In triangle PQR, let X be the intersection of the angle bisector of angle P with side QR, and let Y be the foot of the perpendicular from X to line PR. If PQ = 9, QR = 9, and PR = 9, then compute the length of XY.
Because PQR is an isosceles triangle with PQ = PR, we have that the angle bisector of angle P is equivalently the altitude from P to QR. Thus, we see that angles XYP and XPQ are 90-degree angles, so triangles XYP and XPQ are similar. Furthermore, we have $\triangle PQR \sim \triangle XPQ \sim \triangle XYP.$
This means that $PQ/PR = QX/YR$ and $PQ/PX = XY/YP$, so we can plug in the values $PQ=9/RQ=9$ and get the equations \[\frac{9}{18} = \frac{QX}{YR} \quad \text{and} \quad \frac{9}{PX} = \frac{XY}{PY} .\] We need to solve for $PY$ and $PX$ to get $XY$. To do so, note that \[\tan\angle PQR = \tan \frac{\angle PQX}{2}=\frac{PX}{QX}\] and \[\tan \angle QPR = \frac{PY}{QX}.\] Both equal 1/2 (because $\triangle PQR$ is isosceles), so we find $PX = 2QX$ and $PY = QX/2$. Substituting these into the equations for similarity gives \[\frac{9}{PX} = \frac{9}{18} \cdot \frac{XY}{PY} \implies XY = \frac{9}{18} \cdot \frac{PX^2}{PY} =\boxed{\frac{27}{4}}.\]