Conference rusure::math

     Hint:
     
     I thought that this would be some geometry student's
     nightmare, but it isn't.  When I sat down to do it, the
     first thing I tried was to use the law of cosines in the
     obvious way, and that works.
     
     As they say on those news teasers, more at eleven. :-)
     
     Dan

     In .1 I referred only to part (b) of the problem in .0.
     
>>     (b) If the lengths of the medians are m1, m2, and m3 prove that
>>         (a^2 + b^2 + c^2)/(m1^2 + m2^2 + m3^2) = 4/3.
     
     Let a, b, c be the sides of the triangle; A, B, C the
     opposite angles; and m1, m2, m3 the median lengths from
     vertex A, B, C to side a, b, c.
     
     Use the Law of Cosines in turn on each side:
     
          a^2 = b^2 + c^2 - 2bc cos A
          b^2 = a^2 + c^2 - 2ac cos B
          c^2 = a^2 + b^2 - 2ab cos C
     
     Adding these three together,
     
     a^2 + b^2 + c^2 = 2(a^2 + b^2 + c^2) - 2(bc cos A + ac cos B + ab cos C)
     
     or, after a few minor manipulations,
     
        bc cos A + ac cos B + ab cos C = (a^2 + b^2 + c^2) / 2
     
     Now write out the Law of Cosines formulas for the medians. 
     You have a choice of two triangles to use for each median;
     make the choice so that each angle is represented once and
     each half-side is represented once.
     
          m1^2 = (a/2)^2 + b^2 - 2(a/2)b cos C
          m2^2 = (b/2)^2 + c^2 - 2(b/2)c cos A
          m3^2 = a^2 + (c/2)^2 - 2a(c/2) cos B
     
     Adding these,
     
          m1^2 + m2^2 + m3^3 = (5/4)(a^2 + b^2 + c^2) - 
                               - (ab cos C + bc cos A + ac cos B)
     
               = (5/4)(a^2 + b^2 + c^2) - (1/2)(a^2 + b^2 + c^2)
     
               = (3/4)(a^2 + b^2 + c^2)
     
     So (a^2 + b^2 + c^2) / (m1^2 + m2^2 + m3^2) = 4/3.
     
     Dan

     re part (a)
     
>>     (a) Given the three lengths of the medians of a triangle,
>>         construct the triangle (using a straight edge and a compass).
     
     Don't the medians all meet at a point 2/3 the way (along any
     of the medians) from the vertex to the other side?
     
     Let the lengths of the medians be m1, m2, and m3, with
     m1 >= m2 >= m3.
     
     Take the longest median (m1) and draw it.  Call one end A (for
     vertex A of the triangle) and the other A' (which will be on
     the opposite side of the triangle, i.e., side a).
     
     Now the other two medians will meet the segment AA' 2/3 of
     the way from A to A'.  Construct that point and call it O.
     
     Draw a circle centered at O with radius m2/3.  Side b must
     pass through this circle.  Draw a second circle centered at
     O with radius twice m2/3.  Vertex B must be on this larger
     circle.
     
     Now, once B is determined, the triangle can be completed
     because BA' is one half of BC; just double the length of BA'
     in the B -> A' direction and that is where vertex C is.
     
     Likewise, if B' (the midpoint of AC, the point on the inner
     circle around O where median m2 hits the side opposite B) is
     given, extend AB' to find point C and extend B'O to find
     point B.
     
     So it is only necessary to find either of B (on the outer
     circle) or B' (on the inner circle) to finish the
     construction of triangle ABC with the given medians.
     
     When I figure out how to do that, I'll post it here. :-)
     
     Dan

    
    Hello,
    
    I had a different solution in mind than the one given in .2
    It is easy to prove that in a triangle 
    
    b^2 + c^2 = 2 ( (a/2)^2 + m1^2).
    
    Write the other two equations that follow from the symmetry of
    the problem.
    
    Add the three equations to get the result.
    
    Murali.