tag:blogger.com,1999:blog-6933544261975483399.post3923581173081884820..comments2024-03-26T19:10:02.918-07:00Comments on Go Geometry (Problem Solutions): Elearn Geometry Problem 144Antonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger6125tag:blogger.com,1999:blog-6933544261975483399.post-53623675031876130262015-04-26T06:28:23.861-07:002015-04-26T06:28:23.861-07:00Peter, DE/AC = (h−2r)/h = HG/A'C'Peter, DE/AC = (h−2r)/h = HG/A'C'Anonymoushttps://www.blogger.com/profile/01035278723944920530noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-50632814746591104502012-10-09T18:27:27.194-07:002012-10-09T18:27:27.194-07:00http://imageshack.us/a/img138/9641/problem144.png
...http://imageshack.us/a/img138/9641/problem144.png<br /><br />Extend MH,DE and FG to form triangle A’B’C’ ( see attached sketch)<br />∆ABC similar to ∆A’B’C’…. ( case AA)<br />Ratio of similarity =incircle radius of ∆ABC/incircle radius of ∆A’B”C’= r/ r=1<br />So ∆ABC congruent to ∆A’B’C’∆<br />Let h is the altitude of ∆ABC and A’B’C’ from B and B’<br />In ∆BDE we have DE/AC=(h-r)/h<br />In ∆B’HG we have HG/A’C’= (h-r)/h<br />Bu AC=A’C’ so DE=HG<br />Follow the solution of yassin-NAugust 23, 2009 7:25 AM for the remaining prove<br />Peter Tranhttps://www.blogger.com/profile/02320555389429344028noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-88060499978013048722012-10-07T18:07:31.980-07:002012-10-07T18:07:31.980-07:00Solution of problem 144.
Let P, P1, P2, P3 be the ...Solution of problem 144.<br />Let P, P1, P2, P3 be the perimeters or triangles ABC, AMH, DBE, GFC, respectively.<br />We have P = P1 + P2 + P3 (as proved in problem 141).<br />Triangles AMH and ABC are similar with ratio P1/P.<br />Triangles DBE and ABC are similar with ratio P2/P.<br />Triangles GFC and ABC are similar with ratio P3/P.<br />So r1/r + r2/r + r3/r = P1/P + P2/P + P3/P = (P1 + P2 + P3)/P = P/P = 1<br />and r = r1 + r2 + r3.<br />Nilton Lapanoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-36419003131119976662012-10-07T12:50:43.118-07:002012-10-07T12:50:43.118-07:00I just wonder if there is any information about th...I just wonder if there is any information about the history of this problem.Hasannoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-9412047692020888482009-08-23T08:08:19.832-07:002009-08-23T08:08:19.832-07:00Very good.
What are the proof reasons that you use...Very good.<br />What are the proof reasons that you use?Antonio Gutierrezhttps://www.blogger.com/profile/04521650748152459860noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-83494891192871054532009-08-23T07:25:42.006-07:002009-08-23T07:25:42.006-07:00we have : DE=HG
r1/r=AH/AC
r2/r=DE/AC
r3/r=CG/AC
r...we have : DE=HG<br />r1/r=AH/AC<br />r2/r=DE/AC<br />r3/r=CG/AC<br />r1/r+r2/r+r3/r=AH/AC+DE/AC+CG/AC<br />(r1+r2+r3)/r=(AH+HG+CG)/AC<br />(r1+r2+r3)/r=AC/AC=1<br />r1+r2+r3=r<br />yass_ghaz_57@hotmail.fryassin-Nhttps://www.blogger.com/profile/02621920005785165269noreply@blogger.com