QA-Co3: Gergonne-Steiner Conic
In a random quadrangle it is generally impossible to circumscribe a circle through all 4 points. But it is always possible to circumscribe a conic (ellipse, parabola or hyperbola) through these 4 points. Technically, the conic with least eccentricity is the conic that deviates least from a circle.
The Gergonne-Steiner Conic is the conic through the vertices of the Reference Quadrangle with least eccentricity. It is the conic with center QA-P3 (Gergonne-Steiner Point). The problem of the Quadrangle conic with least eccentricity was initially posed in the “Annales de Gergonne” and solved by J. Steiner. See Ref-7 page 231.
(a2 p q r (2p+q+r) - (b2 r +c2 q) p2 (q+r)) y z
+ (b2 p q r (p+2q+r) - (c2 p +a2 r) q2 (p+r)) z x
+ (c2 p q r (p+q+2r) - (a2 q+b2 p) r2 (p+q)) x y = 0
(2 a2 q2 r2 - b2 r2 (p2+q2-r2) - c2 q2 ( p2 - q2+r2)) x2
+(2 b2 p2 r2 - a2 r2 (p2+q2-r2) - c2 p2 (-p2+q2+r2)) y2
+(2 c2 p2 q2 - a2 q2 (p2-q2+r2) - b2 p2 (-p2+q2+r2)) z2 = 0
- Jean-Pierre Ehrmann commented at Hyacinthos wrt this conic that when e = eccentricity,
then for this conic e2 = √ [2/(k+1)],
where k = O1P1*/R1 = O2P2*/R2 = O3P3*/R3 = O4P4*/R4,
Oi = circumcenter PjPkPl,
Pi*= isogonal conjugate of Pi wrt PjPkPl,
Ri = circumradius PjPkPl.
(See Hyacinthos (Ref-11) message #19970)
- Let d1=QA-P3.QA-P2 and d2=QA-P3.QA-P4. Now d1/d2 = (a2 + b2)/(a2 – b2) when QA-Co3 is an ellipse and d1/d2=(a2 – b2)/(a2 + b2) when QA-Co3 is a hyperbola, where a and b are resp. the lengths of the major and minor axes of QA-Co3 (Benedetto Scimemi, November 12, 2014).
- The axes of the Gergonne-Steiner Conic can be constructed as the angle bisectors of the angle QA-P2.QA-P3.QA-P4 (Benedetto Scimemi, November 12, 2014).
- The axes of the Gergonne-Steiner Conic are parallel to:
– the axes of the QA-Nine-point Conic (QA-Co1),
– the asymptotes of the QA-Orthogonal Hyperbola (QA-Co2).
– the reflection axes of the QA-Orthopole Transformation (QA-Tf3).