## 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.

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