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Homothety Aditya Ghosh

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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 re- spectively 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 ! be a circle tangent to chord AB at K and internally tangent to ! 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 ! are tangent at T; it follows there is a homothety at T taking ! to Ω: Clearly, the tangent to ! 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 ver- tices/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 respec- tively. 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 in- circle to the A-excircle. The line segment DT is the diameter of the incircle that is perpen- dicular to BC; and therefore its image under the dilation must be the diameter of the ex- circle 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; F I. 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.
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