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 Bisect The Segment (Posted on 2005-01-29)
Show how to bisect a line segment with a compass (no straight-edge).

 See The Solution Submitted by David Shin Rating: 4.0000 (4 votes)

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 re: solution | Comment 4 of 8 |
(In reply to solution by Alec)

Re: "Keeping the compasses at AB mark off the 120deg & 180deg points on the circle centred at B.  Call the 180deg mark C.  "

This does not specify how to mark those points, nor specify what the angular (arc) measurements are made from.  In fact, if, as is likely, we had A on the left and B on the right, and we take the usual convention of zero degrees on the right and measure counterclockwise, the 180 deg. mark would actually be point A, and C would coincide with A.  Trying out the subsequent details, namely that AC equals twice AB, it's apparent that the 180 degree mark, C, is in the opposite direction from A, about B.  The instructions still do not explicitly state how to mark that point.  I think the following will be a better description:

Call the line segment AB.

Set the compass to AB and make 2 circles--one around A and one around B.

Call one of the intersections of these two circles C.  It lies at 60 degrees from A on circle B.  Place the point of the compass at point C, leaving the radius the length AB.  Mark off another segment of the circle centered on B with this setting; call that point D, which is then 120 degrees away from A on circle B. Again, place the point of the compass on point D, and mark off another intersection with circle B; call it point E.  This point will be directly opposite point A on circle B.

Place the point of the compass on point E, and set the length to AE (which is twice AB). Swing an arc to cut the circle centered at A and call this intersection F.

Reset the compass to length AB and place its point at F. Inscribe an arc to cut AB.  Call the intersection point M, which is the midpoint of AB.

Proof: Call the unit of length the length of AB. Triangle AEF is isosceles with a base AF of length 1 and sides of length 2. Triangle AFM is also isosceles, with sides AF and FM of length 1. Its angle at the base, FAM, is coincident with a base angle of the larger isosceles triangle, FAE, and so the triangles are similar, and the base AM is 1/2; that is, 1/2 of AB.

Edited on January 30, 2005, 4:15 pm
 Posted by Charlie on 2005-01-30 16:13:38

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