I never particularly enjoyed geometric constructions when I was younger. It may be because I had a tendency to press down too hard, and mess up the width my compass was set to. Like so many things, technology has a fix for that.

You can do geometric constructions on a number of sites, and in geogebra. My favorite site, which I discovered about 5 years ago, is sciencevsmagic.net. I having been telling students about it for years, and decided this summer to play with it again, to see how much I still like it. (A lot, it turns out.)

There are 40 challenges, in sets of 4. This time around I got every shape but the pentagon. I asked for hints (from math friends on facebook) which I then avoided reading because I wanted to do it on my own. I got nowhere that way, and finally followed instructions for how to construct it. Every other shape I have constructed makes sense on its own, but the construction of a regular pentagon must be proved with algebra. Ok, I don't feel so bad about 'cheating'.

I kept constructing it over and over, as I tried to understand it better. The first time (which you see above) I used these instructions (which I found by following links from wikipedia) and constructed it at sciencevsmagic, achieving my final shape challenge. Then I did it again using geogebra. I needed to figure out how it worked, and was at a loss. Cut-the-knot, a site I've gone to many times with my math questions, had a different construction - which I followed on both sciencevsmagic and geogebra - and an explanation (by Scott Brodie) that I kept working through as I wrote this.

[Sadly, Alexander Bogomolny, the creator of the cut-the-knot site, has died. I hope the math blogging community can find a way to maintain his amazing site. We miss you, Alexander.]

I will explain this as I understand it. My explanation will mostly just restate what I learned from these sites. But it may help some people, since some of the reasoning steps in Scott's explanation were hard for me to follow.

The constructions are worth doing first, so you can get a feel for what's happening as you do it. But when you're done, the question is still open - is this

*really*a regular pentagon? The following explanation proceeds in two major steps. First it looks at relationships in the regular pentagon and the pentagram (which is a regular pentagon with a five-sided star inside it). Then it looks at the relationships built by the construction method, and we finally see that the side length given in the two ways is the same. Bingo!

**The Construction**

If you're ready to follow this, I trust that you already know how to construct the perpendicular bisector of a line segment. I will leave out the construction details for each of those.

- Starting with a circle with center O, construct a horizontal diameter, AB.
- Now construct its perpendicular bisector, CD.
- Construct the perpendicular bisector of AO, intersecting it at E.
- Construct the circle with center at E, through C, labeling its intersection with AB as F.
- Construct the circle with center at C, through F, labeling its intersection with the original circle (above B) as G.
- Go around the original circle, constructing this same size circle from each new point. (Center at G, through C, gives H at the other intersection with the original circle. Etc.)

**Relationships in a regular pentagram**

- The angles in a regular pentagon are each 108°, so each external angle, like ∠HMN, is 72°. That makes each angle at a point, like ∠NHM, 36°. Since ∠HMG is 108°, ∠MHG and ∠MGH must each be 36°.
- Since angles ∠GLH and ∠LGH are each 72°, ΔHGL has two equal sides, HL and HG (which also equal CM, HI, etc).
- ΔCHI is similar to ΔIHM. So corresponding sides are in proportion. We get CH/HI=HI/HM, giving us HI
^{2}= CH*HM = CH(CH-CM) = CH(CH-HI). - Let CH = x*HI. Then the equation above gives us HI
^{2}=x*HI(x*HI-HI), which gives us a (perhaps) familiar equation: 1 = x^{2}-x, whose solution (in this case, where x is clearly greater than 1) is x = (1+`√5`

)/2 aka φ. (At this point, I get excited. φ, aka the golden ratio, shows up in such different contexts!) - So CH = φ*HI, and (similar triangles) HI = φ*HM and HM = φ*MN.
- We now have relationships between all the sides of the pentagon, star, and line segments on the star, but these are not yet connected to the radius of the circle.
- Now consider ΔODH. Two sides are radii, and the angle at O is 36°. Another triangle similar to all the others we've found. So OD = φ*DH.
- Also note that ∠DHC (inscribed angle on the diameter) is a right angle.
- Let's switch to simple variables names now. Let r=OD, s=CG, t=DH, and d=CI=CH.
- Then d= φ*s, r= φ*t, and t
^{2}+d^{2}= (2r)^{2}, giving (r/φ)^{2}+(φ*s)^{2}= 4r^{2}. - This gives us s
^{2}= 1/φ^{2}*(4r^{2}- (r/φ)^{2}) = (4/φ^{2}- 1/φ^{4})*r^{2}.

**Analyzing our construction**

- We will find the length of s = CG = CF.
- OE=1/2*r. And EC =
`√(OE`

^{2}+OC^{2}) =`√(5/4)*r =`

/2*r = EF. OF = (`√5`

/2- 1/2)*r.`√5`

- So s
^{2}= CG^{2}= CF^{2}= OC^{2}+ OF^{2}= (1+(

/2- 1/2)`√5`

^{2})*r^{2}. - We have two very messy expressions for s
^{2}, one from the regular pentagon/pentagram and one from our construction. Are they equal? It may help to write out positive and negative powers of φ to help simplify the first expression. Yes! They are both equal to (5-

)/2*r`√5`

^{2}, making the side length of the regular pentagon inscribed in a circle of radius 1 equal the square root of (5-

)/2.`√5`

- And that proves that our construction created a regular pentagon. (Whew!)

I've completed 32 of the 40 challenges at sciencevsmagic. I still haven't figured out how to make some of the shapes within the original circle, and still haven't found the least moves for some of the shapes. I'm so glad there's still something to work toward the next time I revisit this.

[Blogger doesn't deal well with superscripts and square roots. Once again, I am noticing that I ought to learn some LaTex. Sigh.]

Thanks for sharing the sciencevsmagic website. It's a perfect addition to my arsenal for Fun Math Fridays. And it's great fun to play with, regardless of your age!

ReplyDeleteMy friend, Barbara, decided to try this in autocad. She had trouble commenting. Here's what she says:

ReplyDeleteI figured out how to construct a pentagram in AutoCad using only lines and circles (equiv. of straight edge and compass). (I'm actually using Draftsight which is a free clone of 2d AutoCad.) I figured out that the key internal element is the 36-72-72 triangle (apparently called the Golden Triangle). Once I figured out how to construct a 36-72-72 triangle the pentagon was easy. But I didn't find the key move at all intuitive. I had to cheat and look stuff up. But I learned stuff I didn't know (not much call to construct pentagons in civil engineering). I made a Quicktime movie for you. It's big so I put it on Google Drive. Here's the link:

https://drive.google.com/file/d/1jZVa01-aS1ZGuqcfySz5iRkh3qLJOe_6/view?usp=sharing

Euclidea is also awesome.

ReplyDeleteYes! I played at it years ago, and thought I mentioned it in a post. But I don't find any previous mention of it on my blog. Thanks! I am playing now. The sciencevsmagic is smaller, and easier to conceive of finishing. I liked it better before. But this one is great too, and I just figured out the rhombus in a rectangle. Yay.

ReplyDeleteBy coincidence I was working on this too back in December: https://mymathclub.blogspot.com/2018/12/proof-of-pentagon-construction.html and used some ideas from cut the knot as well. +1 on mathjax/Latex for blogging (I use it in the above post)

ReplyDelete