2009 AMC 10A Problem 11

P11-15~2 minPrint

Algebraic Manipulation Solid Geometry Apply an identity: SFFT, sum of squares, difference of cubes, Vieta Use the answer choices (mod checks, size, form) to eliminate or select

One dimension of a cube is increased by $1$ , another is decreased by $1$ , and the third is left unchanged. The volume of the new rectangular solid is $5$ less than that of the cube. What was the volume of the cube?

Show solution without answering

Key insight

The new box has volume s(s+1)(s-1), which is s cubed minus s, so it is always s less than the cube. A drop of 5 means s = 5.

Solution

1. Try one cube first to see what happens. Side $4$ : the cube has volume $64$ . The new box is $5 \times 3 \times 4 = 60$ . The drop is $4$ , which is the side length.
2. Now in general. Call the side $s$ , so the cube has volume $s^3$ . The new box measures $s + 1$ , $s - 1$ and $s$ .
3. First multiply $(s + 1)(s - 1) = s^2 - s + s - 1 = s^2 - 1$ .
4. So the new volume is
$s(s^2 - 1) = s^3 - s.$
It is always exactly $s$ less than the cube.
5. The problem says the drop is $5$ , so $s = 5$ , and the cube's volume is $5^3 = 125$ .
6. Check: $6 \times 4 \times 5 = 120$ , and $125 - 120 = 5$ .

The answer is $\boxed{\textbf{(D)}\ 125}$ .

Why this works

Making one side $1$ longer and another $1$ shorter multiplies them to $s^2 - 1$ instead of $s^2$ . That "minus $1$ " times the third side $s$ is the whole loss. Spotting $(s+1)(s-1) = s^2 - 1$ turns a cube problem into a one-step equation.

Alternative approach

Every choice is a cube: $2^3, 3^3, 4^3, 5^3, 6^3$ . Test the sides. Side $5$ gives a box $6 \times 4 \times 5 = 120$ , which is $5$ less than $125$ . The other sides give drops of $2, 3, 4, 6$ , never $5$ .

Concise solution

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

Let the cube have side $s$ , so its volume is $s^3$ . The new solid has dimensions $s + 1$ , $s - 1$ and $s$ , so its volume is
$s(s+1)(s-1) = s(s^2 - 1) = s^3 - s.$

This is exactly $s$ less than the cube's volume. The problem says the difference is $5$ , so $s = 5$ and the cube's volume is
$5^3 = 125.$

Check: the new box is $6 \times 4 \times 5 = 120 = 125 - 5$ .

The answer is $\boxed{\textbf{(D)}\ 125}$ .

Review in the AoPS books

Introduction to Algebra, Chapter 11: Special Factorizations (§11.2, §11.3, §11.5)
Introduction to Geometry, Chapter 14: Three-Dimensional Geometry
Introduction to Algebra, Chapter 22: Special Manipulations
Introduction to Geometry, Chapter 15: Curved Surfaces