Past exam of the mathematics course of the University of Cambridge / 2025 / ia / Paper 1 / 1B (Vectors and Matrices)

Translate and scale the circle to $|z|=1$. For distinct points $u,v,w$ on it, the angle at $w$ is right exactly whenis purely imaginary. Since $\bar{u}=u^{-1}$ and similarly for $v,w$, direct simplification gives $\bar{R}=(v/u)R$. Thus $\bar{R}=-R$ exactly when $v=-u$, which says that the opposite side joins antipodal points and is a diameter. The converse is the same calculation in reverse.

The vertices satisfyAfter removing the root $z=0$, the product of the other roots has modulus equal to the constant term of $((z+1{)}^N-1)/z$, namely $N$. Hence the product of the $N-1$ chord lengths from one vertex of a unit regular polygon is $N$. Multiplying this identity over all $N$ vertices counts every chord twice, so the product of all chord lengths is $N^{N/2}$. Scaling the circle by $R$ scales each of its $N(N-1)/2$ chords by $R$, giving