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?
(In reply to Solution
At each corner of the boundary triangle, there is one of the small tesselating triangles, pointed inward, but 2 one-quarter sized triangles missing, leaving only a diamond within the tesselating triangle. It has half the area of a tesselating triangle but is missing 2/3 of the ball coverage of an ordinary (interior) tesselating triangle, so instead of pi/(2*sqrt(3)) of that piece's area being covered, only pi/(3*sqrt(3)) is covered, or .60459978807807....
Along the edges (other than their ends, at the corners mentioned above), the triangle of height sqrt(3) is truncated to a trapezoid of height 1 (being the radius of each circle). So the area is the sqrt(3) area of the original triangle minus the area of the small triangle that's cut off: ((sqrt(3)-1)^2)/sqrt(3) = 2 - 1/sqrt(3). Of this, pi/3 is covered by two circular sectors, making the covered fraction pi/(3*(2-1/sqrt(3))), or .736089515583....
But the boundary triangle also has pieces of other tesellating triangles intruding between these trapezoids, so it's best to calculate the area of the entire boundary triangle.
The height consists primarily of four rows of tesellating triangles, of height sqrt(3) each. At the top is the height of one of the corner diamonds: also sqrt(3). At the bottom is a border of height 1, the radius of the circles. That adds up to 5*sqrt(3)+1, making the total area (5*sqrt(3)+1)^2/sqrt(3), or 53.8786204584....
The 15 balls add up to 15*pi = 47.1238898038.... The ratio is .8746305937848....
So the percentage of the triangle covered increases from about 87.463% with 15 balls to about 90.690% asymptotically as the number of balls increases without limit.
Posted by Charlie
on 2004-09-17 14:17:18