2006 AMC 10B Problem 5

Attempt Problem 5 of the 2006 AMC 10B below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2006 AMC 10B solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

5.

A 2×32 \times 3 rectangle and a 3×43 \times 4 rectangle are contained within a square without overlapping at any interior 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?

1616

2525

3636

4949

6464

Answer: B
Concepts:square (geometry)rectangleoptimization
Difficulty rating: 1060
Solution:

Place the rectangles side by side with their 33-length sides vertical. Their widths add to 2+3=5,2+3=5, and the heights 33 and 44 both fit within 5.5.

Because the rectangles are axis-aligned and their interiors do not overlap, their horizontal projections or their vertical projections must be disjoint. In either direction, the first rectangle spans at least 22 and the second spans at least 3,3, so the square's side is at least 2+3=5.2+3=5. The smallest area is therefore 52=25.5^2=25.

Thus, the correct answer is B.

← Problem 4#4
Full Exam

Problem 5 in Other Years