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Start with the set {3, 4, 12}. You are allowed to perform a sequence of replacements, each time replacing two numbers a and b from your set with the new pair 0.6 a - 0.8b and 0.8 a + 0.6b. Can you transform the set into {4, 6, 12}? Look for an invariant.

I am having trouble determining what the invariant is. Any suggestions would be appreciated.

amWhy
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Jebediah
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2 Answers2

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$$(0.6a-0.8b)^2+(0.8a+0.6b)^2+c^2=a^2+b^2+c^2$$

The squaresum of the numbers remains constant. The start triple $\{3,4,12\}$ has the squaresum $3^2+4^2+12^2=169$ . The next triple you generate has also the squaresum $169$ and so on. You will never reach a triple with squaresume $6^2+4^2+12^2=196$

miracle173
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  • Can I simply make the comparison between a^2 + b^2 and (0.6a−0.8b)^2 + (0.8a+0.6b)^2? Since those two sums are not equal, it is not possible to transform the set. – Jebediah Oct 22 '13 at 01:00
  • @David: What are $a$ and $b$ in the case of ${3,4,12}$ and in the case of ${4,6,12}$ – miracle173 Oct 22 '13 at 01:29
  • In the case of {3, 4, 12}, a = 3 and b = 4 and in the case of {4, 6, 12}, a = 4 and b = 6. – Jebediah Oct 23 '13 at 12:27
  • Let's start with the triple $(3,4,12)$. I select two of these numbers (call them $a$ and $b$) and replace them by $0.6a-0.8b$ and $0.6a+0.8b$. Call the new triple $(x,y,z)$. Which property has the new triple? After you have posted your answer I will tell you which numbers I selected. – miracle173 Oct 23 '13 at 13:57
  • I'm not quite sure what you mean by "Which property has the new triple?" – Jebediah Oct 23 '13 at 14:41
  • Better: "which property does not change?". The squaresum $x^2+y^2+z^2$ will does not chanage, it will remain $169$. – miracle173 Oct 23 '13 at 18:25
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The operation you are performing is equivalent to multiplying by one of the matrices $$ \begin{bmatrix} 0.6&-0.8&0\\ 0.8&0.6&0\\ 0&0&1 \end{bmatrix}, \begin{bmatrix} 0.6&0&0.8\\ 0&0&1\\ -0.8&0&0.6 \end{bmatrix},\text{ or} \begin{bmatrix} 1&0&0\\ 0&0.6&-0.8\\ 0&0.8&0.6 \end{bmatrix} $$ or their inverses. These are rotation matrices that rotate by $\sin^{-1}(0.8)$ around one of the coordinate axes. Since they are orthogonal, they are isometries and they preserve distances; in particular the length of a vector.

Since $|(3,4,12)|=13$ and $|(4,6,12)|=14$, the length of the vector is changed, so there is no way to get from one to the other using these matrices.

Note that this is not really different than miracle173's observation, just put into matrix/rotation form.

robjohn
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