tag:blogger.com,1999:blog-6500033298667240354.post4287431990049102674..comments2024-02-19T00:34:12.578-08:00Comments on Always Creative Geometry Problems plus Occasionally Annoying Philosophy: Fox 1758foxeshttp://www.blogger.com/profile/09567328431908997738noreply@blogger.comBlogger2125tag:blogger.com,1999:blog-6500033298667240354.post-46139801603021497602009-11-06T07:18:38.926-08:002009-11-06T07:18:38.926-08:00Your approach of analytic geometry is interesting ...Your approach of analytic geometry is interesting and very good!<br />Thanks for the solution.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-6500033298667240354.post-32047412705679254622009-11-06T04:27:42.608-08:002009-11-06T04:27:42.608-08:00Let the circle with radius R having its centre at ...Let the circle with radius R having its centre at O be x^2+y^2=R^2. Now OP=RS=R. Since areas A and B are equal, we've Pi*R^2/4 -(Commonn Area)=R*OS -(Common Area). In other words, OS = RT = Pi*R/4 and OR^2=(PiR/4)^2+R^2 Hence, OR =(R/4)V(18+Pi^2). Now let T be (a,b). We've R as (Pi*R/4,R) Since T is a point on the circle we can say: a^2 + b^2 = R^2 & ((pR/4)-a)^2+(R-b)^2=p^2*R^2/16 On solving we can obtain a & b and since S is (PiR/4,0) we can show that ST =R(16-Pi^2)/(4V(16+Pi^2). Hence ST/OR =(16-Pi^2)/(16+Pi^2)<br />AjitAjithttps://www.blogger.com/profile/00611759721780927573noreply@blogger.com