Unit VI: Multiplication of Curves & The Method of Similitude¶
Julius Petersen — Methods and Theories (Authorized Translation by Sophus Haagensen, 1879 · pp. 22–33)
Chapter I (Subsections) — Multiplication of Curves & The Method of Similitude¶
Multiplication of Curves (Homothety, pp. 22–27, Problems 117–145)¶
[pp. 22–24] Take a fixed point \(P\) and join it to an arbitrary point \(a\) of a given curve \(k\). If along the straight line \(Pa\) we determine \(A\) such that \(PA : Pa = m : n\)—either on the same side of \(P\) (multiplication by \(+\frac{m}{n}\)) or on the opposite side (multiplication by \(-\frac{m}{n}\))—the locus of \(A\) is a curve \(K\) similar and similarly situated (homothetic) to \(k\), with \(P\) as centre of similitude (a pure specialization of Chapter III’s revolution with \(v = 0^\circ\) or \(180^\circ\)):
Multiplying a Straight Line: Yields a parallel straight line whose distance from \(P\) is multiplied by \(\pm \frac{m}{n}\).
Multiplying a Circle: Yields a circle whose centre and radius are both multiplied from \(P\) by \(\pm \frac{m}{n}\). Conversely, any two unequal circles have two centres of similitude—the external centre (dividing the line of centres externally in ratio \(+R_1 : R_2\)) and the internal centre (dividing it internally in ratio \(-R_1 : R_2\)).
Intersections With Scaled Curves (Problems 117–145): To draw through \(P\) a secant cut by two curves \(k, K_1\) in segments \(Pa : PA = n : m\), multiply \(k\) from \(P\) by \(\pm \frac{m}{n}\) and intersect with \(K_1\)!
The Method of Similitude (pp. 27–33, Problems 146–191)¶
[pp. 27–29] The Method of Similitude ranks alongside the Method of Two Loci as the most natural and fertile heuristic in elementary geometry. It applies under two distinct structural regimes:
Case a — Given Only One Linear Magnitude Together With Angles and Ratios (pp. 27–28, Problems 146–153): If all conditions save one determine only the form (shape) of the sought figure—specifying angles and ratios of lines—while a single given length fixes its size, set aside the given length, adopt any convenient arbitrary length in its stead, construct a prototype figure similar to the sought one, and finally scale the prototype by fourth-proportional similarity (Euclid VI.12)! - Problem 152 (p. 28): Construct a triangle given its three altitudes \(h_a, h_b, h_c\). Since \(a h_a = b h_b = c h_c = 2\Delta\), the sides satisfy \(a : b : c = \frac{1}{h_a} : \frac{1}{h_b} : \frac{1}{h_c} = h_b h_c : h_c h_a : h_a h_b\) (or secants from an external point to a circle of products). Construct any prototype triangle \(\triangle A'B'C'\) with sides in those ratios, measure its altitude \(h'_a\), and scale from \(A'\) by \(h_a / h'_a\)!
Case b — Positional Similitude by Omitting One Boundary Condition (pp. 28–33, Problems 154–191): [pp. 28–29] Even when the sought figure must be inscribed in or placed upon a given figure, if omitting one condition leaves a system of figures that remain similar and similarly situated with respect to a fixed centre of similitude, then every point of the variable figure describes a straight line (or scaled curve) through that centre of similitude! Drawing any one convenient prototype figure fixes that straight-line locus, whose intersection with the omitted boundary immediately gives the true figure!
Figure 6.1 — Petersen’s Problem 155 (p. 29): Inscribing a Square in a Given Triangle by Homothety. Suppressing the condition that the upper-right vertex lie on BC, we erect prototype square D0E0F0C of side \(h_c = 12\) resting on base AB with upper-left vertex at C on AC. Taking A as centre of similitude, the locus of the free vertex is ray AF0, which intersects BC at F \(= (119/13, 84/13)\), yielding exact inscribed square DEFG of side \(84/13\).
Flagship Homothety Constructions (Problems 154–191)¶
Problem 155 (p. 29): To inscribe a square in a given triangle, sector, or segment. - Analysis & Construction for \(\triangle ABC\): Require square \(DEFG\) with \(D, E \in AB\), \(G \in AC\), and \(F \in BC\). Omit the condition \(F \in BC\)! Every square \(D_0 E_0 F_0 G_0\) resting on \(AB\) with \(G_0 \in AC\) is homothetic with respect to vertex \(A\) as centre of similitude, because \(AD_0 : D_0 G_0 = \cot A\) is constant. Hence the free fourth vertex \(F_0\) moves along the straight ray \(AF_0\)! Construct any one prototype square (for instance, with \(G_0 = C\) and side \(h_c\)), connect \(A F_0\) to meet \(BC\) in \(F\), and complete \(DEFG\) by parallels and perpendiculars. - For a circular sector: Take the centre of the circle as centre of similitude (if two vertices lie on the radii) or the midpoint of the base chord (for a segment)!
Problem 166 (p. 30): In a given triangle \(ABC\) to inscribe another \(XYZ\), knowing the direction of \(YZ\), the fixed point \(X \in BC\)? — or knowing the ray \(AX\) on which \(X\) falls, the direction of \(YZ\), and the ratio \(XY : XZ\). Remove \(BC\) (retaining ray \(AX\)) and take \(A\) as centre of similitude!
Problem 170 (p. 30): In a triangle to inscribe a rhombus, one vertex of which coincides with one of the vertices of the triangle (homothety from \(A\), coinciding with the internal angle bisector \(v_A\)).
Problem 181 (p. 31): Construct a circle passing through a given point \(P\) and touching two given intersecting straight lines — take the intersection of the two lines as centre of similitude, draw any prototype circle touching both lines, intersect ray \(OP\) with the prototype circle at \(P_1, P_2\), and scale by \(OP / OP_1\) and \(OP / OP_2\) (two solutions)!