## Tuesday, January 26, 2016

### the triangle of triangles

Is there a triangle that contains every possible triangle? In one sense, yes there is.

In "The Stiener-Lehmus Theorem" (which appears in The Best Writing on Mathematics, 2015, edited by Mircea Pitici), John Conway and Alex Ryba introduce a clever nomogram which they call "the triangle of triangles."

As the authors describe it,
each point corresponds to a triple of numbers A, B, and C, that add to 180, and so to a shape of a triangle. In the figure, A is constant on downward sloping lines, B on upward sloping ones, and C on horizontals.
In the diagram below, the point P provides us with a triple (36, 60, 84), which corresponds to the shape of a triangle ABC with those angles. Note that A and B are easier to read off the figure than C, so it may help to add an additional axis.

To say P corresponds to the "shape" of a triangle, it is meant that P represents an infinite family of similar triangles whose angles (measured in degrees) equal the numbers given by the triple described by P.

Let's see where some favorite triangle families live on the triangle of triangles. In the simplified figure below, green lines are drawn at 90 and light blue lines are drawn at 60. These intersect at the special 45-45-90 right triangles.

And here are some more favorites. With the light green lines drawn at 90, the dark green drawn at 60, and the orange lines drawn at 30, we get the 30-60-90 special triangles, and equilateral.

Maybe it is a little surprising that the triple you get from the points within the triangle should sum to 180. In the diagram below, those values correspond to the lengths of RC, QC and PS, with appropriate scaling factors applied

If for the sake of simplicity, we assume that the lengths of AP and BC are 1, things simplify quite a bit. The relationship to be proven reduces to what is shown in the figure below.

One way to show this is to start with the median DC, and show how the required lengths are related to it.

Of course, it makes sense if you are drawing a triangle-of-triangles, you'd likely want to draw it as an equilateral, but as you might notice, any old triangle seems to do (with the right scaling). The figure below shows one set of measurements taken in GSP, where you can distort the triangle and convince yourself that the relationship holds generally. Feel free to re-do the proof above for the general case :)