What is the Point?

Given a reference triangle $ABC$ , construct similar isosceles triangles $A'BC$ , $B'AC$ and $C'AB$ externally with the edges $BC$ , $AC$ and $AB$ as respective bases, where the base angle of these isosceles triangles is $\theta$ . Then the lines $AA'$ , $BB'$ , $CC'$ are concurrent at the point with trilinear coordinates $\csc(A + \theta) \,:\, \csc(B+ \theta)\;:\; \csc(C+\theta)$ When $\theta=0$ , we obtain the centroid $X(2)$ . When $\theta = \tfrac16\pi$ , we have the first Napoleon point $X(17)$ . When $\theta = \tfrac12\pi$ , we obtain the orthocentre $X(4)$ . As $\theta$ varies, all these points lie on the so-called Kiepert hyperbola .

Constructing a regular pentagon externally on the side of a triangle, and finding its centroid, is essentially constructing an isosceles triangle with base angle $\theta = \tfrac12\pi - \tfrac15\pi$ , and so this problem is concerned with the point with triangle center function $\csc(A + \tfrac12\pi - \tfrac15\pi) \; = \; \sec(\tfrac15\pi - A) \; =\; \sec(A - \tfrac15\pi)$ While the Encyclopedia for Triangle Centers cannot, obviously, list the uncountably infinite number of points that lie on the Kiepert hyperbola, it does list the so-called $\sec(A - \tfrac15\pi)$ point $X(\boxed{3381})$ . For those with access to Geometer's Sketchpad, here is a visualisation of this triangle centre.

To search the ETC for a point that you suspect is a triangle center, construct the point in one of the three "search" triangles shown near the bottom of this page . I chose the 6,9,13 triangle and constructed the figure in Geogebra . Notice the perpendicular distance between $X$ and $CB \approx 0.4081575$ . Look for 0.4081575 in the fourth column of the table. Then find $\boxed{3381}$ in the third column of the same row. Now find $X(3381)$ in Part 3 . It is given the name "X(3381) = SEC(A - π/5) POINT" as well as other information about the point.