Newsgroups: sci.math From: ramsay@unixg.ubc.ca (Keith Ramsay) Subject: Re: Trisect an angle??? Date: Thu, 17 Sep 1992 17:26:11 GMT In article <19a4g0INNah4@bigboote.WPI.EDU> cubanski@tensor4.WPI.EDU (David J Cubanski) writes: >I have heard that such a thing is possible, but have never seen it >done. Is the technique straightforward enough to be posted?? If not, >could you supply a reference? Suppose that we have an angle. Draw a circle around the vertex, and extend one of the rays of the angle, so that we have a circle centered at O with points A, B, and C on the circumference, an angle AOB which we want to trisect, and a diameter AC. Mark the ruler (or use some equivalent trick, like holding the compass next to it) with marks at an interval congruent to the radius OA. Slide the ruler along, with one of the marks on the ray OC beyond C, and the other mark on the circle, until the ruler passes through the point B. Draw a line there. So now we have a point D on the circle, and a point E on the ray OC, such that the length of DE is equal to the radius of the circle, and E, D, and B are collinear in that order. The triangles DOB and EDO are isoceles, since two of the sides of each are congruent to the radius of the circle. Using the fact that the corresponding angles are equal, and that the sum of the angles of a triangle is 180 degrees, you can show that the angle AOB is three times that of OEB. Keith Ramsay ramsay@unixg.ubc.ca ============================================================================== Newsgroups: sci.math From: alopez-o@neumann.uwaterloo.ca (Alex Lopez-Ortiz) Subject: Re: NEW sci.math FAQ: The Trisection of an Angle Date: Thu, 10 Nov 1994 01:56:19 GMT In article <39r5t8\$hb1@dartvax.dartmouth.edu>, Benjamin.J.Tilly@dartmouth.edu (Benjamin J. Tilly) writes: > ... > I think that the solution using a compass and a ruler marked in 2 > places should be put in the FAQ. Unfortunately I do not remember the > construction off of the top of my head. Does anyone else remember it? I do. It is due to Archimedes amd consists in drawing a semicircle of arbitrary radius with base on one of the (extended) lines that form the angle. Then you draw a segment joining the point of intersection of the other line with the semicircle (point marked Z in the drawing) and the extended base of the line in such a way that the distance from the base to the first intersection point (point Y in diagram below) is the radius of the circle. __ - Z / / \ Y / j / / \ | / | -X------------------------------ I'm writing this from memory. Any one smart enough to check the angles using trigonometry? Alex ============================================================================== From: jurjus@kub.nl (H. Jurjus) Newsgroups: sci.math Subject: Re: NEW sci.math FAQ: The Trisection of an Angle Date: Thu, 10 Nov 1994 11:48 MET In Article <39r5t8\$hb1@dartvax.dartmouth.edu> "Benjamin.J.Tilly@dartmouth.edu (Benjamin J. Tilly)" says: (...) > I think that the solution using a compass and a ruler marked in 2 > places should be put in the FAQ. Unfortunately I do not remember the > construction off of the top of my head. Does anyone else remember it? > > Ben Tilly I'll have a go. (The originator of this proof is Archimedes, BTW.) _ ------ _ 'X" . " / " . 'B'. / . .o / . . / . --o---------------(-1,0)---------(0,0)-----------(1,0)---- 'A' 'O' 'E' (A, B, X on one line) Suppose X is a point on the unit circle such that /_ XOE is the angle we would like to 'trisect'. Draw a line AX through a point A on the x-axis such that AB = 1 (= radius of the circle = BO ), where B is the intersection-point of the line AX with the circle. Let a be /_ BAO Then /_ BOA = a, and /_ XBO = 2a (= /_ BXO ) Therefore: /_ AOB + /_ BOX + /_ XOE = pi/2 = /_ XBO + /_ BXO + /_ BOX a + /_ BOX + /_ XOE = 2a + 2a + /_ BOX Therefore: /_ XOE = 3a QED. (I'm not experienced enough in drawing lines in ASCII: I think one should draw the line AX.) H.Jurjus ============================================================================== From: Benjamin.J.Tilly@dartmouth.edu (Benjamin J. Tilly) Newsgroups: sci.math Subject: Re: Need angle trisection disproof help. Date: 24 Jan 1995 02:15:09 GMT In article <19950123143411.jurjus@pi0220.kub.nl> jurjus@kub.nl (H. Jurjus) writes: > In Article <3fp2k1\$a8n@dartvax.dartmouth.edu> "Benjamin.J.Tilly@dartmouth.edu (Benjamin J. Tilly)" says: > > > ... In fact the impossible > > problems are called that for a reason, namely they have been proven to > > be impossible with compass and unmarked ruler. If you check the FAQ I > > believe that they have the method for trisections that uses a compass > > and a ruler marked at 2 points... > > I didn't find it; maybe I looked at the wrong place. It used to be there, they may have changed it. Here is the construction... Suppose that the angle to cut is ABC. Continue the line BC. Now draw a circle around B with radius equal to the distance between the two points on the marked straight-edge. Let D be the point where the ray from B to A intersects this circle. Now take the straightedge, and slide it along to the point where it is touching D, one of the marks is touching another point of the circle, call it E, and the other mark is touching the line BC outside of the circle at a point that we will call F. Then EBF is exactly 1/3 of the original angle. (If you draw the diagram, note which edges are the same length as the distance between the two points so you have isocoles triangles, and then you just work out angles then it is easy to show that this works.) Of course the act of laying your marked straight-edge down in that fashion cannot be done unless the straight-edge has marks on it... ... Ben Tilly ============================================================================== From: Benjamin.J.Tilly@dartmouth.edu (Benjamin J. Tilly) Newsgroups: sci.math Subject: Re: trisect arbitrary angles HELP! Date: 26 Apr 1995 05:00:13 GMT In article <3nk17q\$so@falcon.abare.gov.au> Greg Griffiths writes: > As I recall, the proof that it is impossible to trisect an angle using > ruler and compass assumes that the ruler is used merely form the line > connecting two previously constructed points. {I think that there's a > proof that all ruler and compass constructions are possible with compass > alone (assuming that you don't actually want to draw lines just find > their points of intersection etc.) At the risk of inviting flames (please > don't, I'll freely admit that my memory is poor) I believe that there is > a method (know in pre-medieval times) for trisecting angles using ruler > and compass that involves using the ruler to record a length and then to > find the points AB such that A and B lie on preconstructed lines or rays > and AB has the length desired. However I may merely be another victing of > CRAFT disease (Can't Remember A Flamin' Thing). Sorry I can't provide a > reference. In fact this is true, and I have run across at least 2 ways to trisect an arbitrary angle using compass and a ruler marked in two places. The one that turns up on sci.math more often goes as follows. Suppose that a, b, c are three points and your angle is abc. Draw a circle centered at b with radius equal to the distance between your markings on the ruler. Extend the edges in the angle so they both intersect the circle. Extend the line from a to b into a line that intersects the circle at another point a'. Now slide the ruler along keeping the ruler touching where the ray from b to c touches the circle (call that poin c'), and keeping the far mark on the ruler touching the line from a to b, until the intermediate mark touches the circle at a point d (and the line at point d'). Then angle a'bd is exactly 1/3 of angle abc. Here is why. Draw the diagram and call angle a'bd (= angle d'bd) alpha. (None of this will make sense without a diagram present in front of you.) Note that edges d'd and db are the same length so angle dd'b is also alpha. Therefore angle d'db must be 180-2alpha. Therefore angle bdc' must be 2alpha. But bd and bc' are both the same length so angle dc'b is 2alpha. Therefore angle dbc' is 180-4alpha. However we have 180 = angle d'bd + angle dbc' + angle c'ba so angle c'ba is 3 alpha. However angle c'ba is angle cba, so that completes the demonstration. The other construction is in _A History Of Mathematics_ by Boyer, as well as a number of other solutions by the Greeks which arise if you allow yourself to use some special curve or other. (With those curves you can also double the cube and square the circle, which are other famous problems.) Ben Tilly