EMAT 6680 :: Clay Kitchings
:: Assignment 9 :: Pedal Triangles

Problem 9: Find
all conditions for which the vertices of the pedal triangle are collinear (that
is, a degenerate triangle). Note: This line segment is called the Simson Line.

How do you
construct the Pedal Triangle?

It is obviously
important to first know how to construct a Pedal Triangle if we are to
investigate all conditions for which the vertices of the pedal triangle are
collinear.

First, construct a
triangle ABC. It is helpful to construct lines rather than segments. Construct any point P anywhere in the
plane. Then construct perpendicular lines through P to Lines AC, AB, and
BC. Below is a representative
figure:

The Pedal Triangle
is formed by the intersections of the perpendiculars to the lines AC, AB, and
BC. The triangle (EFG) in yellow below represents the Pedal Triangle.

Now that we
understand how to construct a Pedal Triangle, let us investigate cases for
which the Pedal Triangle degenerates into a line segment. For a GSP File, click here.

At first glance,
we can visually conjecture that when the Pedal Point (P) lies on any vertex of
triangle ABC, then triangle EFG degenerates into a segment. However, this seems
to be more of a ÒspecialÓ case. It
appears in this case that two vertices of the pedal triangle converge and
become the same point. Are there
other occasions for which the vertices of the pedal triangle are collinear and
even perhaps distinct points? I
found one approximate case within about 15-20 seconds by moving P about the
plane to obtain the following figure:

We can see that
points E, F, and G are approximately collinear, and it is reasonable to assume
(while exploring) collinearity near this region. Further movement of P about the plane could lead one to
suspect some sort of circular pattern P such that if P travels this path,
points E, F, and G are collinear.

I decided to construct the circumcircle of
triangle ABC. Then, I used GSPÕs ÒmergeÓ
tool to merge point P onto the circumcircle. I then animated P so that it would travel the path of the
circle. Sure enough, points E, F,
and G are collinear here.