## Problem 6

Find the smallest value $\mathrm{x}$ such that, given any point inside an equilateral triangle of side 1 , we can always choose two points on the sides of the triangle, collinear with the given point and a distance $x$ apart.


## Solution

Answer: $2 / 3$

Let $\mathrm{O}$ be the center of $\mathrm{ABC}$. Let $\mathrm{AO}$ meet $\mathrm{BC}$ at $\mathrm{D}$, let $\mathrm{BO}$ meet $\mathrm{CA}$ at $\mathrm{E}$, and let $\mathrm{CO}$ meet $\mathrm{AB}$ at $\mathrm{F}$. Given any point $\mathrm{X}$ inside $\mathrm{ABC}$, it lies in one of the quadrilaterals $\mathrm{AEOF}, \mathrm{CDOE}, \mathrm{BFOD}$. Without loss of generality, it lies in AEOF. Take the line through X parallel to BC. It meets
$\mathrm{AB}$ in $\mathrm{P}$ and $\mathrm{AC}$ in $\mathrm{Q}$. Then $\mathrm{PQ}$ is shorter than the parallel line $\mathrm{MON}$ with $\mathrm{M}$ on $\mathrm{AB}$ and $\mathrm{N}$ on $A C$, which has length $2 / 3$. If we twist the segment PXQ so that it continues to pass through $X$, and $\mathrm{P}$ remains on $\mathrm{AB}$ and $\mathrm{Q}$ on $\mathrm{AC}$, then its length will change continuously. Eventually, one end will reach a vertex, whilst the other will be on the opposite side and hence the length of the segment will be at least that of an altitude, which is greater than $2 / 3$. So at some intermediate position its length will be $2 / 3$.

To show that no value smaller than $2 / 3$ is possible, it is sufficient to show that any segment POQ with $\mathrm{P}$ and $\mathrm{Q}$ on the sides of the triangle has length at least $2 / 3$. Take $\mathrm{P}$ on $\mathrm{MB}$ and $\mathrm{Q}$ on AN with $\mathrm{P}, \mathrm{O}, \mathrm{Q}$ collinear. Then $\mathrm{PQ} \cos \mathrm{POM}=\mathrm{MN}-\mathrm{QN} \operatorname{cospi} / 3+\mathrm{PM} \cos \mathrm{pi} / 3$. But $\mathrm{PM}>\mathrm{QN}$ (using the sine rule, $\mathrm{PM}=\mathrm{OM} \sin \mathrm{POM} / \sin \mathrm{OPM}$ and $\mathrm{QN}=\mathrm{ON} \sin \mathrm{QON} /$ sinOQN, but $\mathrm{OM}=\mathrm{ON}$, angle $\mathrm{POM}=$ angle $\mathrm{QON}$, and angle $\mathrm{OQN}=$ angle $\mathrm{OPM}+\mathrm{pi} / 3>$ angle $\mathrm{OPM}$ ), and hence $\mathrm{PQ}>\mathrm{MN} \sec \mathrm{POM}>\mathrm{MN}$.

