When Does Area Equal Perimeter?

A slightly different approach to the other solutions is to start from Euclid's formula for Pythagorean triples.

Any integer sided right-angled triangle has sides that can be written as $t(u^2-v^2)$ , $2tuv$ and $t(u^2+v^2)$ for some integers $u>v>0$ and $t>0$ .

These give an area of $t^2 uv(u^2-v^2)$ and perimeter $2tu(u+v)$ . Equating these, we find

$tv(u-v)=2$

It's pretty quick to check cases; if $t=2$ then $v=u-v=1$ so $u=2$ , $v=1$ and the sides are $(6,8,10)$ .

If $v=2$ we need $t=u-v=1$ , so $u=3$ ; this gives the sides $(5,12,13)$

If $u-v=2$ , $t=v=1$ and $u=3$ ; this again gives sides $(8,6,10)$ (which we've already had).

So the only two triangles are $(6,8,10)$ and $(5,12,130$ , giving the answer $\boxed{54}$ .

Incidentally, shapes with the property of having (numerically) equal area and perimeter are known as " equable ", and seem to have been around for quite a while (according to the linked article, an extension of this problem to general integer sided triangles was discussed in 1858 and solved in 1904).

Let the sides of the right triangle be $x,y$ and $\sqrt{x^2+y^2}$ . Then the perimeter of the triangle is $x+y+\sqrt{x^2+y^2}$ and its area is $\frac{xy}{2}$ .

Equating two expressions gives, $x+y+\sqrt{x^2+y^2}=\frac{xy}{2} \implies x^2+y^2=\left(\frac{xy}{2}-x-y\right)^2$ which on simplification yields $x+y=\frac{xy}{4}+2$ It follows that 4 divides $xy$ , which leads to two cases, either 2 divides both $x$ and $y$ , or 4 divides $x$ or $y$ .

Assume 2 divides both $x$ and $y.$ Let $x=2a$ and $y=2b.$ We get $2a+2b=ab+2 \implies 2\vert ab$ . Therefore 2 divides $a$ or $b$ , so either $x$ or $y$ is divisible by 4. Thus both our cases are the same!

Suppose that $y=4c$ for some positive integer $c.$ Substituting this in our equation gives, $x+4c=xc+2$ . Rearrange the equation as $(x-4)+2=c(x-4) \implies (x-4)\vert 2$ .

Thus the only solutions are $x=5$ and $x=6$ which leads to $y=12$ and $y=8$ respectively. $\boxed{(x,y,z)=(5,12,13), (6, 8, 10)}$

Answer: $(6+8+10)+(5+12+13)=\boxed{54}$

From the given conditions of the problem, we get x=4(2-y)/(4-y). y can not be equal to 2 or 4. y<2 has no solution. y>4 has only 2 solutions. They are (i) y=5, x=12, z=13 and (ii) y=6, x=8, z=10. Therefore the required sum is (5+12+13)+(6+8+10)=54