Posted by: Dave Richeson | December 30, 2008

## 2009 Times Square New Year’s Eve Ball

We can look forward to a very mathematical New Year’s eve ball in Times Square this year.

According to the Times Square website:

The new Times Square New Year’s Eve Ball is a 12 foot geodesic sphere, double the size of previous Balls, and weighs 11,875 pounds. Covered in 2,668 Waterford Crystals and powered by 32,256 Philips Luxeon Rebel LEDS, the new Ball is capable of creating a palette of more than 16 million vibrant colors and billions of patterns producing a spectacular kaleidoscope effect atop One Times Square.

Mathematical fact:

Think of this ball as a giant polyhedron with triangular faces (and who wouldn’t!). Notice that the vertices all have degree 5 or 6 (the number of edges meeting at each corner). Without knowing anything else about this shape we can conclude that there are exactly twelve vertices of degree 5. (OK, I am assuming it is truly a sphere and doesn’t have a pole going through its axis.)

General result: if a spherical polyhedron has triangular faces and 5 or 6 edges meet at each vertex, then there are exactly twelve vertices of degree 5.

This theorem has a dual formulation too: if a spherical polyhedron has pentagonal and hexagonal faces and three edges meet at each vertex, then there are exactly 12 pentagons (just think of a traditional soccer ball with twelve black pentagonal patches).

How to prove this? I’ll leave it as an exercise, but here’s a hint. It uses my favorite theorem.

Euler’s polyhedron formula: a spherical polyhedron with $V$ vertices, $E$ edges, and $F$ faces satisfies $V-E+F=2$.