IMO 2016 Problem 1

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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.
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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.
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Thanic Nur Samin
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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 \$CD||AB\$ we get \$FD\perp CD\$ and so \$BCDF\$ cyclic with center \$M\$, so \$MF=MD\$. And \$AEMB\$ is an isosceles trapezoid, so \$ME||AB\$ 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.
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