Problem 1170

Solution:

Let $\triangle PQR$ have sides $p=7$, $q=5$, $r=8$. Of the four sectors determined by the diameters of $C$ that are parallel to $PQ$ and $PR$, two have angles equal to $P$ and the other two have angles equal to $\pi - P$. We first find $P$ using the law of cosines:

$$49 = 25 + 64 - 2(5)(8) \cos P$$
which implies
$$\cos P = \frac{1}{2}$$
so
$$P = \frac{\pi}{3}$$
Thus the two smaller sectors will have angle $\frac{\pi}{3}$.

Next we find the circumradius of $\triangle PQR$ using the formula
$$R = \frac{pqr}{4[PQR]}$$
where $[PQR]$ is the area of $\triangle PQR$. By Heron's Formula we have
$$[PQR] = \sqrt{10 \cdot 5 \cdot 3 \cdot 2} = 10 \sqrt{3}$$
thus
$$R = \frac{5 \cdot 7 \cdot 8}{4(10 \sqrt{3})} = \frac{7}{\sqrt{3}}$$
The area of a smaller sector is thus
$$\frac{\pi/3}{2\pi} (\pi R^2) = \frac{\pi}{6} \left(\frac{7}{\sqrt{3}}\right)^2 = \frac{49}{18} \pi$$