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Nice ques to solve

  1. Aug 18, 2011 #1
    Nice ques to solve !!!

    Given a triangle ABC find a point M on its circumcircle such that MA=MB+MC....It is easy for an equilateral triangle... I got it... but I couldn't get how to find for any arbitrary triangle....
     
  2. jcsd
  3. Aug 18, 2011 #2
    Re: Nice ques to solve !!!

    Hhhmmm...in other words, you have 3 points on a circle and you need to find a 4th one that meets such condition.

    I would first find out the angle that the radius makes to each point A,B,C; let's call these angles THA, THB, THC (TH for theta).

    Then, let's have a point M and its corresponding THM.

    Let's call the center of the circle O.

    By taking the center of the circle, the point M and one point (A,B,C) at a time, we have 3 isosceles triangles where the base-lines (chords) of those triangles are those lines you want to match the equation MA=MB+MC. Also, it is true that MA/2 = MB/2 + MC/2.

    Those base-lines are chords whose lengths are 2Sin(TH/2), where TH is the angle the two sides make at the center of the circle (radius=1).

    For triangle AOM, we have one TH, THAOM = THA - THM

    and

    THBOM = THB - THM
    THCOM = THC - THM

    And so, we want:

    Sin(THAOM/2) = Sin(THBOM/2) + Sin(THCOM/2)

    I think it now gets a bit hairy to isolate THM from here; so, you could simply apply brut force and substitute values between 0 and 2Pi and that's it.
     
  4. Aug 18, 2011 #3
    Re: Nice ques to solve !!!

    The three sides of the triangle as well as the chords MA, MB and MC will form the sides and diagonals of a cyclic quadrilateral. Now if you apply Ptolemy's theorem to the quadrilateral, you should be able to obtain a condition relating the three sides. With that condition, you should be able to derive an expression for the length of either MA, MB or MC and the position of M can be found with that.
     
    Last edited: Aug 18, 2011
  5. Aug 18, 2011 #4

    DaveC426913

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    Gold Member

    Re: Nice ques to solve !!!

    Guys? Help him to the answer. Spoonfeeding is not allowed on PF.
     
  6. Aug 18, 2011 #5
    Re: Nice ques to solve !!!

    Oh my apologies, since this wasn't posted in the homework forums, I though he just wanted some form of solution. I'll edit my answer to provide hints instead.
     
  7. Aug 19, 2011 #6
    Re: Nice ques to solve !!!

    Now we just do THAOM/2=THA/2-THM/2
    Expand out the sines using sin(A-B)=sinAcosB-sinBcosA
    Pull out the sin(THM/2) to one side and the cos(THM/2) to the other.
    Divide for tan(THM/2)
     
  8. Aug 21, 2011 #7
    Re: Nice ques to solve !!!

    i got it mysely...just say whether its r8.....Barycentric coordinates as a function of the side lengths. The circumcenter has trilinear coordinates (cos α, cos β, cos γ) where α, β, γ are the angles of the triangle. The circumcenter M has coordinates:

    M = (a^2(-a^2 + b^2 + c^2), b^2(a^2 - b^2 + c^2), c^2(a^2 + b^2 - c^2))

    where a, b, c are edge lengths (BC, CA, AB respectively) of the triangle.

    and i found the three co-ordinates....now we can find the answer easily....!!!...is it correct.....
     
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