When Area = Perimeter

If the dimensions are $x$ by $y$ with $x \ge y \gt 0$ then we require that $xy = 2x + 2y \Longrightarrow xy - 2x - 2y = 0 \Longrightarrow (x - 2)(y - 2) = 4$ .

As $4$ can be factored as $4 \times 1$ or $2 \times 2$ we obtain the side-length solutions $(x,y) = (6,3)$ and $(x,y) = (4,4)$ , with respective areas $18$ and $16$ .

The desired sum is then $18 + 16 = \boxed{34}$ .

The area of the rectangle with sides x, and y, A=xy = 2(x+y). We can write y=2+4/(x-2), If x, y are integers the possible solutions are (3,6) and (4,4). So the sum of the areas 18+16=34

Suppose that the length of a brilliant rectangle is $a$ and the width is $b$ . Since $a$ and $b$ are interchangeable, add the restriction $a≥b$ without any loss of generality.

The area of the rectangle is given by $ab$ , and the perimeter is given by $2(a+b)$ .

Then, $ab=2(a+b)$ .

Divide both sides by $2ab$ to get that $\frac{1}{a}+\frac{1}{b}=\frac{1}{2}$ .

We need to find possible values for $a$ . First, since $b$ is positive and thus $\frac{1}{b}>0$ , we know that $\frac{1}{a}<\frac{1}{2}$ , or $a>2$ .

Then, since $a≥b$ , $\frac{2}{a}≤\frac{1}{a}+\frac{1}{b}=\frac{1}{2}$ , or $a≤4$ . Thus, possible values of $a$ that work are $3$ and $4$ .

Substituting $a=3$ and $a=4$ both reveal integer solutions for $b$ , $6$ and $4$ respectively.

Thus, we can find the answer $3\cdot6+4\cdot4=34$ .