Three particles, weighing 2 kg, 2 kg and 3 kg respectively, sit on the perimeter of a light disc of radius 1, keeping the disc balanced horizontally on its centre. Find the area of the triangle formed by the particles.

How could you model the positions of the particles mathematically?

Why not assign the positions of the particles and the centre of the disc using Cartesian coordinates?

How about assigning a fixed position for one particle, then finding the positions of the other two?

Assuming the centre of the disc is at $(0,0)$. For the disc to balance at its centre, the particles must satisfy the centre of mass equation $\sum w_i\mathbf{x}_i = \mathbf{0},$ where $w_i, \mathbf{x}_i$ denote the weight and the vector position of particle $i$ respectively.

The centre of mass must be at the centre of the disc, or else it tips. Without loss of generality, place the centre of the disc at $(0,0),$ the 3 kg point at $(0,-1)$ and the two 2 kg points at $(-x,y)$ and $(x,y)$ respectively. Writing the centre of mass equation on the $y$-axis, we get $0 = {-3+4y\over 3},$ which gives $y={3\over4}.$ Using Pythagoras’ Theorem, we then have $x={\sqrt{7}\over4}.$ The total area of the triangle is thus $\frac{2x(1+y)}{2}={7\sqrt{7}\over16}.$