APMO 2011 - P2: A combinatorial geometry

[@skylerplayz3713]

For any finite number of points, I can always find a convex polygon such that all points are on/inside the polygon. This is done by induction in which we alter two of the adjacent to include the new point, then reduce at most one concave angle into a convex one if that process forms one concave angles (cannot be more than one since we just focus on the quadrilateral involving 3 original point and one new point)... Not sure if it is well known. For 5 points, the polygon can be triangle or convex quadrilateral/pentagon. By PHP, I can select one of the polygon vertices

[@ShefsofProblemSolving]

Succinctly put !

[@mohamedboukra498]

Your content is very wonderful, thank you for your effort, but there is a small problem that always bothers me. The table do not appear well, perhaps because of the lighting.

[@AndreyGoryainov-k7o]

Isnt there a simpler solution, which doesnt require going through possible cases. Consider a point A_1, number all of the other points in clockwise order, if all of the angles A_iA_1A_i+1 are >36, then the angle A_2A_1A_5 is > 36*3 = 108, then one of the angles in the triangle A_2A_1A_5 is

[@Yasaah_M]

Loved that!

[@yasser_elmoussaed]

Nice problem

[@sithlordbinks]

Thank you sir!

[@ShefsofProblemSolving]

You are welcome!

[@Samyak_2008]

nice

[@ShefsofProblemSolving]

Thanks

[@USAUSA-g8j]

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