2010 AMC 10B Problem 19

1. Radius. The area is $\pi r^2 = 156\pi$, so $r^2 = 156$. Since $B$ is on the circle, $OB^2 = 156$.
2. Find a line that holds everything. Let $M$ be the midpoint of $BC$, and let $s$ be the side of the triangle. The points that are equally far from $B$ and from $C$ form the line through $M$ perpendicular to $BC$.
- $O$ is equally far from $B$ and $C$ (both are radii), so $O$ is on this line.
- $A$ is equally far from $B$ and $C$ (the triangle is equilateral), so $A$ is on this line too.
So $O$, $A$ and $M$ all lie on one line, which meets $BC$ at a right angle.
3. Lengths along the line. $MB = \frac{s}{2}$, and $AM$ is the height of the equilateral triangle, $AM = \frac{s\sqrt3}{2}$.
4. Use "O is outside". Picture the line: $M$ is on $BC$, $A$ is above it. If $O$ were between $M$ and $A$, it would be inside the triangle. So $O$ is beyond $A$, and
$$OM = OA + AM = 4\sqrt3 + \frac{s\sqrt3}{2}.$$
5. Pythagoras in the right triangle $OMB$: $OM^2 + MB^2 = OB^2$, so
$$\left(4\sqrt3 + \frac{s\sqrt3}{2}\right)^2 + \left(\frac{s}{2}\right)^2 = 156.$$
6. Expand the first square: $48 + 12s + \frac{3s^2}{4}$. Adding $\frac{s^2}{4}$ gives
$$48 + 12s + s^2 = 156 \quad\Longrightarrow\quad s^2 + 12s - 108 = 0 \quad\Longrightarrow\quad (s + 18)(s - 6) = 0.$$
A side length is positive, so $s = 6$.
7. Check: $AM = 3\sqrt3$, so $OM = 7\sqrt3$ and $OM^2 = 147$. With $MB = 3$, $147 + 9 = 156$. It works.

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

With a chord of a circle, the first line to draw goes from the center to the chord's midpoint; it meets the chord at a right angle. Here the triangle's top corner lies on that same line, so the whole picture turns into one right triangle. The phrase "$O$ is outside the triangle" tells you whether lengths along the line add or subtract. (Placing $O$ on the other side of $BC$ and subtracting leads to $s = 18$, but then $AM = 9\sqrt3$ is longer than $OA = 4\sqrt3$, which would put $O$ inside the triangle.)

Test the choices with the same picture. For a side $s$, compute $OM = 4\sqrt3 + \frac{s\sqrt3}{2}$ and check whether $OM^2 + \left(\frac{s}{2}\right)^2 = 156$. For $s = 6$: $OM = 7\sqrt3$, and $147 + 9 = 156$. For $s = 12$: $OM = 10\sqrt3$, and $300 + 36 = 336$ is far too big. The left side keeps growing as $s$ grows, so only one side length can work, and $s = 6$ is it.

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 $\boxed{\textbf{(B)}\ 6}$.