Triangle
Imagine you have an equilateral triangle 2 inches to a side. Prove that no matter how you draw 5 points within the triangle (anywhere in the interior or on the perimeter), there will be two points an inch or less away from each other.
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Triangle
Imagine you have an equilateral triangle 2 inches to a side. Prove that no matter how you draw 5 points within the triangle (anywhere in the interior or on the perimeter), there will be two points an inch or less away from each other. 
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If you choose any 5 points in the triangle, at least two of the points must be in the same 1" equilateral triangle, and therefore 1" or less away from each other.
Some of these bits could use more rigorous argument, I suppose.
I'd use a compass to draw a 1" radius line from each point of the equilateral within the triangle. That would leave a "concave triangle" of an infinite # of points more than an inch away from all of the end points.
Once you have this "concave triangle" of points, then you can draw the interior equilateral that the previous ocmmenters mentioned (which completely surrounds the smaller "concave triangle", though it shares all 3 points). Because all points within the smaller equilateral triangle are within 1" of each other, all points within or on the "concave triangle" must be within 1" of each other. Thus, having 2 points all more than 1" away from any other point on the triangle.
Using a compass to determine the points' radii (rather than just subdividing into smaller equilateral triangles) will show visually that there is no arrangement in which 5 circles of 1" radius that can fit at least their center points into an equilateral triangle of 2" on a side.
Incidentally, is there any importance to the fact that so many of the puzzles recently have basically hinged on the Dirichlet drawer principle?
And this is a good example of why!
Divide the larger triangle into four smaller 1 inch equilateral triangles by connecting the midpoints of each of the sides. Note that these 4 triangles completely partition the larger triangle.
Now begin placing your 5 circles over the larger triangle. The center of each circle placed with necessarily fall within one of the smaller triangles, and will completely cover the triangle. Since the circle defines the area within which another circle's center cannot be placed, and there are at most four triangles within which we can place these circles, the fifth circle placed must overlap one of the previous circles. This overlap means that two points will be within 1 inch of each other.
"At least two of the five points must be contained within or on the perimeter of one of the four small triangle, hence are less than one inch apart."(emphasis added)
But this isn't true. If you draw points only at the vertices of the smaller triangles, then there are six points that are exactly 1 inch from the nearest other points. It is only because the puzzle asks for proof that "there will be two points an inch or less" away from one another that it can be solved. The configuration posited shows that the proposition isn't true if you must show that two points are less than an inch away from one another.
His solution is still wrong; according to his solution, we couldn't fit 5 points >1 inch apart each into a rectangle 2 by sqrt(3), where it's easily possible (1 at each corner and 1 in the center). His solution goes astray when he implicitly assumes that the 1 inch circles can't overlap. They can; they just can't contain the centers of other circles.
Of course, MY triangle lies in a nonEuclidean space. You DID leave that loophole.