Rolling Stones + Math

Christina found a math course at MIT called Street Fighting Math. That's what we need at Tunxis! Contact math. (Plenty of elbow/knee pads though.) Until we offer it, you can follow the link and get all of the MIT course's course materials: lecture notes, video, etc.

What Would a Spectator Friendly Math Competition Look Like?

John and Steve have given me the challenge of creating a math competition that is actually fun to watch. There are tons of competitions out there. Rob runs one every year at Tunxis. However, almost all of these involve students sitting at a desk solving problems. What would a spectator friendly math competition look like? I have been thinking in two directions on this.

1) Taking math problems and making them physical. Classic bucket problems could be fun to watch if you took away paper and pencil and gave competitors actual buckets. (Think Die Hard with a Vengeance minus the bomb.) Or, here's a pack of M&M's, add these two numbers in base 6. Compass and protractor constructions would be interesting, particularly because hardly anyone knows how to do them anymore.

2) Structure the competition so that the audience can evaluate or judge the participants. Math contests are usually only judged for correctness and speed by experts. John mentioned the example of the poetry slam. How could this happen for math? Here are some ideas to get started with. Contestants solve a problem at home before the contest, come to the contest, and explain their solutions in 3 minutes or less. Then the audience judges them on the clarity/creativity of their explanations. Or, contestants prepare a 3 minute lecture/demonstration about math and the audience judges. Another wilder idea is to have them perform a creative task that doesn't have a correct answer. I use an enrichment activity in Math for Liberal Arts that might work. I have students create a poem using the power set structure on a set of words. The audience would definitely be able to evaluate this.

I'm not sure exactly what John and Steve had in mind for the bigger picture. I think they plan to have contests like this in all disciplines. They definitely gave me an interesting thought problem though.

Solving The Rectangles Problem

I keep posting problems and you do all the solving. I figured I better walk the walk, so I worked on Lee's problem from last week's post. It was really a nice one. The idea was to give the dimensions of five rectangles such that
1) the sides have unique integer dimensions between 1 and 10
2) they can be arranged into a square.

The first thing I did was narrow the size of the possible squares. I did it originally with an area argument. The minimum area of 5 rectangles that meet criteria 1 would be 1*10+2*9+3*8+4*7+5*6 = 110. The max would be 10*9+8*7+6*5+4*3+2*1= 190. So, 110<11^2, 12^2, 13^2<190. That limits the squares I was considering to only three different dimensions. (Later I found another way to argue for these as the only three possibilities.)

I have to admit I had a little more information than you guys did on the blog. Lee is a puzzle maker and he brought in a wood version of one of the solutions to show me. I was playing around with it on my desk one day and found it surprisingly hard to arrange into a square. It took me around 5 minutes to arrange 5 rectangles into a square.



Seeing the correct configuration was the insight that broke the problem for me. The rectangles looked something like this when assembled.



Note that this isn't drawn to scale, but essentially you have one rectangle in the middle with the others spiraling around it. The dimensions of the middle depend on the dimensions you choose for your outer rectangles.


That's when I started playing with me three possible areas.

1) If the overall square is 11 by 11, then what are my options for the inner rectangle. 10 can't be one of the sides of the inner triangle because I have to add two numbers to it to get 11. 9 also can't be it because I'd have to add two one's to it to get 11, and I'm only allowed one one. 8 is the highest possible side of the inner rectangle, because the outer rectangles could have 2 and 1 as sides.

But then since the sides of the whole rectangle are 11, we can fill in...

Now only 3,4,5,6,7 are left, and the only pairs that add to 11 are (5,6) and (4,7). That means 3 has to be the other dimension of the center rectangle.

What two remaining numbers could you add to 3 to get 11? There aren't any. Conclusion: 8 can't be the largest dimension of the inner rectangle. Try 7. And so the process repeats through all possible cases. I really didn't do it that systemattically, but this makes for a better demonstration.

When I got through with the process I only found two solutions: An 11x11 rectangle like this

and a 13x13 like this

At this point, I freaked out. I thought my logic was impeccable but Lee promised 4 solutions. At some point I thought to go back and look at the solution he gave me and saw that the sides were identical to mine, but his solution had flipped two of my sides giving different dimensions to the surrounding four rectangles. I'll show you with the 13x13,

So that's it, Lee also has a solution page on his website check it out. What I want now is a proof that 5 rectangles with unique sides cannot be arranged into a square in any other way (than the way I described above). There has to be a simple topological proof of this. Help?

Fun Sent in by Steve

How many seconds are there in 6 weeks. Write the answer in 3 symbols only.

This is a another problem neat problem sent in by Steve L. Thanks!

(pic by in touch)

Happy Square Root Day 3/3/09

I bet you didn't know it was Square Root Day. Don't feel bad. I didn't either. Rob clued me in. Here is the Wikipedia entry explaining the holiday. There are only 9 in each century. The next Square Root Day is not until the year 2016, so party really hard. That might begin by learning how to take square roots by hand, extract responsibly.

pic by _ahn

Question: Anyone seen these numbers before?

There is a long history of identifying subsets of the natural numbers particularly based on their divisibility properties. There are friendly numbers, extravagant numbers, deficient, perfect... The other day in class we were doing prime factorizations and I wound up with something like 2^2*3^3. I started thinking about all numbers n such that if n has a prime factor p, p^p divides n. Even more, I was thinking about adding the stipulation of sequential prime factors. So here are the first 4 numbers...
2^2=4
2^2*3^3=108
2^2*3^3*5^5=337,500
2^2*3^3*5^5*7^7=277,945,762,500
Anyone ever heard of numbers like this? What are they called? Are they used for anything?

Graph of the Day

We're snowed out today! (except for maybe evening classes). Enjoy the day off, solve a math problem.