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 $$\,\triangle ABC.$$

• When $A$ is a point, what does $MA^2$ denote? And what kind of object is $M$? A real number? A matrix? Something else? – John Hughes Feb 3 at 16:12
• M is a point, MA the distance between M and A – DINEDINE Feb 3 at 16:13
• Thanks. Sounds like a cool question, but I have no idea, offhand, how to approach it. Good luck. – John Hughes Feb 3 at 16:20

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, $$\alpha = \frac {\cos {A}}{AC×AB}$$

• If $I$ denotes the circumcenter $m_A$ should be the area of the triangle $IBC$ which is $\frac{1}{2}R^2\cos{2A}$. Is that correct? – DINEDINE Feb 3 at 16:54
• Area of $IBC$ is $\frac {1}{2} R^2 \sin {2A}$ – Dhvanit Feb 3 at 17:19
• Sorry! $\sin2A$. – DINEDINE Feb 3 at 17:21
• Your proof is correct and beautiful. – DINEDINE Feb 3 at 18:01
• Thanks, btw what is the source of this problem? Is it from an Olympiad/book or did you come up with it? – Dhvanit Feb 3 at 18:10

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

$$(\alpha,\beta,\gamma) = (\frac{\cos A}{bc},\frac{\cos B}{ac}, \frac{\cos C}{ab}). \tag{3}$$