April Fool's Day, and the Pythagorean Theorem

      Here is a post in honor of April Fools' Day, which unlike March 14, is celebrated by countries that use either the month-day format or the day-month format.

      Not everyone is a math person, but nearly everyone remembers the Pythagorean Theorem.

      Well, not everyone. The Scarecrow got it wrong in the 1939 movie, "The Wizard of Oz". It is arguable whether the writers accidentally or deliberately got it wrong, or whether the actor Ray Bolger flubbed the line. (See also Singh, "The Simpsons and Their Mathematical Secrets", pp. 119-121.)

      So here is a little Pythagorean Theorem problem. In the following right triangle, calculate C.   A² + B² = C², and then take the square root of C. How hard can it be?

      Of course you have to know if A = (√-1), then A² = (√-1)² = -1, but that seems reasonable.

      But does something seem wrong with the resulting value of C?

      I plotted the triangle in a complex number plane, not a real number plane.   (√-1) is not a "real" number, and perhaps it should be thought of as a non-real number. (I am intentionally avoiding the dreaded "i" word.)   Perhaps what is happening is that in the complex plane, the hypotenuse C is really a line segment of magnitude zero, and hence a point? No, not "really".

      So April Fool's. You calculated a hypotenuse of a right triangle and concluded it has length zero.

      A better answer is that a point (x, y) in the complex plane is represented as x + yi, and the Euclidean distance between two complex points (x, y) and (u, v) equals √( (u - x)² + (v - y)² ) which here is the expected √2.

      Here is the R code for the plot, followed by the Python code for the plot. Note that in R, the shading is done with geom_polygon, while in Python it is done with plt.fill_between:

library(ggplot2)

# Define the vertices of the triangle
triangle_data <- data.frame(
    x = c(0, 0, 1),    # x-coordinates of vertices
    y = c(0, 1, 0)     # y-coordinates of vertices
)

ggplot(data = NULL) +
    geom_polygon(data = triangle_data, aes(x = x, y = y), fill = "#9B111E", color = "black") +
    geom_segment(aes(x = 0, y = 0, xend = 1, yend = 0), color = "black", size = 2) +
    geom_segment(aes(x = 0, y = 0, xend = 0, yend = 1), color = "black", size = 2) +
    geom_segment(aes(x = 0, y = 1, xend = 1, yend = 0), color = "black", size = 2) +
    ggtitle("Use the Pythagorean Theorem to calculate C") +
    xlab(expression(bold(A == sqrt(-1)))) +  # Square root symbol for x-axis label
    ylab("B = 1") +     # Label for y-axis
    geom_text(aes(x = 0.55, y = 0.55, label = "C = ?"), fontface = "bold", size = 6) +
    theme_classic() +
    theme(
        legend.position = "none",
        axis.line = element_blank(),
        axis.text = element_blank(),
        axis.ticks = element_blank(),
        axis.title.x = element_text(size = 20, face = "bold"),
        axis.title.y = element_text(size = 20, face = "bold"),  
        plot.title = element_text(size = 20, face = "bold")
    )       

---

import numpy as np
import matplotlib.pyplot as plt

# Create figure and axes
fig, ax = plt.subplots()
ax.set_title('Use the Pythagorean Theorem to calculate C', fontsize=15, fontweight='bold')

# Define the points for the line
x = [0, 1]
y = [1, 0]

# Plot the horizontal and vertical lines
ax.axhline(y=0, color='black')  # x-axis
ax.axvline(x=0, color='black')  # y-axis

# Plot the diagonal line
ax.plot(x, y, color='black', linewidth=1.5)

# Fill the region enclosed by the axes and the line
plt.fill_between(x, y, 0, color='#9B111E', alpha=0.5)

# Add axis labels with bold and enlarged fonts
ax.text(0.5, -0.05, r"$\mathbf{A = \sqrt{-1}}$", ha='center', va='center', fontsize=14, transform=ax.transAxes)
ax.text(-0.05, 0.5, r"$\mathbf{B = 1}$", ha='center', va='center', fontsize=14, rotation=90, transform=ax.transAxes)

# Add diagonal line label with bold text, enlarged font, and moved farther off the line
ax.text(.6, .5, r"$\mathbf{C}$", ha='center', va='center', fontsize=14, color='black')

# Turn off the axes' tick marks and spines
ax.spines['top'].set_color('none')
ax.spines['right'].set_color('none')
ax.spines['bottom'].set_color('none')
ax.spines['left'].set_color('none')
ax.tick_params(left=False, bottom=False, labelleft=False, labelbottom=False)

# Display the plot
plt.show()

# end

Pi and billiard balls; a different application of π for Pi Day

      So suppose there is some sort of physical experiment, and the first time you do it, the answer comes out as 3. Then you change the experiment to make it a little more complicated, and the answer comes out as 31. Then you make it even more complicated, and the answer comes out 314. Then you ... . Of course you see where I'm going with this.

      It's March 14, or 3 14 (in the US date format), and because 314 are the first digits of π, many people use this day to share interesting and unusual appearances of π. I think you will find Pi and the Billiard Balls to be a little different!

      But first, if your statistics course did not require calculus as a prerequisite, you may be unaware that π is contained within the formula for the probability density of the Normal distribution, f(x) = $\frac{1}{\sqrt{2\pi {\sigma }^2}}{}^{\exp }$ where the 2π is necessary for the integral of the pdf to equal 1.

      A common application of π on Pi Day is the Buffon's needle problem: Given a needle of length l dropped on a plane ruled with parallel lines t units apart, what is the probability that the needle will lie across a line upon landing? There are many references to this such as Buffon, so the solution will not be repeated here.

      π appears in many places in math. One reason is that π is defined in reference to a circle. The trigonometric functions can be defined in terms of triangles within a circle. As a point traverses the circumference of the circle more than once, the trigonometric functions repeat in cycles of 2𝜋. Therefore, phenomena that repeat and can be represented by trigonometric functions are likely to have π somewhere within them. This is because π inherently relates to the periodic nature of these functions, making it indispensable in modeling cyclical behaviors.

      I recently discovered a physics problem called Pi and Billiard Balls. The original article is by G. Galperin, and I will try to summarize it. His paper is not an easy read.

      Suppose we have the first quadrant of an xy coordinate system. Suppose we have a vertical wall at x = 0, y ⪰ 0. We have ball 1 having mass m1 at initial position x1 > 0, and ball 2 having mass m2 ⪰ m1 at initial position x2 > x1. Let the ratio of the masses be a multiple of 100:   m2 / m1 = 100^N for a fixed non-negative integer N, including N = 0. Suppose ball 2 moves from right to left along the x-axis and collides with ball 1, ball 1 moves from right to left and collides with the vertical wall, and assume all collisions will be perfectly elastic. This perfect elasticity assumption implies the balls will satisfy the law of conservation of momentum, and the law of conservation of kinetic energy. We also assume the balls only move along the x-axis; there is no y movement.

      The amazing conclusion of all this is: The total number of collisions C(N) is a number equal to the first N decimal digits of the number π (starting with 3) !

      For case 1, let N = 0 so m2 = m1. Now push ball 2 from right to left at initial velocity v2 until it hits ball 1. This is collision 1. Ball 1 will move from right to left at the same velocity v1' = v2, while ball 2 will now be at rest. Eventually ball 1 hits the wall for collision 2 and now moves from left to right at velocity -v1', until it hits ball 2 for collision 3. Now ball 1 is at rest, and ball 2 is moving to the right with velocity −v1'. There have been 3 collisions (the first digit of π), and there will be no more.

      For case 2, let N = 1 so m2 / m1 = 100. This is more complicated because after collision 1, v1' will not equal v2, and ball 2 will not be at rest. The result of the two conservation equations m1v1 + m2v2 = m1v1' + m2v2' and .5m1(v1^2) + .5m2(v2^2) = .5m1(v1' ^2) + .5m2(v2' ^2), give v1' = [(m1 - m2)v1 + 2m2v2] / (m1 + m2) and v2' = [(m2 - m1)v2 + 2m1v1] / (m1 + m2). After collision 1, substituting v1 = 0 and m2 / m1 = 100 gives v1' = 200v2/101 ≈ 1.98v2 (nearly twice as fast as initial velocity v2) and v2' = 99v2/101 ≈ .98v2 (slightly less than initial velocity v2).

      Ball 1 hits the wall for collision 2 and now moves from left to right at velocity -v1', until it hits ball 2 for collision 3.

      After collision 3, substituting primes into the two conservation equations, gives v1''' ≈ -0.96v2 and v2''' ≈ 1.96v2. Ball 1 will bounce back and forth between ball 2 and the wall many times. After each collision between the two balls, the velocity of each ball changes with ball 1 decreasing in velocity and ball 2 increasing in velocity. The relative velocity between ball 1 and ball 2 decreases with each collision, as the heavier ball will slowly transfer its momentum to the lighter ball. Eventually the relative velocity will be so small that they will effectively move together after the collision. At this point, no more collisions will occur. There will be 31 collisions (the first two digits of π).

      For case 3, let N = 2 and m2 / m1 = 100², there are 314 collisions. And so on.

      Not surprisingly there is a circle lurking under all of this, due to the conservation of kinetic energy equation .5m1(v1^2) + .5m2(v2^2) = constant, and there is a trigonometric function and a calculation involving an angle and π. Galperin proves the conclusion in general: The total number of collisions C(N) is a number equal to the first N decimal digits of the number π (starting with 3) !

      In addition to the Galperin paper, a good explanation is here, but this is not an easy read.

      You are welcome to try this experiment yourself, if you can create the condition of perfect elasticity. Happy Pi Day.