Thread: Math question
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Old January 7th, 2009, 05:06 AM
HallsofIvy HallsofIvy is offline
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Let me see if I understand this. If you set the ant down at (x,y), it immediately walks to (-3x-y, 7x+ ky) and those then become the "(x,y)" for the next step- it now walks to (-3(-3x-y)- (7x+ ky), 7(-3x-y)+ k(7x+ky))= (2x+(3-k)y, (-21+ 7k)x+ (-7+k^2)y)

I think we can best do this as a matrix problem: From point (x,y) the ant goes to A(x,y) where A is the matrix
\begin{bmatrix}-3 & -1 \\ 7 & k\end{bmatrix}
In order that the ant eventually return to the original point, no matter what that point is, we must have k such that, for some finite n, A^n= I or
\begin{bmatrix}-3 & -1 \\ 7 & k\end{bmatrix}^n= \begin{bmatrix}1 & 0 \\ 0 & 1\end{bmatrix}

Since the product of a two matrices has determinant equal to the product of the determinants of the two matrices, If A had determinant larger than 1, it would get bigger and bigger and could never be 1, as it must in order to be the identity matrix. If A had determinant less than 1, it would get smaller and smaller. A must have determinant 1 or -1.
The determinant of A is -3k+ 7 so either -3k+ 7= 1, which gives -3k= -6 and k= 2, or -3k+ 7= -1 so -3k= -8 and k= 8/3. I think I would take k= 2.

Sure enough, with k= 2, [math]A^3= I/[math] so starting at any (x, y), the ant would return to (x,y) after going around a triangle.
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