Le Monde puzzle [#1002]

For once and only because it is part of this competition, a geometric Le Monde mathematical puzzle:

Given both diagonals of lengths p=105 and q=116, what is the parallelogram with the largest area? and when the perimeter is furthermore constrained to be L=290?

This made me jump right away to the quadrilateral page on Wikipedia, which reminds us that the largest area occurs when the diagonals are orthogonal, in which case it is A=½pq. Only the angle between the diagonals matters. Imposing the perimeter 2s in addition is not solved there, so I wrote an R code looking at all the integer solutions, based on one of the numerous formulae for the area, like ½pq sin(θ), where θ is the angle between both diagonals, and discretising in terms of the fractions of both diagonals at the intersection, and of the angle θ:

p=105
q=116
s=145
for (alpha in (1:500)/1000){
 ap=alpha*p;ap2=ap^2;omap=p-ap;omap2=omap^2
 for (beta in (1:999)/1000){
  bq=beta*q;bq2=bq^2;ombq=q-bq;ombq2=ombq^2
  for (teta in (1:9999)*pi/10000){
   d=sqrt(ap2+bq2-2*ap*bq*cos(teta))
   a=sqrt(ap2+ombq2+2*ap*ombq*cos(teta))
   b=sqrt(omap2+ombq2-2*omap*ombq*cos(teta))
   c=sqrt(omap2+bq2+2*omap*bq*cos(teta))
   if (abs(a+b+c+d-2*s)<.01){
     if (p*q*sin(teta)<2*maxur){
       maxur=p*q*sin(teta)/2
       sole=c(a,b,c,d,alpha,beta,teta)}}}}

This code returned an area of 4350, to compare with the optimal 6090 (which is recovered by the above R code when the diagonal lengths are identical and the perimeter is the one of the associated square). (As Jean-Louis Foulley pointed out to me, this area can be found directly by assuming the quadrilateral is a parallelogram and maximising in the length of one side.)

3 Responses to “Le Monde puzzle [#1002]”

  1. FOULLEY Jean-Louis Says:

    See more details about an analytical solution at
    https://www.researchgate.net/publication/315779927_Le_Monde_Puzzle_1002

  2. […] article was first published on R – Xi'an's Og, and kindly contributed to […]

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