## Monday, July 27, 2015

### The Geometric Mean

I've been thinking a bit about the geometric mean this week after it turned up in two separate
problems I've been looking at. In the first which is fairly simple given 2 circles that are tangent to each other the segment from the their intersection to the tangents on the sides is equal to the geometric mean of their radii i.e. $\overline{CD} = \sqrt{\overline{CB} * \overline{AC}}$.  Hint: it helps that triangle ADB is a right triangle.

In the second there is a triangle with an altitude that has two more perpendicular segments drawn from the other sides to it (point D) and then 3 inscribed circles are added in.  Again the middle circle's radius is the geometric mean of the other two.  $\overline{MN} = \sqrt{\overline{KL} * \overline{HI}}$. This figure has a lot of similar triangles.

I can prove both problems separately but it seems like they are connected and I don't yet see how to do so.

## Further Riff

http://blog.ifem.co.uk/am-gm-inequality/ has a discussion of the arithmetic - geometric mean inequality theorem I happened to read this morning.  Its very handy for solving the following: $(a+b)(b+c)(a+c) \geq 8abc$