SOLUTION TO PROBLEM 4:


Start by dividing the corral into four equal sub-squares as shown in the figure.


\begin{picture}(120,120) \put(0,0){\line(0,1){100}} \put(0,0){\line(1,0){100}} \... ...1,0){100}} \put(0,50){\line(1,0){100}} \put(50,0){\line(0,1){100}} \end{picture}


By the pigeonhole principle, when we place 5 horses in the corral, at least two of them will be in the same sub-square. (A sub-square includes both the interior and the boundary.) The maximum distance between any two points in a sub-square is the diagonal which has length

\begin{displaymath}\sqrt{\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^2}= \frac{\sqrt{2}}{2}.\end{displaymath}

It follows that the two horses in the same sub-square must be within a distance of $\sqrt{2}/2$ units from each other.


Questions and/or comments should be directed to Judy Downey or Griff Elder


[Back]    BACK
 
 

Last modified:   Fri Sep 15 16:51:48 CDT 2000