Squaring The Circle Insights

We have some new insights to share about an age old technique in mathematics called Squaring The Circle. This research was partly inspired by an illustration we found in a pdf publication written about Viktor Schauberger (1885 – 1958). In Figure 1, we can see how they oriented the squares, circles, and triangles in this particular model.

We don’t know if this was Schauberger’s own illustration or if he got it from somewhere else. For that matter, we don’t even know if he ever saw it, or if the author just used it as an illustration for his pdf when he was writing the paper. We do know that Schauberger was interested in sacred geometry however, so it’s probably safe to assume that he was familiar with the concept.

The “Hidden Nature” (2003) pdf is cited at the end of this article, so you can download it and read it for yourself. Schauberger was a fascinating character who’s contributions were right up there with the likes of Nikola Tesla, so it’s an interesting read if you want to learn more about him.

We’ll be quoting from the pdf directly where it relates to the illustration in Figure 1 (also seen on page 57 of the pdf). The author has some interesting things to say about it, and he provides a good introduction for what it means and what it represents. Beyond that, we’ll get into our own analysis of it and explore the math.

Some of this might be speculative, and we might bend the rules a little here and there. But if we need to take a leap of faith at some points along the way, we’ll be sure to point this out beforehand and explain why we’re doing it. Some of these discoveries were quite shocking to us at least, so fasten your seat belts!

We first needed to recreate the old image in Figure 1 as a graphic model, and we’re showing that below in Figure 2.

1

Squaring the Circle And Sacred Geometry

From “Hidden Nature” (2003)

“Viktor Schauberger saw patterns and rhythms as the heartbeat of the Universe, and was fascinated by the traditional use of the language of number and form to codify how they are repeated and in what form. It is hard for us, schooled in a rationalist worldview which separates form from the natural order to see that they are part of one whole. The ancients regarded mathematics and geometry as the tools to understand patterns in Nature and in the Universe. The religious leaders of old, who were also the scientists and mathematicians, did not make our mistake of putting different phenomena in separate compartments. To them, the world of matter and reason and the world of spirit and the awareness of God were all one. In the context of myth and symbol, they used numbers and forms in a way that would satisfy the spiritual sense of meaning and the scientific need for structure and reason. Out of this process arose the traditions of numerology and sacred geometry.

Sacred geometry is based on observations of cosmic relationship. The Great Pyramid’s base straddles the Earth’s equatorial diameter; its apex is at the centre of the Moon, which is in true proportion to the Earth, and held to the square by a Pythagorean 3-4-5 right-angled triangle. See Figure 2.

In all the ancient cultures, the square symbolized the Earth of matter and rationalization, and the circle the encompassing world of spirit and feeling. How to bring them into balance was called ‘squaring the circle’ and was the pursuit both of architecture and philosophy. The sum of the sides of the square was equal to the circumference of the circle, so they come into harmony by enclosing the same area. This is sometimes used as a metaphor for the balanced personality. So ‘circling the square’ indicates someone whose rationality is greater than his/her sense of feeling.

As with other problems in sacred geometry, though it is not possible to draw this relationship by simple measurement,* because it is part of the natural order, that is where the solution is to be found — in fact in the relationship of the size of the Moon to the size of the Earth. You draw a square around the circle of the Earth (each side of which will equal the Earth’s diameter). Then you draw the Moon on the same scale, sitting on top of the Earth. A circle with its centre as the centre of the Earth, and its circumference passing through the centre of the Moon will have a circumference equal to the sum of the sides of the square enclosing the Earth.

Figure 2 also contains the 3-4-5 Pythagorean triangle which connects the corners of the Earth and Moon squares. It was from such relationships that the ‘Pythagorean canon of proportions’ was created. The basics of musical harmony depend on intervals created by these divine proportions. There were canons of architecture, of painting and of musical harmony taught in the medieval mystery schools, and partly revived in the Renaissance.”

*We should point out that from what we’ve seen, everything in sacred geometry can actually be measured very precisely with perfect mathematical validity. So we’re not sure which “other problems” the author is referring to.

Some theories about the Great Pyramid of Giza suggest that other proportions might have been used. In particular, check out our research on the Kepler Triangle. This article presented one such possibility known as the Golden Pyramid, which is formed by using a doubled Kepler Triangle as the two perpendicular cross sections of the base.

2

New Squaring the Circle Diagram

The first thing we need to do is rebuild the original Squaring the Circle Diagram in Figure 1. We’ve done this in Figure 3, where we’ve given each of the shapes new names so we can analyze these elements separately.

We can now define these as different sets of shapes. For the Circles, we have ${\mathcal{C}}_1, {\mathcal{C}}_2$, and ${\mathcal{C}}_3$. For the Squares, we have ${\mathcal{S}}_1$ and ${\mathcal{S}}_2$. And for the Triangles, we have ${\mathcal{T}}_1, {\mathcal{T}}_2$, and ${\mathcal{T}}_3$.

The one addition we made to the original diagram in Figure 1 is the top Rectangle ${\mathcal{R}}$. This shape turns out to be one of the bigger missing pieces of the puzzle, and we’ll be discussing that more later.

The Viktor Schauberger document only gave us just enough information to reconstruct all of the dimensions we’ll need before we can see how squaring the circle works. The first thing to notice is the two triangles ${\mathcal{T}}_2$ and ${\mathcal{T}}_3$ are both Pythagorean 3-4-5 right triangles, and this gives us the dimensions for all of the other elements.

According to the article segment we cited above, “The sum of the sides of the square was equal to the circumference of the circle.” We think it’s an interesting choice of words to say “was,” instead of “is” equal to the circumference of the circle. Because this is math, so either it is or it isn’t. Right? So what went wrong?

3

Revisiting the Number Pi

It turns out the problem we mentioned above with Squaring the Circle is encountered the minute you introduce the number pi. The sum of sides of the square ${\mathcal{S}}_2$ is supposed to be equal to the circumference of the circle ${\mathcal{C}}_3$. But what do we need to calculate the circumference of a circle? Well, pi of course, since the circumference of ${\mathcal{C}}_3$ is given by,

\begin{align}
{\mathcal{C}}_{3_c} = 2\pi r_2 \\
\end{align}

Where Does Pi Come From?

So where do we get pi? How does it work and where does it come from? When you think about it, what pi is actually doing is quite fascinating. In radians, pi is considered to be 180 degrees or half of a circle. This is why you’re multiplying by 2 to get the full 360 degrees for the whole circle. So basically all you’re doing is multiplying pi by the radius of a circle, and it magically produces a measurement for half of the circumference!

So at very least, it seems to us like pi should be some kind of a number with harmonic or oscillating properties. And you would expect it to at least be a repeating decimal. In other words, wouldn’t we expect it to be something intrinsically beautiful or meaningful, in at least some small way, when it’s representing the most fundamentally important shape in all of geometry? But no, the generally accepted value for pi doesn’t give us this satisfaction at all, in any way shape or form!

In fact, we have a video referenced below called “Mile of Pi” where the presenter has printed the currently accepted value of pi on a long narrow sheet of brown paper. They rolled this out on a local runway, and it was about a mile long! What’s worse is, they couldn’t find any repeating patterns anywhere in the number whatsoever!

Pi is an Approximation

We hope you can see from this that pi is only ever an approximation at best, and it isn’t terribly useful to us when we need to memorize a number that’s over a mile long! But putting all of that aside, let’s go ahead and use the value for pi that our calculator gives us, and see how squaring the circle was supposed to work.

First, we need the perimeter of the square ${\mathcal{S}}_2$.

\begin{align}
{\mathcal{S}}_{2_{{\mathcal{P}}}} = 4 \times 11 = 44 \\
\end{align}

This should be equal to the circumference of circle ${\mathcal{C}}_3$. From Figure 3, we can see that $ \; r_2 = 11/ 2 + 3 / 2 = 7$. So plugging this into our calculator, we get something that looks like the following,

\begin{align}
{\mathcal{C}}_{3_c} = 2\pi 7 = 14\pi \cong 43.9822971503 \\
\end{align}

So now we can see that ${\mathcal{C}}_{3_c}$ is indeed actually very close to ${\mathcal{S}}_{2_{{\mathcal{P}}}}$. In fact it’s so close, could we go ahead and solve for a new version of $\pi$ based on our Squaring the Circle diagram? If $\pi$ itself is only ever an approximation anyway, why not give it a try and see what that looks like?

New Value of Pi with Harmonic Properties

If we can pretend for a moment that ${\mathcal{S}}_{2_{{\mathcal{P}}}}$ is exactly equal to ${\mathcal{C}}_{3_c} \; $, then,

\begin{align}
44 = 14\pi \\\\
\pi = \frac{44}{14} = \frac{22}{7}
\end{align}

So there we have it, and at least for the purposes of our diagram, we’ll plan on using this version of pi to see if anything else interesting shakes out from the model. The numbers 22 and 7 have fascinating properties, and these are more like the numbers we would expect from something as wholesome and unified as a circle!

If we look at these numbers on a calculator, let’s give the generally accepted pi its usual symbol $\pi$, and we’ll call the new pi, $\pi_h$.

\begin{align}
\pi = 3.14159265359 \\\\
\pi_h = 3.14285714286 \\
\end{align}

So these numbers are quite close, and the new pi is giving us a repeating decimal pattern of 142857. So we could think of this number as being harmonic or oscillating. With that in mind, we wrote the $h$ subscript in $\pi_h$ to mean “pi harmonic.”

Unusual Characteristics of 7 and 22

We’re posting a video below called “Why 7 is Weird,” and it shows an interesting trick for figuring out whether a number is divisible by 7. Then he uses a different technique for the number 11 (which is a multiple of 22). And determining whether a number is divisible by 11 is done with “alternating sums.” So this suggests that the number 11 has oscillating properties, and both of these numbers have unusual characteristics compared with other numbers.

Archimedes Upper Bound For Pi

When we first discovered this, we wondered if anyone else had come across the number $22/7$ for $\pi$. And the astonishing answer is yes. In fact it was Archimedes no less, so we were in good company! And it’s interesting that we found his exact same value using a completely different method.

Archimedes calculated that $\pi$ lies between the values of $22/7$ (approximately 3.142857) as an upper bound, and $223/71$ (approximately 3.140845) as a lower bound, essentially proving that $3.140845 \leq \pi \leq 3.142857$.

He achieved this by inscribing and circumscribing regular polygons with increasing numbers of sides around a circle, then calculating the perimeters of these polygons to determine upper and lower limits for the circle’s circumference, which directly relates to the value of pi.

Pi is still being calculated the same way to this day, using super computers to crunch out smaller and smaller polygons. But doesn’t that seem like it could be kind of a mess? After summing up all of those tiny little slices, what kind of a round off error are you getting? And the fact is, you’ll never really get there anyway because the curve of a circle has an infinite number of points!

Narrowing Down What Pi Isn’t

One thing that is fairly easy to prove is what Pi isn’t. If you can place a large circle on a grid with tiny squares, you can easily calculate the sum of these tiny square areas. Using this method with a sufficiently large number of tiny squares, you can get a sense what pi couldn’t possibly be. Because if your squares will more than cover the area of the circle, you’ll know that any value greater than that needs to be thrown out. Also notice that this technique is attempting to narrow down the upper bound for pi as opposed to the lower bound.

The video below shows how you can see this for yourself using a free online calculator application to create your own circle and grid. The squares have a scaling slider, so you can adjust the number of squares you want to calculate.

Formula For Estimating Lower Bound of Pi

This next video uses a similar technique for evaluating pi, but this time it’s attempting to narrow down the lower bound. This approach is more accurate since it’s showing an actual formula that can be used recursively to find smaller slices of the circle. The shapes of these slices are also giving us a better approximation than the tiny squares in the previous video. This formula can be programmed into an Excel spreadsheet to show the output depending on the number of slices.

How Archimedes Established Upper & Lower Bounds

Then we finally have an explanation for how Archimedes established both an upper and lower bound for pi. Keep in mind that he was doing all of this by hand, and he had to use estimates for square roots. So with our modern technology we’ve been able to greatly improve upon these results.

Improving Archimedes’ Approximation

Since evaluating $\pi$ is still done by narrowing down the upper and lower bounds, we should be taking a look at what sits right in the middle of those boundaries. Using Archimedes’ approximation for example, we can take the average of those two numbers to see what’s right in the middle,

\begin{align}
\frac{\frac{223}{7} + \frac{22}{7}}{2} = 3.141851 \\\\
\end{align}

This is getting close to the standard $\pi$, but we were able to get it much closer than that in our Gabriel’s Horn article — in the section about Improving Archimedes’ Approximation.

4

Further Squaring The Circle Analysis

We’ve seen how squaring the circle works according to the limited instructions we were given from the article cited above. This was based on the circumference of circle ${\mathcal{C}}_3$ being equal to the perimeter of square ${\mathcal{S}}_2$. But rather than the perimeters, what about looking at the areas of these shapes? And what about exploring the shapes we haven’t looked at yet?

If we use our new definition of $\pi_h$ for pi, we would be examining the original diagram in its pure mathematical form. And this could have been the way we were intended to see it from the beginning. In addition, our new pi is a more manageable number than the generally accepted pi (the mile long version discussed above). So it’s a perfect approximation to use for further analysis, and we know it’s very close to the generally accepted version of pi. If anything interesting happens in the pure form using $\pi_h$, we already know it’s very close to reality. That is, if we consider reality to be the generally accepted version of pi in the first place!

We can start by looking at the areas of ${\mathcal{S}}_2$ and ${\mathcal{C}}_3$, and see if any relationships still hold true,

\begin{align}
{\mathcal{S}}_{2_{{\mathcal{a}}}} = 11 \times 11 = 121 \\\\
{\mathcal{C}}_{3_{{\mathcal{a}}}} = \pi_h(r_2)^2 = \frac{22}{7} \times 49 = 154 \\
\end{align}

So ${\mathcal{S}}_{2_{{\mathcal{a}}}}$ is not equal to ${\mathcal{C}}_{3_{{\mathcal{a}}}}$, and we can’t say the area of the square is equal to the area of the circle like we can with the perimeters of these shapes. But does that area relationship still hold true in some other form?

Using Rectangles Instead of Squares

Mathematicians like reducing problems to squares since their areas are so easy to calculate by squaring one of the sides. But another handy shape they like to use is a rectangle, since the area of a rectangle is also easy to manage by simply multiplying the length by the width.

You’ll notice in Figure 5 that we added the rectangle ${\mathcal{R}}$ at the top of the diagram. If we add this to the square ${\mathcal{S}}_2$, we’ve created a new rectangle which we can call ${\mathcal{R}}_2$. If we let ${\mathcal{R}}_{{\mathcal{a}}}$ be the area of the top rectangle ${\mathcal{R}}$, this would be $11 \times 3 = 33$.

To get the area of ${\mathcal{R}}_2$, we can simply add the areas of ${\mathcal{S}}_2$ and ${\mathcal{R}}$, or we can multiply the length times the width of ${\mathcal{R}}_2$,

\begin{align}
{\mathcal{R}}_{2_{{\mathcal{a}}}} = {\mathcal{R}}_a + {\mathcal{S}}_{2_{{\mathcal{a}}}} = 33 + 121 = 154 \\\\
{\mathcal{R}}_{2_{{\mathcal{a}}}} = 14 \times 11 = 154 \\\\
\end{align}

So by “Rectangling the Circle” rather than “Squaring the Circle,” we’re getting another very interesting relationship showing how the areas of ${\mathcal{R}}_2$ and ${\mathcal{C}}_3$ are also equal!

\begin{align}
{\mathcal{R}}_{2_{{\mathcal{a}}}} = \; {\mathcal{C}}_{3_{{\mathcal{a}}}} = 154 \\
\end{align}

5

Squaring The Circle Areas of Shapes

Now that we’ve established a relationship between the areas of shapes in the squaring the circle diagram, things become much more interesting. The relationship between the perimeter of a square being equal to the circumference of a circle is valuable because it sets the stage. But there is only so much you can do with peripheral outlines of shapes.

When you can look at areas however, more things begin to open up. As we illustrated in the section above, adding one area to another creates a new shape with a new area. So we added a rectangle ${\mathcal{R}}$ to a square ${\mathcal{S}}_2$ to create a new rectangle ${\mathcal{R}}_2$. We can also subtract areas of shapes to form new shapes as well. This creates a nice way to visualize what’s going on, in addition to providing visual proofs in some cases.

Areas of Squares

We should start by jotting down some of these areas to see what their numbers look like. First, the areas of squares ${\mathcal{S}}_1$ and ${\mathcal{S}}_2$ are defined below,

\begin{align}
{\mathcal{S}}_{1_{{\mathcal{a}}}} = 3 \times 3 = 9 \\\\
{\mathcal{S}}_{2_{{\mathcal{a}}}} = 11 \times 11 = 121 \\\\
\end{align}

Areas of Circles

The areas of circles ${\mathcal{C}}_1$, ${\mathcal{C}}_2$, and ${\mathcal{C}}_3$ are,

\begin{align}
{\mathcal{C}}_{1_{{\mathcal{a}}}} = \pi_hr^2 = \frac{22}{7} \times \frac{9}{4} = \frac{99}{14} \\\\
{\mathcal{C}}_{2_{{\mathcal{a}}}} = \pi_h(r_1)^2 = \frac{22}{7} \times \frac{121}{4} = \frac{1331}{14} \\\\
{\mathcal{C}}_{3_{{\mathcal{a}}}} = \pi_h(r_2)^2 = \frac{22}{7} \times 49 = 154 \\\\
\end{align}

Areas of Rectangles

We already have the areas of rectangles ${\mathcal{R}}$ and ${\mathcal{R}}_2$, but we’ll include them again here so we have the list of areas in one place,

\begin{align}
{\mathcal{R}}_{2_{{\mathcal{a}}}} = 154 \\\\
{\mathcal{R}}_a = 33 \\\\
\end{align}

We can also look at the rectangles defined by doubling the Pythagorean 3-4-5 right triangles ${\mathcal{T}}_2$ and ${\mathcal{T}}_3$. We can call one of these rectangles $2{\mathcal{T}}_{2_{{\mathcal{a}}}}$ or $2{\mathcal{T}}_{3_{{\mathcal{a}}}}$ since the two triangles are equal.

\begin{align}
2{\mathcal{T}}_{2_{{\mathcal{a}}}} = 2{\mathcal{T}}_{3_{{\mathcal{a}}}} = 3 \times 4 = 12 \\\\
\end{align}

Checking the Diagonals of ${\mathcal{R}}_2$

We should also check the diagonals of ${\mathcal{R}}_2$ to see if anything interesting falls out. If we let the diagonal of ${\mathcal{R}}_2$ be ${\mathcal{R}}_{2_{{\mathcal{d}}}}$, then,

\begin{align}
{\mathcal{R}}_{2_d}^2 = 11^2 + 14^2 = 317 \\\\
{\mathcal{R}}_{2_d} = \sqrt{317} \\\\
\end{align}

We also checked the side lengths of triangle ${\mathcal{T}}_1$, and found these to be $\sqrt{317}/ 2$. The number 317 is prime, so it can’t be reduced any further. And if we look at the numerology on 317, we have,

\begin{align}
3 + 1 + 7 = 11 \\\\
\end{align}

So there is our oscillating number 11 again. We’re not sure if there’s any real significance to that or not, but it’s kind of interesting.

Areas of Triangles

To find ${\mathcal{T}}_{2_{{\mathcal{a}}}}$ and ${\mathcal{T}}_{3_{{\mathcal{a}}}}$, all we have to do is divide one of the doubled Pythagorean triangles defined earlier in half,

\begin{align}
{\mathcal{T}}_{2_{{\mathcal{a}}}} = {\mathcal{T}}_{3_{{\mathcal{a}}}} = 2{\mathcal{T}}_{2_{{\mathcal{a}}}} \div 2 = 12 \div 2 = 6 \\\\
\end{align}

For ${\mathcal{T}}_{1_{{\mathcal{a}}}}$, we can just use the standard formula for the area of a triangle,

\begin{align}
{\mathcal{T}}_{1_{{\mathcal{a}}}} = \frac{1}{2}bh = \frac{11}{2} \times 7 = \frac{77}{2} \\\\
\end{align}

Checking the Base Angles of ${\mathcal{T}}_1$

If we let $\theta$ be one of the base angles of ${\mathcal{T}}_1$, this is easy to check by taking the sine inverse of the opposite side, or $r_2$ in this case, divided by the hypotenuse or side length of ${\mathcal{T}}_1$,

\begin{align}
\theta = \sin^{- 1}\left(\frac{14}{\sqrt{317}}\right) = 51.84277341^{\circ} \\\\
\end{align}

Triangle Area Relationship With Circle Area

At first glance, we didn’t see the significance of triangle ${\mathcal{T}}_1$. It’s not equilateral, and it’s even more scrunched in the vertical direction than the facing triangles of the golden pyramid in our article about Kepler’s Triangle.

In fact, ${\mathcal{T}}_1$ is closer to the pentagonal triangles we discussed in our article about the Five Pointed Star Reference Shapes, since one of those has a base angle of 54 degrees. So ${\mathcal{T}}_1$ is more like a pentagonal triangle slightly scrunched down in the vertical direction.

Then we noticed another astonishing relationship with the areas of ${\mathcal{T}}_1$ and ${\mathcal{C}}_3$.

\begin{align}
4{\mathcal{T}}_{1_{{\mathcal{a}}}} = 4\frac{77}{2} = 154 = {\mathcal{R}}_{2_{{\mathcal{a}}}} = {\mathcal{C}}_{3_{{\mathcal{a}}}} \\\\
\end{align}

You can visualize this as we’re showing in Figure 7, by looking at rectangle ${\mathcal{R}}_2$ and triangle ${\mathcal{T}}_1$. By doubling and inverting ${\mathcal{T}}_1$, you have a diamond shape from the center top to bottom of ${\mathcal{R}}_2$. Then dividing ${\mathcal{T}}_1$ in half leaves us with two right triangles. Each of these right triangles will fit into the corners of ${\mathcal{R}}_2$.

So, $4{\mathcal{T}}_{1_{{\mathcal{a}}}}$ gives us everything we need to perfectly fill the space of ${\mathcal{R}}_{2_{{\mathcal{a}}}}$, thus making it equal to ${\mathcal{C}}_{3_{{\mathcal{a}}}}$.

Triangling a Circle

We can also see from Figure 7 how we could construct a single similar triangle to ${\mathcal{T}}_1$, formed by reorienting the two right triangles on either side of the diagonal of ${\mathcal{R}}_2$. And this would also give us the area of ${\mathcal{C}}_3$. This single combined triangle would have a base length of $22$, side lengths of $\sqrt{317}$, and a height of $14$. So if we wanted this relationship reduced to a single triangle, we could call that ${\mathcal{T}}_4$, and our new relationship would be the following,

\begin{align}
{\mathcal{T}}_{4_{{\mathcal{a}}}} = {\mathcal{R}}_{2_{{\mathcal{a}}}} = {\mathcal{C}}_{3_{{\mathcal{a}}}} \\\\
\end{align}

By now, we can Square a Circle, Rectangle a Circle, and even Triangle a Circle!

6

Squaring The Circle Area Exploration

This next section is where things get even more interesting, as we begin to drill down into the areas of these shapes. By adding or removing areas we can quickly visualize what’s going on, and some of the results are quite surprising.

The first thing we need to add is a third square positioned on the outside of ${\mathcal{C}}_3$. We can call this square ${\mathcal{S}}_3$ as shown in Figure 8. Its side lengths measure $14 \times 14$, so the area is ${\mathcal{S}}_{3_{{\mathcal{a}}}} = 196$.

Subtracting ${\mathcal{S}}_2$ From ${\mathcal{S}}_3$

One of the things we’re interested in is the area outside of ${\mathcal{S}}_2$ but inside of ${\mathcal{S}}_3$,

\begin{align}
{\mathcal{S}}_{3_{{\mathcal{a}}}} – \; {\mathcal{S}}_{2_{{\mathcal{a}}}} = 196 \; – \; 121 = 75 \\\\
\end{align}

The small white squares in the corners of ${\mathcal{S}}_3$ have an area of $3/2 \times 3/2 = 9/4$, and there are 4 of these with a total area of $9$. So we can remove these from the result above,

\begin{align}
{\mathcal{S}}_{3_{{\mathcal{a}}}} – \; {\mathcal{S}}_{2_{{\mathcal{a}}}} – \; 9 = 75 \; – \; 9 = 66 \\\\
\end{align}

The small white squares in the corners of ${\mathcal{S}}_3$ have a total area of $9$, and this is also equal to the area of ${\mathcal{S}}_1$. So the above expression can be written as,

\begin{align}
{\mathcal{S}}_{3_{{\mathcal{a}}}} – \; {\mathcal{S}}_{2_{{\mathcal{a}}}} – \; {\mathcal{S}}_{1_{{\mathcal{a}}}} = 75 \; – \; 9 = 66 \\\\
\end{align}

This is beginning to make sense when you consider that rectangle ${\mathcal{R}}$ was added to ${\mathcal{S}}_2$ to get the area of ${\mathcal{C}}_3$. What we have now are 4 narrow strips on each side of ${\mathcal{S}}_2$ measuring $3/2 \times 11 = 33/2$. But only 2 of these strips are needed to give us the area of ${\mathcal{R}}$ which is $33$.

Subtracting ${\mathcal{C}}_3$ From ${\mathcal{S}}_3$

Next, we can look at what happens when we subtract circle ${\mathcal{C}}_3$ from our new square ${\mathcal{S}}_3$. Also remembering to remove the small white square areas from the corners of ${\mathcal{S}}_3$, which are also equal to the total area of ${\mathcal{S}}_1$,

\begin{align}
{\mathcal{S}}_{3_{{\mathcal{a}}}} – \; {\mathcal{C}}_{3_{{\mathcal{a}}}} = 196 \; – \; 154 = 42 \\\\
{\mathcal{S}}_{3_{{\mathcal{a}}}} – \; {\mathcal{C}}_{3_{{\mathcal{a}}}} – \; {\mathcal{S}}_{1_{{\mathcal{a}}}} = 42 \; – \; 9 = 33 \\\\
\end{align}

This result is shown in Figure 9. What we have now are the narrow strips on each side of ${\mathcal{S}}_3$ as before, but with ${\mathcal{C}}_3$ removed. This time all 4 modified strips are adding up to the area of ${\mathcal{R}}$ or $33$. But notice that subtracting ${\mathcal{C}}_3$ has also left a small remainder (shown in blue) in the corners of ${\mathcal{S}}_2$.

The key thing to notice is how subtracting ${\mathcal{S}}_2$ from ${\mathcal{S}}_3$ and subtracting ${\mathcal{C}}_3$ from ${\mathcal{S}}_3$ are both giving us the same relationship to the area of our top rectangle ${\mathcal{R}}$. We’re subtracting $9$ or ${\mathcal{S}}_{1_{{\mathcal{a}}}}$ from both expressions to remove the corner squares of ${\mathcal{S}}_3$, and we’re dividing the first expression by two. The first expression is giving us solid rectangular sections (which means we’re getting twice the area we need), while the second expression has the circular sections removed along with the blue remainders.

\begin{align}
{\mathcal{R}}_a = \frac{{\mathcal{S}}_{3_{{\mathcal{a}}}}- {\mathcal{S}}_{2_{{\mathcal{a}}}}- {\mathcal{S}}_{1_{{\mathcal{a}}}}}{2} = \; {\mathcal{S}}_{3_{{\mathcal{a}}}}- {\mathcal{C}}_{3_{{\mathcal{a}}}}- {\mathcal{S}}_{1_{{\mathcal{a}}}} = 33 \\\\
\end{align}

Algorithmic Expression For Red, Green, & Blue Areas

If we combine the results of Figure 8 and Figure 9, we can define the red ares as ${\mathcal{Red}}$, the green areas of the circular segments as ${\mathcal{Green}}$, and the little blue remainders in the corners of ${\mathcal{S}}_2$ as ${\mathcal{Blue}}$. Then we can create an algorithmic expression and solve for each shape in various ways,

\begin{align}
4({\mathcal{Red}} + {\mathcal{Blue}}) = 2({\mathcal{Red}} + {\mathcal{Green}}) = 33 \\\\
2{\mathcal{Red}} + 2{\mathcal{Blue}} = {\mathcal{Red}} + {\mathcal{Green}} = 33/2 \\\\
{\mathcal{Red}} = 33/2 \; – \; {\mathcal{Green}}\\\\
{\mathcal{Green}} = 33/2 \; – \; {\mathcal{Red}}\\\\
2{\mathcal{Blue}} = 33/2 \; – \; 2{\mathcal{Red}} \\\\
{\mathcal{Red}} = {\mathcal{Green}} \; – \; 2{\mathcal{Blue}} \\\\
{\mathcal{Green}} = {\mathcal{Red}} + 2{\mathcal{Blue}} \\\\
\end{align}

This shows the relationships between these three shapes, but we would need to get a number for at least one of them before we could solve for the other two. We calculated the side lengths of the blue remainders and found them to be $1.169872981$. This tells us we might not find anything too exciting about calculating these actual numbers.

What’s probably more interesting in this case is just the equalities we’re seeing between the red/green, and red/blue combinations. But it’s an intriguing problem to tackle, and we should have the numbers for those areas anyway. Because you never know, they might pop up in another relationship somewhere else.

Calculating The Green ${\mathcal{C}}_3$ Circle Section Area

Fortunately, we can see a nice way to solve for one of the green sections of the circle ${\mathcal{C}}_3$.

In Figure 10, the idea is to get the length $a$ of right triangle ${\mathcal{T}}$, and then find the area of $2{\mathcal{T}}$. As you can see, the complete sector area would be defined by $2{\mathcal{T}}_a$ plus one ${\mathcal{Green}}$ section. So to find the ${\mathcal{Green}}$ area, all we need to do is subtract $2{\mathcal{T}}_a$ from the full sector.

We can call the full sector area ${\mathcal{Sect}}_a$, and calculate that as follows,

\begin{align}
{\mathcal{Sect}}_a = \frac{2(\theta) }{360}\pi_hr_2^2 \\\\
\end{align}

Then we can call the green area $ \; {\mathcal{Green}}_a$, and we would get that by subtracting $2{\mathcal{T}}_a$ from ${\mathcal{Sect}}_a$,

\begin{align}
{\mathcal{Green}}_a = {\mathcal{Sect}}_a \; – \; 2{\mathcal{T}}_a \\\\
\end{align}

So first we need to find $\theta$ in Figure 10, since twice that angle will show us what percentage of a full circle ${\mathcal{Sect}}_a$ represents. We can use the Cosine rule for that, by dividing the adjacent length $r_1$ by the hypotenuse $r_2$,

\begin{align}
\cos \left(\theta \right) = \frac{r_1}{r_2} = \frac{11}{14} \\\\
\theta = \cos^{- 1}\left(\frac{11}{14}\right) = 38.2132107^{\circ} \\\\
\end{align}

Now we can plug that into our formula above for ${\mathcal{Sect}}_a$,

\begin{align}
{\mathcal{Sect}}_a = \frac{76.4264214}{360}\left(\frac{22}{7}\right)49 = 32.69352471 \\\\
\end{align}

The next step is to find length $a$ so we can find the area of $2{\mathcal{T}}_a$. From Figure 10 we can see that,

\begin{align}
r_2^2 = r_1^2 + a^2 \\\\
a^2 = r_2^2 \; – \; r_1^2 \\\\
a^2 = 49 \; – \; \frac{121}{4} = \frac{75}{4} \\\\
a = \frac{\sqrt{75}}{\sqrt{4}} = \frac{\sqrt{3\left(25\right)}}{2} = \frac{5\sqrt{3}}{2} \\\\
\end{align}

Now we can find the area of $2{\mathcal{T}}$ with a standard formula for the area of a triangle,

\begin{align}
2{\mathcal{T}}_a = \frac{1}{2}bh = \left(\frac{5\sqrt{3}}{2}\right)\frac{11}{2} = \frac{55\sqrt{3}}{4} \\\\
\end{align}

Solving For Red, Green, & Blue Section Areas

Then finally, we can solve our formula for ${\mathcal{Green}}_a$, and we’ll use that result to also find ${\mathcal{Red}}_a$ and ${\mathcal{Blue}}_a$

\begin{align}
{\mathcal{Green}}_a = {\mathcal{Sect}}_a \; – \; 2{\mathcal{T}}_a = 32.69352471 \; – \; \frac{55\sqrt{3}}{4} \\\\
{\mathcal{Green}}_a = 8.87782611 \\\\
{\mathcal{Red}}_a = \frac{33}{2} \; – \; {\mathcal{Green}}_a = 7.62217389 \\\\
2{\mathcal{Blue}}_a = \frac{33}{2} \; – \; 2{\mathcal{Red}}_a = 1.25565222 \\\\
{\mathcal{Blue}}_a = \frac{1\ldotp 25565222}{2} = 0.62782611 \\\\
\end{align}

Subtracting ${\mathcal{C}}_2$ From ${\mathcal{S}}_2$

We get another interesting relationship when we subtract circle ${\mathcal{C}}_2$ from square ${\mathcal{S}}_2$,

\begin{align}
{\mathcal{S}}_{2_{{\mathcal{a}}}} – \; {\mathcal{C}}_{2_{{\mathcal{a}}}} = 121 \; – \; \frac{1331}{14} = \frac{363}{14} \\\\
\end{align}

Something unexpected happens when we subtract this result from the area of our top rectangle ${\mathcal{R}}$,

\begin{align}
{\mathcal{R}}_{{\mathcal{a}}} \; – \; ({\mathcal{S}}_{2_{{\mathcal{a}}}} – \; {\mathcal{C}}_{2_{{\mathcal{a}}}}) = 33 \; – \; \frac{363}{14} = \frac{462 \; – \; 363}{14} = \frac{99}{14} \\\\
\end{align}

If you’ll recall from our earlier area calculations, this result is also equal to the area of our smaller top circle ${\mathcal{C}}_1$,

\begin{align}
{\mathcal{C}}_{1_{{\mathcal{a}}}} = {\mathcal{R}}_{{\mathcal{a}}} \; – \; ({\mathcal{S}}_{2_{{\mathcal{a}}}} – \; {\mathcal{C}}_{2_{{\mathcal{a}}}}) = \frac{99}{14} \\\\
\end{align}

In other words, the remainder of ${\mathcal{S}}_{2_{{\mathcal{a}}}} \; – \; {\mathcal{C}}_{2_{{\mathcal{a}}}}$ is equal to our top rectangle ${\mathcal{R}}$ minus the top circle ${\mathcal{C}}_1$!

\begin{align}
{\mathcal{S}}_{2_{{\mathcal{a}}}} – \; {\mathcal{C}}_{2_{{\mathcal{a}}}} = {\mathcal{R}}_{{\mathcal{a}}} \; – \; {\mathcal{C}}_{1_{{\mathcal{a}}}} = \frac{363}{14} \\\\
\end{align}

We’re illustrating this in Figure 11, where we’re inverting and stacking the $ \; 2({\mathcal{Red}} + {\mathcal{Green}}) = 33 \; $ relationship in our top rectangle ${\mathcal{R}}$. We’re not sure if the eye shape is significant or not, but the math uncovered that as well. And the same is true with all of these relationships. If the math didn’t tell us what was there, we wouldn’t be able to produce any of these results!

Conclusion

We have to conclude that this research has had us wondering about how we could be seeing this many precise relationships in our Squaring the Circle model, only because of using just a slightly different version of pi. The generally accepted version would have had almost exactly the same results, and we would be looking at very close outcomes in every case.

Could it be that Archimedes was right, and there actually is an upper and lower boundary in the number pi? We think it’s possible, and our article about the Multiplicative Inverse And Gabriel’s Horn hints at that possibility as well. We get the “transcendental number” argument, and we’ve seen the proofs stating that pi is irrational. But the curvature of a circle has an infinite number of points, and we would be cautious about making too many assumptions about the nature of infinity.

Shortly after Charles Hermite proved that $e$ was transcendental in 1873, Georg Cantor proved that “infinity comes in different sizes.” So what classification of infinity does a circle belong in, and have we discovered that yet? We think this could still be a potentially wide open problem, so we’re not investing in any final conclusions about pi quite yet.

We also discovered an approximation for pi which is even closer than the Archimedes upper bound in our Heptagon Seven Pointed Star Geometry article. Some might think the search for a true value of pi is a waste of time. But even if you never find it, we still think it’s an invaluable exercise. Because just the process of searching itself will most likely lead you to discovering many other fascinating relationships that you weren’t even looking for!

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Cite This Page As:

The Math Zone. “Squaring The Circle Insights.” From MathZone.io — A Modern Exploration of Ancient Mathematics. https://mathzone.io/squaring-the-circle-insights/

Work Cited

Bartholomew, Alick (2003). Hidden Nature, The Startling Insights of Viktor Schauberger. Retrieved 1:11 PM, October 1, 2024, from United Diversity website: https://library.uniteddiversity.coop/Water_and_Sanitation/Hidden_Nature-The_Startling_Insights_of_Viktor_Schauberger.pdf

Video Reference

Mile of Pi – Numberphile
https://www.youtube.com/watch?v=0r3cEKZiLmg&ab_channel=Numberphile

Why 7 is Weird – Numberphile
https://www.youtube.com/watch?v=UDQjn_-pDSs&ab_channel=Numberphile