## Thursday, 18 October 2012

### Variations on A Mathematical Problem with some Historical Notes

I recently finished Steven Strogatz's "The Joy of X, A guided Tour of Math, from One to Infinity". The book has been well reviewed by a number of bloggers from various backgrounds, and I do not want to do that here; but the one thing I have not seen them say is what a great gift this would be for a good High School student. The book presents math they would well know in a way they may never have thought about it, and then probably touches on areas that will tempt and inspire them to explore farther, and then provides some additional references in both text and on-line at the end. After more than 30 years working with high school students, I think they will love his writing style and presentation.

Now on to my main purpose.....

One of the things I love about mathematics is that there are many seemingly different problems which seem to have a common thread that is not at all obvious at first. I was reminded of one such set of related ideas that I had meant to comment on from a chapter in Joy of X, the very old Cistern Problem.

In Professor Strogatz's book, the problem is presented as filling a bath tub. Suppose the cold-water tap can fill the tub in 1/2 hour, and the hot water tap takes an hour to fill the same tub. How long would it take with both taps opened?
The Professor gives two solution approaches which both work well, I will cover both, and a third way which came to me intuitively as a youth when I was first presented with the problem, so if you want to work it out you might stop reading for a moment and have some fun.

So here are the solutions from the book(Prof Strogatz's book, not Erdos' book).
The first is to find common rate units to compare. He uses 1/30 th of a tub per minute for one, and 1/60 th of a tub per minute for the other. Adding these rates gives the simple combined rate of 3/60th of a tub per minute, so the two together would take 20 minutes to fill the tub.
His second solution is similar but based on using hours, one fills two tubs per hour, the other fills one tub per hour, so together they fill three tubs per hour, or one tub in 1/3 of an hour.
Perhaps the reason this is so difficult for people is also a reflection on the level to which we prepare out students for fractional and proportional mathematics, a point that is suggested again in the Professor's chapter about data and statistics.

My reasoning when I first me the problem was to think of it in terms of work rates. One tap worked twice as fast as the other, so in equal time, it would do twice as much work, or 2/3 of the total. Since the faster could fill the tub in 30 minutes, it would take it 2/3 of 30, or 20 minutes working with the slower tap.

This type of problem and lots of variations originated as the cistern problem. The first record of the problem I have learned of was from David Singmaster's Chronology of Recreational Mathematics in China by Chiu Chang Suan Shu around 150 BC. Within a couple of hundred years it showed up in Heron's Metre'seis around the year fifty of the common era. (I recently read that Heron's method or solution was wrong but have not seen the actual document so if someone will share details, I would love to include them). The problem became a staple in arithmetics and problem books and was used by Alcuin (775) and appears in the Lilavati of Bhaskara (1150). It also appeared in Oughtred's Whetsone of Witte in several variations, and the illustration I use below of the cistern problem was from The First Illustrated Arithmetic, the 1492 arithmetic, Trattato di aritmetica by Filippo Calandri.

There are thousands of similar variations which add more taps, have pipes filling and others draining, or involve animals eating a prey, two men painting at different speeds, etc. Almost all those variations seem immediately recognizable as the same kine of problem, but I want to talk about three types that might not be so easily recognizable. To make the similarities more evident, I will use the same 60, 30 from the book.

Variation 1
The first is a geometric approach. Suppose two vertical poles are erected on horizontal ground and wires are stretched tightly from the top of each to the bottom of the other. How high are the wires when they cross? You may already know what the answer is from all the suggestive connections I have made between this problem and the cistern problem, but take a moment to justify how it is done. One of the first things that made me believe that this was perceived much differently by students was that they would feel they needed to know how far apart the poles were standing, but none ever asked the capacity of the tub. Before I give a solution, or two, I will go on with the other variations that (beware casual reader) do not both have the same solution.

Variation 2
The second can be stated like this: Joe drives to his Mom's for Sunday Dinner and drove 60 mile per hour. Returning home he drives more leisurely and goes only 30 miles per hour. What is his average speed for the whole trip?
If you see how this ties to the previous problems, you are well on the way to a solution.

Variation 3
I regularly read a blog by Dave Marain Math Notations and he once posed a fourth variation of this problem that I did not immediately recognize myself was a variant. In a form matching the problems above, given a square inscribed in a right triangle whose legs are 30 ft and 60 feet (with one corner of the square at the right angle of the triangle), what is the length of a side of the square?

I have read that there is a version of these in the Rhind Papyrus about making loaves of bread from different substances. I was told it was problem 76, but can find no translation that spells out the problem yet. Any assistance would be appreciated.

*** addendum:  I just got an answer from Milo Gardner who pointed out to me that this problem is given in Richard J Gillings' "Mathematics in the Time of the Pharaohs". The  problem as given by Gillings is :"1000 loaves of Peso 10 is to be exchanged for a number of loaves of Peso 20, and the same number of Peso 30. How many of each kind will there be?"

Peso 20 means (if I read the explanation correctly) that you can make 20 loaves from a hekat of grain.  The scribe's solution is first to find that since 1/20 + 1/30 = 1/12, and so 1/12 hekat will produce a loaf of 20 Peso and a loaf of 30 Peso,  so 1 hekat will make 12 loaves of each type or 24 loaves in all. If the original loaves had been 60 peso and 30 peso loaves,  then the answer would be 20 of each type of loaf, or 40 loaves in all.

Ok, So I will give at least a single solution method for each, but the fun part is fleshing out all the details that make the problems alike.

For variation 1, the answer is indeed 20 feet high. My approach to this one was similar to my solution for the other. Since one pole is twice the other over the same horizontal distance, the slope of that wire must be twice as great as the other, so it will only go 1/3 of the way across the span (and up) to reach the height that the smaller slope will reach in twice the distance. Alternately, using what I always called the most important SAT tip of all, if they don't tell you what it is, it probably doesn't matter (make it an easy number). Set up the two right triangles and proceed with similar triangles and algebra using x and width minus x for the two smaller legs.

Variation 2
These seem to be a nightmare for some folks, and they give wildly varying answer. The answer asks for the average speed, so we just need to take the whole distance there and back, see how long it takes, and do the division. Re-invoking the SAT tip, let's say Mom lives 60 miles away (any number would work, but it's a good idea to check the answer by using two distances when you are unsure). So it took one hour to drive there at 60 mph, and two hours to return at 30 mph, for a total of three hours (are we starting to sound like the tub filling problem yet??) So the total trip is 120 miles and the time is 3 hours, the average speed is 40 miles per hour (the fact that it is twice 20 will be explained when I put one more detail in place later.

Variation 3
If the square is to fit in the right triangle, then the corner opposite the one in the right angle of the triangle must be touching the hypotenuse (If it wasn't, we could make the square bigger). I actually solve this with coordinate geometry. Since the square has the same width as height, the opposite corner will be at coordinates (s,s) where s is the length of the sides of the square. If it is on the hypotenuse, we can assume the longer leg is on the x-axis, so one vertex of the hypotenuse is at (60,0) and the other is at (0,30). The equation of the line containing the hypotenuse will then be y= -x/2 + 30 . Since ever point on the line must make this true, it must be true for (s,s) the corner of the square......sooooo,,
s = -s/2 + 30 .

Combining the unknowns on the left tels us that 3s/2 = 30 and once more we see that the sides of the square come out to be 20.

So how are all of these tied together? Interestingly, when Professor Strogatz related his first approach to this problem, his incorrect guess was the average of the two values, 45. I say interestingly because the actual thing that ties them all together is an ancient type of average called the harmonic mean. To the early Greeks, if Nichomachus can be believed, all the means were descriptive of musical relations. Much is often made of the Harmonic Mean in relation to a musical sense, but this may not represent the Greek view. Euclid used the word enarmozein to describe a segment that just fits in a given circle. The word is a form of the word Harmozein which the more competent Greek Scholars tell me means to join or to fit together. Jeff Miller's Web site on the first use of Mathematical terms contains a reference to the very early origin of the harmonic mean, 'A surviving fragment of the work of Archytas of Tarentum (ca. 350 BC) states, 'There are three means in music: one is the arithmetic, the second is the geometric, and the third is the subcontrary, which they call harmonic.' The term harmonic mean was also used by Aristotle. "
The subcontrary comes from the fact that the harmonic mean is the reciprocal of the arithmetic average of the reciprocals of the numbers. for 60 and 30 we get

1/{[(1/30)+(1/60)]/2}

That simplifies to be 2(60)(30)?(60+30) which gives the harmonic average of 40.
The answer of 20 for the three problems that had answers of 20 comes from the fact that the two things were working together. If they average 40 each, it will only take 20 working together. The idea that the problems are the same comes from the occurrence of the 2/1 (or 2/3 to 1/3) proportions in other places. The faster tap fills 2/3 the tub. The wires intersect 2/3 of the way from post to post. (and by the way, if you connected a wire between the tops of the poles, at the point where it is directly over the crossing it will be 40 ft high. The sides of the square will have corners along the two legs of the triangle that are 2/3 of the way along the shorter, and 1/3 of the way along the longer.
As my student's learned eventually to respond when I asked , "Coincidence?"..... "I think Not!"

I recently came across a tub-filling problem in 1921 article from the High School Journal of the University of North Carolina. The title is The Origins of Some Problems in Algebra. It mentions a collection of problems from Metrodorus about 310 A. D. :
We three Loves stand here pouring out water for the bath, sending streams into the fair-flowing tank. I on the right, from my long-winged feet, fill it in the sixth part of a day; I on the left, from my jar, fill it in four hours ; and I in the middle, from my bows in just half a day. Tell me in what a short time we should fill it, pouring water from wings, bows, and jar all at once.

Just as a footnote and plea for help, I have been unable to find very early sources of the variations after some considerable searches. If you would notify me of the earliest printed use of any of these that you know of, or any other historical information, I would be grateful. Thanks.