A pool rack is an equilateral triangle, filled with 15 equal-sized balls. Seen from above, we'd see a triangle, with 15 circles within.

Imagine we used smaller and smaller balls. The more the balls, more area of the triangle would be covered.

In the limit, with infinite balls, would all of the triangle be covered?

It's going to be hard to explain this without a picture, but I'll try to be as clear as possible.

First, consider n layers of balls, each with a radius of r. The top layer would be 1, next 2, etc. There will be a total of n(n+1)/2 balls that cover an area of pi*r^2*n(n+1)/2.

The hard part to visualize is how to find the side of the triangle. The side of the big triangle consists of n*2r plus 2 little pieces left over.

Consider looking only at the top layer with 1 circle. If you draw a line through the vertex and the center of the circle, it will bisect the angle from 60 degrees to 30 degrees (by symmetry). Consider the 30-60-90 triangle formed by the vertex of the big triangle, the center of the circle, and the point of tangency.

One of the legs will be r. The other one of the legs will be r/(tan 30) = r*sqrt(3). This is exactly r + the little piece left over. Therefore the little piece left over is r*(sqrt(3)-1). Meaning the triangle has a side of n*2r+2*r(sqrt(3)-1).

The formula for the area of an equilateral triangle of side s is s^2*sqrt(3)/4. Plugging this in, the area becomes r^2*sqrt(3)*(sqrt(3)-1+n)^2.

Taking the ratio of the area covered by the circles to the area of the triangle, we get [n(n+1)*pi]/[2*sqrt(3)*(sqrt(3)-1+n)^2], which does not depend on r. Taking the limit as n approaches infinity, we get pi/(2*sqrt(3)), which equates to 0.9069 which is shown in Charlies solution.

Posted by np_rt on 2004-09-17 21:33:55