Geometry Problem
Click the figure below to see the complete problem 520 about Triangle, Six Squares, Areas, Ratio.
See also:
Complete Problem 520
Level: High School, SAT Prep, College geometry
Sunday, September 12, 2010
Problem 520: Triangle, Six Squares, Areas, Ratio
See also:
Complete Problem 520
Level: High School, SAT Prep, College geometry
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Draw three median lines on triangle ABC, AA', BB', and CC'.
ReplyDeleteLength 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
Let c=AB, a=BC, b=AC
ReplyDeleteS1=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
To Emil Ekker
ReplyDeleteIt is not clear to me how do you have "Length AA' is (side of S4)/2" in your comment. Please explain
Hello Peter,
ReplyDeleteSee problem 502, and c.t.e.o's solution.
BM=AC/2, in addition, DM=median line from B (=BB' of my solution).
Hello Peter,
ReplyDeleteIs my last explanation enough for you, or not?
To Emil
ReplyDeleteThank you for your explanation.
It is nice if you can give reference in your comment so that people can follow your logic.