Unit 4: Book VI Extensions, Part B & Book III Props. 25, 27 — Inversion, Casey’s & Feuerbach’s Theorems, Poles & Polars¶
John Casey — A Sequel to the First Six Books of the Elements of Euclid (5th ed., 1888 / 6th ed., 1892)
Book VI, Section IV — Theory of Circular Inversion & Pappus’s Arbelos¶
Definition (:term:`Circular Inversion <inversion>`). Given a fixed circle of centre $O$ and radius $k$ (the Circle of Inversion), if to every point $P \neq O$ there corresponds the point $P'$ on ray $OP$ satisfying $OP \cdot OP' = k^2$, the figure formed by $P'$ is called the inverse of the figure formed by $P$.
Fundamental Properties of Inversion (Propositions 1–9, pp. 95–102):
Lines & Circles (Props. 1–3): A straight line not passing through $O$ inverts into a circle passing through $O$ (with diameter perpendicular to the line); a circle passing through $O$ inverts into a straight line not through $O$; a circle not passing through $O$ inverts into a circle not passing through $O$, with $O$ as a centre of similitude of the two circles; and any circle cutting the circle of inversion orthogonally inverts into itself!
Distance Distortion Formula (Prop. 4): By similar triangles $\triangle OAB \sim \triangle OB'A'$ (Euclid
VI.6),$$ |A'B'| = \frac{k^2 \cdot |AB|}{|OA| \cdot |OB|}. $$
Conformal Angle & Cross-Ratio Preservation (Props. 5–8): Two curves intersect at the same angle as their inverse curves (in opposite orientation), tangency is preserved, and the anharmonic ratio of four concyclic or collinear points is preserved under inversion.
Inversion Invariance of Normalized Bitangents (Prop. 9): If two circles of radii $r_1, r_2$ and common direct tangent $t_{12}$ invert into two circles of radii $r_1', r_2'$ and common direct tangent $t_{12}'$, then
$$ \frac{t_{12}^2}{r_1 r_2} = \frac{t_{12}'^{\,2}}{r_1' r_2'}. $$
Proposition 9, Corollary 1 (:person:`Pappus’s <pappus>` Ancient Arbelos Chain Theorem). In the shoemaker’s knife (Arbelos) bounded by three mutually tangent semicircles on collinear diameters $AB, AC, CB$, let a chain of circles $O_1, O_2, \dots, O_n$ be inscribed, each touching the two fixed boundary semicircles on $AB$ and $AC$ and touching the preceding circle of the chain. Then the perpendicular height $h_n$ of the centre of the $n$-th circle above the base line $AB$ is equal to $2n$ times its radius ($h_n = 2 n r_n$).
(Inverting from $A$, the two boundary semicircles through $A$ become parallel straight lines, and the chain $O_1, \dots, O_n$ becomes a column of equal congruent spheres/circles stacked between two parallel lines!)
Casey’s Generalized Ptolemy Theorem, Feuerbach’s Theorem & Hart’s Circle¶
Book VI, Section IV, Proposition 10 (:person:`Casey’s <casey>` :term:`Generalized Ptolemy Theorem <caseys-theorem>`, 1866). If four circles $O_1, O_2, O_3, O_4$ be all tangent to a fifth circle $\Sigma$ (or straight line), the six common tangents joining the four circles in pairs (direct tangents when two circles touch $\Sigma$ on the same side, transverse tangents when on opposite sides) satisfy the relation
$$ \overline{12} \cdot \overline{34} + \overline{14} \cdot \overline{23} = \overline{13} \cdot \overline{24}. $$
Demonstration. Invert the configuration from any point $O$ taken on the circumference of the fifth circle $\Sigma$. Because $\Sigma$ passes through the centre of inversion $O$, $\Sigma$ inverts into a straight line $L$, while $O_1, O_2, O_3, O_4$ invert into four circles $O_1', O_2', O_3', O_4'$ tangent to $L$ at four collinear points $A', B', C', D'$! Along the straight line $L$, Euler's collinear identity (Book II, Prop. 7) holds for the contact segments:
$$ A'B' \cdot C'D' + A'D' \cdot B'C' = A'C' \cdot B'D'. $$
Dividing through by $2\sqrt{r_1' r_2' r_3' r_4'}$ and applying Proposition 9 ($A'B' / \sqrt{r_1' r_2'} = \overline{12} / \sqrt{r_1 r_2}$) multiplies back by $2\sqrt{r_1 r_2 r_3 r_4}$ to yield $\overline{12}\cdot\overline{34} + \overline{14}\cdot\overline{23} = \overline{13}\cdot\overline{24}$ identically! Conversely, when the relation holds, the fourth circle touches the circle touching the first three. When $r_1 = r_2 = r_3 = r_4 = 0$, the theorem reduces to Ptolemy's Theorem for cyclic quadrilaterals (Euclid VI.D). Q.E.D.
First Proof (Immediate Corollary of Casey’s Theorem, pp. 104–105). Regard the three side midpoints Ma, Mb, Mc as three point-circles ($r_1 = r_2 = r_3 = 0$) and the Incircle (I Ta) as the fourth circle $O_4$. Their mutual “common tangents” are:
Midpoint chord lengths: $\overline{23} = |M_b M_c| = \frac{1}{2}a$, $\overline{31} = |M_c M_a| = \frac{1}{2}b$, $\overline{12} = |M_a M_b| = \frac{1}{2}c$.
Tangents from $M_a, M_b, M_c$ to the Incircle touching $BC, CA, AB$ at Ta, Tb, Tc: since $|B T_a| = s - b$ and $|B M_a| = \frac{1}{2}a$, we have
$$ \overline{14} = |M_a T_a| = \tfrac{1}{2}|b - c|, \qquad \overline{24} = |M_b T_b| = \tfrac{1}{2}|c - a|, \qquad \overline{34} = |M_c T_c| = \tfrac{1}{2}|a - b|. $$
Assuming without loss of generality $a \ge c \ge b$ (as in our canonical $(15, 13, 14)$ triangle!),
$$ \overline{23}\cdot\overline{14} + \overline{31}\cdot\overline{24} - \overline{12}\cdot\overline{34} = \frac{a(c - b) + b(a - c) - c(a - b)}{4} \equiv 0! $$
By the converse of Casey’s Theorem, the circle circumscribing the three point-circles $M_a, M_b, M_c$—which is the Nine-Points Circle (N Ma)—is tangent to the Incircle (I Ta) at Fe! Replacing $(s-b)$ by $s$ for the $A$-excircle gives $|M_a T_a'| = \frac{1}{2}(b+c)$, $|M_b T_b'| = \frac{1}{2}(a+c)$, and $|M_c T_c'| = \frac{1}{2}(a+b)$, satisfying $\frac{1}{4}[a(c-b) + c(a+b) - b(a+c)] \equiv 0$, proving external tangency with all three excircles.
Second Proof (:person:`Casey’s <casey>` 1861 Orthogonal Inversion Proof, pp. 105–106). Let Ma be the midpoint of $BC$. The Incircle $(I)$ and $A$-Excircle $(I_a)$ touch $BC$ at Ta and $T_a'$ symmetrically about Ma at equal distance $|M_a T_a| = |M_a T_a'| = \frac{1}{2}|b - c|$. Therefore the circle (Ma Ta) of centre Ma and radius $\frac{1}{2}|b - c|$ cuts both the Incircle and the $A$-Excircle orthogonally, and hence inverts each of them into itself!
Now consider the fourth common tangent [C Ts] to $(I)$ and $(I_a)$ (the reflection of $BC$ across the internal/external bisector line $I I_a$, which on the $(15, 13, 14)$ triangle passes directly through $C$ and touches $(I)$ at Ts!). Under inversion in (Ma Ta), this straight line—being tangent to both $(I)$ and $(I_a)$—inverts into a circle passing through the centre of inversion Ma and through Mb and Mc, namely the Nine-Points Circle (N Ma). Because inversion preserves tangency while leaving $(I)$ and $(I_a)$ fixed, the Nine-Points Circle touches $(I)$ internally at the inverse Fe of Ts, and touches $(I_a)$ externally! Q.E.D.
Proposition 11 (:person:`Hart’s <hart>` :term:`Circle <harts-circle>`, 1861). Given any circular triangle formed by three arbitrary circles, the eight Apollonius circles touching all three partition into groups of four (analogous to the incircle and three excircles of a rectilinear triangle) such that each quartet of four circles is touched by a common circle—Hart’s Circle.
Book III (Props. 25, 27) & Book VI, Section VII — Poles, Polars & Reciprocation¶
Definition (:term:`Pole and Polar <pole-and-polar>`). If $D'$ be the inverse of a point D with respect to a circle (O G1), the straight line [U V] drawn through $D'$ perpendicular to $OD$ is called the Polar of D, and D is called the Pole of [U V].
Book III, Propositions 25 & 27 (:person:`La Hire’s <lahire>` Reciprocity & Harmonic Secant Property).
Prop. 25 (La Hire’s Reciprocity, 1685): If a point Q lie upon the polar [U V] of a point D with respect to a circle, then D lies upon the polar of Q.
Prop. 27 (Harmonic Secant Theorem): Any secant drawn from a pole D cutting the circle in G1, H1 and cutting its polar [U V] in J1 is divided harmonically: $(G_1, H_1; J_1, D) = -1$.
Demonstration of La Hire’s Reciprocity. Let $D'$ be the inverse of D on $OD$ ($OD \cdot OD' = r^2$, $D'Q \perp OD$), and drop $DK \perp OQ$. Because $\angle OD'Q = \angle OKD = 90^\circ$, the quadrilateral $D'DKQ$ is cyclic (Euclid III.22); therefore by Euclid III.36 from $O$, $OQ \cdot OK = OD \cdot OD' = r^2$. Hence $K$ is the inverse of Q, and $DK$—being perpendicular to $OQ$ at $K$—is the polar of Q, passing through D! Q.E.D.
Book VI, Section VII, Propositions 1–6 (Inscribed Quadrangle Polar Construction, Self-Conjugate Triangles & :term:`Reciprocation <reciprocation>`).
Prop. 2 (Pure-Straightedge Polar & Self-Conjugate Triangle): Draw any two secants from D cutting the circle in G1, H1 and G2, H2. For the inscribed quadrangle $G_1 H_1 H_2 G_2$, let opposite chord pairs meet at U and V. By the Complete Quadrilateral theorem (Unit 3) and Prop. 27, the line [U V] is the exact polar of D—constructed using straightedge alone—and $\triangle DUV$ is a Self-Conjugate Triangle in which each vertex is the pole of the opposite side!
Prop. 3 (:person:`Salmon’s <salmon>` Theorem): The distances of any two points from the centre of a circle are proportional to the perpendicular distances of each point from the polar of the other.
Props. 4–6 (:person:`Brianchon’s <brianchon>` Theorem by Polar Reciprocation): Replacing every point on a circle by its tangent (polar) and every inscribed chord by the intersection of tangents (pole) reciprocates Pascal's inscribed hexagon theorem directly into Brianchon’s Theorem: In any hexagon circumscribed about a circle, the three diagonals joining opposite vertices concur at a single point (the Brianchon Point).