Let A(n,θ) be the maximum number of times angle θ can be formed by n points. What are the values of A(4,θ)? How about A(n,θ) for larger n? Can you prove your answers?

John Hoffman gave many bounds for A(n,θ). For example, he found that A(2n+1,θ) ≥ n² and A(2n,θ) ≥ n(n–1) by putting one point at the intersection of two lines and evenly dividing the other points on the two lines.

Hoffman can show that for large n, A(n,π/4) ≥ (1–ε)n². I can show the same thing for θ=π/2, using a square grid of points. Hoffman and I conjecture that if 0<θ<π, then log A(n,θ)/ log n → 2.

Hoffman and Devincentis showed that A(n,0)=2n(n–1)(n–2)/6 and A(n,π)=n(n–1)(n–2)/6. No more than two angles in any triangle can be θ=0 and no more than one angle can be θ=π. Arranging points in a line achieves both these bounds.

Generalizing this, one can show A(n,θ) ≤ A(n–1,θ) n / (n–3). To see this, consider all n collections of n–1 points out of n points. In each collection, at most A(n–1,θ) angles θ occur, and each such angle appears in (n-3) collections.

There are some general constructions which provide lower bounds. A(n,2π/n) ≥ n(n–2) by putting the points at the vertices of an regular n-gon. Similarly, if θ ≤ 2π/(n–1), then A(n,θ) ≥ (n–1)(n–2) because we can put the points at equally spaced points on the circumference of a circle.

Here are the best known bounds for n=3, 4 and 5.

θ A(3,θ) [ 0 , π/2 ) 2 [ π/2 , π ] 1 π/3 3

θ A(4,θ) ( 0 , π/3 ] 6 ( π/3 , π/2 ] 4 ( π/2 , 2π/3 ] 3 ( 2π/3 , π ) 2 0 8 π/4 8 2π/5 5 π 4

θ A(5,θ) ( 0 , π/4 ) 12 , 15 ( π/4 , π/2 ] 8 , 15 ( π/2 , π ) 5 , 7 0 20 π/5 15 π/4 16 , 20 π/3 11 , 15 2π/5 10 , 12 π 10

Here are the conjectured values of A(n,θ) for some special angles. Can you verify or continue any of these sequences?

If you can extend any of these results, please e-mail me. Click here to go back to Math Magic. Last updated 1/28/99.