## 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 A_{0}, A_{1}, and A_{2} 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-1}A_{k+1}, let C_{k} be the foot of the altitude from A_{k}, and let D_{k} be the midpoint of A_{k}H.

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-1}C_{k+1} and B_{k+1}C_{k-1}, Q_{k} the intersection of C_{k-1}D_{k+1} and C_{k+1}D_{k-1}, and R_{k} the intersection of C_{k-1}C_{k+1} and D_{k-1}D_{k+1}.

Let e be the line through {P_{0}, P_{1}, P_{2}}, and f be the line through _{0}, Q_{1}, Q_{2}}._{0}, R_{1}, R_{2}};

(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.

|Contact| |Front page| |Contents| |Up|

Copyright © 1996-2018 Alexander Bogomolny

Let A_{0}, A_{1}, and A_{2} 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-1}A_{k+1}, let C_{k} be the foot of the altitude from A_{k}, and let D_{k} be the midpoint of A_{k}H.

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-1}C_{k+1} and B_{k+1}C_{k-1}, Q_{k} the intersection of C_{k-1}D_{k+1} and C_{k+1}D_{k-1}, and R_{k} the intersection of C_{k-1}C_{k+1} and D_{k-1}D_{k+1}.

Let e be the line through {P_{0}, P_{1}, P_{2}}, and f be the line through _{0}, Q_{1}, Q_{2}}._{0}, R_{1}, R_{2}};

(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_{k}A_{n} and to points B_{n}, C_{n}, A_{m} on line A_{k}A_{m}, we get that the three points P_{k}, G, and H, defined by _{k} = B_{m}C_{n} ∩ B_{n}C_{m},_{m}B_{m} ∩ A_{n}B_{n},_{m}C_{m} ∩ A_{n}C_{n},_{k} lie on the Euler line GH.

(b) Let N be the nine-point circle. Consider the cyclic quadrilateral C_{m}C_{n}D_{m}D_{n}. Because _{m}D_{m} ∩ C_{n}D_{n},_{k} = C_{m}D_{n} ∩ C_{n}D_{m},_{k} = C_{m}C_{n} ∩ D_{m}D_{n},_{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

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

|Contact| |Front page| |Contents| |Up|

Copyright © 1996-2018 Alexander Bogomolny

64668980 |