tag:blogger.com,1999:blog-6933544261975483399.post9091059702996400354..comments2024-03-26T19:10:02.918-07:00Comments on Go Geometry (Problem Solutions): Problem 16: Triangle, Cevian, Perpendicular, Angles, CongruenceAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger8125tag:blogger.com,1999:blog-6933544261975483399.post-11505380630500570662021-07-23T14:32:52.033-07:002021-07-23T14:32:52.033-07:00Extend CE to F such that BF || AC. Then FBC is iso...Extend CE to F such that BF || AC. Then FBC is isosceles with base FC=AC and base angles 2x. Extend AC to G such that CG=CB. Then BFCG is parallelogram and BG=FC=AC. Therefore, triangle ABG and its cevian BC display the same configuration as Problem 5, and therefore x=10 Guzhttps://www.blogger.com/profile/06355506727751211930noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-82975782509835590392017-12-27T05:59:45.136-08:002017-12-27T05:59:45.136-08:00Let CE=1 so that AC=2.
Hence BC= (1 / cos(2x))
In ...Let CE=1 so that AC=2.<br />Hence BC= (1 / cos(2x))<br />In △ABC,<br />sin(180-7x)/AC = sin(3x)/BC<br />sin(7x)/2 = sin(3x)cos(2x)<br />sin(7x) =2sin(3x)cos(2x)<br />sin(7x) = sin(5x) + sin(x) (trigonometric identity)<br />sin(x) = sin(7x) - sin(5x)<br />sin(x) = 2cos(6x)sin(x) (trigonometric identity)<br />0 = 2cos(6x)sin(x)-sin(x)<br />0 = sin(x) (2cos(6x)-1)<br />sin(x)=0 or (2cos(6x)-1)=0<br />x=0 (rejected) or 2cos(6x)=1<br />cos(6x)=0.5<br />6x=60<br />x=10Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-18517416104499715162012-07-21T18:02:27.609-07:002012-07-21T18:02:27.609-07:00Video solution http://youtu.be/3luEqMUwzXM (in spa...Video solution http://youtu.be/3luEqMUwzXM (in spanish :P)<br /><br />GreetingsEder Contreras Ordeneshttps://www.blogger.com/profile/08915905860162020945noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-92022983282083462552012-04-15T06:22:53.111-07:002012-04-15T06:22:53.111-07:00The solution is uploaded to the following link:
h...The solution is uploaded to the following link:<br /><br />https://docs.google.com/open?id=0B6XXCq92fLJJcDNRdVlFR3VKNlkAnonymoushttps://www.blogger.com/profile/07812499400423119847noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-68914636440328871502010-07-02T02:57:12.258-07:002010-07-02T02:57:12.258-07:00http://ahmetelmas.files.wordpress.com/2010/05/cozu...http://ahmetelmas.files.wordpress.com/2010/05/cozumlu-ornekler.pdfAnonymousnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-23323965409743037342009-05-13T09:13:00.000-07:002009-05-13T09:13:00.000-07:00This comment has been removed by a blog administrator.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-21318802131039176622009-05-08T23:59:00.000-07:002009-05-08T23:59:00.000-07:00Since a plane geometry solution hasn't appeared so...Since a plane geometry solution hasn't appeared so far here's a simple trigonometric solution:<br />By applying sine rule to Tr. ABC, we get <br />sin(7x)/sin(3x) = AC/BC. But AC = 2CE; hence <br />sin(7x)/sin(3x)= 2CE/BC. Now CE/BC = cos(2x) from Tr. BCE. Thus, sin(7x)/sin(3x)=2cos(2x) or sin(7x)/cos(2x)= 2sin(3x). By inspection if x=10 then sin(7x)=sin(70)=cos(20)=cos(2x) while sin(3x)=sin(30)= 1/2 which in turn makes LHS = RHS. Therefore, x = 10 is a solution to our equation.<br />Ajit: ajitathle@gmail.comAjithttps://www.blogger.com/profile/00611759721780927573noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-22681339357606050142009-02-18T17:44:00.000-08:002009-02-18T17:44:00.000-08:00It is possible to provide a proof using only eleme...It is possible to provide a proof using only elementary geometry!<BR/>Hints:<BR/>Auxiliary lines, Perpendicular bisector, Congruence, Problem 4.Antonio Gutierrezhttps://www.blogger.com/profile/04521650748152459860noreply@blogger.com