Sunday, September 12, 2010

Problem 520: Triangle, Six Squares, Areas, Ratio

Geometry Problem
Click the figure below to see the complete problem 520 about Triangle, Six Squares, Areas, Ratio.

Complete Problem 520

Level: High School, SAT Prep, College geometry

1. Draw three median lines on triangle ABC, AA', BB', and CC'.
Length AA' is (side of S4)/2, BB'=S6 side/2, CC'=S5 side/2.

Let AB=c, BC=a, CA=b, and AA'=d, BB'=e, CC'=f.
Par median theorem:
a^2+b^2=2*(f^2+(c/2)^2),
b^2+c^2=2*(d^2+(a/2)^2),
c^2+a^2=2*(e^2+(b/2)^2).
--->
2*(a^2+b^2+c^2)=2*(d^2+e^2+f^2)+(a^2+b^2+c^2)/2
--->
3*(a^2+b^2+c^2)=4*(d^2+e^2+f^2)=(2d)^2+(2e)^2+(2f)^2
--->
3*(S1+S2+S3)=S4+S5+S6

2. Let c=AB, a=BC, b=AC
S1=c^2 , S2=b^2, S3=a^2
S4=b^2+c^2-2.b.c.Cos(obtuse A)=b^2+c^2+2.b.c.Cos(A)
Replace 2.b.c.Cos(A)=b^2+c^2-a^2
We get S4=2.b^2+2.c^2-a^2
Similarly S5=2.a^2+2.b^2-c^2 and S6=2.a^2+2.c^2-b^2
S4+S5+S6=3.a^2+3.b^2+3.a^2 and (S4+S5+S6)/(S1+S2+S3)=3

Peter Tran

3. To Emil Ekker

It is not clear to me how do you have "Length AA' is (side of S4)/2" in your comment. Please explain

4. Hello Peter,

See problem 502, and c.t.e.o's solution.
BM=AC/2, in addition, DM=median line from B (=BB' of my solution).

5. Hello Peter,

Is my last explanation enough for you, or not?

6. To Emil