tag:blogger.com,1999:blog-6933544261975483399.post68308922001894067..comments2024-03-26T19:10:02.918-07:00Comments on Go Geometry (Problem Solutions): Problem 909: Bicentric Quadrilateral, Incircle, Circumcircle, Circumscribed, Inscribed, Tangent, Incenter, Inradius, DistanceAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger3125tag:blogger.com,1999:blog-6933544261975483399.post-44904910187479100152014-01-10T20:02:33.662-08:002014-01-10T20:02:33.662-08:00a/r=c/sqrt(c^2-r^2)=csc(A/2), solving for r^2 give...a/r=c/sqrt(c^2-r^2)=csc(A/2), solving for r^2 gives r^2=(c^2a^2)/(c^2+a^2). The same reasoning gives r^2=(b^2d^2)/(b^2+d^2). The result follows immediately Ivan Bazarovnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-33928239243393759282013-08-07T17:59:39.329-07:002013-08-07T17:59:39.329-07:00A+C=180
sin(A/2) = cos(C/2)
1/a^2 + 1/c^2
= sin^2...A+C=180<br />sin(A/2) = cos(C/2)<br /><br />1/a^2 + 1/c^2<br />= sin^2(A/2)/r^2 + sin^2(C/2)/r^2<br />= [cos^2(C/2) + sin^2(C/2)]/r^2<br />= 1/r^2<br /><br />Similarly for 1/b^2+1/d^2=1/r^2. Jacob HA (EMK2000)https://www.blogger.com/profile/17238561555526381028noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-84029853912748842762013-08-07T16:24:13.406-07:002013-08-07T16:24:13.406-07:00∠IHA = ∠CGI = 90
∠IAH = 1/2∠BAD = 1/2(180-∠BCD) = ...∠IHA = ∠CGI = 90<br />∠IAH = 1/2∠BAD = 1/2(180-∠BCD) = 90-1/2∠BCD = 90-∠ICG = ∠CIG<br />So, ∆IAH ~ ∆ICG<br />(a/c) = r/(sqrt(c^2 - r^2)) = (sqrt(a^2 - r^2))/r<br />=> a^4/c^4 = (a^2 - r^2)/(c^2 - r^2)<br />=> r^2 = (a^4*c^2 - c^4*a^2)/(a^4 - c^4)<br />=> 1/r^2 = (a^4 - c^4)/[a^2*c^2*(a^2 - c^2)]<br />=> 1/r^2 = (a^2+c^2) / a^2*c^2nakahttps://www.blogger.com/profile/11277356476170372732noreply@blogger.com