ABC is a triangle, right-angled at A, and D is the foot of the altitude from A. The straight line joining the incenters of the triangles ABD and ACD intersects the sides AB, AC at K, L respectively. Show that the area of the triangle ABC is at least twice the area of the triangle AKL.
Solution
The key is to show that AK = AL = AD. We do this indirectly. Take K' on AB and L' on AC so that AK' = AL' = AD. Let the perpendicular to AB at K' meet the line AD at X. Then the triangles AK'X and ADB are congruent. Let J be the incenter of ADB and let r be the in-radius of ADB. Then J lies on the angle bisector of angle BAD a distance r from the line AD. Hence it is also the incenter of AK'X. Hence JK' bisects the right angle AK'X, so ∠AK'J = 45o and so J lies on K'L'. An exactly similar argument shows that I, the incenter of ADC, also lies on K'L'. Hence we can identify K and K', and L and L'.
The area of AKL is AK·AL/2 = AD2/2, and the area of ABC is BC·AD/2, so we wish to show that 2AD ≤ BC. Let M be the midpoint of BC. Then AM is the hypoteneuse of AMD, so AM ≥ AD with equality if and only if D = M. Hence 2AD ≤ 2AM = BC with equality if and only if AB = AC.
Solutions are also available in István Reiman, International Mathematical Olympiad 1959-1999, ISBN 189-8855-48-X.
© John Scholes
jscholes@kalva.demon.co.uk
17 Nov 1998