IMO Shortlist 1969 problem 10
Dodao/la:
arhiva2. travnja 2012. ![(BUL 4)](/media/m/a/e/7/ae7d3a8aa430228426e88cc9e6437c84.png)
Let
![M](/media/m/f/7/f/f7f312cf6ba459a332de8db3b8f906c4.png)
be the point inside the right-angled triangle
![ABC (\angle C = 90^{\circ})](/media/m/f/d/c/fdcf41581e0482e451e170a215811064.png)
such that
![\angle MAB = \angle MBC = \angle MCA =\phi.](/media/m/1/1/2/112399f372ff3cf3ee9e4d8d43adfd5a.png)
Let
![\Psi](/media/m/1/f/5/1f517132fa7a1100019fe936a467b661.png)
be the acute angle between the medians of
![AC](/media/m/6/4/7/647ef3a5d68f07d59d84afe03a9dc655.png)
and
![BC.](/media/m/c/8/0/c808a4699a08ca37e026a4f20c00723c.png)
Prove that
%V0
$(BUL 4)$ Let $M$ be the point inside the right-angled triangle $ABC (\angle C = 90^{\circ})$ such that $\angle MAB = \angle MBC = \angle MCA =\phi.$ Let $\Psi$ be the acute angle between the medians of $AC$ and $BC.$ Prove that $\frac{\sin(\phi+\Psi)}{\sin(\phi-\Psi)}= 5.$
Izvor: Međunarodna matematička olimpijada, shortlist 1969