The Kariya Problem and Related Constructions

Given a point Q other than the incenter I of a reference triangle, we give a simple conic construction of a homothety mapping I into Q so that the image of the intouch triangle is perspective with the reference triangle. This is a generalization of the Kariya theorem in the case Q = I that the homothety can be arbitrary. The ratio of the homothety (the Kariya factor) is a unique nonzero finite number except when Q lies on the Feuerbach hyperbola or the line joining the incenter to the orthocenter of the reference triangle. For each nonzero real number t, we show that the locus of Q with Kariya factor t is a rectangular hyperbola. We give two simple constructions of this hyperbola. 1. The Kariya problem This note presents several constructions related to the Kariya problem. Given a triangle T := ABC with incenter I , let the incircle be tangent to the sides BC, CA, AB at A1, B1, C1 respectively. A1B1C1 is the intouch triangle of ABC. For a real number t, let Ia(t), Ib(t), Ic(t) be points on the lines IA1, IB1, IC1 respectively, such that as vectors, IIa(t) = tIA1, IIb(t) = tIB1, IIc(t) = tIC1. Theorem (Kariya). For every real number t, the triangle TI(t) := Ia(t)Ib(t)Ic(t) is perspective with T at a point on the Feuerbach hyperbola, the rectangular circum-hyperbola through I and H , the orthocenter of T (see Figure 1). The Kariya problem studies the case when the incenter is replaced by an arbitrary point. We begin with a sign convention for distances along lines perpendicular to the sidelines of T. For two points Y and Z on a line perpendicular to BC, the distance Y Z is reckoned positive or negative according as the vector YZ is directly or oppositely parallel to IA1; similarly for points on lines perpendicular to CA and AB respectively. Given a point Q and a real number t, we denote by Qa(t), Qb(t), Qc(t) the unique points on the perpendiculars from Q to BC, CA, AB respectively with QQa(t) = QQb(t) = QQc(t) = tr, where r is the inradius of T (see Figure 2). In absolute barycentric coordinates, Qa(t) = Q+t(A1−I), Qb(t) = Q+t(B1−I), Qc(t) = Q+t(C1−I). Lemma 1. Triangle TQ(t) is homothetic to the intouch triangle TI(1) = A1B1C1. Proof. Let T be the point dividing QI in the ratio QT : TI = −t : 1. It is clear that TQa(t) : TA1 = TQb(t) : TB1 = TQc(t) : TC1 = TQ : TI = t : 1. Publication Date: October 7, 2015. Communicating Editor: Nikolaos Dergiades.