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.

Posted by Antonio Gutierrez at 4:26 PM
Labels: , , ,

11 comments:

  1. AjitMay 6, 2009 at 7:47 PM

    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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  2. AnonymousMay 19, 2009 at 11:56 AM

    This comment has been removed by a blog administrator.

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  3. AnonymousJuly 2, 2010 at 2:54 AM

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

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  4. Eder Contreras OrdenesMarch 24, 2012 at 11:14 PM

    Solution by cristian, my partnert >:)

    http://youtu.be/neHBBXZiaRI

    Go Go Geometry!

    Geo-Greetings ^^!

    ReplyDelete
  5. geoclidOctober 15, 2016 at 8:59 AM

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

    ReplyDelete
  6. Benjamin LeisAugust 11, 2017 at 2:57 PM

    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. Benjamin LeisAugust 11, 2017 at 9:03 PM

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

      Delete
  7. UnknownAugust 19, 2017 at 2:41 AM

    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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  8. Sumith PeirisFebruary 23, 2021 at 10:50 AM

    Complete isosceles trapezoid ABEC with BE//AC and AB = DC = CE = c
    Let DB and CE extended meet at F, so DF = c (since < ADF = 8x)

    So BF = BE = d say

    Since CE = CD = c & BE//DC, DE bisects < BEC, complete kite BEGD with G on CE and GE = BE =d so that GC = c-d = BG (< GCB = < GBC = x and < EBG = < EGB =2x)

    But BD = DG = c-d hence Tr. BGD is equilateral
    Therefore < DBG = 6x = 60 and x = 10

    Sumith Peiris
    Moratuwa
    Sri Lanka

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    Replies
    1. Sumith PeirisFebruary 28, 2021 at 7:28 AM

      Or simply, complete isosceles trapezoid ABEC (BE//AC) and kite DBGE with G on CE

      < DGB = 6x so < CDG = 4x
      Hence GD = GC = GB. But GD = DB so Tr. BDG is equilateral and 6x = 60 and x = 10

      Sumith Peiris
      Moratuwa
      Sri Lanka

      Delete
  9. AnonymousJuly 1, 2021 at 5:13 AM

    Extend BD to F such that BF=AB
    m(BFA)=6x=2.m(BCA)
    Considering that F is the circumcenter of ABC
    m(CFB)=2.m(CAB)=8x=m(CDF)=> CD=CF (which validates the given that AB=BF=FC=CD=Circum Radius of ABC)
    hence 18x=180=> x=10 Degrees (ABF is an equilateral Tr.)

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