AMC10 2018 A

AMC10 2018 A · Q15

AMC10 2018 A · Q15. It mainly tests Similarity, Circle theorems.

Two circles of radius 5 are externally tangent to each other and are internally tangent to a circle of radius 13 at points A and B, as shown in the diagram. The distance AB can be written in the form $\frac{m}{n}$, where m and n are relatively prime positive integers. What is m + n?

如图所示，两个半径为5的圆外部相切，并且分别与半径为13的大圆在点A和B处内部相切。AB的距离可以写成 $\frac{m}{n}$ 的形式，其中m和n互质，问m + n？

(A) 21 21

(B) 29 29

(D) 69 69

(E) 93 93

Answer

Correct choice: (D)

正确答案：（D）

Solution

Answer (D): Let $C$ be the center of the larger circle, and let $D$ and $E$ be the centers of the two smaller circles, as shown. Points $C$, $D$, and $A$ are collinear because the radii are perpendicular to the common tangent at the point of tangency, and so are $C$, $E$, and $B$. These points form two isosceles triangles that share a vertex angle. Thus $\triangle CAB \sim \triangle CDE$, and therefore $\dfrac{AB}{DE}=\dfrac{CA}{CD}$, so $$ AB=\dfrac{DE\cdot CA}{CD}=\dfrac{(5+5)\cdot 13}{13-5}=\dfrac{65}{4}, $$ and the requested sum is $65+4=69$.

答案（D）：设 $C$ 为大圆的圆心，$D$ 和 $E$ 为两个小圆的圆心，如图所示。点 $C$、$D$、$A$ 共线，因为半径在切点处与公共切线垂直；同理，$C$、$E$、$B$ 也共线。这些点构成两个共享一个顶角的等腰三角形。因此 $\triangle CAB \sim \triangle CDE$，从而 $\dfrac{AB}{DE}=\dfrac{CA}{CD}$，所以 $$ AB=\dfrac{DE\cdot CA}{CD}=\dfrac{(5+5)\cdot 13}{13-5}=\dfrac{65}{4}, $$ 所求的和为 $65+4=69$。

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