Difference between revisions of "2016 IMO Problems/Problem 1"
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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. | 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. | ||
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==Solution== | ==Solution== |
Revision as of 12:21, 20 April 2024
Problem
Triangle has a right angle at
. Let
be the point on line
such that
and
lies between
and
. Point
is chosen so that
and
is the bisector of
. Point
is chosen so that
and
is the bisector of
. Let
be the midpoint of
. Let
be the point such that
is a parallelogram. Prove that
and
are concurrent.
Solution
See Also
2016 IMO (Problems) • Resources | ||
Preceded by First Problem |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 2 |
All IMO Problems and Solutions |