 # 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
Geometry
##### Difficulty Level
Medium
This problem obviously points towards an application of trilinear coordinates (see more here).

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 $ABC$.

# Connected Program at Cheenta

Math Olympiad is the greatest and most challenging academic contest for school students. Brilliant school students from over 100 countries participate in it every year. Cheenta works with small groups of gifted students through an intense training program. It is a deeply personalized journey toward intellectual prowess and technical sophistication.

# Similar Problems

## Sum of Sides of Triangle | PRMO-2018 | Problem No-17

Try this beautiful Problem on Geometry from PRMO -2018.You may use sequential hints to solve the problem.

## Recursion Problem | AMC 10A, 2019| Problem No 15

Try this beautiful Problem on Algebra from AMC 10A, 2019. Problem-15, You may use sequential hints to solve the problem.

## Roots of Polynomial | AMC 10A, 2019| Problem No 24

Try this beautiful Problem on Algebra from AMC 10A, 2019. Problem-24, You may use sequential hints to solve the problem.

## Set of Fractions | AMC 10A, 2015| Problem No 15

Try this beautiful Problem on Algebra from AMC 10A, 2015. Problem-15. You may use sequential hints to solve the problem.

## Indian Olympiad Qualifier in Mathematics – IOQM

Due to COVID 19 Pandemic, the Maths Olympiad stages in India has changed. Here is the announcement published by HBCSE: Important Announcement [Updated:14-Sept-2020]The national Olympiad programme in mathematics culminating in the International Mathematical Olympiad...

## Positive Integers and Quadrilateral | AMC 10A 2015 | Sum 24

Try this beautiful Problem on Rectangle and triangle from AMC 10A, 2015. Problem-24. You may use sequential hints to solve the problem.

## Rectangular Piece of Paper | AMC 10A, 2014| Problem No 22

Try this beautiful Problem on Rectangle and triangle from AMC 10A, 2014. Problem-23. You may use sequential hints to solve the problem.

## Probability in Marbles | AMC 10A, 2010| Problem No 23

Try this beautiful Problem on Probability from AMC 10A, 2010. Problem-23. You may use sequential hints to solve the problem.

## Points on a circle | AMC 10A, 2010| Problem No 22

Try this beautiful Problem on Number theory based on Triangle and Circle from AMC 10A, 2010. Problem-22. You may use sequential hints to solve the problem.

## Circle and Equilateral Triangle | AMC 10A, 2017| Problem No 22

Try this beautiful Problem on Triangle and Circle from AMC 10A, 2017. Problem-22. You may use sequential hints to solve the problem.