I spent this weekend working on and off a very interesting IMO problem – problem 6 from IMO 2006. You can find that here . It was a real hard nut and only led to a crushed ego :(. Let me first describe the problem and a result I managed to prove in this direction, which was however much weaker than the problem statement.
IMO 2006 Problem 6 : Assign to each side of a convex polygon the maximum area of a triangle that has as a side and is contained in . Show that the sum of the areas assigned to the sides of is at least twice the area of .
I managed to prove the following result.
Theorem: Assign to each side of a convex polygon the maximum area of a triangle that has as a side and is contained in . Then the sum of the areas assigned to the sides of is at least the area of .
The proof is as follows. We prove first a small lemma.
Lemma 1 : Let be a convex polygon with sides. Then there is a quadrilateral or a triangle with vertices among the vertices of that has area at least the area of .
Proof: Let the polygon have vertices . Without loss of generality assume that is the longest diagonal. The angles are all less than since is the longest diagonal. See the figure below for an example. For any vertex denote by the distance from the line . We will call the number the height of .
If or then let be the vertex with maximum height. By our earlier observation regarding the angles at vertices with the line , is enclosed inside the rectangle with base and height . As such the area of the triangle is at least the area of .
If consider the vertex that has the maximum height among all vertices and the vertex that has maximum height among all vertices . Clearly as before is enclosed inside the rectangle with a side parallel to through and the other sides through and . Now
We can now prove the theorem as follows.
Proof of Theorem: Let be a maximum area quadrilateral whose vertices are also vertices of . We will call this quadrilateral . Clearly as we saw in the
previous lemma. The lemma actually showed that in few cases there
is actually a triangle with vertices among those of with area at least of . We will not deal with the case of a triangle here since it is similar but we remark that if there is such a triangle, the theorem actually gets stronger and we can in fact prove what was required in the IMO problem.
Consider the triangles in the following four sets
The sum of the areas of the triangles is at most the sum of the numbers on the edges of . However, the sum of the areas is easily seen to be at least
and we are done.
I found the solution to the IMO problem here and it is a real gem. It was a relief to see the solution!