Geometry Problem

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

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Complete Problem 520

Level: High School, SAT Prep, College geometry

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## Sunday, September 12, 2010

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Problem 520: Triangle, Six Squares, Areas, Ratio

See also:

Complete Problem 520

Level: High School, SAT Prep, College geometry

Online Geometry theorems, problems, solutions, and related topics.

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

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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.