Worked examples — Symmetry — line symmetry, rotational symmetry, order of symmetry

This page is a workout. In the parent note you met two ideas:

• Line symmetry — a figure that folds onto itself along a line.
• Rotational symmetry — a figure that spins onto itself before a full turn, with an order (how many "identical positions" there are in one $360°$ turn).

Here we hit every kind of question those ideas can throw at you. First we lay out a map of all the cases; then we solve one example per case.

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The scenario matrix

Every symmetry question is one of these cells. Together they fill the full $2\times 2$ grid of "has line symmetry? / has rotational symmetry?" plus the tricky counting and formula cells. The last column names the worked example that covers it.

#	Case class	What's tricky about it	Example
0	Neither line nor rotational symmetry (scalene triangle)	the empty cell: 0 lines, order 1	Ex 0
1	Regular polygon — both even and odd side counts	line families differ for odd $n$	Ex 1
2	Reflecting a point across vertical, horizontal, $y=x$ and general lines	pick the right rule per mirror	Ex 2
3	Has rotation but no line symmetry	must test the two types separately	Ex 3
4	Has line symmetry but no rotation (order 1)	order-1 = "none" trap	Ex 4
5	Degenerate / lowest case: a plain line segment or circle	order $=2$ vs order $=\infty$ (limiting value)	Ex 5
6	Order-vs-rotations counting trap	"4 clicks" ≠ order 4 done wrong	Ex 6
7	Real-world word problem (ceiling fan / logo)	translate turning into an angle	Ex 7
8	Exam twist: given order, find the angle and back-check	reverse the formula	Ex 8

The tools we lean on the whole way:

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Example 0 — Neither symmetry: the empty cell (matrix cell 0)

Forecast: Guess before reading — is it symmetric at all?

Figure s07 shows a lavender scalene triangle with sides labelled $3$, $5$, $6$ units; a dashed coral attempt at a fold-line simply misses on the other side.

1. Fold test. Any fold-line would have to match one side onto another, but all three sides differ in length, so no fold makes the halves coincide.

   • Why this step? A mirror line pairs equal-length edges; with all sides unequal there is no such pairing.
   • 0 lines of symmetry.

2. Spin test. Turning the triangle by anything less than $360°$ moves a short side to where a long side was — instant mismatch.

   • Why this step? Rotational symmetry needs interchangeable parts; unequal sides are not interchangeable.
   • Order $1$ (no rotational symmetry).

Verify: $0$ lines and order $1$ — this is the only cell of the $2\times 2$ grid with neither type. Every earlier example fills another corner: Ex 1 (both), Ex 3 (rotation only), Ex 4 (line only). ✓

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Example 1 — Regular polygons, even AND odd (matrix cell 1)

Forecast: Guess: how many fold-lines for each, and does the type of line change when $n$ is odd?

Figure s01 shows, on the left, the octagon with two families of mirror lines (lavender through opposite vertices, dashed coral through opposite side-midpoints); on the right, the pentagon where each single line runs from a vertex straight to the midpoint of the opposite side.

1. Octagon — count the fold-lines (even $n$). When $n$ is even, opposite vertices face each other and opposite side-midpoints face each other, giving two families: $\frac{n}{2}$ vertex-to-vertex lines and $\frac{n}{2}$ midpoint-to-midpoint lines.

   • Why this step? Each mirror passes through the centre and must hit matching features on both ends; for even $n$ a vertex is opposite a vertex, so both families exist. Here $\frac{8}{2}+\frac{8}{2}=4+4=8$ lines.

2. Pentagon — count the fold-lines (odd $n$). When $n$ is odd, a vertex sits opposite the middle of a side, so there is only one family: each of the $n$ lines runs from a vertex to the midpoint of the opposite side.

   • Why this step? With odd $n$ there is no "opposite vertex" to pair up, so the two families of the even case merge into a single family of $n$ lines. Pentagon: $5$ lines.
   • Key nuance: for any regular $n$-gon the total is still $n$ lines, but the description differs — even $n$ mixes vertex- and midpoint-lines, odd $n$ uses vertex-to-opposite-midpoint lines only.

3. Order and angle for each. Both have order $=n$ (corners interchangeable). Angles: $\text{octagon: }\frac{360°}{8}=45°,\text{pentagon: }\frac{360°}{5}=72°.$

Verify: $8\times 45°=360°$ ✓ and $5\times 72°=360°$ ✓. Lines $=$ order $=n$ in both cases: $8$ and $5$ ✓. See Regular Polygons for why the parity of $n$ changes the line description but never the count.

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Example 2 — Reflection across every kind of mirror (matrix cell 2)

Forecast: Guess which coordinate changes — and what happens to signs — in each case.

Figure s02 shows point $P(2,5)$ in lavender and its images: across the mint diagonal $y=x$, the slate anti-diagonal $y=-x$, the butter vertical $x=1$, and a coral horizontal $y=3$, each with a dotted perpendicular link.

1. Across $y=x$ — swap coordinates. $(x,y)\to (y,x)\Rightarrow P^{\prime }=(5,2)$

   • Why this step? The diagonal treats horizontal and vertical alike, so across-ness and up-ness trade places.

2. Across $y=-x$ — swap and negate. $(x,y)\to (-y,-x)\Rightarrow P^{\prime }=(-5,-2)$

   • Why this step? The anti-diagonal $y=-x$ is the mirror running top-left to bottom-right. Reflecting across it swaps the roles of $x$ and $y$ (like $y=x$) but flips their signs, sending a point from the upper-right into the lower-left.

3. Across $x=1$ — flip $x$ only. Use $x^{\prime }=2a-x$ with $a=1$: $P^{\prime }=(2\cdot 1-2,5)=(0,5)$

   • Why this step? A vertical mirror leaves height alone; $P$ is $1$ unit right of the line, so $P^{\prime }$ is $1$ unit left.

4. Across $y=3$ — flip $y$ only. Use $y^{\prime }=2b-y$ with $b=3$: $P^{\prime }=(2,2\cdot 3-5)=(2,1)$

   • Why this step? A horizontal mirror leaves left/right alone; $P$ is $2$ up from the line, so $P^{\prime }$ is $2$ down.

Verify:

• $y=x$: midpoint $\left(\frac{2+5}{2},\frac{5+2}{2}\right)=(3.5,3.5)$ lies on $y=x$ ✓.
• $y=-x$: midpoint $\left(\frac{2-5}{2},\frac{5-2}{2}\right)=(-1.5,1.5)$ satisfies $y=-x$ ✓.
• $x=1$: $|2-1|=|0-1|=1$ ✓.
• $y=3$: $|5-3|=|1-3|=2$ ✓.

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Example 3 — Rotation but no line symmetry (matrix cell 3)

Forecast: Guess yes/no for each before continuing.

Figure s03 shows a lavender S with its centre of rotation (the middle of its bounding box) marked in coral, and a curved arrow indicating the $180°$ turn that lands it back on itself.

1. Test folding (line symmetry). Try a vertical fold: the top curve bulges right, the bottom curve bulges left — they don't match. Try horizontal, try diagonal: none match.

   • Why this step? Line symmetry demands a fold where every point meets its twin. S's curves twist in opposite senses, so no fold works.
   • No line symmetry.

2. Test spinning (rotational symmetry). Turn S by $180°$ about its centre (the pinned midpoint shown in the figure). The top-right curve rotates into where the bottom-left curve was — and they coincide.

   • Why this step? A $180°$ turn about the centre sends "up-right" to "down-left"; S is built exactly with that twist, so it matches.
   • It matches at $0°$ and $180°$: order $2$, smallest angle $\frac{360°}{2}=180°$.

Verify: $2\times 180°=360°$ ✓. This proves the parent's point: line and rotational symmetry are independent — S has one without the other.

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Example 4 — Line symmetry but order 1 (matrix cell 4)

Forecast: Does A spin onto itself?

Figure s04 shows a lavender letter A with a single mint dashed vertical fold-line down its middle.

1. Fold test. A vertical fold maps left leg onto right leg exactly.

   • Why this step? A's two legs and the crossbar are mirror-images across the centre column, so one vertical mirror works.
   • 1 line of symmetry.

2. Spin test. Turn A about its centre by any angle less than $360°$ — the apex points down instead of up, so it never looks like an upright A until the full turn.

   • Why this step? Rotational symmetry needs a sub-$360°$ turn that matches. A has none.
   • Order $1$ (the "no rotational symmetry" case).

Verify: Only $0°$ and $360°$ match, and those are the same position — so exactly $1$ distinct position ⇒ order $1$ ✓. Compare with Congruence and Similarity: the folded halves are congruent, confirming the line.

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Example 5 — Degenerate & limiting shapes (matrix cell 5)

Forecast: One order is a small number; one is "infinity". Which is which?

Figure s05 shows, left, a lavender segment with its two mirror lines drawn; right, a mint circle with several diameters drawn to hint at infinitely many.

1. The segment — its lines. A segment has exactly 2 lines of symmetry.

   • Why this step? Line 1 — the perpendicular through the midpoint: folding here swaps the two ends, which are equidistant from the midpoint, so they coincide. Line 2 — the line the segment lies along: this one feels strange, so here is the justification. A "line of symmetry" is any line whose reflection sends the shape exactly onto itself. Reflect the segment across the very line it sits on: every point of the segment is already on that mirror, and a point on a mirror maps to itself. So every point stays put and the whole segment lands exactly on itself — the test passes. No other line works, because any tilted line through the midpoint sends one end off to a point not on the segment.

2. The segment — its order. Rotate about its midpoint. At $180°$ the two ends swap and it looks identical; any other angle tilts it.

   • Why this step? Only the half-turn preserves the segment, giving positions at $0°$ and $180°$.
   • Order $=2$, angle $=\frac{360°}{2}=180°$.

3. The circle — the limiting case. Rotate a circle by any angle $\theta$, however tiny — it looks unchanged.

   • Why this step? Every point of the circle is the same distance from the centre, so no rotation can be "noticed". Now the crucial point about why order is infinite: order counts the matching angles, but for a circle every angle matches, so there is no smallest nonzero matching angle — whatever tiny angle you name, half of it also works. Because the formula $n=\frac{360°}{\text{smallest angle}}$ has no smallest angle to divide by, no finite $n$ can describe the circle; we write order $=\infty$ to mean "matches at infinitely many — in fact all — angles".
   • Order $=\infty$; it also has infinitely many lines of symmetry (every diameter).

Verify: Segment: $2\times 180°=360°$ ✓. Circle: for any candidate smallest angle $\theta >0$, the angle $\theta /2>0$ also matches and is smaller — so no smallest exists, confirming no finite order ✓.

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Example 6 — The counting trap (matrix cell 6)

Forecast: Where does the $3$ come from, and why is it wrong?

1. List the matching positions. Turning a square by $90°$ repeatedly gives matches at $0°,90°,180°,270°$.

   • Why this step? Order counts distinct identical positions in one full turn, including the start. Those are four angles.
   • Order $=4$.

2. Where the "3" came from. Beyond the starting position there are $3$ further rotations ($90°,180°,270°$) before returning. The student counted movements, not positions.

   • Why this step? Movements $=$ order $-1$. Always count positions.

3. The angle. $\text{angle}=\frac{360°}{4}=90°$

Verify: $4\times 90°=360°$ ✓, and positions $\{0,90,180,270\}$ are $4$ in number ✓.

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Example 7 — Real-world word problem (matrix cell 7)

Forecast: Think "how many blades" → order.

Figure s06 shows three coloured arrows (lavender, coral, mint) spaced evenly around a slate centre dot, with a $120°$ arc marked between two of them.

1. Read the order off the arrows. Three identical arrows equally spaced ⇒ each arrow can take the next arrow's place.

   • Why this step? Identical, equally-spaced parts means the number of parts is the order.
   • Order $=3$.

2. Compute the angle. $\text{angle}=\frac{360°}{3}=120°$

   • Why this step? Divide the circle into $3$ equal slices; one slice is the smallest matching turn.

3. Line symmetry? Because each arrow curves one way (like S), no fold matches — no line symmetry, only rotational. Same lesson as Example 3.

Verify: $3\times 120°=360°$ ✓. A $3$-blade fan is exactly this: it looks unchanged every $120°$. This turning-into-itself is a transformation; see Transformations in Geometry.

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Example 8 — Exam twist: reverse the formula (matrix cell 8)

Forecast: Divide $360$ by something.

1. Find the order. Rearrange $\text{angle}=\frac{360°}{n}$ to $n=\frac{360°}{\text{angle}}$: $n=\frac{360°}{40°}=9$

   • Why this step? If the smallest matching turn is $40°$, then $40°$ must fit a whole number of times into $360°$. It fits $9$ times, so order $=9$.

2. List all matching angles. Multiples of $40°$ up to $360°$: $40°,80°,120°,160°,200°,240°,280°,320°,360°$

   • Why this step? Every rotation that matches is a whole-number multiple of the smallest one. That's $9$ angles (the last, $360°$, is the same as $0°$).

3. Could the order be 8? If order were $8$, the smallest angle would be $\frac{360°}{8}=45°\ne 40°$.

   • Why this step? Order and smallest angle are locked together by the formula; you can't have $40°$ and order $8$.
   • So no, order must be $9$.

Verify: $9\times 40°=360°$ ✓; and $\frac{360}{8}=45\ne 40$ confirms order $8$ is impossible ✓.

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Recall

Recall Order of a regular

$n$-gon? Regular $n$-gon: order and number of lines of symmetry both equal $n$ ::: $n$ each; smallest angle $\frac{360°}{n}$.

Recall Do the mirror lines of an odd regular polygon differ from an even one?

For odd $n$ each line goes vertex-to-opposite-midpoint (one family); for even $n$ there are two families (vertex-vertex and midpoint-midpoint) ::: yes — same count $n$, different description.

Recall Reflect

$(x,y)$ across $y=x$? across $y=-x$? across $y=b$? Diagonal swaps; anti-diagonal swaps and negates; horizontal flips $y$ ::: $y=x:(y,x)$; $y=-x:(-y,-x)$; $y=b:(x,2b-y)$.

Recall How do you locate the centre of rotation of a shape?

The fixed point that doesn't move — intersection of perpendicular bisectors of point-to-image segments ::: for regular shapes it's the geometric centre.

Recall Order 1 means?

Order 1 has only the full-turn match ::: it means no rotational symmetry.

Recall Why does a circle have infinite order?

Every angle matches, so there is no smallest nonzero matching angle to plug into $360°/\text{angle}$ ::: no finite order exists — write order $=\infty$.

Recall Scalene triangle symmetry?

All sides unequal ::: $0$ lines, order $1$ — the "neither" cell.

Related deep structure: the set of symmetries of a shape forms a group — see Group Theory. For why regular polygons are so symmetric, see Regular Polygons.