Rotation

Let $O$ be a fixed point (the centre of rotation) and $\varphi$ a signed angle (positive = counterclockwise). The rotation by angle $\varphi$ around $O$ maps each point $A$ to the point $A'$ such that:

$|OA'| = |OA| \qquad \text{and} \qquad \angle AOA' = \varphi \text{ (measured counterclockwise)}.$

The centre $O$ is the only fixed point (it maps to itself).

The rotation by angle $\varphi$ around the origin maps $A(x, y)$ to:

$\boxed{x' = x\cos\varphi - y\sin\varphi, \qquad y' = x\sin\varphi + y\cos\varphi}$

Special cases:

Let $R_{O,\varphi}$ denote the rotation by $\varphi$ around $O$. Then:

1. Isometry: $|A'B'| = |AB|$ for all $A, B$.

2. Direct isometry: $R_{O,\varphi}$ preserves orientation.

3. Angle preservation: the angle between any two lines is preserved.

4. Circles map to circles: a circle of radius $r$ maps to a circle of radius $r$.

5. Composition: $R_{O,\varphi_2} \circ R_{O,\varphi_1} = R_{O,\varphi_1+\varphi_2}$ (rotations about the same centre add up).

6. Special cases:

   • $\varphi = 0°$: the identity transformation.
   • $\varphi = 180°$: central symmetry with centre $O$.

Proof of (1). The distance $|OA|$ is preserved (by definition), and likewise $|OB|$. The angle $\angle AOB$ is preserved (both are rotated by the same $\varphi$, so $\angle A'OB' = \angle AOB$). By the Law of Cosines in triangles $AOB$ and $A'OB'$: $|AB|^2 = |OA|^2 + |OB|^2 - 2|OA||OB|\cos(\angle AOB) = |A'B'|^2. \quad \blacktriangleleft$

Find the image of $A(3, 1)$ under the rotation by $90°$ (counterclockwise) around the origin.

Solution. Apply the formula $(x, y) \to (-y, x)$:

$A' = (-1,\; 3).$

Verification: $|OA| = \sqrt{9+1} = \sqrt{10}$; $|OA'| = \sqrt{1+9} = \sqrt{10}$ ✓. The angle between $\overrightarrow{OA} = (3,1)$ and $\overrightarrow{OA'} = (-1,3)$: $\cos\theta = \frac{(3)(-1)+(1)(3)}{\sqrt{10}\cdot\sqrt{10}} = \frac{0}{10} = 0 \implies \theta = 90° \checkmark.$

Answer: $A'(-1, 3)$. $\blacktriangleleft$

Triangle $ABC$ has vertices $A(2, 0)$, $B(4, 0)$, $C(4, 2)$. Find the image of the triangle under rotation by $90°$ (counterclockwise) around centre $O(2, 1)$.

Solution. Use the formula with $O = (2,1)$, $\varphi = 90°$ ($\cos 90° = 0$, $\sin 90° = 1$):

$x' = 2 + (x-2)\cdot 0 - (y-1)\cdot 1 = 2 - (y-1) = 3 - y$ $y' = 1 + (x-2)\cdot 1 + (y-1)\cdot 0 = 1 + (x-2) = x - 1$

Apply to each vertex: $A(2,0): \quad x' = 3-0 = 3, \quad y' = 2-1 = 1. \quad A' = (3, 1).$ $B(4,0): \quad x' = 3-0 = 3, \quad y' = 4-1 = 3. \quad B' = (3, 3).$ $C(4,2): \quad x' = 3-2 = 1, \quad y' = 4-1 = 3. \quad C' = (1, 3).$

Verification: $|AB| = 2$; $|A'B'| = \sqrt{0+4} = 2$ ✓.

Answer: $A'(3, 1)$, $B'(3, 3)$, $C'(1, 3)$. $\blacktriangleleft$

(a) Find the image of $P(0, 5)$ under rotation by $270°$ (counterclockwise) around the origin.

(b) A rotation around the origin maps $A(1, 0)$ to $A'(0, 1)$. What is the angle of rotation? Is there another rotation (by a different angle) that achieves the same result?

Square $ABCD$ has vertices $A(1,1)$, $B(3,1)$, $C(3,3)$, $D(1,3)$. Show that rotation by $90°$ around the centre of the square $O(2,2)$ maps the square to itself. What is the order of this rotation symmetry?