Friday, July 22, 2016

Geometry Problem 1237: IMO 2016, Problem 1, Triangle, Congruence, Parallel Lines, Midpoint, Concurrency

Geometry Problem. Post your solution in the comment box below.
Level: Mathematics Education, High School, Honors Geometry, College.

Click the figure below to view more details of problem 1237.

Geometry Problem 1237: IMO 2016, Problem 1, Triangle, Congruence, Parallel Lines, Midpoint, Concurrency

Posted by Antonio Gutierrez at 11:06 AM
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2 comments:

  1. Peter TranJuly 23, 2016 at 11:58 PM

    https://goo.gl/photos/YYPhTktUP43uz8Pt5

    Denote (XYZ) as angle (XYZ)
    Let θ= (FBA)=(FAB)=(FAD)=(DAE)=(DCF)=(EDA) ( see sketch)
    So (BFC)=2 θ ( external angle)
    (CMX)=2 θ … ( MX//AE)
    (MXD)=2 θ
    (CDX)= θ
    Let H is the projection of E over AF
    Let u=BF=FA , v= DA=DC , w=ED=EA and t=MC=MF
    In triangle BCF we have FC= u/cos(2 θ)= 2.t
    AD=v= AC/2.cos(θ)= u/(2.cos(θ)) x ( 1+1/cos(2. θ))….. (1)
    AE=w= v/(2.cos(θ))…. (2)
    Replace (1) into (2) and simplifying we get
    AE=w= u/(2.cos(2θ))= t=AF/(2.cos(2θ)) => AF= 2.AE.cos(2. θ)= 2.AH
    So H is the midpoint of AF and triangle AEF is isosceles
    (EAF)= (EFA)= 2. θ => B, F, E are collinear
    we also have EF=EA=FM=MX => triangle MFE is isosceles => ME is an angle bisector of ( BEX)
    In parallelogram MXEA , since DE=FM= w=t => XD=FA=FB
    So XE=BE => triangle XEB is isosceles
    X and D are symmetric points of B and F with respect to symmetric axis EM
    So XF will intersect BD at a point on the symmetric axis EM

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  2. APOSTOLIS MANOLOUDISJuly 24, 2016 at 1:18 AM

    Problem 1237
    Let's say that <FBA=<FAB=<FAD=<DCA=<EAD=<EDA=x
    The straight CD intersects the extension of BF in point K.Then <CFB=2x=<FCK+<FKC
    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
    then BA/AC=FA/DA so triangles CBA and DFA are similar.Then <FDA=<BCA=y.But <FDC=180-
    <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.
    Ιs <EDA=<CAD=x so XE//MA therefore MXEA is parallelogram with <ΜΧD=<MAE=2x=<MDX.
    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.
    APOSTOLIS MANOLOUDIS 4 HIGH SHCOOL OF KORYDALLOS PIRAEUS GREECE

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