Heptagonal triangle

{{short description|Obtuse triangle formed by the side and diagonals of a regular heptagon}} [[Image:Heptagrams.svg|200px|thumb|right| {{legend-line|solid red 2px|Regular heptagon}} {{legend-line|solid green 2px| Longer diagonals}} {{legend-line|solid blue 2px|Shorter diagonals}} Each of the fourteen congruent '''heptagonal triangles''' has one green side, one blue side, and one red side.]]

In Euclidean geometry, a '''heptagonal triangle''' is an obtuse, scalene triangle whose vertices coincide with the first, second, and fourth vertices of a regular heptagon (from an arbitrary starting vertex). Thus its sides coincide with one side and the adjacent shorter and longer diagonals of the regular heptagon. All heptagonal triangles are similar (have the same shape), and so they are collectively known as ''the'' heptagonal triangle. Its angles have measures <math>\pi/7, 2\pi/7,</math> and <math>4\pi/7,</math> and it is the only triangle with angles in the ratios 1:2:4. The heptagonal triangle has various remarkable properties.

==Key points==

The heptagonal triangle's nine-point center is also its first Brocard point.<ref name=Yiu>{{cite journal |last=Yiu |first=Paul |year=2009 |title=Heptagonal Triangles and Their Companions |journal=Forum Geometricorum |volume=9 |pages=125–148 |url=http://forumgeom.fau.edu/FG2009volume9/FG200912.pdf |archive-date=2017-02-12 |access-date=2016-06-15 |archive-url=https://web.archive.org/web/20170212214508/http://forumgeom.fau.edu/FG2009volume9/FG200912.pdf |url-status=dead }}</ref>{{rp|Propos. 12}}

The second Brocard point lies on the nine-point circle.<ref name=BG>{{cite journal | jstor=2688574 | title=The Heptagonal Triangle | last1=Bankoff | first1=Leon | last2=Garfunkel | first2=Jack | journal=Mathematics Magazine | date=1973 | volume=46 | issue=1 | pages=7–19 | doi=10.2307/2688574 }}</ref>{{rp|p. 19}}

The circumcenter and the Fermat points of a heptagonal triangle form an equilateral triangle.<ref name=Yiu/>{{rp|Thm. 22}}

The distance between the circumcenter ''O'' and the orthocenter ''H'' is given by<ref name=BG/>{{rp|p. 19}}

:<math>OH=R\sqrt{2},</math>

where ''R'' is the circumradius. The squared distance from the incenter ''I'' to the orthocenter is<ref name=BG/>{{rp|p. 19}}

:<math>IH^2=\frac{R^2+4r^2}{2},</math>

where ''r'' is the inradius.

The two tangents from the orthocenter to the circumcircle are mutually perpendicular.<ref name=BG/>{{rp|p. 19}}

==Relations of distances==

===Sides===

The heptagonal triangle's sides ''a'' < ''b'' < ''c'' coincide respectively with the regular heptagon's side, shorter diagonal, and longer diagonal. They satisfy<ref name=Altintas>{{cite journal |last=Altintas |first=Abdilkadir |year=2016 |title=Some Collinearities in the Heptagonal Triangle |journal=Forum Geometricorum |volume=16 |pages=249–256 |url=http://forumgeom.fau.edu/FG2016volume16/FG201630.pdf |archive-date=2016-08-13 |access-date=2016-06-15 |archive-url=https://web.archive.org/web/20160813071253/http://forumgeom.fau.edu/FG2016volume16/FG201630.pdf |url-status=dead }}</ref>{{rp|Lemma 1}}

:<math> \begin{align} a^2 & =c(c-b), \\[5pt] b^2 & =a(c+a), \\[5pt] c^2 & =b(a+b), \\[5pt] \frac 1 a & =\frac 1 b + \frac 1 c \end{align} </math>

(the latter<ref name=BG/>{{rp|p. 13}} being the optic equation) and hence

:<math> ab+ac=bc,</math>

and<ref name=Altintas/>{{rp|Coro. 2}}

:<math>b^3+2b^2c-bc^2-c^3=0, </math> :<math>c^3-2c^2a-ca^2+a^3=0, </math> :<math>a^3-2a^2b-ab^2+b^3=0.</math>

Thus –''b''/''c'', ''c''/''a'', and ''a''/''b'' all satisfy the cubic equation

:<math>t^3-2t^2-t + 1=0.</math>

However, no algebraic expressions with purely real terms exist for the solutions of this equation, because it is an example of casus irreducibilis.

The approximate relation of the sides is

:<math>b\approx 1.80193\cdot a, \qquad c\approx 2.24698\cdot a.</math>

We also have<ref name=Wang2>{{cite journal |last=Wang |first=Kai |title=Heptagonal Triangle and Trigonometric Identities |journal=Forum Geometricorum |volume=19 |year=2019 |pages=29–38}}</ref><ref name="Wang2a">{{cite web |last=Wang |first=Kai |date=August 2019 |url=https://www.researchgate.net/publication/335392159 |title=On cubic equations with zero sums of cubic roots of roots |via=ResearchGate}}</ref>

:<math>\frac{a^2}{bc}, \quad -\frac{b^2}{ca}, \quad -\frac{c^2}{ab} </math> satisfy the cubic equation :<math>t^3+4t^2+3t-1=0.</math>

We also have<ref name=Wang2/> :<math>\frac{a^3}{bc^2}, \quad -\frac{b^3}{ca^2}, \quad \frac{c^3}{ab^2} </math> satisfy the cubic equation :<math>t^3-t^2-9t+1=0.</math>

We also have<ref name=Wang2/> :<math>\frac{a^3}{b^2c}, \quad \frac{b^3}{c^2a}, \quad -\frac{c^3}{a^2b} </math> satisfy the cubic equation :<math>t^3+5t^2-8t+1=0.</math>

We also have<ref name=BG/>{{rp|p. 14}}

:<math>b^2-a^2=ac,</math>

:<math>c^2-b^2=ab,</math>

:<math>a^2-c^2=-bc,</math>

and<ref name=BG/>{{rp|p. 15}}

:<math>\frac{b^2}{a^2}+\frac{c^2}{b^2}+\frac{a^2}{c^2}=5.</math>

We also have<ref name=Wang2/> :<math>ab-bc+ca=0, </math> :<math>a^{3}b-b^{3}c+c^{3}a=0, </math> :<math>a^{4}b+b^{4}c-c^{4}a=0, </math> :<math>a^{11}b^{3}-b^{11}c^{3}+c^{11}a^{3}=0. </math>

===Altitudes===

The altitudes ''h''_''a'', ''h''_''b'', and ''h''_''c'' satisfy

:<math>h_a=h_b+h_c</math><ref name=BG/>{{rp|p. 13}}

and

:<math>h_a^2+h_b^2+h_c^2=\frac{a^2+b^2+c^2}{2}.</math><ref name=BG/>{{rp|p. 14}}

The altitude from side ''b'' (opposite angle ''B'') is half the internal angle bisector <math>w_A</math> of ''A'':<ref name=BG/>{{rp|p. 19}}

:<math>2h_b=w_A.</math>

Here angle ''A'' is the smallest angle, and ''B'' is the second smallest.

===Internal angle bisectors===

We have these properties of the internal angle bisectors <math>w_A, w_B,</math> and <math> w_C</math> of angles ''A, B'', and ''C'' respectively:<ref name=BG/>{{rp|p. 16}}

:<math>w_A=b+c,</math>

:<math>w_B=c-a,</math>

:<math>w_C=b-a.</math>

===Circumradius, inradius, and exradius===

The triangle's area is<ref name=Weisstein/>

:<math>A=\frac{\sqrt{7}}{4}R^2,</math>

where ''R'' is the triangle's circumradius.

We have<ref name=BG/>{{rp|p. 12}}

:<math>a^2+b^2+c^2=7R^2.</math>

We also have<ref name=Wang1>{{cite web |last=Wang |first=Kai |date=September 2018 |url=https://www.researchgate.net/publication/327825153 |title=Trigonometric Properties For Heptagonal Triangle |via=ResearchGate}}</ref> :<math>a^4+b^4+c^4=21R^4.</math> :<math>a^6+b^6+c^6=70R^6.</math>

The ratio ''r'' /''R'' of the inradius to the circumradius is the positive solution of the cubic equation<ref name=Weisstein/>

:<math>8x^3+28x^2+14x-7=0.</math>

In addition,<ref name=BG/>{{rp|p. 15}}

:<math>\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}=\frac{2}{R^2}.</math>

We also have<ref name=Wang1/> :<math>\frac{1}{a^4}+\frac{1}{b^4}+\frac{1}{c^4}=\frac{2}{R^4}.</math> :<math>\frac{1}{a^6}+\frac{1}{b^6}+\frac{1}{c^6}=\frac{17}{7R^6}.</math>

In general for all integer ''n'', :<math>a^{2n}+b^{2n}+c^{2n}=g(n)(2R)^{2n}</math> where :<math>g(-1) = 8, \quad g(0)=3, \quad g(1)=7 </math> and :<math>g(n)=7g(n-1)-14g(n-2)+7g(n-3). </math>

We also have<ref name=Wang1/> :<math>2b^2-a^2=\sqrt{7}bR, \quad 2c^2-b^2=\sqrt{7}cR, \quad 2a^2-c^2=-\sqrt{7}aR. </math>

We also have<ref name=Wang2/> :<math>a^{3}c + b^{3}a - c^{3}b = -7R^{4}, </math> :<math>a^{4}c - b^{4}a + c^{4}b = 7\sqrt{7}R^{5}, </math> :<math>a^{11}c^{3}+b^{11}a^{3} - c^{11}b^{3} = -7^{3}17R^{14}. </math>

The exradius ''r''_''a'' corresponding to side ''a'' equals the radius of the nine-point circle of the heptagonal triangle.<ref name=BG/>{{rp|p. 15}}

==Orthic triangle==

The heptagonal triangle's orthic triangle, with vertices at the feet of the altitudes, is similar to the heptagonal triangle, with similarity ratio 1:2. The heptagonal triangle is the only obtuse triangle that is similar to its orthic triangle (the equilateral triangle being the only acute one).<ref name=BG/>{{rp|pp. 12–13}}

==Hyperbola== The rectangular hyperbola through <math>A,B,C,G=X(2),H=X(4)</math> has the following properties: * first focus <math>F_1 = X(5) </math> * center <math>U</math> is on Euler circle (general property) and on circle <math>(O, F_1)</math> * second focus <math>F_2</math> is on the circumcircle

==Trigonometric properties==

===Trigonometric identities===

The various trigonometric identities associated with the heptagonal triangle include these:<ref name=BG/>{{rp|pp. 13–14}}<ref name="Weisstein">{{Cite web |last=Weisstein |first=Eric W. |title=Heptagonal Triangle |url=https://mathworld.wolfram.com/ |access-date=2024-08-02 |website=mathworld.wolfram.com |language=en}}</ref><ref name=Wang1/> <math display=block>\begin{align} A &= \frac{\pi}{7} \\[6pt] \cos A &= \frac{b}{2a} \end{align} \quad\begin{align} B &= \frac{2\pi}{7} \\[6pt] \cos B &= \frac{c}{2b} \end{align} \quad\begin{align} C &= \frac{4\pi}{7} \\[6pt] \cos C &= -\frac{a}{2c} \end{align}</math><ref name=Wang2/>{{rp|Proposition 10}}

<math display=block>\begin{array}{rcccccl} \sin A \!&\! \times \!&\! \sin B \!&\! \times \!&\! \sin C \!&\! = \!&\! \frac{\sqrt{7}}{8} \\[2pt] \sin A \!&\! - \!&\! \sin B \!&\! - \!&\! \sin C \!&\! = \!&\! -\frac{\sqrt{7}}{2} \\[2pt] \cos A \!&\! \times \!&\! \cos B \!&\! \times \!&\! \cos C \!&\! = \!&\! -\frac{1}{8} \\[2pt] \tan A \!&\! \times \!&\! \tan B \!&\! \times \!&\! \tan C \!&\! = \!&\! -\sqrt{7} \\[2pt] \tan A \!&\! + \!&\! \tan B \!&\! + \!&\! \tan C \!&\! = \!&\! -\sqrt{7} \\[2pt] \cot A \!&\! + \!&\! \cot B \!&\! + \!&\! \cot C \!&\! = \!&\! \sqrt{7} \\[8pt]

\sin^2\!A \!&\! \times \!&\! \sin^2\!B \!&\! \times \!&\! \sin^2\!C \!&\! = \!&\! \frac{7}{64} \\[2pt] \sin^2\!A \!&\! + \!&\! \sin^2\!B \!&\! + \!&\! \sin^2\!C \!&\! = \!&\! \frac{7}{4} \\[2pt] \cos^2\!A \!&\! + \!&\! \cos^2\!B \!&\! + \!&\! \cos^2\!C \!&\! = \!&\! \frac{5}{4} \\[2pt] \tan^2\!A \!&\! + \!&\! \tan^2\!B \!&\! + \!&\! \tan^2\!C \!&\! = \!&\! 21 \\[2pt] \sec^2\!A \!&\! + \!&\! \sec^2\!B \!&\! + \!&\! \sec^2\!C \!&\! = \!&\! 24 \\[2pt] \csc^2\!A \!&\! + \!&\! \csc^2\!B \!&\! + \!&\! \csc^2\!C \!&\! = \!&\! 8 \\[2pt] \cot^2\!A \!&\! + \!&\! \cot^2\!B \!&\! + \!&\! \cot^2\!C \!&\! = \!&\! 5 \\[8pt]

\sin^4\!A \!&\! + \!&\! \sin^4\!B \!&\! + \!&\! \sin^4\!C \!&\! = \!&\! \frac{21}{16} \\[2pt] \cos^4\!A \!&\! + \!&\! \cos^4\!B \!&\! + \!&\! \cos^4\!C \!&\! = \!&\! \frac{13}{16} \\[2pt] \sec^4\!A \!&\! + \!&\! \sec^4\!B \!&\! + \!&\! \sec^4\!C \!&\! = \!&\! 416 \\[2pt] \csc^4\!A \!&\! + \!&\! \csc^4\!B \!&\! + \!&\! \csc^4\!C \!&\! = \!&\! 32 \\[8pt] \end{array}</math>

<math display=block>\begin{array}{ccccl} \tan A \!&\! - \!&\! 4\sin B \!&\! = \!&\! -\sqrt{7} \\[2pt] \tan B \!&\! - \!&\! 4\sin C \!&\! = \!&\! -\sqrt{7} \\[2pt] \tan C \!&\! + \!&\! 4\sin A \!&\! = \!&\! -\sqrt{7} \end{array}</math><ref name="Wang1"/><ref name="Moll">{{cite arXiv |last=Moll |first=Victor H. |title=An elementary trigonometric equation |date=2007-09-24 |class=math.NT |eprint=0709.3755}}</ref>

<math display=block>\begin{align} \cot^2\! A &= 1 -\frac{2 \tan C}{\sqrt{7}} \\[2pt] \cot^2\! B &= 1 -\frac{2 \tan A}{\sqrt{7}} \\[2pt] \cot^2\! C &= 1 -\frac{2 \tan B}{\sqrt{7}} \end{align}</math><ref name=Wang2/>

<math display=block>\begin{array}{rcccccl} \cos A \!&\! = \!&\! \frac{-1}{2} \!&\! + \!&\! \frac{4}{\sqrt{7}} \!&\! \times \!&\! \sin^3\! C \\[2pt] \sec A \!&\! = \!&\! 2 \!&\! + \!&\! 4 \!&\! \times \!&\! \cos C \\[4pt] \sec A \!&\! = \!&\! 6 \!&\! - \!&\! 8 \!&\! \times \!&\! \sin^2\! B \\[4pt] \sec A \!&\! = \!&\! 4 \!&\! - \!&\! \frac{16}{\sqrt{7}} \!&\! \times \!&\! \sin^3\! B \\[2pt] \cot A \!&\! = \!&\! \sqrt{7} \!&\! + \!&\! \frac{8}{\sqrt{7}} \!&\! \times \!&\! \sin^2\! B \\[2pt] \cot A \!&\! = \!&\! \frac{3}{\sqrt{7}} \!&\! + \!&\! \frac{4}{\sqrt{7}} \!&\! \times \!&\! \cos B \\[2pt] \sin^2\! A \!&\! = \!&\! \frac{1}{2} \!&\! + \!&\! \frac{1}{2} \!&\! \times \!&\! \cos B \\[2pt] \cos^2\! A \!&\! = \!&\! \frac{3}{4} \!&\! + \!&\! \frac{2}{\sqrt{7}} \!&\! \times \!&\! \sin^3\! A \\[2pt] \cot^2\! A \!&\! = \!&\! 3 \!&\! + \!&\! \frac{8}{\sqrt{7}} \!&\! \times \!&\! \sin A \\[2pt] \sin^3\! A \!&\! = \!&\! \frac{-\sqrt{7}}{8} \!&\! + \!&\! \frac{\sqrt{7}}{4} \!&\! \times \!&\! \cos B \\[2pt] \csc^3\! A \!&\! = \!&\! \frac{-6}{\sqrt{7}} \!&\! + \!&\! \frac{2}{\sqrt{7}} \!&\! \times \!&\! \tan^2\! C \end{array}</math><ref name=Wang2/>

<math display=block>\sin A\sin B - \sin B\sin C + \sin C\sin A = 0 </math> <math display=block>\begin{align} \sin^3\!B\sin C - \sin^3\!C\sin A - \sin^3\!A\sin B &= 0 \\[3pt] \sin B\sin^3\!C - \sin C\sin^3\!A - \sin A\sin^3\!B &= \frac{7}{2^4\!} \\[2pt] \sin^4\!B\sin C - \sin^4\!C\sin A + \sin^4\!A\sin B &= 0 \\[2pt] \sin B\sin^4\!C + \sin C\sin^4\!A - \sin A\sin^4\!B &= \frac{7\sqrt{7}}{2^{5}} \end{align}</math> <math display=block>\begin{align} \sin^{11}\!B\sin^3\!C - \sin^{11}\!C\sin^3\!A - \sin^{11}\!A\sin^3\!B &= 0 \\[2pt] \sin^3\!B\sin^{11}\!C - \sin^3\!C\sin^{11}\!A - \sin^3\!A\sin^{11}\!B &= \frac{7^3\cdot17}{2^{14}} \end{align}</math><ref name="Wang3">{{cite web |last=Wang |first=Kai |date=October 2019 |url=https://www.researchgate.net/publication/336813631 |title=On Ramanujan Type Identities For PI/7 |via=ResearchGate}}</ref>

===Cubic polynomials===

The cubic equation <math>64y^3-112y^2+56y-7=0</math> has solutions<ref name=BG/>{{rp|p. 14}} <math>\sin^2\! A,\ \sin^2\! B,\ \sin^2\! C.</math>

The positive solution of the cubic equation <math>x^3+x^2-2x-1=0</math> equals <math>2\cos B.</math><ref name="Gleason">{{cite journal|last=Gleason|first=Andrew Mattei|title=Angle trisection, the heptagon, and the triskaidecagon |journal=The American Mathematical Monthly|date=March 1988|volume=95|issue=3 |pages=185–194|url=http://apollonius.math.nthu.edu.tw/d1/ne01/jyt/linkjstor/regular/1.pdf#3 |archiveurl=https://web.archive.org/web/20151219180208/http://apollonius.math.nthu.edu.tw/d1/ne01/jyt/linkjstor/regular/7.pdf#3 |doi= 10.2307/2323624|jstor=2323624 |archive-date=2015-12-19 |url-status=dead}}</ref>{{rp|p. 186–187}}

The roots of the cubic equation <math>x^3 - \tfrac{\sqrt 7}{2}x^2 + \tfrac{\sqrt 7}{8} = 0 </math> are<ref name=Wang2/> <math>\sin 2A,\ \sin 2B,\ \sin 2C.</math>

The roots of the cubic equation <math>x^3 - \tfrac{\sqrt 7}{2} x^2 + \tfrac{\sqrt 7}{8} = 0</math> are <math>-\sin A,\ \sin B,\ \sin C.</math>

The roots of the cubic equation <math>x^3 + \tfrac{1}{2}x^2 - \tfrac{1}{2}x - \tfrac{1}{8} = 0</math> are <math>-\cos A,\ \cos B,\ \cos C.</math>

The roots of the cubic equation <math>x^3 + \sqrt{7}x^2 - 7x + \sqrt{7} = 0</math> are <math> \tan A,\ \tan B,\ \tan C.</math> The roots of the cubic equation <math>x^3 - 21x^2 + 35x - 7 = 0</math> are <math>\tan^2\! A,\ \tan^2\! B,\ \tan^2\! C.</math>

===Sequences===

For an integer {{mvar|n}}, let <math display=block>\begin{align} S(n) &= (-\sin A)^n + \sin^n\! B + \sin^n\! C \\[4pt] C(n) &= (-\cos A)^n + \cos^n\! B + \cos^n\! C \\[4pt] T(n) &= \tan^n\! A + \tan^n\! B + \tan^n\! C \end{align}</math>

<div style="overflow-x:auto"> {|class=wikitable style="text-align: center;" ! Value&nbsp;of&nbsp;{{mvar|n}}: !! 0 !! 1 !! 2 !! 3 !! 4 !! 5 !! 6 !! 7 !! 8 !! 9 !! 10 !! 11 !! 12 !! 13 !! 14 !! 15 !! 16 !! 17 !! 18 !! 19 !! 20 |- ! <math>S(n)</math> | <math>\ 3\ </math> || <math> \tfrac{\sqrt 7}{2}</math> || <math> \tfrac{7}{2^2}</math> || <math> \tfrac{\sqrt 7}{2}</math> || <math> \tfrac{7\cdot3}{2^4}</math> || <math> \tfrac{7\sqrt 7}{2^4}</math> || <math> \tfrac{7\cdot5}{2^5}</math> || <math> \tfrac{7^2\sqrt 7}{2^7}</math> || <math> \tfrac{7^2\cdot5}{2^8}</math> || <math> \tfrac{7\cdot25\sqrt 7}{2^9}</math> || <math> \tfrac{7^2\cdot9}{2^9}</math> || <math> \tfrac{7^2\cdot13\sqrt 7}{2^{11}}</math> || <math> \tfrac{7^2\cdot33}{2^{11}}</math> || <math> \tfrac{7^2\cdot3\sqrt 7}{2^9}</math> || <math> \tfrac{7^4\cdot5}{2^{14}}</math> || <math> \tfrac{7^2\cdot179\sqrt 7}{2^{15}}</math> || <math> \tfrac{7^3\cdot131}{2^{16}}</math> || <math> \tfrac{7^3\cdot3\sqrt 7}{2^{12}}</math> || <math> \tfrac{7^3\cdot493}{2^{18}}</math> || <math> \tfrac{7^3\cdot181\sqrt 7}{2^{18}}</math> || <math> \tfrac{7^5\cdot19}{2^{19}} </math> |- ! <math>S(-n)</math> | <math>3</math> || <math> 0</math> || <math> 2^3</math> || <math> -\tfrac{2^3\cdot3\sqrt 7}{7}</math> || <math> 2^5</math> || <math> -\tfrac{2^5\cdot5\sqrt 7}{7}</math> || <math> \tfrac{2^6\cdot17}{7}</math> || <math> -2^7\sqrt{7}</math> || <math> \tfrac{2^9\cdot11}{7}</math> || <math> -\tfrac{2^{10}\cdot33\sqrt 7}{7^2}</math> || <math> \tfrac{2^{10}\cdot29}{7}</math> || <math> -\tfrac{2^{14}\cdot11\sqrt 7}{7^2}</math> || <math> \tfrac{2^{12}\cdot269}{7^2}</math> || <math> -\tfrac{2^{13}\cdot117\sqrt 7}{7^2}</math> || <math> \tfrac{2^{14}\cdot51}{7}</math> || <math> -\tfrac{2^{21}\cdot17\sqrt 7}{7^3}</math> || <math> \tfrac{2^{17}\cdot237}{7^2}</math> || <math> -\tfrac{2^{17}\cdot1445\sqrt 7}{7^3}</math> || <math> \tfrac{2^{19}\cdot2203}{7^3}</math> || <math> -\tfrac{2^{19}\cdot1919\sqrt 7}{7^3}</math> || <math> \tfrac{2^{20}\cdot5851}{7^3} </math> |- ! <math>C(n)</math> | <math>3</math> || <math> -\tfrac{1}{2}</math> || <math> \tfrac{5}{4}</math> || <math> -\tfrac{1}{2}</math> || <math> \tfrac{13}{16}</math> || <math> -\tfrac{1}{2}</math> || <math> \tfrac{19}{32}</math> || <math> -\tfrac{57}{128}</math> || <math> \tfrac{117}{256}</math> || <math> -\tfrac{193}{512}</math> || <math> \tfrac{185}{512}</math> |- ! <math>C(-n)</math> | <math>3</math> || <math> -4</math> || <math> 24</math> || <math> -88</math> || <math> 416</math> || <math> -1824</math> || <math> 8256</math> || <math> -36992</math> || <math> 166400</math> || <math> -747520</math> || <math> 3359744</math> |- ! <math>T(n)</math> | <math>3</math> || <math> -\sqrt{7}</math> || <math> 7\cdot3</math> || <math> -31\sqrt{7}</math> || <math> 7\cdot53</math> || <math> -7\cdot87\sqrt{7}</math> || <math> 7\cdot1011</math> || <math> -7^2\cdot239\sqrt{7}</math> || <math> 7^2\cdot2771</math> || <math> -7\cdot32119\sqrt{7}</math> || <math> 7^2\cdot53189</math> |- ! <math>T(-n)</math> | <math>3</math> || <math> \sqrt{7}</math> || <math> 5</math> || <math> \tfrac{25\sqrt 7}{7}</math> || <math> 19</math> || <math> \tfrac{103\sqrt 7}{7}</math> || <math> \tfrac{563}{7}</math> || <math> 7\cdot9\sqrt{7}</math> || <math> \tfrac{2421}{7}</math> || <math> \tfrac{13297\sqrt 7}{7^2}</math> || <math> \tfrac{10435}{7}</math> |} </div>

===Ramanujan identities=== We also have Ramanujan type identities,<ref name="Wang1"/><ref name="WS1">{{cite journal |last1=Witula |first1=Roman |last2=Slota |first2=Damian |title=New Ramanujan-Type Formulas and Quasi-Fibonacci Numbers of Order 7 |url=https://www.emis.de/journals/JIS/VOL10/Slota/witula13.pdf |journal=Journal of Integer Sequences |volume=10 |year=2007 |issue=5 |bibcode=2007JIntS..10...56W |article-number=07.5.6}}</ref>

<math display=block>\begin{array}{ccccccl} \sqrt[3]{2\sin 2A} \!&\! + \!&\! \sqrt[3]{2\sin 2B} \!&\! + \!&\! \sqrt[3]{2\sin 2C} \!&\! = \!&\! -\sqrt[18]{7} \times \sqrt[3]{-\sqrt[3]{7} + 6 + 3\left(\sqrt[3]{5 - 3 \sqrt[3]{7}} + \sqrt[3]{4 - 3 \sqrt[3]{7}}\right)} \\[2pt]

\sqrt[3]{2\sin 2A} \!&\! + \!&\! \sqrt[3]{2\sin 2B} \!&\! + \!&\! \sqrt[3]{2\sin 2C} \!&\! = \!&\! -\sqrt[18]{7} \times \sqrt[3]{-\sqrt[3]{7} + 6 + 3\left(\sqrt[3]{5 - 3 \sqrt[3]{7}} + \sqrt[3]{4 - 3 \sqrt[3]{7}}\right)} \\[2pt]

\sqrt[3]{4\sin^2 2A} \!&\! + \!&\! \sqrt[3]{4\sin^2 2B} \!&\! + \!&\! \sqrt[3]{4\sin^2 2C} \!&\! = \!&\! \sqrt[18]{49} \times \sqrt[3]{ \sqrt[3]{49} + 6 + 3\left(\sqrt[3]{12 + 3( \sqrt[3]{49} + 2\sqrt[3]{7})} + \sqrt[3]{11 + 3( \sqrt[3]{49} + 2\sqrt[3]{7})}\right)} \\[6pt]

\sqrt[3]{2\cos 2A} \!&\! + \!&\! \sqrt[3]{2\cos 2B} \!&\! + \!&\! \sqrt[3]{2\cos 2C} \!&\! = \!&\! \sqrt[3]{5 - 3\sqrt[3]{7}} \\[8pt]

\sqrt[3]{4\cos^2 2A} \!&\! + \!&\! \sqrt[3]{4\cos^2 2B} \!&\! + \!&\! \sqrt[3]{4\cos^2 2C} \!&\! = \!&\! \sqrt[3]{11 + 3(2\sqrt[3]{7} + \sqrt[3]{49})} \\[6pt]

\sqrt[3]{\tan 2A} \!&\! + \!&\! \sqrt[3]{\tan 2B} \!&\! + \!&\! \sqrt[3]{\tan 2C} \!&\! = \!&\! -\sqrt[18]{7} \times \sqrt[3]{\sqrt[3]{7} + 6 + 3\left(\sqrt[3]{5 + 3(\sqrt[3]{7} - \sqrt[3]{49})} + \sqrt[3]{- 3 + 3(\sqrt[3]{7} - \sqrt[3]{49})}\right)} \\[2pt] \sqrt[3]{\tan^2 2A} \!&\! + \!&\! \sqrt[3]{\tan^2 2B} \!&\! + \!&\! \sqrt[3]{\tan^2 2C} \!&\! = \!&\! \sqrt[18]{49} \times \sqrt[3]{3\sqrt[3]{49} + 6 + 3\left(\sqrt[3]{89 + 3(3\sqrt[3]{49} + 5\sqrt[3]{7})} + \sqrt[3]{25 + 3(3\sqrt[3]{49} + 5\sqrt[3]{7})}\right)} \end{array}</math>

<math display=block>\begin{array}{ccccccl} \frac{1}{\sqrt[3]{2\sin 2A}} \!&\! + \!&\! \frac{1}{\sqrt[3]{2\sin 2B}} \!&\! + \!&\! \frac{1}{\sqrt[3]{2\sin 2C}} \!&\! = \!&\! -\frac{1}{\sqrt[18]{7}} \times \sqrt[3]{6 + 3\left(\sqrt[3]{5 - 3 \sqrt[3]{7}} + \sqrt[3]{4 - 3 \sqrt[3]{7}}\right)} \\[2pt]

\frac{1}{\sqrt[3]{4\sin^2 2A}} \!&\! + \!&\! \frac{1}{\sqrt[3]{4\sin^2 2B}} \!&\! + \!&\! \frac{1}{\sqrt[3]{4\sin^2 2C}} \!&\! = \!&\! \frac{1}{\sqrt[18]{49}} \times \sqrt[3]{ 2\sqrt[3]{7} + 6 + 3\left(\sqrt[3]{12 + 3( \sqrt[3]{49} + 2\sqrt[3]{7})} + \sqrt[3]{11 + 3( \sqrt[3]{49} + 2\sqrt[3]{7})}\right)} \\[2pt]

\frac{1}{\sqrt[3]{2\cos 2A}} \!&\! + \!&\! \frac{1}{\sqrt[3]{2\cos 2B}} \!&\! + \!&\! \frac{1}{\sqrt[3]{2\cos 2C}} \!&\! = \!&\! \sqrt[3]{4 - 3\sqrt[3]{7}} \\[6pt]

\frac{1}{\sqrt[3]{4\cos^2 2A}} \!&\! + \!&\! \frac{1}{\sqrt[3]{4\cos^2 2B}} \!&\! + \!&\! \frac{1}{\sqrt[3]{4\cos^2 2C}} \!&\! = \!&\! \sqrt[3]{12 + 3(2\sqrt[3]{7} + \sqrt[3]{49})} \\[2pt]

\frac{1}{\sqrt[3]{\tan 2A}} \!&\! + \!&\! \frac{1}{\sqrt[3]{\tan 2B}} \!&\! + \!&\! \frac{1}{\sqrt[3]{\tan 2C}} \!&\! = \!&\! -\frac{1}{\sqrt[18]{7}} \times \sqrt[3]{-\sqrt[3]{49} + 6 + 3\left(\sqrt[3]{5 + 3(\sqrt[3]{7} - \sqrt[3]{49})} + \sqrt[3]{- 3 + 3(\sqrt[3]{7} - \sqrt[3]{49})}\right)} \\[2pt]

\frac{1}{\sqrt[3]{\tan^2 2A}} \!&\! + \!&\! \frac{1}{\sqrt[3]{\tan^2 2B}} \!&\! + \!&\! \frac{1}{\sqrt[3]{\tan^2 2C}} \!&\! = \!&\! \frac{1}{\sqrt[18]{49}} \times \sqrt[3]{5\sqrt[3]{7} + 6 + 3\left(\sqrt[3]{89 + 3(3\sqrt[3]{49} + 5\sqrt[3]{7})} + \sqrt[3]{25 + 3(3\sqrt[3]{49} + 5\sqrt[3]{7})}\right)} \end{array}</math>

<math display=block>\begin{array}{ccccccl} \sqrt[3]{\frac{\cos 2A}{\cos 2B}} \!&\! + \!&\! \sqrt[3]{\frac{\cos 2B}{\cos 2C}} \!&\! + \!&\! \sqrt[3]{\frac{\cos 2C}{\cos 2A}} \!&\! = \!&\! -\sqrt[3]{7} \\[2pt]

\sqrt[3]{\frac{\cos 2B}{\cos 2A}} \!&\! + \!&\! \sqrt[3]{\frac{\cos 2C}{\cos 2B}} \!&\! + \!&\! \sqrt[3]{\frac{\cos 2A}{\cos 2C}} \!&\! = \!&\! 0 \\[2pt]

\sqrt[3]{\frac{\cos^4 2B}{\cos 2A}} \!&\! + \!&\! \sqrt[3]{\frac{\cos^4 2C}{\cos 2B}} \!&\! + \!&\! \sqrt[3]{\frac{\cos^4 2A}{\cos 2C}} \!&\! = \!&\! -\frac{\sqrt[3]{49}}{2} \\[2pt]

\sqrt[3]{\frac{\cos^5 2A}{\cos^2 2B}} \!&\! + \!&\! \sqrt[3]{\frac{\cos^5 2B}{\cos^2 2C}} \!&\! + \!&\! \sqrt[3]{\frac{\cos^5 2C}{\cos^2 2A}} \!&\! = \!&\! 0 \\[2pt]

\sqrt[3]{\frac{\cos^5 2B}{\cos^2 2A}} \!&\! + \!&\! \sqrt[3]{\frac{\cos^5 2C}{\cos^2 2B}} \!&\! + \!&\! \sqrt[3]{\frac{\cos^5 2A}{\cos^2 2C}} \!&\! = \!&\! -3\times \frac{\sqrt[3]{7}}{2} \\[2pt]

\sqrt[3]{\frac{\cos^{14}2A}{\cos^5 2B}} \!&\! + \!&\! \sqrt[3]{\frac{\cos^{14}2B}{\cos^5 2C}} \!&\! + \!&\! \sqrt[3]{\frac{\cos^{14}2C}{\cos^5 2A}} \!&\! = \!&\! 0 \\[2pt]

\sqrt[3]{\frac{\cos^{14}2B}{\cos^5 2A}} \!&\! + \!&\! \sqrt[3]{\frac{\cos^{14}2C}{\cos^5 2B}} \!&\! + \!&\! \sqrt[3]{\frac{\cos^{14}2A}{\cos^5 2C}} \!&\! = \!&\! -61\times \frac{\sqrt[3]{7}}{8}. \end{array}</math><ref name=Wang3/>

<math display=block> </math>

==References== {{reflist}}

Category:Types of triangles