OFF ON A TANGENT
A Fortnightly Electronic Newsletter from the Hope College Department of Mathematics
September 20, 2006 Vol. 5, No. 2
http://www.math.hope.edu/newsletter.html


Upcoming Colloquia
  Join us for Tea Time on Thursdays before colloquia

As part of our colloquium series this year, the mathematics department will host a "tea time" in the Reading Room (VWF 222) at 3:45 p.m.  If tea isn't really your cup of tea, have no fear -- we'll provide some other beverages and snacks, too.  So please join us for a little food and fellowship before you go to the colloquia.  It'll be a great time to chat with the speaker, your professors and other students.

There are three ways to tile a plane using a single type of congruent regular polygons.  Do you remember what they are?  If we allow the use of two or more kinds of regular polygons, there are other possibilities. Tiling, both polygonal and non-polygonal, has played an important role in art and design.

We can also consider tiling a polygonal region using polygons. For example, a square can be tiled using 2 congruent triangles, or 4 congruent triangles, or 6 congruent triangles, etc. Can a square be tiled using an odd number of congruent triangles?  Perhaps requiring the triangular tiles to be congruent is too strong.  Can a square be tiled using an odd number of triangles of equal area? In tomorrow’s presentation Prof. Tom Jager from Calvin College will provide an answer to the latter question.


In 1801, Karl Gauss announced the "Class Number Problem," which he was never able to solve.  In fact, it was not solved until 1983.  But what IS the class number problem and what was so unusual about its proof?  What's special about the number 163?  More philosophically, what is a number?

Join us next week as Prof. John Stoughton poses these and other questions.

 

Grand Valley State University is hosting a series of four lectures this year that will present the beauty of mathematics to a general audience. Using images to convey mathematical ideas, these talks will highlight the aesthetic qualities, diversity, and relevance of mathematics. All of the lectures are accessible to a wide audience, including students at all levels.

Tomorrow’s lecture is titled “Playing Penrose’s Tile Game” and will be presented by David Austin of GVSU’s Mathematics Department.  In this talk, Prof. Austin will describe Penrose tilings.  These tiles can also be put together to cover an area like your kitchen floor. However, the patterns they form are not created by repeating a single pattern, and this is what makes them so interesting to mathematicians and others. Because they are not formed by repetition, there is a disorderliness to the patterns. Yet at the same time, there is a principle, which will be explained in this talk, that leads to a new type of order and allows us to investigate these patterns.

Roger Penrose discovered these tiles in the mid 1970's while doodling during a visit to a relative in the hospital. About ten years later, a radically new phenomenon was observed in crystallography; after further investigation, it was found that Penrose's tiles provided an explanation of this phenomenon. This illustrates something fundamental about the nature of mathematics: new mathematics is usually created out of simple curiosity, yet it often provides clear, elegant explanations of important features of our natural world.

Note:  If you are planning to attend this talk, please allow plenty of time for parking as there is a football game on campus at the same time.  For directions to Padnos Hall of Science Hall at GVSU, please consult the map.  For more information on the Art of Mathematics Lectures at GVSU, please visit their information page.


Upcoming Events at Hope

The 13th annual Michigan Autumn Take-Home (MATH) Challenge will be held on the Hope College campus on Saturday, October 28.  Sign-up information will be available in the next issue of Off on a Tangent.  So keep reading your favorite fortnightly electronic mathematics department newsletter, dear reader, and be sure to meet the Challenge head on!


Hope Floats Kegger Starts Colloquium Season with a Splash!

On the covered walkway outside Van der Werf last Thursday, about 80 math students and faculty gathered to enjoy frosty and frothy root beer floats, as well as each other's good company.  Students and faculty alike pushed their artistic skills to the limit in playing a picture-drawing game, and the assembled throng ooh-ed and aah-ed at the decidedly unmathematical, but nevertheless spectacular, "soda fountains" constructed from 2L bottles of Diet Coke, Diet Root Beer, Mentos and a little fishing line.  Departmental celebrity Elvis made a cameo appearance, which delighted the crowd, but he    appeared more interested in the left over ice cream than in signing autographs.     



Problem Solvers of the Fortnight

Congratulations to Allie Hartley, Anonymous mathematician, Ben Johnson, Bill Buckman, Bryan McMahon, Carleen Dykstra, Chad Rector, Clint Jepkema, Dan Halma, Dayna Waters, Grace Olson, James Daly, Jeff Meyers, Jeff Minkus, Martha Precup and Karena Schroeder for correctly surmising that the ten integers producing the sums of 82, 83, 84, 85, 87, 89, 90, 91, 92 are:

5,6,7,7,8,10,12,13,14,15

For a complete solution to the problem, just ask Elvis!  (See his barkings below.)


  
Problem of the Fortnight
The Hamilton family wanted to cross Broom Bridge at night, but they had only one lantern and the bridge was too weak for more than two to cross at a time.  William, the father, could cross the bridge in 1 minute, and his wife Helen could cross in 2 minutes.  Their eldest son Edwin could cross the bridge in 5 minutes, but the youngest son Archibald took 10 minutes to cross the bridge.  Given that anyone crossing the bridge must have the lantern in order to see the way across, what is the fastest way for the Hamilton family to cross Broom Bridge, and how do they do it?

Write your solution on the back of a picture of Broom Bridge and drop it in the Problem of the Fortnight slot outside Dr. Pearson's office (VWF 212) by 3 pm on Friday, September 29.  As always, show all your work for full credit, and please put your name, the name of your professor(s) and your math class(es) -- e.g. I.M. Student, Math 972, Professor Carl Friedrich Gauss -- on the top of your solution.

Historical note: On the evening of October 16, 1843, as William Rowan Hamilton was walking along the Royal Canal in Dublin with his wife Helen, thinking about autumn evenings, strolls with his loved one, and a way to generalize the complex numbers from two dimensions into higher dimensions, he had a flash of insight and scrawled the equations
i2 = j2 = k2 = ijk =  -1
into the Broom Bridge.  Please visit Wikipedia to read more about Hamilton, and see what Elvis has to say about Hamilton's discovery below.
Got a Math Question?

Ask Elvis ...

... email him at elvis@hope.edu

Dear Friends,

I didn't receive any questions from on campus in the last fortnight.  I did, however, receive one from Al at Fresno State in California and one from Pat in Australia.  I answer both below.  The first question is, of course, our problem of the fortnight.  The second question is rather complex!  So, if it is over your head, don't worry.  After all, I am only about one foot tall so there are a lot of things that are over my head!

If you have any questions (hopefully easier than the one here), don't hesitate to write me at elvis@hope.edu.

Dear Elvis,

I've been trying to figure out this puzzle but I couldn't make ends meet...please help.

Ten (not necessarily distinct) integers have the property that if all but one of them are added the possible results are: 82, 83, 84, 85, 87, 89, 90, 91, 92. What is the smallest of the integers?

Al

Dear Al,

I'm so glad you asked about this one!  When it was posed as the problem of the fortnight in the last issue, I chewed on it for a while before realizing I was at the end of my leash, so to speak.  After all, I know a lot of calculus (and even learned a bit about bifurcations this summer), but I never learned anything like this at The Training Academy.   So, I went to Van Wylen Library the other day and started sniffing around in the stacks.  I found this great book, The Wohascum County Problem Book by Gilbert, Krusemeyer and Larson (which is where I think the Problems Editors found this problem in the first place), and the authors (on p. 41) offered the following solution:

The integers are 5,6,7,7,8,10,12,13,14,15.

Let n1 ≤ n2 ≤ ... ≤ n10 be the ten integers, and let S = n1 + ... + n10 be their sum.  If all but n are added, the result is S - n1; similarly, if all but n2 are added, the result is S - n2; and so forth.  If the ten results are added, we get (S - n1) + (S - n2) + ... + (S - n10) = 10S - S = 9S.  Now we are given that there are only nine possible results: 82, 83, 84, 85, 87, 89, 90, 91, 92.  Let x denote the sum that occurs twice.  Then the sum of the ten results is 82 + 83 + 84 + 85 + 87 + 89 + 90 + 91 + 92 + x; by the above, this sum is also 9S.  Therefore, 783 + x = 9S; equivalently, x = 9(S - 87).  This shows that x must be divisible by 9.  However, of the given results 82, 83, ..., 92, only 90 is divisible by 9, so x = 90, and S = 97.  Subtracting 82, 83, ..., 92, with 90 occurring twice, from 97, we find the ten integers listed above.

Pretty neat!  I guess I should stop chasing squirrels so much and hit the books a little more!

Al's pal,
Elvis


Dear Elvis,
I was flicking through the net looking for an answer to my math question and I came across this website.  I'm a student studying extension math in senior high school and I have an assignment on quaternions (which I've never heard of but have done a bit of research and sort of understand them now). So anyway, my question is:

Assuming ij = a + bi + cj where a, b, c are real numbers, show c2 = -1 which is of course impossible over reals. Explain carefully what conclusions may be drawn.

Now this is my attempt but I don't know where to go from here.

ij = a + bi + cj
(multiply equation by i)
i2j = ai + bi2 + cij
-j = ai - b + cij
(sub in value of ij given in the question)
-j = ai - b + c(a + bi + cj)
-j = ai - b + ca + cbi + c2j

Now how do I get it to equal c2 = -1? And what conclusions may be drawn?

Thanks
Evil Patty

Dear Evil Patty,

I am just a humble dog that has a knack for calculus (which I use to make my life easier), so I don't really know that much about quaternions.  I thought they might be inhabitants of the planet Quaternoid.  But I guess that is not the case.  There are a lot of smart people in the hallway where I hang out every day.  With their help, I think I have an answer to your question. 

Most math students know that a complex number can be written as a + bi.  We can think of this as a number in two dimensions, the real part, a, and the imaginary part, bi, where i2 = -1.  But why limit ourselves to two dimensions!  We can let j2 also equal -1 (but that doesn't mean that i = j).  We can then write a number out like the right side of your first equation, a + bi + cj.  We will see, however, that that leads to problems.

It seems you are on the right track, and in fact, almost done.  You need to take your last equation and reorder it as follows.

-j = (b + ca) + (a+ cb)i + c2j

Since -j can only be written as a linear combination of  i and j in one way, this implies that (b + ca) + (a+ cb)i = 0.  So we therefore have

-j = c2j.

This then implies, of course that c2 = -1.  However, c is suppose to be a real number and this gives us our contradiction.  We can then conclude that our assumption was wrong and that ij cannot equal a + bi + cj.  However, in this algebra we want i, -i, j, -j, 1, and -1 to be closed under multiplication.   We have shown that i and j are not closed under multiplication.  This means we need to add a fourth dimension to our quaternions so that it forms a group.  To do that, we make ij = k where k2 = -1. 

Hence, a quaternion can be written as

H = a + bi + cj +dk

where a, b, c and d are real numbers and i2 = j2 = k2 = -1.

Sometimes two or even three dimensions is not enough.  The beauty of mathematics is we can just add another one if we need to.  Kind of like the dimension of smell that I enjoy and you humans don't quite seem to understand. 

Elvis


Who would not rather have the fame of Archimedes than that of his conqueror Marcellus?  ~ William Rowan Hamilton