2016 IMO Problems/Problem 1
Contents
[hide]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
The Problem shows that
And
Finally
Solution 2
Let . And WLOG, . Hence, ,
,
and
.
So which means , , and are concyclic. We know that and , so we conclude is parallelogram. So . That means is isosceles trapezoid. Hence, . By basic angle chasing,
and and we have seen that , so is isosceles trapezoid. And we know that bisects , so is the symmetrical axis of .
and , and are symmetrical respect to . Hence, the symmetry of with respect to is . And we are done .
~EgeSaribas
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 |