tag:blogger.com,1999:blog-6933544261975483399.post887264206614127913..comments2024-03-26T19:10:02.918-07:00Comments on Go Geometry (Problem Solutions): Problem 461: Three Circles, Tangent, Right Angle, Center, Distance, MeasurementAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger2125tag:blogger.com,1999:blog-6933544261975483399.post-57008857901701654702010-05-24T08:16:55.087-07:002010-05-24T08:16:55.087-07:00BC = x; proof:
x²=CN²+BC²=(2√(ac)+a-b)²+(2√(ab)+...BC = x; proof: <br /><br />x²=CN²+BC²=(2√(ac)+a-b)²+(2√(ab)+a-c)²<br /><br />Draw a line from B to OG and a line from C to OF.<br /><br />The intersection of these two lines is N.<br /><br />Draw a line from A to OG, the intersection with CN is P. Draw a line from A to OF, the intersection with BN is Q.<br /><br />Now BN = BQ + NQ and CN = CP + NP.<br /><br />BQ²=AB²-AQ²=(a+b)²-(a-b)²=a²+2ab+b²-(a²-2ab+b²)=4ab <br />BQ=2√(ab)<br /><br />CP²=AC²-AP²=(a+c)²-(a-c)²=a²+2ac+c²-(a²-2ac+c²)=4ac <br />CP=2√(ac)<br /><br />NQ=a-c<br /><br />NP=a-b<br /><br />BN=2√(ab)+a-c<br /><br />CN=2√(ac)+a-b<br /><br />x²=CN²+BC²=(2√(ac)+a-b)²+(2√(ab)+a-c)²<br /><br />QEDAnonymousnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-75414082524292296362010-05-22T18:00:29.124-07:002010-05-22T18:00:29.124-07:00From C draw a perpendicular to OF and from B draw ...From C draw a perpendicular to OF and from B draw a perpendicular to OG, Let the two intersect in P. Now CP = GM + OM - b. GM is the common tangent between A & C and hence = 2√(ac)while OM=a and thus, CP=(2√(ac)+a-b). Similarly, BP = HF + OH – c = (2√(ab)+a-c) while x^2 = BP^2 + CP^2. Hence etc.<br />VihaanVihaanhttps://www.blogger.com/profile/14122321143712979648noreply@blogger.com