Question

一个 $2 \times 3$ 矩形和一个 $3 \times 4$ 矩形放在一个正方形内，内部不重叠，且正方形的边与两个矩形的边平行。这个正方形最小可能面积是多少？

A $2 \times 3$ rectangle and a $3 \times 4$ 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?

$16$

$25$

$36$

$49$

$64$

答案：B

知识点：正方形（几何）矩形最优化

难度评级：1060

解答：

将两个矩形并排放置，使长度为 $3$ 的边竖直。它们宽度相加为 $2+3=5$ ，高度 $3$ 和 $4$ 都能放进边长 $5$ 的正方形。

边长不能小于 $2+3=5$ ，因为两个较小尺寸 $2$ 和 $3$ 必须同时容纳。最小面积为 $5^2=25$ 。

所以正确答案是 B。

Place the rectangles side by side with their $3$ -length sides vertical. Their widths add to $2+3=5,$ and the heights $3$ and $4$ both fit within $5.$

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 $2$ and the second spans at least $3,$ so the square's side is at least $2+3=5.$ The smallest area is therefore $5^2=25.$

Thus, the correct answer is B.

2006 AMC 10B 第 5 题

答案：B

难度评级：1060

解答：

其他年份的第 5 题