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Compression of a Geometric Figure — Definition & Examples

Compression of a Geometric Figure
Contraction of a Geometric Figure

A transformation in which all distances are shortened by a common factor. This is done by contracting all points toward some fixed point P.

Note: The common factor is less than 1 for a contraction. When the common factor is greater than 1 the transformation is called a dilation.

A large pentagon ABCDE (pre-image) and a smaller, vertically-compressed pentagon A′B′C′D′E′ (image), with dashed rays from a centre point P showing the compression transformation.

Key Formula

P′=O+k(P−O)P' = O + k(P - O)
Where:
  • PP = Any point on the original (pre-image) figure
  • P′P' = The corresponding point on the compressed (image) figure
  • OO = The fixed center point of the compression
  • kk = The scale factor, where 0 < k < 1 for a compression

Worked Example

Problem: A triangle has vertices A(2, 6), B(8, 4), and C(4, 0). Apply a compression with scale factor k = 0.5 centered at the origin O(0, 0). Find the vertices of the compressed triangle.
Step 1: Write down the compression formula. Since the center is the origin, the formula simplifies to P' = kP.
P′=k⋅PP' = k \cdot P
Step 2: Apply the formula to vertex A(2, 6) with k = 0.5.
A′=0.5⋅(2,6)=(1,3)A' = 0.5 \cdot (2, 6) = (1, 3)
Step 3: Apply the formula to vertex B(8, 4) with k = 0.5.
B′=0.5⋅(8,4)=(4,2)B' = 0.5 \cdot (8, 4) = (4, 2)
Step 4: Apply the formula to vertex C(4, 0) with k = 0.5.
C′=0.5⋅(4,0)=(2,0)C' = 0.5 \cdot (4, 0) = (2, 0)
Step 5: Verify by checking a distance. The original distance AB and the new distance A'B' should satisfy A'B' = k · AB.
AB=(8−2)2+(4−6)2=40,A′B′=(4−1)2+(2−3)2=10=0.5⋅40  ✓AB = \sqrt{(8-2)^2 + (4-6)^2} = \sqrt{40}, \quad A'B' = \sqrt{(4-1)^2 + (2-3)^2} = \sqrt{10} = 0.5 \cdot \sqrt{40} \; \checkmark
Answer: The compressed triangle has vertices A'(1, 3), B'(4, 2), and C'(2, 0). Every distance in the figure is halved.

Another Example

This example uses a center of compression that is not the origin, showing how the figure shrinks toward an arbitrary fixed point. The center O(3, 3) is the center of the original square, so the compressed square is concentric with the original.

Problem: A square has vertices at P(1, 1), Q(5, 1), R(5, 5), and S(1, 5). Apply a compression with scale factor k = 0.25 centered at the point O(3, 3). Find the image vertices.
Step 1: Use the full compression formula since the center is not the origin. For each point, compute P' = O + k(P − O).
P′=(3,3)+0.25⋅((x,y)−(3,3))P' = (3, 3) + 0.25 \cdot \bigl((x, y) - (3, 3)\bigr)
Step 2: Compress vertex P(1, 1). Subtract the center, scale, then add the center back.
P′=(3,3)+0.25(−2,−2)=(3,3)+(−0.5,−0.5)=(2.5,  2.5)P' = (3, 3) + 0.25(-2, -2) = (3, 3) + (-0.5, -0.5) = (2.5,\; 2.5)
Step 3: Compress vertex Q(5, 1).
Q′=(3,3)+0.25(2,−2)=(3,3)+(0.5,−0.5)=(3.5,  2.5)Q' = (3, 3) + 0.25(2, -2) = (3, 3) + (0.5, -0.5) = (3.5,\; 2.5)
Step 4: Compress vertices R(5, 5) and S(1, 5) the same way.
R′=(3,3)+0.25(2,2)=(3.5,  3.5),S′=(3,3)+0.25(−2,2)=(2.5,  3.5)R' = (3, 3) + 0.25(2, 2) = (3.5,\; 3.5), \quad S' = (3, 3) + 0.25(-2, 2) = (2.5,\; 3.5)
Step 5: Check the side length. The original side was 4 units; the compressed side should be 0.25 × 4 = 1 unit.
P′Q′=3.5−2.5=1  ✓P'Q' = 3.5 - 2.5 = 1 \; \checkmark
Answer: The compressed square has vertices P'(2.5, 2.5), Q'(3.5, 2.5), R'(3.5, 3.5), and S'(2.5, 3.5), with side length 1.

Frequently Asked Questions

What is the difference between compression and dilation of a geometric figure?
Both are the same type of transformation — scaling from a fixed center point. The only difference is the scale factor. A compression uses a factor k with 0 < k < 1, which makes the figure smaller. A dilation uses a factor k > 1, which makes the figure larger. When k = 1, the figure stays unchanged.
Does compression change the shape of a geometric figure?
No. Compression is a similarity transformation, meaning it preserves the shape, angle measures, and proportions of the figure. Only the size changes — all lengths are multiplied by the same factor k. The image is always similar to the pre-image.
What happens to the area when you compress a 2D figure?
The area of the compressed figure equals the original area multiplied by k². For example, if k = 0.5, the new area is 0.25 (one-quarter) of the original area. This is because area depends on two dimensions, and each dimension is scaled by k.

Compression (Contraction) vs. Dilation (Expansion)

Compression (Contraction)Dilation (Expansion)
Scale factor0 < k < 1k > 1
Effect on sizeFigure gets smallerFigure gets larger
FormulaP' = O + k(P − O) with 0 < k < 1P' = O + k(P − O) with k > 1
Effect on shapePreserved (similar figure)Preserved (similar figure)
Effect on area (2D)Area decreases by factor k²Area increases by factor k²
Fixed pointCenter point O stays in placeCenter point O stays in place

Why It Matters

Compression appears throughout geometry courses whenever you study similarity transformations, scale drawings, and coordinate geometry problems. It is also essential in computer graphics and animation, where objects must be resized smoothly relative to a reference point. Understanding compression helps you solve problems about similar figures, map scales, and the relationship between linear scale factors and area or volume changes.

Common Mistakes

Mistake: Forgetting to subtract and then re-add the center point when the center of compression is not the origin.
Correction: Always use the full formula P' = O + k(P − O). Only when O is the origin does this simplify to P' = kP. If you skip the translation steps, your image will be placed in the wrong location.
Mistake: Applying the scale factor to the area directly instead of squaring it.
Correction: If the scale factor is k, lengths multiply by k, areas multiply by k², and volumes multiply by k³. For instance, a compression with k = 0.5 reduces the area to 0.25 of the original, not 0.5.

Related Terms