AMC10 2009 B

AMC10 2009 B · Q16

AMC10 2009 B · Q16. It mainly tests Triangles (properties), Circle theorems.

Points A and C lie on a circle centered at O, each of $\overline{BA}$ and $\overline{BC}$ are tangent to the circle, and $\triangle ABC$ is equilateral. The circle intersects BO at D. What is BD/BO?

点 A 和 C 位于以 O 为圆心的圆上，$\overline{BA}$ 和 $\overline{BC}$ 各切于该圆，且 $\triangle ABC$ 是等边三角形。该圆与 BO 相交于 D。求 BD/BO？

(A) $\frac{\sqrt{2}}{3}$ $\frac{\sqrt{2}}{3}$

(B) $\frac{1}{2}$ $\frac{1}{2}$

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

(E) $\frac{\sqrt{3}}{2}$ $\frac{\sqrt{3}}{2}$

Answer

Correct choice: (B)

正确答案：（B）

Solution

Answer (B): Let the radius of the circle be $r$. Because $\triangle BCO$ is a right triangle with a $30^\circ$ angle at $B$, the hypotenuse $BO$ is twice as long as $OC$, so $BO=2r$. It follows that $BD=2r-r=r$, and \[ \frac{BD}{BO}=\frac{r}{2r}=\frac{1}{2}. \]

答案（B）：设圆的半径为$r$。因为$\triangle BCO$是一直角三角形，且在$B$处的角为$30^\circ$，所以斜边$BO$是$OC$的两倍，因此$BO=2r$。由此可得$BD=2r-r=r$，并且 \[ \frac{BD}{BO}=\frac{r}{2r}=\frac{1}{2}. \]

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