Another Variation on the Steiner-Lehmus Theme

Transcription

1 Forum Geometricorum Volume 8 (2008) FORUM GOM ISSN Another Variation on the Steiner-Lehmus Theme Sadi Abu-Saymeh, Mowaffaq Hajja, and Hassan Ali ShahAli Abstract. Let the internal angle bisectors and of angles and of triangle A be extended to meet the circumcircle at and. The Steiner- Lehmus theorem states that if =, then A = A. In this article, we investigate those triangles for which = and we address several issues that arise within this investigation. 1. Introduction The celebrated Steiner-Lehmus theorem states that if the internal angle bisectors of two angles of a triangle are equal, then the triangle is isosceles. In terms of triangle centers and cevians, it states that if two cevians through the incenter of a triangle are equal, then the triangle is isosceles. Variations on the theme can be obtained by replacing the incenter by any of the hundreds of centers known in the literature; see [6] and the website [7]. Other variations on this theme are obtained by letting the cevians of A through a center P meet the circumcircle of A at A,, and and asking whether the equality = implies that A = A, where XY denotes the length of the line segment XY. This variation, together with several others, is investigated in [5] where it is proved that if P is the incenter, the orthocenter, or the Fermat-Torricelli point, then = if and only if A = A or A = π 3. When P is the centroid, the triangles for which = are proved, in Theorem 9 below, to be the ones whose side lengths satisfy the relation a 4 = b 4 b 2 c 2 + c 4, a relation that has no geometric interpretation and cannot be fitted into a traditional geometry context such as uclid s lements. Using geometric arguments, we show that if the centroid P of a scalene triangle A is such that =, then A must lie in the interval [ π 3, π 2 ] and that to every θ in [ π 3, π 2 ] there is essentially a unique scalene triangle with A = θ and with =. The proof uses a generalization of Proposition 7 of ook III of uclid s lements, in brief uclid III.7 1, and deserves recording on its own. Publication Date: June 16, ommunicating ditor: Paul Yiu. The first and second named authors are supported by a research grant from Yarmouk University and would like to express their thanks for this support. The authors would also like to thank the referee for his valuable remarks and for providing the construction given at the very end of this note, and to Mr. ssam Darabseh for drawing the figures. 1 Throughout, the symbol uclid. designates Proposition of ook in uclid s lements.

2 132 S. Abu-Saymeh, M. Hajja, and H. A. ShahAli 2. uclid III.7 and a generalization uclid III.7, not that well known, states that if Ω is a circle centered at O, if M O is a point inside Ω, and if the intersection of a ray MX with Ω is denoted by X, then (i) the maximum value of MX is attained when the ray MX passes through O and the minimum is attained when the ray MX is the opposite ray MO, (ii) as the ray MX rotates from the position MO to the opposite position MO, the quantity MX changes monotonically. We restate this proposition in Theorem 1 as a preparation for the generalization that is made in Theorem 5. Theorem 1 (uclid III.7). Let be a chord in a circle Ω, let M be the mid-point of, and let the line perpendicular to through M meet Ω at and F.As a point P moves from to F along the arc F of Ω, the length changes monotonically. It increases or decreases according as is closer or farther than F from M. P O θ Q M F Figure 1. Proof. Referring to Figure 1, we shall show that if M > MF, i.e., if the center O of Ω is between and M, and if P and Q are any points on the arc F such that P is closer to than Q, then > MQ. Under these assumptions, MQP > OQP = OPQ > Q. Thus MQP > Q, and therefore > MQ, as desired. Remark. The proof above uses the fairly simple-minded fact that in a triangle, the greater angle is subtended by the greater side. This is uclid I.19. It is interesting that uclid s proof uses the more sophisticated uclid I.24. This theorem, referred to in [8, Theorem 6.3.9, page 140] as the Open Mouth Theorem, states that if triangles A and A are such that A = A,A= A, A > A, then >. Quoting [8], this says that the wider you open your mouth, the farther apart your lips are. Although this follows immediately from the

3 Another variation on the Steiner-Lehmus theme 133 law of cosines, the intricate proofs given by uclid and in [8] have the advantage of showing that the theorem is a theorem in neutral geometry. Theorem 5 below generalizes Theorem 1. In fact Theorem 1 follows from Theorem 5 by taking to be a diameter of one of the circles Ω and Ω. For the proof of Theorem 5, we need the following simple lemmas. Lemma 2. Let A be a triangle and let D and be points on the sides A and A respectively (see Figure 2). Then AD A A is greater than, less than, or equal to A according as A is greater than, less than, or equal to AD, respectively. A D Figure 2 Proof. Let be the point on A such that A A A ; i.e., D is parallel to. If A A = AD A, then = and A = AD. If A A > AD A, then lies between and, and A = AD < AD. Similarly for the case A A < AD A. = AD Ω P P S M Ω F F Figure 3 Lemma 3. Two circles Ω and Ω intersect at and, and the line perpendicular to through the midpoint M of meets Ω and Ω at and, respectively, such that is inside Ω (see Figure 3). IfP is any point on the arc F of Ω and if the ray meets Ω at P, then > M M.

4 134 S. Abu-Saymeh, M. Hajja, and H. A. ShahAli Proof. Let S be the point of intersection of FP and F P. Since PF = π 2 = P F, it follows that M P + MF P = π 2 = MP + MFP. ut MFP > MF P, by the exterior angle theorem. Hence M P > MP. y Lemma 2, we have > M M, as desired. Lemma 4. Let be an isosceles triangle having =. Let M be the midpoint of and let be the circumcenter of (see Figure 4). Then M M is greater than, equal to, or less than 1 3 according as is less than, equal to, or greater than π 3, respectively. R x R M Figure 4. Proof. Let θ =, x = M, and let R be the circumradius of. Then M = θ and M M 1 3 = x x + R 1 3 = R cos θ R cos θ + R 1 3 = 2cosθ 1 3(cos θ +1). cos θ is greater than, equal to, or less than 1 2. Theorem 5. Two circles Ω and Ω intersect at and and the line perpendicular to through the midpoint M of meets Ω at and F and meets Ω at and F. For every point P on Ω, let P be the point where the ray meets Ω. As a point P moves from to F along the arc F, the ratio changes monotonically. It decreases or increases according as is inside or outside Ω. Proof. Referring to Figure 5, suppose that lies inside Ω and let P and Q be two points on the arc F of Ω such that P is closer to than Q. we are to show that < MQ MQ. xtend QM to meet Ω at U and Ω at U. Let T be the point of intersection of U and U. Since the quadrilaterals PQU and P Q U are cyclic, it follows that UQP + UP = π = U Q P + U P. (1)

5 Another variation on the Steiner-Lehmus theme 135 Ω P P Q T Q M U U F F Ω Figure 5 ut U P = U M + M P > UM + M P (by the exterior angle theorem) > UM + MP (by Lemmas 3 and 2) = UP. From this and (1) it follows that U Q P > UQP. y Lemma 2, we conclude that < MQ MQ, as desired. Note that if P is on the arc and Q is on the arc F, then < 1 < MQ MQ. 3. onditions of equality of two chords through a given point The next simple lemma exhibits the relation between two geometric properties of a point P inside a triangle A. It will be used in the proof of Theorem 9. Lemma 6. Let P be a point inside triangle A and let the rays P and P meet the circumcircle of A at and respectively (see Figure 6). Then (a) = if and only if P = P or P =2 A; (b) P =2 A P = P. Moreover, if P is the centroid, then (c) P = P A = A.

6 136 S. Abu-Saymeh, M. Hajja, and H. A. ShahAli Proof. (a) It is clear that = A = A or A + A = π A = A or A + A +2 A = π = or + +2 A = π P = P or P =2 A. A P Figure 6. (b) Also, P =2 A P + P =2 P P = P P = P. This proves the first part of (b). The implication P = P is easy. (c) Let the lengths of the medians from and be β and γ, respectively. y Apollonius theorem, we have b β2 = a 2 + c 2, c2 2 +2γ2 = a 2 + b 2. The rest follows from the facts that P = 2β 3 and P = 2γ hords of circumcircle through the centroid In Theorem 7, we focus on triangles A whose centroid G has the property that G =2 A. Interest in this property stems from Lemma 6. Note that Part (i) provides a solution of the problem in [4]. Theorem 7. (i) If A is a triangle whose centroid G has the property that G =2 A, then π 3 A < π 2 with A = π 3 if and only if A is equilateral.

7 Another variation on the Steiner-Lehmus theme 137 (ii) If θ is any angle in the interval ( π 3, π 2 ) and if is any line segment, then there is a triangle A, unique up to reflection about and about the perpendicular bisector of, having A = θ and whose centroid G has the property G =2 A. Proof. (i) Let Ω be the circumcircle of A and let be its circumcenter. Let Ω be the circumcircle of. Let M be the midpoint of and let the perpendicular bisector of meet Ω at and F and meet Ω at ( and) F, where is on the arc A of Ω (see Figure 7). Let A = θ, and let G be the centroid of A. Also, for every P on Ω, let P be the point where the ray meets Ω. Suppose that G =2 A. Since =2 A, it follows that G lies on the arc of Ω. Also, G lies on the median AM of A. Therefore, G is the point A where the ray MA meets Ω. In particular, MA MA = 1 3. As P moves from to F along the arc F, the ratio increases by Theorem 5. Therefore M M MA MA = 1 3. y Lemma 4, θ π 3, with equality if and only if A =, or equivalently if and only if A is equilateral. The possibility that A π 2 is ruled out since it would lead to the contradiction G π. Ω A θ Ω θ A = G M M F F Ω Ω F F Figure 7 Figure 8 (ii) Suppose that θ is a given angle such that π 3 θ<π 2 and that isagiven segment. Let be an isosceles triangle with = and with = θ. Let Ω be the circumcircle of and let be its circumcenter. Let Ω be the circumcircle of. Let M be the midpoint of and let the perpendicular bisector of meet Ω at ( and) F and meet Ω at ( and) F ; see Figure 8. For

8 138 S. Abu-Saymeh, M. Hajja, and H. A. ShahAli every P on Ω, we let P be the point where the ray meets Ω. Let t = M M. Since θ π 3, it follows from Lemma 4 that t 1 3. Also, = and M M =1. Thus as P moves from to along one of the arcs of Ω, the ratio increases from t 1 3 to 1. y continuity and the intermediate value theorem, there is a unique point A on that arc for which MA MA = 1 3. If we think of M as the x-axis and of M as the y-axis, then the point A is the only point in the first quadrant for which A has the desired property. Points in the other quadrants are obtained by reflection about the x- and y-axes. This is precisely the point A on the arc F for which A is the centroid of A. This triangle A is the unique triangle (up to reflection about and about the perpendicular bisector of ) whose vertex angle at A is θ and whose centroid G has the property that G =2 A. Theorem 9 characterizes those triangles whose centroid has the property =. For the proof, we need the following simple lemma. Lemma 8. Let A be a triangle with side-lengths a, b, and c (in the standard order) and with centroid G. Let the rays G and G meet the circumcircle of A at and respectively. 2 = (a2 + c 2 ) 2 2a 2 +2c 2 b 2. Proof. Let m =, x =. y Apollonius theorem, m 2 = 2(a2 +c 2 ) b 2 4. Since and A are diagonals of a cyclic quadrilateral, m(x m) = b2 4. It follows that mx = a2 +c 2 2 and x 2 = (a2 +c 2 ) 2 = (a2 +c 2 ) 2. 4m 2 2a 2 +2c 2 b 2 Theorem 9. Let A be a triangle with side-lengths a, b, and c (in the standard order) and with centroid G. Let the rays G and G meet the circumcircle of A at and, respectively. If b c, then the following are equivalent: (i) =, (ii) G =2 A, (iii) a 4 = b 4 + c 4 b 2 c 2. Proof. Since b c, it follows that G G. y Lemma 6, (i) is equivalent to (ii). To see that (i) is equivalent to (iii), let x =, y =, and let s = a 2 + b 2 + c 2. y Lemma 8, x 2 = (s b2 ) 2 2s 3b 2, y2 = (s c2 ) 2 2s 3c 2.

9 Another variation on the Steiner-Lehmus theme 139 Therefore x = y (s b2 ) 2 2s 3b 2 = (s c2 ) 2 2s 3c 2 (s 2 2b 2 s + b 4 )(2s 3c 2 )=(s 2 2c 2 s + c 4 )(2s 3b 2 ) s 2 (c 2 b 2 ) 2s(c 2 b 2 )(c 2 + b 2 )+3c 2 b 2 (c 2 b 2 )=0 s 2 2s(c 2 + b 2 )+3c 2 b 2 =0 (because b c) (s (c 2 + b 2 )) 2 =(c 2 + b 2 ) 2 3c 2 b 2 as claimed. a 4 = c 4 + b 4 c 2 b 2, It follows from [1, Theorem , page 83] (or [9, page 20]) that the only positive solutions of the diophantine equation a 4 + b 4 a 2 b 2 = c 4 (2) are given by a = b = c. Thus there are no non-isosceles triangles A with integer side-lengths whose centroid G has the property =. We end this note by a uclidean construction, provided by a referee, of triangles A whose centroid has the property =. We start with any segment. (i) Take any point A on the major arc A 0 of an equilateral triangle A 0. (ii) xtend A and A to Y and Z respectively such that Y = Z =. (iii) onstruct a circle with diameter A Z and the perpendicular at to A Z, intersecting the circle at (iii ) onstruct a circle with diameter A Y and the perpendicular at to A Y, intersecting the circle at (iv) onstruct the circles centered at and and passing through and, respectively. Letting A be a point of intersection of the two circles in (iv), one can verify that triangle A satisfies =. In this regard, one may ask whether one can construct a triangle A having the property = and having preassigned side and angle A (in [ π 3, π 2 ]). The answer is affirmative as seen below. Without loss of generality, assume = 1. Let b = A, c = A, and t =cosa. We are to show that b and c are constructible. These are defined by b 4 + c 4 b 2 c 2 =1,b 2 + c 2 =2bct +1. Subtracting the square of the second from the first and simplifying, we obtain bc = 4t 3 4t 2. Thus bc is constructible. Since b 2 + c 2 =2bct +1, it follows that b 2 + c 2 is constructible. Thus both b 2 c 2 and b 2 + c 2 are constructible, and hence b 2 and c 2, being the zeros of f(t ):=T 2 (b 2 + c 2 )T + b 2 c 2, are constructible. This shows that b and c are constructible, as desired. The restriction A [60, 90 ], i.e., t [ 0, 1 2], guarantees that the zeros of f(t ) are real (and positive).

10 140 S. Abu-Saymeh, M. Hajja, and H. A. ShahAli References [1] T. Andreescu and D. Andrica, An Introduction to Diophantine quations, GIL Publishing House, Zalau, Romania, [2] uclid, The lements, Sir Thomas L. Heath, editor, Dover Publications, Inc., New York, [3] uclid s lements, aleph0.clarku.edu/ djoyce/mathhist/alexandria.html [4] M. Hajja, Problem 1767, Math. Mag., 80 (2007), 145; solution, ibid., 81 (2008), 137. [5] M. Hajja, yril F. Parry s variations on the Steiner-Lehmus theme, Preprint. [6]. Kimberling, Triangle centers and central triangles, ongressus Numerantium, 129 (1998) [7]. Kimberling, ncyclopaedia of Triangle enters, [8] R. S. Millman and G. D. Parker, Geometry A Metric Approach with Models, second edition, Springer-Verlag, New York, [9] L. J. Mordell, Diophantine quations, Academic Press, New York, [10]. M. Stewart, Theory of Numbers, second edition, The Macmillan o., New York, Sadi Abu-Saymeh: Mathematics Department, Yarmouk University, Irbid, Jordan -mail address: sade@yu.edu.jo, ssaymeh@yahoo.com Mowaffaq Hajja: Mathematics Department, Yarmouk University, Irbid, Jordan -mail address: mhajja@yu.edu.jo, mowhajja@yahoo.com Hassan Ali ShahAli: Fakultät für Mathematik und Physik, Leibniz Universität, Hannover, Welfengarten 1, Hannover, Germany