图中显示一个圆，圆心为 $A$，半径为 $r$，并经过点 $B$、$C$ 和 $D$。一个较大的圆半径为 $s$，圆心为 $C$，并经过点 $B$ 和 $D$。线段 $BD$ 的长度也为 $s$。

(a) 证明 $s=\sqrt{3}r$。

(b) 求阴影区域面积的表达式。答案写成 $(a+b\pi)r^2$ 的形式，其中 $a$ 和 $b$ 是待求常数。

大圆的三条长度 $BC$、$CD$ 和 $BD$ 都是 $s$，所以 $BCD$ 是等边三角形。小圆中同一弦 $BD$ 所对的圆心角是圆周角的两倍。

Since $BC=CD=BD=s$, angle $BCD$ is $\pi/3$. By the centre-angle theorem in the smaller circle, angle $BAD$ is $2\pi/3$. Applying the cosine rule in triangle $BAD$ gives

As $r,s>0$, $s=\sqrt3r$.

也可用弦长与圆心角的关系。过 $A$ 作弦 $BD$ 的垂线，得到一个直角三角形。

The equilateral triangle $BCD$ gives angle $BCD=\pi/3$. Therefore the smaller circle subtends angle $BAD=2\pi/3$ at its centre. The perpendicular from $A$ to $BD$ bisects both the chord and this angle. Its right triangle has hypotenuse $r$ and opposite side $s/2$, so

Thus $s=\sqrt3r$.

两条弧有同一条弦 $BD$。阴影等于小圆的弓形减去大圆的弓形，每个弓形都等于扇形减三角形。

The smaller circle has sector angle $2\pi/3$ and radius $r$, so its segment on the shaded side of $BD$ has area

For the larger circle, the sector angle is $\pi/3$ and $s^2=3r^2$, so its segment on that side has area

Subtracting the second segment from the first gives the shaded area

Hence $a=\frac{\sqrt3}2$ and $b=-\frac16$.