tag:blogger.com,1999:blog-6933544261975483399.post1000595403260562762..comments2024-03-26T19:10:02.918-07:00Comments on Go Geometry (Problem Solutions): Problem 429: Circumscribed and Inscribed Regular Pentagon, Perpendicular, MeasurementAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger4125tag:blogger.com,1999:blog-6933544261975483399.post-51439235949574468642010-06-23T09:51:53.514-07:002010-06-23T09:51:53.514-07:00The sweetest proof seems to be this...
Let x, y, z...The sweetest proof seems to be this...<br />Let x, y, z be the perpendicular distances from O to AE, A'E', A''B'’ respectively.<br />Note that x = the radius of circle O, so x=OA', and also that z=b/2.<br />Since all regular pentagons are similar,<br /> x:y:z= a:b:c.<br />But x^2 = y^2 + z^2 (Pythagoras), <br />so a^2 = b^2 + c^2. QED<br /><br />Corollary: In fact, z=b/2 =(b/2c)×c; <br />so y=(b/2c)×b=b^2/2c and x=(b/2c)×a = ab/2c.<br />So if d=diameter of circle O, <br />then d=2(ab/2c)=ab/c. ∴ ab=cd. How elegant!MasterStreamnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-69224358374475065522010-04-09T04:31:53.700-07:002010-04-09T04:31:53.700-07:00rotate A'B'C'D'E' so that A...rotate A'B'C'D'E' so that A' is on AB<br />mE'A'O=mE"A"O => O,A",A',E' are concyclic => mA"OE'=pi/2=mEE'O => A"O//E'E, note that mE'A"O=mOEE' so A"OEE' is a //gram and A"E'=OE<br />consider OA"E' we have OE^2=A"E'^2=OE'^2+OA"^2<br />as three regular pentagons are similar to each others, it yields a^2=b^2+c^2Jankonyexnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-39109018867720103232010-03-07T07:38:51.629-08:002010-03-07T07:38:51.629-08:00in [0;90[ 18 is the only root of
(1-sinx)-cosx.tan...in [0;90[ 18 is the only root of<br />(1-sinx)-cosx.tan2x=0<br />4sin18=sqr(5)-1<br />4cos18=sqr(10+2sqr5)<br />4tan36=(sqr(5)-1).sqr(10-2sqr5)<br />this numbers are related with the golden ratio<br />.-.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-79933383998134750962010-03-06T18:14:26.596-08:002010-03-06T18:14:26.596-08:00From figure geometry: (b/2)/sin(36) =(a/2)/tan(36)...From figure geometry: (b/2)/sin(36) =(a/2)/tan(36) or b=acos(36). Moreover, (c+btan(18))cos(18)=b which gives us: c =b(1-sin(18))/cos(18)=btan(36) or c = a*cos(36)*tan(36)=a*sin(36)<br />Thus: b^2+ c^2 = a^2[(cos(36))^2 +sin(36))^2]=a^2 QED<br />PS: Will someone pl. explain why (1-sin(18))/cos(18)= tan(36)?Ajithttps://www.blogger.com/profile/00611759721780927573noreply@blogger.com