Finding Barycentric coordinates Let $\,\triangle ABC\,$ be a triangle, and $\alpha, \beta, \gamma$ 3 real numbers such that $\alpha+\beta+\gamma \neq 0.\,$ We define $\Gamma$ as the set of all points $\,M\,$, such that $\,\alpha\,\overline{MA}^2+\beta\,\overline{MB}^2+\gamma\,  \overline{MC}^2=1.\,$ Find $\,\alpha, \beta, \gamma\,$ when $\Gamma$ is the circumcircle of $\,\triangle ABC.$
 A: Disclaimer : I am new to this forum and this is my first answer... Please let me know if anything I write here can be improved.
We look at this problem from a physical perspective.
We let three masses $m_A$ ,$m_B$, $m_C$ be placed on the vertices of the triangle such that the centre of mass lies on the circumcentre of the triangle.
We can assume the total mass to be $1$ unit.
Then, if the moment of inertia of the system about an axis perpendicular to the plane of the triangle passing through $M$ be $I$
$$I = I_c + mD^2$$
 where $I_c$ is the moment of inertia about a parallel axis passing through the centre of mass, $m$ is the total mass and $D$ is distance between centre of mass and  point $M$. This is the 'parallel axis theorem'.
Simplifying, we get the relation :
$$(m_A)MA^2  +  (m_B)MB^2 + (m_C)MC^2 = 2 R^2$$ 
where $R$ is the circumradius of the triangle.
Now, we let $\alpha  =  \frac {m_A}{2R^2}$ and similarly others
Note:
To prove that such masses $m_A$ ,$m_B$, $m_C$ will exist, we note that this is equivalent to 
$$(m_A)\mathbf {A} + (m_B)\mathbf {B} + (m_C)\mathbf {C} = 0$$ 
Or 
$$(m_A)\mathbf {A} + (m_B)\mathbf {B}= (m_C)(- \mathbf {C})$$
here $\mathbf {A}$, $\mathbf {B}$, $\mathbf {C}$  denote the vectors with tails at circumcentre and heads at respective vertices. This will always have a solution as:
In a coordinate system with axes parallel to $\mathbf {A}$ and $\mathbf {B}$, and origin at circumcentre, the coordinates of $\mathbf {C}$ will be $$(\frac {-m_A}{m_C},\frac {-m_B}{m_C})$$
Alternatively,
We can use Lami's theorem and the fact that $\mathbf {A}$, $\mathbf {B}$, $\mathbf {C}$ have same magnitude to calculate :
$$\frac {m_A}{m_A + m_B + m_C}  =  \frac {\sin{2A}}{\sin{2A} + \sin{2B} + \sin{2C} }$$
Which simplifies to :
$$m_A  =  \frac {\cos{A}}{2 \sin{B} \sin{C}}$$
Since we assumed total mass to be $1$ unit. And so, 
$$\alpha = \frac {\cos {A}}{AC×AB}$$
A: By definition the circumcircle of the triangle passes through the three vertices. Thus when $\,M\,$ becomes the points $A,B,C$ we get three equations in the three unknown distances $\,a,b,c.\,$ That is:
$$ \alpha\, 0^2+\beta\, c^2+\gamma\, b^2=1, \quad
   \alpha\, c^2+\beta\, 0^2+\gamma\, a^2=1, \quad
   \alpha\, b^2+\beta\, a^2+\gamma\, 0^2=1. \tag{1}$$
The solution is
$$ (\alpha,\beta,\gamma) = (a^2(-a^2+b^2+c^2),b^2(a^2-b^2+c^2),c^2(a^2+b^2-c^2))/(2a^2b^2c^2). \tag{2}$$
If $\,A,B,C\,$ are the three angles at the three points, then using the law of cosines we get the result
$$ (\alpha,\beta,\gamma) = (\frac{\cos A}{bc},\frac{\cos B}{ac},
\frac{\cos C}{ab}). \tag{3}$$
