## Necklace of circles

### Problem 428

Published on Sunday, 19th May 2013, 01:00 am; Solved by 133Let `a`, `b` and `c` be positive numbers.

Let W, X, Y, Z be four collinear points where |WX| = `a`, |XY| = `b`, |YZ| = `c` and |WZ| = `a` + `b` + `c`.

Let C_{in} be the circle having the diameter XY.

Let C_{out} be the circle having the diameter WZ.

The triplet (`a`, `b`, `c`) is called a *necklace triplet* if you can place `k` ≥ 3 distinct circles C_{1}, C_{2}, ..., C_{k} such that:

- C
_{i}has no common interior points with any C_{j}for 1 ≤`i`,`j`≤`k`and`i`≠`j`, - C
_{i}is tangent to both C_{in}and C_{out}for 1 ≤`i`≤`k`, - C
_{i}is tangent to C_{i+1}for 1 ≤`i`<`k`, and - C
_{k}is tangent to C_{1}.

For example, (5, 5, 5) and (4, 3, 21) are necklace triplets, while it can be shown that (2, 2, 5) is not.

Let T(`n`) be the number of necklace triplets (`a`, `b`, `c`) such that `a`, `b` and `c` are positive integers, and `b` ≤ `n`.
For example, T(1) = 9, T(20) = 732 and T(3000) = 438106.

Find T(1 000 000 000).