Notes — Book II: Algebraic Audit of Dürer’s Pentagon, Heptagon & $\pi$

1. Exact Radical Audit of Dürer’s Rusty-Compass Pentagon (Fig. 16)

Why did Dürer present two pentagon constructions back-to-back on pages 058–059 (Figs. 15 and 16)? While Figure 15 gives Ptolemy's exact golden-ratio chord over $\mathbb{Q}(\phi)$, Figure 16 requires never changing the compass opening (auß vnuerruckten zirckel)—a classic medieval stonemasons’ and Islamic workshop protocol going back to Abū al-Wafāʾ (abulwafa:craftsmen) and Matthäus Roriczer (roritzer:1487), which Hermann Staigmüller (staigmuller:1891) and Émile Lemoine (lemoine:1902) celebrated as a milestone of constructive simplicity.

In construct_durer_vs_euclid_pentagon(), let the given base be $A = (0, 0)$ and $B = (1, 0)$ with fixed unit aperture $R = 1$:

  • The vesica intersections are $C = \left(\frac{1}{2}, \frac{\sqrt{3}}{2}\right)$ and $D = \left(\frac{1}{2}, -\frac{\sqrt{3}}{2}\right)$.

  • The third unit circle centered at $D$ meets $\bigcirc(A, 1)$ at $E = \left(\frac{1}{2} - \frac{\sqrt{3}}{2},\, -\frac{\sqrt{3}}{2} + \frac{1}{2}\right)$ and meets the vertical axis $CD$ at $G = \left(\frac{1}{2},\, 1 - \frac{\sqrt{3}}{2}\right)$.

  • Ray $EG$ has slope $m = \frac{1}{\sqrt{3}}(2 - \sqrt{3})$? Let us read off $Delta x = frac{sqrt{3}}{2}$ and $\Delta y = \frac{1}{2}$, so ray $EG$ makes an exact $30^\circ$ angle with the horizontal! Intersecting ray $EG$ with $\bigcirc(B, 1)$ yields the closed-form nested quadratic coordinates of vertex $H_{\text{Dürer}}$:

$$H_{\text{Dürer}} = \left(\frac{5}{8} + \frac{\sqrt{3}\,\sqrt{6\sqrt{3} - 4}}{8},\; -\frac{\sqrt{3}}{2} + \frac{7}{8} + \frac{\sqrt{6\sqrt{3} - 4}}{8}\right),$$

and by bilateral reflection $I_{\text{Dürer}} = (1 - x_H, y_H)$. Stepping unit circles from $I_{\text{Dürer}}$ and $H_{\text{Dürer}}$ closes at apex $K_{text{Dürer}} = left(frac{1}{2}, y_Kright)$.

Symbolic verification in geometor.durer proves:

  1. 100% Exact Equilateral Side Quadrances: Every edge $AB, BH_{text{Dürer}}, H_{text{Dürer}}K_{text{Dürer}}, K_{text{Dürer}}I_{text{Dürer}}, I_{text{Dürer}}A$ has exact quadrance $Q = 1$.

  2. Non-Equiangularity (+21.97 Arcminutes): Whereas Euclid’s $\mathbb{Q}(\phi)$ pentagon has all five interior angles equal to $108^\circ$ (interior line spread $s = \frac{5 + \sqrt{5}}{8}$), Dürer’s pentagon has interior angles:

    • At base vertices $A, B$: $108.36612^\circ$ ($+0.36612^\circ = +21.97'$ too wide),

    • At shoulder vertices $I_{\text{Dürer}}, H_{\text{Dürer}}$: $107.03899^\circ$ ($-57.66'$),

    • At apex $K_{\text{Dürer}}$: $109.18978^\circ$ ($+1^\circ 11.39'$).

Within twenty-five years of Dürer’s book, Lodovico Ferrari and Gerolamo Cardano proved that even the exact $\mathbb{Q}(\phi)$ pentagon can be constructed with straightedge and a single fixed compass opening, albeit requiring many more circles than Dürer’s ultra-fast five-circle workshop layout.

2. Eisenstein Irreducibility & Residual of the Workshop Heptagon (Figs. 11 & 15)

For a regular heptagon inscribed in the unit circle $R = 1$, setting $\theta = \frac{2\pi}{7}$ gives chord quadrance $q = 4\sin^2\frac{\pi}{7} = 2 - 2\cos\frac{2\pi}{7}$. Since $2\cos\theta$ satisfies $t^3 + t^2 - 2t - 1 = 0$, substituting $t = 2 - q$ yields the minimal polynomial of the regular heptagon’s side quadrance (construct_durer_heptagon_workshop()):

$$P_7(q) = q^3 - 7q^2 + 14q - 7 = 0.$$

By Eisenstein’s criterion at the prime $p = 7$ ($7 \mid -7, 14, -7$ and $7^2 \nmid -7$), $P_7(q)$ is irreducible over $\mathbb{Q}$ of degree $3 \neq 2^k$. By the Gauss–Wantzel theorem (Gauss gauss:1801; Wantzel wantzel:1837), no unmarked ruler-and-compass construction can ever construct $q_{text{exact}} approx 0.753020396$, vindicating Dürer’s explicit Mechanice label! Evaluating $P_7$ at Dürer’s half-equilateral quadrance $Q_{\text{Dürer}} = \left(\frac{\sqrt{3}}{2}\right)^2 = \frac{3}{4}$ gives the extraordinarily tiny exact rational residual:

$$P_7\!\left(\frac{3}{4}\right) = \frac{27}{64} - \frac{63}{16} + \frac{21}{2} - 7 = -\frac{1}{64},$$

corresponding to a relative chord error of $-0.2008\%$ (less than $0.2\text{ mm}$ on a $100\text{ mm}$ radius circle!). Similarly, Dürer’s hendecagon rule $s_{11} = frac{9}{16}R = 0.5625,R$ misses $2\sin\frac{\pi}{11} \approx 0.563465\,R$ by just $-0.1713\%$.

Philological Note on the 1525 Errata (SLUB N1525 p. 182)

In the 1525 first printing of p. 058 (Fig. 15), the typesetter omitted the clause bisecting a.c. at e before drawing segment e.d., and the woodblock cutter omitted line d.e. inside the first circle. Dürer caught this himself on the final Errata leaf (p. 182): “Im .E. am vierten plat in der fünftzeheten figur im ersten cirkel soll .d.e. mit einer lini zusamen gezogen werden.”

3. Dürer’s Circle Quadrature & $\pi_{\text{Dürer}} = \frac{25}{8}$ (Fig. 34)

For a square of side $a$ and area $a^2$, the diagonal (ortstrich) has length $a\sqrt{2}$. Taking $d = \frac{8}{10}\,a\sqrt{2} = \frac{4\sqrt{2}}{5}a$ gives radius $r = frac{2sqrt{2}}{5}a = sqrt{frac{8}{25}},a$. Equating $\pi_{\text{Dürer}}\,r^2 = a^2$ yields

$$\pi_{\text{Dürer}} = \frac{25}{8} = 3.125,$$

the classical Roman-Vitruvian value ($3\frac{1}{8}$, relative error $-0.528\%$), which Dürer rightly refuses to pass off as demonstrative—anticipating Lindemann’s 1882 proof of the transcendence of $\pi$ (klein:1895).