tag:blogger.com,1999:blog-6933544261975483399.post1571743202789425823..comments2024-03-26T19:10:02.918-07:00Comments on Go Geometry (Problem Solutions): Geometry Problem 1237: IMO 2016, Problem 1, Triangle, Congruence, Parallel Lines, Midpoint, ConcurrencyAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger2125tag:blogger.com,1999:blog-6933544261975483399.post-63604451201535412022016-07-24T01:18:57.592-07:002016-07-24T01:18:57.592-07:00Problem 1237
Let's say that <FBA=<FAB=&...Problem 1237<br />Let's say that <FBA=<FAB=<FAD=<DCA=<EAD=<EDA=x<br />The straight CD intersects the extension of BF in point K.Then <CFB=2x=<FCK+<FKC<br />so <FKC=x and CF=FK.Then triangle FAK=triangle FBC.So <FAK=90. Passes the midpoint E’ of FK,is triangleFBM=triangleFAE’ so <E’AF=2x or E’AD=x.But triangles FBA and DCA are similar<br />then BA/AC=FA/DA so triangles CBA and DFA are similar.Then <FDA=<BCA=y.But <FDC=180-<br /><DCF-<DFC=180-(x+<FAD+<FDA)=180-(2x+y)=180-90=90.So MD=MF=MB=MC=FE’=AE’=KE’=DE’.Is triangleADE=triangleADE’,so the points E, E’ coincides.<br />Ιs <EDA=<CAD=x so XE//MA therefore MXEA is parallelogram with <ΜΧD=<MAE=2x=<MDX.<br />Then triangleMXD=triangleMBF soXD=BF and triangleFDX=triangleFDB.Si BD intersect FX in point P then the point P It lies on the perpendicular bisector of FD.Byt ME is perpendicular bisector of FD.Therefore lines BD,FX and ME are concurreant.<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-71642133935273251452016-07-23T23:58:23.457-07:002016-07-23T23:58:23.457-07:00https://goo.gl/photos/YYPhTktUP43uz8Pt5
Denote (X...https://goo.gl/photos/YYPhTktUP43uz8Pt5<br /><br />Denote (XYZ) as angle (XYZ)<br />Let θ= (FBA)=(FAB)=(FAD)=(DAE)=(DCF)=(EDA) ( see sketch)<br />So (BFC)=2 θ ( external angle)<br />(CMX)=2 θ … ( MX//AE)<br />(MXD)=2 θ<br />(CDX)= θ<br />Let H is the projection of E over AF<br />Let u=BF=FA , v= DA=DC , w=ED=EA and t=MC=MF<br />In triangle BCF we have FC= u/cos(2 θ)= 2.t<br />AD=v= AC/2.cos(θ)= u/(2.cos(θ)) x ( 1+1/cos(2. θ))….. (1)<br />AE=w= v/(2.cos(θ))…. (2)<br />Replace (1) into (2) and simplifying we get <br />AE=w= u/(2.cos(2θ))= t=AF/(2.cos(2θ)) => AF= 2.AE.cos(2. θ)= 2.AH<br />So H is the midpoint of AF and triangle AEF is isosceles <br />(EAF)= (EFA)= 2. θ => B, F, E are collinear <br />we also have EF=EA=FM=MX => triangle MFE is isosceles => ME is an angle bisector of ( BEX)<br />In parallelogram MXEA , since DE=FM= w=t => XD=FA=FB <br />So XE=BE => triangle XEB is isosceles<br /> X and D are symmetric points of B and F with respect to symmetric axis EM <br />So XF will intersect BD at a point on the symmetric axis EM<br />Peter Tranhttps://www.blogger.com/profile/02320555389429344028noreply@blogger.com