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BdMO National Higher Secondary 2009/4

Posted: Sun Feb 06, 2011 11:33 pm
Problem 4:
Triangle $ABC$ is acute and $M$ is its circumcenter. Determine what point $P$ inside the triangle satisfy $1\le \frac {\angle APB}{\angle ACB} \le 2,\ 1\le \frac{\angle BPC}{\angle BAC}\le 2,\ 1\le \frac {\angle CPA}{\angle CBA} \le 2$

Re: BdMO National Higher Secondary 2009/4

Posted: Wed Oct 19, 2011 12:02 am
AM, BM ,CM are the locus of p that satisfies the given inequality. the proof is very trivial.....

Re: BdMO National Higher Secondary 2009/4

Posted: Tue Feb 07, 2012 2:42 pm
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.

Re: BdMO National Higher Secondary 2009/4

Posted: Tue Feb 07, 2012 4:46 pm
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