## Sunday, September 11, 2011

### Problem 668: Right triangle, Incircles, Parallel, Common Tangent, Mind Map

Geometry Problem
Level: Mathematics Education, High School, Honors Geometry, College.

Click the figure below to see the complete problem 668.

1. http://img819.imageshack.us/img819/5083/problem668.png
Connect CO and AO
Note that CO and CO2 are angle bisector of angle (BCA) so
C, O, O2 are collinear
Triangle OEC congruence with triangle OHC ( right triangles with common hypotenuse and CO is angle bisector)
So OC is the angle bisector of angle (HOE) >>Incircle O2 tangent to OE also tangent to OH.
Similarly AO is angle bisector of angle HOF) and incircle O1 tangent to both OF and OH
Peter Tran

2. Let circle O₁ touch AB at J, AC at K & FG at L.
Complete the square O₁JHM
O₁M = KH = AH - AK = AF - AJ = KF = radius of O₁ (note FJ O₁K is a square)
follows OH touches circle O₁.
Similarly we can show OH touches circle O₂

3. Refer my solution to Problem 668:
"FJO₁L is a square" (not FJO₁K as previously typed)

4. another point of vue
in ABC r=s-b
triangles AFG and ABC are homothetic,center A ratio (s-a)/c
the incircle of AFG is the image of the incircle of ABC;r1=(s-a)r/c=(s-b)(s-a)/c
line OH becomes the parallel line O1H1
distance H1H=(s-a)-(s-a)²/c=(s-a)(s-b)c=r1
incircle O1 tangent to OH
The same way CDE and ABC are homothetic with center C and ratio (s-c)/a