AMC12 2003 A

AMC12 2003 A · Q17

AMC12 2003 A · Q17. It mainly tests Circle theorems, Coordinate geometry.

Square $ABCD$ has sides of length $4$, and $M$ is the midpoint of $\overline{CD}$. A circle with radius $2$ and center $M$ intersects a circle with radius $4$ and center $A$ at points $P$ and $D$. What is the distance from $P$ to $\overline{AD}$?

正方形 $ABCD$ 的边长为 $4$，$M$ 是 $\overline{CD}$ 的中点。以 $M$ 为圆心、半径为 $2$ 的圆与以 $A$ 为圆心、半径为 $4$ 的圆相交于点 $P$ 和 $D$。点 $P$ 到 $\overline{AD}$ 的距离是多少？

(A) 3 3

(B) \frac{16}{5} \frac{16}{5}

(D) 2\sqrt{3} 2\sqrt{3}

(E) \frac{7}{2} \frac{7}{2}

Answer

Correct choice: (B)

正确答案：（B）

Solution

$APMD$ obviously forms a kite. Let the intersection of the diagonals be $E$. $AE+EM=AM=2\sqrt{5}$ Let $AE=x$. Then, $EM=2\sqrt{5}-x$. By Pythagorean Theorem, $DE^2=4^2-AE^2=2^2-EM^2$. Thus, $16-x^2=4-(2\sqrt{5}-x)^2$. Simplifying, $x=\frac{8}{\sqrt{5}}$. By Pythagoras again, $DE=\frac{4}{\sqrt{5}}$. Then, the area of $ADP$ is $DE\cdot AE=\frac{32}{5}$. Using $4$ instead as the base, we can drop a altitude from P. $\frac{32}{5}=\frac{bh}{2}$. $\frac{32}{5}=\frac{4h}{2}$. Thus, the horizontal distance is $\frac{16}{5} \implies \boxed{\textbf{(B)}\frac{16}{5}}$

$APMD$ 显然构成一个风筝形。设两条对角线的交点为 $E$。$AE+EM=AM=2\sqrt{5}$。令 $AE=x$，则 $EM=2\sqrt{5}-x$。由勾股定理，$DE^2=4^2-AE^2=2^2-EM^2$。因此，$16-x^2=4-(2\sqrt{5}-x)^2$。化简得 $x=\frac{8}{\sqrt{5}}$。再由勾股定理，$DE=\frac{4}{\sqrt{5}}$。于是，$\triangle ADP$ 的面积为 $DE\cdot AE=\frac{32}{5}$。以 $4$ 为底，从 $P$ 作高，则 $\frac{32}{5}=\frac{bh}{2}$，即 $\frac{32}{5}=\frac{4h}{2}$。因此所求距离为 $\frac{16}{5} \implies \boxed{\textbf{(B)}\frac{16}{5}}$。

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