# Understand the problem

[/et_pb_text][et_pb_text _builder_version="3.22.4" text_font="Raleway||||||||" background_color="#f4f4f4" box_shadow_style="preset2" custom_margin="10px||10px" custom_padding="10px|20px|10px|20px"]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.

[/et_pb_text][et_pb_tabs active_tab_background_color="#0c71c3" inactive_tab_background_color="#000000" _builder_version="3.23.3" tab_text_color="#ffffff" tab_font="||||||||" background_color="#ffffff"][et_pb_tab title="Hint 0" _builder_version="3.22.4"]Do you really need a hint? Try it first!

[/et_pb_tab][et_pb_tab title="Hint 1" _builder_version="3.23.3"]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.[/et_pb_tab][et_pb_tab title="Hint 2" _builder_version="3.23.3"]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.[/et_pb_tab][et_pb_tab title="Hint 3" _builder_version="3.23.3"]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.[/et_pb_tab][et_pb_tab title="Hint 4" _builder_version="3.23.3"]

.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 $ABC$.