2009 AMC 10B Problem 16

Below is the professionally curated solution for Problem 16 of the 2009 AMC 10B, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2009 AMC 10B solutions, or check the answer key.

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Concepts:tangent line special right triangle circle

Difficulty rating: 1420

16.

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 $\overline{BO}$ at $D.$ What is $\dfrac{BD}{BO}?$

$\dfrac{\sqrt2}{3}$

$\dfrac12$

$\dfrac{\sqrt3}{3}$

$\dfrac{\sqrt2}{2}$

$\dfrac{\sqrt3}{2}$

Solution:

Let the radius be $r.$ By symmetry $BO$ bisects the $60^\circ$ angle $ABC,$ so $\angle OBC=30^\circ.$ Since $OC\perp BC,$ triangle $BCO$ is a $30$ - $60$ - $90$ triangle with hypotenuse $BO=2\,OC=2r.$

Then $BD=BO-OD=2r-r=r,$ so $\dfrac{BD}{BO}=\dfrac{r}{2r}=\dfrac12.$

Thus, the correct answer is B.

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