Mediterranean Mathematical Competition

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