Problem 39: IMO 2016, Day 1, Problem 1

by henderson, Jul 13, 2016, 4:08 PM

$\color{red}\bf{Problem \ 39}$ 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.
$\color{red}\bf{My \ solution}$ Let $\angle FAB=\angle FBA=\alpha.$ Then $\angle DAC=\angle DAE=\angle ADE=\alpha,$ so $AM\parallel DE.$
Thus $X$ lies on $DE.$

$\color{green}\textbf{Claim 1.}$ $D$ is the circumcenter of $\triangle ABC.$
$\color{green}\textbf{Proof.}$ Let's consider the circle $\omega$ centered at $D$ and containing the points $C$ and $A.$ If $B$ is outside the circle, then $\angle CBA<90^\circ+\alpha,$ if $B$ is inside the circle, then $\angle CBA>90^\circ+\alpha.$ But with the given conditions, we have $\angle CBA=90^\circ+\alpha,$ which means that $B$ is on $\omega.$ $\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \square$
So, $\angle DBA=\angle DAB$ $\color{blue}\Longrightarrow$ $\angle DBF+\angle FBA=2\alpha$ $\color{blue}\Longrightarrow$ $\angle DBF=\alpha.$ $\quad ({\color{blue}{\clubsuit}})$

Let $BD\cap FX=Y.$ Observe that $\angle CDB=\angle CFB=2\alpha,$ so $CBFD$ is a cyclic quadrilateral with center $M$ $($ since $CF$ is a diameter and $M$ is its midpoint. $)$ Therefore $\angle DMF=\angle DMA=2\alpha.$ Note that we also have $\angle EAM=2\alpha,$ so $DMAE$ is an isosceles trapezoid. Since $AD$ is an angle-bisector, $ME$ is an angle-bisector, too $\color{blue}\Longrightarrow$ $\angle AME=\alpha.$ $\quad ({\color{red}{\clubsuit}})$

$\color{green}\textbf{Claim 2.}$ $BMYF$ is a cyclic quadrilateral.
$\color{green}\textbf{Proof.}$ Because $\angle FBA=\angle FAB=\alpha$ $\color{red}\Longrightarrow$ $\angle BFM=2\alpha.$ $\quad ({\color{blue}{\bigstar}})$
Since $CBFD$ is a cyclic quadrilateral with center $M,$ it follows that $MC=MD=MF.$ From isosceles $DMAE$ and parallelogram $AMXE$ $\color{red}\Longrightarrow$ $MX=AE=MD$ $\color{red}\Longrightarrow$ $CXDF$ is a cyclic quadrilateral with circumcenter $M.$ Notice that it's also a parallelogram, thus is an isoscelec trapezoid. It's easy to see that $\triangle CMD \cong \triangle FMX$ $\color{red}\Longrightarrow$ $\angle MXF=\angle MXY=\alpha.$ Also from the cyclic quadrilateral $CBFD$ and the isosceles triangle $CMD$ $\color{red}\Longrightarrow$ $\angle MDY=\angle MDB=\angle CDB-\angle CDM=2\alpha-\alpha=\alpha.$ Therefore, $\angle MXY=\alpha=\angle MDY,$ so, it follows that $MXDY$ is a cyclic quadrilateral. Then $\angle MYB=\angle MXD=\angle MDX=\angle DMA=\angle EAM=2\alpha.$ $\quad ({\color{red}{\bigstar}})$
Combining $({\color{blue}{\bigstar}})$ and $({\color{red}{\bigstar}}),$ we obtain that $BMYF$ is a cyclic quadrilateral. $\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \square$

The last step: The cyclic quadrilateral $BMYF$ gives $\angle AMY=\angle FMY=\angle FBY=\angle FBD=\alpha$ (because of $({\color{blue}{\clubsuit}})$ ). Comparing this fact with $({\color{red}{\clubsuit}}),$ we conclude that the points $M,Y$ and $E$ are collinear, as desired. $\square$

This post has been edited 41 times. Last edited by henderson, Feb 3, 2017, 8:14 PM

Math Path

Problem 39: IMO 2016, Day 1, Problem 1

by henderson, Jul 13, 2016, 4:08 PM

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