Question

一个圆桌半径为 $4$ 。桌上放置六个矩形餐垫。每个餐垫宽 $1$ 、长 $x$ ，如图。每个餐垫有两个角在桌边上，这两个角是同一条长为 $x$ 的边的端点。此外，餐垫的位置使得每个内侧角都与相邻餐垫的一个内侧角接触。求 $x$ ？

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 endpoints 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：圆等腰三角形勾股定理

难度评级：2120

解答：

取一个餐垫，外侧两角为 $P$ 和 $Q$ ，令 $R$ 是圆桌边上与 $P$ 直径相对的点。则 $PR = 8$ 为直径，所以 $\triangle PQR$ 在 $Q$ 处为直角，且 $PQ = x$ 。

沿 $QR$ 方向，相邻餐垫的内角形成两边长为 $x$ 、顶角为 $120^\circ$ 的等腰三角形，其底边为 $\sqrt{3}\,x$ 。因此 $QR = \sqrt{3}\,x + 2$ 。

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

取正根， $x = \dfrac{-\sqrt{3} + \sqrt{63}}{2} = \dfrac{3\sqrt{7} - \sqrt{3}}{2}.$

所以正确答案是 C。

Take one mat with outer corners $P$ and $Q,$ and let $R$ be the point of the table's edge diametrically opposite $P.$ Then $PR = 8$ is a diameter, so $\triangle PQR$ has a right angle at $Q,$ with $PQ = x.$

Along $QR,$ the inner corners of neighboring mats meet in an isosceles triangle with two sides of length $x$ and vertex angle $120^\circ,$ whose base is $\sqrt{3}\,x.$ Hence $QR = \sqrt{3}\,x + 2.$

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

Taking the positive root, $x = \dfrac{-\sqrt{3} + \sqrt{63}}{2} = \dfrac{3\sqrt{7} - \sqrt{3}}{2}.$

Thus, C is the correct answer.

2008 AMC 12A 第 22 题

答案：C

难度评级：2120

解答：

其他年份的第 22 题