The Sliding Triangle

If you read the ReuleauxTriangle article, you may have wondered about the curves traced by the centroid and vertices of the reuleaux as it rotated within the rhombus. Essentially, what we had was a rigid figure with two of its points sliding along intersecting lines. I want to examine the locus of a third point on the figure. My generalization of the question does not fit the reuleaux exactly, as in that case it is not always the same two vertices in contact with sides of the rhombus, and it is not always the same two sides. Those conditions change as the figure turns. Still the question interested me.

In my model, there are only three points in motion. Two of them are the points tethered to intersecting lines. The third point forms a rigid frame with those two, and traces out a locus. It was easy enough to model the configuration on Sketchpad. Since we are interested in only three points of the figure, substitute a triangle and have two of the vertices attached to intersecting lines. When I move the triangle along the lines, the trace of the third vertex appears to be an ellipse, but that has not been proved.

What I found here was satisfying, but it came as no great surprise that I was not the first to discover it. Several years after publishing the first edition of this article, it came to my attention that Frans van Schooten (1615‑1660) had reached the same conclusions more than three centuries earlier. I commend him for it, but, frankly, I am not convinced that he was the first either.

To understand this problem, it might be helpful to review the definitions of the ellipse. It has several equivalent definitions. I tend to avoid the Cartesian plane when discussing conics, but here I will use it to get things started. The parametric equations below represent an ellipse in standard position, having semi‑major axis a and semi‑minor b.

This may seem unnecessarily restrictive. This definition applies only to an ellipse centered on the origin and having the x‑axis as its major axis. It will do here though, because the origin itself can be arbitrarily placed anywhere on a given plane. The axes must be perpendicular to each other, but they may be tilted in any direction we please.

The image above is an interactive Web Sketch. The line segement lengths OP and PQ are fixed. The direction of OP is θ, as measure counterclockwise from the positive x direction. The direction of PQ is θ measured clockwise. Drag point P and observe the trace left by point Q. It sure looks like an ellipse, but is it? With a little thought, these parametric equations can be derived for Q:

Now try these substitutions:

So it is an ellipse. Notice that the directions of OP and PQ change at the same rate, but in opposite rotation directions. Also notice that the lengths of the segments do not matter. That is, we never did say what a and b were. We can choose arbitrary segment lengths and use them to define a and b. From this we can make a general statement. Given a point P tracing a circle about some fixed center O, if point Q revolves about P at the same rate, but opposite direction, and if OP and PQ have constant lengths, then Q must trace an ellipse.

I skipped a few steps there, so maybe you can fill in the gaps. Since OP and PQ are rotating in opposite directions, they must be collinear for some orientation, but not necessarily on the x- and y‑axes, as in the example above. In fact, there is no reason that O has to be the origin. If you start with any points O, P, and Q, you should be able to just set them in motion. Point P orbits O, and Q orbits P, like a star, a planet, and a moon. If they have the same orbital period, but opposite direction, then point Q will trace an ellipse. Try it.

Now what about that triangle? In the stated problem, two vertices of a triangle are each tethered to one of two intersecting lines. The lines are not necessarily perpendicular, so these are not coordinate axes, and those parametric equations will no longer serve us well. A dynamic geometry sketch can still be constructed though. Triangle ABQ is defined off to the side. A congruent image is transformed in such a way that points A and B remain on their respective lines. To understand the motion of Q, use the action button to show it in a different frame of reference. Suppose the triangle is stationary and it is the two lines that are in motion. They still have the constant angle of intersection, so point O must trace a circle. It is the circumcircle of ΔABO. Let P be its center, and its distance from O must be constant.

Moving back to the original frame of reference, the triangle is moving, and P moves with it. As P rotates about O, Q rotates about P with the same angular speed and opposite direction. Yes, its locus is an ellipse.

Questions to Consider

There is an action button in the sketch allowing you to show the axes and respective vertices of the ellipse. Given the intersecting lines and ΔABQ, could you construct these objects?

What conditions for the triangle would result in a circle locus?

Another action button allows you to set the linear paths perpendicular, essentially returning to the Cartesian plane. Under what conditions would the axes of the resulting ellipse be the coordinate axes?

Generally, excepting certain special cases, there is a second distinct triangle that will trace this same ellipse. When this was originally written, I called it the ghost for some reason. In the Web Sketch there is an action button revealing the ghost as ΔCDR. How could this triangle be constructed from the original ΔABQ and the intersecting lines? What are the special conditions that would make these two triangles congruent, hence not distinct at all?

Web Sketchpad Files

The animated sketches on this page were created with Web Sketchpad, a developmental version of The Geometer’s Sketchpad.