AMC12 2020 B

AMC12 2020 B · Q10

AMC12 2020 B · Q10. It mainly tests Triangles (properties), Coordinate geometry.

In unit square ABCD, the inscribed circle $\omega$ intersects CD at M, and AM intersects $\omega$ at a point P different from M. What is AP?

在单位正方形 ABCD 中，内接圆 $\omega$ 与 CD 交于 M，AM 交 $\omega$ 于不同于 M 的点 P。AP 长为？

(A) $\sqrt{5}/12$ $\sqrt{5}/12$

(B) $\sqrt{5}/10$ $\sqrt{5}/10$

(D) $\sqrt{5}/8$ $\sqrt{5}/8$

(E) $2\sqrt{5}/15$ $2\sqrt{5}/15$

Answer

Correct choice: (B)

正确答案：（B）

Solution

Let Q be the midpoint of $\overline{AB}$. By the Power of a Point Theorem, $AP \cdot AM = AQ^2 = \frac{1}{4}$. By the Pythagorean Theorem, $AM = \frac{1}{2}\sqrt{5}$. Thus $$AP = \frac{\frac{1}{4}}{\frac{1}{2}\sqrt{5}} = \frac{\sqrt{5}}{10}.$$

设 Q 为 $\overline{AB}$ 中点。由点到圆的幂定理，$AP \cdot AM = AQ^2 = \frac{1}{4}$。由勾股定理，$AM = \frac{1}{2}\sqrt{5}$。故 $$AP = \frac{\frac{1}{4}}{\frac{1}{2}\sqrt{5}} = \frac{\sqrt{5}}{10}.$$

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