A nest of Euler Inequalities

Transcription

1 31 nest of Euler Inequalities Luo Qi bstract For any given BC, we define the antipodal triangle. Repeating this construction gives a sequence of triangles with circumradii R n and inradii r n obeying a generalized form of Euler s inequality n R n R R 1 R 0 r 0 r 1 n+1 r n, (n 1,, ), with equalities iff BC is equilateral. Key words: Euler inequality; antipodal triangle Let R,r be the radius of circumcircle and inscribed circle of a triangle; then R r, with equalitiesiff the triangleis equilateral([1], p.50). This is the famous Euler inequality. In this note, we are going to build a nest of Euler inequalities for a certain family of related triangle. Definition 1 If a vertex of a triangle BC and another point on the perimeter divide the perimeter into two equal parts (that is, B + B C + C ) we call the antipode of, and the triangle B C of which three vertices are antipodes of, B, C respectively the antipodal triangle of BC. Note that is necessarily on the (non-extended) edge BC and in fact it is the point where that edge touches the appropriate escribed circle.[] Thus, we can easily find a way to draw an antipodal triangle B C of a given triangle BC. Lemma 1 Denote by a,b,c,a 1,b 1,c 1,s,s 1,, 1 the sides, semiperimeters, and areas of BC and its antipodal triangle 1 B 1 C 1, and let R and r be the circumradius and inradius of BC. Then 1. B 1 B 1 s c, C 1 C 1 s b, BC 1 CB 1 s a; r R ; a 1 b 1 c 1 abc r 4R (with equality iff BC is equilateral ); 4. s 1 s( with equality iff BC is equilateral ). Copyright c 011 Canadian Mathematical Society Crux Mathematicorum with Mathematical Mayhem, Volume 37, Issue 5

2 313 Proof (See the figure 1) (1) We have B 1 1 ( B + BC + C ) B, B 1 1 ( B + BC + C ) B and so B 1 B 1 s c. In the same way, we have C 1 C 1 s b, BC 1 CB 1 s a. B 1 C 1 B 1 Figure 1 C () Denote by B1C 1, B1C 1, C1B 1, the areas of B 1 C 1, B 1 C 1, C 1 B 1. Because B1C 1 B1C 1 C1B 1 B 1 C 1 B C B 1 BC 1 B BC C 1 CB 1 C CB (s c)(s b), c b (s c)(s a), c a (s b)(s a) b a we have 1 B 1C 1 B1C 1 C1B 1 1 B 1C 1 B 1C 1 C 1B 1 (s c)(s b) (s c)(s a) (s b)(s a) 1 c b c a b a (s a)(s b)(s c) a b c

3 314 Thus By Heron s and other well-known formulas, 1 s(s a)(s b)(s c) abc 4R sr (s a)(s b)(s c) a b c (3) Using the law of sines on B 1 C 1, we have a 1 sin Therefore we have s b sin B 1 C 1 s 4 R sr r R s c s b + s c sin C 1 B 1 sin B 1 C 1 + sin C 1 B 1 a sin B1C1+ C1B1 cos B1C1 C1B1 a sin π In the same way, we have a cos (with equality iff B 1 C 1 C 1 B 1 ). a 1 a sin cos sin (with equality iff B 1 C 1 C 1 B 1 ). b 1 b sin B (with equality iff B 1 C 1 BC 1 1 ), and c 1 c sin C (with equality iff CB 1 1 C 1 B 1 ). Multiplying these three inequalities, we obtain lso, because we have a 1 b 1 c 1 abc sin sin B sin C (with equality iff BC is equilateral). 1 abc absinc 4R r(a+b+c) (R) 3 sinsinbsinc 4R Rr(sin + sinb + sinc).

4 315 So r R sinsinbsinc sin + sinb + sinc 16sin sin B sin C cos cos B cos C sin +B cos B + sin C cos C 16sin sin B sin C cos cos B cos C cos C B (cos + sin C ) 8sin sin B sin C cos cos B cos cos B + sin sin B +B + cos 8sin sin B sin C cos cos B cos cos B + sin sin B + cos cos B sin sin B 4sin sin B sin C Thus a 1 b 1 c 1 abc r 4R (with equality iff BC is equilateral). (4) Construct perpendiculars B 1 E and C 1 D to BC at E and D, respectively (See Figure ). Then a 1 DE (with equality iff BC B 1 C 1.) But DE a BD CE a (s a)cosb (s a)cosc, so a 1 a (s a)(cosb + cosc) with equality iff BC B 1 C 1. B 1 C 1 B 1 D Figure E C Similarly, we have: b 1 b (s b)(cosc + cos) with equality iff C C 1 1, c 1 c (s c)(cos + cosb) with equality iff B 1 B 1.

5 316 dding up these three inequalities yields s 1 s (acos + bcosb + ccosc) s 1 (acosb + acosc + bcos + bcosc + ccos + ccosb) s 1 (a + b + c) s Therefore s 1 s with equality iff BC is equilateral. Studying these two triangles we can also find other interesting properties. The reader may verify that the antipodal triangle is less equilateral in that the ratio between longest and shortest side is always greater than in the original triangle. Theorem 1 Denote by R,R 1,r,r 1, the circumradii and inradii of BC and its antipodal triangle 1 B 1 C 1. Then R 1 R r 4r 1, with equalities iff BC is equilateral. Proof By () and (3) of lemma 1, we have r R 1 a 1b 1 c 1 /4R 1 abc/4r Ra 1b 1 c 1 R 1 abc R r R 1 4R r 4R 1 So R 1 R, with equality iff BC is equilateral. By () and (4) of lemma 1, we have 1 4 r R 1 r 1s 1 rs r 1s 1 r s 1 r 1 r and so r 4r 1 with equality iff BC is equilateral. Hence, we get R 1 R r 4r 1, again with equalities iff BC is equilateral. Using mathematical induction and the theorem we immediately get: Corollary 1 (See Figure 3) Let 0 B 0 C 0 be given, and let R 0,R 1,,R n ; r 0,r 1,,r n denote the circumradii andinradii of 0 B 0 C 0, 1 B 1 C 1,, n B n C n respectively, and i B i C i is the antipodal triangle of i 1 B i 1 C i 1, (i 1,, ). Then n R n R R 1 R 0 r 0 r 1 n+1 r n, with equalities iff 0 B 0 C 0 is equilateral.

6 317 0 C n 1 n B n 1 C n B n C 0 B n 1 0 Figure 3 So, we build a nest of Euler inequalities. Open question: In [3] Yang derives Euler-type inequalities for tetrahedra in 3-dimensional space. Can we define antipodal points for a tetrahedron in such a way that the results of this paper generalize? References [1] O. Bottema et al. Geometric Inequalities, Wolter-Noordhoff, Groningen, [] Weisstein, Eric W. Semiperimeter. FromMathWord Wolfram Web Resource. ccessed 10 January 01. [3] Yang Shiguo. Several improvements of tetrahedral Euler inequality. China collection of studies of elementary mathematics [M]. Zhengzhou province: Henan education press, Mathematics Department, Guilin Normal College, Guilin, Luo Qi Mathematics Department Guilin Normal College Guilin, Guangxi China luoqi67@163.com