## 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.

#### 2 comments:

1. 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

2. 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