13. (2 marks) Let $A B C D$ be a square of side $5, E$ a point on $B C$ such that $B E=3, E C=2$. Let $P$ be a variable point on the diagonal $B D$. Determine the length of $P B$ if $P E+P C$ is smallest.
(2 分) 正方形 $A B C D$ 邊長為 $5, E$ 為 $B C$ 上的點使得 $B E=3, E C=2$ 。若 $P$ 是對角線 $B D$ 上的點使 $P E+P C$ 的值最小, 求 $P B$ 的長度。
13. Apply cosine rule to $\triangle B C P$, we have $\cos \angle P E B=\frac{B E^{2}+P E^{2}-P B^{2}}{2 B E \cdot P E}$,
$\cos \angle P E C=\frac{P E^{2}+E C^{2}-P C^{2}}{2 P E \cdot E C}$. Using the fact $\cos \angle P E B=-\cos \angle P E C \quad, \quad$ and after
simplifications, get $P E^{2}=P B^{2} \cdot \frac{E C}{B C}$
$+P C^{2} \cdot \frac{B E}{B C}-E C \cdot B E \quad, \quad$ or $\quad P E^{2}=\frac{2}{5} P B^{2}$
$+\frac{3}{5} P C^{2}-6 \ldots$ (1) Apply cosine rule to
$\triangle B C D$, we have $\cos \angle C P B=\frac{B P^{2}+P C^{2}-B C^{2}}{2 B P \cdot P C}, \cos \angle C P D=\frac{D P^{2}+P C^{2}-C D^{2}}{2 D P \cdot P C}$. Again by $\cos \angle C P B=-\cos \angle C P D \quad, \quad B C=C D=5 \quad, \quad B P=5 \sqrt{2} \quad, \quad$ after simplifications, get $P C^{2}=B C^{2}-B P \cdot D P=B C^{2}-B P(B D-B P)=25-B P(5 \sqrt{2}-B P) \ldots$ (2). Put (2) in (1), get $P E^{2}=P B^{2}-3 \sqrt{2} B P+9 \ldots$ (3). Hence $P E+P C=\sqrt{P B^{2}-3 \sqrt{2} B P+9}+\sqrt{P B^{2}-5 \sqrt{2} B P+25}$ $=\sqrt{\left(P B-\frac{3}{2} \sqrt{2}\right)^{2}+\frac{9}{2}}+\sqrt{\left(P B-\frac{5 \sqrt{2}}{2}\right)^{2}+\frac{25}{2}}$. Take $P B=x$, then $P E+P C$ corresponds to the sum of distances from $(x, 0)$ to $\left(\frac{3}{2} \sqrt{2}, \frac{3}{2} \sqrt{2}\right)$, and $\left(\frac{5 \sqrt{2}}{2}, \frac{-5 \sqrt{2}}{2}\right)$. For $P E+P C$ to be smallest $(x, 0)$ lies on the line formed by these two points. Hence $\frac{\frac{3}{2} \sqrt{2}-\left(-\frac{5 \sqrt{2}}{2}\right)}{\frac{3}{2} \sqrt{2}-\frac{5 \sqrt{2}}{2}}=\frac{\frac{5 \sqrt{2}}{2}}{x-\frac{5 \sqrt{2}}{2}}$, solve to get $x=\underline{\underline{\frac{15 \sqrt{2}}{8}}}$. (Corresponds to $\left.P E+P C=\sqrt{34}\right)$.

Alternate Solution

Clearly $P C=P A$, hence $P E+P C$ is smallest when $P E+P A$ is smallest, i.e. when $A, P, E$ are collinear.

Assign the coordinates $A(0,5), B(5,5), C(5,0), D(0,0)$,
then $E=(5,2)$, and the equation of the line $A E$ is $y=-\frac{3}{5} x+5$.
This means $P=\left(\frac{25}{8}, \frac{25}{8}\right)$ and $P B^{2}=\left(5-\frac{25}{8}\right)^{2}+\left(5-\frac{25}{8}\right)^{2}$
$=\frac{450}{64}$ or $P B=\underline{\underline{\frac{15}{8}} \sqrt{2}}$