Unit III: Circles, Tangency, Golden Section, Projective Polarity & Modern Triangle Centers

Émile Lemoine — Géométrographie, ou Art des constructions géométriques (English Translation, Problems XXI & XXV–LXIII, pp. 12–51)

[Folio Problèmes XXI, XXV–XXVII | p. 15]

§ 1. Circumcircle (XXI), Circle Tangents (XXV–XXVI), and Incircle \(X(1)\) (XXVII)

Problem XXI (Euclid IV.5 — Circumcircle and Circumcenter :math:`X(3)`, :math:`S = 15`). Given three non-collinear points \(A, B, C\), open the compass to one arbitrary radius \(R\) greater than half of \(AB\) and \(BC\). Sweep \(B(R)\) once (\(C_1 + C_3\)), then \(A(R)\) (\(C_1 + C_3\)) and \(C(R)\) (\(C_1 + C_3\)). Reusing \(B(R)\) across both perpendicular bisectors cuts the auxiliary circle count from four to three; drawing the two mediatrices (\(4R_1 + 2R_2\)) locates the circumcenter \(O = X(3)\), and sweeping \(O(A)\) (\(2C_1 + C_3\)) completes the circumcircle in \(4R_1 + 2R_2 + 5C_1 + 4C_3\) (\(S = 15, E = 9\), 2 lines, 4 circles vs. classical \(S = 17\)).

Problems XXV & XXVI (Euclid III.16 Por. & III.17 — Tangents to a Circle).

  • Tangent at a Point :math:`A` on Traced Circle :math:`O` (XXV): Place one leg at \(A\) (\(C_1\)), pivot around an arbitrary exterior point \(P\) to draw \(P(PA)\) (\(C_3\)) cutting circle \(O\) again at \(B\), draw diameter \(BPC\) (\(2R_1 + R_2\))—when \(P\) is chosen on ray \(OA\) or via Thales semicircle on \(OA\)—yielding \(4R_1 + 2R_2 + C_1 + C_3\) (\(S = 8\), 1 circle).

  • Tangents from an External Point :g:`A` to Circle :g:`O(R)` (XXVI): Classical Elements III.17 uses concentric circles (\(8R_1 + 4R_2 + 4C_1 + 3C_3, S = 19\)). Lemoine reduces the cost to \(S = 18\) via two distinct syntheses—either \(5R_1 + 3R_2 + 7C_1 + 3C_3\) (\(E = 12\), 3 circles) or bisecting OA at M to sweep MMO cutting OR at T, T and joining AT, AT (\(5R_1 + 3R_2 + 6C_1 + 4C_3, S = 18, E = 11\)).

A M O T T' Problems XXVI & XXXIX — Tangents (S = 18) & Straightedge Polar TT' (12R₁ + 7R₂)

Problem XXVII (Euclid IV.4 — Incircle and Incenter :math:`I = X(1)`, :math:`S = 23` vs. :math:`S = 30`). Textbook execution bisects two angles of ABC (6 circles, 2 lines), drops a full perpendicular line from incenter I to BC (3 circles, 1 line), and finally draws the incircle (1 circle): 3 lines and 10 circles (\(6R_1 + 3R_2 + 11C_1 + 10C_3, S = 30, E = 17\)). Lemoine cuts seven operations and four circles by: (1) sharing one compass radius across vertices B and C to draw bisectors BI and CI meeting at I (\(X(1)\) in ), and (2) determining the perpendicular touch-radius ID to BC via symmetric compass arcs without tracing the third straight line before sweeping IID:

\[\mathrm{Op.} : \left(4R_1 + 2R_2 + 11C_1 + 6C_3\right), \qquad S = 23, \quad E = 15 \quad (2\text{ lines}, 6\text{ circles}).\]
A B C I = X(1) D E F Problem XXVII (Euclid IV.4) — Incircle : 4R₁ + 2R₂ + 11C₁ + 6C₃ (S = 23 vs. S = 30)
[Folio Problèmes XXXIII–XXXVI | p. 24]

§ 2. Proportions (XXXIII–XXXV) & Division in Extreme and Mean Ratio \(\phi\) (XXXVI)

  • Fourth & Third Proportionals (XXXIII–XXXIV, Euclid VI.12 & VI.11): Constructing \(x = bc/a\) drops from \(S = 25\) to \(2R_1 + 2R_2 + 11C_1 + C_2 + 6C_3\) (\(S = 22, E = 14\)), while \(x = b^2/a\) drops from \(S = 22\) to \(4R_1 + 2R_2 + 5C_1 + C_2 + 3C_3\) (\(S = 15, E = 10\)).

  • Mean Proportional :math:`sqrt{ab}` (XXXV, Euclid II.14 / VI.13 — Lemoine–Gouzy Construction, :math:`S = 14`): Classical semicircle erection on \(a + b\) requires 3 lines and 6 circles (\(S = 24\)). With Auguste Gouzy, Lemoine constructs \(\sqrt{ab}\) in half the circles (\(2R_1 + 2R_2 + 6C_1 + C_2 + 3C_3\), \(S = 14, E = 9\), 2 lines, 3 circles).

Problem XXXVI (Euclid II.11 & VI.30 — Golden Section :math:`phi`, :math:`S = 18` in One Single Line!). To divide a given segment AB in extreme and mean ratio (\(AB / AX = AX / XB = \phi = \frac{1+\sqrt{5}}{2}\)), classical constructions draw three auxiliary straight lines—erecting \(BC \perp AB\) of length \(AB/2\), joining hypotenuse \(AC\), and transferring \(AC - BC\) onto \(AB\) (\(6R_1 + 3R_2 + 9C_1 + 7C_3, S = 25, E = 15\)).

Lemoine eliminates two of the three straight lines! Sweeping equal-radius arcs from A and B traces the single straight line MC—the perpendicular bisector of AB, simultaneously giving midpoint M (\(MB = AB/2\)) and the right angle at M. Laying off \(MC = AB\) along that perpendicular produces right triangle \(\triangle BMC\) with legs \(AB/2\) and \(AB\), so the un-drawn distance BC equals \(\frac{\sqrt{5}}{2}\,AB\)! Sweeping circle \(B(C)\) cuts the carrier line beyond A at E, so segment \(ME = BC - MB = \frac{\sqrt{5}-1}{2}\,AB = AB/\phi\), and arc AME marks the golden point X (AX \(= AB/\phi\)):

\[\mathrm{Op.} : \left(2R_1 + R_2 + 9C_1 + 6C_3\right), \qquad S = 18, \quad E = 11 \quad (1\text{ line}, 6\text{ circles}).\]
E A M X (φ⁻¹) B C Problem XXXVI (Euclid II.11) — Golden Section in 1 Line & 6 Circles (S = 18 vs. S = 25) MC = AB, MB = AB/2 ⇒ BC = (√5/2)·AB ⇒ ME = AX = AB / φ
[Folio Problèmes XXXVIII–XLVII | p. 29]

§ 3. Inversion (XXXVIII), Polarity (XXXIX), Radical Axis (XLI) & Harmonic Conjugate (XLVI)

Problems XXXVIII–XLVI reveal that Géométrographie naturally discovers minimal-instrument constructions whenever compass-only or straightedge-only syntheses beat mixed methods:

  • Problem XXXVIII (:term:`Circle Inversion <inversion>` — Compass Alone, :math:`5C_1 + 3C_3, S = 8`): For a point \(A\) with \(OA > R/2\), sweep \(A(O)\) (\(2C_1 + C_3\)) cutting \(O(R)\) at \(M, N\); since \(MO = NO = R\), sweeping \(M(O)\) and \(N(O)\) adds only \(3C_1 + 2C_3\). In total, zero lines and three circles (\(5C_1 + 3C_3, S = 8, E = 5\)) locate \(A'\) with \(OA \cdot OA' = R^2\)—the exact lemma by which August Adler (adler-1890) proved Mascheroni’s theorem!

  • Problem XXXIX (:term:`Polar of a Point <polaire>` — Straightedge Alone, :math:`12R_1 + 7R_2, S = 19`): Drawing two secants through A cutting circle \(O\) in four vertices of an inscribed complete quadrangle and joining diagonal intersections constructs the polar TT in seven lines and zero circles (\(S = 19, E = 12\), beating metric ruler-and-compass \(S = 25\)), identically to Steiner’s § 10 (steiner-1833).

  • Problem XLI (:term:`Radical Axis <linie-der-gleichen-potenzen>` of Two Non-Intersecting Circles — :math:`S = 17` vs. Mannheim’s :math:`S = 58`): Answering Amédée Mannheim (pp. 57–58), Lemoine contrasts Mannheim’s common-tangent midpoint construction (\(18R_1 + 9R_2 + 19C_1 + 12C_3, S = 58\)) with sweeping two completely arbitrary secant circles (\(2C_3\), \(0\) alignment cost!) cutting both given circles and joining chord pairs to two radical centers (\(10R_1 + 5R_2 + 2C_3, S = 17, E = 10\)).

  • Problem XLVI (:term:`Harmonic Conjugate <conjugue-harmonique>` — Straightedge Alone, :math:`9R_1 + 6R_2, S = 15`): Following Pappus (Synagoge Book VII, pappus-synagoge), the complete quadrilateral constructs \(D\) with \((A, B; C, D) = -1\) in 6 lines and 0 circles (\(9R_1 + 6R_2, S = 15, E = 9\)).

  • Problem XLVII (:person:`Eugène Catalan <catalan>`’s Square on :math:`AB`, :math:`S = 19`): Constructs vertices \(C, D\) in \(2R_1 + R_2 + 6C_1 + 4C_3\) (\(S = 13\)) and the full square in \(6R_1 + 3R_2 + 6C_1 + 4C_3\) (\(S = 19\) vs. classical \(S = 21\)).

[Folio Problèmes L–LXIII | p. 42]

§ 4. Modern Triangle Geometry: From \(X(2)\) and Lemoine’s \(X(6)\) to Brocard and Nagel

As co-founder of modern triangle geometry (lemoine-1873, altshiller-court-1952; cross-referenced to Clark Kimberling’s ETC (kimberling-etc) in ), Lemoine culminates his treatise by benchmarking the remarkable points of triangle ABC:

  • Problem L — Centroid :g:`G` :math:`= X(2)` (:math:`S = 17`): Sweeping \(B(AC)\) and \(C(AB)\) forms parallelogram \(ABDC\) whose diagonal \(AD\) bisects BC at \(M_a\) without tracing a mediatrix; intersecting two medians gives \(6R_1 + 3R_2 + 6C_1 + 2C_3\) (\(S = 17\), 2 circles vs. classical 3 circles, \(S = 18\)).

  • Problem LI — :term:`Lemoine Symmedian Point <point-de-lemoine>` :g:`K` :math:`= X(6)` (:math:`S = 27` vs. :math:`S = 54`): Constructing \(K = (a : b : c)\) via general trilinear coordinates (Problem XLIX) costs \(S = 54\). Lemoine halves the cost by constructing an antiparallel PaQa to BC and bisecting it by symmetric arcs so only the two symmedian lines AK and BK are ever drawn:

\[\mathrm{Op.} : \left(4R_1 + 2R_2 + 13C_1 + 8C_3\right), \qquad S = 27, \quad E = 17 \quad (2\text{ lines}, 8\text{ circles}).\]
A B C Pₐ Qₐ M'ₐ Mₐ G = X(2) K = X(6) Problem LI — Lemoine Symmedian Point K = X(6) : 4R₁ + 2R₂ + 13C₁ + 8C₃ (S = 27) Antiparallel PₐQₐ bisected by symmedian AK (isogonal to median AMₐ through G = X(2))

  • Problems LII–LIII — :term:`Lemoine Line <droite-de-lemoine>` (:math:`S = 19`) & Simultaneous :math:`(G, K)` (:math:`S = 32`): Trilinear polar of \(X(6)\) in \(6R_1 + 3R_2 + 6C_1 + 4C_3\) (\(S = 19\)); joint construction of \(X(2)\) and \(X(6)\) in \(12R_1 + 6R_2 + 9C_1 + 5C_3\) (\(S = 32\)).

  • Problems LIV–LV — :term:`Brocard Points <points-de-brocard>` :math:`omega, omega’` (:math:`S = 28` and :math:`S = 38` vs. Brocard’s :math:`S = 74`): Single Brocard point in \(8R_1 + 4R_2 + 10C_1 + 6C_3\) (\(S = 28\)); both \(\omega\) and \(\omega'\) simultaneously in \(12R_1 + 6R_2 + 12C_1 + 8C_3\) (\(S = 38\)), cutting Henri Brocard’s own letter construction (\(22R_1 + 11R_2 + 24C_1 + 17C_3, S = 74\), pp. 58–59) nearly in half!

  • Problems LVI–LXIII — Steiner :math:`X(99)` (:math:`S = 24`), Tarry :math:`X(98)` (:math:`S = 28`), Gergonne :math:`X(7)` (:math:`S = 27`), Spieker :math:`X(10)` (:math:`S = 29`), Nagel :math:`X(8)` (:math:`S = 16`), Neuberg :math:`Phi` (:math:`S = 37`), and Point :math:`W` (:math:`S = 148 to 62`): Notably, Nagel’s point \(X(8)\) requires only \(4R_1 + 2R_2 + 7C_1 + 3C_3\) (\(S = 16, E = 11\), 2 lines, 3 circles) by direct semiperimeter transfer without constructing excircles!