Chapter 19 — Transformations
Transformational geometry is a beautiful and elegant subject with many applications in physics and optics. When we transform a figure, we use rules to move each point to a new point. The output is the image of the original, and the process is a mapping.
In this chapter
- 19.1 Translation
- 19.2 Rotation
- 19.3 Reflection
- 19.4 Distortion
- 19.5 Dilation
- 19.6 The More Things Change… (Preserved Properties)
- 19.7 Transformation Proofs
- Transformation Summary Table
- Exercises & Problems
Transformation Summary Table
| Transformation | Preserves Distance? | Preserves Angles? | Preserves Area? | Preserves Shape? | Isometry? |
|---|---|---|---|---|---|
| Translation | Yes | Yes | Yes | Yes | Yes |
| Rotation | Yes | Yes | Yes | Yes | Yes |
| Reflection | Yes | Yes | Yes | Yes | Yes |
| Dilation (factor $k$) | No ($\times k$) | Yes | No ($\times k^2$) | Yes (similar) | No |
| Distortion (factor $k$) | No | No | No ($\times k$) | No | No |
19.1 Translation
A translation slides every point the same distance in the same direction, without rotation or distortion. In coordinates:
\[(x, y) \to (x + a,\; y + b)\]where $a$ is the horizontal shift and $b$ is the vertical shift.
The image is always congruent to the original figure. Orientation and shape are preserved.
Example 19-1. What translation moves a point 3 units left and 5 units down?
Solution. Subtract 3 from $x$, subtract 5 from $y$:
\((x', y') = (x - 3,\; y - 5)\)
Exercise 19-1. What translation maps $(3, 4)$ to $(5, -3)$?
Exercise 19-2. A fixed point of a transformation is a point whose image is itself. Which translations have fixed points?
19.2 Rotation
A rotation fixes one point (the center of rotation) and turns everything else by a specified angle $\theta$ in a specified direction (clockwise or counterclockwise).
The image of point $B$ under rotation by $\theta$ about center $O$ is the point $B’$ such that:
- $OB = OB’$ (distance preserved)
- $\angle BOB’ = \theta$
Key properties:
- Rotations map figures to congruent figures.
- $AB = A’B’$ and $OA = OA’$ for any point $A$ and its image $A’$.
- In polar coordinates centered at the center of rotation: $(r, \theta) \to (r, \theta + \alpha)$.
Example 19-2. Which of $(2,5)$, $(5,8)$, $(7,7)$ could be the image of $(5,3)$ rotated about $(2,7)$?
Solution. Distance from $(5,3)$ to $(2,7)$: $\sqrt{9+16} = 5$. Check which candidate is 5 units from $(2,7)$:
- $(2,5)$: distance $= \sqrt{0 + 4} = 2$ ✗
- $(5,8)$: distance $= \sqrt{9+1} = \sqrt{10}$ ✗
- $(7,7)$: distance $= \sqrt{25+0} = 5$ ✓
Answer: $(7, 7)$.
Example 19-3. How many nonnegative clockwise rotations less than 360° about the center of a regular pentagon map the pentagon to itself? (MATHCOUNTS 1984)
Solution. The central angle subtended by each side is $\frac{360°}{5} = 72°$. Rotations of $0°$, $72°$, $144°$, $216°$, and $288°$ each map the pentagon to itself. That’s $\boxed{5}$ rotations.
Exercise 19-3. Show that if line $\ell$ is rotated about point $O$ by angle $\alpha$ to line $\ell’$, then $\ell$ and $\ell’$ intersect at angle $\alpha$. (Mandelbrot #3)
Exercise 19-4. If $ABCDEF$ is a regular hexagon, what rotation about $D$ maps $B$ to $F$?
19.3 Reflection
A reflection in a line $\ell$ (the line of reflection / mirror line) maps each point $P$ to the point $P’$ such that:
- $\ell$ is the perpendicular bisector of $PP’$.
- Equivalently, $P$ and $P’$ are equidistant from $\ell$ on opposite sides.
If the image of a reflection of a figure in a line is the figure itself, then $\ell$ is a line of symmetry.
Reflection in a point. To reflect point $A$ in point $O$: the image $A’$ lies on line $AO$, on the opposite side of $O$, with $AO = A’O$. This is equivalent to a $180°$ rotation about $O$.
If the image of reflection of a figure about a point is the figure itself, the point is a point of symmetry (center of symmetry).
Useful properties:
- A point and its image are equidistant from the line of reflection.
- The line of reflection is perpendicular to the segment connecting a point and its image.
- Reflections preserve distances: every segment maps to a segment of equal length.
Example 19-4. Find the image of $(4, 3)$ reflected in the point $(2, 0)$.
Solution. Point $O = (2, 0)$ is 2 left and 3 below $A = (4,3)$. So $A’$ is 2 left and 3 below $O$:
\(A' = (2 - 2,\; 0 - 3) = \boxed{(0, -3)}\)
Example 19-5. If line $\ell$ meets line $m$ at angle $\alpha$, show that $\ell’$ (the reflection of $\ell$ in $m$) also meets $m$ at angle $\alpha$.
Proof. The intersection point $O$ of $\ell$ and $m$ maps to itself under reflection in $m$. Take $A$ on $\ell$ with image $A’$ on $\ell’$. Since $A’X = AX$ (equal distances from $m$) and $\triangle A’XO \cong \triangle AXO$ by Leg-Leg, we get $\angle A’OX = \angle AOX = \alpha$. $\square$
Exercise 19-5. Show that reflection through a point in the plane is the same as a $180°$ rotation about that point.
Exercise 19-6. Draw a figure with a line of symmetry but no point of symmetry. Draw a figure with a point of symmetry. Can a figure have a point of symmetry but no line of symmetry?
Exercise 19-7. How many lines of symmetry does a regular hexagon have?
19.4 Distortion
A distortion (stretching/squishing) multiplies one dimension by a factor $k$ while leaving other dimensions unchanged. For example, stretching vertically by factor $k$:
\((x, y) \to (x,\; ky)\)
Key difference from other transformations: Distortions do not preserve angles. A figure is not similar to its image.
- An isosceles triangle can be distorted into an equilateral triangle.
- A circle distorted along one axis becomes an ellipse.
Although distortions don’t preserve area, they do have simple behavior: area is multiplied by $k$ (the stretching factor). This means distortions preserve ratios of areas.
Example 19-8. Prove that area is multiplied by the stretching factor $k$ under a distortion.
Proof sketch. Think of the area as the sum of tiny parallel rectangles. Aligned with the distortion, each rectangle is stretched by $k$ in one direction and unchanged in the other, so each rectangle’s area is multiplied by $k$. Thus the total area is multiplied by $k$. $\square$
Exercise 19-8. Give a rigorous proof (without “tiny rectangles”) that distortions multiply the area of triangles and rectangles by $k$.
19.5 Dilation
A dilation (similitude) maps each point $A$ to a point $A’$ on ray $OA$ such that:
\[OA' = k \cdot OA\]where $O$ is the center and $k$ is the scale factor (ratio of dilation).
Properties of dilation:
- The image is similar (not congruent unless $k = 1$) to the original.
- Angles are preserved; distances are multiplied by $k$.
- Areas are multiplied by $k^2$.
- Two similar figures oriented the same way in the plane are homothetic — lines through corresponding points all pass through the center of dilation.
Example 19-6. Square $ABCD$ with $AB = 4$ is dilated about its center $O$ with factor 2 to form $A’B’C’D’$. Find $A’C$.
Solution. Diagonal $AC$ passes through $O$, so $A’$ lies on line $AC$ extended past $A$. Since $OA’ = 2 \cdot OA$:
\(A'C = A'O + OC = 2 \cdot OA + OC = 2 \cdot \frac{4\sqrt{2}}{2} + \frac{4\sqrt{2}}{2} = 4\sqrt{2} + 2\sqrt{2} = \boxed{6\sqrt{2}}\)
Example 19-7. Show that if segment $AB$ is dilated about any center with ratio 2, then the image $A’B’$ is twice as long as $AB$.
Proof. Let $O$ be the center. Then $\frac{OA’}{OA} = \frac{OB’}{OB} = 2$ and $\angle AOB = \angle A’OB’$ (same angle). By SAS Similarity, $\triangle AOB \sim \triangle A’OB’$, so $\frac{A’B’}{AB} = 2$, i.e., $A’B’ = 2AB$.
This also shows $\angle OAB = \angle OA’B’$, hence $A’B’ | AB$. $\square$
19.6 The More Things Change…
What each transformation preserves:
| Property | Translation | Rotation | Reflection | Dilation | Distortion | |—|—|—|—|—|—| | Collinearity | ✓ | ✓ | ✓ | ✓ | ✓ | | Distance | ✓ | ✓ | ✓ | ✗ | ✗ | | Angles | ✓ | ✓ | ✓ | ✓ | ✗ | | Area | ✓ | ✓ | ✓ | ✗ ($\times k^2$) | ✗ ($\times k$) | | Shape (similarity) | ✓ | ✓ | ✓ | ✓ | ✗ | | Orientation | ✓ | Changed | Changed | ✓ | ✓ |
Isometries (distance-preserving transformations) are: translations, rotations, and reflections. They produce congruent images.
Similarity transformations (angle-preserving): isometries + dilations. They produce similar images.
Distortions preserve neither distance nor angles, but they do preserve collinearity and ratios of areas.
Under distortion, circles don’t stay circles, equilateral triangles don’t stay equilateral (though triangles stay triangles since lines map to lines).
19.7 Transformation Proofs
Approach: Use the definition of each transformation plus congruence/similarity theorems. Be careful to prove things rigorously — “obvious” properties still need justification.
Example 19-9. Show that if three points are collinear, their images under any rotation are also collinear.
Proof. Let $A$, $B$, $C$ be collinear with images $A’$, $B’$, $C’$ under rotation about $O$. We need $\angle OB’A’ + \angle OB’C’ = \pi$.
Since rotation preserves angles and distances:
- $\angle A’OB’ = \angle AOB$ and $A’O = AO$, $B’O = BO$. By SAS: $\triangle A’B’O \cong \triangle ABO$.
- Similarly $\triangle C’B’O \cong \triangle CBO$.
Using corresponding angles: $\angle OB’A’ + \angle OB’C’ = \angle OBA + \angle OBC = \pi$ (since $A$, $B$, $C$ are collinear). $\square$
Example 19-10. Assuming reflections preserve segment length, show that the reflection of a circle is a circle of the same radius.
Proof. Let circle have center $O$, radius $r$, with $O’$ the reflection of $O$.
Part 1 (image points are on a circle): For any $P$ on the original circle, $OP = r$. Its image $P’$ satisfies $O’P’ = OP = r$ (distances preserved). So $P’$ lies on the circle of radius $r$ centered at $O’$.
Part 2 (every point on the new circle is an image): For any $P’$ with $O’P’ = r$, reflect $P’$ back to get $P$. Then $OP = O’P’ = r$, so $P$ is on the original circle. Hence $P’$ is the image of a point on the original circle.
Both parts together show the image is exactly the circle of radius $r$ centered at $O’$. $\square$
Exercise 19-9. Show that any segment is equal in length to its image upon reflection (“reflections preserve distances”).
Problems to Solve for Chapter 19
Problem 329. Prove that given any two points $A$ and $A’$, there exists a point $O$ such that $A’$ is the image of $A$ under reflection through $O$.
Problem 330. Let $y = mx + b$ be the image when $x + 3y + 11 = 0$ is reflected across the $x$-axis. Find $m + b$. (AHSME 1992)
Problem 331. Given $P_0 = (0,0)$ and $P_1 = (3,4)$. Reflect $P_1$ in $P_0$ to get $P_2$, reflect $P_2$ in $P_1$ to get $P_3$, etc. If $P_4 = (a,b)$, find $a + b$. (MAΘ 1991)
Problem 332. Two successive clockwise rotations about the origin of angles $x$ and $y$ ($0 < x, y < \pi$) result in a reflection through the origin. Find $x + y$. (MAΘ 1992)
Problem 333. Prove that if a rotation maps $A$ to $C$ and $B$ to $D$, then it maps segment $AB$ to segment $CD$. (Mandelbrot #3)
Problem 334. Let $\triangle ABC$ be isosceles with $AB = AC$, $\angle BAC = \alpha$, and circumcenter $O$. Prove rotations about both $A$ and $O$ carry segment $AB$ to $AC$. (Mandelbrot #3)
Problem 335. With the same triangle, suppose $M$ on $AB$ and $N$ on $AC$ satisfy $BM = AN$. Find the angle of rotation (in terms of $\alpha$) about $O$ mapping $AB$ to $AC$, and show $\angle MON$ equals this angle. (Mandelbrot #3)
Problem 336. Find the reflection of $(2, 2)$ in the line $x + 2y = 4$.
Problem 337. A circle centered at $(6, -5)$ is reflected about $y = x$. Find the center of the image. (MATHCOUNTS 1992)
Problem 338. Which does not have point symmetry: equilateral triangle, square, or regular hexagon? (MATHCOUNTS 1984)
Problem 339. If $ABCDE$ is a regular pentagon, find the smallest rotation about $E$ mapping $A$ to $D$. (MATHCOUNTS 1984)
Problem 340. Quadrilateral $ABCD$ reflected over line $\ell$ gives $A’B’C’D’$. $E$ is the intersection of lines $AB$ and $A’B’$. If $AA’ = 10$ and $A’E = 13$, find the area of $\triangle AEA’$. (MATHCOUNTS 1984)
Problem 341. If $y = 3x + 2$ is reflected with respect to the $y$-axis, what is the equation of the resulting graph? (MATHCOUNTS 1989)
Problem 342. Points $Q(9, 14)$ and $R(a, b)$ are symmetric with respect to $P(5, 3)$. Find $R$. (MATHCOUNTS 1989)