Central Symmetry

Let $O$ be a fixed point called the centre. The central symmetry with centre $O$ is the transformation that maps each point $A$ to the point $A'$ such that $O$ is the midpoint of segment $AA'$:

$\overrightarrow{OA'} = -\overrightarrow{OA}$

If $O = (a, b)$ and $A = (x, y)$, then the midpoint condition gives:

$\boxed{A' = (2a - x,\; 2b - y)}$

Special case — centre at the origin: $(x, y) \to (-x,\; -y)$.

The central symmetry with centre $O$ is identical to the rotation by $180°$ around $O$.

Proof. Under rotation by $180°$ about $O$, each point $A$ maps to the unique point $A'$ with $|OA'| = |OA|$ and $\angle AOA' = 180°$. This means $O$ lies on segment $AA'$ and $|OA| = |OA'|$, i.e.\ $O$ is the midpoint of $AA'$. This is exactly the definition of central symmetry. $\blacktriangleleft$

Let $S_O$ denote the central symmetry with centre $O$. Then:

1. Isometry: $|A'B'| = |AB|$ for all $A, B$.
2. Direct isometry: $S_O$ preserves orientation.
3. Involution: $S_O \circ S_O = \text{id}$.
4. Lines map to parallel lines: every line $\ell$ maps to a line $\ell' \parallel \ell$ (or $\ell' = \ell$ if $\ell$ passes through $O$).
5. Segments map to parallel, equal segments: $A'B' \parallel AB$ and $|A'B'| = |AB|$.

Proof of (1). With centre $O$ at the origin: $|A'B'| = |(-x_2)-(-x_1),\, (-y_2)-(-y_1)| = |(x_1-x_2,\, y_1-y_2)| = |AB|. \quad \blacktriangleleft$

A point $O$ is called a centre of symmetry of a figure $F$ if the central symmetry $S_O$ maps $F$ to itself: $S_O(F) = F$.

Examples of figures with a centre of symmetry:

• Parallelogram — centre is the intersection of its diagonals.
• Circle — centre of the circle.
• Regular $n$-gon with $n$ even — centre of the polygon.
• Line segment — midpoint of the segment.

Examples of figures without a centre of symmetry: any triangle (except degenerate cases), a regular polygon with odd $n$ (e.g.\ equilateral triangle, regular pentagon).

Find the image of triangle $ABC$ with $A(1, 2)$, $B(3, -1)$, $C(-2, 4)$ under the central symmetry with centre $O(2, 1)$.

Solution. Apply the formula $A' = (2\cdot2 - x,\; 2\cdot1 - y) = (4-x,\; 2-y)$ to each vertex:

$A' = (4-1,\; 2-2) = (3,\; 0)$ $B' = (4-3,\; 2-(-1)) = (1,\; 3)$ $C' = (4-(-2),\; 2-4) = (6,\; -2)$

Verification: $|AB| = \sqrt{(3-1)^2+(-1-2)^2} = \sqrt{4+9} = \sqrt{13}$; $|A'B'| = \sqrt{(1-3)^2+(3-0)^2} = \sqrt{4+9} = \sqrt{13}$ ✓.

Answer: $A'(3, 0)$, $B'(1, 3)$, $C'(6, -2)$. $\blacktriangleleft$

Parallelogram $ABCD$ has vertices $A(0,0)$, $B(4,0)$, $C(5,2)$, $D(1,2)$. Find its centre of symmetry and verify.

Solution. In a parallelogram the diagonals bisect each other. The centre of symmetry is the midpoint of diagonal $AC$:

$O = \left(\frac{0+5}{2},\, \frac{0+2}{2}\right) = \left(\frac{5}{2},\, 1\right).$

Verify: apply $S_O$ to each vertex using $(4 - x \cdot 2/2\ldots)$, i.e.\ $A' = (5-0,\; 2-0) = (5,2) = C$ ✓; $C' = (5-5,\; 2-2) = (0,0) = A$ ✓; $B' = (5-4,\; 2-0) = (1,2) = D$ ✓; $D' = (5-1,\; 2-2) = (4,0) = B$ ✓.

Answer: The centre of symmetry is $O\!\left(\dfrac{5}{2},\, 1\right)$. $\blacktriangleleft$

Find the image of point $P(3, -2)$ under the central symmetry with centre $O(-1, 4)$. Then find the preimage of the point $Q(0, 0)$ under the same symmetry.

A central symmetry maps segment $AB$ ($A(1,3)$, $B(5,1)$) to segment $A'B'$ ($A'(7,5)$, $B'(3,7)$). Find the centre of symmetry.