Sunday, December 23, 2012

Problem 835: Isosceles Triangle, Double Angle, Triple Angle, Auxiliary Lines

Geometry Problem
Level: Mathematics Education, High School, Honors Geometry, College.

Click the figure below to see the complete problem 835.

Online Geometry Problem 835: Isosceles Triangle, Double Angle, Triple Angle, Auxiliary Lines

Posted by Antonio Gutierrez at 8:12 AM
Labels: , , , ,

7 comments:

  1. AnonymousDecember 23, 2012 at 11:32 AM

    Result: 150

    Solution: Let BC=AC=1 a:alfa, b:beta
    From sinus theorem in BDC
    DC=sina/sin(a+b)
    Sinus theorem in ADC using DC we get
    sina*sin(2a+3b)=sin2a*sin(a+b) since sin2a=2*sina*cosa we get
    2*sin(a+b)*cosa=sin(2a+b)+sinb=sin(2a+3b)
    sinb=sin(2a+3b)-sin(2a+b)=2*sinb*cos(2a+2b)=sinb
    eliminating sinb(b can't be zero) we get
    cos(2a+2b)=1/2 which means 2a+2b=60, a+b=30
    So x=180-30=150 degrees.

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  2. Peter TranDecember 23, 2012 at 1:35 PM

    http://img38.imageshack.us/img38/3597/problem835.png

    Make drawing per attached sketch
    1. Draw altitude CF and bisector AF of angle DAC
    Make angle ACE= beta
    2. Due to symmetric properties , CDE and AEDB are isosceles
    And BE, AD and CF are concurrent at point G
    3. In triangle AGC , GE is an angle bisector => angle BGF=angle AGF=angle AGE=angle CGE=60
    In triangle ABC we have 60+ 4alpha+4.beta=180 => alpha+ beta=30
    In triangle BDC x=180-alpha-beta= 150

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    Replies
    1. UnknownJanuary 1, 2020 at 7:32 PM

      not clear at all. Your proposition is to
      1) Drop an altitude from C to the base, intercepting the base at F ( I think) that will bisect the vertex angle at C into 2 equal angles of 2beta each
      2) The second construction I am not sure I understand. but it seems that you are proposing and angle bisector from A intersecting BC at E or intersecting DC At E.
      the rest pointing to concurrency at G etc. I have no clue at all.
      do you have a sketch of your construction?

      Delete
  3. Peter TranDecember 24, 2012 at 6:45 PM

    Minor typo correction.
    Line 3 of my solution should read as "1. Draw altitude CF of triangle ABC and bisector AE of angle DAC "
    Peter Tran

    ReplyDelete
  4. Prof Radu Ion,Sc.Gim.Bozioru,BuzauDecember 25, 2012 at 9:04 AM

    Fie P punctul de pe segmentul AD astfel incat m(: PC este mediatoarea segmetului AB si bisectoarea unghiului =>m(In ∆BPC
    m(6α+6β=180 =>α+β=30 de unde in ∆BDC x=180-(α+β)=150

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  5. Sumith PeirisSeptember 7, 2015 at 10:37 AM

    Let CX be the bisector of < ACB, X being on AD. Now thro' congruence or symmetry < AXC = < BXC. But D is the in centre of Tr. ACX, so XD bisects < BXC. It thus follows that < BXD = 60.
    So 2 alpha + 2 beta + 120 = 180 in Tr. ADC, implying that alpha + beta = 30.
    So from Tr. BDC x = 150

    Sumith Peiris
    Moratuwa
    Sri Lanka

    ReplyDelete

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