tag:blogger.com,1999:blog-6933544261975483399.post4747754114016795834..comments2024-03-26T19:10:02.918-07:00Comments on Go Geometry (Problem Solutions): Problem: 525 Circles, Diameter, Tangent, Radius, Congruence, MeasurementAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger4125tag:blogger.com,1999:blog-6933544261975483399.post-79321997157684600002010-10-10T07:12:24.214-07:002010-10-10T07:12:24.214-07:00yes you are rightyes you are rightAnonymousnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-7758040469122851932010-10-05T02:47:51.067-07:002010-10-05T02:47:51.067-07:00(2*r1-r3)/r3=(d-r2)/r2 ---> 2*r1*r2=d*r3
(2*r2-...(2*r1-r3)/r3=(d-r2)/r2 ---> 2*r1*r2=d*r3<br />(2*r2-r4)/r4=(d-r1)/r1 ---> 2*r1*r2=d*r4 ---> r3=r4=2*r1*r2/dAnonymousnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-23846520858755718852010-10-05T01:14:31.100-07:002010-10-05T01:14:31.100-07:00▲AGD ~ ▲CND => DG/DN = d/2r2
▲DOG ~ ▲DNM => ...▲AGD ~ ▲CND => DG/DN = d/2r2<br />▲DOG ~ ▲DNM => DG/DN = r1/r4<br />=> r4 = (2r1r2)/d<br />about r3 see ▲ADE, ▲AKBc .t . e. onoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-72706838738458342172010-10-04T21:01:13.225-07:002010-10-04T21:01:13.225-07:00Note that triangle AKJ similar to triangle AEO’ ...Note that triangle AKJ similar to triangle AEO’ (case AA)<br />So KJ/O’E=AJ/AO’ or r3/r2= (2r1-r3)/(d-r2)<br />Develop above expression we get r3=2.r1.r2/d<br /><br />Similarly we also have triangle DNM similar to triangle DGO<br />So r4/r1=(2.r2-r4)/(d-r1) and r4=2.r1.r2/d<br /><br />Peter TranPeter Tranhttps://www.blogger.com/profile/02320555389429344028noreply@blogger.com