AMC10 2004 A

AMC10 2004 A · Q23

AMC10 2004 A · Q23. It mainly tests Pythagorean theorem, 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、C两两外部相切，且都与圆D内部相切。圆B与圆C全等。圆A半径为1且经过圆D的圆心。求圆B的半径。

(A) 2/3 $\frac{2}{3}$

(B) $\sqrt{3}/2$ $\frac{\sqrt{3}}{2}$

(D) 8/9 $\frac{8}{9}$

(E) $(1 + \sqrt{3})/3$ $\frac{1 + \sqrt{3}}{3}$

Answer

Correct choice: (D)

正确答案：（D）

Solution

(D) Let $E$, $H$, and $F$ be the centers of circles $A$, $B$, and $D$ respectively, and let $G$ be the point of tangency of circles $B$ and $C$. Let $x=FG$ and $y=GH$. Since the center of circle $D$ lies on circle $A$ and the circles have a common point of tangency, the radius of circle $D$ is $2$, which is the diameter of circle $A$. Applying the Pythagorean Theorem to right triangles $EGH$ and $FGH$ gives $$(1+y)^2=(1+x)^2+y^2 \quad \text{and} \quad (2-y)^2=x^2+y^2,$$ from which it follows that $$y=x+\frac{x^2}{2} \quad \text{and} \quad y=1-\frac{x^2}{4}.$$ The solutions of this system are $(x,y)=(2/3,8/9)$ and $(x,y)=(-2,0)$. The radius of circle $B$ is the positive solution for $y$, which is $8/9$.

（D）设 $E$、$H$、$F$ 分别为圆 $A$、圆 $B$、圆 $D$ 的圆心，并设 $G$ 为圆 $B$ 与圆 $C$ 的切点。令 $x=FG$，$y=GH$。由于圆 $D$ 的圆心在圆 $A$ 上，且两圆有一个公共切点，所以圆 $D$ 的半径为 $2$，这等于圆 $A$ 的直径。对直角三角形 $EGH$ 与 $FGH$ 应用勾股定理得 $$(1+y)^2=(1+x)^2+y^2 \quad \text{且} \quad (2-y)^2=x^2+y^2,$$ 由此推出 $$y=x+\frac{x^2}{2} \quad \text{且} \quad y=1-\frac{x^2}{4}.$$ 该方程组的解为 $(x,y)=(2/3,8/9)$ 和 $(x,y)=(-2,0)$。圆 $B$ 的半径取 $y$ 的正解，即 $8/9$。

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