Understand the problem

Let $ ABC$ be an equilateral triangle and $ P$ in its interior. The distances from $ P$ to the triangle’s sides are denoted by $ a^2, b^2,c^2$respectively, where $ a,b,c>0$. Find the locus of the points $ P$ for which $ a,b,c$ can be the sides of a non-degenerate triangle.

Source of the problem
Romanian Master in Mathematics, 2008
Difficulty Level
This problem obviously points towards an application of trilinear coordinates (see more here).

Start with hints

Do you really need a hint? Try it first!

If there exists a non-degenerate triangle with sides a,b,c then the area of the triangle must be positive. Make use of this fact employing Heron’s formula.
Hint 1 will give you a locus in terms of trilinear coordinates. However, it is easier to work in cartesian coordinates as we are already familiar with many curves in them.
Take A=(1,0,0), B= (0,1,0), C= (0,0,1). There is a simple relationship between the cartesian and trilinear coordinates for this choice.

.Note that all points in the plane of ABC satisfy x+y+z=1 (why?). For any P in the interior, Let Q be the foot of the perpendicular from P to AB and R be the foot of the perpendicular from P to the XY plane. It is possible to show that \frac{PR}{PQ}= \sqrt{\frac{2}{3}} (it is because the angle between the plane and the Z axis is \arccos \sqrt{\frac{2}{3}}). Hence, \frac{z}{c^2}=\sqrt{\frac{3}{2}}. By symmetry, \frac{x}{a^2}=\sqrt{\frac{2}{3}}=\frac{y}{b^2}. This relates the two coordinate systems. For a triangle with sides a,b,c, the square of the area is \frac{1}{16}(a+b+c)(-a+b+c)(a-b+c)(a+b-c). For this to be positive, we must have (after simplification) a^4+b^4+c^4< 2(a^2b^2+b^2c^2+c^2a^2). In cartesian coordinates, this translates to \frac{3}{2}(x^2+y^2+z^2)<3(xy+yz+zx), which is equivalent to (x+y+z)^2>2(x^2+y^2+z^2). As P lies on the plane x+y+z=1, this means that x^2+y^2+z^2<\frac{1}{2}. This last equation is that of the interior of a solid sphere. Hence, our desired locus is the intersection of this solid sphere with the plane x+y+z=1, which is precisely the interior of the circumcircle of ABC.  


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