Mathematics Education

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Centers of Triangles Learning Tasks


Jim Wilson

Instructional materials and instructional tasks dealing with centers of a triangle occur in many places in published material and on the internet. I have my own contribution to this at, a web page with its title of "Centers of a Triangle" in which definitions and explorations of four concurrencies -- the centroid, the orthocenter, the incenter, and the circumcenter -- are put forward and students are challenged to prove the concurrencies and investigate properties of each "center." Further, many of the problems from my problem solving course tie into these and other concurrencies of the triangle. Likewise in my Foundations of Geometry course, there are discussions of other points of concurrency relative to the triangle. In the paper, "Extended Concurrencies of a Triangle" I show a set of concurrencies for a triangle that lie along a locus that is the hyperbola. In other words, there are not four concurrencies, but rather infinitely many. Kimberling (1994) list a very large number of concurrencies of a triangle and the use of trilinear coordinates to classify them. A list of about 30 named triangle centers can be found at Triangles Centers and that web page includes links to references as well.

My first point here is that it amazes me that so many lessons for presenting students with elementary explorations explicitly state that there are FOUR triangle centers or triangle concurrencies. A Google Search for Centers of Triangles Learning Tasks will uncover many of them.

That is not the only message of this paper, however. The imprecise language on many of these introductions is serious. For example, there is rampant indiscrimint use of "line" and "segment" in these materials.

The centroid is the point of concurrency of the three medians of a triangle. A median is the segment from a vertex of a triangle to the midpoint of the opposite side. Segments have a length and thus one corollary of the proof of the existence of the centroid is that it divides each median into two parts in a ratio of 2:1. We properly discuss the concurrence of these segments.

The circumcenter, on the other hand, is the concurrence of the perpendicular bisectors of the sides of a triangle. Thes are lines, not segments. We prove that the three lines are concurrent and along the way observe that the circumcenter is equidistant from each end of a side, therefore it is equidistant from the ends of each side, and so it is equidistant from the three points. It is NOT the perpendicular bisector of a side that has a distance. Amazingly, many sets of instructional materials will instruct the students to consider the segments that are perpendicular bisectors. It is hard to know what the intent might be. Perhaps it refers to segments from the midpoint of the side (the foot of the perpendicular) to the circumcenter. Regardless, the circumcenter is the concurrency of three lines. For obtuse triangles, the circumcenter is outside of the triangle.

The orthocenter, perhaps, does the most violence to the line vs. segment distinction. Curriculum writers are not consistent in their definition of an altitude -- some define it as a line from a vertext perpendicular to the line of the opposite side; some define it as a segment from a vertex of a triangle to the foot of the perpendicular. Almost always, those who call the altitude a line will then inconsistently attribute a distance to it corresponding to the height of the triangle from the vertex to the foot of the perpendicular. I prefer to define the altitude as a segment. This means that the orthocenter is the concurrence of the lines from the vertices and perpendicular to the lines of the opposite sides. For obtuse triangles, the orthocenter is outside of the triangle; it is sloppy language to say the orthocenter is the concurrency of the altitudes. The othocenter is contained in the altitudes (defined as segments) only when the triangle is not obtuse.

The incenter is a point equidistant from each of the sides. Points equidistant from two sides of an angle will lie on the angle bisector. So the incenter is the point of concurrency of the three angle bisectors. There is not too much violence done to precision and language to observe that the incenter will always be inside the triangle, regardless of its shape, and so the segments from the vertices to the point of intersection of the opposite sides are often called the angle bisectors. In reality, however, the angle bisector is a ray that extends beyond the opposite and when we explore the bisectors of the external angles of a triangle, we may be interested in the concurrency of the two external bisectors on a side of the triangle and the internal angle bisector of the vertex opposte that side. This point of concurrency will be outside the triangle; there are three of them and they are equidistant from the lines of the sides of the triangle (i.e. the sides extended) and thus the center of an excircle.

Using the centers to "Solve Problems"

So how are students led to use triangle centers to solve real world problems. In many of the materials one can find on the web or in printed instructional materials, a 'map' is constructed with three towns and problems posed about locating some facility with relation to the three towns (vertices of the triangle). It is usually explicit that one can expect to use one of the centroid, circumcenter, orthocenter, or incenter to solve such problems. Unfortunately, many misconceptions and errors are brought about.

If the facility to be built means building new roads either from the three towns or from the existing highways along the sides of the triangle, then the cost of the construction of the roads is a function of the combined length of the roads.

A. If highways are built from each town to the facilty, on the map this is the SUM of the distances from each vertex to the point locating the facility. To minimize the cost, it would mean finding the point inside the triangle so that the sum of the distances is minimized. This is a standard problem in geometry but it IS not at this introductory level. THAT IS, none of the four centers (centroid, incenter, circumcenter, or orthocenter) would be the the point of minimum distance unless the three cities were at the vertices of an equilateral triangle. To lead students through an instructional sequence that leads to an erroneous conclusion is unacceptable.

That minimum sum of the distances from an internal point is the first Fermat point of the triangle when no angle is greater than 120 degrees. The first Fermat point is the concurrency of lines drawn from each vertex of the triangle to the external vertex of an equilateral triangle constructed on the opposite side of the triangle.

If the problem was restated as "Which of these four centers would be the best, that is, involve the least total distance?" then perhaps it is a little better.

If the problem was stated "Where should a facility be located that is equidistant from the three towns?" it is a straightforward exercise and not much problem solving. More important would be a discussion of "Why do this?" Looking at an example where the three towns are at the vertices of an obtuse triangle should raise issues about "equidistant" not being a solution to "minimum total distance.

B. If the problem is to build roads from the existing highways to minimize the cost, then the point to locate the facility would have the sum of the distances from the point along perpendiculars to the sides. This minimum turns out to be the length of the shortest altitude (yes a distance, a segment) from a vertex to its opposite side.


This map is taken from Centers of Triangles Learning Task from GeorgiaStandards.Org. The students are asked to locate Orthocenter (H), the Incenter (I), the Centroid (G), and the Circumcenter (C). After several exercises with constructing those four points, the problem is stated:

The president of the company building the park is concerned about the cost of building roads from the towns to the parks. What recommendation would you give him?


Careful statement of questions and their interpretation is important. What is being asked here? Is it reasonable to assume that this question wants to get at the conditions for minimal cost? I think so and that would mean the SUM of the lengths of the roads from the park to the cities.

In the map above I have constructed the points H, I, G, and C for this triangle. I have also located the first Fermat Point (F) and the map below the roads that would be built are shown. The Fermat point is also a concurrency of certain lines related to the triangle. One of the corollaries of this is that the three line segments from F all are separated by an angle of 120 degrees.

The argument here is NOT that the Fermat point is a better answer. Rather, the task itself is flawed. The intent was to give students experience with some pretty basic ideas of geometry -- which it does. But the illusion of an "application" leads to misconceptions. The tools of geometry provided by the first four centers do not suffice for answering the question posed yet the students are explicitly told all through the investigation that one of those center will be the location best for the builder.

The tasks posed in investigations like this must allow the available tools to provide answers that will not have to be revised when more advanced tools are available.

There can be other questions. For example, the location equally convenient to the people in all three towns might be the circumcenter. Here, though, the question is not about minimizing the cost of construction. Yet, if the three towns were Albany, Valdosta, and Waycross the obtuse triangle makes the circumcenter less appealing for any criteria other than "same distance for all" even if that is further than any of they would need to travel. If the question was where to locate a new trauma center, it would make sense to locate it at the midpoint of the longest side between Albany and Waycross.

Real world decisions would bring in many other factors and cost of roads may not be of the highest importance.