2021 AMC 10B Problem 11

P11-15~4 minPrint

Diophantine Equations Algebraic Manipulation Apply an identity: SFFT, sum of squares, difference of cubes, Vieta Substitute to simplify (u = x+1/x, shifting, scaling)

Grandma has just finished baking a large rectangular pan of brownies. She is planning to make rectangular pieces of equal size and shape, with straight cuts parallel to the sides of the pan. Each cut must be made entirely across the pan. Grandma wants to make the same number of interior pieces as pieces along the perimeter of the pan. What is the greatest possible number of brownies she can produce?

Show solution without answering

Key insight

The inside pieces must be half of all pieces: 2(m − 2)(n − 2) = mn. Only two pans work, 5 by 12 and 6 by 8.

Solution

Step 1: count inside and edge pieces.

1. Say the cuts make $m$ rows and $n$ columns of pieces, so there are $mn$ pieces in all.
2. Try a small case first: a $4 \times 5$ pan has $20$ pieces. Peeling off the outer ring leaves a $2 \times 3$ block inside, which is $6$ pieces, so $14$ are on the edge.
3. In general, removing the outer ring takes one row off the top and bottom and one column off each side, so the inside is $(m - 2) \times (n - 2)$ .
4. "Inside equals edge" means the inside is exactly half of all the pieces:
$2(m-2)(n-2) = mn.$

Step 2: simplify.

5. To make the numbers smaller, write $a = m - 2$ and $b = n - 2$ (the inside's length and width). Then $m = a + 2$ and $n = b + 2$ , and the equation becomes
$2ab = (a+2)(b+2) = ab + 2a + 2b + 4.$
6. Subtract $ab$ from both sides: $ab - 2a - 2b = 4$ .
7. The left side is almost a product. Notice that $(a - 2)(b - 2) = ab - 2a - 2b + 4$ . So add $4$ to both sides:
$(a - 2)(b - 2) = 8.$

Step 3: list the possibilities.

Both brackets are whole numbers that multiply to $8$ . Say $a \le b$ . The positive options are $1 \times 8$ and $2 \times 4$ . (Two negative numbers would make $a$ or $b$ zero or negative, which is not a real pan.)

$a - 2$ , $b - 2$	$a$ , $b$	$m$ , $n$	pieces $mn$
$1, 8$	$3, 10$	$5, 12$	$60$
$2, 4$	$4, 6$	$6, 8$	$48$

Step 4: check and choose.

Check $5 \times 12$ : the inside is $3 \times 10 = 30$ pieces, and the edge is $60 - 30 = 30$ . Equal, so $60$ is possible.
These are the only two shapes that work, so no pan can give more than $60$ .

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

Why this works

The condition turns into an equation with a product and a few extra terms. Adding the right number to both sides turns it into "one bracket times another bracket equals $8$ ," and $8$ can be split into factors in only a few ways. Whenever you see something like $ab - 2a - 2b$ , look for the product it almost is.

Alternative approach

Check the choices from the largest down, trying shapes with an inside.

$64$ : $8 \times 8$ has inside $6 \times 6 = 36$ and edge $28$ . $4 \times 16$ has inside $2 \times 14 = 28$ and edge $36$ . Neither works, and other shapes such as $2 \times 32$ have no inside at all.
$60$ : $5 \times 12$ has inside $3 \times 10 = 30$ and edge $30$ . It works, so $60$ is the answer.

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

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 $\boxed{\textbf{(D)}\ 60}$ .

Review in the AoPS books

Introduction to Number Theory, Chapter 7: Algebra With Integers
Introduction to Algebra, Chapter 11: Special Factorizations (§11.2, §11.3, §11.5)
Intermediate Algebra, Chapter 20: Some Advanced Strategies (§20.2)
Introduction to Algebra, Chapter 22: Special Manipulations