Unit II: Chapter I, Part A — Loci of Points & The Method of Two Loci¶
Julius Petersen — Methods and Theories (Authorized Translation by Sophus Haagensen, 1879 · pp. 7–22, 37–40)
Chapter I — Loci · A. Loci of Points¶
[p. 7] In applying the Method of Two Loci, everything depends on recognizing instantly which elementary curve corresponds to a single detached condition on an unknown point \(P\). Every proof of a locus requires two reciprocal demonstrations (or one chain of reversible equivalences): first, that every point satisfying the condition lies upon the stated curve; second, that every point of the curve (or specified arc) satisfies the condition.
Fundamental Point Loci a – k (pp. 7–14)¶
Locus a (p. 7): The locus of all points situated at a given distance from a given point is a circle, having the given point for its centre and the given distance for its radius. - App. 1: The locus of the centres of circles of given radius touching a given circle \((O, R)\) consists of the two concentric circles \((O, R + r)\) and \((O, |R - r|)\). - App. 2: The locus of points from which a given circle \((O, R)\) is seen under a given angle \(v\) (or from which tangents to the circle have a given length) is a circle concentric with the given one.
Locus b (p. 8): The locus of all points situated at a given distance from a given straight line consists of two straight lines, parallel to the given line at the given distance on either side. - App.: The locus of vertices of triangles having a fixed base \(AB\) and a given area (hence a given altitude \(h_c\)) consists of two parallels to \(AB\).
Locus c (p. 8): The locus of all points equidistant from two given points A and B is the straight line bisecting AB perpendicularly.
Locus d (p. 8): The locus of all points equidistant from two given intersecting straight lines consists of the two perpendicular straight lines bisecting the angles formed by the given lines. (If the two given lines are parallel, the locus reduces to the single midline parallel to both.)
Locus e (pp. 9–11): The locus of all points, from which a given finite line AB is seen under a given angle, consists of two circular arcs passing through A and B symmetric across AB (Euclid III.21 & III.33). When the given angle is a right angle, the two arcs unite into the single circle having \(AB\) for diameter (Euclid III.31). - App. 1: As a chord of fixed direction or fixed angle moves in a circle, the locus of the middle point of chords passing through a fixed point \(P\) is the circle on diameter \(OP\) (\(O\) being the centre of the given circle). - Remark on Directed Angles (p. 10): If we distinguish the sense of rotation from \(PA\) to \(PB\), points on opposite sides of \(AB\) belonging to supplementary arcs satisfy the same modulo-\(180^\circ\) directed angle relation, unifying the two pieces into a single complete circle. - Problem 19 — Pothenot’s Problem (p. 11, Pothenot): Determine a station P in the plane from which three known landmarks A, B, C are seen under known angles \(\angle APB = u\) and \(\angle BPC = v\). Intersect the locus
earc on \(AB\) for angle \(u\) with the locusearc on \(BC\) for angle \(v\) (indeterminate precisely when \(P\) lies on circle \(ABC\)).
Locus f — The Circle of Apollonius (pp. 11–12)¶
[p. 11] f. The locus of the points, the distances of which from two given points have a given ratio (\(m : n\)), is a circle (Apollonius, Circle of Apollonius).
Figure 2.1 — Petersen’s Locus f (Circle of Apollonius) & the Method of Two Loci. Points C and C1 divide AB internally and externally in ratio \(m : n = 2 : 1\) so \((A, B; C, C_1) = -1\). The Apollonius circle MC centered at midpoint M meets parallel locus H0LocusB (\(h = 8\)) in exact integer solutions P1 and P2 with \(P_1 A : P_1 B = 2 : 1\), while bisectors P1C and P1C1 meet orthogonally.
Let \(A\) and \(B\) be the given points, and \(P\) any point such that \(PA : PB = m : n\) (\(m \neq n\)). Draw the internal and external bisectors \(PC\) and \(PC_1\) of \(\angle APB\) meeting the straight line \(AB\) in \(C\) and \(C_1\). By Euclid VI.3 and its external counterpart,
The points \(C\) and \(C_1\) are therefore fixed points dividing \(AB\) harmonically in the given ratio \(m : n\), independently of the position of \(P\). Moreover, because \(PC\) and \(PC_1\) bisect supplementary adjacent angles, \(\angle CPC_1 = 90^\circ\)! Consequently \(P\) lies on the circle described upon \(CC_1\) as diameter (by e). Conversely, for every point \(P\) on this circle, if we draw \(A U \parallel B V \parallel P C\) or reflect across the bisector, \(PA : PB = CA : CB = m : n\). (When \(m = n\), \(C_1\) recedes to infinity and f degenerates into the perpendicular bisector c.)
Locus g (p. 12): The locus of the points, the distances of which from two given straight lines have a given ratio (\(m : n\)), consists of two straight lines passing through the point of intersection of the given lines. Together with the two given lines, these two locus lines form four harmonical rays, intersecting every transversal in four harmonic points.
Locus h (pp. 12–13): The locus of the points, the squares of the distances of which from two given points A and B have a given constant difference (\(\overline{PB}^2 - \overline{PA}^2 = a^2\)), is a straight line perpendicular to AB. - App. 1 — Radical Axis & Radical Centre (p. 13): Given two circles \((A, r_1)\) and \((B, r_2)\), the tangents drawn from an external point \(P\) satisfy \(t_1^2 = \overline{PA}^2 - r_1^2\) and \(t_2^2 = \overline{PB}^2 - r_2^2\). Setting \(t_1 = t_2\) yields \(\overline{PB}^2 - \overline{PA}^2 = r_2^2 - r_1^2 = \text{const.}\), so the locus of points of equal tangent power is a straight line perpendicular to the line of centres, called the radical axis of the two circles (containing their common chord if they intersect). For three circles, the three pairwise radical axes concur in one point, the radical centre (Problem 30, p. 14). - App. 3 (p. 13): The locus of centres of circles cutting two given circles orthogonally is their radical axis.
Locus i (p. 13): The locus of the points, the sum of the squares of whose distances from two given points A and B equals a given square (\(\overline{PA}^2 + \overline{PB}^2 = a^2\)), is a circle whose centre is the middle point C of AB. By the median identity, \(\overline{PA}^2 + \overline{PB}^2 = 2\overline{PC}^2 + \tfrac{1}{2}\overline{AB}^2\), whence \(\overline{PC}^2 = \tfrac{1}{2}a^2 - \tfrac{1}{4}\overline{AB}^2\) (real precisely when \(a^2 \ge \tfrac{1}{2}\overline{AB}^2\)).
Locus k (pp. 13–14): The locus of points whose distances from two given intersecting lines have a given sum or difference \(a\) consists of four straight line segments forming a rectangle (for the sum) and their eight infinite prolongations (for the difference).
Selected Canonical Problems on Point Loci (Problems 1–116 & 202–219)¶
Problem 21 (p. 14): To find a point, the distances of which from three given lines have given ratios (g — four solutions).
Problem 22 (p. 14): In a triangle to find a point, the distances of which from the three vertices have given ratios (f — intersection of two Apollonius circles; for \(PA : PB : PC = bc : ca : ab\), yields the isodynamic points).
Problem 30 (p. 14): To find a point from which the tangents to three given circles are of equal length (h App. 1 — radical centre).
Problem 31 (p. 14): To construct a triangle of \(A, a\) and \(b^2 + c^2\) (e and i).
Problem 32 (p. 14): In a triangle ABC to find a point O, from which lines drawn to the three vertices divide the triangle into three equal parts. The condition \(\triangle AOB = \triangle AOC\) determines the locus of \(O\) as the median through \(A\); the sought point is the common intersection of the three medians (centroid).
Problem 37 (p. 15): Construct a triangle of \(a, A\) and \(b^2 - c^2\) (e and h).
Problem 42 (p. 15): Divide a given line into two parts having a given line as a mean proportional (e and b).
Problem 45 (p. 15): Find a point from which three contiguous segments \(AB, BC, CD\) of a straight line are seen under equal angles (f, since the internal bisector of \(\angle BPC\) divides \(BC\) in the ratio \(PB : PC\)!).
Problem 47 (p. 15): To find a point from which three given circles appear under equal visual angles (their distances from the centres must stand in the same ratios as the radii; solved by f).
Loci in General (pp. 37–40, Problems 202–219): [p. 38] Even when a locus is not immediately listed under a–k, removing one positional condition often permits drawing the rigid figure in any convenient position first, and then bringing it into the required position by parallel translation or revolution around a fixed centre—the loci of its vertices being parallel lines or concentric circles (Problems 202–209).