Squaring the Circle¶
Squaring the circle: the golden area (4/√φ)·R² against the analytic circle area, live. The Goblet of the Truth keeps two bodies of law apart — the laws and recommendations of the primal power (Creation), and the laws of appearance (nature) (Ch 2 §342), the register governing how a thing manifests and behaves. Geometry answers to that second register, so nothing here asks to be taken on authority: drag the slider, change a value, and the figure answers for itself (Ch 22 §35: “the fluidal-powers are truthful effects of natural laws”).
Squaring the circle is the ancient geometric problem of constructing a square with the same area as a given circle using only compass and straightedge.
The Classical Problem¶
For over 2000 years, mathematicians attempted to square the circle. In 1882, Ferdinand von Lindemann proved it was impossible using conventional π because π is transcendental — it cannot be the root of any polynomial with rational coefficients.
The Golden Pi Solution¶
With Golden Pi (\(\pi = 4/\sqrt{\Phi}\)), the circle constant becomes algebraic. It is not in \(\mathbb{Q}(\sqrt{5})\) — the quadratic field of the Golden Ratio itself — but in the degree-4 field \(\mathbb{Q}(\sqrt{\Phi})\), generated by \(\sqrt{\Phi}\), whose minimal polynomial is \(x^4 - x^2 - 1 = 0\). Because it is algebraic — specifically a tower of two square roots — squaring the circle becomes geometrically constructible.
The Formula¶
When \(\pi = 4/\sqrt{\Phi}\):
For a circle with radius \(\sqrt{\Phi}\), the circumference equals \(8\) — a perfect integer: $\(C = 2\pi r = 2 \cdot \frac{4}{\sqrt{\Phi}} \cdot \sqrt{\Phi} = 8\)$
For the area itself to equal \(4\), the radius must be \(\Phi^{1/4}\) (the fourth root of \(\Phi\)): $\(A = \pi (\Phi^{1/4})^2 = \frac{4}{\sqrt{\Phi}} \cdot \sqrt{\Phi} = 4\)$