See complete Problem 61

Triangle, Trisection of Sides. Level: High School, SAT Prep, College geometry

Post your solutions or ideas in the comments.

## Monday, May 19, 2008

### Elearn Geometry Problem 61

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## Monday, May 19, 2008

###
Elearn Geometry Problem 61

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

See complete Problem 61

Triangle, Trisection of Sides. Level: High School, SAT Prep, College geometry

Post your solutions or ideas in the comments.

Subscribe to:
Post Comments (Atom)

hi good day.uhm you had a great project. i just wnt to ask where did u get the values 3/20? thank you

ReplyDeleteIt can be easily demonstrated that line AMF meeting AC in F1 is a median of the triangle ABC. Likewise AGD and CEH extended are also medians of Tr. ABC.

ReplyDeleteNow we apply Menelaus's theorem repetitively to various triangles and their transversals. First we can see that A'M/MA*AF1/F1C*3/1=1 or A'M/MA*1/1*3/1=1 or A'M/MA=1/3 or A'M/AA'=1/4.

AE/EA"*A"B/BC*CB"/B"A=1 or AE/EA"*2/3*1/2=1 or AE/EA"=3/1. Thus, A'H/HA*1/2*3/1=1 or A'H/HA=2/3 or A'H/AA'=2/5 or AH/AA'=3/5

Now, HM = AA' - AH - MA' = AA'(1-3/5-1/4)=3/20*AA'(This the genesis of the magic fraction "3/20"). Similarly, it is possible to prove that: MD = 3/20 * CC", DE = 3/20 * BB", EF = 3/20*AA", FG = 3/20*CC', GH = 3/20*BB'.

Hence, P=(3/20)(AA'+AA"+BB'+BB"+CC'+CC")

Ajit: ajitathle@gmail.com

FA" = 4/10 AA"

ReplyDeleteEA" = 1/4 AA" ( see P122, P123 )

EF = 4/10 - 1/4

EF = 3/20 AA"

in the same way others