2009 AMC 10B Problem 16

Attempt Problem 16 of the 2009 AMC 10B below, then check your answer against the professionally curated solution 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.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

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}

Answer: B

Concepts:tangent line special right triangle circle

Difficulty rating: 1420

Small hint:

Equilateral

\triangle ABC

gives

\angle OBC=30^\circ,

and

OC\perp BC

Big hint:

In a

30

-

60

-

90

triangle the hypotenuse

BO=2\,OC

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.