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

(Created page with "Pr♦❜❧❡♠ ✶✳ ❚r✐❛♥❣❧❡ BCF ❤❛s ❛ r✐❣❤t ❛♥❣❧❡ ❛t B✳ ▲❡t A ❜❡ t❤❡ ♣♦✐♥t ♦♥ ❧✐♥❡ CF s✉❝❤...")
 
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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 BCF has a right angle at B.Let A be the point on line CF such that FA=FB and F lies between C and A. Point D is chosen such that DA=DC and AC is the bisector of ∠DAB. Point E is chosen such that EA=ED and AD is the bisector of ∠EAC. Let M be the midpoint of CF . Let X be the point such that AMXE is a parallelogram (where AM||EX and AE||MX). Prove that the lines BD,FX and ME are concurrent.

Revision as of 02:29, 8 June 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 C and A. Point D is chosen such that DA=DC and AC is the bisector of ∠DAB. Point E is chosen such that EA=ED and AD is the bisector of ∠EAC. Let M be the midpoint of CF . Let X be the point such that AMXE is a parallelogram (where AM||EX and AE||MX). Prove that the lines BD,FX and ME are concurrent.