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Why the diagonal of a cube isn't its symmetry line?

We should find a point that its reflection is out of cube but I can't find that.Please don't use softwares or so.Because it was a math olympiads question.

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    Hold a cube on a diagonal and rotate it to see what happens. – 2017-02-09
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    What happens to the vertices of the cube? – 2017-02-09
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    There is no such thing as symmetry line. If you mean an axis of symmetry, then, well, it is. – 2017-02-09
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    For a **space** diagonal, it's an axis of $3$-fold symmetry. – 2017-02-09

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Consider the standard unit cube in $\mathbb{R}^3$.

Let $d$ be the diagonal line through the points $(0,0,0)$, $(1,1,1)$.

Let $A = (1,1,0)$.

A generic point on $d$ has the form $(t,t,t)$.

The square of the distance from $A$ to $(t,t,t)$ is $3t^2-4t+2$, which is a quadratic function which achieves its minimum value at $t=2/3$.

It follows that the point on $d$ which is closest to $A$ is the point $B=(2/3,2/3,2/3)$.

Let $C$ be the reflection of $A$ about $d$.

Then $C$ must be the reflection of $A$ through the point $B$.

It follows that $B$ is the midpoint of the line segment $AC$.

Applying the midpoint formula, we get $C = (1/3,1/3,4/3)$.

Since the $z$-coordinate of $C$ is greater than $1$, $C$ is outside the cube.

It follows that the diagonal line $d$ is not a line of symmetry for the cube.

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    How can You find that $B=(2/3,2/3,2/3)$? – 2017-02-09
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    @Taha Akbari: I just added an explanation in the answer. – 2017-02-09