Hofstadter points

File:HofstadterPoint.svg

Let {{math|△ABC}} be a given triangle. Let {{mvar|r}} be a positive real constant.

Rotate the line segment {{mvar|{{overline|BC}}}} about {{mvar|B}} through an angle {{mvar|rB}} towards {{mvar|A}} and let {{mvar|L_BC}} be the line containing this line segment. Next rotate the line segment {{mvar|{{overline|BC}}}} about {{mvar|C}} through an angle {{mvar|rC}} towards {{mvar|A}}. Let {{mvar|L'_BC}} be the line containing this line segment. Let the lines {{mvar|L_BC}} and {{mvar|L'_BC}} intersect at {{math|A(r)}}. In a similar way the points {{math|B(r)}} and {{math|C(r)}} are constructed. The triangle whose vertices are {{math|A(r), B(r), C(r)}} is the Hofstadter {{mvar|r}}-triangle (or, the {{mvar|r}}-Hofstadter triangle) of {{math|△ABC}}.{{cite web|last=Weisstein|first=Eric W.|title=Hofstadter Triangle|url=http://mathworld.wolfram.com/HofstadterTriangle.html|publisher=MathWorld--A Wolfram Web Resource|accessdate=11 May 2012}}

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

The trilinear coordinates of the vertices of the Hofstadter {{mvar|r}}-triangle are given below:

$$\backslash begin\{array\}\{ccccccc\}$$

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}

File:HofstadterPointAnimation.gif {{mvar|I}} of the triangle.]]

For a positive real constant {{math|r > 0}}, let {{math|A(r), B(r), C(r)}} be the Hofstadter {{mvar|r}}-triangle of triangle {{math|△ABC}}. Then the lines {{math|AA(r), BB(r), CC(r)}} are concurrent.{{cite journal|last=C. Kimberling|title=Hofstadter points|journal=Nieuw Archief voor Wiskunde|year=1994|volume=12|pages=109–114}} The point of concurrence is the Hofstdter {{mvar|r}}-point of {{math|△ABC}}.

The trilinear coordinates of the Hofstadter {{mvar|r}}-point are given below.

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

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

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

$$\backslash begin\{array\}\{rccccc\}$$

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

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

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

$$\backslash begin\{array\}\{rccccc\}$$

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

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

{{reflist}}

{{Douglas Hofstadter}}

Category:Triangle centers

Category:1992 introductions