Lost theorem about angular proportions

Last week, I came across a so called missing theorem about angular proportions in a triangle, discovered (or rediscovered) by Leon Romain. This triangle construction is presented by user Linelites on youtube.

For such a "Romain triangle", the missing theorem states that, if one of the inner angles is twice another inner angle, one has the property a² = bc +c², where a, b and c are outlined in Figure 1.

The sine chord pattern presented in the Teaching sine function with spaghetti provides helpful insights for the missing theorem. In this pattern, all angles at the intersection of chords are integer multiples of a chosen unit angle or its complement, modulo 90°. All segment lengths in this pattern can therefore be written as sums or differences of cosine and sine products and ratios of that angle. Figure 2 pictures some measures of sine chords in a circle of unit diameter, for an arbitrary angle θ.

Figure 3 shows a Romain triangle for angle α in this pattern. There are numerous other Romain triangles in this pattern. Can you figure them out? With the help of the measures pictured in Figure 2, one can follow visually the elements of a proof for the missing theorem.

Side b is the sine of the chosen angle θ (Figure 4):

b = sinθ

Side c times sin3θ equals b times sinθ (Figure 5):

c = sin²θ/sin3θ

Side a equals b cosθ minus c cos3θ (Figure 6)

a = sinθ cosθ - sin²θ cos3θ /sin3θ

Working out a² and bc +c², one finds that they are both equal to 2 sin²θ cosθ / sin3θ.

The interesting thing is that the sine chord pattern hosts plenty of "missing theorems" about angular proportions, which are only waiting to be (re)discovered.

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Update (June 7, 2010): I corrected Figure 6, which held a wrong term.

Teaching sine function with spaghetti

Sine functions appear everywhere in physics and mathematics. This seems to be related to the circular symmetry of space and to the periodic behavior of dynamical processes. The usual definition is that the sine of an angle is the ratio of the length of the opposite side to the length of the hypotenuse in a right triangle (definition at wikipedia). The sine is also often represented as the ordinate of a point running on a unit circle centered at the origin. Any right triangle can be put in that setting. The length of the blue segment on Figure 1 illustrates this definition for an arbitrary angle α, in a circle of unit radius.

It is of interest to have in mind other geometric representations of the sine function. This helps to associate figures with sine and cosine operations or identities. The red segment in Figure 1 shows such an alternative representation, because the sine is also a chord of a circle of unit diameter. When we increment the angle by steps α, one endpoint of the chord advances on the little circle, while the other endpoint remains fixed at the origin, as illustrated by the animation in Figure 2.

When it represents the sine of angle α, the chord in the little circle spans an angle 2α. It is therefore a direct way to the angle multiplication and division by 2. Moreover, the endpoints of this chord can be placed two by two in any other direction on this circle. This enables one to find other variations on the recurrence pattern of the angle in the circle. During one of my circle drawing sessions, I was surprised by the stepwise zigzagging pattern of Figure 3.

Teaching the sine function with this zigzag pattern is particularly suited for daily life situations. Children like it, especially when you explain it at the table.

Keeping track of the circle for integral representations of π

The fundamental constant π is characterized in many ways. Historically, it all began from tentative measurements of the circle's perimeter or area and gradually shifted into more advanced mathematics, in such a way that the link between the circle and modern characterizations of π faded away, see for example the formulas presented at Wolfram MathWorld or Wikipedia.

In the preceding posts, I mentioned infinite products as approximations for π. These may be seen geometrically as exhaustion methods, where the area of a polygon approaches the circular area alternately from above, from below, from above, from below, etc.

There are also integral representations of pi. In such integral representations, π appears in the quantitative value of the integral of a mathematical function. Visually, this is often represented as the area delimited by the bounds of the function. However, the relation with the circle is lost, when viewed under Cartesian coordinates. For example, the graph of the simplest instance of the Cauchy-Lorentz distribution, f(x)=1/(1+x²), "has nothing at all to do with circles or geometry in any obvious way" as quoted from last Pi-day Sunday function from Matt Springer's Built on Facts blog.

In order to view the role of the circle in integral representations of π, we need to switch to alternative ways to visualize math functions. As an example, let's take the constant function y=f(x)=2. The function f maps an element x from a domain to the element y of the target. In this case, for every x, the target y has the constant value 2. With Cartesian coordinates, we are used to represent this function as a horizontal straight line, like in Figure 1a (click on the figure to view it enlarged). If however we write it as R=f(r)=2, where the function f maps any circle of radius r of the domain to a target circle of radius R=2, the same function can be viewed as a circle of constant radius, like in Figure 1b. So the same function f can be equally well viewed as a straight line or as a circle (x, y, r or R are only dummy variables).

Now if we take another example, the linear function, y=f(x)=2x, we are often used to view it in Cartesian coordinates as a straight line with slope 2, like in Figure 1c. In the circular representation R=f(r)=2r, this works however differently. Because we are relating circles of the input domain to other circles of the target, for each circle of radius r, we need to draw the target circle of radius 2r. A single line won't do. For one value of r, we need to draw two circles. If we use blue circles for elements of the input domain and red circles for elements of the target, we could visualize it for successive values of r as an animation like in Figure 1d. In that way, we view the progression of the target circle as the input circle becomes larger.

Unlike the Cartesian representation which shows the progression of a function in a static graph, this circular representation needs a dynamic or recurrent process to get grip of the progression of the function. Therefore it isn't very adapted for illustrations in print media. On the other hand, it has the advantage of keeping track of the geometrical form of the circle. And that's exactly what we need in order to perceive the circular nature when π shows up in mathematical functions. The relation of the integral of the Cauchy-Lorentz distribution f(r)=1/(1+r²) with the circle can then be seen with the help of the geometric counterparts of arithmetic operations like addition, squaring and dividing. A convenient procedure is illustrated in the successive steps of Figure 2.

Step 1. Draw the input circle of radius r and the reference circle of radius unity.

Step 2. Determine r².

Step 3. Add 1 to r².

Step 4. Invert (1+r²). We now have the target circle of radius R=1/(1+r²).

Step 5. Find the target ring related to the input ring ranging over [r, r + dr]. This yields a ring of width dr/(1+r²). The location of this ring depends on the relative progression rates of r and r² (I've not yet found a straightforward explanation for this determination).

Step 6. Integrate dr/(1+r²) for r running over all space. For r becoming larger and larger, the summed area tends towards the area of a circle of radius 1. For the positive half plane, this corresponds to the π/2 value found analytically.

The tricky step seems to be the way how to relate the progression between r and 1/(1+r²) in steps 5 and 6. One can verify for example the value of the integral at intermediate steps. For the integral from r=0 to 1, the value in the positive half plane must be π/4, which can be verified on the figure below.

In order to gain more insight on π, it could be of interest to develop skills for this circular representation.

Wallis product for nested equilateral triangles

I'm discovering the Wallis product thanks to an approach involving geometry of areas. In order to relate that to what has already been written on the subject, here are some works which I would like to study in the coming time:

• the appropriate parts of John Wallis' Arithmetica Infinitorum (is there a translation online ?)
• Some generalizations of the Wallis product (1992), by Les Short, containing a general formula for different starting ratios
• An Elementary Proof of the Wallis Product Formula for pi (2005), by Johan Wästlund

Maybe you have other suggestions on this subject.

Meanwhile, I tried to draw equilateral triangles in the exhaustion scheme of nested circles. We start from the concentric rings of same area as described in Figure 4 of the Variations of circular area division into equal parts post. Remember the inner circle has radius 1 and area π. Each subsequent Nth circle has radius square root of N (noted √N) and area Nπ. Then we can draw an equilateral triangle whose sides are tangent to the first circle. The neat thing about this procedure is that we can read out the pertaining numerical values right on the figure. Figure 1 shows us that the area of the equilateral triangle is 3√3. So the ratio between the outscribed equilateral triangle area and the circle area is equal to 3√3/π.

By drawing equilateral triangles tangent to circles of incrementing area Nπ, we discover the nested equal area division for equilateral triangles: the area between each equilateral triangle is equal to 3√3 (see Figure 2, you can click on the figure to view it enlarged). Among other interesting features, this figure contains the table of 4. For each triangle outscribing a circle of radius √N, its vertices are located on the circle of order 4×N.

If we outscribe alternately a circle, a triangle, a circle, a triangle, a circle, etc., the circles (and the vertices of the triangles) go through the subsequent powers of 4, see Figure 3. By changing the form of the triangle, we have access to other tables and powers. It seems there are plenty of properties that are hidden in this structure.

Now, we can try the Wallis exhaustion scheme on the nested circle and triangle structure. The procedure apparently goes something like:

Step 1. Draw the initial circle and triangle with area ratio 3√3/4π.

Step 2. A better fit would be to inscribe both figures in higher orders of the nested structure. But the area of the triangle is too small with respect to the circle. So we must enlarge the area of the triangle more than the circle. We enlarge the area of the triangle by 3 while leaving unchanged the area of the circle. This is illustrated in the figure below.

Step 3. But now the area of the triangle is too large with respect to the circular area. So for this fit, we must enlarge the area of the triangle a little less than the area of the circle. We therefore enlarge the area of the triangle by 3, while enlarging the area of the circle by 4, as illustrated below.

Step 4. But now the area of the triangle is too small with respect to the circular area. So for the next fit, we must enlarge the area of the triangle a little more than the area of the circle. We therefore enlarge the area of the circle by 6, while enlarging the area of the circle by 5, as illustrated below.

This exhaustion scheme that makes the area of equilateral triangles alternately larger, smaller, larger, smaller, etc... than the circular area can be written as:

This formula corresponds to an infinite product approximating π, which I've found in a French book Le fascinant nombre π by Jean-Paul Delahaye but I couldn't find it mentioned elsewhere. As approximation method, it is about twice as fast as Wallis' original product. This equilateral triangle exhaustion product is probably a special case in a large family of Wallis products, each of which pertaining to a particular geometrical configuration from which one tries to approximate π. The starting and recurrence conditions depend on the form and the initial area ratio between the polygon and the circle. The rules governing this exhaustion scheme in the general case still elude me.

Exhaustion of nested squares and Wallis product

When one is asked to divide a square into N equal area parts, we commonly think of dividing a square into N equal rectangles. Figure 1 shows an example for N=11.

There is another general solution to that problem, in particular we could wrap it up in an enigma: is it possible to divide a square into N square-shaped parts of equal area, N being any integer?

The nested circles structure described in Figure 4 of the Variations of circular area division into equal parts post suggests the following solution.

Step 1. Starting from a circle of radius 1, area π, draw circles of incrementing area 2π, 3π, 4π, 5π, 6π, and so on... The resulting concentric rings have all the same area π.

Step 2. For each circle, circumscribe a square. The resulting squares have area 4, 8, 12, 16, 20, 24, and so on...

Step 3. Conclusion: the areas between each nested squares are equal. The figure illustrates the case for N=11.

So here we have a compass and ruler solution for dividing a square into N equal area parts. An interesting corollary is that this division into N equal parts can be applied to any 2D-figure, whether it be a triangle, a pentagon, any polygon, or even arbitrary drawings, provided that it be circumscribed (or inscribed) in a circle. As an answer to a comment on the preceding post, why not divide Leonardo da Vinci's Vitruvian man into N nested Vitruvian men? There are certainly other applications, one of them has been suggested to me by another comment on the preceding post.

In order to calculate an approximate value of π, Archimedes used a method of exhaustion which consists in inscribing a circle between two identically shaped polygons. Increasing gradually the number of sides of the polygon, the shape of the polygon approaches the shape of a circle. This geometrical approach is at the origin of his numerical approximation of π. Many other ways have been developed to approximate π, especially using infinite series. Geometrical representations of these series are rarely available, probably because no direct link has been found between them and the circle. The benefit of the nested circle and square structure of Figure 2 is that it suggests alternative methods of exhaustion.

For example, we could try to equate the area of a circle to the area of a square in this nested structure. This yields an approximation for squaring the circle (or circling the square). For a circle of radius 1, the area of the circle is π, the area of the circumscribed square is 4. The ratio between both areas is therefore π/4. The procedure to let that ratio go to 1 then follows the following steps.

Step 1. Draw the initial circle and square with ratio π/4.

Step 2. A better fit would be to inscribe both figures in higher orders of the nested structure. But the area of the circle is too small with respect to the square. So, we must enlarge the area of the circle a little more than the square. We therefore enlarge the area of the circle by 3 while enlarging the area of the square by 2, which is illustrated in the figure below.

Step 3. But now the area of the circle is too large with respect to the square. So for this fit, we must enlarge the area of the circle a little less than the area of the square. We therefore enlarge the area of the circle by 3, while enlarging the area of the square by 4, as illustrated below.

Step 4. But now the area of the circle is again too small with respect to the square. So for the next fit, we must enlarge the area of the circle a little more than the area of the square. We therefore enlarge the area of the circle by 5, while enlarging the area of the square by 4, as illustrated below.

Step 5. As after step 2, the area of the circle is too large with respect to the square... and I become a bit exhausted by drawing at exponentially growing scales;-) but I hope I've shown enough to suggest the continuation of this exhaustion scheme. You can of course rescale the figures, keeping track of the successive numbers and their roots on the figures.

The described method of exhaustion searches for a limit by alternately letting the circular area become larger, smaller, larger, smaller, etc... than the square area, but each time with a reduced fraction. As a limit it can be expressed by the following equation:

It seems we have gone through Wallis' product, which is usually disconnected from any geometrical approach.

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Physics quote of the day :

Profound study of nature is the most fertile source of mathematical discoveries. ~ Joseph Fourier

Variations on dividing circular area into equal parts

Today is π-day. Investigating the properties of a circle is one of my favorite hobbies, so this is an appropriate event for some discussion on the circle. I love drawing circles and lines on a piece of paper. It's fascinating to discover natural laws in the circle. At each of my drawings, I gain some new insghts. Recently I came across a Cut The Knot post presenting an inventive way to divide a circular area into any number of equal parts, with compass and ruler. I googled a little more over it and curiously, it seems that the presented yin and yang-like solution is the only one that has been presented (if you know of other published references, I would be eager to hear about). It appeared in a Mathematical Association book in 1995 but I didn't find any other reference to solutions of this problem. This is surprising, because area division is a fundamental problem with broad applications in this world where sharing physical resources equally is a necessary guarantee for peace. Playing a bit with the elementary components of the yin and yang pattern gives related solutions, of which a random one is sketched in Figure 1.

I convinced myself that the most general solution is in fact a quite natural and intuitive one, because it figures the way how circular patterns often grow in nature, like the annual rings in wood. Other patterns also give some insight in various ways to divide a circular area into parts, like the beautiful quintuple emulsions formed with microfluidics of the figured image (courtesy of A. Abate/ Harvard University).

The general rule to have a circular area divided equally into N of such nested circles is that, for a starting circle of radius 1, the surrounding Nth circle radius is the square root of N (noted √N).

The first circle has radius 1, area π.
The second circle has radius √2, area 2π.
The third circle has radius √3, area 3π.
The fourth circle has radius √4 (= 2), area 4π.
And so on...

Square roots can be easily drawn with compass and ruler, so the same for nested circles with radius equal to √N, where N is incremented from 1 to any positive integer value. Figure 4 shows a convenient procedure for N=1 to 9:

Step 1. Draw nested circles with diameter 1 to 9 whose left diameter endpoints coincide.
Step 2. Draw a perpendicular to the diameter direction at the intersection of the right endpoint of the circle with diameter 1.
Step 3. Draw circles of radius equal to √N thanks to intersections of first circles with perpendicular at N=1.
Step 4. Erase lines and circles of intermediate steps. The obtained outer circle is divided into N equal parts each of area π.

Now, from this pattern, it is possible to start a set of variations on circular area division, because the nested circles are whole, contrarily to the yin and yang pattern. We may for example start to divide the circle of area 2π into two equal half-circles and shift those components randomly (see Figure 5). Or we may start to divide the circle of area 3π into equal parts, etc. I'm sure anyone can find original new ways to arrange and transform all those equal parts, in order to gain some new intuitions on the circle.

Happy π-day!

One year of Physics Quotes of the Day

It's been exactly one year since I left my job as IT project manager at French telecom operator SFR. In that job, I missed the physics, the photons, the electrons, the atoms... So when SFR gave the opportunity to change course, I eagerly applied for a year Master of Science training at the Institut d'Optique, which could give me an upgrade relative to almost 20 years out of the professional physics world. This back to school period is extremely satisfying. It's a pleasure to learn new things, or learn them again and under different circumstances, to meet instructors, researchers and students who share the same interest. My last exam was past Friday and the last term of the MSc year is a 4-month long internship in a lab. So I'll be teaming up with a group at the ESPCI who's growing semiconductor nanocrystals.

It has also been a year ago since I started to tweet daily quotes from physisicts. I needed a speed course in all the fields of physics. It seemed that looking for quotes from diverse physicists is a great way to achieve that goal. Monte Zerger, a mathematics professor, wrote an interesting paper: "A quote a day educates". It explains how a daily quote can "instill in students an appreciation for the human in the mathematician as well as the mathematician in the human".

From March 2 last year, I challenged myself to quote only physicists (or physics related scientists) who where born on the day I quoted them, and to provide the reference for that quote, in order that one can check the context in which it was written or said. For about a tenth of the days in the year, such quotes could already be found easily on the web, but the 90% other ones needed a lot of reading, of searching in oral histories, in archives or in online parts of books or papers with Google Books or Scholar. I now have a year long physics calendar, a bit in the same trend as the catholic saints calendar. So if anyone is interested in publishing such a calendar, I'm the man;-) I currently have about 1300 sourced physics related quotes from 800 different scientists in my collection, part of which may be found on Wikiquote, or in my past posts or tweets.

Here are the two last quotes in the Physics Quotes of the Day series on this blog. I hope, you've enjoyed it. And I'm looking for another challenge...

"Your waistline may be spreading but you can't blame it on the expansion of the universe." Richard H. Price, born 1 March 1943.

"The atoms become like a moth, seeking out the region of higher laser intensity." Steven Chu, born 28 February 1948.