Howard Phillips, on his blog “Saving School Math” proposed a real life geometry problem that proved harder to solve than it first appeared. I love to banter about mathematics education with Howard. We agree on some things about math education and disagree on other things, most likely because his background is in higher ed and mine is in K-12 ed. Nevertheless, the back and forth is always interesting.
Howard hails from from England and was a senior lecturer in mathematics and more at the University of Huddersfield. He retired from teaching in 2004 and moved to Puerto Rico with his wife, who is the director of the Western Ballet Theater.
I suspect that this real life geometry problem was somehow related to his building a stage prop for the next ballet performance.
Here’s Howard’s problem:
While designing a system for connecting “educational” cubes together I figured that the holes in the faces had to be positioned very carefully. To achieve what I wanted the holes had to be positioned with length a equal to length b, and length c had to be twice the length a.
So what is length a, as a fraction of the side of the cube ?
Howard wrote back and challenged me: “And now without trigonometry…”
I took the challenge. After several messy hand-drawn diagrams, I created the shape on GeoGebra and messed around for a long time, getting nowhere. I concluded that the length of the side of the square consisted of 6 copies of length “a” plus 2 copies of a length shorter than “a”. I couldn’t figure out the value of the shorter length nor how it was related to “a” . I was wondering if somehow phi was involved, but this was late at night, so visions of sugarplums and phi were dancing in my head. It would not make sense for phi to be involved with something so static and “non-growthy”. Phi is often involved in nature, but rarely in stage props.
To get an idea of how far flung my messing around took me, check out the image below of the GeoGebra mess. This is where I noticed the 6 copies of length “a” plus 2 copies of a length shorter than “a”. If you look at the horizontal lines cutting the vertical sides of the square, you will see along the vertical sides the 6 equal lengths and the 2 smaller equal lengths. You will also notice the octagon that informed a later solution.
Howard wrote me back with a BIG HINT. He used the fact that an angle bisector in a triangle divides the opposite side into lengths proportional to the two sides. He wrote down the first few lines. I would like to say that I immediately jumped on his idea and came up with my own solution. However, if I said that I would be lying! I labored over hand-drawn diagrams. Then I decided that I would again use GeoGebra to create a a simpler diagram than the one you see above, but one that I could print out and mess around with.
Creating the clean GeoGebra diagram was helpful in allowing me to really wrap myself around the problem. It made it easier to follow Howard’s approach and finally get the same answer that I got with trig. Rather than let the side length of the square be x, I set the length of the side of the square to be 1 unit (whatever unit that may be). This allowed me to not worry about yet another variable.
Here’s my work using Howard’s approach of the angle bisector:
I had not forgotten about the octagon formed from the 8 holes, so decided to try out a solution using it. I got the same answer as I got using trig and using Howard’s angle bisector: If the side of the square is 1 unit, then “a” = 0.14644 units.
While solving it this way, I was reminded of my discovery the night before. I had noticed that the side of the square had a length that consisted of 6 copies of length “a” plus two shorter lengths, but I had no idea what the shorter lengths were or how they were related to a.
The relationship became clear when I broke up the top of the square into different lengths. It’s easy to see on the top edge that there are 4 copies of length “a” plus 2 copies of length “(sq rt 2)a”. Square root 2 is approximately 1.414. So the 2 copies could be expressed as having length “1.414 a” units, which could also be written as “(1 + .414) a”, which is the same a “1a + .414a”. We have two of these lengths of “1a + .414a”, so now we have the other 2 copies of “a” to make “6a” and we have the 2 copies of the shorter length, which we now know each have a length of .414 units.
This is a great example of showing how dead-ends like I thought I had during my late night playing around with GeoGebra actually informed my final geometric solution. Not only that, but my finally geometric solution verified my observation that the side consisted of 6 copies of length “a” plus two shorter lengths. I figured out the shorter lengths!
Now I am inspired to pull off the cobwebs of a post I started writing in 2013 about a “multiplication machine” at MoMath in NYC. Thank you, Howard, for helping me get my blog post mojo back!