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 Hofstadter 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 Template:Math be a given triangle. Let Template:Mvar be a positive real constant.

Rotate the line segment Template:Mvar about Template:Mvar through an angle Template:Mvar towards Template:Mvar and let Template:Mvar be the line containing this line segment. Next rotate the line segment Template:Mvar about Template:Mvar through an angle Template:Mvar towards Template:Mvar. Let Template:Mvar be the line containing this line segment. Let the lines Template:Mvar and Template:Mvar intersect at Template:Math. In a similar way the points Template:Math and Template:Math are constructed. The triangle whose vertices are Template:Math is the Hofstadter Template:Mvar-triangle (or, the Template:Mvar-Hofstadter triangle) of Template:Math.

Special case

• The Hofstadter 1/3-triangle of triangle Template:Math is the first Morley's triangle of Template:Math. 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 Template:Mvar-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

Animation showing various Hofstadter points. Template:Math is the Hofstadter zero-point. Template:Math is the Hofstadter one-point. The little red arc in the center of the triangle is the locus of the Hofstadter Template:Mvar-points for Template:Math. This locus passes through the incenter Template:Mvar of the triangle.

For a positive real constant Template:Math, let Template:Math be the Hofstadter Template:Mvar-triangle of triangle Template:Math. Then the lines Template:Math are concurrent. The point of concurrence is the Hofstdter Template:Mvar-point of Template:Math.

Trilinear coordinates of Hofstadter Template:Mvar-point

The trilinear coordinates of the Hofstadter Template:Mvar-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 Template:Mvar in the expressions for the trilinear coordinates for the Hofstadter Template:Mvar-point.

The Hofstadter zero-point is the limit of the Hofstadter Template:Mvar-point as Template:Mvar 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 Template:Mvar-point as Template:Mvar 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 $\underset{r\to 1}{lim}\frac{(1-r)A}{\sin (A-rA)}=\underset{r\to 1}{lim}\frac{(1-r)B}{\sin (B-rB)}=\underset{r\to 1}{lim}\frac{(1-r)C}{\sin (C-rC)}=1,$

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