tag:blogger.com,1999:blog-6933544261975483399.post1415868295311675011..comments2022-09-27T03:11:10.165-07:00Comments on Go Geometry (Problem Solutions): Problem 286: Square, Midpoints, OctagonAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger7125tag:blogger.com,1999:blog-6933544261975483399.post-17245251166148749732021-03-13T05:59:36.787-08:002021-03-13T05:59:36.787-08:00The previous comment should be improved as follows...The previous comment should be improved as follows to make it completely accurate:<br /><br />Another way to prove that the octagon is equilateral is to see that each vertice Pn has a predecessor Pn-1 and a successor Pn+1 that are such that Pn-1 and Pn+1 are the symetric of one another accross either AC, BD, EG or FH <b>to which Pn belongs</b>. Therefore the length of each segment Pn-1_Pn is equal to that of segment Pn_Pn+1.Greghttps://www.blogger.com/profile/14941282981782772132noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-90215968303620758412021-02-27T08:31:22.005-08:002021-02-27T08:31:22.005-08:00Another way to prove that the octagon is equilater...Another way to prove that the octagon is equilateral is to see that each vertice Pn has a predecessor Pn-1 and a successor Pn+1 that are such that Pn-1 and Pn+1 are the symetric of one another accross either AC, BD, EG or FH. Therefore the length of each segment Pn-1_Pn is equal to that of segment Pn_Pn+1.Greghttps://www.blogger.com/profile/14941282981782772132noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-50154717962658961972021-01-08T07:15:05.421-08:002021-01-08T07:15:05.421-08:00Let EG and HF meet at O. Now EG goes thro' 2 v...Let EG and HF meet at O. Now EG goes thro' 2 vertices of the octagon, name them U and V<br /><br />OV = VG since Tr.s CVG & HOG are congruent. And so it can easily be proved that all the lines which connect O to the vertices of the octagon are of equal length x. <br /><br />Now < HOD = < COD since Tr.s HOD & COD are congruent & each and angle = 45<br />Therefore we have the octagon broken down into 8 congruent Tr.s SAS and all the sides are = x and the octagon is therefore a regular octagon which is the 1st required result<br /><br />Now HD = a/2 = AH, VG = a/4 from the midpoint theorem and so VH = x + 2x = 3x from similar triangles<br /><br />Hence CH = V5a/2 = 6x from which x = V5a/12 which is the 2nd required result<br /><br />Sumith Peiris<br />Moratuwa<br />Sri LankaSumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-5020809035858656582021-01-07T10:32:59.654-08:002021-01-07T10:32:59.654-08:00For the first part see congruent forms
https://pho...For the first part see congruent forms<br />https://photos.app.goo.gl/JDrBZ1sPdG864iBE6c.t.e.o.https://www.blogger.com/profile/16937400830387715195noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-70937616426048056782019-01-04T13:05:15.742-08:002019-01-04T13:05:15.742-08:00It is an equilateral octagon because all the verti...It is an equilateral octagon because all the vertices lie on a line of symmetry for the given square. Using the horizontal symmetry line and vertical symmetry line only and facts of rotational symmetry, one can prove that we have 8 pairs of congruent triangles via one of the four commonly used triangle congruences. Via CPCTC, we can show that all sides of the octagon are congruent thus proving the figure equilateral. A similar series of arguments could made using the diagonal symmetry lines as starting points. Equilateral doesn't necessarily mean regular here.Geek37https://www.blogger.com/profile/12171388277139538068noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-58294268201867759212010-11-22T12:47:11.085-08:002010-11-22T12:47:11.085-08:00Sorry, the octagon can't be regular (equilater...Sorry, the octagon can't be regular (equilateral) because if it would be, the points A,H,D,G,C,F,B,E of the intersections would lay on a circle!<br />but it is approximative an equilateral octagon.<br /><br />sincerely<br />gwenglerAnonymousnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-74154735438616153192009-04-27T20:39:00.000-07:002009-04-27T20:39:00.000-07:00One may do this by using Menelaus's Theorem or...One may do this by using Menelaus's Theorem or by analytical geom. Following the latter, let A be (0,0)and name the polygon vertex near H as P and then go clockwise with the vertices as Q,R,S,T,U,V & W. Then B:(0,a),C:(a,a) and D:(a,0).Now AG is y=x/2 and DE is x/a+2y/a=1. Thus, P is (a/2,a/4). Likewise, BH is 2x/a+y/a=1. Hence, Q is (a/3,a/3) and PQ^2 =(a/2-a/3)^2+(a/4-a/3)^2 or PQ =V5*a/12. Likewise, one may show QR=RS=ST=TU=UV=VW=WP=V5*a/12. Hence PQRSTUVW is a regular polygon of side = V5*a/12<br />Ajit: ajitathle@gmail.comAjithttps://www.blogger.com/profile/00611759721780927573noreply@blogger.com