Harmonic Division and its Applications

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Dwain Gilbert
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1 Harmonic ivision and its pplications osmin Pohoata Let d be a line and,,, and four points which lie in this order on it. he four-point () is called a harmonic division, or simply harmonic, if =. If is a point not lying on d, then we say that pencil () (which consists of the four lines,,, ) is harmonic if () is harmonic. In this note, we show how to use harmonic division as a tool in solving some difficult Euclidean geometry problems. We begin by stating two very useful lemmas without proof. he first lemma shows one of the simplest geometric characterizations of harmonic divisions, based on the theorems of Menelaus and eva. Lemma 1. In a triangle consider three points, Y, on the sides,, respective. If is the point of intersection of Y with the extended side, then the four-point ( ) forms and harmonic division if and only if the cevians, Y and are concurrent. Y P ' he second lemma is a consequence of the ppollonius circle property. It can be found in [1] followed by several interpretations. Lemma 2. Let four points,, and, in this order, lying on d. hen, if two of the following three propositions are true, then the third is also true: Mathematical Reflections 4 (2007) 1

2 (1) he division () is harmonic. (2) is the internal angle bisector of. (3). We begin our journey with a problem from the IMO 1995 Shortlist. Problem 1. Let be a triangle, and let, E, F be the points of tangency of the incircle of triangle with the sides, and respectively. Let be in the interior of such that the incircle of touches, and in, Y and respectively. Prove that EF Y is cyclic. E F Y Solution. enote = EF. ecause of the concurency of the lines, E, F in the Gergonne point of triangle, we deduce that the division ( ) is harmonic. Similarly, the lines, Y and are concurrent in the Gergonne point of triangle, so Y as a consequence of Lemma 1. Now expressing the power of point with respect to the incircle of triangle and the incircle of triangle we have that 2 = E F and 2 = Y. So E F = Y, therefore the quadrilateral EF Y is cyclic. For our next application, we present a problem given at the hinese IMO eam Selection est in Problem 2. Let be a convex quadrilateral. Let E =, F =, P =, and let O the foot of the perpendicular from P to the line EF. Prove that O = O. Mathematical Reflections 4 (2007) 2

3 S O F P E Solution. enote S = EF and = EF. s from Lemma 1, we deduce that the division (E F S) is harmonic. Furthermore, the division (P S) is also harmonic, due to the pencil (E F S). ut now, the pencil E(P S) is harmonic, so by intersecting it with the line, it follows that the four-point (P ) is harmonic. herefore, the pencil O(P S) is harmonic and OP OS, thus by Lemma 2, P O = P O. Similarly, the pencil O(P ) is harmonic and OP O, thus again by Lemma 2, P O = P O. It follows that O = O. We continue with an interesting problem proposed by inu Serbanescu at the Romanian Junior alkan MO 2007, eam Selection est. Problem 3. Let be a right triangle with = 90 and let be a point on side. enote by E the reflection of across the line and F the intersection point of E with the perpendicular to at. Prove that F, E and are concurrent. Y E F Solution. enote the points = E, Y = E, = E F and = F. From Lemma 1, applied to triangle E and for the cevians F Mathematical Reflections 4 (2007) 3

4 and E, we observe that the lines F, E and are concurrent if and only if the division (Y E) is harmonic. Since the quadrilateral Y is cyclic, tan Y = tan, which is equivalent to /Y = /. So = Y. Since triangles and are similar, we have that 2 =, so 2 = Y. Using = E, we obtain that Y Y E = E, and thus the division (Y E) is harmonic. he next problem was proposed by the author and given at the Romanian IMO eam Selection est in Problem 4. Let be a triangle, let E, F be the tangency points of the incircle Γ(I) to the sides, respectively, and let M be the midpoint of the side. Let N = M EF, let γ(m) be the circle of diameter, and let lines I and I meet γ again at and Y, respectively. Prove that N NY =. Y F N I E M Mathematical Reflections 4 (2007) 4

5 Solution. We will assume, so the solution matches a possible drawing. Let = EF (for =, = ), and the tangency point of Γ to. laim 1. In the configuration described above, for = I EF, one has. Proof. he fact that I effectively intersects EF follows from F E = 1 2 ( + ) = 1 2 π 1 2 < 1 2π, and I F (similarly, I effectively intersects EF ). he division ( ) is harmonic, and triangles F and are congruent, therefore =, which is equivalent to (similarly, for Y = I EF, one has Y Y ). laim 2. In the configuration described above, one has N = I EF. Proof. It is enough to prove that NI. Let d be the line through, parallel to. Since the pencil (M ) is harmonic, it follows the division (F NE) is harmonic, where = d EF. herefore N lies on the polar of relative to circle Γ, and as N EF (the polar of ), it follows that is the polar of N relative to circle Γ, hence NI d, so NI. In conclusion, since I, one has N I. It follows, according to laim 1, that = and Y = Y, therefore, Y EF. Since the division ( ) is harmonic, it follows that lies on the polar p of relative to circle γ. ut M p, so p, and since I, it follows that p is, in fact, I. Now, according to laim 2, it follows that, I, N are collinear. Since N is the polar, it means the division ( Y N) is harmonic, thus the pencil ( Y N) is harmonic. ut N, so N is the angle bisector of Y, hence N NY = Y = sin Y sin Y. s quadrilaterals IY and I are cyclic (since pairs of opposing angles are right angles), it follows that 1 2 = I = Y I = 1 2 Y (triangles Y and EY are congruent), so Y =. Similarly, Y =. herefore N NY = sin Y sin = = Y sin Y sin =. he following problem was posted on the MathLinks forum [2]: Problem 5. Let be a triangle and ρ(i) its incircle., E and F are the points of tangency of ρ(i) with, and respectively. enote M = ρ(i), N the intersection of the circumcircle of M with F and G = N. Prove that = 3F G. Mathematical Reflections 4 (2007) 5

6 F M G N E Solution. enote = EF G and = EF. Now because the four-point ( ) forms an harmonic division, so does the pencil F ( ) and now by intersecting it with the line G, we obtain that the division (GN) is harmonic. ccording to the Menelaus theorem applied to G for the transversal NF, we find that = 3GF is equivalent to N = 3NG. N Since (GN ) is harmonic, NG = G, so it suffices to show that N is the midpoint of. Observe that ME = MF = M, therefore the quadrilateral ME is cyclic, which implies that M = ME = E and M = F. lso, MN = F and MN = M MN = EF F = E. Using the above angle relations and the equation N N we obtain that N = N, so sin M sin MN = sin M sin MN, sin F sin F = sin E sin E. On other hand, coincides with a symmedian of triangle EF, so sin F sin E = F E herefore, N is the midpoint of. = sin EF sin F E = sin F sin E. Mathematical Reflections 4 (2007) 6

7 Let be a cyclic quadrilateral and a point on the circle. hen, the is called harmonic if the pencil () is harmonic. For a list of properties regarding the harmonic quadrilateral, interested readers may can consult [1] and [3]. he following problem was given at an IMO eam Preparation ontest, held in acau, Romania, in Problem 6. Let be a convex quadrilateral, for which denote O =. If O is a symmedian of triangle and O is a symmedian of triangle, prove that O is a symmedian of triangle. S O Solution. enote 1 =, 2 =, =, where, respective represents the tangent in to the circumcircle of and the tangent in to the circumcircle of. Since O is a symmedian of triangle and O is a symmedian of triangle, the divisions (O 1 ) and (O 2 ) are harmonic, so 1 = 2 =. Hence, is the polar of 1 with respect to the circumcircle of and also the polar of 2 with respect to the circumcircle of. ut because 1 = 2, we deduce that the circles and coincide, i.e. the quadrilateral is cyclic, and since the division (O ) is harmonic, the pencil (O ) is, and by intersecting it by the circle, it follows that the quadrilateral is also harmonic. hen, the pencil () is harmonic. y intersecting it with the line, we see that the division (OS) is harmonic, where S =. It follows that O is a symmedian of triangle. he next problem was also given in an IMO eam Preparation est, at the IMR ontest, held in ucharest in Problem 7. Let be an isosceles triangle with =, and M the midpoint of. Find the locus of the point P interior to the triangle for which P M + P = π. Mathematical Reflections 4 (2007) 7

8 P I S M Solution. enote the point as the intersection of the line P with the circumcircle of P and S = P. Since SP = 180 P, it follows that P S = P M. From the Steiner theorem applied in to triangle P for the isogonals P S and P M, S S = P 2 P 2. On other hand, using Sine Law, we obtain S S = sin S sin S = sin P sin P = P P. hus by the above relations, it follows that = P P, i.e. the quadrilateral P is harmonic, therefore the point = lies on the line P. If =, then lines and are always tangent to the circle P, and so the locus of P is the circle I, where I is the incircle of. Otherwise, if, then = M P S, due to the fact that P and and = P S M, therefore by maintaining the condition that, we obtain that P S = M, therefore P lies on M. he next problem was selected in the Senior MO 2007 Shortlist, proposed by the author. Problem 8. Let ρ(o) be a circle and a point outside it. enote by, the points where the tangents from with respect to ρ(o) meet the circle, the point on ρ(o), for which O, the foot of the perpendicular from to, Y the midpoint of the line segment and by the second intersection of Y with ρ(o). Prove that. Mathematical Reflections 4 (2007) 8

9 Y O H Solution. Let us call H = O ρ(o). hus H, so H. ecause Y is the midpoint of, we deduce that the division (Y ) is harmonic, so also is the pencil (Y H) and by intersecting it with ρ(o), it follows that the quadrilateral H is harmonic. hen, the pencil (H) is harmonic, so by intersecting it with the line H, it follows that the division ( H) is harmonic, where = H and = H. So, the line H is the polar of with respect to ρ(o), but H = is the polar of as well, so =, hence the points H,, are collinear, therefore. he last problem is a generalization of a problem by Virgil Nicula [4]. solution covers all concepts and methods presented throughout this paper. he N O P S 1 2 S 2 1 M N' Mathematical Reflections 4 (2007) 9

10 Problem 9. Let d be a line and,,, four points in this order on it such that the division () is harmonic. enote by M the midpoint of the line segment. Let ω be a circle passing through and M. Let NP be the diameter of ω perpendicular to M. Let lines N, N, P, P meet ω again at S 1, 1, S 2, 2, respectively. Prove that = S 1 1 S 2 2. Solution. Since the four-point () is harmonic, so is the pencil N() and by intersecting it with ω, it follows that the quadrilateral S 1 N 1 is harmonic, hence the lines S 1 S 1, 1 1 and N are concurrent, where N = N ω. ecause the tangent in N to ω is parallel with the line M and since M is the midpoint of, the division (M ) is harmonic, therefore the pencil N(NM) also is, and by intersecting it with ω, it follows that the quadrilateral N 1 MS 1 is harmonic, hence the lines S 1 S 1, 1 1 and MN are concurrent. From the above two observations, we deduce that the lines S 1 S 1, 1 1, MN, N are concurrent at a point. On the other hand, since the pencils (S 1 N 1 ) and (N 1 MS 1 ) are harmonic, by intersecting them with ω, it follows that the quadrilaterals N 3 MS 3 and S 3 N 3 harmonic, where S 3 = S 1 ω and 3 = 1 ω. Similarly, we deduce that the lines S 3 S 3, 3 3, MN and N are concurrent in the same point. herefore, S 3 3 is the polar of with respect to ω, but so is S 1 1, thus S 1 1 = S 3 3, so S 1 = S 3 and 1 = 3, therefore the points S 1,, 1 are collinear. Similarly, the points S 2,, 2 are collinear, from which it follows that = S 1 1 S 2 2. References [1] Virgil Nicula, osmin Pohoata. iviziunea armonica. GIL, [2] [3] [4] osmin Pohoata pohoata_cosmin2000@yahoo.com Mathematical Reflections 4 (2007) 10

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