tag:blogger.com,1999:blog-6933544261975483399.post7849799658808680971..comments2024-03-26T19:10:02.918-07:00Comments on Go Geometry (Problem Solutions): Geometry Problem 1199: Equilateral Triangle, Square, Altitude, Circle, Incircle, Inradius, Metric RelationsAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger4125tag:blogger.com,1999:blog-6933544261975483399.post-20025472362081419122016-03-14T07:27:16.401-07:002016-03-14T07:27:16.401-07:00From my proof above x = q/(1+m) which upon simplif...From my proof above x = q/(1+m) which upon simplifying gives<br /><br />(6+sqrt3)/(11+ sqrt(160+12sqrt3))<br /><br />Thanks Antonio Sumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-62661473516084904732016-03-14T05:42:19.331-07:002016-03-14T05:42:19.331-07:00Dear Sumith,
Please review the algebra simplificat...Dear Sumith,<br />Please review the algebra simplification<br />ThanksAntonio Gutierrezhttps://www.blogger.com/profile/04521650748152459860noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-74687686936119208672016-03-14T01:39:56.558-07:002016-03-14T01:39:56.558-07:00배덕락(Bae deokrak)
We see that two triangles BDF, C...배덕락(Bae deokrak)<br /><br />We see that two triangles BDF, CEF are 30-60-90 triangle and so <br />CE=1, EF=1/sqrt3, FC 2/sqrt3, DF=1-1/sqrt3 , BD=sqrt3 –1 and BF=2sqrt3 –1<br />Furthermore, since two triangles FDG and AHG are similar, we have<br />DF/DG=AM/(1-DG) <=> DG=(5-sqrt3)/11 <=> HG= 1-DG=(6+sqrt3)/11<br />The area of triangle ACG =[tri_ACG]= HG= x(1+GA) =x(1+sqrt{1+HG^2}), that is, <br />x=HG/(1+sqrt{1+HG^2}). where HG=(6+sqrt3)/11.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-71740885892719178762016-03-13T19:35:06.592-07:002016-03-13T19:35:06.592-07:00Let DF = p, GH =q and AG = m
So BD = sqrt3 -1 and...Let DF = p, GH =q and AG = m<br /><br />So BD = sqrt3 -1 and from similar triangles p = 1 - sqrt(1/3)<br /><br />and p/1 = (1-q)/q from which q = (6+sqrt3)/11 upon simplifying the algebra<br /><br />m^2 = q^2 + 1 and so simplifying again m = (1/11) sqrt(149-10sqrt3)<br /><br />x = S(AGC)/s = (1/2)q X 2/{(1/2) (2m+2)} = 2q / (2m+2)<br /><br />Substituting for q and m and again simplifying the algebra <br /><br />x = (5 - sqrt3)/{11 + sqrt(149-10sqrt3)}<br /><br />I have not tried to simplify this unwieldy expression but it can be done by rationalizing the denominator.<br /><br />Sumith Peiris<br />Moratuwa<br />Sri LankaSumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.com