Points Generated by the Nine Points

The following engaging problem has been proposed by Michael Goldenberg and Mark Kaplan, with their solution published by American Mathematical Monthly, v 117 (2010), p. 281. The attraction of the problem to me is twofold. Together, the formulation and the solution rely in a very immediate way on several - I would even say, on an excessive number of - well known theorems and, with that, the solution appears noticeably shorter than the formulation. The editors of the Monthly found it necessary to observe that Most solvers proceeded analytically. Some solvers simplified the algebra by using complex numbers or determinants. Some used Maple to help.

Let A0, A1, and A2 be the vertices of a nonequilateral triangle T. Let G and H be the centroid and orthocenter of T, respectively. Treating all indices modulo 3, let Bk be the midpoint of Ak-1Ak+1, let Ck be the foot of the altitude from Ak, and let Dk be the midpoint of AkH.

The nine-point circle of T is the circle through all Bk, Ck, and Dk. We now introduce nine more points, each obtained by intersecting a pair of lines. (The intersection is not claimed to occur between the two points specifying a line.) Let Pk be the intersection of Bk-1Ck+1 and Bk+1Ck-1, Qk the intersection of Ck-1Dk+1 and Ck+1Dk-1, and Rk the intersection of Ck-1Ck+1 and Dk-1Dk+1.

Let e be the line through {P0, P1, P2}, and f be the line through {Q0, Q1, Q2}. (By Pascal's theorem, these triples of points are collinear.) Let g be the line through {R0, R1, R2}; by Desargues' theorem, these points are also collinear.

(a) Show that the line e is the Euler line of T .
(b) Show that g coincides with f .
(c) Show that f is perpendicular to e.
(d) Show that the intersection S of e and f is the inverse of H with respect to the nine-point circle.

Solution

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Copyright © 1996-2017 Alexander Bogomolny

Let A0, A1, and A2 be the vertices of a nonequilateral triangle T. Let G and H be the centroid and orthocenter of T, respectively. Treating all indices modulo 3, let Bk be the midpoint of Ak-1Ak+1, let Ck be the foot of the altitude from Ak, and let Dk be the midpoint of AkH.

The nine-point circle of T is the circle through all Bk, Ck, and Dk. We now introduce nine more points, each obtained by intersecting a pair of lines. (The intersection is not claimed to occur between the two points specifying a line.) Let Pk be the intersection of Bk-1Ck+1 and Bk+1Ck-1, Qk the intersection of Ck-1Dk+1 and Ck+1Dk-1, and Rk the intersection of Ck-1Ck+1 and Dk-1Dk+1.

Let e be the line through {P0, P1, P2}, and f be the line through {Q0, Q1, Q2}. (By Pascal's theorem, these triples of points are collinear.) Let g be the line through {R0, R1, R2}; by Desargues' theorem, these points are also collinear.

(a) Show that the line e is the Euler line of T .
(b) Show that g coincides with f .
(c) Show that f is perpendicular to e.
(d) Show that the intersection S of e and f is the inverse of H with respect to the nine-point circle.

Solution

(a) Let k, m, n be 1, 2, 3 in some order. Applying Pappus's theorem to points Bm, Cm, An on line AkAn and to points Bn, Cn, Am on line AkAm, we get that the three points Pk, G, and H, defined by Pk = BmCn ∩ BnCm, G = AmBm ∩ AnBn, and H = AmCm ∩ AnCn, are collinear. So all Pk lie on the Euler line GH.

(b) Let N be the nine-point circle. Consider the cyclic quadrilateral CmCnDmDn. Because H = CmDm ∩ CnDn, Qk = CmDn ∩ CnDm, and Rk = CmCn ∩ DmDn, we conclude that points Qk and Rk are on the polar of H with respect to N. So f and g coincide.

(c) By the definition of polar, we have NH ⊥ f or e⊥ f.

(d) This also follows from the definition of polar.

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Copyright © 1996-2017 Alexander Bogomolny

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