Difference between revisions of "Orthocenter"
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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]]s. It is [[mathematical convention | conventionally]] denoted <math>H</math>. | ||
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''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.'' | ''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 <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 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. | ||
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==Properties== | ==Properties== | ||
| − | *The orthocenter is collinear with the [[circumcenter]] and [[de Longchamps point]]. | + | * The orthocenter is collinear with the [[circumcenter]] and [[de Longchamps point]]. |
| − | *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 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. |
| + | * 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> | ||
| + | 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> | ||
==See Also== | ==See Also== | ||
Revision as of 08:30, 22 July 2009
The orthocenter of a triangle is the point of intersection of its altitudes. It is conventionally denoted 
.
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]](http://latex.artofproblemsolving.com/6/7/d/67d616b35fc9702763fcb0b034d0d91933a6d069.png)
Consider a triangle 
with circumcenter 
and centroid 
. Let 
be the midpoint of 
. Let be the point such that is between and and 
. Then the triangles 
, 
are similar by angle-side-angle similarity. It follows that 
is parallel to 
and is therefore perpendicular to ; i.e., it is the altitude from 
. Similarly, 
, 
, are the altitudes from 
, 
. Hence all the altitudes pass through . 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.
Properties
- The orthocenter is collinear with the circumcenter and de Longchamps point.
- If the orthocenter's triangle is acute, then the orthocenter is on the triangle, it 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
be a triangle and 
its orthocenter. Then the reflections of over 
, 
, and 
are on the circumcircle of :
, 
are on the circumcircle of ![[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]](http://latex.artofproblemsolving.com/d/4/3/d43dfa0aa5888fc748071e9fb210a3deee6c8212.png)
