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

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Pr♦❜❧❡♠ ✶✳ ❚r✐❛♥❣❧❡ BCF ❤❛s ❛ r✐❣❤t ❛♥❣❧❡ ❛t B✳ ▲❡t A ❜❡ t❤❡ ♣♦✐♥t ♦♥ ❧✐♥❡ CF s✉❝❤ t❤❛t FA = FB ❛♥❞ F ❧✐❡s ❜❡t✇❡❡♥ A ❛♥❞ C✳ P♦✐♥t D ✐s ❝❤♦s❡♥ s✉❝❤ t❤❛t DA = DC ❛♥❞ AC ✐s t❤❡ ❜✐s❡❝t♦r ♦❢ ∠DAB✳ P♦✐♥t E ✐s ❝❤♦s❡♥ s✉❝❤ t❤❛t EA = ED ❛♥❞ AD ✐s t❤❡ ❜✐s❡❝t♦r ♦❢ ∠EAC✳ ▲❡t M ❜❡ t❤❡ ♠✐❞♣♦✐♥t ♦❢ CF✳ ▲❡t X ❜❡ t❤❡ ♣♦✐♥t s✉❝❤ t❤❛t AMXE ✐s ❛ ♣❛r❛❧❧❡❧♦❣r❛♠ ✭✇❤❡r❡ AM k EX ❛♥❞ AE k MX✮✳ Pr♦✈❡ t❤❛t ❧✐♥❡s BD✱ FX✱ ❛♥❞ ME ❛r❡ ❝♦♥❝✉rr❡♥t✳
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Triangle <math>BCF</math> has a right angle at <math>B</math>. Let <math>A</math> be the point on line <math>CF</math> such that <math>FA=FB</math> and <math>F</math> lies between <math>A</math> and <math>C</math>. Point <math>D</math> is chosen so that <math>DA=DC</math> and <math>AC</math> is the bisector of <math>\angle{DAB}</math>. Point <math>E</math> is chosen so that <math>EA=ED</math> and <math>AD</math> is the bisector of <math>\angle{EAC}</math>. Let <math>M</math> be the midpoint of <math>CF</math>. Let <math>X</math> be the point such that <math>AMXE</math> is a parallelogram. Prove that <math>BD,FX</math> and <math>ME</math> are concurrent.

Revision as of 20:00, 26 December 2019

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.