## Math Tutoring Service

See my Mathematics Tutoring Service on Thumbtack

### The Four Kinds of Students

Richard Feynman is generally regarded as a great teacher, and I'd agree, based on my experience when I had him in sophomore physics at Caltech. However, he sometimes despaired of the teaching enterprise. He said something to the effect that teaching a concept is either unsuccessful (in the case of a poor student) or unnecessary (in the case of a good student who can pick it up by reading). I'd like to suggest that there are not just two, but four kinds of students.

The poor student is unable or unwilling to learn the material.

The mediocre student will learn as much as the teacher presents, but no more.

The good student will continue to learn after leaving the teacher.

The excellent student will surpass his or her teacher.

### A Sangaku Problem

I will be giving a talk Saturday at the NES-MAA meeting in Bridgewater MA in which I will talk about sangaku, or Japanese temple geometry problems. These problems, created by people from a wide walk of life, were beautifully drawn on wooden tablets which were then placed in Buddhist temples or Shinto shrines. Hundreds have been discovered, and probably thousands existed at one time. I became involved in this when I was challenged to solve one of the difficult sangaku. Here I present one of the easier ones, from the delightful book Sacred Geometry: Japanese Temple Geometry by Fukagawa Hidetoshi and Tony Rothman.

I chose this problem because the diagram is so beautiful, the solution is fairly simple, yet satisfying, and it is one of the few sangaku created by a woman (Okuda Tsume).

In a circle of diameter AB = 2R, draw two arcs of radius R with centers A and B respectively, and 10 inscribed circles, two green circles of diameter R, four red circles of radius t, and four blue circles of radius t'. Show that t = t' = R/6.

In the diagram below, we follow the convention of labeling the center of a circle with the radius of that circle.

### E22. A simple probability problem

How many rolls of one fair die does it take, on average, before all six numbers show up? Make a guess and then see if you can figure it out.

### A6. Counting Triangulations

Here is a counting problem that was solved a long time ago. Feel free to try your hand at it.

Given P, a convex n-gon, a triangulation of P is a subdivision of P into n - 2 non-overlapping triangles. A triangulation is obtained by drawing n - 3 non-intersecting diagonals. Let f(n) be the number of different triangulations. Clearly, f(3) = 1, f(4) = 2, and f(5) = 5. Careful counting shows f(6) = 14. Find an expression for f(n).

### The Waitress and the Mathematicians

I recently heard two stories on a LinkedIn math forum, under the topic of humor in mathematics. The first is a funny little story, which I first heard years ago. The second is a cute logic puzzle.

Story
Two mathematicians, Tom and Joe, are in a restaurant, discussing the state of mathematical illiteracy in the general public. Tom goes to the restroom, and Joe calls over the waitress and says, "I'd like to play a trick on my friend. I'll call you over and ask you a question. I'll give you ten dollars if you answer my question with 'x squared'". She agrees, and takes the money. Tom returns, and sometime later Joe says to Tom, "I'll bet most people know how to find the antiderivative of a simple function". Tom disagrees strongly, and Joe says. "OK, I'll bet that our waitress knows the antiderivative of 2x. If I'm wrong, I'll pay for lunch. If I'm right, you pay." Tom says "You're on."

The waitress comes over, and Joe asks her, "Excuse me, miss, but do you happen to know the antiderivative of 2x." The waitress replies "Sure. It's x squared ... plus C".

Puzzle
Four mathematicians come into a restaurant together, and a waitress comes over, and asks "Would you all like coffee?". The first mathematician says, "I don't know". The second mathematician says "I don't know". The third mathematician says "I don't know". The fourth mathematician says "no".

The waitress, who is no slouch at logic, comes back with the correct number of coffees. How many coffees did she bring?

### E21. Equiangular and Equilateral Polygons

A polygon is equiangular if all of its angles are equal. In particular, if the polygon has n sides, each angle measures (n - 2) * 180 / n degrees. A polygon is equilateral if all of its sides have the same length. It can be shown very easily that every equiangular triangle is equilateral. Of course, it is not true that every equiangular quadrilateral is equilateral. Any rectangle that is not a square provides a counterexample. Show that for every n > 3 there exists an equiangular n-gon that is not equilateral.

### A4. Five Circles Theorem

Harold Jacobs presents the fascinating Five Circles Theorem on page 568 of his excellent high school text: Geometry: Seeing, Doing, Understanding (3rd ed.). It states if one starts with a cyclic quadrilateral ABCD, draws the diagonals AC and BD, inscribes a circle in each of the 4 triangles produced, and connects the centers of these circles, then the quadrilateral EFGH produced is a rectangle!

Peter Renz, an editor of Jacobs, called this theorem to my attention and mentioned that the proof that Jacobs gives in the Teacher's Guide uses transformational geometry. He asked if I could find a more elementary proof.

I struggled a bit with this, but finally came up a proof which I have posted at http://www.scribd.com/doc/98720253. I found the Geometer's Sketchpad computer program to be invaluable in helping me discovering geometric truths which I was able to prove and put together to create the proof.

If you are good at geometry, you may want to see if you can come up with a proof on your own.

### E20. The 5-5-9-9 Problem

Introduction: A problem that has been well publicized on the Internet is the "Four Fours" problem. It asks for a representation of all integers from 1 to N (where N is as large as possible) using the digit 4 exactly four times, in addition to basic arithmetic symbols as needed. See, for example, http://www.wheels.org/math/44s.html. I have found this problem to be excellent for group work for students at the middle school or high school level, because it allows students of many different ability levels to participate. However, when I gave this problem to a group of teachers in an online course, one teacher responded by listing the answers available from an online source, verbatim, even though it was clear she did not even understand some of the symbols used in "her" solution. I thought this was a clear case of cheating, but the student felt that using the Internet was a legitimate way to solve a mathematics problem. Maybe she had a point. In any case, it became clear to me that it was time to modify the problem to make it one whose solutions would not be available on line. See the following.

Problem: Show how to represent each of the integers from 1 to 50 using the digit 5 exactly twice and the digit 9 exactly twice, in addition to basic mathematics symbols as needed. Allowed symbols are plus, minus, times, divides (including fraction bar), parentheses, square root, exponentiation, decimal point, factorial, floor function, and ceiling function.

1. The numbers 5, 9, and 50 in the problem are all arbitrary, and the teacher is free to change them to customize the problem.
2. The definition of "basic mathematics symbols" is crucial to the problem. For example, if you allow the successor function (++ suffix in the C programming language), the problem is trivial, and uninteresting. On the other hand, disallowing the decimal point, floor function, and ceiling function would make the problem much more difficult.
3. There are some ambiguities that will come up that must be resolved, such as whether (9-5)(9-5) may be used to represent 44. A question like this would make for good class discussion.
4. The following lemma makes the problem easier to solve: If a number can be represented using a subset of the 4 allowed digits, it can be represented using all 4 allowed digits. Proofs involve showing how to use any number of the allowed digits to produce either a factor with value 1, or a term with value 0. I think most students at the high school level could arrive at this lemma without being told about it. In any case, this lemma might make the idea of using a simplifying lemma, so important in mathematics, something students would remember.

### Simple but remarkable math facts

I'm putting together a list of simple but remarkable math facts that could be used to spice up a math class, more-or-less at the level of high school math. These should be presentable in the form of questions that could be asked of a class, using some showmanship to draw students into them. I'll give a starter list of questions. Most are from geometry, a couple from probability. Happy to hear from anyone of similar questions.

(1) A length of railroad track is two miles long. To account for expansion in hot weather, the track has been joined in the middle. In hot weather the track expands, with each one-mile long piece expanding by one foot. It will form a shallow isosceles triangle, with base exactly two miles long, and sides 2 miles + 1 foot. How high will the track be at the middle? Less than 1 inch? More than one inch but less than one foot? More than one foot but less than 10 feet? More than 10 feet but less than 20 ft? More than 20 ft?

(2) Tennis balls are frequently sold in packages of 3, with the 3 balls packed into a cylinder. Which is greater, the height of the can or the circumference of the can? By how much? The answer to this question will be much more surprising if the teacher displays such a can. Most students will guess from looking that the height of the can is considerably greater than the circumference.

(3) This is a very common optical illusion. Draw two lines exactly the same length. At the end of one place arrowheads, and at the end of the other reverse arrowheads, something like this:
<-------------------->     >--------------------<
Which line is shorter? By what percentage? While not strictly a math problem, it does show that we cannot always depend on the evidence of our eyes, and that in this case measurement with a ruler is more reliable than vision.

(4) Display two statues. The statues are similar (in the sense of geometry, same shape but different size). Tell the class that both statues are made of the steel (or gold?) that the smaller statue weights 10 pounds, and the larger statue is exactly twice the height of the other. How much would they guess it would weigh?  You could tie this into the use of a scale model of the Titanic in James Cameron's movie. This is also related to the misuse of 3-d figures in statistical graphs, where typically oil consumption might be indicated by the size of an oil barrel, with the height of the barrel proportional to a country's consumption, greatly exaggerating the difference in consumption between different countries.

(5) A rope has been stretched around the Equator of the Earth (25,000 miles, approximately). How much longer does the rope need to be if it is to be raised two feet off the ground and remain a closed loop?

(6) How many people, picked at random, must be in a room before the probability that two or more people share the same birthday is greater than 1/2?

(7) The Monty Hall problem. You are a contestant in Monty's TV show. You are presented with three doors. There is a prize behind just one of them. You choose one door. Without opening that door, Monty opens one of the other two doors that does not hide the prize. (If your first pick was correct so both unpicked doors have no prize, Monty chooses randomly which of those to open.) He says you may stay with your original pick, or switch to the other unopened door. Should you switch or not? How would your choice affect the probability of your winning?

### Changes in Mathematics Research Methods

The following math problem appeared in a math forum that I follow:
You have a deck of 56 cards, labeled 1 through 56. All the cards are drawn randomly one at a time. What is the probability that for exactly one card, the face value of the card is one less than the face value of the card before it.
The reason for considering the number 56 was not specified, but that doesn't matter because it is clear that to solve the problem in any kind of satisfying way we have to solve it in a more general case, where 56 is replaced by n. In this case, the probability is P(n)/n!, where P(n) is the number of permutations p of (1, … ,n) for which p(k – 1) – p(k) = 1 for exactly one k. The problem reduces to finding P(n).
The answer is rather neat, and (SPOILER ALERT) I give it below. However what I find most interesting is the way my finding the solution to this problem depended so much on technology. To solve it I did not need to know much mathematics at all. If I had had to solve the problem 30 years ago, it would have taken much more knowledge and/or much more time.
The first thing I did was to generate some numerical evidence. By writing out the acceptable permutations and counting them, I found that P(2) = 1, P(3) = 2, P(4) = 9, and P(5) = 43. I sent the problem to a friend of mine, Ken Cutter, who is a MATLAB guru, and he wrote a short program in MATLAB, using perms(1 … n) to list all permutations and the diff function to compute the differences p(k) – p(k-1). He discovered that I had missed one acceptable permutation of (1 … 5) so that actually P(5) = 44. He also gave me the list of a few more values of P, including P(6) = 265.
I now knew I was looking for a sequence 1 , 2, 9, 44, 265, … , so I went to the Online Encyclopedia of Integer Sequences (OEIS, https://oeis.org/) and entered 1 , 2, 9, 44, 265 into the search field, and out came two known sequences. Only the first of these two sequences seemed to relate to permutations, and sure enough, one the descriptions of this sequence identified it as what I was looking for.
The answer turns out to be that P(n) number of derangements of P(n), that is, the number of permutations with no fixed points. The formula for P(n) is given on OEIS as P(n) = n!*Sum((-1)^k/k!, k=0..n). On the Wikipedia page for derangements, it is noted that a common notation for P(n) is !n, so the desired probability is !n/n! = Sum((-1)^k/k!, k=0..n) which, as Wikipedia notes, converges rapidly to 1/e. In fact, it is the first n terms of the Taylor expansion for exp(x) about 0, evaluated at x = -1.  For the original problem with n = 56, the answer for all practical purposes is 1/e.
Looking back, the most crucial tool in my finding the solution was The Online Encyclopedia of Integer Sequences, started by Neil J. A. Sloane, and a wonderful resource for anyone involved in mathematical work. MATLAB was also very helpful. Without it, I might have still gotten the answer if I had gone over my work and discovered my mistake which made me get P(5) off by one. If I had entered the correct first 4 terms, Online Encyclopedia of Integer Sequences would have returned 17 sequences, but it would still have been pretty easy to find the correct sequence from amongst these 17. But MATLAB (and Ken Cutter) saved a lot of time. Lastly, Wikipedia was a help in explaining the result.
When I was in school, none of these resources were available. If I knew more about combinatorial mathematics, I might have known the answer immediately, or at least recognized it once I generated the numerical examples. Otherwise, I would have had to look at my lists of acceptable permutations more carefully and perhaps derived a recursion relation or induction step that would enable me to find the solution. I would be interested in hearing from anyone would like to send me a solution to this problem from scratch, imagining that they do not know the answer already.
This experiences brings home to me how much the methods of mathematics research have changed since I was in school. I wonder what changes we should be making in K-12 math education that take into account these changes.

### Reed Magazine Profile

My undergraduate Alma Mater, Reed College, just published a profile of me focusing on Math for the Rest of Us at http://www.reed.edu/reed_magazine/march2012/articles/alumni_profiles/ash.html. I feel particularly honored because not only do I have great fondness for Reed College, but Reed Magazine is one publication I make a point of reading. The writing is superior to most college alumni magazines I have seen, and the articles describe varied, idiosyncratic work by fascinating people in many intellectual areas.

### Toastmaster Project Followup

Sources mentioned in my presentation at Lexington ISBN Toastmasters on February 23:

The Online Encyclopedia of Integer Sequences which allows you to search for previously studied sequences by simply entering the first several numbers of the sequence: http://oeis.org. This is an indispensable resource for mathematicians and math students. You can find out a lot about Alcuin's sequence here.

The paper I used to prepare my presentation was by Donald J Bindner and Martin Erickson. It appeared in the February 2012 American Mathematical Monthly, which can be obtained at JSTOR:
http://www.jstor.org/pss/10.4169/amer.math.monthly.119.02.115
Cost is \$12 unless you have a JSTOR subscription for this journal, and you can read the abstract and first page for free. It is very much worth reading for those wanting an in-depth treatment of Alcuin's Sequence.

### The Toastmaster Project

I am using a puzzle problem to illustrate the way in which I teach math to adult students who may not have much experience in the subject, which is the purpose of my company, Math for the Rest of Us.
I will introduce the puzzle and some problems related to it in this blog, and then will talk about its solution during the meeting of Lexington (MA) Toastmasters at 12 noon on February 23. (The source of my presentation and information relating to the solution will be listed on the blog after February 23.)
A rich merchant died and left an inheritance to his three sons. The inheritance consisted of 30 identical silver boxes, of which 10 were full of gold, 10 were exactly half-full of gold, and 10 were empty. He instructed that the inheritance be divided equally, so that each of the three sons received the same amount of silver and the same amount of gold. For the purpose of the division, it is impossible to remove any gold from any box. How can the treasure be divided equally?
This puzzle goes back to the early middle ages. Before following the link, try your hand at solving the puzzle. It can be solved by a little trial and error, and some elementary algebra may be helpful.