Center of Gravity (Posted on 2008-02-06)

	Submitted by Bractals
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Assume that each of the four points has a mass of one unit. Procedure: Replace points I and J with a point EJ having two units of mass and located at the midpoint of line segment IJ. Replace points K and L with a point OJ having two units of mass and located at the midpoint of line segment KL. Replace points EJ and OJ with a point O having four units of mass and located at the midpoint of line segment EJOJ. If we repeat the above procedure with the letters "JKL" replaced with the letters "KLJ" and then with "LJK", we see that the point O (which is the center of gravity of points I, J, K , and L) is the midpoint of line segments EJOJ, EKOK, and ELOL. Since I is the orthocenter of triangle JKL, the points EJ, OJ, EK, OK, EL, and OL are six of the nine points which define the nine-point circle for triangle JKL. Therefore, O is the center of the nine-point circle of triangle JKL. The other three points defining the nine-point circle are the feet of the altitudes of triangle JKL, but those feet are the vertices of triangle ABC. Therefore, the nine-point circle of triangle JKL is the circumcircle of triangle ABC. Therefore, the center of gravity of points I, J, K, and L is the circumcenter of triangle ABC. NOTE: Since the four points form an orthocentric system, the above proof is valid nomatter what permutation of the letters I, J, K, and L are applied to the four points.

	Subject	Author	Date
	Puzzle Answer	K Sengupta	2022-09-03 12:55:11
	I think I am centered on the solution now...	Dej Mar	2008-02-08 08:34:51
	re: Solution ???	Bractals	2008-02-07 13:10:29
	Solution ???	Dej Mar	2008-02-07 12:10:47
	re(2): ????	Dej Mar	2008-02-07 00:32:48
	re: ????	Charlie	2008-02-06 16:42:48
	????	J	2008-02-06 16:20:25
	playing with Geometer's Sketchpad (spoiler)	Charlie	2008-02-06 12:45:46