AMC12 2017 A

AMC12 2017 A · Q16

AMC12 2017 A · Q16. It mainly tests Circle theorems, Coordinate geometry.

In the figure below, semicircles with centers at $A$ and $B$ and with radii 2 and 1, respectively, are drawn in the interior of, and sharing bases with, a semicircle with diameter $JK$. The two smaller semicircles are externally tangent to each other and internally tangent to the largest semicircle. A circle centered at $P$ is drawn externally tangent to the two smaller semicircles and internally tangent to the largest semicircle. What is the radius of the circle centered at $P$?

下图中，直径为 $JK$ 的大半圆内部绘制了以 $A$ 和 $B$ 为圆心、半径分别为 2 和 1 的半圆，且它们与大半圆共用底边。两个小半圆外切于彼此，并内切于最大半圆。以 $P$ 为圆心的一个圆外切于两个小半圆，并内切于最大半圆。以 $P$ 为圆心的圆的半径是多少？

(A) \frac{3}{4} \frac{3}{4}

(B) \frac{6}{7} \frac{6}{7}

(D) \frac{5}{8}\sqrt{2} \frac{5}{8}\sqrt{2}

(E) \frac{11}{12} \frac{11}{12}

Answer

Correct choice: (B)

正确答案：（B）

Solution

Answer (B): Let $C$ be the center of the largest semicircle, and let $r$ denote the radius of the circle centered at $P$. Note that $PA=2+r$, $PC=3-r$, $PB=1+r$, $AC=1$, $BC=2$, and $AB=3$. Let $F$ be the foot of the perpendicular from $P$ to $\overline{AB}$, let $h=PF$, and let $x=CF$. The Pythagorean Theorem in $\triangle PAF$, $\triangle PCF$, and $\triangle PBF$ gives $$ h^2=(2+r)^2-(1+x)^2=(3-r)^2-x^2=(1+r)^2-(2-x)^2. $$ This reduces to two linear equations in $r$ and $x$, whose solution is $r=\frac{6}{7}$, $x=\frac{9}{7}$.

答案（B）：设 $C$ 为最大半圆的圆心，设 $r$ 表示以 $P$ 为圆心的圆的半径。注意 $PA=2+r$，$PC=3-r$，$PB=1+r$，$AC=1$，$BC=2$，以及 $AB=3$。设 $F$ 为从 $P$ 到 $\overline{AB}$ 的垂足，令 $h=PF$，并令 $x=CF$。在 $\triangle PAF$、$\triangle PCF$ 与 $\triangle PBF$ 中使用勾股定理可得 $$ h^2=(2+r)^2-(1+x)^2=(3-r)^2-x^2=(1+r)^2-(2-x)^2. $$ 这可化为关于 $r$ 与 $x$ 的两个线性方程，其解为 $r=\frac{6}{7}$，$x=\frac{9}{7}$。

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