Cyclic Quadrilateral 2018

Suppose that $\angle AOB = 2\theta$ and $\angle AOC =2\phi$ . Let $R$ be the outradius. Then $49 = AB = 2R\sin\theta$ and $73 = AC = 2R\sin\phi$ , and hence $73\sin\theta \; = \; 49\sin\phi$ Since $IJ$ is bisected by $BD$ , we deduce that the triangles $ABD$ and $CBD$ have the same inradius $r$ . Using Carnot's Theorem that the sum of the inradius and the outradius of a triangle is equal to the sum of the (directed) distances to the three sides of the outcentre, we deduce (considering the two ways to express $R+r$ in the triangles $ABD$ and $CBD$ ), $\begin{aligned} R\cos\theta + R\cos\theta + R\cos(\pi-2\theta) & = \; R\cos2\theta + R\cos(\phi-\theta) + R\cos(\pi-\phi-\theta) \\ 2\cos\theta - \cos2\theta & = \; \cos2\theta + \cos(\phi-\theta) - \cos(\phi+\theta) \; = \; \cos2\theta + 2\sin\phi\sin\theta \\ \cos\theta - \cos2\theta & = \; \sin\phi\sin\theta \\ 49(\cos\theta - \cos2\theta) & = \; 73\sin^2\theta \\ 25\cos^2\theta -49\cos\theta + 24 & = \; 0 \\ (25\cos\theta - 24)(\cos\theta - 1) & = \; 0 \end{aligned}$ Since the case $\theta=0$ is not possible, we deduce that $\cos\theta = \tfrac{24}{25}$ , $\sin\theta = \tfrac{7}{25}$ , and hence $\sin\phi = \tfrac{73}{175}$ and $R = \tfrac{175}{2}$ .

The rest is just calculation. We use the fact that the incentre $I$ of a triangle with vertices $A,B,C$ and side lengths $a,b,c$ (named in the normal manner) as has position vector $\overrightarrow{OI} \; = \; \frac{1}{a+b+c}\big[a \overrightarrow{OA} + b\overrightarrow{OB} + c\overrightarrow{OC}\big]$ Setting up a coordinate system with the outcentre $O$ as the origin, and $A$ lying on the positive $x$ -axis, we have $A\;: \;(\tfrac{175}{2},0) \hspace{1cm} B\;:\; (\tfrac{3689}{50},\tfrac{1176}{25}) \hspace{1cm} C\;:\; (\tfrac{19967}{350}, \tfrac{292 \sqrt{1581}}{175})\hspace{1cm} D\;:\;(\tfrac{3689}{50},-\tfrac{1176}{25})$ and so we obtain the incentres $I \;:\; ( \tfrac{161}{2},0) \hspace{2cm} J \;:\; (\tfrac{3353}{50}, \tfrac{28 \sqrt{1581}}{25})$ and hence $IJ \; = \; \frac{28 \sqrt{69}}{5} \; = \; \boxed{46.51709363}$ There may be a neater way to do this using the Japanese Theorems about cyclic quadrilaterals...