IMO 2016 Problem 1
 Phlembac Adib Hasan
 Posts: 1016
 Joined: Tue Nov 22, 2011 7:49 pm
 Location: 127.0.0.1
 Contact:
IMO 2016 Problem 1
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.
Welcome to BdMO Online Forum. Check out Forum Guides & Rules
 ahmedittihad
 Posts: 181
 Joined: Mon Mar 28, 2016 6:21 pm
Re: IMO 2016 Problem 1
We will show the result by the** radical axis theorem**. So, we will show that $BFDX, BMDE, FEMX$ is concyclic.
$ \angle ADF = \angle ACB$ by **spiral similarity**.
$\angle FDC = 180  \angle ADE  \angle CDX = 180  2\angle ADE  \angle ADF = 180  \angle BFC  \angle ACB = 90 $.
So, $BCDF$ is a cyclic quadrilateral and $M$ is the center of this circle. So, $\angle AMD = 2\angle ACD = \angle MAE$. Along with the fact that $ AC \parallel DE$ we get that $AMDE$ is an isosceles trapizium. Meaning also that it is cyclic.
Now, $ \angle BAD + \angle BMD = 2\angle BAC + 2\angle ACD + 2\angle BCF = 2( 2\angle BAC + \angle BCF) = 2(\angle BFC + \angle BCF) = 180 $. So,$ABMD$ is cyclic. Also, $ \angle ABD = \angle ABF + \angle FBD = \angle ABF + \angle ACD = 2\angle BAC$. So, $ \angle ABD + \angle AED = 2\angle BAC +180  2\angle DAE = 180$.
So we get, $ABMDE$ is cyclic.
Now, we prove that $BDFX$ is cyclic. $\angle AMD = \angle EAM = \angle EXM = \angle MDX$. So, $\triangle MDX$ is isosceles. So, $X$ lies on the circle $BFDC$.
At last, we need to prove that $FEXM$ is cyclic. See that. $A$ is the center of similitude that sends $BD$ to $FE$. So, $ \angle ABD = \angle AFE = 2\angle BAC$.
We already have that $\angle MXE = 2\angle BAC$. And the result follows.
$ \angle ADF = \angle ACB$ by **spiral similarity**.
$\angle FDC = 180  \angle ADE  \angle CDX = 180  2\angle ADE  \angle ADF = 180  \angle BFC  \angle ACB = 90 $.
So, $BCDF$ is a cyclic quadrilateral and $M$ is the center of this circle. So, $\angle AMD = 2\angle ACD = \angle MAE$. Along with the fact that $ AC \parallel DE$ we get that $AMDE$ is an isosceles trapizium. Meaning also that it is cyclic.
Now, $ \angle BAD + \angle BMD = 2\angle BAC + 2\angle ACD + 2\angle BCF = 2( 2\angle BAC + \angle BCF) = 2(\angle BFC + \angle BCF) = 180 $. So,$ABMD$ is cyclic. Also, $ \angle ABD = \angle ABF + \angle FBD = \angle ABF + \angle ACD = 2\angle BAC$. So, $ \angle ABD + \angle AED = 2\angle BAC +180  2\angle DAE = 180$.
So we get, $ABMDE$ is cyclic.
Now, we prove that $BDFX$ is cyclic. $\angle AMD = \angle EAM = \angle EXM = \angle MDX$. So, $\triangle MDX$ is isosceles. So, $X$ lies on the circle $BFDC$.
At last, we need to prove that $FEXM$ is cyclic. See that. $A$ is the center of similitude that sends $BD$ to $FE$. So, $ \angle ABD = \angle AFE = 2\angle BAC$.
We already have that $\angle MXE = 2\angle BAC$. And the result follows.
Frankly, my dear, I don't give a damn.
 Thanic Nur Samin
 Posts: 176
 Joined: Sun Dec 01, 2013 11:02 am
Re: IMO 2016 Problem 1
Let $\angle FAB=x$.
Since $\triangle AFB \sim \triangle ADC$, $\triangle AFD \sim \triangle ABC$ [sprial similarity]
Thus $\angle AFD = 90^{\circ}+x$, and so $FD\perp AB$. Since $FB=FA$, $FD$ is the perpendicular biscector of $AB$. Now, $\angle DBF = x$. Since $\angle AED=180^{\circ}2x$, $ABDE$ is cyclic and since $BF$ is the angle biscector of $\angle ABD$ and $E$ is the midpoint of arc $AD$ not containing $B$, we get that $B,F,E$ are collinear.
$\angle AMB=2\angle FCB=180^{\circ}4x=\angle AEB$, and consequently $AEMB$ is cyclic. Note that since $FA=FB$ and $FA.FM=FB.FE$, $FE=FM$ and $AM=BE$ and $BM=AE$.
Now, since $CDAB$ we get $FD\perp CD$ and so $BCDF$ cyclic with center $M$, so $MF=MD$. And $AEMB$ is an isosceles trapezoid, so $MEAB$ implying $FD\perp ME$.
Now, $MB=AE=MX$ and $EB=AM=EX$.
So $BD$ and $XF$ are reflections of each other wrt $ME$ and hence they are collinear.
Since $\triangle AFB \sim \triangle ADC$, $\triangle AFD \sim \triangle ABC$ [sprial similarity]
Thus $\angle AFD = 90^{\circ}+x$, and so $FD\perp AB$. Since $FB=FA$, $FD$ is the perpendicular biscector of $AB$. Now, $\angle DBF = x$. Since $\angle AED=180^{\circ}2x$, $ABDE$ is cyclic and since $BF$ is the angle biscector of $\angle ABD$ and $E$ is the midpoint of arc $AD$ not containing $B$, we get that $B,F,E$ are collinear.
$\angle AMB=2\angle FCB=180^{\circ}4x=\angle AEB$, and consequently $AEMB$ is cyclic. Note that since $FA=FB$ and $FA.FM=FB.FE$, $FE=FM$ and $AM=BE$ and $BM=AE$.
Now, since $CDAB$ we get $FD\perp CD$ and so $BCDF$ cyclic with center $M$, so $MF=MD$. And $AEMB$ is an isosceles trapezoid, so $MEAB$ implying $FD\perp ME$.
Now, $MB=AE=MX$ and $EB=AM=EX$.
So $BD$ and $XF$ are reflections of each other wrt $ME$ and hence they are collinear.
 Attachments

 IMOgeo.png (22.75 KiB) Viewed 6217 times
Hammer with tact.
Because destroying everything mindlessly isn't cool enough.
Because destroying everything mindlessly isn't cool enough.

 Posts: 1
 Joined: Thu Sep 29, 2016 5:32 pm
 Contact:
Re: IMO 2016 Problem 1
Thanks for sharing the Sample Question Papers for <a href="https://www.olympiadsuccess.com/courses ... mpiad">IMO Olympiad Exam</a>. It is very useful to understand the question pattern of the exam. Please keep posting with the latest updates on IMO and other Olympiad Exams.