2022 AMC 10A Problem 10

1. Name the sides. Let the card be $a$ by $b$. The diagonal is $8$, so by the Pythagorean theorem
$$a^2 + b^2 = 64.$$
2. Place the card on a grid with corners $(0,0)$, $(a,0)$, $(a,b)$, $(0,b)$. Cut the $1 \times 1$ squares from the bottom-left and top-right corners. The two closest corners of the cut squares are their inner corners, $(1,1)$ and $(a-1,\ b-1)$.
3. Distance between them. Going from one to the other you move $a - 2$ across and $b - 2$ up. By the Pythagorean theorem again,
$$(a-2)^2 + (b-2)^2 = (4\sqrt2)^2 = 32.$$
4. Find $a + b$. Multiply out: $a^2 - 4a + 4 + b^2 - 4b + 4 = 32$. Replace $a^2 + b^2$ by $64$:
$$64 - 4(a + b) + 8 = 32, \quad\text{so}\quad 4(a + b) = 40 \quad\text{and}\quad a + b = 10.$$
5. Find the area $ab$. We know $a + b$ and $a^2 + b^2$, and the square of a sum connects them:
$$(a + b)^2 = a^2 + 2ab + b^2, \qquad 100 = 64 + 2ab.$$
So $2ab = 36$ and $ab = 18$.

The answer is $\boxed{\textbf{(E)}\ 18}$.

The question asks for the product $ab$, not the two sides. Knowing $a + b$ and $a^2 + b^2$ is enough, because $(a+b)^2 = a^2 + 2ab + b^2$ links them. Whenever the target treats $a$ and $b$ the same way, look for a shortcut like this before solving for each one.

If you want the actual sides: two numbers with sum $10$ and product $18$ are $a = 5 + \sqrt7$ and $b = 5 - \sqrt7$. Check: $(5+\sqrt7)(5-\sqrt7) = 25 - 7 = 18$, and $a^2 + b^2 = (32 + 10\sqrt7) + (32 - 10\sqrt7) = 64$.

Let the card have side lengths $a$ and $b$. The diagonal gives
$$a^2 + b^2 = 8^2 = 64.$$

Place the card with corners $(0,0)$, $(a,0)$, $(a,b)$, $(0,b)$ and cut the unit squares at $(0,0)$ and $(a,b)$. The two vertices closest to each other are the inner corners $(1,1)$ and $(a-1, b-1)$, so
$$(a-2)^2 + (b-2)^2 = (4\sqrt{2})^2 = 32.$$
Expanding: $a^2 + b^2 - 4(a+b) + 8 = 32$, and using $a^2 + b^2 = 64$ gives $4(a+b) = 40$, so $a + b = 10$.

The area is $ab$, and
$$2ab = (a+b)^2 - (a^2+b^2) = 100 - 64 = 36,$$
so $ab = 18$.

The answer is $\boxed{\textbf{(E)}\ 18}$.