2006 AMC 10B Problem 5

P1-10~2 minPrint

Geometric Optimization Quadrilaterals & Polygon Areas Bound the quantity above/below or estimate to pin it down

A $2 \times 3$ rectangle and a $3 \times 4$ rectangle are contained within a square without overlapping at any point, and the sides of the square are parallel to the sides of the two given rectangles. What is the smallest possible area of the square?

Show solution without answering

Key insight

The rectangles cover 18 square units, more than a 4 by 4 square holds. Side by side they make a 5 by 4 block, which fits.

Solution

This asks for the smallest square, so there are two halves: show a size-5 square works, then show a size-4 square cannot.

Part 1: a $5 \times 5$ square is enough.

Stand the $3 \times 4$ rectangle up so it is $3$ wide and $4$ tall.
Put the $2 \times 3$ rectangle right beside it, standing $2$ wide and $3$ tall.
Together they are $3 + 2 = 5$ wide, and the taller one is $4$ tall. A block $5$ wide and $4$ tall fits inside a $5 \times 5$ square, with room to spare.

Part 2: a $4 \times 4$ square is too small.

The rectangles have areas $2 \cdot 3 = 6$ and $3 \cdot 4 = 12$ , so together they need $18$ square units of space.
A $4 \times 4$ square has only $16$ square units. Since they cannot overlap, they do not fit. Any smaller square is even worse.

Among the answer choices, $16$ is too small and $25$ works, so the smallest area is $5^2 = 25$ .

The answer is $\boxed{\textbf{(B)}\ 25}$ .

Why this works

"Smallest possible" needs both an example that works and a reason nothing smaller works. Comparing areas is the fastest way to rule out small squares. To build the example, line the rectangles up so their short sides add together, keeping the overall shape as close to square as possible.

Concise solution

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

Lower bound: the rectangles have areas $6$ and $12$ , so the square must have area at least $18$ . A $4 \times 4$ square has area $16 < 18$ , so side $4$ is impossible, and any smaller side is too.

Construction for side $5$ : stand the $3 \times 4$ rectangle with its $3$ -side horizontal and $4$ -side vertical, and stand the $2 \times 3$ rectangle next to it with its $2$ -side horizontal and $3$ -side vertical. Together they span width $3 + 2 = 5$ and height $\max(4, 3) = 4$ , which fits inside a $5 \times 5$ square.

So the smallest square has area $5^2 = 25$ .

The answer is $\boxed{\textbf{(B)}\ 25}$ .

Review in the AoPS books

Introduction to Geometry, Chapter 10: Geometric Inequalities
Introduction to Geometry, Chapter 8: Quadrilaterals
Introduction to Geometry, Chapter 9: Polygons (§9.3)