part 3 - LaTeX does not like spoilers

[spoiler]Suppose that B throws at A who is R away, and that C is to throw at D such that CD crosses BA.[/spoiler]

[spoiler]Let us view this geometricly, creating a circle of radius R about B[/spoiler]
[spoiler](which nobody except B can be within) centered at (0,0) with A at (R,0).[/spoiler]

[spoiler]Suppose that C is at a position (X1,Y1) with Y1>0 outside the circle,[/spoiler]
[spoiler]and D is at a position (X2,Y2) with Y2<0 outside the circle.[/spoiler]

[spoiler]We then break this into three cases for C.[/spoiler]

[spoiler]1) If X1<0 then C is closer to B than D (smaller dx, dy)[/spoiler]

[spoiler]2) If 0<X1<R then C is closer to A than D,[/spoiler]
[spoiler]which can be seen by noting D is on the opposite side of the line through A[/spoiler]
[spoiler]perpendicular to AC (or the arc through A with vertex C and radius AC).[/spoiler]

[spoiler]3) If R<X1 then C is closer to A than D (smaller dx, dy)[/spoiler]

[spoiler]Edit: Note that the 1) and 3) reasoning hold because CD is said to intersect BA[/spoiler]

[spoiler]I have a picture to better illustrate the middle case,[/spoiler]
[spoiler]which can be posted at a later time if there is an interest.[/spoiler]

I got a little confused by your second case.
[spoiler]Since all you say about C and D is that they are both outside[/spoiler]
[spoiler]the circle centred on B, then they could both be equidistant[/spoiler]
[spoiler]from A and B so that a perpendicular to AC could in fact be AD.[/spoiler]
[spoiler]Which means that D is not on the opposite side of this line as[/spoiler]
[spoiler]you claim. But this is irrelevant to your answer because in that[/spoiler]
[spoiler]case CD would be a perpendicular bisector of AB at the midpoint,[/spoiler]
[spoiler]which by definition is the points equidistant from A and B. So if[/spoiler]
[spoiler]either (or both) C and D are on this line then they are closer to[/spoiler]
[spoiler]either A or B than they are to each other. I'm sure you considered[/spoiler]
[spoiler]this and simply failed to include it in your post.[/spoiler]

[spoiler]For my solution I considered AB to be a cord of a circle centred on C[/spoiler]
[spoiler]from where it was pretty straightforward to show that D (who has to be[/spoiler]
[spoiler]inside the circle) must be closer to one end of the cord than he is to[/spoiler]
[spoiler]any point on the opposite side of it, unless he is sufficiently far away[/spoiler]
[spoiler]that C must suffer the same fate.[/spoiler]

[QUOTE=Numbers]I got a little confused by your second case.
[spoiler]....in that case CD would be a perpendicular bisector of AB at the midpoint,[/spoiler]
[spoiler]which by definition is the points equidistant from A and B.[/spoiler]
[/QUOTE]

[spoiler]From here we know this cannot be the case since we are told that[/spoiler]
[spoiler]"If any two players are n feet apart, they are the only two players who are n feet apart." (Rule 3)[/spoiler]
[spoiler]so this need not be considered further - but was an oversight originally on my part when posting[/spoiler]