BdMO Online Forum

Problem 4:
Triangle $ABC$ is acute and $M$ is its circumcenter. Determine what point $P$ inside the triangle satisfy \[1\le \frac {\angle APB}{\angle ACB} \le 2,\ 1\le \frac{\angle BPC}{\angle BAC}\le 2,\ 1\le \frac {\angle CPA}{\angle CBA} \le 2\]

AM, BM ,CM are the locus of p that satisfies the given inequality. the proof is very trivial.....

I'm not sure about it.But I'm trying to solve it.If it is wrong plz send the write answer.
I think the point p & M is same.because \[\angle APB> AMB> ACB.\] When P & M are different point.
notice that,\[ \angle AMB= 2 \left (\angle ACB \right )\] in similar fashion we can prove the other eqation also.
So,M & P are same point.

I also did the same solution.and, i believe it is OK.we can say it like this, if $P$ is any other point but $M$,at least one of the equation can not be fulfilled