## Le Monde puzzle [#1067]

Posted in Books, Kids, R with tags , , , , , , , on September 19, 2018 by xi'an The second Le Monde mathematical puzzle in the new competition is sheer trigonometry: When in the above figures both triangles ABC are isosceles and the brown segments are all of length 25cm, find the angle in A and the value of DC², respectively. This could have been solved by R coding the various possible angles of the three segments beyond BC until the isosceles property is met, but it went much faster by pen and paper, leading to an angle of π/9 in the first case and a square of 1250 in the second case. The third puzzle is basic arithmetic that only seems solvable by enumeration…

## Le Monde puzzle [#1012]

Posted in Books, Kids with tags , , , , , on June 14, 2017 by xi'an 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.