ICOT:
1) Yes.
2) Yes. Also, I meant to say that Y stays on the boundary of the circle.
Yes also to your square question – in fact, this strategy works in any compact metric space. (My square visualization hint at the beginning was correct given the assumption on X and Y’s optimal strategies. It turns out, though, that Y has an optimal strategy that does not necessarily maximize the distance from X.)
Radius vector – this is just the line drawn from the center of the circle to Y.
{t_1, t_2, t_3, …} – I don’t know if it’s possible to do math type here, but the “_number” is supposed to mean subscript. The number just identifies the time period, so t_1 is the first period of movement and so on.
ith – “i” just means that it refers to any number. It could refer to t_1 or t_2 or t_557 or anything you want. So using “i” instead of a number is a generalization. So if the ith time period happens to be t_557, then the (i+1)th time period is t_558. But “i” can be any number.
half-plane – in our example of a closed circle, we were using R^2 space, i.e. your standard 2-dimensional metric space. R^2 is the metric space that children do math in for all of grade school; when you draw a graph with an x-axis and y-axis, this is R^2 space. The whole R^2 space makes up a plane. If I only consider everything above the x-axis, then I am considering a “half-plane.”
So in the context of the problem, we could consider Y’s radius vector as dividing the “world” in half. By imagining that the vector actually is a line rather than a line segment, everything to one side of the line would constitute a half-plane whereas everything to the other side of the line would constitute the other half-plane.
The Pythagorean Theorem is actually quite easy; draw a line from the center of the circle to where Y is located. This is Y’s radius vector. Now Y is moving perpendicular to his radius vector, so draw a small line segment perpendicular to the radius vector that begins at Y’s current location. This line segment represents Y’s movement. Now connect the other end of the segment to the center of the circle. This will be Y’s new radius vector after moving. Note that since the original radius vector is perpendicular to the line segment of movement, we now have a right triangle, where the new radius vector after moving is the hypotenuse of the triangle.
The boundedness of the radii is a bit more difficult, but basically the idea is that in order for the strategy to work, Y cannot make “too big” of moves or else the strategy will require him to leave the circle, which he is not allowed to do.
frummy in the tummy: If I understand your kasha (which I might not), you are concerned that Y does not know where X is heading necessarily, so he cannot know which half-plane to choose.
Y does not need to know X’s direction; Y only needs to know X’s location.