Hofstadter points

In plane geometry, a Hofstadter point is a special point associated with every plane triangle. In fact there are several Hofstadter points associated with a triangle. All of them are triangle centers. Two of them, the Hofstadter zero-point and Hofstadter one-point, are particularly interesting. They are two transcendental triangle centers. Hofstadter zero-point is the center designated as X(360) and the Hofstafter one-point is the center denoted as X(359) in Clark Kimberling's Encyclopedia of Triangle Centers. The Hofstadter zero-point was discovered by Douglas Hofstadter in 1992.

Hofstadter triangles

Let △ABC be a given triangle. Let r be a positive real constant.

Rotate the line segment BC about B through an angle rB towards A and let L_BC be the line containing this line segment. Next rotate the line segment BC about C through an angle rC towards A. Let L'_BC be the line containing this line segment. Let the lines L_BC and L'_BC intersect at A(r). In a similar way the points B(r) and C(r) are constructed. The triangle whose vertices are A(r), B(r), C(r) is the Hofstadter r-triangle (or, the r-Hofstadter triangle) of △ABC.

Special case

• The Hofstadter 1/3-triangle of triangle △ABC is the first Morley's triangle of △ABC. Morley's triangle is always an equilateral triangle.
• The Hofstadter 1/2-triangle is simply the incentre of the triangle.

Trilinear coordinates of the vertices of Hofstadter triangles

The trilinear coordinates of the vertices of the Hofstadter r-triangle are given below:

$${\begin{array}{ccccccc}A(r)&=&1&:&{\frac {\sin rB}{\sin(1-r)B}}&:&{\frac {\sin rC}{\sin(1-r)C}}\\[2pt]B(r)&=&{\frac {\sin rA}{\sin(1-r)A}}&:&1&:&{\frac {\sin rC}{\sin(1-r)C}}\\[2pt]C(r)&=&{\frac {\sin rA}{\sin(1-r)A}}&:&{\frac {\sin(1-r)B}{\sin rB}}&:&1\end{array}}$$

Hofstadter points

Animation showing various Hofstadter points. H₀ is the Hofstadter zero-point. H₁ is the Hofstadter one-point. The little red arc in the center of the triangle is the locus of the Hofstadter r-points for 0 < r < 1. This locus passes through the incenter I of the triangle.

For a positive real constant r > 0, let A(r), B(r), C(r) be the Hofstadter r-triangle of triangle △ABC. Then the lines AA(r), BB(r), CC(r) are concurrent. The point of concurrence is the Hofstdter r-point of △ABC.

Trilinear coordinates of Hofstadter r-point

The trilinear coordinates of the Hofstadter r-point are given below.

$${\frac {\sin rA}{\sin(A-rA)}}\ :\ {\frac {\sin rB}{\sin(B-rB)}}\ :\ {\frac {\sin rC}{\sin(C-rC)}}$$

Hofstadter zero- and one-points

The trilinear coordinates of these points cannot be obtained by plugging in the values 0 and 1 for r in the expressions for the trilinear coordinates for the Hofstadter r-point.

The Hofstadter zero-point is the limit of the Hofstadter r-point as r approaches zero; thus, the trilinear coordinates of Hofstadter zero-point are derived as follows:

$${\begin{array}{rccccc}\displaystyle \lim _{r\to 0}&{\frac {\sin rA}{\sin(A-rA)}}&:&{\frac {\sin rB}{\sin(B-rB)}}&:&{\frac {\sin rC}{\sin(C-rC)}}\\[4pt]\implies \displaystyle \lim _{r\to 0}&{\frac {\sin rA}{r\sin(A-rA)}}&:&{\frac {\sin rB}{r\sin(B-rB)}}&:&{\frac {\sin rC}{r\sin(C-rC)}}\\[4pt]\implies \displaystyle \lim _{r\to 0}&{\frac {A\sin rA}{rA\sin(A-rA)}}&:&{\frac {B\sin rB}{rB\sin(B-rB)}}&:&{\frac {C\sin rC}{rC\sin(C-rC)}}\end{array}}$$

Since $\lim _{r\to 0}{\tfrac {\sin rA}{rA}}=\lim _{r\to 0}{\tfrac {\sin rB}{rB}}=\lim _{r\to 0}{\tfrac {\sin rC}{rC}}=1,$

$$\implies {\frac {A}{\sin A}}\ :\ {\frac {B}{\sin B}}\ :\ {\frac {C}{\sin C}}\quad =\quad {\frac {A}{a}}\ :\ {\frac {B}{b}}\ :\ {\frac {C}{c}}$$

The Hofstadter one-point is the limit of the Hofstadter r-point as r approaches one; thus, the trilinear coordinates of the Hofstadter one-point are derived as follows:

$${\begin{array}{rccccc}\displaystyle \lim _{r\to 1}&{\frac {\sin rA}{\sin(A-rA)}}&:&{\frac {\sin rB}{\sin(B-rB)}}&:&{\frac {\sin rC}{\sin(C-rC)}}\\[4pt]\implies \displaystyle \lim _{r\to 1}&{\frac {(1-r)\sin rA}{\sin(A-rA)}}&:&{\frac {(1-r)\sin rB}{\sin(B-rB)}}&:&{\frac {(1-r)\sin rC}{\sin(C-rC)}}\\[4pt]\implies \displaystyle \lim _{r\to 1}&{\frac {(1-r)A\sin rA}{A\sin(A-rA)}}&:&{\frac {(1-r)B\sin rB}{B\sin(B-rB)}}&:&{\frac {(1-r)C\sin rC}{C\sin(C-rC)}}\end{array}}$$

Since $\lim _{r\to 1}{\tfrac {(1-r)A}{\sin(A-rA)}}=\lim _{r\to 1}{\tfrac {(1-r)B}{\sin(B-rB)}}=\lim _{r\to 1}{\tfrac {(1-r)C}{\sin(C-rC)}}=1,$

$$\implies {\frac {\sin A}{A}}\ :\ {\frac {\sin B}{B}}\ :\ {\frac {\sin C}{C}}\quad =\quad {\frac {a}{A}}\ :\ {\frac {b}{B}}\ :\ {\frac {c}{C}}$$