By the angle bisector theorem, $PQ/PS = QR/SR$. Re-arranging, $\frac{QR}{SR} = \frac{5}{3} = \frac{PS}{PQ}$.
By the angle bisector theorem, $RS/RQ = PS/PQ$. Combining this with $\frac{QR}{SR} = \frac{PS}{PQ}$, we have $RS/RQ = QR/RP$, so triangle $PQR$ is isosceles, and $m\angle QPR = m\angle PQR = \theta$.
By the Law of Sines on triangle $RPQ$, $\frac{PQ}{\sin \theta} = \frac{RP}{\sin (180^\circ - 2 \theta)}$, or
\[\frac{PQ}{\sin \theta} = \frac{RP}{\sin 2 \theta} = 2 \cdot \frac{RP}{2 \sin \theta \cos \theta}.\]Then, $\frac{PQ}{2 \sin^2 \theta} = \frac{PR}{\cos \theta} = \frac{PS}{1 - \cos^2 \theta} = 3$. Hence,
\begin{align*}
PR &= 2 \sin^2 \theta, \\
PS &= 3 - 3 \cos^2 \theta.
\end{align*}(We know that $PS < PQ$, so $3 - 3 \cos^2 \theta < 2 \sin^2 \theta$.)
By the Law of Sines on triangle $QSR$, $\frac{RS}{\sin \theta} = \frac{QS}{\sin 2 \theta}$, so
\[RS = \frac{\sin \theta \cdot QS}{\sin 2 \theta} = \frac{\sin \theta \cdot \frac{5}{3} PQ}{2 \sin \theta \cos \theta} = \frac{5}{6} PQ \cot \theta.\]Then $SR = RQ + RS$, so
\[SR = PQ - PR + \frac{5}{6} PQ \cot \theta.\]Hence,
\[PQ - PR + \frac{5}{6} PQ \cot \theta = 8.\]From the equation $PR = 2 \sin^2 \theta,$ $\cot \theta = \frac{1 - \cos^2 \theta}{2 \sin \theta \cos \theta} = \frac{1 - \cos^2 \theta}{\sin 2 \theta}$.
Thus,
\[SR = PQ - 2 \sin^2 \theta + \frac{5}{6} PQ \cdot \frac{1 - \cos^2 \theta}{\sin 2 \theta}.\]Since $PQ = 3 \sin^2 \theta,$
\[SR = 3 \sin^2 \theta - 2 \sin^2 \theta + \frac{5}{6} \cdot 3 \sin^2 \theta \cdot \frac{1 - \cos^2 \theta}{\sin 2 \theta}.\]This reduces to
\[SR = \frac{19}{6} \sin^2 \theta + \frac{5}{6} \cdot \frac{1 - \cos^2 \theta}{\sin 2 \theta}.\]From the equation $3 - 3 \cos^2 \theta < 2 \sin^2 \theta$, $1 - \cos^2 \theta < 5 \sin^2 \theta$, so
\[SR < \frac{19}{6} \sin^2 \theta + \frac{5}{6} \cdot \frac{5 \sin^2 \theta}{\sin 2 \theta}.\]We can write $\sin 2 \theta$ in terms of $\sin \theta$ and $\cos \theta$: $\sin 2 \theta = 2 \sin \theta \cos \theta$. Hence,
\[SR < \frac{19}{6} \sin^2 \theta + \frac{25 \cos \theta}{6}.\]Recall that $PQ = 3 \sin^2 \theta = 2 - 2 \cos^2 \theta$. Then $2 \cos^2 \theta = 2 - PQ$, so $\cos \theta = \pm \sqrt{1 - \frac{PQ}{2}} = \pm \frac{\sqrt{5 - PQ}}{2}$.
Since $0^\circ < \theta < 180^\circ$, $0 < PQ < 4$. Therefore, $\cos \theta = \frac{\sqrt{5 - PQ}}{2}$.
Since $\sqrt{5 - PQ} > 0$, $SR$ is an increasing function of $\sin^2 \theta$, subject to $0 < PQ < 4$. Therefore, $SR$ is an increasing function of $\cos \theta$, or an increasing function of $\frac{\sqrt{5 - PQ}}{2}$.
Both $\frac{\sqrt{5 - PQ}}{2}$ and $\frac{\sqrt{5 - PQ}}{2}$ are positive. If $\frac{\sqrt{5 - PQ}}{2} > \cos \theta$, then
\[SR < \frac{19}{6} \sin^2 \theta + \frac{25 \sqrt{5 - PQ}}{6}.\]Otherwise, $SR \ge \frac{19}{6} \sin^2 \theta + \frac{25 \sqrt{5 - PQ}}{6}$.
\[
\begin{array}{c|c}
PQ & SR \\ \hline
0 & 6 \\
1 & \frac{25 \sqrt{4}}{6} = \frac{25}{3} \\
2 & \frac{19}{3} \cdot \frac{4}{9} + \frac{25 \sqrt{3}}{6} > \frac{25}{3} \\
3 & \frac{19}{3} \cdot \frac{9}{9} + \frac{25 \sqrt{2}}{6} > \frac{29}{3} \\
4 & \frac{19}{3} \cdot \frac{16}{9} + \frac{25 \sqrt{1}}{6} > \frac{32}{3}
\end{array}
\]We cannot have $SR$ greater than $\frac{32}{3}$, so $SR < \frac{32}{3}$.
Therefore, $SR < \frac{32}{3}$. But $PQSR$ is a quadrilateral inscribed in a circle, so $SR = RQ$. Hence, the largest angle is $\boxed{\frac{32}{3}}$.
In cyclic quadrilateral PQRS, P/3 = Q/5 = R/8 + 60. Find the largest angle in quadrilateral PQRS, in degrees.
1 answer