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.

26 January 2016, Created with GeoGebra

Solution

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Copyright © 1996-2017 Alexander Bogomolny

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.

Circle through the incenter 2 - problem

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.

Circle through the incenter 2 - solution

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:

What Is Red Herring

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