## Friday, April 10, 2009

### Problem 27: Right triangle, Incenter, Parallel, Incircles, Inradius

1. Hints: Two tangent segments to a circle are congruent. Square...

2. http://img98.imageshack.us/img98/3497/problem27.png
Connect CO and AO
Note that CO and CO2 are angle bisector of angle (BCA) so
C, O, O2 are collinear
Triangle OGC congruence with triangle OKC ( right triangles with common hypotenuse and CO is angle bisector)
So OC is the angle bisector of angle (GOK) >>Incircle O2 tangent to OG also tangent to OK and O2N=r2
Similarly AO is angle bisector of angle HOK) and incircle O1 tangent to both OH and OK and O1P=r1
DE=DK+KE=O1P+O2N=r1+r2
Peter Tran

3. Using Antonio's hint above....

If X is the point of tangency of circle O with AC

AK = c - r and KC = a - r

Similarly AD = c-r-r1 and CE = a-r-r2

So AC - AD - EC = r1+r2 + (b-c-a + 2r) from using the above expressions for AD and CE.

The expression within the bracket = 0 and the result follows

Sumith Peiris
Moratuwa
Sri Lanka

4. Define O1 and O2 the center of incirles with radius r1 and r2
Points A, O1 and O are aligned on the angle bisector AD=AD’ and CE=CE’
D’H=r1, E’G=r2 and HB=HG=r