For $1)$
Rotate $\triangle{AMC}$ with center so that side $AC$ becomes $BC$, get $\triangle{BM'C}$
Then $\triangle{BM'C}$ is Pompeiu's Triangle
We have $\angle{BM'M}=\angle{BMC}-60^{\circ}=\angle{AMC}-60^{\circ}$
And $\angle{BMM'}=\angle{BMC}-60^{\circ}$
And then we get $\angle{MBM'}=180^{\circ}-\angle{BMM'}-\angle{BM'M}=\angle{AMB}-60^{\circ}$
And let $x$ denote area of that triangle
Consider other $2$ rotations get $3$ triangles outside $\triangle{ABC}$
Then we have $2[ABC]=3x+\frac{\sqrt{3}}{4}(MA^2+MB^2+MC^2)$
Then we use well-known identity that $MA^2+MB^2+MC^2=3MG^2+\frac{1}{3}(AB^2+BC^2+CA^2)$ for any triangle $ABC$ with centroid $G$ (in this case $G=O$)
The result follow immediately
|