# Homework Help: Unit Disc, Dyadic Squares

1. Nov 9, 2009

### JG89

1. The problem statement, all variables and given/known data

Given $$\epsilon > 0$$, show that the unit disc contains finitely many dyadic squares whose total area exceeds $$\pi - \epsilon$$, and which intersect each other only along their boundaries.

2. Relevant equations

3. The attempt at a solution

I've tried to solve this two ways. First, by finding a series of dyadic squares whose area converges to pi. But I realized it would be very complicated to show that these squares are all contained in the unit disc, and in addition, intersect each other only along their boundaries.

For my second method, I thought that if I could prove the original statement, but for the portion of the semi-circle in the first-quadrant, which is described by the equation $$\sqrt{1 - x^2}$$. I thought I could take a look at the Riemann sums and choose my partition cleverly so that it's only dyadic squares that are approximating the area under the graph. But then I realized that these dyadic squares would also intersect some other dyadic square at only a vertex, which isn't allowed by this problem. So I dropped this idea too.

One thing that I proved, which may help with this, is that for any two dyadic squares of the same size, they are either identical, intersect eachother along an edge, intersect at a vertex, or are completely distinct.

Any ideas?
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Nov 9, 2009

### fantispug

Surely intersecting only at a vertex is allowed by "intersecting only along their boundaries", since the vertex is part of the boundary. From a practical perspective you'd be doing this to find the area of the circle, so you'd just want the area of overlap to be zero.

I don't see what's wrong with your second method.

3. Nov 9, 2009

### JG89

Nice! I had a lot of hope in my second method, as that was turning out some interesting results.

Time to get back to work...thanks for the reply :)