tag:blogger.com,1999:blog-6933544261975483399.post7373217908821292714..comments2024-03-26T19:10:02.918-07:00Comments on Go Geometry (Problem Solutions): Problem 410: Two Regular Pentagons, AngleAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger7125tag:blogger.com,1999:blog-6933544261975483399.post-61996099978541933572016-06-17T13:28:59.256-07:002016-06-17T13:28:59.256-07:00Problem 410
The circle described in both pentagon...Problem 410<br /><br />The circle described in both pentagons intersect at point Ν , then <ANB=36=<ANE=<br />=<END=<DNJ=<JNH ,but <BNG=36+36+36+36+36=180.Therefore section B, N, C and A, N, H is collinear so points M, N coincide.So <BMG=144.(<GMH=36).<br />APOSTOLIS MANOLOUDIS 4 HIGH SHCOOL OF KORYDALLOS PIRAEUS GREECE<br />APOSTOLIS MANOLOUDIShttps://www.blogger.com/profile/15561495997090211148noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-67043731321448534772012-03-19T22:08:44.112-07:002012-03-19T22:08:44.112-07:00http://img577.imageshack.us/img577/1528/problem410...http://img577.imageshack.us/img577/1528/problem410.png<br /><br />Note that triangles ADG and BDH are congruence ( case SAS) <br />∠ (XMD)= ∠ (MHD) => quadrilateral MGHD is cyclic<br />And ∠ ( GMX)= ∠ (GDH) = 36 ----- ( both angles face same arc GH)<br />And x= 180-36=180Peter Tranhttps://www.blogger.com/profile/02320555389429344028noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-54365447088964411792010-07-06T21:22:29.731-07:002010-07-06T21:22:29.731-07:00The problem is simple if you assume that there is ...The problem is simple if you assume that there is an answer to the problem. It is posed in a way that suggests that the answer is independent of the angle between the sides of the pentagon. If I assume that there is a unique answer to the problem, then I can orient the pentagons any way I choose without loss of generality. So I choose to orient them so that angle x occurs at D. Then x is simply the exterior angle of the well-know 36-72-72 triangle, and thus must measure 144 degrees.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-65360748701278742562010-01-11T01:11:23.074-08:002010-01-11T01:11:23.074-08:00in the complex plane we can choose O center from t...in the complex plane we can choose O center from the circumcircle of ABCDE as origin,affix of D<br />zD=1,then:<br />zC= exp(i2pi/5);zB= exp(i4pi/5);zA= exp(i6pi/5)<br />the similarity s can be written: z'=az+b<br />s(D)=D b=1-a<br />zG=azA+1-a;zH=azB+1-a<br />ang(HMG)=arg((zG-zA)/(zH-zB))<br />=arg[(exp(i4pi/5)-1)/(exp(i6pi/5)-1)]<br />=arg[exp(ipi/5).sin(2pi/5)/sin(3pi/5)]=pi/5<br />in degrees<br />x=180-36=144<br />.-.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-79982909477867675762010-01-07T10:45:49.574-08:002010-01-07T10:45:49.574-08:00To Anonymous:
The answer is wrong.To Anonymous:<br />The answer is wrong.Antonio Gutierrezhttps://www.blogger.com/profile/04521650748152459860noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-45281477029463298832010-01-07T10:22:57.625-08:002010-01-07T10:22:57.625-08:00the two pentagons are similar ,by the similarity s...the two pentagons are similar ,by the similarity s,with center D,angle (108+m(CDF)),ratio GH/AB<br />s(B)=H;s(A)=G<br />x=108+m(CDF)<br />.-.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-59471398577118057622009-12-27T10:20:09.487-08:002009-12-27T10:20:09.487-08:00http://geometri-problemleri.blogspot.com/2009/12/p...http://geometri-problemleri.blogspot.com/2009/12/problem-58-ve-cozumu.htmlAnonymousnoreply@blogger.com