AMC10 2023 A

AMC10 2023 A · Q22

AMC10 2023 A · Q22. It mainly tests Pythagorean theorem, Circle theorems.

Circle $C_1$ and $C_2$ each have radius $1$, and the distance between their centers is $\frac{1}{2}$. Circle $C_3$ is the largest circle internally tangent to both $C_1$ and $C_2$. Circle $C_4$ is internally tangent to both $C_1$ and $C_2$ and externally tangent to $C_3$. What is the radius of $C_4$?

圆 $C_1$ 和 $C_2$ 半径均为 $1$，两圆心距离为 $\frac{1}{2}$。圆 $C_3$ 是与 $C_1$ 和 $C_2$ 都内切的最大的圆。圆 $C_4$ 与 $C_1$ 和 $C_2$ 都内切，且与 $C_3$ 外切。求 $C_4$ 的半径。

(A) \frac{1}{14} \frac{1}{14}

(B) \frac{1}{12} \frac{1}{12}

(D) \frac{3}{28} \frac{3}{28}

(E) \frac{1}{9} \frac{1}{9}

Answer

Correct choice: (D)

正确答案：（D）

Solution

Let $O$ be the center of the midpoint of the line segment connecting both the centers, say $A$ and $B$. Let the point of tangency with the inscribed circle and the right larger circles be $T$. Then $OT = BO + BT = BO + AT - \frac{1}{2} = \frac{1}{4} + 1 - \frac{1}{2} = \frac{3}{4}.$ Since $C_4$ is internally tangent to $C_1$, center of $C_4$, $C_1$ and their tangent point must be on the same line. Now, if we connect centers of $C_4$, $C_3$ and $C_1$/$C_2$, we get a right angled triangle. Let the radius of $C_4$ equal $r$. With the pythagorean theorem on our triangle, we have \[\left(r+\frac{3}{4}\right)^2+\left(\frac{1}{4}\right)^2=(1-r)^2\] Solving this equation gives us \[r = \boxed{\textbf{(D) } \frac{3}{28}}\]

设 $O$ 是连接两圆心 $A$ 和 $B$ 的线段中点的中心。设内切圆与右侧大圆的切点为 $T$。则 $OT = BO + BT = BO + AT - \frac{1}{2} = \frac{1}{4} + 1 - \frac{1}{2} = \frac{3}{4}$。由于 $C_4$ 与 $C_1$ 内切，$C_4$ 的中心、$C_1$ 的中心及其切点必须共线。现在，连接 $C_4$、$C_3$ 和 $C_1$/$C_2$ 的中心，得到一个直角三角形。设 $C_4$ 的半径为 $r$。利用勾股定理，有 \[\left(r+\frac{3}{4}\right)^2+\left(\frac{1}{4}\right)^2=(1-r)^2\] 解此方程得到 \[r = \boxed{\textbf{(D) } \frac{3}{28}}\]

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