## Archive for December, 2008

### 3, 2, 1, 1 Happy New Year!

December 31, 2008

Ahh, the perennial problem:  what to do with the pesky fact that the year isn’t exactly as long as we want.   The Romans initially handled this by having the season of “winter” be of varying length.  Of course, those early Romans weren’t exactly known for their accurate record keeping: the pontifex maximus (Calendar Guy) would add an extra month from time to time to keep things sort of on track, but since a calendar year was also the same as the term of office in politics, he would also add days to a year when his allies were in power and subtract them when his opponents were in charge.  In other words, it pays to be on good terms with the person in charge of the calendar.

So we probably shouldn’t turn to the ancient Romans for calendar advice, though to be fair Julius Caesar is the one who authorized giving the calendar a much-needed overhaul and began the regular adding of leap days by having a second February 24 every 4 years (seriously:  February 29 didn’t happen for more than a thousand years after that).

But an extra day every 4 years isn’t exactly right either:  you’re off by a day every hundred or so years.  So with the Gregorian calendar the leap day is skipped at the turn of a new century.  Well, it’s skipped 3 out of 4 centuries:  1700, 1800, 1900, 2100, etc. skip their leap days, although 2000 (and 1600 and 2400) still have them.  And that got the alignment mostly on track, but it’s not perfect.

The problem is that the imperfection is inconsistent.  The length of a year actually varies a bit:  According to The Guardian:

The snag is that the rotation of the Earth is not so reliable. It is gradually slowing down and factors such as disruptions in the Earth’s core, extreme weather, volcanic eruptions and earthquakes can all influence the precise length of the astronomical day. From time to time, the rotation-based clock — UT1 time — and UTC [Coordinated Universal Time] need to be brought back into line.

So occasionally, like tonight, an extra second is added [or in theory subtracted, although that’s never actually happened].  Really, it’s like a Roman Intercalary month second since it doesn’t happen on a regular or predictable fashion.  And for most people it means nothing except perhaps an extra kiss at midnight, but people who pay very close attention to time get some extra work in by making sure that everything is aligned perfectly world-wide.

And that causes its own problems, because what with new-fangled technology like computers, being off by a second requires all sorts of overtime and problems.  The extra second is actually added at midnight Greenwich Time: according to a bulletin from the U.S.  Navy here, the official sequence will be

31 DEC 2008 23 HOURS 59 MINUTES 59 SECONDS
31 DEC 2008 23 HOURS 59 MINUTES 60 SECONDS
01 JAN 2009 00 HOURS 00 MINUTES 00 SECONDS

but that’s actually 7pm here in New York and is mid-day in other parts of the world.  In other words, if you’re working 9-5 in California, you might want to ask for that extra second in overtime pay.

Note 1: Did you notice that UTC, not CUT, is the abbreviation for Coordinated Universal Time and assume that the abbreviation must be from another language? If so, you’d be wrong but for an amusing reason:   according to the National Institute of Standards and Technology, the abbreviation was specifically chosen not to stand for anything.  The English Coordinated Universal Time would be reduced to CUT and the French Temps Universel Coordonné would be abbreviated as TUC, so the International Telecommunication Union decided that everyone in the whole world should use the initials UTC as a compromise.

Note 2: The photo of the Times Square Ball is actually from last year, and is licensed under Creative Commons Attribution 2.0 by Clare Cridland.  But if you want to be current, Dave Richeson has a really cool post about the 2009 Times Square New Year’s Eve ball here on his blog Division by Zero.

### The Last Carnival…

December 29, 2008

Don’t worry, it’s just the last Carnival of the year, not forever!  The 46th Carnival of Mathematics is up at Walking Randomly, one of our favorite blog reads.

This was a short carnival since some people (insert embarrassed face here) got all distracted with sending out Christmas cards New Year cards Happy Winter cards, and forgot to submit anything.  But fortunately those submissions are good reads, and Mike supplemented the Carnival by adding 12 additional posts of his own choosing, one for each month of the year.

### Mathy Fails

December 26, 2008

The other day I noticed that the Fail Blog had several math-related fails on the site.  “I’ll post this!” I thought.  Then I noticed that QED had already done exactly that.  Oh, well, so much for originality!  [Then again, since I was already taking them from somewhere, I suppose originality was never a risk.]

But without further ado, here are a few pictures for laughs:

see more pwn and owned pictures

see more pwn and owned pictures

see more pwn and owned pictures

### Oh Christmas Tree!

December 25, 2008

If you have a Christmas Tree, you might be wondering where it came from.  Well, not the tree itself (though come to think of it, I’m not at all sure where ours came from other than a store on the highway, but it’s big and fabulous and hasn’t dropped too many needles.) but the custom of bringing a tree into the house and decorating it.

My research first led to the story that the Christmas tree dated back to a 7th century monk who used the triangular shape of the tree to represent the Holy Trinity.   This story is false, which should come as no surprise to anyone who has actually looked at an evergreen and realized that they are in fact three-dimensional and not flat triangles (and it’s hard to see how a cone could represent a trinity, although that does lead to interesting speculation about just what it would represent).

According to History.com, the Christmas tree actually has a much longer tradition, dating all the way back to the Roman Winter Solstice holiday of Saturnalia.  Saturnalia sounds a bit like baccanalia, and indeed both involved enjoying food and drink.  There were plenty of other feasts at this same time, including Juvenalia (honoring children) and the December 25 birthday of the infant god Mithra, which was viewed by some devout Romans as the most sacred day of the year.  During this season of celebrations, the Romans would bring in evergreen boughs to represent that the days were starting to get longer.

But the Romans weren’t the only ones who used evergreens in this way.  Egyptians and Druids also decorated with evergreens, and the Vikings thought this was the special plant of their Sun God Balder [no relation to the word balderdash].

But still, bringing in tree branches is a little different than bringing in the entire tree.  Although this site says that in the middle ages  Paradise Trees (evergreens decorated with apples) were used to celebrate/symbolize the Feast of Adam and Even on December 24, it’s the Germans who were primarily responsible some 500 years ago for the modern Christmas tree.  The Germans are also the ones who decorated the trees, and Martin Luther apparently is the first person to put candles on the tree.   (I haven’t seen too many candle-lit trees in the United States, but my German roommate in grad school said that her family still put little candles all over the tree and and lit them on Christmas Eve.)

Folk in the Colonies were much more skeptical of this whole tree thing and its pagan origins.    Many Puritan folk viewed decorated trees, Christmas carols, and the like as a distraction from the sacred aspect of the day, and a 1659 Massachusetts Law made it illegal to decorate or do anything to celebrate Christmas other than go to church.  Bah, humbug!  But the Irish and particularly the Germans were still all rah rah in favor of really celebrating Christmas, and in the mid 1700s the  German folk in Pennsylvania still had community trees.  Of course, many of those were actually wooden pyramids decorated with branches, making artificial trees more traditional to the US than actual trees.

However, the biggest influence on the acceptability of Christmas Trees in the United States was England’s Queen Victoria.  In 1846 she tacitly gave her approval to the tradition by being featured with her family in front of a big tree on the cover of the  Illustrated London News.

That made it The Thing to Do, and so the concerns about the Pagan Origins of the Christmas tree were put aside.   Ironically, it was about this time that the Germans started to get really concerned about conserving the fir forests, and encouraging people to use artificial trees.  Plastic wasn’t readily available, so the trees of those days used feathers, dyed for a more realistic look.  And from there, whether artificial or real, aluminum, wooden, or living, Christmas trees stuck with the mainstream Christmas celebrations of many countries.

### Puzzler from Car Talk

December 23, 2008

Any of you listen to Car Talk, the auto show by Tom and Ray?  It’s a great weekend listen (although thanks to the glory of the Internet, you can hear episodes at pretty much any time from pretty much any place with an internet connection).  During each show they have a Puzzler, the answer to which they reveal the following week.  Some of the puzzlers have a mathematical bent, like the one from Ray for the week of December 8:

This puzzler is from my mathematical series. Every two-digit number can be represented as AB, where B is the ones digit and A is the tens digit. Right? So for example the number 43, A is 4 and B is 3.

Imagine then that you took this two-digit number and you squared it, AB x AB, and when you did that the result was a three-digit number, CAB.

Here’s the question: What’s the value of C? So, for example if AB is 43, CAB might be 943. Of course this is a totally bogus answer, but you get the idea.

So again, what is the value of C, so that AB(squared)= CAB?

This can be done by brute force, but examples of more elegant solutions are certainly welcome!  (Indeed, since the answer has already been posted online, there’s no harm is sharing partial or complete solutions).

Thanks to Ted for bringing this puzzler to my attention!

### The 2008 Edublog Winners!

December 22, 2008

They were announced this weekend here.  There are plenty of great blogs on the list, including the winner of the Best Group Blog was SCC English:  a blog of the English Department from St. Columbia’s College in Dublin, Ireland.  One other favorite of mine is Extreme Biology:  a class blog where the contrubutors are primarily 9th grade students and AP biology students.  (And now I have to make a conscious choice to stop listing favorites or I might as well just reproduce the entire winers list!)  Congratulations to all!

### Star of David Theorem

December 21, 2008

Hanukkah starts today at sundown, so in honor of the holiday here is the Star of David Theorem.  In simplest terms, this theorem says that the greatest common divisor of  ${{n-1} \choose k}$, ${n \choose {k-1}}$, and ${{n+1} \choose {k+1}}$ is equal to the greatest common divisor of  ${{n-1} \choose {k-1}}$, ${n \choose {k+1}}$, and ${{n+1} \choose k}$.  To see why it’s called the Star of David, look at the following visual.  the greatest common divisor of the blue corners and the greatest common divisor of the purple corners are equal.  Together, the two triangles form the Star of David.

For example, if $n=4$ and $k=2$, this says that the greatest common divisor of  ${3 \choose 2}$, ${4 \choose 1}$, and ${5 \choose 3}$ is equal to the greatest common divisor of  ${3 \choose 1}$, ${4\choose 3}$, and ${5 \choose 2}$.  As it turns out, this is one of the less interesting examples because both sides simplify to gcd(3,4,10), which is 1.  So let’s look for another example.

Where there are binomial coefficients, Pascal’s triangle can’t be too far behind.  Sadly, when the above Theorem is placed visually into Pascal’s triangle, it ends up looking kind of turned and squooshed.

Visually, the top star illustrates the previous not-so-interesting example of how gcd(3,4,10) is equal to gcd(3,4,10).  But the lower star illustrates that that gcd(36,210,165) is equal to gcd(84,45,330), and this is a little less obvious.  In this second case, but greatest common divisors are equal to 3.

According to Wolfram’s Mathworld, the Star of David Theorem was first stated by H. W. Gould in 1972, and there were several generalization in the years immediately following.  Apprently the association with Pascal’s triangle wasn’t noticed until 6 years ago, however, by B. Butterworth in this article (which is originally about using Pascal’s triangle to illustrate the song “The Twelve Days of Christmas”).

December 20, 2008

So there are a lot of cool math things out on the web, that’s clear with a quick google.  But suppose you want to try something homemade?  One option is to make cookie ornaments.  But not with real cookies, because there’s pretty much a 100% chance that the cookies would start sporting bite marks, and then they’d disapper entirely leaving only little crumb covered ornament hooks, and that would just be sad.

No, it’s better to make ornaments out of something a little less tasty.  My favorite is Penzey’s Cinnamon Applesauce Holiday Ornaments:  instructions formally written out in a .pdf file here.  Essentially, you take a jar of applesauce, drain it overnight by putting it on a dishtowel or cheesecloth over a bowl (supposedly this is unnecessary, but enough water drains out that it seems to be quite a useful step).  Then you add a bunch of cinnamon, which is pretty cheap if you buy it at a dollar store.  Stir in enough for it to become a stiff dough (it’s edible, but about as tasty as pure cinnamon),  roll it out thin, cut out shapes, and poke out a hole for yarn or ribbon  Then you cook them for 175° [assuming your oven goes that low] for 6-8 hours.  That’s right, 6-8 hours:  you pretty much need to play on being home all day.  On the bright side, your house will smell GOOD!], decorate them, and add something to hang them from like yarn.

You can use your favorite cookie cutter (perhaps in the shape of pi).   In this case I used a butterfly to represent the butterfly effect from chaos theory [an idea gathered from the Halloween Costume suggestion here].

That symbol is gamma because I thought that γ deserved to step out of π’s shadow for the holidays.

So that’s Homemade Holiday Gift #1.  For 1.5, I was going to show you how to make Gem Magnets, but it turns out that with their glass tops, photographing them requires a lot more skill than I have.  Or maybe a better camera.  They look something like this:

See what I mean about the photography?  But they’re still pretty cool to make:  you get magnets, then cut out circles either from pictures or from plain paper that you can draw on (perhaps your favorite math symbol, which is the faint tie in to mathematics with this post).  Then you glue one of those clear stones on top, and lo and behold you have a homemade present!  Woo hoo!  (And if you like directions that are a little more precise, you could try here.  If words like “silicone” scare you, just use an all purpose adhesive that sticks to metal and glass).

Happy Holidays!

December 16, 2008

What to get the math folk on your holiday list? There are already a few suggestions on holiday lists (e.g. Walking Randomly has some great books listed), so we’ll just add to the fray.

Originally this was going to combine store-bought and homemade presents, but what with the hecticness (that’s a word, right?) of this time of year, the homemade math things are only partially made. Godzilla has grand plans to show a couple in the next day or so, but for today it’s a sample of What’s On The Internet.

You’re probably first wondering about ornaments. These could hang in your window, on a tree, or even be attached to some other lovely gift. There’s a 72 pages of math ornaments here from The Library of Math (e.g. a round ornament with HO3 on it). Cost: $7.49-$9.99. They appear to have roughly a million other objects (buttons, shirts, etc.) with the same kinds of mathy statements, and that’s a lot of browsing.

But if it doesn’t appeal? Perhaps you’d like to do some baking. In that case, of course, you’ll need some cookie cutters shaped like numbers ($23.95 for the whole set of digits) or like the symbol π (at$12.95).

Or a pie dish with the digits of pi. Cost: $24.95. Or maybe you want an original shirt. Then this isn’t the one for you, because TwoPi already created the design from a public domain photo of Iwan Stranski and put it on CafePress so he could order himself a copy. But you can be one of only a few people who has a shirt or mug with the following image on it. Cost:$17.99 — $21.99, or$11.99 for the mug.

(Note: We find the whole money thing confusing, what with this being a group blog and all, so to make it simple we don’t get any profit from this. Which, sadly, doesn’t make the shirts actually free.)

(Another note: Since TwoPi originally uploaded this so he could order one for himself, he put our address https://threesixty360.wordpress.com on the back of some of the shirts.)

Coming up next: things a lot closer to free!

### How tall was the murderer?

December 13, 2008

In a recent episode (“The Bone that Blew”) of the TV series Bones — and yes, I am thinking of creating an entire category on this blog just for brief mentions of math in Bones — a body was discovered and it turned out to have been dragged by a choke chain. Temperance Brennan points out that the angle of the fracture was 18°, indicating that the leash was being pulled at that angle. The blond assistant guy then added “Assuming a standard 4 foot leash, the person who dragged the victim is at most 5’5″.” They showed a picture like the following, only it had a skull and broken neck at the bottom. I spared you that.

How did they come up with the 5’5″ height? Since the leash was 4 feet long and being held at 18°, the vertical height from the body to the hand that was dragging it must have been 4·sin(18°)=1.24 feet. The leash probably didn’t touch the ground, though: it would have been attached to the neck, and therefore something like 6″ above the ground. Let’s assume the vertical height is therefore 1.74 feet above the ground.

Presumably Blond Guy knows some formula that relates how tall one’s hand is above the ground with their total height, although my extensive searching (i.e. Google) didn’t reveal any such formula. Never one to be deterred by a lack of actual facts, I forged ahead and found that my height is 2.9 times as big as the distance from the floor to my hand. Assuming that ratio holds constant for everyone [which is certainly doesn’t], this would mean that the person dragging the body was (1.74·2.9=5.05) just over 5 feet tall.

So I still have no idea where the 5’5″ came from, assuming it wasn’t just made up out of thin air (which it wasn’t, right? Because that would make me really sad.) Maybe the person was shorter. Maybe my adding 6″ was an underestimate and it should have been closer to 9″ (which would mean I’d add 0.75 to 1.24, getting 1.99, which yields 5’9″ when multiplied by 2.9). Maybe the 2.9 really varies quite a bit person by person. Or maybe the person was tall, but leaning over, and Blond Guy took that into account.

So alas, I really don’t have an answer. But fortunately for the folk at the Jeffersonian, they found additional evidence (identifying the trajectory of the bullets that killed the man, and used that to get a different height that led them to the killer). If you’re dying to know who did it — so to speak — you can see the episode at least for the time being here on Fox. It’s Season 4, Episode 10, and will probably only be up for a few more weeks.

### Recovering from the recession

December 11, 2008

I have a Modest Proposal for how to get the US (and World) economy out of the current recession.

It came to me last Friday, on our last day of classes, as I was walking across campus to Calculus II.  I wanted to talk about cool applications of the course content, as a way of summing up (so to speak) the semester: a course on techniques of integration, applications of integration, and infinite series.  I decided to go with something simple:  applications of geometric series.  And one of my favorites is computing the total amount of economic activity that ensues from a single injection of economic stimulus from the government.

Here’s the (fairly standard) story:  Suppose that each individual saves 10% of each dollar they receive, and they spend (recirculate) the other 90%.  Then for each $1000 (say) of government stimulus,$100 gets saved, and $900 is spent, becoming additional income for other individuals in the economy. But now consider that$900 on the rebound.  10% of it ($90) goes into savings, and the other 90% ($810) gets spent again, this time by its second owners.  And now of the $810, 10% is saved, 90% spent. And so it goes… This story leads to the following calculation. A stimulus of$1000 will lead to a total amount of economic activity equal to

$1000 + 1000(.9) + 1000(.9)^2 + 1000(.9)^3 + \cdots$

an infinite geometric sum, where each summand is 90% of the previous term.  Now a geometric series $a + ar + ar^2 + ar^3 + \cdots$ with common ratio $r$ converges to a finite sum provide $|r|<1$, and in that case the limiting value is $\frac{a}{1-r}$.   In the case of the $1000 above, the total amount of economic activity is $\frac{1000}{1 - .9}$, or$10,000.  [So the $700 billion stimulus package, under these assumptions, could lead to$7 trillion in economic activity, or roughly half of the US Gross National Product.]

The brilliant thought I had whilst crossing campus:  what if we drop the savings rate from 10% to a smaller number?  If instead of recirculating 90% of our income, each of us went out and spent 95%?  or 99%, or even… more???  This model predicts that the total amount of economic activity from a given stimulus is the amount of that stimulus divided by 1 – (the proportion recirculated).

So, what would happen if no one put any money into savings, and ALL of our income went directly into consumer spending?  In that case, we have a common ratio of $r=1$ in the geometric series, which now diverges, and the model predicts that any amount of government stimulus leads to an infinite amount of economic activity!

Woo hoo!  Let’s send out 5 cents in economic stimulus, and watch the American electorate spend us out of recession!!!

[inhale, exhale]

Ok, so that’s obviously wrong, which means the original discussion (standard fodder for all the calculus and precalculus texts I’ve seen in recent years) is also flawed.  And seeing what’s wrong is easier now in this extreme case.  Suppose the government gives Joe the Drummer a $10 stimulus check. Joe goes out and buys new drumsticks; the music store spends all of Joe’s$10 on rent; the landlord spends all of the $10 on utilities; the utility company etc…. What is missing in the series is the issue of time. Joe might take a day or two between getting his government rebate check and actually spending it. The music store owner won’t spend the$10 until the utilities are due on the 15th of the month; and so it goes.   $10 in stimulus in theory leads to an unending succession of financial activity, but it cannot do so in a finite period of time. What can happen in a finite period of time is a finite number of transactions. If we assume that over the course of the month (say), each dollar spent in stimulus changes hands a total of ten times, we get a truncated geometric series (again, with a$1000 initial stimulus, and 90% recirculation rate):

$1000 + 1000(.9) + 1000(.9)^2 + \cdots +1000(.9)^9 = \frac{1000( 1 - (.9)^{10})}{1-.9} \simeq \ 6513$

This is quite a bit less than the infinite sum ($10,000). However, to some extent the predicted multiplier effect is real, if not quite as dramatic as one gets with infinite series. Now if the savings rate drops toward 0, and the recirculation rate increases toward 100%, the simplification on the right-hand side of the equation no longer works, and we instead decide that after 10 transactions, a stimulus of$1000 with all of it being recirculated 10 times leads to a total of \$10,000 in total economic activity.

So much for my 5 cent solution.

[Although it *does* optimize the amount of economic activity in a fixed number of transactions.  Just sayin’.  That Vox AC-30 amp would look mighty good under the tree this year….  Spending more this season just might be patriotic!]

### The Day the Earth Stood Still

December 10, 2008

The Earth first stood still over 50 years ago, when from out in space came a Warning and an Ultimatum.   And on December 12, it’s halting again for a remake.

So what does this have to do with math? Two things.  The first (which is the one that initially led to this post) is that one of the people in this Friday’s film is 10-year old Jaden Smith.  And do you know what is favorite subject is?  That’s right, math.  Thus says People magazine, after an interview with him last night at the New York City premiere:

“It’s math,” Jaden told PEOPLE at the New York premiere of Earth on Tuesday. Why math? “Because I’m good at it,” he added.

And the second reason, which I only discovered while trying to find out if anyone else in the movie also had math as a favorite subject, is that there is an Actual Math Equation in this movie.  Apparently there’s a chalkboard scene with John Cleese (who plays Professor Barnhardt), which Director Scott Derrickson referred to in this interview on SciFi Weekly:

Derrickson: It’s a real math equation about a real significant high-physics theory about the universe, and we tried to be truthful to the scientific aspects. I just did an interview with Discover, and the interviewer was really surprised at little things in the movie, but that being the biggest one. We had an astrophysicist who worked with Keanu [Reeves] and John. I still remember watching them for quite a long time. I don’t remember where we were in the production, but for quite a long time in a room working out the back-and-forth of that, and then we added material to make it longer at one point … just to get that kind of flow and rhythm to it. I didn’t have much to do with that. I’m just remembering, because it was really Keanu and the math guy, the theoretical physicist, and John Cleese. The three of them just kind of figured that out, and I saw it and thought it was fantastic.

And fantastic it is.

### How Fast Could They Have Gone?

December 9, 2008

I love the show Top Gear. In particular, I am a big fan of the “challenges” they have. Recently (yesterday afternoon, when I should have been grading), there was an episode on BBC America from 2006 in which host Richard Hammond was at London’s ExCeL Centre for the British Motor Show, and before the exhibits were set up, he had The Stig (the show’s “tame racing driver”) test two vehicles to see how fast they could go in the 385-metre hall. The first was a Chevy Lacetti, their “Star in a Reasonably-Priced Car” car, which reached 70 mph. The second was a Toyota F1 car (the TF105, I think), which reached…

only 81 mph?

This led me to wonder what the top speed of a car could be on a 385-metre stretch. Let’s find out.

For simplicity, I will assume constant acceleration a (I said simplicity, not accuracy), and constant deceleration from braking (again, not particularly realistic). Let $v_T$ be the top speed. Then the distance covered in accelerating to top speed is

$\displaystyle\frac{1}{2}a \left( \frac{v_T}{a} \right)^2$

and the braking distance is

$\displaystyle \frac{v_T^2}{2\mu g},$

where μ is the coefficient of friction between the tires and the floor, and g is gravity. We then have the following equation:

$\displaystyle \frac{v_T^2}{2} \left( \frac{1}{a} + \frac{1}{\mu g} \right) = 385$

and thus

$v_T = \displaystyle \sqrt{\frac{770a\mu g}{\mu g+a}}.$

At this point, I plead ignorance. I tried (not very hard) to find a reasonable coefficient of friction for racing tires, and to find a 0-60 time for a Formula 1 car (the McLaren F1 does it in 3.9 seconds). In the end, I made a spreadsheet for μ between 0.4 and 0.7, and a between 6 and 7 m/s2. Given an ideal setup—starting at one end of the hall and stopping perfectly at the other end—the top speed is somewhere between 95.6 mph (μ=0.4, a=6) and 115.5 mph (μ=0.7, a=7).

What does this mean? I don’t know, but it makes The Stig’s 81 mph sound pretty good, given the initial burnout, the nonconstant acceleration and braking, driver reaction time, and an interest in personal safety (cf. Hammond’s “Formula 1 car-shaped hole” comment).

### McMorran’s Math

December 8, 2008

Last week I had two guest speakers in my Thinking Mathematically class.  They came to talk about how math is used in the fiber arts, particularly clothing design.  For example, shawls made in different places combine simple geometric shapes in different ways to come up with styles that serve different purposes, depending on the styles and needs of the people involved (only they said it without using the word “different” three times in a single sentence because they’re professionals). Or how to use proportions to design a sweater, or even how to use the relationship between the circumference of a circle and the radius to create a design for a circular shawl by doubling the number of stitches whenever the number of rows has been doubled.

At the end, they showed me a doohickey called a McMorran Yarn Balance. Both Mary Louise and Meg are spinners as well as knitters and dyers, and when you spin your own yarn, this McMorran Balance can tell you how many yards of yarn you’re getting per pound!

Godzilla is here to show you how it works.

First, you take the Yarn Balance out of the box and set it up. It’s a rectangular prism with grooves for the balance part. There’s a notch in the balance part itself, and that has to be face up because it’s where the yarn will go.

Then you take a piece of your yarn and drape it in the slot.  Get out your scissors, because you’ll need those next!

Now carefully trim the yarn until the balance part is completely balanced.

Now you take that little piece of yarn that is perfectly balanced, and you measure it in Inches. Because this in an Imperial Yarn Balance (which sounds a bit like Imperial Storm Troopers, but there are none of those around. And if there were, Godzilla could totally take them.). They also sell Metric Yarn Balances.

This piece of yarn measures 4 inches. Multiply that by 100 {that’s the mathy part of this post} to get 400, and that’s how many yards of this yarn there are in one pound! Most yarns actually have a higher number of yards per pound, but you can see that this yarn is kind of thick so it’s a little heavier (meaning that it doesn’t take as much length until you have a pound).

I totally wish that I were the person who invented this balance and figured out exactly where the notch would have to be so that you only have to multiply by 100. Sadly, I can’t find anything about the history of this device online, so I don’t know who to thank for this little creation!

Thanks to Mary Louise and Meg for coming to my class again and sharing your talents!

### 2008 Putnam

December 7, 2008

The 2008 William Lowell Putnam Mathematical Competition officially took place this weekend! Yup, six hours of grueling math problems. We had a record number of students take it this year: 21! [That’s just 21, not 21 factorial. That would be an impressive number.]

Here’s one of the videos that our students watched while they were gathered around my dining room table Friday night eating dinner (because we totally bribe our students with food: dinner the night before at my house, bagels for breakfast, and lunch at the local pub in between the two three-hour sessions). It’s “I Will Derive” and I know it’s made its way around the internet, but I still think it’s fabulous:

And here’s the problem (A2) that caused the most discussion over lunch:

Alan and Barbara play a game in which they take turns filling entries of an initially empty 2008×2008 array. Alan plays first. At each turn, a player chooses a real number and places it in a vacant entry. The game ends when all the entries are filled. Alan wins if the determinant of the resulting matrix is nonzero; Barbara wins if it is zero. Which player has a winning strategy?

Here’s my favorite problem (A1) because I was able to solve it right away and do you know how often that happens? Not very.

Let f : R2 -> R be a function such that f(x,y)+f(y,z)+f(z,x)=0 for all real numbers x,y, and z. Prove that there exists a function g : R->R such that f(x,y) = g(x)-g(y) for all real numbers x and y.

You can see the rest of the problems here, and pretty soon you should be able to find the answers hereEdited Monday 12/8 to add: yes, the answers are posted!

No, of course Godzilla didn’t really use a calculator on an exam. He’s a stickler for following the rules.