tag:blogger.com,1999:blog-6933544261975483399.post6980044113620360481..comments2022-09-27T03:11:10.165-07:00Comments on Go Geometry (Problem Solutions): Archimedes' Book of Lemmas, Proposition #1Antonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger13125tag:blogger.com,1999:blog-6933544261975483399.post-22489819298486227782018-04-09T11:49:12.033-07:002018-04-09T11:49:12.033-07:00Join A to O1 and O, => A, O1, O collinear =>...Join A to O1 and O, => A, O1, O collinear => CO1A = EOA <br />=> O1CA = OEA c.t.e.onoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-20507114212650387582015-11-04T23:57:10.612-08:002015-11-04T23:57:10.612-08:00Let O be the centre of the larger circle and O1 th...Let O be the centre of the larger circle and O1 that of the smaller circle<br /><br />Now OA and O1A are both perpendicular to the common tangent at A hence O, O1 and A are collinear <br /><br />Now O1A = O1C and OA = OE so < O1AC = < O1CA = < OEA and since BC// DE this can only happen if A,C,E are collinear <br /><br />Similarly A,B,D are collinear <br /><br />This is also similarly proved for externally tangent circles<br /><br />Sumith Peiris<br />Moratuwa<br />Sri LankaSumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-47625571779448034742010-01-17T13:33:14.921-08:002010-01-17T13:33:14.921-08:00bjhvash44@sbcglobal,net
Draw diameter from A thru ...bjhvash44@sbcglobal,net<br />Draw diameter from A thru centers F smaller G other AnglesAFC AGE are equal Corr angles parallel lines arc AC =AE. Angle between AC and tangent at A=1/2 arc AC.Same angle must be 1/2 of AE This equality cannot be maintainted unless ACE is a common chord Give the line a bend and there will be no equalityAnonymousnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-65238346717951036422009-05-28T14:28:20.212-07:002009-05-28T14:28:20.212-07:00Oh boy, I did not intend to reduce every statement...Oh boy, I did not intend to reduce every statement I make to Euklid's axioms. There are some facts that are well-known, and it makes communication about geometry a lot easier, if the others accept that.<br /><br />But ok, here is the proof, that A, M and N lie on a common line: Draw the tangent t to the big circle at the tangency point A. This must also be a tangent to the small circle, as if there was a second point of intersection, there also had to be points of the small circle on the other side of t, which is impossible, as the small circle remains completely inside the big one.<br /><br />Then the radii of both circles, that end in A, are perpendicular to t, which means that they both lie on the same line. Right?Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-54360497635101603052009-05-28T05:20:05.570-07:002009-05-28T05:20:05.570-07:00In mathematics you cannot assume anything is 'clea...In mathematics you cannot assume anything is 'clear'. Do I have a proof for this? Yes, I do.<br /><br />Suppose that more than one radial axis exists and arrive at a contradiction.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-79622461869467638732009-05-27T13:57:10.475-07:002009-05-27T13:57:10.475-07:00I thought it was clear, that, if two circles are t...I thought it was clear, that, if two circles are tangent, their centers and the point of tangency lie on the same line. Did you reclaim a proof for this?Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-36516562990408119332009-05-26T16:26:34.326-07:002009-05-26T16:26:34.326-07:00No, your proof would still be incorrect. You have ...No, your proof would still be incorrect. You have to establish that AMN is the radial axis of both circles. Without this, you cannot be sure that E and C lie on a straight line.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-82813021231397893612009-05-26T03:04:44.886-07:002009-05-26T03:04:44.886-07:00Alright, one can clarify some details, bu that doe...Alright, one can clarify some details, bu that does not change, that the proof is correct.<br /><br />First, that BC and DE are diameters is explicitly given (and the proposition would not hold if we dropped that premise).<br /><br />That M and N are the midpoints of BC and DE, respectively (and thus the centers of the circles), is my definition, because they were not named and I needed them for the proof.<br /><br />The main argument then was that E is the image of C, and the property of dilatation is, that a point, its image and the center always lie on the same line. That E is the image of C follows from the facts, that it has to lie on the big circle (the image of te small one) and on DE (the image of BC). This should do the proof. (Thomas)Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-86116950737494703972009-05-23T06:22:39.777-07:002009-05-23T06:22:39.777-07:00To: Anonymous of Dilation Proof
By assuming that M...To: Anonymous of Dilation Proof<br />By assuming that M is the midpoint of BC and N the midpoint of DE, you have in fact assumed quite a lot that is central to the proof. For example, how do you know that BC will intersect AE at C and DN will intersect AE at E? Furthermore, you also assume that BC and DE are diameters. It is true that they are, but you have not given a reason. Is it self-evident or is there a proof for this?<br /><br />Once this is established, the proof is simple.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-63460275921853873122009-05-19T03:59:00.000-07:002009-05-19T03:59:00.000-07:00Could you tell me your problem with my solution by...Could you tell me your problem with my solution by dilatation? What in particular seems uncorrect to you?Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-35206734028559380532009-05-18T04:17:00.000-07:002009-05-18T04:17:00.000-07:00? The first two solutions on Cut-the-knot are corr...? The first two solutions on Cut-the-knot are correct. Where do you see a problem? The first solution assumes you can construct a radial axis that includes both the circle diameters. However, this is axiomatic.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-79906882985522293932009-05-14T15:21:00.000-07:002009-05-14T15:21:00.000-07:00I don't think so. Your dilation is possible, but i...I don't think so. Your dilation is possible, but it does not prove the proposition. <br /><br />Aside:<br />I have looked at the 'solutions' on 'Cut the knot' and the first two are in error.<br /><br />The author of this lovely site (Guttierez) claims these are nice 'high school' problems. Although most of them are easy to solve, proposition 1 is very difficult given that diameters are parallel and point of intersection is internal. I wonder why he does not post his solutions?Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-83788489101845800352009-05-13T13:00:00.000-07:002009-05-13T13:00:00.000-07:00Well, it may be different from Archimedes' proof, ...Well, it may be different from Archimedes' proof, but here is my proposition:<br />Let the midpoint of BC be M and the Midpoint of DE be N.<br />Consider a dilatation with center A, that maps M to N. As by dilatation, every line is mapped to a parallel, BC (through M) has to be mapped to DE (through N). Obviously, the small circle is mapped to the big one.<br /><br />Therefore the points of intersection of the parallels with the circles are mapped to each other, more precisely C is mapped to E, and so they lie on the same line through the center A.Anonymousnoreply@blogger.com