2008 AMC 12A 第 22 题

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

小提示：

对一个餐垫，设

P, Q

为外侧角，

R

为与

P

直径相对的点；则

\triangle PQR

是直角三角形且

PR = 8

。

For one mat, let

P, Q

be its outer corners and

R

the point diametrically opposite

P;

then

\triangle PQR

is right-angled with

PR = 8

大提示：

内侧角形成顶角

120^\circ

的等腰三角形，贡献长度

\sqrt{3}\,x

。

The inner corners form isosceles triangles with vertex angle

120^\circ,

contributing a segment of length

\sqrt{3}\,x

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

x^2 + \left(\sqrt{3}\,x + 2\right)^2 = 64\text{，}

化简为

x^2 + \sqrt{3}\,x - 15 = 0

。

取正根，

x = \dfrac{-\sqrt{3} + \sqrt{63}}{2} = \dfrac{3\sqrt{7} - \sqrt{3}}{2}\text{。}

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

小提示：

大提示：

解答：

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