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Title: Cartesian Game Post by THUDandBLUNDER on Dec 23rd, 2006, 8:41pm Starting at a point (x,y), you can move to either (x+2y,y), (x-2y,y), (x,y+2x), or (x,y-2x). This becomes the new (x,y), and you have the same linear transformations for your next move, except that you are not allowed to move directly back to the previous point. This can continue indefinitely. Prove that if you begin at the point (1, sqrt(2)), it is impossible to return there. |
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Title: Re: Cartesian GAme Post by Eigenray on Dec 24th, 2006, 3:14pm Define the linear transformations A(x, y) = (x+2y, y) and B(x, y) = (x, y+2x). Then any valid sequence of steps is a linear transformation given by some reduced word W in L={A, A-1, B, B-1}. Since the matrices of A and B have rational entries, W does too, so if W fixes the point (1, sqrt(2)), then it must be the identity. Hence it suffices to show that no nontrivial reduced word in L is the identity. Let S(A) = {(x,y) : |x| > |y|, xy>0}, S(A-1) = {(x,y) : |x| > |y|, xy<0}, S(B) = {(x,y) : |x| < |y|, xy>0}, S(B-1) = {(x,y) : |x| < |y|, xy<0}. Suppose that (x,y) is in S(A) or S(A-1), so that |x|>|y|. Then |y+2x| > 2|x|-|y| > |x|, and 2x+y has the same sign as x, so B(x,y) is in S(B). And if (x,y) is in S(B), then since x,y have the same sign, y+2x has the same sign as x and is larger in absolute value, so again B(x,y) is in S(B). Thus B maps every region other than S(B-1) into S(B). Similarly, we find that for any X in L, X maps everything other than S(X-1) into S(X). Now suppose W = Xn...X1 is a nontrivial reduced word in L. That is, each Xi is in L, and no Xi = Xi+1-1. Pick v=(x,y) not in S(X1-1) or in S(Xn). Then X1(v) is in S(X1). Inductively, if Xi...X1(v) is in S(Xi), then because Xi is not Xi+1-1, we have Xi+1...X1(v) is in S(Xi+1). It follows W(v) is in S(Xn). Since we chose v not in S(Xn), it follows W is not the identity. |
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