Mathematics Education

J. W. Wilson

Consider the triangle ABC with base AB fixed and vertex C moving along some path. What will be the locus of the orthocenter? (The orthocenter is the intersection point of the altitudes of a triangle.) The path for vertex C could be along a straight line, along the arc of a circle, along a parabola, or along any function.

Consider the vertex C to move along the path of a function given by f(x). Let A and B be arranged on the x-axis at (-5,0) and (5,0) respectively. Then if the orthocenter is at (x,y) then vertex C is at (x,f(x)).The slope of BC is given by

Therefore, the equation of the line through A and the orthocenter is

When the vertex C of the triangle with base on the segment AB, with A(-5,0) and B(5,0), moves along the line y = 5, the locus of the orthocenter is given by the equation

which is clearly a parabola. If other lines parallel to the x-axis are used, the locus of the orthocenter will remain a parabola. For example, when y = 3 or y = 8

Now, what if the line for the vertex is not parallel to the x-axis? For example, take y = x. The equation of the locus is

and the graph is given at the right. If the lin is y = -x or y = 3x + 4, the graphs below will result.

There is no restriction, of course, to move the vertex C along a line. For example, consider the locus of the orthocenter as the vertex C is moved along various parabolas:

There is a special case where the third vertex of the triangle moves along a parabola that crosses the x-axis at -5 and 5. In this case ther structure of the function would lead to the path of the orthocenter being a constant, y = 1 or y = -1.

For moving the vertex around a circle, the limitations of being a function require investigating two functions, the upper half-circle and the lower half-circle.

We can examine moving the vertex C around the entire circle by overlaying the upper and the lower halves of the circle

An interesting case results when the circle is

so that the base AB of the triangle is a

radiusof the circle. This pattern generates aStrophoidand the base can be any portion of the diameter of the circle as long as one vertex is on the circle. If the base AB is the diameter, the locus of the orthocenter is the circle (the triangles are all right triangles). If the base AB is achordbut not a diameter, the locus of the orthocenter is a circle of the same radius but with a different center. Clickherefor a GSP sketch to experiment with these various configurations and others.

The circles

provide some contrast to this last graph:

Some other interesting cases can be defined for other locations of a circle relative to the triangle with base on the x-axis from -5 to 5. I will use a GSP sketch to show some of these.

Let's try moving vertex C along a cosine curve. The cosine curve shown here is

y = cos(5x) +3

The red curve is the locus of the orthocenter of a triangle with base AB between - 5 and 5 on the x-axis and vertex C moving along the cosine curve.

This locus of the orthocenter is outlined by a pair of parabolas as shown below:

If the vertex of the triangle moves along the rectangular hyperbola

the locus of the orthocenter follows a cubic equation. The orthocenter approaches the origin, where it is undefined, as the third vertex of the triangle approaches an infinite length from the base. Thus, the orthocenter at the origin is defined in the limit.

The travels with the orthocenter presents an interesting context to explore interchange between algebraic and geometric depiction of situations. With technology, some situations are easier to depict with dynamic geometry software than with a function grapher, and some are vice versa.

Further investigations could explore moving a vertex along segmented paths such as the sides of a polygonal figure or a path made up of arcs of conics.

The combination of algebraic tools and geometric tools facilitate the definition and proof of locus paths at a more elementary level. For example, we began with the locus of the orthocenter as the third vertex is moved along a line parallel to the base. This appears to be a parabola. The algebra of the situation makes it much easier to prove that the locus is a parabola and to determine such things as the location of the focus and the directrix, although all of this could be done with geometric arguments alone.