Question

A convex polygon $\Gamma$ is such that the distance between any two vertices of $\Gamma$ does not exceed $1$.
(i) Prove that the distance between any two points on the boundary of $\Gamma$ does not exceed $1.$
(ii) If $X$ and $Y$ are two distinct points inside $\Gamma$, prove that there exists a point $Z$ on the boundary of $\Gamma$ such that $XZ+YZ\le 1$.

Answer 1

(i) Let S and T be two points on the boundary of $\Gamma$, with S lying on the side AB and T lying on the side PQ of $\Gamma$ (See Fig).
Join TA, TB, TS.
ST lies between TA and TB in triangle TAB.
One of $\angle AST$ and $\angle BST$ is at least ${90}^o$, say $\angle AST\ge {90}^o$.
$\therefore AT\ge TS$.
But AT lies inside triangle APQ and one of $\angle ATP$ and $\angle ATQ$ is at least ${90}^o$, say $\angle ATP\ge {90}^o$.
Then $AP\ge AT$.
$\therefore TS\le AT\le AP\le 1$
(ii) Let X and Y be points in the interior $\Gamma$.
Join XY and produce them on either side to meet the sides CD and EF of $\Gamma$ at $Z_1$ and $Z_2$ respectively.
We have, $(X{Z}_1+Y{Z}_1)+(X{Z}_2+Y{Z}_2)=(X{Z}_1+X{Z}_2)+(Y{Z}_1+Y{Z}_2)$
$=2Z_1{Z}_2\le 2$, by the first part.
Therefore one of the sums $X{Z}_1+Y{Z}_1$ and $X{Z}_2+Y{Z}_2$ is at most 1.
Z can be chosen accordingly as $Z_1$ or $Z_2$.