Unexpectedly easy,isn't it? Suppose that these sides are $ AB,AC$ and for convenience denote them $ c,b$ respectively.Let $ \alpha$ be the angle between these two sides.Our goal is to prove that $ \alpha < 60^{\circ}$.Assume the contrary,i.e $ \alpha\geq 60^{\circ}$,therefore,$ \cos{\alpha}\leq\frac {1}{2}$ By the well-known formula of area of $ \triangle ABC$,it follows that: \[ \frac {abc}{4R} = a\cdot 2r = pr = \frac {a + b + c}{2}\cdot r \] It means that $ b + c = 3a$.Squaring the last relation gives us: $ b^2 + 2bc + c^2 = 9a^2 = 9(b^2 + c^2 - 2bc\cos{\alpha})$,or $ 8(b^2 + c^2) = bc(2 + 18\cos{\alpha})\leq 11bc$,what is clearly impossible,as it contradicts to am-gm inequality... P.S: It seems to me that I missed something,because if my solution would be correct we can even prove that $ \alpha\leq 45^{\circ}$ But I can't figure out where I did the mistake

everytihng is ok This is the first problem of second day

also we have $ \alpha \leq \arcsin(\frac {3\sqrt3}{8})$

in fact from the fact that $ b+c=3a$ it follows that $ a=s-a<\min\{b,c\}$, which implies that $ \angle A<60^{\circ}$..

Here is my solution: $ bc=8Rr\Rightarrow \frac{b}{2R}\frac{c}{2R}=2\frac{r}{R}$ $ \Rightarrow \sin B\sin C=8\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}$ $ \Rightarrow \cos \frac{B}{2}\cos \frac{C}{2}=2\sin\frac{A}{2}$ $ \Rightarrow\frac{1}{2}\ge \sin\frac{A}{2}$ $ \Rightarrow 60\ge A$