Rioplatense Mathematical Olympiad

Let $\angle CAB=2\alpha$, $\angle ABC=2\beta$ and $\angle BCA=2\gamma$. Notice that $\alpha +\beta +\gamma =90^{\circ }$, and if $I$ is the incenter of $△ABC$ we can compute $\angle AIB$, $\angle BIC$, $\angle CIA$ in terms of these variables.

Fact 1: The number of $60^{\circ }$ angles in any step of the process is equal to the number of $120^{\circ }$ angles in the next step. Proof: It suffices to prove that each $60^{\circ }$ angle generates a $120^{\circ }$ angle, and that every $120^{\circ }$ angle is generated by a $60^{\circ }$ angle. The first statement is clear: if, say, $2\alpha =60^{\circ }$, then $90^{\circ }+\alpha =120^{\circ }$. For the second statement, observe that the only way we can obtain a $120^{\circ }$ angle is if one of the angles at I measures $120^{\circ }$, because those are the only obtuse angles generated. For this to be true, one of our variables must be equal to $30^{\circ }$, which in turn means one of the angles of ABC must be $60^{\circ }$. Fact 2: The number of $120^{\circ }$ angles in any step of the process is half the number of $60^{\circ }$ angles in the next step. Proof: When performing a step, every $120^{\circ }$ angle gets divided into two $60^{\circ }$ angles. Moreover, this is the only way of obtaining a $60^{\circ }$ angle. $\square$ Using these two facts we can readily complete the following table, which shows the answer is $32$.

Step	1	2	3	4	5	6	7	8	9	10
Measure	60°	120°	60°	120°	60°	120°	60°	120°	60°	120°
Angles	2	2	4	4	8	8	16	16	32	32