This easy via the binomial theorem for the quantity is just 
,
and 
.
and 
.
Where 
Y by Adventure10, Mango247
A point 
is chosen on the side 
of the triangle 
in such a way that the radii of the circles inscribed in the triangles 
and 
are equal. Prove that
![\[ BM^{2} = X \cot \left( \frac {B}{2}\right)
\]](//latex.artofproblemsolving.com/f/7/4/f74aae0bfb024a98ba92cacc78824f11a0ab1eb8.png)
where X is the area of triangle
is chosen on the side 
of the triangle 
in such a way that the radii of the circles inscribed in the triangles 
and 
are equal. Prove that![\[ BM^{2} = X \cot \left( \frac {B}{2}\right)
\]](http://latex.artofproblemsolving.com/f/7/4/f74aae0bfb024a98ba92cacc78824f11a0ab1eb8.png)
where X is the area of triangle
This post has been edited 1 time. Last edited by orl, Sep 12, 2008, 11:54 PM
,
is the semiperimeter of triangle ABC. Thus, 
.
,
.
,
