a.m.-g.m.-inequalitycirclescoordinate systemsgeometryquadrilateral
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.
Let $\mathcal{C}$ be a circle wichi is tangent to $BC,CD$ and $DA$ and let it center be $O$. Then $O$ must lie on angle bisector of angle $\angle ADC$ and angle $\angle DCB$. So these angle bisectors meet on $AB$.
Now let $E$ be on $AB$ so that $AE=AD$. We have to prove $BC=BE$ i.e. triangle $BCE$ is isosceles. 
If $\angle ADE =\alpha$ then $\angle AED =\alpha$ and $\angle ADE =180^{\circ}-2\alpha$. So $\angle BCD =2\alpha$ and thus $\angle OCD =\alpha$. Since $\angle DEB =180^{\circ}-\alpha$ we have $CDEO$ is cyclic.
Let $\angle OEC =\beta$ then $\angle ODC =\beta$ and $\angle ADO = \beta $ so $\angle ABC =180^{\circ}-2\beta$ and thus $\angle BCE =\beta$.
You found $AC$ using the cosine rule. With another version of that same law it is possible to find any angle given all three sides of a triangle. You now know all three sides of $\triangle ABC$ and $\triangle ADC$. With that you can derive $\angle DAC$ and $\angle CAB$, hence $\angle DAB$.
Now, consider $\triangle DAB$. You have sides $AD$ and $AB$, and $\angle DAB$. Use the cosine rule once more to solve for $BD$.
Best Answer
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!