2010 AMC 10A Problem 17

P16-20~4 minPrint

Solid Geometry Cut the figure into known shapes (triangles, rectangles, sectors)

A solid cube has side length $3$ inches. A $2$ -inch by $2$ -inch square hole is cut into the center of each face. The edges of each cut are parallel to the edges of the cube, and each hole goes all the way through the cube. What is the volume, in cubic inches, of the remaining solid?

Show solution without answering

Key insight

Slice the cube into lengths 1/2, 2, 1/2 each way. Only the twelve edge bars and eight corner pieces survive the holes: 6 + 1 = 7.

Solution

The holes overlap in the middle, so subtracting them is tricky. Instead, cut the cube into boxes and count which boxes survive.

Cut the cube. A $2$ -inch hole centered on a $3$ -inch face leaves a strip $\tfrac12$ inch wide on each side, because $\tfrac12 + 2 + \tfrac12 = 3$ . So slice the cube in each of the three directions into lengths $\tfrac12$ , $2$ , $\tfrac12$ . This makes $3 \times 3 \times 3 = 27$ boxes.
Which boxes does a hole remove? A box is removed if, in two of the three directions, it is in the middle $2$ -inch band, because then it lines up with a hole going through in the third direction.
Sort the 27 boxes.

Kind of box	Size	How many	Survives?
Center	$2 \times 2 \times 2$	$1$	no
Center of a face	$2 \times 2 \times \tfrac12$	$6$	no
Along an edge	$\tfrac12 \times \tfrac12 \times 2$	$12$	yes
Corner	$\tfrac12 \times \tfrac12 \times \tfrac12$	$8$	yes

( $1 + 6 + 12 + 8 = 27$ , so every box is counted.)
4. Add the survivors.
- Edge boxes: each has volume $\tfrac12 \times \tfrac12 \times 2 = \tfrac12$ , and $12 \times \tfrac12 = 6$ .
- Corner boxes: each has volume $\tfrac18$ , and $8 \times \tfrac18 = 1$ .
5. The remaining solid has volume $6 + 1 = 7$ cubic inches. It is just a thin frame along the edges of the cube.

The answer is $\boxed{\textbf{(A)}\ 7}$ .

Why this works

When several cut-outs overlap, it is often easier to describe what is left than what is removed. Slicing the cube along the edges of the holes breaks it into a few kinds of boxes, like a Rubik's cube, and each kind is simple to measure and count.

Alternative approach

Count the removed part carefully. Each hole is a $2 \times 2 \times 3$ tunnel, volume $12$ . Three tunnels give $36$ , but they overlap: all three pass through the same central $2 \times 2 \times 2$ cube (volume $8$ ). That cube was counted three times instead of once, so subtract it twice: $36 - 2 \times 8 = 20$ removed. What remains is $27 - 20 = 7$ .

Concise solution

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

Three tunnels of cross-section $2 \times 2$ run through the cube in the three coordinate directions. A $2$ -inch hole centered on a $3$ -inch face leaves a border of width $\tfrac12$ around it, so a point survives only if at least two of its three coordinates lie in that outer half-inch border (if two coordinates were both in the middle $2$ -inch band, the point would sit inside the tunnel running along the third direction).

Count what survives:

Corners: all three coordinates in the border. These are the $8$ corner cubes of side $\tfrac12$ , total volume $8 \cdot \tfrac18 = 1$ .
Edge bars: exactly two coordinates in the border, the third free along the middle $2$ -inch band. Along each of the $12$ edges this is a $\tfrac12 \times \tfrac12 \times 2$ bar of volume $\tfrac12$ , total $12 \cdot \tfrac12 = 6$ .

The remaining solid has volume $1 + 6 = 7$ .

The answer is $\boxed{\textbf{(A)}\ 7}$ .

Review in the AoPS books

Introduction to Geometry, Chapter 14: Three-Dimensional Geometry
Introduction to Geometry, Chapter 4: Perimeter and Area (§4.2, §4.3)
Introduction to Geometry, Chapter 15: Curved Surfaces
Introduction to Geometry, Chapter 11: Circles (§11.3)