Homothety and Similarity of Figures

Let $O$ be a fixed point (the centre of homothety) and $k \neq 0$ a real number (the ratio or coefficient of homothety). The homothety with centre $O$ and ratio $k$ maps each point $A$ to the point $A'$ such that:

$\overrightarrow{OA'} = k\,\overrightarrow{OA}$

If $O = (a, b)$ and $A = (x, y)$, this gives:

$\boxed{A'= \bigl(a + k(x-a),\; b + k(y-b)\bigr)}$

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

Let $H_{O,k}$ denote the homothety with centre $O$ and ratio $k$. Then:

1. Distances are scaled: $|A'B'| = |k|\cdot|AB|$ for all points $A, B$.

2. Angles are preserved: the angle between any two lines equals the angle between their images.

3. Lines map to parallel lines: if line $\ell$ does not pass through $O$, its image $\ell'$ is parallel to $\ell$; if $\ell$ passes through $O$, then $\ell' = \ell$.

4. Circles map to circles: a circle with centre $C$ and radius $r$ maps to a circle with centre $C'$ (image of $C$) and radius $|k|r$.

5. Area is scaled by $k^2$: $S_{A'B'C'} = k^2 \cdot S_{ABC}$.

Proof of (1). With $O$ at the origin: $A' = (kx_1, ky_1), \quad B' = (kx_2, ky_2).$ $|A'B'| = \sqrt{(kx_2-kx_1)^2+(ky_2-ky_1)^2} = |k|\sqrt{(x_2-x_1)^2+(y_2-y_1)^2} = |k|\cdot|AB|. \quad \blacktriangleleft$

A similarity transformation (or similarity) with similarity ratio $k > 0$ is a transformation that can be expressed as the composition of a homothety with ratio $k$ (or $-k$) and an isometry. Equivalently, $f$ is a similarity with ratio $k$ if:

$|A'B'| = k\cdot|AB| \quad \text{for all points } A, B.$

Every isometry is a similarity with ratio $k = 1$.

If a similarity transformation with ratio $k$ maps figure $F$ to figure $F'$, then:

$\boxed{\dfrac{S_{F'}}{S_F} = k^2}$

Proof sketch. Divide $F$ into small triangles. Each triangle maps to a similar triangle with ratio $k$. The area of each small image triangle is $k^2$ times the area of the original, so the total area scales by $k^2$. $\blacktriangleleft$

Triangle $ABC$ has vertices $A(1, 1)$, $B(3, 1)$, $C(2, 3)$. Find the image of the triangle under the homothety with centre $O(0, 0)$ and ratio $k = 2$.

Solution. Apply $(x, y) \to (2x, 2y)$:

$A' = (2, 2), \quad B' = (6, 2), \quad C' = (4, 6).$

Check: $|AB| = 2$, $|A'B'| = 4 = 2\cdot|AB|$ ✓; $|BC| = \sqrt{1+4} = \sqrt{5}$, $|B'C'| = \sqrt{4+16} = 2\sqrt{5}$ ✓.

Area of $ABC$: $S = \frac{1}{2}|2\cdot(1-3)+1\cdot(3-1)+3\cdot(1-1)| = \frac{1}{2}|-4+2+0| = 1$.

Area of $A'B'C'$: $S' = 4\cdot 1 = 4 = k^2 S$ ✓.

Answer: $A'(2, 2)$, $B'(6, 2)$, $C'(4, 6)$. $\blacktriangleleft$

A homothety maps circle $\omega_1$ with centre $C_1(1, 2)$ and radius $r_1 = 2$ to circle $\omega_2$ with centre $C_2(5, 6)$ and radius $r_2 = 6$. Find the centre and ratio of the homothety.

Solution. The ratio is:

$k = \pm\dfrac{r_2}{r_1} = \pm\dfrac{6}{2} = \pm 3.$

Case $k = 3$ (external centre): The centre $O$ divides $C_1C_2$ in the ratio $|OC_1| : |OC_2| = 1 : 3$ externally such that $C_2 - O = 3(C_1 - O)$:

$5 - O_x = 3(1 - O_x) \implies 5 - O_x = 3 - 3O_x \implies 2O_x = -2 \implies O_x = -1.$ $6 - O_y = 3(2 - O_y) \implies 6 - O_y = 6 - 3O_y \implies 2O_y = 0 \implies O_y = 0.$

So $O = (-1, 0)$ with $k = 3$.

Case $k = -3$ (internal centre): $C_2 - O = -3(C_1 - O)$ gives $O = (2, 3)$ with $k = -3$.

Answer: Centre $O(-1, 0)$, ratio $k = 3$; or centre $O(2, 3)$, ratio $k = -3$. $\blacktriangleleft$

Triangle $PQR$ has area $S = 12$. It is mapped by a homothety with ratio $k = \frac{3}{2}$ to triangle $P'Q'R'$. Find the area of $P'Q'R'$. How does the answer change if $k = -\frac{3}{2}$?

A homothety with centre $O(2, -1)$ maps point $A(4, 3)$ to $A'(8, 11)$. Find the ratio $k$ and the image of point $B(0, -1)$ under the same homothety.