There are 5 points in a rectangle (including its boundary) with area 1, no three of them are in the same line. Find the minimum number of triangles with the area not more than $\frac 1{4}$, vertex of which are three of the five points.
Problem
Source: china mathematical olympiad cmo 2005 final round - Problem 5
Tags: geometry, rectangle, combinatorial geometry, combinatorics unsolved, combinatorics
Myth
25.01.2005 20:00
we are to analize few cases, aren't we?
pengshi
26.01.2005 01:21
minimum is 2? (If I didn't miss any case.) Divide the rectangle into 2 equal parts using a line parallel to one of the sides. One can show that if 3 points lie inside one of the 2 equal parts, then the area of the triangle with these 3 points as vertice as area <=1/4. Do this for the 2 orientations, then either 3 points lie in a quater rectangle, or there are 2 triangles with area <=1/4, in which case we are done. If 3 points lie in a quater rectangle, let them be A,B,C, and let another point be D. One can show that the area of the convex hull of A,B,C,D is at most 1/2. Then since there exists 2 triangles out of ABD,ACD,BCD that add up to at most this convex hull, so another triangle has area <=1/4. Thus, 2 is minimum. Finally, note that 2 is possible. (see diagram)
Attachments:
Lopes
25.02.2005 01:33
your solution is not correct. this diagram not contains 2 trianles with area not more than 1/4. This problem is not easy. One Idea is Complex geometry. (crazy idea)
Myth
25.02.2005 07:15
You are wrong, Lopes! The diagram contains exactly two small triangles.