My son took a pretty computational approach to the problem and the calculations gave him a little difficulty. Today my main idea was to share Ian Agol’s neat solution, but I wanted to review one piece of yesterday’s calculation with him as an arithmetic review:

Now we moved on to looking at Ian Agol’s solution (using the picture made by Greg Egan):

1/3. Rotate this picture around the bottom vertex v to get 12 tangent circles equidistant to v. Since the circle picture is symmetric, we can rotate by π/6 about v, preserving the circle configuration. This sends the aqua hexagon to one whose vertices lie at the centers.