2010 AMC 10B Problem 19

Question

A circle with center

O

has area

156\pi.

Triangle

ABC

is equilateral,

\overline{BC}

is a chord on the circle,

OA = 4\sqrt{3},

and point

O

is outside

\triangle ABC.

What is the side length of

\triangle ABC?

2\sqrt{3}

6

4\sqrt{3}

12

18

Let

s

be the side length and drop the altitude from

A

Use the circle radius

\sqrt{156}

and the equilateral triangle altitude

Consider the following diagram:

Using the formula for the area of a circle, we have that

BO = \sqrt{156}

since it is a radius.

Extend

\overline{AO}

to intersect

\overline{BC}

at

X.

Let

s

be the side length of

\triangle ABC.

Then we have that

BX = \dfrac{s}{2}

and

AX = \dfrac{s\sqrt3}{2}.

We have that

\triangle OXB

is right, which means that we can apply the Pythagorean Theorem. This gives us

(\sqrt{156})^2 = \left(\dfrac{s}{2}\right)^2

+ \left(\dfrac{s\sqrt3}{2} + 4\sqrt3\right)^2.

Simplifying, we get

156 = s^2 + 12s + 48

s^2 + 12s - 108 = 0

(s - 6)(s + 18) = 0.

Since

s

is positive, we must have that

s = 6.

Thus, B is the correct answer.

Problem 19 in other years