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27 Mar

Weekly report, March 21-27, 2016

This week a discussion on tweeter, brought to mind a quote by Underwood Dudley I used years ago

Can you recall why you fell in love with mathematics? It was not, I think, because of its usefulness in controlling inventories. Was it not because of the delight, the feeling of power and satisfaction it gave; the theorems that inspired awe, or jubilation, or amazement; the wonder and glory of what I think is the human race's supreme intellectual achievement? Mathematics is more important than jobs. It transcends them, it does not need them.

Is mathematics necessary? No. But it is sufficient.

With nine pages - solved problems and proved math statements - added to my site this week, I truly have a very good reason to mention that quote.

In the prior week, Galina Gavrilenko - a Russian mathematics teacher -has informed me of a theorem that she discovered, and that long searches on the web made her believe that the theorem was new.

Let similarly oriented squares $ABGH,$ $BCIJ,$ $CDKL,$ and $ADEF$ be erected on the sides of a parallelogram $ABCD.$ Assume $U,V,W,Z$ are the midpoints of segments $FH,$ $GJ,$ $IL,$ and $KE,$ respectively.

Prove that the quadrilateral UVWZ is a square.

The theorem was obviously related to a few well-known and popular results in geometry: Finsler-Hadwiger theorem, van Aubel's theorem, Vecten's configuration, Thebault's first problem. With such a variety of possible connections, it seemed rather implausible that the statement has been overlooked so far. But my web searches also failed to find a precedent. With that, I set out to prove that theorem, found three simple but independent proofs, with confirmed connections to the better known problems. Finding the proofs was a delight in itself; it was augmented by the realization that, although the connections to the older problems have been really very strong and direct, Galina's discovery was in all likelihood indeed new.

Another enjoyable piece of mathematics has been supplied by Miguel Ochoa Sanchez from Peru.

He formulated a statement for a square, but his beautiful proof worked equally well for trapezoids. It was a pleasure to come up with this realization. Sometimes problem posers insert into their formulations unnecessary details that may obscure the real focus of the problem. These are commonly referred to as red herrings. Miguel is a great inventor of geometric problems; I am quite confident he chose a more specific configuration because it suggested applicability of analytic methods thus leading away from his wonderfully simple and general geometric argument.

I also received an attractive statement from Teo Lopez Puccio from Argentina - an aspiring student of mathematics. The statement is related to the configuration of arbelos, the shoemaker's knife, made famous yet by Archimedes.

In the two diagrams, the blue and yellow figures have the same areas, to check which is a simple exercise.

My Romanian correspondents continued the stream of various inequalities. Dorin Marghidanu offered an inequality involving the three altitudes of a triangle:

Solutions came from Japan (Kunihiko Chikaya); Greece (Vaggelis Stamatiadis); Romania (Dorin Marghidanu's and separately by Leo Giugiuc.) Leo Giugiuc was also a partner with Daniel Sitaru in coming up with another inequality

The two solutions (one by Imad Zak from Lebanon) were both based on the famous and very useful Schur's inequality.

23 Feb

A First Look at "The Population Explosion" Book

The book in fact has a longer title: The Population Explosion and Other Mathematical Puzzles. The title warrants an observation.

I once wrote of the difference in attitude of mathematicians and puzzlists to solving problems. While, for a puzzlist, solving a problem is a goal in itself, for a mathematician it may serve as a starting point for further investigation, for turning the problem each other way, for trying to generalize, and learn something from. As Murray S. Klamkin once wrote

...small solved and unsolved problems lead to larger solved and unsolved problems which in turn lead to important mathematical results.

At first glance, the author of the book, Dick Hess, is fully justified to refer to the book as a collection of mathematical puzzles. And this is not only because the problems in the book - the puzzles - need some kind of mathematics to be solved successfully, but also because the author exhibits a remarkably "mathematician's attitude" in his approach to generating the puzzles. Many a problem in the book come as modifications of each other or of various better known puzzles. This is pretty uncommon for a puzzle book. Some readers might want to have a greater puzzle variety, but in my view, having problems turned around, seen from different angles, even munched under slightly modified conditions, makes the book a valuable resource not just for puzzle fans, but for teachers of mathematics who may want to introduce their students to problem solving strategies and instill in them the right kind of attitude to problem solving in general.

Here, for example, the opening of Chapter 2, Geometric Puzzles that come as a modification of a well-known geometric conundrum:

Mining on Rigel IV An amazing thing about the planet Rigel IV is that it is a perfectly smooth sphere of radius 4,000 miles. Like the earth it rotates about a north pole so a latitude and longitude system of coordinates referenced to the poles serves to locate positions on Rigel IV just as it does on earth. Three prospectors make the following reports to headquarters.

(a) Prospector A: "From my base camp I faced north and went 1 mile in that direction without turning. Then I went east for 1 mile. I rested for lunch before facing north again and going 1 mile in that direction without turning. Finally, I went west for 1 mile and arrived precisely at my base camp." What are the possible locations for base camp A?

(b) Prospector B: "From my base camp I went 1 mile north; then I went 1 mile east. I next went 1 mile south and, finally, I went 1 mile west and arrived precisely at my base camp:" What are the possible locations for base camp B?

(c) Prospector C: "From my base camp I went 1 mile north; then I went 1 mile east. I next went 1 mile south and, finally, I went 1 mile west and arrived at a point the most distance from my base camp under these conditions." What are the possible locations for base camp C and how far from base camp C does the prospector end up?

Chapter 9 is devoted in its entirety to an enormous number of variations on a single puzzle:

Jeeps in the Desert The problems in this chapter deal with fleets of jeeps in the desert initially located at point A where there is a fuel depot with unlimited fuel supply. All jeeps end up at either point A or at a delivery point B, as far from point A as possible. The jeeps are all identical, can go a unit distance on a tank of fuel and consume fuel at a constant rate per mile. Jeeps may not tow each other or carry more than a tankful of fuel.

One-Way Trip with a Single Jeep Your fleet consists of one jeep, which is trying to get to point B as far away as possible from the depot and finish at point B. It is permitted to cache fuel unattended in the desert for later use.

(a) How far can you get if you may use only 2 tanks of fuel?
(b) How far can you get if you may use only 1.9 tanks of fuel?
(c) How much fuel is required to go 1.33 units of distance?
(d) How far can you get if you may use only 3 tanks of fuel?
(e) How far can you get if you may use only 2.5 tanks of fuel?
(f) How much fuel is required to go 1.5 units of distance?
(g) How much fuel is required to go 2 units of distance?

These are followed with subsections Round trip with a single jeep, One One-Way Trip with Two Jeeps, and so on. The chapter is stretched over five full pages of questions. There are 10 chapters in all: Playful puzzles, Geometric Puzzles, Digital Puzzles, Logical Puzzles, Probability Puzzles, Analytical Puzzles, Physical Puzzles, Trapezoid Puzzles, Jeeps in the Desert, and MathDice Puzzles.

Among the "Analytical puzzles" one caught my eye. Rate Race:

Three cats and a rat are confined to the edges of a tetrahedron. The cats are blind but catch the rat if any cat meets the rat. One cat can travel 1% faster than the rat's top speed and the other two cats can travel 1% faster than half the rat's speed. Devise a strategy for cats to catch the rat.

The essential point in the cats being blind is of course their inability to detect the location of the mouse. Thus, when solving the problem, the reader should assume that the cats also have dysfunctional olfactory and hearing faculties.

The book is titled after Problem 6 The Population Explosion, in the first Chapter:

In March 2015 the estimated population of the earth reached 7.3 billion people. The average pewrson is estimated to occupy a volume of 0.063 m3, s the volume ov the total population is 0.4599 km3.

(a) Model the earth as a sphere with a radius of 6,371 km and spread the volume of people over the surface of the earth in a shell of constant thickness. How thick is the shell?

(b) The population currently grows geometrically at 1.14% a year. How long will it take at this rate for the population to fill a shell one meter thick covering the earth? What will the population be then?

(c) At the 1.14% geometric rate how long will it take and what will the population be to occupy a sphere whose radius is expanding at the speed of light (= 9.4605284 x 1012 km/yr)? Ignore relativistic effects.

The framework of the question is rather unexpected and funny, although I'd prefer comparing the total volume of the population to that of the Grand Canyon than to smearing the people all over the surface of the Earth.

The book is small but offers a rich collection of interesting problems - puzzles, if you will - not requiring math knowledge beyond high, and many middle, school level.

15 Feb

A Problem in Complex Numbers Sets 3 Aside

Kunihiko Chikaya‎ has posted an elegant problem on a facebook page which is a simple exercise in complex numbers that also submits - due to an insightful observation - to a solution with a very little computational effort.

The problem is

Let $a,b$ be complex numbers that satisfy $\displaystyle\frac{a}{b}+\frac{b}{a}+1=0.$
Find the value of the expression $\displaystyle\frac{a^{2016}}{b^{2016}}+\frac{b^{2016}}{a^{2016}}+1.$

Denoting $\displaystyle x=\frac{a}{b}$ leads to a quadratic equation $x^2+x+1=0,$ with roots $\displaystyle x_{1,2}=-\frac{1}{2}\pm i\frac{\sqrt{3}}{2}=\cos 120^{\circ}\pm i\sin 120^{\circ}.$ It is now hard to miss the fact that, according to De Moivre's formula, $\displaystyle x_{1,2}^3=\cos 360^{\circ}\pm i\sin 360^{\circ}=1.$ From here, since $2016=3\times 672,$ $\displaystyle\frac{a^{2016}}{b^{2016}}+\frac{b^{2016}}{a^{2016}}+1=1+1+1=3.$

However, one solution - submitted by Rozeta Atanasova - obtained the same result without so much as even writing the quadratic equation, let alone solving it. Rozeta Atanasova just observed that $(x-1)(x^2+x+1)=x^3-1,$ immediately implying $x^3=1.$

This led to the inquiry of answering Kunihiko Chikaya question without solving any equation. Here's one such method. I'd be curious to learn of any other. Algebraically, it takes more steps than that of Rozeta Atanasova, but is still based on a simple insight into the implied properties of the original condition. Let's rewrite it as $\displaystyle\frac{a}{b}+\frac{b}{a}=-1.$ Squaring gives $\displaystyle\left(\frac{a}{b}\right)^2+2+\left(\frac{b}{a}\right)^2=1,$ or $\displaystyle\frac{a^2}{b^2}+\frac{b^2}{a^2}=-1.$ Next,

$\displaystyle(-1)(-1)=\left(\frac{a^2}{b^2}+\frac{b^2}{a^2}\right)\left(\frac{a}{b}+\frac{b}{a}\right)=\left(\frac{a^3}{b^3}+\frac{b^3}{a^3}\right)+\left(\frac{a}{b}+\frac{b}{a}\right)=\left(\frac{a^3}{b^3}+\frac{b^3}{a^3}\right)-1,$ so that $\displaystyle\frac{a^3}{b^3}+\frac{b^3}{a^3}=2.$ The same way, $\displaystyle\left(\frac{a^3}{b^3}+\frac{b^3}{a^3}\right)\left(\frac{a^3}{b^3}+\frac{b^3}{a^3}\right)=\left(\frac{a^6}{b^6}+\frac{b^6}{a^6}\right)+\left(\frac{a^3}{b^3}+\frac{b^3}{a^3}\right)=\left(\frac{a^6}{b^6}+\frac{b^6}{a^6}\right)+2=4.$ Thus $\displaystyle\frac{a^6}{b^6}+\frac{b^6}{a^6}=2,$ and, by induction, $\displaystyle\frac{a^{3n}}{b^{3n}}+\frac{b^{3n}}{a^{3n}}=2.$ It follows that $\displaystyle\frac{a^{2016}}{b^{2016}}+\frac{b^{2016}}{a^{2016}}=2,$ implying $3$ as the answer to Kunihiko Chikaya's question.

With other exponents the situation is more involved. For example,

$\displaystyle\frac{a^2}{b^2}+\frac{b^2}{a^2}=-1,$ $\displaystyle\frac{a^4}{b^4}+\frac{b^4}{a^4}=-1,$ $\displaystyle\frac{a^6}{b^6}+\frac{b^6}{a^6}=2,$ $\displaystyle\frac{a^8}{b^8}+\frac{b^8}{a^8}=-1,$ $\displaystyle\frac{a^{10}}{b^{10}}+\frac{b^{10}}{a^{10}}=-1,$ $\displaystyle\frac{a^{12}}{b^{12}}+\frac{b^{12}}{a^{12}}=2,$ etc.

The same behavior is observed for the factors of $5$ and $7$:

$\displaystyle\frac{a^{5n}}{b^{5n}}+\frac{b^{5n}}{a^{5n}}=2,$ if n is divisible by 3, and $\displaystyle\frac{a^{5n}}{b^{5n}}+\frac{b^{5n}}{a^{5n}}=-1,$ otherwise. And

$\displaystyle\frac{a^{7n}}{b^{7n}}+\frac{b^{7n}}{a^{7n}}=2,$ if n is divisible by 3, and $\displaystyle\frac{a^{7n}}{b^{7n}}+\frac{b^{7n}}{a^{7n}}=-1,$ otherwise. The latter confirms the answer to the original problem.

In general,

Which clearly makes $3$ to stand out.

26 May

The Jeweler’s Observation, a look back

Paul Brown, an Australian math teacher and author of Proof, a book that I may characterize as a well-written guided introduction into that most fundamental activity, has brought to my attention a recent post at the Futility Closet blog, The Jeweler’s Observation, which I fully reproduce below:

Prove that every convex polyhedron has at least two faces with the same number of sides.

The solution is initially hidden but becomes available on a button click:

Consider the face with the largest number of sides. If that face has $m$ sides, then it’s surrounded by $m$ faces. Any face must have at least $3$ sides. So altogether in this group we have $m + 1$ faces, and each face must have between $3$ and $m$ sides. At least two faces must have the same number of sides.

From Arthur Engel’s Problem-Solving Strategies, Springer Verlag, 1999.

This example provides nice references to two powerful tools in the arsenal of mathematical problem solving: the Extremal and Pigeonhole Principles. However, that was not the reason I decided to write this blog. One of the most (if not the most) important strategy in problem solving, namely, G. Polya's Looking Back step.

So, looking back at the quote from A. Engel's book, what has been actually proved? From the proof we can extract a stronger assertion than that claimed in the problem. This can be formulated in a couple of equivalent ways. But first note that the convexity of the polyhedron was not absolutely essential and that condition can be weakened. For example, the argument appears to work for any polyhedron in which two faces may only share an entire edge (I am not sure even this is necessary). Thus, thinking of such polyhedrons, the original problem can be so generalized:

• Every polyhedron has a face with two adjacent faces that have the same number of edges.
• Every polyhedron has two faces with the same number of edges adjacent to the same face.

On a second look back, what is essential for the proof is the requirement that all faces of the polyhedron be convex polygons. If that condition is not satisfied, there are 3D objects with a "skeleton" formed by straight line segments, and "faces" bounded by these edges, with no two faces having the same number of edges. Can you give an example? (That would be a counterexample to the original claim.) I can think of a 3D object that consists of four pieces that have $3,4,5,$ and $6$- sided faces, the ones with $4$ and $5$ sides not being convex, or even flat.

01 May

A wrapping surprise

Jim Henle's book "The Proof and the Pudding" that I have recently reviewed contains a good deal of surprises. Across several chapters of the book the author looks into the billiard problem (that is very much like the Two Pails puzzle) and its modifications. One of these involves wrapping a ribbon around a 3D box. Start at a corner and move within one of the faces along the diagonal of that corner. When you hit an edge, just bend the ribbon around the edge to the adjacent face, keeping the same angle $(45^{\circ}).$ Here's an example with an incredible box of dimensions $\sqrt{2}\times e\times\pi:$

Can you figure out how that path will proceed? In a 2D billiard with the crazy dimensions like $\sqrt{2}$ or $\pi$, the path will circle without end. Most likely, this is what your intuition may suggest happens in the 3D case at hand. My intuition did. So here comes the first surprise:

The path that consists of five legs ends up right at the corner where it started. There is an easy explanation why this is so. As in the study of the regular billiard, we may - instead of bending around an edge - flatten two adjacent faces of the cube and let the path pass on a straight line:

The sequence of five successive face traversals snugly fit into a square of size $\sqrt{2}+e+\pi,$ with the (image of the) path going from one corner of the square to its opposite along the diagonal. Note one face outside the square not touched by the path.

Now, that you may think that you understand and can explain why the path returns to the starting corner, consider wrapping exactly same box but starting in a different face. Before, our first leg lay in a $\sqrt{2}\times\pi$ face. Now, let's start within a $\sqrt{2}\times e$ face. Can you predict what will happen to the path?

As you may surmise, the path will behave - if I may say so - in a more rational way. Given the incommensurate dimensions of the box it was rational to expect an endless path. This is what you get on the second attempt. But there remains a question to ponder: Why was the first path so short? Jim Henle leaves it to his readers to find the answer. So do I, though I'd like to suggest also starting the wrapping in the $e\times\pi$ face.

02 Apr

An impossible building, at least this is how it looks

I had very little time driving through Newark, NJ when I caught the sight of the Panasonic building:

It absolutely appears an impossible structure that reminded me of an old applet Structural Constellation one example of which you can see below:

I'll have to visit this place again to learn what creates the illusion.

01 Apr

Why Learn Mental Math Tricks?

"Why Learn Mental Math Tricks?" is the question author Presh Talwalkar tries to answer in the introduction to his new book "The Best Mental Math Tricks". He gives several reasons, starting with

For one, math skills are needed for regular tasks like calculating the tip in a restaurant or comparison shopping to find the best deal. Second, mental math tricks are one of the few times people enjoy talking about math. Third, mental math methods can help students build confidence with math and numbers.

Mental math tricks are fun to share.

I absolutely agree with all of the above and wish only to remark that his book makes a strong stand against the common meaning of the word "trick" whose connotations include magic, deceit, and disapproval.

The book is not a collection of disparate math facts but rather a textbook of well organized and well explained methods of handling arithmetic operations on classes of numbers grouped by their size, divisibility properties, last digits, etc. A whole section that is devoted to each of the "tricks" contains practice problems with complete solutions to illustrate the method. Every method is accompanied by a mathematical proof that sheds the mystery of why the method works. And this is what places the contents of the book in contraposition with the most likely interpretation of the book's title.

In a recent book "Mathematics without Apologies" Michael Harris devotes a whole chapter to the role played by the so called "math tricks" in the body of mathematics.

While capital-M Mathematics is neatly divided among definitions, axioms, theorems, and proofs, the mathematics of mathematicians blurs taxonomical boundaries. ... A mathematical trick is a notorious crosser of conventional borders.

While it is most unlikely that Michael Harris was ever thinking of tricks described and explained by Presh Talwalkar I have no qualms of making this association. Presh Talwalkar has included in his book

... tricks that are relatively easy to learn, are fun, or have educational value.

I'd go further as I strongly believe that those tricks are more than anything else convey to the early learners the essence of practicing mathematics. Presh Talwalkar's book may also open the eyes of an older generation on what they missed in the early grades.

07 Oct

Review of "Zombies and Calculus" by Colin Adams

Colin Adams, author of the unique book "Zombies and Calculus", opens the book with a warning that "if you are squeamish you should not read the book." I venture an additional warning: if you lack a sense of humor, you should not read it either. As an afterthought, the author considers that, given the title of the book, there's little likelihood a squeamish person wold pick it up, in the first place. In my view, such thinking is a mistake - the first of two mistakes the author has committed. Many a student felt like a zombie in a calculus class. These might have shown interest in the book in the hope of finding an explanation to their experience. To these students I say, no, the book is about hard-core zombies - stiff-legged cannibals, entirely devoid of high mental functionality. The one thing on their mind is feeding on anyone yet alive and thinking. But, unlike in the movies featuring Milla Jovovich, in the book zombies have been observed to be tearing clothes off zombies of the opposite sex - in public and with an obvious intention. Unfortunately, the narrator - a college math professor - being followed by his former students and colleagues who now saw him only as a potential meal had not the time to further dwell on his observations.

One really needs a well developed sense of humor to read about a talented student who, with a chunk of her neck missing and the head at a weird angle, tries to grab her professor of a few minutes ago, and so had to be knocked down. On the other hand, every one would laugh at a delinquent student, oblivious of the surrounding dangers, who hands to the narrator (concerned with escaping a crowd of zombies in hot pursuit around the corner) his late homework. One again needs to summon one's sense of humor to read about the sad fate that befell the unthoughtful student short time after the encounter.

At the outset, I have mentioned that the author committed two mistakes. Here's the second one. Early in the story, several survivors of the initial attack, found themselves locked up in an office, with zombies moaning and banging into(?) each other just outside the door. Naturally there arose a question of active protection, and a girl removed her stockings to be filled with heavy objects that, when swung, could bring a zombie down. The narrator then cuts the stockings with scissors into two legs. However, as every grownup knows, stockings are already a two-piece item. Pantyhose is the one dress item that needs separation.

 Stockings Pantyhose

The episode with the stockings serves to develop a conversation (and related mathematical tools) about speed, force, and the strength of human (i.e., zombie's) skull. As the story evolves, other mathematics comes in handy: equations of pursuit, logistic equation, predator-prey model, stationary point of a system of ODEs, Newton's Law of Cooling. Mathematics in the book is impeccable. However, from the incomplete list of topics, one may start suspecting (and justly) that the book should be more properly titled "Zombies and ODE" or "Zombies and ODE Modeling". (The author admits as much in the introduction by pointing to those who already learned calculus as his primary audience.) I do not count this as a third mistake, for this is a common marketing ploy to rely on a sound bite title to attract broader readership. And the book well deserves attention from not squeamish math instructors and a wider audience of intelligent readers, curious of a new literary genre that mixes storytelling with gentle mathematical instruction.

14 Sep

Review of Coffee, Love and Matrix Algebra

I am admittedly a compulsive reader. I either stop reading a book if I do not like it, or I continue reading until finished - with only obligatory interruptions. Gary Davis' book brought me an entirely new and tantalizing experience. Gary used daily tweets on twitter.com to announce new installation at his popular blog Republic of Mathematics (http://www.blog.republicofmath.com/). So there was no other way but to do the reading a chapter a day. Had it been my choice - and I can candidly say that after the fact - I would have gobbled the book in one setting.

The story is about a year long episode in a life of a college math department. Any one, I believe, who ever held a position in an academic department would easily identify the traits of Gary's protagonists as shared by some of their colleagues. The characters were authentic, evolution of events realistic; it took me a while to realize that the book was entirely a work of fiction.

Naturally, while there are similarities, not all math departments are the same; Gary's no different in this respect. It is painted with its own problems and peculiarities. Although a mathematics professor, Gary navigates his story with the skill of a professional writer. He narrates his story that takes several imaginative turns with confidence of a participant and kind humor of life's keen observer. That's a great story, masterfully and engagingly told. Read and enjoy.

01 Aug

Distance to the Horizon on the Fourth of July

I had the luck to celebrate the past 4th of July with our friends in their newly acquired home just above the marina in Atlantic Highlands, NJ. The view from their backyard was absolutely breathtaking. The ambient light that appeared to blur the background made the view even more enchanting.

Here is a map that would help you identify parts of the panorama.

In the middle above the center line, across the Sandy Hook and Low bays the part of the Interstate 278 is the famous Verrazano Bride, past which there are visible Manhattan tall buildings; on the right, that's Brooklyn whose buildings seem taller, but only because of relative proximity.

The sunset was spectacular.

From the beginning (above) and to the end (below).

With the dusk, came fireworks. We were able to simultaneously see the bursts of illumination over Manhattan (that appeared just above the Verrazano bridge) and Brooklyn. (You can see these too if you click on the photo below.)

And next came the fireworks just above our heads shot from the marina below.

Next day I was downloading my 4th of July photos to my computer from the camera. In the zoomed-in version of the Verrazano bridge I noticed a short upward stick under the bridge which - to my great surprise - happened to be the Stature of Liberty.

To inquire the distance to the stature has now appeared quite natural. This can be estimated with the following map.

It is also possible to use the well known formulas for such estimates, I refer to the wikipedia. The distance is about 22 miles, Verrazano bridge being at about 2/3 of the way. Using the reversed formula it is possible to estimate how high above the marina have I been celebrating the 4th of July this year.