AMC12 2005 A

AMC12 2005 A · Q16

AMC12 2005 A · Q16. It mainly tests Pythagorean theorem, Coordinate geometry.

Three circles of radius $s$ are drawn in the first quadrant of the $xy$-plane. The first circle is tangent to both axes, the second is tangent to the first circle and the $x$-axis, and the third is tangent to the first circle and the $y$-axis. A circle of radius $r > s$ is tangent to both axes and to the second and third circles. What is $r/s$?

在 $xy$ 平面的第一象限中画三个半径为 $s$ 的圆。第一个圆与两条坐标轴都相切，第二个圆与第一个圆和 $x$ 轴相切，第三个圆与第一个圆和 $y$ 轴相切。一个半径为 $r > s$ 的圆与两条坐标轴以及第二、第三个圆都相切。求 $r/s$。

(A) 5 5

(B) 6 6

(D) 9 9

(E) 10 10

Answer

Correct choice: (D)

正确答案：（D）

Solution

Error creating thumbnail: Unable to save thumbnail to destination Set $s =1$ so that we only have to find $r$. Draw the segment between the center of the third circle and the large circle; this has length $r+1$. We then draw the radius of the large circle that is perpendicular to the x-axis, and draw the perpendicular from this radius to the center of the third circle. This gives us a right triangle with legs $r-3,r-1$ and hypotenuse $r+1$. The Pythagorean Theorem yields: Quite obviously $r > 1$, so $r = 9$.

Error creating thumbnail: Unable to save thumbnail to destination 令 $s =1$，这样我们只需要求 $r$。连接第三个圆的圆心与大圆的圆心，这条线段长度为 $r+1$。再作大圆上一条与 $x$ 轴垂直的半径，并从该半径向第三个圆的圆心作垂线。这样得到一个直角三角形，其两条直角边分别为 $r-3,r-1$，斜边为 $r+1$。由勾股定理得： Quite obviously $r > 1$, so $r = 9$.

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