## Problem 6

(a) A and B move clockwise with equal angular speed along circles center $\mathrm{P}$ and $\mathrm{Q}$ respectively. $\mathrm{C}$ moves continuously so that $\mathrm{AB}=\mathrm{BC}=\mathrm{CA}$. Establish $\mathrm{C}$ 's locus and speed. *(b) $\mathrm{ABC}$ is an equilateral triangle and $\mathrm{P}$ satisfies $\mathrm{AP}=2, \mathrm{BP}=3$. Establish the maximum possible value of CP.


## Solution

(a) Represent A, B as complex numbers $\mathrm{z}_{1}+\mathrm{w}_{1} \mathrm{e}^{\mathrm{it}}, \mathrm{z}_{2}+\mathrm{w}_{2} \mathrm{e}^{\mathrm{it}}$. Then $\mathrm{C}$ is $\left(\mathrm{z}_{1}+\mathrm{w}_{1} \mathrm{e}^{\mathrm{it}}\right)+\left(\mathrm{z}_{2}+\right.$ $\left.\mathrm{w}_{2} \mathrm{e}^{\mathrm{it}}-\mathrm{z}_{1}-\mathrm{w}_{1} \mathrm{e}^{\mathrm{it}}\right) \mathrm{e}^{\mathrm{i} \pi / 3}$, which is also of the form $\mathrm{z}+\mathrm{w} \mathrm{e}^{\mathrm{it}}$.

However, there is one subtlety. There are actually two circles possible for $\mathrm{C}$ depending on which side of $\mathrm{AB}$ we place it. The continuity requirement means that $\mathrm{C}$ is normally confined to one of the circles. However, if A and B ever coincide then C may be able to switch to the other circle.

If we regard "moves continuously" as allowing a discontinuous velocity, then a switch is always possible (provided A and B coincide).

(b) Answer: 5.

P must be the opposite side of $\mathrm{AB}$ to $\mathrm{C}$ (or we could increase $\mathrm{CP}$, whilst keeping $\mathrm{AP}$ and $\mathrm{BP}$ the same, by reflecting in $A B$ ). Similarly it must be on the same side of $A C$ as $B$, and on the same side of BC as A. For any P in this region the quadrilateral APBC is convex and hence satisfies Ptolemy's inequality $\mathrm{CP} \cdot \mathrm{AB} \leq \mathrm{AP} \cdot \mathrm{BC}+\mathrm{BP} \cdot \mathrm{AC}$, with equality iff $\mathrm{APBC}$ is cyclic. But $\mathrm{AB}=\mathrm{BC}=\mathrm{CA}$, so we have $\mathrm{CP} \leq \mathrm{AP}+\mathrm{BP}=5$ with equality iff $\mathrm{P}$ lies on the arc $\mathrm{AB}$ of the circle $\mathrm{ABC}$. Note that there is just one such point, because the locus of $\mathrm{P}$ such that $\mathrm{BP}=1.5$ $\mathrm{AP}$ is a circle which cuts the arc just once.

Ptolemy's inequality for 4 points $\mathrm{A}, \mathrm{B}, \mathrm{C}, \mathrm{D}$ : $\mathrm{AB} \cdot \mathrm{CD}+\mathrm{BC} \cdot \mathrm{AD} \geq \mathrm{AC} \cdot \mathrm{BD}$ with equality iff $\mathrm{ABCD}$ is a cyclic quadrilateral (meaning $\mathrm{A}, \mathrm{B}, \mathrm{C}, \mathrm{D}$ lie on a circle in that order).

## Proof

Take $\mathrm{E}$ inside $\mathrm{ABCD}$ such that $\angle \mathrm{DAE}=\angle \mathrm{CAB}$ and $\angle \mathrm{ADE}=\angle \mathrm{ACB}$. Then $\mathrm{ADE}$ and $\mathrm{ACB}$ are similar, so $\mathrm{DE} / \mathrm{CB}=\mathrm{AD} / \mathrm{AC}$ and hence $\mathrm{BC} \cdot \mathrm{AD}=\mathrm{AC} \cdot \mathrm{DE}$. It also follows that $\mathrm{AE} / \mathrm{AB}=$ $\mathrm{AD} / \mathrm{AC}$. But we also have $\angle \mathrm{EAB}=\angle \mathrm{DAC}$ and hence $\mathrm{AEB}$ and $\mathrm{ADC}$ are also similar. So $\mathrm{EB} / \mathrm{AB}=\mathrm{DC} / \mathrm{AC}$, and hence $\mathrm{AB} \cdot \mathrm{CD}=\mathrm{AC} \cdot \mathrm{EB}$. Adding, we have: $\mathrm{AB} \cdot \mathrm{CD}+\mathrm{BC} \cdot \mathrm{AD}=$ $\mathrm{AC}(\mathrm{BE}+\mathrm{ED}) \geq \mathrm{AC} \cdot \mathrm{BD}$ with equality iff $\mathrm{E}$ lies on $\mathrm{BD}$, or equivalently $\mathrm{ABCD}$ is cyclic.

This glosses over one point. It only follows that $\angle \mathrm{EAB}=\angle \mathrm{DAC}$ if $\mathrm{ABCD}$ is convex. For the convex case, we have that $\angle \mathrm{EAB}=\angle \mathrm{CAB}+\angle \mathrm{EAC}$ and $\angle \mathrm{DAC}=\angle \mathrm{DAE}+\angle \mathrm{EAC}$, or $\angle \mathrm{EAB}=\angle \mathrm{CAB}-\angle \mathrm{EAC}$ and $\angle \mathrm{DAC}=\angle \mathrm{DAE}-\angle \mathrm{EAC}$. Either way $\angle \mathrm{EAB}=\angle \mathrm{DAC}$. But in the non-convex case, we can have $\angle \mathrm{EAB}=\angle \mathrm{CAB}+\angle \mathrm{EAC}$ and $\angle \mathrm{DAC}=\angle \mathrm{DAE}-\angle \mathrm{EAC}$ (or $-\ldots+$ ) and hence the angles $\angle \mathrm{EAB}$ and $\angle \mathrm{DAC}$ are not necessarily equal.

