Yay, it’s another Carnival of Mathematics (This one #56)! It’s up at Reasonable Deviations, and includes computers, statistics, geometry, primes, and even tinkertoys [which match the previous post on RD about a marble adding machine]. Thanks, Rod, for putting together so many posts!
(Incidentally, we’ll be hosting the Carnival of Math #57 in two weeks time. Would anyone like to volunteer for #58?)
There was an interesting article in ScienceNOW last week about how a study has shown that people walk in circles if they are blindfolded and set loose in a field (unless it’s sunny out; then they stay in a straight line). The article referred to various novels and movies, indicating that as a plot devise walking in circles isn’t unrealistic.
As I read it, I kept thinking, “Circles? Really? But wouldn’t maintaining constant curvature be about as remarkable as walking in a straight line?” Yet no mention was made of this, and when I looked at the accompanying picture it turns out that circle was used in a more literary sense: the circles were really just closed loops occurring within the path. And although that too is interesting because I’d be freaked out by crossing my own path again, my mathematician self was sorry that they weren’t actually circles.
This squiggle picture actually has a lofty pedigree: it appeared on page 1 of La Peau de Chagrin by Honore de Balzac in 1901.
Yup, the title pretty much sums it up!
Math Teachers at Play #14 is up at Math Mama Writes… and, as always, is full of great posts. One of my favorite is a chart illustrating how few quadratics have integer roots. It’s a little daunting (although I suppose you can increase the number by considering negatives).
Speaking of Carnivals, there will be a Carnival of Mathematics #56 next Friday up at Reasonable Deviations. The official Carnival submission page doesn’t seem to work anymore, but there are instructions at the link above to email or post a submission.
It seems to be the month for odd pricing schemes. First, Michael at God Plays Dice questioned why two items sometimes cost more than twice the cost of one: there were lots of examples in the comments, but one of my favorites is Chicken McNuggets at McDonalds, where a 4-pack costs 99 cents but a 10-pack typically costs over $3.
Then JD2718 shared a story of how he went to buy a bottle of water, only to discover that one bottle cost $1.09 but you could get two bottles for a total of only $1.
I followed up in the comments of JD’s post with a similar experience from earlier this week. We have a cheapo movie theater in town, which is really quite a bargain: big plush seats, second run movies, and only $2 per person for matinees. Except on Mondays, when groups of 3 or more can get in for 50 cents each.
That means that two of us had to pay $4 to escape the heat for a few hours, but if we’d though to bring along Godzilla and buy him his very own seat, it would have cost a grand total of $1.50.
Guess who’se coming to the theater with us next time?
The photo above isn’t our theater, it’s one in Victoria, Australia that Fernando de Sousa took (CC license). But our cheapo theater is pretty similar, except I think the seats are purple.
You know how some math problems start off simple and then, the more you look at them, the more complex they appear, with hidden depths and meanings?
This isn’t one of those. This one needs a sign with “No diving allowed.” Nonetheless, it was surprising to me in its shallowness (like the joke about the smallest not-interesting whole number: hey, isn’t that interesting?).
I wore my hair down yesterday, but I had to braid it at night because I knew it it would get tangled up something fierce if I slept with it down. That got me wondering about the mathematical reason for the number of tangles. In particular, how does the number of possible tangles decrease as you partition hair into three segments for braiding, assuming that tangles could still occur within a segment but not between the three segments?
Now with braids, a big reason that there isn’t much tangling is because of the added tension on the strands of hair. So let’s switch to something else instead: spaghetti.
Suppose you put 
If you look at the ratio:
you get a lot of canceling, and it simplifies to
which, since spaghetti and hair both have relatively large values of 
Why isn’t there a square root? There should be, I thought, because the number of pairs increases like a quadratic as you add more strands. But as TwoPi pointed out, when I interrupted his reading to discuss this problem, each pot or group of 
So in the end, since I end up with far fewer than 1/3 as many tangles by braiding, it turns out that it has very little to do with the mathematics and pretty much everything to do with the tension in the braid. I’m afraid that in the great Math-Physics Match, this round goes to Physics.
Isn’t that a great picture from Harper’s Bazaar? It’s from 1868. The spaghetti is published under GNU-FDL by Tim ‘Avatar’ Bartel.
(There’s got to be a chemistry joke in there somewhere about humours and aluminium.)
So I’m cleaning my office, and I’m cleaning out my Inbox too (I was briefly down to under 200 messages — woo hoo! Then the summer ended.), and I ran across a couple of pieces that a couple of our alumni had sent me over the past year.
Here’s the first:
Proof that Dating is Evil:
First we state that dating requires time and money.
And we all know “time is money”.
Therefore:
And because “money is the root of all evil”:
Therefore:
And we are forced to conclude that:
.
The second piece is a link to a page with a lot of comics and quotes, which seems to be updated at least periodically (since there is one piece from the Monthly in 2009). For lots of distractions and ways to avoid ever getting that Inbox cleaned out, check out this site.
Thanks Lacey and Sarah! And while I’m at it, the original piece was a proof that Girls are Evil, sent by a female alumna to me, a female professor of hers, but after going back and forth a bit I decided to change it to be gender-neutral.
This should be subtitled, “Cleaning my office, Part I” since I ran across it yesterday when I was going through the piles and folders and boxes of papers that all came out of hiding when I moved offices recently.
Anyway, it was a copy of an article in The Onion from right about 10 years ago: “Nobody Really Understands Me” by Fermat’s Last Theorem. It began:
Look, I really don’t have all that much to complain about. I’m well respected. I’m considered quite elegant in my own way. And, in certain circles, I’m seen as quite a romantic and mysterious set of figures. But despite all this, sometimes I still can’t help feeling like no one truly understands me.
You can find the rest of the article here.
Given that there are no Sonic restaurants withing a 150-mile radius of my home, I spend a surprising amount of time talking about them, or at least their commercials (which they show on local TV stations for…no reason?). I recently showed a friend the Food Math commerical, and he responded by showing me something several orders of magnitude better (you’ll have to watch it on YouTube):
So. Good.
Last Friday, right on schedule, Math Teachers at Play #13 appeared at Blog, She Wrote, the blog of homeschooler, science lover, scrapbook lover Heather. As usual, there is a great collection of materials, including one of my favorite card tricks. 😀
I am, right the second, baking chocolate chip cookies for a bunch of teachers who are at a workshop this week. And since me and cookies are pretty much Best Buds I’m not actually confused by the following instructions (which I’m only loosely following anyway), but still, how many ounces of chocolate chips would you add if you were following the recipe below:
(butter, sugar, all that good stuff, then):
1/2 pkg (24 oz) Wegmans Semi-Sweet Chocolate Morsels
[Incidentally, when did Chocolate Chips become Chocolate Morsels?]
Chocolate chip photo published under CC by Editor at Large.
(No, it’s not Sudoku.)
After a many-year hiatus, I just re-subscribed to GAMES Magazine, and in my first issue (September 2009), I was pleased to discover several puzzles with a mathematical slant. One of them was Strimko, a puzzle based on Latin squares, and developed by the Grabarchuk family. Here’s an example (click to solve online):
The idea is simple: each row and column of an nxn grid must contain the number 1, 2, …, n exactly once (that is, the grid must form a Latin square), and each “stream” (connected path in the grid) must also contain the numbers 1, 2, …, n exactly once.
The official site claims that the minimum number of clues required for an nxn grid is n-1 for n=4, 5, 6, and 7, and also says, “This is another unique feature of Strimko.” They do not provide a proof, though, so here’s an opportunity for a nice exercise. (On a related note, a MathSciNet search for “Strimko” returned 0 results, while “latin square” returned 1888 results. It is left to the reader to determine if there’s anything relevant there.)
There are a few sites that provide weekly (here) or monthly (here, here) puzzle sets. So in addition to your daily Sudoku fix, maybe a crossword puzzle, and checking your email, you now have yet another way to avoid doing work.
I’m moving offices: just next door, but I have about 20% less shelf space so I figured it was a good opportunity to see if there were any books that I didn’t really use (answer: yes). In looking through one (Man and Number by Donald Smeltzer, which is a distracting title even though I can see that this book was published 50 years ago) I found another way to multiply! It’s not anything dramatic, but was apparently described by Nichomachus of Alexandria over 1900 years ago in his Introduction to Arithmetic. It uses the following fact:
Here’s the method: suppose first that a and b are both odd or both even. Let x be the average of the two numbers (so x is a whole number, because their sum must be even) and let y be the positive difference between x and a or b. The radius, as it were. Then:
The example given in the book is:
You still have to know your perfect squares, but if you happen to have a table of squares like the Babylonians, that’s no trouble at all. (Indeed, this is basically Formula (4) here, so I don’t know if this should count at all. But I like having an actual citation for the method.)
But what if you have an even and an odd number? Never fear, just ignore that pesky odd bit and add it on at the end. For example, if you have 24·15, you know that this is 24·14 plus an additional 24. So you find 24·14 as above, and add 24 to get 360.
We adopted kittens this week, within 24 hours of arriving home. This was the culmination of several months of deciding whether or not it was a good idea given that we already have two adult cats who weren’t actually asking for younger siblings. So far, the kittens are happy, one adult cat is curious and I think will be fine, and the other adult cat has at least progressed to the point where she’ll eat snacks right by the doorway to the room the kittens are primarily staying in.
Anyway, the place we got them from advertises that one adult female can produce 420,000 offspring over 7 years. That seemed a little high, so I decided to check the math.
The first thing to wonder about was how to count offspring. Clearly this was more than just kittens: it must be counting all future generations. But predicting how many offspring males can sire was difficult, because it really only takes a few males to father a LOT of kittens. So I decided to initially look only at the female lines.
In looking around, it seems that cat pregnancies last about two months, and a female can get pregnant again about a month after that. In theory this would mean 4 litters a year, but The Internet implies that most cats have 2-3 litters per year. I decided to go with 3, since I was trying to see if that 420,000 was even resonable (as opposed to typical).
There’s also the question of how old cats have to be until they are able to mate. It turns out to be around 6 months, though for convenience (since I was going by 4 month intervals) I decided to go with 8. Or, looking at it another way, I figured that kittens would have their own litter on their 1st birthday (and every 4 months after that), so that’s assuming mating at about 10 months old.
Finally, there’s the size of the little. I decided to go with 4 kittens per litter, two of which were female.
Then I made a table. It started off like this:
0 months: 1 adult female
4 months: 1 adult, 2 (female) newborn kittens
8 months: 1 adult, 2 newborns, 2 kid-kittens
12 months: 3 adults, 2 newborns, 2 kids
16 months: 5 adults, 6 newborns, 2 kids
20 months: 7 adults, 10 newborns, 6 kids
By the end of 7 years, there were 35951 adults, 42410 newborns, and 25006 kid-kittens, leading to a grand total of 103,367 cats, including the original one, all along female-descendent lines. If you double it in order to count the males in each litter (but no separate offspring of those males), you get around 206, 732 offspring.
Initially it sounds like the 420,000 is an overestimate, but the wording only is that cats can have that that many youngsters, not that they normally do. If you assume that there are 3 females in each litter (all of whom procreate, etc, since this is just a ballpark estimate) you actually get 898798 offspring along the all-female lines, or over 1.5 million offspring in 7 years. So while I doubt that having half a million offspring within 7 years is the norm, it certainly seems possible in extreme cases.
360 
