A Beautiful Triangle Relation, with an Unexpected Twist.

  From an exploration a few years back:

It started with a tweet from John Golden@mathhombre who put me on to this beautiful recent discovery by Lee Sallows that he found at the Futility Closet blog, a veritable cornucopia of interesting posts.
So first the details, and then my discovery (although I can't imagine I was the first, but then things like the neat little idea below just lay there for centuries waiting to be found, so who knows?)

Lee Sallows recently published (Dec 2014)in Mathematics Magazine his discovery of the following:

It starts with the standard high school construction of the medians of a triangle meeting in a point, a regular feature of most high school classes. If you then take the six small triangles in pairs as shown, and rotate them about the three midpoints of the original sides, you get three triangles.... three CONGRUENT triangles, but in general, none of which is similar to the original.

Now if the original is equilateral, then so will the three triplets, and if they are isosceles, then so will the triplets... but NOT with the same vertex angle. For example if you start with a 4,4,6 isosceles triangle (vertex angle = appx. 97^o), then the newly created triangles will have a vertex angle of about 37.5^o, and both base angles are much larger.
I harbored a conjecture when I began that each new generation of triangles would work themselves toward an equilateral triangle as a limiting case, but this seemed to make that unlikely, one angle had jumped from being obtuse to well smaller than a 60 degree angle.
Confused, I decided to pursue what would happen if we took one of these smaller congruent triangles and applied this process again.
I will give some of the process here in the hopes that it will make it easier to follow.

It is easy to show they are congruent, although it fooled me for a minute because they are all in different orientations.   If you are familiar with the properties of the medians, you are well aware that the medians intersect each other at a point which is 2/3 of the way from the vertex to the midpoint of the opposite side. Studying the diagram, you can see that each side of the new triangle will be 2/3 of one of the medians (two copies of the 1/3 of a median next to the midpoint, and two sides of 2/3 of a median from the vertices). For convenience we will call the median to the midpoint of a, m_a and similarly the other two will be m_b and m_c.

There is actually a somewhat easily remembered formula for the lengths of each median based on the lengths of the three sides. \( m_a = 1/2 (\sqrt{2b^2 + 2c^2-a^2}) \) and similarly for each of the others. Using 6 as side a we arrive at \( m_a = \sqrt{7} \). The other two medians are both \( \sqrt{22} \) . This would make the three triplets, have sides of  \( ( \frac{2 \sqrt{7}}{3} , \frac{2 \sqrt{22}}{3}, and \frac{2 \sqrt{22}}{3} ) \).

Now if we use these as sides a', b', and c' we could run them through the same process, and get the medians of \( \Delta A'B'C'\) , which I proceeded to do. And the base of the isosceles triangle turns out to be ..... 3... yep, .....but not just plain old 3, but 3 like 1/2 of six. And when I did the isosceles sides they turn out to be 2. These are the medians of the first generation of triplets, so 2/3 of these medians should be the sides of our second generation triplets. Sides of 2, 4/3 and 4/3.... all exactly one third of our original triangle. The second generation of triplets were not only congruent to each other, but they were also similar to the original triangle with sides 1/3 as long.

Now there was only one question left to figure out... WHY?

After a break for some thinking (and a piece of chocolate cream pie) it hit me.

If you look at the figure above, (or John's illustrations below which I just received as I'm writing) focus for a moment on the two blue triangles and call the side of the triangle below them side a. Now when the hinge is opened, the two halves of this edge come together to make the seam of this one of the triplet. But look hard, these two a/2 halves form a median of the triplet.
Now look again at the original triangle (so I don't have to draw so many figures) and think of the median from the hinge between the two blue triangles up to the opposite vertex as a/2. When that is broken into second generation triplets, the part of the median between the two blue triangles running up to the concurrency point, is 1/3 of a/2, or a/6 in length. Now when the hinge is opened out, two copies of this will form a side of the second generation triplet, twice as long as a/6, or a/3.
Now we could have done all this hand waving starting with side b or c, and arrived at b/3 or c/3.

As I was working on this, I exchanged a few tweets with John and he was working on some Geogebra sketches, one of which I think makes the evolution of each generation really clear.

 The original pink \( \Delta \) ABC with the blue triangle  \(\Delta \)BGG' , made up of BGF and the rotation of  \(\Delta \)AGF  to become \(\Delta \)AG'F  as the first generation triplet on side c.  This is then trisected and the next generation is \(\Delta \) G'I I' which is a 1/3 dilation of the original rotated 180^o

It had to be the chocolate pie (and John's help). I would never have figured it out otherwise.  My thanks to John for both the intro to the problem, and the neat Geogebra graphics.