(a)Prove that every pentagon with integral coordinates has at least two vertices , whose respective coordinates have the same parity. (b)What is the smallest area possible of pentagons with integral coordinates. Albanian National Mathematical Olympiad 2010---12 GRADE Question 3.
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Tags: calculus, integration, analytic geometry, geometry, pigeonhole principle, combinatorics unsolved, combinatorics
06.03.2011 02:26
For a: There are only four possible parities for a point in the Cartesian plane: (odd, odd) (odd, even) (even, odd) (even, even) By the Pigeonhole Principle, there must be at least two points with the same parity in both coordinates. For b: The smallest area I can find is $ 2.5 $, but I'm not sure if this is the smallest possible area. This is in a national olympiad? : : It seems really easy -- and I'm only in eighth grade!
06.03.2011 03:42
Well, assuming nonconvex pentagons are okay, $(0,0), (1,1), (2,1), (1,2), (0,1)$ has area $3/2$. This is clearly the minimum by Pick's theorem.
09.05.2012 17:36
(a) The couple of coordinates of any lattice point can have at most four different parity patterns by the pigeonhole principle, the assertion follows. (b) Any triangulation of the pentagon has three triangles with all integer coordinates. Since any lattice triangle has area at least $\frac 1 2$, the minimum possible area for our pentagon is $\frac 3 2$, which is attained for the set of vertices $(0, 0)$, $(0, 1)$, $(1, 0)$, $(1, 1)$, $(2, -1)$.
15.05.2012 11:24
well ,for (1) there can be 4 types of pts.,if we classify them in terms of the parity of their co ordinates. namely (odd,odd),(even,even) , (odd,even) ,(even ,odd). then applying PHP gives our desired result.
20.05.2012 15:33
For convex pentagons... http://www.artofproblemsolving.com/Forum/viewtopic.php?f=47&t=403513