Unit VI: Multiplication of Curves & The Method of Similitude — Notes

Historical, Mathematical & Cross-Repository Commentary (pp. 22–33)

Historical, Mathematical & GEOMETOR Commentary

1. Why Positional Homothety Eliminates Quadratic Degree (Problem 155)

Compare solving Problem 155 (inscribing square \(DEFG\) in \(\triangle ABC\), p. 29) algebraically by coordinates versus synthetically by Petersen’s Method of Similitude:

  • Every square \(D_0 E_0 F_0 G_0\) resting on \(AB\) with \(G_0 \in AC\) satisfies \(\overrightarrow{A F_0} = \lambda \overrightarrow{v}\) because dilations centered at \(A\) preserve both the line \(AB\), the line \(AC\), right angles, and the aspect ratio \(1 : 1\).

  • Therefore the locus of the free fourth vertex \(F_0\) is linear (a ray through \(A\)), meaning the inscribed square is rationally constructible in the base field \(\mathbb{Q}(A, B, C)\) whenever altitude direction onto \(AB\) is rational—requiring zero square roots beyond erecting a perpendicular!

  • By contrast, in Problem 181 (p. 31: circle through \(P\) tangent to two intersecting lines \(l_1, l_2\)), omitting \(P\) gives a homothetic family of circles centered on the angle bisector; intersecting ray \(OP\) with a prototype circle yields two points \(P_1, P_2\), cleanly explaining why Problem 181 has two solutions governed by a single quadratic radical!

2. Exact geometor.model Realization: inscribe_square_by_homothety

In geometor.petersen.constructions.inscribe_square_by_homothety(), we choose the classic Heronian \((13, 14, 15)\) acute triangle \(A = (0, 0)\), \(B = (14, 0)\), \(C = (5, 12)\) (altitude \(h_c = 12\) onto \(AB\)):

  • Dropping perpendicular \(C D_0\) onto \(AB\) at \(D_0 = (5, 0)\) and laying off \(h_c = 12\) along \(AB\) constructs prototype square \(D_0 E_0 F_0 C\) with \(E_0 = (17, 0)\) and \(F_0 = (17, 12)\).

  • Homothetic ray \([A, F_0]\) (\(12x - 17y = 0\)) intersects side \([B, C]\) (\(4x + 3y = 56\)) at the exact rational vertex

    \[F = \left(\frac{119}{13},\, \frac{84}{13}\right),\]

    matching the classical harmonic-mean formula \(s = \frac{c \cdot h_c}{c + h_c} = \frac{14 \cdot 12}{14 + 12} = \frac{168}{26} = \frac{84}{13}\)!

  • Completing \(DEFG\) gives \(E = (\frac{119}{13}, 0)\), \(D = (\frac{35}{13}, 0)\), and \(G = (\frac{35}{13}, \frac{84}{13})\), all four sides having exact quadrance \((\frac{84}{13})^2 = \frac{7056}{169}\) and orthogonal corner spreads 1.