**F**or 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?

**T**his 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.)*