Let ABC be an arbitrary triangle with side lengths a = BC, b = CA, and c = AB.
Let X, Y, and Z be the points of tangency of the incircle with the sides BC, CA, and AB respectively.
Let X', Y', and Z' be the points of tangency of the excircles with "sides" BC, CA, and AB respectively.
What is the value of XX' + YY' + ZZ' in terms of a, b, and c?
let t=BX by similar triangles we have BZ=BX=t we also have CX=CY=at AZ=AY=ba+t thus AZ+BZ=AB
t+ba+t=c
2t=a+cb
t=(a+cb)/2
let u=BX'
let Z,W be the centers of the excircle from side BC and incircle respectivly
let Rx and Ri be the radii of the excircle and incircle respectivly
ZBX' and BWX are right triangles and since angles BZX' and BWX are congruent and ZBX'=BXW=90 the we have that
u=Rx*Ri/t
now if A is the area of triangle ABC then
ri=2A/(a+b+c)
rx=2A/(b+ca)
thus
u=4A^2/[(a+b+c)(b+ca)t]
substituting in our know value for t we get
u=8A^2/[(a+b+c)(b+ca)(a+cb)]
now using Herons formula for area we get
u=(1/2)(a+b+c)(a+bc)(a+cb)(b+ca)/[(a+b+c)(b+ca)(a+cb)]
simplfying we get
u=(a+bc)/2
now we can get XX' with tu which is
(1/2)a+cbab+c
(1/2)2c2b
cb
now using similar methods we can get YY' and ZZ' to be ac and ab respectivly thus the desired sum is
cb+ac+ab

Posted by Daniel
on 20080118 02:26:11 