Question bank — Symmetry — line symmetry, rotational symmetry, order of symmetry

True. A $360°$ spin is a do-nothing move that always maps a figure onto itself, so the smallest possible order is $1$, which we read as "no real rotational symmetry".

False. The letter A folds vertically (one line of symmetry) but spinning it by any angle under $360°$ never matches it — order $1$. Fold-ability and spin-ability are independent tests.

False. The letter S has order-$2$ rotational symmetry (looks the same upside-down) yet no fold line matches its two halves. A pinwheel or the letter Z are the same story.

True. Two through opposite side-midpoints (vertical and horizontal) and two along the diagonals — $4$ folds all match, matching the general rule "regular $n$-gon has $n$ lines".

True. The $n$ identical positions are equally spaced around the circle, so the smallest turn between neighbours is the full circle split into $n$ equal slices, $\frac{360°}{n}$.

True. Every diameter is a fold line, and any rotation about the centre (however tiny) leaves the circle unchanged — both counts are infinite.

False as worded. A $90°$ match only guarantees the order is a multiple of $4$. A regular octagon also matches at $90°$ yet first matches at $45°$, giving order $8$. It is order $4$ only if $90°$ is the smallest non-trivial matching angle.

True. No fold matches its slanted sides, but a $180°$ turn about its centre swaps opposite corners and leaves it unchanged — order $2$.

True. A regular $n$-gon has $n$ fold lines and order $n$; the two counts coincide precisely because the shape is fully regular.

False on the order. It has exactly one fold line (through the apex), but its rotational order is $1$ — turning it any angle under $360°$ never matches, so it has no real rotational symmetry.

The number is right but the reasoning is wrong: order counts identical positions (at $0°,90°,180°,270°$), not the number of moves. There are only $3$ turns beyond the start, yet $4$ positions.

Folding and spinning are separate tests. E folds top-to-bottom but a $180°$ turn makes it face backward (Ǝ), which does not match — so E has rotational order $1$, no rotational symmetry.

Order $1$ is the technical label for no rotational symmetry — the only "match" is the trivial full turn. Symmetry begins at order $2$.

That counts only half of them. There are also $3$ lines through the midpoints of opposite sides, giving $3+3=6$ — the full count for a regular $6$-gon.

It is backwards. A vertical mirror line $x=4$ leaves $y$ untouched and flips $x$ via $x^{\prime }=2(4)-x=8-x$. The coordinate that changes is the one perpendicular to the mirror.

Visual balance is not fold-matching. No fold of N lands its two halves on each other; its diagonal only survives a $180°$ rotation, so N has rotational order $2$ and zero lines of symmetry.

Adding vertices and sides is not how order works. Only vertices-to-vertices rotations count: turning by $120°$ maps vertex to vertex, giving $3$ identical positions — order $3$, matching $\frac{360°}{120°}$.

Because it makes the tidy formula $\text{order}=\frac{360°}{\text{smallest match angle}}$ true: a $90°$-symmetric square then has order $\frac{360}{90}=4$, and every figure honestly has order $\ge 1$.

A fold line must map the polygon onto itself, so it must swap the polygon's corners in pairs while pinning the centre — meaning it passes through the centre. Trace where it hits the boundary: either straight through a corner (which must stay fixed) or through the exact midpoint of a side (splitting a pinned pair). Between one corner and the next there is exactly one such direction, and with $n$ corners you get exactly $n$ of them — any other tilt would send a corner to a non-corner point, which cannot match.

Rotation and reflection are different moves. A "swirl" shape (pinwheel, S, Z) is unchanged by a turn but a mirror flip reverses its chirality — the swirl now spins the other way — so no fold matches.

The mirror only needs equal distance on opposite sides: $P^{\prime }$ is as far right of $a$ as $P$ is left. That gives $x^{\prime }-a=-(x-a)$, i.e. $x^{\prime }=2a-x$ — pure distances, no squaring involved.

Larger matching angles are just repeats of the smallest one. The smallest angle is the true "step", and all others (its multiples) fall out automatically; that step alone fixes $n$.

The fold line reflects the triangle onto itself, so the two half-triangles are congruent mirror images — that forces the line to bisect the base perpendicularly and bisect the apex angle, merging all three special lines.

It is order $1$, not $0$ — because the $360°$ full turn always maps a figure to itself, guaranteeing at least one identical position.

Yes: two fold lines (along it and its perpendicular bisector) and rotational order $2$ (a $180°$ flip about its midpoint), making it as symmetric as a rectangle.

The smallest is $1$, meaning the figure matches itself only after a complete turn — the technical way of saying "no rotational symmetry".

No — for one connected figure spun onto itself the centre must be its own balance point (inside it or on its boundary), because a point far outside would sweep the figure into a different location instead of back onto itself.

Test the smaller candidate first: if $90°$ does not match, order cannot be $4$. Since only $180°$ matches, the smallest winning angle is $180°$ and the order is $\frac{360°}{180°}=2$.

Yes — the circle. Any angle matches, so there is no smallest nonzero angle; its order is infinite, the limiting case of the $\frac{360°}{n}$ rule as $n\to \infty$.