Berkeley Math Circle: Monthly Contest 4

Solution:

WLOG, assume $O_1$ is at least as close to $BC$ as is $O_2$. Let $M, N$, and $Q$ be the respective midpoints of $\overline{BP}$, $\overline{CP}$, and $\overline{AP}$, and let $T$ be the foot of the altitude from $O_1$ to $\overline{O_2N}$.

As $\overline{O_1M}$ and $\overline{O_2N}$ are the perpendicular bisectors of $\overline{BP}$ and $\overline{PC}$, respectively, $O_1TNM$ is a rectangle, so

$$O_1T = MN = MP + PN = \frac{BP}{2} + \frac{CP}{2} = \frac{BC}{2} = \frac{O_1O_2}{2},$$
and $\triangle O_2O_1T$ is a respective $30^\circ$-$60^\circ$-$90^\circ$ triangle. Therefore, looking at the sum of the angles of quadrilateral $NPQO_2$, $\triangle ACP$, and $\triangle ABC$, we find

- $\angle NPQ = 360^\circ - \angle PQO_2 - \angle QO_2N - \angle O_2NP = 360^\circ - 90^\circ - 30^\circ - 90^\circ = 150^\circ$,

- $\angle ACP = 180^\circ - \angle CPA - \angle PAC < 180^\circ - \angle CPA = 180^\circ - 150^\circ = 30^\circ$, and

- $\angle A + \angle B = 180^\circ - \angle C > 180^\circ - 30^\circ = 150^\circ$.

This is impossible unless either $\angle A$ or $\angle B$ is

$>75^\circ$, as desired.