AMC12 2002 A

AMC12 2002 A · Q18

AMC12 2002 A · Q18. It mainly tests Circle theorems, Coordinate geometry.

Let $C_1$ and $C_2$ be circles defined by $(x-10)^2 + y^2 = 36$ and $(x+15)^2 + y^2 = 81$ respectively. What is the length of the shortest line segment $PQ$ that is tangent to $C_1$ at $P$ and to $C_2$ at $Q$?

设圆 $C_1$ 和 $C_2$ 分别由 $(x-10)^2 + y^2 = 36$ 和 $(x+15)^2 + y^2 = 81$ 定义。与 $C_1$ 在点 $P$ 相切且与 $C_2$ 在点 $Q$ 相切的最短线段 $PQ$ 的长度是多少？

(A) 15 15

(B) 18 18

(D) 21 21

(E) 24 24

Answer

Correct choice: (C)

正确答案：（C）

Solution

First examine the formula $(x-10)^2+y^2=36$, for the circle $C_1$. Its center, $D_1$, is located at (10,0) and it has a radius of $\sqrt{36}$ = 6. The next circle, using the same pattern, has its center, $D_2$, at (-15,0) and has a radius of $\sqrt{81}$ = 9. So we can construct this diagram: Line PQ is tangent to both circles, so it forms a right angle with the radii (6 and 9). This, as well as the two vertical angles near O, prove triangles $D_2QO$ and $D_1PO$ similar by AA, with a scale factor of $6:9$, or $2:3$. Next, we must subdivide the line $D_2D_1$ in a 2:3 ratio to get the length of the segments $D_2O$ and $D_1O$. The total length is $10 - (-15)$, or $25$, so applying the ratio, $D_2O$ = 15 and $D_1O$ = 10. These are the hypotenuses of the triangles. We already know the length of $D_2Q$ and $D_1P$, 9 and 6 (they're radii). So in order to find $PQ$, we must find the length of the longer legs of the two triangles and add them. $15^2 - 9^2 = (15-9)(15+9) = 6 \times 24 = 144$ $\sqrt{144} = 12$ $10^2-6^2 = (10-6)(10+6) = 4 \times 16 = 64$ $\sqrt{64} = 8$ Finally, the length of PQ is $12+8=\boxed{20}$, or (C).

先看圆 $C_1$ 的方程 $(x-10)^2+y^2=36$。其圆心 $D_1$ 在 $(10,0)$，半径为 $\sqrt{36}=6$。同理，另一个圆的圆心 $D_2$ 在 $(-15,0)$，半径为 $\sqrt{81}=9$。直线 $PQ$ 与两圆都相切，因此它与两条半径（6 和 9）都成直角。再结合点 $O$ 附近的两对对顶角，可知三角形 $D_2QO$ 与 $D_1PO$ 由 AA 相似，相似比为 $6:9=2:3$。接下来，将线段 $D_2D_1$ 按 $2:3$ 分点以得到 $D_2O$ 与 $D_1O$ 的长度。总长度为 $10-(-15)=25$，按比例得 $D_2O=15$，$D_1O=10$。它们分别是两个直角三角形的斜边。又已知 $D_2Q=9$、$D_1P=6$（半径）。因此要找 $PQ$，只需分别求出两三角形较长的直角边并相加。 $15^2 - 9^2 = (15-9)(15+9) = 6 \times 24 = 144$ $\sqrt{144} = 12$ $10^2-6^2 = (10-6)(10+6) = 4 \times 16 = 64$ $\sqrt{64} = 8$ 所以 $PQ=12+8=\boxed{20}$，即 (C)。

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