tag:blogger.com,1999:blog-6933544261975483399.post6591721015203454323..comments2024-03-26T19:10:02.918-07:00Comments on Go Geometry (Problem Solutions): Archimedes' Book of Lemmas, Proposition #15Antonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger3125tag:blogger.com,1999:blog-6933544261975483399.post-2239461872153988542015-11-04T23:29:42.171-08:002015-11-04T23:29:42.171-08:00We first note that chord BC subtends an angle of 3...We first note that chord BC subtends an angle of 36 at the circumference and thus 72 at the centre. CD and DB subtend 18 at the circumference and AC subtends 54. <br /><br />Hence Tr.s ACF and AGF are congruent ASA. Hence AC = AG and thus < AGC = 72 = < COG. Hence CO = CG = R the radius. <br /><br />Now < CBO = 54 and < BCD = 18 so < CEB = 36 and so CG = GE = R<br /><br />Sumith Peiris<br />Moratuwa<br />Sri LankaSumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-74394129547720774712011-08-03T07:54:26.614-07:002011-08-03T07:54:26.614-07:00Sketch:
OG = AG - R = AC - R = 2R cos 36 - R
BE = ...Sketch:<br />OG = AG - R = AC - R = 2R cos 36 - R<br />BE = BD = 2R sin 18<br />OG - BE = 2R (cos 36 - sin 18) - R<br />=2R(1/2)- R = 0<br />So OG = BE, OG + GB = GB + BE, <br />EG = R <br />(Center is O & radius is R)Pravinhttps://www.blogger.com/profile/05947303919973968861noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-59629509380083131142011-08-03T01:21:28.276-07:002011-08-03T01:21:28.276-07:00http://img839.imageshack.us/img839/3698/problem655...http://img839.imageshack.us/img839/3698/problem655.png<br />Let H is the projection of C over AB<br />Observe that (CAD)=(DAB)=(DCB)=18<br />And (OBC)=(OCB)=54<br />(CEB)=36<br />AF is angle bisector of (CAB)<br />So we have CF/FB=AC/AB=HG/GB=HC/BC<br />So CG is angle bisector of (HCB) and (HCG)=(GCB)=18=(OCH)<br />And OCG and ACE are isosceles triangles and OH=HG, HA=HE<br />So GE=OA= radius of the circle<br />Peter TranPeter Tranhttps://www.blogger.com/profile/02320555389429344028noreply@blogger.com