Symmetry

Match the whole figure

A motion can move points while leaving a figure in the same place. Consider an unmarked rectangle 6 units wide and 2 units tall. A rectangle here is a four-sided figure with four right angles. Put its center at the origin and its corners at $(-3,-1)$, $(3,-1)$, $(3,1)$, and $(-3,1)$. A half-turn negates both coordinates, exchanging opposite corners. It takes each joining side onto the opposite side, so the whole boundary matches. A $90^\circ$ counterclockwise turn instead sends $(3,1)$ to $(-1,3)$, absent from the boundary. This turn fails.

A symmetry of a figure is a rigid motion that maps the whole figure exactly onto itself. Every image point must belong to the original figure, and every original position must be covered. Test edges, curves, interiors if included, and any marks that belong to the figure. A point label used only to name a corner is not a decoration unless the task says it is.

For the triangle, rectangle, and square boundaries here, a whole-boundary match must send corners to corners. A rigid motion preserves the angle where two straight sides meet. At a corner this angle is less than $180^\circ$; at an interior point of a straight side, the two boundary directions make $180^\circ$. A corner therefore cannot land in the middle of a straight side.

Reflections gives the flip rules, and Rotations gives the turn rules. Here we use those rules to test whether the image is the same figure. One failed point disproves a candidate; one successful point does not establish it.

Line symmetry is a fold-and-match test

Fold the same 6-by-2 rectangle along its vertical center line. The left half covers the right half, including the top and bottom edges. Folding along the horizontal center line also works. Folding along a diagonal fails: a long edge would have to cover a short edge, although reflection preserves length.

A line of symmetry is a mirror line whose reflection maps the whole figure onto itself. Points on the line stay fixed; other points can exchange places. A line that merely cuts an area into two equal amounts need not be a line of symmetry.

A square is a rectangle with four equal side lengths. Consider the unmarked square with corners $A(2,2)$, $B(-2,2)$, $C(-2,-2)$, and $D(2,-2)$. It has four lines of symmetry: the two center lines $x=0$ and $y=0$, and the two diagonals $y=x$ and $y=-x$.

The axis and diagonal reflection rules check the whole boundary:

Each rule maps every side onto the side joining its image endpoints, so every boundary point matches. This checks the folds using coordinates and straight-segment mapping, rather than assuming a fold works merely because adjacent sides have equal lengths.

There are no other mirrors. A corner $A$ must land on $A$, $B$, $C$, or $D$. Sending it to $B$, $C$, or $D$ forces, respectively, $x=0$, $y=-x$, or $y=0$ by the perpendicular-midpoint reflection rule. If $A$ stays fixed, the two sides at $A$ must map onto the two sides at $A$. Their other endpoints, the adjacent corners $B,D$, must therefore stay or exchange. Keeping both fixed would fix three points not on one line, whereas a reflection fixes only its mirror. Thus $B,D$ exchange, forcing $y=x$.

Rotational symmetry, order, and identity

For the coordinate square above, the counterclockwise quarter-turn rule $(-y,x)$ sends $A$ to $B$, $B$ to $C$, $C$ to $D$, and $D$ to $A$. Each joining side therefore maps onto the next side, matching the whole boundary. Its interior matches too, if included. The matching angles in one full turn are $0^\circ$, $90^\circ$, $180^\circ$, and $270^\circ$.

Rotational symmetry usually means that some turn strictly between $0^\circ$ and $360^\circ$ maps the whole figure onto itself. Specify its center. The identity rotation leaves every point fixed; $0^\circ$ and $360^\circ$ describe the same mapping.

For counting, this lesson includes identity exactly once. The rotational order about a stated center is the number of distinct matching rotations in $0^\circ\leq\theta<360^\circ$, when that number is finite. Thus a square has order $4$. Order $1$ means only identity matches: the figure has no nontrivial rotational symmetry about that center. We do not count $360^\circ$ again.

A proposed least turn must fit a full turn exactly. Suppose a finite-order figure were said to have least positive matching angle $50^\circ$. Seven repeats would also match:

$7\times50^\circ=350^\circ.$

The inverse of that $350^\circ$ counterclockwise turn is equivalent to the remaining $10^\circ$ counterclockwise turn:

$360^\circ-350^\circ=10^\circ.$

An inverse of a matching turn also matches. This $10^\circ$ match contradicts the claim that $50^\circ$ was least.

For any figure with finite rotational order $n$ about a fixed center, its least positive matching angle $\alpha$ satisfies

$n\alpha=360^\circ.$

$\alpha=\frac{360^\circ}{n}.$

Why are the matching turns evenly spaced? Repeating the least turn gives matching positions because each step matches the whole figure. Take the last repeat before a full turn. If a positive gap smaller than the least turn remained, that remaining turn would also match: it is the inverse of the last matching repeat, up to a full turn. This contradicts the choice of the least positive angle. Likewise, any extra match between consecutive repeats would give a smaller matching turn by undoing the earlier repeat. Thus the repeats account for all matches and divide a full turn exactly. This uses inverses and compositions already taught, not a rule about numbers of polygon sides.

For order $1$, the least positive matching angle is $360^\circ$, an identity turn. It does not supply a nontrivial symmetry.

Point symmetry and exact outlines

The 6-by-2 rectangle matches after a half-turn, even though most points move. A figure has point symmetry about $O$ when a $180^\circ$ turn about $O$ maps it onto itself. Thus point symmetry requires a half-turn, rather than a fold across a line. A quarter-turn alone is not the definition.

Letters are useful test figures only when their exact outlines are supplied. A curved font, unequal stroke widths, or a decoration can change the answer. Judge the drawn boundary below, not another printed version of the letter.

Triangles and circles need different tests

We can establish a particular triangle’s symmetries from how it is constructed, without a future triangle theorem. Join three points equally spaced around a circle: the three successive central turns are each $120^\circ$. The circle and radii are guides, not part of the triangular figure.

A $120^\circ$ turn about $O$ cycles the three vertices and their connecting sides. Any smaller positive turn misses all three vertices. Thus the triangle has order $3$ and least angle $120^\circ$, with no point symmetry. Each line through $O$ and one vertex also reflects the other two vertices onto one another: their rays are equally far to either side and have equal radii. It fixes that vertex and swaps the joined sides, so there are three symmetry lines. These conclusions are for this constructed triangle. The triangle in Example 1 has a different symmetry pattern.

A circle is the set of points at one fixed positive distance from its center. Any line through the center reflects that distance condition onto itself. Any rotation about the center does the same, because each point retains its radius. Therefore a circle has infinitely many lines of symmetry and infinitely many rotational symmetries, including point symmetry. A line not through the center does not work: it sends the center to a different location and the circle with it.

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