From HMMT:
Triangle $\triangle PQR$, with $PQ=PR=5$ and $QR=6$, is inscribed in circle $\omega$. Compute the radius of the circle with center on $QR$ which is tangent to both $\omega$ and $PQ$.
I haven't made much progress. I've set $QS=x$ and $SR=y$ to try for Law of Cosines, since we know $\cos\angle QSR$, but that really hasn't lead anywhere. With Ptolemy's, I 've found that $PS=\displaystyle\frac{5(x+y)}{6}$, but unfortunately $PS$ isn't colinear with anything useful (like the line connecting the centers). I also haven't really been able to use the tangent properties.
Hints beyond what I've done or any useful insights would be greatly appreciated!
Given that $|PQ|=|PR|=5,\ |QR|=6$, the area, the height and the circumradius of $\triangle PQR$ are $S=12$, $|PF|=4$ and $R_0=\tfrac{25}8$, respectively. Let $\angle PQR=\alpha$, $\angle FOE=\phi$.
Assuming that the center of the circle $O_t\in QR$, we must have $|DQ_t|=|EQ_t|=r$.
\begin{align} \sin\alpha&=\frac{|PF|}{|PQ|} =\frac45 ,\\ |OF|&=|PF|-R_0=\tfrac78 \tag{1}\label{1} . \end{align}
We have two conditions for $r$, $\phi$:
\begin{align} |QO_t|+|FO_t|&=\tfrac12\,|QR| \tag{2}\label{2} ,\\ \frac r{\sin\alpha} + (R_0-r)\sin\phi &=3 \tag{3}\label{3} ,\\ (R_0-r)\cos\phi&=|OF| \tag{4}\label{4} . \end{align}
Excluding $\phi$ from \eqref{3},\eqref{4} and using known values, we get
\begin{align} r&=\frac{20}9 . \end{align}