, 2016

Solution:

Let $O$ be circumcenter, $R$ the circumradius and $r$ the common inradius. We have $IO^{2} = JO^{2} = R(R-2r)$ by a result of Euler; denote $x$ for the common value of $IO$ and $JO$. Additionally, we know $AJ = AB = AD = 49$ (angle chase to find that $\angle BJA = \angle JBA$). Since $A$ is the midpoint of the arc $\widehat{BD}$ not containing $C$, both $J$ and $A$ lie on the angle bisector of angle $\angle BCD$, so $C, J, A$ are collinear. So by Power of a Point we have
$$R^{2} - x^{2} = 2Rr = AJ \cdot JC = 49 \cdot 24$$
Next, observe that the angle bisector of angle $BAD$ contains both $I$ and $O$, so $A, I, O$ are collinear. Let $M$ be the midpoint of $IJ$, lying on $\overline{BD}$. Let $K$ be the intersection of $IO$ and $BD$. Observing that the right triangles $\triangle IMO$ and $\triangle IKM$ are similar, we find $IM^{2} = IK \cdot IO = r x$, so $IJ^{2} = 4 r x$. Now apply Stewart's Theorem to $\triangle AOJ$ to derive
$$R(x(R-x) + 4 r x) = 49^{2} x + x^{2}(R-x)$$
Eliminating the common factor of $x$ and rearranging gives
$$49^{2} - (R-x)^{2} = 4 R r = 48 \cdot 49$$
so $R-x = 7$. Hence $R+x = \frac{49 \cdot 24}{7} = 168$, and thus $2R = 175, 2x = 161$. Thus $r = \frac{49 \cdot 24}{175} = \frac{168}{25}$.
Finally, $IJ = 2 \sqrt{r x} = 2 \sqrt{\frac{84 \cdot 161}{25}} = \frac{28 \sqrt{69}}{5}$.