Difference between revisions of "Orthocenter"

Revision as of 20:52, 3 May 2020

The orthocenter of a triangle is the point of intersection of its altitudes. It is conventionally denoted $H$ . $[asy] size(15cm); markscalefactor = 0.01; dot((0,0)); dot((4,0)); dot((3,3)); dot((3,0)); dot((2,2)); dot((3,1)); dot((3.6,1.2)); draw((0,0)--(4,0)--(3,3)--cycle); draw((0,0)--(3.6,1.2),red); draw((2,2)--(4,0),red); draw((3,0)--(3,3), red); draw(rightanglemark((0,0),(3,0),(3,1))); draw(rightanglemark((4,0),(3.6,1.2),(0,0))); draw(rightanglemark((0,0),(2,2),(4,0))); label("Orthocenter",(2.85,1.1),W); [/asy]$ The lines highlighted are the altitudes of the triangle, they meet at the orthocenter.

Proof of Existence

Note: The orthocenter's existence is a trivial consequence of the trigonometric version Ceva's Theorem; however, the following proof, due to Leonhard Euler, is much more clever, illuminating and insightful.

$[asy] defaultpen(fontsize(8)); pair A=(8,7), B=(0,0), C=(10,0), A1 = (B+C)/2, O = circumcenter(A,B,C), G = (A+B+C)/3, H = 3*G-2*O; draw(A--B--C--cycle); draw(A--G--H--cycle); draw(A1--G--O--cycle); label("A",A,(0,1));label("B",B,(0,-1));label("C",C,(0,-1));label("G",G,(1,-1));label("H",H,(0,-1));label("O",O,(-1,1));label("$A'$",A1,(0,-1));dot(H); [/asy]$ Consider a triangle $ABC$ with circumcenter $O$ and centroid $G$ . Let $A'$ be the midpoint of $BC$ . Let $H$ be the point such that $G$ is between $H$ and $O$ and $HG = 2 GO$ . Then the triangles $AGH$ , $A'GO$ are similar by side-angle-side similarity. It follows that $AH$ is parallel to $OA'$ and is therefore perpendicular to $BC$ ; i.e., it is the altitude from $A$ . Similarly, $BH$ , $CH$ , are the altitudes from $B$ , ${C}$ . Hence all the altitudes pass through $H$ . Q.E.D.

This proof also gives us the result that the orthocenter, centroid, and circumcenter are collinear, in that order, and in the proportions described above. The line containing these three points is known as the Euler line of the triangle, and also contains the triangle's de Longchamps point and nine-point center.

Easier proof

That seems somewhat overkill to prove the existence of the orthocenter. We use a much easier (and funnier) way.

Let the line through $B$ parallel to $AC$ and the line through $C$ parallel to $AB$ intersect at $D.$ Define $E,F$ similarly. Note that $FA=BC=AE,$ so the $A$ altitude of $\triangle ABC$ is the perpendicular bisector of $DE.$ Since the circumcenter exists, the altitude must too.

$[asy] import olympiad; size(4cm); pair A=dir(110), B=dir(200), C=dir(-20), D,E,F,H; H=orthocenter(A,B,C); D=B+C-A; E=C+A-B; F=A+B-C; draw(A--B--C--A); draw(D--E--F--D); draw(A--H,dotted); draw(B--H,dotted); draw(C--H,dotted); label("A",A,N); label("B",B,dir(190)); label("C",C,dir(-10)); label("D",D,S); label("E",E,dir(45)); label("F",F,dir(140)); label("H",H,dir(50)); [/asy]$

Properties

The orthocenter and the circumcenter of a triangle are isogonal conjugates.
If the orthocenter's triangle is acute, then the orthocenter is in the triangle; if the triangle is right, then it is on the vertex opposite the hypotenuse; and if it is obtuse, then the orthocenter is outside the triangle.
Let $ABC$ be a triangle and $H$ its orthocenter. Then the reflections of $H$ over $AB$ , $BC$ , and $CA$ are on the circumcircle of $ABC$ :

$[asy] defaultpen(fontsize(8)); pair A=(8,7), B=(0,0), C=(10,0), H=orthocenter(A,B,C), A1, B1, C1; A1 = 2*foot(A,B,C)-H; B1 = 2*foot(B,C,A)-H; C1 = 2*foot(C,A,B)-H; draw(A--B--C--cycle,black+1); draw(A--A1);draw(B--B1);draw(C--C1); draw(A1--B--C1--A--B1--C--cycle); draw(circumcircle(A,B,C)); dot(A1^^B1^^C1^^H); label("$A$",A,(0,1));label("$B$",B,(-1,0));label("$C$",C,(1,0)); label("$A'$",A1,(0,-1));label("$B'$",B1,(1,1));label("$C'$",C1,(-1,1)); label("$H$",H,(-1,-1)); [/asy]$

Even more interesting is the fact that if you take any point $P$ on the circumcircle and let $M$ to be the midpoint of $HP$ , then $M$ is on the nine-point circle.

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Difference between revisions of "Orthocenter"

Revision as of 20:52, 3 May 2020

Contents

Proof of Existence

Easier proof

Properties

See Also

@@ Line 1: / Line 1: @@
-The '''orthocenter''' of a [[triangle]] is the point of intersection of its [[altitude]]s.  It is [[mathematical convention | conventionally]] denoted <math>H</math>.
+The '''orthocenter''' of a [[triangle]] is the point of intersection of its [[altitude|altitudes]].  It is [[mathematical convention | conventionally]] denoted <math>H</math>.
+<asy>
+size(15cm);
+markscalefactor = 0.01;
+dot((0,0));
+dot((4,0));
+dot((3,3));
+dot((3,0));
+dot((2,2));
+dot((3,1));
+dot((3.6,1.2));
+draw((0,0)--(4,0)--(3,3)--cycle);
+draw((0,0)--(3.6,1.2),red);
+draw((2,2)--(4,0),red);
+draw((3,0)--(3,3), red);
+draw(rightanglemark((0,0),(3,0),(3,1)));
+draw(rightanglemark((4,0),(3.6,1.2),(0,0)));
+draw(rightanglemark((0,0),(2,2),(4,0)));
+label("Orthocenter",(2.85,1.1),W);
+</asy>
+The lines highlighted are the [[altitude|altitudes]] of the triangle, they meet at the orthocenter.
 == Proof of Existence ==
@@ Line 12: / Line 32: @@
 label("A",A,(0,1));label("B",B,(0,-1));label("C",C,(0,-1));label("G",G,(1,-1));label("H",H,(0,-1));label("O",O,(-1,1));label("$A'$",A1,(0,-1));dot(H);
 </asy>
-Consider a triangle <math>ABC</math> with [[circumcenter]] <math>O</math> and [[centroid]] <math>G</math>.  Let <math>A'</math> be the midpoint of <math>BC</math>.  Let <math>H</math> be the point such that <math>G</math> is between <math>H</math> and <math>O</math> and <math>HG = 2 GO</math>.  Then the triangles <math>AGH</math>, <math>A'GO</math> are [[similar]] by angle-side-angle similarity.  It follows that <math>AH</math> is parallel to <math>OA'</math> and is therefore perpendicular to <math>BC</math>; i.e., it is the altitude from <math>A</math>.  Similarly, <math>BH</math>, <math>CH</math>, are the altitudes from <math>B</math>, <math>{C}</math>.  Hence all the altitudes pass through <math>H</math>.  Q.E.D.
+Consider a triangle <math>ABC</math> with [[circumcenter]] <math>O</math> and [[centroid]] <math>G</math>.  Let <math>A'</math> be the midpoint of <math>BC</math>.  Let <math>H</math> be the point such that <math>G</math> is between <math>H</math> and <math>O</math> and <math>HG = 2 GO</math>.  Then the triangles <math>AGH</math>, <math>A'GO</math> are [[similar]] by side-angle-side similarity.  It follows that <math>AH</math> is parallel to <math>OA'</math> and is therefore perpendicular to <math>BC</math>; i.e., it is the altitude from <math>A</math>.  Similarly, <math>BH</math>, <math>CH</math>, are the altitudes from <math>B</math>, <math>{C}</math>.  Hence all the altitudes pass through <math>H</math>.  Q.E.D.
+This proof also gives us the result that the orthocenter, centroid, and circumcenter are [[collinear]], in that order, and in the proportions described above.  The line containing these three points is known as the [[Euler line]] of the triangle, and also contains the triangle's [[de Longchamps point]] and [[nine-point center]].
+=== Easier proof ===
+That seems somewhat overkill to prove the existence of the orthocenter. We use a much easier (and funnier) way.
+Let the line through <math>B</math> parallel to <math>AC</math> and the line through <math>C</math> parallel to <math>AB</math> intersect at <math>D.</math> Define <math>E,F</math> similarly. Note that <math>FA=BC=AE,</math> so the <math>A</math> altitude of <math>\triangle ABC</math> is the perpendicular bisector of <math>DE.</math> Since the circumcenter exists, the altitude must too.
-This proof also gives us the result that the orthocenter, centroid, and circumcenter are [[collinear]], in that order, and in the proportions described above.  The line containing these three points is known as the [[Euler line]] of the triangle.
+<asy>
+    import olympiad;
+    size(4cm);
+    pair A=dir(110), B=dir(200), C=dir(-20), D,E,F,H;
+    H=orthocenter(A,B,C);
+    D=B+C-A;
+    E=C+A-B;
+    F=A+B-C;
+    draw(A--B--C--A);
+    draw(D--E--F--D);
+    draw(A--H,dotted);
+    draw(B--H,dotted);
+    draw(C--H,dotted);
+    label("A",A,N);
+    label("B",B,dir(190));
+    label("C",C,dir(-10));
+    label("D",D,S);
+    label("E",E,dir(45));
+    label("F",F,dir(140));
+    label("H",H,dir(50));
+</asy>
 ==Properties==
-* The orthocenter is collinear with the [[circumcenter]] and [[de Longchamps point]].
+* The orthocenter and the circumcenter of a triangle are [[isogonal conjugate]]s.
-* If the orthocenter's triangle is [[acute triangle|acute]], then the orthocenter is on the triangle, it the triangle is [[right triangle|right]], then it is on the vertex opposite the [[hypotenuse]], and if it is [[obtuse triangle|obtuse]], then the orthocenter is outside the triangle.
+* If the orthocenter's triangle is [[acute triangle|acute]], then the orthocenter is in the triangle; if the triangle is [[right triangle|right]], then it is on the vertex opposite the [[hypotenuse]]; and if it is [[obtuse triangle|obtuse]], then the orthocenter is outside the triangle.
 * Let <math>ABC</math> be a triangle and <math>H</math> its orthocenter. Then the reflections of <math>H</math> over <math>AB</math>, <math>BC</math>, and <math>CA</math> are on the circumcircle of <math>ABC</math>:
 <asy>
@@ Line 35: / Line 82: @@
 label("$H$",H,(-1,-1));
 </asy>
+*Even more interesting is the fact that if you take any point <math>P</math> on the circumcircle and let <math>M</math> to be the midpoint of <math>HP</math>, then <math>M</math> is on the nine-point circle.
 ==See Also==
 *[[Centroid]]
+*[[Incenter]]
 *[[Altitude]]
 *[[Circumcenter]]