Solution: playing with triangles

We are publishing here the solution to the divertimento Playing with triangles. 

The fun

A figure consisting of an equilateral triangle and a line parallel to one side passing through its centre is given. A game is set up between two players, the rules of which are as follows:

1. Each player chooses a point on the triangle, including the sides. The winner is the one who obtains the highest value by adding the distances from the chosen point to the sides of the triangle.
2. In case of a tie, the winner is the player who has the greatest distance from the point to the side of the triangle.
3. In case of a new tie, the winner is the one with the greatest distance from the point to the line through the centre.

How should the point be chosen to win?

Solution

Let \(P\) be a point of our equilateral triangle. If \(P\) is joined to the three vertices, the triangle is decomposed in three others if \(P\) is interior, in two if \(P\) is on a side and is not a vertex or it remains invariant if \(P\) is a vertex. In any case, if we equal the area of the original triangle and the sum of those of the triangles formed, we can verify that the sum of the distances from \(P\) to the sides is always equal to the height of the triangle (Viviani’s Theorem). Therefore, the first criterion of the game does not discern which move is better.

According to the second tie-breaker criterion, in order to have the greatest possible distance to a side, one must obviously choose a vertex of the side. Therefore a tie can occur again if both players choose both vertices.

Finally, of the three vertices, there is one that gives the greatest distance to the line and it is the one that does not belong to the side parallel to the line (this distance being double that of the other two). This is therefore the optimal choice.