Calculating area of quadrilateral when distance of vertices from an arbitrary point is known Given a convex quadrilateral $ABCD$ circumscribed about a circle of diameter $1$. Inside $ABCD$ there is a point $M$ such that $MA^2 + MB^2 +MC^2 + MD^2 =2$. Find the area of the quadrilateral. 


My attempt at a solution:
I tried to solve it using coordinate geometry. I guessed $M$ to be on centre of circle and taking $ABCD$ as a square, the sides come out to be $1$. Hence, area comes out to be $1$ unit sq.

However I was looking for a proper solution for the question.
By AM-GM

$$\sum_{cyc}MA^2=\frac{1}{2}\sum_{cyc}\left(MA^2+MB^2\right)\geq\sum_{cyc}MA\cdot MB\geq$$
$$\geq2\sum_{cyc}S_{\Delta AMB}=2S_{ABCD}=r\sum_{cyc}AB,$$
where $r$ is a radius of our circle. 

In another hand, by AM-GM again 
$$\left(\sum_{cyc}AB\right)^2\geq4(AB+CD)(AD+BC)=4\sum_{cyc}AB\cdot AD=$$
$$=2\sum_{cyc}(AB\cdot AD+BC\cdot CD)\geq2\cdot4\cdot2S_{ABCD}=16S_{ABCD},$$
which gives 
$$\sum_{cyc}AB\geq4\sqrt{S_{ABCD}}=4\sqrt{\frac{1}{2}r\sum_{cyc}AB}$$ or
$$\sum_{cyc}AB\geq8r.$$
Thus, $$\sum_{cyc}MA^2\geq r\sum_{cyc}AB\geq8r^2=2.$$
The equality occurs for $$\measuredangle DAB=\measuredangle ABC=\measuredangle BCD=\measuredangle CDA=90^{\circ}$$ and 
$$\measuredangle AMB=\measuredangle BMC=\measuredangle CMD=\measuredangle DMA=90^{\circ},$$
which says that $ABCD$ is a square. 

Id est, $$S_{ABCD}=1\cdot1=1.$$
Done!
You can't do that, as there is no single solution.

Take the degenerate quadrilateral $ABCD$ where $A=B$ and $C=D$, with $M=C$, you have fulfilled the equation, but the area is $0$.