Wednesday, February 18, 2009

Problem 13: Triangle, Angles, Congruence, Cevian

Proposed Problem
Problem 13: Triangle, Angles, Congruence.

See complete Problem 13 at:
http://gogeometry.com/problem/problem013.htm

Level: High School, SAT Prep, College geometry

Post your solutions or ideas in the comments.

11 comments:

  1. Solucion:
    http://forogeometras.com/index.php?topic=884.0
    http://img25.imageshack.us/img25/7149/rsolmarzo8.jpg

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  2. trazar DE//AB
    el angulo EDC=4x
    Los triangulos ABD y DCE son congruentes
    entonces ABD=3x
    en el triangulo ABC: 3x+4x+3x+5x=180
    x=12º

    atte:tino!!!

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  3. Dear Tino,
    The answer is not correct.
    Why triangles ABD and DCE are congruent?

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  4. We've CD/sin(5x)=BC/sin(8x) and AB/sin(3x)=BC/sin(4x). Thus, sin(5x)/sin(8x)=sin(3x)/sin(4x) or
    sin(5x)/2sin(4x)cos(4x)=sin(3x)/sin(4x) or
    sin(5x)/2cos(4x)=sin(3x)
    By inspection if x=10 then sin(5x)=cos(4x) and sin(3x)=1/2. Thus, x=10 is clearly a solution to the equation.
    Ajit: ajitathle@gmail.com

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

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

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  7. Solution by cristian, my partnert >:)

    http://youtu.be/neHBBXZiaRI

    Go Go Geometry!

    Geo-Greetings ^^!

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  8. https://www.youtube.com/watch?v=-i4RlIl4B0s

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  9. 1. Draw E on AB such that BE = BD and F on CD such that DF = BD also. This forms 2 isosceles triangles BDE (with 2 6x angles) and BDF (with 2 4x angles).
    2. By angle chasing <BFA = 4x = <BAD so tr. ABF is isosceles also and
    BF = AB = CD.
    3. By angle chasing <BDC = <EBF = 180 - 8x so tr BEF = tr BCD by SAS.
    4. So <BEF = <CBD = 5x and by angle subtraction <DEF = x
    then also <BFE = <BCD = 3x and by angle subtraction <EFD = x also.
    5. So tr. EDF is also isosceles and DE = DF = BD = BE.
    6. This implies tr. BDE is actually equilateral and 6x = 60. X = 10.

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    Replies
    1. Btw: compare this construction to problem #9. They are the same.

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  10. I derived the following equation in the same way that Joe did in 2009:
    sin(5x) / 2cos(4x) = sin(3x)
    At this point, I rearranged the terms and apply the product-to-sum formula:
    sin(5x) = 2cos(4x)sin(3x) = sin(7x) - sin(x)
    Now, rearrange the terms again and apply the sum-to-product formula,
    sin(x) = sin(7x) - sin(5x) = 2cos(6x)sin(x)
    sin(x)(2cos(6x) - 1) = 0
    This suggests either sin(x) = 0 or cos(6x) = 1/2.
    The former is invalid (which would imply x is a multiple of 180°), whereas the latter suggest 6x = 60°, 300°, ..., that is x = 10°, 50°, ...
    In order to keep ∠DBC = 5x under 180°, we must choose x = 10°.

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