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Orthocenter

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 consequence of Ceva's Theorem (proof). 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.


[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]


Let the line through $B$ parallel to $AC$ and the line through $C$ parallel to $AB$ intersect at $D.$ Define $E,F$ similarly.

(Alternatively, $\triangle DEF$ can be constructed by reflecting $\triangle ABC$ across each of its edges' midpoints, respectively.)

Note that $FA=BC=AE$ (and likewise for the other sides $AB$ and $CA$ of $\triangle ABC$), and so each altitude of $\triangle ABC$ is a perpendicular bisector of $\triangle DEF$.

Since the perpendicular bisectors of $\triangle DEF$ intersect (at its circumcenter), this intersection point is also the the intersection of altitudes of $\triangle ABC$, its orthocenter.

Properties

  • The orthocenter and the circumcenter of a triangle are isogonal conjugates.
  • If a triangle is acute, then its orthocenter $H$ is in the triangle; if the triangle is right, then $H$ is on the vertex opposite the hypotenuse; and if it is obtuse, then $H$ 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]

  • The reflections of $H$ over the midpoints of the sides $AB$, $BC$, and $CA$ are also on the circumcircle of $ABC$:
  • The circumcircle and the nine-point circle are [homethy|homethetic] with respect to the orthocenter $H$, with scaling factor equal to 2. That is, the midpoint of $H$ and a point on the circumcircle is a point on the nne-point circle.

Resources

Art of Problem Solving Volume 2 - Example 21-4 Euclidean Geometry in Mathematical Olympiads by Evan Chen - Section 1.3

See Also

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