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2009 AMC 10B Problems/Problem 16

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Contents

1 Problem
2 Solution
- 2.1 Solution 1
- 2.2 Solution 2
3 See Also

Problem

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 $\frac{BD}{BO}$ ?

$\text{(A) } \frac {\sqrt2}{3}\qquad\text{(B) } \frac {1}{2}\qquad\text{(C) } \frac {\sqrt3}{3}\qquad\text{(D) } \frac {\sqrt2...$

Solution

Solution 1

As $\triangle ABC$ is equilateral, we have $\angle BAC = \angle BCA = 60^\circ$ , hence $\angle OAC = \angle OCA = 30^\circ$ . Then $\angla AOC = 120^\circ$ , and from symmetry we have $\angle AOB = \angle COB = 60^\circ$ . Finally this gives us $\angle ABO = \angle CBO = 30^\circ$ .

We know that $DO = AO$ , as $D$ lies on the circle. From $\triangle ABO$ we also have $AO = BO \sin 30^\circ = \frac{BO}2$ , Hence $DO = \frac{BO}2$ , therefore $BD = BO - DO = \frac{BO}2$ , and $\frac{BD}{BO} = \boxed{\frac 12}$ .

Solution 2

As in the previous solution, we find out that $\angle AOB = \angle COB = 60^\circ$ . Hence $\triangle AOD$ and $\triangle COD$ are both equilateral.

We then have $\angle SCD = \angle SAD = 30^\circ$ , hence $D$ is the incenter of $\triangle ABC$ , and as $\triangle ABC$ is equilateral, $D$ is also its centroid. Hence $2 \cdot SD = BD$ , and as $SD = SO$ , we have $2\cdot SD = SD + SO = OD$ , therefore $BD=OD$ , and as before we conclude that $\frac{BD}{BO} = \boxed{\frac 12}$ .

See Also

2009 AMC 10B (Problems • Resources)
Preceded by Problem 15	Followed by Problem 17
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25

Retrieved from "http://www.artofproblemsolving.com/Wiki/index.php/2009_AMC_10B_Problems/Problem_16"

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