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E18. Spider and Bug

A room is in the shape of a rectangular prism, 12 feet high, 12 feet wide, and 30 feet long. A spider is in the center of one of the 12 x 12 walls, one foot from the ceiling. A bug is in the center of the opposite 12 x 12 wall, one foot from the floor. The spider wants to reach the bug by the shortest possible route, and can only travel on the surface. What is the shortest distance, and what is the route? (Hint. The shortest route is NOT the obvious one of going straight up to the ceiling, straight across the middle of the ceiling, and straight down the opposite wall for a total of 42 feet.)

I remember this problem from my school days, and managed to find it again in the Math Forum archives (1995). I would appreciate hearing from anyone who knows the original source. I expect it may be due to Ernest Dudeney.

Teaching Partial Fractions

Students commonly encounter the method of partial fractions for the first time (without proofs) in Calculus II, as a method to aid in integrating rational functions. These days, partial fractions are sometimes not taught at all, since students can determine most any common indefinite integral by using a CAS. Without taking sides in the debate over how much methods of integration should be taught, I would like to make a case that partial fractions should be taught in high school or below.

Of course, partial fractions are a technique that comes up when discussing the algebra of rational functions. However, they also come up very naturally in arithmetic. I propose introducing them in the context of solving a problem that students might find interesting. I call this the problem of the base-p rulers.

The smallest distance measurable by an ordinary English-units ruler is 1/2^n inch, where n is typically 5 (32nds) or 6 (64ths). Define a base-2 ruler to be an idealized version of this ruler, where all coordinates of the form a/2^n are marked, where a and n are non-negative integers. It's clear that not all rational distances are measurable with such a ruler, for example 1/3 is not. To measure all rational distances, we can create an infinite number of base-p rulers, where p varies over the prime numbers. A base-p ruler has all co-ordinates of the form a/p^n, where a and n are non-negative integers. A length of length a/b can be laid with base-p rulers, provided a/b can be expressed as a sum of signed base-p numbers a/p^n. For example, the length 1/6 can be laid out by measuring 1/2, and then backing up 1/3: 1/6 = 1/2 - 1/3.

We want to have students discover that every rational number length a/b (in lowest terms) can be expressed using base-p rulers, where p varies over the primes that divide b.

Providing a proof requires some number theory. Clearly, it is necessary and sufficient to show that every number of the form 1/b can be represented in the required form, and the number theory involves finding a generalization of the fact that if (a, b) = 1 there is a solution in integers to ax + by = 1.

The mystic pentagram and the discovery of irrationals

According to one legend, the Pythagorean Hippasus of Metapontum first discovered that not all numbers are rational by proving that the square root of two is irrational, and he was murdered by other Pythagoreans who believed that all numbers are rational.

However, some people believe that a Pythagorean, possibly Hippasus, discovered the existence of irrational numbers in a different way, by considering the mystic pentagram. Since this figure was sacred to the Pythagoreans, they must have been curious about determining its dimensions. And it is not too difficult to imagine that one of them was led to the discovery of irrationals this way. Indeed, if the five diagonals of a regular pentagon are drawn, forming the mystic pentagram, the ratio of the length of a diagonal of the pentagon to the length of a side is the irrational number phi, the golden ratio.

It makes a great exercise for good beginning geometry students to prove, as some early Greek geometer must have, that the ratio mentioned above cannot be a rational number, using what is essentially Fermat's method of infinite descent. I'll sketch an outline of the proof below.

Consider a regular pentagon of side length s and let the length of each diagonal be d. A regular pentagon is formed inside the original one. Let its length be s', and the length of its diagonals be d'. After drawing the diagram, the student needs to make repeated use of the following elementary facts:
(1) The interior angle in a regular pentagon is 3 * 180 / 5 = 108 degrees.
(2) The sum of the angles in a triangle is 180 degrees.
(3) Two sides of a triangle are equal iff the two angles opposite the sides are equal.
Using these facts it becomes apparent that the diagram has 36 degree angles all over the place [ 36 = (180 - 108)/2] and lots of isosceles triangles. Using this information, the following simple relations can be determined:

(4) s' = 2s - d
(5) d' = d -s

If the ratio d:s is rational, then (by scaling) we may assume that d and s are positive integers. But according to formulas (4) and (5), this means that d' and s' are integers too, and obviously from the diagram d' < d and s' < s. Now, we can imagine continuing the process of drawing diagonals and producing smaller and smaller nested pentagons over and over. Each time the length of the side and the length of the diagonal is a smaller positive integer. But after (much) less than s iterations, the length of a side will be less than 1, not an integer. So d and s can not both be integers.

To see where phi arises, use (4) and (5) to write

(6) d'/s' = (d - s)/(2s - d)

Since the original pentagon and the nested one are similar, we can replace the left hand side by d/s, and then by dividing the numerator and denominator of the right hand side by s we obtain a quadratic equation for (d/s). The positive solution is phi = (1 + sqrt(5))/2.