AMC12 2004 A

AMC12 2004 A · Q19

AMC12 2004 A · Q19. It mainly tests Triangles (properties), Circle theorems.

Circles $A, B$ and $C$ are externally tangent to each other, and internally tangent to circle $D$. Circles $B$ and $C$ are congruent. Circle $A$ has radius $1$ and passes through the center of $D$. What is the radius of circle $B$?

圆 $A, B$ 和 $C$ 两两外切，并且都与圆 $D$ 内切。圆 $B$ 和 $C$ 全等。圆 $A$ 的半径为 $1$，且经过圆 $D$ 的圆心。圆 $B$ 的半径是多少？

(A) $\dfrac{2}{3}$ $\dfrac{2}{3}$

(B) $\dfrac{\sqrt{3}}{2}$ $\dfrac{\sqrt{3}}{2}$

(D) $\dfrac{8}{9}$ $\dfrac{8}{9}$

(E) $1 + \dfrac{\sqrt{3}}{3}$ $1 + \dfrac{\sqrt{3}}{3}$

Answer

Correct choice: (D)

正确答案：（D）

Solution

Let $O_{i}$ be the center of circle $i$ for all $i \in \{A,B,C,D\}$ and let $E$ be the tangent point of $B,C$. Since the radius of $D$ is the diameter of $A$, the radius of $D$ is $2$. Let the radius of $B,C$ be $r$ and let $O_{D}E = x$. If we connect $O_{A},O_{B},O_{C}$, we get an isosceles triangle with lengths $1 + r, 2r$. Then right triangle $O_{D}O_{B}O_{E}$ has legs $r, x$ and hypotenuse $2-r$. Solving for $x$, we get $x^2 = (2-r)^2 - r^2 \Longrightarrow x = \sqrt{4-4r}$. Also, right triangle $O_{A}O_{B}O_{E}$ has legs $r, 1+x$, and hypotenuse $1+r$. Solving, \begin{eqnarray*} r^2 + (1+\sqrt{4-4r})^2 &=& (1+r)^2\\ 1+4-4r+2\sqrt{4-4r}&=& 2r + 1\\ 1-r &=& \left(\frac{6r-4}{4}\right)^2\\ \frac{9}{4}r^2-2r&=& 0\\ r &=& \frac 89 \end{eqnarray*} So the answer is $\boxed{\mathrm{(D)}\ \frac{8}{9}}$.

对所有 $i \in \{A,B,C,D\}$，设 $O_{i}$ 为圆 $i$ 的圆心，并设 $E$ 为 $B,C$ 的切点。由于 $D$ 的半径是 $A$ 的直径，所以 $D$ 的半径为 $2$。设 $B,C$ 的半径为 $r$，并令 $O_{D}E = x$。连接 $O_{A},O_{B},O_{C}$，得到一个等腰三角形，其边长为 $1 + r, 2r$。则直角三角形 $O_{D}O_{B}O_{E}$ 的两直角边为 $r, x$，斜边为 $2-r$。解得 $x$： $x^2 = (2-r)^2 - r^2 \Longrightarrow x = \sqrt{4-4r}$。另外，直角三角形 $O_{A}O_{B}O_{E}$ 的两直角边为 $r, 1+x$，斜边为 $1+r$。解得 \begin{eqnarray*} r^2 + (1+\sqrt{4-4r})^2 &=& (1+r)^2\\ 1+4-4r+2\sqrt{4-4r}&=& 2r + 1\\ 1-r &=& \left(\frac{6r-4}{4}\right)^2\\ \frac{9}{4}r^2-2r&=& 0\\ r &=& \frac 89 \end{eqnarray*} 所以答案是 $\boxed{\mathrm{(D)}\ \frac{8}{9}}$。

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