Results 1 to 4 of 4

Math Help - in a triangle

  1. #1
    Member grgrsanjay's Avatar
    Joined
    May 2010
    From
    chennai,tamil nadu
    Posts
    143
    Thanks
    1

    in a triangle

    in a triangle ABC and R is the radius of the circle into which the triangle is inscribed(in other words it is the circum radius of the triangle)

    \frac{a\cos A + b\cos B + c\cos C}{a\sin B + b\sin C + c\sin A} = \frac{a+b+c}{9R}

    then prove it is an equilateral triangle
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    Joined
    Jun 2009
    Posts
    806
    Thanks
    4
    When I simplified the left hand side, I got

    \frac{abc}{ab+bc+ca} = \frac{a+b+c}{9}

    By observation you can see the the equation is true only when a = b = c. But how to prove it?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    Opalg's Avatar
    Joined
    Aug 2007
    From
    Leeds, UK
    Posts
    4,041
    Thanks
    7
    Quote Originally Posted by sa-ri-ga-ma View Post
    When I simplified the left hand side, I got

    \frac{abc}{ab+bc+ca} = \frac{a+b+c}{9}

    By observation you can see the the equation is true only when a = b = c. But how to prove it?
    It took me quite a while to see how to simplify the left hand side to get \frac{abc}{ab+bc+ca} = \frac{a+b+c}{9}. (It's correct, but I'll leave it to sa-ri-ga-ma to provide the details. )

    So we know that (a+b+c)(ab+bc+ca) = 9abc. Multiply out the left side to get a(b^2+c^2) + b(c^2+a^2) + c(a^2+b^2) + 3abc. Thus the equation becomes a(b^2+c^2) + b(c^2+a^2) + c(a^2+b^2) = 6abc, which you can write as a(b-c)^2 + b(c-a)^2 + c(a-b)^2 = 0. That obviously only holds if a=b=c.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member grgrsanjay's Avatar
    Joined
    May 2010
    From
    chennai,tamil nadu
    Posts
    143
    Thanks
    1

    i think i got it

    i can simplify the denominator first
    a\sin B + b\sin C + c\sin A  =  a(b/2R) + b(c/2R) + c(a/2R)
                                         =  \frac{ab +bc+ca}{2R}
    now about the numerator
    a\cos A + b\cos B + c\cos C = R(2\sin a\cos a+ 2\sin b\cos b + 2\sin c\cos c)
                                           = R(\sin 2a+\sin 2b+\sin 2c)
                                           = 4R(\sin a\sin b\sin c)
                                           = \frac{abc}{2 R^2}

    then thats it you can get from there
    Last edited by grgrsanjay; September 2nd 2010 at 08:27 AM.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 3
    Last Post: April 23rd 2011, 09:10 AM
  2. Replies: 3
    Last Post: April 30th 2009, 08:41 AM
  3. Replies: 1
    Last Post: October 28th 2008, 08:02 PM
  4. Replies: 7
    Last Post: July 19th 2008, 07:53 AM
  5. Replies: 27
    Last Post: April 27th 2008, 11:36 AM

Search Tags


/mathhelpforum @mathhelpforum