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 $\{b_1,b_2,\ldots b_n\}$ ?

#Algebra #Generalize #InvariantPrinciple #ProblemSolving #Transformation

Easy Math Editor

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`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
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Comments

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

ww margera - 6 years, 10 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 .

Abdullah Ahmed - 5 years, 8 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$