$e, \varphi$ and $\pi$ in a right angle

Question

\begin{align} e &= 2.7182\ldots \\ \varphi &= 1.618 \ldots \\ \pi &= 3.1415\ldots \\ \end{align}

\begin{align} e^2 &= 7.38905\dots \\ \varphi^2 &= 2.61803\ldots \\ \implies e^2 + \varphi^2 &= 10.00708 \ldots \\ &> \pi^2 \\ &= 9.8696 \ldots \end{align}

$10.00708 - 9.8696 = 0.137485 \Rightarrow 0.1375 \Rightarrow 137.5$deg is the Golden Angle

How come $\pi$ is short by the value of the Golden Angle?

Answer 1

It is not a right triangle in your diagram above precisely because $\pi^2 \neq \varphi^2 + e^2$.

There is the Kepler triangle, which has sides $(1, \varphi, \sqrt{\varphi})$ for which $$1+\varphi = \varphi^2$$

Answer 2

Hunt for near misses for long enough and you will find them. In this case, the sum is out by more than 1% which seems too much to be interesting. The coincidental similarity to an angle in degrees is also uninteresting since it depends on the arbitrary choice of 360 degrees in a circle.