2013 AMC 10A Problem 14

P11-15~4 minPrint

Solid Geometry Set up the equation/formula and compute; no special trick needed Cut the figure into known shapes (triangles, rectangles, sectors)

A solid cube of side length $1$ is removed from each corner of a solid cube of side length $3$ . How many edges does the remaining solid have?

Show solution without answering

Key insight

Each of the 12 old edges keeps its middle piece, and each of the 8 corner holes adds 9 new edges: 12 + 72 = 84.

Solution

Picture the big cube as a $3 \times 3 \times 3$ stack of little cubes, like a Rubik's cube. Removing the $8$ corner cubes leaves a square hole at each corner. Count the edges in two groups.

1. What is left of the old edges. The big cube had $12$ edges, each $3$ units long. Each corner cube bites $1$ unit off each end, so the middle $1$ unit of every edge is still there. That is $12$ edges.
2. Edges around one corner hole. The hole is shaped like the inside corner of a box: three little square walls that meet at one inner point.
- Three squares have $3 \times 4 = 12$ sides.
- But each of the $3$ edges running into the inner point is shared by two walls, so it was counted twice.
- Distinct edges: $12 - 3 = 9$ .

Another way to see it: $3$ edges meet at the inner point, and $6$ more edges lie on the faces of the big cube, outlining the hole. That is $9$ .
3. All eight holes. $8 \times 9 = 72$ edges. None of these is one of the $12$ middle pieces from Step 1.
4. Total. $12 + 72 = 84$ .

The answer is $\boxed{\textbf{(D)}\ 84}$ .

Why this works

Split a hard count into groups you can trust: what is left of the old shape, and what each change adds. All eight corners are the same, so one careful count, times eight, finishes the job. The only tricky part is not counting a shared edge twice.

Alternative approach

Use Euler's formula for solids: vertices $-$ edges $+$ faces $= 2$ .
- Faces: the $6$ big faces (now shaped like plus signs) plus $3$ little squares in each of the $8$ holes: $6 + 24 = 30$ .
- Vertices: each hole has $7$ corners (the inner point and $6$ on the big faces), and holes do not share any: $8 \times 7 = 56$ .
- Edges: $56 + 30 - 2 = 84$ .

Concise solution

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

Think of the big cube as a $3 \times 3 \times 3$ block of unit cubes; removing the eight corner unit cubes leaves a shape with a square notch at every corner.

Old edges. Each of the $12$ edges of the original cube loses a unit length at both ends but keeps its middle unit segment: $12$ edges.

New edges from one notch. A notch is bounded by three unit squares meeting at an inner corner. Those squares have $3 \cdot 4 = 12$ sides in total, but the three sides that run into the inner corner are each shared by two of the squares, so the notch has $12 - 3 = 9$ distinct edges. (Equivalently: $3$ inner edges plus $6$ edges lying on the original faces.)

Eight notches contribute $8 \cdot 9 = 72$ edges, and none of these coincide with the surviving middle segments of the old edges.

Total: $12 + 72 = 84$ .

The answer is $\boxed{\textbf{(D)}\ 84}$ .

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)