Math Tutoring Service

See my Mathematics Tutoring Service on Thumbtack

Ha-ha is like Aha!

Yesterday there was an interview on NPR Science Friday with a writer from The Simpsons and Futurama who sneaks in lots of references to advanced mathematics and physics into the cartoons. The references would not be recognized by 99% of the audience. Of course, those who get the references feel really good about being in on a secret. The references are often connected to the plots of the episodes.

For example, an episode in which the theme is "everything becomes easy" has a board filled with mathematical equations including "NP = P". In another episode there is an equation written down which, if true, would be a counterexample to Fermat's Last Theorem. The left and right hand sides of the equation evaluate as equal on a 10-digit calculator, but of course they are not exactly equal.

The interviewer asked the screenwriter if he could guess why there seemed to be some connection between comedy and mathematics and he made a suggestion that I don't remember exactly. I gave my own answer to this question in this blog a couple of years ago.

Back then I mentioned that the "ha-ha" moment in a joke is somewhat like the "aha!" moment in discovering, or appreciating, a mathematical proof, in that both often depend on recognizing the likeness between seemingly dissimilar things. In humor, if one doesn't understand a reference and therefore doesn't understand why something is funny, detailed explanations will almost never make the joke funny. Similarly in mathematics, a proof is very satisfying when it ties together ideas that are well known but seemingly unrelated. Imagine presenting the standard proof that 2 is irrational to a group of intelligent but mathematically ignorant college freshmen. A candid student might respond, "I follow what you did, but what is 2 and what is an irrational number, and why does it matter?" (Unfortunately, this kind of classroom experience is all too common.) By the time that the professor answers all these questions, the aha! moment that the professor was trying to elicit has disappeared forever.

A8. Acute triangles with given side lengths

This is a neat problem from last year's Putnam Examination. It was published (with answer given) in the most recent MAA Monthly.

Given 12 real numbers d1, ... , d12 on the interval (1, 12), show that there exist distinct indices i, j, k such that there is an acute triangle with side lengths di, dj, dk.

I will post a hint as a comment in a few days.

E 24. Egyptian Area Formula

Let ABCD be a quadrilateral with area K and let w = AB, x = BC, y = CD, z = DA. The ancient Egyptians used the expression K' = [(w + y)/2][(x + z)/2] for the area. Since they used this formula to compute the area of fields for taxing purposes, the government probably didn't mind that this formula overestimates the area for all quadrilaterals except rectangles. Prove this fact, that is K <= K', with K = K' iff ABCD is a rectangle. The proof is surprisingly easy, and only requires high school mathematics.

Open Letter to Harper's

Harper's Magazine published a long essay by Nicholson Baker in the current (September 2013) issue, entitled: "Wrong Answer: the Case against Algebra II". Baker is taking up the same anti-algebra banner that was waved by Andrew Hacker in last year's NY Times. Baker rightly points out many problems with the current math curriculum in the schools, but I think he makes many errors of logic and interpretation.

Baker starts off by establishing the difficulty of algebra by quoting a solution of Cardano to the partial cubic equation from 1545, stating that "the algebraic rules that Cardano described and codified are variants of the techniques that students are taught, with varying degrees of success, today". It is as if I argued that  the reading of English is too difficult for students to learn by giving a quotation from Chaucer, whose language is a "variant" of what students are taught today. In point of fact, (1) the notation that we use today has been refined over the last 5 centuries or so, precisely to make the techniques of algebra as easy as possible, (2) Cardano did not "codify" algebra in any reasonable meaning of the term, and (3) the solution of the cubic equation in radicals was no longer being taught in Algebra II when I took it 50 years ago and is, practically speaking, a fairly useless algorithm today.

It is no secret that large numbers of students dislike high school mathematics. Baker seems to feel that this proves that high school mathematics is too difficult for most children to master. It never seems to have crossed his mind that perhaps it is the way that algebra is taught is the problem, not the fact that it is taught at all. In fact, advances in understanding how people learn make it possible to teach high school mathematics to anyone except the most learning disabled. The problem is that these methods are not being used because teaching mathematics well is not an easy task. It is one that we entrust to teachers who are overworked, underpaid, and inadequately trained. In the elementary grades, where the foundations need to be set that will allow the child to master algebra, teachers often themselves have a misunderstanding and dislike of mathematics, which is all too easily passed on to children.

All this might not be so bad if we subscribe to Baker's theory that most students only need a one-year survey course of mathematics in high school: they need to learn "courtesy and kindness, the times tables, fractions, percentages, ratios, reading, writing, some history -- the rest is gravy, really". He cites the fact that American technology ruled the world in the 1950s when only a fourth of students took high school algebra, and seems to think this proves that, as a country, it is fine if most students don't learn algebra. I find it impossible to follow this logic. Haven't a few things changed since the 1950s that might make American technological supremacy less sure and teaching our children mathematics a tad more important?

One of the things that has changed is that we now at least pay lip service to the ideal of equity in education. The idea that it is fine that 3/4 of students (mostly female and people of color) will never be able to enter technical fields that require college mathematics is no longer acceptable to most of us. When algebra, geometry, and trigonometry were not required subjects, the lower classes went to the lower classes where they learned a little shop math or "general math". I don't believe we want to go back to those days. I don't think that students should be allowed to have their future position in society cemented in ninth grade because they could opt out of a subject they might not like. As a nation, we need to teach mathematics better and make it more relevant, not make it optional. Anything less is a cop-out.

A7. Comparing areas

In the diagram below, ABCD is a square, DCE is an equilateral triangle, F is the intersection of AE with CD, G is the intersection of BE with CD, IFJ and HGK are perpendicular to CD, and L is on CB so CF = CL. Prove that the area of triangle AFL is equal to the area of rectangle FGKJ.

E 23. Problem from an insurance exam

Victor Gutenmacher sent me this great puzzle. It appeared on a job interview in an insurance company.

In each equation below, insert one or more mathematical symbols to make the equation correct. You may use the symbols +, -, *, / sqrt, (, ), and the decimal point (e.g. you may insert a decimal point before 1 to get .1.)

1 1 1 = 3
2 2 2 = 3
3 3 3 = 3
4 4 4 = 3
5 5 5 = 3
6 6 6 = 3
7 7 7 = 3
8 8 8 = 3
9 9 9 = 3

For example, 1 + 1 + 1 = 3 and sqrt9 – sqrt9 + sqrt9 = 3.

Victor's friend, Benjy, came up with a solution which, amazingly, is given by one formula, f(n) where n = 1, 2, … , 9. See if you can find it.

Sketchpad - Jacobs problems now available

I have finished writing 23 pages of problems for Jacobs' high school geometry book that use GSP. It is posted on Scribd as a pdf.

http://www.scribd.com/doc/146440260/Extra-Problems-for-Jacobs-Using-GSP

My Jacobs - GSP Project

I am using the textbook Geometry: Seeing, Doing, Understanding (Third Edition) by Harold R. Jacobs in my course for prospective middle school math teachers. I love many things about this book, but I also am a strong believer in having my students use Geometer's Sketchpad (GSP) software. Jacobs has lots of hands-on work, but he made a decision not to use geometry software, so I have been supplementing his book with Sketchpad work. Mostly, I have taken a number of exercises from Jacobs and made very similar Sketchpad exercises out of them.

So far, I have written out seven pages of exercises, and I imagine I will end up with 20 - 30 pages. I key each of my exercise sets to section of Jacobs. I intend to class-test my exercises, and once they are complete to post them on Scribd. In the meantime, I am happy to send my work-in-progress to anyone who wants to send me an email for it. My address is peter.ash@MathForTheRestOfUs.com, and I appreciate feedback.

Sangaku, Harold Jacobs, and Geometer's Sketchpad

As I mentioned at the time, I delivered a talk at the New England Section of the Mathematical Association of America meeting in Bridgewater, Massachusetts, in November. I decided that it would be good to make the talk available online. The talk was about my adventures in trying to prove a difficult theorem mentioned in Harold Jacobs' Geometry. After finding a proof with the aid of Geometer's Sketchpad I happened to discover through Wikipedia that the theorem has a name: The Japanese Theorem For Quadrilaterals. Then Peter Renz, one of Jacobs' editors, suggested I look at the book Sacred Geometry: Japanese Temple Geometry by Fukagawa Hidetoshi and Tony Rothman, which allowed me to place the theorem in a rich cultural and mathematical context.

The talk is at http://www.scribd.com/doc/129968316/NES-MAA-Presentation. This consists of the slides that I used, put in portrait page orientation, but otherwise unchanged. It is a bit terse, but I hope some find it interesting.

The Power of Cryptograms

When I was a boy, a relative bought me and my siblings a copy of a book about solving cryptograms, which described techniques for solving substitution cyphers and gave a lot of cryptograms of varying difficulty to play with. I really enjoyed the book. At the time, it seems like many newspapers would publish a daily cryptogram, though I don't see them very much any more. My local newspaper, the Boston Globe, publishes a daily crossword, a Sudoku, a Kenken, and a few other puzzles, but no cryptograms. I think that's a loss. The kind of thinking used to solve cryptograms is very similar to what is used in solving mathematical problems of all kinds, and cryptograms is something that can be enjoyed both by mathematically-oriented people and literature-oriented ones, since the quotations that are encrypted can be quite memorable. Also, an interest in simple cryptograms could lead to a long-term interest in cryptography, the importance of which in today's Internet-driven world can scarcely be overestimated.

I'd suggest anyone teaching math in the middle grades think about challenging their students with cryptograms. As a starting point, I found Simon Singh's web page at http://simonsingh.net/cryptography/cryptograms/ to be a nice introduction.

Constructible Angles

A student asked what angles are constructible, considering only angles that measure a whole number of degrees (which I'll call integral angles). The answer is very simple: The only constructible integral angles are those which measure 3n degrees, where n is any natural number.

To see this, first note that angles of measure 60 and 72 (vertex angles in a regular pentagon) are constructible. Since angles may be bisected repeatedly by construction, this means that angles of measure 15 = 60/4 and 9 = 72/8 degrees may be constructed. Since 3 = 9*2 – 15, an angle of 3 degrees may be constructed by constructing two adjacent 9 degree angles, and then a 15 degree angle inside the resulting 18 degree angle. By adding n 3-degree angles together, a 3n degree angle can be constructed.

The proof that only these integral angles can be constructed seems to require the more advanced result that there is at least one integral angle that is not constructible. For example, an angle of 20 degrees is non-constructible. (This is the standard proof, using Galois theory, that demonstrates the impossibility of angle trisection.) If there were an integral angle of measure n that is not a multiple of 3 that is constructible, then since (n,3) = 1, there is a linear combination with integral coefficients of n and 3 that gives 1, and so an angle of 1 degree would be constructible, and hence an angle of 20 degrees would be constructible, which it is not..

Wallace Feurzeig 1927 - 2013

Computer pioneer, mathematician, educator, mentor, and friend Wally Feurzeig passed away on January 4. Wally was a key member of the small  team that created the LOGO educational programming languages in the 60s. I' was privileged to know him and his lovely wife Nanni for over 15 years. Wally would listen patiently anc carefully to my various enthusiasms about math education, and always find something incredibly helpful to say, or introduce me to others from his circle such as Victor Gutenmacher and the late Oliver Selfridge who really expanded my understanding of what mathematics education could be. My wife Leslie and I were grateful for invitations to a Monday evening dinner that was cooked by his lovely wife Nanni, and featured some of their fascinating friends.

You can read about some of Wally's accomplishments on Wikipedia, but I just wanted to record a few personal remembrances here.

I am feeling the loss of a kind, gentle, and most intelligent man. We will not see Wally's like again.