2009 AMC 10B Problem 16

Problem 16 of 25IntermediateGeometry

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}?

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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.

More practice

Concepts: tangent line · special right triangle · circle

Problem text and solution from the LIVE past-contest archive. See also the AoPS wiki page for community solutions.