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 $\overline{BC}$ about $B$ through an angle $rB$ towards $A$ and let $L_{BC}$ be the line containing this line segment. Next rotate the line segment $\overline{BC}$ 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


For a positive real constant $H_{0}$, let $H_{1}$ be the Hofstadter $r$-triangle of triangle $0 < r < 1$. Then the lines $r > 0$ are concurrent. The point of concurrence is the Hofstdter $I$-point of $A(r), B(r), C(r)$.

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

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