Homothety Aditya Ghosh

An Introduction to Homothety Aditya Ghosh

A homothety (or dilation) is just an expansion or contraction, in which a figure is ‘zoomed’ (in/out) from a center. For instance, figure1 shows the effect of a homothety centred at O. In the left panel, the triangle ABC is magnified by a factor of 2 to become triangle A0B0C0. We say that 4A0B0C0 is the image of 4ABC under a homothety centred at O with scale factor 2. Mathematically we have OA0 = 2OA, OB0 = 2OB etc. In contrast, figure2 shows the effect of a homothety centred at O with scale factor −2, i.e. OP 0/OP = −2 for each point P . The minus sign indicates that P 0 is outside the segment OP for any point P (e.g. A0 is outside the segment OA).

Figure 1: A homothety centred at O with Figure 2: A homothety centred at O with scale factor 2 that sends ABC to A0B0C0 scale factor −2 that sends ABC to A0B0C0

More formally, a homothety h is a transformation on the plane defined by a center O and a real number k, which sends any point P to another point P 0 = h(P ), such that OP 0 OP = k. The number k is the scale factor. Note that the transformation is meaningful for any point other than O itself; we either leave h(O) as undefined, or set h(O) = O as a convention. The homothety centred at O with scale factor k is sometimes denoted by H(O, k). We emphasize again on the fact that this k can be negative, indicating that the image P 0 lies on the ‘other side’ of O w.r.t. point P. Here is an applet where you can use the slider to change the scale factor k and see the effect of the homothety: https://www.geogebra.org/m/DMkPrGTR Let us consider another simple example. Suppose that D,E,F are the midpoints of the sides BC, CA, AB respectively of 4ABC (see the adjacent diagram). The homothety with centre A and ratio 2 sends 4AF E to 4ABC. On the other hand, the homothety with centre at the centroid G and ratio −1/2 sends 4ABC to 4DEF.

If two triangles are similar, is it necessary that they are homothetic (i.e. there exists a homothety that maps one of them to the other)? No. For instance, draw two arbitrary equilateral triangles ABC and DEF such that AD, BE, CF does not concur.

1 Following are some properties of homothety. 1. The image of any object under a homothety is similar to the original object. 2. A homothety is deﬁned uniquely by where any 2 points are mapped to. 3. The slope of a line is preserved under homothety. Hence, parallel lines are preserved.

0 0 0 4. Angles are preserved, meaning that ∠ABC = ∠A B C . 5. H(O, −1) is a reﬂection through the center, and is also a 180◦ rotation about the center. 6. For k 6= 1, the set of lines that remain invariant under H(O, k) are precisely the lines that pass through the center O.

7. If k1k2 6= 1, then H(O1, k1) ◦ H(O2, k2) is a homothety.

8. If k1k2 = 1, then H(O1, k1) ◦ H(O2, k2) is a translation.

9. In general, it need not be true that H(O1, k1) ◦ H(O2, k2) = H(O2, k2) ◦ H(O1, k1).

This is true only if O1 = O2, or if either homothety is the identity. 10. If 3 non-congruent sets of points are pairwise homothetic, then the (direct) centers of homothety are collinear. Some of the above results can be proved using complex numbers. Note that the image of point z under the homothety H(a, k) is given by z0 = k(z − a) + a. You may also visit the following webpage: https://brilliant.org/wiki/euclidean-geometry-homothety/ Let us now dive into some applications of the idea of homothety.

Example 1. Chord AB is given in a circle Ω. Let ω be a circle tangent to chord AB at K and internally tangent to ω at T. Then show that ray TK passes through the midpoint M of arc AB of Ω, not containing T.

Solution. Let us assume that TK meets the arc AB at N and we shall show that AN = NB. Since Ω and ω are tangent at T, it follows there is a homothety at T taking ω to Ω. Clearly, the tangent to ω at K (which is line AB) will be mapped to the tangent to Ω at N. Hence it follows that the tangent to Ω at N is parallel to AB. Now AN = NB follows by observing that the three colored angles in the adjacent diagram are equal.

For an animation of the homothety, visit https://www.geogebra.org/geometry/g4prfqcm (use the slider to adjust the scale factor).

2 Remark. There are some more interesting results in the above conﬁguration, e.g. 4TMB is similar to 4BMK, which in turn implies that MK · MT = MB2. Look up section 4.7 of the book EGMO (by Evan Chen).

Example 2. (Nine point circle) The three midpoints of the sides of the triangle, the three feet of the altitudes of the triangle, and the three midpoints from the vertices to the orthocenter of the triangle are concyclic.

Let us first recall the following facts: (i) The reflection of H in the feet of the altitude K falls on the circumcircle (ABC). (ii) The reflection of H in the midpoint D of BC is nothing but the point A0, which is the reflection of A in the circumcentre O. Figure3 illustrates these two facts. These two to- gether tell us that the homothety centred at H with scale factor 2 sends K and D on the circumcircle of ABC. 0 We can also say that the points H1 and A lying on (ABC) are sent to K and D respectively, under 1 Figure 3: Two famous facts the homothety centred at H with scale factor 2 .

Clearly, similar argument applies to other vertices/sides as well. Let us now look at ﬁgure4. 1 The homothety centred at H with scale factor 2 0 0 0 sends the points H1,H2,H3,A ,B ,C , A, B, C to the nine points K, L, M, D, E, F, P, Q, S respectively. Since the former set of 9 points lie on (ABC), their images must also lie on a circle (which is in fact the image of (ABC) under this homothety). If N be the centre of the smaller circle (the nine- point circle) then the homothety also yields that N is the midpoint of OH. Furthermore, we can tell that the radius of the nine-point circle is half of the Figure 4: The nine-point circle radius of the circumcircle.

For an animation of this homothety, visit https://www.geogebra.org/geometry/mnzykf2n and use the slider to adjust the scale factor.

3 Example 3. (Diameter of the incircle lemma) Let the incircle of triangle ABC touch side BC at D, and let DT be a diameter of the incircle. If line AT meets BC at X, then show that BD = CX.

Assume w.l.o.g that AB ≤ AC. Consider the dilation with center A that carries the incircle to the A-excircle. The line segment DT is the diameter of the incircle that is perpendicular to BC, and therefore its image under the dilation must be the diameter of the excircle that is perpendicular to BC. It follows that T must get mapped to the point of tan- gency between the excircle and BC. In addition, the image of T must lie on the line AT, and hence T gets mapped to X. Thus, the excircle is tangent to BC at X. From here it is easy to show that BD = CX, I leave that to the reader. For an animation of the homothety, visit https://www.geogebra.org/geometry/fgrbjqhz The above lemma is important. We shall see few of its applications in this note itself.

Example 4. Three circles of equal radius have a common point O and lie inside a given triangle. Each circle touches a pair of sides of the triangle. Prove that the incentre and the circumcentre of the triangle are collinear with the point O.

Solution. Let A0,B0,C0 be the centres of the circles. Since the radii are the same, so A0B0 is parallel to AB, B0C0 is parallel to BC,C0A0 is parallel to CA. Since AA0,BB0,CC0 bisect ∠A, ∠B, ∠C respectively, they concur at the incentre I of 4ABC. Note O is the circumcentre of 4A0B0C0 as it is equidistant from A0,B0,C0. Then the homothety with center I sending 4A0B0C0 to 4ABC will send O to the circumcentre S of 4ABC (not shown in the above diagram). Therefore, I,O,S are collinear.

4 Example 5. Let I,O be the incentre and the circumcentre of triangle ABC, and D,E,F be the circumcentres of triangles the BIC, CIA, AIB respectively. Let P, Q, R be the midpoints of segments DI,EI,FI. Prove that the circumcentre of triangle P QR, say M, is the midpoint of segment IO.

Solution. The key idea here is that the circumcentres D,E,F lie on (ABC). To see why, 0 note that if the internal bisector of ∠BAC intersect (ABC) at D , then simple angle chasing gives D0B = D0C = D0I which tells us that D0 is nothing but the circumcentre D of 4BIC. Thus the circumcentre D lies on (ABC). Similar argument holds for E and F as well.

Now consider the homothety centred at I with factor 2 that sends the midpoints P, Q, R to D,E,F respectively. Since this homothety sends (P QR) to (ABC), it must send IM IP 1 M to O. Furthermore, MO = ID = 2 , which completes the proof.

5 Example 6. In a triangle ABC we have AB = AC. A circle which is internally tangent with the circumcircle of the triangle is also tangent to the sides AB, AC in the points P, respectively Q. Prove that the midpoint of PQ is the centre of the incircle of the triangle ABC.

Solution. Let O be the centre of the circle. Let the circle be tangent to the circumcircle of 4ABC at D. Let I be the midpoint of PQ. Then A, I, O, D are collinear by symmetry. Consider the homothety with centre A that sends 4ABC to 4AB0C0 such that D is on B0C0. The scaling factor here is k = AB0/AB. Now since the right triangles AIP, ADB0, ABD, and AP O are similar,

AI AI AP AD AB AB 1 = = = = . AO AP AO AB0 AD AB0 k This tells us that the homothety sends I to O. Then O being the incentre of 4AB0C0 implies that I is the incentre of 4ABC. Alternate approach: Example 1 tells us that DP and DQ must meet the arcs AB and AC at their midpoints, say at X and Y. Now applying Pascal’s theorem one can show that CX and BY (which are angle bisectors of 4ABC) must meet on P Q.

Example 7. Triangle ABC has orthocentre H, incentre I and circumcentre O. Let K be the point where the incircle touches BC. If IO is parallel to BC, then prove that AO is parallel to HK.

6 Solution: Let KE be a diameter of the incircle, and let line AE meet BC at D. Let M be the midpoint of BC.

By the diameter of the incircle lemma, M is also the midpoint of KD. Since IO is parallel to BC, we can say that KMOI is a rectangle. Since I is the midpoint of KE and M is the midpoint of KD, we see that1 O must be the midpoint of ED. Thus lines AE and AO coincide.

The rest is easy. Since AH = 2OM = EK and AH and EK are both perpendicular to BC, it follows that AHKE is a parallelogram. Hence HK is parallel to AE, which coincides with line AO.

1In view of midpoint theorem, the line through midpoint M parallel to KE must pass through the midpoint of DE. Similarly, the line through midpoint I parallel to KD must also pass through the midpoint of DE. In other words, these two lines must intersect each other at the midpoint of DE. But they intersect at O! So not only O has to lie on DE, it must also be the midpoint of DE.

7 Problems for the reader

1. (Monge’s thm) Consider disjoint circles ω1, ω2, ω3 in the plane, no two congruent. For each pair of circles, we construct the intersection of their common external tangents. Prove that these three intersections are collinear. 2. (IMO 1983) Let A be one of the two distinct points of intersection of two unequal

coplanar circles C1 and C2 with centers O1 and O2 respectively. One of the common

tangents to the circles touches C1 at P1 and C2 at P2, while the other touches C1 at

Q1 and C2 at Q2. Let M1 be the midpoint of P1Q1 and M2 the midpoint of P2Q2. Prove that ∠O1AO2 = ∠M1AM2. 3. (IMO 1992) In the plane let C be a circle, ` a line tangent to the circle C, and M a point on `. Find the locus of all points P with the following property: there exists two points Q, R on ` such that M is the midpoint of QR and C is the inscribed circle of 4P QR. 4. Let ABC be a triangle. Let the excircle of ABC opposite to A touch side BC at D. Similarly deﬁne E on AC and F on AB. First show that AD, BE, CF concur at a point N (known as the Nagel point). Let G be the centroid of ABC and I the incentre of ABC. Show that I, G, N lie in that order on a line (known as the Nagel line), and GN = 2IG.

5. (USAMO 2001) Let ABC be a triangle and let ω be its incircle. Denote by D1 and

E1 the points where ω is tangent to sides BC and AC, respectively. Denote by D2

and E2 the points on sides BC and AC, respectively, such that CD2 = BD1 and

CE2 = AE1, and denote by P the point of intersection of segments AD2 and BE2.

Circle ω intersects segment AD2 at two points, the closer of which to the vertex A is

denoted by Q. Prove that AQ = D2P.

6. (IMO 1999) Two circles Ω1 and Ω2 touch internally the circle Ω in M and N and the

centre of Ω2 is on Ω1. The common chord of the circles Ω1 and Ω2 intersects Ω in A

and B. MA and MB intersects Ω1 in C and D. Prove that Ω2 is tangent to CD.

7. (IMO 1982) A non-isosceles triangle A1A2A3 has sides a1, a2, a3 with the side ai lying

opposite to the vertex Ai. Let Mi be the midpoint of the side ai, and let Ti be the

point where the inscribed circle of triangle A1A2A3 touches the side ai. Denote by Si

the reﬂection of the point Ti in the interior angle bisector of the angle Ai. Prove that

the lines M1S1, M2S2 and M3S3 are concurrent. 8. (APMO 2000) Let ABC be a triangle. Let M and N be the points in which the median and the angle bisector, respectively, at A meet the side BC. Let Q and P be the points in which the perpendicular at N to NA meets MA and BA, respectively. And O the point in which the perpendicular at P to BA meets AN produced. Prove that QO is perpendicular to BC.

8 Solutions (Sorry for not providing the ﬁgures.)

1. Let the centers of the circles be A, B, C and denote the radii by ra, rb, rc. Let the tangents for the circles centered at B and C meet at X. Deﬁne Y and Z analogously.

Note that X lies outside BC. Use homothety to show that XB/XC = rb/rc. Now use Menelaus’ theorem.

2. By symmetry, the lines P1P2,Q1Q2,O1O2 concur at a point, call it O. Consider the

homothety with center O which sends C1 to C2. Let OA intersect C1 at B. Then

under the homothety A is the image of B. Since 4BM1O1 is sent to 4AM2O2, so OO1 M1BO1 = M2AO2. Now 4OP1O1 is similar to 4OM1P1, which implies = ∠ ∠ OP1 OP1 2 . Then OO1 · OM1 = OP = OA · OB, which implies A, M1,B,O1 are concyclic. OM1 1 Then ∠M1NO1 = ∠M1AO1. Hence ∠M1AO1 = ∠M2AO2. Adding ∠O1AM2 to both sides, we have ∠O1AO2 = ∠M1AM2. 3. Let the line ` touch the circle C at D and let T,X be the the reﬂection of D with respect to the centre of C and M respectively. Note that the points T,X are independent of P. But the diameter of the incircle lemma tells us that P must lie on the ray XT beyond T. Conversely, given a point P lying on the ray XT beyond T, let the tangents from P to C meet ` at Q and R. By the lemma we must have QD = XR, from which it follows that M is the midpoint of QR. Therefore, the locus is the ray XT beyond T. 4. Let the incircle of ABC touch BC at X, and let XY be a diameter of the incircle. By the lemma, A, Y, D are collinear. Let M be the midpoint of BC. Then MI is a midline of triangle XYD, so IM and YD are parallel. The dilation centered at G with ratio −2 takes M to A, and thus it takes line IM to the line through A parallel to IM, namely the line AD. Hence the image of I under the dilation lies on the line AD. Analogously, it must also lie on BE and CF, and therefore the image of I is precise N. This proves that I, G, N are collinear in that order with GN = 2IG.

5. From the lemma we know that D1Q is a diameter of the incircle. Let the incenter of ABC be I, its centroid be G, and the midpoint of BC be M. Note that P is the Nagel point of ABC. From the previous problem, we know that the dilation centered at G with ratio −2 sends M to A and I to P, and hence sends segment IM to P A, thus

PA = 2IM. On the other hand, a dilation centered at D1 with ratio 2 sends IM to

QD2, so QD2 = 2IM = P A. Therefore, AQ = PA − QP = QD2 − QP = D2P. 6. (copying verbatim from Mathematical Excalibur v9n4) Let EF be the chord of Γ which

is the common tangent to Γ1 and Γ2 on the same side of line O1O2 as A. Let EF 0 touch Γ1 at C (see the ﬁgure). The homothety with center M that sends Γ1 to Γ will send C0 to some point A0 and line EF to the tangent line L of Γ at A0. Since 0 0 0 0 lines EF and L are parallel, A must be the midpoint of arc FA E. Then ∠A EC = 0 0 0 0 0 0 ∠A FC = ∠A ME. So 4A EC is similar to 4A ME. Then the power of A with

9 0 0 0 0 2 0 respect to Γ1 is A C · A M = A E . Similarly, the power of A with respect to Γ2 is 0 2 0 0 0 0 A F . Since A E = A F,A has the same power with respect to Γ1 and Γ2. So A lies

on the radical axis AB. Hence, A0 = A. Then C0 = C and C is on EF. Similarly, the

other common tangent to Γ1 and Γ2 passes through D. Let Oi be the center of Γi. By

symmetry with respect to O1O2, we see that O2 is the midpoint of arc CO2D. Then ∠DCO2 = ∠CDO2 = ∠FCO2. This implies O2 is on the angle bisector of ∠FCD. Since CF is tangent to Γ2, therefore CD is tangent to Γ2.

7. Some hints: show that (i) S1,S2,S3 all lie on the incircle, (ii) S1S2 k M1M2,S2S3 k

M2M3,S3S1 k M3M1, (iii) 4M1M2M3 ∼ 4S1S2S3, (iv) M1S1,M2S2,M3S3 concur at the centre of homothety. For full solution, see Mathematical Excalibur v9n4. 8. The case AB = AC is clear. W.l.o.g, we may assume AB > AC. Let AN intersect 1 (ABC) at D. Then ∠DBC = ∠DAC = 2 ∠BAC = ∠DAB = ∠DCB. So DB = DC and MD is perpendicular to BC. Consider the homothety with center A that sends 4DBC to 4OB0C0. Then OB0 = OC0 and BC k B0C0. Now let B0C0 intersect PN at K. Show that K lies on MA. Hence prove that K = Q and ﬁnish. (For full solution, see Mathematical Excalibur v9n4.)

References

• Mathematical Excalibur v9n4 • Euclidean Geometry in Mathematical Olympiads, by Evan Chen • Three Lemmas in Geometry, Yufei Zhao