A drawing doubles in width and height. Why does its area quadruple without making the depicted plot larger? Separate a figure’s transformation, a change to its picture and the measurement units.
A homothety is specified by center O and factor k≠0: multiply every vector from O to a point by k. Lengths and perimeter multiply by |k|; area multiplies by k². A negative k takes the image across the center rather than reflecting it in a line. We use similarity, areas and isometries.
The center determines where measurement starts
For k>0, X and X′ lie on the same ray from O. For 0<k<1, the image is nearer the center; for k>1 it is farther away. For k<0, use the opposite ray. Distance OX′=|k|·OX remains nonnegative. The center itself stays fixed.
For O(a;b), the formulas are:
x′ = a + k(x − a), y′ = b + k(y − b).
Only for O(0;0) do they simplify to x′=kx,y′=ky. For example, center (2;1),k=2 and point (3;2) give image (4;3), rather than (6;4). Scale the displacement from the center, rather than from an arbitrary coordinate origin.
The difference between two image coordinates is multiplied by k; the distance formula therefore multiplies distance by |k|. The similarity factor is positive s=|k|. This extends G013, now specifying the center and image position too. AMSI, scale and similarity.
The same shape on a grid
Our rectangle has vertices A(1;1),B(4;1),C(4;3),D(1;3). Its sides are 3 and 2, perimeter 10 and area 6. Lengths use one chosen unit; area uses its square.
Same center, different scale
k = 2: S = 6 → S′ = 24
k = −1/2: S = 6 → S′ = 1.5
About O(0;0),k=2 gives sides 6 and 4, perimeter 20 and area 24. Each original grid cell becomes a 2×2 block, or four cells. This shows where the square of the scale factor comes from.
Why area multiplies by k²
In a triangle, both base and perpendicular altitude multiply by s=|k|. From S=ah/2 we get S′=(sa)(sh)/2=s²S=k²S. Partitioning a simple polygon into triangles without overlap gives the same multiplier for the sum of their areas. For a disk, S=πr² shows it directly: radius multiplies by s and area by s².
Angles and ratios of corresponding lengths are preserved. A line through the center maps onto itself; a line not through the center maps to a parallel line. Homothety is one kind of similarity. To superimpose arbitrary similar figures, a further isometry may be necessary: translation, rotation or reflection. Similarity does not guarantee a single common homothety center for corresponding vertices already placed in the plane.
A negative factor does not make negative area
For k=−1/2, our rectangle maps to (−1/2;−1/2),(−2;−1/2),(−2;−3/2),(−1/2;−3/2). Its sides are 1.5 and 1, perimeter 5 and area 1.5. Lengths halve and area becomes one-quarter.
A negative homothety is positive scaling by |k| together with a half-turn about O. In the plane this preserves the orientation of vertex traversal, unlike reflection in a line. For k=−1 we obtain the central symmetry of G022; for k=1, the identity. Both are isometries. Other values of |k| change distances. Transformations teaching notes, section4.
We exclude k=0: all points would collapse to O, shape would be lost and the operation could not be reversed. For k≠0, the inverse homothety has the same center and factor 1/k.
From area back to length
If a similar figure’s area is 9 times larger, its lengths are √9=3 times larger, rather than 9 times. Area alone does not tell us the sign of k: both k=3 and k=−3 give area factor 9. Locating a homothety also requires corresponding point positions relative to the center.
For our rectangle, initial area 6 and desired area 24 give s=√(24/6)=2. This agrees with perimeter 20, twice the original 10. Volumes of similar solids need a separate argument; we return to them in the spatial part of the course.
The drawing and the plot are different objects
Our own example: on a 1:250 plan a plot occupies 5.6 cm². Its real area is 5.6·250²=350,000 cm²=35 m². Apply the factor twice, then convert square centimetres to square metres.
If printing actually doubles both linear dimensions of the plan, the paper area becomes 22.4 cm² and the scale becomes 1:125. Calculating 22.4·125² again gives 350,000 cm², or 35 m². The plot has not changed. Browser zoom likewise does not prove any change in the object’s physical dimensions; check the actual print scale before measuring paper.
Using coordinatesOptional · safe to skip
Find the center when k is known
For k≠1 and known X(x;y),X′(x′;y′), the formulas give a=(x′−kx)/(1−k),b=(y′−ky)/(1−k). With X(3;2),X′(4;3),k=2 we obtain O(2;1). Other corresponding point pairs must give the same center. One correct pair alone does not prove a homothety of the whole figure. At k=1 the center formula fails: the identity does not single out one center.