2013 AMC 10A Problem 14

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 $\boxed{\textbf{(D)}\ 84}$.

Decompose the boundary into pieces you can count reliably: what survives from the original solid, and what each identical modification adds. Because all eight corners are alike, one careful count times eight does the job. The shared inner edges are the only subtlety.

Euler's formula. Faces: the $6$ original faces (now cross-shaped) plus $3 \cdot 8 = 24$ notch squares, $F = 30$. Vertices: each notch has $7$ vertices (one inner corner and six on the original faces), and no vertex is shared between notches, so $V = 56$. Then $E = V + F - 2 = 56 + 30 - 2 = 84$.

• Counting each notch as 12 edges (three squares times four sides) without noticing the three inner edges are shared by two squares each.
• Forgetting the $12$ surviving middle segments of the original edges and answering $72$.