AMC12 2008 A

AMC12 2008 A · Q22

AMC12 2008 A · Q22. It mainly tests Triangles (properties), Circle theorems.

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 $x$?

一个圆桌的半径为 $4$。在桌上放置了六个矩形餐垫。每个餐垫的宽为 $1$，长为 $x$，如图所示。它们的位置使得每个餐垫有两个角在桌边上，这两个角是同一条长度为 $x$ 的边的端点。此外，餐垫的位置使得每个餐垫的内侧角都与相邻餐垫的一个内侧角相接触。求 $x$。

(A) $2\sqrt{5} - \sqrt{3}$ $2\sqrt{5} - \sqrt{3}$

(B) 3 3

(D) $2\sqrt{3}$ $2\sqrt{3}$

(E) $\frac{5 + 2\sqrt{3}}{2}$ $\frac{5 + 2\sqrt{3}}{2}$

Answer

Correct choice: (C)

正确答案：（C）

Solution

Let one of the mats be $ABCD$, and the center be $O$ as shown: Since there are $6$ mats, $\Delta BOC$ is equilateral (the hexagon with side length $x$ is regular). So, $BC=CO=x$. Also, $\angle OCD = \angle OCB + \angle BCD = 60^\circ+90^\circ=150^\circ$. By the Law of Cosines: $4^2=1^2+x^2-2\cdot1\cdot x \cdot \cos(150^\circ) \Rightarrow x^2 + x\sqrt{3} - 15 = 0 \Rightarrow x = \frac{-\sqrt{3}\pm 3\sqrt{7}}{2}$. Since $x$ must be positive, $x = \frac{3\sqrt{7}-\sqrt{3}}{2} \Rightarrow (C)$.

设其中一个餐垫为 $ABCD$，圆心为 $O$，如图所示：由于共有 $6$ 个餐垫，$\Delta BOC$ 为等边三角形（边长为 $x$ 的六边形是正六边形）。因此 $BC=CO=x$。并且 $\angle OCD = \angle OCB + \angle BCD = 60^\circ+90^\circ=150^\circ$。由余弦定理：$4^2=1^2+x^2-2\cdot1\cdot x \cdot \cos(150^\circ) \Rightarrow x^2 + x\sqrt{3} - 15 = 0 \Rightarrow x = \frac{-\sqrt{3}\pm 3\sqrt{7}}{2}$。因为 $x$ 必须为正，故 $x = \frac{3\sqrt{7}-\sqrt{3}}{2} \Rightarrow (C)$。

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