2010 AMC 10B Problem 19

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.