HMMT February

Geometry Difficulty 5.3 AIME, harder Prove it United States

Problem:

Triangle $ABC$ has incenter $I$ . Let $D$ be the foot of the perpendicular from $A$ to side $BC$ . Let $X$ be a point such that segment $AX$ is a diameter of the circumcircle of triangle $ABC$ . Given that $ID = 2$ , $IA = 3$ , and $IX = 4$ , compute the inradius of triangle $ABC$ .

Solution

Solution:

Let $R$ and $r$ be the circumradius and inradius of $ABC$ , let $AI$ meet the circumcircle of $ABC$ again at $M$ , and let $J$ be the $A$ -excenter. We can show that $\triangle AID \sim \triangle AXJ$ (e.g. by $\sqrt{bc}$ inversion), and since $M$ is the midpoint of $IJ$ and $\angle AMX = 90^\circ$ , $IX = XJ$ . Thus, we have
$\frac{2R}{IX} = \frac{XA}{XJ} = \frac{IA}{ID},$
so $R = \frac{IX \cdot IA}{2 ID} = 3$ . But we also know
$R^2 - 2Rr = IO^2 = \frac{2 IX^2 + 2 IA^2 - AX^2}{4}.$
Thus, we have
$r = \frac{1}{2R} \left(R^2 - \frac{2 IX^2 + 2 IA^2 - 4R^2}{4}\right) = \frac{11}{12}$

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