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# A Nice Invariant Problem

I recently found a problem that I found was really cool. Here it is:

(Tom Rike) Start with the set $$\{3,4,12\}$$. You are then allowed to replace any two numbers $$a$$ and $$b$$ with the numbers $$0.6a-0.8b$$ and $$0.8a+0.6b$$. Can you transform the set into $$\{4,6,12\}$$?

Generalize: given a set $$\{a_,a_2,\ldots a_n\}$$, what rule(s) determine if you can transform it into the set $$\{b_1,b_2,\ldots b_n\}$$?

Note by Daniel Liu
3 years, 5 months ago

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The sum of squares of the numbers is invariant as $$(0.6a - 0.8b)^2 + (0.8a + 0.6b)^2 = a^2 + b^2$$. Therefore, as $$3^2 + 4^2 + 12^2 = 169$$ while $$4^2 + 6^2 + 12^2 = 196$$, one cannot be transformed into the other

- 3 years, 5 months ago

if we dive into a little deep, we find that the important invariant is , the distance of the point (x,y,z) from midpoint O

here (0.6a-0.8b)²+(0.8a+0.6b)²=a²+b² Since 3²+4²+12²=169=13² assume the point lies on the sphere around O with the radius of 13 . Again 4²+6²+12²=196=14² and also assume the point lies on the sphere around I with the radius of 14 .

Then it Will never be transformed .

- 2 years, 3 months ago