XVI OBM

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