UNIT V — DISTINCTIO VII–VIII, FLOS (1225) & LIBER ABACI CONVERGENCE TO $\phi$¶
Optical Surveying with the Quadrans, Geometrical Subtleties of the Pentagon, the Flos Cubic, and the Rabbit Recurrence
Source Text: fibonacci:pg-7, fibonacci:pg-8, fibonacci:flos, boncompagni:1857 (Boncompagni Vol. II, pp. 202–247 & Vol. I, pp. 283–284) · Latin (1220 / Boncompagni 1862) · Summary · GEOMETOR Notes
Distinctio VII — Surveying Heights & Distances with the Quadrans¶
Latin Locus: Distinctio VII — De Altitudinibus et Quadrante (pp. 202–207) · Commentary: Notes on Optical Similar Triangles
Without an Astrolabe or Quadrant: Set up a vertical gnomon or rod of known height, or place a horizontal plane mirror (or vessel of water) upon the ground and step backward until the summit of the tower appears at the center of reflection; by equal angles of incidence and reflection (or parallel rays of the sun), the observer’s eye-height and tower height form equiangular right triangles, [VI. 4] giving $\text{Tower Height} : \text{Tower Distance} = \text{Eye Height} : \text{Mirror Distance}$.
With the *Quadrans* Shadow Square: Within the quarter-circle limb (DI) centered at sighting apex D, inscribe the square DHAI divided along both its horizontal umbra recta side IA and vertical umbra versa side AH into $12$ equal points (puncta).
When the weighted plumb line hangs along the diagonal DA (meeting the arc at B at $45^\circ$), the horizontal distance from the observer’s foot to the base of the tower equals the tower’s height above the observer’s eye.
When sighting a steeper tower, the plumb line DP (extended as DC to C) cuts umbra recta IA at P (for example, at $IP = 8$ puncta out of $DI = 12$). By similar triangles [VI. 4], the ratio of horizontal distance to vertical elevation is $8 : 12 = 2 : 3$; hence at distance $40$ perticae (and eye-height $1$ pertica), the tower height is $\frac{12}{8} \times 40 + 1 = 61$ perticae. Conversely, for shallow angles where the plumb line cuts umbra versa AH at $v$ puncta, elevation to distance is $v : 12$.
Distinctio VIII — Subtilitates Geometricae: Pentagon, Decagon & Al-Jabr¶
Latin Locus: Distinctio VIII — De Subtilitatibus Geometricis (pp. 207–226) · Commentary: Notes on Abū Kāmil’s Quartic & Euclid XIII.10
Euclidean Construction in Circle Diameter $10$ ([XIII. 9–10]): In circle (IA) of diameter $CA = 10$ through C, center I, and A, erect perpendicular radius ID $= 5$ to apex D, and bisect $IC$ at H ($IH = 2\frac{1}{2}$). Then hypotenuse HD has square $25 + 6\frac{1}{4} = 31\frac{1}{4}$. Turning circle (HD) onto the diameter at K along IK divides radius $IA = 5$ (the inscribed hexagon side) in extreme and mean ratio, yielding the decagon side IK equal to chord DQ1 (an apotome):
$$s_{10} = IK = \sqrt{31\frac{1}{4}} - 2\frac{1}{2} = \frac{5(\sqrt{5} - 1)}{2}.$$
By Euclid’s theorem that the square on the pentagon side equals the sum of the squares on the hexagon and decagon sides [XIII. 10], hypotenuse DK is the exact side of the regular pentagon DP1P2P3P4 (with vertices P1, P2, P3, P4):
$$s_5^2 = ID^2 + IK^2 = 25 + \left(37\frac{1}{2} - \sqrt{781\frac{1}{4}}\right) = 62\frac{1}{2} - \sqrt{781\frac{1}{4}}, \qquad s_5 = \frac{5\sqrt{10 - 2\sqrt{5}}}{2}.$$
Algebraic Derivation via *Res*, *Census* & *Census Census* (ptolemy:almagest): Following al-Khwārizmī and Abū Kāmil, let the unknown pentagon side be one thing (res $= x$), so its square is one census ($x^2$) and its fourth power is census census ($x^4$). Combining Ptolemy’s cyclic-quadrilateral theorem [VI. D] with diameter $d = 10$ produces the biquadratic equation:
$$x^4 + 3{,}125 = 125\,x^2.$$
Completing the square in census ($x^2 = 62\frac{1}{2} - \sqrt{3{,}906\frac{1}{4} - 3{,}125} = 62\frac{1}{2} - \sqrt{781\frac{1}{4}}$) recovers the identical Euclid X minor irrational!
Bridge I — Flos (1225): The Imperial Cubic & Book X Irrationals¶
Latin Locus: Flos super Solutionibus Quarundam Questionum (pp. 227–234) · Commentary: Notes on the Flos Cubic & Congruum
Five years after Practica Geometriae, at the Pisan court of Emperor Frederick II, Master John of Palermo proposed two celebrated challenges preserved in Leonardo’s Flos (fibonacci:flos, dedicated to Cardinal Raniero Capocci):
The Congruum of Five (Liber Quadratorum & Flos): Find a rational square $x^2$ which, increased or diminished by $5$, remains a rational square ($x^2 \pm 5 = \square$). Scaling by $12^2 = 144$ (since $5 \times 144 = 720$ is a congruum arising from $m = 5, n = 4$: $4mn(m^2 - n^2) = 720$), Leonardo exhibits $31^2 + 720 = 41^2$ and $41^2 + 720 = 49^2$, whence $x = 3\frac{5}{12} = \frac{41}{12}$:
$$\left(\frac{41}{12}\right)^2 - 5 = \left(\frac{31}{12}\right)^2, \qquad \left(\frac{41}{12}\right)^2 + 5 = \left(\frac{49}{12}\right)^2.$$
The Irreducible Cubic Equation: Find a line $x$ such that a cube plus two squares and ten roots equals twenty:
$$x^3 + 2x^2 + 10x = 20.$$
Testing every species of rational fraction and every quadratic irrational classified in Euclid’s Book X (binomiale, recisum, medial, minor, major), Leonardo proves rigorously that $x$ cannot be expressed by any square-root tower of Book X, and then exhibits its astonishing six-place sexagesimal root:
$$x \approx 1^{\circ}\ 22'\ 7''\ 42'''\ 33''''\ 4'''''\ 40'''''' = \frac{1{,}596{,}577{,}777}{1{,}166{,}400{,}000} \approx 1.36880810785322\dots$$
whose error from the true real root $1.36880810782137\dots$ is less than $3.2 \times 10^{-11}$!
Bridge II — Liber Abaci Cap. XII: The Rabbit Sequence & $\phi$¶
Latin Locus: Quot Paria Coniculorum in Uno Anno (Vol. I, pp. 283–284) · Commentary: Notes on Cassini’s Identity & Golden Convergence
A certain man placed one pair of rabbits in a place surrounded on all sides by a wall, to find how many pairs would be bred from it in one year, when by nature every mature pair produces another pair every month, and each newborn pair begins to bear in the second month after its birth. Starting with $1$ pair, in the first month there are $2$ pairs, in the second $3$, in the third $5$, and continuing month by month through the twelfth:
$$1,\; 2,\; 3,\; 5,\; 8,\; 13,\; 21,\; 34,\; 55,\; 89,\; 144,\; 233,\; 377.$$