MHB Problem of the Week # 220 - Jun 14, 2016

[Ackbach]

Here is this week's POTW:

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Find the least number $A$ such that for any two squares of combined area $1$, a rectangle of area $A$ exists such that the two squares can be packed into the rectangle (without interior overlap). You may assume that the sides of the squares are parallel to the sides of the rectangle.

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[Ackbach]

Re: Problem Of The Week # 220 - Jun 14, 2016

This was Problem A-1 in the 1996 William Lowell Putnam Mathematical Competition.

No one answered this week's POTW correctly, though I'm happy to give a shout-out to kiwi for a valiant attempt. The solution, attributed to Kiran Kedlaya and his associates, follows:

Spoiler

If $x$ and $y$ are the sides of two squares with combined area 1, then
$x^2 + y^2 = 1$. Suppose without loss of generality that $x \geq y$.
Then the shorter side of a rectangle containing both squares without
overlap must be at least $x$, and the longer side must be at least
$x+y$. Hence the desired value of $A$ is the maximum of $x(x+y)$.

To find this maximum, we let $x = \cos \theta, y = \sin \theta$ with
$\theta \in [0, \pi/4]$. Then we are to maximize
\begin{align*}
\cos^2 \theta + \sin \theta \cos \theta
&= \frac 12 (1 + \cos 2\theta + \sin 2\theta) \\
&= \frac 12 + \frac{\sqrt{2}}{2} \cos (2\theta - \pi/4) \\
&\leq \frac{1 + \sqrt{2}}{2},
\end{align*}
with equality for $\theta = \pi/8$. Hence this value is the desired
value of $A$.