tag:blogger.com,1999:blog-6933544261975483399.post8117785851466324737..comments2024-03-26T19:10:02.918-07:00Comments on Go Geometry (Problem Solutions): Geometry Problem 1276 Equilateral Triangle, Circumcircle, Internally Tangent Circles, Tangent Line, MeasurementAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger4125tag:blogger.com,1999:blog-6933544261975483399.post-8980493151014197622016-11-20T08:00:47.406-08:002016-11-20T08:00:47.406-08:00Pradyumna thank you very much and your own solut...Pradyumna thank you very much and your own solution is very good.APOSTOLIS MANOLOUDIShttps://www.blogger.com/profile/15561495997090211148noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-68083155285514215712016-10-24T10:20:43.223-07:002016-10-24T10:20:43.223-07:00Great!!!
Very nice solution.
-PradyumnaGreat!!!<br />Very nice solution. <br /><br />-PradyumnaPradyumna Agashehttps://www.blogger.com/profile/10300531209692781145noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-28460348408028856852016-10-24T06:34:22.021-07:002016-10-24T06:34:22.021-07:00Problem 1276
We apply the Casey’s theorem for ...Problem 1276<br />We apply the Casey’s theorem for zero circles A, B, C, and the circle with center O_1.Then<br />AC.AA_1=AC.BB_1+AB.CC_1 or AA1=BB1+CC1.(AB=BC=AC).<br />APOSTOLIS MANOLOUDIS 4 HIGH SCHOOL KORYDALLOS PIRAEUS GREECE<br /><br /><br />APOSTOLIS MANOLOUDIShttps://www.blogger.com/profile/15561495997090211148noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-88662717609748929102016-10-24T01:11:46.238-07:002016-10-24T01:11:46.238-07:00ABTC is cyclic quadrilateral, By Ptolemy's the...ABTC is cyclic quadrilateral, By Ptolemy's theorem AT*BC = BT*AC + CT*AB, <br />Since AB=BC=AC, we have AT = BT + CT, <br />Lets assume AT,BT and CT meets inner circle O1 at X,Y and Z. We can see that Angle XTY = Angle XTZ = 60 Deg, also Angle YTZ=120 Deg, hence Triangle XYZ is equilateral. <br /><br />Also Angle ATO = Angle XTO1, hence angle subtended by AT and XT on circle O and O1 are equal. We have XT/AT=YT/BT=ZT/CT=r/R (r is radius of inner circle O1, and R is radius of outer circle O). <br />Subtracting each term from 1. We get AX/AT=BY/BT=CZ/CT=1-(r/R)=k <br />AA1^2=AX.AT=k.AT^2 or AA1=sqrt(k).AT,similarly BB1=sqrt(k).BT,CC1=sqrt(k).CT <br />Since AT=BT+CT we get sqrt(k).AT=sqrt(k).BT + sqrt(k).CT or AA1=BB1+CC1 <br />Pradyumna Agashehttps://www.blogger.com/profile/10300531209692781145noreply@blogger.com