1973 IMO Problems/Problem 4
A soldier needs to check on the presence of mines in a region having the shape of an equilateral triangle. The radius of action of his detector is equal to half the altitude of the triangle. The soldier leaves from one vertex of the triangle. What path shouid he follow in order to travel the least possible distance and still accomplish his mission?
Let our triangle be , let the midpoint of be , and let the midpoint of be . Let the height of the triangle be . Draw circles around points and with radius , and label them and . Let the intersection of and be .
The path that is the solution to this problem must go from to a point on to a point on . Let us first find the shortest possible path. We will then prove that this path fits the requirements.
The shortest path is in fact the path from to to . We will prove this as follows:
Suppose that a different point on is the optimal point to go to. Let this point be . Then, the optimal point on would be the intersection of and (let this point be ). Let the line through parallel to be , and the intersection of and be . Then, we have
The second inequality is true due to the triangle inequality, and we can prove the first to be true as follows:
Suppose we have a point on , . Reflect across to get . Then, , which is obviously minimized at the intersection of the two lines, .
By , we have
and we have proved the claim.
This path easily covers the whole triangle. This is because if you draw a line perpendicular to from any point in the triangle, this line will hit the path in a distance less than or equal to . We can compute the length of the path to be
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