All vertices of a regular 2016-gon are initially white. What is the least number of them that can be painted black so that: (a) There is no right triangle (b) There is no acute triangle having all vertices in the vertices of the 2016-gon that are still white?
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Tags: geometry, combinatorics
ThE-dArK-lOrD
04.05.2017 21:20
We must paint at least one point of each diagonal to be black. So the answer is exactly $1008$ by painting one point from each diagonal.
Note that we can paint $1007$ consecutive vertices. Then assume that one can paint at most $1006$ points to avoid all acute triangles.
Since there are still at least $1010$ white points remain, so there are two distinct diagonals whose vertices are all white.
It's not hard to deduce (by some angle chasing) that for any four positions of other white points of the polygon, we can form an acute triangle using that vertex and other two from the two diagonals.
This complete the prove and so the answer is $1007$.
PROF65
04.05.2017 22:15
b) the answer is $ 1008 \because$ we can leave at most all the vertices over one diagonal plus one of the point in the diagonal we have to prove that necessary all the points are ove one diagonal .consider the greatest chord $PQ$ that we can have with the points as ends and let $P',Q'$ theirsymmetric ibn the circumcenter if there exists any point in the minor arcs $PA$ and $P'Q'$ then we ll be able to form a chord geater than $PQ$ which is impossible more over if we have any vertices on the circle in opposite side of $P$ wrt $P'Q'$ then we can form an acute triangle so aal the points leaved white must be ove the diagonal . RH HAS
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