## Tuesday, March 31, 2009

### Math Boo-yah

I just got done with a round of midterms. I'm always fascinated by the scribbles students write on their test. My favorite scribble among the recently corrected is a student that got a little over zealous when checking his answer. I'm always harping on students to check. I say it so many times I'm even sick of hearing it. Well, I finally reached someone. Boo-yah!

## Monday, March 30, 2009

### Probability Problem

Lee found this nice problem in the Minitab manual that accompanies "Elementary Statistics" by Mario F. Triola. The problem posed is:

Two points along a straight stick are randomly selected. The stick is then broken at these two points. Find the probability that the three pieces can be arranged to form a triangle.

pic by flowers & machinery

## Friday, March 27, 2009

## Thursday, March 26, 2009

### 100 Greatest Theorems

Jay sent me an unbelievable link. Paul and Jack Abad presented this list of 100 greatest theorems at a math conference in 1999. Their criteria for the greatness of a theorem was

1) the place the theorem holds in the literature

2) the quality of the proof

3) the unexpectedness of the result

I had fun looking for my favorites: denumerability of the rationals, the non-denumerability of the continuum, and the incompleteness theorems. Jeff and Lee will be happy to see that the infinitude of the primes made it (#11). The list has dates for all the theorems, and links to proofs for some of them. Any guess as to what's number 1?

pic from Boston Bill

## Wednesday, March 25, 2009

### Using Selectricity

I'm entering a unit on elections in my Math for Liberal Arts class. There are a lot of slick tools out there to run online polls like PollDaddy. I was having trouble finding any tool that allowed user's to fully order their preferences... until Selectricity. It's a really great tool, particularly if you are studying elections. Check out this poll, it allows you to see current results, and the current winner under a number of common election systems: Borda, Condorcet, etc. Anyone can create an election without registering. For more extensive security measures, register for a full account. All free!

pic by Enrico Fuente

Labels:
146,
Math Courseware,
mathcrawl

## Tuesday, March 24, 2009

### 19th Annual Tunxis Math Contest

Rob Clark is hosting his annual Math melee. You can problem solve for money. Here are the details.

*Tunxis Annual Math Contest: Friday, April 3, 2009 from 12:00 – 2:00 p.m. in Room 6-201. Cash Prizes Awarded! $250, $150, $100 for 1st, 2nd, and 3rd.*

OPEN TO ANY STUDENT CURRENTLY ENROLLED IN ANY COURSE AT TUNXIS. QUESTIONS ON THE EXAM ARE MOSTLY ALGEBRAIC. THERE IS NO TRIG OR CALCULUS ON THE EXAM. SIGN UP WITH R. CLARK IN OFFICE F-32. SAMPLE EXAMS AVAILABLE IN OFFICE F-32.

OPEN TO ANY STUDENT CURRENTLY ENROLLED IN ANY COURSE AT TUNXIS. QUESTIONS ON THE EXAM ARE MOSTLY ALGEBRAIC. THERE IS NO TRIG OR CALCULUS ON THE EXAM. SIGN UP WITH R. CLARK IN OFFICE F-32. SAMPLE EXAMS AVAILABLE IN OFFICE F-32.

## Monday, March 23, 2009

### My New Favorite Puzzle: Tentaizu

I found a new puzzle. New to me anyway. It's called Tentaizu. I found it in the in-flight magazine on the way home from vacation. Btw, Southwest's magazine has the best puzzle section that I've ever seen. Not a reason to buy a ticket, but a definite perk. Anyway, I was completely charmed by this game. I thought we would play one this week. Maybe if we get excited we can try to automate a solution technique. I also tried to find a website that generates these like those I've found for Sudoku and KenKen. No luck.

Here are the instructions from Southwest's magazine:

*In each of the grids on this page, stars are*

hiding in 10 of the 49 squares. Your task is

to determine the positions of the stars. The

numbers in the grid provide clues: A number

in a square indicates how many stars lie next

to the square—in other words, how many

adjacent squares (including diagonally adjacent

squares) contain stars. No square with

a number in it contains a star, but a star may

appear in a square with no adjacent numbers.

hiding in 10 of the 49 squares. Your task is

to determine the positions of the stars. The

numbers in the grid provide clues: A number

in a square indicates how many stars lie next

to the square—in other words, how many

adjacent squares (including diagonally adjacent

squares) contain stars. No square with

a number in it contains a star, but a star may

appear in a square with no adjacent numbers.

## Friday, March 13, 2009

### YofX on Spring Break

Tunxis is going on spring break this week. I'll be enjoying the vacation, but will be back posting on 3/23. Until then, stay geeky, and solve some good problems.

Already, bored? Check out this Geeks Gone Wild Vid

### Jean-Marc's Angle of Fire

Jean-Marc passed on this problem to me. Here's the story. You're on one side of a pit of fire. You see two walls protruding from it. You must figure out how to build a wall so that if it were continued past the pit of fire it would bisect the angle. I solved this with trig the other day during a meeting and told Jean-Marc. He said the fun solution is purely geometric. So I'm back at work on this problem with my weak geometry. Help!

## Thursday, March 12, 2009

### Understanding Instant Runoff

I've been crawling the web, gearing up to teach voting systems in my Math for Liberal Arts course. I want to pass on this interesting site.

The voting that we do in the US is a modified plurality vote. One problem with our system is that often the winning canidate has not garnered a majority. One popular alternative used frequently in other countries is run-off voting, or what we call the Hare method in class. I found a video (below) and a great site, Choiceranker to explain how run-off voting works. Choiceranker even gives you the chance to participate in opinion polls that use the run-off method. Check them out.

The voting that we do in the US is a modified plurality vote. One problem with our system is that often the winning canidate has not garnered a majority. One popular alternative used frequently in other countries is run-off voting, or what we call the Hare method in class. I found a video (below) and a great site, Choiceranker to explain how run-off voting works. Choiceranker even gives you the chance to participate in opinion polls that use the run-off method. Check them out.

## Wednesday, March 11, 2009

### Sign Up for Tunxis Pi Contest

Just wanted to let you know that the 3rd annual Tunxis Pi recitation competition is coming up Friday, April 3rd at 10:00 am. Sue has even designed M&M's for it. The sign up sheet is outside her door (office F-7). There are cash prizes for the top three finishers: $150, $100, $50. Come eat chocolate, make money, be geeky!

## Tuesday, March 10, 2009

### 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.

## Monday, March 9, 2009

### 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.

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.

## Friday, March 6, 2009

### 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?

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?

## Thursday, March 5, 2009

### 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)

## Tuesday, March 3, 2009

### 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?

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?

## Monday, March 2, 2009

### Graph of the Day

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