2017 AMC 10B Problem 6

P1-10~2 minPrint

Solid Geometry Bound the quantity above/below or estimate to pin it down

What is the largest number of solid $2\text{ in.}$ by $2\text{ in.}$ by $1\text{ in.}$ blocks that can fit in a $3\text{ in.}$ by $2\text{ in.}$ by $3\text{ in.}$ box?

Show solution without answering

Key insight

Volume allows at most 4 blocks, since 5 blocks need 20 cubic inches and the box has 18. Three flat blocks plus one on its side fit.

Solution

This is a "largest number" question, so it has two halves: show that some number of blocks fits, then show that one more cannot.

Part 1: five blocks cannot fit.

The box has volume $3 \cdot 2 \cdot 3 = 18$ cubic inches.
Each block has volume $2 \cdot 2 \cdot 1 = 4$ cubic inches.
Five blocks would need $5 \cdot 4 = 20$ cubic inches, more than the box has. So at most $4$ blocks fit.

Part 2: four blocks do fit.

Think of the box as $3$ inches wide, $2$ inches deep and $3$ inches tall.
Lay a block flat: $2$ wide, $2$ deep, $1$ tall. Stack three of them in the left part of the box. Together they fill a space $2$ wide, $2$ deep and $3$ tall.
What is left is a slab $1$ inch wide, $2$ deep and $3$ tall.
Turn the fourth block on its side so it is $1$ wide, $2$ deep and $2$ tall. It fits in the slab, with $1$ inch of height to spare.

Four blocks fit and five cannot, so the answer is $\boxed{\textbf{(B)}\ 4}$ .

Why this works

Packing questions always have two halves: a volume count that says "no more than this," and an actual arrangement that reaches that number. When the two agree, you are done. The blocks do not all have to face the same way; turning one of them is often what uses up the leftover space.

Concise solution

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

Upper bound: the box has volume $3 \cdot 2 \cdot 3 = 18$ cubic inches and each block has volume $2 \cdot 2 \cdot 1 = 4$ . Since $4 \cdot 5 = 20 > 18$ , at most $4$ blocks fit.

Construction: think of the box as $3$ wide, $2$ deep, $3$ tall. Lay a block flat so its $2 \times 2$ face covers a $2$ wide by $2$ deep footprint and it is $1$ tall; stack three of these to fill a $2 \times 2 \times 3$ column. The remaining space is a $1 \times 2 \times 3$ slab. Stand the fourth block on its edge: $1$ wide, $2$ deep, $2$ tall. It fits in the slab with one inch of height to spare.

Four blocks fit and five cannot, so the answer is $\boxed{\textbf{(B)}\ 4}$ .

Review in the AoPS books

Introduction to Geometry, Chapter 14: Three-Dimensional Geometry
Introduction to Geometry, Chapter 15: Curved Surfaces