Problem

Source: Romanian TST 2007, Day 6, Problem 2

Tags: geometry, trigonometry, AMC, USA(J)MO, USAJMO, perpendicular bisector, geometry proposed



Let $ ABC$ be a triangle, let $ E, F$ be the tangency points of the incircle $ \Gamma(I)$ to the sides $ AC$, respectively $ AB$, and let $ M$ be the midpoint of the side $ BC$. Let $ N = AM \cap EF$, let $ \gamma(M)$ be the circle of diameter $ BC$, and let $ X, Y$ be the other (than $ B, C$) intersection points of $ BI$, respectively $ CI$, with $ \gamma$. Prove that \[ \frac {NX} {NY} = \frac {AC} {AB}. \] Cosmin Pohoata