## Monday, May 19, 2008

### Geometry Problem 97

See complete Problem 97
Similar Triangles, Areas. Level: High School, SAT Prep, College geometry

Post your solutions or ideas in the comments.

#### 2 comments:

1. We've AC = AM + MC = DF+ MC = DE + EF + MC
or 1 = DE/AC + EF/AC + MC/AC ---(1). Noting that triangles GMC, GEF, BDE & BAC are all similar we can say that their areas are proportional to the squares of their corresponding sides. Hence DE/AC = V(S1/S) where V(x)=square root of x. Similarly, EF/AC=V(S2/S) and MC/AC=V(S3/S). BY (1) we can say that 1 = V(S1/S)+V(S2/S)+V(S3/S) or VS= V(S1)+ V(S2)+ V(S3) QED
Ajit

2. Solution to problem 97.
Let h, h1, h2 and h3 be the altitudes of triangles ABC, DBE, FGE and MGC, corresponding to sides BC, BE, EG and GC, respectively.
As these four triangles are similar,
BC/h = BE/h1 = EG/h2 = GC/h3 = k.
For the areas we have
S = BC.h/2 = (k/2).h*2,
S1 = BE.h1/2 = (k/2).h1*2,
S2 = EG.h2/2 = (k/2).h2*2
and S3 = GC.h3/2 = (k/2).h3*2.
Then sqr(S) = sqr(k/2).h,
sqr(S1) = sqr(k/2).h1,
sqr(S2) = sqr(k/2).h2
and sqr(S3) = sqr(k/2).h3.
So sqr(S1) + sqr(S2) + sqr(S3) =
= sqr(k/2).(h1 + h2 + h3) =
= sqr(k/2).h = sqr(S).
We have used here the result of problem 91,
h = h1 + h2 + h3.