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.

Problem 520: Triangle, Six Squares, Areas, Ratio


See also:
Complete Problem 520

Level: High School, SAT Prep, College geometry

6 comments:

  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

    ReplyDelete
  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

    ReplyDelete
  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

    ReplyDelete
  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).

    ReplyDelete
  5. Hello Peter,

    Is my last explanation enough for you, or not?

    ReplyDelete
  6. To Emil

    Thank you for your explanation.
    It is nice if you can give reference in your comment so that people can follow your logic.

    ReplyDelete