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

File:HofstadterPoint.svg

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:

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

Hofstadter points

File:HofstadterPointAnimation.gif

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:

$$\begin{array}{cccccc}\underset{r\to 0}{\lim }& \frac{\sin rA}{\sin (A-rA)}& :& \frac{\sin rB}{\sin (B-rB)}& :& \frac{\sin rC}{\sin (C-rC)}\\ ⟹\underset{r\to 0}{\lim }& \frac{\sin rA}{r\sin (A-rA)}& :& \frac{\sin rB}{r\sin (B-rB)}& :& \frac{\sin rC}{r\sin (C-rC)}\\ ⟹\underset{r\to 0}{\lim }& \frac{A\sin rA}{rA\sin (A-rA)}& :& \frac{B\sin rB}{rB\sin (B-rB)}& :& \frac{C\sin rC}{rC\sin (C-rC)}\end{array}$$

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

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

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:

$$\begin{array}{cccccc}\underset{r\to 1}{\lim }& \frac{\sin rA}{\sin (A-rA)}& :& \frac{\sin rB}{\sin (B-rB)}& :& \frac{\sin rC}{\sin (C-rC)}\\ ⟹\underset{r\to 1}{\lim }& \frac{(1-r)\sin rA}{\sin (A-rA)}& :& \frac{(1-r)\sin rB}{\sin (B-rB)}& :& \frac{(1-r)\sin rC}{\sin (C-rC)}\\ ⟹\underset{r\to 1}{\lim }& \frac{(1-r)A\sin rA}{A\sin (A-rA)}& :& \frac{(1-r)B\sin rB}{B\sin (B-rB)}& :& \frac{(1-r)C\sin rC}{C\sin (C-rC)}\end{array}$$

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

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