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
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
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
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