"Polyominoes are formed by joining congruent squares...."
This is the first page of a new section, "Some Mathematical Recreations." This first page also has the subtitle "Playing with Polyhexes."
But before explaining what a polyhex is, Pappas reminds us about polyominoes. There is only one monomino, a single square, as well as only one domino. There are exactly two trominoes (up to isometry) -- three squares in a row, and an L shape. There are five tetrominoes (up to isometry) -- and all are familiar to players of Tetris, a game that is named after the tetromino. We could continue on with pentominoes and so on, but we won't.
Now polyhexes are just like polyominoes, except they consist of congruent regular hexagons, instead of squares. Naturally, there is only one monohex and one "bihex." (This is the name that Pappas gives it, but I suppose it should be called "dihex" to be linguistically consistent.)
But since hexagons have more sides than squares, there are more ways to join them, and so for a given n > 2, there are more n-hexes than n-ominoes. Indeed, there are three trihexes -- just as with trominoes, we can have all hexes in one row or an L-shape, except that the third hex to form the "L" could touch either both or just one of the other two hexes. And there are seven tetrahexes -- two more of these than tetrominoes.
The number of n-ominoes for a given n follow the sequence mentioned above -- 1, 1, 2, 5, .... This seems like a sequence I should be able to find on the OEIS -- and sure enough, here it is:
https://oeis.org/A000105
This sequence goes 1, 1, 1, 2, 5, 12, .... There's an extra 1 there because there is one "zeromino" -- the empty set. The number 12 refers to the number of pentominoes. The OEIS mentions something interesting about polyominoes and the classroom (albeit the elementary school classroom):
Polyominoes are worth exploring in the elementary school classroom. Students in grade 2 can reproduce the first 6 terms. Grade 3 students can explore area and perimeter. Grade 4 students can talk about polyomino symmetries.
The pentominoes should be singled out for special attention: 1) they offer a nice, manageable set that a teacher can commercially acquire without too much expense. 2) There are also deeply strategic games and perplexing puzzles that are great for all students. 3) A fraction of students will become engaged because of the beautiful solutions.
But Pappas is writing here about polyhexes, not polyominoes. Here is the OEIS polyhex sequence:
https://oeis.org/A000228
Now here's the question that Pappas asks:
"Which of the three following shapes can be tessellated using one of each of the seven tetrahexes shown above? Have fun!"
There are seven tetrahexes, and so they can cover 28 hexes in all. The three shapes for us to fill are easily described here on the blog. The first is a 4 * 7 Hex board, so that's clearly 28 hexes. The second has the hexes in rows of 5-6-7-6-5. That adds up to 29, so we remove the center hex. The last one is a triangle, 1-2-3-4-5-6-7. In all cases, the hexes are oriented so that a vertical line bisects two of its angles.
But you can't solve these puzzles unless you see what the puzzle pieces -- the seven tetrahexes themselves -- look like. Well, here's a link to them:
http://www.recmath.com/PolyPages/PolyPages/index.htm?Polyhexes.html
The seven valid tetrahexes appear at the top of the page. (Ignore "one-sided tetrahexes," as these are invalid in the Pappas puzzles.)
As usual, I'll give the solution tomorrow.
Chapter 1 of Paul Hoffman's The Man Who Loved Only Numbers is "Straight from the Book." As usual, he opens the chapter with a quote, which I post here straight from the book:
"I have wished to understand the hearts of men. I have wished to know why the stars shine. And I have tried to apprehend the Pythagorean power by which number holds sway above the flux."
-- Bertrand Russell
(Note: I've mentioned the mathematician Bertrand Russell several times on the blog -- most recently in my August 8th post, since he's mentioned on Pappas page 220.)
In this chapter, Hoffman writes about the Paul Erdos and his conception of God. Actually, Erdos refers to God as "SF" -- the Supreme Fascist. The mathematician provides us with two limericks:
There was a young man who said, "God,
It has always struck me as odd,
That the sycamore tree
Simply ceases to be
When there's no one about in the quad."
"Dear Sir, Your astonishment's odd;
I am always about in the quad:
And that's why the tree
Will continue to be,
Since observed by, Yours faithfully, God."
To Erdos, this plainly settles the debate whether mathematics is created or discovered. All we know about math is discovered. In fact, Erdos believes that God -- er, the SF -- has an infinite book with the proofs of all possible theorems. All mathematicians do is rediscover these proofs, which ultimately come "straight from the book." This explains the chapter title. Indeed, Erdos later on says that he doesn't necessarily believe in God, but he definitely believes in God's holy math book.
Hoffman tells us that many young students believe that math is solely arithmetic -- so mathematicians simply spend time solving harder arithmetic problems. Indeed, the author writes more about his own mathematical journey. As a child, he excels in math throughout elementary school and often wins his in-class math bees. In middle school, he becomes interested in chess instead. After he earns his degree, he works with Martin Gardner on his games columns. It is through Gardner that Hoffman eventually meets several famous mathematicians, ultimately including Erdos. By now, the author knows that there's much more to math than arithmetic.
Despite this, much of today's chapter is all about numbers -- or number theory, to be precise. Many of the results Hoffman mentions in this chapter are already familiar to us, since they appeared in Ogilvy's book from last month. (Once again, we keep returning to Ogilvy's book!)
And as we found out in Ogilvy, number theory begins with the primes. Hoffman tells us that prime numbers remain prime no matter what base we use. Even if we had 26 fingers (and presumably used base 26), we'd still have the same primes. Later on, he comments that the best base for housewives would be octal, since there are 16 ounces in a pound and 16 (actually 32) ounces in a quart. (It's still Metric Week, so let me point out that instead of changing our base to octal, we can keep decimal and change our measurements to the metric system.)
After giving Euclid's infinitude of primes book (which I omit, since we already saw it in Ogilvy), Hoffman write about large primes, including the Mersenne primes. This book was written in 1998, and so more Mersenne primes had been discovered since Ogilvy's book. The largest Mersenne prime known at the time of Hoffman's book is 2^3021377-1. It was discovered by the famous GIMPS project, where primes are discovered by computer. I already know about GIMPS -- in fact, here's a link to that website:
https://www.mersenne.org/
But what I didn't know before reading Hoffman is that the prime is discovered by Roland Clarksen, a 19-year-old sophomore at California State University, Dominguez Hills. That is actually the school where I earned my teaching credential!
Hoffman lists the first few Mersenne primes:
For n = 2, 2^2 - 1 = 3
For n = 3, 2^3 - 1 = 7
For n = 5, 2^5 - 1 = 31
For n = 7, 2^7 - 1 = 127
But, as we already know, for n = 11, 2^11 - 1 = 2047 = 23 * 89. In fact, Hoffman tells us the story of a mathematician who proves, entirely by hand, that 2^67 - 1 factors as:
193,707,721 * 761,838,257,287
Hoffman proceeds to state Goldbach's conjecture, with the first few examples:
4 = 2 + 2
6 = 3 + 3
8 = 5 + 3
10 = 5 + 5
12 = 7 + 5
14 = 7 + 7
What does any of this have to do with Erdos? Well, Hoffman quotes him as saying, "Descartes actually discovered this before Goldbach, but it's better that the conjecture was named for Goldbach because, mathematically speaking, Descartes was infinitely rich and Goldbach was very poor."
Erdos himself provides an elementary proof of a result already proved by Chebyshev. We don't have a full limerick, but Erdos writes a short ditty about it:
Chebyshev said it, and I say it again
There is always a prime between n and 2n.
I skip around in this chapter to results not proved by Ogilvy. After retelling the story of Pythagoras and the discovery that sqrt(2) is irrational (oh, and a more Freudian reason for why the Greek sage refused to eat beans), Hoffman tells us of another Pythagorean fascination -- friendly numbers (also amicable numbers). Two numbers are called friendly if the proper divisors of one add up to the other (and these are different from perfect numbers, whose own proper divisors add up to the number.) The first friendly pair is 220 and 284. The next smallest pair, discovered by an Italian teen in the 19th century, is 1184 and 1210.
Hoffman continues by describing Ramsey theory. This theory starts with a simple problem:
Prove that at a party of six people either there are three mutual acquaintances or there are three mutual strangers.
Rather than give a complete description of what Erdos proves about this question, let's just give a link to Cut the Knot:
https://www.cut-the-knot.org/Curriculum/Combinatorics/ThreeOrThree.shtml
Hoffman tells us that if we change "three" to "five," it's unknown how many people must be at the party to guarantee either five mutual acquaintances or five mutual strangers -- except that the answer must be between 43 and 49. Based on the Cut the Knot link above, no progress has been made. We notice that the proof in the (3, 3) case depends on graph theory, which I mentioned yesterday.
In the final example, we show that no matter how we order the natural numbers 1 to 101, there must be either an increasing subsequence or a decreasing subsequence of length at least 11. Basically, we try as hard as we can to go from 1 to 100 with no such sequence -- first go 91-100, then 81-90, then 71-80, all the way down to 1-10. Then no matter where we put the 101, there must be a subsequence of either 11 increasing or decreasing. Basically, if we put it after the 100 (or later), then 91-101 is an increasing subsequence of length 11. And if we place it before the 100 (or earlier), then 101, 100, 90, 80, ..., 10 is a decreasing subsequence of length 11.
Because the universal book of math is so long, we know almost nothing about the integers. Hoffman closes by quoting Erdos: "It will be millions of years before we'll have any understanding, and even then it won't be a complete understanding because we're up against the infinite."
Lesson 4-1 of the U of Chicago text is called "Reflecting Points." This is what I wrote two years ago about today's lesson:
At last, we have reached what makes Common Core Geometry different from traditional geometry -- transformations, including reflections, rotations, and translations. So far, what I posted in September is not much different from a traditional course. But I had to give all that preliminary material first -- after all, the Common Core Standards demand it:
CCSS.MATH.CONTENT.HSG.CO.A.4
Develop definitions of rotations, reflections, and translations in terms of angles, circles, perpendicular lines, parallel lines, and line segments.
And that's exactly what had to do in the first three chapters of the U of Chicago text --define angles, circles, perpendicular lines, parallel lines, and line segments. Only now after defining those basic terms can we actually define rotations, reflections, and translations so that we can finally do Common Core Geometry.
Lesson 4-1 of the U of Chicago text deals with reflections. As I mentioned last year, we do reflections first because the text defines rotations and translations in terms of reflections!
The definition of reflection is so important that I repeat it here. (Remember that I use a strikethrough to represent the segment symbol, since I can't reproduce the vinculum here.)
For a point P not on a line m, the reflection image of P over line m is the point Q if and only if m is the perpendicular bisector of
For a point P on m, the reflection image of P over line m is P itself.
This definition is highly intuitive -- after all, suppose I gave a student a line m and two points P and Q such that the reflection image of P over m is the Q. Now suppose that I drew in segment
(This is a trick that I often do with students -- whenever I ask a student whether
For this section, I'll repeat my first worksheet on reflections. Then I follow it with some exercises. Keep in mind that the method I suggested to generate reflection images is folding -- and it may be hard to fold when there is writing on both sides. As much as I want to save paper and not tie up the Xerox machine, this lesson, and the ones that follow, are very intensive on drawing and folding.
Still, there are a few more things that I want to include here. As I mentioned earlier, one way to generate reflection images is folding. Another method suggested in the U of Chicago text is utilizing a protractor. And once again, this is an important lesson, so let me restate the method for those of you who don't have the U of Chicago text:
1. Place your protractor so that its 90-degree mark and the center of the protractor are onm.
2. Slide the protractor along m so that the edge line (the line through the 0- and 180-degree marks) goes through P.
3. Measure the distance d from P to m along the edge line. You may wish to draw the line lightly.
4. Locate P' on the other side of m along the edge, the same distance from P.
We can see why this works. The first two steps take care of the "perpendicular" part of the definition -- as m is on the 90-degree mark and
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