Newton–Gauss line

{{main|Complete quadrilateral}}

Any four lines in general position (no two lines are parallel, and no three are concurrent) form a complete quadrilateral. This configuration consists of a total of six points, the intersection points of the four lines, with three points on each line and precisely two lines through each point.{{Cite web|url=https://www.researchgate.net/publication/266061112|title=Gauss–Newton Lines and Eleven Point Conics|last=Alperin|first=Roger C.|author-link=Roger C. Alperin|date=6 January 2012|website=Research Gate}} These six points can be split into pairs so that the line segments determined by any pair do not intersect any of the given four lines except at the endpoints. These three line segments are called diagonals of the complete quadrilateral.

File:Labeling for a complete quadrilateral.svg

It is a well-known theorem that the three midpoints of the diagonals of a complete quadrilateral are collinear.{{harvnb|Johnson|2007|loc=p. 62}}

There are several proofs of the result based on areas or wedge products{{citation|first=Dan|last=Pedoe|title=Geometry A Comprehensive Course|year=1988|orig-year=1970|publisher=Dover|pages=46–47|isbn=0-486-65812-0}} or, as the following proof, on Menelaus's theorem, due to Hillyer and published in 1920.{{harvnb|Johnson|2007|loc=p. 152}}

Let the complete quadrilateral {{mvar|ABCA'B'C'}} be labeled as in the diagram with diagonals {{mvar|{{overline|AA'}}, {{overline|BB'}}, {{overline|CC'}}}} and their respective midpoints {{mvar|L, M, N}}. Let the midpoints of {{mvar|{{overline|BC}}, {{overline|CA'}}, {{overline|A'B}}}} be {{mvar|P, Q, R}} respectively. Using similar triangles it is seen that {{mvar|QR}} intersects {{mvar|{{overline|AA'}}}} at {{mvar|L}}, {{mvar|RP}} intersects {{mvar|{{overline|BB'}}}} at {{mvar|M}} and {{mvar|PQ}} intersects {{mvar|{{overline|CC'}}}} at {{mvar|N}}. Again, similar triangles provide the following proportions,

:$\backslash frac\{\backslash overline\{RL\}\}\{\backslash overline\{LQ\}\}\; =\; \backslash frac\{\backslash overline\{BA\}\}\{\backslash overline\{AC\}\},\; \backslash quad\; \backslash frac\{\backslash overline\{QN\}\}\{\backslash overline\{NP\}\}\; =\; \backslash frac\{\backslash overline\{A\text{'}C\text{'}\}\}\{\backslash overline\{C\text{'}B\}\},\; \backslash quad\; \backslash frac\{\backslash overline\{PM\}\}\{\backslash overline\{MR\}\}\; =\; \backslash frac\{\backslash overline\{CB\text{'}\}\}\{\backslash overline\{B\text{'}A\text{'}\}\}.$

However, the line {{math|A'B'C }} intersects the sides of triangle {{math|△ABC}}, so by Menelaus's theorem the product of the terms on the right hand sides is −1. Thus, the product of the terms on the left hand sides is also −1 and again by Menelaus's theorem, the points {{mvar|L, M, N}} are collinear on the sides of triangle {{math|△PQR}}.

The following are some results that use the Newton–Gauss line of complete quadrilaterals that are associated with cyclic quadrilaterals, based on the work of Barbu and Patrascu.{{Cite web|url=http://forumgeom.fau.edu/FG2012volume12/FG201212.pdf|title=Some Properties of the Newton–Gauss Line|last=Patrascu|first=Ion|date=|website=Forum Geometricorum|access-date=29 April 2019|archive-date=29 March 2023|archive-url=https://web.archive.org/web/20230329151523/https://forumgeom.fau.edu/FG2012volume12/FG201212.pdf|url-status=dead}}

File:Newton-Gauss_Line_Figure_1_subtext.png

Given any cyclic quadrilateral {{mvar|ABCD}}, let point {{mvar|F}} be the point of intersection between the two diagonals {{mvar|{{overline|AC}}}} and {{mvar|{{overline|BD}}}}. Extend the diagonals {{mvar|{{overline|AB}}}} and {{mvar|{{overline|CD}}}} until they meet at the point of intersection, {{mvar|E}}. Let the midpoint of the segment {{mvar|{{overline|EF}}}} be {{mvar|N}}, and let the midpoint of the segment {{mvar|{{overline|BC}}}} be {{mvar|M}} (Figure 1).

If the midpoint of the line segment {{mvar|{{overline|BF}}}} is {{mvar|P}}, the Newton–Gauss line of the complete quadrilateral {{mvar|ABCDEF}} and the line {{mvar|PM}} determine an angle {{math|∠PMN}} equal to {{math|∠EFD}}.

First show that the triangles {{math|△NPM, △EDF}} are similar.

Since {{math|BE ∥ PN}} and {{math|FC ∥ PM}}, we know {{math|1=∠NPM = ∠EAC}}. Also, $\backslash tfrac\{\backslash overline\{BE\}\}\{\backslash overline\{PN\}\}\; =\; \backslash tfrac\{\backslash overline\{FC\}\}\{\backslash overline\{PM\}\}\; =\; 2.$

In the cyclic quadrilateral {{mvar|ABCD}}, these equalities hold:

:$\backslash begin\{align\}$

\angle EDF &= \angle ADF + \angle EDA, \\

&= \angle ACB + \angle ABC, \\

&= \angle EAC.

\end{align}

Therefore, {{math|1=∠NPM = ∠EDF}}.

Let {{math|R{{sub|1}}, R{{sub|2}}}} be the radii of the circumcircles of {{math|△EDB, △FCD}} respectively. Apply the law of sines to the triangles, to obtain:

: $\backslash frac\; \{\backslash overline\{BE\}\}\{\backslash overline\{FC\}\}=\backslash frac\; \{2R\_1\backslash sin\; \backslash angle\; EDB\}\{2R\_2\backslash sin\; \backslash angle\; FDC\}=\backslash frac\; \{R\_1\}\{R\_2\}=\backslash frac\; \{2R\_1\backslash sin\; \backslash angle\; EBD\}\{2R\_2\backslash sin\; \backslash angle\; FCD\}\; =\; \backslash frac\{\backslash overline\{DE\}\}\{\backslash overline\{DF\}\}.$

Since {{math|1={{overline|BE}} = 2 · {{overline|PN}}}} and {{math|1={{overline|FC}} = 2 · {{overline|PM}}}}, this shows the equality $\backslash tfrac\{\backslash overline\{PN\}\}\{\backslash overline\{PM\}\}\; =\; \backslash tfrac\{\backslash overline\{DE\}\}\{\backslash overline\{DF\}\}.$ The similarity of triangles {{math|△PMN, △DFE}} follows, and {{math|1=∠NMP = ∠EFD}}.

If {{mvar|Q}} is the midpoint of the line segment {{mvar|{{overline|FC}}}}, it follows by the same reasoning that {{math|1=∠NMQ = ∠EFA}}.

File:Newton-Gauss_Line_Figure_2_Subtext.png

The line through {{mvar|E}} parallel to the Newton–Gauss line of the complete quadrilateral {{mvar|ABCDEF}} and the line {{mvar|EF}} are isogonal lines of {{math|∠BEC}}, that is, each line is a reflection of the other about the angle bisector. (Figure 2)

Triangles {{math|△EDF, △NPM}} are similar by the above argument, so {{math|1=∠DEF = ∠PNM}}. Let {{mvar|E'}} be the point of intersection of {{mvar|BC}} and the line parallel to the Newton–Gauss line {{mvar|NM}} through {{mvar|E}}.

Since {{math|PN ∥ BE}} and {{math|NM ∥ EE',}} {{math|1=∠BEF = ∠PNF}}, and {{math|1=∠FNM = ∠E'EF}}.

Therefore,

:$\backslash begin\{align\}$

\angle CEE' &= \angle DEF - \angle E'EF, \\

&= \angle PNM - \angle FNM, \\

&= \angle PNF = \angle BEF.

\end{align}

File:Newton-Gauss_Line_Subtext_3.1.png

Let {{mvar|G}} and {{mvar|H}} be the orthogonal projections of the point {{mvar|F}} on the lines {{mvar|AB}} and {{mvar|CD}} respectively.

The quadrilaterals {{mvar|MPGN}} and {{mvar|MQHN}} are cyclic quadrilaterals.

{{math|1=∠EFD = ∠PMN}}, as previously shown. The points {{mvar|P}} and {{mvar|N}} are the respective circumcenters of the right triangles {{math|△BFG, △EFG}}. Thus, {{math|1=∠PGF = ∠PFG}} and {{math|1=∠FGN = ∠GFN}}.

Therefore,

: $\backslash begin\{align\}$

\angle PGN + \angle PMN

& = (\angle PGF + \angle FGN) + \angle PMN \\[4pt]

& = \angle PFG + \angle GFN + \angle EFD \\[4pt]

&= 180^\circ

\end{align}.

Therefore, {{mvar|MPGN}} is a cyclic quadrilateral, and by the same reasoning, {{mvar|MQHN}} also lies on a circle.

File:Newton-Gauss_Line_Figure_3_subtext_2.png

Extend the lines {{mvar|GF, HF}} to intersect {{mvar|EC, EB}} at {{mvar|I, J}} respectively (Figure 4).

The complete quadrilaterals {{mvar|EFGHIJ}} and {{mvar|ABCDEF}} have the same Newton–Gauss line.

The two complete quadrilaterals have a shared diagonal, {{mvar|{{overline|EF}}}}. {{mvar|N}} lies on the Newton–Gauss line of both quadrilaterals. {{mvar|N}} is equidistant from {{mvar|G}} and {{mvar|H}}, since it is the circumcenter of the cyclic quadrilateral {{mvar|EGFH}}.

If triangles {{math|△GMP, △HMQ}} are congruent, and it will follow that {{mvar|M}} lies on the perpendicular bisector of the line {{mvar|HG}}. Therefore, the line {{mvar|MN}} contains the midpoint of {{mvar|{{overline|GH}}}}, and is the Newton–Gauss line of {{mvar|EFGHIJ}}.

To show that the triangles {{math|△GMP, △HMQ}} are congruent, first observe that {{mvar|PMQF}} is a parallelogram, since the points {{mvar|M, P}} are midpoints of {{mvar|{{overline|BF}}, {{overline|BC}}}} respectively.

Therefore,

:$\backslash begin\{align\}$

& \overline{MP} = \overline{QF} = \overline{HQ}, \\

& \overline{GP} = \overline{PF} = \overline{MQ}, \\

& \angle MPF = \angle FQM.

\end{align}

Also note that

: $\backslash angle\; FPG\; =\; 2\; \backslash angle\; PBG\; =\; 2\; \backslash angle\; DBA\; =\; 2\; \backslash angle\; DCA\; =\; 2\; \backslash angle\; HCF\; =\; \backslash angle\; HQF.$

Hence,

: $\backslash begin\{align\}$

\angle MPG &= \angle MPF + \angle FPG, \\

&= \angle FQM + \angle HQF, \\

&= \angle HQF + \angle FQM, \\

&= \angle HQM.

\end{align}

Therefore, {{math|△GMP}} and {{math|△HMQ}} are congruent by SAS.

Due to {{math|△GMP, △HMQ}} being congruent triangles, their circumcircles {{mvar|MPGN, MQHN}} are also congruent.

The point at infinity along the Newton–Gauss line is the isogonal conjugate of the Miquel point.

Dao Thanh Oai showed a generalization of the Newton–Gauss line.{{Cite journal |title=Generalizations of some famous classical Euclidean geometry theorems |journal=International Journal of Computer Discovered Mathematics |last=Thanh Oai |first=Dao |url=https://journal-1.eu/2016-3/Dao-Thanh-Oai-Generalizations-pp.12-20.pdf |volume=3}}

For a triangle {{Mvar|ABC}}, let {{Mvar|l}} an arbitrary line and {{Math|A{{sub|0}}B{{sub|0}}C{{sub|0}}}} the Cevian triangle of an arbitrary point {{Mvar|P}}. {{Mvar|l}} intersects {{Mvar|BC, CA}}, and {{Mvar|AB}} at {{Math|A{{sub|1}}, B{{sub|1}},}} and {{Math|C{{sub|1}}}} respectively. Then {{Math|AA{{sub|1}}∩B{{sub|0}}C{{sub|0}}, BB{{sub|1}}∩C{{sub|0}}A{{sub|0}}}}, and {{Math|CC{{sub|1}}∩A{{sub|0}}B{{sub|0}}}} are colinear.

If {{Mvar|P}} is the centroid of the triangle {{Mvar|ABC}}, the line is Newton–Gauss line of the quadrilateral composed of {{Mvar|AB, BC, CA,}} and {{Mvar|l}}.

The Newton–Gauss line proof was developed by the two mathematicians it is named after: Sir Isaac Newton and Carl Friedrich Gauss.{{citation needed|date=June 2019}} The initial framework for this theorem is from the work of Newton, in his previous theorem on the Newton line, in which Newton showed that the center of a conic inscribed in a quadrilateral lies on the Newton–Gauss line.{{citation|first=David|last=Wells|title=The Penguin Dictionary of Curious and Interesting Geometry|year=1991|publisher=Penguin Books|page=[https://archive.org/details/penguindictionar0000well/page/36 36]|isbn=978-0-14-011813-1|url=https://archive.org/details/penguindictionar0000well/page/36}}

The theorem of Gauss and Bodenmiller states that the three circles whose diameters are the diagonals of a complete quadrilateral are coaxal.{{harvnb|Johnson|2007|loc=p. 172}}

{{reflist}}

• {{citation|first=Roger A.|last=Johnson|title=Advanced Euclidean Geometry|publisher=Dover|year=2007|orig-year=1929|isbn=978-0-486-46237-0}}
• (available on-line as) {{Cite web|url=https://babel.hathitrust.org/cgi/pt?id=wu.89043163211;view=1up;seq=14|title=Modern Geometry: An Elementary Treatise on the Geometry of the Triangle and the Circle|last=Johnson|first=Roger A.|date=1929|website=HathiTrust|access-date=28 May 2019}}

• {{Cite web|url=https://www.cut-the-knot.org/Curriculum/Geometry/Quadri.shtml#explanation|title=Theorem of Complete Quadrilateral: What is it?|last=Bogomonly|first=Alexander|access-date=11 May 2019}}

{{DEFAULTSORT:Newton-Gauss line}}

Category:Geometry

Category:Quadrilaterals