Le Monde puzzle [#1012]

A basic geometric Le Monde mathematical puzzle:

Take a triangle ABC such that the side AB is c=42 long, each side has an integer length, and the area is 756. Given an inner point D, draw three lines parallel to the three sides of ABC through D in order to construct three triangles with common summit D and bases supported by these three sides.

  1. How far is D from the base AB when all three triangles have perimeters equal to the sides that support their basis?
  2. How far is D from the previous solution when the sum of the areas of the three triangles is minimal?

Since the puzzle is purely geometric, I was quite tempted to bypass it and to watch instead the British elections and the Comey audition! However, the sides a and b are easily found by an exhaustive search, a=39 and b=45 (or the reverse). From there, the problem resolution proceeds by a similar triangles argument, since all triangles constructed by the game rule have the same angles, hence proportional sides. For the first question, this leads to a straightforward determination of the basis of each triangle by the perimeter equation, meaning that D is then 12 units away from AB. The second question is not harder in that the surface of a triangle with basis a and opposite angles β and γ can be written as

a²sin(β)sin(γ)/2sin(β+γ)

meaning it suffices to minimise a²+a’²+a”² under the constraint that the sum of the three sides parallel to BC is the complete length of BC, a²+a’²+a”²=39. The solution is then that all triangles are identical, leading to a summit D’ at a distance 12 from AB, again!, but in the middle of the segment, hence distance to the earlier D equal to one.

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