Let ABC be an arbitrary triangle and O, r, and R its circumcenter, inradius, and circumradius respectively.
Let O_A, O_B, and O_C be the midpoints of sides BC, CA, and AB respectively.

For X = A, B, and C let

     SX = |OOX|  if the line segment OOX intersects
                 the interior of triangle ABC,

        = -|OOX| otherwise.

Prove that r + R = S_A + S_B + S_C.

Hi!

First let prove that :

In a triangle : cos(A)+cos(B)+cos(C)=1+r/R

From cos theorem cos(A)=(b^2+c^2-a^2)/(2bc)

But cos(A)=1-2sin^2(A/2) and so we find that

sin^2(A/2) = (p-b)*(p-c)/(b*c)  relation (1)

where p=(a+b+c)/2 of course.

So

cos(A)+cos(B)+cos(C) = after some trigonometry transformation

= 1+4*sin(A/2)sin(B/2)*sin(C/2)  and using relation (1) =

1+4*S/(p*a*b*c).

But R=a*b*c/(4*S) and r=S/p,  where S is the area of the triangle.

So

cos(A)+cos(B)+cos(C) = 1 + r/R (a nice relation !)

Is very interesting that some time ago i post on this site a problem to prove that cos(A)+cos(B)+cos(C)>1 (this is a new and nice demonstration!) 

Coming back to our problem we have

S(A) = R*cos(A) ; S(B) = R*cos(B) ; S(C) = R*cos(C)

So S(A)+S(B)+S(C) = R*(cos(A)+cos(B)+cos(C)) = R*(1+r/R) = R+r.

This is the situation when line segment OO_X intersects the interior of triangle ABC!

Edited on March 13, 2008, 4:15 pm

Posted by Chesca Ciprian on 2008-03-13 16:11:17