 2$ integers, then the property in the statement of the problem does not hold. Indeed, in this case, let $a_{k}=a$ for $1 \leqslant k \leqslant n-1$ and $a_{n}=0$. The sum $a_{1}+a_{2}+\cdots+a_{n}=a \cdot(n-1)$ is not divisible by $n$. Let $i$ with $1 \leqslant i \leqslant n$ be an arbitrary index. Taking $j=b$ if $1 \leqslant i \leqslant n-b$, and $j=b+1$ if $n-b<i \leqslant n$, we have $$ a_{i}+a_{i+1}+\cdots+a_{i+j-1}=a \cdot b=n \equiv 0 \quad(\bmod n) $$ It follows that the given example is indeed a counterexample to the property of the statement. Now let $n$ be a prime number. Suppose by contradiction that the property in the statement of the problem does not hold. Then there are integers $a_{1}, a_{2}, \ldots, a_{n}$ whose sum is not divisible by $n$ such that for each $i, 1 \leqslant i \leqslant n$, there is $j, 1 \leqslant j \leqslant n$, for which the number $a_{i}+a_{i+1}+$ $\cdots+a_{i+j-1}$ is divisible by $n$. Notice that, in any such case, we should have $1 \leqslant j \leqslant n-1$, since $a_{1}+a_{2}+\cdots+a_{n}$ is not divisible by $n$. So we may construct recursively a finite sequence of integers $0=i_{0}<i_{1}<i_{2}<\cdots<i_{n}$ with $i_{s+1}-i_{s} \leqslant n-1$ for $0 \leqslant s \leqslant n-1$ such that, for $0 \leqslant s \leqslant n-1$, $$ a_{i_{s}+1}+a_{i_{s}+2}+\cdots+a_{i_{s+1}} \equiv 0 \quad(\bmod n) $$ (where we take indices modulo $n$ ). Indeed, for $0 \leqslant s<n$, we apply the previous observation to $i=i_{s}+1$ in order to define $i_{s+1}=i_{s}+j$. In the sequence of $n+1$ indices $i_{0}, i_{1}, i_{2}, \ldots, i_{n}$, by the pigeonhole principle, we have two distinct elements which are congruent modulo $n$. So there are indices $r, s$ with $0 \leqslant r<s \leqslant n$ such that $i_{s} \equiv i_{r}(\bmod n)$ and $$ a_{i_{r}+1}+a_{i_{r}+2}+\cdots+a_{i_{s}}=\sum_{j=r}^{s-1}\left(a_{i_{j}+1}+a_{i_{j}+2}+\cdots+a_{i_{j+1}}\right) \equiv 0 \quad(\bmod n) $$ Since $i_{s} \equiv i_{r}(\bmod n)$, we have $i_{s}-i_{r}=k \cdot n$ for some positive integer $k$, and, since $i_{j+1}-i_{j} \leqslant$ $n-1$ for $0 \leqslant j \leqslant n-1$, we have $i_{s}-i_{r} \leqslant(n-1) \cdot n$, so $k \leqslant n-1$. But in this case $$ a_{i_{r}+1}+a_{i_{r}+2}+\cdots+a_{i_{s}}=k \cdot\left(a_{1}+a_{2}+\cdots+a_{n}\right) $$ cannot be a multiple of $n$, since $n$ is prime and neither $k$ nor $a_{1}+a_{2}+\cdots+a_{n}$ is a multiple of $n$. A contradiction.
|
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5289e4d8-5dcf-51f1-97ae-835d64fa95a3
| 25,074
|
Given any set $S$ of positive integers, show that at least one of the following two assertions holds: (1) There exist distinct finite subsets $F$ and $G$ of $S$ such that $\sum_{x \in F} 1 / x=\sum_{x \in G} 1 / x$; (2) There exists a positive rational number $r<1$ such that $\sum_{x \in F} 1 / x \neq r$ for all finite subsets $F$ of $S$. (Luxembourg)
|
Argue indirectly. Agree, as usual, that the empty sum is 0 to consider rationals in $[0,1)$; adjoining 0 causes no harm, since $\sum_{x \in F} 1 / x=0$ for no nonempty finite subset $F$ of $S$. For every rational $r$ in $[0,1)$, let $F_{r}$ be the unique finite subset of $S$ such that $\sum_{x \in F_{r}} 1 / x=r$. The argument hinges on the lemma below. Lemma. If $x$ is a member of $S$ and $q$ and $r$ are rationals in $[0,1)$ such that $q-r=1 / x$, then $x$ is a member of $F_{q}$ if and only if it is not one of $F_{r}$. Proof. If $x$ is a member of $F_{q}$, then $$ \sum_{y \in F_{q} \backslash\{x\}} \frac{1}{y}=\sum_{y \in F_{q}} \frac{1}{y}-\frac{1}{x}=q-\frac{1}{x}=r=\sum_{y \in F_{r}} \frac{1}{y} $$ so $F_{r}=F_{q} \backslash\{x\}$, and $x$ is not a member of $F_{r}$. Conversely, if $x$ is not a member of $F_{r}$, then $$ \sum_{y \in F_{r} \cup\{x\}} \frac{1}{y}=\sum_{y \in F_{r}} \frac{1}{y}+\frac{1}{x}=r+\frac{1}{x}=q=\sum_{y \in F_{q}} \frac{1}{y} $$ so $F_{q}=F_{r} \cup\{x\}$, and $x$ is a member of $F_{q}$. Consider now an element $x$ of $S$ and a positive rational $r<1$. Let $n=\lfloor r x\rfloor$ and consider the sets $F_{r-k / x}, k=0, \ldots, n$. Since $0 \leqslant r-n / x<1 / x$, the set $F_{r-n / x}$ does not contain $x$, and a repeated application of the lemma shows that the $F_{r-(n-2 k) / x}$ do not contain $x$, whereas the $F_{r-(n-2 k-1) / x}$ do. Consequently, $x$ is a member of $F_{r}$ if and only if $n$ is odd. Finally, consider $F_{2 / 3}$. By the preceding, $\lfloor 2 x / 3\rfloor$ is odd for each $x$ in $F_{2 / 3}$, so $2 x / 3$ is not integral. Since $F_{2 / 3}$ is finite, there exists a positive rational $\varepsilon$ such that $\lfloor(2 / 3-\varepsilon) x\rfloor=\lfloor 2 x / 3\rfloor$ for all $x$ in $F_{2 / 3}$. This implies that $F_{2 / 3}$ is a subset of $F_{2 / 3-\varepsilon}$ which is impossible. Comment. The solution above can be adapted to show that the problem statement still holds, if the condition $r<1$ in (2) is replaced with $r<\delta$, for an arbitrary positive $\delta$. This yields that, if $S$ does not satisfy (1), then there exist infinitely many positive rational numbers $r<1$ such that $\sum_{x \in F} 1 / x \neq r$ for all finite subsets $F$ of $S$.
|
proof
|
Yes
|
Yes
|
proof
|
Combinatorics
|
Given any set $S$ of positive integers, show that at least one of the following two assertions holds: (1) There exist distinct finite subsets $F$ and $G$ of $S$ such that $\sum_{x \in F} 1 / x=\sum_{x \in G} 1 / x$; (2) There exists a positive rational number $r<1$ such that $\sum_{x \in F} 1 / x \neq r$ for all finite subsets $F$ of $S$. (Luxembourg)
|
Argue indirectly. Agree, as usual, that the empty sum is 0 to consider rationals in $[0,1)$; adjoining 0 causes no harm, since $\sum_{x \in F} 1 / x=0$ for no nonempty finite subset $F$ of $S$. For every rational $r$ in $[0,1)$, let $F_{r}$ be the unique finite subset of $S$ such that $\sum_{x \in F_{r}} 1 / x=r$. The argument hinges on the lemma below. Lemma. If $x$ is a member of $S$ and $q$ and $r$ are rationals in $[0,1)$ such that $q-r=1 / x$, then $x$ is a member of $F_{q}$ if and only if it is not one of $F_{r}$. Proof. If $x$ is a member of $F_{q}$, then $$ \sum_{y \in F_{q} \backslash\{x\}} \frac{1}{y}=\sum_{y \in F_{q}} \frac{1}{y}-\frac{1}{x}=q-\frac{1}{x}=r=\sum_{y \in F_{r}} \frac{1}{y} $$ so $F_{r}=F_{q} \backslash\{x\}$, and $x$ is not a member of $F_{r}$. Conversely, if $x$ is not a member of $F_{r}$, then $$ \sum_{y \in F_{r} \cup\{x\}} \frac{1}{y}=\sum_{y \in F_{r}} \frac{1}{y}+\frac{1}{x}=r+\frac{1}{x}=q=\sum_{y \in F_{q}} \frac{1}{y} $$ so $F_{q}=F_{r} \cup\{x\}$, and $x$ is a member of $F_{q}$. Consider now an element $x$ of $S$ and a positive rational $r<1$. Let $n=\lfloor r x\rfloor$ and consider the sets $F_{r-k / x}, k=0, \ldots, n$. Since $0 \leqslant r-n / x<1 / x$, the set $F_{r-n / x}$ does not contain $x$, and a repeated application of the lemma shows that the $F_{r-(n-2 k) / x}$ do not contain $x$, whereas the $F_{r-(n-2 k-1) / x}$ do. Consequently, $x$ is a member of $F_{r}$ if and only if $n$ is odd. Finally, consider $F_{2 / 3}$. By the preceding, $\lfloor 2 x / 3\rfloor$ is odd for each $x$ in $F_{2 / 3}$, so $2 x / 3$ is not integral. Since $F_{2 / 3}$ is finite, there exists a positive rational $\varepsilon$ such that $\lfloor(2 / 3-\varepsilon) x\rfloor=\lfloor 2 x / 3\rfloor$ for all $x$ in $F_{2 / 3}$. This implies that $F_{2 / 3}$ is a subset of $F_{2 / 3-\varepsilon}$ which is impossible. Comment. The solution above can be adapted to show that the problem statement still holds, if the condition $r<1$ in (2) is replaced with $r<\delta$, for an arbitrary positive $\delta$. This yields that, if $S$ does not satisfy (1), then there exist infinitely many positive rational numbers $r<1$ such that $\sum_{x \in F} 1 / x \neq r$ for all finite subsets $F$ of $S$.
|
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|
c58676f5-244c-5ff8-a8cc-d604eeb3942e
| 23,543
|
Given any set $S$ of positive integers, show that at least one of the following two assertions holds: (1) There exist distinct finite subsets $F$ and $G$ of $S$ such that $\sum_{x \in F} 1 / x=\sum_{x \in G} 1 / x$; (2) There exists a positive rational number $r<1$ such that $\sum_{x \in F} 1 / x \neq r$ for all finite subsets $F$ of $S$. (Luxembourg)
|
A finite $S$ clearly satisfies (2), so let $S$ be infinite. If $S$ fails both conditions, so does $S \backslash\{1\}$. We may and will therefore assume that $S$ consists of integers greater than 1 . Label the elements of $S$ increasingly $x_{1}<x_{2}<\cdots$, where $x_{1} \geqslant 2$. We first show that $S$ satisfies (2) if $x_{n+1} \geqslant 2 x_{n}$ for all $n$. In this case, $x_{n} \geqslant 2^{n-1} x_{1}$ for all $n$, so $$ s=\sum_{n \geqslant 1} \frac{1}{x_{n}} \leqslant \sum_{n \geqslant 1} \frac{1}{2^{n-1} x_{1}}=\frac{2}{x_{1}} $$ If $x_{1} \geqslant 3$, or $x_{1}=2$ and $x_{n+1}>2 x_{n}$ for some $n$, then $\sum_{x \in F} 1 / x<s<1$ for every finite subset $F$ of $S$, so $S$ satisfies (2); and if $x_{1}=2$ and $x_{n+1}=2 x_{n}$ for all $n$, that is, $x_{n}=2^{n}$ for all $n$, then every finite subset $F$ of $S$ consists of powers of 2 , so $\sum_{x \in F} 1 / x \neq 1 / 3$ and again $S$ satisfies (2). Finally, we deal with the case where $x_{n+1}<2 x_{n}$ for some $n$. Consider the positive rational $r=1 / x_{n}-1 / x_{n+1}<1 / x_{n+1}$. If $r=\sum_{x \in F} 1 / x$ for no finite subset $F$ of $S$, then $S$ satisfies (2). We now assume that $r=\sum_{x \in F_{0}} 1 / x$ for some finite subset $F_{0}$ of $S$, and show that $S$ satisfies (1). Since $\sum_{x \in F_{0}} 1 / x=r<1 / x_{n+1}$, it follows that $x_{n+1}$ is not a member of $F_{0}$, so $$ \sum_{x \in F_{0} \cup\left\{x_{n+1}\right\}} \frac{1}{x}=\sum_{x \in F_{0}} \frac{1}{x}+\frac{1}{x_{n+1}}=r+\frac{1}{x_{n+1}}=\frac{1}{x_{n}} $$ Consequently, $F=F_{0} \cup\left\{x_{n+1}\right\}$ and $G=\left\{x_{n}\right\}$ are distinct finite subsets of $S$ such that $\sum_{x \in F} 1 / x=\sum_{x \in G} 1 / x$, and $S$ satisfies (1).
|
proof
|
Yes
|
Yes
|
proof
|
Combinatorics
|
Given any set $S$ of positive integers, show that at least one of the following two assertions holds: (1) There exist distinct finite subsets $F$ and $G$ of $S$ such that $\sum_{x \in F} 1 / x=\sum_{x \in G} 1 / x$; (2) There exists a positive rational number $r<1$ such that $\sum_{x \in F} 1 / x \neq r$ for all finite subsets $F$ of $S$. (Luxembourg)
|
A finite $S$ clearly satisfies (2), so let $S$ be infinite. If $S$ fails both conditions, so does $S \backslash\{1\}$. We may and will therefore assume that $S$ consists of integers greater than 1 . Label the elements of $S$ increasingly $x_{1}<x_{2}<\cdots$, where $x_{1} \geqslant 2$. We first show that $S$ satisfies (2) if $x_{n+1} \geqslant 2 x_{n}$ for all $n$. In this case, $x_{n} \geqslant 2^{n-1} x_{1}$ for all $n$, so $$ s=\sum_{n \geqslant 1} \frac{1}{x_{n}} \leqslant \sum_{n \geqslant 1} \frac{1}{2^{n-1} x_{1}}=\frac{2}{x_{1}} $$ If $x_{1} \geqslant 3$, or $x_{1}=2$ and $x_{n+1}>2 x_{n}$ for some $n$, then $\sum_{x \in F} 1 / x<s<1$ for every finite subset $F$ of $S$, so $S$ satisfies (2); and if $x_{1}=2$ and $x_{n+1}=2 x_{n}$ for all $n$, that is, $x_{n}=2^{n}$ for all $n$, then every finite subset $F$ of $S$ consists of powers of 2 , so $\sum_{x \in F} 1 / x \neq 1 / 3$ and again $S$ satisfies (2). Finally, we deal with the case where $x_{n+1}<2 x_{n}$ for some $n$. Consider the positive rational $r=1 / x_{n}-1 / x_{n+1}<1 / x_{n+1}$. If $r=\sum_{x \in F} 1 / x$ for no finite subset $F$ of $S$, then $S$ satisfies (2). We now assume that $r=\sum_{x \in F_{0}} 1 / x$ for some finite subset $F_{0}$ of $S$, and show that $S$ satisfies (1). Since $\sum_{x \in F_{0}} 1 / x=r<1 / x_{n+1}$, it follows that $x_{n+1}$ is not a member of $F_{0}$, so $$ \sum_{x \in F_{0} \cup\left\{x_{n+1}\right\}} \frac{1}{x}=\sum_{x \in F_{0}} \frac{1}{x}+\frac{1}{x_{n+1}}=r+\frac{1}{x_{n+1}}=\frac{1}{x_{n}} $$ Consequently, $F=F_{0} \cup\left\{x_{n+1}\right\}$ and $G=\left\{x_{n}\right\}$ are distinct finite subsets of $S$ such that $\sum_{x \in F} 1 / x=\sum_{x \in G} 1 / x$, and $S$ satisfies (1).
|
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c58676f5-244c-5ff8-a8cc-d604eeb3942e
| 23,543
|
Let $n \geqslant 3$ be an integer. Prove that there exists a set $S$ of $2 n$ positive integers satisfying the following property: For every $m=2,3, \ldots, n$ the set $S$ can be partitioned into two subsets with equal sums of elements, with one of subsets of cardinality $m$. (Iceland)
|
We show that one of possible examples is the set $$ S=\left\{1 \cdot 3^{k}, 2 \cdot 3^{k}: k=1,2, \ldots, n-1\right\} \cup\left\{1, \frac{3^{n}+9}{2}-1\right\} $$ It is readily verified that all the numbers listed above are distinct (notice that the last two are not divisible by 3 ). The sum of elements in $S$ is $$ \Sigma=1+\left(\frac{3^{n}+9}{2}-1\right)+\sum_{k=1}^{n-1}\left(1 \cdot 3^{k}+2 \cdot 3^{k}\right)=\frac{3^{n}+9}{2}+\sum_{k=1}^{n-1} 3^{k+1}=\frac{3^{n}+9}{2}+\frac{3^{n+1}-9}{2}=2 \cdot 3^{n} $$ Hence, in order to show that this set satisfies the problem requirements, it suffices to present, for every $m=2,3, \ldots, n$, an $m$-element subset $A_{m} \subset S$ whose sum of elements equals $3^{n}$. Such a subset is $$ A_{m}=\left\{2 \cdot 3^{k}: k=n-m+1, n-m+2, \ldots, n-1\right\} \cup\left\{1 \cdot 3^{n-m+1}\right\} . $$ Clearly, $\left|A_{m}\right|=m$. The sum of elements in $A_{m}$ is $$ 3^{n-m+1}+\sum_{k=n-m+1}^{n-1} 2 \cdot 3^{k}=3^{n-m+1}+\frac{2 \cdot 3^{n}-2 \cdot 3^{n-m+1}}{2}=3^{n} $$ as required. Comment. Let us present a more general construction. Let $s_{1}, s_{2}, \ldots, s_{2 n-1}$ be a sequence of pairwise distinct positive integers satisfying $s_{2 i+1}=s_{2 i}+s_{2 i-1}$ for all $i=2,3, \ldots, n-1$. Set $s_{2 n}=s_{1}+s_{2}+$ $\cdots+s_{2 n-4}$. Assume that $s_{2 n}$ is distinct from the other terms of the sequence. Then the set $S=\left\{s_{1}, s_{2}, \ldots, s_{2 n}\right\}$ satisfies the problem requirements. Indeed, the sum of its elements is $$ \Sigma=\sum_{i=1}^{2 n-4} s_{i}+\left(s_{2 n-3}+s_{2 n-2}\right)+s_{2 n-1}+s_{2 n}=s_{2 n}+s_{2 n-1}+s_{2 n-1}+s_{2 n}=2 s_{2 n}+2 s_{2 n-1} $$ Therefore, we have $$ \frac{\Sigma}{2}=s_{2 n}+s_{2 n-1}=s_{2 n}+s_{2 n-2}+s_{2 n-3}=s_{2 n}+s_{2 n-2}+s_{2 n-4}+s_{2 n-5}=\ldots, $$ which shows that the required sets $A_{m}$ can be chosen as $$ A_{m}=\left\{s_{2 n}, s_{2 n-2}, \ldots, s_{2 n-2 m+4}, s_{2 n-2 m+3}\right\} $$ So, the only condition to be satisfied is $s_{2 n} \notin\left\{s_{1}, s_{2}, \ldots, s_{2 n-1}\right\}$, which can be achieved in many different ways (e.g., by choosing properly the number $s_{1}$ after specifying $s_{2}, s_{3}, \ldots, s_{2 n-1}$ ). The solution above is an instance of this general construction. Another instance, for $n>3$, is the set $$ \left\{F_{1}, F_{2}, \ldots, F_{2 n-1}, F_{1}+\cdots+F_{2 n-4}\right\}, $$ where $F_{1}=1, F_{2}=2, F_{n+1}=F_{n}+F_{n-1}$ is the usual Fibonacci sequence.
|
proof
|
Yes
|
Yes
|
proof
|
Combinatorics
|
Let $n \geqslant 3$ be an integer. Prove that there exists a set $S$ of $2 n$ positive integers satisfying the following property: For every $m=2,3, \ldots, n$ the set $S$ can be partitioned into two subsets with equal sums of elements, with one of subsets of cardinality $m$. (Iceland)
|
We show that one of possible examples is the set $$ S=\left\{1 \cdot 3^{k}, 2 \cdot 3^{k}: k=1,2, \ldots, n-1\right\} \cup\left\{1, \frac{3^{n}+9}{2}-1\right\} $$ It is readily verified that all the numbers listed above are distinct (notice that the last two are not divisible by 3 ). The sum of elements in $S$ is $$ \Sigma=1+\left(\frac{3^{n}+9}{2}-1\right)+\sum_{k=1}^{n-1}\left(1 \cdot 3^{k}+2 \cdot 3^{k}\right)=\frac{3^{n}+9}{2}+\sum_{k=1}^{n-1} 3^{k+1}=\frac{3^{n}+9}{2}+\frac{3^{n+1}-9}{2}=2 \cdot 3^{n} $$ Hence, in order to show that this set satisfies the problem requirements, it suffices to present, for every $m=2,3, \ldots, n$, an $m$-element subset $A_{m} \subset S$ whose sum of elements equals $3^{n}$. Such a subset is $$ A_{m}=\left\{2 \cdot 3^{k}: k=n-m+1, n-m+2, \ldots, n-1\right\} \cup\left\{1 \cdot 3^{n-m+1}\right\} . $$ Clearly, $\left|A_{m}\right|=m$. The sum of elements in $A_{m}$ is $$ 3^{n-m+1}+\sum_{k=n-m+1}^{n-1} 2 \cdot 3^{k}=3^{n-m+1}+\frac{2 \cdot 3^{n}-2 \cdot 3^{n-m+1}}{2}=3^{n} $$ as required. Comment. Let us present a more general construction. Let $s_{1}, s_{2}, \ldots, s_{2 n-1}$ be a sequence of pairwise distinct positive integers satisfying $s_{2 i+1}=s_{2 i}+s_{2 i-1}$ for all $i=2,3, \ldots, n-1$. Set $s_{2 n}=s_{1}+s_{2}+$ $\cdots+s_{2 n-4}$. Assume that $s_{2 n}$ is distinct from the other terms of the sequence. Then the set $S=\left\{s_{1}, s_{2}, \ldots, s_{2 n}\right\}$ satisfies the problem requirements. Indeed, the sum of its elements is $$ \Sigma=\sum_{i=1}^{2 n-4} s_{i}+\left(s_{2 n-3}+s_{2 n-2}\right)+s_{2 n-1}+s_{2 n}=s_{2 n}+s_{2 n-1}+s_{2 n-1}+s_{2 n}=2 s_{2 n}+2 s_{2 n-1} $$ Therefore, we have $$ \frac{\Sigma}{2}=s_{2 n}+s_{2 n-1}=s_{2 n}+s_{2 n-2}+s_{2 n-3}=s_{2 n}+s_{2 n-2}+s_{2 n-4}+s_{2 n-5}=\ldots, $$ which shows that the required sets $A_{m}$ can be chosen as $$ A_{m}=\left\{s_{2 n}, s_{2 n-2}, \ldots, s_{2 n-2 m+4}, s_{2 n-2 m+3}\right\} $$ So, the only condition to be satisfied is $s_{2 n} \notin\left\{s_{1}, s_{2}, \ldots, s_{2 n-1}\right\}$, which can be achieved in many different ways (e.g., by choosing properly the number $s_{1}$ after specifying $s_{2}, s_{3}, \ldots, s_{2 n-1}$ ). The solution above is an instance of this general construction. Another instance, for $n>3$, is the set $$ \left\{F_{1}, F_{2}, \ldots, F_{2 n-1}, F_{1}+\cdots+F_{2 n-4}\right\}, $$ where $F_{1}=1, F_{2}=2, F_{n+1}=F_{n}+F_{n-1}$ is the usual Fibonacci sequence.
|
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|
93c0c0a5-8c34-584e-aea7-3c10cd4a4bdb
| 23,564
|
Let $n$ be a given positive integer. Sisyphus performs a sequence of turns on a board consisting of $n+1$ squares in a row, numbered 0 to $n$ from left to right. Initially, $n$ stones are put into square 0, and the other squares are empty. At every turn, Sisyphus chooses any nonempty square, say with $k$ stones, takes one of those stones and moves it to the right by at most $k$ squares (the stone should stay within the board). Sisyphus' aim is to move all $n$ stones to square $n$. Prove that Sisyphus cannot reach the aim in less than $$ \left\lceil\frac{n}{1}\right\rceil+\left\lceil\frac{n}{2}\right\rceil+\left\lceil\frac{n}{3}\right\rceil+\cdots+\left\lceil\frac{n}{n}\right\rceil $$ turns. (As usual, $\lceil x\rceil$ stands for the least integer not smaller than $x$.) (Netherlands)
|
The stones are indistinguishable, and all have the same origin and the same final position. So, at any turn we can prescribe which stone from the chosen square to move. We do it in the following manner. Number the stones from 1 to $n$. At any turn, after choosing a square, Sisyphus moves the stone with the largest number from this square. This way, when stone $k$ is moved from some square, that square contains not more than $k$ stones (since all their numbers are at most $k$ ). Therefore, stone $k$ is moved by at most $k$ squares at each turn. Since the total shift of the stone is exactly $n$, at least $\lceil n / k\rceil$ moves of stone $k$ should have been made, for every $k=1,2, \ldots, n$. By summing up over all $k=1,2, \ldots, n$, we get the required estimate. Comment. The original submission contained the second part, asking for which values of $n$ the equality can be achieved. The answer is $n=1,2,3,4,5,7$. The Problem Selection Committee considered this part to be less suitable for the competition, due to technicalities.
|
proof
|
Yes
|
Yes
|
proof
|
Combinatorics
|
Let $n$ be a given positive integer. Sisyphus performs a sequence of turns on a board consisting of $n+1$ squares in a row, numbered 0 to $n$ from left to right. Initially, $n$ stones are put into square 0, and the other squares are empty. At every turn, Sisyphus chooses any nonempty square, say with $k$ stones, takes one of those stones and moves it to the right by at most $k$ squares (the stone should stay within the board). Sisyphus' aim is to move all $n$ stones to square $n$. Prove that Sisyphus cannot reach the aim in less than $$ \left\lceil\frac{n}{1}\right\rceil+\left\lceil\frac{n}{2}\right\rceil+\left\lceil\frac{n}{3}\right\rceil+\cdots+\left\lceil\frac{n}{n}\right\rceil $$ turns. (As usual, $\lceil x\rceil$ stands for the least integer not smaller than $x$.) (Netherlands)
|
The stones are indistinguishable, and all have the same origin and the same final position. So, at any turn we can prescribe which stone from the chosen square to move. We do it in the following manner. Number the stones from 1 to $n$. At any turn, after choosing a square, Sisyphus moves the stone with the largest number from this square. This way, when stone $k$ is moved from some square, that square contains not more than $k$ stones (since all their numbers are at most $k$ ). Therefore, stone $k$ is moved by at most $k$ squares at each turn. Since the total shift of the stone is exactly $n$, at least $\lceil n / k\rceil$ moves of stone $k$ should have been made, for every $k=1,2, \ldots, n$. By summing up over all $k=1,2, \ldots, n$, we get the required estimate. Comment. The original submission contained the second part, asking for which values of $n$ the equality can be achieved. The answer is $n=1,2,3,4,5,7$. The Problem Selection Committee considered this part to be less suitable for the competition, due to technicalities.
|
{
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|
5dbe36f2-5d24-552c-b9e6-9745e81192ef
| 23,569
|
Let $a$ and $b$ be distinct positive integers. The following infinite process takes place on an initially empty board. (i) If there is at least a pair of equal numbers on the board, we choose such a pair and increase one of its components by $a$ and the other by $b$. (ii) If no such pair exists, we write down two times the number 0 . Prove that, no matter how we make the choices in $(i)$, operation (ii) will be performed only finitely many times. (Serbia)
|
We may assume $\operatorname{gcd}(a, b)=1$; otherwise we work in the same way with multiples of $d=\operatorname{gcd}(a, b)$. Suppose that after $N$ moves of type (ii) and some moves of type $(i)$ we have to add two new zeros. For each integer $k$, denote by $f(k)$ the number of times that the number $k$ appeared on the board up to this moment. Then $f(0)=2 N$ and $f(k)=0$ for $k<0$. Since the board contains at most one $k-a$, every second occurrence of $k-a$ on the board produced, at some moment, an occurrence of $k$; the same stands for $k-b$. Therefore, $$ f(k)=\left\lfloor\frac{f(k-a)}{2}\right\rfloor+\left\lfloor\frac{f(k-b)}{2}\right\rfloor $$ yielding $$ f(k) \geqslant \frac{f(k-a)+f(k-b)}{2}-1 $$ Since $\operatorname{gcd}(a, b)=1$, every integer $x>a b-a-b$ is expressible in the form $x=s a+t b$, with integer $s, t \geqslant 0$. We will prove by induction on $s+t$ that if $x=s a+b t$, with $s, t$ nonnegative integers, then $$ f(x)>\frac{f(0)}{2^{s+t}}-2 . $$ The base case $s+t=0$ is trivial. Assume now that (3) is true for $s+t=v$. Then, if $s+t=v+1$ and $x=s a+t b$, at least one of the numbers $s$ and $t$ - say $s$ - is positive, hence by (2), $$ f(x)=f(s a+t b) \geqslant \frac{f((s-1) a+t b)}{2}-1>\frac{1}{2}\left(\frac{f(0)}{2^{s+t-1}}-2\right)-1=\frac{f(0)}{2^{s+t}}-2 . $$ Assume now that we must perform moves of type (ii) ad infinitum. Take $n=a b-a-b$ and suppose $b>a$. Since each of the numbers $n+1, n+2, \ldots, n+b$ can be expressed in the form $s a+t b$, with $0 \leqslant s \leqslant b$ and $0 \leqslant t \leqslant a$, after moves of type (ii) have been performed $2^{a+b+1}$ times and we have to add a new pair of zeros, each $f(n+k), k=1,2, \ldots, b$, is at least 2 . In this case (1) yields inductively $f(n+k) \geqslant 2$ for all $k \geqslant 1$. But this is absurd: after a finite number of moves, $f$ cannot attain nonzero values at infinitely many points.
|
proof
|
Yes
|
Yes
|
proof
|
Combinatorics
|
Let $a$ and $b$ be distinct positive integers. The following infinite process takes place on an initially empty board. (i) If there is at least a pair of equal numbers on the board, we choose such a pair and increase one of its components by $a$ and the other by $b$. (ii) If no such pair exists, we write down two times the number 0 . Prove that, no matter how we make the choices in $(i)$, operation (ii) will be performed only finitely many times. (Serbia)
|
We may assume $\operatorname{gcd}(a, b)=1$; otherwise we work in the same way with multiples of $d=\operatorname{gcd}(a, b)$. Suppose that after $N$ moves of type (ii) and some moves of type $(i)$ we have to add two new zeros. For each integer $k$, denote by $f(k)$ the number of times that the number $k$ appeared on the board up to this moment. Then $f(0)=2 N$ and $f(k)=0$ for $k<0$. Since the board contains at most one $k-a$, every second occurrence of $k-a$ on the board produced, at some moment, an occurrence of $k$; the same stands for $k-b$. Therefore, $$ f(k)=\left\lfloor\frac{f(k-a)}{2}\right\rfloor+\left\lfloor\frac{f(k-b)}{2}\right\rfloor $$ yielding $$ f(k) \geqslant \frac{f(k-a)+f(k-b)}{2}-1 $$ Since $\operatorname{gcd}(a, b)=1$, every integer $x>a b-a-b$ is expressible in the form $x=s a+t b$, with integer $s, t \geqslant 0$. We will prove by induction on $s+t$ that if $x=s a+b t$, with $s, t$ nonnegative integers, then $$ f(x)>\frac{f(0)}{2^{s+t}}-2 . $$ The base case $s+t=0$ is trivial. Assume now that (3) is true for $s+t=v$. Then, if $s+t=v+1$ and $x=s a+t b$, at least one of the numbers $s$ and $t$ - say $s$ - is positive, hence by (2), $$ f(x)=f(s a+t b) \geqslant \frac{f((s-1) a+t b)}{2}-1>\frac{1}{2}\left(\frac{f(0)}{2^{s+t-1}}-2\right)-1=\frac{f(0)}{2^{s+t}}-2 . $$ Assume now that we must perform moves of type (ii) ad infinitum. Take $n=a b-a-b$ and suppose $b>a$. Since each of the numbers $n+1, n+2, \ldots, n+b$ can be expressed in the form $s a+t b$, with $0 \leqslant s \leqslant b$ and $0 \leqslant t \leqslant a$, after moves of type (ii) have been performed $2^{a+b+1}$ times and we have to add a new pair of zeros, each $f(n+k), k=1,2, \ldots, b$, is at least 2 . In this case (1) yields inductively $f(n+k) \geqslant 2$ for all $k \geqslant 1$. But this is absurd: after a finite number of moves, $f$ cannot attain nonzero values at infinitely many points.
|
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|
6cb94fc8-f971-5d56-b53d-9a64e5849bc6
| 23,579
|
Consider 2018 pairwise crossing circles no three of which are concurrent. These circles subdivide the plane into regions bounded by circular edges that meet at vertices. Notice that there are an even number of vertices on each circle. Given the circle, alternately colour the vertices on that circle red and blue. In doing so for each circle, every vertex is coloured twice once for each of the two circles that cross at that point. If the two colourings agree at a vertex, then it is assigned that colour; otherwise, it becomes yellow. Show that, if some circle contains at least 2061 yellow points, then the vertices of some region are all yellow. (India)
|
The first two lemmata in Call the circles from the two classes white and black, respectively. Call a region yellow if its vertices are all yellow. Let $w$ and $b$ be the numbers of white and black circles, respectively; clearly, $w+b=n$. Assume that $w \geqslant b$, and that there is no yellow region. Clearly, $b \geqslant 1$, otherwise each region is yellow. The white circles subdivide the plane into $w(w-1)+2$ larger regions - call them white. The white regions (or rather their boundaries) subdivide each black circle into black arcs. Since there are no yellow regions, each white region contains at least one black arc. Consider any white region; let it contain $t \geqslant 1$ black arcs. We claim that the number of points at which these $t$ arcs cross does not exceed $t-1$. To prove this, consider a multigraph whose vertices are these black arcs, two vertices being joined by an edge for each point at which the corresponding arcs cross. If this graph had more than $t-1$ edges, it would contain a cycle, since it has $t$ vertices; this cycle would correspond to a closed contour formed by black sub-arcs, lying inside the region under consideration. This contour would, in turn, define at least one yellow region, which is impossible. Let $t_{i}$ be the number of black arcs inside the $i^{\text {th }}$ white region. The total number of black arcs is $\sum_{i} t_{i}=2 w b$, and they cross at $2\left(\begin{array}{l}b \\ 2\end{array}\right)=b(b-1)$ points. By the preceding, $$ b(b-1) \leqslant \sum_{i=1}^{w^{2}-w+2}\left(t_{i}-1\right)=\sum_{i=1}^{w^{2}-w+2} t_{i}-\left(w^{2}-w+2\right)=2 w b-\left(w^{2}-w+2\right) $$ or, equivalently, $(w-b)^{2} \leqslant w+b-2=n-2$, which is the case if and only if $w-b \leqslant\lfloor\sqrt{n-2}\rfloor$. Consequently, $b \leqslant w \leqslant(n+\lfloor\sqrt{n-2}\rfloor) / 2$, so there are at most $2(w-1) \leqslant n+\lfloor\sqrt{n-2}\rfloor-2$ yellow vertices on each circle - a contradiction.
|
proof
|
Yes
|
Yes
|
proof
|
Combinatorics
|
Consider 2018 pairwise crossing circles no three of which are concurrent. These circles subdivide the plane into regions bounded by circular edges that meet at vertices. Notice that there are an even number of vertices on each circle. Given the circle, alternately colour the vertices on that circle red and blue. In doing so for each circle, every vertex is coloured twice once for each of the two circles that cross at that point. If the two colourings agree at a vertex, then it is assigned that colour; otherwise, it becomes yellow. Show that, if some circle contains at least 2061 yellow points, then the vertices of some region are all yellow. (India)
|
The first two lemmata in Call the circles from the two classes white and black, respectively. Call a region yellow if its vertices are all yellow. Let $w$ and $b$ be the numbers of white and black circles, respectively; clearly, $w+b=n$. Assume that $w \geqslant b$, and that there is no yellow region. Clearly, $b \geqslant 1$, otherwise each region is yellow. The white circles subdivide the plane into $w(w-1)+2$ larger regions - call them white. The white regions (or rather their boundaries) subdivide each black circle into black arcs. Since there are no yellow regions, each white region contains at least one black arc. Consider any white region; let it contain $t \geqslant 1$ black arcs. We claim that the number of points at which these $t$ arcs cross does not exceed $t-1$. To prove this, consider a multigraph whose vertices are these black arcs, two vertices being joined by an edge for each point at which the corresponding arcs cross. If this graph had more than $t-1$ edges, it would contain a cycle, since it has $t$ vertices; this cycle would correspond to a closed contour formed by black sub-arcs, lying inside the region under consideration. This contour would, in turn, define at least one yellow region, which is impossible. Let $t_{i}$ be the number of black arcs inside the $i^{\text {th }}$ white region. The total number of black arcs is $\sum_{i} t_{i}=2 w b$, and they cross at $2\left(\begin{array}{l}b \\ 2\end{array}\right)=b(b-1)$ points. By the preceding, $$ b(b-1) \leqslant \sum_{i=1}^{w^{2}-w+2}\left(t_{i}-1\right)=\sum_{i=1}^{w^{2}-w+2} t_{i}-\left(w^{2}-w+2\right)=2 w b-\left(w^{2}-w+2\right) $$ or, equivalently, $(w-b)^{2} \leqslant w+b-2=n-2$, which is the case if and only if $w-b \leqslant\lfloor\sqrt{n-2}\rfloor$. Consequently, $b \leqslant w \leqslant(n+\lfloor\sqrt{n-2}\rfloor) / 2$, so there are at most $2(w-1) \leqslant n+\lfloor\sqrt{n-2}\rfloor-2$ yellow vertices on each circle - a contradiction.
|
{
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|
100edfe4-41a0-537f-bde4-258da827da2c
| 23,582
|
Let $A B C$ be an acute-angled triangle with circumcircle $\Gamma$. Let $D$ and $E$ be points on the segments $A B$ and $A C$, respectively, such that $A D=A E$. The perpendicular bisectors of the segments $B D$ and $C E$ intersect the small arcs $\overparen{A B}$ and $\overparen{A C}$ at points $F$ and $G$ respectively. Prove that $D E \| F G$. (Greece)
|
In the sequel, all the considered arcs are small arcs. Let $P$ be the midpoint of the $\operatorname{arc} \overparen{B C}$. Then $A P$ is the bisector of $\angle B A C$, hence, in the isosceles triangle $A D E, A P \perp D E$. So, the statement of the problem is equivalent to $A P \perp F G$. In order to prove this, let $K$ be the second intersection of $\Gamma$ with $F D$. Then the triangle $F B D$ is isosceles, therefore $$ \angle A K F=\angle A B F=\angle F D B=\angle A D K $$ yielding $A K=A D$. In the same way, denoting by $L$ the second intersection of $\Gamma$ with $G E$, we get $A L=A E$. This shows that $A K=A L$.  Now $\angle F B D=\angle F D B$ gives $\overparen{A F}=\overparen{B F}+\overparen{A K}=\overparen{B F}+\overparen{A L}$, hence $\overparen{B F}=\overparen{L F}$. In a similar way, we get $\widehat{C G}=\widehat{G K}$. This yields $$ \angle(A P, F G)=\frac{\overparen{A F}+\overparen{P G}}{2}=\frac{\overparen{A L}+\overparen{L F}+\overparen{P C}+\overparen{C G}}{2}=\frac{\overparen{K L}+\overparen{L B}+\overparen{B C}+\overparen{C K}}{4}=90^{\circ} . $$
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
Let $A B C$ be an acute-angled triangle with circumcircle $\Gamma$. Let $D$ and $E$ be points on the segments $A B$ and $A C$, respectively, such that $A D=A E$. The perpendicular bisectors of the segments $B D$ and $C E$ intersect the small arcs $\overparen{A B}$ and $\overparen{A C}$ at points $F$ and $G$ respectively. Prove that $D E \| F G$. (Greece)
|
In the sequel, all the considered arcs are small arcs. Let $P$ be the midpoint of the $\operatorname{arc} \overparen{B C}$. Then $A P$ is the bisector of $\angle B A C$, hence, in the isosceles triangle $A D E, A P \perp D E$. So, the statement of the problem is equivalent to $A P \perp F G$. In order to prove this, let $K$ be the second intersection of $\Gamma$ with $F D$. Then the triangle $F B D$ is isosceles, therefore $$ \angle A K F=\angle A B F=\angle F D B=\angle A D K $$ yielding $A K=A D$. In the same way, denoting by $L$ the second intersection of $\Gamma$ with $G E$, we get $A L=A E$. This shows that $A K=A L$.  Now $\angle F B D=\angle F D B$ gives $\overparen{A F}=\overparen{B F}+\overparen{A K}=\overparen{B F}+\overparen{A L}$, hence $\overparen{B F}=\overparen{L F}$. In a similar way, we get $\widehat{C G}=\widehat{G K}$. This yields $$ \angle(A P, F G)=\frac{\overparen{A F}+\overparen{P G}}{2}=\frac{\overparen{A L}+\overparen{L F}+\overparen{P C}+\overparen{C G}}{2}=\frac{\overparen{K L}+\overparen{L B}+\overparen{B C}+\overparen{C K}}{4}=90^{\circ} . $$
|
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4a218847-d7e3-5c87-887e-7737b56d546e
| 23,586
|
Let $A B C$ be an acute-angled triangle with circumcircle $\Gamma$. Let $D$ and $E$ be points on the segments $A B$ and $A C$, respectively, such that $A D=A E$. The perpendicular bisectors of the segments $B D$ and $C E$ intersect the small arcs $\overparen{A B}$ and $\overparen{A C}$ at points $F$ and $G$ respectively. Prove that $D E \| F G$. (Greece)
|
Let $Z=A B \cap F G, T=A C \cap F G$. It suffices to prove that $\angle A T Z=\angle A Z T$. Let $X$ be the point for which $F X A D$ is a parallelogram. Then $$ \angle F X A=\angle F D A=180^{\circ}-\angle F D B=180^{\circ}-\angle F B D, $$ where in the last equality we used that $F D=F B$. It follows that the quadrilateral $B F X A$ is cyclic, so $X$ lies on $\Gamma$.  Analogously, if $Y$ is the point for which $G Y A E$ is a parallelogram, then $Y$ lies on $\Gamma$. So the quadrilateral $X F G Y$ is cyclic and $F X=A D=A E=G Y$, hence $X F G Y$ is an isosceles trapezoid. Now, by $X F \| A Z$ and $Y G \| A T$, it follows that $\angle A T Z=\angle Y G F=\angle X F G=\angle A Z T$.
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
Let $A B C$ be an acute-angled triangle with circumcircle $\Gamma$. Let $D$ and $E$ be points on the segments $A B$ and $A C$, respectively, such that $A D=A E$. The perpendicular bisectors of the segments $B D$ and $C E$ intersect the small arcs $\overparen{A B}$ and $\overparen{A C}$ at points $F$ and $G$ respectively. Prove that $D E \| F G$. (Greece)
|
Let $Z=A B \cap F G, T=A C \cap F G$. It suffices to prove that $\angle A T Z=\angle A Z T$. Let $X$ be the point for which $F X A D$ is a parallelogram. Then $$ \angle F X A=\angle F D A=180^{\circ}-\angle F D B=180^{\circ}-\angle F B D, $$ where in the last equality we used that $F D=F B$. It follows that the quadrilateral $B F X A$ is cyclic, so $X$ lies on $\Gamma$.  Analogously, if $Y$ is the point for which $G Y A E$ is a parallelogram, then $Y$ lies on $\Gamma$. So the quadrilateral $X F G Y$ is cyclic and $F X=A D=A E=G Y$, hence $X F G Y$ is an isosceles trapezoid. Now, by $X F \| A Z$ and $Y G \| A T$, it follows that $\angle A T Z=\angle Y G F=\angle X F G=\angle A Z T$.
|
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4a218847-d7e3-5c87-887e-7737b56d546e
| 23,586
|
Let $A B C$ be an acute-angled triangle with circumcircle $\Gamma$. Let $D$ and $E$ be points on the segments $A B$ and $A C$, respectively, such that $A D=A E$. The perpendicular bisectors of the segments $B D$ and $C E$ intersect the small arcs $\overparen{A B}$ and $\overparen{A C}$ at points $F$ and $G$ respectively. Prove that $D E \| F G$. (Greece)
|
As in the first solution, we prove that $F G \perp A P$, where $P$ is the midpoint of the small arc $\overparen{B C}$. Let $O$ be the circumcentre of the triangle $A B C$, and let $M$ and $N$ be the midpoints of the small arcs $\overparen{A B}$ and $\overparen{A C}$, respectively. Then $O M$ and $O N$ are the perpendicular bisectors of $A B$ and $A C$, respectively.  The distance $d$ between $O M$ and the perpendicular bisector of $B D$ is $\frac{1}{2} A B-\frac{1}{2} B D=\frac{1}{2} A D$, hence it is equal to the distance between $O N$ and the perpendicular bisector of $C E$. This shows that the isosceles trapezoid determined by the diameter $\delta$ of $\Gamma$ through $M$ and the chord parallel to $\delta$ through $F$ is congruent to the isosceles trapezoid determined by the diameter $\delta^{\prime}$ of $\Gamma$ through $N$ and the chord parallel to $\delta^{\prime}$ through $G$. Therefore $M F=N G$, yielding $M N \| F G$. Now $$ \angle(M N, A P)=\frac{1}{2}(\overparen{A M}+\overparen{P C}+\overparen{C N})=\frac{1}{4}(\overparen{A B}+\overparen{B C}+\overparen{C A})=90^{\circ} $$ hence $M N \perp A P$, and the conclusion follows.
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
Let $A B C$ be an acute-angled triangle with circumcircle $\Gamma$. Let $D$ and $E$ be points on the segments $A B$ and $A C$, respectively, such that $A D=A E$. The perpendicular bisectors of the segments $B D$ and $C E$ intersect the small arcs $\overparen{A B}$ and $\overparen{A C}$ at points $F$ and $G$ respectively. Prove that $D E \| F G$. (Greece)
|
As in the first solution, we prove that $F G \perp A P$, where $P$ is the midpoint of the small arc $\overparen{B C}$. Let $O$ be the circumcentre of the triangle $A B C$, and let $M$ and $N$ be the midpoints of the small arcs $\overparen{A B}$ and $\overparen{A C}$, respectively. Then $O M$ and $O N$ are the perpendicular bisectors of $A B$ and $A C$, respectively.  The distance $d$ between $O M$ and the perpendicular bisector of $B D$ is $\frac{1}{2} A B-\frac{1}{2} B D=\frac{1}{2} A D$, hence it is equal to the distance between $O N$ and the perpendicular bisector of $C E$. This shows that the isosceles trapezoid determined by the diameter $\delta$ of $\Gamma$ through $M$ and the chord parallel to $\delta$ through $F$ is congruent to the isosceles trapezoid determined by the diameter $\delta^{\prime}$ of $\Gamma$ through $N$ and the chord parallel to $\delta^{\prime}$ through $G$. Therefore $M F=N G$, yielding $M N \| F G$. Now $$ \angle(M N, A P)=\frac{1}{2}(\overparen{A M}+\overparen{P C}+\overparen{C N})=\frac{1}{4}(\overparen{A B}+\overparen{B C}+\overparen{C A})=90^{\circ} $$ hence $M N \perp A P$, and the conclusion follows.
|
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4a218847-d7e3-5c87-887e-7737b56d546e
| 23,586
|
Let $A B C$ be a triangle with $A B=A C$, and let $M$ be the midpoint of $B C$. Let $P$ be a point such that $P B<P C$ and $P A$ is parallel to $B C$. Let $X$ and $Y$ be points on the lines $P B$ and $P C$, respectively, so that $B$ lies on the segment $P X, C$ lies on the segment $P Y$, and $\angle P X M=\angle P Y M$. Prove that the quadrilateral $A P X Y$ is cyclic. (Australia)
|
Since $A B=A C, A M$ is the perpendicular bisector of $B C$, hence $\angle P A M=$ $\angle A M C=90^{\circ}$.  Now let $Z$ be the common point of $A M$ and the perpendicular through $Y$ to $P C$ (notice that $Z$ lies on to the ray $A M$ beyond $M$ ). We have $\angle P A Z=\angle P Y Z=90^{\circ}$. Thus the points $P, A, Y$, and $Z$ are concyclic. Since $\angle C M Z=\angle C Y Z=90^{\circ}$, the quadrilateral $C Y Z M$ is cyclic, hence $\angle C Z M=$ $\angle C Y M$. By the condition in the statement, $\angle C Y M=\angle B X M$, and, by symmetry in $Z M$, $\angle C Z M=\angle B Z M$. Therefore, $\angle B X M=\angle B Z M$. It follows that the points $B, X, Z$, and $M$ are concyclic, hence $\angle B X Z=180^{\circ}-\angle B M Z=90^{\circ}$. Finally, we have $\angle P X Z=\angle P Y Z=\angle P A Z=90^{\circ}$, hence the five points $P, A, X, Y, Z$ are concyclic. In particular, the quadrilateral $A P X Y$ is cyclic, as required. Comment 1. Clearly, the key point $Z$ from the solution above can be introduced in several different ways, e.g., as the second meeting point of the circle $C M Y$ and the line $A M$, or as the second meeting point of the circles $C M Y$ and $B M X$, etc. For some of definitions of $Z$ its location is not obvious. For instance, if $Z$ is defined as a common point of $A M$ and the perpendicular through $X$ to $P X$, it is not clear that $Z$ lies on the ray $A M$ beyond $M$. To avoid such slippery details some more restrictions on the construction may be required. Comment 2. Let us discuss a connection to the Miquel point of a cyclic quadrilateral. Set $X^{\prime}=$ $M X \cap P C, Y^{\prime}=M Y \cap P B$, and $Q=X Y \cap X^{\prime} Y^{\prime}$ (see the figure below). We claim that $B C \| P Q$. (One way of proving this is the following. Notice that the quadruple of lines $P X, P M, P Y, P Q$ is harmonic, hence the quadruple $B, M, C, P Q \cap B C$ of their intersection points with $B C$ is harmonic. Since $M$ is the midpoint of $B C, P Q \cap B C$ is an ideal point, i.e., $P Q \| B C$.) It follows from the given equality $\angle P X M=\angle P Y M$ that the quadrilateral $X Y X^{\prime} Y^{\prime}$ is cyclic. Note that $A$ is the projection of $M$ onto $P Q$. By a known description, $A$ is the Miquel point for the sidelines $X Y, X Y^{\prime}, X^{\prime} Y, X^{\prime} Y^{\prime}$. In particular, the circle $P X Y$ passes through $A$.  Comment 3. An alternative approach is the following. One can note that the (oriented) lengths of the segments $C Y$ and $B X$ are both linear functions of a parameter $t=\cot \angle P X M$. As $t$ varies, the intersection point $S$ of the perpendicular bisectors of $P X$ and $P Y$ traces a fixed line, thus the family of circles $P X Y$ has a fixed common point (other than $P$ ). By checking particular cases, one can show that this fixed point is $A$. Comment 4. The problem states that $\angle P X M=\angle P Y M$ implies that $A P X Y$ is cyclic. The original submission claims that these two conditions are in fact equivalent. The Problem Selection Committee omitted the converse part, since it follows easily from the direct one, by reversing arguments.
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
Let $A B C$ be a triangle with $A B=A C$, and let $M$ be the midpoint of $B C$. Let $P$ be a point such that $P B<P C$ and $P A$ is parallel to $B C$. Let $X$ and $Y$ be points on the lines $P B$ and $P C$, respectively, so that $B$ lies on the segment $P X, C$ lies on the segment $P Y$, and $\angle P X M=\angle P Y M$. Prove that the quadrilateral $A P X Y$ is cyclic. (Australia)
|
Since $A B=A C, A M$ is the perpendicular bisector of $B C$, hence $\angle P A M=$ $\angle A M C=90^{\circ}$.  Now let $Z$ be the common point of $A M$ and the perpendicular through $Y$ to $P C$ (notice that $Z$ lies on to the ray $A M$ beyond $M$ ). We have $\angle P A Z=\angle P Y Z=90^{\circ}$. Thus the points $P, A, Y$, and $Z$ are concyclic. Since $\angle C M Z=\angle C Y Z=90^{\circ}$, the quadrilateral $C Y Z M$ is cyclic, hence $\angle C Z M=$ $\angle C Y M$. By the condition in the statement, $\angle C Y M=\angle B X M$, and, by symmetry in $Z M$, $\angle C Z M=\angle B Z M$. Therefore, $\angle B X M=\angle B Z M$. It follows that the points $B, X, Z$, and $M$ are concyclic, hence $\angle B X Z=180^{\circ}-\angle B M Z=90^{\circ}$. Finally, we have $\angle P X Z=\angle P Y Z=\angle P A Z=90^{\circ}$, hence the five points $P, A, X, Y, Z$ are concyclic. In particular, the quadrilateral $A P X Y$ is cyclic, as required. Comment 1. Clearly, the key point $Z$ from the solution above can be introduced in several different ways, e.g., as the second meeting point of the circle $C M Y$ and the line $A M$, or as the second meeting point of the circles $C M Y$ and $B M X$, etc. For some of definitions of $Z$ its location is not obvious. For instance, if $Z$ is defined as a common point of $A M$ and the perpendicular through $X$ to $P X$, it is not clear that $Z$ lies on the ray $A M$ beyond $M$. To avoid such slippery details some more restrictions on the construction may be required. Comment 2. Let us discuss a connection to the Miquel point of a cyclic quadrilateral. Set $X^{\prime}=$ $M X \cap P C, Y^{\prime}=M Y \cap P B$, and $Q=X Y \cap X^{\prime} Y^{\prime}$ (see the figure below). We claim that $B C \| P Q$. (One way of proving this is the following. Notice that the quadruple of lines $P X, P M, P Y, P Q$ is harmonic, hence the quadruple $B, M, C, P Q \cap B C$ of their intersection points with $B C$ is harmonic. Since $M$ is the midpoint of $B C, P Q \cap B C$ is an ideal point, i.e., $P Q \| B C$.) It follows from the given equality $\angle P X M=\angle P Y M$ that the quadrilateral $X Y X^{\prime} Y^{\prime}$ is cyclic. Note that $A$ is the projection of $M$ onto $P Q$. By a known description, $A$ is the Miquel point for the sidelines $X Y, X Y^{\prime}, X^{\prime} Y, X^{\prime} Y^{\prime}$. In particular, the circle $P X Y$ passes through $A$.  Comment 3. An alternative approach is the following. One can note that the (oriented) lengths of the segments $C Y$ and $B X$ are both linear functions of a parameter $t=\cot \angle P X M$. As $t$ varies, the intersection point $S$ of the perpendicular bisectors of $P X$ and $P Y$ traces a fixed line, thus the family of circles $P X Y$ has a fixed common point (other than $P$ ). By checking particular cases, one can show that this fixed point is $A$. Comment 4. The problem states that $\angle P X M=\angle P Y M$ implies that $A P X Y$ is cyclic. The original submission claims that these two conditions are in fact equivalent. The Problem Selection Committee omitted the converse part, since it follows easily from the direct one, by reversing arguments.
|
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|
f5925bff-4299-5bf1-b2d5-357df2b50075
| 23,590
|
Let $A B C$ be a triangle with circumcircle $\omega$ and incentre $I$. A line $\ell$ intersects the lines $A I, B I$, and $C I$ at points $D, E$, and $F$, respectively, distinct from the points $A, B, C$, and $I$. The perpendicular bisectors $x, y$, and $z$ of the segments $A D, B E$, and $C F$, respectively determine a triangle $\Theta$. Show that the circumcircle of the triangle $\Theta$ is tangent to $\omega$. (Denmark)
|
Claim 1. The reflections $\ell_{a}, \ell_{b}$ and $\ell_{c}$ of the line $\ell$ in the lines $x, y$, and $z$, respectively, are concurrent at a point $T$ which belongs to $\omega$.  Proof. Notice that $\Varangle\left(\ell_{b}, \ell_{c}\right)=\Varangle\left(\ell_{b}, \ell\right)+\Varangle\left(\ell, \ell_{c}\right)=2 \Varangle(y, \ell)+2 \Varangle(\ell, z)=2 \Varangle(y, z)$. But $y \perp B I$ and $z \perp C I$ implies $\Varangle(y, z)=\Varangle(B I, I C)$, so, since $2 \Varangle(B I, I C)=\Varangle(B A, A C)$, we obtain $$ \Varangle\left(\ell_{b}, \ell_{c}\right)=\Varangle(B A, A C) . $$ Since $A$ is the reflection of $D$ in $x$, $A$ belongs to $\ell_{a}$; similarly, $B$ belongs to $\ell_{b}$. Then (1) shows that the common point $T^{\prime}$ of $\ell_{a}$ and $\ell_{b}$ lies on $\omega$; similarly, the common point $T^{\prime \prime}$ of $\ell_{c}$ and $\ell_{b}$ lies on $\omega$. If $B \notin \ell_{a}$ and $B \notin \ell_{c}$, then $T^{\prime}$ and $T^{\prime \prime}$ are the second point of intersection of $\ell_{b}$ and $\omega$, hence they coincide. Otherwise, if, say, $B \in \ell_{c}$, then $\ell_{c}=B C$, so $\Varangle(B A, A C)=\Varangle\left(\ell_{b}, \ell_{c}\right)=\Varangle\left(\ell_{b}, B C\right)$, which shows that $\ell_{b}$ is tangent at $B$ to $\omega$ and $T^{\prime}=T^{\prime \prime}=B$. So $T^{\prime}$ and $T^{\prime \prime}$ coincide in all the cases, and the conclusion of the claim follows. Now we prove that $X, X_{0}, T$ are collinear. Denote by $D_{b}$ and $D_{c}$ the reflections of the point $D$ in the lines $y$ and $z$, respectively. Then $D_{b}$ lies on $\ell_{b}, D_{c}$ lies on $\ell_{c}$, and $$ \begin{aligned} \Varangle\left(D_{b} X, X D_{c}\right) & =\Varangle\left(D_{b} X, D X\right)+\Varangle\left(D X, X D_{c}\right)=2 \Varangle(y, D X)+2 \Varangle(D X, z)=2 \Varangle(y, z) \\ & =\Varangle(B A, A C)=\Varangle(B T, T C), \end{aligned} $$ hence the quadrilateral $X D_{b} T D_{c}$ is cyclic. Notice also that since $X D_{b}=X D=X D_{c}$, the points $D, D_{b}, D_{c}$ lie on a circle with centre $X$. Using in this circle the diameter $D_{c} D_{c}^{\prime}$ yields $\Varangle\left(D_{b} D_{c}, D_{c} X\right)=90^{\circ}+\Varangle\left(D_{b} D_{c}^{\prime}, D_{c}^{\prime} X\right)=90^{\circ}+\Varangle\left(D_{b} D, D D_{c}\right)$. Therefore, $$ \begin{gathered} \Varangle\left(\ell_{b}, X T\right)=\Varangle\left(D_{b} T, X T\right)=\Varangle\left(D_{b} D_{c}, D_{c} X\right)=90^{\circ}+\Varangle\left(D_{b} D, D D_{c}\right) \\ =90^{\circ}+\Varangle(B I, I C)=\Varangle(B A, A I)=\Varangle\left(B A, A X_{0}\right)=\Varangle\left(B T, T X_{0}\right)=\Varangle\left(\ell_{b}, X_{0} T\right) \end{gathered} $$ so the points $X, X_{0}, T$ are collinear. By a similar argument, $Y, Y_{0}, T$ and $Z, Z_{0}, T$ are collinear. As mentioned in the preamble, the statement of the problem follows. Comment 1. After proving Claim 1 one may proceed in another way. As it was shown, the reflections of $\ell$ in the sidelines of $X Y Z$ are concurrent at $T$. Thus $\ell$ is the Steiner line of $T$ with respect to $\triangle X Y Z$ (that is the line containing the reflections $T_{a}, T_{b}, T_{c}$ of $T$ in the sidelines of $X Y Z$ ). The properties of the Steiner line imply that $T$ lies on $\Omega$, and $\ell$ passes through the orthocentre $H$ of the triangle $X Y Z$.  Let $H_{a}, H_{b}$, and $H_{c}$ be the reflections of the point $H$ in the lines $x, y$, and $z$, respectively. Then the triangle $H_{a} H_{b} H_{c}$ is inscribed in $\Omega$ and homothetic to $A B C$ (by an easy angle chasing). Since $H_{a} \in \ell_{a}, H_{b} \in \ell_{b}$, and $H_{c} \in \ell_{c}$, the triangles $H_{a} H_{b} H_{c}$ and $A B C$ form a required pair of triangles $\Delta$ and $\delta$ mentioned in the preamble. Comment 2. The following observation shows how one may guess the description of the tangency point $T$ from Solution 1. Let us fix a direction and move the line $\ell$ parallel to this direction with constant speed. Then the points $D, E$, and $F$ are moving with constant speeds along the lines $A I, B I$, and $C I$, respectively. In this case $x, y$, and $z$ are moving with constant speeds, defining a family of homothetic triangles $X Y Z$ with a common centre of homothety $T$. Notice that the triangle $X_{0} Y_{0} Z_{0}$ belongs to this family (for $\ell$ passing through $I$ ). We may specify the location of $T$ considering the degenerate case when $x, y$, and $z$ are concurrent. In this degenerate case all the lines $x, y, z, \ell, \ell_{a}, \ell_{b}, \ell_{c}$ have a common point. Note that the lines $\ell_{a}, \ell_{b}, \ell_{c}$ remain constant as $\ell$ is moving (keeping its direction). Thus $T$ should be the common point of $\ell_{a}, \ell_{b}$, and $\ell_{c}$, lying on $\omega$.
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
Let $A B C$ be a triangle with circumcircle $\omega$ and incentre $I$. A line $\ell$ intersects the lines $A I, B I$, and $C I$ at points $D, E$, and $F$, respectively, distinct from the points $A, B, C$, and $I$. The perpendicular bisectors $x, y$, and $z$ of the segments $A D, B E$, and $C F$, respectively determine a triangle $\Theta$. Show that the circumcircle of the triangle $\Theta$ is tangent to $\omega$. (Denmark)
|
Claim 1. The reflections $\ell_{a}, \ell_{b}$ and $\ell_{c}$ of the line $\ell$ in the lines $x, y$, and $z$, respectively, are concurrent at a point $T$ which belongs to $\omega$.  Proof. Notice that $\Varangle\left(\ell_{b}, \ell_{c}\right)=\Varangle\left(\ell_{b}, \ell\right)+\Varangle\left(\ell, \ell_{c}\right)=2 \Varangle(y, \ell)+2 \Varangle(\ell, z)=2 \Varangle(y, z)$. But $y \perp B I$ and $z \perp C I$ implies $\Varangle(y, z)=\Varangle(B I, I C)$, so, since $2 \Varangle(B I, I C)=\Varangle(B A, A C)$, we obtain $$ \Varangle\left(\ell_{b}, \ell_{c}\right)=\Varangle(B A, A C) . $$ Since $A$ is the reflection of $D$ in $x$, $A$ belongs to $\ell_{a}$; similarly, $B$ belongs to $\ell_{b}$. Then (1) shows that the common point $T^{\prime}$ of $\ell_{a}$ and $\ell_{b}$ lies on $\omega$; similarly, the common point $T^{\prime \prime}$ of $\ell_{c}$ and $\ell_{b}$ lies on $\omega$. If $B \notin \ell_{a}$ and $B \notin \ell_{c}$, then $T^{\prime}$ and $T^{\prime \prime}$ are the second point of intersection of $\ell_{b}$ and $\omega$, hence they coincide. Otherwise, if, say, $B \in \ell_{c}$, then $\ell_{c}=B C$, so $\Varangle(B A, A C)=\Varangle\left(\ell_{b}, \ell_{c}\right)=\Varangle\left(\ell_{b}, B C\right)$, which shows that $\ell_{b}$ is tangent at $B$ to $\omega$ and $T^{\prime}=T^{\prime \prime}=B$. So $T^{\prime}$ and $T^{\prime \prime}$ coincide in all the cases, and the conclusion of the claim follows. Now we prove that $X, X_{0}, T$ are collinear. Denote by $D_{b}$ and $D_{c}$ the reflections of the point $D$ in the lines $y$ and $z$, respectively. Then $D_{b}$ lies on $\ell_{b}, D_{c}$ lies on $\ell_{c}$, and $$ \begin{aligned} \Varangle\left(D_{b} X, X D_{c}\right) & =\Varangle\left(D_{b} X, D X\right)+\Varangle\left(D X, X D_{c}\right)=2 \Varangle(y, D X)+2 \Varangle(D X, z)=2 \Varangle(y, z) \\ & =\Varangle(B A, A C)=\Varangle(B T, T C), \end{aligned} $$ hence the quadrilateral $X D_{b} T D_{c}$ is cyclic. Notice also that since $X D_{b}=X D=X D_{c}$, the points $D, D_{b}, D_{c}$ lie on a circle with centre $X$. Using in this circle the diameter $D_{c} D_{c}^{\prime}$ yields $\Varangle\left(D_{b} D_{c}, D_{c} X\right)=90^{\circ}+\Varangle\left(D_{b} D_{c}^{\prime}, D_{c}^{\prime} X\right)=90^{\circ}+\Varangle\left(D_{b} D, D D_{c}\right)$. Therefore, $$ \begin{gathered} \Varangle\left(\ell_{b}, X T\right)=\Varangle\left(D_{b} T, X T\right)=\Varangle\left(D_{b} D_{c}, D_{c} X\right)=90^{\circ}+\Varangle\left(D_{b} D, D D_{c}\right) \\ =90^{\circ}+\Varangle(B I, I C)=\Varangle(B A, A I)=\Varangle\left(B A, A X_{0}\right)=\Varangle\left(B T, T X_{0}\right)=\Varangle\left(\ell_{b}, X_{0} T\right) \end{gathered} $$ so the points $X, X_{0}, T$ are collinear. By a similar argument, $Y, Y_{0}, T$ and $Z, Z_{0}, T$ are collinear. As mentioned in the preamble, the statement of the problem follows. Comment 1. After proving Claim 1 one may proceed in another way. As it was shown, the reflections of $\ell$ in the sidelines of $X Y Z$ are concurrent at $T$. Thus $\ell$ is the Steiner line of $T$ with respect to $\triangle X Y Z$ (that is the line containing the reflections $T_{a}, T_{b}, T_{c}$ of $T$ in the sidelines of $X Y Z$ ). The properties of the Steiner line imply that $T$ lies on $\Omega$, and $\ell$ passes through the orthocentre $H$ of the triangle $X Y Z$.  Let $H_{a}, H_{b}$, and $H_{c}$ be the reflections of the point $H$ in the lines $x, y$, and $z$, respectively. Then the triangle $H_{a} H_{b} H_{c}$ is inscribed in $\Omega$ and homothetic to $A B C$ (by an easy angle chasing). Since $H_{a} \in \ell_{a}, H_{b} \in \ell_{b}$, and $H_{c} \in \ell_{c}$, the triangles $H_{a} H_{b} H_{c}$ and $A B C$ form a required pair of triangles $\Delta$ and $\delta$ mentioned in the preamble. Comment 2. The following observation shows how one may guess the description of the tangency point $T$ from Solution 1. Let us fix a direction and move the line $\ell$ parallel to this direction with constant speed. Then the points $D, E$, and $F$ are moving with constant speeds along the lines $A I, B I$, and $C I$, respectively. In this case $x, y$, and $z$ are moving with constant speeds, defining a family of homothetic triangles $X Y Z$ with a common centre of homothety $T$. Notice that the triangle $X_{0} Y_{0} Z_{0}$ belongs to this family (for $\ell$ passing through $I$ ). We may specify the location of $T$ considering the degenerate case when $x, y$, and $z$ are concurrent. In this degenerate case all the lines $x, y, z, \ell, \ell_{a}, \ell_{b}, \ell_{c}$ have a common point. Note that the lines $\ell_{a}, \ell_{b}, \ell_{c}$ remain constant as $\ell$ is moving (keeping its direction). Thus $T$ should be the common point of $\ell_{a}, \ell_{b}$, and $\ell_{c}$, lying on $\omega$.
|
{
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|
447d7e84-425b-55ee-b286-7fae4ff79645
| 23,599
|
Let $A B C$ be a triangle with circumcircle $\omega$ and incentre $I$. A line $\ell$ intersects the lines $A I, B I$, and $C I$ at points $D, E$, and $F$, respectively, distinct from the points $A, B, C$, and $I$. The perpendicular bisectors $x, y$, and $z$ of the segments $A D, B E$, and $C F$, respectively determine a triangle $\Theta$. Show that the circumcircle of the triangle $\Theta$ is tangent to $\omega$. (Denmark)
|
As mentioned in the preamble, it is sufficient to prove that the centre $T$ of the homothety taking $X Y Z$ to $X_{0} Y_{0} Z_{0}$ belongs to $\omega$. Thus, it suffices to prove that $\Varangle\left(T X_{0}, T Y_{0}\right)=$ $\Varangle\left(Z_{0} X_{0}, Z_{0} Y_{0}\right)$, or, equivalently, $\Varangle\left(X X_{0}, Y Y_{0}\right)=\Varangle\left(Z_{0} X_{0}, Z_{0} Y_{0}\right)$. Recall that $Y Z$ and $Y_{0} Z_{0}$ are the perpendicular bisectors of $A D$ and $A I$, respectively. Then, the vector $\vec{x}$ perpendicular to $Y Z$ and shifting the line $Y_{0} Z_{0}$ to $Y Z$ is equal to $\frac{1}{2} \overrightarrow{I D}$. Define the shifting vectors $\vec{y}=\frac{1}{2} \overrightarrow{I E}, \vec{z}=\frac{1}{2} \overrightarrow{I F}$ similarly. Consider now the triangle $U V W$ formed by the perpendiculars to $A I, B I$, and $C I$ through $D, E$, and $F$, respectively (see figure below). This is another triangle whose sides are parallel to the corresponding sides of $X Y Z$. Claim 2. $\overrightarrow{I U}=2 \overrightarrow{X_{0} X}, \overrightarrow{I V}=2 \overrightarrow{Y_{0} Y}, \overrightarrow{I W}=2 \overrightarrow{Z_{0} Z}$. Proof. We prove one of the relations, the other proofs being similar. To prove the equality of two vectors it suffices to project them onto two non-parallel axes and check that their projections are equal. The projection of $\overrightarrow{X_{0} X}$ onto $I B$ equals $\vec{y}$, while the projection of $\overrightarrow{I U}$ onto $I B$ is $\overrightarrow{I E}=2 \vec{y}$. The projections onto the other axis $I C$ are $\vec{z}$ and $\overrightarrow{I F}=2 \vec{z}$. Then $\overrightarrow{I U}=2 \overrightarrow{X_{0} X}$ follows. Notice that the line $\ell$ is the Simson line of the point $I$ with respect to the triangle $U V W$; thus $U, V, W$, and $I$ are concyclic. It follows from Claim 2 that $\Varangle\left(X X_{0}, Y Y_{0}\right)=\Varangle(I U, I V)=$ $\Varangle(W U, W V)=\Varangle\left(Z_{0} X_{0}, Z_{0} Y_{0}\right)$, and we are done. 
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
Let $A B C$ be a triangle with circumcircle $\omega$ and incentre $I$. A line $\ell$ intersects the lines $A I, B I$, and $C I$ at points $D, E$, and $F$, respectively, distinct from the points $A, B, C$, and $I$. The perpendicular bisectors $x, y$, and $z$ of the segments $A D, B E$, and $C F$, respectively determine a triangle $\Theta$. Show that the circumcircle of the triangle $\Theta$ is tangent to $\omega$. (Denmark)
|
As mentioned in the preamble, it is sufficient to prove that the centre $T$ of the homothety taking $X Y Z$ to $X_{0} Y_{0} Z_{0}$ belongs to $\omega$. Thus, it suffices to prove that $\Varangle\left(T X_{0}, T Y_{0}\right)=$ $\Varangle\left(Z_{0} X_{0}, Z_{0} Y_{0}\right)$, or, equivalently, $\Varangle\left(X X_{0}, Y Y_{0}\right)=\Varangle\left(Z_{0} X_{0}, Z_{0} Y_{0}\right)$. Recall that $Y Z$ and $Y_{0} Z_{0}$ are the perpendicular bisectors of $A D$ and $A I$, respectively. Then, the vector $\vec{x}$ perpendicular to $Y Z$ and shifting the line $Y_{0} Z_{0}$ to $Y Z$ is equal to $\frac{1}{2} \overrightarrow{I D}$. Define the shifting vectors $\vec{y}=\frac{1}{2} \overrightarrow{I E}, \vec{z}=\frac{1}{2} \overrightarrow{I F}$ similarly. Consider now the triangle $U V W$ formed by the perpendiculars to $A I, B I$, and $C I$ through $D, E$, and $F$, respectively (see figure below). This is another triangle whose sides are parallel to the corresponding sides of $X Y Z$. Claim 2. $\overrightarrow{I U}=2 \overrightarrow{X_{0} X}, \overrightarrow{I V}=2 \overrightarrow{Y_{0} Y}, \overrightarrow{I W}=2 \overrightarrow{Z_{0} Z}$. Proof. We prove one of the relations, the other proofs being similar. To prove the equality of two vectors it suffices to project them onto two non-parallel axes and check that their projections are equal. The projection of $\overrightarrow{X_{0} X}$ onto $I B$ equals $\vec{y}$, while the projection of $\overrightarrow{I U}$ onto $I B$ is $\overrightarrow{I E}=2 \vec{y}$. The projections onto the other axis $I C$ are $\vec{z}$ and $\overrightarrow{I F}=2 \vec{z}$. Then $\overrightarrow{I U}=2 \overrightarrow{X_{0} X}$ follows. Notice that the line $\ell$ is the Simson line of the point $I$ with respect to the triangle $U V W$; thus $U, V, W$, and $I$ are concyclic. It follows from Claim 2 that $\Varangle\left(X X_{0}, Y Y_{0}\right)=\Varangle(I U, I V)=$ $\Varangle(W U, W V)=\Varangle\left(Z_{0} X_{0}, Z_{0} Y_{0}\right)$, and we are done. 
|
{
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|
447d7e84-425b-55ee-b286-7fae4ff79645
| 23,599
|
Let $A B C$ be a triangle with circumcircle $\omega$ and incentre $I$. A line $\ell$ intersects the lines $A I, B I$, and $C I$ at points $D, E$, and $F$, respectively, distinct from the points $A, B, C$, and $I$. The perpendicular bisectors $x, y$, and $z$ of the segments $A D, B E$, and $C F$, respectively determine a triangle $\Theta$. Show that the circumcircle of the triangle $\Theta$ is tangent to $\omega$. (Denmark)
|
Let $I_{a}, I_{b}$, and $I_{c}$ be the excentres of triangle $A B C$ corresponding to $A, B$, and $C$, respectively. Also, let $u, v$, and $w$ be the lines through $D, E$, and $F$ which are perpendicular to $A I, B I$, and $C I$, respectively, and let $U V W$ be the triangle determined by these lines, where $u=V W, v=U W$ and $w=U V$ (see figure above). Notice that the line $u$ is the reflection of $I_{b} I_{c}$ in the line $x$, because $u, x$, and $I_{b} I_{c}$ are perpendicular to $A D$ and $x$ is the perpendicular bisector of $A D$. Likewise, $v$ and $I_{a} I_{c}$ are reflections of each other in $y$, while $w$ and $I_{a} I_{b}$ are reflections of each other in $z$. It follows that $X, Y$, and $Z$ are the midpoints of $U I_{a}, V I_{b}$ and $W I_{c}$, respectively, and that the triangles $U V W$, $X Y Z$ and $I_{a} I_{b} I_{c}$ are either translates of each other or homothetic with a common homothety centre. Construct the points $T$ and $S$ such that the quadrilaterals $U V I W, X Y T Z$ and $I_{a} I_{b} S I_{c}$ are homothetic. Then $T$ is the midpoint of $I S$. Moreover, note that $\ell$ is the Simson line of the point $I$ with respect to the triangle $U V W$, hence $I$ belongs to the circumcircle of the triangle $U V W$, therefore $T$ belongs to $\Omega$. Consider now the homothety or translation $h_{1}$ that maps $X Y Z T$ to $I_{a} I_{b} I_{c} S$ and the homothety $h_{2}$ with centre $I$ and factor $\frac{1}{2}$. Furthermore, let $h=h_{2} \circ h_{1}$. The transform $h$ can be a homothety or a translation, and $$ h(T)=h_{2}\left(h_{1}(T)\right)=h_{2}(S)=T $$ hence $T$ is a fixed point of $h$. So, $h$ is a homothety with centre $T$. Note that $h_{2}$ maps the excentres $I_{a}, I_{b}, I_{c}$ to $X_{0}, Y_{0}, Z_{0}$ defined in the preamble. Thus the centre $T$ of the homothety taking $X Y Z$ to $X_{0} Y_{0} Z_{0}$ belongs to $\Omega$, and this completes the proof.
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
Let $A B C$ be a triangle with circumcircle $\omega$ and incentre $I$. A line $\ell$ intersects the lines $A I, B I$, and $C I$ at points $D, E$, and $F$, respectively, distinct from the points $A, B, C$, and $I$. The perpendicular bisectors $x, y$, and $z$ of the segments $A D, B E$, and $C F$, respectively determine a triangle $\Theta$. Show that the circumcircle of the triangle $\Theta$ is tangent to $\omega$. (Denmark)
|
Let $I_{a}, I_{b}$, and $I_{c}$ be the excentres of triangle $A B C$ corresponding to $A, B$, and $C$, respectively. Also, let $u, v$, and $w$ be the lines through $D, E$, and $F$ which are perpendicular to $A I, B I$, and $C I$, respectively, and let $U V W$ be the triangle determined by these lines, where $u=V W, v=U W$ and $w=U V$ (see figure above). Notice that the line $u$ is the reflection of $I_{b} I_{c}$ in the line $x$, because $u, x$, and $I_{b} I_{c}$ are perpendicular to $A D$ and $x$ is the perpendicular bisector of $A D$. Likewise, $v$ and $I_{a} I_{c}$ are reflections of each other in $y$, while $w$ and $I_{a} I_{b}$ are reflections of each other in $z$. It follows that $X, Y$, and $Z$ are the midpoints of $U I_{a}, V I_{b}$ and $W I_{c}$, respectively, and that the triangles $U V W$, $X Y Z$ and $I_{a} I_{b} I_{c}$ are either translates of each other or homothetic with a common homothety centre. Construct the points $T$ and $S$ such that the quadrilaterals $U V I W, X Y T Z$ and $I_{a} I_{b} S I_{c}$ are homothetic. Then $T$ is the midpoint of $I S$. Moreover, note that $\ell$ is the Simson line of the point $I$ with respect to the triangle $U V W$, hence $I$ belongs to the circumcircle of the triangle $U V W$, therefore $T$ belongs to $\Omega$. Consider now the homothety or translation $h_{1}$ that maps $X Y Z T$ to $I_{a} I_{b} I_{c} S$ and the homothety $h_{2}$ with centre $I$ and factor $\frac{1}{2}$. Furthermore, let $h=h_{2} \circ h_{1}$. The transform $h$ can be a homothety or a translation, and $$ h(T)=h_{2}\left(h_{1}(T)\right)=h_{2}(S)=T $$ hence $T$ is a fixed point of $h$. So, $h$ is a homothety with centre $T$. Note that $h_{2}$ maps the excentres $I_{a}, I_{b}, I_{c}$ to $X_{0}, Y_{0}, Z_{0}$ defined in the preamble. Thus the centre $T$ of the homothety taking $X Y Z$ to $X_{0} Y_{0} Z_{0}$ belongs to $\Omega$, and this completes the proof.
|
{
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|
447d7e84-425b-55ee-b286-7fae4ff79645
| 23,599
|
A convex quadrilateral $A B C D$ satisfies $A B \cdot C D=B C \cdot D A$. A point $X$ is chosen inside the quadrilateral so that $\angle X A B=\angle X C D$ and $\angle X B C=\angle X D A$. Prove that $\angle A X B+$ $\angle C X D=180^{\circ}$. (Poland)
|
Let $B^{\prime}$ be the reflection of $B$ in the internal angle bisector of $\angle A X C$, so that $\angle A X B^{\prime}=\angle C X B$ and $\angle C X B^{\prime}=\angle A X B$. If $X, D$, and $B^{\prime}$ are collinear, then we are done. Now assume the contrary. On the ray $X B^{\prime}$ take a point $E$ such that $X E \cdot X B=X A \cdot X C$, so that $\triangle A X E \sim$ $\triangle B X C$ and $\triangle C X E \sim \triangle B X A$. We have $\angle X C E+\angle X C D=\angle X B A+\angle X A B<180^{\circ}$ and $\angle X A E+\angle X A D=\angle X D A+\angle X A D<180^{\circ}$, which proves that $X$ lies inside the angles $\angle E C D$ and $\angle E A D$ of the quadrilateral $E A D C$. Moreover, $X$ lies in the interior of exactly one of the two triangles $E A D, E C D$ (and in the exterior of the other).  The similarities mentioned above imply $X A \cdot B C=X B \cdot A E$ and $X B \cdot C E=X C \cdot A B$. Multiplying these equalities with the given equality $A B \cdot C D=B C \cdot D A$, we obtain $X A \cdot C D$. $C E=X C \cdot A D \cdot A E$, or, equivalently, $$ \frac{X A \cdot D E}{A D \cdot A E}=\frac{X C \cdot D E}{C D \cdot C E} $$ Lemma. Let $P Q R$ be a triangle, and let $X$ be a point in the interior of the angle $Q P R$ such that $\angle Q P X=\angle P R X$. Then $\frac{P X \cdot Q R}{P Q \cdot P R}<1$ if and only if $X$ lies in the interior of the triangle $P Q R$. Proof. The locus of points $X$ with $\angle Q P X=\angle P R X$ lying inside the angle $Q P R$ is an arc $\alpha$ of the circle $\gamma$ through $R$ tangent to $P Q$ at $P$. Let $\gamma$ intersect the line $Q R$ again at $Y$ (if $\gamma$ is tangent to $Q R$, then set $Y=R)$. The similarity $\triangle Q P Y \sim \triangle Q R P$ yields $P Y=\frac{P Q \cdot P R}{Q R}$. Now it suffices to show that $P X<P Y$ if and only if $X$ lies in the interior of the triangle $P Q R$. Let $m$ be a line through $Y$ parallel to $P Q$. Notice that the points $Z$ of $\gamma$ satisfying $P Z<P Y$ are exactly those between the lines $m$ and $P Q$.
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
A convex quadrilateral $A B C D$ satisfies $A B \cdot C D=B C \cdot D A$. A point $X$ is chosen inside the quadrilateral so that $\angle X A B=\angle X C D$ and $\angle X B C=\angle X D A$. Prove that $\angle A X B+$ $\angle C X D=180^{\circ}$. (Poland)
|
Let $B^{\prime}$ be the reflection of $B$ in the internal angle bisector of $\angle A X C$, so that $\angle A X B^{\prime}=\angle C X B$ and $\angle C X B^{\prime}=\angle A X B$. If $X, D$, and $B^{\prime}$ are collinear, then we are done. Now assume the contrary. On the ray $X B^{\prime}$ take a point $E$ such that $X E \cdot X B=X A \cdot X C$, so that $\triangle A X E \sim$ $\triangle B X C$ and $\triangle C X E \sim \triangle B X A$. We have $\angle X C E+\angle X C D=\angle X B A+\angle X A B<180^{\circ}$ and $\angle X A E+\angle X A D=\angle X D A+\angle X A D<180^{\circ}$, which proves that $X$ lies inside the angles $\angle E C D$ and $\angle E A D$ of the quadrilateral $E A D C$. Moreover, $X$ lies in the interior of exactly one of the two triangles $E A D, E C D$ (and in the exterior of the other).  The similarities mentioned above imply $X A \cdot B C=X B \cdot A E$ and $X B \cdot C E=X C \cdot A B$. Multiplying these equalities with the given equality $A B \cdot C D=B C \cdot D A$, we obtain $X A \cdot C D$. $C E=X C \cdot A D \cdot A E$, or, equivalently, $$ \frac{X A \cdot D E}{A D \cdot A E}=\frac{X C \cdot D E}{C D \cdot C E} $$ Lemma. Let $P Q R$ be a triangle, and let $X$ be a point in the interior of the angle $Q P R$ such that $\angle Q P X=\angle P R X$. Then $\frac{P X \cdot Q R}{P Q \cdot P R}<1$ if and only if $X$ lies in the interior of the triangle $P Q R$. Proof. The locus of points $X$ with $\angle Q P X=\angle P R X$ lying inside the angle $Q P R$ is an arc $\alpha$ of the circle $\gamma$ through $R$ tangent to $P Q$ at $P$. Let $\gamma$ intersect the line $Q R$ again at $Y$ (if $\gamma$ is tangent to $Q R$, then set $Y=R)$. The similarity $\triangle Q P Y \sim \triangle Q R P$ yields $P Y=\frac{P Q \cdot P R}{Q R}$. Now it suffices to show that $P X<P Y$ if and only if $X$ lies in the interior of the triangle $P Q R$. Let $m$ be a line through $Y$ parallel to $P Q$. Notice that the points $Z$ of $\gamma$ satisfying $P Z<P Y$ are exactly those between the lines $m$ and $P Q$.
|
{
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|
ec4e8976-6e0a-5c42-87b9-7bec8492ca6e
| 23,604
|
A convex quadrilateral $A B C D$ satisfies $A B \cdot C D=B C \cdot D A$. A point $X$ is chosen inside the quadrilateral so that $\angle X A B=\angle X C D$ and $\angle X B C=\angle X D A$. Prove that $\angle A X B+$ $\angle C X D=180^{\circ}$. (Poland)
|
The solution consists of two parts. In Part 1 we show that it suffices to prove that $$ \frac{X B}{X D}=\frac{A B}{C D} $$ and $$ \frac{X A}{X C}=\frac{D A}{B C} $$ In Part 2 we establish these equalities. Part 1. Using the sine law and applying (1) we obtain $$ \frac{\sin \angle A X B}{\sin \angle X A B}=\frac{A B}{X B}=\frac{C D}{X D}=\frac{\sin \angle C X D}{\sin \angle X C D} $$ so $\sin \angle A X B=\sin \angle C X D$ by the problem conditions. Similarly, (2) yields $\sin \angle D X A=$ $\sin \angle B X C$. If at least one of the pairs $(\angle A X B, \angle C X D)$ and $(\angle B X C, \angle D X A)$ consists of supplementary angles, then we are done. Otherwise, $\angle A X B=\angle C X D$ and $\angle D X A=\angle B X C$. In this case $X=A C \cap B D$, and the problem conditions yield that $A B C D$ is a parallelogram and hence a rhombus. In this last case the claim also holds. Part 2. To prove the desired equality (1), invert $A B C D$ at centre $X$ with unit radius; the images of points are denoted by primes. We have $$ \angle A^{\prime} B^{\prime} C^{\prime}=\angle X B^{\prime} A^{\prime}+\angle X B^{\prime} C^{\prime}=\angle X A B+\angle X C B=\angle X C D+\angle X C B=\angle B C D . $$ Similarly, the corresponding angles of quadrilaterals $A B C D$ and $D^{\prime} A^{\prime} B^{\prime} C^{\prime}$ are equal. Moreover, we have $$ A^{\prime} B^{\prime} \cdot C^{\prime} D^{\prime}=\frac{A B}{X A \cdot X B} \cdot \frac{C D}{X C \cdot X D}=\frac{B C}{X B \cdot X C} \cdot \frac{D A}{X D \cdot D A}=B^{\prime} C^{\prime} \cdot D^{\prime} A^{\prime} $$  Now we need the following Lemma. Lemma. Assume that the corresponding angles of convex quadrilaterals $X Y Z T$ and $X^{\prime} Y^{\prime} Z^{\prime} T^{\prime}$ are equal, and that $X Y \cdot Z T=Y Z \cdot T X$ and $X^{\prime} Y^{\prime} \cdot Z^{\prime} T^{\prime}=Y^{\prime} Z^{\prime} \cdot T^{\prime} X^{\prime}$. Then the two quadrilaterals are similar. Proof. Take the quadrilateral $X Y Z_{1} T_{1}$ similar to $X^{\prime} Y^{\prime} Z^{\prime} T^{\prime}$ and sharing the side $X Y$ with $X Y Z T$, such that $Z_{1}$ and $T_{1}$ lie on the rays $Y Z$ and $X T$, respectively, and $Z_{1} T_{1} \| Z T$. We need to prove that $Z_{1}=Z$ and $T_{1}=T$. Assume the contrary. Without loss of generality, $T X>X T_{1}$. Let segments $X Z$ and $Z_{1} T_{1}$ intersect at $U$. We have $$ \frac{T_{1} X}{T_{1} Z_{1}}<\frac{T_{1} X}{T_{1} U}=\frac{T X}{Z T}=\frac{X Y}{Y Z}<\frac{X Y}{Y Z_{1}} $$ thus $T_{1} X \cdot Y Z_{1}<T_{1} Z_{1} \cdot X Y$. A contradiction.  It follows from the Lemma that the quadrilaterals $A B C D$ and $D^{\prime} A^{\prime} B^{\prime} C^{\prime}$ are similar, hence $$ \frac{B C}{A B}=\frac{A^{\prime} B^{\prime}}{D^{\prime} A^{\prime}}=\frac{A B}{X A \cdot X B} \cdot \frac{X D \cdot X A}{D A}=\frac{A B}{A D} \cdot \frac{X D}{X B} $$ and therefore $$ \frac{X B}{X D}=\frac{A B^{2}}{B C \cdot A D}=\frac{A B^{2}}{A B \cdot C D}=\frac{A B}{C D} $$ We obtain (1), as desired; (2) is proved similarly. Comment. Part 1 is an easy one, while part 2 seems to be crucial. On the other hand, after the proof of the similarity $D^{\prime} A^{\prime} B^{\prime} C^{\prime} \sim A B C D$ one may finish the solution in different ways, e.g., as follows. The similarity taking $D^{\prime} A^{\prime} B^{\prime} C^{\prime}$ to $A B C D$ maps $X$ to the point $X^{\prime}$ isogonally conjugate of $X$ with respect to $A B C D$ (i.e. to the point $X^{\prime}$ inside $A B C D$ such that $\angle B A X=\angle D A X^{\prime}$, $\left.\angle C B X=\angle A B X^{\prime}, \angle D C X=\angle B C X^{\prime}, \angle A D X=\angle C D X^{\prime}\right)$. It is known that the required equality $\angle A X B+\angle C X D=180^{\circ}$ is one of known conditions on a point $X$ inside $A B C D$ equivalent to the existence of its isogonal conjugate. This page is intentionally left blank
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
A convex quadrilateral $A B C D$ satisfies $A B \cdot C D=B C \cdot D A$. A point $X$ is chosen inside the quadrilateral so that $\angle X A B=\angle X C D$ and $\angle X B C=\angle X D A$. Prove that $\angle A X B+$ $\angle C X D=180^{\circ}$. (Poland)
|
The solution consists of two parts. In Part 1 we show that it suffices to prove that $$ \frac{X B}{X D}=\frac{A B}{C D} $$ and $$ \frac{X A}{X C}=\frac{D A}{B C} $$ In Part 2 we establish these equalities. Part 1. Using the sine law and applying (1) we obtain $$ \frac{\sin \angle A X B}{\sin \angle X A B}=\frac{A B}{X B}=\frac{C D}{X D}=\frac{\sin \angle C X D}{\sin \angle X C D} $$ so $\sin \angle A X B=\sin \angle C X D$ by the problem conditions. Similarly, (2) yields $\sin \angle D X A=$ $\sin \angle B X C$. If at least one of the pairs $(\angle A X B, \angle C X D)$ and $(\angle B X C, \angle D X A)$ consists of supplementary angles, then we are done. Otherwise, $\angle A X B=\angle C X D$ and $\angle D X A=\angle B X C$. In this case $X=A C \cap B D$, and the problem conditions yield that $A B C D$ is a parallelogram and hence a rhombus. In this last case the claim also holds. Part 2. To prove the desired equality (1), invert $A B C D$ at centre $X$ with unit radius; the images of points are denoted by primes. We have $$ \angle A^{\prime} B^{\prime} C^{\prime}=\angle X B^{\prime} A^{\prime}+\angle X B^{\prime} C^{\prime}=\angle X A B+\angle X C B=\angle X C D+\angle X C B=\angle B C D . $$ Similarly, the corresponding angles of quadrilaterals $A B C D$ and $D^{\prime} A^{\prime} B^{\prime} C^{\prime}$ are equal. Moreover, we have $$ A^{\prime} B^{\prime} \cdot C^{\prime} D^{\prime}=\frac{A B}{X A \cdot X B} \cdot \frac{C D}{X C \cdot X D}=\frac{B C}{X B \cdot X C} \cdot \frac{D A}{X D \cdot D A}=B^{\prime} C^{\prime} \cdot D^{\prime} A^{\prime} $$  Now we need the following Lemma. Lemma. Assume that the corresponding angles of convex quadrilaterals $X Y Z T$ and $X^{\prime} Y^{\prime} Z^{\prime} T^{\prime}$ are equal, and that $X Y \cdot Z T=Y Z \cdot T X$ and $X^{\prime} Y^{\prime} \cdot Z^{\prime} T^{\prime}=Y^{\prime} Z^{\prime} \cdot T^{\prime} X^{\prime}$. Then the two quadrilaterals are similar. Proof. Take the quadrilateral $X Y Z_{1} T_{1}$ similar to $X^{\prime} Y^{\prime} Z^{\prime} T^{\prime}$ and sharing the side $X Y$ with $X Y Z T$, such that $Z_{1}$ and $T_{1}$ lie on the rays $Y Z$ and $X T$, respectively, and $Z_{1} T_{1} \| Z T$. We need to prove that $Z_{1}=Z$ and $T_{1}=T$. Assume the contrary. Without loss of generality, $T X>X T_{1}$. Let segments $X Z$ and $Z_{1} T_{1}$ intersect at $U$. We have $$ \frac{T_{1} X}{T_{1} Z_{1}}<\frac{T_{1} X}{T_{1} U}=\frac{T X}{Z T}=\frac{X Y}{Y Z}<\frac{X Y}{Y Z_{1}} $$ thus $T_{1} X \cdot Y Z_{1}<T_{1} Z_{1} \cdot X Y$. A contradiction.  It follows from the Lemma that the quadrilaterals $A B C D$ and $D^{\prime} A^{\prime} B^{\prime} C^{\prime}$ are similar, hence $$ \frac{B C}{A B}=\frac{A^{\prime} B^{\prime}}{D^{\prime} A^{\prime}}=\frac{A B}{X A \cdot X B} \cdot \frac{X D \cdot X A}{D A}=\frac{A B}{A D} \cdot \frac{X D}{X B} $$ and therefore $$ \frac{X B}{X D}=\frac{A B^{2}}{B C \cdot A D}=\frac{A B^{2}}{A B \cdot C D}=\frac{A B}{C D} $$ We obtain (1), as desired; (2) is proved similarly. Comment. Part 1 is an easy one, while part 2 seems to be crucial. On the other hand, after the proof of the similarity $D^{\prime} A^{\prime} B^{\prime} C^{\prime} \sim A B C D$ one may finish the solution in different ways, e.g., as follows. The similarity taking $D^{\prime} A^{\prime} B^{\prime} C^{\prime}$ to $A B C D$ maps $X$ to the point $X^{\prime}$ isogonally conjugate of $X$ with respect to $A B C D$ (i.e. to the point $X^{\prime}$ inside $A B C D$ such that $\angle B A X=\angle D A X^{\prime}$, $\left.\angle C B X=\angle A B X^{\prime}, \angle D C X=\angle B C X^{\prime}, \angle A D X=\angle C D X^{\prime}\right)$. It is known that the required equality $\angle A X B+\angle C X D=180^{\circ}$ is one of known conditions on a point $X$ inside $A B C D$ equivalent to the existence of its isogonal conjugate. This page is intentionally left blank
|
{
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ec4e8976-6e0a-5c42-87b9-7bec8492ca6e
| 23,604
|
Let $O$ be the circumcentre, and $\Omega$ be the circumcircle of an acute-angled triangle $A B C$. Let $P$ be an arbitrary point on $\Omega$, distinct from $A, B, C$, and their antipodes in $\Omega$. Denote the circumcentres of the triangles $A O P, B O P$, and $C O P$ by $O_{A}, O_{B}$, and $O_{C}$, respectively. The lines $\ell_{A}, \ell_{B}$, and $\ell_{C}$ perpendicular to $B C, C A$, and $A B$ pass through $O_{A}, O_{B}$, and $O_{C}$, respectively. Prove that the circumcircle of the triangle formed by $\ell_{A}, \ell_{B}$, and $\ell_{C}$ is tangent to the line $O P$. (Russia)
|
As usual, we denote the directed angle between the lines $a$ and $b$ by $\Varangle(a, b)$. We frequently use the fact that $a_{1} \perp a_{2}$ and $b_{1} \perp b_{2}$ yield $\Varangle\left(a_{1}, b_{1}\right)=\Varangle\left(a_{2}, b_{2}\right)$. Let the lines $\ell_{B}$ and $\ell_{C}$ meet at $L_{A}$; define the points $L_{B}$ and $L_{C}$ similarly. Note that the sidelines of the triangle $L_{A} L_{B} L_{C}$ are perpendicular to the corresponding sidelines of $A B C$. Points $O_{A}, O_{B}, O_{C}$ are located on the corresponding sidelines of $L_{A} L_{B} L_{C}$; moreover, $O_{A}, O_{B}$, $O_{C}$ all lie on the perpendicular bisector of $O P$.  Claim 1. The points $L_{B}, P, O_{A}$, and $O_{C}$ are concyclic. Proof. Since $O$ is symmetric to $P$ in $O_{A} O_{C}$, we have $$ \Varangle\left(O_{A} P, O_{C} P\right)=\Varangle\left(O_{C} O, O_{A} O\right)=\Varangle(C P, A P)=\Varangle(C B, A B)=\Varangle\left(O_{A} L_{B}, O_{C} L_{B}\right) . $$ Denote the circle through $L_{B}, P, O_{A}$, and $O_{C}$ by $\omega_{B}$. Define the circles $\omega_{A}$ and $\omega_{C}$ similarly. Claim 2. The circumcircle of the triangle $L_{A} L_{B} L_{C}$ passes through $P$. Proof. From cyclic quadruples of points in the circles $\omega_{B}$ and $\omega_{C}$, we have $$ \begin{aligned} \Varangle\left(L_{C} L_{A}, L_{C} P\right) & =\Varangle\left(L_{C} O_{B}, L_{C} P\right)=\Varangle\left(O_{A} O_{B}, O_{A} P\right) \\ & =\Varangle\left(O_{A} O_{C}, O_{A} P\right)=\Varangle\left(L_{B} O_{C}, L_{B} P\right)=\Varangle\left(L_{B} L_{A}, L_{B} P\right) . \end{aligned} $$ Claim 3. The points $P, L_{C}$, and $C$ are collinear. Proof. We have $\Varangle\left(P L_{C}, L_{C} L_{A}\right)=\Varangle\left(P L_{C}, L_{C} O_{B}\right)=\Varangle\left(P O_{A}, O_{A} O_{B}\right)$. Further, since $O_{A}$ is the centre of the circle $A O P, \Varangle\left(P O_{A}, O_{A} O_{B}\right)=\Varangle(P A, A O)$. As $O$ is the circumcentre of the triangle $P C A, \Varangle(P A, A O)=\pi / 2-\Varangle(C A, C P)=\Varangle\left(C P, L_{C} L_{A}\right)$. We obtain $\Varangle\left(P L_{C}, L_{C} L_{A}\right)=$ $\Varangle\left(C P, L_{C} L_{A}\right)$, which shows that $P \in C L_{C}$. Similarly, the points $P, L_{A}, A$ are collinear, and the points $P, L_{B}, B$ are also collinear. Finally, the computation above also shows that $$ \Varangle\left(O P, P L_{A}\right)=\Varangle(P A, A O)=\Varangle\left(P L_{C}, L_{C} L_{A}\right) $$ which means that $O P$ is tangent to the circle $P L_{A} L_{B} L_{C}$. Comment 1. The proof of Claim 2 may be replaced by the following remark: since $P$ belongs to the circles $\omega_{A}$ and $\omega_{C}, P$ is the Miquel point of the four lines $\ell_{A}, \ell_{B}, \ell_{C}$, and $O_{A} O_{B} O_{C}$. Comment 2. Claims 2 and 3 can be proved in several different ways and, in particular, in the reverse order. Claim 3 implies that the triangles $A B C$ and $L_{A} L_{B} L_{C}$ are perspective with perspector $P$. Claim 2 can be derived from this observation using spiral similarity. Consider the centre $Q$ of the spiral similarity that maps $A B C$ to $L_{A} L_{B} L_{C}$. From known spiral similarity properties, the points $L_{A}, L_{B}, P, Q$ are concyclic, and so are $L_{A}, L_{C}, P, Q$. Comment 3. The final conclusion can also be proved it terms of spiral similarity: the spiral similarity with centre $Q$ located on the circle $A B C$ maps the circle $A B C$ to the circle $P L_{A} L_{B} L_{C}$. Thus these circles are orthogonal. Comment 4. Notice that the homothety with centre $O$ and ratio 2 takes $O_{A}$ to $A^{\prime}$ that is the common point of tangents to $\Omega$ at $A$ and $P$. Similarly, let this homothety take $O_{B}$ to $B^{\prime}$ and $O_{C}$ to $C^{\prime}$. Let the tangents to $\Omega$ at $B$ and $C$ meet at $A^{\prime \prime}$, and define the points $B^{\prime \prime}$ and $C^{\prime \prime}$ similarly. Now, replacing labels $O$ with $I, \Omega$ with $\omega$, and swapping labels $A \leftrightarrow A^{\prime \prime}, B \leftrightarrow B^{\prime \prime}, C \leftrightarrow C^{\prime \prime}$ we obtain the following Reformulation. Let $\omega$ be the incircle, and let $I$ be the incentre of a triangle $A B C$. Let $P$ be a point of $\omega$ (other than the points of contact of $\omega$ with the sides of $A B C$ ). The tangent to $\omega$ at $P$ meets the lines $A B, B C$, and $C A$ at $A^{\prime}, B^{\prime}$, and $C^{\prime}$, respectively. Line $\ell_{A}$ parallel to the internal angle bisector of $\angle B A C$ passes through $A^{\prime}$; define lines $\ell_{B}$ and $\ell_{C}$ similarly. Prove that the line $I P$ is tangent to the circumcircle of the triangle formed by $\ell_{A}, \ell_{B}$, and $\ell_{C}$. Though this formulation is equivalent to the original one, it seems more challenging, since the point of contact is now "hidden".
|
proof
|
Yes
|
Yes
|
proof
|
Geometry
|
Let $O$ be the circumcentre, and $\Omega$ be the circumcircle of an acute-angled triangle $A B C$. Let $P$ be an arbitrary point on $\Omega$, distinct from $A, B, C$, and their antipodes in $\Omega$. Denote the circumcentres of the triangles $A O P, B O P$, and $C O P$ by $O_{A}, O_{B}$, and $O_{C}$, respectively. The lines $\ell_{A}, \ell_{B}$, and $\ell_{C}$ perpendicular to $B C, C A$, and $A B$ pass through $O_{A}, O_{B}$, and $O_{C}$, respectively. Prove that the circumcircle of the triangle formed by $\ell_{A}, \ell_{B}$, and $\ell_{C}$ is tangent to the line $O P$. (Russia)
|
As usual, we denote the directed angle between the lines $a$ and $b$ by $\Varangle(a, b)$. We frequently use the fact that $a_{1} \perp a_{2}$ and $b_{1} \perp b_{2}$ yield $\Varangle\left(a_{1}, b_{1}\right)=\Varangle\left(a_{2}, b_{2}\right)$. Let the lines $\ell_{B}$ and $\ell_{C}$ meet at $L_{A}$; define the points $L_{B}$ and $L_{C}$ similarly. Note that the sidelines of the triangle $L_{A} L_{B} L_{C}$ are perpendicular to the corresponding sidelines of $A B C$. Points $O_{A}, O_{B}, O_{C}$ are located on the corresponding sidelines of $L_{A} L_{B} L_{C}$; moreover, $O_{A}, O_{B}$, $O_{C}$ all lie on the perpendicular bisector of $O P$.  Claim 1. The points $L_{B}, P, O_{A}$, and $O_{C}$ are concyclic. Proof. Since $O$ is symmetric to $P$ in $O_{A} O_{C}$, we have $$ \Varangle\left(O_{A} P, O_{C} P\right)=\Varangle\left(O_{C} O, O_{A} O\right)=\Varangle(C P, A P)=\Varangle(C B, A B)=\Varangle\left(O_{A} L_{B}, O_{C} L_{B}\right) . $$ Denote the circle through $L_{B}, P, O_{A}$, and $O_{C}$ by $\omega_{B}$. Define the circles $\omega_{A}$ and $\omega_{C}$ similarly. Claim 2. The circumcircle of the triangle $L_{A} L_{B} L_{C}$ passes through $P$. Proof. From cyclic quadruples of points in the circles $\omega_{B}$ and $\omega_{C}$, we have $$ \begin{aligned} \Varangle\left(L_{C} L_{A}, L_{C} P\right) & =\Varangle\left(L_{C} O_{B}, L_{C} P\right)=\Varangle\left(O_{A} O_{B}, O_{A} P\right) \\ & =\Varangle\left(O_{A} O_{C}, O_{A} P\right)=\Varangle\left(L_{B} O_{C}, L_{B} P\right)=\Varangle\left(L_{B} L_{A}, L_{B} P\right) . \end{aligned} $$ Claim 3. The points $P, L_{C}$, and $C$ are collinear. Proof. We have $\Varangle\left(P L_{C}, L_{C} L_{A}\right)=\Varangle\left(P L_{C}, L_{C} O_{B}\right)=\Varangle\left(P O_{A}, O_{A} O_{B}\right)$. Further, since $O_{A}$ is the centre of the circle $A O P, \Varangle\left(P O_{A}, O_{A} O_{B}\right)=\Varangle(P A, A O)$. As $O$ is the circumcentre of the triangle $P C A, \Varangle(P A, A O)=\pi / 2-\Varangle(C A, C P)=\Varangle\left(C P, L_{C} L_{A}\right)$. We obtain $\Varangle\left(P L_{C}, L_{C} L_{A}\right)=$ $\Varangle\left(C P, L_{C} L_{A}\right)$, which shows that $P \in C L_{C}$. Similarly, the points $P, L_{A}, A$ are collinear, and the points $P, L_{B}, B$ are also collinear. Finally, the computation above also shows that $$ \Varangle\left(O P, P L_{A}\right)=\Varangle(P A, A O)=\Varangle\left(P L_{C}, L_{C} L_{A}\right) $$ which means that $O P$ is tangent to the circle $P L_{A} L_{B} L_{C}$. Comment 1. The proof of Claim 2 may be replaced by the following remark: since $P$ belongs to the circles $\omega_{A}$ and $\omega_{C}, P$ is the Miquel point of the four lines $\ell_{A}, \ell_{B}, \ell_{C}$, and $O_{A} O_{B} O_{C}$. Comment 2. Claims 2 and 3 can be proved in several different ways and, in particular, in the reverse order. Claim 3 implies that the triangles $A B C$ and $L_{A} L_{B} L_{C}$ are perspective with perspector $P$. Claim 2 can be derived from this observation using spiral similarity. Consider the centre $Q$ of the spiral similarity that maps $A B C$ to $L_{A} L_{B} L_{C}$. From known spiral similarity properties, the points $L_{A}, L_{B}, P, Q$ are concyclic, and so are $L_{A}, L_{C}, P, Q$. Comment 3. The final conclusion can also be proved it terms of spiral similarity: the spiral similarity with centre $Q$ located on the circle $A B C$ maps the circle $A B C$ to the circle $P L_{A} L_{B} L_{C}$. Thus these circles are orthogonal. Comment 4. Notice that the homothety with centre $O$ and ratio 2 takes $O_{A}$ to $A^{\prime}$ that is the common point of tangents to $\Omega$ at $A$ and $P$. Similarly, let this homothety take $O_{B}$ to $B^{\prime}$ and $O_{C}$ to $C^{\prime}$. Let the tangents to $\Omega$ at $B$ and $C$ meet at $A^{\prime \prime}$, and define the points $B^{\prime \prime}$ and $C^{\prime \prime}$ similarly. Now, replacing labels $O$ with $I, \Omega$ with $\omega$, and swapping labels $A \leftrightarrow A^{\prime \prime}, B \leftrightarrow B^{\prime \prime}, C \leftrightarrow C^{\prime \prime}$ we obtain the following Reformulation. Let $\omega$ be the incircle, and let $I$ be the incentre of a triangle $A B C$. Let $P$ be a point of $\omega$ (other than the points of contact of $\omega$ with the sides of $A B C$ ). The tangent to $\omega$ at $P$ meets the lines $A B, B C$, and $C A$ at $A^{\prime}, B^{\prime}$, and $C^{\prime}$, respectively. Line $\ell_{A}$ parallel to the internal angle bisector of $\angle B A C$ passes through $A^{\prime}$; define lines $\ell_{B}$ and $\ell_{C}$ similarly. Prove that the line $I P$ is tangent to the circumcircle of the triangle formed by $\ell_{A}, \ell_{B}$, and $\ell_{C}$. Though this formulation is equivalent to the original one, it seems more challenging, since the point of contact is now "hidden".
|
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|
c9cab443-dedf-5d37-bd12-8c6b92990272
| 23,608
|
Let $n>1$ be a positive integer. Each cell of an $n \times n$ table contains an integer. Suppose that the following conditions are satisfied: (i) Each number in the table is congruent to 1 modulo $n$; (ii) The sum of numbers in any row, as well as the sum of numbers in any column, is congruent to $n$ modulo $n^{2}$. Let $R_{i}$ be the product of the numbers in the $i^{\text {th }}$ row, and $C_{j}$ be the product of the numbers in the $j^{\text {th }}$ column. Prove that the sums $R_{1}+\cdots+R_{n}$ and $C_{1}+\cdots+C_{n}$ are congruent modulo $n^{4}$. (Indonesia)
|
Let $A_{i, j}$ be the entry in the $i^{\text {th }}$ row and the $j^{\text {th }}$ column; let $P$ be the product of all $n^{2}$ entries. For convenience, denote $a_{i, j}=A_{i, j}-1$ and $r_{i}=R_{i}-1$. We show that $$ \sum_{i=1}^{n} R_{i} \equiv(n-1)+P \quad\left(\bmod n^{4}\right) $$ Due to symmetry of the problem conditions, the sum of all the $C_{j}$ is also congruent to $(n-1)+P$ modulo $n^{4}$, whence the conclusion. By condition $(i)$, the number $n$ divides $a_{i, j}$ for all $i$ and $j$. So, every product of at least two of the $a_{i, j}$ is divisible by $n^{2}$, hence $R_{i}=\prod_{j=1}^{n}\left(1+a_{i, j}\right)=1+\sum_{j=1}^{n} a_{i, j}+\sum_{1 \leqslant j_{1}<j_{2} \leqslant n} a_{i, j_{1}} a_{i, j_{2}}+\cdots \equiv 1+\sum_{j=1}^{n} a_{i, j} \equiv 1-n+\sum_{j=1}^{n} A_{i, j} \quad\left(\bmod n^{2}\right)$ for every index $i$. Using condition (ii), we obtain $R_{i} \equiv 1\left(\bmod n^{2}\right)$, and so $n^{2} \mid r_{i}$. Therefore, every product of at least two of the $r_{i}$ is divisible by $n^{4}$. Repeating the same argument, we obtain $$ P=\prod_{i=1}^{n} R_{i}=\prod_{i=1}^{n}\left(1+r_{i}\right) \equiv 1+\sum_{i=1}^{n} r_{i} \quad\left(\bmod n^{4}\right) $$ whence $$ \sum_{i=1}^{n} R_{i}=n+\sum_{i=1}^{n} r_{i} \equiv n+(P-1) \quad\left(\bmod n^{4}\right) $$ as desired. Comment. The original version of the problem statement contained also the condition (iii) The product of all the numbers in the table is congruent to 1 modulo $n^{4}$. This condition appears to be superfluous, so it was omitted.
|
proof
|
Yes
|
Yes
|
proof
|
Number Theory
|
Let $n>1$ be a positive integer. Each cell of an $n \times n$ table contains an integer. Suppose that the following conditions are satisfied: (i) Each number in the table is congruent to 1 modulo $n$; (ii) The sum of numbers in any row, as well as the sum of numbers in any column, is congruent to $n$ modulo $n^{2}$. Let $R_{i}$ be the product of the numbers in the $i^{\text {th }}$ row, and $C_{j}$ be the product of the numbers in the $j^{\text {th }}$ column. Prove that the sums $R_{1}+\cdots+R_{n}$ and $C_{1}+\cdots+C_{n}$ are congruent modulo $n^{4}$. (Indonesia)
|
Let $A_{i, j}$ be the entry in the $i^{\text {th }}$ row and the $j^{\text {th }}$ column; let $P$ be the product of all $n^{2}$ entries. For convenience, denote $a_{i, j}=A_{i, j}-1$ and $r_{i}=R_{i}-1$. We show that $$ \sum_{i=1}^{n} R_{i} \equiv(n-1)+P \quad\left(\bmod n^{4}\right) $$ Due to symmetry of the problem conditions, the sum of all the $C_{j}$ is also congruent to $(n-1)+P$ modulo $n^{4}$, whence the conclusion. By condition $(i)$, the number $n$ divides $a_{i, j}$ for all $i$ and $j$. So, every product of at least two of the $a_{i, j}$ is divisible by $n^{2}$, hence $R_{i}=\prod_{j=1}^{n}\left(1+a_{i, j}\right)=1+\sum_{j=1}^{n} a_{i, j}+\sum_{1 \leqslant j_{1}<j_{2} \leqslant n} a_{i, j_{1}} a_{i, j_{2}}+\cdots \equiv 1+\sum_{j=1}^{n} a_{i, j} \equiv 1-n+\sum_{j=1}^{n} A_{i, j} \quad\left(\bmod n^{2}\right)$ for every index $i$. Using condition (ii), we obtain $R_{i} \equiv 1\left(\bmod n^{2}\right)$, and so $n^{2} \mid r_{i}$. Therefore, every product of at least two of the $r_{i}$ is divisible by $n^{4}$. Repeating the same argument, we obtain $$ P=\prod_{i=1}^{n} R_{i}=\prod_{i=1}^{n}\left(1+r_{i}\right) \equiv 1+\sum_{i=1}^{n} r_{i} \quad\left(\bmod n^{4}\right) $$ whence $$ \sum_{i=1}^{n} R_{i}=n+\sum_{i=1}^{n} r_{i} \equiv n+(P-1) \quad\left(\bmod n^{4}\right) $$ as desired. Comment. The original version of the problem statement contained also the condition (iii) The product of all the numbers in the table is congruent to 1 modulo $n^{4}$. This condition appears to be superfluous, so it was omitted.
|
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69aea495-5c1a-5784-9692-9cbd8ec943c3
| 23,614
|
Let $n>1$ be a positive integer. Each cell of an $n \times n$ table contains an integer. Suppose that the following conditions are satisfied: (i) Each number in the table is congruent to 1 modulo $n$; (ii) The sum of numbers in any row, as well as the sum of numbers in any column, is congruent to $n$ modulo $n^{2}$. Let $R_{i}$ be the product of the numbers in the $i^{\text {th }}$ row, and $C_{j}$ be the product of the numbers in the $j^{\text {th }}$ column. Prove that the sums $R_{1}+\cdots+R_{n}$ and $C_{1}+\cdots+C_{n}$ are congruent modulo $n^{4}$. (Indonesia)
|
We present a more straightforward (though lengthier) way to establish (1). We also use the notation of $a_{i, j}$. By condition (i), all the $a_{i, j}$ are divisible by $n$. Therefore, we have $$ \begin{aligned} P=\prod_{i=1}^{n} \prod_{j=1}^{n}\left(1+a_{i, j}\right) \equiv 1+\sum_{(i, j)} a_{i, j} & +\sum_{\left(i_{1}, j_{1}\right),\left(i_{2}, j_{2}\right)} a_{i_{1}, j_{1}} a_{i_{2}, j_{2}} \\ & +\sum_{\left(i_{1}, j_{1}\right),\left(i_{2}, j_{2}\right),\left(i_{3}, j_{3}\right)} a_{i_{1}, j_{1}} a_{i_{2}, j_{2}} a_{i_{3}, j_{3}}\left(\bmod n^{4}\right), \end{aligned} $$ where the last two sums are taken over all unordered pairs/triples of pairwise different pairs $(i, j)$; such conventions are applied throughout the solution. Similarly, $$ \sum_{i=1}^{n} R_{i}=\sum_{i=1}^{n} \prod_{j=1}^{n}\left(1+a_{i, j}\right) \equiv n+\sum_{i} \sum_{j} a_{i, j}+\sum_{i} \sum_{j_{1}, j_{2}} a_{i, j_{1}} a_{i, j_{2}}+\sum_{i} \sum_{j_{1}, j_{2}, j_{3}} a_{i, j_{1}} a_{i, j_{2}} a_{i, j_{3}} \quad\left(\bmod n^{4}\right) $$ Therefore, $$ \begin{aligned} P+(n-1)-\sum_{i} R_{i} \equiv \sum_{\substack{\left(i_{1}, j_{1}\right),\left(i_{2}, j_{2}\right) \\ i_{1} \neq i_{2}}} a_{i_{1}, j_{1}} a_{i_{2}, j_{2}} & +\sum_{\substack{\left(i_{1}, j_{1}\right),\left(i_{2}, j_{2}\right),\left(i_{3}, j_{3}\right) \\ i_{1} \neq i_{2} \neq i_{3} \neq i_{1}}} a_{i_{1}, j_{1}} a_{i_{2}, j_{2}} a_{i_{3}, j_{3}} \\ & +\sum_{\substack{\left(i_{1}, j_{1}\right),\left(i_{2}, j_{2}\right),\left(i_{3}, j_{3}\right) \\ i_{1} \neq i_{2}=i_{3}}} a_{i_{1}, j_{1}} a_{i_{2}, j_{2}} a_{i_{3}, j_{3}}\left(\bmod n^{4}\right) . \end{aligned} $$ We show that in fact each of the three sums appearing in the right-hand part of this congruence is divisible by $n^{4}$; this yields (1). Denote those three sums by $\Sigma_{1}, \Sigma_{2}$, and $\Sigma_{3}$ in order of appearance. Recall that by condition (ii) we have $$ \sum_{j} a_{i, j} \equiv 0 \quad\left(\bmod n^{2}\right) \quad \text { for all indices } i $$ For every two indices $i_{1}<i_{2}$ we have $$ \sum_{j_{1}} \sum_{j_{2}} a_{i_{1}, j_{1}} a_{i_{2}, j_{2}}=\left(\sum_{j_{1}} a_{i_{1}, j_{1}}\right) \cdot\left(\sum_{j_{2}} a_{i_{2}, j_{2}}\right) \equiv 0 \quad\left(\bmod n^{4}\right) $$ since each of the two factors is divisible by $n^{2}$. Summing over all pairs $\left(i_{1}, i_{2}\right)$ we obtain $n^{4} \mid \Sigma_{1}$. Similarly, for every three indices $i_{1}<i_{2}<i_{3}$ we have $$ \sum_{j_{1}} \sum_{j_{2}} \sum_{j_{3}} a_{i_{1}, j_{1}} a_{i_{2}, j_{2}} a_{i_{3}, j_{3}}=\left(\sum_{j_{1}} a_{i_{1}, j_{1}}\right) \cdot\left(\sum_{j_{2}} a_{i_{2}, j_{2}}\right) \cdot\left(\sum_{j_{3}} a_{i_{3}, j_{3}}\right) $$ which is divisible even by $n^{6}$. Hence $n^{4} \mid \Sigma_{2}$. Finally, for every indices $i_{1} \neq i_{2}=i_{3}$ and $j_{2}<j_{3}$ we have $$ a_{i_{2}, j_{2}} \cdot a_{i_{2}, j_{3}} \cdot \sum_{j_{1}} a_{i_{1}, j_{1}} \equiv 0 \quad\left(\bmod n^{4}\right) $$ since the three factors are divisible by $n, n$, and $n^{2}$, respectively. Summing over all 4 -tuples of indices $\left(i_{1}, i_{2}, j_{2}, j_{3}\right)$ we get $n^{4} \mid \Sigma_{3}$.
|
proof
|
Yes
|
Yes
|
proof
|
Number Theory
|
Let $n>1$ be a positive integer. Each cell of an $n \times n$ table contains an integer. Suppose that the following conditions are satisfied: (i) Each number in the table is congruent to 1 modulo $n$; (ii) The sum of numbers in any row, as well as the sum of numbers in any column, is congruent to $n$ modulo $n^{2}$. Let $R_{i}$ be the product of the numbers in the $i^{\text {th }}$ row, and $C_{j}$ be the product of the numbers in the $j^{\text {th }}$ column. Prove that the sums $R_{1}+\cdots+R_{n}$ and $C_{1}+\cdots+C_{n}$ are congruent modulo $n^{4}$. (Indonesia)
|
We present a more straightforward (though lengthier) way to establish (1). We also use the notation of $a_{i, j}$. By condition (i), all the $a_{i, j}$ are divisible by $n$. Therefore, we have $$ \begin{aligned} P=\prod_{i=1}^{n} \prod_{j=1}^{n}\left(1+a_{i, j}\right) \equiv 1+\sum_{(i, j)} a_{i, j} & +\sum_{\left(i_{1}, j_{1}\right),\left(i_{2}, j_{2}\right)} a_{i_{1}, j_{1}} a_{i_{2}, j_{2}} \\ & +\sum_{\left(i_{1}, j_{1}\right),\left(i_{2}, j_{2}\right),\left(i_{3}, j_{3}\right)} a_{i_{1}, j_{1}} a_{i_{2}, j_{2}} a_{i_{3}, j_{3}}\left(\bmod n^{4}\right), \end{aligned} $$ where the last two sums are taken over all unordered pairs/triples of pairwise different pairs $(i, j)$; such conventions are applied throughout the solution. Similarly, $$ \sum_{i=1}^{n} R_{i}=\sum_{i=1}^{n} \prod_{j=1}^{n}\left(1+a_{i, j}\right) \equiv n+\sum_{i} \sum_{j} a_{i, j}+\sum_{i} \sum_{j_{1}, j_{2}} a_{i, j_{1}} a_{i, j_{2}}+\sum_{i} \sum_{j_{1}, j_{2}, j_{3}} a_{i, j_{1}} a_{i, j_{2}} a_{i, j_{3}} \quad\left(\bmod n^{4}\right) $$ Therefore, $$ \begin{aligned} P+(n-1)-\sum_{i} R_{i} \equiv \sum_{\substack{\left(i_{1}, j_{1}\right),\left(i_{2}, j_{2}\right) \\ i_{1} \neq i_{2}}} a_{i_{1}, j_{1}} a_{i_{2}, j_{2}} & +\sum_{\substack{\left(i_{1}, j_{1}\right),\left(i_{2}, j_{2}\right),\left(i_{3}, j_{3}\right) \\ i_{1} \neq i_{2} \neq i_{3} \neq i_{1}}} a_{i_{1}, j_{1}} a_{i_{2}, j_{2}} a_{i_{3}, j_{3}} \\ & +\sum_{\substack{\left(i_{1}, j_{1}\right),\left(i_{2}, j_{2}\right),\left(i_{3}, j_{3}\right) \\ i_{1} \neq i_{2}=i_{3}}} a_{i_{1}, j_{1}} a_{i_{2}, j_{2}} a_{i_{3}, j_{3}}\left(\bmod n^{4}\right) . \end{aligned} $$ We show that in fact each of the three sums appearing in the right-hand part of this congruence is divisible by $n^{4}$; this yields (1). Denote those three sums by $\Sigma_{1}, \Sigma_{2}$, and $\Sigma_{3}$ in order of appearance. Recall that by condition (ii) we have $$ \sum_{j} a_{i, j} \equiv 0 \quad\left(\bmod n^{2}\right) \quad \text { for all indices } i $$ For every two indices $i_{1}<i_{2}$ we have $$ \sum_{j_{1}} \sum_{j_{2}} a_{i_{1}, j_{1}} a_{i_{2}, j_{2}}=\left(\sum_{j_{1}} a_{i_{1}, j_{1}}\right) \cdot\left(\sum_{j_{2}} a_{i_{2}, j_{2}}\right) \equiv 0 \quad\left(\bmod n^{4}\right) $$ since each of the two factors is divisible by $n^{2}$. Summing over all pairs $\left(i_{1}, i_{2}\right)$ we obtain $n^{4} \mid \Sigma_{1}$. Similarly, for every three indices $i_{1}<i_{2}<i_{3}$ we have $$ \sum_{j_{1}} \sum_{j_{2}} \sum_{j_{3}} a_{i_{1}, j_{1}} a_{i_{2}, j_{2}} a_{i_{3}, j_{3}}=\left(\sum_{j_{1}} a_{i_{1}, j_{1}}\right) \cdot\left(\sum_{j_{2}} a_{i_{2}, j_{2}}\right) \cdot\left(\sum_{j_{3}} a_{i_{3}, j_{3}}\right) $$ which is divisible even by $n^{6}$. Hence $n^{4} \mid \Sigma_{2}$. Finally, for every indices $i_{1} \neq i_{2}=i_{3}$ and $j_{2}<j_{3}$ we have $$ a_{i_{2}, j_{2}} \cdot a_{i_{2}, j_{3}} \cdot \sum_{j_{1}} a_{i_{1}, j_{1}} \equiv 0 \quad\left(\bmod n^{4}\right) $$ since the three factors are divisible by $n, n$, and $n^{2}$, respectively. Summing over all 4 -tuples of indices $\left(i_{1}, i_{2}, j_{2}, j_{3}\right)$ we get $n^{4} \mid \Sigma_{3}$.
|
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69aea495-5c1a-5784-9692-9cbd8ec943c3
| 23,614
|
Define the sequence $a_{0}, a_{1}, a_{2}, \ldots$ by $a_{n}=2^{n}+2^{\lfloor n / 2\rfloor}$. Prove that there are infinitely many terms of the sequence which can be expressed as a sum of (two or more) distinct terms of the sequence, as well as infinitely many of those which cannot be expressed in such a way. (Serbia)
|
Call a nonnegative integer representable if it equals the sum of several (possibly 0 or 1) distinct terms of the sequence. We say that two nonnegative integers $b$ and $c$ are equivalent (written as $b \sim c$ ) if they are either both representable or both non-representable. One can easily compute $$ S_{n-1}:=a_{0}+\cdots+a_{n-1}=2^{n}+2^{\lceil n / 2\rceil}+2^{\lfloor n / 2\rfloor}-3 . $$ Indeed, we have $S_{n}-S_{n-1}=2^{n}+2^{\lfloor n / 2\rfloor}=a_{n}$ so we can use the induction. In particular, $S_{2 k-1}=2^{2 k}+2^{k+1}-3$. Note that, if $n \geqslant 3$, then $2^{[n / 2]} \geqslant 2^{2}>3$, so $$ S_{n-1}=2^{n}+2^{\lceil n / 2\rceil}+2^{\lfloor n / 2\rfloor}-3>2^{n}+2^{\lfloor n / 2\rfloor}=a_{n} . $$ Also notice that $S_{n-1}-a_{n}=2^{[n / 2]}-3<a_{n}$. The main tool of the solution is the following claim. Claim 1. Assume that $b$ is a positive integer such that $S_{n-1}-a_{n}<b<a_{n}$ for some $n \geqslant 3$. Then $b \sim S_{n-1}-b$. Proof. As seen above, we have $S_{n-1}>a_{n}$. Denote $c=S_{n-1}-b$; then $S_{n-1}-a_{n}<c<a_{n}$, so the roles of $b$ and $c$ are symmetrical. Assume that $b$ is representable. The representation cannot contain $a_{i}$ with $i \geqslant n$, since $b<a_{n}$. So $b$ is the sum of some subset of $\left\{a_{0}, a_{1}, \ldots, a_{n-1}\right\}$; then $c$ is the sum of the complement. The converse is obtained by swapping $b$ and $c$. We also need the following version of this claim. Claim 2. For any $n \geqslant 3$, the number $a_{n}$ can be represented as a sum of two or more distinct terms of the sequence if and only if $S_{n-1}-a_{n}=2^{[n / 2]}-3$ is representable. Proof. Denote $c=S_{n-1}-a_{n}<a_{n}$. If $a_{n}$ satisfies the required condition, then it is the sum of some subset of $\left\{a_{0}, a_{1}, \ldots, a_{n-1}\right\}$; then $c$ is the sum of the complement. Conversely, if $c$ is representable, then its representation consists only of the numbers from $\left\{a_{0}, \ldots, a_{n-1}\right\}$, so $a_{n}$ is the sum of the complement. By Claim 2, in order to prove the problem statement, it suffices to find infinitely many representable numbers of the form $2^{t}-3$, as well as infinitely many non-representable ones. Claim 3. For every $t \geqslant 3$, we have $2^{t}-3 \sim 2^{4 t-6}-3$, and $2^{4 t-6}-3>2^{t}-3$. Proof. The inequality follows from $t \geqslant 3$. In order to prove the equivalence, we apply Claim 1 twice in the following manner. First, since $S_{2 t-3}-a_{2 t-2}=2^{t-1}-3<2^{t}-3<2^{2 t-2}+2^{t-1}=a_{2 t-2}$, by Claim 1 we have $2^{t}-3 \sim S_{2 t-3}-\left(2^{t}-3\right)=2^{2 t-2}$. Second, since $S_{4 t-7}-a_{4 t-6}=2^{2 t-3}-3<2^{2 t-2}<2^{4 t-6}+2^{2 t-3}=a_{4 t-6}$, by Claim 1 we have $2^{2 t-2} \sim S_{4 t-7}-2^{2 t-2}=2^{4 t-6}-3$. Therefore, $2^{t}-3 \sim 2^{2 t-2} \sim 2^{4 t-6}-3$, as required. Now it is easy to find the required numbers. Indeed, the number $2^{3}-3=5=a_{0}+a_{1}$ is representable, so Claim 3 provides an infinite sequence of representable numbers $$ 2^{3}-3 \sim 2^{6}-3 \sim 2^{18}-3 \sim \cdots \sim 2^{t}-3 \sim 2^{4 t-6}-3 \sim \cdots \text {. } $$ On the other hand, the number $2^{7}-3=125$ is non-representable (since by Claim 1 we have $125 \sim S_{6}-125=24 \sim S_{4}-24=17 \sim S_{3}-17=4$ which is clearly non-representable). So Claim 3 provides an infinite sequence of non-representable numbers $$ 2^{7}-3 \sim 2^{22}-3 \sim 2^{82}-3 \sim \cdots \sim 2^{t}-3 \sim 2^{4 t-6}-3 \sim \cdots . $$
|
proof
|
Yes
|
Yes
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proof
|
Number Theory
|
Define the sequence $a_{0}, a_{1}, a_{2}, \ldots$ by $a_{n}=2^{n}+2^{\lfloor n / 2\rfloor}$. Prove that there are infinitely many terms of the sequence which can be expressed as a sum of (two or more) distinct terms of the sequence, as well as infinitely many of those which cannot be expressed in such a way. (Serbia)
|
Call a nonnegative integer representable if it equals the sum of several (possibly 0 or 1) distinct terms of the sequence. We say that two nonnegative integers $b$ and $c$ are equivalent (written as $b \sim c$ ) if they are either both representable or both non-representable. One can easily compute $$ S_{n-1}:=a_{0}+\cdots+a_{n-1}=2^{n}+2^{\lceil n / 2\rceil}+2^{\lfloor n / 2\rfloor}-3 . $$ Indeed, we have $S_{n}-S_{n-1}=2^{n}+2^{\lfloor n / 2\rfloor}=a_{n}$ so we can use the induction. In particular, $S_{2 k-1}=2^{2 k}+2^{k+1}-3$. Note that, if $n \geqslant 3$, then $2^{[n / 2]} \geqslant 2^{2}>3$, so $$ S_{n-1}=2^{n}+2^{\lceil n / 2\rceil}+2^{\lfloor n / 2\rfloor}-3>2^{n}+2^{\lfloor n / 2\rfloor}=a_{n} . $$ Also notice that $S_{n-1}-a_{n}=2^{[n / 2]}-3<a_{n}$. The main tool of the solution is the following claim. Claim 1. Assume that $b$ is a positive integer such that $S_{n-1}-a_{n}<b<a_{n}$ for some $n \geqslant 3$. Then $b \sim S_{n-1}-b$. Proof. As seen above, we have $S_{n-1}>a_{n}$. Denote $c=S_{n-1}-b$; then $S_{n-1}-a_{n}<c<a_{n}$, so the roles of $b$ and $c$ are symmetrical. Assume that $b$ is representable. The representation cannot contain $a_{i}$ with $i \geqslant n$, since $b<a_{n}$. So $b$ is the sum of some subset of $\left\{a_{0}, a_{1}, \ldots, a_{n-1}\right\}$; then $c$ is the sum of the complement. The converse is obtained by swapping $b$ and $c$. We also need the following version of this claim. Claim 2. For any $n \geqslant 3$, the number $a_{n}$ can be represented as a sum of two or more distinct terms of the sequence if and only if $S_{n-1}-a_{n}=2^{[n / 2]}-3$ is representable. Proof. Denote $c=S_{n-1}-a_{n}<a_{n}$. If $a_{n}$ satisfies the required condition, then it is the sum of some subset of $\left\{a_{0}, a_{1}, \ldots, a_{n-1}\right\}$; then $c$ is the sum of the complement. Conversely, if $c$ is representable, then its representation consists only of the numbers from $\left\{a_{0}, \ldots, a_{n-1}\right\}$, so $a_{n}$ is the sum of the complement. By Claim 2, in order to prove the problem statement, it suffices to find infinitely many representable numbers of the form $2^{t}-3$, as well as infinitely many non-representable ones. Claim 3. For every $t \geqslant 3$, we have $2^{t}-3 \sim 2^{4 t-6}-3$, and $2^{4 t-6}-3>2^{t}-3$. Proof. The inequality follows from $t \geqslant 3$. In order to prove the equivalence, we apply Claim 1 twice in the following manner. First, since $S_{2 t-3}-a_{2 t-2}=2^{t-1}-3<2^{t}-3<2^{2 t-2}+2^{t-1}=a_{2 t-2}$, by Claim 1 we have $2^{t}-3 \sim S_{2 t-3}-\left(2^{t}-3\right)=2^{2 t-2}$. Second, since $S_{4 t-7}-a_{4 t-6}=2^{2 t-3}-3<2^{2 t-2}<2^{4 t-6}+2^{2 t-3}=a_{4 t-6}$, by Claim 1 we have $2^{2 t-2} \sim S_{4 t-7}-2^{2 t-2}=2^{4 t-6}-3$. Therefore, $2^{t}-3 \sim 2^{2 t-2} \sim 2^{4 t-6}-3$, as required. Now it is easy to find the required numbers. Indeed, the number $2^{3}-3=5=a_{0}+a_{1}$ is representable, so Claim 3 provides an infinite sequence of representable numbers $$ 2^{3}-3 \sim 2^{6}-3 \sim 2^{18}-3 \sim \cdots \sim 2^{t}-3 \sim 2^{4 t-6}-3 \sim \cdots \text {. } $$ On the other hand, the number $2^{7}-3=125$ is non-representable (since by Claim 1 we have $125 \sim S_{6}-125=24 \sim S_{4}-24=17 \sim S_{3}-17=4$ which is clearly non-representable). So Claim 3 provides an infinite sequence of non-representable numbers $$ 2^{7}-3 \sim 2^{22}-3 \sim 2^{82}-3 \sim \cdots \sim 2^{t}-3 \sim 2^{4 t-6}-3 \sim \cdots . $$
|
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4b44c906-35d4-53d8-9d7a-5da8557f075e
| 23,618
|
Let $a_{1}, a_{2}, \ldots, a_{n}, \ldots$ be a sequence of positive integers such that $$ \frac{a_{1}}{a_{2}}+\frac{a_{2}}{a_{3}}+\cdots+\frac{a_{n-1}}{a_{n}}+\frac{a_{n}}{a_{1}} $$ is an integer for all $n \geqslant k$, where $k$ is some positive integer. Prove that there exists a positive integer $m$ such that $a_{n}=a_{n+1}$ for all $n \geqslant m$. (Mongolia)
|
The argument hinges on the following two facts: Let $a, b, c$ be positive integers such that $N=b / c+(c-b) / a$ is an integer. (1) If $\operatorname{gcd}(a, c)=1$, then $c$ divides $b$; and (2) If $\operatorname{gcd}(a, b, c)=1$, then $\operatorname{gcd}(a, b)=1$. To prove (1), write $a b=c(a N+b-c)$. Since $\operatorname{gcd}(a, c)=1$, it follows that $c$ divides $b$. To prove (2), write $c^{2}-b c=a(c N-b)$ to infer that $a$ divides $c^{2}-b c$. Letting $d=\operatorname{gcd}(a, b)$, it follows that $d$ divides $c^{2}$, and since the two are relatively prime by hypothesis, $d=1$. Now, let $s_{n}=a_{1} / a_{2}+a_{2} / a_{3}+\cdots+a_{n-1} / a_{n}+a_{n} / a_{1}$, let $\delta_{n}=\operatorname{gcd}\left(a_{1}, a_{n}, a_{n+1}\right)$ and write $$ s_{n+1}-s_{n}=\frac{a_{n}}{a_{n+1}}+\frac{a_{n+1}-a_{n}}{a_{1}}=\frac{a_{n} / \delta_{n}}{a_{n+1} / \delta_{n}}+\frac{a_{n+1} / \delta_{n}-a_{n} / \delta_{n}}{a_{1} / \delta_{n}} . $$ Let $n \geqslant k$. Since $\operatorname{gcd}\left(a_{1} / \delta_{n}, a_{n} / \delta_{n}, a_{n+1} / \delta_{n}\right)=1$, it follows by (2) that $\operatorname{gcd}\left(a_{1} / \delta_{n}, a_{n} / \delta_{n}\right)=1$. Let $d_{n}=\operatorname{gcd}\left(a_{1}, a_{n}\right)$. Then $d_{n}=\delta_{n} \cdot \operatorname{gcd}\left(a_{1} / \delta_{n}, a_{n} / \delta_{n}\right)=\delta_{n}$, so $d_{n}$ divides $a_{n+1}$, and therefore $d_{n}$ divides $d_{n+1}$. Consequently, from some rank on, the $d_{n}$ form a nondecreasing sequence of integers not exceeding $a_{1}$, so $d_{n}=d$ for all $n \geqslant \ell$, where $\ell$ is some positive integer. Finally, since $\operatorname{gcd}\left(a_{1} / d, a_{n+1} / d\right)=1$, it follows by (1) that $a_{n+1} / d$ divides $a_{n} / d$, so $a_{n} \geqslant a_{n+1}$ for all $n \geqslant \ell$. The conclusion follows.
|
proof
|
Yes
|
Yes
|
proof
|
Number Theory
|
Let $a_{1}, a_{2}, \ldots, a_{n}, \ldots$ be a sequence of positive integers such that $$ \frac{a_{1}}{a_{2}}+\frac{a_{2}}{a_{3}}+\cdots+\frac{a_{n-1}}{a_{n}}+\frac{a_{n}}{a_{1}} $$ is an integer for all $n \geqslant k$, where $k$ is some positive integer. Prove that there exists a positive integer $m$ such that $a_{n}=a_{n+1}$ for all $n \geqslant m$. (Mongolia)
|
The argument hinges on the following two facts: Let $a, b, c$ be positive integers such that $N=b / c+(c-b) / a$ is an integer. (1) If $\operatorname{gcd}(a, c)=1$, then $c$ divides $b$; and (2) If $\operatorname{gcd}(a, b, c)=1$, then $\operatorname{gcd}(a, b)=1$. To prove (1), write $a b=c(a N+b-c)$. Since $\operatorname{gcd}(a, c)=1$, it follows that $c$ divides $b$. To prove (2), write $c^{2}-b c=a(c N-b)$ to infer that $a$ divides $c^{2}-b c$. Letting $d=\operatorname{gcd}(a, b)$, it follows that $d$ divides $c^{2}$, and since the two are relatively prime by hypothesis, $d=1$. Now, let $s_{n}=a_{1} / a_{2}+a_{2} / a_{3}+\cdots+a_{n-1} / a_{n}+a_{n} / a_{1}$, let $\delta_{n}=\operatorname{gcd}\left(a_{1}, a_{n}, a_{n+1}\right)$ and write $$ s_{n+1}-s_{n}=\frac{a_{n}}{a_{n+1}}+\frac{a_{n+1}-a_{n}}{a_{1}}=\frac{a_{n} / \delta_{n}}{a_{n+1} / \delta_{n}}+\frac{a_{n+1} / \delta_{n}-a_{n} / \delta_{n}}{a_{1} / \delta_{n}} . $$ Let $n \geqslant k$. Since $\operatorname{gcd}\left(a_{1} / \delta_{n}, a_{n} / \delta_{n}, a_{n+1} / \delta_{n}\right)=1$, it follows by (2) that $\operatorname{gcd}\left(a_{1} / \delta_{n}, a_{n} / \delta_{n}\right)=1$. Let $d_{n}=\operatorname{gcd}\left(a_{1}, a_{n}\right)$. Then $d_{n}=\delta_{n} \cdot \operatorname{gcd}\left(a_{1} / \delta_{n}, a_{n} / \delta_{n}\right)=\delta_{n}$, so $d_{n}$ divides $a_{n+1}$, and therefore $d_{n}$ divides $d_{n+1}$. Consequently, from some rank on, the $d_{n}$ form a nondecreasing sequence of integers not exceeding $a_{1}$, so $d_{n}=d$ for all $n \geqslant \ell$, where $\ell$ is some positive integer. Finally, since $\operatorname{gcd}\left(a_{1} / d, a_{n+1} / d\right)=1$, it follows by (1) that $a_{n+1} / d$ divides $a_{n} / d$, so $a_{n} \geqslant a_{n+1}$ for all $n \geqslant \ell$. The conclusion follows.
|
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71c0fa0d-95e0-581c-87e5-053f159121c3
| 23,622
|
Let $f:\{1,2,3, \ldots\} \rightarrow\{2,3, \ldots\}$ be a function such that $f(m+n) \mid f(m)+f(n)$ for all pairs $m, n$ of positive integers. Prove that there exists a positive integer $c>1$ which divides all values of $f$. (Mexico)
|
For every positive integer $m$, define $S_{m}=\{n: m \mid f(n)\}$. Lemma. If the set $S_{m}$ is infinite, then $S_{m}=\{d, 2 d, 3 d, \ldots\}=d \cdot \mathbb{Z}_{>0}$ for some positive integer $d$. Proof. Let $d=\min S_{m}$; the definition of $S_{m}$ yields $m \mid f(d)$. Whenever $n \in S_{m}$ and $n>d$, we have $m|f(n)| f(n-d)+f(d)$, so $m \mid f(n-d)$ and therefore $n-d \in S_{m}$. Let $r \leqslant d$ be the least positive integer with $n \equiv r(\bmod d)$; repeating the same step, we can see that $n-d, n-2 d, \ldots, r \in S_{m}$. By the minimality of $d$, this shows $r=d$ and therefore $d \mid n$. Starting from an arbitrarily large element of $S_{m}$, the process above reaches all multiples of $d$; so they all are elements of $S_{m}$. The solution for the problem will be split into two cases. Case 1: The function $f$ is bounded. Call a prime $p$ frequent if the set $S_{p}$ is infinite, i.e., if $p$ divides $f(n)$ for infinitely many positive integers $n$; otherwise call $p$ sporadic. Since the function $f$ is bounded, there are only a finite number of primes that divide at least one $f(n)$; so altogether there are finitely many numbers $n$ such that $f(n)$ has a sporadic prime divisor. Let $N$ be a positive integer, greater than all those numbers $n$. Let $p_{1}, \ldots, p_{k}$ be the frequent primes. By the lemma we have $S_{p_{i}}=d_{i} \cdot \mathbb{Z}_{>0}$ for some $d_{i}$. Consider the number $$ n=N d_{1} d_{2} \cdots d_{k}+1 $$ Due to $n>N$, all prime divisors of $f(n)$ are frequent primes. Let $p_{i}$ be any frequent prime divisor of $f(n)$. Then $n \in S_{p_{i}}$, and therefore $d_{i} \mid n$. But $n \equiv 1\left(\bmod d_{i}\right)$, which means $d_{i}=1$. Hence $S_{p_{i}}=1 \cdot \mathbb{Z}_{>0}=\mathbb{Z}_{>0}$ and therefore $p_{i}$ is a common divisor of all values $f(n)$. Case 2: $f$ is unbounded. We prove that $f(1)$ divides all $f(n)$. Let $a=f(1)$. Since $1 \in S_{a}$, by the lemma it suffices to prove that $S_{a}$ is an infinite set. Call a positive integer $p$ a peak if $f(p)>\max (f(1), \ldots, f(p-1))$. Since $f$ is not bounded, there are infinitely many peaks. Let $1=p_{1}<p_{2}<\ldots$ be the sequence of all peaks, and let $h_{k}=f\left(p_{k}\right)$. Notice that for any peak $p_{i}$ and for any $k<p_{i}$, we have $f\left(p_{i}\right) \mid f(k)+f\left(p_{i}-k\right)<$ $2 f\left(p_{i}\right)$, hence $$ f(k)+f\left(p_{i}-k\right)=f\left(p_{i}\right)=h_{i} . $$ By the pigeonhole principle, among the numbers $h_{1}, h_{2}, \ldots$ there are infinitely many that are congruent modulo $a$. Let $k_{0}<k_{1}<k_{2}<\ldots$ be an infinite sequence of positive integers such that $h_{k_{0}} \equiv h_{k_{1}} \equiv \ldots(\bmod a)$. Notice that $$ f\left(p_{k_{i}}-p_{k_{0}}\right)=f\left(p_{k_{i}}\right)-f\left(p_{k_{0}}\right)=h_{k_{i}}-h_{k_{0}} \equiv 0 \quad(\bmod a), $$ so $p_{k_{i}}-p_{k_{0}} \in S_{a}$ for all $i=1,2, \ldots$ This provides infinitely many elements in $S_{a}$. Hence, $S_{a}$ is an infinite set, and therefore $f(1)=a$ divides $f(n)$ for every $n$. Comment. As an extension of the solution above, it can be proven that if $f$ is not bounded then $f(n)=a n$ with $a=f(1)$. Take an arbitrary positive integer $n$; we will show that $f(n+1)=f(n)+a$. Then it follows by induction that $f(n)=a n$. Take a peak $p$ such that $p>n+2$ and $h=f(p)>f(n)+2 a$. By (1) we have $f(p-1)=$ $f(p)-f(1)=h-a$ and $f(n+1)=f(p)-f(p-n-1)=h-f(p-n-1)$. From $h-a=f(p-1) \mid$ $f(n)+f(p-n-1)<f(n)+h<2(h-a)$ we get $f(n)+f(p-n-1)=h-a$. Then $$ f(n+1)-f(n)=(h-f(p-n-1))-(h-a-f(p-n-1))=a . $$ On the other hand, there exists a wide family of bounded functions satisfying the required properties. Here we present a few examples: $$ f(n)=c ; \quad f(n)=\left\{\begin{array}{ll} 2 c & \text { if } n \text { is even } \\ c & \text { if } n \text { is odd; } \end{array} \quad f(n)= \begin{cases}2018 c & \text { if } n \leqslant 2018 \\ c & \text { if } n>2018\end{cases}\right. $$
|
proof
|
Yes
|
Yes
|
proof
|
Number Theory
|
Let $f:\{1,2,3, \ldots\} \rightarrow\{2,3, \ldots\}$ be a function such that $f(m+n) \mid f(m)+f(n)$ for all pairs $m, n$ of positive integers. Prove that there exists a positive integer $c>1$ which divides all values of $f$. (Mexico)
|
For every positive integer $m$, define $S_{m}=\{n: m \mid f(n)\}$. Lemma. If the set $S_{m}$ is infinite, then $S_{m}=\{d, 2 d, 3 d, \ldots\}=d \cdot \mathbb{Z}_{>0}$ for some positive integer $d$. Proof. Let $d=\min S_{m}$; the definition of $S_{m}$ yields $m \mid f(d)$. Whenever $n \in S_{m}$ and $n>d$, we have $m|f(n)| f(n-d)+f(d)$, so $m \mid f(n-d)$ and therefore $n-d \in S_{m}$. Let $r \leqslant d$ be the least positive integer with $n \equiv r(\bmod d)$; repeating the same step, we can see that $n-d, n-2 d, \ldots, r \in S_{m}$. By the minimality of $d$, this shows $r=d$ and therefore $d \mid n$. Starting from an arbitrarily large element of $S_{m}$, the process above reaches all multiples of $d$; so they all are elements of $S_{m}$. The solution for the problem will be split into two cases. Case 1: The function $f$ is bounded. Call a prime $p$ frequent if the set $S_{p}$ is infinite, i.e., if $p$ divides $f(n)$ for infinitely many positive integers $n$; otherwise call $p$ sporadic. Since the function $f$ is bounded, there are only a finite number of primes that divide at least one $f(n)$; so altogether there are finitely many numbers $n$ such that $f(n)$ has a sporadic prime divisor. Let $N$ be a positive integer, greater than all those numbers $n$. Let $p_{1}, \ldots, p_{k}$ be the frequent primes. By the lemma we have $S_{p_{i}}=d_{i} \cdot \mathbb{Z}_{>0}$ for some $d_{i}$. Consider the number $$ n=N d_{1} d_{2} \cdots d_{k}+1 $$ Due to $n>N$, all prime divisors of $f(n)$ are frequent primes. Let $p_{i}$ be any frequent prime divisor of $f(n)$. Then $n \in S_{p_{i}}$, and therefore $d_{i} \mid n$. But $n \equiv 1\left(\bmod d_{i}\right)$, which means $d_{i}=1$. Hence $S_{p_{i}}=1 \cdot \mathbb{Z}_{>0}=\mathbb{Z}_{>0}$ and therefore $p_{i}$ is a common divisor of all values $f(n)$. Case 2: $f$ is unbounded. We prove that $f(1)$ divides all $f(n)$. Let $a=f(1)$. Since $1 \in S_{a}$, by the lemma it suffices to prove that $S_{a}$ is an infinite set. Call a positive integer $p$ a peak if $f(p)>\max (f(1), \ldots, f(p-1))$. Since $f$ is not bounded, there are infinitely many peaks. Let $1=p_{1}<p_{2}<\ldots$ be the sequence of all peaks, and let $h_{k}=f\left(p_{k}\right)$. Notice that for any peak $p_{i}$ and for any $k<p_{i}$, we have $f\left(p_{i}\right) \mid f(k)+f\left(p_{i}-k\right)<$ $2 f\left(p_{i}\right)$, hence $$ f(k)+f\left(p_{i}-k\right)=f\left(p_{i}\right)=h_{i} . $$ By the pigeonhole principle, among the numbers $h_{1}, h_{2}, \ldots$ there are infinitely many that are congruent modulo $a$. Let $k_{0}<k_{1}<k_{2}<\ldots$ be an infinite sequence of positive integers such that $h_{k_{0}} \equiv h_{k_{1}} \equiv \ldots(\bmod a)$. Notice that $$ f\left(p_{k_{i}}-p_{k_{0}}\right)=f\left(p_{k_{i}}\right)-f\left(p_{k_{0}}\right)=h_{k_{i}}-h_{k_{0}} \equiv 0 \quad(\bmod a), $$ so $p_{k_{i}}-p_{k_{0}} \in S_{a}$ for all $i=1,2, \ldots$ This provides infinitely many elements in $S_{a}$. Hence, $S_{a}$ is an infinite set, and therefore $f(1)=a$ divides $f(n)$ for every $n$. Comment. As an extension of the solution above, it can be proven that if $f$ is not bounded then $f(n)=a n$ with $a=f(1)$. Take an arbitrary positive integer $n$; we will show that $f(n+1)=f(n)+a$. Then it follows by induction that $f(n)=a n$. Take a peak $p$ such that $p>n+2$ and $h=f(p)>f(n)+2 a$. By (1) we have $f(p-1)=$ $f(p)-f(1)=h-a$ and $f(n+1)=f(p)-f(p-n-1)=h-f(p-n-1)$. From $h-a=f(p-1) \mid$ $f(n)+f(p-n-1)<f(n)+h<2(h-a)$ we get $f(n)+f(p-n-1)=h-a$. Then $$ f(n+1)-f(n)=(h-f(p-n-1))-(h-a-f(p-n-1))=a . $$ On the other hand, there exists a wide family of bounded functions satisfying the required properties. Here we present a few examples: $$ f(n)=c ; \quad f(n)=\left\{\begin{array}{ll} 2 c & \text { if } n \text { is even } \\ c & \text { if } n \text { is odd; } \end{array} \quad f(n)= \begin{cases}2018 c & \text { if } n \leqslant 2018 \\ c & \text { if } n>2018\end{cases}\right. $$
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0170eb5a-2761-5378-8837-1c45f79bd6a6
| 23,629
|
Let $f:\{1,2,3, \ldots\} \rightarrow\{2,3, \ldots\}$ be a function such that $f(m+n) \mid f(m)+f(n)$ for all pairs $m, n$ of positive integers. Prove that there exists a positive integer $c>1$ which divides all values of $f$. (Mexico)
|
Let $d_{n}=\operatorname{gcd}(f(n), f(1))$. From $d_{n+1} \mid f(1)$ and $d_{n+1}|f(n+1)| f(n)+f(1)$, we can see that $d_{n+1} \mid f(n)$; then $d_{n+1} \mid \operatorname{gcd}(f(n), f(1))=d_{n}$. So the sequence $d_{1}, d_{2}, \ldots$ is nonincreasing in the sense that every element is a divisor of the previous elements. Let $d=\min \left(d_{1}, d_{2}, \ldots\right)=\operatorname{gcd}\left(d_{1} \cdot d_{2}, \ldots\right)=\operatorname{gcd}(f(1), f(2), \ldots)$; we have to prove $d \geqslant 2$. For the sake of contradiction, suppose that the statement is wrong, so $d=1$; that means there is some index $n_{0}$ such that $d_{n}=1$ for every $n \geqslant n_{0}$, i.e., $f(n)$ is coprime with $f(1)$. Claim 1. If $2^{k} \geqslant n_{0}$ then $f\left(2^{k}\right) \leqslant 2^{k}$. Proof. By the condition, $f(2 n) \mid 2 f(n)$; a trivial induction yields $f\left(2^{k}\right) \mid 2^{k} f(1)$. If $2^{k} \geqslant n_{0}$ then $f\left(2^{k}\right)$ is coprime with $f(1)$, so $f\left(2^{k}\right)$ is a divisor of $2^{k}$. Claim 2. There is a constant $C$ such that $f(n)<n+C$ for every $n$. Proof. Take the first power of 2 which is greater than or equal to $n_{0}$ : let $K=2^{k} \geqslant n_{0}$. By Claim 1, we have $f(K) \leqslant K$. Notice that $f(n+K) \mid f(n)+f(K)$ implies $f(n+K) \leqslant$ $f(n)+f(K) \leqslant f(n)+K$. If $n=t K+r$ for some $t \geqslant 0$ and $1 \leqslant r \leqslant K$, then we conclude $f(n) \leqslant K+f(n-K) \leqslant 2 K+f(n-2 K) \leqslant \ldots \leqslant t K+f(r)<n+\max (f(1), f(2), \ldots, f(K))$, so the claim is true with $C=\max (f(1), \ldots, f(K))$. Claim 3. If $a, b \in \mathbb{Z}_{>0}$ are coprime then $\operatorname{gcd}(f(a), f(b)) \mid f(1)$. In particular, if $a, b \geqslant n_{0}$ are coprime then $f(a)$ and $f(b)$ are coprime. Proof. Let $d=\operatorname{gcd}(f(a), f(b))$. We can replicate Euclid's algorithm. Formally, apply induction on $a+b$. If $a=1$ or $b=1$ then we already have $d \mid f(1)$. Without loss of generality, suppose $1<a<b$. Then $d \mid f(a)$ and $d|f(b)| f(a)+f(b-a)$, so $d \mid f(b-a)$. Therefore $d$ divides $\operatorname{gcd}(f(a), f(b-a))$ which is a divisor of $f(1)$ by the induction hypothesis. Let $p_{1}<p_{2}<\ldots$ be the sequence of all prime numbers; for every $k$, let $q_{k}$ be the lowest power of $p_{k}$ with $q_{k} \geqslant n_{0}$. (Notice that there are only finitely many positive integers with $q_{k} \neq p_{k}$.) Take a positive integer $N$, and consider the numbers $$ f(1), f\left(q_{1}\right), f\left(q_{2}\right), \ldots, f\left(q_{N}\right) $$ Here we have $N+1$ numbers, each being greater than 1 , and they are pairwise coprime by Claim 3. Therefore, they have at least $N+1$ different prime divisors in total, and their greatest prime divisor is at least $p_{N+1}$. Hence, $\max \left(f(1), f\left(q_{1}\right), \ldots, f\left(q_{N}\right)\right) \geqslant p_{N+1}$. Choose $N$ such that $\max \left(q_{1}, \ldots, q_{N}\right)=p_{N}$ (this is achieved if $N$ is sufficiently large), and $p_{N+1}-p_{N}>C$ (that is possible, because there are arbitrarily long gaps between the primes). Then we establish a contradiction $$ p_{N+1} \leqslant \max \left(f(1), f\left(q_{1}\right), \ldots, f\left(q_{N}\right)\right)<\max \left(1+C, q_{1}+C, \ldots, q_{N}+C\right)=p_{N}+C<p_{N+1} $$ which proves the statement.
|
proof
|
Yes
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Yes
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proof
|
Number Theory
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Let $f:\{1,2,3, \ldots\} \rightarrow\{2,3, \ldots\}$ be a function such that $f(m+n) \mid f(m)+f(n)$ for all pairs $m, n$ of positive integers. Prove that there exists a positive integer $c>1$ which divides all values of $f$. (Mexico)
|
Let $d_{n}=\operatorname{gcd}(f(n), f(1))$. From $d_{n+1} \mid f(1)$ and $d_{n+1}|f(n+1)| f(n)+f(1)$, we can see that $d_{n+1} \mid f(n)$; then $d_{n+1} \mid \operatorname{gcd}(f(n), f(1))=d_{n}$. So the sequence $d_{1}, d_{2}, \ldots$ is nonincreasing in the sense that every element is a divisor of the previous elements. Let $d=\min \left(d_{1}, d_{2}, \ldots\right)=\operatorname{gcd}\left(d_{1} \cdot d_{2}, \ldots\right)=\operatorname{gcd}(f(1), f(2), \ldots)$; we have to prove $d \geqslant 2$. For the sake of contradiction, suppose that the statement is wrong, so $d=1$; that means there is some index $n_{0}$ such that $d_{n}=1$ for every $n \geqslant n_{0}$, i.e., $f(n)$ is coprime with $f(1)$. Claim 1. If $2^{k} \geqslant n_{0}$ then $f\left(2^{k}\right) \leqslant 2^{k}$. Proof. By the condition, $f(2 n) \mid 2 f(n)$; a trivial induction yields $f\left(2^{k}\right) \mid 2^{k} f(1)$. If $2^{k} \geqslant n_{0}$ then $f\left(2^{k}\right)$ is coprime with $f(1)$, so $f\left(2^{k}\right)$ is a divisor of $2^{k}$. Claim 2. There is a constant $C$ such that $f(n)<n+C$ for every $n$. Proof. Take the first power of 2 which is greater than or equal to $n_{0}$ : let $K=2^{k} \geqslant n_{0}$. By Claim 1, we have $f(K) \leqslant K$. Notice that $f(n+K) \mid f(n)+f(K)$ implies $f(n+K) \leqslant$ $f(n)+f(K) \leqslant f(n)+K$. If $n=t K+r$ for some $t \geqslant 0$ and $1 \leqslant r \leqslant K$, then we conclude $f(n) \leqslant K+f(n-K) \leqslant 2 K+f(n-2 K) \leqslant \ldots \leqslant t K+f(r)<n+\max (f(1), f(2), \ldots, f(K))$, so the claim is true with $C=\max (f(1), \ldots, f(K))$. Claim 3. If $a, b \in \mathbb{Z}_{>0}$ are coprime then $\operatorname{gcd}(f(a), f(b)) \mid f(1)$. In particular, if $a, b \geqslant n_{0}$ are coprime then $f(a)$ and $f(b)$ are coprime. Proof. Let $d=\operatorname{gcd}(f(a), f(b))$. We can replicate Euclid's algorithm. Formally, apply induction on $a+b$. If $a=1$ or $b=1$ then we already have $d \mid f(1)$. Without loss of generality, suppose $1<a<b$. Then $d \mid f(a)$ and $d|f(b)| f(a)+f(b-a)$, so $d \mid f(b-a)$. Therefore $d$ divides $\operatorname{gcd}(f(a), f(b-a))$ which is a divisor of $f(1)$ by the induction hypothesis. Let $p_{1}<p_{2}<\ldots$ be the sequence of all prime numbers; for every $k$, let $q_{k}$ be the lowest power of $p_{k}$ with $q_{k} \geqslant n_{0}$. (Notice that there are only finitely many positive integers with $q_{k} \neq p_{k}$.) Take a positive integer $N$, and consider the numbers $$ f(1), f\left(q_{1}\right), f\left(q_{2}\right), \ldots, f\left(q_{N}\right) $$ Here we have $N+1$ numbers, each being greater than 1 , and they are pairwise coprime by Claim 3. Therefore, they have at least $N+1$ different prime divisors in total, and their greatest prime divisor is at least $p_{N+1}$. Hence, $\max \left(f(1), f\left(q_{1}\right), \ldots, f\left(q_{N}\right)\right) \geqslant p_{N+1}$. Choose $N$ such that $\max \left(q_{1}, \ldots, q_{N}\right)=p_{N}$ (this is achieved if $N$ is sufficiently large), and $p_{N+1}-p_{N}>C$ (that is possible, because there are arbitrarily long gaps between the primes). Then we establish a contradiction $$ p_{N+1} \leqslant \max \left(f(1), f\left(q_{1}\right), \ldots, f\left(q_{N}\right)\right)<\max \left(1+C, q_{1}+C, \ldots, q_{N}+C\right)=p_{N}+C<p_{N+1} $$ which proves the statement.
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0170eb5a-2761-5378-8837-1c45f79bd6a6
| 23,629
|
Let $n \geqslant 2018$ be an integer, and let $a_{1}, a_{2}, \ldots, a_{n}, b_{1}, b_{2}, \ldots, b_{n}$ be pairwise distinct positive integers not exceeding $5 n$. Suppose that the sequence $$ \frac{a_{1}}{b_{1}}, \frac{a_{2}}{b_{2}}, \ldots, \frac{a_{n}}{b_{n}} $$ forms an arithmetic progression. Prove that the terms of the sequence are equal.
|
Suppose that (1) is an arithmetic progression with nonzero difference. Let the difference be $\Delta=\frac{c}{d}$, where $d>0$ and $c, d$ are coprime. We will show that too many denominators $b_{i}$ should be divisible by $d$. To this end, for any $1 \leqslant i \leqslant n$ and any prime divisor $p$ of $d$, say that the index $i$ is $p$-wrong, if $v_{p}\left(b_{i}\right)<v_{p}(d)$. $\left(v_{p}(x)\right.$ stands for the exponent of $p$ in the prime factorisation of $x$.) Claim 1. For any prime $p$, all $p$-wrong indices are congruent modulo $p$. In other words, the $p$-wrong indices (if they exist) are included in an arithmetic progression with difference $p$. Proof. Let $\alpha=v_{p}(d)$. For the sake of contradiction, suppose that $i$ and $j$ are $p$-wrong indices (i.e., none of $b_{i}$ and $b_{j}$ is divisible by $\left.p^{\alpha}\right)$ such that $i \not \equiv j(\bmod p)$. Then the least common denominator of $\frac{a_{i}}{b_{i}}$ and $\frac{a_{j}}{b_{j}}$ is not divisible by $p^{\alpha}$. But this is impossible because in their difference, $(i-j) \Delta=\frac{(i-j) c}{d}$, the numerator is coprime to $p$, but $p^{\alpha}$ divides the denominator $d$. Claim 2. $d$ has no prime divisors greater than 5 . Proof. Suppose that $p \geqslant 7$ is a prime divisor of $d$. Among the indices $1,2, \ldots, n$, at most $\left\lceil\frac{n}{p}\right\rceil<\frac{n}{p}+1$ are $p$-wrong, so $p$ divides at least $\frac{p-1}{p} n-1$ of $b_{1}, \ldots, b_{n}$. Since these denominators are distinct, $$ 5 n \geqslant \max \left\{b_{i}: p \mid b_{i}\right\} \geqslant\left(\frac{p-1}{p} n-1\right) p=(p-1)(n-1)-1 \geqslant 6(n-1)-1>5 n, $$ a contradiction. Claim 3. For every $0 \leqslant k \leqslant n-30$, among the denominators $b_{k+1}, b_{k+2}, \ldots, b_{k+30}$, at least $\varphi(30)=8$ are divisible by $d$. Proof. By Claim 1, the 2 -wrong, 3 -wrong and 5 -wrong indices can be covered by three arithmetic progressions with differences 2,3 and 5 . By a simple inclusion-exclusion, $(2-1) \cdot(3-1) \cdot(5-1)=8$ indices are not covered; by Claim 2, we have $d \mid b_{i}$ for every uncovered index $i$. Claim 4. $|\Delta|<\frac{20}{n-2}$ and $d>\frac{n-2}{20}$. Proof. From the sequence (1), remove all fractions with $b_{n}<\frac{n}{2}$, There remain at least $\frac{n}{2}$ fractions, and they cannot exceed $\frac{5 n}{n / 2}=10$. So we have at least $\frac{n}{2}$ elements of the arithmetic progression (1) in the interval $(0,10]$, hence the difference must be below $\frac{10}{n / 2-1}=\frac{20}{n-2}$. The second inequality follows from $\frac{1}{d} \leqslant \frac{|c|}{d}=|\Delta|$. Now we have everything to get the final contradiction. By Claim 3, we have $d \mid b_{i}$ for at least $\left\lfloor\frac{n}{30}\right\rfloor \cdot 8$ indices $i$. By Claim 4 , we have $d \geqslant \frac{n-2}{20}$. Therefore, $$ 5 n \geqslant \max \left\{b_{i}: d \mid b_{i}\right\} \geqslant\left(\left\lfloor\frac{n}{30}\right\rfloor \cdot 8\right) \cdot d>\left(\frac{n}{30}-1\right) \cdot 8 \cdot \frac{n-2}{20}>5 n . $$ Comment 1. It is possible that all terms in (1) are equal, for example with $a_{i}=2 i-1$ and $b_{i}=4 i-2$ we have $\frac{a_{i}}{b_{i}}=\frac{1}{2}$. Comment 2. The bound $5 n$ in the statement is far from sharp; the solution above can be modified to work for $9 n$. For large $n$, the bound $5 n$ can be replaced by $n^{\frac{3}{2}-\varepsilon}$. The activities of the Problem Selection Committee were supported by DEDEMAN LA FANTANA vine la tine
|
proof
|
Yes
|
Yes
|
proof
|
Algebra
|
Let $n \geqslant 2018$ be an integer, and let $a_{1}, a_{2}, \ldots, a_{n}, b_{1}, b_{2}, \ldots, b_{n}$ be pairwise distinct positive integers not exceeding $5 n$. Suppose that the sequence $$ \frac{a_{1}}{b_{1}}, \frac{a_{2}}{b_{2}}, \ldots, \frac{a_{n}}{b_{n}} $$ forms an arithmetic progression. Prove that the terms of the sequence are equal.
|
Suppose that (1) is an arithmetic progression with nonzero difference. Let the difference be $\Delta=\frac{c}{d}$, where $d>0$ and $c, d$ are coprime. We will show that too many denominators $b_{i}$ should be divisible by $d$. To this end, for any $1 \leqslant i \leqslant n$ and any prime divisor $p$ of $d$, say that the index $i$ is $p$-wrong, if $v_{p}\left(b_{i}\right)<v_{p}(d)$. $\left(v_{p}(x)\right.$ stands for the exponent of $p$ in the prime factorisation of $x$.) Claim 1. For any prime $p$, all $p$-wrong indices are congruent modulo $p$. In other words, the $p$-wrong indices (if they exist) are included in an arithmetic progression with difference $p$. Proof. Let $\alpha=v_{p}(d)$. For the sake of contradiction, suppose that $i$ and $j$ are $p$-wrong indices (i.e., none of $b_{i}$ and $b_{j}$ is divisible by $\left.p^{\alpha}\right)$ such that $i \not \equiv j(\bmod p)$. Then the least common denominator of $\frac{a_{i}}{b_{i}}$ and $\frac{a_{j}}{b_{j}}$ is not divisible by $p^{\alpha}$. But this is impossible because in their difference, $(i-j) \Delta=\frac{(i-j) c}{d}$, the numerator is coprime to $p$, but $p^{\alpha}$ divides the denominator $d$. Claim 2. $d$ has no prime divisors greater than 5 . Proof. Suppose that $p \geqslant 7$ is a prime divisor of $d$. Among the indices $1,2, \ldots, n$, at most $\left\lceil\frac{n}{p}\right\rceil<\frac{n}{p}+1$ are $p$-wrong, so $p$ divides at least $\frac{p-1}{p} n-1$ of $b_{1}, \ldots, b_{n}$. Since these denominators are distinct, $$ 5 n \geqslant \max \left\{b_{i}: p \mid b_{i}\right\} \geqslant\left(\frac{p-1}{p} n-1\right) p=(p-1)(n-1)-1 \geqslant 6(n-1)-1>5 n, $$ a contradiction. Claim 3. For every $0 \leqslant k \leqslant n-30$, among the denominators $b_{k+1}, b_{k+2}, \ldots, b_{k+30}$, at least $\varphi(30)=8$ are divisible by $d$. Proof. By Claim 1, the 2 -wrong, 3 -wrong and 5 -wrong indices can be covered by three arithmetic progressions with differences 2,3 and 5 . By a simple inclusion-exclusion, $(2-1) \cdot(3-1) \cdot(5-1)=8$ indices are not covered; by Claim 2, we have $d \mid b_{i}$ for every uncovered index $i$. Claim 4. $|\Delta|<\frac{20}{n-2}$ and $d>\frac{n-2}{20}$. Proof. From the sequence (1), remove all fractions with $b_{n}<\frac{n}{2}$, There remain at least $\frac{n}{2}$ fractions, and they cannot exceed $\frac{5 n}{n / 2}=10$. So we have at least $\frac{n}{2}$ elements of the arithmetic progression (1) in the interval $(0,10]$, hence the difference must be below $\frac{10}{n / 2-1}=\frac{20}{n-2}$. The second inequality follows from $\frac{1}{d} \leqslant \frac{|c|}{d}=|\Delta|$. Now we have everything to get the final contradiction. By Claim 3, we have $d \mid b_{i}$ for at least $\left\lfloor\frac{n}{30}\right\rfloor \cdot 8$ indices $i$. By Claim 4 , we have $d \geqslant \frac{n-2}{20}$. Therefore, $$ 5 n \geqslant \max \left\{b_{i}: d \mid b_{i}\right\} \geqslant\left(\left\lfloor\frac{n}{30}\right\rfloor \cdot 8\right) \cdot d>\left(\frac{n}{30}-1\right) \cdot 8 \cdot \frac{n-2}{20}>5 n . $$ Comment 1. It is possible that all terms in (1) are equal, for example with $a_{i}=2 i-1$ and $b_{i}=4 i-2$ we have $\frac{a_{i}}{b_{i}}=\frac{1}{2}$. Comment 2. The bound $5 n$ in the statement is far from sharp; the solution above can be modified to work for $9 n$. For large $n$, the bound $5 n$ can be replaced by $n^{\frac{3}{2}-\varepsilon}$. The activities of the Problem Selection Committee were supported by DEDEMAN LA FANTANA vine la tine
|
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|
d9877fd8-9b07-53be-875c-197042beed9d
| 23,633
|
Given any set $S$ of positive integers, show that at least one of the following two assertions holds: (1) There exist distinct finite subsets $F$ and $G$ of $S$ such that $\sum_{x \in F} 1 / x=\sum_{x \in G} 1 / x$; (2) There exists a positive rational number $r<1$ such that $\sum_{x \in F} 1 / x \neq r$ for all finite subsets $F$ of $S$. ## (Luxembourg)
|
Argue indirectly. Agree, as usual, that the empty sum is 0 to consider rationals in $[0,1)$; adjoining 0 causes no harm, since $\sum_{x \in F} 1 / x=0$ for no nonempty finite subset $F$ of $S$. For every rational $r$ in $[0,1)$, let $F_{r}$ be the unique finite subset of $S$ such that $\sum_{x \in F_{r}} 1 / x=r$. The argument hinges on the lemma below. Lemma. If $x$ is a member of $S$ and $q$ and $r$ are rationals in $[0,1)$ such that $q-r=1 / x$, then $x$ is a member of $F_{q}$ if and only if it is not one of $F_{r}$. Proof. If $x$ is a member of $F_{q}$, then $$ \sum_{y \in F_{q} \backslash\{x\}} \frac{1}{y}=\sum_{y \in F_{q}} \frac{1}{y}-\frac{1}{x}=q-\frac{1}{x}=r=\sum_{y \in F_{r}} \frac{1}{y} $$ so $F_{r}=F_{q} \backslash\{x\}$, and $x$ is not a member of $F_{r}$. Conversely, if $x$ is not a member of $F_{r}$, then $$ \sum_{y \in F_{r} \cup\{x\}} \frac{1}{y}=\sum_{y \in F_{r}} \frac{1}{y}+\frac{1}{x}=r+\frac{1}{x}=q=\sum_{y \in F_{q}} \frac{1}{y} $$ so $F_{q}=F_{r} \cup\{x\}$, and $x$ is a member of $F_{q}$. Consider now an element $x$ of $S$ and a positive rational $r<1$. Let $n=\lfloor r x\rfloor$ and consider the sets $F_{r-k / x}, k=0, \ldots, n$. Since $0 \leqslant r-n / x<1 / x$, the set $F_{r-n / x}$ does not contain $x$, and a repeated application of the lemma shows that the $F_{r-(n-2 k) / x}$ do not contain $x$, whereas the $F_{r-(n-2 k-1) / x}$ do. Consequently, $x$ is a member of $F_{r}$ if and only if $n$ is odd. Finally, consider $F_{2 / 3}$. By the preceding, $\lfloor 2 x / 3\rfloor$ is odd for each $x$ in $F_{2 / 3}$, so $2 x / 3$ is not integral. Since $F_{2 / 3}$ is finite, there exists a positive rational $\varepsilon$ such that $\lfloor(2 / 3-\varepsilon) x\rfloor=\lfloor 2 x / 3\rfloor$ for all $x$ in $F_{2 / 3}$. This implies that $F_{2 / 3}$ is a subset of $F_{2 / 3-\varepsilon}$ which is impossible. Comment. The solution above can be adapted to show that the problem statement still holds, if the condition $r<1$ in (2) is replaced with $r<\delta$, for an arbitrary positive $\delta$. This yields that, if $S$ does not satisfy (1), then there exist infinitely many positive rational numbers $r<1$ such that $\sum_{x \in F} 1 / x \neq r$ for all finite subsets $F$ of $S$.
|
proof
|
Yes
|
Yes
|
proof
|
Combinatorics
|
Given any set $S$ of positive integers, show that at least one of the following two assertions holds: (1) There exist distinct finite subsets $F$ and $G$ of $S$ such that $\sum_{x \in F} 1 / x=\sum_{x \in G} 1 / x$; (2) There exists a positive rational number $r<1$ such that $\sum_{x \in F} 1 / x \neq r$ for all finite subsets $F$ of $S$. ## (Luxembourg)
|
Argue indirectly. Agree, as usual, that the empty sum is 0 to consider rationals in $[0,1)$; adjoining 0 causes no harm, since $\sum_{x \in F} 1 / x=0$ for no nonempty finite subset $F$ of $S$. For every rational $r$ in $[0,1)$, let $F_{r}$ be the unique finite subset of $S$ such that $\sum_{x \in F_{r}} 1 / x=r$. The argument hinges on the lemma below. Lemma. If $x$ is a member of $S$ and $q$ and $r$ are rationals in $[0,1)$ such that $q-r=1 / x$, then $x$ is a member of $F_{q}$ if and only if it is not one of $F_{r}$. Proof. If $x$ is a member of $F_{q}$, then $$ \sum_{y \in F_{q} \backslash\{x\}} \frac{1}{y}=\sum_{y \in F_{q}} \frac{1}{y}-\frac{1}{x}=q-\frac{1}{x}=r=\sum_{y \in F_{r}} \frac{1}{y} $$ so $F_{r}=F_{q} \backslash\{x\}$, and $x$ is not a member of $F_{r}$. Conversely, if $x$ is not a member of $F_{r}$, then $$ \sum_{y \in F_{r} \cup\{x\}} \frac{1}{y}=\sum_{y \in F_{r}} \frac{1}{y}+\frac{1}{x}=r+\frac{1}{x}=q=\sum_{y \in F_{q}} \frac{1}{y} $$ so $F_{q}=F_{r} \cup\{x\}$, and $x$ is a member of $F_{q}$. Consider now an element $x$ of $S$ and a positive rational $r<1$. Let $n=\lfloor r x\rfloor$ and consider the sets $F_{r-k / x}, k=0, \ldots, n$. Since $0 \leqslant r-n / x<1 / x$, the set $F_{r-n / x}$ does not contain $x$, and a repeated application of the lemma shows that the $F_{r-(n-2 k) / x}$ do not contain $x$, whereas the $F_{r-(n-2 k-1) / x}$ do. Consequently, $x$ is a member of $F_{r}$ if and only if $n$ is odd. Finally, consider $F_{2 / 3}$. By the preceding, $\lfloor 2 x / 3\rfloor$ is odd for each $x$ in $F_{2 / 3}$, so $2 x / 3$ is not integral. Since $F_{2 / 3}$ is finite, there exists a positive rational $\varepsilon$ such that $\lfloor(2 / 3-\varepsilon) x\rfloor=\lfloor 2 x / 3\rfloor$ for all $x$ in $F_{2 / 3}$. This implies that $F_{2 / 3}$ is a subset of $F_{2 / 3-\varepsilon}$ which is impossible. Comment. The solution above can be adapted to show that the problem statement still holds, if the condition $r<1$ in (2) is replaced with $r<\delta$, for an arbitrary positive $\delta$. This yields that, if $S$ does not satisfy (1), then there exist infinitely many positive rational numbers $r<1$ such that $\sum_{x \in F} 1 / x \neq r$ for all finite subsets $F$ of $S$.
|
{
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|
6a3a6868-9c5b-562e-8168-7c9764e8298e
| 23,643
|
Given any set $S$ of positive integers, show that at least one of the following two assertions holds: (1) There exist distinct finite subsets $F$ and $G$ of $S$ such that $\sum_{x \in F} 1 / x=\sum_{x \in G} 1 / x$; (2) There exists a positive rational number $r<1$ such that $\sum_{x \in F} 1 / x \neq r$ for all finite subsets $F$ of $S$. ## (Luxembourg)
|
A finite $S$ clearly satisfies (2), so let $S$ be infinite. If $S$ fails both conditions, so does $S \backslash\{1\}$. We may and will therefore assume that $S$ consists of integers greater than 1 . Label the elements of $S$ increasingly $x_{1}<x_{2}<\cdots$, where $x_{1} \geqslant 2$. We first show that $S$ satisfies (2) if $x_{n+1} \geqslant 2 x_{n}$ for all $n$. In this case, $x_{n} \geqslant 2^{n-1} x_{1}$ for all $n$, so $$ s=\sum_{n \geqslant 1} \frac{1}{x_{n}} \leqslant \sum_{n \geqslant 1} \frac{1}{2^{n-1} x_{1}}=\frac{2}{x_{1}} $$ If $x_{1} \geqslant 3$, or $x_{1}=2$ and $x_{n+1}>2 x_{n}$ for some $n$, then $\sum_{x \in F} 1 / x<s<1$ for every finite subset $F$ of $S$, so $S$ satisfies (2); and if $x_{1}=2$ and $x_{n+1}=2 x_{n}$ for all $n$, that is, $x_{n}=2^{n}$ for all $n$, then every finite subset $F$ of $S$ consists of powers of 2 , so $\sum_{x \in F} 1 / x \neq 1 / 3$ and again $S$ satisfies (2). Finally, we deal with the case where $x_{n+1}<2 x_{n}$ for some $n$. Consider the positive rational $r=1 / x_{n}-1 / x_{n+1}<1 / x_{n+1}$. If $r=\sum_{x \in F} 1 / x$ for no finite subset $F$ of $S$, then $S$ satisfies (2). We now assume that $r=\sum_{x \in F_{0}} 1 / x$ for some finite subset $F_{0}$ of $S$, and show that $S$ satisfies (1). Since $\sum_{x \in F_{0}} 1 / x=r<1 / x_{n+1}$, it follows that $x_{n+1}$ is not a member of $F_{0}$, so $$ \sum_{x \in F_{0} \cup\left\{x_{n+1}\right\}} \frac{1}{x}=\sum_{x \in F_{0}} \frac{1}{x}+\frac{1}{x_{n+1}}=r+\frac{1}{x_{n+1}}=\frac{1}{x_{n}} $$ Consequently, $F=F_{0} \cup\left\{x_{n+1}\right\}$ and $G=\left\{x_{n}\right\}$ are distinct finite subsets of $S$ such that $\sum_{x \in F} 1 / x=\sum_{x \in G} 1 / x$, and $S$ satisfies (1).
|
proof
|
Yes
|
Yes
|
proof
|
Combinatorics
|
Given any set $S$ of positive integers, show that at least one of the following two assertions holds: (1) There exist distinct finite subsets $F$ and $G$ of $S$ such that $\sum_{x \in F} 1 / x=\sum_{x \in G} 1 / x$; (2) There exists a positive rational number $r<1$ such that $\sum_{x \in F} 1 / x \neq r$ for all finite subsets $F$ of $S$. ## (Luxembourg)
|
A finite $S$ clearly satisfies (2), so let $S$ be infinite. If $S$ fails both conditions, so does $S \backslash\{1\}$. We may and will therefore assume that $S$ consists of integers greater than 1 . Label the elements of $S$ increasingly $x_{1}<x_{2}<\cdots$, where $x_{1} \geqslant 2$. We first show that $S$ satisfies (2) if $x_{n+1} \geqslant 2 x_{n}$ for all $n$. In this case, $x_{n} \geqslant 2^{n-1} x_{1}$ for all $n$, so $$ s=\sum_{n \geqslant 1} \frac{1}{x_{n}} \leqslant \sum_{n \geqslant 1} \frac{1}{2^{n-1} x_{1}}=\frac{2}{x_{1}} $$ If $x_{1} \geqslant 3$, or $x_{1}=2$ and $x_{n+1}>2 x_{n}$ for some $n$, then $\sum_{x \in F} 1 / x<s<1$ for every finite subset $F$ of $S$, so $S$ satisfies (2); and if $x_{1}=2$ and $x_{n+1}=2 x_{n}$ for all $n$, that is, $x_{n}=2^{n}$ for all $n$, then every finite subset $F$ of $S$ consists of powers of 2 , so $\sum_{x \in F} 1 / x \neq 1 / 3$ and again $S$ satisfies (2). Finally, we deal with the case where $x_{n+1}<2 x_{n}$ for some $n$. Consider the positive rational $r=1 / x_{n}-1 / x_{n+1}<1 / x_{n+1}$. If $r=\sum_{x \in F} 1 / x$ for no finite subset $F$ of $S$, then $S$ satisfies (2). We now assume that $r=\sum_{x \in F_{0}} 1 / x$ for some finite subset $F_{0}$ of $S$, and show that $S$ satisfies (1). Since $\sum_{x \in F_{0}} 1 / x=r<1 / x_{n+1}$, it follows that $x_{n+1}$ is not a member of $F_{0}$, so $$ \sum_{x \in F_{0} \cup\left\{x_{n+1}\right\}} \frac{1}{x}=\sum_{x \in F_{0}} \frac{1}{x}+\frac{1}{x_{n+1}}=r+\frac{1}{x_{n+1}}=\frac{1}{x_{n}} $$ Consequently, $F=F_{0} \cup\left\{x_{n+1}\right\}$ and $G=\left\{x_{n}\right\}$ are distinct finite subsets of $S$ such that $\sum_{x \in F} 1 / x=\sum_{x \in G} 1 / x$, and $S$ satisfies (1).
|
{
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"problem_match": null,
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|
6a3a6868-9c5b-562e-8168-7c9764e8298e
| 23,643
|
An anti-Pascal pyramid is a finite set of numbers, placed in a triangle-shaped array so that the first row of the array contains one number, the second row contains two numbers, the third row contains three numbers and so on; and, except for the numbers in the bottom row, each number equals the absolute value of the difference of the two numbers below it. For instance, the triangle below is an anti-Pascal pyramid with four rows, in which every integer from 1 to $1+2+3+4=10$ occurs exactly once: $$ \begin{array}{ccccc} & & & \\ & 2 & 6 & \\ & 5 & 7 & \\ 8 & 3 & 10 & 9 . \end{array} $$ Is it possible to form an anti-Pascal pyramid with 2018 rows, using every integer from 1 to $1+2+\cdots+2018$ exactly once? (Iran) Answer: No, it is not possible.
|
Let $T$ be an anti-Pascal pyramid with $n$ rows, containing every integer from 1 to $1+2+\cdots+n$, and let $a_{1}$ be the topmost number in $T$ (Figure 1). The two numbers below $a_{1}$ are some $a_{2}$ and $b_{2}=a_{1}+a_{2}$, the two numbers below $b_{2}$ are some $a_{3}$ and $b_{3}=a_{1}+a_{2}+a_{3}$, and so on and so forth all the way down to the bottom row, where some $a_{n}$ and $b_{n}=a_{1}+a_{2}+\cdots+a_{n}$ are the two neighbours below $b_{n-1}=a_{1}+a_{2}+\cdots+a_{n-1}$. Since the $a_{k}$ are $n$ pairwise distinct positive integers whose sum does not exceed the largest number in $T$, which is $1+2+\cdots+n$, it follows that they form a permutation of $1,2, \ldots, n$.  Figure 1  Figure 2 Consider now (Figure 2) the two 'equilateral' subtriangles of $T$ whose bottom rows contain the numbers to the left, respectively right, of the pair $a_{n}, b_{n}$. (One of these subtriangles may very well be empty.) At least one of these subtriangles, say $T^{\prime}$, has side length $\ell \geqslant\lceil(n-2) / 2\rceil$. Since $T^{\prime}$ obeys the anti-Pascal rule, it contains $\ell$ pairwise distinct positive integers $a_{1}^{\prime}, a_{2}^{\prime}, \ldots, a_{\ell}^{\prime}$, where $a_{1}^{\prime}$ is at the apex, and $a_{k}^{\prime}$ and $b_{k}^{\prime}=a_{1}^{\prime}+a_{2}^{\prime}+\cdots+a_{k}^{\prime}$ are the two neighbours below $b_{k-1}^{\prime}$ for each $k=2,3 \ldots, \ell$. Since the $a_{k}$ all lie outside $T^{\prime}$, and they form a permutation of $1,2, \ldots, n$, the $a_{k}^{\prime}$ are all greater than $n$. Consequently, $$ \begin{array}{r} b_{\ell}^{\prime} \geqslant(n+1)+(n+2)+\cdots+(n+\ell)=\frac{\ell(2 n+\ell+1)}{2} \\ \geqslant \frac{1}{2} \cdot \frac{n-2}{2}\left(2 n+\frac{n-2}{2}+1\right)=\frac{5 n(n-2)}{8}, \end{array} $$ which is greater than $1+2+\cdots+n=n(n+1) / 2$ for $n=2018$. A contradiction. Comment. The above estimate may be slightly improved by noticing that $b_{\ell}^{\prime} \neq b_{n}$. This implies $n(n+1) / 2=b_{n}>b_{\ell}^{\prime} \geqslant\lceil(n-2) / 2\rceil(2 n+\lceil(n-2) / 2\rceil+1) / 2$, so $n \leqslant 7$ if $n$ is odd, and $n \leqslant 12$ if $n$ is even. It seems that the largest anti-Pascal pyramid whose entries are a permutation of the integers from 1 to $1+2+\cdots+n$ has 5 rows.
|
proof
|
Yes
|
Yes
|
proof
|
Combinatorics
|
An anti-Pascal pyramid is a finite set of numbers, placed in a triangle-shaped array so that the first row of the array contains one number, the second row contains two numbers, the third row contains three numbers and so on; and, except for the numbers in the bottom row, each number equals the absolute value of the difference of the two numbers below it. For instance, the triangle below is an anti-Pascal pyramid with four rows, in which every integer from 1 to $1+2+3+4=10$ occurs exactly once: $$ \begin{array}{ccccc} & & & \\ & 2 & 6 & \\ & 5 & 7 & \\ 8 & 3 & 10 & 9 . \end{array} $$ Is it possible to form an anti-Pascal pyramid with 2018 rows, using every integer from 1 to $1+2+\cdots+2018$ exactly once? (Iran) Answer: No, it is not possible.
|
Let $T$ be an anti-Pascal pyramid with $n$ rows, containing every integer from 1 to $1+2+\cdots+n$, and let $a_{1}$ be the topmost number in $T$ (Figure 1). The two numbers below $a_{1}$ are some $a_{2}$ and $b_{2}=a_{1}+a_{2}$, the two numbers below $b_{2}$ are some $a_{3}$ and $b_{3}=a_{1}+a_{2}+a_{3}$, and so on and so forth all the way down to the bottom row, where some $a_{n}$ and $b_{n}=a_{1}+a_{2}+\cdots+a_{n}$ are the two neighbours below $b_{n-1}=a_{1}+a_{2}+\cdots+a_{n-1}$. Since the $a_{k}$ are $n$ pairwise distinct positive integers whose sum does not exceed the largest number in $T$, which is $1+2+\cdots+n$, it follows that they form a permutation of $1,2, \ldots, n$.  Figure 1  Figure 2 Consider now (Figure 2) the two 'equilateral' subtriangles of $T$ whose bottom rows contain the numbers to the left, respectively right, of the pair $a_{n}, b_{n}$. (One of these subtriangles may very well be empty.) At least one of these subtriangles, say $T^{\prime}$, has side length $\ell \geqslant\lceil(n-2) / 2\rceil$. Since $T^{\prime}$ obeys the anti-Pascal rule, it contains $\ell$ pairwise distinct positive integers $a_{1}^{\prime}, a_{2}^{\prime}, \ldots, a_{\ell}^{\prime}$, where $a_{1}^{\prime}$ is at the apex, and $a_{k}^{\prime}$ and $b_{k}^{\prime}=a_{1}^{\prime}+a_{2}^{\prime}+\cdots+a_{k}^{\prime}$ are the two neighbours below $b_{k-1}^{\prime}$ for each $k=2,3 \ldots, \ell$. Since the $a_{k}$ all lie outside $T^{\prime}$, and they form a permutation of $1,2, \ldots, n$, the $a_{k}^{\prime}$ are all greater than $n$. Consequently, $$ \begin{array}{r} b_{\ell}^{\prime} \geqslant(n+1)+(n+2)+\cdots+(n+\ell)=\frac{\ell(2 n+\ell+1)}{2} \\ \geqslant \frac{1}{2} \cdot \frac{n-2}{2}\left(2 n+\frac{n-2}{2}+1\right)=\frac{5 n(n-2)}{8}, \end{array} $$ which is greater than $1+2+\cdots+n=n(n+1) / 2$ for $n=2018$. A contradiction. Comment. The above estimate may be slightly improved by noticing that $b_{\ell}^{\prime} \neq b_{n}$. This implies $n(n+1) / 2=b_{n}>b_{\ell}^{\prime} \geqslant\lceil(n-2) / 2\rceil(2 n+\lceil(n-2) / 2\rceil+1) / 2$, so $n \leqslant 7$ if $n$ is odd, and $n \leqslant 12$ if $n$ is even. It seems that the largest anti-Pascal pyramid whose entries are a permutation of the integers from 1 to $1+2+\cdots+n$ has 5 rows.
|
{
"resource_path": "IMO/segmented/en-IMO2018SL.jsonl",
"problem_match": null,
"solution_match": null
}
|
41a73402-3458-5c33-b7d8-69152895cc34
| 23,661
|
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