Euclidean Geometry - Homothety
A homothety, also known as a dilation, is an affine transformation of the plane, determined by a point and a ratio that sends any point to a point called the image of such that
- If this transformation is known as an expansion.
- If the homothety is the identity transformation.
- If this transformation is known as a contraction.
Homothety is a simple concept with surprisingly sophisticated consequences. It is useful for proving collinearity, concurrency, determining ratios, and constructing points.
Contents
Representations
There are several ways to express a homothety. Switching between these different representations confer different benefits and can greatly simplify the problem, as will be observed later.
1) Geometric interpretation
A homothety is just an expansion or contraction. The facts which apply to those situations also apply here. Here are examples of
- applied to a triangle
- applied to a circle , where is a point on the circumference:
- applied to the line , where does not lie on
2) Euclidean geometry (simplified)
- In the case that the center of homothety is the origin, an expansion by would bring the point to . This is the simplest case and often makes calculations very direct.
- In the case that the center of homothety is the point an expansion by would bring the point to . This expression is slightly more complicated, which is why setting is often more helpful.
3) Complex numbers
In the argand plane, if the center of homothety is represented by the complex number , then an expasion by would bring the complex number to .
4) Vectors
As stated above, if the center of homothety is represented by the vector , then an expansion by would bring the vector to .
5) Abstract identification
Given familiarity with the applications of homothety, we can identify a homothetic transformation and apply the results.
Important Results
Proof of the results can be viewed by clicking on the "Proof" button.
The points and form a straight line.
This follows directly from the geometric interpretation.
The slope of a line is preserved under homothety. Hence, parallel lines are preserved.
We will use the simplified Euclidean geometry approach, where we let the center be the origin.
Suppose we have the points and .
Then, and .
We can verify that the slopes are equal:
Note: The general case can easily be proven using vectors or complex numbers to keep the mathematics simple.
Corollary: Since the slope of a line is preserved, if two lines are parallel, then their images are parallel to the original lines, and thus the images are parallel.Angles are preserved, meaning that .
Since the slopes of lines are preserved (2), the angle between and is equal to the angle between and .
Every homothety has an inverse: .
This follows directly from the geometric interpretation.
The ratio of lengths satisfy .
We will use the simplified Euclidean geometry approach, where we let the center be the origin.
Suppose we have the points and .
Then, and .
Hence, the distances obey
Note: Using the interpretation of directed lengths, we have .The image is similar to the original object.
Since angles are preserved (3) and the ratio of lengths is a constant (5), the image is similar.
The ratio of areas satisfy .
Since the image is similar and the ratio of lengths is , the ratio of areas is .
Note: Using the interpretation of directed area, we have . In particular, orientation is preserved.
is the identity transformation.
This follows directly from the geometric interpretation.
is a reflection through the center, and is also a rotation about the center.
This follows directly from the geometric interpretation.
has 1 fixed point.
We will use the complex number interpretation. Suppose that is a fixed point, meaning that . Then we have . Since , is the only fixed point.
The set of lines that remain invariant under are lines through the center .
We will use the vector interpretation with homothety . Suppose that the line is preserved. Since slopes are preserved, this means that the point must be mapped onto this line, so there exists some such that . Hence, is parallel to , meaning that lies on the line.
Conversely, given any line through the center of homothety, if follows from the geometric interpretation that the line is preserved.A homothety is defined uniquely by where any 2 points are mapped to.
Geometric interpretation: Suppose that the points are mapped to . Then, since the center lies on both and , this can be determined by the intersection of the lines. After that, the scaling factor can be determined by .
Complex number interpretation: Suppose that the complex numbers are mapped to . Then, we need to find a complex number and a real number such that
If this system has no solutions, then we do not have a homothety.
If this system has a unique solution, then that uniquely defines the homothety.
If this system has infinitely many solutios, then this must be the identity, which is unique.
Note: We can solve this system to obtain and .Any 2 (non-congruent) circles have 2 centers of homothety. The positive scaling factor corresponds to the direct homothety, and the negative scaling factor corresponds to the indirect homothety.
We know that the centers of the circle must be mapped onto each other, and thus the scaling factor must satisfy . The center of homothety lies on the line containing these circles. For the negative ratio, the center of homothety will lie on the line segment between the 2 centers. For the positive ratio, the center of homothety will lie outside of the line segment between the 2 centers.
If , then is a homothety.
We will use the complex number interpretation: In order to make this a homothety, we must find .
Equating coefficients of , we obtain .
Equating the constant term, we obtain . Since , we can set .
Note: If we used the geometric interpretation, then this (and the following) result would not be immediately obvious.If , then is a translation.
We will use the complex number interpretation. Since , the previous expression looks like . Thus, this is a translation by the complex number .
In general, it need not be true that . This is true only if , or if either homothety is the identity.
We will use the complex number interpretation: The coefficient of is equal to . Comparing the constant terms, we have equality if and only if (after some simplification) . This correspeonds to 1) the 2 centers of homothety are identical, 2) the first homothety is the identity, 3) the second homothety is the identity,
If 3 non-congruent sets of points are pairwise homothetic, then the (direct) centers of homothety are collinear.
Suppose the homotheties are for .
By (12), it follows that .
The line is invariant under , and again under , and hence it is invariant under .
By (11), it follows that lies on the line .
Thus these 3 points are collinear.
Hint: Find an invariant line.
Problems
Consider 2 non-intersecting, non-congruent circles.
Show that the external tangents intersect at the external center of homothety.
Show that the internal tangents intersect at the internal center of homothety.
Let the external tangents intersect at .
Let an external tangent touch circles at points respectively.
Consider the external homothety applied to point .
It has to be mapped to some point on .
Consider the external tangent at By (11), since the line passes through , it is invariant under the map. Hence, the only possible point that it gets mapped to is . Hence, by (1), points lie on a straight line.
Note: The internal version of the problem follows using a similar argument.Circles and are tangential at . Draw a line through cutting the circles at and respectively.
Show that the tangent at is parallel to the tangent at .
Let the center of be and the center of be . Because is the center of homothety between the two circles, : = : = : = : . Therefore, =. Since the angles the radii make with the tangents are both 90 degrees, the angles that line makes with the tangents to both and are equal by simple angle subtraction. By transversal properties, this shows that the tangents at and are parallel.
Using a straightedge and compass, in triangle where and are acute, inscribe a square whose edge lies on .
Construct square such that is constructed on the side of opposite to point . Let be the point where intersects and let be the point where intersects . Then, construct rectangle inscribed in triangle . Since triangles and are similar with the same orientation, points and can be obtained by applying to points and respectively. Similarly, since triangles and are similar with the same orientation, points and can be obtained by applying the same homothety, , to points and respectively, noting that = and =. Since rectangle is homothetic with square , is a square, and we are done.
Any 2 non-congruent triangles whose pairs of sides are parallel to each other and are oriented similarly can be mapped by a homothety.
Hint: First show that these triangles are similar. Then guess the center of homothety.
Corollary: [Centroid] Draw the medians of triangle , where points lie on the opposite sides.
Show that the medians of a triangle are concurrent at .
Show that .
Consider 3 non-congruent circles.
Show that the (possibly indirect) centers of homothety are collinear if and only if the scaling factors satisfy .
Corollary: [Monge's theorem] Given 3 non-congruent, non-overlapping circles, the 3 intersection points of the external tangents taken pairwise are collinear.
In circle , draw chord . Circle is internally tangent to at , and tangential to at . Let be the midpoint of arc that does not contain .
Show that is a straight line.
Call the center of , the radius of , the center of , and the radius of . Since is the midpoint of arc , the tangent at is parallel to . Also, it is clear that the homothety maps circle to circle . Therefore, , , and are collinear. Since and are perpendicular to parallel lines, and are parallel. Using line as a transversal, =. Also, : = : = : . By SAS similarity, triangle is similar to triangle . Since , , and are collinear and = , , , and must be collinear as well.
In a large circle , draw 2 smaller circles and that are internally tangent to at points and respectively. Draw an external tangent of and intersecting these circles at and respectively.
Show that and intersect on the circumference of , which we will denote by .
Show that is a cyclic quadrilateral and that . In fact, is the radical center of which 3 circles?
Let be the center of , be the center of , and be the center of . Since is internally tangent to , , , and are collinear. Construct a tangent line to that is parallel to . Let the point of tangency be . Since the constructed tangent line and are parallel, = . Also, it is clear that :=:, since = and =. Therefore, by SAS similarity, is similar to . Therefore, =, and it follows that , , and are collinear by transversal properties. We can similarly show that , , and are collinear, and therefore and intersect on the circumference of , and is . Afterthought: Another way to show that is similar to would be through homothety.
Given 2 non-overlapping circles, let be the internal tangent with on and on . Let be the indirect center of homothety. Draw diameters and .
Show that is a straight line.
Let be the center of and be the center of . If is the radius of and is the radius of , the ratio of all homothetic transformations of points between the two circles is :. Since is the center of homothecy, : = : = : = :. Therefore, is similar to . Therefore, = . Therefore, is a straight line.
Consider a semicircle with diameter . A circle is internally tangent to the circumference at and touches the diameter at .
Calculate .
Call the midpoint of , also the center of the semicircle, . Call the center of the internally tangent circle . Let us call the measure of . We know that , and that . Since is right, . Since the smaller circle is internally tangent to the larger semicircle at point , , , and are collinear. Therefore, . Since triangle is an isosceles triangle, . Finally, degrees.
In a large circle , draw 2 smaller circles and that are internally tangent to at points and . Let the direct center of homothety of and be .
Show that is a straight line.
Note: Does the indirect center of homothety lie on this line?
Let the center of be , the center of be , and the center of be . Without loss of generality, let be closer to point than . Draw ray until it intersects for the first time. Let us call this intersection point . Since is the direct center of homothety, . Since circles and are internally tangent to , points , , and , and points , , and are collinear. By transversal properties, which contains is parallel to . Therefore, . Therefore, by SAS similarity, triangle is similar to triangle . Therefore, . Since , , and therefore ray contains .
is a quadrilateral. The line through parallel to meets at , and the line through parallel to meets at . Prove that is parallel to .
Hint: Let the homothety that maps to be , and the homothety that maps to be . Show that .
Let the intersection of the diagonals of the quadrilateral be . Then, triangles and are similar, and straight lines connect corresponding points, so there exists a homothety that maps to and to . In the same way, through the similarity of triangles and , we can find a homothety that maps to and to . It follows that the homothety maps to and to . Since homothetic transformations of lines end up parallel to the original lines, is parallel to .
Two circles are externally tangent at . Their common external tangents meet at , and the circles are homothetic via . Take a point on the circumference of the circle, not on the line .
Calculate .
Let us call the center of the circle closer to , and the center of the other circle . Without loss of generality, let point be on the circle closer to , and be on the other circle. First, we know that , , and are collinear. Properties of homothetic transformations tell us that is parallel to . Set and . Since triangle is isosceles, . Therefore, by transversal properties, . Since triangle is isosceles, . Since , by transversal properties, , and therefore by simple angle subtraction. We now know two angles in triangle , and we can easily find that measures 90 degrees.
In a large circle , there are 3 circles that are internally tangent at points respectively, where lie on the circumference in that order. Suppose that are externally tangent with tangent and are externally tangent with tangent .
Show that and intersect on if and only if share a common exterior tangent.
In triangle , let the incircle be tangential to at . Let be the diameter of the incircle. Let intersect at .
Show that .
(IMO 1978) In isosceles triangle , . A circle is tangent internally to the circumcircle of , and also to sides and at and respectively.
Show that the midpoint of segment is the incenter of .
In a large circle of diameter 4, we draw two circles that are each internally tangent at and We then draw the common interior tangent If intersects the large circle at and intersects the large circle at what is the distance