Suppose $AOCN$ is cyclic. This means that $\angle ANC=\pi-\angle AOC$. The triangles $ACN,ABM$ are directly congruent, so $\angle ANC=\angle AMB=\frac{\angle AOB}2$. We thus get $\angle AOC=\pi-\frac{\angle AOB}2$, or, in other words, the line $OC$ is the internal bisector of $\angle AOB$. Since $OA=OB$, this immediately implies $AC=BC$, and we find $ABC$ to be equilateral. For the converse, we just reverse the reasoning.

In the complex plane let $A=0$ and $B=1$ then $N=CM$. If $O'$ is the circumcenter of $ACN$ then since $ABM$ and $ACN$ are similar by a factor of $C$ we have $O'=OC$. Then the assumption becomes that $O'$ has equal distances to $A$ and $O$. That is \[ |O'-O|=|O'| \] Then \[ |OC-O|=|OC| \] By canceling $|O|$ we get \[ |C-1|=|C| \] that is the length of $BC$ and $AC$ are equal.

Tell me if anything is wrong with the following: We take a rotation of center $A$ and angle $\angle BAC$. This takes the circumcenter $O$ of $ABM$ into the circumcenter $P$ of $ACN$. Thus, $AO=AP$. The condition that $O,C,N,A$ are cyclic is equivalent to $PA=PO \, \, \Longleftrightarrow \, \, \triangle OAP$ equilateral $\, \, \Longleftrightarrow \, \, \angle BAC = \frac{\pi}{3}$.

Remark

There is nothing wrong with it (In fact it is exactly the same as mine but you expressed it without complex numbers) And I agree that it is very trivial

perfect_radio wrote: If the above is correct, then this problem is one of the most trivial problems given at a Romanian TST, in the last three years, at least. Thanks for the appreciation (the problem is mine ). However, let me point out that for example Micnea Dragos (IMO Gold medalist, JBMO multiple medalist, etc. etc.) didn't solve it. And in fact less than 50% of the contestants solved it. Only $4$ (!!) solved it using the rotations (or directed angles) the rest did either complex numbers or other types of calculatorial arguments. So I really do not know what trivial means ... For problem 1 I belive it was a good choice.

This is a lovely little problem. Congrats, Valentin!

Valentin Vornicu wrote: Thanks for the appreciation (the problem is mine ). Sorry if I offended you in any way. This was not my intention. Valentin Vornicu wrote: And in fact less than 50% of the contestants solved it. This was "the most solved" problem, along with your problem from the other day. (I didn't manage to solve that one, yet) Valentin Vornicu wrote: So I really do not know what trivial means ... Trivial means here that I could solve a geometry problem in not so much time using synthetic arguments, which is really a rarity. Probably I should have specified that. PS: The fact that Michnea Dragos didn't solve it is really strange. I knew he was quite good.

perfect_radio wrote: The fact that Michnea Dragos didn't solve it is really strange. I knew he was quite good. He is, it is just that the problem is not as trivial as you think it is Of course, once you see a solution, and/or solve it yourself you might call it easy, but it's not that easy as it seems. Don't get me wrong: it's not hard, and it's not ment to be hard: it is the first problem. We wanted to give the kids from 9th grade a chance (since all the other problems were not accesible to them).

This problem isn't trivial, but is simply and nicely (using a rotation - see the Perfect_Radio's proof). Can solve it without rotation (the level $7^{\mathrm{th}}$ class) and through a chain of equivalencies. Denote the circumcentre $O'$ of the triangle $ACN\ .$ Then, $\triangle ABM\equiv\triangle ACN$ $\Longrightarrow$ $OA=O'A$ and $\widehat {BAO}\equiv\widehat {CAO'}\ ,$ i.e. $\widehat {BAC}\equiv \widehat {OAO'}\ .$ Therefore, the points $O,A,C,N$ are concyclically $\Longleftrightarrow$ $O'A=O'O$ $\Longleftrightarrow$ the triangle $OAO'$ is equilateral $\Longleftrightarrow$ $m(\widehat{OAO'})=60^{\circ}$ $\Longleftrightarrow$ $m(\widehat {BAC})=60^{\circ}\ .$

"with the same orientation" what does it mean?