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2009 AMC 10B Problem 16

2009 AMC 10B Problem 16AMC 10 Step by Step

Name: Date: September 13, 2026

Circles · difficulty 2 of 5 · about 3 min · position P16-20

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 $\frac{BD}{BO}$ ?

A)
$\frac {\sqrt2}{3}$
B)
$\frac {1}{2}$
C)
$\frac {\sqrt3}{3}$
D)
$\frac {\sqrt2}{2}$
E)
$\frac {\sqrt3}{2}$

Answer

Key insight

BO bisects the 60-degree angle at B and OA is perpendicular to the tangent, so triangle OAB is 30-60-90: BO = 2r and BD = r.

Solution

Draw the radius $OA$ . Because $BA$ is tangent at $A$ , $OA \perp BA$ , so triangle $OAB$ has a right angle at $A$ .

The two tangent segments from $B$ are symmetric about the line $BO$ , so $BO$ bisects $\angle ABC = 60^\circ$ and $\angle ABO = 30^\circ$ . Triangle $OAB$ is therefore a $30$ - $60$ - $90$ triangle with the short leg $OA = r$ opposite the $30^\circ$ angle, which makes the hypotenuse $BO = 2r$ .

Point $D$ lies on the circle and on segment $BO$ , so $OD = r$ and $BD = BO - OD = 2r - r = r$ . Hence
$\frac{BD}{BO} = \frac{r}{2r} = \frac12.$

The answer is $\boxed{\textbf{(B)}\ \frac12}$ .

Why this works

Two facts about tangents do all the work: the radius to the point of tangency is perpendicular to the tangent, and the two tangents from an external point make equal angles with the line to the center. Together they produce a right triangle with a known angle, and then $D$ is located simply by subtracting the radius.

The trap

Assuming D is the midpoint of BO by symmetry without reason, or using the 60-degree angle at B instead of the 30-degree half-angle.

Common mistakes

Assuming D is the midpoint of BO by symmetry without reason, or using the 60-degree angle at B instead of the 30-degree half-angle.
Confusing $D$ with the point where the circle meets $AC$ , or measuring $OD$ along $OA$ instead of along $BO$ .

Techniques

Add construction lines/points (drop altitudes, extend segments, connect centers) · Set up the equation/formula and compute; no special trick needed