Precariously Balanced (Posted on 2009-12-22)

	Submitted by brianjn
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Firstly the sum of the lengths of the shorter arms must be greater than the length of the longest arm. The task can be simply accomplished with a little geometry and without knowing any angles. For ease of explanation consider the arms of the "Y" to be the sides of a triangle of lengths in the ratio of a 3-4-5 triangle. Move the "3" and "4" sides to the same end of the "5" side but keep them parallel to their original position ensuring that they form a "Y". Consider the extremities of the "Y" as vertices of a triangle. This triangle is also of the ratio of the original triangle with the 'centre' of the "Y" being the centroid of the newly formed triangle. The requirement then is that the centre point of the arms is to be the centroid of a triangle which can be formed by the extremities of the arms.

	Subject	Author	Date
	Simplicity?	brianjn	2009-12-27 23:15:11
	re: Three solutions	Kenny M	2009-12-27 10:48:19
	Three solutions	Tony	2009-12-27 04:00:42
	Another general soution method	Kenny M	2009-12-24 11:17:55
	Somthing missing in "simple soultion"?	Kenny M	2009-12-24 10:59:05
	Rather Simple Solution	Brian Smith	2009-12-23 22:59:57
	Is this the solution intended?	Kenny M	2009-12-22 22:25:03
	re: Insufficient info to answer?	brianjn	2009-12-22 22:12:03
	Antoehr question - alternative?	Kenny M	2009-12-22 22:11:02
	Insufficient info to answer?	Kenny M	2009-12-22 18:45:34