Finding side of a triangle

Question

$ST$ is the perpendicular bisector of $PR$ and $SP$ is the angle bisector of $\angle QPR$. If $QS=9cm$ and $SR=7cm$ then $PR$ = $x/y$ where $x$, $y$ are coprimes. $x$ + $y$ = ?

I tried to use the angle bisector theorem and found that the ratio of $PR/PQ=SR/QS=7/9$. But afterwards could not make any progress. Can anybody help me with this?

Answer 1

Let $PT=TR=x$ and $\angle{SPR}=\theta$. Then note that $\angle{PSQ}=2\angle{QPS}=2\theta.$ By the angle bisector theorem,

$$PQ=\frac{18x}{7}.$$

Observe that $\cos{\theta}=\frac{x}{7}$. By the law of sines on $\triangle{PQR}$ we have,

\begin{align*} \frac{PQ}{\sin{\theta}} &=\frac{16}{\sin{2\theta}} \\ \implies{} \dfrac{18}{7}x &=\dfrac{8}{\cos{\theta}} \\ \implies{}x\cos{\theta} &=\dfrac{28}{9} \end{align*}

Thus $x=14/3$ and $PR=2x=28/3$ which gives us the answer $m+n=31.$

Answer 2

Hint: since $\frac{PQ}{PR}=\frac{SQ}{SR}=\frac{9}{7}$, $P$ belongs to a circle of Apollonius. Since $ST$ is the perpendicular bisector of $PR$, $PS=SR$ and $P$ belongs to the circle having centre at $S$ and radius $SR$. Both circles are symmetric with respect to the $QR$ line.