8. $A B C D$ is a square of side length 1. $X$ and $Y$ are points on $B C$ and $C D$ respectively such that $C X=C Y=m$. When extended, $A B$ meets $D X$ at $P ; A D$ meets $B Y$ at $Q$; $A X$ meets $D C$ at $R ; A Y$ meets $B C$ at $S$. If $P, Q, R, S$ are collinear, find $m$.
(1 mark)
$A B C D$ 是正方形, 邊長爲 $1 \circ X$ 和 $Y$ 分別是 $B C$ 和 $C D$ 上的點, 使得 $C X=C Y=m$ 。經延長後, $A B$ 交 $D X$ 於 $P ; A D$ 交 $B Y$ 於 $Q ; A X$ 交 $D C$ 於 $R$; $A Y$ 交 $B C$ 於 $S$ 。若 $P 、 Q 、 R 、 S$ 共線, 求 $m$ 。
(1 分)
8. $\frac{3-\sqrt{5}}{2}$
8. Let $C R=C S=n$ and $D Q=k$.
Since $\triangle R C X \sim \triangle R D A$, we have $\frac{n}{m}=\frac{n+1}{1}$ or $n=\frac{m}{1-m}$. Since $\triangle Q D Y \sim \triangle Q A B$, we have $\frac{k}{1-m}=\frac{k+1}{1}$ or $k=\frac{1-m}{m}$. Since $\triangle R C S \sim \triangle R D Q$, we have $\frac{n}{k}=\frac{n}{n+1}$ or $k=n+1$. From the above, we get $\frac{1-m}{m}=\frac{m}{1-m}+1$. Solving this equation subject to the condition $0<m<1$, we get $m=\frac{3-\sqrt{5}}{2}$.