1996 IMO Problems/Problem 5
Problem
Let be a convex hexagon such that is parallel to , is parallel to , and is parallel to . Let , , denote the circumradii of triangles , , , respectively, and let denote the perimeter of the hexagon. Prove that
Solution
Let
Let
Let
From the parallel lines on the hexagon we get:
[Equations 1]
So now we look at . We construct a perpendicular from to and a perpendicular from to .
We find out the length of these two perpendiculars and add them to get the distance between parallel lines and and because of the triangle inequality the distance is greater or equal to tha the distance between parallel lines and :
This provides the following inequality:
Using the [Equations 1] we simplify to:
[Equation 2]
We now construct a perpendicular from to and a perpendicular from to . Then we find out the length of these two perpendiculars and add them to get the distance between parallel lines and and get:
Using the [Equations 1] we simplify to:
[Equation 3]
We now add [Equation 2] and [Equation 3] to get:
[Equation 4]
We now use the Extended law of sines on with to get:
[Equation 5]
Substitute [Equation 5] into [Equation 4]:
[Equation 6]
To find the equivalent inequality for and we just need shift the indexes by two. That is to add to each of the indexes of and and adjust the indexes so that for the indexes are 1 through 6, and for the indexes are 1 through 3.
[Equation 7]
[Equation 8]
Adding [Equation 6], [Equation 7], and [Equation 8] we get:
From AM-GM inequality we get:
Therefore, for any index and
Therefore,
Since , then
~ Tomas Diaz. orders@tomasdiaz.com
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.
See Also
1996 IMO (Problems) • Resources | ||
Preceded by Problem 4 |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 6 |
All IMO Problems and Solutions |