Overview of Section 5.4 Similarity transformations and constructions

A transformation Sk of the plane that multiplies all distances by the same positive constant k is a similarity of the plane.

The constant k is called a scale factor, or similarity ratio, or similarity coefficient, or . . .

If k = 1, the transformation is an isometry. If 0 < k and k ≠ 1, then the transformation Sk is not an isometry.

A dilation or dilatation is a similarity transformation with center O, k > 0, assigns to each point P a point P' such that OP' = kOP (Note typo in book where this is given as OP' = kOA). If k = 1, P is mapped to itself.

O is called the center of similitude or the center of dilation.

Why do you suppose the scale factor was defined to be a positive number? Note that we are dealing with distance.

For a dilation (or homothety or homothecy) the transformation is usually defined for k < 0 as well. When k < 0, P is mapped to P' on the opposite side of point O such that OP' = |k|OP.


Now Solve This 5.10. Justify.

1. A dilation is a similarity transformation.

2. With a dilation:

a. A segment AB not belonging to the line through the center O is transformed on to the segment A'B' parallel to AB

b. A line through O is transformed onto itself, and line not through O is transformed into a parallel line.

c. Every triangle is transformed to a similar triangle, and consequently angle are transformed into congruent angles.

d. A circle is transformed into a circle. The image of the center of the circle is the center of the transformed circle, and the ratio of the radius of the transformed circle to the radius of the given circle is equal to the scale factor of the dilation.

e. In a coordinate plane, if the center of the dilation is the origin, then the image of point (x,y) is (kx, ky).


Now Solve This 5.11

1. a. Let Sk be a similarity translation with a scale factor k, and let C1/k be a dilation about some point O.

Prove that Sk º C1/k is an isometry.

C1/k maps every point P to a point on ray OP' such that OP' = OP. Since the point O for the dilation can be chosen at any point in the plane, let O be at A. Then by C 1/k AB is mapped to A'B' such that A'B' =AB. Now Sk maps that image so that distances are multiplied by the constant k. Therefore Sk º C1/k maps AB to an image with the same length as AB and hence it is an isometry.

Hence any similarity transformation is a composition of a dilation followed by an isometry?

1. b. Is it true that a similarity transformation is a composition of an isometry followed by a dilation? Justify.

2. Prove that the centers of dilation of three triangles that are pairwise centrally similar are collinear. See proof process in book.










Theorem: Any Two Circles with Different Radii are similar.

Essentially the proof is to show that dilations exist between any two given circles.

Given two circles and their centers P1 and P2.

Construct diameter A2B2.Construct parallel to A2B2 through P1. The intersection O1 of lines P1P2 and A1A2 will be the center of dilation for similar triangles O1A1P1 and O1A2P2.

In a like manner, we could find another center of similitude by considering the point O2 at the intersection of P1P2 and A1B2. In this case triangles O2A1P1 and O2B2P2 are similar.

In each case the scale factor or similarity coefficient is the ratio of the two radii.




The cases when the circles are not disjoint need to be explored.


Construction of a common tangent to two circles.

Simply, locate the center of dilation. Then construct a tangent from that center to one of the circles. The constructed tangent must be tangent to the second circle. The second tangent can be found by a reflection in P1P2 or by constructing the other intersection point.


If O2 is the center of similitude, construct a tangent to one of the circles, as above, but now we have the internal common tangent.







A GSP file is available if you need it.

The usual 'textbook' proof of the Euler Line Theorem uses the following construction. The proof makes use of similar triangles.












Problem Set 5.4