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.

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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 $\frac{2x(1+y)}{2}={7\sqrt{7}\over16}.$

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