2024 IMO Problems/Problem 4
Let be a triangle with . Let the incentre and incircle of triangle be and , respectively. Let be the point on line different from such that the line through parallel to is tangent to . Similarly, let be the point on line different from such that the line through parallel to is tangent to . Let intersect the circumcircle of triangle again at . Let and be the midpoints of and , respectively. Prove that .
Contents
[hide]Video Solution
Video Solution (AI solution)
Video Solution(In Chinese)
Video Solution
Part 1: Derive tangent values and with trig values of angles , ,
Part 2: Derive tangent values and with side lengths , , , where is the midpoint of
Part 3: Prove that and .
Comments: Although this is an IMO problem, the skills needed to solve this problem have all previously tested in AMC and its system math contests, such as HMMT.~ also proved by Kislay Kai
Evidence 1: 2020 Spring HMMT Geometry Round Problem 8
I used the property that because point is on the angle bisector , is isosceles. This is a crucial step to analyze . This technique was previously tested in this HMMT problem.
Evidence 2: 2022 AMC 12A Problem 25
The technique in this AMC problem can be easily and directly applied to this IMO problem to quickly determine the locations of points and . If you read my solutions to both this AMC problem and this IMO problem, you will find that I simply took exactly the same approach to solve both.
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
Video Solution with discussion of a generalized case
https://youtu.be/NJc79Ccg82E?si=J0YdHAz-46miJIO2
Solution simple
Let point be
is the parallelogram with equal heights, so is rhomb points and are concyclic.
Therefore
Similarly,
vladimir.shelomovskii@gmail.com, vvsss
Video Solution
See Also
2024 IMO (Problems) • Resources | ||
Preceded by Problem 3 |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 5 |
All IMO Problems and Solutions |