I tried very much but since tomorrow is my exam, i cannot risk it.
The following is a geometry problem, which i have tried very much but could not grasp a solution.
I think that i require pythagoras' theorem here, but can not understand how to apply it. I first thought of extending a line congruent to BP such that it intersects CP at a point. But that did not help me. I also have the hint that all the other angles are 60 in measure, but if P were the incentre, in that case it would not have been difficult to show to P is the subtending angle of measure 150 degrees. Any help would be helpful!
Let in $\Delta ABC$ vertexes by the rotation similar to the clock.
The first problem.
Let $K\in AM$ such that $BM=KM$.
Thus, since $$\measuredangle BMK=\measuredangle BMA=\measuredangle BCA=60^{\circ},$$ we obtain that $\Delta BMK$ is equilateral triangle.
Now, take $\Delta BAK$ and rotate around point $B$ by $60^{\circ}.$
We obtain that $A$ goes to $C$ and $K$ goes to $M$,
which says $AK=MC$ and $$BM+MC=MK+AK=AM$$ and we are done!
The second problem.
Take the rotation around $C$ by $60^{\circ}.$
We see that $B$ goes to $A$ and let $M$ goes to $M'$.
Thus, $BM=AM'$ and $MC=MM'$ and we got $\Delta AMM'$ with needed properties.