Judging from the early reviews, my math blog seems to have failed in

what it was attempting to do. And I can see why: the trouble is that you

can't just read mathematics; you have to get involved and work on it

yourself. It's understandable that even those who might be inclined to

invest a little effort don't have the time to do so. At any rate, sadly,

I'm going to abandon the attempt to introduce mathematics systematically

in blog form.

On the other hand, there has been a burst of mathematical activity in

recent meetings of the NKPS (North Kirkland Philosophical Society),

including very interesting contributions from Jay and Oliver, as well as

an old contribution from Abby that I'd like to resurrect. All of these

ideas were new to me, and I've had a lot of fun learning about

them. The problems involved are easy to state and self-contained, and so

perhaps are more suitable for the blog.

PROCEEDINGS OF THE NORTH KIRKLAND PHILOSOPHICAL SOCIETY, JUNE 2015

Jay ``Rocket Man'' Foster called our attention to a beautifully simple

proof of the Pythagorean theorem (the good old ``a squared + b squared

=c squared for right triangles). Oddly enough, this proof is due to

President James Garfield, who--tragically--is best known for being

assassinated in 1881, after just six months in office. An excellent

video of the proof can be found on YouTube, as pointed out by

Mr. Foster. In fact there are several that come up if you search

``garfield proof of pythagorean theorem'', but the one found by Jay is

from the ``Khan Academy''. It's exceptionally well done, and highly

recommended. The idea is to compute the area of a certain trapezoid

related to the right triangle, in two different ways. Comparing the two

answers yields the theorem. By the way, in the course of the proof the

speaker assumes one basic formula for the area of the trapezoid, involving

the height times the average of the top and bottom sides. All

self-respecting readers will, of course, want a proof of the

aforementioned formula. And should supply such a proof themselves!

Oliver ``Count Almaviva'' Henderson reported on a fascinating problem

that originated as a brain-teaser in Victorian times, but in its general

form was only solved in 2014. See

http://www.wired.com/2015/06/answer-150-year-old-math-conundrum-brings-mystery\

/

The original version was published by the Reverend Thomas Kirkman in

1850, in the ``Lady's and Gentleman's Diary'', and requires arranging

fifteen young ladies in a school in certain groups. One abstract

generalization asks the following: Suppose you have numbers r<k<n, and a

set with n elements (school girls, for instance). When is it possible to

choose subsets of size k in such a way that every subset of size r

occurs in exactly one of the chosen subsets of size k? For example, in

the school girl version n=15, k=3 and r=2, although an extra wrinkle is

thrown in as well. The problem is known as the ``combinatorial design

problem''. Peter Keevash of Oxford University recently surprised workers

in the field by finding a complete solution. It involves ``randomised

algebraic constructions'' and ``clique decompositions of hypergraphs''

among other sophisticated techniques. Simple-looking problems can turn out to

have complicated solutions!

At the end of the above-cited article you'll find another recreational

problem, similar in spirit to the schoolgirl problem: the Prisoner

Problem. Several members of the Society have worked on it, without

success as yet.

Some time ago Resident Diva Abigail M. Mitchell discovered the ``3n+1''

problem (while searching for interesting properties of the number 27 for

a friend's upcoming birthday). The fascinating thing about this problem

is that it's very easy to understand, and yet it remains unsolved, by

anyone.

Here it is: Start with any positive whole number. If it's even, divide

it by 2. If it's odd, multiply by 3 and then add 1. Repeat the process

on the new number. If the output is 1, stop. The question is whether or

not you always get back to 1.

Let's try some small numbers to illustrate. 2-->1, stop. Not very

interesting.

3-->10-->5-->16-->8-->4-->2-->1, stop. More interesting!

No need to even try starting with 4, because it already occurs in the

previous example. Similarly for 5. The next interesting case is 7, try

it! In fact you can easily check by hand (i.e. computer-free) that it

works for all numbers less than 27: You always get back to 1. If I

didn't make any dumb arithmetic errors (always a possibility!), the

longest sequence in this range starts from 25 and takes 23 iterations to

reach 1. On the other hand the highest number you ever hit in the

process is 160, which you hit when you start from 15.

But when you start from 27, it takes 111 steps to reach 1 and a high

point of 9,232 is reached! It's been checked by computer that for

starting values up to about a billion billion, the sequence always

returns to 1. This isn't a proof though (see the Wikipedia article for

more info). It's an open problem; no one has proved that it's true for

ALL numbers. And maybe it isn't, who knows?

Finally, as chief editor of the Proceedings I'd like to add a problem to

the list. It's fascinating and fun for anyone who likes

geometry. Suppose we want to tile the plane (think of the floor in your

bathroom, say) with tiles of the same uniform shape (``regular

polygons'', to be precise, which means the sides all have equal length

and the interior angles are all equal). You can do it with squares,

obviously, and with equilateral triangles. You can also do it with

regular hexagons, a fact that is well-known to bees. But these are the

only regular polygons that work! You can't do it with regular

pentagons. And you can't do it with regular n-sided polygons for any

number n greater than 6. Why? See if you can prove this! You can prove

it using little more than the fact that the interior angles in a triangle sum to 180 degrees, plus ingenuity (the ingenuity being the fun

part).

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