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 B_k be the midpoint of A_k-1A_k+1, let C_k be the foot of the altitude from A_k, and let D_k be the midpoint of A_kH.

The nine-point circle of T is the circle through all B_k, C_k, and D_k. 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 P_k be the intersection of B_k-1C_k+1 and B_k+1C_k-1, Q_k the intersection of C_k-1D_k+1 and C_k+1D_k-1, and R_k the intersection of C_k-1C_k+1 and D_k-1D_k+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 © 1996-2018 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 B_k be the midpoint of A_k-1A_k+1, let C_k be the foot of the altitude from A_k, and let D_k be the midpoint of A_kH.

The nine-point circle of T is the circle through all B_k, C_k, and D_k. 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 P_k be the intersection of B_k-1C_k+1 and B_k+1C_k-1, Q_k the intersection of C_k-1D_k+1 and C_k+1D_k-1, and R_k the intersection of C_k-1C_k+1 and D_k-1D_k+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 B_m, C_m, A_n on line A_kA_n and to points B_n, C_n, A_m on line A_kA_m, we get that the three points P_k, G, and H, defined by P_k = B_mC_n ∩ B_nC_m, G = A_mB_m ∩ A_nB_n, and H = A_mC_m ∩ A_nC_n, are collinear. So all P_k lie on the Euler line GH.

(b) Let N be the nine-point circle. Consider the cyclic quadrilateral C_mC_nD_mD_n. Because H = C_mD_m ∩ C_nD_n, Q_k = C_mD_n ∩ C_nD_m, and R_k = C_mC_n ∩ D_mD_n, we conclude that points Q_k and R_k 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 © 1996-2018 Alexander Bogomolny