, 2014

Solution:
WLOG, suppose $OA \geq OB$. Let $\omega'$ be the circle of radius $OA$ centered at $O$. We have that $B$ lies inside $\omega'$. Thus, it is possible to scale $\omega'$ down about the point $A$ to get a circle $\omega''$ passing through both $A$ and $B$. Since $\omega''$ lies inside $\omega'$ and $\omega'$ lies inside $\omega$, $\omega''$ lies inside $\omega$.

Alternative solution 1:
WLOG, suppose $OA \geq OB$. Since $OA \geq OB$, the perpendicular bisector of $AB$ intersects segment $OA$ at some point $C$. We claim that the circle $\omega'$ passing through $A$ and $B$ and centered at $C$ lies entirely in $\omega$. Let $x = OA$ and $y = AC = BC$. Note that $y$ is the length of the radius of $\omega'$. By definition, any point $P$ contained in $\omega'$ is of distance at most $y$ from $C$. Applying the triangle inequality to $OCP$, we see that $OP \leq OC + CP \leq (x - y) + y = x$, so $P$ lies in $\omega$. Since $P$ was arbitrary, it follows that $\omega'$ lies entirely in $\omega$.

Alternative solution 2:
Draw line $AB$, and let it intersect $\omega$ at $A'$ and $B'$, where $A$ and $A'$ are on the same side of $B$. Choose $X$ inside the segment $AB$ so that $A'X / AX = B'X / BX$; such a point exists by the intermediate value theorem. Notice that $X$ is the center of a dilation taking $A'B'$ to $AB$—the same dilation carries $\omega$ to $\omega'$ which goes through $A$ and $B$. Since $\omega'$ is $\omega$ dilated with respect to a point in its interior, it's clear that $\omega'$ must be contained entirely within $\omega$, and so we are done.