2010 AMC 10B Problem 19

The radius satisfies $\pi r^2 = 156\pi$, so $r^2 = 156$.

Let $s$ be the side length and $M$ the midpoint of $BC$. The center $O$ lies on the perpendicular bisector of the chord $BC$, and so does the apex $A$ of the equilateral triangle. Hence $O$, $A$, $M$ are collinear, with $AM = \frac{s\sqrt3}{2}$ and $MB = \frac{s}{2}$.

Since $O$ is outside the triangle and on line $AM$, either $O$ is beyond $A$ (so $OM = OA + AM$) or $O$ is on the other side of $BC$ (so $OM = OA - AM$, requiring $AM < 4\sqrt3$, i.e. $s < 8$).

Take the first case. In right triangle $OMB$:
$$\left(4\sqrt3 + \frac{s\sqrt3}{2}\right)^2 + \left(\frac{s}{2}\right)^2 = 156.$$
Expanding: $48 + 12s + \frac{3s^2}{4} + \frac{s^2}{4} = 156$, so $s^2 + 12s - 108 = 0$, i.e. $(s + 18)(s - 6) = 0$ and $s = 6$.

The second case gives $s^2 - 12s - 108 = 0$, whose positive root $s = 18$ violates $s < 8$, so it is impossible.

The answer is $\boxed{\textbf{(B)}\ 6}$.

Chord problems almost always start with the perpendicular from the center to the chord; here that line also carries the triangle's apex, collapsing the picture to one right triangle. The phrase "O is outside the triangle" is not decoration: it decides whether distances add or subtract along the line.

Coordinates: put $M$ at the origin, $B = (3, 0)$ for $s = 6$, $A = (0, 3\sqrt3)$, $O = (0, 7\sqrt3)$. Then $OB^2 = 9 + 147 = 156$, confirming the choice; testing each answer this way is quick.

• Placing O on the far side of BC (OM = OA - AM), which gives s = 18 but is inconsistent because AM would exceed OA.
• Assuming $A$ lies on the circle or that $OA$ is a radius, which contradicts $OA = 4\sqrt3 \ne \sqrt{156}$.