, 2021

Suppose that $PQ=PR = 2x$ and $QR=2y$. The semi-circles with diameters $PQ$ and $PR$ thus have radii $x$ and the radius of the semi-circle with diameter $QR$ is $y$. The area of each semi-circle with radius $x$ is $\frac{1}{2}\pi x^2$ and the area of the semi-circle with radius $y$ is $\frac{1}{2}\pi y^2$. Since the sum of the areas of the three semi-circles equals 5 times the area of the semi-circle with diameter $QR$, then $$\tfrac{1}{2}\pi x^2 + \tfrac{1}{2}\pi x^2 + \tfrac{1}{2}\pi y^2 = 5 \cdot \tfrac{1}{2}\pi y^2$$ which gives $x^2 + x^2 + y^2 = 5y^2$ and so $2x^2 = 4y^2$ which gives $x^2 = 2y^2$ and so $x = \sqrt{2}y$. Suppose that $M$ is the midpoint of $QR$ and that $P$ is joined to $M$. [[IMAGE0]] Since $\triangle PQR$ is isosceles with $PQ = PR$, then $PM$ is perpendicular to $QR$. In other words, $\triangle PMQ$ is right-angled at $M$. Therefore, $\cos(\angle PQR) = \cos(\angle PQM) = \dfrac{QM}{PQ} = \dfrac{\tfrac{1}{2}QR}{PQ} = \dfrac{y}{2x} = \dfrac{y}{2\sqrt{2}y} = \dfrac{1}{2\sqrt{2}} = \dfrac{1}{\sqrt{4\cdot 2}} = \dfrac{1}{\sqrt{8}}$.