tag:blogger.com,1999:blog-6933544261975483399.post2355710216198061853..comments2024-03-26T19:10:02.918-07:00Comments on Go Geometry (Problem Solutions): Archimedes' Book of Lemmas, Proposition #6Antonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger1125tag:blogger.com,1999:blog-6933544261975483399.post-16809590215308322232011-07-28T23:41:02.450-07:002011-07-28T23:41:02.450-07:00Let AC = 2a, CB = 2b, DE = 2k .
Denote the midpoi...Let AC = 2a, CB = 2b, DE = 2k . <br />Denote the midpoints of AC, CB, AB, DE <br />by X, Y, O, Z respectively.<br />(These are respectively the centres of the circles o AC, CB,AB, DE as diameters) <br />Clearly AO = OB = a + b,<br />AX = XC = a, CY = YB = b and so<br />XO = b - a, XY = a + b, DE / AB = k / (a + b). <br />XZ = a + k, YZ = b + k, OZ = a + b -k <br />by tangency considerations<br />Also note a /b = 2a / 2b = AC / CB = r.<br />Let angle ZXY = x<br />Apply Cosine Rule for x in triangles XYZ, XOZ<br />2(a + k)(a + b)cos x <br />= (a + k)^2 + (a + b)^2 - (b + k)^2<br />= 2k(a - b)+ 2a(a + b) <br />In the same manner<br />2(a + k)(b)cos x <br />= (a + k)^2 + b^2 - (a + b - k)^2<br />= 2k(2a + b)+ a(a - 2b)<br />Dividing respective sides;<br />(a + b )/ b<br />= [2k(a - b)+ 2a(a + b)]/[2k(2a + b)- 2ab]<br />implies<br />2bk(a - b)+ 2ab(a + b) <br />= 2k(a + b)(2a + b) - 2ab(a + b)<br />k(2ab - 2b^2 - 4a^2 - 6ab - 2b^2 ) = - 4ab(a + b)<br />k(- 4ab – 4a^2 – 4b^2) = -4ab(a + b)<br />k = ab(a + b ) / (a^2 + ab + b^2)<br />DE / AB = k / (a + b) <br />= ab / (a^2 + ab + b^2) <br />= 1 /[(a/b) + 1 + (b/a)]<br />= 1 / [r + 1 + (1/r)] <br />= r / (r^2 + r +1)Pravinhttps://www.blogger.com/profile/05947303919973968861noreply@blogger.com