Brazilian Mathematical Olympiad

Let $a \ge b \ge c$ be the sides of the triangle. If the triangle is right-angled, then
$r = \frac{b+c-a}{2}$, $2R = a$ and $s - r = \frac{a+b+c}{2} - \frac{b+c-a}{2} = a = 2R$.

If $2R = s - r$, consider
$$E = (2R-a)(2R-b)(2R-c) = 8R^3 - 4R^2(a+b+c) + 2R(ab+bc+ca) - abc$$
Let $\Delta$ be the area of the triangle. We have $a+b+c = 2s$ and $R = \frac{abc}{4\Delta} \iff abc = 4R\Delta$. Finally, since $\Delta = sr$, $abc = 4Rsr$ and
$$\begin{align*} sr &= \Delta = \sqrt{s(s-a)(s-b)(s-c)} \\ \iff sr^2 &= s^3 - (a+b+c)s^2 + (ab+bc+ca)s - abc \\ \iff ab + bc + ca &= r^2 - s^2 + 2s \cdot s + 4Rr = r^2 + s^2 + 4Rr \end{align*}$$
Thus
$$\begin{align*} E &= 8R^3 - 4R^2 \cdot 2s + 2R(r^2 + s^2 + 4Rr) - 4Rsr \\ &= 2R(4R^2 - 4Rs + r^2 + s^2 + 4Rr - 2sr) \\ &= 2R(4R^2 + (r-s)^2 + 4Rr - 4Rs) \\ &= 2R(4R^2 + 4R^2 + 4Rr - 4Rs) \\ &= 8R^2(2R - s + r) = 0 \end{align*}$$
and so one of the sides of the triangle is equal to $2R$, completing the proof.