IMO Shortlist 1995 Geometry #3

by KingSmasher3, Apr 3, 2013, 5:29 AM

The incircle of triangle $ \triangle ABC$ touches the sides $ BC$, $ CA$, $ AB$ at $ D, E, F$ respectively. $ X$ is a point inside triangle of $ \triangle ABC$ such that the incircle of triangle $ \triangle XBC$ touches $ BC$ at $ D$, and touches $ CX$ and $ XB$ at $ Y$ and $ Z$ respectively.
Show that $ E, F, Z, Y$ are concyclic.
________

Solution 1: Consider the circle centered at $B$ passing through $F, Z,$ and $D$ and the circle centered at $C$ passing through $E, Y,$ and $D.$ Their radical axis is the line through $D$ perpendicular to $BC.$ This line contains $I_1$ and $I_2,$ the incenters of $\triangle ABC$ and $\triangle XBC$ respectively. Let $EY$ intersect this line at the point $P.$ Let $Q=I_2Y \cap EI_1$ and $R = I_2Z \cap FI_1.$ Using Menelaus on $\triangle I_1I_2Q$ and $\triangle I_1I_2R,$ see find that $F, Z,$ and $P$ are collinear if and only if $EQ/QY=FR/RZ$ which is clearly true since $EQ=QY$ and $FR=RZ.$ This, in addition to the fact that $P$ is on the radical axis of the two circles centered at $B$ and $C,$ shows that $PY \cdot PE = PZ \cdot PF,$ so by power of a point, $EFZY$ is cyclic.

Solution 2: Extend $FE$ to meet $BC$ at $M.$ Using Menelaus of $\triangle ABC$ and $\triangle XBC,$ we see that $Z, Y,$ and $M$ are collinear. By power of a point, it follows that $ME \cdot MF = MD^2 = MY \cdot MZ,$ so $EFZY$ is cyclic as desired.

Inversion Solution Hint

Main Point(s): Menelaus can be used to prove collinearity (or concurrency) when applying power of a point.
This post has been edited 2 times. Last edited by KingSmasher3, Apr 3, 2013, 7:32 PM

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