## Saturday, January 15, 2011

### Problem 567: Right triangle, Incircle, Excircle, Collinear tangency points

Geometry Problem
Click the figure below to see the complete problem 567.

Zoom at: Geometry Problem 567

1. http://img843.imageshack.us/img843/4514/problem567.png

Let 2p=perimeter of tri. ABC ; r= radius of incenter circle
We have CF=p and AG= p-BC
FB=CF-BC=p-BC
So AG=FB
Triangle FBG congruent to tri. AGD ( case SAS)
So F, G, H are collinear

Peter Tran

2. EA//GH ( EAH + GHA = 90 + A/2 + 90 - A/2 )
EF//AG, EF = AG = R => EFGA paralelogram
=> EA//FG
from G two // to AE => F, G, H collinear

3. Other,
ADGH são cocíclicos => DGH=a/2 (i)
M é a interseção de AD com GH, então AMBF tb é cocíclico e BFG = a/2 (ii)
De (i) e (ii) BGF = 90 - a/2 = AGH => F, G, H colineares.

4. If the circle centered on the point D intersects the BC in K and s is half-perimeter of triangle ABC.(AH=AG=s-a=BF, CH=CK=s-c, BG=BK=s-b).Then CH/HA.AG/GB.BK/KC=1 and from the inverse theorem Menelaus, the points F,Gand H are collinear.
APOSTOLIS MANOLOUDIS 4 HIGH SHCOOL OF KORYDALLOS PIRAEUS GREECE