Diary

The spiral of Theodorus is a spiral composed of right triangles. Hundreds of years ago, Theodorus of Cyrene constructed continuous right triangles and got a beautiful spiral. He used that spiral to prove that all non-square integers from 3–17 are irrational.

How would you plot this spiral? At each step, you need to draw a segment of length 1, perpendicular to the hypotenuse of the previous triangle. There are two perpendicular directions, and you want to choose the one that moves counterclockwise.

If we step outside the xy plane, we can compute the cross product of the unit vector in the z direction with the vector (x, y). The cross product will be perpendicular to both, and by the right-hand rule, it will point in the counterclockwise direction.

The cross product of (0, 0, 1) and (x, y, 0) is (-y, x, 0), so the direction we want to go in the xy plane is (-y, x). We divide this vector by its length to get a vector of length 1, then add it to our previous point.

Here is a code written in Python to plot the spiral

import matplotlib.pyplot as plt
    
def vertex(x, y):
    h = (x**2 + y**2)**0.5
    return (x - y/h, y + x/h)
    
plt.axes().set_aspect(1)
plt.axis('off')
    
# base of the first triangle
plt.plot([0, 1], [0, 0])
    
N = 17
x_old, y_old = 1, 0

for n in range(1, N):
    x_new, y_new = vertex(x_old, y_old)
    # draw short side
    plt.plot([x_old, x_new], [y_old, y_new])
    # draw hypotenuse
    plt.plot([0, x_new], [0, y_new])
    x_old, y_old = x_new, y_new

plt.show()

The following proof was discovered by Tom M. Apostol, and was published as “Irrationality of the Square Root of Two – A Geometric Proof” in the American Mathematical Monthly, November 2000, pp. 841–842.

In short, if \(\sqrt{2}\) were rational, we could construct an isosceles right triangle with integer sides. Given one such triangle, it is possible to construct another that is smaller. Repeating the construction, we could construct arbitrarily small integer triangles. But this is impossible since there is a lower limit on how small a triangle can be and still have integer sides. Therefore no such triangle could exist in the first place, and \(\sqrt{2}\) is irrational.

In oher words, suppose that \(\sqrt{2}\) is rational. Then by scaling up the isosceles right triangle with sides 1, 1, and \(\sqrt{2}\) appropriately, we obtain the smallest possible isosceles right triangle whose sides are all integers.
Note: If \(\sqrt{2}=a/b\), where \(a/b\) is in lowest terms, then the desired tringle has catheti with length b and hypotenuse a. Indeed, for the pythagorean theorem, we have:

\(b^2+b^2=a^2 \Leftrightarrow 2b^2=a^2\)

but, for hypothesis, \(a=\sqrt{2}b\). So: \(2b^2=2b^2\).
This is \(\Delta OAB\) in the figure below:

By hypothesis, OA, OB, and AB are all integers.

Now construct arc BC, whose center is at A.

Note: \(AC=AB\), and thus AC is an integer. Since \(OC = OA – CA\), OC is also an integer.

Let CD be the perpendicular to OA at point C. Then \(\Delta OCD\) is also an isosceles right triangle, so OC = CD, and CD is an integer.
\(\overline {CD}\) and \(\overline {BD}\) are tangents to the same arc from the same point D, so \(CD = BD\), and BD is an integer. Since OB and BD are both integers, so is OD.
Since OC, CD, and OD are all integers, \(\Delta OCD\) is another isosceles right triangle with integer sides, which contradicts the assumption that \(\Delta OAB\) was the smallest such.

The logical contradiction we have arrived at shows that \(\sqrt{2}\) is a irrational number.

so, we can convince ourselves that:

\(\cos \left( {2x} \right) = {\cos ^2}\left( x \right) – {\sin ^2}\left( x \right)\)

\(\sin \left( {2x} \right) = 2\sin \left( x \right)\cos \left( x \right)\)

We have the hypothesis of having a watch with infinitely thin hour and minute hands (i.e. with continuous motion, and not jerky as usually, it happens). The ideal clock has a perfect mechanics, so that there is exactly 1 to 12 ratio between the angular speeds of the two hands. For convenience, we will measure the angles in sexagesimal degrees and time in minutes.

If we indicate with \(\omega_m\) the angular speed of the minute hand and \(\omega_h\) that of the hour hand, the angular motion laws of the two hands are obviously given by:

In these relations, \({\vartheta_m}\) and \({\vartheta _h}\) are the angles formed by the two hands with a predetermined reference direction, while \({\vartheta_{m,0}}\) and \({\vartheta _{h,0}}\) are the values of these angles at the initial time of our observation (when \(t=0\)).

For simplicity, but without losing in generalities, we can assume that the initial instant corresponds exactly at noon (or midnight), when the two hands are perfectly superimposed, and agree to measure the corners precisely from the direction of the twelve hours; in this way the two equations become:

\({\vartheta _m} = {\omega _m}t\)

\({\vartheta _h} = \frac{{{\omega _m}}}{{12}}t\)

In particular, if the clock is accurate, the two angular speeds are respectively \({\omega_m} = {6^o}/{\text{min}}\) (for the minute hand) and \({\omega _h} = {0.5^o}/\min \) (for that of the hours).

Therefore:

\({\vartheta _m} = 6t\)

\({\vartheta _h} = 0.5t\)

The question: from the beginning of the observation, after how long the two hands are for the first time in an orthogonal position?
The answer:

\(6t – 0.5t = {90^o}\)

or:

\(t = \frac{{90}}{{5.5}} \simeq 16\min {\text{ e }}20{\text{sec}}.\)