---
title: "Central Symmetry and Parallelograms"
book: "Primary & Middle School Mathematics"
subject: math
language: en
chapter: 52
exercises: 11
source: https://one-course.com/books/math/1/en/chapter/52-central-symmetry-and-parallelograms
---

# Chapter 52 — Central Symmetry and Parallelograms

Turn a figure by [half](https://one-course.com/books/math/1/en/chapter/17-sharing-and-division#def-g3-division-half) a turn around a point: this is *[central symmetry](#def-g7-central-def)*, the second transformation of the book after the reflections of [Chapter 42](https://one-course.com/books/math/1/en/chapter/42-axial-symmetry#ch-g6-symmetry). Its star figure is the [parallelogram](#def-g7-central-parallelogram) — the [quadrilateral](https://one-course.com/books/math/1/en/chapter/26-lines-and-polygons#def-g4-geometry-polygon) that is [symmetric](#def-g7-central-def) about the crossing point of its own diagonals.

## 52.1 Central symmetry

**Definition 52.1 (Symmetric about a point).**

The *symmetric* of a point $M$ about a point $O$ is the point $M'$ such that $O$ is the *midpoint* of $[MM']$. The symmetric figure is what the original becomes after a half-turn ($180^\circ$) around $O$.

![Symmetry about O: each point and its image are aligned with O, at equal distances on both sides (tick marks).](https://one-course.com/images/onecourse/chapters/math-1/g7-central/fig-3a5d10b2de5e.svg)

*Symmetry about $O$: each point and its image are aligned with $O$, at equal distances on both sides (tick marks).*

**Method 52.2 (Constructing the symmetric of a point).**

To construct the [symmetric](#def-g7-central-def) $M'$ of $M$ about $O$:

1. draw the line $(MO)$ and extend it beyond $O$ ;
2. measure $MO$ (ruler or compass);
3. place $M'$ on the extension with $OM' = OM$ .

On grid paper: count the horizontal and vertical steps from $M$ to $O$, and repeat the *same* steps from $O$ to reach $M'$.

**Example 52.3 (On a grid).**

If $M$ is $3$ squares left and $1$ square up from $O$, its [symmetric](#def-g7-central-def) $M'$ is $3$ squares *right* and $1$ square *down* from $O$: [central symmetry](#def-g7-central-def) reverses both directions at once (a reflection reverses only one).

**Proposition 52.4 (What a half-turn preserves).**

[Central symmetry](#def-g7-central-def) preserves lengths, angles, [perimeters](https://one-course.com/books/math/1/en/chapter/43-perimeter-area-volume#def-g6-measure-perimeter) and [areas](https://one-course.com/books/math/1/en/chapter/43-perimeter-area-volume#def-g6-measure-area); the image of a line is a *[parallel](https://one-course.com/books/math/1/en/chapter/40-lines-circles-and-angles#def-g6-lines-perp)* line, the image of a [segment](https://one-course.com/books/math/1/en/chapter/40-lines-circles-and-angles#def-g6-lines-objects) is a [parallel](https://one-course.com/books/math/1/en/chapter/40-lines-circles-and-angles#def-g6-lines-perp) [segment](https://one-course.com/books/math/1/en/chapter/40-lines-circles-and-angles#def-g6-lines-objects) of the same length, the image of a circle is a circle of the same [radius](https://one-course.com/books/math/1/en/chapter/18-shapes-and-right-angles#def-g3-shapes-shapes) (centered at the [symmetric](#def-g7-central-def) of the center).

**Proof.** *Admitted at this level.* ∎

**Remark 52.5.**

Compare with reflections ([Proposition 42.4](https://one-course.com/books/math/1/en/chapter/42-axial-symmetry#prop-g6-symmetry-preserved)): both preserve lengths and angles; but the reflection flips figures over (a “b” becomes a “d”), while the half-turn keeps them readable — upside down (a “b” becomes a “q”). And unlike a reflection, a half-turn sends every line to a line [parallel](https://one-course.com/books/math/1/en/chapter/40-lines-circles-and-angles#def-g6-lines-perp) to it.

**Definition 52.6 (Center of symmetry).**

A point $O$ is a *center of symmetry* of a figure when the half-turn around $O$ sends the figure exactly onto itself. For instance, the letters S, N, Z have a center of symmetry; a circle has one (its center); a triangle never has one.

## 52.2 Parallelograms

**Definition 52.7 (Parallelogram).**

A *parallelogram* is a [quadrilateral](https://one-course.com/books/math/1/en/chapter/26-lines-and-polygons#def-g4-geometry-polygon) whose diagonals cross at their common midpoint. That crossing point is then a [center of symmetry](#def-g7-central-center) of the figure: each [vertex](https://one-course.com/books/math/1/en/chapter/35-cubes-and-boxes#def-g5-solids-def) is the half-turn image of the opposite [vertex](https://one-course.com/books/math/1/en/chapter/35-cubes-and-boxes#def-g5-solids-def).

**Theorem 52.8 (Properties of a parallelogram).**

In a [parallelogram](#def-g7-central-parallelogram) $ABCD$ with center $O$:

1. opposite sides are [parallel](https://one-course.com/books/math/1/en/chapter/40-lines-circles-and-angles#def-g6-lines-perp) : $(AB) \parallel (CD)$ and $(BC) \parallel (AD)$ ;
2. opposite sides have the same length: $AB = CD$ and $BC = AD$ ;
3. opposite angles are equal: $\widehat A = \widehat C$ and $\widehat B = \widehat D$ .

**Proof.** The half-turn around $O$ sends $A$ to $C$ and $B$ to $D$ (definition: $O$ is the midpoint of both diagonals). So it sends the [segment](https://one-course.com/books/math/1/en/chapter/40-lines-circles-and-angles#def-g6-lines-objects) $[AB]$ to the [segment](https://one-course.com/books/math/1/en/chapter/40-lines-circles-and-angles#def-g6-lines-objects) $[CD]$; by [Proposition 52.4](#prop-g7-central-preserved) the image is [parallel](https://one-course.com/books/math/1/en/chapter/40-lines-circles-and-angles#def-g6-lines-perp) to $[AB]$ and of the same length: points 1 and 2. It also sends the angle at $A$ to the angle at $C$, and angles are preserved: point 3. ∎

![A parallelogram and its center O: the diagonals (red) cut each other at their midpoints, opposite sides are parallel and equal (tick marks).](https://one-course.com/images/onecourse/chapters/math-1/g7-central/fig-f7a6741d6cd6.svg)

*A [parallelogram](#def-g7-central-parallelogram) and its center $O$: the diagonals (red) cut each other at their midpoints, opposite sides are [parallel](https://one-course.com/books/math/1/en/chapter/40-lines-circles-and-angles#def-g6-lines-perp) and equal (tick marks).*

**Theorem 52.9 (Recognizing a parallelogram).**

A [quadrilateral](https://one-course.com/books/math/1/en/chapter/26-lines-and-polygons#def-g4-geometry-polygon) (with no crossing sides) is a [parallelogram](#def-g7-central-parallelogram) as soon as *one* of the following holds:

1. its diagonals have the same midpoint (the definition);
2. its opposite sides are [parallel](https://one-course.com/books/math/1/en/chapter/40-lines-circles-and-angles#def-g6-lines-perp) two by two;
3. two opposite sides are [parallel](https://one-course.com/books/math/1/en/chapter/40-lines-circles-and-angles#def-g6-lines-perp) *and* of the same length.

**Proof.** *Admitted at this level.* ∎

**Example 52.10.**

$ABCD$ has $(AB) \parallel (CD)$ and $AB = CD = 5$ cm. Criterion 3 applies: $ABCD$ is a [parallelogram](#def-g7-central-parallelogram) — so without any further measuring we also know $AD = BC$, $(AD) \parallel (BC)$, and that its diagonals cut each other in their middle.

**Remark 52.11 (Special parallelograms).**

The rectangle, [rhombus](https://one-course.com/books/math/1/en/chapter/41-triangles-and-quadrilaterals#def-g6-shapes-quadrilaterals) and square of [Chapter 41](https://one-course.com/books/math/1/en/chapter/41-triangles-and-quadrilaterals#ch-g6-shapes) are exactly the [parallelograms](#def-g7-central-parallelogram) with an extra property: equal diagonals (rectangle), [perpendicular](https://one-course.com/books/math/1/en/chapter/40-lines-circles-and-angles#def-g6-lines-perp) diagonals ([rhombus](https://one-course.com/books/math/1/en/chapter/41-triangles-and-quadrilaterals#def-g6-shapes-quadrilaterals)), both (square). The angle argument promised in [Example 41.6](https://one-course.com/books/math/1/en/chapter/41-triangles-and-quadrilaterals#ex-g6-shapes-recognize) now works: in a [parallelogram](#def-g7-central-parallelogram), consecutive angles are supplementary (alternate/corresponding angles with the [parallels](https://one-course.com/books/math/1/en/chapter/40-lines-circles-and-angles#def-g6-lines-perp), [Theorem 51.2](https://one-course.com/books/math/1/en/chapter/51-triangles-and-angles#thm-g7-triangles-equalities)), so one [right angle](https://one-course.com/books/math/1/en/chapter/18-shapes-and-right-angles#def-g3-shapes-rightangle) forces all four.

## 52.3 Exercises

**Exercise 52.1 ★.**

On grid paper, place $O$, then $M$ four squares right of $O$ and one square up. Construct the [symmetric](#def-g7-central-def) $M'$ of $M$ about $O$, and the [symmetric](#def-g7-central-def) of $M$ across the *vertical grid line* through $O$. Compare the two images.

**Solution of Exercise 52.1.**

Half-turn image: $M'$ is four squares *left* of $O$ and one square *down* (both directions reversed). Reflection image across the vertical line: four squares left but still one square *up*. The two images differ: they are vertical mirror images of each other.

**Exercise 52.2 ★.**

Draw a triangle $ABC$ and a point $O$ outside it. Construct its image $A'B'C'$ by the half-turn around $O$. What can you say about the lengths $A'B'$ and $AB$? About the lines $(A'B')$ and $(AB)$?

**Solution of Exercise 52.2.**

$A'B' = AB$ (half-turns preserve lengths) and $(A'B') \parallel (AB)$ (the image of a line is a [parallel](https://one-course.com/books/math/1/en/chapter/40-lines-circles-and-angles#def-g6-lines-perp) line, [Proposition 52.4](#prop-g7-central-preserved)).

**Exercise 52.3 ★.**

Which digits, written as on a calculator display ($0$ to $9$), have a [center of symmetry](#def-g7-central-center)? Which capital letters among H, A, N, S, T, Z, O?

**Solution of Exercise 52.3.**

Calculator digits with a [center of symmetry](#def-g7-central-center): $0$, $2$, $5$, $8$ (and $1$ if drawn as a plain bar). Letters: H, N, S, Z, O have one; A and T do not (they have an axis instead).

**Exercise 52.4 ★.**

$KLMN$ is a [parallelogram](#def-g7-central-parallelogram) of center $O$ with $KL = 7$ cm, $LM = 4$ cm and $\widehat K = 115^\circ$. Give, with reasons: $MN$, $NK$, $\widehat M$, and the midpoint of $[LN]$.

**Solution of Exercise 52.4.**

Opposite sides equal: $MN = KL = 7$ cm and $NK = LM = 4$ cm. Opposite angles equal: $\widehat M = \widehat K = 115^\circ$. The diagonals cut each other at their common midpoint, and that point is $O$: the midpoint of $[LN]$ is $O$.

**Exercise 52.5 ★.**

Construct a [parallelogram](#def-g7-central-parallelogram) $ABCD$ from its diagonals: $AC = 8$ cm and $BD = 5$ cm, crossing at their midpoints with an angle of $60^\circ$ between them. (Draw the diagonals first!)

**Solution of Exercise 52.5.**

Draw a [segment](https://one-course.com/books/math/1/en/chapter/40-lines-circles-and-angles#def-g6-lines-objects) $[AC]$ of $8$ cm, mark its midpoint $O$, draw through $O$ a line making $60^\circ$ with $(AC)$, and place $B$ and $D$ on it at $2.5$ cm from $O$ on either side. Join $A$, $B$, $C$, $D$: the diagonals cut at their midpoints, so $ABCD$ is a [parallelogram](#def-g7-central-parallelogram) by definition.

**Exercise 52.6 ★.**

In a [parallelogram](#def-g7-central-parallelogram), one angle measures $115^\circ$. Using the remark on supplementary consecutive angles, give the three other angles.

**Solution of Exercise 52.6.**

Consecutive angles are supplementary: next to $115^\circ$ sits $180 - 115 = 65^\circ$. Opposite angles are equal, so the four angles are $115^\circ$, $65^\circ$, $115^\circ$, $65^\circ$ ([sum](https://one-course.com/books/math/1/en/chapter/2-addition-first-steps#def-g1-addition-def) $360^\circ$).

**Exercise 52.7 ★★.**

Plot $A(1, 1)$, $B(4, 2)$ and the point $O(2.5, 2.5)$. Compute the coordinates of the images $A'$ and $B'$ by the half-turn around $O$ (use the grid counting of [Example 52.3](#ex-g7-central-grid)).

**Solution of Exercise 52.7.**

From $A(1,1)$ to $O(2.5, 2.5)$: $1.5$ right and $1.5$ up; continuing the same: $A'(4, 4)$. From $B(4,2)$ to $O$: $1.5$ left and $0.5$ up; continuing: $B'(1, 3)$.

**Exercise 52.8 ★★.**

$ABCD$ is a [quadrilateral](https://one-course.com/books/math/1/en/chapter/26-lines-and-polygons#def-g4-geometry-polygon) in which $AB = CD$. Is that enough to make it a [parallelogram](#def-g7-central-parallelogram)? If not, sketch a counterexample (an [isosceles](https://one-course.com/books/math/1/en/chapter/41-triangles-and-quadrilaterals#def-g6-shapes-triangles) trapezoid helps), and state what must be added to the hypothesis.

**Solution of Exercise 52.8.**

No. An [isosceles](https://one-course.com/books/math/1/en/chapter/41-triangles-and-quadrilaterals#def-g6-shapes-triangles) trapezoid has its two slanted sides equal without being a [parallelogram](#def-g7-central-parallelogram) (the two [parallel](https://one-course.com/books/math/1/en/chapter/40-lines-circles-and-angles#def-g6-lines-perp) sides have different lengths). What suffices is criterion 3 of [Theorem 52.9](#thm-g7-central-recognize): two opposite sides equal *and [parallel](https://one-course.com/books/math/1/en/chapter/40-lines-circles-and-angles#def-g6-lines-perp)*.

**Exercise 52.9 ★★.**

Let $ABC$ be a triangle and $O$ the midpoint of $[BC]$. Construct the [symmetric](#def-g7-central-def) $A'$ of $A$ about $O$. Show that $ABA'C$ is a [parallelogram](#def-g7-central-parallelogram). (Which criterion of [Theorem 52.9](#thm-g7-central-recognize) is free of charge here?)

**Solution of Exercise 52.9.**

$O$ is the midpoint of $[BC]$ by choice, and it is the midpoint of $[AA']$ by construction of the [symmetric](#def-g7-central-def). The [quadrilateral](https://one-course.com/books/math/1/en/chapter/26-lines-and-polygons#def-g4-geometry-polygon) $ABA'C$ has diagonals $[AA']$ and $[BC]$ sharing the midpoint $O$: criterion 1 (the definition) makes it a [parallelogram](#def-g7-central-parallelogram) at no extra cost.

**Exercise 52.10 ★★.**

True or false, with a reason: “a [quadrilateral](https://one-course.com/books/math/1/en/chapter/26-lines-and-polygons#def-g4-geometry-polygon) whose opposite angles are equal two by two is a [parallelogram](#def-g7-central-parallelogram)”; “a [parallelogram](#def-g7-central-parallelogram) with [perpendicular](https://one-course.com/books/math/1/en/chapter/40-lines-circles-and-angles#def-g6-lines-perp) diagonals is a [rhombus](https://one-course.com/books/math/1/en/chapter/41-triangles-and-quadrilaterals#def-g6-shapes-quadrilaterals)”; “a [parallelogram](#def-g7-central-parallelogram) with one [right angle](https://one-course.com/books/math/1/en/chapter/18-shapes-and-right-angles#def-g3-shapes-rightangle) is a rectangle”.

**Solution of Exercise 52.10.**

“Opposite angles equal $\Rightarrow$ [parallelogram](#def-g7-central-parallelogram)”: *true* (this is another recognition criterion; with the angle [sum](https://one-course.com/books/math/1/en/chapter/2-addition-first-steps#def-g1-addition-def) $360^\circ$, equal opposite pairs force supplementary consecutive angles, hence [parallel](https://one-course.com/books/math/1/en/chapter/40-lines-circles-and-angles#def-g6-lines-perp) sides by [Theorem 51.2](https://one-course.com/books/math/1/en/chapter/51-triangles-and-angles#thm-g7-triangles-equalities)).

“[Perpendicular](https://one-course.com/books/math/1/en/chapter/40-lines-circles-and-angles#def-g6-lines-perp) diagonals $\Rightarrow$ [rhombus](https://one-course.com/books/math/1/en/chapter/41-triangles-and-quadrilaterals#def-g6-shapes-quadrilaterals)”: *true* for a [parallelogram](#def-g7-central-parallelogram) (the diagonals already cut at their midpoints).

“One [right angle](https://one-course.com/books/math/1/en/chapter/18-shapes-and-right-angles#def-g3-shapes-rightangle) $\Rightarrow$ rectangle”: *true* — consecutive angles are supplementary, so all four become right.

**Exercise 52.11 ★★★.**

Let $ABCD$ be any [quadrilateral](https://one-course.com/books/math/1/en/chapter/26-lines-and-polygons#def-g4-geometry-polygon) and $I$, $J$, $K$, $L$ the midpoints of its sides $[AB]$, $[BC]$, $[CD]$, $[DA]$. Draw several examples. What does $IJKL$ always seem to be? (The proof, with the midpoint theorem, comes in [Chapter 59](https://one-course.com/books/math/1/en/chapter/59-midpoints-and-parallels#ch-g8-midpoints) — and again with vectors, in the High School volume.)

**Solution of Exercise 52.11.**

Whatever the [quadrilateral](https://one-course.com/books/math/1/en/chapter/26-lines-and-polygons#def-g4-geometry-polygon) — even a very irregular one — the midpoints $I$, $J$, $K$, $L$ always form a *[parallelogram](#def-g7-central-parallelogram)*. The proof with the midpoint theorem is in [Chapter 59](https://one-course.com/books/math/1/en/chapter/59-midpoints-and-parallels#ch-g8-midpoints) ([Exercise 59.9](https://one-course.com/books/math/1/en/chapter/59-midpoints-and-parallels#exo-g8-midpoints-9)).

## 52.4 Problem: The cake, the median, and the impossible second center

**Problem 52.1.**

Weekend problem — half-turns at work: fair cake cuts, a bound on medians, and why no shape can have two centers of symmetry

The half-turn looks like the simplest of transformations — yet it cuts cakes into provably fair halves, measures the medians of a triangle, and hides a small pearl of pure mathematics: a bounded figure can have one [center of symmetry](#def-g7-central-center), but *never two*. The proof uses a discovery you will make in Part I: two half-turns, performed one after the other, add up to a slide.

**Part I — Half-turn gymnastics.**

1. On grid paper, take the origin $O(0, 0)$ as center. Using the grid counting of [Example 52.3](#ex-g7-central-grid) , give the images of $M(3, 1)$ , $N(-2, 4)$ and $P(0, -5)$ by the half-turn about $O$ . What is the general rule for $(x, y)$ ?
2. Now the center is $C(2, 1)$ . Compute the image of $M(5, 3)$ , and check the general recipe: each coordinate of the image is *twice the center’s minus the point’s* .
3. Playing cards are designed to read the same upside down — a [central symmetry](#def-g7-central-def) , so neither player sees the card “wrongly”. Explain why, on a card with an *[odd](https://one-course.com/books/math/1/en/chapter/14-numbers-up-to-10-000#def-g3-numbers-evenodd)* number of pips, one pip must sit exactly at the card’s center.
4. Two half-turns in a row: take the centers $O_1(2, 0)$ and $O_2(5, 0)$ . Send the point $A(1, 1)$ through the half-turn about $O_1$ , then send the image through the half-turn about $O_2$ . Compare start and finish. Repeat with $B(0, 3)$ . What single, simple transformation did the two half-turns amount to?
5. Measure the slide of question 4 against the distance $O_1 O_2$ , and state the discovery. (Compare with the two [parallel](https://one-course.com/books/math/1/en/chapter/40-lines-circles-and-angles#def-g6-lines-perp) mirrors of [Problem 42.1](https://one-course.com/books/math/1/en/chapter/42-axial-symmetry#pb-g6-symmetry-1) : same phenomenon, new actors.)

**Part II — Cakes and medians.**

6. Let $O$ be the center of a [parallelogram](#def-g7-central-parallelogram) . Explain why any line through $O$ meets the boundary at two points that are [symmetric](#def-g7-central-def) about $O$ , and why the line cuts the [parallelogram](#def-g7-central-parallelogram) into two pieces that are exact half-turn copies of each other — hence of equal [areas](https://one-course.com/books/math/1/en/chapter/43-perimeter-area-volume#def-g6-measure-area) .
7. The baker’s theorem: a rectangular cake is shared perfectly fairly by *any* straight cut through its center. Better: a rectangular cake has a rectangular hole (any size, any position, [even](https://one-course.com/books/math/1/en/chapter/14-numbers-up-to-10-000#def-g3-numbers-evenodd) tilted). Describe the one straight cut that gives two parts with equally much cake — and justify it with question 6.
8. Three [vertices](https://one-course.com/books/math/1/en/chapter/35-cubes-and-boxes#def-g5-solids-def) of a [parallelogram](#def-g7-central-parallelogram) $ABCD$ are $A(0, 0)$ , $B(6, 1)$ and $C(8, 5)$ . Compute the coordinates of $D$ and of the center $O$ .
9. [Exercise 52.9](#exo-g7-central-9) built, from a triangle $ABC$ and the midpoint $O$ of $[BC]$, the [parallelogram](#def-g7-central-parallelogram) $ABA'C$ where $A'$ is the [symmetric](#def-g7-central-def) of $A$ about $O$. Use it, together with the triangle inequality ([Theorem 51.6](https://one-course.com/books/math/1/en/chapter/51-triangles-and-angles#thm-g7-triangles-inequality)) in the triangle $ABA'$, to prove the *median bound*: the median $[AO]$ satisfies $$AO < \frac{AB + AC}{2} .$$ (Note that $AA' = 2\,AO$ and $BA' = AC$.)
10. In a triangle with $AB = 5$ cm and $AC = 7$ cm, between which two values must the length of the median from $A$ lie? (Use both directions of the triangle inequality in $ABA'$ .)

**Part III — How many centers can a figure have?**

11. For each figure, say whether it has a [center of symmetry](#def-g7-central-center) , and where: a [segment](https://one-course.com/books/math/1/en/chapter/40-lines-circles-and-angles#def-g6-lines-objects) ; a full line; an [equilateral triangle](https://one-course.com/books/math/1/en/chapter/41-triangles-and-quadrilaterals#def-g6-shapes-triangles) ; a square; a circle; the letters S and Z.
12. Prove your claim for the [equilateral triangle](https://one-course.com/books/math/1/en/chapter/41-triangles-and-quadrilaterals#def-g6-shapes-triangles) : a half-turn sending the triangle to itself would pair up the three [vertices](https://one-course.com/books/math/1/en/chapter/35-cubes-and-boxes#def-g5-solids-def) — but three is [odd](https://one-course.com/books/math/1/en/chapter/14-numbers-up-to-10-000#def-g3-numbers-evenodd) , so some [vertex](https://one-course.com/books/math/1/en/chapter/35-cubes-and-boxes#def-g5-solids-def) would be sent to itself. What would that force the [vertex](https://one-course.com/books/math/1/en/chapter/35-cubes-and-boxes#def-g5-solids-def) to be, and why is it impossible?
13. Generalize question 12: explain why *no* [polygon](https://one-course.com/books/math/1/en/chapter/26-lines-and-polygons#def-g4-geometry-polygon) with an [odd number](https://one-course.com/books/math/1/en/chapter/14-numbers-up-to-10-000#def-g3-numbers-evenodd) of [vertices](https://one-course.com/books/math/1/en/chapter/35-cubes-and-boxes#def-g5-solids-def) has a [center of symmetry](#def-g7-central-center) .
14. The pearl: suppose a figure had *two* distinct centers of symmetry $O_1$ and $O_2$ . Perform the two half-turns one after the other and use the discovery of question 5: what motion must send the figure exactly onto itself? Explain why a figure that fits on a sheet of paper cannot survive that, and conclude.
15. Question 14’s conclusion leaves a loophole: *unbounded* figures. Verify on an infinite frieze — say, an endless strip of footprints, left, right, left, right … — that it has (at least) two different centers of symmetry, and identify the translation that question 14 predicted.

**Solution of Problem 52.1.**

**1.** $M(3, 1) \to (-3, -1)$; $N(-2, 4) \to (2, -4)$; $P(0, -5) \to (0, 5)$. Rule: the half-turn about the origin sends $(x, y)$ to $(-x, -y)$ — both coordinates change sign.

**2.** From $M(5, 3)$: three squares right of $C$ becomes three squares left, two up becomes two down: image $(-1, -1)$. Recipe check: twice the center minus the point, $(2 \times 2 - 5,\ 2 \times 1 - 3) = (-1, -1)$.

**3.** The half-turn about the card’s center swaps the pips in pairs. With an [odd number](https://one-course.com/books/math/1/en/chapter/14-numbers-up-to-10-000#def-g3-numbers-evenodd) of pips, one pip has no partner: it must be its own image, and the only point that is its own image is the center itself. Hence the middle pip of the 3, the 5, the 7 … sits dead center (look at a real card: it does).

**4.** About $O_1(2,0)$: $A(1,1) \to (3,-1)$; about $O_2(5,0)$: $(3,-1) \to (7,1)$. Start $A(1,1)$, finish $(7,1)$: slid $6$ to the right, not flipped. Same for $B(0,3) \to (4,-3) \to (6,3)$: slid $6$ right. The two half-turns amount to a *translation*.

**5.** The slide is $6$, and $O_1 O_2 = 3$: the translation covers *twice* the distance between the centers (in the direction from $O_1$ to $O_2$) — just as two [parallel](https://one-course.com/books/math/1/en/chapter/40-lines-circles-and-angles#def-g6-lines-perp) mirrors translated by twice their gap in [Problem 42.1](https://one-course.com/books/math/1/en/chapter/42-axial-symmetry#pb-g6-symmetry-1).

**6.** The half-turn about $O$ sends the [parallelogram](#def-g7-central-parallelogram) onto itself (that is what “[center of symmetry](#def-g7-central-center)” means for it, [Theorem 52.8](#thm-g7-central-properties)). A line through $O$ is also sent onto itself, so the two points where it crosses the boundary swap with each other: they are [symmetric](#def-g7-central-def) about $O$. The two pieces of the [parallelogram](#def-g7-central-parallelogram) on either side of the line swap too, so they are exact copies ([Proposition 52.4](#prop-g7-central-preserved)) — equal [areas](https://one-course.com/books/math/1/en/chapter/43-perimeter-area-volume#def-g6-measure-area).

**7.** Cut along the line through the *two centers*: the center of the cake rectangle and the center of the hole. By question 6 applied to the cake rectangle, the cut halves the cake; applied to the hole rectangle (the line passes through its center too), it halves the hole. Each part gets [half](https://one-course.com/books/math/1/en/chapter/17-sharing-and-division#def-g3-division-half) the cake minus [half](https://one-course.com/books/math/1/en/chapter/17-sharing-and-division#def-g3-division-half) the hole: perfectly fair.

**8.** $D = (0 + 8 - 6,\ 0 + 5 - 1) = (2, 4)$ (the diagonals share their midpoint, so $D$ is the half-turn image of $B$ about the center). Center: midpoint of $[AC]$, $O(4,\ 2.5)$.

**9.** In the [parallelogram](#def-g7-central-parallelogram) $ABA'C$, opposite sides are equal: $BA' = AC$. The triangle inequality in $ABA'$ gives $AA' < AB + BA' = AB + AC$. Since $O$ is the midpoint of $[AA']$, $AA' = 2\,AO$, so $2\,AO < AB + AC$, i.e. $AO < \frac{AB + AC}{2}$.

**10.** Upper bound: $AO < \frac{5 + 7}{2} = 6$ cm. Lower bound: in $ABA'$, $AA' > BA' - AB = 7 - 5 = 2$, so $AO > 1$ cm. The median from $A$ lies strictly between $1$ and $6$ cm.

**11.** [Segment](https://one-course.com/books/math/1/en/chapter/40-lines-circles-and-angles#def-g6-lines-objects): yes, its midpoint. Line: yes — every one of its points is a center. [Equilateral triangle](https://one-course.com/books/math/1/en/chapter/41-triangles-and-quadrilaterals#def-g6-shapes-triangles): none. Square: yes, the crossing of the diagonals. Circle: yes, its center. S and Z: yes, their middle point (turn the page upside down: they read the same).

**12.** The half-turn would pair the three [vertices](https://one-course.com/books/math/1/en/chapter/35-cubes-and-boxes#def-g5-solids-def) among themselves; three being [odd](https://one-course.com/books/math/1/en/chapter/14-numbers-up-to-10-000#def-g3-numbers-evenodd), one [vertex](https://one-course.com/books/math/1/en/chapter/35-cubes-and-boxes#def-g5-solids-def) $A$ would be its own image, forcing $A$ to be the center of the half-turn. The two other [vertices](https://one-course.com/books/math/1/en/chapter/35-cubes-and-boxes#def-g5-solids-def) $B$ and $C$ would then be [symmetric](#def-g7-central-def) about $A$ — so $A$ would be the midpoint of $[BC]$, putting the three [vertices](https://one-course.com/books/math/1/en/chapter/35-cubes-and-boxes#def-g5-solids-def) on one line. A triangle has no three aligned [vertices](https://one-course.com/books/math/1/en/chapter/35-cubes-and-boxes#def-g5-solids-def): contradiction. No center exists.

**13.** The same parity argument: a half-turn preserving the [polygon](https://one-course.com/books/math/1/en/chapter/26-lines-and-polygons#def-g4-geometry-polygon) pairs up its [vertices](https://one-course.com/books/math/1/en/chapter/35-cubes-and-boxes#def-g5-solids-def); an [odd number](https://one-course.com/books/math/1/en/chapter/14-numbers-up-to-10-000#def-g3-numbers-evenodd) of [vertices](https://one-course.com/books/math/1/en/chapter/35-cubes-and-boxes#def-g5-solids-def) forces a fixed [vertex](https://one-course.com/books/math/1/en/chapter/35-cubes-and-boxes#def-g5-solids-def), which must be the center and the midpoint of the [segment](https://one-course.com/books/math/1/en/chapter/40-lines-circles-and-angles#def-g6-lines-objects) joining two other [vertices](https://one-course.com/books/math/1/en/chapter/35-cubes-and-boxes#def-g5-solids-def) — three aligned [vertices](https://one-course.com/books/math/1/en/chapter/35-cubes-and-boxes#def-g5-solids-def), impossible in a [polygon](https://one-course.com/books/math/1/en/chapter/26-lines-and-polygons#def-g4-geometry-polygon) … so no [polygon](https://one-course.com/books/math/1/en/chapter/26-lines-and-polygons#def-g4-geometry-polygon) with an [odd number](https://one-course.com/books/math/1/en/chapter/14-numbers-up-to-10-000#def-g3-numbers-evenodd) of [vertices](https://one-course.com/books/math/1/en/chapter/35-cubes-and-boxes#def-g5-solids-def) has a [center of symmetry](#def-g7-central-center).

**14.** Performing the half-turn about $O_1$, then about $O_2$, sends the figure onto itself both times — so their combination does too. By question 5, that combination is a translation by twice the distance $O_1 O_2$, which is not [zero](https://one-course.com/books/math/1/en/chapter/1-counting-to-20#def-g1-counting-zero) since the centers differ. A figure fitting on a sheet cannot equal itself slid by a fixed nonzero amount (slide it enough times and it leaves the sheet entirely). So a bounded figure has at most one [center of symmetry](#def-g7-central-center).

**15.** On the endless strip, a half-turn about the point midway between a left footprint and the following right one sends the pattern onto itself; the point one full step further does too: two distinct centers. Composing the two half-turns gives the translation by twice the half-step — exactly the one-step slide that visibly maps the infinite frieze onto itself, as question 14 predicts. Unbounded patterns live by different rules: that is the mathematics of wallpaper.