AMC10 2022 A

AMC10 2022 A · Q15

AMC10 2022 A · Q15. It mainly tests Circle theorems, Area & perimeter.

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 $\frac{a\pi-b}{c},$ where $a,b,$ and $c$ are positive integers such that $a$ and $c$ have no common prime factor. What is $a+b+c?$

边长 $AB=7$，$BC=24$，$CD=20$，$DA=15$ 的四边形 $ABCD$ 内接于一个圆。圆内四边形外的面积可写成 $\frac{a\pi-b}{c}$ 的形式，其中 $a,b,c$ 为正整数，且 $a$ 与 $c$ 无公质因数。求 $a+b+c$？

(A) 260 260

(B) 855 855

(D) 1565 1565

(E) 1997 1997

Answer

Correct choice: (D)

正确答案：（D）

Solution

Opposite angles of every cyclic quadrilateral are supplementary, so \[\angle B + \angle D = 180^{\circ}.\] We claim that $AC=25.$ We can prove it by contradiction: - If $AC<25,$ then $\angle B$ and $\angle D$ are both acute angles. This arrives at a contradiction. - If $AC>25,$ then $\angle B$ and $\angle D$ are both obtuse angles. This arrives at a contradiction. By the Inscribed Angle Theorem, we conclude that $\overline{AC}$ is the diameter of the circle. So, the radius of the circle is $r=\frac{AC}{2}=\frac{25}{2}.$ The area of the requested region is \[\pi r^2 - \frac12\cdot AB\cdot BC - \frac12\cdot AD\cdot DC = \frac{625\pi}{4}-\frac{168}{2}-\frac{300}{2}=\frac{625\pi-936}{4}.\] Therefore, the answer is $a+b+c=\boxed{\textbf{(D) } 1565}.$

任意内接四边形的对角互补，即 $\angle B + \angle D = 180^{\circ}$。我们断言 $AC=25$。可用反证法证明： - 若 $AC<25$，则 $\angle B$ 和 $\angle D$ 均为锐角。矛盾。 - 若 $AC>25$，则 $\angle B$ 和 $\angle D$ 均为钝角。矛盾。由内接角定理，得出 $\overline{AC}$ 为圆的直径。故圆半径 $r=\frac{AC}{2}=\frac{25}{2}$。所求区域面积为 \[\pi r^2 - \frac12\cdot AB\cdot BC - \frac12\cdot AD\cdot DC = \frac{625\pi}{4}-\frac{168}{2}-\frac{300}{2}=\frac{625\pi-936}{4}.\] 因此答案为 $a+b+c=\boxed{\textbf{(D) } 1565}$。

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