# 1983 IMO Problems/Problem 2

## Problem

Let $A$ be one of the two distinct points of intersection of two unequal coplanar circles $C_1$ and $C_2$ with centers $O_1$ and $O_2$ respectively. One of the common tangents to the circles touches $C_1$ at $P_1$ and $C_2$ at $P_2$, while the other touches $C_1$ at $Q_1$ and $C_2$ at $Q_2$. Let $M_1$ be the midpoint of $P_1Q_1$ and $M_2$ the midpoint of $P_2Q_2$. Prove that $\angle O_1AO_2=\angle M_1AM_2$.

## Solution 1

Let $A$ be one of the two distinct points of intersection of two unequal coplanar circles $C_1$ and $C_2$ with centers $O_1$ and $O_2$ respectively. Let $S$ be such point on line $O_1O_2$ so that tangents on $C_1$ touches it at $P_1$ and $Q_1$ and tangents on $C_2$ touches it at $P_2$ and $Q_2$. Let $M_1$ be the midpoint of $P_1Q_1$ and $M_2$ the midpoint of $P_2Q_2$. Prove that $\angle O_1AO_2 = \angle M_1AM_2$.

Proof: Since $S$ is image of $M_1$ under inversion wrt circle $C_1$ we have:$$\angle O_1AM_1 = \angle O_1M_1'A'= \angle O_1SA$$Since $S$ is image of $M_2$ under inversion wrt circle $C_2$ we have:$$\angle O_2SA= \angle O_2A'S'= \angle O_2AM_2$$Image of $A$ is in both cases $A$ itself, since it lies on both circles. Since $\angle O_1SA = \angle O_2SA$ we have:$$\angle M_1AO_1=\angle M_2AO_2$$Now:$$\angle O_1AO_2 = \angle M_1AM_2-\angle M_1AO_1+\angle M_2AO_2 = \angle M_1AM_2$$

This solution was posted and copyrighted by Number1. The original thread for this problem can be found here: [1]

## Solution 2

Let $P_1P_2$ and $Q_1Q_2$ meet at $R$. Let $RA$ meet $C_2$ at $B$. Now, it is well-known that $O_1O_2$, $P_1P_2$, and $Q_1Q_2$ are concurrent at $R$, the center of homothety between $C_1$ and $C_2$. Now, it is well-known that $O_1O_2$ bisects $\angle P_1RQ_1$. Since $P_1R = Q_1R$, we have that $O_1O_2R$ meets $P_1Q_1$ at its midpoint, $M_1$, and $P_1Q_1$ is perpendicular to $O_1O_2R$. Similarly, $O_1O_2R$ passes through $M_2$ and is perpendicular to $P_2Q_2$. Since $O_2P_2\perp P_2R$, we have that $RA\cdot RB = RP_2^2 = RM_2\cdot RO_2$, which implies that $ABM_2O_2$ is cyclic. Yet, since $A$ and $B$ lie on $C_1$ and $C_2$ respectively and are collinear with $R$, we see that the homothety that maps $C_1$ to $C_2$ about $R$ maps $A$ to $B$. Also, $O_1$ is mapped to $O_2$ by this homothety, and since $M_1$ and $M_2$ are corresponding parts in these circles, $M_1$ is mapped to $M_2$ by this homothety, so $\angle O_1AM_1 = \angle O_2BM_2 = \angle O_2AM_2$, from which we conclude that $\angle O_2AO_1 = \angle M_2AOM_1$.

This solution was posted and copyrighted by The QuattoMaster 6000. The original thread for this problem can be found here: [2]

## Solution 3

Let $B$ be the other intersection point of these two circles. Let $P_1P_2$, $Q_1Q_2$, $O_1O_2$ meet at $Q$. Let $AB$ meet $P_1P_2$ at $P$. Clearly, $P_1Q_1$ and $O_1O_2$ are perpendicular at $M_1$; $P_2Q_2$ and $O_1O_2$ are perpendicular at $M_2$. Since $\triangle O_1P_1M_1 \sim \triangle O_2P_2M_2$,$$\dfrac{O_1P_1}{O_1M_1} = \dfrac{O_2P_2}{O_2M_2} \Rightarrow \dfrac{O_1A}{O_1M_1} = \dfrac{O_2A}{O_2M_2} \qquad{(*)}$$Since $P$ is on the radical axis, $PP_1 = PP_2$, so in the right trapezoid $P_1P_2M_2M_1$, $AB$ is the midsegment. So we have $M_1A = AM_2$ and $\angle AM_1O_1 = \angle AM_2O_1$. Let $M$ be a point on $O_1O_2$ such that $\triangle AM_1O_1 \cong \triangle AM_2M$ (which means $AM=AO_1$ ve $MM_2 = M_1O_1$). So, from $(*)$, we get$$\dfrac{AM}{MM_2} = \dfrac{O_2A}{O_2M_2}$$This means, $AM_2$ is the angle bisector of $\angle MAO_2$. So,$$\angle O_2AM_2 = \angle M_2AM= \angle M_1AO_1 \Longrightarrow \angle M_1AM_2 = \angle O_1AO_2.$$

This solution was posted and copyrighted by matematikolimpiyati. The original thread for this problem can be found here: [3]

## Solution 4

Let, $P_1P_2,Q_1Q_2,O_1O_2$ concur at $Q$.Then a homothety with centre $Q$ that sends $C_1$ to $C_2$.Let $QA \cap C_1 =B$.Under the homothety $A$ is the image of $B$.So, $\angle M_1BO_1 =\angle M_2AO_2$ and $\triangle QP_1O_1 \sim \triangle QM_1P_1 \implies \frac{QO_{1}}{QP_{1}} = \frac{QP_{1}}{QM_{1}}$.so $QO_1.QM_1 = Q.P^2_1 = QA .QB$ Hence $A,M_1,B,O_1$ are concyclic.so $\angle O_1BM_1=\angle O_1AM_1 \implies \angle O_1AM_1= \angle O_2AM_2 \implies \angle O_1AO_2=\angle M_1AM_2$

This solution was posted and copyrighted by spikerboy. The original thread for this problem can be found here: [4]

## Solution 5

It can be easily solved with this lemma: Let $A$-symmedian of $\triangle ABC$ intersect its circumscribed circle $u$ at $D, P$ on $AD$ satisfying $AP = PD$. Let us define center of $u$ as $O$. Then $(B, C, P, O)$ are concyclic. Now let $S$ be intersection of tangents to circles from the problem, $AS$ intersects $C_1$ at $B$. One can prove that $\square AP_1BQ_1$ is harmonic qudrilateral, so $P_1Q_1$ is symmedian of $\triangle AP_1B$. By lemma we get that $(B, A, O_1, M_1)$ are concyclic, thus $\angle O_1AM_1$ is equivalent to $\angle O_1BM_1$. By homothety we get $\angle O_1BM_1 = \angle O_2AM_2$. $Q. E. D.$

This solution was posted and copyrighted by LeoLeon. The original thread for this problem can be found here: [5]

 1983 IMO (Problems) • Resources Preceded byProblem 1 1 • 2 • 3 • 4 • 5 • 6 Followed byProblem 3 All IMO Problems and Solutions