35. Suppose the three sides of a triangular field are all integers, and its area equals the perimeter (in numbers). What is the largest possible area of the field?
35. Ans. 60 .
Let the three sides of the triangle be $a, b, c$ respectively. Then
$$
\sqrt{s(s-a)(s-b)(s-c)}=a+b+c=2 s,
$$
where $s=\frac{a+b+c}{2}$. Note that $s$ is an integer; otherwise $s(s-a)(s-b)(s-a)$ is a non-integer. Let $x=s-a, y=s-b$ and $z=s-c$. Then $x, y, z$ are positive integers satisfying
$$
x y z=4(x+y+z) .
$$

Assume that $x \geq y \geq z$. Then $x y z \leq 12 x$, i.e., $y z \leq 12$, and thus $z \leq 3$.
If $z=1, x y=4(x+y+1)$ implies $(x-4)(y-4)=20=20 \cdot 1=10 \cdot 2=5 \cdot 4$. So $(x, y)=(24,5),(14,6),(9,8)$.
If $z=2,2 x y=4(x+y+2)$ implies $(x-2)(y-2)=8=8 \cdot 1=4 \cdot 2$. So $(x, y)=(10,3),(6,4)$.
If $z=3,3 x y=4(x+y+3)$ implies $(3 x-4)(3 y-4)=52$, which has no solution $x \geq y \geq 3$.
The area is $60,42,36,30,24$, respectively; and the largest possible value is 60 .