Saturday, September 8, 2018

Geometric Construction of the Regular Pentagon


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.)
The question remains: Does this really construct a regular pentagon, or is it maybe just pretty close?






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 HI2 = CH*HM = CH(CH-CM) = CH(CH-HI).
  • Let CH = x*HI. Then the equation above gives us HI2 =x*HI(x*HI-HI), which gives us a (perhaps) familiar equation: 1 = x2-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 t2+d2 = (2r)2, giving (r/φ)2 +(φ*s)2 = 4r2.
  • This gives us s2 = 1/φ2*(4r2 - (r/φ)2) = (4/φ2 - 1/φ4)*r2.



Analyzing our construction
  • We will find the length of s = CG = CF.
  • OE=1/2*r. And EC = √(OE2+OC2) = √(5/4)*r = √5/2*r = EF. OF = (√5/2- 1/2)*r.
  • So s2 = CG2 = CF2 = OC2 + OF2 = (1+(√5/2- 1/2)2)*r2.
  • We have two very messy expressions for s2, 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-√5)/2*r2, making the side length of the regular pentagon inscribed in a circle of radius 1 equal the square root of (5-√5)/2.
  • 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.]

Sunday, July 1, 2018

Math Teachers at Play #118


In two more months, we'll hit MTAP #120, which should be the ten-year anniversary mark. Is it?



The number 118 ...
... factors to 59*2.
... is 1110110 in base two, 11101 in base three, 1312 in base four, 433 in base five, and 226 in base 7.
Not particularly exciting...



Aha! Can you make 118 using four 4's? (I did.) I wonder if you can make it using five 5s, or ...?








Summertime Learning
Summer Math Resources from Math Mammoth's Maria Miller. 

Denise Gaskins offers us one of her FAQs, on forgetting what they learned. My favorite of all her suggestions? Play games!

On Routines and Lessons
I like Geoff's perspective: "What’s the shortest amount of time you could possibly do the talking? Go with that. And maybe subtract a few more minutes." After 30 years of teaching, I am still learning, and I still get excited when I see something that might make a difference for my students. This looks helpful, and perfect for my summer meditations on teaching.



Young Math
Can you imagine doing a college math lesson with 2nd graders, and having it work out well? Manan Shah did it! His lesson was on set theory.

Ahh... A whole blog on magical math books. And she wrote about Christopher Danielson's new book, How Many? If you like this, she also has a list of every book she has posted about. Thank you, Kelly Darke!

Remember playing the card game War? If you're a parent, those memories may not be so fond. Kids love it and parents can get sooo bored. Kent wrote a great post about his interactions with his son around Addition, Subtraction, and Multiplication War. (If you want a good summary of even more mathy variations on war, check out Denise Gaskins' classic post.)

The first ever Global Math Week went well, with its exploding dots exploding around the world. Thank you, James Tanton, for getting people all over the globe into math.

How do you talk about numbers with young children? There are so many ways! Here's another: Counting with Dice, from Dave Martin.




Calculus
Which is bigger, eπ or πe? Sure, you can check it on a calculator. Or you can use areas to see why it's true. Lovely! (from Glenn Waddell)

The Fundamental Theorem of calculus says that you can figure out areas by using anti-derivatives. I do a project to help students understand it. Sam Shah does even more. This is Part III of his lessons for this topic. I recommend clicking on the links to Parts I and II first.





A Few More Goodies...
Dan McKinnon shares his notes from his Origami Workshop. For those who enjoy doing origami, can you find something new here?

Logic and Math go together so well. Check out this blog full of Venn Diagrams to fill in.

In his post From Surds to Ab-surds, Pat Bellew looks at an interesting relationship, which looks like bad simplifying but is still correct. He suggests it as a challenge in an algebra course to produce more correct 'bad simplifying' equations. Hmm, I want to think about how to do that.




Until Next Month...
Our sister blog carnival, The Carnival of Mathematics, often includes posts that are above my head. This month, our big sister is quite approachable. Enjoy!

If you have suggestions for next month's MTAP, share via the form at the Carnival home page.  Sharing in the carnival, or hosting, is a great way to increase connections in the #mtbos/#iteachmath community.

Sunday, January 7, 2018

Logic Puzzle - What Does Your Friend See?

I love logic puzzles, and was drawn by the title saying this was a hard one. Usually Nautilus is well-written, but their version of this puzzle isn't as good (in my opinion) as the original, blogged about by Presh Talwalkar.
 
The Nautilus version of the puzzle says to imagine your brightest friend. I imagined a friend I know likes logic puzzles (Sharon), but since I wanted a different initial than mine for notation purposes, I imagined another smart friend who might like logic puzzles (Linda). And I began scribbling away with S's (for Sue) and L's.
The answer I got is different than the answer the author got because we made different assumptions. Mine were based on his wording, his were based on Presh's wording.

Puzzle #1 (with Sue's interpretation):
You’ve been caught snooping around a spooky graveyard with your best friend. The caretaker, a bored old man fond of riddles (and not so fond of trespassers), imprisons each of you in a different room inside the storage shed, and, taking your phones, says, “Only your mind can set you free.”

To you, he gestures toward a barred window. Through it, you can see 12 statues. Out of your friend’s window, which overlooks the opposite side of the graveyard, she can see eight. Neither of you know the other’s count.The caretaker tells you each, individually, that together you can see either 18 or 20 statues. Unfortunately, there’s no way to tell your friend how many you can spot.

The only way for you both to escape is for one of you to give the total number of visible statues. Get it wrong, and neither of you ever leave. The caretaker asks you each once a day [Sue assumed neither person knew who was asked first], and you can choose to answer or to pass. If you both pass on a given day, the question—are there 18 or 20?—is posed to each of you again the next day, and the next, and so on, until you get it right or wrong.
Puzzle #2 is at Presh's blog. (No way to copy-paste that one.) It's now Alice and Bob, and they know that Alice gets asked first, so they'd both be released before Bob is asked if she has it right.
I think I have found a solution to (my version of) Puzzle #1. I would love to hear other folks reasoning before posting anything.
 
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