tag:blogger.com,1999:blog-6933544261975483399.post8020849779150501924..comments2023-02-06T18:39:11.207-08:00Comments on Go Geometry (Problem Solutions): Elearn Geometry Problem 82Antonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger3125tag:blogger.com,1999:blog-6933544261975483399.post-81795943421854351032016-10-11T23:18:45.025-07:002016-10-11T23:18:45.025-07:00 Reference my proof of Problem 81, S(DEF) = ar^2/... Reference my proof of Problem 81,<br /><br />S(DEF) = ar^2/4R + br^2/4R + cr^2/4R<br /><br />Hence S1 = (r/2R)(ra/2 + rb/2 + rc/2) = (r/2R).S<br /><br />Sumith Peiris<br />Moratuwa<br />Sri LankaSumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-52909718445097329362012-04-23T19:29:26.538-07:002012-04-23T19:29:26.538-07:00Alternative solution to problem 82. It was proved ...Alternative solution to problem 82.<br />It was proved in problem 081 that S(DIE) = (a.r*2)/(4R). Similarly, S(DIF) = (b.r*2)/(4R) and S(EIF) = (c.r*2)/(4R). The area S1 of tr. DEF is the sum of those three areas, so S1 = (a+b+c).r*2/(4R).<br />We also know that the area of tr. ABC is given by S = (a+b+c).r/2. Hence<br />S1/S = ((a+b+c).r*2).2)/(4R.(a+b+c).r) = r/2RNilton Lapanoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-27488710025347365192009-03-25T22:18:00.000-07:002009-03-25T22:18:00.000-07:00S1 = Tr.IDE + Tr.IEF + Tr.IFD = [r^2*S/bc + r^2*...S1 = Tr.IDE + Tr.IEF + Tr.IFD<BR/> = [r^2*S/bc + r^2*S/ca + r^2*S/ab] by previous problem.<BR/>Thus, S1= S*r^2[1/bc + 1/ca + 1/ab]<BR/>and S = abc/4R where a,b &amp; c have their usual meaning.<BR/>Hence. S1 = (abcr^2/4R)[1/bc + 1/ca + 1/ab]<BR/> = (r^2/4R)(a+b+c)<BR/> = (r/2R)[r*(a+b+c)/2]<BR/>We know that S = Inradius*Semisum of sides<BR/> = [r*(a+b+c)/2]<BR/>Hence, S1 = (r/2R)* S or S1/S = r/2R<BR/>Ajit: ajitathle@gmail.comAjithttps://www.blogger.com/profile/00611759721780927573noreply@blogger.com