## Monday, May 19, 2008

### Geometry Problem 86 See complete Problem 86
Intouch and Extouch Triangles, Area. Level: High School, SAT Prep, College geometry

Post your solutions or ideas in the comments.

1. AF=AE=p-a=BM=CH
BF=BD=p-b=AM=CG
CD=CE=p-c=BG=AH
[DEF]=[ABC]-0.5(p-a)²sinA-0.5(p-b)²sinB
-0.5(p-c)²sinC
[MGH]=[ABC]-0.5(p-b)(p-c)sinA-0.5(p-a)(p-b)sinC
-0.5(p-a)(p-c)sinB
sinA=a/(2R);sinB=b/(2R);sinC=c/(2R)
with a "little" algebra
S_i=S_e

2. In the solution of pr. 86 posted by Anonymous, could anyone explain me why is it true that
AF = BM = CH = p - a ?
Thanks for the help.
Nilton Lapa

1. To Nilton,
Problem 86, See

Semiperimeter s, Incircle and Excircles
" or

Semiperimeter Index
" or

use "two tangent segments to a circle from an external point are congruent."

3. To Antonio: thanks for the hints. My doubt is solved now.

4. of the above solution, why it can be said that the triangle DEF = triangle MGH??

5. In the solution of pr. 86 posted by Anonymous, could anyone please explain me how to do this ''little'' algebra?