Swimming in Three-Dimensional Space

Algebra Level 3

For an ordered triple $(a, b, c),$ define a transformation as follows: choose any two of the three numbers, say $a$ and $b,$ and replace them with $\frac{a+b}{\sqrt2}$ and $\frac{a-b}{\sqrt2},$ respectively.

For instance, if we chose to transform the first two elements of $(4,2,4)$ as above, a transformation would result in $\Big(\frac{4+2}{\sqrt2}, \frac{4-2}{\sqrt2}, 4\Big)= \big(3\sqrt2, \sqrt2, 4\big).$

Can $\big(2,2\sqrt2, 3\sqrt2\big)$ transform into $(2,1,1)$ after finitely many transformations defined above?

Yes No

2 solutions

Kelvin Hong
Aug 28, 2018

For any transformation $(a,b,c)$ into $\bigg(\dfrac{a+b}{\sqrt2},\dfrac{a-b}{\sqrt2}, c\bigg)$ , we calculate its sum of squares:

$\bigg(\dfrac{a+b}{\sqrt2}\bigg)^2+\bigg(\dfrac{a-b}{\sqrt2}\bigg)^2+c^2=\dfrac{a^2+2ab+b^2}{2}+\dfrac{a^2-2ab+b^2}{2}+c^2=a^2+b^2+c^2$

Note that the sum of squares before and after transformation remains the same, hence we can now compute the sum of squares of these two triples:

$2^2+(2\sqrt2)^2+(3\sqrt2)^2=30$

$2^2+1^2+1^2=6$

Since the sum of squares are different, the triple is not transformable, the answer is $\boxed{\text{NO}}$ .

Brian Moehring
Aug 29, 2018

We can recognize each possible transformation as a rotation $45^\circ$ about one of the axes. Then we note the following facts:

Each of these rotations preserve its respective axis, which contains the origin, so all of them preserve the origin.
Rotations, in general, preserve distance between two points.

Combined, this means the distance between the origin and any point is preserved by each transformation.

We can check that the distances from the origin for the two points are different, so the answer is $\boxed{\text{no}}$

Swimming in Three-Dimensional Space

2 solutions

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