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Consider a parallelogram which is neither a rhombus nor a rectangle. It is well-known that such shapes do not have an axis of symmetry.

Is there a simple proof for this?

I prefer a proof that I can give to a kid (~12 year old), but more involved proofs will do as well.

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It has a twofold axis around the line perpendicular to the plane and through the center. –  Ross Millikan Dec 4 '10 at 17:58
    
@Ross: I don't want to go 3D. Stickling to the 2D plane, it doesn't have any symmetry axes, right? –  Sadeq Dousti Dec 4 '10 at 18:11

4 Answers 4

up vote 3 down vote accepted

If a symmetry pairs two vertices, then the angles at those vertices must be congruent. A symmetry that pairs adjacent vertices of a parallelogram requires that the figure be a rectangle: adjacent angles are supplementary; if they're also congruent, then they must be right angles.

Having ruled out rectangles, your proposed symmetry must pair each vertex (a) with its non-adjacent counterpart, or (b) with itself. In (a), the axis would be the perpendicular bisector of the diagonal between the paired vertices; in (b) the axis would contain the vertex. If both (a) and (b) occur, then we have a vertex (from (b)) on the perpendicular bisector of the diagonal joining two non-adjacent others (from (a)); this implies that we have two congruent adjacent edges, which in turn implies that the figure is a rhombus.

To avoid both rectangles and rhombi, we must have that the axis provides only-(a) symmetry, or only-(b) symmetry. An axis of only-(a) symmetry must be the perpendicular bisector of both diagonals, which (as perpendiculars of that axis, and lying in the same plane of that axis) must therefore lie on the same line: the vertices of the figure are collinear. An axis of only-(b) symmetry contains all the vertices, making them collinear (but in a different arrangement than required by (a)).

If you disallow "degenerate" parallelograms with all vertices collinear, then you have your impossibility argument. However, I prefer to recognize degenerate figures as legitimate, and I try to avoid stigmatizing them whenever possible. (After all, they come in handy as intermediate steps in transforming one figure smoothly into another. Besides, it's not like the "Parallelogram Law of Vector Addition" becomes invalid when the vectors involved are linearly dependent.) So, I would go so far as to observe that an axis of only-(a) symmetry implies the existence of a (separate) axis of only-(b) symmetry, and vice versa; that is, we've determined a third class of parallelograms --along with rectangles and rhombi-- that have two axes of symmetry.

(Consideration of parallelograms whose vertices coincide is left to the reader.)

Note: An only-(a) axial symmetry can also be realized with an axis perpendicular to the plane of the parallelogram (which need not be degenerate), but you've ruled this out.

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Perhaps this will be convincing?

1) The axis of symmetry intersects one of the sides at point which is not the vertex.

In this case, by considering the perpendicular to the axis at the point of intersection of the axis and the side, we see that the axis has to be perpendicular to the side and intersects it at the midpoint. In which case it is not an axis of symmetry as it won't intersect the opposite side at the midpoint.

2) The axis of symmetry intersects a side at a vertex. Then by 1) it also intersects the opposite side at the vertex, and hence has to be a diagonal, in which case, it is not an axis of symmetry either.

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Yes, this is how I would do it. –  Qiaochu Yuan Dec 4 '10 at 20:07

Let's prove the converse: if a parallelogram $ABCD$ has a symmetry axis, it's either a rectangle or a rhombus.

Clearly the image of a vertex is a vertex.

If some of the vertices (say $A$) is mapped to itself, than there must be another vertex mapped to itself (because otherwise the rest of vertices must be split into pairs by mapping). This another vertex must be $C$, because if it's adjacent to $A$, than the symmetry axis is one of the edges, which is impossible. So, $AC$ is the axis, $B$ maps to $D$, thus $BD$ is perpendicular to $AC$, therefore the parallelogram $ABCD$ is a rhombus.

Consider the case when every vertex is mapped to some other vertex. If $A$ maps to $C$, than $B$ must map to $D$, so the axis should be a line perpendicular to both $AC$ and $BD$, which is impossible since $AC$ and $BD$ are parallel. So, $A$ maps to an adjacent vertex (let it be $B$), $C$ maps to $D$, so the axis connects the midpoints of $AB$ and $CD$ and is perpendicular to them. This implies that $AB$ is perpendicular to $CD$.

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I would start by pointing out that parallelogram has two points farther apart than any other pair-two of the vertices. Any symmetry must leave them in place or exchange them. To leave them in place, the axis has to go through them. To exchange them, the axis has to be the perpendicular bisector of the segment between them. By inspection, neither is a symmetry.

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