Aryan geometry challenge 1

Suppose that $AC = 2R$ and that $\angle BAC = \theta$ .

Let $X$ be the foot of the perpendicular from the incentre $I$ to the hypotenuse $AC$ , so that $IX = r$ , the inradius. The triangle $AEF$ is right-angled, and similar to $ABC$ , and so the circumcircle of $AEF$ is the one with $AF$ as diameter. Since $I$ lies on this circle, this means that $\angle AIF$ is right-angled. Thus the right-angled triangles $AIF$ and $AXI$ are similar. Indeed $\frac{AX}{AI} \; = \; \frac{AI}{AF} \; = \; \cos\tfrac12\theta \;,$ so that $AX \,=\, AF\,\cos^2\tfrac12\theta \,=\, R\cos^2\tfrac12\theta \,=\, \tfrac12R(\cos\theta + 1)$ .

By standard results concerning the incentre, the length $AX \,=\, s-a$ , where $s$ is the semiperimeter and $a = BC$ . In terms of $\theta$ , this becomes $\begin{array}{rcl} \tfrac12R(\cos\theta + 1) & = & AX \; =\; R(1 + \cos\theta - \sin\theta) \\ 2\sin\theta & = & 1 + \cos\theta \\ 4\sin^2\theta & = & 1 + 2\cos\theta + \cos^2\theta \\ 5\cos^2\theta + 2\cos\theta - 3 & = & 0 \\ (5\cos\theta - 3)(\cos\theta + 1) & = & 0 \end{array}$ and hence $\cos\theta = \tfrac35$ . Thus the triangle $ABC$ is a $(3,4,5)$ -triangle, with $AB = \tfrac65R$ , $BC = \tfrac85R$ , and hence $\frac{BC}{AB} = \boxed{\frac43}$ .