Mediterranean Mathematical Competition

According to the cosine rule we have $$b^2+c^2-2bc\cos A=a^2.(1)$$ If $\tau =\frac{a+b+c}{2}$ the given condition can be written as $$\begin{gathered} a^2=rR=\frac{(ABC)}{\tau }\cdot \frac{a}{\sin A}=\frac{bc\sin A}{(a+b+c)\sin A}\iff a^2=\frac{abc}{a+b+c} \\ \iff bc-2a(b+c)=2a^2.(2) \end{gathered}$$ We write (1) in the form $$(b+c{)}^2-(1+\cos A)bc=a^2.(3)$$ Since $b+c>0$, from (2) and (3) we find $$b+c=a\left(1+8{\cos }^2\frac{A}{2}\right)(4)$$ $$bc=4a^2\left(1+4{\cos }^2\frac{A}{2}\right)(5)$$ From (3) and (4) it follows that $b$ and $c$ are the solutions of the quadratic equation $$\begin{array}{rl}& t^2-a\left(1+8{\cos }^2\frac{A}{2}\right)t+4a^2\left(1+4{\cos }^2\frac{A}{2}\right)=0\\ \iff & t^2-a(5+4\cos A)t+4a^2(3+2\cos A)=0\\ \iff & t=\frac{a}{2}\left[5+4\cos A\pm \sqrt{16{\cos }^{2}A+8\cos A-23}\right],\end{array}$$ provided that $16{\cos }^{2}A+8\cos A-23\ge 0$. Considering the trinomial $f(x)=16x^2+8x-23$ we have $$f(x)\ge 0\iff x\le \frac{-1-\sqrt{24}}{4}\text{ or }x\ge \frac{\sqrt{24}-1}{4}.$$ The first condition cannot be satisfied. From the second condition we have $$\cos A\ge \frac{\sqrt{24}-1}{4}\iff 0<A\le \arccos \frac{\sqrt{24}-1}{4}.$$