Earlier today Patrick Honner posted an interesting piece about a famous triangle:

His post sparked a perhaps odd connection in my mind with a blog post about solving inequalities that Bob Lochel had posted earlier in the week:

The connection had to do with how you describe a triangle using inequalities. I thought that creating 3D prints of the triangles in Patrick Honner’s blog would be an interesting exercise in inequalities, so I gave it a shot.

First, though, I was pretty surprised by the 2nd picture in the triangle blog, so I used a ruler and compass to draw the triangles (yes, a ruler – I measured out lengths of 13 cm, 14 cm, and 15 cm. Sorry construction purists!!)

Amazing how close the 13-13-13 equilateral triangle is to fitting perfectly into the 13-14-15 triangle – wow!

Next I went to Mathematica to see how tough it would be to describe a 3D triangle. This is only the third shape hat I’ve created on Mathematica to 3D print, so I am by no means an expert. I wouldn’t be surprised at all to learn that there is an easier way to do what I did. Actually, I’d be stunned if there wasn’t, but I wanted to use inequalities specifically for this project even if there was a better way.

Though my code is clumsy, it wasn’t too difficult to create the inequalities that would describe the 3D triangles:

The code above was for the equilateral triangle. It assumes that the three verticies of the triangle are at points (0,0), (13,0), and (6.5, 13 / 2) in the x-y plane. The 4 variables after the a,b,c, and d are just giving me the slope and y intercept of the 3rd side of the triangle (that I used “xint” for the y intercept says more about the quality of the code than any other words I could speak .. . . ).

The next line defines the triangle and gets to the part about inequalities. We’ll define this triangle by all of the points satisfying:

(i) the y coordinate is greater than 0. (y = 0 describes the line segment at the bottom of the triangle.)

(ii) the point is underneath the line y = (a/b) * x. (This formula describes the line segment on the left side of the triangle.)

(iiI) the point is under the line y = mx + xint. (This formula describes the line segment on the right hand side of the triangle.)

(iv) and finally, to make the triangle three dimensional, I’ll take the z coordinate be between -0.25 and 0.25.

That’s it for the simple equilateral triangle. The 13-14-15 triangle and the third triangle I made – the 13-14-2 triangle – are a tiny bit harder since they require a little trigonometry to get the third coordinate (I assumed (0,0) and (13,0) were points in all three triangles). If you are working through this same exercise with students, I suggest also using (0,0) and (13,0) for two of the points in all three triangles. The reason is that you have to switch up one of the inequalities to describe the 13-14-2 triangle, and thinking through that switch of inequalities is a nice exercise for students.

So, a fun connection between two blog posts and a neat printing exercise which doubles as an interesting inequality exercise. Made for a fun Sunday afternoon recreational math adventure. Even got to show my 8 year old how to make an equilateral triangle using a ruler and compass!

As usual, always happy to see and play around with the math ideas that people share on Twitter.

Which reminds me, I’d previously written a post about Mr. Honner’s square, so I guess this post was just a matter of time!!

https://mikesmathpage.wordpress.com/2013/12/21/numberphiles-pebbling-the-chessboard-game-and-mr-honners-square/