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Given the halfcircle equation of the unit circle

$$y = \sqrt{1 - x^2}$$

and its derivative

$$y' = \frac{-x}{\sqrt{1-x^2}} $$

why does $y' = y$ yield a solution that contains the golden ratio. Is this random or is there a simple relation here which I am overcomplicating.

Taking the full circle equation $x^2 + y^2 = 1$ and finding the intersections with the full derivative equation (equation that would describe the derivative for the whole circle) there are $4$ intersections each with a relation of $\phi - 1$.

EDIT:

I came so far by just solving the equation $$\frac{-x}{\sqrt{1-x^2}} = \sqrt{1-x^2}$$

Simplifying this yields the quadratic equation $$x^2 - x - 1 = 0$$

and a solution $$\frac{1}{2}\pm\frac{\sqrt{5}}{2}$$

Yet I still don't know why exactly the golden ratio is a solution and not just a random number leaves me clueless. Maybe there is some geometric relation...

Greg Martin
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Murad Babayev
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    Sorry, I swear I know how to write equations in Latex I just don't know how to embed them here :P – Murad Babayev Jul 25 '17 at 22:19
  • The golden ratio is the positive solution of $x^2-x-1=0$ by definition, so it just pops up when solving your equation because of simple algebra, I guess. – Angelo Rendina Jul 25 '17 at 22:25
  • Note: Since $y\sqrt{1-x^2}\ge0$, the equation $y=y'$ is satisfied only by the negative root of $x^2-x-1=0$. – Barry Cipra Jul 25 '17 at 22:42
  • Yes that is why I suggested using the full circle equation and as derivative $\frac{x}{\sqrt{1-x^2}}$ and $\frac{-x}{\sqrt{1-x^2}}$ – Murad Babayev Jul 25 '17 at 22:45

1 Answers1

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This is not a coincidence, but rather a consequence of simple geometry. Within a unit circle there will be right triangle with sides $x=1/\varphi, y=1/\sqrt{\varphi}$ and hypotenuse $R=1$. Thus, the condition $y=-y'=x/y$ is automatically satisfied at this point. What you missed was that at $x=1/\varphi$, we also find $y=1/\sqrt{\varphi}$. This has been noted previously in the literature: Gray SB, et al. (2016) The Method of Archimedes: Propositions 13 and 14. Notices of the AMS 62(9) 1036-1040.

But there is more to be seen here. If $x=1/\varphi, y=1/\sqrt{\varphi}$ has a unit hypotenuse, the so must $x=1/\sqrt{\varphi}, y=1/\varphi$. Here, we find that $y'=-1/\sqrt{y}$.

Similarly, we find that for the parabola $y=1-x^2$, that $y(1/\varphi)=1/\varphi$ and $y(1/\sqrt{\varphi})=1/\varphi^2$. But I don't find any interesting relation between $y'$ and $y$.

Cye Waldman
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