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Iranian Mathematical Olympiad

Geometry Difficulty 5.4 AIME, harder Prove it Iran

Given two intersecting circles $\omega_1, \omega_2$ . Points $A, C$ lie on $\omega_1$ and points $B, D$ lie on $\omega_2$ so that $AB, CD$ are common tangents of $\omega_1, \omega_2$ . Let $M$ be the midpoint of $AB$ . Tangent lines through $M$ to $\omega_1, \omega_2$ (other than $AB$ ) intersect $CD$ at $Y, X$ . If $I$ be the incenter of triangle $MXY$ , prove that $IC = ID$ .

Solution

Suppose that $MX, MY$ touch $\omega_2, \omega_1$ at $E, F$ also let $N$ be the foot of perpendicular line through $I$ to $CD$ .

Notice that $MF = MA = MB = ME$ . $N$ is the intersection of $XY$ and incircle of triangle $MXY$ so
$\begin{align*} NY &= \frac{YX + YM - MX}{2} \\ &= \frac{YX + YF - XE}{2} \\ &= \frac{YX + YC - XD}{2} \\ &= \frac{YD + YC}{2} \\ &= YD + \frac{CD}{2}. \end{align*}$
Therefore
$\frac{CD}{2} = NY - YD = ND,$
hence $N$ is the midpoint of $CD$ so $IC = ID$ and the result follows. ■

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