See complete Problem 123

Area of triangle, Similarity. Level: High School, SAT Prep, College geometry

Post your solutions or ideas in the comments.

## Monday, June 16, 2008

### Elearn Geometry Problem 123

Labels:
area,
similarity,
triangle,
trisection

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see P122

ReplyDeleteSAGEC = SAGFEC + SFGE

using the fact h is the same for AGF and GFE,

and AF = 6/10 AA", FE = 3/20 AA" ( 4/10 - 1/4 )

SFGE = 1/80

=>

SAGEC = 1/20 + 1/5 + 1/80 + 1/20

SAGEC = 5/16

for three sides of tr

3∙5/16 = 15/16

SGME = 16/16 - 15/16

SGME = 1/16

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Apply Menelaus’s theorem to ABA’ and secant CMC”

ReplyDeleteCA’/CB x C”B/C”A x MA/MA’= 1

Replace value of CA’/CB and C”B/C”A in above we get MA/MA’= 3 so AM/AA’=3/4

Similarly apply Menelaus’s theorem to triangle CAA’ and secant BEB’

We get EA/EA”=3 => AE/AA’=>3/4

ME//BC and triangle MGE similar to triangle BAC with the factor ME/BC= ¼

So S1/S= (1/4)^2= 1/16

Solution to problem 123.

ReplyDeleteLet K be the center of gravity of triangle ABC. Let S be the area of ABC. As proved in problem 121, KE/KC = 1/4, and also KG/KA = 1/4. So triangles KGE and KAC are similar with

S(KGE) = (1/16).S(KAC). Similarly,

S(KEM) = (1/16).S(KCB) and S(KMG) = (1/16).S(KBA).

Adding these three equalities we get S1 = S/16.