Question

一个长方形的边长为 $a$ 和 $36$ 。在长方形的每个顶点，以及每条长为 $36$ 的边的中点处都安装铰链。把两条长为 $a$ 的边彼此压近，同时保持这两条边平行，于是长方形变成如图所示的凸六边形。当图形成为一个六边形，且两条长为 $a$ 的边平行并相距 $24$ 时，该六边形的面积与原长方形相同。求 $a^2$ 。

A rectangle has sides of length $a$ and $36.$ A hinge is installed at each vertex of the rectangle and at the midpoint of each side of length $36.$ The sides of length $a$ can be pressed toward each other keeping those two sides parallel so the rectangle becomes a convex hexagon as shown. When the figure is a hexagon with the sides of length $a$ parallel and separated by a distance of $24,$ the hexagon has the same area as the original rectangle. Find $a^2.$

答案

答案：720

知识点：勾股定理梯形面积

难度评级：2110

解答：

在六边形中，每条长为 $36$ 的边都在中点折成两根长为 $18$ 的杆。两条长为 $a$ 的边相距 $24$ ，所以每根杆的竖直跨度为 $12$ ，水平跨度为 $\sqrt{18^2 - 12^2} = \sqrt{180} = 6\sqrt{5}$ 。

过两个中点铰链的直线把六边形分成两个全等梯形，梯形的平行边为 $a$ 和 $a + 12\sqrt{5}$ ，高为 $12$ ，因此六边形面积为 $\begin{aligned} &2 \cdot \frac{a + (a + 12\sqrt{5})}{2} \cdot 12 \\ &= 24a + 144\sqrt{5}. \end{aligned}$

令它等于长方形面积 $36a$ ，得到 $12a = 144\sqrt{5}$ ，所以 $a = 12\sqrt{5}$ ，且 $a^2 = 720$ 。

In the hexagon, each side of length $36$ has folded at its midpoint into two bars of length $18.$ The two sides of length $a$ are $24$ apart, so each bar spans a vertical distance of $12$ and hence a horizontal distance of $\sqrt{18^2 - 12^2} = \sqrt{180} = 6\sqrt{5}.$

The line through the two midpoint hinges splits the hexagon into two congruent trapezoids with parallel sides $a$ and $a + 12\sqrt{5}$ and height $12,$ so the hexagon has area $\begin{aligned} &2 \cdot \frac{a + (a + 12\sqrt{5})}{2} \cdot 12 \\ &= 24a + 144\sqrt{5}. \end{aligned}$

Setting this equal to the rectangle's area $36a$ gives $12a = 144\sqrt{5},$ so $a = 12\sqrt{5}$ and $a^2 = 720.$

2014 AIME II 第 3 题

答案：720

难度评级：2110

解答：

其他年份的第 3 题