The Two Squaring-the-Circle Graphs: What the Instrumentum Identity Really Returns¶

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Two nearly identical Desmos graphs of the squaring-the-circle area match differ in exactly one number — the circle's radius — and that one number changes everything. Decoding them side by side shows precisely where the Golden Pi construction is exact, and what happens when its defining identity is fed a foreign radius.

The construction (shared by both graphs)¶

A square of half-side \(s\) (area \(A_s = (2s)^2\), perimeter \(P = 8s\)) is drawn together with a circle of radius \(r\). The circle constant is not hard-coded; it is computed from the instrumentum identity

\[ \pi_t \;=\; \frac{4\,s^2 r^4}{(r^4 + s^4)\,\sqrt{r^4 - s^4}} . \]

For \(s = 1\), this is the two-variable instrumentum form evaluated at \((s^2, r^2)\). The circle area is \(A_c = \pi_t \, r^2\), and the two figures are "squared" when \(A_c = A_s = 4\).

Graph 1 — the golden radius: the match is exact¶

In the first graph the radius is set to the golden fourth root

\[ r \;=\; s\,\varphi^{1/4} \;=\; 1.12783848556, \qquad r^4 = \varphi . \]

Feeding \(r^4 = \varphi\), \(s^4 = 1\) into the instrumentum identity:

\[ \pi_t \;=\; \frac{4\varphi}{(\varphi+1)\sqrt{\varphi-1}} \;=\; \frac{4}{\sqrt{\varphi}} \;=\; 3.14460551103 \;=\; \hat{\pi}. \]

Then the circle area is

\[ A_c \;=\; \hat{\pi}\,r^2 \;=\; \frac{4}{\sqrt{\varphi}}\cdot\sqrt{\varphi} \;=\; 4 \;=\; A_s. \]

The circle and square have exactly equal area (verified to full machine precision — this is a true equality, not a rounded display). The circumference is \(C = 2\hat{\pi}r = 8/\varphi^{1/4} = 7.0932142345\). Under the conventional \(\pi\), the same circle would have area \(\pi r^2 = 3.99617 \neq 4\) — short by ~0.096%. So this squaring is achieved only by Golden Pi.

Graph 2 — the classical radius: the match breaks¶

The second graph is a copy of the first, but the radius is changed to the classical squaring radius

\[ r \;=\; s\,\sqrt{\frac{4}{\pi}} \;=\; \frac{2}{\sqrt{\pi}} \;=\; 1.1283791671, \]

the value chosen so that conventional \(\pi\) makes the areas equal (\(\pi \cdot (4/\pi) = 4\)).

But the graph keeps the instrumentum identity for the circle constant. At this radius, the identity no longer returns Golden Pi — it returns

\[ \pi_t \;=\; 3.13903247529, \]

a number that is neither conventional \(\pi\) (3.14159…) nor Golden Pi (3.14461…). Consequently

\[ A_c \;=\; \pi_t\,r^2 \;=\; 3.99674 \;\neq\; 4 \;=\; A_s . \]

The circle and square no longer have equal area. The "squaring" fails.

What this demonstrates¶

The two constructions are mutually exclusive — you cannot mix their parts:

Framework	Radius	Circle constant	Area match?
Golden Pi	\(r = \varphi^{1/4}\)	\(\pi_t = 4/\sqrt{\varphi} = 3.14460551103\)	yes (\(A_c = 4\))
Classical	\(r = 2/\sqrt{\pi}\)	\(\pi = 3.14159265\ldots\)	yes (\(A_c = 4\))
Mixed (the copy)	\(r = 2/\sqrt{\pi}\)	\(\pi_t = 3.13903247529\)	no (\(A_c = 3.99674\))

The instrumentum identity only returns Golden Pi at its own characteristic ratio (\(r^4 = \varphi\)). At any other input it is just a different number — feeding it a radius derived from conventional \(\pi\) yields the orphan value 3.13903, which belongs to neither framework.

The honest boundary¶

This pair of graphs establishes a self-consistency, not a physical proof. It shows that \(4/\sqrt{\varphi}\) is the circle constant that makes this area-match construction exact, and that the construction is tuned to that value. It does not overturn the analytic fact that the unit circle's Euclidean arc-length integral converges to \(2\pi = 6.283185\ldots\); the conventional \(\pi\) remains the constant of the measured/analytic world. Both columns stay on the table — the constructed world where \(4/\sqrt{\varphi}\) is exact, and the analytic world where the series evaluate to \(3.14159\ldots\).