Geometry Problem 1237:
IMO 2016, Problem 1, Triangle, Congruence, Parallel
Lines,
Midpoint, Concurrency. Level:
College, High School.
|
Triangle CBF has a right angle at B. A
is a point on line CF extended such that FA = FB. Point D is chosen such DA = DC and AC is the bisector of angle DAB. Point E is chosen such that EA = ED and AD is the bisector of angle EAC. Let M be the midpoint of CF. Let X be
a point on ED extended such that MX is parallel to AE. Prove
that lines BD, FX, and ME are concurrent.
Reference:
International Mathematical Olympiad. IMO 2016,
Problem 1.
|
Art of Geometry Problem 1237 using Mobile Apps. Light Patterns.
Geometric art is a form of art based on the use and application of geometric figures. A geometric figure is any set or combination of points, lines, surfaces and solids.
A mobile app or mobile application software is a computer program designed to run on smartphones and tablet computers.
See also
kaleidoscope of problem 1237.
Home
|
Search | Geometry
|
Problems |
All
Problems |
Open Problems
|
Visual Index
|
10 Problems
|
Problems Art Gallery
|
Art |
1231-1240
|
Triangles
|
Concurrent lines
|
Congruence
|
Midpoint
|
Parallel lines
|
Email
Post or view a solution to the problem 1237
Last updated: Jul 22, 2016
|
|