2022 AMC 10A Problem 10

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 $\boxed{\textbf{(E)}\ 18}$.

Two distance measurements produce two equations in $a$ and $b$, but the question asks for $ab$, a symmetric expression. Symmetric targets are reached through $a + b$ and $a^2 + b^2$ without ever finding $a$ and $b$ themselves; the identity $(a+b)^2 = a^2 + 2ab + b^2$ is the bridge.

Notice the segment between the inner corners is the diagonal of an $(a-2) \times (b-2)$ rectangle, giving the same second equation. If you do solve, $a + b = 10$ and $ab = 18$ give $a, b = 5 \pm \sqrt{7}$, and $(5+\sqrt7)(5-\sqrt7) = 18$ confirms.

• Trying to solve for a and b individually (they are 5 plus or minus sqrt 7) instead of going straight for ab through (a+b)^2 - (a^2+b^2).
• Using $(a-1)^2 + (b-1)^2$ for the inner-corner distance, forgetting that a unit square is removed at both ends of the segment.