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In a documentation about the Great Pyramid of Giza, I heared following three theses about its measurements and the numbers $\pi$ and $\phi$ (the golden ratio).

Pyramid with symbols

Measurement

The Great Pyramid of Giza had originally following measurements:

$$s = 230.33\textrm{m}$$

$$h_p = 146.59\textrm{m}$$

$$\alpha = 51°50'$$

Thesis 1:

If you divide the perimeter of the pyramid $4s$ by the height $h_p$, you will get $2\pi$ (error < 0.01%)

$$\frac{4s}{h_p} = 2\pi$$

Thesis 2:

If you divide the surface of the base $s^2$ by the of the rest of the lateral surface $4A_m$, you will get the major of the golden ratio (0.618..., error < 0.01%):

$$\frac{s^2}{4A_m} = maj. a = \phi-1$$

Thesis 3:

The cosine of the angle of the Great Pyramid $\alpha$, will result in the major of the golden ratio, too (error unknown):

$$cos(\alpha) = maj. a = \phi-1$$

Putting together thesis 1 and thesis 2, I get following formula:

$$\phi = f(\pi) = 1 + \pi \frac{\sqrt{16+\pi^2}}{16+\pi^2}$$

The very interesting part is that $s$, $A_m$ and $h_p$ have been cancelled out, so that this formula is valid for all square pyramids with $2\frac{s}{h_p} = \pi$.

  • Question: How can I prove or falsify this formula? Using my calculator, it seems that both terms are not equal, but it could be simply a rounding error of the calculator.

  • If it is true, is there a way to express $\pi = f(\phi)$ using this formula?

  • Side-question about thesis 3: Is it possible that $\alpha = arccos(\phi-1) = arccos(\frac{\sqrt{5}-1}{2})$ can be expressed as a non-trigonometrial formula or as non-transcendental number?

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    $\begingroup$ I always wonder how to even measure the dimensions of the pyramid with such great precision as interpreted into it by pyramidologists after thousands of years of erosion and uneven coverage by sand ... $\endgroup$ – Hagen von Eitzen Nov 6 '14 at 18:55
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    $\begingroup$ @HagenvonEitzen You can't prove the stated values are wrong after so much erosion, and you need that sort of "precision" to get impressive values for the approximations to $\pi$ or whatever. What's not to understand? $\endgroup$ – Daniel Fischer Nov 6 '14 at 18:57
  • $\begingroup$ I'm sorry, I have written the wrong value. The correct value is $\alpha = 51°50'$ . $\endgroup$ – Daniel Marschall Nov 6 '14 at 19:04
  • $\begingroup$ Theses 2 and 3 are the same claim, since $$\frac{s^2}{4A_m} = \cos\alpha = \frac{1}{\sqrt{1+(\frac{h_p}{s/2})^2}}$$ in every square pyramid. $\endgroup$ – Henning Makholm Nov 6 '14 at 19:09
  • $\begingroup$ ... and the claim that this equals $\phi-1$ is equivalent to simply claiming that $\frac{h_p}{\frac12 s}=\sqrt{\phi}$. $\endgroup$ – Henning Makholm Nov 6 '14 at 19:16
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$\phi$ is algebraic while $\pi$ is transcendental, so an equation such as $$ \phi = 1 + \pi \dfrac{\sqrt{16 + \pi^2}}{16 + \pi^2}$$ can't possibly be correct. Indeed, your calculator should show you that it isn't even very close.

$\arccos(\phi - 1)$ is a transcendental number by Lindemann's theorem that $e^x$ is transcendental whenever $x$ is nonzero and algebraic.

EDIT: $\arccos(\phi - 1)$ is also not a rational multiple of $\pi$. If $t = \arccos(\phi - 1)$, we have $w = \exp(it)$ satisfying $w + 1/w = 2 z$, and then $p(w) = w^4 + 2 w^3 - 2 w^2 + 2 w + 1 = 0$. This polynomial is irreducible over the rationals. Now if $t$ were a rational multiple of $\pi$, $w$ would be a root of unity, so $p(w)$ would have to be a cyclotomic polynomial, which it is not.

It is also not an algebraic multiple of $\pi$, by the Gelfond-Schneider theorem.

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  • $\begingroup$ Thank you very much for your answers. Indeed, I was assuming that this formula $\phi = f(\pi)$ was wrong, but I am still not 100% sure if it could be some rounding error, since I don't know with which precision the calculator is working. I wonder, is there a possibility to falsify this formula without using a calculator? $\endgroup$ – Daniel Marschall Nov 6 '14 at 19:51
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    $\begingroup$ I did falsify the formula without using a calculator. That's exactly what the "$\phi$ is algebraic while $\pi$ is transcendental" does. $\endgroup$ – Robert Israel Nov 6 '14 at 20:25
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You don't need to trust the eggheads on the true value of $\pi$. You can easily calculate it yourself, the same way as Archimedes did it around 2000 years ago.

Draw a circle, and then draw some regular polygon inscribed inside and circumscribed around it. (The simplest one to start with is a hexagon, since it's enough to measure the radius and mark it on a circumference six times with a pair of compasses. You can also start with a square.) Then calculate their perimeters. (It's quite easy, a bit of trigonometry would suffice.) This won't give you the value of $\pi$ yet, but it will give you a rough estimate of the boundaries in which $\pi$ has to fit. That's because the circumference of the circle is definitely smaller than the perimeter of the circumscribed polygon, but greater than the perimeter of the inscribed one. By doubling the number of sides of both polygons, you narrow the boundary in which $\pi$ has to fit, giving you better and better precision of your estimate. You can double the number of sides as many times as you please, assuring more and more decimal places of $\pi$. So you still don't know the final value of $\pi$ (and never will), but you can at least tell which numbers aren't $\pi$ for sure.

When you apply this technique to the value calculated from the infamous formula involving $\Phi$, the golden ratio (an inverse of its square root actually, multiplied by four), you'll get $3.144605511...$, and you can easily prove that this value lies far off from the allowed range. In other words, it doesn't fit between the inscribed and circumscribed polygon, so it cannot be equal to the circumference of the circle more still.

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