AMC10 2008 A

AMC10 2008 A · Q25

AMC10 2008 A · Q25. It mainly tests Quadratic equations, 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 $x$?

一张半径为4的圆桌上有六个矩形餐垫放置。每张餐垫宽度为1，长度为$x$，如图所示。它们定位使得每张餐垫有两个角在桌边上，这两个角是长度为$x$的同一边的端点。此外，餐垫定位使得内角每个都接触相邻餐垫的一个内角。$x$是多少？

(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

Answer (C): Select one of the mats. Let $P$ and $Q$ be the two corners of the mat that are on the edge of the table, and let $R$ be the point on the edge of the table that is diametrically opposite $P$ as shown. Then $R$ is also a corner of a mat and $\triangle PQR$ is a right triangle with hypotenuse $PR=8$. Let $S$ be the inner corner of the chosen mat that lies on $QR$, $T$ the analogous point on the mat with corner $R$, and $U$ the corner common to the other mat with corner $S$ and the other mat with corner $T$. Then $\triangle STU$ is an isosceles triangle with two sides of length $x$ and vertex angle $120^\circ$. It follows that $ST=\sqrt{3}x$, so $QR=QS+ST+TR=\sqrt{3}x+2$. Apply the Pythagorean Theorem to $\triangle PQR$ to obtain $(\sqrt{3}x+2)^2+x^2=8^2$, from which $x^2+\sqrt{3}x-15=0$. Solve for $x$ and ignore the negative root to obtain \[ x=\frac{3\sqrt{7}-\sqrt{3}}{2}. \]

答案（C）：选择其中一张垫子。设 $P$ 和 $Q$ 为该垫子位于桌边的两个角点，设 $R$ 为桌边上与 $P$ 关于圆直径相对的位置（如图）。则 $R$ 也是某张垫子的一个角点，并且 $\triangle PQR$ 是一直角三角形，其斜边 $PR=8$。设 $S$ 为所选垫子落在 $QR$ 上的内角点，$T$ 为以 $R$ 为角点的那张垫子上的对应点，$U$ 为分别与角点 $S$、$T$ 相邻的另外两张垫子所共有的角点。则 $\triangle STU$ 为等腰三角形，两边长为 $x$，顶角为 $120^\circ$。因此 $ST=\sqrt{3}x$，从而 $QR=QS+ST+TR=\sqrt{3}x+2$。对 $\triangle PQR$ 使用勾股定理得 $(\sqrt{3}x+2)^2+x^2=8^2$，由此得到 $x^2+\sqrt{3}x-15=0$。解出 $x$ 并舍去负根，得到 \[ x=\frac{3\sqrt{7}-\sqrt{3}}{2}. \]

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