Difference between revisions of "2016 IMO Problems/Problem 1"

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[[File:2016IMOQ1Solution.jpg|600px]]
 
[[File:2016IMOQ1Solution.jpg|600px]]
  
Given DAC=DCA=CAD, it follows that ABCD. Extend DC to intersect AB at G, making GFA=GFB=CFD. Triangles CDF and AGF are similar. Also, FDC=FGA=90 and FBC=90, which implies points D, C, B, and F are concyclic.
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The Problem shows that DAC=DCA=CAD, it follows that ABCD. Extend DC to intersect AB at G, we get GFA=GFB=CFD. Making triangles CDF and AGF similar. Also, FDC=FGA=90 and FBC=90, which points D, C, B, and F are concyclic.
  
Furthermore, BFC=FBA+FAB=FAE=AFE. Triangle AFE is congruent to FBM, and AE=EF=FM=MB. Let MX=EA=MF, then points B, C, D, F, and X are concyclic.
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And BFC=FBA+FAB=FAE=AFE. Triangle AFE is congruent to FBM, and AE=EF=FM=MB. Let MX=EA=MF, then points B, C, D, F, and X are concyclic.
  
It is also given that AD=DB and DAF=DBF=FXD. Additionally, MFX=FXD=FXM and FEMD with EF=FM=MD=DE, making EFMD a rhombus. Consequently, FBD=MBD=MXF=DXF and triangle BEM is congruent to XEM, while MFX is congruent to MBD which is congruent to FEM, and EM=FX=BD.
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Finally AD=DB and DAF=DBF=FXD. MFX=FXD=FXM and FEMD with EF=FM=MD=DE, making EFMD a rhombus. And FBD=MBD=MXF=DXF and triangle BEM is congruent to XEM, while MFX is congruent to MBD which is congruent to FEM, so EM=FX=BD.
  
 
~Athmyx
 
~Athmyx

Revision as of 11:32, 20 April 2024

Problem

Triangle $BCF$ has a right angle at $B$. Let $A$ be the point on line $CF$ such that $FA=FB$ and $F$ lies between $A$ and $C$. Point $D$ is chosen so that $DA=DC$ and $AC$ is the bisector of $\angle{DAB}$. Point $E$ is chosen so that $EA=ED$ and $AD$ is the bisector of $\angle{EAC}$. Let $M$ be the midpoint of $CF$. Let $X$ be the point such that $AMXE$ is a parallelogram. Prove that $BD,FX$ and $ME$ are concurrent.

2016IMOQ1.jpg

Solution

2016IMOQ1Solution.jpg

The Problem shows that DAC=DCA=CAD, it follows that ABCD. Extend DC to intersect AB at G, we get GFA=GFB=CFD. Making triangles CDF and AGF similar. Also, FDC=FGA=90 and FBC=90, which points D, C, B, and F are concyclic.

And BFC=FBA+FAB=FAE=AFE. Triangle AFE is congruent to FBM, and AE=EF=FM=MB. Let MX=EA=MF, then points B, C, D, F, and X are concyclic.

Finally AD=DB and DAF=DBF=FXD. MFX=FXD=FXM and FEMD with EF=FM=MD=DE, making EFMD a rhombus. And FBD=MBD=MXF=DXF and triangle BEM is congruent to XEM, while MFX is congruent to MBD which is congruent to FEM, so EM=FX=BD.

~Athmyx

See Also

2016 IMO (Problems) • Resources
Preceded by
First Problem
1 2 3 4 5 6 Followed by
Problem 2
All IMO Problems and Solutions