Library / Rioplatense Mathematical Olympiad /30 of 41

Rioplatense Mathematical Olympiad

Geometry Difficulty 7.0 National Olympiad, round 2 Prove it Argentina

Initially we have a paper triangle $ABC$ such that $\angle BAC = 120^\circ$ . In the first step, we draw the angle bisectors of the three angles of the triangle, which intersect in $I$ , and then using a pair of scissors we cut along segments $AI$ , $BI$ and $CI$ , obtaining 3 triangles: $ABI$ , $BCI$ and $CAI$ . In the second step we repeat the same procedure with the three triangles, that is: each one of those is cut into three smaller triangles cutting along the angle bisectors. At the end of the second step we have 9 triangles in total. This procedure continues in the same way until we complete 10 steps.
How many of the triangles at the end of the process have an angle of $120^\circ$ ?

Solution

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 $\triangle ABC$ 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

Want a route through all this instead of an archive? The track puts 2,604 problems in a working order, from Junior Challenge level to the IMO shortlist.

Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.