Bariz - Problem

Source: 1st Romanian Master in Mathematics (RMIM) 2008, Bucharest, Problem 1

Tags: conics, geometry unsolved, geometry

freemind

09.02.2008 22:27

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.

Using areal (or trilinear - both the same here) co-ordinates we are looking for the a region on which $ (\sqrt{x}+\sqrt{y}-\sqrt{z})(\sqrt{x}-\sqrt{y}-\sqrt{z})(-\sqrt{x}+\sqrt{y}+\sqrt{z}) > 0$ and on which $ \sqrt{x}+\sqrt{y}+\sqrt{z}$ is strictly positive. Multiplying it all together and factorising we get $ 2(xy+yz+zx)-(x^2+y^2+z^2)>0$. It's easy to show that this is the interior of a conic which is a circle and passes through the three midpoints of the sides, and that it is therefore the interior of the incircle. So this is our locus.