New Problems to Think About ...

Last night, I attended the January meeting of the Metropolitan Mathematics Club of Chicago (MMC), at which Steve Viktora talked about word problems through the ages,  He showed examples of word problems across cultures and times (going back 6000 years!), and highlighted some common themes he found.  Here's an example of a type of problem Steve called a "hydraulic problem"; these types of problems deal with some sort of public works, like building dikes, storing grain, or marking out land, and they appear in documents from Mesopotamia, Egypt, and China going back thousands of years.  This one is from the 4th millennium BCE from the Sumerian city of Shuruppog, and is the oldest known word problem:

A granary of barley.  One man received 7 sila [of grain].  What are its men?  [i.e. How many men can be given a ration?]  

Of course, to solve this, one needs to know that capacity of a granary was 2400 gur, and one gur was equal to 480 sila.  (Any Sumerian bureaucrat could tell you that.)

Steve presented other hydraulic problems.  This one got everyone talking, and there were a number of different, interesting, and elegant ways that people solved it.  There were also some good questions about what happens if the base is not a leg of the triangle.  I don't remember the  time or place for this problem:

A triangular piece of land [in the form of a right triangle] is divided among six brothers by equidistant lines constructed perpendicular to the base of the triangle.  The length of the base is 390 units and the area of the triangle is 40950 square units.  What is the difference in area between adjacent plots of land?

There were abstract problems from Islamic cultures, temple problems from Japan, puzzles from Medieval Europe, and many others.  They were fun to see, and we got to work on a few, which is always a good time.  (Yes, I like discussing and solving problems.  I am a Math Geek, but you already knew that.)

After Steve's presentation, a few of us were talking, and Carol said she was teaching a Geometry class, and asked the students to come up with four integers that could represent the sides of a rectangular solid and the length of one of its interior diagonals.  Nice problem, and if you are studying the Pythagorean Theorem and Triples, you can figure out that if the dimensions are 3, 4, and 12, the diagonal is 13.  One of her students said, "Sure, and if you pick and two consecutive numbers for the base, and multiply them to get the height, then the diagonal will be one more."  Carol thought that was an interesting hypothesis, and matched the expected solutions, but cautioned that it might not really be a general solution.  Then she did the algebra ...  So what do we have?  A way to find "Pythagorean Quadruples"?  I love math!  Paul added that when he was teaching the triples, he listed several out for his students:  3,4,5;  5,12,13;  7,24,25;  9,40,41;  11,60,61 ... and hoped they would see some patterns that could help generate more triples.  (There are all sorts of patterns here.)  Paul said one of his students noticed this one, which I had not seen before:  4 = (1/2)(3+5); 12 = (2/3)(5+13); 24 = (3/4)(7+25);  40 = (4/5)(9+41);  60 = (5/6)(11+61).  Okay, mind blown.  Did I mention that I love math?  MMC Dinners are always worth the price of admission.