Difference between revisions of "2016 IMO Problems/Problem 1"
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− | Let <math>\angle FBA = \angle FAB = \angle FAD = \angle FCD = \alpha</math>. And WLOG, <math>MF = 1</math>. Hence, <math>CF = 2</math>, | + | Let <math>\angle FBA = \angle FAB = \angle FAD = \angle FCD = \angle DAE = \angle ADE = \alpha</math>. And WLOG, <math>MF = 1</math>. Hence, <math>CF = 2</math>, |
<math>\implies</math> <math>BF = CF.cos(2\alpha) = 2.cos(2\alpha) = FA</math>, | <math>\implies</math> <math>BF = CF.cos(2\alpha) = 2.cos(2\alpha) = FA</math>, |
Revision as of 09:04, 27 May 2024
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
~Athmyx
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 |