1990 APMO Problems/Problem 5

Not a perfect solution, but I think it works. \\ The construction would obviously depend on the parity of n, let's denote m to be , so either or . \\ By the condition, we know that , then we construct some rectangles with side lengths and . We can observe an obvious fact that these rectangles can be dissected into exactly small triangles with side length . One simply cut those into 1 by stuff small rectangles and then cut each small rectangle with a diagnonal. \\ Now we do a small trick. For those even , we add two of those to the top and bottom of the rectangle with side length so that the longest side of the triangle coincide with the longer edge of the rectangle. Adding each triangle now adds a vertex so we have six vertices, a hexagon. \\ We can verify the hexagon is convex this way because at the four vertices that was vertices of the rectangle, we are adding an acute angle to a right angle while in the two added vertices we just have two right angles. \\ It is slightly trickier with the odd case. We add one instead of two triangle to the top of the rectangle. This would make a pentagon instead of hexagon. \\ WLOG we can make the right angle vertex of the newly added triangle pointing left, and at the same time we make sure the rightmost small rectangle gets an upperright-lowerleft diagonal to get a small triangle sitting on the lower right corner that only shares a vertex with the newly added triangle. We now flip it across the perpendicular bisector of its longest edge. \\ We again can verify that this creates a hexagon, since in order to travel from the lower left vertex of the upper right vertex of the original rectangle one now have to pass three edges instead of two. We can also verify it is a convex hexagon. \\ Let's denote the two acute angles of the small triangle as and assume . The angle of the vertices can be thus calculated to be . Since we know that both are acute both angles are less than 180, while , so the angle is also less than 180. \\ Q.E.D \\ Note 1: Someone please try to read this and add some pictures. This gets very very absurd just by reading. \\ Note 2: I don't know why starting with 6 because from this construction 5 also seems doable from this exact construction. It's just unfortunate we cannot use this construction to get 4, but we can just generate a regular 100-gon and cut it into 100 isosceles pieces containing two adjacent vertices, pick four adjacent triangles and they make a perfect convex hexagon. [Solution by krosa1910]