orthocenter

Let {{mvar|A, B, C}} denote the vertices and also the angles of the triangle, and let $a\; =\; \backslash left|\backslash overline\{BC\}\backslash right|,\; b\; =\; \backslash left|\backslash overline\{CA\}\backslash right|,\; c\; =\; \backslash left|\backslash overline\{AB\}\backslash right|$ be the side lengths. The orthocenter has trilinear coordinatesClark Kimberling's Encyclopedia of Triangle Centers {{cite web|url=http://faculty.evansville.edu/ck6/encyclopedia/ETC.html |title=Encyclopedia of Triangle Centers |access-date=2012-04-19 |url-status=dead |archive-url=https://web.archive.org/web/20120419171900/http://faculty.evansville.edu/ck6/encyclopedia/ETC.html |archive-date=2012-04-19 }}

$$\backslash begin\{align\}$$

& \sec A:\sec B:\sec C \\

&= \cos A-\sin B \sin C:\cos B-\sin C \sin A:\cos C-\sin A\sin B,

\end{align}

and barycentric coordinates

$$\backslash begin\{align\}$$

& (a^2+b^2-c^2)(a^2-b^2+c^2) : (a^2+b^2-c^2)(-a^2+b^2+c^2) : (a^2-b^2+c^2)(-a^2+b^2+c^2) \\

&= \tan A:\tan B:\tan C.

\end{align}

Since barycentric coordinates are all positive for a point in a triangle's interior but at least one is negative for a point in the exterior, and two of the barycentric coordinates are zero for a vertex point, the barycentric coordinates given for the orthocenter show that the orthocenter is in an acute triangle's interior, on the right-angled vertex of a right triangle, and exterior to an obtuse triangle.

In the complex plane, let the points {{mvar|A, B, C}} represent the numbers {{mvar|z{{sub|A}}, z{{sub|B}}, z{{sub|C}}}} and assume that the circumcenter of triangle {{math|△ABC}} is located at the origin of the plane. Then, the complex number

:$z\_H=z\_A+z\_B+z\_C$

is represented by the point {{mvar|H}}, namely the altitude of triangle {{math|△ABC}}.Andreescu, Titu; Andrica, Dorin, "Complex numbers from A to...Z". Birkhäuser, Boston, 2006, {{ISBN|978-0-8176-4326-3}}, page 90, Proposition 3 From this, the following characterizations of the orthocenter {{mvar|H}} by means of free vectors can be established straightforwardly:

:$\backslash vec\{OH\}=\backslash sum\backslash limits\_\{\backslash scriptstyle\backslash rm\; cyclic\}\backslash vec\{OA\},\backslash qquad2\backslash cdot\backslash vec\{HO\}=\backslash sum\backslash limits\_\{\backslash scriptstyle\backslash rm\; cyclic\}\backslash vec\{HA\}.$

The first of the previous vector identities is also known as the problem of Sylvester, proposed by James Joseph Sylvester.Dörrie, Heinrich, "100 Great Problems of Elementary Mathematics. Their History and Solution". Dover Publications, Inc., New York, 1965, {{ISBN|0-486-61348-8}}, page 142

Let {{mvar|D, E, F}} denote the feet of the altitudes from {{mvar|A, B, C}} respectively. Then:

• The product of the lengths of the segments that the orthocenter divides an altitude into is the same for all three altitudes:{{harvnb|Johnson|2007|loc=p. 163, Section 255}}{{Cite web |url=http://www.pballew.net/orthocen.html |title="Orthocenter of a triangle" |access-date=2012-05-04 |archive-url=https://web.archive.org/web/20120705102919/http://www.pballew.net/orthocen.html |archive-date=2012-07-05 |url-status=usurped }}

:$\backslash overline\{AH\}\; \backslash cdot\; \backslash overline\{HD\}\; =\; \backslash overline\{BH\}\; \backslash cdot\; \backslash overline\{HE\}\; =\; \backslash overline\{CH\}\; \backslash cdot\; \backslash overline\{HF\}.$

:The circle centered at {{mvar|H}} having radius the square root of this constant is the triangle's polar circle.{{harvnb|Johnson|2007|loc=p. 176, Section 278}}

• The sum of the ratios on the three altitudes of the distance of the orthocenter from the base to the length of the altitude is 1:[http://jwilson.coe.uga.edu/EMAT6680Fa06/Panapoi/assignment_8/assignment8.htm Panapoi, Ronnachai, "Some properties of the orthocenter of a triangle"], University of Georgia. (This property and the next one are applications of a more general property of any interior point and the three cevians through it.)

:$\backslash frac\{\backslash overline\{HD\}\}\{\backslash overline\{AD\}\}\; +\; \backslash frac\{\backslash overline\{HE\}\}\{\backslash overline\{BE\}\}\; +\; \backslash frac\{\backslash overline\{HF\}\}\{\backslash overline\{CF\}\}\; =\; 1.$

• The sum of the ratios on the three altitudes of the distance of the orthocenter from the vertex to the length of the altitude is 2:

:$\backslash frac\{\backslash overline\{AH\}\}\{\backslash overline\{AD\}\}\; +\; \backslash frac\{\backslash overline\{BH\}\}\{\backslash overline\{BE\}\}\; +\; \backslash frac\{\backslash overline\{CH\}\}\{\backslash overline\{CF\}\}\; =\; 2.$

• The isogonal conjugate of the orthocenter is the circumcenter of the triangle.{{harvnb|Smart|1998|loc=p. 182}}
• The isotomic conjugate of the orthocenter is the symmedian point of the anticomplementary triangle.Weisstein, Eric W. "Isotomic conjugate" From MathWorld--A Wolfram Web Resource. http://mathworld.wolfram.com/IsotomicConjugate.html
• Four points in the plane, such that one of them is the orthocenter of the triangle formed by the other three, is called an orthocentric system or orthocentric quadrangle.

{{excerpt|Orthocentric system}}

Denote the circumradius of the triangle by {{mvar|R}}. Then[http://mathworld.wolfram.com/Orthocenter.html Weisstein, Eric W. "Orthocenter." From MathWorld--A Wolfram Web Resource.]{{harvnb|Altshiller-Court|2007|loc=p. 102}}

:$a^2\; +\; b^2\; +\; c^2\; +\; \backslash overline\{AH\}^2\; +\; \backslash overline\{BH\}^2\; +\; \backslash overline\{CH\}^2\; =\; 12R^2.$

In addition, denoting {{mvar|r}} as the radius of the triangle's incircle, {{mvar|r{{sub|a}}, r{{sub|b}}, r{{sub|c}}}} as the radii of its excircles, and {{mvar|R}} again as the radius of its circumcircle, the following relations hold regarding the distances of the orthocenter from the vertices:[http://forumgeom.fau.edu/FG2006volume6/FG200639.pdf Bell, Amy, "Hansen's right triangle theorem, its converse and a generalization", Forum Geometricorum 6, 2006, 335–342.]

:$\backslash begin\{align\}$

& r_a + r_b + r_c + r = \overline{AH} + \overline{BH} + \overline{CH} + 2R, \\

& r_a^2 + r_b^2 + r_c^2 + r^2 = \overline{AH}^2 + \overline{BH}^2 + \overline{CH}^2 + (2R)^2.

\end{align}

If any altitude, for example, {{mvar|{{overline|AD}}}}, is extended to intersect the circumcircle at {{mvar|P}}, so that {{mvar|{{overline|AD}}}} is a chord of the circumcircle, then the foot {{mvar|D}} bisects segment {{mvar|{{overline|HP}}}}:

:$\backslash overline\{HD\}\; =\; \backslash overline\{DP\}.$

The directrices of all parabolas that are externally tangent to one side of a triangle and tangent to the extensions of the other sides pass through the orthocenter.Weisstein, Eric W. "Kiepert Parabola." From MathWorld--A Wolfram Web Resource. http://mathworld.wolfram.com/KiepertParabola.html

A circumconic passing through the orthocenter of a triangle is a rectangular hyperbola.Weisstein, Eric W. "Jerabek Hyperbola." From MathWorld--A Wolfram Web Resource. http://mathworld.wolfram.com/JerabekHyperbola.html

{{main|Nine-point circle}}

The orthocenter {{mvar|H}}, the centroid {{mvar|G}}, the circumcenter {{mvar|O}}, and the center {{mvar|N}} of the nine-point circle all lie on a single line, known as the Euler line.{{harvnb|Berele|Goldman|2001|loc=p. 123}} The center of the nine-point circle lies at the midpoint of the Euler line, between the orthocenter and the circumcenter, and the distance between the centroid and the circumcenter is half of that between the centroid and the orthocenter:{{harvnb|Berele|Goldman|2001|loc=pp. 124-126}}

:$\backslash begin\{align\}$

& \overline{OH} = 2\overline{NH}, \\

& 2\overline{OG} = \overline{GH}.

\end{align}

The orthocenter is closer to the incenter {{mvar|I}} than it is to the centroid, and the orthocenter is farther than the incenter is from the centroid:

:$\backslash begin\{align\}$

\overline{HI} &< \overline{HG}, \\

\overline{HG} &> \overline{IG}.

\end{align}

In terms of the sides {{mvar|a}}, {{mvar|b}}, {{mvar|c}}, inradius {{mvar|r}} and circumradius {{mvar|R}},Marie-Nicole Gras, "Distances between the circumcenter of the extouch triangle and the classical centers", Forum Geometricorum 14 (2014), 51-61. http://forumgeom.fau.edu/FG2014volume14/FG201405index.htmlSmith, Geoff, and Leversha, Gerry, "Euler and triangle geometry", Mathematical Gazette 91, November 2007, 436–452.{{rp|p. 449}}

:$\backslash begin\{align\}$

\overline{OH}^2 &= R^2 -8R^2 \cos A \cos B \cos C \\

&= 9R^2-(a^2+b^2+c^2), \\

\overline{HI}^2 &= 2r^2 -4R^2 \cos A \cos B \cos C.

\end{align}

==Orthic triangle==

File:Altitudes and orthic triangle SVG.svg

If the triangle {{math|△ABC}} is oblique (does not contain a right-angle), the pedal triangle of the orthocenter of the original triangle is called the orthic triangle or altitude triangle. That is, the feet of the altitudes of an oblique triangle form the orthic triangle, {{math|△DEF}}. Also, the incenter (the center of the inscribed circle) of the orthic triangle {{math|△DEF}} is the orthocenter of the original triangle {{math|△ABC}}.

{{cite book |title=Continuous symmetry: from Euclid to Klein |author= William H. Barker, Roger Howe |chapter-url=https://books.google.com/books?id=NIxExnr2EjYC&pg=PA292 |chapter=§ VI.2: The classical coincidences |isbn=978-0-8218-3900-3 |publisher=American Mathematical Society|year=2007|page= 292}} See also: Corollary 5.5, p. 318.

Trilinear coordinates for the vertices of the orthic triangle are given by

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

D =& 0 &:& \sec B &:& \sec C \\

E =& \sec A &:& 0 &:& \sec C \\

F =& \sec A &:& \sec B &:& 0

\end{array}

The extended sides of the orthic triangle meet the opposite extended sides of its reference triangle at three collinear points.{{harvnb|Johnson|2007|loc=p. 199, Section 315}}{{harvnb|Altshiller-Court|2007|loc=p. 165}}

In any acute triangle, the inscribed triangle with the smallest perimeter is the orthic triangle.{{harvnb|Johnson|2007|loc=p. 168, Section 264}} This is the solution to Fagnano's problem, posed in 1775.{{harvnb|Berele|Goldman|2001|loc=pp. 120-122}} The sides of the orthic triangle are parallel to the tangents to the circumcircle at the original triangle's vertices.{{harvnb|Johnson|2007|loc = p. 172, Section 270c}}

The orthic triangle of an acute triangle gives a triangular light route.Bryant, V., and Bradley, H., "Triangular Light Routes," Mathematical Gazette 82, July 1998, 298-299.

The tangent lines of the nine-point circle at the midpoints of the sides of {{math|△ABC}} are parallel to the sides of the orthic triangle, forming a triangle similar to the orthic triangle.{{citation|first=David C.|last=Kay|title=College Geometry / A Discovery Approach|year=1993|publisher=HarperCollins|isbn=0-06-500006-4|page=6}}

The orthic triangle is closely related to the tangential triangle, constructed as follows: let {{mvar|L{{sub|A}}}} be the line tangent to the circumcircle of triangle {{math|△ABC}} at vertex {{mvar|A}}, and define {{mvar|L{{sub|B}}, L{{sub|C}}}} analogously. Let $A$ = L_B \cap L_C, $B$ = L_C \cap L_A, $C$ = L_C \cap L_A. The tangential triangle is {{math|△A"B"C"}}, whose sides are the tangents to triangle {{math|△ABC''}}'s circumcircle at its vertices; it is homothetic to the orthic triangle. The circumcenter of the tangential triangle, and the center of similitude of the orthic and tangential triangles, are on the Euler line.{{rp|p. 447}}

Trilinear coordinates for the vertices of the tangential triangle are given by

$$\backslash begin\{array\}\{rrcrcr\}$$

A'' =& -a &:& b &:& c \\

B'' =& a &:& -b &:& c \\

C'' =& a &:& b &:& -c

\end{array}

The reference triangle and its orthic triangle are orthologic triangles.

For more information on the orthic triangle, see here.

The theorem that the three altitudes of a triangle concur (at the orthocenter) is not directly stated in surviving Greek mathematical texts, but is used in the Book of Lemmas (proposition 5), attributed to Archimedes (3rd century BC), citing the "commentary to the treatise about right-angled triangles", a work which does not survive. It was also mentioned by Pappus (Mathematical Collection, VII, 62; {{c.}} 340).

{{cite book |last=Newton |first=Isaac |author-link=Isaac Newton |editor-last=Whiteside |editor-first=Derek Thomas |year=1971 |title=The Mathematical Papers of Isaac Newton |volume=4 |publisher=Cambridge University Press |chapter=3.1 The 'Geometry of Curved Lines' |pages=454–455 |chapter-url=https://archive.org/details/mathematicalpape0004newt/page/454/ |chapter-url-access=limited }} Note Whiteside's footnotes 90–92, pp. 454–456.

The theorem was stated and proved explicitly by al-Nasawi in his (11th century) commentary on the Book of Lemmas, and attributed to al-Quhi ({{floruit|10th century}}).{{cite journal |last1=Hajja |first1=Mowaffaq |last2=Martini |first2=Horst |year=2013 |title=Concurrency of the Altitudes of a Triangle |journal=Mathematische Semesterberichte |volume=60 |number=2 |pages=249–260 |url=https://www.researchgate.net/publication/257442911 |doi=10.1007/s00591-013-0123-z

}}

{{cite journal |last=Hogendijk |first=Jan P. |title=Two beautiful geometrical theorems by Abū Sahl Kūhī in a 17th century Dutch translation |journal=Tārīk͟h-e ʾElm: Iranian Journal for the History of Science |volume=6 |pages=1–36 |year=2008 |url=https://www.sid.ir/paper/146602/en }}

This proof in Arabic was translated as part of the (early 17th century) Latin editions of the Book of Lemmas, but was not widely known in Europe, and the theorem was therefore proven several more times in the 17th–19th century. Samuel Marolois proved it in his Geometrie (1619), and Isaac Newton proved it in an unfinished treatise Geometry of Curved Lines {{nobr|({{c.}} 1680).}} Later William Chapple proved it in 1749.{{cite journal |last=Davies |first=Thomas Stephens |author-link=Thomas Stephens Davies |year=1850 |title=XXIV. Geometry and geometers |journal=Philosophical Magazine |series=3 |volume=37 |number=249 |pages=198–212 |url=https://zenodo.org/record/1919807 |doi=10.1080/14786445008646583}} [https://archive.org/details/londonedinburg3371850lond/page/207 Footnote on pp. 207–208]. Quoted by {{cite web |first=Alexander |last=Bogomolny |author-link=Alexander Bogomolny |year=2010 |url=https://www.cut-the-knot.org/triangle/Chapple.shtml |title=A Possibly First Proof of the Concurrence of Altitudes |work=Cut The Knot |access-date=2019-11-17}}

A particularly elegant proof is due to François-Joseph Servois (1804) and independently Carl Friedrich Gauss (1810): Draw a line parallel to each side of the triangle through the opposite point, and form a new triangle from the intersections of these three lines. Then the original triangle is the medial triangle of the new triangle, and the altitudes of the original triangle are the perpendicular bisectors of the new triangle, and therefore concur (at the circumcenter of the new triangle).

{{cite book |last=Servois |first=Francois-Joseph |author-link=Francois-Joseph Servois |title=Solutions peu connues de différens problèmes de Géométrie-pratique |language=fr |trans-title=Little-known solutions of various Geometry practice problems |publisher=Devilly, Metz et Courcier |year=1804 |page=15

}}

{{cite book |contributor-last=Gauss |contributor-first=Carl Friedrich |year=1810 |contributor-link=Carl Friedrich Gauss |last=Carnot |first=Lazare |translator-last=Schumacher |title=Geometrie der Stellung |contribution=Zusätze |language=de }} republished in {{cite book |last=Gauss |first=Carl Friedrich |title=Werke |volume=4 |chapter-url=https://archive.org/details/werkecarlf04gausrich/page/n405/ |chapter=Zusätze |page=396 |publisher= Göttingen Academy of Sciences |year=1873

}}

See {{cite journal |last=Mackay |first=John Sturgeon |author-link=John Sturgeon Mackay |year=1883 |title=The Triangle and its Six Scribed Circles §5. Orthocentre |journal=Proceedings of the Edinburgh Mathematical Society |volume=1 |pages=60–96 |doi=10.1017/S0013091500036762 |doi-access=free}}

• Triangle center

{{reflist}}

• {{citation |last=Altshiller-Court |first=Nathan |author-link=Nathan Altshiller Court |year=2007 |orig-year=1952 |title=College Geometry |publisher=Dover }}
• {{citation |last1=Berele |first1=Allan |last2=Goldman |first2=Jerry |year=2001 |title=Geometry: Theorems and Constructions |publisher=Prentice Hall |isbn=0-13-087121-4 }}
• {{citation |last=Johnson |first=Roger A. |year=2007 |orig-year=1960 |title=Advanced Euclidean Geometry |publisher=Dover |isbn=978-0-486-46237-0 }}
• {{citation |last=Smart |first=James R. |year=1998 |title=Modern Geometries |edition=5th |publisher=Brooks/Cole |isbn=0-534-35188-3}}

• {{MathWorld|title=Altitude|urlname=Altitude}}
• [http://www.mathopenref.com/triangleorthocenter.html Orthocenter of a triangle] With interactive animation
• [http://www.mathopenref.com/constorthocenter.html Animated demonstration of orthocenter construction] Compass and straightedge.
• [http://demonstrations.wolfram.com/FagnanosProblem/ Fagnano's Problem] by Jay Warendorff, Wolfram Demonstrations Project.

Category:Triangle centers