Results 1 to 10 of 10

Math Help - Prove that triangle ABC is isosceles

  1. #1
    Newbie
    Joined
    Mar 2011
    Posts
    9

    Prove that triangle ABC is isosceles

    Triangle ABC has the following properties: there is an interior point P such that angle PAB = 10 degrees, angle PBA = 20 degrees, angle PCA = 30 degrees and angle PAC = 40 degrees. Prove that triangle ABC is isosceles.

    Very difficult problem don't know where to start.
    Any help appreciated.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Master Of Puppets
    pickslides's Avatar
    Joined
    Sep 2008
    From
    Melbourne
    Posts
    5,236
    Thanks
    28
    Have you drawn a picture to help solve this problem? After doing this label the angles with the information given. Use the fact that a triangle has 180 degrees and the sum of angles around P is 360. Label your any unknowns (there should be 2) and solve them by setting up some simultaneous equations
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Mar 2011
    Posts
    9
    i dont think i can use simultaneous equations to solve this.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,713
    Thanks
    633
    Hello, ahm0605!

    \text}Triangle }ABC\text{ has the following properties:}

    \text{There is an interior point }P\text{ such that:}
    . . \angle PAB = 10^o,\;\angle PBA = 20^o, \;\angle PCA = 30^o,\;\angle PAC = 40^o.

    \text{Prove that triangle }ABC\text{ is isosceles.}

    Very difficult problem, don't know where to start.
    Of course you do . . . Make a sketch!

    I don't think the proof is possible.

    Code:
    
                    A
                    o
                    *
                   ***
                    *
                  * * *
                    *
                 *  *  *
                 10 * 40
                *   *   *
                    *
               *    *    *
                    *
              *     *     *
                    *
             *      o      *
                150 * 110
            *     *   *     *
             20 *  100  * 30
           *  *           *  *
            *  x         y  *
        B o   *   *   *   *   o C

    And all we have is: . x + y \:=\:80^o

    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Mar 2011
    Posts
    9
    I have drawn a sketch but don't know where to go from there. I think it's something to do with the cosine rule
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Super Member TheChaz's Avatar
    Joined
    Nov 2010
    From
    Northwest Arkansas
    Posts
    600
    Thanks
    2
    This has been bothering me! If we don't crack it soon, I might seek "outside assistance"...
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,713
    Thanks
    633

    Is anyone listening?

    With the given information, ANY triangle is possible.

    (Well, with the exception of an equilateral.)

    Here is a possible right triangle.

    Code:
    
          A
          ♥
          *◊*
          * ◊ *
          *  ◊  *
          *10 ◊ 40*
          *    ◊    *
          *     ♥     *
          *    ◊ p      *
          *20 ◊       ◊ 30*
          *  ◊              *
          * ◊               ◊ *
          *◊ 70           10    *
        B ♥  *  *  *  *  *  *  *  ♥ C
    Follow Math Help Forum on Facebook and Google+

  8. #8
    MHF Contributor

    Joined
    Apr 2005
    Posts
    15,579
    Thanks
    1418
    Soroban has already told you all you need to know- you can't prove it- it is NOT true!
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Super Member TheChaz's Avatar
    Joined
    Nov 2010
    From
    Northwest Arkansas
    Posts
    600
    Thanks
    2
    I just wanted to see him draw another picture...

    Follow Math Help Forum on Facebook and Google+

  10. #10
    Member
    Joined
    Feb 2010
    From
    in the 4th dimension....
    Posts
    122
    Thanks
    9

    Re: Prove that triangle ABC is isosceles

    Sorry to open the old thread, but a user has just posted this problem recently, and I think i should give a link to the solution, as the problem is both famous, interesting and not wrong...
    mathhelpforum.com/new-users/212009-mongolias-province-math-competition-9th-grade.html#post765883

    the link does not seem to work:
    consider \angle PCB = x and we know \angle PBC + \angle PCB =80, hence  \angle PBC= 80-x
    now we have to use the trigonometric version of ceva's theorem:
    \frac{\sin \angle PBA}{\sin \angle PAB}*\frac{\sin \angle PCB}{\sin  \angle PBC}*\frac{\sin \angle PAC}{\sin \angle PCA} =1
    => \frac{\sin 20 \sin x \sin 40}{\sin 10 \sin (80-x) \sin 30}=1
    => \frac{4\sin x \sin 40 \cos 10}{\sin (80-x)}=1
    => \frac{2\sin x(\sin 30+\sin 50)}{\sin (80-x)}=1 [#using product-sum formula]
    => \frac{\sin x(1+2\cos 40)}{\sin (80-x)}=1
    => 2\sin x \cos 40 = \sin (80-x)-\sin x = 2\sin(40-x)\cos 40
    so as 2\sin x \cos 40 = 2\sin(40-x)\cos 40, we can write x=40-x or x=20
    hence \angle ACB=\angle ABC = 50,therefore triangle ABC is isosceles
    Last edited by earthboy; January 25th 2013 at 10:52 AM.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. value of x in an isosceles triangle
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: April 6th 2011, 08:43 AM
  2. Isosceles triangle
    Posted in the Trigonometry Forum
    Replies: 3
    Last Post: February 28th 2010, 05:44 AM
  3. isosceles triangle
    Posted in the Geometry Forum
    Replies: 3
    Last Post: November 9th 2009, 05:59 AM
  4. Replies: 2
    Last Post: March 20th 2009, 09:21 AM
  5. Replies: 27
    Last Post: April 27th 2008, 10:36 AM

Search Tags


/mathhelpforum @mathhelpforum