12. Let $A B C D$ be a square of side length 3. $P$ is a point on the plane such that each of $\angle A P B, \angle B P C, \angle C P D$ and $\angle D P A$ is at least $60^{\circ}$. If each possible position of $P$ is painted red, find the area of the red region.
(2 marks)設 $A B C D$ 為正方形, 邊長為 $3 \circ P$ 是平面上的一點, 使得 $\angle A P B 、 \angle B P C$ 、 $\angle C P D$ 和 $\angle D P A$ 均不小於 $60^{\circ}$ 。若把每個 $P$ 的可能位置均塗上紅色, 求紅色區域的面積。
(2 分)
12. $2 \pi+3-3 \sqrt{3}$
12. As shown in the left hand figure below, the set of points $P$ for which $\angle A P B \geq 60^{\circ}$ is the union of two segments (the major segment formed by $A B$ in the circumcircle of $\triangle A B P$, where $P$ may take one of the two positions for which $\triangle A B P$ is equilateral). Hence the red region $R$ is the intersection of the four such congruent shapes (denoted by $X Y W Z$ in the centre figure below).
Let $O$ be the centre of major arc $A X W B$. By symmetry, $\angle X B A=45^{\circ}$, and so $\angle X O A=90^{\circ}$. Similarly, $\angle W O B=90^{\circ}$. Also, $\angle A O B=120^{\circ}$ as angle at centre is twice angle at circumference. Thus $\angle X O W=360^{\circ}-90^{\circ}-120^{\circ}-90^{\circ}=60^{\circ}$.
As shown in the right hand figure below, $R$ consists of four identical minor segments (like $X P W$ ) and a square $X Y Z W$. Since $A B=3$, each segment has radius $\sqrt{3}$. Thus the square $X Y Z W$ has area 3 and each segment has area $\frac{1}{6} \pi(\sqrt{3})^{2}-\frac{\sqrt{3}}{4}(\sqrt{3})^{2}=\frac{\pi}{2}-\frac{3 \sqrt{3}}{4}$. It follows that the answer is $3+4\left(\frac{\pi}{2}-\frac{3 \sqrt{3}}{4}\right)=2 \pi+3-3 \sqrt{3}$.