Rotational and point symmetry
Some shapes look exactly the same after you turn them part of the way round. Which ones, and by how much can you turn them?
Where you are 11 in Transformations
Why this exists
Turn a square a quarter of the way round its center and it lands exactly on its own outline. If you looked away while it turned, you could not tell that anything had happened. A pinwheel does the same, and so does a regular hexagon, after a sixth of a turn.
Line symmetry does not describe this. A mirror line catches shapes whose two halves match, but a pinwheel has no mirror line at all, and it is still plainly symmetric. What it has is a turn that leaves it unchanged. This lesson names that kind of symmetry, counts it, and finds the one special turn, the half turn, that shows up in every parallelogram and in the graph of every odd function.
Symmetry also pays. A turn is a rigid motion, so it keeps every length and angle. When a turn carries a figure onto itself, each side is equal to the side it lands on and each angle to the angle it lands on. A symmetry hands you equal parts without measuring anything.
Definition
A rotation turns every point of the plane about one fixed point O through the same angle. A figure has rotational symmetry when some rotation about a point O, through an angle strictly between 0^\circ and 360^\circ, maps the figure exactly onto itself: every point of the figure lands on a point of the figure, and every point of the figure is landed on.
The point O is the center of the symmetry, and each angle that works is an angle of rotational symmetry. A turn through 360^\circ puts every figure back where it started, so it is left out of the definition: it says nothing about the shape.
The order of the rotational symmetry is the number of angles \theta with 0^\circ < \theta \le 360^\circ that map the figure onto itself. Put another way, it is how many times the figure sits exactly on itself during one full turn. A figure with no rotational symmetry has order 1, since only the full turn works.
A figure has point symmetry when a half turn, a rotation through 180^\circ about some point O, maps it onto itself. Then every point P of the figure has a partner P' in the figure with O the midpoint of \overline{PP'}: straight through the center, the same distance out on the other side.
Angles here are counterclockwise unless a problem says otherwise. The direction never changes the answer: a clockwise turn through \theta ends in the same place as a counterclockwise turn through 360^\circ - \theta, and a figure that survives one turn survives the turn that undoes it.
Intuition
Trace the figure on a sheet of tracing paper, pin the tracing at the center, and turn it. Each time the tracing sits exactly on the figure again, you have found an angle of rotational symmetry.
For the regular hexagon in Figure 1 that happens at 60^\circ, 120^\circ, 180^\circ, 240^\circ, 300^\circ and finally at 360^\circ: six times in one full turn, so the order is 6. Notice that each angle is a whole number of 60^\circ steps. Once one turn works, doing it twice works, and three times, and so on, which is why the angles of a figure come in an evenly spaced list.
Point symmetry is the same game with one half turn. Pick any point of the figure, walk in a straight line to the center, and keep walking the same distance. If you always land on the figure again, it has point symmetry.
Results
Turns combine. If turns through a and b about the same center both map a figure onto itself, so do turns through a + b and a - b. A turn through more than 360^\circ ends where a turn through 360^\circ less would, and a negative turn is a clockwise one.
The smallest angle divides 360^\circ. If a figure has rotational symmetry and \alpha is its smallest angle of rotational symmetry, then its order n is a whole number with
and the angles that work are exactly \alpha, 2\alpha, 3\alpha, \ldots, (n-1)\alpha.
Regular polygons. A regular polygon with n sides has rotational symmetry of order n about its center, with smallest angle \dfrac{360^\circ}{n}.
Point symmetry and order. A figure with rotational symmetry of order n has point symmetry about the same center exactly when n is even. In particular a regular polygon has point symmetry exactly when its number of sides is even.
On coordinates. The half turn about the origin sends (x, y) to (-x, -y), and the half turn about the point (a, b) sends (x, y) to (2a - x,\ 2b - y). A figure has point symmetry about the origin exactly when (-x, -y) is in the figure whenever (x, y) is.
Why it is true
Turns combine. Turning through a and then through b about the same center is a single turn through a + b (doing one transformation after another). If each turn alone carries the figure onto itself, doing one after the other does too. A turn through b that carries the figure onto itself is undone by the turn through -b, and the undoing turn carries the figure onto itself as well, since it simply sends every point back where it came from. Turning through a and then through -b gives a - b.
The smallest angle divides 360^\circ. Let \theta be any angle that works, with 0^\circ < \theta \le 360^\circ. Divide \theta by \alpha with a remainder: \theta = q\alpha + r, where q is a whole number and 0^\circ \le r < \alpha. A turn through q\alpha is q turns through \alpha in a row, so it works. Then r = \theta - q\alpha works too, by the subtraction rule. If r were bigger than 0^\circ it would be a working angle smaller than \alpha, which is impossible, so r = 0^\circ and \theta is a multiple of \alpha. The full turn always works, so 360^\circ is a multiple too: 360^\circ = n\alpha for some whole number n. The working angles up to 360^\circ are then \alpha, 2\alpha, \ldots, n\alpha, exactly n of them, so n is the order.
Regular polygons. The vertices of a regular polygon are all the same distance from its center O. Join O to every vertex. Each triangle formed by O and one side has two sides equal to that distance and a third side equal to the polygon’s side, so all n triangles are congruent by SSS, and their angles at O are equal. Those n angles fill the full turn, so each is \dfrac{360^\circ}{n}. A turn through that angle carries each vertex to the next one along, so each side to the next side and the polygon onto itself. No smaller turn works: a rotation that maps the polygon onto itself must send corners to corners, and every vertex is \dfrac{360^\circ}{n} round the center from its neighbors, so a smaller turn puts a vertex at a point between two corners.
Point symmetry and order. By the second result the angles that work are the multiples of \dfrac{360^\circ}{n}. So 180^\circ works exactly when
for some whole number k, which says n = 2k: n is even.
On coordinates. The half turn about (a, b) sends (x, y) to the point (x', y') that has (a, b) as the midpoint of the segment joining them. So \dfrac{x + x'}{2} = a and \dfrac{y + y'}{2} = b, which give x' = 2a - x and y' = 2b - y. With a = b = 0 that is (-x, -y).
Worked examples
The equilateral triangle
Find every angle of rotational symmetry of an equilateral triangle, give its order, and decide whether it has point symmetry.
An equilateral triangle is a regular polygon with 3 sides, so it has rotational symmetry of order 3 about its center. The smallest angle is \dfrac{360^\circ}{3} = 120^\circ, and the angles of rotational symmetry are 120^\circ and 2 \cdot 120^\circ = 240^\circ.
The order 3 is odd, so the triangle has no point symmetry. Figure 2 shows why directly. The center O lies on the median from A, two thirds of the way down it. If the median has length h, then A is \tfrac{2}{3}h above O, while the triangle reaches only \tfrac{1}{3}h below O, where the base crosses the median at a right angle. The half turn sends A to A', \tfrac{2}{3}h below O, which is outside the triangle. The dashed image does not match the original.
The regular hexagon
For the regular hexagon ABCDEF in Figure 1, list the angles of rotational symmetry, give the order, and say where the half turn sends each vertex. What does the half turn tell you about the sides \overline{AB} and \overline{DE}?
SolutionA regular hexagon has order 6, and its smallest angle is \dfrac{360^\circ}{6} = 60^\circ. The angles of rotational symmetry are the multiples of 60^\circ below 360^\circ:
The order is even, so the hexagon has point symmetry. The half turn is 180^\circ = 3 \cdot 60^\circ, three steps round, so it sends each vertex three places on: A to D, B to E and C to F. Each vertex and its partner lie on a line through O, at the same distance on opposite sides.
The half turn carries \overline{AB} onto \overline{DE}. A rotation keeps lengths, so AB = DE. A half turn also sends every line to a line parallel to it, so \overline{AB} \parallel \overline{DE}. The same holds for \overline{BC} and \overline{EF}, and for \overline{CD} and \overline{FA}: opposite sides of the hexagon are parallel.
A pinwheel
Figure 3 is made of four congruent right-angled triangles, each one a quarter turn on from the last about the center. Find its order, decide whether it has point symmetry, and decide whether it has a line of symmetry.
A quarter turn about the center moves each blade onto the next one, so 90^\circ works. By the combining rule 180^\circ and 270^\circ work too. No smaller turn works. A turn keeps every distance from the center, so it must send the points of the figure farthest from the center to points equally far. Those are the four outer tips of the blades, and they sit a quarter turn apart, so a turn of less than 90^\circ sends a tip to a point that is not a tip. So the smallest angle is 90^\circ, and the order is \dfrac{360^\circ}{90^\circ} = 4. Since 4 is even, the pinwheel has point symmetry: the half turn swaps opposite blades.
It has no line of symmetry. Each blade has two edges running out from the center: a shorter one that ends at the right angle, and the longer hypotenuse that ends at the tip. Going counterclockwise round the center, every blade begins with its shorter edge and ends with its hypotenuse. A reflection keeps lengths but reverses the direction of travel round any point, so in the mirror image every blade would begin with its hypotenuse. The mirror image is a pinwheel spinning the other way, and it does not sit on the original. Rotational symmetry and line symmetry are separate properties: a figure can have either one without the other.
The parallelogram
Show that every parallelogram has point symmetry about the point M where its diagonals cross. Then decide when a turn of 90^\circ about M also maps it onto itself.
The diagonals of a parallelogram bisect each other (properties of a parallelogram), so M is the midpoint of \overline{AC} and of \overline{BD}. The half turn about M therefore sends A to C, C to A, B to D and D to B. A rotation sends segments to segments, so \overline{AB} goes to \overline{CD}, \overline{BC} to \overline{DA}, and so on round the figure. The four sides land on the four sides, and the parallelogram maps onto itself.
Now suppose a turn of 90^\circ about M also works. It must send A to a vertex. Not to A itself, since a turn moves every point except its center, and not to C, which is 180^\circ round from A. So A goes to B or to D, and either way that vertex is the same distance from M as A, a right angle round. That makes the diagonals equal in length and perpendicular. A parallelogram whose diagonals are equal is a rectangle, and one whose diagonals are perpendicular is a rhombus (what each special quadrilateral guarantees), so it is a square.
So a parallelogram that is not a square has rotational symmetry of order 2, and a square has order 4.
Completing a figure on the coordinate plane
A hexagon ABCDEF has point symmetry about the origin. Three of its vertices, in order, are A(5, 1), B(2, 4) and C(-3, 3). Find D, E and F.
The half turn about the origin sends (x, y) to (-x, -y), and it sends vertices to vertices. It also keeps neighbors as neighbors and never reverses the order round the hexagon, because a rotation does not reverse direction. So it moves every label the same number of places round. That number is not 0, since only the center stays put and A is not the origin. Doing the half turn twice brings every point back, so two such moves make a whole lap of six places, and each move is three places. A goes to D, B to E and C to F:
Check. Each pair has the origin as its midpoint, for example \left(\dfrac{5 + (-5)}{2}, \dfrac{1 + (-1)}{2}\right) = (0, 0). And the half turn should make \overline{AB} parallel to \overline{DE}. The slope of \overline{AB} is \dfrac{4 - 1}{2 - 5} = -1 and the slope of \overline{DE} is \dfrac{-4 - (-1)}{-2 - (-5)} = -1. They match.
Working back to the number of sides
A regular polygon maps onto itself under a turn of 40^\circ about its center. What is the fewest sides it can have, and what other numbers of sides are possible?
SolutionThe angles of rotational symmetry of a regular n-sided polygon are the multiples of \dfrac{360^\circ}{n}, so 40^\circ = k \cdot \dfrac{360^\circ}{n} for some whole number k. Multiply both sides by n:
The fewest sides come from k = 1: n = 9, and then \dfrac{360^\circ}{9} = 40^\circ is the smallest angle itself. Any multiple of 9 also works. For n = 18 the smallest angle is \dfrac{360^\circ}{18} = 20^\circ, and 40^\circ is two of those steps.
Two turns force a third
A figure maps onto itself under a turn of 90^\circ and under a turn of 150^\circ, both about the same center. Show that a turn of 30^\circ must also work, and find the smallest order the figure can have.
SolutionTurns about the same center can be subtracted. 150^\circ - 90^\circ = 60^\circ, so a turn of 60^\circ works. Then 90^\circ - 60^\circ = 30^\circ, so a turn of 30^\circ works.
The smallest angle \alpha divides every angle that works, so it divides 30^\circ, and the order is \dfrac{360^\circ}{\alpha}. The largest \alpha can be is 30^\circ, which gives the smallest order, \dfrac{360^\circ}{30^\circ} = 12. A regular 12-sided polygon has exactly this: 90^\circ = 3 \cdot 30^\circ and 150^\circ = 5 \cdot 30^\circ.
The subtraction did the same job as finding the greatest common factor: 30 is the greatest common factor of 90 and 150, and the turns a figure has are always the multiples of one angle.
Boundaries
Exactly, not nearly. The figure must land on itself point for point. A drawing that looks almost the same after a turn has no rotational symmetry unless the description says the parts are equal. A letter or a digit has the symmetry its particular drawing has, which depends on how it is drawn.
The full turn does not count. Every figure survives a turn of 360^\circ, which is why the definition asks for an angle strictly between 0^\circ and 360^\circ, and why a figure with no rotational symmetry still has order 1 rather than 0.
There may be no smallest angle. A circle maps onto itself under every turn about its center, so no angle is the smallest and it has no whole-number order. The results about \alpha need a smallest angle. Every polygon has one, because its finitely many vertices can be shuffled in only finitely many ways.
A figure that fits on a page has at most one center. A turn about one point followed by the reverse turn about a different point is a slide, and a figure that fits on a page cannot slide onto itself. Patterns that go on for ever, such as a tessellation of the plane, can have many centers.
Turns and mirrors are separate. The pinwheel has rotational symmetry and no line of symmetry; an isosceles triangle that is not equilateral has a line of symmetry and no rotational symmetry. A parallelogram that is not a rectangle or a rhombus has point symmetry and no line of symmetry at all. The two meet in one place: two lines of symmetry crossing at an angle \beta always produce rotational symmetry through 2\beta, because two reflections in crossing lines make a rotation.
Point symmetry is about a point, in the plane. In space, turning about a point is not enough to describe a turn; a solid turns about a line, its axis, and a half turn about an axis is a different motion from sending every point straight through a center.
Where this goes next
The angle \dfrac{360^\circ}{n} at the center of a regular polygon is the central angle, and the n congruent triangles in the proof above are how the area of a regular polygon is found. The same triangles give the size of each interior angle.
Point symmetry about the origin, (x, y) \mapsto (-x, -y), is exactly what makes a graph the graph of an odd function. Testing whether an equation survives replacing both variables by their opposites is testing a graph for symmetry with algebra.
Example 7 showed the symmetries of a figure combining with each other: any two give a third, and each can be undone. A collection of motions that behaves this way is what later mathematics calls a group, and the symmetries of a regular polygon, its turns together with its mirror lines, are one of the first groups anyone meets. Seen from the other side, a symmetry is a congruence between a figure and itself.
Exercises
Finding the turns
For Exercises 1 to 3, list every angle of rotational symmetry, give the order, and say whether the figure has point symmetry.
- A square.
- A regular pentagon.
- A regular octagon.
- Find the smallest angle of rotational symmetry of a regular heptagon (seven sides) exactly, and say whether it has point symmetry.
- Give the order of a rectangle that is not a square, and its angles of rotational symmetry.
- Give the order of a rhombus that is not a square, and the center of its rotational symmetry.
- A five-pointed star is drawn by joining every second vertex of a regular pentagon. Find its smallest angle of rotational symmetry and its order, and say whether it has point symmetry.
Working backwards
- The smallest angle of rotational symmetry of a regular polygon is 24^\circ. How many sides does it have?
- A figure has rotational symmetry of order 9. Find its smallest angle of rotational symmetry, and list the angles of rotational symmetry less than 180^\circ.
- Can a regular polygon have a smallest angle of rotational symmetry of 50^\circ? Explain.
- A regular polygon maps onto itself under a turn of 135^\circ about its center. What is the fewest sides it can have?
- A regular polygon maps onto itself under a turn of 100^\circ about its center. What is the fewest sides it can have?
Point symmetry on the coordinate plane
- A figure has point symmetry about the origin and contains the point (3, -7). Which other point must it contain?
- Does the quadrilateral with vertices A(1, 4), B(-3, 2), C(-1, -4) and D(3, -2) have point symmetry about the origin? Explain.
- A parallelogram has vertices (0, 0), (4, 1), (6, 5) and (2, 4). Find its center of point symmetry.
- A figure has point symmetry about (2, 1) and contains the point (5, -3). Which other point must it contain?
- Three vertices of a quadrilateral, in order, are (2, 5), (-1, 3) and (-2, -5). The quadrilateral has point symmetry about the origin. Find the fourth vertex.
Combining turns
- A figure maps onto itself under a turn of 72^\circ. Which other angles between 0^\circ and 360^\circ must also work?
- A figure maps onto itself under a turn of 80^\circ. Show that a turn of 40^\circ also works, and find the smallest order the figure can have.
- A figure maps onto itself under turns of 120^\circ and 135^\circ about the same center. Find the smallest order it can have.
Reasoning
- A figure has point symmetry and also maps onto itself under a turn of 120^\circ about the same center. Show that its order is a multiple of 6.
- A kite has two short sides meeting at one vertex and two long sides meeting at the opposite vertex. Explain why it has no rotational symmetry.
- Explain why a circle has no whole-number order, and say which step in the proof that the smallest angle divides 360^\circ fails for it.
- Show that a figure that fits on a page cannot have point symmetry about two different points.
- Using the circle through the vertices, explain why a regular polygon with an odd number of sides has no point symmetry.
Answers
- 90^\circ, 180^\circ and 270^\circ; order 4; it has point symmetry. A square is a regular polygon with 4 sides, so the smallest angle is \dfrac{360^\circ}{4} = 90^\circ, and 180^\circ = 2 \cdot 90^\circ is on the list.
- 72^\circ, 144^\circ, 216^\circ and 288^\circ; order 5; no point symmetry. The smallest angle is \dfrac{360^\circ}{5} = 72^\circ, and \dfrac{180}{72} = \dfrac{5}{2} is not a whole number, as the odd order says.
- 45^\circ, 90^\circ, 135^\circ, 180^\circ, 225^\circ, 270^\circ and 315^\circ; order 8; it has point symmetry. The smallest angle is \dfrac{360^\circ}{8} = 45^\circ, and 180^\circ = 4 \cdot 45^\circ.
- \dfrac{360^\circ}{7} = 51\tfrac{3}{7}^\circ, and no point symmetry, since the order 7 is odd.
- Order 2; the only angle is 180^\circ, about the point where the diagonals cross. Every parallelogram has this half turn (Example 4), and a turn of 90^\circ would make it a square. A figure on a page has only one center, so no other point helps.
- Order 2, about the point where the diagonals cross. As in Example 4, a quarter turn about that point would need equal diagonals, and a rhombus with equal diagonals is a square.
- Smallest angle 72^\circ, order 5, no point symmetry. A turn of 72^\circ about the center moves each pentagon vertex to the next, so each segment of the star, joining a vertex to the one two along, moves to the next such segment. The five tips are the points of the star farthest from the center and sit 72^\circ apart, so no smaller turn sends tips to tips. The order 5 is odd.
- 15 sides. The smallest angle is \dfrac{360^\circ}{n}, so 24n = 360 and n = 15.
- 40^\circ; the angles below 180^\circ are 40^\circ, 80^\circ, 120^\circ and 160^\circ. The smallest angle is \dfrac{360^\circ}{9} = 40^\circ.
- No. The smallest angle of a regular n-sided polygon is \dfrac{360^\circ}{n}, so n = \dfrac{360}{50} = \dfrac{36}{5} = 7.2, which is not a whole number of sides.
- 8 sides. 135 = k \cdot \dfrac{360}{n} gives 3n = 8k, so k must be a multiple of 3, and k = 3 gives n = 8. Check: \dfrac{360^\circ}{8} = 45^\circ and 135^\circ is 3 steps of 45^\circ.
- 18 sides. 100 = k \cdot \dfrac{360}{n} gives 5n = 18k, so k must be a multiple of 5, and k = 5 gives n = 18. Check: \dfrac{360^\circ}{18} = 20^\circ and 100^\circ is 5 steps of 20^\circ.
- (-3, 7). The half turn about the origin sends (x, y) to (-x, -y).
- Yes. The half turn sends A(1, 4) to (-1, -4), which is C, and B(-3, 2) to (3, -2), which is D; it sends C and D back to A and B. Vertices go to vertices in the same order round, so \overline{AB} lands on \overline{CD} and \overline{BC} on \overline{DA}.
- \left(3, \tfrac{5}{2}\right), where the diagonals cross. The midpoint of the diagonal from (0, 0) to (6, 5) is \left(\tfrac{0 + 6}{2}, \tfrac{0 + 5}{2}\right), and the midpoint of the one from (4, 1) to (2, 4) is \left(\tfrac{4 + 2}{2}, \tfrac{1 + 4}{2}\right), the same point.
- (-1, 5). The half turn about (a, b) sends (x, y) to (2a - x, 2b - y), here (2 \cdot 2 - 5,\ 2 \cdot 1 - (-3)).
- (1, -3). Opposite vertices of a figure with point symmetry are partners, and (2, 5) and (-2, -5) already are. The fourth vertex is the partner of (-1, 3), which is (1, -3).
- 144^\circ, 216^\circ and 288^\circ: two, three and four turns of 72^\circ in a row. Other angles may work as well, if the figure’s smallest angle is a smaller divisor of 72^\circ, but these three must.
- Five turns of 80^\circ make 400^\circ = 360^\circ + 40^\circ, which ends where a turn of 40^\circ does, so 40^\circ works. The smallest angle divides 40^\circ, so the order is a multiple of \dfrac{360}{40} = 9. The smallest order is 9, which a regular nonagon has.
- 24. The turn 135^\circ - 120^\circ = 15^\circ works, so the smallest angle divides 15^\circ and the order is a multiple of \dfrac{360}{15} = 24. A regular 24-sided polygon has both turns: 120^\circ = 8 \cdot 15^\circ and 135^\circ = 9 \cdot 15^\circ.
- Point symmetry means 180^\circ works, so 180^\circ - 120^\circ = 60^\circ works too. The smallest angle divides 60^\circ, so the order, 360^\circ divided by the smallest angle, is a multiple of \dfrac{360}{60} = 6.
- A rotation keeps lengths, so if it mapped the kite onto itself it would send short sides to short sides and long sides to long sides. The vertex where the two short sides meet is the only such vertex, so it would have to stay put, and so would the vertex where the two long sides meet. A turn through an angle strictly between 0^\circ and 360^\circ moves every point except its center, so it cannot keep two different points still. No such turn works, and the order is 1.
- Every turn about the center maps a circle onto itself. Whatever angle you try, half of it also works, so there is no smallest angle, and the number of working angles in one turn is not finite. The proof begins by taking the smallest angle \alpha and dividing by it; for a circle there is no \alpha to divide by.
- Put one center P at the origin and call the other Q(a, b), with (a, b) \ne (0, 0). The half turn about P sends (x, y) to (-x, -y), and the half turn about Q then sends that to (2a + x,\ 2b + y). Doing both is a slide by (2a, 2b), and it maps the figure onto itself. Repeating it k times slides by (2ka, 2kb), which for large enough k carries a point of the figure farther from P than the figure reaches. That is impossible for a figure that fits on a page.
- The vertices lie on a circle about the center, spaced \dfrac{360^\circ}{n} apart. A half turn sends a vertex V to the point of the circle directly opposite, 180^\circ round. That point is a vertex only if 180^\circ is a whole number of steps, 180 = k \cdot \dfrac{360}{n}, which needs n = 2k. When n is odd the image of V falls between two vertices, so the half turn does not send corners to corners and cannot map the polygon onto itself.