Question

一个圆外接于边长为 $20, 21$ 和 $29$ 的三角形，从而将圆内部划分为四个区域。令 $A$ 、 $B$ 、 $C$ 为三个非三角形区域的面积，其中 $C$ 最大。则

A circle is circumscribed about a triangle with sides $20, 21,$ and $29,$ thus dividing the interior of the circle into four regions. Let $A,$ $B,$ and $C$ be the areas of the non-triangular regions, with $C$ being the largest. Then

$A + B = C$

$A + B + 210 = C$

$A^2 + B^2 = C^2$

$20A + 21B = 29C$

$\dfrac{1}{A^2} + \dfrac{1}{B^2} = \dfrac{1}{C^2}$

答案：B

知识点：直角三角形圆面积面积分割

难度评级：1810

解答：

因为 $20^2 + 21^2 = 841 = 29^2$ ，该三角形为直角三角形，斜边 $29$ 是外接圆直径。因此最大区域 $C$ 是该直径一侧的半圆。

另一半圆由三角形以及区域 $A$ 、 $B$ 组成。两个半圆面积相等，而三角形面积为 $\tfrac12 \cdot 20 \cdot 21 = 210$ ，所以 $A + B + 210 = C.$

所以正确答案是 B。

Since $20^2 + 21^2 = 841 = 29^2,$ the triangle is right-angled, and its hypotenuse of length $29$ is a diameter of the circle. Thus the largest region $C$ is the semicircle on one side of that diameter.

The other semicircle consists of the triangle together with regions $A$ and $B.$ Since the two semicircles are congruent and the triangle has area $\tfrac12 \cdot 20 \cdot 21 = 210,$ we get $A + B + 210 = C.$

Thus, the correct answer is B.

1999 AMC 12 第 21 题

答案：B

难度评级：1810

解答：

其他年份的第 21 题