✔ 最佳答案
Working on it.
Couple hints for others:
1.) Consider Pick's theorem - I think it is relevant here.
2.) The GCD constraint is really simply removing lines with exactly same slope.
2010-06-24 13:25:20 補充：
We can think of this problem as a maximin problem - for a particular point, minmize by finding a point nearby that minimize the angle, and maximize by finding the point that maximize the quantity above.
2010-06-24 13:25:29 補充：
From this perspective - we can approach the problem this way. From a point, try to find an upper bound of the minimized angle. If this upper bound is less than the one between P1 and P2, we are done.
2010-06-24 13:25:39 補充：
By Pick's Theorem - we have the area of the triangle = 0.5, I wonder how can we make use of this fact. It sounds like a constrainted maximization problem - find a triangle with with integer coordinates, area = 0.5, with maximum angle.
2010-06-24 13:25:46 補充：
kwanhimshek - if you have a simulation program, can you remove the constraint that the points are in the quarter circle - does that break the conjecture?
2010-06-29 11:06:52 補充：
I don't have a complete answer yet, but I want to share what I think about the problem and hopefully we can answer that together.
First of all, we can leverage Pick's theorem. For all the triangles that are formed by three lattice point on the vertexs and no lattice point on the edge in interior, the area must be 0.5.
We can also calculate the area in a different way using vector cross product,
suppose the points are (a, b) and (c, d) respectively, we have
2A = |ad - bc| = 1
We can also calculate the area in a third way, in particular, we have
2A = |(a,b)| |(c,d)| sin theta = 1
Now since sine is an increasing function, therefore the larger the angle, the larger the sine, therefore, if we want big angles, we will have small length products.
Unfortunately, Pick's theorem does not help us to eliminate all wrong triangles, in particular, we can have triangles with no lattice points in the interior and boundary but yet not a valid triangle, (0,1)-(0,0)-(1,0) is a counter example. In fact, it gives the largest possible angle = 90 degree.
2010-06-30 12:16:45 補充：
I think we need more explanation here:
If we join P(m), O, P(m + 1) into a triangle, it will be a triangle that
1) Have no points in the interior (for if there is one, then there is a line with smaller angle and P(m) and P(m+1) could not be next to each other), and
2010-06-30 12:16:49 補充：
2) Have no point in the edge (or otherwise GCD(a,b) > 1 or GCD(c,d) > 1)
Therefore it will satisfy the Pick's theorem, let us call these right triangles.
There are wrong triangles that also satisfy Pick's theorem, as I mention in the last paragraph.
2010-07-02 10:36:30 補充：
I have more findings here.
For ad-bc must equals to 1, and gcd(a, b) = 1.
We can guarentee the existence of (c,d) by the extended euclidean algorithm.
Suppose we get some (c,d) here such that ad-bc=1, another pair of (c',d') that also satisfy the relation::
ad' - bc' = 1
ad - bc = 1
a(d'-d)-b(c'-c)=0
