tag:blogger.com,1999:blog-6933544261975483399.post8597082132934356066..comments2024-03-26T19:10:02.918-07:00Comments on Go Geometry (Problem Solutions): Problem 14: Triangle, Cevian, Midpoint, AnglesAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger12125tag:blogger.com,1999:blog-6933544261975483399.post-10391777761357422702021-02-22T12:40:52.863-08:002021-02-22T12:40:52.863-08:00Geometry Solution 2
Let the bisector of <CBD m...Geometry Solution 2<br /><br />Let the bisector of <CBD meet AC at E. Let the foot of the perpendicular from B to AC be F.<br /><br />Then AB = BE = CE = c, DE = b/2-c and DF = c/2<br /><br />Tr.s BCD & ABC are similar so a/(b/2) = b/a = c/d so b = a.srqt2 and d = c.sqrt2<br /><br />Hence Tr. BFD is right isosceles and < ADB = 3x = 45 <br /><br />Therefore x = 15<br /><br />Sumith Peiris<br />Moratuwa<br />Sri LankaSumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-29495523619739276942021-02-13T02:31:06.916-08:002021-02-13T02:31:06.916-08:00Consider triangle BCD
sin2x/CD=sinx/BD
CD=BDsin2x/...Consider triangle BCD<br />sin2x/CD=sinx/BD<br />CD=BDsin2x/sinx<br /><br />Consider triangle ABD<br />sin2x/BD=sin5x/AD<br />AD=BDsin5x/sin2x<br /><br />CD=AD<br />sin2x/sinx=sin5x/sin2x<br />2cosxsin2x=sin5x<br />sin3x+sinx=sin5x<br />sin5x-sin3x=sinx<br />2cos4xsinx=sinx<br />2cos4x=1<br />cos4x=1/2<br />x=15Marcohttps://www.blogger.com/profile/04632526355171968456noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-53878717066728002992021-02-10T06:50:49.099-08:002021-02-10T06:50:49.099-08:00Let BE be perpendicular to AC, E on AC. Extend CA ...Let BE be perpendicular to AC, E on AC. Extend CA to F so that BF = BC. Let BE = h<br /><br />Then 2a^2 = b^2 ......(1) since BC is a tangent to circle ABD<br /><br />Also AF = AB = c and so AE = (b-c)/2 and hence CE = (b+c)/2 ......(2)<br /><br />Therefore <br />h^2 = a^2 - {(b+c)/2}^2 = c^2 - {(b-c)/2)^2} and so 2a^2 = 2c^2 + 2bc = b^2 ...(2) upon substituting for 2a^2 from (1) <br /> <br />Therefore h^2 = c^2 - {(b-c)/2)^2} = c^2 - (1/4)(b^2 - 2bc + c^2) = c^2 - 3c^2/4 = c^2/4<br /><br />So h = c/2 and hence ABE is a 30-60-90 Triangle and 2x = 30 and x = 15<br /><br />Sumith Peiris<br />Moratuwa<br />Sri LankaSumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-18972294779722866462019-02-20T05:16:40.597-08:002019-02-20T05:16:40.597-08:00See original solution on this video<a title="See the video" href="https://youtu.be/WDyCMASKDSI" rel="nofollow"> See original solution on this video</a><br />rv.littlemanhttps://www.blogger.com/profile/05572092955468280791noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-12431986266937638172017-09-01T01:53:04.928-07:002017-09-01T01:53:04.928-07:00Assume length of AC=2
Note triangle ABC and BDC a...Assume length of AC=2 <br />Note triangle ABC and BDC are similar. <br />AC/BC=BC/DC<br />2/BC=BC/1<br />Hence BC=√2<br /><br />Construct point E which lies on BC and <EDC=90. By joining AE, triangle EBA is again similar to ABC and BDC, and EA=EC.Let length of AE and EC=a<br />AC/BC=AE/AB<br />2/√2=AE/AB<br />AB=a/√2<br /><br />BC/AB=AB/BE<br />√2/AB=AB/BE<br />BE=AB^2/√2<br />BE=(a/√2)^2/√2<br />BE=(√2)a^2/4<br /><br />BE+EC=√2<br />(√2)a^2/4 + a - √2=0 (quadratic equation)<br />a=1.03527618<br />In triangle EDC,<br />cos x=1/a<br />x=15<br />Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-5761551108864795012016-10-15T09:00:35.188-07:002016-10-15T09:00:35.188-07:00https://www.youtube.com/watch?v=ikWNS9tpYJYhttps://www.youtube.com/watch?v=ikWNS9tpYJYgeoclidhttps://www.blogger.com/profile/07989522895596673545noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-77992278672521510072015-12-09T21:11:05.041-08:002015-12-09T21:11:05.041-08:00Excellent solution MichaelExcellent solution MichaelSumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-26189558525357383802013-02-01T14:29:38.065-08:002013-02-01T14:29:38.065-08:00 CE perpendicular to AB and BE=EH. So , angle AHC... CE perpendicular to AB and BE=EH. So , angle AHC=3x. Because, angle BDA=3x ,<br />BDCH , is a cyclic quadrilateral , therefore , angle BHD=x. The triangle AEC, is rectangular and AD=DC. So, ED=DA=DC, and angle BED=2x, therefore, angle EDH=x. So ,DE=EH=EB .Then angle BDH=90 degrees and 3x+2x+x=90 .Therefore x=15 degrees.<br />image : http://img248.imageshack.us/img248/8017/geogebra4.pngMichael Tsourakakishttps://www.blogger.com/profile/01337344627083642570noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-28090729839588522682012-04-14T09:43:50.106-07:002012-04-14T09:43:50.106-07:00the solution is uploaded to the following link:
ht...the solution is uploaded to the following link:<br />https://docs.google.com/open?id=0B6XXCq92fLJJS2NfODE5ZmEwNkUAnonymoushttps://www.blogger.com/profile/07812499400423119847noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-64401129314833499232010-07-02T02:55:15.605-07:002010-07-02T02:55:15.605-07:00http://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-55643706040771445792009-05-13T20:18:00.000-07:002009-05-13T20:18:00.000-07:00http://geometri-problemleri.blogspot.com/2009/05/p...http://geometri-problemleri.blogspot.com/2009/05/problem-21-ve-cozumu.htmlAnonymousnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-53058568055963835792009-03-16T21:53:00.000-07:002009-03-16T21:53:00.000-07:00WLOG one may assume D:(0,0), A:(-1,0), C:(1,0). Le...WLOG one may assume D:(0,0), A:(-1,0), C:(1,0). Let angle BCA = m and angle BAC=angle CBD = 2m<BR/>Thus angle BDA = 3m. Further, let tan(m)=t<BR/>Now, DB: y =(-tan3m)x =-x(3t-t^3)/(1-3t^2)<BR/>CB:y=(1-x)t which give B as((3t^2-1)/2(t2+1),(3t-t^3)/2(t2+1)) while A is (-1,0). hence slope BA=<BR/>(3t-t^3)/(5t^2+1) which as we know is tan(2m)=2t/(1-t^2). The equation obtained: (3t-t^3)/(5t^2+1)=2t/(1-t^2) has five solutions of which the only admissible solution is t=0.2679491924 which gives m=15 deg. Thus, angle BCA = 15 deg.<BR/>Alternatively, in triangle ABD, AD/BD =sin(5m)/sin(2m) while triangle BDC gives, CD/BD=sin(2m)/sin(m). Since AD=CD, we can say that sin(5m)/sin(2m)=sin(2m)/sin(m) which can be solved to get m=15 deg as before.<BR/>Now, having proven my bonafides, can someone please show me how such problems can be solved more elegantly by plane geometry alone?<BR/>Ajit: ajitathle@gmail.comAjithttps://www.blogger.com/profile/00611759721780927573noreply@blogger.com