jmc

Firstly, we note the statement in the problem that "$AD$ is the only chord starting at $A$ and bisected by $BC$" – what is its significance? What is the criterion for this statement to be true? We consider the locus of midpoints of the chords from $A$. It is well-known that this is the circle with diameter $AO$, where $O$ is the center of the circle. The proof is simple: every midpoint of a chord is a dilation of the endpoint with scale factor $\frac{1}{2}$ and center $A$. Thus, the locus is the result of the dilation with scale factor $\frac{1}{2}$ and centre $A$ of circle $O$. Let the center of this circle be $P$. Now, $AD$ is bisected by $BC$ if they cross at some point $N$ on the circle. Moreover, since $AD$ is the only chord, $BC$ must be tangent to the circle $P$. The rest of this problem is straightforward. Our goal is to find $\sin \angle AOB=\sin \left(\angle AOM-\angle BOM\right)$, where $M$ is the midpoint of $BC$. We have $BM=3$ and $OM=4$. Let $R$ be the projection of $A$ onto $OM$, and similarly let $Q$ be the projection of $P$ onto $OM$. Then it remains to find $AR$ so that we can use the addition formula for $\sin$. As $PN$ is a radius of circle $P$, $PN=2.5$, and similarly, $PO=2.5$. Since $OM=4$, we have $OQ=OM-QM=OM-PN=4-2.5=1.5$. Thus $PQ=\sqrt{2.5^2-1.5^2}=2$. Further, we see that $△OAR$ is a dilation of $△OPQ$ about center $O$ with scale factor $2$, so $AR=2PQ=4$. Lastly, we apply the formula:$$\sin \left(\angle AOM-\angle BOM\right)=\sin \angle AOM\cos \angle BOM-\sin \angle BOM\cos \angle AOM=\left(\frac{4}{5}\right)\left(\frac{4}{5}\right)-\left(\frac{3}{5}\right)\left(\frac{3}{5}\right)=\frac{7}{25}$$Thus the answer is $7\cdot 25=\boxed{175}$.