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Orthocenter
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==Formulation== Let {{mvar|A, B, C}} denote the vertices and also the angles of the triangle, and let <math>a = \left|\overline{BC}\right|, b = \left|\overline{CA}\right|, c = \left|\overline{AB}\right|</math> be the side lengths. The orthocenter has [[trilinear coordinates]]<ref name=ck>Clark 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 }}</ref> <math display=block>\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}</math> and [[Barycentric coordinates (mathematics)|barycentric coordinates]] <math display=block>\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}</math> 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|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 [[complex number|numbers]] {{mvar|z{{sub|A}}, z{{sub|B}}, z{{sub|C}}}} and assume that the [[circumscribed circle|circumcenter]] of triangle {{math|△''ABC''}} is located at the origin of the plane. Then, the complex number :<math>z_H=z_A+z_B+z_C</math> is represented by the point {{mvar|H}}, namely the altitude of triangle {{math|△''ABC''}}.<ref name="Andreescu">Andreescu, Titu; [[Dorin Andrica|Andrica, Dorin]], "Complex numbers from A to...Z". Birkhäuser, Boston, 2006, {{ISBN|978-0-8176-4326-3}}, page 90, Proposition 3</ref> From this, the following characterizations of the orthocenter {{mvar|H}} by means of [[euclidean vector|free vectors]] can be established straightforwardly: :<math>\vec{OH}=\sum\limits_{\scriptstyle\rm cyclic}\vec{OA},\qquad2\cdot\vec{HO}=\sum\limits_{\scriptstyle\rm cyclic}\vec{HA}.</math> The first of the previous vector identities is also known as the ''problem of Sylvester'', proposed by [[James Joseph Sylvester]].<ref name=Dorrie>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</ref>
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