Hofstadter_points Knowpia

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.^[1] 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.^[1]

Hofstadter triangles edit

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 $△ ABC$ .^[2]^[1]

Special case edit

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 edit

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 edit

Animation showing various Hofstadter points.

H 0

is the Hofstadter zero-point.

H 1

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.^[3] The point of concurrence is the Hofstdter $r$ -point of $△ ABC$ .

Trilinear coordinates of Hofstadter $r$ -point edit

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 edit

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 Hofstdter $r$ -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}}

References edit

^ ^a ^b ^c Kimberling, Clark. "Hofstadter points". Retrieved 11 May 2012.
^ Weisstein, Eric W. "Hofstadter Triangle". MathWorld--A Wolfram Web Resource. Retrieved 11 May 2012.
^ C. Kimberling (1994). "Hofstadter points". Nieuw Archief voor Wiskunde. 12: 109–114.