20. Let $A B C D$ be the piece of paper, with $A B=a \leq b=B C$. Let $O$ be the center of the rectangle, $\mathrm{B}$ be folded to coincide with $D$, and $P O Q$ be the crease, with $P$ on $B C$ and $Q$ on $D A$. Then $P Q$ is the perpendicular bisector of $B D$, and $\triangle B P O \sim \triangle D B C$. Let $x$ be the length of $P Q$, then $\frac{\frac{1}{2} \sqrt{a^{2}+b^{2}}}{\frac{1}{2} x}=\frac{b}{a}$, or $x=\frac{a}{b} \sqrt{a^{2}+b^{2}}$. Now $x=\frac{a}{b} \sqrt{a^{2}+b^{2}}=65$, thus $b=\frac{a^{2}}{\sqrt{65^{2}-a^{2}}}$. For
$b$ to be an integer, the denominator $\sqrt{65^{2}-a^{2}}$ must be an integer. Hence we try $65^{2}$ as a sum of two squares. Using $5^{2}=3^{2}+4^{2}$ and $13^{2}=5^{2}+12^{2} \quad$ we have $65^{2}=(13 \times 3)^{2}+(13 \times 4)^{2}=(5 \times 5)^{2}+(12 \times 5)^{2}=(3 \times 12+4 \times 5)^{2}+(4 \times 12-3 \times 5)^{2}=(3 \times 12-4 \times 5)^{2}$ $+(4 \times 12+3 \times 5)^{2}$. Consider all these $a$ which may be $16,25,33,39,52,56,60$ or 63 . Put these to find $b$, and bear in mind $b$ is an integer with $b \geq a$, we get $a=60$, and $b=144$ the only possibility. Thus the perimeter is $2 \times(60+144)=\underline{\underline{408}}$.
20. (3 marks) A rectangular piece of paper has integer side lengths. The paper is folded so that a pair of diagonally opposite vertices coincide, and it is found that the crease is of length 65. Find a possible value of the perimeter of the paper.
(3 分) A rectangular piece of paper has integer side lengths. The paper is folded so that a pair of diagonally opposite vertices coincide, and it is found that the crease is of length 65. Find a possible value of the perimeter of the paper.