Math Puzzle: Integer Points

There are four pairs of values modulo two, so if you have five points some two of them must be the same modulo two. Say these points are A and B. Then A + B = (even, even), and so (A + B) / 2 has integer coordinates.

I don't understand. Choose points (0,0) (0,1) (1,0) (1,2) (2,1) (2,2). What line contains three points?<p>Diagram:<p>_OO<p>O_O<p>OO_<p>e: oh, not a point in the original set, any integer point. thanks!

Technically, the problem doesn't specify that the points have to be distinct...

This seems rather trivial.<p>Take two arbitrary points, named A and B, where Ax &#60; Bx (if Ax == Bx, this becomes absurdly simple).<p>Calculate the delta ∂ between the points, such that ∂x = Bx - Ax and ∂y = By - Ay<p>Define a third point C such that Cx = Bx + ∂x and Cy = By + ∂y.<p>C is integral, and is colinear with A and B.

While we're all solving puzzles, can anyone help me with this one?<p>No doubt you've doodled this shape in your graph paper before: <a href="http://i.imgur.com/oY29sBc.png" rel="nofollow">http://i.imgur.com/oY29sBc.png</a><p>What function does this slope approximate? It almost a circle, but not quite.

LOL. For any two integer points P(i, j), Q(l, m). We have the equation of a line:<p>y - j = ((m - j) / (l - i)) * (x - i)<p>Choose x = k * (l - i) + i y = k*(m - j) + j Since i, j, l, m, k are all integers we obtain another integer point. No need 5 points... I may misunderstand the question. :(

I don't see any obvious problem with this counterexample: (0,5), (3,0), (4,6), (7,1), (9,4)<p><a href="http://farm9.staticflickr.com/8114/8661127874_f9269f0ee5_b.jpg" rel="nofollow">http://farm9.staticflickr.com/8114/8661127874_f9269f0ee5_b.j...</a><p>Is there something I'm missing?

back of the napkin: take the first 4 points to be a unit box with the lower left point at the origin. You can draw any of 4 lines to the 5th point (x,y). These 4 lines will have slopes x:y, x-1:y, x:y-1, x-1,y-1. You can always arrive at a slope ratio that is evenly divisible through some combination of -1 and reducing the ratio to simplest form. An evenly divisible slope ratio will pass through another point.<p>example: 5th point (11,4). Choose line with slope 10:4. Divisible by 2, this line passes through the point (5,2). A ratio divisible by 3 would pass through 2 additional points etc.

If I am remembering correctly, this was on the ARML(American Regions Math League) competition "power question" roughly, say 16-17 years ago. The answers given here from nilkn and the children comments are correct.

choose two integers x and y. These represent the first point on a plain. Only adjacent points can be added i.e. any point can only be 1 integer away from this origin point otherwise there will exist an integer point between the origin point the chosen point. so now we have (x, y), (x+1, y), (x, y+1), and (x+1, y+1). Adding a fifth point anywhere on the graph will make it such that an integer point will exists between two of the selected points.

It doesn't hold. Pick (0,1), (0,2), (0,3), (0,4), (0,5) on the plane z = 0

Okay, we have a solution for the plane. The plane is in two dimensions. Now, for any positive integer n, generalize the problem to n dimensions.<p>So, now the problem is, for any positive integer n, given any 2^n + 1 n-tuples of integers, at least two of these n-tuples if added have all even components.