Wednesday, February 18, 2009

Problem 14: Triangle, Cevian, Midpoint, Angles

Proposed Problem
Problem 14: Triangle, Cevian, Midpoint, Angles.

See complete Problem 14 at:
http://gogeometry.com/problem/problem014.htm

Level: High School, SAT Prep, College geometry

Post your solutions or ideas in the comments.

7 comments:

  1. WLOG one may assume D:(0,0), A:(-1,0), C:(1,0). Let angle BCA = m and angle BAC=angle CBD = 2m
    Thus angle BDA = 3m. Further, let tan(m)=t
    Now, DB: y =(-tan3m)x =-x(3t-t^3)/(1-3t^2)
    CB:y=(1-x)t which give B as((3t^2-1)/2(t2+1),(3t-t^3)/2(t2+1)) while A is (-1,0). hence slope BA=
    (3t-t^3)/(5t^2+1) which as we know is tan(2m)=2t/(1-t^2). The equation obtained: (3t-t^3)/(5t^2+1)=2t/(1-t^2) has five solutions of which the only admissible solution is t=0.2679491924 which gives m=15 deg. Thus, angle BCA = 15 deg.
    Alternatively, in triangle ABD, AD/BD =sin(5m)/sin(2m) while triangle BDC gives, CD/BD=sin(2m)/sin(m). Since AD=CD, we can say that sin(5m)/sin(2m)=sin(2m)/sin(m) which can be solved to get m=15 deg as before.
    Now, having proven my bonafides, can someone please show me how such problems can be solved more elegantly by plane geometry alone?
    Ajit: ajitathle@gmail.com

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  2. http://geometri-problemleri.blogspot.com/2009/05/problem-21-ve-cozumu.html

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

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  4. the solution is uploaded to the following link:
    https://docs.google.com/open?id=0B6XXCq92fLJJS2NfODE5ZmEwNkU

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  5. CE perpendicular to AB and BE=EH. So , angle AHC=3x. Because, angle BDA=3x ,
    BDCH , is a cyclic quadrilateral , therefore , angle BHD=x. The triangle AEC, is rectangular and AD=DC. So, ED=DA=DC, and angle BED=2x, therefore, angle EDH=x. So ,DE=EH=EB .Then angle BDH=90 degrees and 3x+2x+x=90 .Therefore x=15 degrees.
    image : http://img248.imageshack.us/img248/8017/geogebra4.png

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  6. https://www.youtube.com/watch?v=ikWNS9tpYJY

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