Bisect The Segment (Posted on 2005-01-29)

	Submitted by David Shin
Rating: 4.0000 (4 votes)
Solution:	(Hide)
(bits and pieces borrowed from Charlie's solution) Let the line segment be AB. Construct two circles of radius AB: one centered at A, and one centered at B. Maintain the compass width and mark off three 60-degree arcs on the circle centered at A to find the point C diametrically opposite B. Place the point of the compass on point C, and set the length to AC (which is twice AB). Swing an arc to cut the circle centered at B and call this intersection D. Reset the compass to length AB and place its point at D. Inscribe an arc to cut AB at a point M. Claim: M is the midpoint of AB. Proof: Assume AB to be of unit length. Then, triangle CBD is isosceles with a base BD of length 1 and sides of length 2. Triangle DMB is also isosceles, with sides DM and BD of length 1. Its angle at the base, DBM, is coincident with a base angle of the larger isosceles triangle, CBD, and so the triangles are similar. Since the ratio is similarity is 1/2, we find that BM = 1/2, implying that M is the midpoint of AB, as desired.

	Subject	Author	Date
	re(3): solution	owl	2005-01-31 17:27:33
	re(3): solution	owl	2005-01-31 16:18:00
	re(2): solution	David Shin	2005-01-31 07:08:36
	re: solution	owl	2005-01-31 05:25:43
	re: solution	Charlie	2005-01-30 16:13:38
	solution	Alec	2005-01-30 09:47:19
	trick answer	Larry	2005-01-30 04:28:17
	Guess	Penny	2005-01-30 01:36:40