7. $\triangle A B C$ is equilateral with side length 4. $D$ is a point on $B C$ such that $B D=1$. If $r$ and $s$ are the radii of the inscribed circles of $\triangle A D B$ and $\triangle A D C$ respectively, find $r s$.
(1 mark) $A B C$ 是等邊三角形, 邊長爲 $4 \circ D$ 是 $B C$ 上的一點, 使得 $B D=1 \circ$ 若 $r$ 和 $\mathrm{s}$ 分別是 $\triangle A D B$ 和 $\triangle A D C$ 的內切圓牛徑, 求 $r s$ 。
7. $4-\sqrt{13}$
7. We have $A D=\sqrt{4^{2}+1^{2}-2(4)(1) \cos 60^{\circ}}=\sqrt{13}$. The area of $\triangle A D B$ is $\frac{1}{2}(4)(1) \sin 60^{\circ}=\sqrt{3}$, and similarly $\triangle A D C$ has area $3 \sqrt{3}$. Recall that the radius of the inscribed circle of a triangle is equal to twice the area divided by the perimeter. (This can be seen by connecting the in-centre to the three vertices and then considering area.) It follows that
$$
r s=\frac{2 \cdot \sqrt{3}}{4+1+\sqrt{13}} \cdot \frac{2 \cdot 3 \sqrt{3}}{4+3+\sqrt{13}}=\frac{36}{48+12 \sqrt{13}}=4-\sqrt{13} .
$$