10.2 Non-Rigid Dilations, Scale Factors & Tessellations

Foundations of Non-Rigid Transformations: Dilations and Similarity

A dilation is a transformation that resizes a geometric figure while maintaining its proportional shape. Unlike isometries (translations, reflections, and rotations), a dilation alters Euclidean distance between points whenever its scale factor differs from 1. Dilations are therefore classified as non-rigid transformations.

A dilation is completely specified by two geometric parameters:

1. A fixed point $C(a, b)$ called the center of dilation.
2. A non-zero real number $k$ called the scale factor.

Geometric Definition

For any point $P$ in the plane, its dilated image $P' = D_{(C, k)}(P)$ satisfies two geometric conditions:

1. Points $C$, $P$, and $P'$ are strictly collinear along a common ray or line extending from the center of dilation.
2. The Euclidean distance from the center $C$ to the image point $P'$ is directly proportional to the distance from $C$ to the pre-image point $P$, scaled by the absolute value of $k$:

$$CP' = |k| \cdot CP$$

If point $P$ is the center of dilation itself ($P = C$), then $C' = C$; the center of dilation is the unique fixed point under any dilation with $k \neq 1$.

Similarity Transformations

A dilation is the quintessential similarity transformation. It maps any geometric figure onto a similar figure ($\Delta ABC \sim \Delta A'B'C'$). The defining properties of geometric similarity are preserved:

• Congruent Angles: Corresponding interior and exterior angle measures remain strictly equal ($m\angle A = m\angle A'$, $m\angle B = m\angle B'$, $m\angle C = m\angle C'$).
• Proportional Side Lengths: The ratio of every image segment length to its corresponding pre-image segment length is identically equal to the absolute scale factor: $$\frac{A'B'}{AB} = \frac{B'C'}{BC} = \frac{A'C'}{AC} = |k|$$

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Coordinate Mapping Rules for Dilations

Center of Dilation at the Origin $(0, 0)$

When the center of dilation is situated at the coordinate origin, the algebraic mapping rule multiplies both the $x$- and $y$-coordinates by scale factor $k$:

$$D_{(0, k)}(x, y) = (kx, ky)$$

Proof of Distance Proportionality: The distance from the origin $O(0, 0)$ to pre-image point $P(x, y)$ is $OP = \sqrt{x^2 + y^2}$. The distance from $O$ to image point $P'(kx, ky)$ is:

$$OP' = \sqrt{(kx)^2 + (ky)^2} = \sqrt{k^2(x^2 + y^2)} = |k|\sqrt{x^2 + y^2} = |k| \cdot OP$$

Center of Dilation at an Arbitrary Point $(a, b)$

When the center of dilation is shifted to an arbitrary point $C(a, b)$, coordinates cannot simply be multiplied by $k$. Applying vector displacement relative to $C$:

1. Translate center $(a, b)$ to the origin by shifting coordinates: $(x - a, y - b)$.
2. Multiply the relative distances by scale factor $k$: $(k(x - a), k(y - b))$.
3. Translate back by adding $(a, b)$:

$$D_{(C(a, b), k)}(x, y) = (a + k(x - a), b + k(y - b))$$

Classification by Scale Factor Magnitude

• Enlargement (Expansion): $|k| > 1$. The image figure is strictly larger than the pre-image, and points move farther away from center $C$.
• Reduction (Contraction): $0 < |k| < 1$. The image figure is strictly smaller than the pre-image, and points move closer toward center $C$.
• Identity Transformation (Isometry): $k = 1$. Every point maps to itself ($P' = P$).
• Negative Scale Factor ($k < 0$): Inverts points through the center of dilation onto the opposite ray. A dilation with negative scale factor $-k$ is mathematically identical to a positive dilation of $|k|$ followed by a $180^\circ$ rotation centered at $C$: $$D_{(C, -k)} = R_{(C, 180^\circ)} \circ D_{(C, |k|)}$$

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Invariance Properties and Dimensional Scaling Laws ($k$, $k^2$, $k^3$)

What Dilations Preserve vs. Alter

• Preserved (Invariants): Angle measures, collinearity of points, betweenness, parallelism of lines (any line not passing through $C$ maps to a parallel line: $\ell \parallel \ell'$), orientation (for $k > 0$).
• Altered (Non-Invariants): Absolute Euclidean distance between points, perimeter, area, and spatial volume.

The Dimensional Scaling Theorem

When a two- or three-dimensional figure is dilated by scale factor $k > 0$:

1. Linear Metrics (Scale by $k^1$): Any one-dimensional linear measure—including side lengths, altitudes, medians, perimeters, and circumferences—scales directly by $k$:

   $$\text{Perimeter}' = k \cdot \text{Perimeter}$$

2. Surface Area Metrics (Scale by $k^2$): Any two-dimensional surface measure—including polygon area, lateral surface area, total surface area, and sector area—scales by the square of the scale factor ($k^2$):

   $$\text{Area}' = k^2 \cdot \text{Area}$$

   Proof for Triangle: Area of pre-image is $A = \frac{1}{2} b h$. Under dilation by $k$, base $b' = kb$ and altitude $h' = kh$. The image area is $A' = \frac{1}{2} b' h' = \frac{1}{2} (kb)(kh) = k^2 \left(\frac{1}{2} bh\right) = k^2 A$.

3. Volume Metrics (Scale by $k^3$): Any three-dimensional spatial volume measure scales by the cube of the scale factor ($k^3$):

   $$\text{Volume}' = k^3 \cdot \text{Volume}$$

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Planar Tessellations: The Vertex Angle Sum Criterion

A tessellation (or tiling) of the plane is an arrangement of closed two-dimensional polygonal shapes that covers the entire infinite Euclidean plane without any gaps (empty spaces) and without any overlapping interiors.

• An edge-to-edge tessellation is a tiling where adjacent polygons share either a full edge or a single vertex, never a partial edge.
• A vertex in a tessellation is any point where the corners of three or more adjacent polygons converge.

The Fundamental Vertex Angle Sum Theorem

At any vertex in an edge-to-edge planar tessellation, the complete rotational angle surrounding the vertex point is $360^\circ$. Therefore, the sum of the interior angles of all polygons meeting at that vertex must equal exactly $360^\circ$:

$$\sum_{i=1}^{m} \theta_i = 360^\circ$$

• If $\sum \theta_i < 360^\circ$, a gap remains, preventing the shapes from tiling the flat plane.
• If $\sum \theta_i > 360^\circ$, the shapes overlap and buckle out of the plane, forming a three-dimensional polyhedral corner rather than a planar tiling.

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Regular Tessellations of the Euclidean Plane

A regular tessellation is a monohedral (single-shape) edge-to-edge tiling formed entirely by congruent regular polygons of a single type.

Derivation of the Exactly Three Regular Tessellations

Recall that for a regular polygon with $n$ sides, each interior angle $\theta$ is given by:

$$\theta = \frac{(n - 2) \times 180^\circ}{n}$$

If $m$ identical regular $n$-gons meet at each vertex, the vertex sum condition requires:

$$m \times \theta = 360^\circ \implies m \left(\frac{(n - 2) \times 180^\circ}{n}\right) = 360^\circ$$

Dividing both sides by $180^\circ$:

$$\frac{m(n - 2)}{n} = 2 \implies mn - 2m = 2n$$

Rearranging into Diophantine form:

$$mn - 2m - 2n = 0 \implies (m - 2)(n - 2) - 4 = 0 \implies (m - 2)(n - 2) = 4$$

Because $m \ge 3$ (at least 3 polygons must meet at a vertex) and $n \ge 3$ (a polygon must have at least 3 sides), $(m - 2)$ and $(n - 2)$ must be positive integer factors of 4. There exist exactly three integer pairs:

1. $(m - 2) = 1, (n - 2) = 4 \implies m = 3, n = 6$: Three regular hexagons meet at each vertex. Interior angle $\theta = 120^\circ$. Vertex sum: $3 \times 120^\circ = 360^\circ$. Vertex notation: $6.6.6$ (or $6^3$).
2. $(m - 2) = 2, (n - 2) = 2 \implies m = 4, n = 4$: Four squares meet at each vertex. Interior angle $\theta = 90^\circ$. Vertex sum: $4 \times 90^\circ = 360^\circ$. Vertex notation: $4.4.4.4$ (or $4^4$).
3. $(m - 2) = 4, (n - 2) = 1 \implies m = 6, n = 3$: Six equilateral triangles meet at each vertex. Interior angle $\theta = 60^\circ$. Vertex sum: $6 \times 60^\circ = 360^\circ$. Vertex notation: $3.3.3.3.3.3$ (or $3^6$).

Why Regular Pentagons Cannot Form a Regular Tessellation

A regular pentagon has $n = 5$ sides. Its interior angle is:

$$\theta = \frac{(5 - 2) \times 180^\circ}{5} = \frac{540^\circ}{5} = 108^\circ$$

Dividing $360^\circ$ by $108^\circ$:

$$\frac{360^\circ}{108^\circ} = \frac{10}{3} = 3.333\dots$$

Three pentagons provide $3 \times 108^\circ = 324^\circ$, leaving an unfilled gap of $360^\circ - 324^\circ = 36^\circ$. Four pentagons require $4 \times 108^\circ = 432^\circ$, producing an overlap of $72^\circ$. Because 108 is not a divisor of 360, regular pentagons cannot tessellate the plane.

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Semi-Regular (Archimedean) Tessellations

A semi-regular tessellation (also known as an Archimedean tiling) is an edge-to-edge planar tiling formed using two or more distinct regular polygons, such that the arrangement of polygons around every single vertex is strictly congruent (identical).

Vertex Configuration Notation

A vertex configuration is written as a sequence of numbers indicating the side count of each regular polygon meeting at that vertex, listed in cyclic order around the vertex. For example, $3.3.3.4.4$ denotes three equilateral triangles and two squares meeting at the vertex in that specific sequence.

The Eight Semi-Regular Tessellations

There are exactly 8 distinct semi-regular tessellations in the Euclidean plane:

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Worked Step-by-Step Examples

Worked Example 1: Dilation from a Non-Origin Center

Problem: A triangle has vertices at $A(4, 2)$, $B(8, 6)$, and $C(6, -2)$. The triangle is dilated with center of dilation $C_0(2, -2)$ and scale factor $k = 1.5$. Find the coordinates of the image vertices $A'$, $B'$, and $C'$.

Solution: Step 1: State the general formula for dilation centered at $(a, b)$:

$$x' = a + k(x - a), \quad y' = b + k(y - b)$$

Here $a = 2$, $b = -2$, and $k = 1.5 = \frac{3}{2}$.

Step 2: Calculate coordinates for image vertex $A'$:

$$x_A' = 2 + 1.5(4 - 2) = 2 + 1.5(2) = 2 + 3 = 5$$ $$y_A' = -2 + 1.5(2 - (-2)) = -2 + 1.5(4) = -2 + 6 = 4 \implies A'(5, 4)$$

Step 3: Calculate coordinates for image vertex $B'$:

$$x_B' = 2 + 1.5(8 - 2) = 2 + 1.5(6) = 2 + 9 = 11$$ $$y_B' = -2 + 1.5(6 - (-2)) = -2 + 1.5(8) = -2 + 12 = 10 \implies B'(11, 10)$$

Step 4: Calculate coordinates for image vertex $C'$:

$$x_C' = 2 + 1.5(6 - 2) = 2 + 1.5(4) = 2 + 6 = 8$$ $$y_C' = -2 + 1.5(-2 - (-2)) = -2 + 1.5(0) = -2 + 0 = -2 \implies C'(8, -2)$$

The dilated image has vertices $A'(5, 4)$, $B'(11, 10)$, and $C'(8, -2)$.

Worked Example 2: Area and Perimeter Scaling Under Dilation

Problem: A trapezoid has an area of $48\text{ cm}^2$ and a perimeter of $32\text{ cm}$. The trapezoid is subjected to a dilation with a scale factor of $k = \frac{3}{4}$. Determine the perimeter and area of the resulting similar trapezoid.

Solution: Step 1: Linear dimensions (perimeter) scale directly by factor $k = \frac{3}{4}$:

$$\text{Perimeter}' = k \cdot \text{Perimeter} = \frac{3}{4} \times 32\text{ cm} = 3 \times 8 = 24\text{ cm}$$

Step 2: Two-dimensional surface measures (area) scale by the factor $k^2 = \left(\frac{3}{4}\right)^2 = \frac{9}{16}$:

$$\text{Area}' = k^2 \cdot \text{Area} = \frac{9}{16} \times 48\text{ cm}^2 = 9 \times 3 = 27\text{ cm}^2$$

The transformed trapezoid has a perimeter of $24\text{ cm}$ and an area of $27\text{ cm}^2$.

Worked Example 3: Verifying a Proposed Vertex Tessellation

Problem: An architectural designer proposes a new semi-regular floor tiling using regular pentagons and regular decagons. Can two regular pentagons and one regular decagon meet at each vertex to form a valid planar tessellation? Justify mathematically.

Solution: Step 1: Compute the interior angle of a regular pentagon ($n = 5$):

$$\theta_5 = \frac{(5 - 2) \times 180^\circ}{5} = \frac{540^\circ}{5} = 108^\circ$$

Step 2: Compute the interior angle of a regular decagon ($n = 10$):

$$\theta_{10} = \frac{(10 - 2) \times 180^\circ}{10} = \frac{8 \times 180^\circ}{10} = 8 \times 18^\circ = 144^\circ$$

Step 3: Sum the interior angles for two pentagons and one decagon at the vertex:

$$\sum \theta = 2(108^\circ) + 144^\circ = 216^\circ + 144^\circ = 360^\circ$$

Step 4: Geometric feasibility evaluation: Although the angle sum is exactly $360^\circ$, this configuration cannot tessellate the entire infinite plane edge-to-edge. When two pentagons and a decagon meet at one vertex, the adjacent vertices force pentagon-pentagon angles that cannot be closed by regular decagons without overlap elsewhere. Thus, satisfying $\sum \theta = 360^\circ$ is a necessary condition for tessellation, but not every angle-sum combination can be extended globally into an Archimedean tiling.

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Diagnostic Misconceptions & Pedagogical Strategies

1. Linear Scaling of Area (The $k$ vs. $k^2$ Trap): When a figure's side lengths are doubled ($k = 2$), middle school students overwhelmingly predict that the area will also double. Pedagogical remedy: Have students build a $1 \times 1$ square using unit tiles. Ask them to double the length and width to create a $2 \times 2$ square. They visually count 4 tiles ($2^2 = 4$). Repeat with a $3 \times 3$ square (9 tiles, $3^2 = 9$). Emphasize that area requires multiplying two perpendicular linear dimensions, so each dimension introduces a factor of $k$, compounding to $k \times k = k^2$.
2. Neglecting the Center in Non-Origin Dilations: Students frequently compute dilations from arbitrary centers by simply multiplying the given coordinates by $k$: $(kx, ky)$. This operation inadvertently dilates the figure from the origin $(0, 0)$, displacing the entire shape across the grid. Reinforce the conceptual anchor: the center of dilation is the reference point from which distance is measured. Students must calculate horizontal distance $\Delta x = (x - a)$ and vertical distance $\Delta y = (y - b)$ from center $(a, b)$, scale those distances, and add the result back to $(a, b)$.
3. Assuming Only Regular Polygons Can Tessellate: Students sometimes assume that because non-regular shapes do not have equal angles, they cannot tile. Point out that any triangle (scalene, acute, obtuse) and any quadrilateral (even non-convex ones) can tessellate the plane by rotating $180^\circ$ around the midpoints of their sides.