2018 AMC 10B 第 12 题

Question

线段

AB

是一个圆的直径，且

AB = 24

。点

C

在圆上，但不等于

A

或

B

。当

C

绕圆运动时，

\triangle ABC

的重心描出一条少了两个点的闭曲线。四舍五入到最接近的正整数，这条曲线围成区域的面积是多少？

Line segment

AB

is a diameter of a circle with

AB = 24.

Point

C,

not equal to

A

or

B,

lies on the circle. As point

C

moves around the circle, the centroid (center of mass) of

\triangle ABC

traces out a closed curve missing two points. To the nearest positive integer, what is the area of the region bounded by this curve?

25

38

50

63

75

答案：C

知识点：重心位似圆面积

难度评级：1530

小提示：

\triangle ABC

的重心是

\tfrac13(A + B + C)

，而

A + B

是圆心的两倍。

The centroid of

\triangle ABC

is

\tfrac13(A + B + C),

and

A + B

is twice the center

大提示：

当

C

在半径为

12

的圆上运动时，重心在半径为其三分之一的圆上运动。

As

C

moves on a circle of radius

12,

the centroid moves on a circle of one third the radius

解答：

把圆心

O

放在原点，则可取

A = (-12, 0)

、

B = (12, 0)

，而

C

在半径为

12

的圆上运动。因为

A + B = 0

，重心为

\tfrac13(A + B + C) = \tfrac13 C

。当

C

绕圆运动时，

\tfrac13 C

描出半径为

\tfrac{12}{3} = 4

的圆，少掉

C = A

或

B

时的两个点不影响面积。面积为

\pi \cdot 4^2 = 16\pi \approx 50

。正确答案是 C。

Put the center

O

at the origin, so

A = (-12, 0)

and

B = (12, 0),

while

C

runs over the circle of radius

12.

Then

A + B = 0,

so the centroid is

\tfrac13(A + B + C) = \tfrac13 C.

As

C

circles,

\tfrac13 C

traces a circle of radius

\tfrac{12}{3} = 4

(minus the two points where

C = A

or

B

). Its area is

\pi \cdot 4^2 = 16\pi \approx 50.

Therefore, the answer is C.

2018 AMC 10B 第 12 题

答案：C

难度评级：1530

小提示：

大提示：

解答：

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