IMO 1973 SL 14

Origin: YUG

Problem

A soldier has to investigate whether there are mines in an area that has the form of an equilateral triangle. The radius of his detector is equal to one-half of an altitude of the triangle. The soldier starts from one vertex of the triangle. Determine the shortest path that the soldier has to traverse in order to check the whole region.

Solution

Suppose that the soldier starts at the vertex A of the equilateral tri- angle ABC of side length a. Let ϕ, \psi be the arcs of circles with centers B and C and radii a \sqrt 3/4 respectively, that lie inside the triangle. In order to check the vertices B, C, he must visit some points D \inϕ and E \in\psi. A B C D E F M N ϕ \psi Thus his path cannot be shorter than the path ADE (or AED) itself. The length of the path ADE is AD +DE \geqAD +DC −a \sqrt 3/4. Let F be the reflection of C across the line MN, where M, N are the midpoints of AB and BC. Then DC \geqDF and hence AD + DC \geqAD + DF \geqAF. Consequently AD + DE \geqAF −a \sqrt = a  \sqrt 2 − \sqrt

, with equality if and only if D is the midpoint of arc ϕ and E = (CD) \cap\psi.

Moreover, it is easy to verify that, in following the path ADE, the soldier will check the whole region. Therefore this path (as well as the one sym- metric to it) is shortest possible path that the soldier can take in order to check the entire field.