Primary & Middle School Mathematics · Grades 1–9

52Central Symmetry and Parallelograms

Turn a figure by half a turn around a point: this is central symmetry, the second transformation of the book after the reflections of Chapter 42. Its star figure is the parallelogram — the quadrilateral that is symmetric 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). — 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 $M'$ of $M$ about $O$ :

draw the line $(MO)$ and extend it beyond $O$ ;
measure $MO$ (ruler or compass);
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 $M'$ is $3$ squares right and $1$ square down from $O$ : central symmetry reverses both directions at once (a reflection reverses only one).

Proposition 52.4 (What a half-turn preserves)

Central symmetry preserves lengths, angles, perimeters and areas; the image of a line is a parallel line, the image of a segment is a parallel segment of the same length, the image of a circle is a circle of the same radius (centered at the symmetric of the center).

Proof. Admitted at this level. ∎

Remark 52.5

Compare with reflections (Proposition 42.4): 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 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 whose diagonals cross at their common midpoint. That crossing point is then a center of symmetry of the figure: each vertex is the half-turn image of the opposite vertex.

Theorem 52.8 (Properties of a parallelogram)

In a parallelogram $ABCD$ with center $O$ :

opposite sides are parallel: $(AB) \parallel (CD)$ and $(BC) \parallel (AD)$ ;
opposite sides have the same length: $AB = CD$ and $BC = AD$ ;
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 $[AB]$ to the segment $[CD]$ ; by Proposition 52.4 the image is parallel 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). — A parallelogram and its center $O$ : the diagonals (red) cut each other at their midpoints, opposite sides are parallel and equal (tick marks).

Theorem 52.9 (Recognizing a parallelogram)

A quadrilateral (with no crossing sides) is a parallelogram as soon as one of the following holds:

its diagonals have the same midpoint (the definition);
its opposite sides are parallel two by two;
two opposite sides are parallel 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 — 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 and square of Chapter 41 are exactly the parallelograms with an extra property: equal diagonals (rectangle), perpendicular diagonals (rhombus), both (square). The angle argument promised in Example 41.6 now works: in a parallelogram, consecutive angles are supplementary (alternate/corresponding angles with the parallels, Theorem 51.2), so one right angle 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 $M'$ of $M$ about $O$ , and the symmetric of $M$ across the vertical grid line through $O$ . Compare the two images.

Solution

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

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 line, Proposition 52.4).

Exercise 52.3 ★

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

Solution

Solution of Exercise 52.3.

Calculator digits with a center of symmetry: $0$ , $2$ , $5$ , $8$ (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 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

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 $ABCD$ 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

Solution of Exercise 52.5.

Draw a segment $[AC]$ 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 by definition.

Exercise 52.6 ★

In a parallelogram, one angle measures $115^\circ$ . Using the remark on supplementary consecutive angles, give the three other angles.

Solution

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 $360^\circ$ ).

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).

Solution

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 in which $AB = CD$ . Is that enough to make it a parallelogram? If not, sketch a counterexample (an isosceles trapezoid helps), and state what must be added to the hypothesis.

Solution

Solution of Exercise 52.8.

No. An isosceles trapezoid has its two slanted sides equal without being a parallelogram (the two parallel sides have different lengths). What suffices is criterion 3 of Theorem 52.9: two opposite sides equal and parallel.

Exercise 52.9 ★★

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

Solution

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. The quadrilateral $ABA'C$ has diagonals $[AA']$ and $[BC]$ sharing the midpoint $O$ : criterion 1 (the definition) makes it a parallelogram at no extra cost.

Exercise 52.10 ★★

True or false, with a reason: “a quadrilateral whose opposite angles are equal two by two is a parallelogram”; “a parallelogram with perpendicular diagonals is a rhombus”; “a parallelogram with one right angle is a rectangle”.

Solution

Solution of Exercise 52.10.

“Opposite angles equal $\Rightarrow$ parallelogram”: true (this is another recognition criterion; with the angle sum $360^\circ$ , equal opposite pairs force supplementary consecutive angles, hence parallel sides by Theorem 51.2).

“Perpendicular diagonals $\Rightarrow$ rhombus”: true for a parallelogram (the diagonals already cut at their midpoints).

“One right angle $\Rightarrow$ rectangle”: true — consecutive angles are supplementary, so all four become right.

Exercise 52.11 ★★★

Let $ABCD$ be any quadrilateral 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 — and again with vectors, in the High School volume.)

Solution

Solution of Exercise 52.11.

Whatever the quadrilateral — even a very irregular one — the midpoints $I$ , $J$ , $K$ , $L$ always form a parallelogram. The proof with the midpoint theorem is in Chapter 59 (Exercise 59.9).

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, 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.

On grid paper, take the origin $O(0, 0)$ as center. Using the grid counting of Example 52.3, 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 $(x, y)$ ?
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.
Playing cards are designed to read the same upside down — a central symmetry, so neither player sees the card “wrongly”. Explain why, on a card with an odd number of pips, one pip must sit exactly at the card’s center.
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?
Measure the slide of question 4 against the distance $O_1 O_2$ , and state the discovery. (Compare with the two parallel mirrors of Problem 42.1: same phenomenon, new actors.)

Part II — Cakes and medians.

Let $O$ be the center of a parallelogram. Explain why any line through $O$ meets the boundary at two points that are symmetric about $O$ , and why the line cuts the parallelogram into two pieces that are exact half-turn copies of each other — hence of equal areas.
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 tilted). Describe the one straight cut that gives two parts with equally much cake — and justify it with question 6.
Three vertices of a parallelogram $ABCD$ are $A(0, 0)$ , $B(6, 1)$ and $C(8, 5)$ . Compute the coordinates of $D$ and of the center $O$ .
Exercise 52.9 built, from a triangle $ABC$ and the midpoint $O$ of $[BC]$ , the parallelogram $ABA'C$ where $A'$ is the symmetric of $A$ about $O$ . Use it, together with the triangle inequality (Theorem 51.6) in the triangle $ABA'$ , to prove the median bound: the median $[AO]$ satisfies

$AO < \frac{AB + AC}{2} .$

(Note that $AA' = 2\,AO$ and $BA' = AC$ .)
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?

For each figure, say whether it has a center of symmetry, and where: a segment; a full line; an equilateral triangle; a square; a circle; the letters S and Z.
Prove your claim for the equilateral triangle: a half-turn sending the triangle to itself would pair up the three vertices — but three is odd, so some vertex would be sent to itself. What would that force the vertex to be, and why is it impossible?
Generalize question 12: explain why no polygon with an odd number of vertices has a center of symmetry.
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.
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

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 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 mirrors translated by twice their gap in Problem 42.1.

6. The half-turn about $O$ sends the parallelogram onto itself (that is what “center of symmetry” means for it, Theorem 52.8). 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 about $O$ . The two pieces of the parallelogram on either side of the line swap too, so they are exact copies (Proposition 52.4) — equal areas.

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 the cake minus 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 $ABA'C$ , 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: yes, its midpoint. Line: yes — every one of its points is a center. Equilateral triangle: 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 among themselves; three being odd, one vertex $A$ would be its own image, forcing $A$ to be the center of the half-turn. The two other vertices $B$ and $C$ would then be symmetric about $A$ — so $A$ would be the midpoint of $[BC]$ , putting the three vertices on one line. A triangle has no three aligned vertices: contradiction. No center exists.

13. The same parity argument: a half-turn preserving the polygon pairs up its vertices; an odd number of vertices forces a fixed vertex, which must be the center and the midpoint of the segment joining two other vertices — three aligned vertices, impossible in a polygon … so no polygon with an odd number of vertices has a center of symmetry.

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 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.

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.