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

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== Solution 2 ==
 
== Solution 2 ==
Let <math>\angle FBA = \angle FAB = \angle FAD = \angle FCD = \alpha</math>. And WLOG, <math>MF = 1</math>. Hence, <math>CF = 2</math>, <math>BF = 2.cos(2\alpha) = FA</math>, <math>DA = \frac{AC}{2cos(\alpha)} = \frac{1+cos(2\alpha}{cos(\alpha)}</math> and <math>DE = AE = \frac{AD}{2cos(\alpha)} = \frac{1+cos(2\alpha)}{2.(cos(\alpha))^2} = 1</math>. So <math>MX = 1</math> which means <math>B</math>, <math>C</math>, <math>X</math> and <math>F</math> are concyclic. We know that <math>DE // MC</math> and <math>DE = 1 = MC</math>, so we conclude <math>MCDE</math> is parallelogram. So <math>\angle AME = \alpha</math>. That means <math>MDEA</math> is isosceles trapezoid. Hence, <math>MD = EA = 1</math>. By basic angle chasing, <math>\angle MBF = \angle MFB = 2\alpha</math> and <math>\angle MXD = \angle MDX = 2\alpha</math> and we have seen that <math>MB = MF = MD = MX</math>, so <math>BFDX</math> is isosceles trapezoid. And we know that <math>ME</math> bisects <math>\angle FMD</math> so <math>ME</math> is the symmetrical axis of <math>BFDX</math>. İt is clear that the symmetry of <math>BD</math> with respect to <math>ME</math> is <math>FX</math>. And we are done <math>\blacksquare</math>.
+
Let <math>\angle FBA = \angle FAB = \angle FAD = \angle FCD = \alpha</math>. And WLOG, <math>MF = 1</math>. Hence, <math>CF = 2</math>,  
 +
 
 +
<math>BF = 2.cos(2\alpha) = FA</math>,  
 +
 
 +
<math>DA = \frac{AC}{2cos(\alpha)} = \frac{1+cos(2\alpha)}{cos(\alpha)}</math> and  
 +
 
 +
<math>DE = AE = \frac{DA}{2cos(\alpha)} = \frac{1+cos(2\alpha)}{2.(cos(\alpha))^2} = 1</math>.  
 +
 
 +
So <math>MX = 1</math> which means <math>B</math>, <math>C</math>, <math>X</math> and <math>F</math> are concyclic. We know that <math>DE || MC</math> and <math>DE = 1 = MC</math>, so we conclude <math>MCDE</math> is parallelogram. So <math>\angle AME = \alpha</math>. That means <math>MDEA</math> is isosceles trapezoid. Hence, <math>MD = EA = 1</math>. By basic angle chasing,  
 +
 
 +
<math>\angle MBF = \angle MFB = 2\alpha</math> and <math>\angle MXD = \angle MDX = 2\alpha</math> and we have seen that <math>MB = MF = MD = MX</math>, so <math>BFDX</math> is isosceles trapezoid. And we know that <math>ME</math> bisects <math>\angle FMD</math>, so <math>ME</math> is the symmetrical axis of <math>BFDX</math>.  
 +
 
 +
İt is clear that the symmetry of <math>BD</math> with respect to <math>ME</math> is <math>FX</math>. And we are done <math>\blacksquare</math>.
  
 
~EgeSaribas
 
~EgeSaribas

Revision as of 10:53, 19 May 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

Solution 2

Let $\angle FBA = \angle FAB = \angle FAD = \angle FCD = \alpha$. And WLOG, $MF = 1$. Hence, $CF = 2$,

$BF = 2.cos(2\alpha) = FA$,

$DA = \frac{AC}{2cos(\alpha)} = \frac{1+cos(2\alpha)}{cos(\alpha)}$ and

$DE = AE = \frac{DA}{2cos(\alpha)} = \frac{1+cos(2\alpha)}{2.(cos(\alpha))^2} = 1$.

So $MX = 1$ which means $B$, $C$, $X$ and $F$ are concyclic. We know that $DE || MC$ and $DE = 1 = MC$, so we conclude $MCDE$ is parallelogram. So $\angle AME = \alpha$. That means $MDEA$ is isosceles trapezoid. Hence, $MD = EA = 1$. By basic angle chasing,

$\angle MBF = \angle MFB = 2\alpha$ and $\angle MXD = \angle MDX = 2\alpha$ and we have seen that $MB = MF = MD = MX$, so $BFDX$ is isosceles trapezoid. And we know that $ME$ bisects $\angle FMD$, so $ME$ is the symmetrical axis of $BFDX$.

İt is clear that the symmetry of $BD$ with respect to $ME$ is $FX$. And we are done $\blacksquare$.

~EgeSaribas

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