1972 AMC 12 第 25 题

Question

一个圆内接四边形的四条边依次长为

25

，

39

，

52

，和

60

。该圆的直径长度为：

Inscribed in a circle is a quadrilateral having sides of lengths

25,

39,

52,

and

60

taken consecutively. The diameter of this circle has length:

62

63

65

66

69

在四条边长中找出两组勾股数的倍数

Notice two scaled Pythagorean triples among the four side lengths

25,60

这一对与

39,52

这一对对应同一条可能的斜边

The

25,60

pair and the

39,52

pair share the same possible hypotenuse

可将连续四边标记为

\begin{gathered} AB=25,\\ BC=39,\\ CD=52,\\ DA=60 \end{gathered}\text{。}

因为

25^2+60^2=65^2=39^2+52^2\text{，}

三角形

ABD

与

BCD

都是直角三角形，且有公共斜边

BD=65

。二者拼成这个圆内接四边形，而直角三角形的斜边是其外接圆的直径。因此直径为

65

。

因此，正确答案是 C。

The consecutive sides may be labeled

\begin{gathered} AB=25,\\ BC=39,\\ CD=52,\\ DA=60. \end{gathered}

Since

25^2+60^2=65^2=39^2+52^2,

triangles

ABD

and

BCD

are right triangles with common hypotenuse

BD=65.

Their union is the cyclic quadrilateral, and a right triangle’s hypotenuse is a diameter of its circumcircle. Thus the diameter is

65.

Therefore, the correct answer is C.

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