1971 AMC 12 第 9 题

Question

一条不交叉且没有松弛的皮带绕在半径分别为

14

英寸和

4

英寸的两个圆形皮带轮上。若皮带与两皮带轮的两个切点之间相距

24

英寸，则两皮带轮圆心之间的距离（英寸）为：

An uncrossed belt is fitted without slack around two circular pulleys with radii of

14

inches and

4

inches. If the distance between the points of contact of the belt with the pulleys is

24

inches, then the distance between the centers of the pulleys in inches is:

24

2\sqrt{119}

25

26

4\sqrt{35}

连接两个圆心，并利用半径之差

Join the centers and use the difference of the radii

圆心距是两直角边为

24

和

10

的直角三角形的斜边

The center distance is the hypotenuse of a right triangle with legs

24

and

10

公外切线与连接圆心和切点的两条半径构成一个直角三角形，其两条直角边长为

24

和

14-4=10

。因此圆心距为

\sqrt{24^2+10^2}=\sqrt{676}=26\text{。}

因此，正确答案为 D。

The common external tangent and the radii to its contact points form a right triangle whose legs are

24

and

14-4=10.

Thus the center distance is

\sqrt{24^2+10^2}=\sqrt{676}=26.

Therefore, the correct answer is D.

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