AMC12 2000 A

AMC12 2000 A · Q24

AMC12 2000 A · Q24. It mainly tests Circle theorems, Geometry misc.

If circular arcs $\widehat{AC}$ and $\widehat{BC}$ have centers at $B$ and $A$, respectively, then there exists a circle tangent to both $\widehat{AC}$ and $\widehat{BC}$, and to $\overline{AB}$. If the length of $\overline{BC}$ is $12$, then the circumference of the circle is

如果圆弧 $\widehat{AC}$ 和 $\widehat{BC}$ 的圆心分别在 $B$ 和 $A$，那么存在一个圆同时与圆弧 $\widehat{AC}$、圆弧 $\widehat{BC}$ 以及线段 $\overline{AB}$ 相切。若线段 $\overline{BC}$ 的长度为 $12$，则该圆的周长为（）

(A) 24 24

(B) 25 25

(D) 27 27

(E) 28 28

Answer

Correct choice: (D)

正确答案：（D）

Solution

First, note the triangle $ABC$ is equilateral. Next, notice that since the arc $BC$ has length 12, it follows that we can find the radius of the sector centered at $A$. $\frac {1}{6}({2}{\pi})AB=12 \implies AB=\frac{36}{\pi}$. Next, connect the center of the circle to side $AB$, and call this length $r$, and call the foot $M$. Since $ABC$ is equilateral, it follows that $MB=\frac{18}{\pi}$, and $OA$ (where $O$ is the center of the circle) is $\frac{36}{\pi}-r$. By the Pythagorean Theorem, you get $r^2+\left(\frac{18}{\pi}\right)^2=\left(\frac{36}{\pi}-r\right)^2 \implies r=\frac{27}{2\pi}$. Finally, we see that the circumference is $2{\pi}\cdot \frac{27}{2\pi}=\boxed{(D)27}$.

首先注意到三角形 $ABC$ 是等边三角形。接着注意到，由于弧 $BC$ 的长度为 12，我们可以求出以 $A$ 为圆心的扇形的半径。$\frac {1}{6}({2}{\pi})AB=12 \implies AB=\frac{36}{\pi}$。接下来，连接该圆的圆心到边 $AB$，记这段长度为 $r$，垂足记为 $M$。由于 $ABC$ 为等边三角形，可得 $MB=\frac{18}{\pi}$，且 $OA$（其中 $O$ 为该圆的圆心）为 $\frac{36}{\pi}-r$。由勾股定理，得到 $r^2+\left(\frac{18}{\pi}\right)^2=\left(\frac{36}{\pi}-r\right)^2 \implies r=\frac{27}{2\pi}$。最后，该圆的周长为 $2{\pi}\cdot \frac{27}{2\pi}=\boxed{(D)27}$。

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