tag:blogger.com,1999:blog-6933544261975483399.post7457615350357583848..comments2024-03-26T19:10:02.918-07:00Comments on Go Geometry (Problem Solutions): Routh's Theorem 4: Triangle, Cevians, Ratio, AreasAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger2125tag:blogger.com,1999:blog-6933544261975483399.post-9474285843397042622010-03-25T12:33:15.365-07:002010-03-25T12:33:15.365-07:00second way, like P2
SA'B'C' = S - ( S...second way, like P2<br /><br />SA'B'C' = S - ( SA'BC' + SA'CB' + SAB'C' )<br /><br />SA'B'C' = S - [n/(n+1)(k+1)+ k/(k+1)(m+1)+ m/(m+1)(n+1)]<br /><br /><br />SA'B'C' = (kmn + 1) / (k+1)(m+1)(n+1)<br />---------------------------------------------c .t . e. onoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-78813763949973557202010-03-24T14:55:33.551-07:002010-03-24T14:55:33.551-07:00SB'A'C/SAA'B' = 1/m => SAA'...SB'A'C/SAA'B' = 1/m => SAA'C/SAA'B' = (1+m)/m<br />SAA'B' = km/(1+k)(1+m) (1) (SAA'C = k/(1+k) S from P1)<br />SAA'B' = AA'∙h3 ( h3 alt from B' to AA')<br />AA' = [km/(1+k)(1+m)]/[m/(1+m)] (h3 = m/(1+m) h1 P1<br />AA' = k/(1+k) h1 (2)<br />SAA'C' = AA'∙h4 ( h4 alt from C' on AA')<br />SAA'C' = AA'∙1/(k(1+n) h1 (h4 related to h from B and<br /> h fro C see P1)<br />AA' = (SAA'C')/[1/(k(1+n)] (3)<br />from (2) and (3)<br />SAA'C' = 1/(1+k)(1+n) (4)<br />from (1) and (4)<br /><br />SAB'A'C' = 1/(1+k)(1+n) + km/(1+k)(1+m)<br />SAB'A'C' =(1+m+kmn+km)/(1+k)(1+m)(1+n)<br /><br />SB'C'A' = SAB'A'C' - SAB'C' (SAB'C'=m/(1+m)(1+n) P3<br />SB'C'A' = (1+m+kmn+km)/(1+k)(1+m)(1+n) - m/(1+m)(1+n)<br /><br />SB'C'A' = (kmn+1)/ (1+k)(1+m)(1+n)<br />---------------------------------------------c .t . e. onoreply@blogger.com