I do not think such a point can exist! Because $M$ is interior, angles AMB, BMC and CMA must add up to 360 degrees, so angle CMA is ninety degrees. But no interior point meets this condition. Is there something wrong?

Good problem. I'll give a vague couple steps of a solution--try to fill in the gaps. Let D and E be points outside the triangle s.t. $\angle ABD = 90, \angle DAB = 30, \angle CBE = 90, \angle BEC = 30$. It suffices to prove that the area of ABC is at most one fourth of the area of ADEC (why?). Note that by the trig formula for areas, triangles DBE and ABC have the same area. We now see that we might as well have $\angle ABC = 90$, so the problem becomes algebraic; prove that $[ABD] + [CBE] \ge 2 [ABC]$ $\frac{x^2}{\sqrt{3}} + y^2\sqrt{3} \ge 2xy$ which is simply AM-GM. Edit--in response to JBL, I think he meant $\angle AMC = 90$.

Whoops, my bad... yeah, that's right $AMC = 90^\circ$, not $ABC$. EDITED.

How do you know that $\angle ABD = 90, \angle DAB = 30, etc.$?

Karth wrote: How do you know that $\angle ABD = 90, \angle DAB = 30, etc.$? probability1.01 constructed $D$ and $E$ to meet those conditions.

solved also here