A soldier needs to sweep a region with the shape of an equilateral triangle for mines. The detector has an effective radius equal to half the altitude of the triangle. He starts at a vertex of the triangle. What path should he follow in order to travel the least distance and still sweep the whole region?
Solution
In particular he must sweep the other two vertices. Let us take the triangle to be ABC, with side 1 and assume the soldier starts at A. So the path must intersect the circles radius √3/4 centered on the other two vertices. Let us look for the shortest path of this type. Suppose it intersects the circle center B at X and the circle center C at Y, and goes first to X and then to Y. Clearly the path from A to X must be a straight line and the path from X to Y must be a straight line. Moreover the shortest path from X to the circle center C follows the line XC and has length AX + XC - √3/4. So we are looking for the point X which minimises AX + XC.
Consider the point P where the altitude intersects the circle. By the usual reflection argument the distance AP + PC is shorter than the distance AP' + P'C for any other point P' on the line perpendicular to the altitude through P. Moreover for any point X on the circle, take AX to cut the line at P'. Then AX + XC > AP' + P'C > AP + PC.
It remains to check that the three circles center A, X, Y cover the triangle. In fact the circle center X covers the whole triangle except for a small portion near A and a small portion near C, which are covered by the triangles center A and Y.
Solutions are also available in: Samuel L Greitzer, International Mathematical Olympiads 1959-1977, MAA 1978, and in István Reiman, International Mathematical Olympiad 1959-1999, ISBN 189-8855-48-X.
© John Scholes
jscholes@kalva.demon.co.uk
10 Oct 1998