Happy New Year to all Hyacinthists.

Paul wrote :

> Dear Clark and friends,

>

> Happy New Year!

> Consider the circle tangent internally to each of the

excircles of

> triangle ABC. This circle can be constructed as the inversive image

of the

> nine-point circle in the radical circle of the excircles. It has

long been

> known that the radius of this circle is (r^2+s^2)/(4r). See, for

example,

> Problem 1864 and solution, Crux Math. 20 (1994) 174--175. What is

interesting

> is that the center is the point X(970) in ETC. As such, this center

lies on

> the Brocard axis. Is there a simple explanation of this fact?

Your circle intersects BC at (in barycentrics)

(0,s-cc,cc) and (0,bb,s-bb) where s = -2abc/(a+b+c).

Thus your circle is a Tucker circle and he is centered on OK.

In fact, as the radical center of the excircles is the Spieker point

S, the center of your circle is the common point of OK and SN

(N=NPcenter), which is, as you've noticed X(970).

Friendly. Jean-Pierre