# What will it look like?

Tried out what I hoped would be a fun Zometool project this morning – what will it look like?

I first asked the kids to make some zome segments using a long blue strut in the middle and a medium blue strut on each end. Then I told them that we’d be making a cube and a dodecahedron out of these pieces.

After we made those shapes, we’d remove the medium pieces (and, as corrected by my older son, connecting up the shape after we removed those pieces!) – what do you think this new shapes will look like?

Here are their initial thoughts:

First up was the cube – here is the cube and their revised thoughts on what the new shape will look like after we removed the medium blue struts:

Here’s what our truncated cube looks like once the corners are removed and re-connected. It was interesting to hear them describe this shape – it was harder for them to describe that I was expecting.

Next up was the dodecahedron – here’s the shape with the medium-long-medium struts:

Finally, here’s the truncated dodecahedron. The surprise here is that the shape has regular faces – so it was an Archimedian solid. There is a truncated cube that is also an Archimedian solid, just not the one we made with our struts.

So, a fun little project working on a little bit of 3d visualization. I really wish I had something similar to the Zometool set when I was a kid!

# A neat tweet from Cilff Pickover

[sorry if this doesn’t read so well – this was my 6:00 am on the morning of daylight savings time exercise. We had to be out the door for something at 7:30, so I was a little rushed for time]

Last night I finally got around to checking out what was going on and ended up getting a fun surprise. Here’s how I approached the problem:

I’ll let $A,$ $B,$ and $C$ be the sides of the triangle and use the law of cosines to find their length. Hopefully those lengths will satisfy the Pythagorean theorem. I’ll use degrees for the angle measures for reasons that will become clear below and let the radius of the three circles be $R$. So . . .

$C^2 = 2R^2 - 2R^2 \cos(36)$,

$B^2 = 2R^2 - 2R^2 \cos(60)$,

$A^2 = 2R2 - 2R^2 \cos(72)$

Setting $A^2 + B^2 = C^2$, a little algebra tells us that we have to have:

$\cos(36) - \cos(72) = 1/2$

This was the pleasant surprise, because I’d seen this exact problem before back in high school. In was problem 30 on the 1975 American High School Mathematics Exam (the AHSME):

Here’s a link to that test on Art of Problem Solving’s website:

The 1975 AHSME on Art of Problem Solving’s website

I remember running across this problem when I was reviewing old contests while studying to take the AHSME. This problem struck me because it taught me something really neat, and something that I thought then (and still do!) would be a really fun thing to show kids in trig class:

$\cos(36) = \frac{\sqrt{5} + 1}{4}$ and $\cos(72) = \frac{\sqrt{5} - 1}{4}$

The proof that I know is a really neat combination of algebra and geometry, and also a really neat illustration of mathematical thinking. For the particular problem of finding the value of $\cos(36) - \cos(72)$ there is an extra bit of fun math, too 🙂

Here’s a sketch of the argument.

Let $x = \cos(36).$ The double angle formula tells us that $\cos(72) = 2x^2 - 1,$ so the difference we are looking for is:

$-2x^2 + x + 1$.

We really don’t know anything about $\cos(36)$ other than that $\cos(5*36) = -1,$ so let’s see what the trig formulas tells us about $cos(5\theta)$:

$\cos(5\theta) = \cos^5(\theta) - 10\cos^3(\theta)\sin^2(\theta) + 5\cos(\theta)\sin^4(\theta).$

A little algebra plus the identity $\sin^2(\theta) = 1 - \cos^2(\theta)$ gives us:

$\cos(5\theta) = 16\cos^5(\theta) - 20\cos^3(\theta) + 5\cos(\theta)$

Substituting $x = \cos(36)$ and $\cos(180) = -1$ we find that:

$16x^5 - 20x^3 + 5x + 1 = 0$

Now, I don’t want to spoil the geometric surprise, so I’ll just say that if you are interested in this problem marking the 5 angles on the unit circle that satisfy $\cos(5\theta) = -1$ is a good exercise. What do you notice?

That geometric exercise leads you to a nice algebraic observation – the above equation has one easy root and two pairs of double roots. This observation means that the above equation is surprisingly easy to factor:

$16x^5 - 20x^3 + 5x + 1 = (x + 1)(4x^2 - 2x - 1)^2$

but that means that $4x^2 - 2x = 1$ which we can re-write as $-2x^2 + x = -1/2$. BUT, we were looking to find the value of:

$-2x^2 + x + 1$, which we now know must be 1/2 !!! 🙂

That means the identity we found by applying the law of cosines to the triangle in Cliff Pickover’s tweet is indeed a right triangle. Yay!! Fun little problem.

As an aside, a couple of years ago I walked my older son through this argument just to show how some basic ideas in math can lead to pretty cool stuff. I didn’t expect him to understand every step, rather, I just wanted him to see the fun result that magically showed up again yesterday!