Discs and tablecloth: Oscar needs more than 10 points to thwart me. The circles can jiggle around too much.

Is that a mathy enough answer for ya'?

No? Ok.

Here's the outline of my proof:

0) Oscar secretly places the points on the table trying to thwart me. I have no idea where they are.

1) I tile the table with infinite number of discs such that they are maximally dense and give the tiling a good spin so that what I've covered is uniformly random.

2) Let Ei be the event that I covered point i (i = 1,2,...,10).

3) I have covered all ten points with probability Pr(E1 & E2 & E3 & ... & E10).

3a) Pr(E1 & E2 & E3 & ... & E10) = 1 - Pr(notE1 or notE2 or notE3 or ... or notE10) >= 1 - Pr(notE1) - Pr(notE2) - ... - Pr(notE10) > 0 (see detail)

4) The probability I have covered the 10 points is positive, so there is some way to cover those 10 points no matter how Oscar places them (pick one and call it the special tiling).

5) What, so we're only allowed to use 10 discs? Remove discs from the special tiling which do not cover a point. We must be left with 10 or fewer discs.

Detail: Pr(notEi) = 1 - pi/(2*sqrt(3)) or approximately .09. To see this, draw equilateral triangles over the circle tiling such that the center of the circles are the vertices of the triangles. Each triangle has area sqrt(3)*r^2. Each triangle is covered by half a circle, area pi*r^2/2. The ratio is pi/(2*sqrt(3)). The ratio of empty space to covered space in the tiling is thus 1 - pi/(2*sqrt(3)).

I should not have spent my morning on this.....