TITLE: Can an arbitrary collection of circles of total area 1/2 fit into a circle of area 1?
QUESTION [53 upvotes]: Assume the circles are actually open disks, otherwise two circles each of area $\frac{1}{4}$ wouldn't fit into the circle of area 1. 
This seems like it should be true, thinking about packing density, but I've not been able to find an algorithm that works in all cases.

REPLY [4 votes]: After weeks of sweating through computations without success, I think I have an auxillary result which I can use to tackle the problem posted above.  However, the path to the conclusion is so surprising that I invite verification, in particular to point out any show-stopper mistake that may or may not be lurking.
The auxillary result involves a region that is the "suitcase" circle with a segment removed.  Let C be a disk of radius 1, and draw a chord of C which has distance 1/2 from the center of C.  Let D be that part of C (including the chord itself) that is on the same side of the chord as the center of C, so D is a disk with a segment chopped off so that the area of D is less than 5/6 the area of C.  Let F and G be two closed disks with unequal radii whose areas sum to half the area of C.
Result:  F and G pack inside D.
Note that if F and G have the same radius, they will not pack inside D for the same reason they won't pack inside C: they have to share a point which is the circle center.
The proof of the result involves placing F and G so that they touch the chord and otherwise are as far apart inside D as possible.  The analogous problem for packing inside C involves placing F and G inside C so that the centers of F, G, and C are on the same line and F and G are as far apart as possible, then use the area sum condition to show the radii of F and of G sum to less than 1, and so they can be pushed together just enough to remain disjoint and keep away from the region boundary.
For the result about D, I orient the coordinates to have the chord parallel to the x axis, and then I show that the x coordinates of the centers of F and G are greater than distance 1 apart.  It took me weeks to realize that was what was needed to be shown.  Even so, the algebra involved led to a surprise, which I want someone else to confirm or deny.
Now for the algebraic verification.  Let $0 \leq s \lt 1/2 \lt r \leq \sqrt{1/2}$ where $s$ and $r$ are the lengths of the radii of F and G.  The area sum condition yields $r^2 + s^2 = 1/2$, and the inequality to be shown is inequality (A): $\sqrt{3/4 - r} + \sqrt{3/4 - s} \gt 1$.  Since this inequality on the x coordinates of the centers of F and G imply that F and G are disjoint, inequality (A) will yield the result.
There may be a slick analytic way to show (A), but I use repeated squaring and rearrangments to yield a quartic polynomial in $r$.  I start by writing $s$ as $\sqrt{1/2 - r^2}$ and isolate the radical with $s$ on one side, and square both sides. I end up with an inequality to prove that has $\sqrt{1/2 - r^2}$ as a subexpression, and so I isolate the summand containing that subexpression on one side and square again.
I collect and rearrange and square again to get the following inequality to be shown:  $49/4 - 42r + 50r^2 - 24r^3 + 4r^4 \gt (1 - 2r + r^2)(12 - 16r)$.  I then collect terms and decide to factor out a term of $(r - 1/2)$ in hopes of getting an easier polynomial to handle to show that it is positive for $r$ in the range given above.
I know the relations above yield 0 when $r$ is $1/2$, so it seems like a nice simplification.
The surprise comes when I find out the above inequality is equivalent to (B): $4(r - 1/2)^4 \gt 0$.  I was not expecting that at all!
If I have it right, then the result is proved, and I can move on after weeks of fruitless labor involving delta and epsilons and sketches that had no $r$'s or $s$'s.  Can someone rederive or otherwise prove inequality A above?  Even better, can someone tell me how I could know inequality B was coming?
Gerhard "Surprised After All These Weeks" Paseman, 2012.05.07