A Putnam Geometry Problem

[jay_shark]

I was looking through some old Putnam problems and came across this gem . I stumbled upon a fascinating solution that doesn't require calculus or trigonometry .
Give it a try and see what you come up with .

What is the smallest alpha such that two squares with total area 1 can always be placed inside a rectangle area alpha with sides parallel to those of the rectangle and with no overlap (of their interiors)?

[bigpooch]

Was this really on the Putnam? [seems too easy]

Answer:

<font color="white">
Such a rectangle must be able to hold two squares of side
1/sqrt(2), so it has to have length AT LEAST twice that, or
sqrt(2). The width must be 1 to hold a square with side
1-epsilon for all epsilon&gt;0, i.e., alpha = sqrt(2).
</font>

[TomCowley]

I read the question differently (on the third try)- namely what is the smallest alpha, such that for any two squares with total area 1, a rectangle of area alpha can be drawn with those two squares placed inside (parallel, no overlap, etc). It's still trivial with calculus, but I'm trying to figure out a non-calculus way to get the same answer.

[jay_shark]

Your solution is off BigPooch .

The problem is 1996 A1 , but the solutions posted on the web are done with calculus /trigonometry .

[LeadbellyDan]

Why is this not just 1?

[jay_shark]

Alpha cannot be 1 because we can always find x,y that satisfy the conditions of the problem and such that the area of the rectangle exceeds 1.

We need to find the greatest lowest bound or the infimum of the area of the rectangle which satisfy the conditions stated in the problem .