 All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars  perplexus dot info  Minimize parallelogram diagonal (Posted on 2021-04-28) Of all parallelograms of a given area find the one with the shortest possible longer diagonal.

 No Solution Yet Submitted by Danish Ahmed Khan No Rating Comments: ( Back to comment list | You must be logged in to post comments.) Solution | Comment 1 of 4

Intuitively it feels like the answer is simply a square with the given area.

Imagine a square with side length b.

Stretching it into a rectangle with the same area increases the length of the long diagonal.

If we keep the base and height fixed and "tilt" it into a parallelogram with the same area, clearly the longer diagonal just keeps getting longer and longer the more we tilt it over.

Geometry is not my strong suit but I took a stab at proving this.  Given an area A, we have two independent variables: the length of the base b (which we'll choose to be the longer of the sides, and thus has a minimum length sqrt(A)), and the size of the larger interior angle C (which can range from 90 to 180 degrees).

With these we can determine the length of the other side of the parallelogram: a = A/(b * sin(C)).

The length of the longest diagonal L = sqrt( a^2 + b^2 - 2ab cos(C) ).

Substituting for a and simplifying I ended up with:

L = sqrt( A^2*b^2*csc^2(C) + b^2 - 2A cot(C) )

csc^2(C) and -cot(C) are both strictly increasing as C increases from 90 to 180 degrees.  So those terms are minimized when C = 90.  In fact csc^2(90) = 1 and cot(90) = 0 so plugging those in we get:

L = sqrt( b^2 * (A^2 + 1) )  = b * sqrt(A^2 + 1)

And since we decided at the outset that b was the longer of the two sides, then this is minimized when b = sqrt(A).

Therefore longest diagonal is minimized when b = sqrt(A) and C = 90, which is a square.

Edited on April 28, 2021, 2:41 pm
 Posted by tomarken on 2021-04-28 08:46:36 Please log in:

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