A Sequence of Triangles and Geometric Inequalities

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1 Forum Geometricorum Volume 9 (009) FORUM GEOM ISSN A Sequence of Triangles and Geometric Inequalities Dan Marinescu, Mihai Monea, Mihai Opincariu, and Marian Stroe Abstract. We construct a sequence of triangles from a given one, and deduce a number of famous geometric inequalities. 1. A geometric construction Throughout this paper we use standard notations of triangle geometry. Given a triangle ABC with sidelengths a, b, c, let s, R, r, and denote the semiperimeter, circumradius, inradius, and area respectively. We begin with a simple geometric construction. Let H be the orthocenter of triangle ABC. Construct a circle, center H, radius R = Rr to intersect the half lines HA, HB, HC at A, B, C respectively (see Figure 1). A A H C B C B Figure 1. If the triangle ABC has a right angle at A with altitude AD (D on the hypotenuse BC), we choose A on the line AD such that A is between D and A. Lemma 1. Triangle A B C has (a) angle measures A = π A, B = π B, C = π C, (b) sidelengths a = a(b + c a), b = b(c + a b), c = c(a + b c), and (c) area =. Publication Date: December 16, 009. Communicating Editor: Paul Yiu.

2 9 M. Marinescu et al Proof. (a) B A C = 1 B HC = 1 (b) By the law of sines, a = R sin A = Rr cos A = similarly for b and c. (c) Triangle A B C has area BHC = π A ; similarly for B and C. abc 4 s(s a) s bc = 1 b c sin A = 1 b c cos A = 1 s(s a) b(c + a b) c(a + b c) bc = s(s a)(s b)(s c) =. = a(b + c a); Proposition. (a) a + b + c = a + b + c (b c) (c a) (a b). (b) a + b + c a + b + c. (c) a + b + c a + b + c. (d) sin A + sinb + sin C sin A + sin B + sinc. (e) R R. (f) r r. In each case, equality holds if and only if ABC is equilateral. Proof. (a) follows from Lemma 1(b); (b) follows from (a). For (c), a + b + c = a(b + c a) + b(c + a b) + c(a + b c) For (d), we have sin A + sin B + sin C b + c + c + a = a + b + c. + a + b = 1 (sinb + sin C + sin C + sina + sina + sin B) = sin B + C cos B C + sin C + A cos C A sin B + C + sin C + A + sin A + B + sin A + B cos A B = cos A + cos B + cos C = sin A + sin B + sin C. (e) R = a +b +c (sin A +sin B +sinc ) (f) r = s s = r. a+b+c (sin A+sinB+sin C) = R.

3 A sequence of triangles and geometric inequalities 93 Remark. The inequality R R certainly follows from Euler s inequality R r. From the direct proof of (e), Euler s inequality also follows (see Theorem 6(b) below).. A sequence of triangles Beginning with a triangle ABC, we repeatedly apply the construction in 1 to obtain a sequence of triangles (A n B n C n ) n N with A 0 B 0 C 0 ABC, and angle measures and sidelengths defined recursively by A n+1 = π A n, B n+1 = π B n, C n+1 = π C n ; a n+1 = a n (b n + c n a n ), b n+1 = b n (c n + a n a n ), c n+1 = c n (a n + b n c n ). Denote by s n, R n, r n, n the semiperimeter, circumradius, inradius, and area of triangle A n B n C n. Note that n = for every n. Lemma 3. The sequences (A n ) n N, (B n ) n N, (C n ) n N are convergent and lim A n = lim B n = lim C n = π n n n 3. Proof. It is enough to consider the sequence (A n ) n N. Rewrite the relation A n+1 = π An as A n+1 π ( 3 = 1 A n π ). 3 It follows that the sequence ( A n π ) 3 is a geometric sequence with common n N ratio 1. It converges to 0, giving lim n A n = π 3. Proposition 4. The sequence (R n ) n N is convergent and lim n R n = 3 3. Proof. Since R n = anbncn 4 n = 8R3 n R n = The result follows from Lemma 3. sinan sin Bn sin Cn 4 n, we have sin A n sinb n sin C n. Proposition 5. The sequences (a n ) n N, (b n ) n N, (c n ) n N are convergent and lim a n = lim b n = lim c n =. n n n 3 Proof. This follows from a n = R n sin A n, Lemma 3 and Proposition 4. From these basic results we obtain a number of interesting convergent sequences. In each case, the increasing or decreasing property is clear from Proposition.

4 94 M. Marinescu et al Sequence Limit Reference (a) n constant Lem.1(c) 3 3 (b) sin A n + sin B n + sin C n increasing Prop.(d), Lem.3 (c) R n decreasing 3 3 Prop.(e),4 (d) s n decreasing 3 3 Prop.(c),4 1 (e) r n increasing 3 3 Prop.(f) R (f) n r n decreasing (g) a n + b n + c n decreasing 4 3 Prop.(b),5 (h) a n + b n + c n (b n c n ) (c n a n ) (a n b n ) decreasing 4 3 Prop.(a,b),5 3. Geometric inequalities The increasing or decreasing properties of these sequences, along with their limits, lead easily to a number of famous geometric inequalities [1, 3]. Theorem 6. The following inequalities hold for an arbitrary angle ABC. (a) sin A + sin B + sin C 3 3. (b) [Euler s inequality] R r. (c) [Weitzenböck inequality] a + b + c 4 3. (d) [Hadwiger-Finsler inequality] a +b +c (b c) (c a) (a b) 4 3. In each case, equality holds if and only if the triangle is equilateral. Remark. Weitzenböck s inequality is usually proved as a consequence of the Hadwiger - Finsler s inequality ([, 4]). Our proof shows that they are logically equivalent. References [1] O. Bottema, R. Z. Djordjević, R. R. Janić, D. S. Mitrinović, and P. M. Vasic, Geometric Inequalities, Wolters-Noordhoe Publishing, Groningen, [] P. von Finsler and H. Hadwiger, Einige Relationen im Dreieck, Comment. Math. Helvetici, 10 (1937) [3] D. S. Mitrinović, J. E. Peĉarić, and V. Volenec, Recent Advances in Geometric Inequalities, Kluwer Academic Publisher, 1989 [4] R. Weitzenböck, Uber eine Ungleichung in der Dreiecksgeometrie, Math. Zeit., 5 (1919)

5 A sequence of triangles and geometric inequalities 95 Dan Marinescu: Colegiul Naţional Iancu de Hunedoara, Hunedoara, Str.l Libertatii No. Bl. 9 Ap.14, 33103, Romania address: marinescuds@yahoo.com Mihai Monea: Colegiul Naţional Decebal Deva, Str. Decebal Bl. 8 Ap.10, 33001, Romania address: mihaimonea@yahoo.com Mihai Opincariu: Colegiul Naţional Avram Iancu Brad, Str. 1 Iunie No.13 Bl. 15 Ap.14, 33500, Romania address: opincariumihai@yahoo.com Marian Stroe: Coleguil Economic Emanoil Gojdu Hunedoara, Str. Viorele No. 4, Bl 10 Ap.10, , Romania address: maricu stroe@yahoo.com