Library / /8 of 11

, 2012

Geometry Difficulty 8.7 Shortlist Prove it Balkan Mathematical Olympiad

Let $M$ be the point of intersection of the diagonals of a cyclic quadrilateral $ABCD$ . Let $I_1$ and $I_2$ be the incenters of triangles $AMD$ and $BMC$ , respectively, and let $L$ be the point of intersection of the lines $DI_1$ and $CI_2$ . The foot of the perpendicular from the midpoint $T$ of $I_1I_2$ to $CL$ is $N$ , and $F$ is the midpoint of $TN$ . Let $G$ and $J$ be the points of intersection of the line $LF$ with $I_1N$ and $I_1I_2$ , respectively. Let $O_1$ be the circumcenter of triangle $LI_1J$ , and let $\Gamma_1$ and $\Gamma_2$ be the circles with diameters $O_1L$ and $O_1J$ , respectively. Let $V$ and $S$ be the second points of intersection of $I_1O_1$ with $\Gamma_1$ and $\Gamma_2$ , respectively. If $K$ is the point where the circles $\Gamma_1$ and $\Gamma_2$ meet again, prove that $K$ is the circumcenter of the triangle $SVG$ .

Solution

The point $L$ is the midpoint of the arc $AB$ and, as $I_1$ , $M$ , $I_2$ are collinear, we have $\angle LI_2I_1 = \angle I_2CM + \angle I_2MC = \angle I_1DM + \angle I_1MD = \angle LI_1I_2$ . Therefore the triangle $LI_1I_2$ is isosceles and $LT \perp I_1I_2$ .

Let $Z$ be the midpoint of $NI_2$ . Then $FZ \parallel I_1I_2$ , so $FZ \perp LT$ and it follows that $F$ is the orthocenter of the triangle $LTZ$ . Therefore we have $LG \perp I_1N$ .

Let $X$ be the orthogonal projection of $O_1$ to $I_1N$ and $H$ be the other end point of diameter of $\Gamma_1$ through $K$ . Since $O_1L$ and $O_1J$ are diameters, $O_1K \perp LG$ and $O_1K \parallel I_1G$ . Since $HVKO_1$ is cyclic, $\angle VHK = \angle VO_1K = \angle O_1I_1X$ . As we also have $\angle HVK = \angle O_1XI_1 = 90^\circ$ and $HK = O_1L = O_1I_1$ , the triangles $HKV$ and $I_1O_1X$ are congruent and $VK = O_1X = KG$ .

On the other hand, since the quadrilaterals $LVKO_1$ and $SKJO_1$ are cyclic, we have
$\angle O_1VK = \angle O_1LK = \angle O_1JK = \angle VSK,$
so the triangle $\Delta SKV$ is isosceles and $VK = SK$ . Hence $K$ is the circumcenter of the triangle $SVG$ .

Want a route through all this instead of an archive? The track puts 2,000 problems in a working order, from AMC 10 level to the IMO shortlist.

Source: MathNet, licensed CC-BY-4.0. Statement and solution reproduced as published; topic and difficulty added by this site.