1982 IMO Problems/Problem 2

Problem

A non-isosceles triangle $A_{1}A_{2}A_{3}$ has sides $a_{1}$ , $a_{2}$ , $a_{3}$ with the side $a_{i}$ lying opposite to the vertex $A_{i}$ . Let $M_{i}$ be the midpoint of the side $a_{i}$ , and let $T_{i}$ be the point where the inscribed circle of triangle $A_{1}A_{2}A_{3}$ touches the side $a_{i}$ . Denote by $S_{i}$ the reflection of the point $T_{i}$ in the interior angle bisector of the angle $A_{i}$ . Prove that the lines $M_{1}S_{1}$ , $M_{2}S_{2}$ and $M_{3}S_{3}$ are concurrent.

Solution

Rename the vertices so that $A_1=A,A_2=B$ and $A_3=C$ . Let $I$ be the incenter of $\triangle ABC$ .

Claim 1: $S_1, S_2, S_3$ all lie on the incircle. Proof: By construction, $IS_1$ is the reflection of $IT_1$ with respect to the $A$ angle bisector since $I$ lies on the angle bisector of $A$ . So, $IS_1=IT_1$ so $S_1$ lies on the incircle. Similarly, $S_2$ and $S_3$ lie on the incircle. $\Box$

Note that $I$ is the circumcenter of $\triangle S_1S_2S_3$ because of equal radii due to Claim 1.

Claim 2: $\angle S_1IS_2=2\angle C.$ Proof: In cyclic quadrilateral $T_1IT_2C$ , $\angle T_1IT_2=180^{\circ}-\angle C$ . Let $D$ be the point where the $A$ angle bisector intersects $BC$ . Due to the construction of $S_1$ , $\angle S_1IT_1=2\angle DIT_1.$ But, $\angle ADC=180^{\circ}-\angle C-\frac{1}{2}\angle A$ so $\angle DIT_1=\frac{1}{2}\angle A+\angle C-90^{\circ}.$ Thus, $\angle S_1IT_1=\angle A+2\angle C-180^{\circ}.$ Similarly, $\angle S_2IT_2=\angle B+2\angle C-180^{\circ}.$ Therefore, $\angle S_1IS_2=\angle S_1IT_1+\angle T_1IT_2+\angle T_2IS_2=2\angle C.$ $\Box$

Hence, we use Claim 2 to get $\angle S_1S_3S_2=\frac{1}{2}\angle S_1IS_2=\angle C$ and we similarly get $\angle S_3S_1S_2=\angle A$ and $\angle S_1S_2S_3=\angle B.$

Claim 3: Let $X_1$ be the midpoint of $S_2S_3.$ Then, $T_1IX_1$ is collinear. Proof: We already derived that $\angle T_1IT_2=180^{\circ}-\angle C$ and $\angle S_2IT_2=\angle B+2\angle C-180^{\circ}.$ Recall that a radius is perpendicular to the midpoint of a chord so $S_2S_3\perp IX_1.$ Then, $\angle X_1IS_2=90^{\circ}-\angle X_1S_2I=90^{\circ}-(90^{\circ}-\angle A)=\angle A.$ Thus, $\angle T_1IX_1=\angle X_1IS_2+\angle S_2IT_2+\angle T_2IT_1=\angle A+\angle B+2\angle C-180^{\circ}+180^{\circ}-\angle C=180^{\circ},$ as desired. $\Box$

Then, $T_1IX_1\perp S_2S_3$ because a diameter is perpendicular to a chord at its midpoint and because $T_1IX_1$ is part of a diameter, by Claim 3. Hence, $S_2S_3\parallel BC.$ But, $M_2M_3\parallel BC$ so $S_2S_3\parallel M_2M_3$ . Similarly, we get $S_1S_2\parallel M_1M_2$ and $S_1S_3\parallel M_1M_3$ so $\triangle M_1M_2M_3\sim \triangle S_1S_2S_3$ with corresponding parallel sides. Thus, there exists a homothety that maps $M_1M_2M_3$ to $S_1S_2S_3$ and $M_1S_1, M_2S_2,$ and $M_3S_3$ intersect at the center of the homothety. $\blacksquare$

This solution was posted and copyrighted by Jzhang21. The original thread for this problem can be found here: [1]

1982 IMO (Problems) • Resources
Preceded by Problem 1	1 • 2 • 3 • 4 • 5 • 6	Followed by Problem 3
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1982 IMO Problems/Problem 2

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