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2010 AMC 10A Problem 17

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2010 AMC 10A Problem 17AMC 10 Step by Step
Name:                             Date: September 13, 2026
Solid Geometry · difficulty 3 of 5 · about 4 min · position P16-20

A solid cube has side length 33 inches. A 22 -inch by 22 -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?

  1. A)

    77

  2. B)

    88

  3. C)

    1010

  4. D)

    1212

  5. E)

    1515

Answer

A

Key insight

Each 2-by-2 hole leaves a border of width 1/2 on its face, so what survives is eight half-inch corner cubes plus twelve 1/2-by-1/2-by-2 edge bars.

Solution

Three tunnels of cross-section 2×22 \times 2 run through the cube in the three coordinate directions. A 22-inch hole centered on a 33-inch face leaves a border of width 12\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 22-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 88 corner cubes of side 12\tfrac12, total volume 8⋅18=18 \cdot \tfrac18 = 1.
  • Edge bars: exactly two coordinates in the border, the third free along the middle 22-inch band. Along each of the 1212 edges this is a 12×12×2\tfrac12 \times \tfrac12 \times 2 bar of volume 12\tfrac12, total 12⋅12=612 \cdot \tfrac12 = 6.

The remaining solid has volume 1+6=71 + 6 = 7.

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

Why this works

Instead of subtracting the tunnels (which overlap), describe the survivors directly: the cube's frame of edges and corners, with thickness 12\tfrac12. The "at least two coordinates in the border" description makes the pieces easy to enumerate and measure.

Alternative approach

Inclusion-exclusion on the removed volume: each tunnel removes 2⋅2⋅3=122 \cdot 2 \cdot 3 = 12, any two tunnels overlap in the central 2×2×22 \times 2 \times 2 cube (volume 88), and so do all three. Removed =3⋅12−3⋅8+8=20= 3 \cdot 12 - 3 \cdot 8 + 8 = 20, leaving 27−20=727 - 20 = 7.

The trap

Treating the surviving corners as unit cubes and answering 8, or not counting the edge bars at all.

Common mistakes

  • Treating the surviving corners as unit cubes and answering 8, or not counting the edge bars at all.
  • Subtracting 3⋅12=363 \cdot 12 = 36 from 2727 without correcting for the overlapping central cube.

Techniques

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