# Why the number of symmetry lines is equal to the number of sides/vertices of a regular polygon?

Considering the/a definition of a regular polygon from Wiki :

In Euclidean geometry, a regular polygon is a polygon that is equiangular (all angles are equal in measure) and equilateral (all sides have the same length).

, my question is how to prove that the number of symmetry lines is equal to the number of sides/vertices of a regular polygon?

To prove the statement (i.e. my question), i.e. by means of some logical 'discussion' to start from the definition to arrive to the conclusion I think a way that:

A line of symmetry is a line that if you fold the polygon by that line each side will fit exactly on each other. So choose a point on the edge of some polygon and start to move from that point to another point again on the edge of that polygon with the condition that the line bisects the perpendicular line joining two points of the boundary of the polygon. And because there are n distinct point to start these lines (why? I guess but can't prove) so there are n lines of symmetry in a regular n-gon.

I don't feel this as a rigorous proof and I don't know any other way to prove. Any simple detailed explanation would much be appreciated.

• Can you use linear algebra? Mar 6, 2016 at 4:51
• @sinbadh - Yes, but how to prove it by use of linear algebra? Thank you
– user200918
Mar 6, 2016 at 5:25
• Don't look too far, or at the end we even have to prove that such polygons exist. Note that you have the axioms (e.g., congruence laws) and basic theorems (e.g., angle sums) of euclidean geometry at your disposal. Jan 19, 2018 at 15:50

I'm not saying you can't refine the proof you attempted, but the first thing that comes to mind when dealing with symmetries of polygons is a group action of the dihedral group on the polygon. In a sense, group theory is the study of symmetry: when a group acts on a set, each group element describes an operation that will rearrange the set in such a way that it looks like it did before, and only the labels have changed. A permutation is precisely a reordering of elements, and Cayley's theorem states that every group is isomorphic to a subgroup of a symmetric group (which is the group of all permutations on a certain number of elements).

I'll use $D_{2n}$ to denote the dihedral group of $2n$ elements. It is generated by two elements: a reflection $R$ (which you can think of as reflecting an $n$-gon across your favorite axis, say the vertical one), and a rotation $r$ (which you can think of as rotating the $n$-gon clockwise). It should be clear that if you reflect across the same axis twice, you're back to where you started (so $R^2 = 1$), and the same is true if you rotate $n$ times in the same direction (so $r^n = 1$). Another relation on these elements is that if you reflect your polygon, and then rotate it clockwise, and then reflect it again, it's the same as rotating it counterclockwise. In fact, the dihedral group can be presented as follows:

$$D_{2n} = <r,R : R^2 = r^n = 1, RrR = r^{-1}>.$$

This is more information than you asked for, but it means that there are only two kinds of symmetries of an $n$-gon: those made of just rotations, and those made of reflections too. Moreover, there are the same number of each, and the number of symmetries with reflections is just the number of lines of symmetry.

• Well my question actually is "Why the number of symmetries with reflections is same as of the number of vertices?" which unfortunately I can't prove.
– user200918
Mar 6, 2016 at 5:22
• The number of rotations is the same as the number of vertices. A clockwise rotation sends one vertex to the next one clockwise. Mar 6, 2016 at 5:23
• Thank you for your answer. I know an elementary level of group theory but the text is not an answer to my question. I wonder to find a reason why there exists n lines of symmetry.
– user200918
Mar 6, 2016 at 5:28
• I don't understand why this is unsatisfactory. Is there a difference between asking "why" and asking "how to prove"? Mar 6, 2016 at 5:32
• OK, can you define a line of symmetry geometrically? and prove that why from the $2n$ possible relabeling only $n$ of them occurs by reflection of the symmetry axis/line? Or please underline the answer to these questions in your text? Thank you.
– user200918
Mar 6, 2016 at 5:44

Let me try to give partial answer to this using a little bit of analysis along with linear algebra.

We are basically finding line of reflection which does not change shape of Polygon at all.

Now any reflection is a linear transformation(these are nothing but orthogonal matrices with deterimant -1). Now any linear transformation in $R^2$ is continuous and its inverse being same reflection also continuous, so it is homeomorphism.

So it fixes boundary of any set, takes convex set to convex and lines to lines(easy to prove).

Now corners are nothing but points on intersection of two line segment which lie on boundary and every other point on boundary is not intersection of two line segments lying on boundary.

Now under required reflection since line goes to line and intersection points of two line will go to intersection point of two line segments. Corner will go to corner.

Hence if $x_1, x_2,\dots, x_n$ corner points then $v= x_1,+x_2+\dots+x_n/n$ will remain invariant under given transformation. And it lies in our polygon being convex linear combination of points, infact it lies in the interior of polygon i.e. there exist a ball around it lying completely inside polygon.

Now under any reflection, only points on a line are fixed and every other point changes their position. Hence that line has to pass through $v$.

Now as point in interior of polygon, every line passing through it will intersect boundary.(shown on Convex Set and an interior point)

Claim: any line passing through $v$ can only intersect boundary at corner or at midpoints of sides(or line segment joining corners).

Proof: If it intersect at any other point, we can transform coordinates to make that point origin and line passing through $v$ as $x-$ axis. Now side which was intersected is now line segment passing through origin. Now it is easy to pove that any line segment if reflected alone $x-$ axis remain unaffected, then midpoint of line lies on origin (use endpoint has to go to endpoint, and being reflection along x-axis), then $x-$ coordinates of endpoint will become zero and line segment would be just perpendicular to $x-$ axis. Now it is obvious that endpoint lies on origin.

Now using claim and fact since there are only $n$ corner points and $n$ sides, we get that line of symmetry can be atmost $2n$.

Now using result mentioned in Convex Set and an interior point

we also get that any line passing through $v$ will intersect boundary at atleast 2 points because $v$ is in interior of polygon.

And as any line passing thtough $v$ is determined by another point on it. Hence we get that under equivalence relation of being on same line of symmetry passing through $v$ on the set consisting of all corner and midpoints through which a symnetry line pass, we get that each equivalence class will have atleast 2 elements. And no. of equivalence class will be equal to no. of symmety lines.

We get since no. of equivalnece classes has to be less than equal to $n$ as set has atmost $2n$ elements while each equivalence class at atleast 2 elements.

We get line of symmetry can be atmost $n$

P.S. We proved for general n-polygon, there are atmost n line of symmetry. If for regular you want me to prove there are exactly $n$, that I don't know.

P.S. We also get by same argument $v$ is only point fixed under rotation, and as each vertex has to go to vertex, there are atmost $n$ rotations.

Let $n\geq3$ be given, and let $P$ be a regular polygon with $n$ vertices and $n$ edges according to the definition in the question. Draw through each vertex angle bisector pointing inwards. Then each edge becomes the base of an isosceles triangle, and all these triangles are congruent. It follows that the bisectors intersect in a single point $O$, and that $P$ can be inscribed in a circle centered at $O$. The angle between two successive bisectors is ${2\pi\over n}$. Congruence axioms then allow to conclude that the ${2\pi\over n}$ rotation with center $O$ is an automorphism of $P$.

Any mirror symmetry of $P$ maps vertices to vertices and edges to edges. Consider a symmetry line $\ell$ of $P$. This line will pass through $O$ (or there would be a second "center") and therefore intersect the boundary $\partial P$ in exactly two points. If such a point $A\in\ell\cap\partial P$ is an interior point of an edge $e$ then $\ell$ intersects $e$ orthogonally in its midpoint. It follows that all points $A\in\ell\cap\partial P$ produced in this way are either vertices of $P$ or midpoints of edges of $P$.

Discussing the cases when $n$ is odd or is even separately one then finds that we have exactly $n$ symmetry lines in both cases.