tag:blogger.com,1999:blog-6933544261975483399.post6048866220098131493..comments2024-03-26T19:10:02.918-07:00Comments on Go Geometry (Problem Solutions): Problem 320: Triangle, Circumcircle, Incenter, Excenter, Collinear pointsAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger1125tag:blogger.com,1999:blog-6933544261975483399.post-44352286135865713762011-04-11T22:30:02.583-07:002011-04-11T22:30:02.583-07:00http://img33.imageshack.us/img33/5290/problem320.p...http://img33.imageshack.us/img33/5290/problem320.png<br />1. Connect BI, EC and AI<br />Note that A,I,E are collinear.<br />We have angle(BCE)= ½ angle (A) + ½ angle (B)<br />And angle (BIF)= ½ angle (A)+1/2 angle (B) so angle (BCE)= angle (BIF) and quad(BICE) is cyclic<br /><br />2. Since H is the midpoint of arc AG >> HI.HM=HN.HE= HA ^2<br />And quad(NIME) is cyclic.<br /><br />3. Note that MN is radical line of circle(ABC) and circle(NIME)<br />In circle(BICE) , F is the intersection of chords BC and IE >> FI.FE=FB.FC <br />Since power of F to circle (NIME)=power of F to circle(ABC)<br />So F must lie on radical axis MN of circles (ABC) and circle (NIME)<br /><br />Peter TranPeter Tranhttps://www.blogger.com/profile/02320555389429344028noreply@blogger.com