Suppose the vertices of the polygon are given by where the n^th set of co-ordinates is equal to the first, then the area of the polygon is given by the nice formula

.

A really nice solution to this problem is using Green’s Theorem. This theorem equates a line integral around the boundary of a simple region to a double integral. The theorem states that if C is a piecewise smooth, simple, closed curve in a plane and it encloses a region D, then

If we can choose L and M such that , then the double integral on the right hand side in Green’s Theorem will just calculate the area of the region D. One option is to take and .

If D is the polygon and C is the boundary, all we need to do to calculate the area of D is to calculate the line integral on the left hand side in Green’s Theorem. Consider the path integral from to of . The path is parametrized by a function from to the line joining to , and is given by

So the path integral becomes

This simplifies to

Source: Area Measurement: Planimeters & Green’s Theorem

Murali told us about this amazing card trick his professor did in his statistics class. The professor asked a student to shuffle a deck of cards and then lay all the cards face up in a straight line. Every card was given a value between 1 and 10. The value of the numbered cards was equal to their number (Aces are given the value 1) and the value of the remaining face cards (Jack, Queen, King) was 2. To start with, the student chooses a secret number n between 1 and 8. He or she then looks at the n^th card and silently figures out its value. If that value is m then the student finds the value of the m^th card after the n^th. This continues till the student reaches the end of the deck. For example, suppose the secret number is 5 and in the 5^th place is the 7 of clubs. Then the student has to look at the 12^th (5 + 7) place and if that is the Jack of Hearts then he or she would have to look at the 14^th (12 + 2) place. At some point the student runs out of cards and cannot do this process anymore. The card that (s)he had just before this process ended is the “magic” card. For example, if the student happens to get to the 47^th card and that is the 9 of spades, then since 9 + 47 is greater than 52, the 9 of spades is the magic card. The professor does not know what secret number the student has chosen but is still able to figure out what the magic card is.

How is this trick done? Well initially we thought that the magic card depends only on the distribution of the cards and not on the secret number. However it is easy to produce a distribution of the cards where two different secret numbers end up giving two different magic cards. This is very fragile, because if the distribution of the cards is changed even slightly it ends up giving the same magic card no matter what secret number is chosen. So it appears to be a probability game where the magician who performs the trick ends up getting the very same magic card almost all the time no matter what secret number is chosen. In the rare occasion that the magician is wrong, the person doing the counting should be ridiculed for being bad at simple arithmetic. It would help to quickly shuffle the deck while doing the ridiculing so that all evidence is erased and the trick can be performed again

The mathematical idea behind this trick is very simple. It is known as the Kruskal Principle after Martin D. Kruskal. The magician goes through the same procedure using 1 as the starting secret value. After a few steps there is a great probability that the two procedures will “hit” the same card and then after that they will be “in step”, thus producing the same magic card. I didn’t realise it at the time but this is principle is very similar to the one used in the Rho Pollard Method of factorizing large integers. The idea being to iterate a function till it falls into a cycle. Here is an arxiv paper which uses coupling in Markov chains to analyse this card trick.

(Kepler Conjecture) No packing of balls of the same radius in three dimensions has density greater than the face-centered cubic packing.

It seems very intuitive that the “shop method” is the optimal way of stacking oranges to take up the most space, however, proving this statement is another matter altogether. The history of this conjecture is very fascinating however it has recently been “solved” by Thomas C. Hales. The only problem is that the “proof” uses a computer program to solve about 100,000 linear programming problems and required 250 pages of notes and 3 gigabytes of data. The Annals of Mathematics agreed to publish the result if a series of 12 referees agreed that the proof was correct. After 4 years Gabor Fejes Tóth, the chief referee, returned a report stating that he was 99% certain of the correctness of the proof, but that the team had been unable to completely certify the proof. However, the annals has decided to publish the theoretical aspects of the proof and mention that they have not verified the computational aspects of the proof.