Mythematics

Solving the Twelve Labors of Hercules

Michael Huber

• Publisher: Princeton University Press
• Year: 2009
• ISBN:
  9781400831906
• Topic: Mathematics
• Subtopic: Mathematics General
• Index: Mathematics

Fun math puzzles based on the Twelve Labors of Hercules How might Hercules, the most famous of the Greek heroes, have used mathematics to complete his astonishing Twelve Labors? From conquering the Nemean Lion and cleaning out the Augean Stables, to capturing the Erymanthean Boar and entering the Underworld to defeat the three-headed dog Cerberus, Hercules and his legend are the inspiration for this book of fun and original math puzzles.While Hercules relied on superhuman strength to accomplish the Twelve Labors, Mythematics shows how math could have helped during his quest. How does Hercules defeat the Lernean Hydra and stop its heads from multiplying? Can Hercules clean the Augean Stables in a day? What is the probability that the Cretan Bull will attack the citizens of Marathon? How does Hercules deal with the terrifying Kraken? Michael Huber's inventive math problems are accompanied by short descriptions of the Twelve Labors, taken from the writings of Apollodorus, who chronicled the life of Hercules two thousand years ago. Tasks are approached from a mathematical modeling viewpoint, requiring varying levels of knowledge, from basic logic and geometry to differential and integral calculus. Mythematics provides helpful hints and complete solutions, and the appendixes include a brief history of the Hercules tale, a review of mathematics and equations, and a guide to the various disciplines of math used throughout the book.An engaging combination of ancient mythology and modern mathematics, Mythematics will enlighten and delight mathematics and classics enthusiasts alike.

CHAPTER 1

The First Labor: The Nemean Lion

From Apollodorus:

First, Eurystheus ordered him to bring the skin of the Nemean lion; now that was an invulnerable beast begotten by Typhon. On his way to attack the lion he came to Cleonae and lodged at the house of a day-laborer, Molorchus; and when his host would have offered a victim in sacrifice, Hercules told him to wait for thirty days, and then, if he had returned safe from the hunt, to sacrifice to Saviour Zeus, but if he were dead, to sacrifice to him as to a hero. And having come to Nemea and tracked the lion, he first shot an arrow at him, but when he perceived that the beast was invulnerable, he heaved up his club and made after him. And when the lion took refuge in a cave with two mouths, Hercules built up the one entrance and came in upon the beast through the other, and putting his arm round its neck held it tight till he had choked it; so laying it on his shoulders he carried it to Cleonae. And finding Molorchus on the last of the thirty days about to sacrifice the victim to him as to a dead man, he sacrificed to Saviour Zeus and brought the lion to Mycenae. Amazed at his manhood, Eurystheus forbade him thenceforth to enter the city, but ordered him to exhibit the fruits of his labours before the gates. They say, too, that in his fear he had a bronze jar made for himself to hide in under the earth, and that he sent his commands for the labours through a herald, Copreus, son of Pelops the Elean. This Copreus had killed Iphitus and fled to Mycenae, where he was purified by Eurystheus and took up his abode.

1.1. The Tasks

The Nemean lion was ravaging the countryside near the town of Nemea, which is northwest of Mycenae. Hercules has three tasks to complete. After he tracks the lion, he shoots an arrow at the beast. First, he determines the speed at which his arrow strikes the invulnerable lion, given an angle of elevation and a distance to the lion (the Shooting an Arrow problem). In the second task, Hercules determines which set of regular polygons will allow him to tile the area of the cave mouth and trap the lion inside the cave (the Closing the Cave Mouth problem). As an associated exercise, Hercules chooses the mouth of the cave which will give him the greatest chance of finding the lion (the Zeus Makes a Deal problem).

1.1.1. Shooting an Arrow

Task: Calculate the speed at which an arrow strikes the lion at a distance of 200 meters given a launch angle of 20°. Assume Hercules aims for the lion’s head and shoulder area, which is the same distance off the ground as the arrow when it leaves Hercules’ bow. Ignoring air resistance, how long does it take the arrow to travel from Hercules’ bow to the lion?

1.1.2. Hercules Closes the Cave Mouth

TASK: To defeat the lion, Hercules must close up one cave entrance and attack the lion through the other. He finds several stacks of tiles nearby, each of which contains sets of regular polygons. There is one stack of equilateral triangles, one stack of squares, one stack of regular pentagons, one stack of regular hexagons, and one stack of regular octagons. Which stack(s) of polygons will allow Hercules to construct an edge-to-edge tiling in order to close up the mouth of the cave with no two tiles overlapping?

1.1.3. Exercise: Zeus Makes a Deal

TASK: Suppose that the cave has three, rather than two, mouths and that the lion is hiding just inside one of the mouths. Hercules selects one of the three cave mouths at random and is about to enter when Zeus, the king of the gods, suddenly tells him that the lion is not in a second cave mouth (not the one Hercules has choosen). Should Hercules change his mind and enter the remaining third mouth to the cave?

1.2. The Solutions

1.2.1. Shooting an Arrow

From Apollodorus:

First, Eurystheus ordered him to bring the skin of the Nemean lion; now that was an invulnerable beast begotten by Typhon. On his way to attack the lion he came to Cleonae and lodged at the house of a day-laborer, Molorchus; and when his host would have offered a victim in sacrifice, Hercules told him to wait for thirty days, and then, if he had returned safe from the hunt, to sacrifice to Saviour Zeus, but if he were dead, to sacrifice to him as to a hero. And having come to Nemea and tracked the lion, he first shot an arrow at him.

TASK: Calculate the speed at which an arrow strikes the lion at a distance of 200 meters given a launch angle of 20°. Assume Hercules aims for the lion’s head and shoulder area, which is the same distance off the ground as the arrow when it leaves Hercules’ bow. Ignoring air resistance, how long does it take the arrow to travel from Hercules’ bow to the lion?

SOLUTION: Let’s place a coordinate axis system so that the point where the arrow leaves Hercules’ bow is at the origin when time t = 0. It is natural to assume constant acceleration and let the positive y-axis be vertically upward. The constant acceleration, directed only downward, is due to gravity and is denoted by −g. Further, think of this as a two-dimensional motion of the arrow through the air. We will neglect any effects the air might have on the arrow (in a simplified model, Hercules can neglect any effects of wind on the arrow’s time of flight).

Figure 1.1. Determining the velocity components of Hercules’ arrow.

Because of our reference system, the initial position is given by x(0) = y(0) = 0. The initial velocity at time t = 0, which is at the exact instant the arrow begins its flight, is given by v(0) = v₀. With constant acceleration, we obtain the velocity by multiplying the acceleration by time t and adding the initial velocity:

Using trigonometry (see Figure 1.1), we can then determine the initial x-and y-components of v₀ as

v_x0 = v₀ cos θ and v_y0 = v₀ sin θ.

Since there is no acceleration in the x-direction, the horizontal component of velocity will remain constant. What does this mean? The horizontal velocity component (v₀ cos θ) will keep its initial value throughout the flight of the arrow. The vertical component, however, will change because of the downward acceleration. The x-and y-components of velocity (at any time t) become

v_x = v₀ cos θ and v_y = v₀ sin θ − gt.

The expression for v_y comes from replacing v₀ in Equation 1.1 with v₀ sin θ. Integrating these velocity components with respect to time, we can now determine the x-and y-components of the arrow’s position at any time. These are given by

We can now calculate the horizontal distance (the range) that the arrow will travel before striking the lion. Setting the y-component to zero (recall that the vertical height where the arrow strikes the lion is the same as the height of the arrow leaving the bow), we obtain

So, solving for t, we find that either t = 0 or

Substituting this expression for t into the x-component, we find that the distance the arrow travels is

We can now solve for the speed at which the arrow leaves Hercules’ bow, v₀. Substituting in the launch angle (θ = 20°), range (200 meters), and gravitational constant (9.8 meters per second squared), we find t...

Table of contents

• Cover
• Title
• Copyright
• Dedication
• List of Figures
• Foreword
• Chapter 1 The First Labor: The Nemean Lion
• Chapter 2 The Second Labor: The Lernean Hydra
• Chapter 3 The Third Labor: The Hind of Ceryneia
• Chapter 4 The Fourth Labor: The Erymanthian Boar
• Chapter 5 The Fifth Labor: The Augean Stables
• Chapter 6 The Sixth Labor: The Stymphalian Birds
• Chapter 7 The Seventh Labor: The Cretan Bull
• Chapter 8 The Eighth Labor: The Horses of Diomedes
• Chapter 9 The Ninth Labor: The Belt of Hippolyte
• Chapter 10 The Tenth Labor: Geryon’s Cattle
• Chapter 11 The Eleventh Labor: The Apples of the Hesperides
• Chapter 12 The Twelfth Labor: Cerberus
• Appendix A: The Labors and Subject Areas of Mathematics
• Appendix B: Hercules before the Labors
• Appendix C: The Authors of the Hercules Myth
• Appendix D: The Laplace Transform
• Appendix E: Solution to the Sudoku Puzzles
• Bibliography
• Index