Berkeley Math Circle

Solution:

We claim that the center $I$ of the triangle's inscribed circle is such a point. To see this, note that
$$\angle BIC = 180 - \angle ICB + \angle CBI = 180 - \frac{\angle C + \angle B}{2} = 180 - \frac{180 - A}{2} = 90 + \frac{A}{2} > 90.$$
Thus $\triangle BIC$ is obtuse and if we drop a perpendicular $BD$ from $B$ to $CI$, then $D$ lies on the extension of ray $CI$. The circle with diameter $BC$ passes through $D$ since $\angle BDC = 90$, and thus $I$, an interior point of chord $CD$, lies inside the circle. By symmetry, we can conclude that $I$ lies inside the other two circles as well.