Question

一张圆桌半径为 $4$ 。桌上放着六个矩形餐垫。每个餐垫宽为 $1$ ，长为 $x$ ，如图所示。每个餐垫都有两个角在桌边上，这两个角是一条长为 $x$ 的边的两个端点。此外，每个内侧角都与相邻餐垫的一个内侧角相接。 $x$ 是多少？

A round table has radius $4.$ Six rectangular place mats are placed on the table. Each place mat has width $1$ and length $x$ as shown. They are positioned so that each mat has two corners on the edge of the table, these two corners being end points of the same side of length $x.$ Further, the mats are positioned so that the inner corners each touch an inner corner of an adjacent mat. What is $x?$

$2\sqrt{5} - \sqrt{3}$

$3$

$\dfrac{3\sqrt{7} - \sqrt{3}}{2}$

$2\sqrt{3}$

$\dfrac{5 + 2\sqrt{3}}{2}$

答案：C

知识点：圆圆周角勾股定理二次方程

难度评级：2150

解答：

取一个餐垫的外侧角为 $P$ 和 $Q$ ，令 $R$ 为圆上与 $P$ 关于直径相对的点。于是 $\triangle PQR$ 在 $Q$ 处为直角，斜边 $PR = 8$ 。

相邻餐垫的内角相接，形成顶角为 $120^\circ$ 、两腰为 $x$ 的等腰三角形，其底边为 $\sqrt{3}\,x$ 。再加上两个餐垫宽度， $QR = \sqrt{3}\,x + 2$ 。

由勾股定理， $\left(\sqrt{3}\,x + 2\right)^2 + x^2 = 64,$ 化简为 $x^2 + \sqrt{3}\,x - 15 = 0$ 。

取正根， $x = \dfrac{3\sqrt{7} - \sqrt{3}}{2}.$

所以正确答案是 C。

Pick a mat with outer corners $P$ and $Q,$ and let $R$ be the point on the circle diametrically opposite $P.$ Then $\triangle PQR$ is right-angled at $Q$ with hypotenuse $PR = 8.$

The inner corners of adjacent mats meet in isosceles triangles with vertex angle $120^\circ$ and sides $x,$ whose base is $\sqrt{3}\,x.$ Together with the two mat widths, $QR = \sqrt{3}\,x + 2.$

By the Pythagorean theorem, $\left(\sqrt{3}\,x + 2\right)^2 + x^2 = 64,$ which simplifies to $x^2 + \sqrt{3}\,x - 15 = 0.$

Taking the positive root, $x = \dfrac{3\sqrt{7} - \sqrt{3}}{2}.$

Thus, the correct answer is C.

2008 AMC 10A 第 25 题

答案：C

难度评级：2150

解答：

其他年份的第 25 题