The area of quadrilateral $P Q R S$ is 48. It is known that $P Q=Q R=6, R S=S P$ and exactly three vertices $P, Q$ and $R$ lie on a circle of radius 5. Find the sides $R S$ and $S P$.

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Let $Q D$ be the diameter of the circle. Then $S_{\mathrm{PQRD}}=48=S_{\mathrm{PQRS}}$. Using this fact, prove that points $Q$ and $S$ lie on the same side of the line $R P$.

## Solution

Since $P Q=Q R$ and $R S=S P$, points $Q$ and $S$ lie on the perpendicular bisector of segment $P R$. Therefore, the line $Q S$ passes through the center of the given circle. Consider the case when points $Q$ and $S$ lie on opposite sides of the line $R P$ (Fig.1). If $Q D$ is the diameter of the circle, then $\angle Q R S=90^{\circ}$. Therefore,

$$
D R=\sqrt{Q D^{2}-Q R^{2}}=\sqrt{100-36}=8
$$

In this case,

$$
S_{\mathrm{PQRD}}=2 \cdot S_{\mathrm{QRD}}=2 \cdot \frac{1}{2} \cdot Q R \cdot R D=48=S_{\mathrm{PQRS}}
$$

This means that point $S$ coincides with point $D$, which is impossible since, by the problem's condition, exactly three vertices of the quadrilateral $P Q R S$ lie on the given circle. The only remaining case is when points $Q$ and $S$ lie on the same side of the line $R P$ (Fig.2). Let $E$ be the midpoint of $P R$. Since triangle $P Q R$ is inscribed in a circle with radius 5, then

$$
\sin \angle Q R P=\frac{Q P}{2 \cdot 5}=\frac{3}{5}
$$

Therefore,

$$
Q E=Q R \cdot \sin \angle Q R P=6 \cdot \frac{3}{5}=\frac{18}{5}, E R=Q R \cdot \cos \angle Q R P=6 \cdot \frac{4}{5}=\frac{24}{5}
$$

Since $S_{P Q R S}=\frac{1}{2} \cdot S Q \cdot P R$, then

$$
S Q=2 \cdot \frac{S_{P Q R S}}{2 \cdot E R}=2 \cdot \frac{48}{\frac{48}{5}}=10
$$

By the cosine theorem in triangle $S Q R$, we find that

$$
\begin{gathered}
S R^{2}=S Q^{2}+Q R^{2}-2 \cdot S Q \cdot Q R \cdot \cos \angle S Q R=100+36-120 \cdot \cos \left(90^{\circ}+\angle Q R P\right)= \\
=136+120 \cdot \sin \angle Q R P=136+120 \cdot \frac{3}{5}=136+72=208
\end{gathered}
$$

Therefore, $S R=\sqrt{208}=4 \sqrt{13}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_96a7587f2ba0a7b3aee2g-04.jpg?height=418&width=734&top_left_y=1126&top_left_x=665)

Answer

$4 \sqrt{13}$.