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 .

[jay_shark]

Bump

[blah_blah]

[ QUOTE ]
Bump

[/ QUOTE ]

nb the area of the rectangle is fixed, but not its dimensions. so we want to maximize x(x+y) subject to the condition that x^2+y^2=1, since the optimal configuration will have the rectangles sharing (part of) a side.

i don't know a way of doing this with am-gm like techniques, so put x = cos t, y = sin t for t\in (0,\pi).

then we want to maximize cos t (cos t + sin t). i believe that you can write this as a single trigonometric function, but it's simple enough to differentiate and get cos 2t - sin 2t, which implies that 2t = pi/4 or t = pi/8. now all that remains is to check that this is optimal.

note that it's a little tricky to plug in pi/8 into cos t(cos t + sin t) unless you know the half angle formulas; it's probably better to use the double angle formulas to convert cos^2 t and cos t sin t into trigonometric functions in 2t.

[Siegmund]

Grrrr, jay.

This was one of the years I took the Putnam, and I felt this was an extremely easy question - until finding out, after I took the test, that I had interpreted the question quite differently than the writers intended it (apparently a majority, but not all, of my fellow test-takers understood it as intended.)

I remain as annoyed today as I was the day after I found out.

[tshort]

[ QUOTE ]
then we want to maximize cos t (cos t + sin t). i believe that you can write this as a single trigonometric function, but it's simple enough to differentiate and get cos 2t - sin 2t, which implies that 2t = pi/4 or t = pi/8. now all that remains is to check that this is optimal.

[/ QUOTE ]

(cos t)^2 + sin(t)cost(t)

= (1+cos(2t))/2 + (sin(2t)+sin(0))/2)

= 1/2 + 1/2 cos2t + 1/2 sin2t

= 1/2 + Sqrt[1/2] cos(2t-Pi/4)

Then obviously this is maximized when 2t-Pi/4 = 0.