Question

四边形 $ABCD$ 内接于圆，边长分别为 $AB = 7, BC = 24, CD = 20$ ， $DA = 15$ 。圆内部但四边形外部的面积可写成 $\dfrac{a \pi - b}{c}$ ，其中 $a, b$ 和 $c$ 为正整数，且 $a$ 与 $c$ 没有公共质因数。求 $a + b + c$ 。

Quadrilateral $ABCD$ with side lengths $AB = 7, BC = 24, CD = 20,$ $DA = 15$ is inscribed in a circle. The area interior to the circle but exterior to the quadrilateral can be written in the form $\dfrac{a \pi - b}{c},$ where $a, b,$ and $c$ are positive integers such that $a$ and $c$ have no common prime factor. What is $a + b + c?$

$260$

$855$

$1235$

$1565$

$1997$

答案：D

知识点：圆内接四边形勾股数圆面积

难度评级：1950

解答：

注意到 $7^2 + 24^2$ 与 $15^2 + 20^2$ 相等。这迫使 $AC = 25$ ，否则 $\angle B$ 与 $\angle D$ 会同时为锐角或同时为钝角，与它们的和为 $180^{\circ}.$ 矛盾。

又因为 $\angle B$ 是直角，所以 $AC$ 是圆的直径。圆的面积为 $\dfrac{625}{4} \pi.$

四边形的面积等于两个三角形的面积之和，即 $\dfrac{1}{2}(7 \cdot 24 + 20 \cdot 15) =$ $84 + 150 = 234.$

圆内而四边形外的面积为 $\dfrac{625}{4} \pi - 234 = \dfrac{625 \pi - 936}{4}.$

因此 $a + b + c = 625 + 936 + 4$ $= 1565.$

所以正确答案是 D。

Notice that $7^2 + 24^2$ and $15^2 + 20^2$ are both the same. This forces $AC = 25$ since otherwise $\angle B$ and $\angle D$ would both be acute or obtuse, violating the fact that their sum is $180^{\circ}.$

Also since $\angle B$ is right, we know that $AC$ is the diameter of the circle. The area of the circle is then $\dfrac{625}{4} \pi.$

To find the area of the quadrilateral, we can find the area of each of the triangles, which is $\dfrac{1}{2}(7 \cdot 24 + 20 \cdot 15) =$ $84 + 150 = 234.$

To find the area outside the quadrilateral, we subtract to get $\dfrac{625}{4} \pi - 234 = \dfrac{625 \pi - 936}{4}.$

Therefore, $a + b + c = 625 + 936 + 4$ $= 1565.$

Thus, D is the correct answer.

2022 AMC 10A 第 15 题

答案：D

难度评级：1950

解答：

其他年份的第 15 题