## Friday, July 25, 2008

### Elearn Geometry Problem 144

See complete Problem 144
Four Triangles, Incircle, Tangent and Parallel to Side, Circumradii. Level: High School, SAT Prep, College geometry

1. we have : DE=HG
r1/r=AH/AC
r2/r=DE/AC
r3/r=CG/AC
r1/r+r2/r+r3/r=AH/AC+DE/AC+CG/AC
(r1+r2+r3)/r=(AH+HG+CG)/AC
(r1+r2+r3)/r=AC/AC=1
r1+r2+r3=r
yass_ghaz_57@hotmail.fr

2. Very good.
What are the proof reasons that you use?

3. I just wonder if there is any information about the history of this problem.

4. Solution of problem 144.
Let P, P1, P2, P3 be the perimeters or triangles ABC, AMH, DBE, GFC, respectively.
We have P = P1 + P2 + P3 (as proved in problem 141).
Triangles AMH and ABC are similar with ratio P1/P.
Triangles DBE and ABC are similar with ratio P2/P.
Triangles GFC and ABC are similar with ratio P3/P.
So r1/r + r2/r + r3/r = P1/P + P2/P + P3/P = (P1 + P2 + P3)/P = P/P = 1
and r = r1 + r2 + r3.

5. http://imageshack.us/a/img138/9641/problem144.png

Extend MH,DE and FG to form triangle A’B’C’ ( see attached sketch)
∆ABC similar to ∆A’B’C’…. ( case AA)
Ratio of similarity =incircle radius of ∆ABC/incircle radius of ∆A’B”C’= r/ r=1
So ∆ABC congruent to ∆A’B’C’∆
Let h is the altitude of ∆ABC and A’B’C’ from B and B’
In ∆BDE we have DE/AC=(h-r)/h
In ∆B’HG we have HG/A’C’= (h-r)/h
Bu AC=A’C’ so DE=HG
Follow the solution of yassin-NAugust 23, 2009 7:25 AM for the remaining prove

6. Peter, DE/AC = (h−2r)/h = HG/A'C'