Circle through the IncenterThe 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,
 |Activities| |Contact| |Front page| |Contents| |Geometry| |Store| Copyright © 1996-2012 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.
Two configurations are possible: one with Y inside BC and X outside AC and the other with Y outside BC and X inside AC. 
Without loss of generality, we shall consider just one of them. This is illustrated below: 
In the diagram, J is the other intersection of the bisector CI with the circle and, for simplicity, the arcs are enumerated, so that, in the following I shall refer to the arcs by those numbers: The problem is solved through angle calculations, commonly referred to as "angle chasing". The fact that I is the incenter of ΔABC provides sufficient reasons and grounds to insure the satisfactory outcome. First of all, in ΔABC 2α + 2β + 2γ = 180° while, in ΔAIB, α + β + δ = 180°. This gives δ = 90° + γ. Since AIB is an inscribed angle we get one arc relation:
Next, ∠ACJ = ∠BCJ = γ. From the properties of the angles between two secants,
Since BI is the angle bisector, ∠ABI = ∠CBI = β, from which
Together with (2) this implies
Further, ∠BAI = α and is subtended by the arc BI so that
Combining this with (3) gives
On the other hand, from (1) and (4),
Together, (6) and (7) show that 3 = 6, as required. (Indeed, the identity checks out: (A different proof has been posted by Vo Duc Dien at the wiki part of the site.) One engaging interpretation of this result is that A circle through two vertices of a triangle and its incenter is always centered on the angle bisector of the remaining vertex and is therefore symmetric in that angle bisector. Another solution makes a better use of this observation. There are five solutions in all: Red Herrings: a Sample List
|Activities| |Contact| |Front page| |Contents| |Geometry| |Store| Copyright © 1996-2012 Alexander Bogomolny |
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