AMC10 2006 A

AMC10 2006 A · Q16

AMC10 2006 A · Q16. It mainly tests Triangles (properties), Circle theorems.

A circle of radius 1 is tangent to a circle of radius 2. The sides of $\triangle ABC$ are tangent to the circles as shown, and the sides $AB$ and $AC$ are congruent. What is the area of $\triangle ABC$?

一个半径为 1 的圆与一个半径为 2 的圆相切。$ riangle ABC$ 的三边如图所示与这些圆相切，且边 $AB$ 和 $AC$ 相等。求 $ riangle ABC$ 的面积。

(A) $\frac{35}{2}$ $\frac{35}{2}$

(B) $15\sqrt{2}$ $15\sqrt{2}$

(D) $16\sqrt{2}$ $16\sqrt{2}$

(E) 24 24

Answer

Correct choice: (D)

正确答案：（D）

Solution

(D) Let $O$ and $O'$ denote the centers of the smaller and larger circles, respectively. Let $D$ and $D'$ be the points on $\overline{AC}$ that are also on the smaller and larger circles, respectively. Since $\triangle ADO$ and $\triangle AD'O'$ are similar right triangles, we have $$ \frac{AO}{1}=\frac{AO'}{2}=\frac{AO+3}{2}, $$ so $AO=3$. As a consequence, $$ AD=\sqrt{AO^2-OD^2}=\sqrt{9-1}=2\sqrt{2}. $$ Let $F$ be the midpoint of $\overline{BC}$. Since $\triangle ADO$ and $\triangle AFC$ are similar right triangles, we have $$ \frac{FC}{1}=\frac{AF}{AD}=\frac{AO+OO'+O'F}{AD}=\frac{3+3+2}{2\sqrt2}=2\sqrt2. $$ So the area of $\triangle ABC$ is $$ \frac12\cdot BC\cdot AF=\frac12\cdot 4\sqrt2\cdot 8=16\sqrt2. $$

（D）设 $O$ 与 $O'$ 分别为小圆与大圆的圆心。设 $D$ 与 $D'$ 为线段 $\overline{AC}$ 上分别与小圆、大圆的交点。由于 $\triangle ADO$ 与 $\triangle AD'O'$ 为相似的直角三角形，我们有 $$ \frac{AO}{1}=\frac{AO'}{2}=\frac{AO+3}{2}, $$ 因此 $AO=3$。因此， $$ AD=\sqrt{AO^2-OD^2}=\sqrt{9-1}=2\sqrt{2}. $$ 设 $F$ 为 $\overline{BC}$ 的中点。由于 $\triangle ADO$ 与 $\triangle AFC$ 为相似的直角三角形，我们有 $$ \frac{FC}{1}=\frac{AF}{AD}=\frac{AO+OO'+O'F}{AD}=\frac{3+3+2}{2\sqrt2}=2\sqrt2. $$ 所以 $\triangle ABC$ 的面积为 $$ \frac12\cdot BC\cdot AF=\frac12\cdot 4\sqrt2\cdot 8=16\sqrt2. $$

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