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.

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.

Another novelty to let your students discover, take an equilateral triangle and circumscribe it in a rectangle. (there are several ways to do that, so they should do a couple each. Now find the area of the three new triangles created inside the rectangle but outside the equilateral triangle. Make an observation. One they may pick up after some thought, if we call the three area A, B, and C, then there will always be one that is the sum of the other two.

Two of my favorite equilateral triangle ideas are the arithmetic triangle and Sierpinski's gasket.

And of course the second can be arrived by coloring in all the odd numbers in one color and the evens in another.

I must not forget Thiebault's theorem(s): Given a square, construct equilateral triangles on two adjacent edges, either both inside or both outside the square. Then the triangle formed by joining the vertex of the square distant from both triangles and the vertices of the triangles distant from the square is equilateral.

*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.

Ok, How about a mathematical quadrilateral you never (probably) heard of, the equlic quadrilateral. A quadrilateral ABCD such that AD and BC are of equal length, and angles A and B have a sum of 120 degrees.

The two lower images show two relations of this shape with Equilateral triangles. If an equilateral PCD is erected outside the quadrilateral, then PAB is also equilateral. The bottom shows the midpoints of diagonals P, and Q, and the midpoint of CD will always form an equilateral triangle at all. Toss "equalic quadrilateral" out at the next math dept meeting and rule the crew.

Two non-associated pieces not from the book, I'm writing this on November 7th and just printed up my post for On This Day in Math for November 8th, and it includes a note about the Death of Gino Fano on that date in 1952. Fano's relation to the equilateral Triangle, of course, is his famous illustration of a finite geometry in which every line had three points and every point lies on three lines.

One last hat tip to John D Cook who pointed out on Linkedin a connection between the Fermat Primes and the Sierpinski Gasket. If you've forgotten, the only Fermat numbers now known to be prime are 3, 5, 17, 257, and 65537. You may also know that these numbers are related to what regular polygons are constructible by straight edge and compass. Only numbers of the form are constructible by classic geometry methods, where F is a product of distinct Fermat Primes .

So that means there can only be values for F if we allow the empty set to be represented by 1, If you print out those values in binary, and he did, you get a perfect Gasket. John aligned his on the left margin so his gasket is a right triangle, but it's the same beast.

So go check out this book, especially if you are a student of geometry, a parent of a geometry student, or a teacher of geometry, or just like me, and you love looking at beautiful math. And check out John Cook's blogsite as well.

Oh, and you might drop back by this blog some time just to see that I'm staying busy and out of trouble.

One application of equilateral triangles is the use of the Warren Truss. The Warren Truss was designed in 1848 by James Warren and Willoughby Theobald Monzani. This truss consists of longitudinal members joined only by angled cross-members, forming alternately inverted equilateral triangle-shaped spaces along its length. One example of the use is in the new Bridge across the Tennessee River between Paducah and Ledbetter Kentucky on Hwy US 60.

How about some other ideas related to equilateral triangles and circles:

One blog I follow regularly is Antonio Gutierrez's gogeometry. If you teach/study/like plane geometry he should be one of your regular references.

Recently among his posts have been a couple with a related theme, circles inscribed or circumscribed about an equilateral triangle. I'm listing these because they are each a wonderful relationship, and together give these otherwise somewhat mundane seeming triangles a luster students?teachers/others might miss.

I will post the problems, but not the proofs, which (if you can't/won't work them out yourself you can find at the links provided to Antonio's site.

So on we go...This one is credited to Dutch mathematician Frans van Schooten, who was the first person to use rectangular coordinates in his extended translation of Descartes work.

1) draw a circle and inscribe an equilateral triangle. Now pick any point on the circumference and construct segments from this point to the three vertices. The sum of the lengths of the two shorter segments will equal the third. The problem, and solution is here.

2) OK:

Same triangle, same circle, but now we sum the square of the three distances ...????? and they sum to twice the square of a side of the equilateral triangle. That proof is here.

3) And now one with the circle on the inside. Again, from any point on the circle construct segments to the three vertices of the equilateral triangle. Again the sum of the squares is related to a side length, but I'll let you chase that down for yourself. Or you can go to the site here.

Addendum: John Golden sent a comment with a link to a GeoGebra sketch showing all three.

With that same diagram, we should point out an interesting and almost trivial relationship that I think is often overlooked. If you draw a median from either side to the opposite vertex, the encircle will trident that median into three equal parts. The proof for any student in elementary geometry involves only two relationships. The first is that the incenter divides each of the medians in a ratio of 2:1. Now use the fact that the distance from the incenter to the encircle is equal on both sides.

And let's add one beautiful proof about infinite series that is approachable to clever students at the geometry level. Analysis by elementary geometry. Try it with your students.

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?

Students will see that there is obviously a single triangle when there is only one matchstick per side. At 2 per side, they get four small triangles and one larger, or four triangles in all. Above I've given away the number of triangles for lengths of 3 on a side, and 9 on a side. Can students determine the relationship for the number of triangles with n toothpicks on a side?

Dissection problems date back to Plato, but one of the newer was the Habadasher's problwm, created by Henry E Dudeney in 1907. Simply take an equilateral triangle and by drawing three straight cuts, produce four pieces that can be reassembled as a square. It was solved by Dudeney, as well. His solutiuo can be made into a tabl that folda and unfolds between the two shapes.

**Solution Below**

**Don't Peek**

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

The Construction of the three lines is here

And in case you wondered, you can dissect a hexagon with four lines to get a square also. Discovered by Harry Lindgren (1961).

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