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Iranian Mathematical Olympiad

Geometry Difficulty 5.6 AIME, harder Prove it Iran

In triangle $ABC$ the points $M$ and $N$ are the midpoints of the sides $AB$ and $AC$ , respectively. The lines $BN$ and $CM$ intersect the circumcircle of $ABC$ , for the second time, at $N'$ and $M'$ , respectively. Points $X$ and $Y$ are on the extension of $BC$ such that, point $B$ is between $X$ and $C$ and point $C$ is between $B$ and $Y$ .

$\angle BXM' = \angle ACM, \angle CYN' = \angle ABN$

Prove that $AY = AX$ .

Solution

Since $\angle BXM' = \angle ACM$ , $\angle M'BX = \angle M'AC$ it follows that $\triangle M'AC \sim \triangle M'BX$ . Hence, $M'B/M'A = BX/AC$ . Yielding, $BX = AC \cdot \frac{M'B}{M'A} = b \cdot \frac{\sin \angle MCB}{\sin \angle MCA}$ . On the other hand, $\frac{\sin \angle MCB}{\sin \angle MCA} = \frac{b}{a}$ . Therefore, $BX = b^2/a$ and analogously $CY = c^2/a$ . Hence, if $D$ is the foot of the altitude from $A$ it suffices to show that $XD = YD$ . We know that $BD = \frac{a^2+c^2-b^2}{2ac}$ and $CD = \frac{a^2+b^2-c^2}{2a}$ . It follows that $XD = BX+BD = \frac{c^2}{a} + \frac{a^2+b^2-c^2}{2a} = CY+CD = YD$ . As desired. ■

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