Orthocenter of a Triangle

The orthocenter of a triangle is the point where the three altitudes meet.  This point may be inside, outside, or on the triangle.  Here are some properties of the orthocenter that I find to be interesting. Many of these are found in Ross Honsberger's wonderful book, Episodes in Nineteenth and Twentieth Century Euclidean Geometry, which I highly recommend.   The figure below will be the general reference for the names of elements.In the figure, ABC is the triangle, with H as the orthocenter at the intersection of altitudes AD, BE, and CH.  P is the point where the Altitude AD extended meets the circumcircle.
   One of the first things we need to show is that the three altitudes always pass through a single point.  We will do this with a simple illustration at the right.  The image shows the same triangle with some lines removed, and the circumcenter, O, and the median AA' through Geocenter G added.  If we rotate point O 180 degrees about G and double its length, we get point H, which we wish to show is the orthocenter.  By a simple property of medians we know that AG is twice the length of A'G, and by our construction we have GH is twice GO.  Since the two angles HGA and OGA' are a vertical pair, we know they are congruent and can assert that the triangles are similar and AH must be parallel to OA'.  Since we already know that OA' is Perpendicular to BC, this means that AH must also be perpendicular, and must lie on the altitude from A to BC.  But A was in no way special. Had we selected either of the other vertices and drawn the same construction, G, O, and H would have been the same three points, thus all three altitudes pass through H, and it is the orthocenter.  A second small gem pops out from the figure.  The distance from the orthocenter to a vertex (AH in figure) is always twice  the distance from the circumcenter to the opposite edge (OA' ).

It is well known that the orthocenter is an isogonal conjugate of the circumcenter. This heavy language means that if the altitudes are reflected in the angle bisectors from the same vertex, the three new lines intersect at the circumcenter of the triangle.

   The next useful little piece of information we can exploit is that the length of HD is equal to the length of DP.  This can be  easily shown from two basic observations. The first is that angles BPA and BCA are congruent, since they are both inscribed angles cutting the same arc.  The second, less obvious fact, is that angle BHD is also equal in measure to both of these. This is made clear by the fact that the two legs of BHD are respectively perpendicular to the two legs of BCA (BH perpendicular to BC and HD perpendicular to CA).  But if BPD and BHD are congruent, then triangle BPH is an isosceles triangle with BD as its perpendicular bisector and HD is the same length as DP.  As small an observation as this seems, it produces several little corollaries.
   The first can be seen when we view AP and BC as chords of the circle containing ABC.  From early geometry we know that when two chords intersect, the products of their parts are equal, so we have AD*DP = BD*DC.  If we substitute DH in place of DP, we get AD*HD=BD*DC, and a general rule that says the altitude divides the base segment into two parts whose product is the product of  the altitude times the distance from the base to the orhtocenter.  A version of this is well known in high school geometry classes when the angle at A is a right angle, "The square of the altitude is equal to the product of the two parts of the hypotenuse created by the foot of the altitude."
   A second fact emerges when we apply the fact that the three altitudes passing through H can all be extended beyond the triangle to be chords of the circumcircle. This means that the products BH*HQ = AH*HP=CH*HR.  Using the already established fact that the distance from the orthocenter to the foot of the altitude (HD for example) is equal to the length from the foot of the altitude to the circumcircle (DP),  we can rewrite these as BH*2*HE = AH*2*HD=CH*2*HF.  Dividing every thing by two will produce BH*HE=AH*HD=CH*HF, which shows that in any triangle, the two pieces of the altitude formed by the orthocenter have the same product for all three altitudes.
   With the knowledge that BC is the perpendicular bisector of HP, we realize that the reflection of the circumcenter in BC will be a congruent circle passing through B, C, and H.  This is the circumcircle of B, C, H .  One thing that may not be immedaitely obvious to the reader, is that if the Point A was moved around the circle holding B and C stationary, the orthocenter would follow the path of this circle.

Since the choice of BC was arbitrary, we could have done the same with any side of the circle and so we see that the circumcenter of the triangles formed by the orthocenter and any two vertices of the original triangle must all have the same radius, and pass through the orthocenter.   This figure also makes it clear that the arc intercepted by angle BHC in the image circle,  is the explement (they add up to 360 degrees) of the arc intercepted by BAC in the original circle, thus the angle formed at the orhtocenter is the supplement of the angle at the vertex (/BAC + /BHC = 180o).
   There is one more "well known but worth repeating" property that is visible from this figure and also the original figure.  If we focus on the triangle BHC we notice that line AC is one altitude (passes through vertex C and is perpendicular to BH).  By similar reasoning we get that BA is also an altitude, as is DA, and the orthocenter of triangle BHC is at A.  It is easy to show that the same thing would hold if we picked any three of ABCH to be the vertices of the triangle.  The remaining point would be the orthocenter of the other three.  Because of this ability to interchange roles, we often refer to the four points as an orthocentric system.

Returning to the fact that the reflection of the orthocenter, H, in any side falls on the circumcenter, we get a second triangle, PQR, (dotted dark in figure) which are the other ends of the chords formed by the altitudes extended to the circumcircle. I have never seen a name for this, so I will call it the double-orthic triangle since, as I will soon show, it is mearly a dilation by two of the orthic triangle. Because each point is a reflection of H in one of the sides of the triangle, we can see that each of points P,Q, and R is the reflection of the others across two of the sides. For example, the reflection of P in BC is H, and H reflected in AC is Q. Since two reflections are twice the angle formed by the lines of reflection, we know that angle PCQ is twice angle ACB. Angle PRQ intercepts the same chord, so we know that Angles PRQ and PCQ are supplementary. Now because PCQ is twice ACB, we see that the angles of the double-orthic triangle are each equal to 180 - twice the opposite angle of ABC.

One more property of the double-orthic triangle may be useful. I wish to show that the points of the double-orthic triangle on each side of a vertex of ABC is the same arc measure from the Vertex. For example, arc AR = arc AQ. To show this we need only examine the two right triangles ABE and ACF. Since the angle at B and the angle at C are both the compliment of the angle at A, they are equal, and they are both inscibed angles cutting out the arcs AR and AQ, thus the arcs are equal.

What use can we make of this knowledge? Well, an easy answer is that we know know that RQ is a chord of the circumcircle, and that the line from A perpendicular to RQ must not only bisect RQ, but pass through the center of the circle. Stated another way, a radius of the circumcircle to a vertex of ABC is perpendicular to the side of the double-orthic triangle. With that done we proceed to extend these two ideas to the Orthic triangle.

The Orthic triangle is a name for the triangle joining the feet of the perpendiculars, D, E, and F. We have already established that HD = DP and similarly for the other three vertices, so the double-orthic triangle we mentioned above is just a similar copy of the orthic triangle with sides twice as long and the same orientation. It should be clear that the sides of the orthic triangle, and the double orthic triangle are respectively parallel. From this we see that the two properties shown for the double-orthic must also be true for the orthic triangle, that is, the orthic trianlge has angles that are the supplement of twice the opposite angle, and the sides are perependicular to the radius of the circumcircle to a vertex.

Of all the triangles that could be inscribed in a given triangle, the one with the smallest perimeter is the orthic triangle. This has sometimes been called Fagnano's Problem since it was first posed and answered by Giovanni Francesco Fagnano dei Toschi. Fagnano also was the first to show that the altitudes of the original triangle are the angle bisectors of the orhtic triangle, so the incenter of the orthic triangle is the orthocenter of the original triangle.

A simple but pretty formula relates the lengths of the sides of the orthic triangle to the sides and angles of the orginal triangle. For example, in the Triangle ABC, the side of the orthic triangle nearest to vertex A is given by a*Cos (A), and likewise for the other two sides. The perimeter of the orthic triangle, then, is equal to a*Cos(A) + b*Cos(B) + c*Cos(C). It can also be shown that the perimeter of the orthic triangle to ABC is equal to twice the area of ABC divided by R, the radius of the circumscibed circle of ABC.

If we pick a random point P2 on the circumcircle and reflect it in the sides a, b, and c, we get three distinct points P2b, P2a, and P2c . The pretty part of this little activity is that all three points lie on a straight line, and, the line always contains the Orhthocenter, H. This line is parallel to the Simson line for P2, and the Simson line is exactly half way from P2 to H.

A recent thread on the Historia-Matematica discussion group pointed out that Euclid did not address the concept of the orthocenter in the Elements. Archimedes did address the concept, although not by that name, in Lemma 5 of the Liber Assumptorum. Dick Tahta suggested that Lemma 12 is a more appropriate source for the Orthocenter.

A posting by Emili Bifit provides a quote from an article by John Satterly that gives credit for the creation of the name "Orthocenter" to Besant and Ferrers in 1865.

"Note_: As a matter of historical interest our readers may be reminded that the term ``Orthocentre'' was invented by two mathematicians, Besant and Ferrers, in 1865, while out for a walk along the Trumpington Road, a road leading out of Cambridge toward London. In those days it was a tree-lined quiet road with a sidewalk, a favourite place for a conversational walk."
From Mathematical Gazette, Feb 1962, pp 51

In Ross Honsberger's Episodes in Nineteenth and Twentieth Century Euclidean Geometry he points out another interesting property about the orthocenter that is called the Droz-Farney Theorem, after the Swedish mathematician Arnold Droz and his wife Lina Farny. If perpendicular lines, neither parallel to a side of the triangle, are drawn through the othocenter of a triangle each line will cut each of the sides (possibly extended) in one place. The Droz-Farny Theorem states that the midpoints of the two intersections on each side of the triangle will be collinear. Here is a link to an interactive Java-script of the Droz-Farny Theorem.

After experimenting with this theorem on the Geometer's Sketchpad I have convinced myself, (which does NOT constitute a proof) that the following is true about the set of lines produced for a given triangle when the perpendiculars through the orthocenter are rotated. If the triangle is acute, the set of lines will describe an ellipse that is tangent to the three sides of the triangle. If the triangle is obtuse, the lines will describe an hyperbola tangent to the longest side of the triangle and to the extended portions of the other two sides. In each case the conics have their foci at the orthocenter and the circumcenter, and the conic center is at the center of the nine point circle. The major diagonal of each of the conics is the Euler Line.

Arnold Droz was born in La Chaux-de-Fonds, Switzerland, on Feb. 12, 1856. Lina was born in the same town on April 12 of 1854. In 1880 he was appointed to be professor of physics at the cantonical school of Porrentruy, Switzerland, where he taught until 1908. He died on 14 January in 1912. (The above biographical data is based on a web posting of Julio Gonzalez Cabillon to the Historia Matematica mailing list in August of 2000)

Monge's Point in a tetrahedron The most general extension of a triangle to 3-space is the tetrahedron, and one is led to wonder if the altitudes of a tetrahedron would also be concurrent. The answer is that they may or may not be, depending on certain conditions of the tetrahedron, but there are six planes in every tetrahedron that intersect in a single point, and the point is related to the concurrency points of a tetrahedron.

If a plane is constructed passing through the midpoint of each of the six edges, and perpendicular to the opposite edge, the planes will intersect in a single point. Read that sentence carefully and note that these are not the planes which are the perpendicular bisectors of each edge. The perpendicular bisecting planes of each edge do contain a common point, but it is the circumcenter of the sphere that contains the tetrahedron. The point of intersection of the six midplanes is the Monge point, named for Gaspard Monge who first discovered the property and published in an 1809 paper. The Monge point is the reflection of the center of gravity of the tetrahedron around the circumcenter. If a tetrahedron has all three sets of opposite sides orthogonal (perpendicular) in pairs, then the altitudes will intersect, and they will intersect in the Monge point. A tetrahedron with this property is called an orthocentric tetrahedron. It is known that if three altitudes of a tetrahedron are concurrent, the fourth will also pass through the same point.

The Monge point seems very closly related to the anti-center of a cyclic quadrilateral and may help in visualziing the 3-D image of tetrahedron and midplanes. If the four edges and two diagonals of a cyclic quadrilateral are viewed as a drawing of a tetrahedron and the four maltitudes of a cyclic quadrilateral and the two segments perpendicular to the diagonals that pass through the midpoints of the opposite diagonal are visualized as segments in the midplanes of the tetrahedron then a pretty good picture of the concurrence may emerge.

I have just seen an outline of the June,2003 issue of the MAA "Mathematics Magazine" which will carry an article on the Monge point, and may be of interest also.