X is a point on the side BC of the triangle ABC. Take the other common tangent (apart from BC) to the incircles of ABX and ACX which intersects the segments AB and AC. Let it meet AX at Y. Show that the locus of Y, as X varies, is the arc of a circle.
Solution
We show that AY = (AB + AC - BC)/2 = constant.
Let the common tangent meet the incircles of ABX, ACX at R, S respectively. Let AX meet them at P, Q respectively and let BC meet them at U, V respectively. Let AB meet the incircle of ABX at K and let AC meet the incircle of ACX at L. We have AY = AQ - QY and AY = AP - PY. So adding 2AY = AP + AQ - (YQ + YP) = AP + AQ - (YS + YR) = AP + AQ - RS. If we reflect about the line of centers of the two incircles R goes to U and S to V. Hence RS = UV. So 2AY = AP + AQ - UV. We have AP = AK = AB - BK = AB - BU and AQ = AL = AC - CL = AC - CV. Hence 2AY = AB + AC - BU - UV - CV = AB + AC - BC.
© John Scholes
jscholes@kalva.demon.co.uk
11 May 2002