IMO Shortlist 1988 problem 30
Dodao/la:
arhiva2. travnja 2012. A point
![M](/media/m/f/7/f/f7f312cf6ba459a332de8db3b8f906c4.png)
is chosen on the side
![AC](/media/m/6/4/7/647ef3a5d68f07d59d84afe03a9dc655.png)
of the triangle
![ABC](/media/m/a/c/7/ac75dca5ddb22ad70f492e2e0a153f95.png)
in such a way that the radii of the circles inscribed in the triangles
![ABM](/media/m/1/0/f/10f2e13fa27b4f7846a6ffc4baab3603.png)
and
![BMC](/media/m/1/3/e/13e8443f7a40fc4d03204aef5b6a1e46.png)
are equal. Prove that
where X is the area of triangle
%V0
A point $M$ is chosen on the side $AC$ of the triangle $ABC$ in such a way that the radii of the circles inscribed in the triangles $ABM$ and $BMC$ are equal. Prove that
$$BM^{2} = X \cot \left( \frac {B}{2}\right)$$
where X is the area of triangle $ABC.$
Izvor: Međunarodna matematička olimpijada, shortlist 1988