IRL_ABooklet_2023

We will use complex numbers and suppose that $\Gamma$ is centred at the origin so that $\Gamma$ is the set of complex numbers $w$ such that $|w|=1$. Moreover, w.l.o.g., we can suppose that $A=1$, $B=\omega$, $C={\omega }^2$, where $\omega =e^{2\pi i/3}$ is a cube root of unity and $1+\omega +{\omega }^2=0$. Let the complex number $z$ stand for $P$. Then $$|PA|=|z-1|,|PB|=|z-\omega |,|PC|=|z-{\omega }^2|,$$ and so $$\begin{array}{rl}|PA|\cdot |PB|\cdot |PC|& =|z-1||z-\omega ||z-{\omega }^2|\\ & =|(z-1)(z-\omega )(z-{\omega }^2)|\\ & =|z^3-1|\\ & \le |z^3|+1(\text{since }|a+b|\le |a|+|b|)\\ & =|z{|}^3+1\\ & \le 2\end{array}$$ if $P$ is within or on $\Gamma$, in which case $|z|\le 1$.

We will now show that equality occurs exactly when $P$ is one of the three points on $\Gamma$ that are obtained by a $60^{\circ }$-rotation of $A,B,C$. Equality occurs above when $|z^3|=1$ and $|z^3-1|=2$, i.e. when $z^3$ is on $\Gamma$ and has distance $2$ from the complex number $1$. This happens exactly when $z^3=-1$. This equation has three solutions: $-1,-\omega ,-{\omega }^2$. These three complex numbers are diametrically opposite $A,B,C$ on $\Gamma$ and so can also be obtained by rotating $A,B,C$ by $60^{\circ }$ about the circumcentre of triangle $ABC$.