CHAPTER 11

HOMOTHETIES

11.1 The Pantograph

Isometries provide a dynamic way of dealing with congruency. In this chapter, we study a transformation which serves the same purpose for the notion of similarity.

Without employing photography, photocopying, or computer graphics, it is not a simple matter to produce an enlarged or reduced copy of a figure. There is, however, a physical instrument called a pantograph that allows us to accomplish this.

A pantograph is formed from four thin fiat rods of equal length that are joined together by four hinge pins P, Q, A, and B so that APBQ is a parallelogram and OA = AP. The instrument lies flat on the drawing board and is fixed to the board at the pivot point O. Pencils are attached to the instrument at points P and P′.

If an enlargement is desired, the pencil at P is used to trace the original feature. As this is being done, the pencil at P′ draws a copy magnified by a factor equal to OQ/OA. If a reduction is desired, the pencil at P′ is used to trace the figure so that the pencil at P draws the reduced copy.

To see why this works, note that OAP and OQP′ are similar triangles by sAs. It follows that O, P, and P′ are collinear. Moreover,

and thus the figure that is traced by P may be considered a “contraction” towards O of the figure that is traced by P′.

It should be noted that the magnification factor OQ/OA for the pantograph illustrated in the figure above is fixed. In an actual pantograph, the positions of the hinge pins at A and B may be adjusted so that OQ/OA can be set as desired. The pins A and B must be adjusted in such a manner that APBQ remains a parallelogram and so that OA = AP.

11.2 Some Basic Properties

A similarity in which a figure contracts towards a point or expands away from a point in this manner is called a homothety.

More formally, let O be a point and let k be a nonzero real number. The homothety centered at O with ratio k, denoted by H_{O, k}, maps O to O and each point P ≠ O onto another point P′ on the line OP such that

Since k ≠ 0, it follows that H_{O, k} is a transformation, and its inverse is easily seen to be H_O,1/k.

Remark. In this chapter, the word parallel includes the case where two lines coincide. In other words, two lines are considered to be parallel if there is a translation that maps one onto the other, including a translation through a distance of magnitude zero. Also, the notation is used to denote the directed segment from A to B.

A similarity is the composition of an isometry and a homothety. Thus, any similarity that is not an isometry is

Theorem 11.2.1. Let A and B be two points whose images under the homothety H_{O, k} are A′ and B′, respectively. Then A′B′ is parallel to AB and = k .

Proof.

Case 1. O, A, and B are collinear.

In this case. A′B′ and AB are both contained in the line OA, so they are parallel.

Moreover,

Case 2. O, A, and B are noncollinear (as in the figure).

If the points O, A, and B do not lie on a line, then angle O is common to both ΔOA′B′ and ΔOAB. Also,

so the triangles are similar by sAs. It follows that

and that the segments A′B′ and AB are parallel.

Corollary 11.2.2. Let A′, B′, and C′ be the images of A, B, and C, respectively, under a homothety.

Proof. Suppose the homothety is H_{O, k}.

A homothety preserves any geometric relationship that can be completely characterized by ratios of distances. For example, midpoints, centroids, and angle bisectors can all be characterized by distance ratios. This means that the midpoint of BC is mapped to the midpoint of B′C′, the centroid of ΔABC is mapped to the centroid of ΔA′B′C′, and the bisector of ∠ABC is mapped to the bisector of ∠A′B′C′. We will use these facts freely throughout this chapter.

All of these facts can be proved using Theorem 11.2.1 and Corollary 11.2.2. For example, here is a proof that the midpoint M of BC is mapped to the midpoint M′ of B′C′ by the homothety H_{O, k}.

Corollary 11.2.2 tells us that B′, C′, and M′ are collinear. Also, by Theorem 11.2.1,

so that

Thus, the image M′ of M is indeed the midpoint of B′C′.

11.2.1 Circles

Theorem 11.2.3. The image of a circle C with center C and radius r under the homothety H_{O, k} is a circle with center D = H_{O, k}(C) and radius |k| r.

Proof. The homothety maps each point X of C to a point Y such that

so the points Y form a circle centered at D with radius |k| r.

If C and are two circles, the center of any homothety that transforms one circle into the other is called a center of similitude.

Example 11.2.4. Given two circles with two different centers and two different radii, explain how to construct all centers of similitude for the circles.

Solution. The process is illustrated in the figure below.

Construct any diameter of one of the circles, say WX. Then construct a radius CY of the other circle that is parallel to this diameter. The points P and Q, where the lines WY and XY intersect the line CD through the centers of the circles, will be the centers of similitude. There are no more centers of similitude for these two circles, for if O is a center of similitude, then O must be on the line CD, and the homothety H_{O, k} that carries C to D must transform the radius CY into a parallel radius, either DW or DX. This leaves only two possible locations for O.

Theorem 11.2.5. Let P be a point on the line joining the centers A₁ and A₂ of two circles with radii r₁ and r₂, respectively. If

then P is a center of similitude of the circles.

Proof. Note that there are at most only two points P on A₁ A₂ such that

For one of them the ratio

is positive, for the other it is negative. Letting

we see that the homothety H_{P, k} maps the circle centered at A₁ to the circle centered at A₂.

Example 11.2.6. Show that any common tangent to two circles of unequal radii passes through a center of similitude.

Solution. The tangent t cannot be parallel to the the line m joining the centers A and B. Suppose that t meets m at P. Let A′ and B′ be the points of tangency, as in the figure on the following page.

Then ΔPA′A ~ ΔPB′B by AAA similarity. Consequently,

and it follows from Theorem 11.2.5 that P is a center of similitude.

11.3 Construction Problems

If one figure is the image of another under a homothety, the figures are said to be homothetic to each other. Homothetic figures are similar, but similar figures need not be homothetic, since similar figures need not be oriented the same way.

Some construction problems can be solved by first constructing a homothetic image of the figure and then enlarging or shrinking the image to get the desired solution.

Example 11.3.1. Given an acute triangle ABC, construct a square PQRS with P on AB, S on AC, and edge QR on BC.

Solution. Let P′ be any point AB. Drop the perpendicular P′Q′ to BC and complete the square P′Q′R′S′, as in figure (a) below.

If we draw the line A′C through S′ parallel to AC, we note that the square P′Q′R′S′, together with the triangle A′B′C, is homothetic to the desired solution, with B being the center of the homothety.

Thus, we need a homothet of P′Q′R′S′ so that the image of S′ is the point S on AC. To accomplish this, draw the line BS′. Then S is the point where BS′ meets AC, as in figure (b) above. Now drop the perpendicular SR to BC and complete the desired square.

Here is another example that uses the same idea.

Example 11.3.2. Construct points X and Y on the sides AB and BC, respectively, of a given triangle ABC such that

Solution, The desired result is shown in figure (a) on the following page. Take a point D on BC and a point F on AB such that AF = CD, as in figure (b) on the following page.

Draw a line l through D parallel to AC. Draw a circle with center F and radius FA, cutting l at a point G inside ΔABC. Complete the parallelogram CDGE, with E on AC. We have

Join A and G and extend the segment AG to cut BC at Y. Draw a line through Y parallel to FG, cutting AB at X. Then AFGE and AXYC are homothetic to each other with A being the center of homothety. Hence,

Example 11.3.3. Given points P and X and line segments AB and CD and letting k = CD/AB, construct the point H_{P, k}(X).

Solution. Construct a line m through P at an angle to PX and construct points D′ and B′ on m so that PB′ = AB and PD′ = CD. Join B′ to X and construct the line through D′ parallel to B′X meeting PX at Y. Therefore,

since triangles PB′X and PD′Y are similar. Thus, Y = H_{P, k}(X).

This shows that given any point X, we can construct the image of X under the homothety H_P, where k = CD/AB.

Example 11.3.3 can be considered a basic construction. It shows us how to construct the homothetic images of all standard geometric figures.

For example, to construct the image of ΔABC under H_{P, k}, we can construct the images A′, B′, and C′ of the vertices A, B, and C, respectively, under the homothety.

Also, to construct the image under H_{P, k} of a circle with center C and radius r, let X be a point on the circle and construct H_{P, k}(C) and H_{P, k}(X). These are, respectively, the center of the image circle and a point on the image circle, and we can now construct the image circle.

Example 11.3.4. Given three points A, B, and C on a circle C, explain how to find all chords through the point C that are bisected by AB.

Solution. There are two ways to solve this problem.

The first way is to construct the image m of the line AB under H_C,2, as in the figure on the right. Then for any point X′ on the line m, the point

is the midpoint of X′C. Thus, the points P and Q, where m meets the circle, will provide the chords PC and QC.

The second way is to apply H_C,1/2 to the circle, obtaining another circle , as in the figure on the right Let P′ and Q′ be the pains where intersects AB. Let P and Q be the points where CP′ and CQ′ meet C; that is,

so that, again, CP and CQ are the desired chords.

Example 11.3.5. Given two lines l and m intersecting at Q and given a point P not on either line, construct a line through P cutting l at L and m at M such that PL = 2PM.

Solution. There are two pairs of points L and M. To find the first pair L₁ and M₁, construct the line

and let M₁ = l₁ ∩ m, and then let L₂ = PM₁ ∩ l.

To find the second pair, construct the line

and let M₂ = l₂ ∩ m, and then let L₁ = PM₂ ∩ l.

Example 11.3.6. Given ∠ABC and a point P on the arm BC, construct a circle passing through P that is tangent to both arms of the angle.

Solution. There are two possible solutions to this problem.

11.4 Using Homotheties in Proofs

Recall that the four major concurrency points of a triangle are:

The incircle is the circle centered at the incenter and is internally tangent to all three sides of the triangle.

The circumcircle is the circle centered at the circumcenter that passes through the three vertices of the triangle.

Despite their suggestive names, neither the centroid nor the orthocenter is the center of any significant circle associated with the triangle.

Theorem 11.4.1. The incircle is the smallest circle that meets all three sides of a given triangle.

Proof. Let C be a circle with center P that intersects all three sides of triangle ABC. If none of the sides are tangent to the circle C, contract it using a homothety centered at P, so that the image circle intersects all sides of ABC but is tangent to at least one side. Let T be the point of tangency.

If the image circle is not tangent to two sides of triangle ABC, then contract it using a homothety centered at T, so that the new image circle still intersects all three sides of ABC and is tangent to at least two sides. Let the second point of tangency be denoted by S.

If this image circle is not tangent to all three sides of triangle ABC, then contract it again using a homothety centered at the vertex that is common to the two sides tangent to the circle.

This image circle is the incircle of triangle ABC. Since it was obtained using only contractions, it is smaller than the original circle C, unless C was already the incircle.

Example 11.4.2. (Euler’s Inequality)

Prove that R ≥ 2r, where R is the circumradius and r the inradius of a triangle. Equality holds if and only if the triangle is equilateral.

Solution. As in the figure below, let ABC be the triangle. Let D, E, and F be the midpoints of BC, CA, and AB, respectively. Let G be the centroid of ΔABC, and let C be the circumcircle.

Recalling that G is a trisection point of each median, we see that H_G,−1/2 maps A, B, and C to D, E, and F, respectively. Hence, C′, the image of C under H_G,−1/2, is the circumcircle of triangle DEF, and the radius of C′ is R/2.

Thus, C′ is a circle that has a point in common with all three sides of ABC, and the smallest such circle is its incircle, so we can conclude that R/2 ≥ r.

For equality to hold, C′ must be the incircle of ABC, touching the sides at D, E, and F. Hence, the circumcenter coincides with the incenter of ABC, which must therefore be equilateral.

Example 11.4.3. (The Euler Line)

The centroid G, the circumcenter O, and the orthocenter H of a triangle are collinear. Moreover, G is between O and H and = .

Solution. This is also proved by applying the homothety H_G,−1/2. Given triangle ABC, let D, E, and F be the midpoints of BC, CA, and AB, respectively. As in the previous example, H_G,−1/2 maps A, B, and C to D, E, and F, respectively.

The homothety maps the altitude AH of ΔABC into a line DH′ that is parallel to AH. In other words, the homothety maps the altitude from A into the right bisector of the opposite side. A similar thing happens to the other two altitudes, so the homothety maps the intersection of the altitudes to the intersection of the right bisectors. In other words, H_G,−1/2 maps the orthocenter H of ABC into the circumcenter O of ABC.

The definition of H_G,−1/2 tells us that G, H, and O are collinear and also that

Example 11.4.4. ABCD is a cyclic quadrilateral. Prove that the centroids of the triangles ABC, BCD, CDA, and DAB are cyclic, that is, that they lie on a circle.

Solution. Let K and L be the midpoints of the diagonals AC and BD, and let M be the midpoint of KL. We will prove the result by showing that the quadrilateral formed by the centroids is the image A′B′C′D′ of ABCD under H_M,−1/3. Since ABCD is a cyclic quadrilateral, it follows that the image is also a cyclic quadrilateral, since the circumcircle C of ABCD will be mapped into the circumcircle of A′B′C′D′.

Let G be the centroid of triangle ABC. We will show that G is actually D′, the image of D under H_m,−1/3.

Let P be the midpoint of BG. Then P and G trisect BK. In triangle BGD. the points P and L are the midpoints of BG and BD, and the segments PL and GD are parallel, with = .

In triangle KPL, the points G and M are the midpoints of KP and KL, and the segments PL and GM are parallel, with = .

Since GD and GM are both parallel to PL, it follows that G, M, and D are collinear. Since = and = , it follows that = or, equivalently, that .

This shows that H_M,−1/3 (D) = G, as claimed. Similarly, A′, B′, and C′ are the centroids of BCD, CDA, and DAB, respectively. Hence, the four points A′, B′, C′, and D′ lie on the circle C′, which is the image of C under H_M,−1/3.

11.5 Dilatation

Any transformation of the plane that transforms a line segment into a parallel line segment is called a dilatation or a dilation. The following theorem follows directly from the definition, and its proof is left to the reader.

Theorem 11.5.1. The collection of all dilatations of the plane is a group.

Examples of dilatations are translations, halfturns, homotheties, and the identity. It turns out that these are the only ones.

Theorem 11.5.2. A dilatation that has two fixed points must be the identity.

Proof. Suppose the dilatation T has A and B as fixed points. Let P be any point not on AB and let T(P) = P′.

Then P′A || PA and P′B || PB. so it follows that P′ is on both PA and PB, Since PA and PB have only one point of intersection, it follows that P′ = P.

Suppose now that Q is a point on AB other than A or B. Then Q is not on the line AP, and since T fixes A and P, the same proof shows that T maps Q onto itself.

This shows that T maps every point of the plane onto itself; that is, T is the identity.

Theorem 11.5.3. A dilatation is completely determined by its action on any two given points.

Proof. Suppose that the dilatations T and S map A and B to A′ and B′, respectively. Then T⁻¹ maps A′ to A and B′ to B, and so

By the previous theorem, T⁻¹S = I, and multiplying by T shows that S = T.

Given parallel lines l and m, it is possible to compare directed segments on them so that if A and B are on I and C and D are on m, the meaning of an equation such as

is unambiguous (see the figure below).

Corollary 11.5.4. Suppose that the dilatation T maps AB to A′B′. Then:

(1) = if and only if T is a translation (or the identity).

(2) = if and only if T is a halfturn.

Theorem 11.5.5. Any dilatation T that is not a translation or the identity is a homothety.

Proof. Since T is not the identity, there is at least one point A that is not fixed. Let its image be A′. Now, let B be a point that is not on the line AA′. The segment A′B′ is parallel to AB, which is not parallel to AA′, so B′ cannot be on AA′.

The lines AA′ and BB′ cannot be parallel, for if they were, AA′B′B would be a parallelogram; that is, = , which would contradict the fact that T is not a translation.

Thus, we may assume that the lines AA′ and BB′ intersect at a unique point P, as in one of the three situations depicted in the figure above. It follows that

and so

Thus, the homothety H_{P, k}, where k = , maps A to A′ and B to B′, and therefore T = H_{P, k} by Theorem 11.5.3.

11.6 Problems

2. Show that two homotheties commute if and only if they have the same center or at least one of the ratios is k = 1.

3. Show that the product of two homotheties with the same center is a homothety and find its center and ratio.

4. Show that the product of two homotheties whose ratios are k and 1/k is a translation.

5. Show that if ΔABC and ΔA′B′C are similar, with AB parallel to A′ B′, AC parallel to A′C, and BC parallel to B′C′, then the lines joining corresponding vertices are concurrent, and there is a homothety H(O, k) such that ΔA′B′C is the image of ΔABC under H(O, k). Find the center O and the ratio k.

6. Show that a homothety preserves angles between lines.

7. Show that the inverse of the homothety H(O, k) is the homothety H(O, 1/k).

8. Show that the center of a homothety of ratio k ≠ 1 is the only fixed point of the homothety and lines through the center are the only fixed lines.

9. Show that a product of three homotheties is a homothety or a translation.

10. Show that a similarity preserves angles.

11. and are two nonparallel segments such that A₁B₁ = 2A₂B₂, as in the figure below.

(a) Find a point O such that may be obtained from by means of a homothety centered at O with ratio 1/2 followed by a rotation about O.

(b) Find a line ℓ and a point O on ℓ such that may be obtained from A₁ B₁ by a homothety centered at O with ratio 1/2 followed by a reflection across ℓ.

12. Using homotheties, show that the figure formed by joining the midpoints of the sides of a square is a square having half the area of the original square.

Hint: Show that the similarity composed of a 45° rotation and a homothety of ratio OA/OA′, both with center O, maps the square ABCD onto the quadrilateral A′B′C′D′.

13. Let P be a fixed point on a circle. Using homotheties, find the locus of midpoints of all chords PA.

Hint: Consider the image of the given circle under the homothety H(P, 1/2).

14. If PT is a tangent and PAB a secant from an external point P to a circle C, show that = PT².

Hint: Reflect ΔPAT in the internal bisector of ∠P and apply the homothety H(P, PB/PT).

15. The point Q is a point inside ∠AOB, as in the figure.

Construct a line through Q intersecting OA and OB at P and Q, respectively, such that PQ = 2QR.

16. Construct ΔABC given its circumcenter O, its orthocenter H, and the vertex A, as in the figure.

17. Construct a line intersecting three given concentric circles ω₁, ω₂, and ω₃ at A, B, and C, respectively, so that AB = BC.

18. Let P be a point on side BC of equilateral triangle ΔABC that is closer to C than to B. Construct a point Q on CA and a point R on AB so that ∠RPQ = 90° and PR = 2PQ.

19. Two rods AD and BC are hinged at fixed points A and B on the ground. They are also connected to each other by means of a third rod CD, as in the figure, so that ABCD is a parallelogram.

As the hinged rods move in a vertical plane, what is the locus of the point P of intersection of AC and BD?

20. The circles ω₁ and ω₂ intersect at M and N, and A₁ is a variable point on ω₂. A₂ is the point of intersection of the line with ω₂. B is the third vertex of an equilateral triangle A₁A₂B, with the vertices in counterclockwise order. Prove that the locus of B is a circle.

21. The circles ω₁ and ω₂ intersect at M and N, and A₁ is a variable point on ω₁. A₂ is the point of intersection of the line with ω₂, and L is the point of intersection of the tangent to ω₁ at A₁ and the tangent to ω₂ at A₂. Let O₁ and O₂ be the respective centers of ω₁ and ω₂. The line through O₁ parallel to LA₁ intersects the line through O₂ parallel to LA₂ at K, as in the figure below.

Prove the following:

(a) ΔA₁ NA₂ and ΔO₁ NO₂ are similar.

(b) A₁LA₂N is acyclic quadrilateral.

(c) O₁ KO₂N is a cyclic quadrilateral.

(d) K, L, and N are collinear.

22. Two circles ω₁ and ω₂ are tangent to each other at the point T. A line through T intersects ω₁ at A₁ and ω₂ at A₂. Prove that the tangent to ω₁ at A₁ is parallel to the tangent to ω₂ at A₂.

23. The incircle of triangle ABC touches BC at D. The excircle of triangle ABC opposite A touches BC at K. The line AK intersects the incircle at two points, and we let H be the one closer to A. Prove that DH is perpendicular to BC.

24. ABCD is a quadrilateral with AB parallel to DC. The extensions of DA and CB intersect at P, and the diagonals AC and BD intersect at Q. Prove that PQ passes through the midpoints of AB and CD.

25. D, E, and F are the respective midpoints of the sides BC, CA, and AB of triangle ABC. P is a point inside ABC, K, L, and M are points such that D, E, and F are also the respective midpoints of PK, PL, and PM. Prove that AK, BL, and CM bisect one another at a common point.