# Mathematics Magazine

### Contents for April 2008

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ARTICLES

Larry W. Cusick
Page 83
Archimedes' use of Eudoxos’ method of exhaustion to determine the area bounded by a parabolic arc and a line segment was a crowning achievement in Greek mathematics. The promise of the method, so apparent to us now, seems to have died with Archimedes, only to rise again in different form some 1900 years later with the modern calculus. Archimedes' result though is not just about computing an area. It is about comparing a parabolic area with a related triangular area. In this article we would like to make the case that Archimedes' area comparisons deserve more attention, not so much because of his methods, but rather because of the interesting geometric content of the comparisons and the new questions they suggests. We feel that there are more results to be had, and present a few here with some speculation on further research directions.

Three Transcendental Numbers From the Last Non-Zero Digits of nn, Fn and n!
Gregory P. Dresden
Page 96
Consider three infinite decimals, A, B, and C. The decimal A = 0.14765... has the property that its nth decimal digit is the last non-zero digit from the sequence nn , the decimal B = 0.1123583... is formed from the last non-zero digits in the Fibonacci sequence Fn , and the decimal C = 0.1264224... arises in the same manner from the last non-zero digit of n! . In this article, we show that all three numbers A, B, and C are transcendental. We use both the Thue-Siegel-Roth theorem and the Adamczewski-Bugeaud-Luca theorem in our paper.

An Exploration of Pick's Theorem in Space
Howard Iseri
Page 106
Pick's theorem describes a surprising relationship between the area of a polygon and the number of lattice points lying in the polygon's interior and on its boundary. In this article we explore an extension of Pick's theorem to flat polygonal disks sitting in space. While it is non-trivial to determine the number of spatial lattice points that lie on a particular polygonal disk, we show how the area of such a disk can still be found using Pick's theorem, and then how the number of spatial lattice points can be found from the area. In other words, we explore solutions to the inverse Pick's problems presented by Agfalvi, Kadar, and Papp.

Revisiting James Watt's Linkage with Implicit Functions and Modern Techniques
Christopher Sangwin
Page 116
This article reexamines a simple linkage mechanism invented by James Watt to constrain the piston of a steam engine to move in a straight line. Such linkages played a key role in the industrial revolution and still find many contemporary uses in robotics and flexible structures. We use implicit functions and apply Gröbner basis techniques to obtain an algebraic expression that represents the path followed by this linkage.

Quasigeometric Distributions and Extra Inning Baseball Games
Darren Glass and Phil Lowry
Page 127
Unlike most sports, there are no tie games in baseball, and games continue to be played until there is a winner. In theory, this means there could be arbitrarily long games, but in practice the longest major league game lasted only twenty-six innings. In this paper, we investigate scoring patterns in baseball games and use a variation on the geometric distribution to construct a model that we use to analyze, among other questions, how likely we are to see this record broken in our lifetimes.

NOTE

A Closer Look at the Crease Length Problem
Sean Ellermeyer
Page 138
A standard calculus optimization problem is to determine the minimum crease length that can be obtained by folding one corner of a rectangular piece of paper to some point on the opposite edge of the paper. Though not always stated as such, the problem that calculus textbook authors consider is actually a restricted version of the problem stated above in which only folds that do not produce a flap that protrudes over one of the edges of the paper are considered to be admissible. By removing this restriction, we discover some surprising aspects of the crease length problem that do not reveal themselves when only the restricted problem is considered. In particular, we find that the minimum crease length is NEVER achieved in the restricted set of admissible creases and we also find that the minimum (and maximum) achievable crease lengths depend critically on the dimensions of the paper being folded. Specifically, we show that the correct constructions of the optimal crease lengths depend on whether the square of the ratio of the paper dimensions exceeds or does not exceed the Golden Ratio.

An Offer You Can’t Refuse
Ron Hirshon and Bob De Simone
Page 146
We deal in part with the following situation: Players X and Y play a gambling game and start with positive bankrolls of x and y respectively. To simplify the game somewhat, we assume that their bankrolls consist of chips without any monetary value. They take turns flipping an unfair coin. The probability of this coin falling heads is p, p>½. Player X wins the flip if the coin falls heads; otherwise Y wins the flip. They continue the flips, each time trading one chip per flip as they win or lose. The first person to go bankrupt (have 0 chips) loses the match and pays the winner \$500. The game is a particular form of the classic Gambler’s Ruin game. We consider various situations in this note. One sample result is given below and describes when “Big Bucks” for Y can help Y. (For the fun of the game, we assume we have a real fast computer to simulate the game and play it for X and Y even if Y has an extraordinarily large bankroll, say for example when x=10, y = one trillion)

WHEN BIG BUCKS REALLY COUNT: Given any positive integer x, there exists a corresponding probability P, P=P(x), such that ½ < P < 1 and such that X’s fate in the game can change depending on whether his probability p of winning on a single flip during the game is less than P or bigger than P. Namely, if ½ < p < P, then Y’s probability of winning the match will be greater than ½ provided that Y starts with a sufficiently large bankroll. If P £ p < 1, then X’s probability of winning the match will always be greater than ½ no matter how much Y starts with. The value of P(x) is given by P = P(x) = [ (½ )1/x + 1 ]-1. As an example, P(10) is approximately 0.51732.

Lazy Student Integrals
Gregory Galperin and Gregory Ronsse
Page 152
A thoughtful teacher intends to induce the students to recognize the use of symmetry to evaluate challenging integrals, but a lazy, yet enterprising, student observes an even simpler “rule” that sidesteps the teacher’s intent. Each modification by the teacher to enforce more careful analysis is countered by a student move that still reaches the right answer while avoiding tedious analysis. Laziness may disguise excellent insight.

PROBLEMS
pp.155

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