tag:blogger.com,1999:blog-6933544261975483399.post1604700883767909938..comments2024-03-26T19:10:02.918-07:00Comments on Go Geometry (Problem Solutions): Elearn Geometry Problem 63Antonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger6125tag:blogger.com,1999:blog-6933544261975483399.post-5327426440932287152020-03-04T13:20:08.251-08:002020-03-04T13:20:08.251-08:00https://www.youtube.com/watch?v=u4m2Ewj1Vakhttps://www.youtube.com/watch?v=u4m2Ewj1VakGeek37https://www.blogger.com/profile/12171388277139538068noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-53809849518700977632018-12-16T04:24:39.117-08:002018-12-16T04:24:39.117-08:00Hello, the graph supporting my proof can be found ...Hello, the graph supporting my proof can be found there:<br /><br />https://drive.google.com/file/d/1MwCoNGh443qVvTIBlmLH4dbLLyRIYGMq/view<br /><br />GregGreghttps://www.blogger.com/profile/14941282981782772132noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-43390747627074519702018-12-15T05:55:47.413-08:002018-12-15T05:55:47.413-08:00Extend segments AB and DC to intersect outside of ...Extend segments AB and DC to intersect outside of the circle in I (see graph).<br />ABCD cyclic quadrilateral implies <IBC = <ADC = <BAD = <ICB = 2π/7, so ΔIAD and ΔIBC are similar and IB/BC = IA/AD (1).<br />Consider ΔCAE and ΔIAD : they are equal (isoceles with same base length d and same base angles 2π/7). So IA = CA = c and from (1) IB = s*c/d.<br />Also, IA = AB+IB = s+IB so IB = c-s.<br />So the following holds :<br />c-s = s*c/d<br />c = s+s*c/d<br />Dividing both terms by s*c : 1/s = 1/c + 1/d QEDGreghttps://www.blogger.com/profile/14941282981782772132noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-10528368838799519122012-11-01T02:11:29.497-07:002012-11-01T02:11:29.497-07:00Apply inversion on A with any inversion radius (A ...Apply inversion on A with any inversion radius (A -> A' ; B -> B' ...etc)<br />Then we can rewrite the statement by proving A'B'=A'C'+A'D'.<br /><br />The inversion image is easy : AB,AC,AD remains unchanged while the circumcircle becomes a line B'C'D' not passing through A.<br />By the conformal properties, ∠BAC=∠B'A'C'=∠C'A'D'=∠A'B'C'=180/7<br /><br />Construct a point E' on A'B' such that A'E' = A'C', so we simply need to prove E'B'=A'D'.<br />This is actually quite obvious. <br /><br />A'C'=B'C'<br />∠E'B'C'=∠C'A'D'=180/7<br />∠B'E'C'=720/7=∠B'D'A'<br />So ΔB'C'E'=ΔA'C'D', and hence E'B'=A'D<br /><br />Q.E.D.<br />W Fungnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-50864648003138823622012-10-31T18:33:58.193-07:002012-10-31T18:33:58.193-07:00Problem 63 solution. This solution was submitted b...Problem 63 solution. This <a href="http://gogeometry.com/problem/p063-geometry-dan-suttin-san-antonio-tx.pdf" rel="nofollow">solution was submitted by Dan Suttin from San Antonio, TX</a> <br><br />Thanks Dan.Antonio Gutierrezhttps://www.blogger.com/profile/04521650748152459860noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-91115279528732994752008-09-08T15:21:00.000-07:002008-09-08T15:21:00.000-07:00Let AB = CD = DE = s, AC = CE = c, AD = AE = d.App...Let AB = CD = DE = s, AC = CE = c, AD = AE = d.<BR/>Applying Ptolemy's Theorem to cyclic quadrilateral ACDE, we obtain cs + ds = cd.<BR/>Then s(c + d) = cd, and so 1/s = (c + d)/cd = 1/d + 1/c.<BR/><BR/>Therefore, in regular heptagon ABCDEFG, 1/s = 1/c + 1/dAnonymousnoreply@blogger.com