# Three Circles through the Incenter

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.

There is already one solution obtained by "angle chasing". While it solved the problem, it left me unsatisfied. There was a feeling that there had to be a more transparent explanation. There I believe is one which is illustrated by the applet below. The idea was suggested by Douglas Rogers in a private correspondence.

### This applet requires Sun's Java VM 2 which your browser may perceive as a popup. Which it is not. If you want to see the applet work, visit Sun's website at https://www.java.com/en/download/index.jsp, download and install Java VM and enjoy the applet.

 What if applet does not run?

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 applet requires Sun's Java VM 2 which your browser may perceive as a popup. Which it is not. If you want to see the applet work, visit Sun's website at https://www.java.com/en/download/index.jsp, download and install Java VM and enjoy the applet.

 What if applet does not run?

Let's complicate the picture by drawing two other circles through the incenter and the remaining pairs of vertices. Let's draw the internal and external angle bisectors of ΔABC. We do not need that many extra elements for the proof, but having them all together may make the argument more transparent.

The internal bisector of angle C and the two external bisectors of angles at A and B intersect at the excenter Ic of ΔABC. The internal and external bisectors at any vertex are perpendicular, implying, in particular, that BI⊥BIc and AI⊥AIc. Thus we have two right triangles IAIc and IBIc. Put together they form a cyclic quadrilateral IAIcB. The circumcircle of IAIcB passes through the points A, B and I and is, therefore, the circle stipulated by the problem.

Since triangles IAIc and IBIc are right, their shared hypotenuse IIc is a diameter of the circle at hand. It follows that the center of the circle lies on the (internal) bisector of angle at C. This means that the whole circle is symmetric in the bisector as are the side lines AC and BC. We conclude that the circle intercepts the two side lines in two mutual reflections in the bisector. But this are AX and BY, with A corresponding to Y and B to X under the reflection. The two segments are therefore equal.

There are five solutions in all: 