# Circle through the Incenter And Antiparallels

The applet below illustrates problem 7 from the 2009 Australian Mathematical Olympiad.

Let I be the incenter of a triangle ABC in which AC ≠ BC. Let Γ be the circle passing through A, I and B. Suppose Γ intersects the line AC at A and X and intersects the line BC at B and Y . Show that AX = BY.

The condition AC ≠ BC is obviously a red herring as, in this case, A = X and B = Y. The applet illustrates a fifth solution (see below) to the problem.

Solution

Let I be the incenter of a triangle ABC in which AC ≠ BC. Let Γ be the circle passing through A, I and B. Suppose Γ intersects the line AC at A and X and intersects the line BC at B and Y . Show that AX = BY.

This is a well known fact that a circle through two vertices of a triangle cuts a chord (XY in the applet) antiparallel to the side joining the two vertices (AB in the applet).

In particular this means that ∠BAX = ∠BYX.

These angles are subtended by the arcs BXA and YBX. For the arcs (depending on the layout and assuming J is the second point of intersection of the circle with CI), either

BXA = BJX + XA,
YBX = YB + BJX,

or

BXA = BIX + XA,
YBX = YB + BIX,

In both cases the arcs XA and YB are equal, implying the identity of the subtended chords.

There are five solutions in all: