Wednesday, February 18, 2009

Problem 16: Triangle, Cevian, Perpendicular, Angles, Congruence

Proposed Problem
Problem 16: Triangle, Cevian, Perpendicular, Angles, Congruence.

See complete Problem 16 at:
http://gogeometry.com/problem/problem016.htm

Level: High School, SAT Prep, College geometry

Post your solutions or ideas in the comments.

7 comments:

  1. It is possible to provide a proof using only elementary geometry!
    Hints:
    Auxiliary lines, Perpendicular bisector, Congruence, Problem 4.

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  2. Since a plane geometry solution hasn't appeared so far here's a simple trigonometric solution:
    By applying sine rule to Tr. ABC, we get
    sin(7x)/sin(3x) = AC/BC. But AC = 2CE; hence
    sin(7x)/sin(3x)= 2CE/BC. Now CE/BC = cos(2x) from Tr. BCE. Thus, sin(7x)/sin(3x)=2cos(2x) or sin(7x)/cos(2x)= 2sin(3x). By inspection if x=10 then sin(7x)=sin(70)=cos(20)=cos(2x) while sin(3x)=sin(30)= 1/2 which in turn makes LHS = RHS. Therefore, x = 10 is a solution to our equation.
    Ajit: ajitathle@gmail.com

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  3. This comment has been removed by a blog administrator.

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  4. http://ahmetelmas.files.wordpress.com/2010/05/cozumlu-ornekler.pdf

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  5. The solution is uploaded to the following link:

    https://docs.google.com/open?id=0B6XXCq92fLJJcDNRdVlFR3VKNlk

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  6. Video solution http://youtu.be/3luEqMUwzXM (in spanish :P)

    Greetings

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  7. Let CE=1 so that AC=2.
    Hence BC= (1 / cos(2x))
    In △ABC,
    sin(180-7x)/AC = sin(3x)/BC
    sin(7x)/2 = sin(3x)cos(2x)
    sin(7x) =2sin(3x)cos(2x)
    sin(7x) = sin(5x) + sin(x) (trigonometric identity)
    sin(x) = sin(7x) - sin(5x)
    sin(x) = 2cos(6x)sin(x) (trigonometric identity)
    0 = 2cos(6x)sin(x)-sin(x)
    0 = sin(x) (2cos(6x)-1)
    sin(x)=0 or (2cos(6x)-1)=0
    x=0 (rejected) or 2cos(6x)=1
    cos(6x)=0.5
    6x=60
    x=10

    ReplyDelete