2021 AMC 10B Problem 11

Suppose the cuts produce an $m \times n$ array of pieces. The interior pieces form an $(m-2)\times(n-2)$ array, and the perimeter pieces are the rest. "Interior equals perimeter" means the interior is exactly half the total:
$$2(m-2)(n-2) = mn.$$

Let $a = m - 2$ and $b = n - 2$. Then $2ab = (a+2)(b+2) = ab + 2a + 2b + 4$, so
$$ab - 2a - 2b = 4 \quad\Longrightarrow\quad (a-2)(b-2) = 8.$$

With $a \le b$, the positive factor pairs of $8$ give $(a-2, b-2) = (1, 8)$ or $(2, 4)$, i.e. $(a, b) = (3, 10)$ or $(4, 6)$, i.e. $(m, n) = (5, 12)$ or $(6, 8)$. (Negative factor pairs would force $a \le 1$ and $b < 0$, impossible.)

The two arrays have $60$ and $48$ pieces. Check $5 \times 12$: interior $3 \times 10 = 30$, perimeter $60 - 30 = 30$. Equal.

The answer is $\boxed{\textbf{(D)}\ 60}$.

The condition is a quadratic Diophantine equation in the grid dimensions; shifting variables and applying Simon's Favorite Factoring Trick turns it into "product of two integers equals $8$," which has only a handful of solutions. Whenever an integer equation has the form $xy + px + qy = r$, complete the product.

Test the choices as rectangles with an interior. $64 = 8 \times 8$: interior $36$, perimeter $28$, no; $4 \times 16$: interior $28$, no. $60 = 5 \times 12$: interior $30$, perimeter $30$, yes. Since $60$ is the largest choice that works, it is the answer.

• Guessing from the choices that a near-square grid like 8 × 8 = 64 works, without checking that its interior 36 ≠ perimeter 28.
• Counting the perimeter pieces as $2m + 2n$ (double-counting the corners) instead of $2m + 2n - 4$.