tag:blogger.com,1999:blog-6933544261975483399.post6949480900018070011..comments2024-03-26T19:10:02.918-07:00Comments on Go Geometry (Problem Solutions): Problem 394: Square, 90 Degree Arc, Diagonal, CongruenceAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger11125tag:blogger.com,1999:blog-6933544261975483399.post-46244295381990124442023-09-10T02:31:49.866-07:002023-09-10T02:31:49.866-07:00AE=(sqrt2)*a/2
AF=AD=a (radii)
EF=AF-AE
=a-(sqrt2)...AE=(sqrt2)*a/2<br />AF=AD=a (radii)<br />EF=AF-AE<br />=a-(sqrt2)*a/2<br />=(2-sqrt2)*a/2<br /><br />In triangle EFD, <FDE=90<br />tan<EDF=EF/ED<br />=(2-sqrt2)/sqrt2<br />=sqrt2-1<br /><EDF=22.5<br /><GDC=45-22.5<br />=22.5<br /><br />In triangle GDC<br />tan<GDC=GC/CD<br />GC=(sqrt2-1)*a<br /><br />BG=BC-CG<br />=a-(sqrt2-1)*a<br />=(2-sqrt2)*a<br />=2EFMarcohttps://www.blogger.com/profile/04632526355171968456noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-18064486389402264302022-12-07T21:35:41.311-08:002022-12-07T21:35:41.311-08:00Thank You Thank You Peter Tranhttps://www.blogger.com/profile/02320555389429344028noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-90878754375221171912022-12-06T19:29:37.840-08:002022-12-06T19:29:37.840-08:00Excellent approach to the solution. Mine was rathe...Excellent approach to the solution. Mine was rather cumbersome trying to express EF and BG in terms of square side. Rogerio de Souzahttps://www.blogger.com/profile/11622594719602267726noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-13426784018864447032018-12-20T09:28:58.814-08:002018-12-20T09:28:58.814-08:00Since GD bisects < BDC,
EP/PC = 1/sqrt2 and BG/...Since GD bisects < BDC,<br />EP/PC = 1/sqrt2 and BG/GC = sqrt2<br /><br />Dividing and since PC = CG, the result follows<br /><br />Sumith Peiris<br />Moratuwa<br />Sri LankaSumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-81562061340609881602016-08-15T05:07:30.629-07:002016-08-15T05:07:30.629-07:00Problem 394
Sulotion 1
Is <FDC=<FBC=<FAD...Problem 394<br />Sulotion 1<br />Is <FDC=<FBC=<FAD/2=22.5=<FDB.If DF intersect AB at point P then <BPG=<FDC=22.5=<BDP so BP=BD=2BE. But triangle BPG and triangle EBF are<br />similar.Then BG/EF=BP/BE=2.Therefore PB=2BE.<br />Sulotion 2<br />Is <CGF=67.5=<AFD=<GFC so CG=CF .In triangle EDC DF is bisector so DE/DC=EF/FC (1)<br /> and in triangle BDC DG is bisector so DC/DB=CG/BG (2). Μultiply by members of (1),(2)<br />σο ½=DE/DB=EF/GB or BG=2EF.<br />APOSTOLIS MANOLOUDIS 4 HIGH SHCOOL OF KORYDALLOS PIRAEUS GREECE<br /><br />APOSTOLIS MANOLOUDIShttps://www.blogger.com/profile/15561495997090211148noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-75595866724764386512015-08-26T03:28:55.972-07:002015-08-26T03:28:55.972-07:00Proved this by writing EF and BG in terms of a the...Proved this by writing EF and BG in terms of a the side of the square and noting that Tr. CGF is isoceles but Peter's proof is far more easy and elegantSumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-88265323046360948772015-04-20T18:18:24.767-07:002015-04-20T18:18:24.767-07:00From E draw EH// BC ( H on GH)
Note that H is the ...From E draw EH// BC ( H on GH)<br />Note that H is the midpoint of GD<br />And triangle EFH is isoceles ( EFH=FDA and FHE=FDA)<br />So EF=EH but EH=1/2. BG<br />So EF= ½. BG<br />Peter Tranhttps://www.blogger.com/profile/02320555389429344028noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-60284702204678251842009-12-08T12:48:44.011-08:002009-12-08T12:48:44.011-08:00suggest to others
(i did not understand anything ...suggest to others<br /><br />(i did not understand anything from third comment)<br /><br />a easy way: use EF as middle line for any trianglec.t.e.onoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-77112568011162314462009-12-04T01:37:55.635-08:002009-12-04T01:37:55.635-08:00let us find the angles of angleBDG,BGD,DFE,EFD,DGC...let us find the angles of angleBDG,BGD,DFE,EFD,DGC,GDC. then we can observe that DG is the angular bisector of angle BDC,and tr DEF, tr DGC are simalar triangles using these two concepts we can prove the requiredvijay9290009015http://vijay.comnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-56374022322146615812009-11-26T07:57:33.690-08:002009-11-26T07:57:33.690-08:00EF = a-a/V2 and FC=V2a-a where V=square root. So A...EF = a-a/V2 and FC=V2a-a where V=square root. So Another way to prove this w/o any construction would be: a=square side. EF/FC=a(1-1/V2)/a(V2-1)=1/V2 and ED/DC=(a/V2)/a =1/V2. Thus,ED/DC=EF/FC or DG bisects angle BDC. Therefore. BG/GC=DB/DC= aV2/a=V2/1 or BG/(BG+GC)=V2*/(V2+1)or rationalizing, BG =V2*a(V2-1) = 2a-aV2. <br />Now, EF=a-a/V2 or 2EF=2a - aV2. Hence, BG=2*EF.<br />AjitAjithttps://www.blogger.com/profile/00611759721780927573noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-48473464491583706612009-11-24T13:37:56.135-08:002009-11-24T13:37:56.135-08:00http://geometri-problemleri.blogspot.com/2009/11/p...http://geometri-problemleri.blogspot.com/2009/11/problem-51-ve-cozumu.htmlAnonymousnoreply@blogger.com