Catching up on some reading lately I came across an interesting book on Academia called Mysteries of the Equilateral Triangle. The only mystery is why he chose to use that word in the title, but it does include a lot of obscure historical and mathematical tidbits about this basic geometric standard. I will share a few I like and add a couple of things I've come across in other places to share as well.

The first is in the historical section where he includes an ancient geometric motif from a temple in Chennai, India. It shows three interlocking triangles, but notice they are interlocked in a Brunnian link, like the famous Borromean Rings. While the three can not be separated, no two are individually linked.

Along the way he enumerates lots of interesting things about history and property of equilateral triangles that you may well teach, Viviani's theorem, and one of my favorites, Morley's theorem

Then he mentions two common recreational challenges. The harder, in my mind, Find the side length and area of the largest square that can be inscribed in an equilateral triangle with unit side lengths. The easier, perhaps, find the side length and area of an the largest equilateral triangle that can be inscribed in a unit square.

The two are interestingly tied together.

The Figure above showed up on the wonderful Futility Closet blog a while back. It shows that two simple maximization problems are curiously related. Problem one, shown in the equilateral triangle at top is the solution to what is the largest square that can be inscribed in an equilateral triangle with unit side lengths. Problem two, shown in the bottom square is the solution to what is the largest equilateral triangle that can be inscribed in a square with unit side lengths. The coincidence? The length of the side of the square in the triangle is units; and the area of the equilateral triangle in the bottom square, is square units. Greg Ross, the mastermind behind the Futility Closet blog credits John Conway with discovering the proof of this relationship.

Digressing a moment from equilateral triangles, another couple of relationships I came across that, somehow, lead me back to the same topic in a roundabout way.

I'm strolling through my twitter feed, and Colin Beveridge AKA @icecolbeveridge posted one of his always entertaining "Math Ninja" blogs and it was about Ailles Rectangle. He just showed that it was a great memory device to figure out the trigonometric values of 15^{o} and 75^{o}. It is easy to construct and a nice way to verify directly the sum and addition formulas.

I had seen the Ailles (pronounced like the beverage, by coincidence) rectangle several years before, (it's been around even before I was a teacher, but seems not very well known) and had to do a little research to figure out why Colin's looked different.

I found an article by Jack S. Calcut at Oberlin College. He gave the original Ailles rectangle from the 1971 article, and sure enough it was different. (*Chris Maslanka pointed out that the segment in the upper left should be*

Where Colin had a 30-60-90 triangle inscribed in the rectangle, Ailles had used an isosceles right triangle. Both however, contain the three essential triangles necessary to demonstrate, what I believe is it's great power as a classroom demonstration, they contain ALL of the right triangles that exist with rational angles and each side length containing at most one square root. There is a 30-60-90, a 45-45-90, and a 15-75-90 triangle. That's it, that's all of them, there are no others. And that seems to be impressive as heck to high school students. "Here they are, memorize this image and you have the whole set!" And all those Pythagorean triples you know how to create.... None of them have rational angles in degree measure or as multiples of .

Since the theme of the day is coincidences, I noticed that the diagrams of Ailles contained another somewhat well known historical result.; If you look remove the 15-75-90 triangle at the top, you are left with a trapezoid that is used in the proof of the Pythagorean Theorem by President James Garfield of the US . Garfield was a professor of mathematics at Hiram College in Ohio for several years before being elected to the Ohio Senate in 1859. He was in congress, not president, when he did the proof which was published in the New England Journal of Education in 1876.

As I was studying Conway's curiosity, I realized that there was one more coincidence between the diagram and Ailles rectangle. Using Colin's illustration, if you reflect the 30-60-90 triangle about its longest leg and extend the other lines it will look like this.

Constructing the Equilateral Triangle at the bottom side of the figure and we have Conway's figure rotated 180 degrees. So Garfield's Pythagorean proof, with an additional triangle becomes the Ailles Rectangle showing all the rational right triangles with sides with a single quadratic root; and reflecting part of that gives us the square that demonstrates a curious equality between two classic maximization problems. I imagine you could assemble the whole thing out of tiles available for the elementary school.

Another little know truth the book reveals is that if you find the Euler Line in a triangle with a 60 degree angle, the Euler line will cut off an equilateral triangle.

Ok, how about creating a parabolic arc with equilateral triangles? Simply done by erecting equilateral triangles along a straight line with progressive side lengths of the odd numbers.

*Futility Closet |

He also had one for parallelograms and squares: Given any parallelogram, construct on its sides four squares external to the parallelogram. The quadrilateralformed by joining the centers of those four squares is a square.

I have stated previously how much I like "napkin" techniques that give quick calculations or estimates of a problem. I also like things like visual displays which essentially prove some mathematical idea. The one at the top of the page is from the cover of Roger Nelsen's "Proofs without words II.." which is way too expensive for a paperback, but I will probably break down and buy it.

The problem, is that, except for mathematicians who already know the proof, it seldom convinces "without words". Most of my high school students will not look at this image and be able to explain easily and clearly why it shows that the limit as n goes to infinity of 1/4 + (1/4)^{2} + (1/4)^{3}+ ... +(1/4)^{n} is equal to 1/3. If I'm wrong, not generally, but in your particular case, then stay with me and read what's written, and tell me if you see it the way I do. The amazing thing in my experience, is that a question about analysis can be illustrated with simple plane geometry.

I think they will be able to see that the triangle is divided into thirds... by the colors, 1/3 purple, 1/3 orange, 1/3 white. But I don't know if they can see the series of powers of 1/4 going off to infinity. That's why they have high school teachers... and so here are some words to help make it more "visual"

Look at the largest white triangle... can you see it is 1/4 of the largest (outside) triangle?... Ok, now look at the line across the top of the biggest white triangle, it connects the midpoints of two legs of an equilateral triangle, so the triangle above this medial segment, the one with multiple smaller triangles in it, is exactly congruent to the Biggest white triangle and is 1/4 of the total area of the outside triangle also. This upper triangle is a scale model of the original outside triangle, with all the same colors in the same positions and the white triangle in it is 1/4 of the area of this upper replica. So the second largest white triangle is 1/4 of the area of the Largest white triangle.... its area is 1/16 or (1/4) ^{2}. Now the line above the second smallest white triangle is a medial segment of the upper triangle, and so the triangle above it, which is also a scale model of the original biggest triangle, is also 1/16 of the total area... and the third smallest white triangle, is 1/4 of that, so its area is 1/4 of 1/16 or (1/4)^{3}... OK, now you see it, and as you move out each white triangle is 1/4 of the previous one... and sure enough, all the white triangles add up to 1/3 of the total area.

And from a slightly different approach, making equilateral triangles from matchsticks. I posed this question on the appropriate day of the year on my "On This Day in Math " blog.

If you build an equilateral triangle with nine matchsticks on each side, then subdivide into additional equilateral triangles, there will be a total of 235 triangles of several different sizes. The image shows the subdivision of a equilateral triangle with three matchsticks on a side. Can you find the thirteen triangles in it?

**Solution Below**

**Don't Peek**

**At Least Try for awhile!!!!!**

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