20. How many different triangles with integer side lengths are there such that the sum of the lengths of any two sides exceeds the length of the third side by at least 5 units, and that the area is numerically twice the perimeter? (Two triangles are regarded to be the same if they are congruent.)
(2 marks)
有多少個不同的三角形各邊的長度均為整數, 任何兩邊的長度之和均比第三邊長 5 單位或以上, 且其面積在數值上是其周界的兩倍?（兩個全等的三角形視為相同。)
20. 8
20. Let $a, b, c$ (where $a \leq b \leq c$ ) be the lengths of the sides of such a triangle. By Heron's formula, we have
$$
\sqrt{\left(\frac{a+b+c}{2}\right)\left(\frac{a+b-c}{2}\right)\left(\frac{c+a-b}{2}\right)\left(\frac{b+c-a}{2}\right)}=2(a+b+c),
$$
which simplifies to
$$
(a+b-c)(c+a-b)(b+c-a)=64(a+b+c) .
$$

Observe that $a+b-c, c+a-b, b+c-a, a+b+c$ have the same parity and hence must be even. Set $a+b-c=2 r, c+a-b=2 s, b+c-a=2 t$, where $r, s, t$ are positive integers with $3 \leq r \leq s \leq t$. The above equation is then reduced to
$$
r s t=16(r+s+t) \text {. }
$$

As $t<r+s+t \leq 3 t$, we have $16<r s \leq 48$. Also, the above equation implies $t=\frac{16(r+s)}{r s-16}$. Hence we need to find $r$ and $s$ so that $16<r s \leq 48$ and $r s-16$ divides $16(r+s)$. We can then list out all such pairs of $(r, s)$ and compute the corresponding $t$, getting 8 different solutions, namely, $(r, s, t)=(3,6,72),(3,7,32),(3,8,22),(3,12,12),(4,5,36),(4,6,20),(4,8,12)$ and $(6,7,8)$. Therefore there are 8 such triangles.