TWO INEQUALITIES FOR A POINT IN THE PLANE OF A TRIANGLE

Transcription

1 INTERNATIONAL JOURNAL OF GEOMETRY Vol. (013), No., 68-8 TWO INEQUALITIES FOR A POINT IN THE PLANE OF A TRIANGLE JIAN LIU Abstract. In this paper we establish two new geometric inequalities involving an arbitrary point in the plane of a triangle. It is interesting that the equalities in both inequalities hold if and only if the point coincide with a special point of the original triangle. We also give a related geometric identity and several new inequalities by the main results. Finally, two elegant conjectures for the Erdös-Mordell inequality are put forward. 1. Introduction and main results Let P be a point in the plane of triangle ABC. Denote by R 1 ; R ; R 3 the distance of P from the vertices A; B; C, and r 1 ; r ; r 3 the distance of P from the sidelines BC; CA; AB respectively. If P lies in the interior of ABC, then we have the following famous Erdös-Mordell inequality: (1) R 1 + R + R 3 (r 1 + r + r 3 ); with equality holds if and only if 4ABC is equilateral and P is its center. This inequality has attracted attention of many researchers since it rst appeared in [10](see, e.g. [1]-[4], [8], [9], [11]-[14], [1]-[5]). In fact, it is well known that the stronger inequality: () R 1 + R + R 3 b c + c b r 1 + c a + a a r + c b + b r 3 a holds for interior point P, where a = BC; b = CA; c = AB. Equality holds if and only if P is the circumcenter of 4ABC. In [8], N.Dergiades generalized () to an arbitrary point in the plane: b (3) R 1 + R + R 3 c + c c ~r 1 + b a + a a ~r + c b + b ~r 3 ; a where ~r 1 ; ~r ; ~r 3 denote the directed (signed) distances from P to the sidelines BC; CA; AB respectively. Keywords and phrases: Erdös-Mordell inequality, pedal triangle, barycentric coordinates. (010)Mathematics Subject Classi cation: 51M04, 51M5. Received: In revised form: Accepted:

2 Two inequalities for a point in the plane of a triangle 69 The directed distance ~r 1 from P to BC is de ned as follows: When the orientation around vertices P; B; C is counterclockwise, then ~r 1 is positive and ~r 1 = r 1 ; Conversely, ~r 1 is negative and ~r 1 = r 1 (If P lies on the line BC then ~r 1 = r 1 = 0, etc). Similarly, we de ne ~r and ~r 3. For example, ~r 1 ; ~r ; ~r 3 are all positive in Figure 1; ~r is negative and ~r 1 ; ~r 3 are positive in Figure. In addition, we shall use directed ares of triangles. The de nition of directed area of a triangle is as follows: Given a triangle XY Z, if the orientation around the vertices X; Y; Z is counterclockwise, then the directed area S ~ 4XY Z of 4XY Z is positive and S ~ 4XY Z = S 4XY Z ; Conversely, ~S 4XY Z is negative and S ~ 4XY Z = S 4XY Z (If P lies on the line BC then ~S 4XY Z = S 4XY Z = 0, etc). For example, we have S ~ 4ABC = S 4ABC and ~S 4DEF = S 4DEF in Figure. Figure 1 Figure D.Nikolaos proof used the following simple inequality: (4) ar 1 b~r + c~r 3 ; (two similar relations are also valid) with equality holds if and only if P A? BC. Recently, the author rst found that the following interesting inequality can be derived by using (4): (5) a R 1 r 1 + b R r + c R 3 r 3 4S ; where S is the area of 4ABC. Equality holds if and only if P is the orthocenter of 4ABC. The proof of (5) is very simple. By (4) and the evident identity (6) a~r 1 + b~r + c~r 3 = ~ S; where S ~ is the directed area of 4ABC, we have a R1 r1 + b R r + c R3 r3 4S (b~r + c~r 3 ) + (c~r 3 + a~r 1 ) + (a~r 1 + b~r ) (a r1 + b r + c r3) (a~r 1 + b~r + c~r 3 ) = a ~r 1 + b ~r + c ~r 3 = 0: (a r 1 + b r + c r 3) Thus, inequality (5) is proved and the equality in (5) holds only when P A? BC; P B? CA; P C? AB, i.e., P is the orthocenter of 4ABC.

3 70 Jian Liu Inequality (5) inspires the author to nd similar results. Finally, we obtain Theorem 1.1. Let P be a point in the plane of triangle ABC with sides a; b; c, the semi-perimeter s and the area S. Then (7) a(s a)(r 1 r 1) + b(s b)(r r ) + c(s c)(r 3 r 3) S : Equality holds if and only if P coincide with the symmetrical point I 0 of the incentre I with respect to the circumcenter O of 4ABC. If we denote by h a ; h b ; h c the altitudes corresponding to sides a; b; c and denote by r a ; r b ; r c the corresponding radii of the excircles of 4ABC, then inequality (7) is equivalent to (8) R 1 r 1 h a r a + R r h b r b + R 3 r 3 h c r c 1; since we have ah a = S; (s a)r a = S, etc.. Theorem 1.1 implies the following interesting conclusion: If 4ABC is given, then the left-hand side of (7) or (8) attains the minimum value at when P coincides with the point I 0. After nding and proving inequality (7), the author noted a succinct formula about point I 0 (see Lemma.6 below). This formula motives the author to nd again a new inequality whose equality condition is the same as in (7). In this respect we prove the following: Theorem 1.. Let P be a point in the plane of triangle ABC with the directed area ~ S and the circumradius R, let ~ S p be the directed area of the pedal triangle DEF of P with respect to triangle ABC. Then (9) ~r 1 + ~r + ~r 3 + R ~ S p ~S R: Equality holds if and only if P coincide with the symmetrical point I 0 of the incentre I with respect to the circumcenter O of 4ABC. In particular, when P lies inside triangle ABC, it follows from (9) that (10) r 1 + r + r 3 + RS p R; S where S p is the area of the pedal triangle DEF. In [18], the author conjectured that the following inequality: (11) r 1 + r + r 3 R p + RS p S holds for any interior point P of 4ABC, where R p is circumradius of the pedal triangle DEF. It is interesting to compare (10) with (11).. Proofs of Theorems.1. Proof of Theorem 1.1. In order to prove Theorem 1.1, we need the following several lemmas in which Lemma 1,, 3 and 4 are all well-known in geometry(cf. [5], [6], [7], [], [6]).

4 Two inequalities for a point in the plane of a triangle 71 Lemma.1. Let P and M be two points in the plane of 4ABC and let (x; y; z) be the barycentric coordinates of M with respect to 4ABC. Then (1) (x+y +z) P M = (x+y +z)(xp A +yp B +zp C ) (yza +zxb +xyc ); where a; b; c are the lengths of the sides BC; CA; AB respectively. In particular, putting M = A in (1) we obtain the following important consequence (e.g. see []): Lemma.. Let P be a point with barycentric coordinates (x; y; z) in the plane of 4ABC. Then (13) (x + y + z) P A = (x + y + z)(yc + zb ) (yza + zxb + xyc ); where a; b; c are the lengths of the sides BC; CA; AB respectively. For the distances from a point to the sides of a triangle, we have the following known formulae: Lemma.3. Let P be a point in the plane of 4ABC with barycentric coordinates (x; y; z) and let ~r 1 ; ~r ; ~r 3 be the directed distances from P to the sidelines BC; CA; AB respectively. Then ~r 1 = xh a x + y + z ; ~r = yh b x + y + z ; ~r 3 = zh c x + y + z ; where h a ; h b ; h c are altitudes corresponding to the sides BC; CA; AB. Lemma.4. Let P i be points in the plane of 4ABC with barycentric coordinates (x i ; y i ; z i ) (i = 1; ; 3). Then P 1 ; P ; P 3 are collinear if and only if the following determinant holds: (14) x 1 y 1 z 1 x y z x 3 y 3 z 3 = 0: Lemma.5. Let p 1 ; p ; p 3 ; q 1 ; q ; q 3 be real numbers, then the following ternary quadratic inequality: (15) p 1 x + p y + p 3 z q 1 yz + q zx + q 3 xy holds for all real numbers x; y; z if and only if p 1 0; p 0; p 3 0; 4p p 3 q1 0; 4p 3p 1 q 0; 4p 1p q3 0; and (16) D 4p 1 p p 3 (q 1 q q 3 + p 1 q 1 + p q + p 3 q 3) 0: The equality conditions of (15) (suppose that p 1 > 0; p > 0; p 3 > 0) are as follows: 1 If D = 0, 4p 1 p q3 = 0; 4p 3p 1 q = 0, then 4p p 3 q1 = 0 and the equality in (15) holds only when p 1 x = q 3 y + q z. If D 0; 4p p 3 q1 > 0; 4p 3p 1 q > 0; 4p 1p q3 = 0, then the equality in (15) holds only when D = 0; z = 0; p 1 x = q 3 y: 3 If D 0; 4p p 3 q1 > 0; 4p 3p 1 q > 0; 4p 1p q3 > 0; x 6= 0; y 6= 0; z 6= 0, then the equality in (15) holds only when D = 0; (p 1 q 1 + q q 3 )x = (p q + q 3 q 1 )y = (p 3 q 3 + q 1 q )z.

5 7 Jian Liu The above lemma not only gives a necessary and su cient condition of (15)(this is well-known, see e.g. [0]) but also give its equality conditions. Here, we o er an elementary proof as follows: Proof. Inequality (15) can be rewritten as (17) p 1 x (q 3 y + q z)x + p y q 1 yz + p 3 z 0: We know this inequality holds for any real number x if and only if p 1 0, and (18) (19) p y q 1 yz + p 3 z 0; (q 3 y + q z) 4p 1 (p y q 1 yz + p 3 z ) 6 0: Again, (18) holds for any real numbers y; z if and only if p 0; p 3 0; q 1 4p p Inequality (18) is equivalent to (0) (4p 1 p q 3)y (p 1 q 1 + q q 3 )yz + (4p 3 p 1 q )z 0; which holds for any real numbers y; z if and only if 4p 1 p q3 0; 4p 3p 1 q 0, and or [ (p 1 q 1 + q q 3 )] 4(4p 1 p q 3)(4p 3 p 1 q ) 6 0; (1) (4p 1 p q 3)(4p 3 p 1 q ) (p 1 q 1 + q q 3 ) 0; which is equivalent to Since p 1 0 we have 16p 1 (p 1 q 1 + p q + p 3 q 3 + q 1 q q 3 4p 1 p p 3 ) 6 0: () 4p 1 p p 3 (q 1 q q 3 + p 1 q 1 + p q + p 3 q 3) 0; i.e. D 0. Combining with the above arguments, we conclude that the necessary and su cient conditions of (15) holds for any real numbers x; y; z are p 1 0; p 0; p 3 0; 4p p 3 q1 0; 4p 3p 1 q 0; 4p 1p q3 0; D 0: In what follows, we are going to discuss the equality condition in (15). Clearly, the equality in (17) and then (15) holds if and only if (3) p 1 x q 3 y q z = 0 and there is equality in (18) or (0), i.e. (4) (4p 1 p q 3)y (p 1 q 1 + q q 3 )yz + (4p 3 p 1 q )z = 0: According to these, we will discuss the equality conditions of (15) under the di erent cases(suppose that p 1 > 0; p > 0; p 3 > 0 for each case): Case 1 D = 0; 4p 1 p q3 = 0; 4p 3p 1 q = 0: In this case, we have 16p p 3 p 1 = q q 3 and (1) becomes an identity. Hence it follows from (1) that p 1 q 1 + q q 3 = 0. Thus(4) holds and 4p p 3 q 1 = q q 3 4p 1 q 1 = 0. Also, the equality in (15) holds only when (3) is valid. Case D 0; 4p p 3 q1 >; 4p 3p 1 q > 0; 4p 1p q3 = 0:

6 Two inequalities for a point in the plane of a triangle 73 By 4p 1 p q3 = 0 and (1), we have p 1q 1 +q q 3 = 0. Then (4) becomes (4p 3 p 1 q )z = 0, hence z = 0 and it follows from (3) that p 1 x = q 3 y. Therefore, there is equality in (15) only when D = 0; z = 0; p 1 x = q 3 y. Case 3 D 0; 4p p 3 q1 > 0; 4p 3p 1 q > 0; 4p 1p q3 > 0; x 6= 0; y 6= 0; z 6= 0: First, it is easy to know that there is equality in (0) only when D = 0 and (4p 1 p q3)y (p 1 q 1 + q q 3 )z = 0; i.e. (5) Using (3) and (5), we easily get y z = p 1q 1 + q q 3 4p 1 p q3 : x (6) z = p q + q 3 q 1 4p 1 p q3 : In addition, using D = 0, it is easy to prove the following two identities: (7) p 1 q 1 + q q 3 4p 1 p q 3 = p 3q 3 + q 1 q p q + q 3 q 1 ; (8) p q + q 3 q 1 4p 1 p q 3 Thus, it follows from (5)-(8) that = p 3q 3 + q 1 q p 1 q 1 + q q 3 : (9) (p 1 q 1 + q q 3 )x = (p q + q 3 q 1 )y = (p 3 q 3 + q 1 q )z: Hence the equality in (15) under the third case if and only if D = 0 and (9) are both valid. This completes the proof of Lemma.5. We now prove Theorem 1.1. Proof. Denote by (x; y; z) the barycentric coordinates of P with respect to 4ABC. By Lemma., it is easy to obtain that (30) R 1 = y c + z b + yz(b + c a ) (x + y + z) : From and Lemma.3, we have (31) R 1 r 1 = y c + z b + yz(b + c a ) (x + y + z) 4x S a (x + y + z) : Therefore X a(s a)(r 1 r1) P a(s a)[y c + z b + yz(b + c a )] = ( P 4S X s a x) ( P x) a x ; where P denote cyclic sums. Hence, the inequality (7) of Theorem 1.1 is equivalent to (3) X a(s a)[y c + z b + yz(b + c a )] 4S X s a X a x x S :

7 74 Jian Liu Replacing x! xa; y! yb; z! zc, again we know (3) is equivalent to (33) abc X (s a) y bc + z bc + yz(b + c a ) 4S X a(s a)x X xa S : Multiplying both sides of (33) by 8, then using s = a + b + c and Heron s formula: (34) 16S = (a + b + c)(b + c a)(c + a b)(a + b c); inequality (33) becomes (35) 4abc X (b + c a) y bc + z bc + yz(b + c a ) X X (a + b + c)(b + c a)(c + a b)(a + b c) a(b + c a)x + xa 0: Expansion and simpli cation give (36) p 1 x + p y + p 3 z (q 1 yz + q zx + q 3 xy) 0; where p 1 = a (b + c)a 4 (b + c)(b c) a + 4abc(b c) + (b c) (b + c) 3 ; p = b (c + a)b 4 (c + a)(c a) b + 4abc(c a) + (c a) (c + a) 3 ; p 3 = c (a + b)c 4 (a + b)(a b) c + 4abc(a b) + (a b) (a + b) 3 ; q 1 = bc(b + c a) 3a 3 + (b + c)a + a(bc 3b 3c ) (b + c)(b c) ; q = ca(c + a b) 3b 3 + (c + a)b + b(ca 3c 3a ) (c + a)(c a) ; q 3 = ab(a + b c) 3c 3 + (a + b)c + c(ab 3a 3b ) (a + b)(a b) : We now apply Lemma.5 to prove inequality (36). First, with the help of the mathematics software for calculating, we can check the following identity: (37) 4p 1 p p 3 q 1 q q 3 p 1 q 1 p q p 3 q 3 = 0: By Lemma.5, it remains to prove that p 1 > 0; 4p p 3 q1 0 and their analogues. Because of symmetry, we only need to prove these two inequalities. To prove p 1 > 0 we need to prove that (38) Q 0 (b + c)a 4 (b + c)(b c) a + 4abc(b c) + (b c) (b + c) 3 > 0: Putting s a = u; s b = v; s c = w, then a = v + w; b = w + u; c = u + v. Furthermore, we can check that Q 0 8(v w) u (v + w)(v w) u + 8(v vw + w )(v + w) u (39) +8vw(v + w)(v + w ): Since u > 0; v > 0; w > 0 and v and p 1 > 0 is proved. vw + w > 0, inequality Q 0 > 0 is valid

8 Two inequalities for a point in the plane of a triangle 75 It remains to prove that 4p p 3 identities: q 1 0: By calculating we get the following (40) (41) (4) 4p p 3 q 1 = 4abcm 1 k 1; 4p 3 p 1 q = 4abcm k ; 4p 1 p q 3 = 4abcm 3 k 3; where (43) (44) (45) (46) (47) (48) m 1 = a 3 (b + c)a (b + c 6bc)a + (b + c)(b c) ; m = b 3 (c + a)b (c + a 6ca)b + (c + a)(c a) ; m 3 = c 3 (a + b)c (a + b 6ab)c + (a + b)(a b) ; k 1 = a 3 + (b + c)a a(b + c) (b + c)(b c) ; k = b 3 + (c + a)b b(c + a) (c + a)(c a) ; k 3 = c 3 + (a + b)c c(a + b) (a + b)(a b) : To show that 4p p 3 rewritten as q 1 0, we have to prove m 1 > 0. But m 1 can be m 1 = 4(v + w)u + 4(v + w )u + 4vw(v + w); where u = s a > 0; v = s b > 0; w = s c > 0. So m 1 > 0 is true and 4p p 3 q1 0 is proved. This completes the proof of inequality (36), and then (33), (3), (7) are proved. Next, we shall discuss the equality conditions in (7). Clearly, the values of k 1 ; k ; k 3 may be zero. In fact, it is easy to prove that one of k 1 ; k ; k 3 at most equal zero(we omit details). Hence by Lemma.5, we will consider two cases below to discuss the equality condition in (36) and further that of (7). Case 1. None of k 1 ; k ; k 3 equals zero. First, we will prove that equality in (7) holds only when the barycentric coordinates of P is (ak 1 ; bk ; ck 3 ). Then we will show that this point just is the symmetrical point I 0 of the incentre I with respect to the circumcenter O of 4ABC: By the hypothesis, three strict inequalities 4p p 3 q1 > 0; 4p 3p 1 q > 0; 4p 1 p q3 > 0 follow from (40), (41), (4) respectively. In addition, we have (49) (50) (51) p 1 q 1 + q q 3 = 4abcu 1 v 1 w 1 ; p q + q 3 q 1 = 4abcu v w ; p 3 q 3 + q 1 q = 4abcu 3 v 3 w 3 ;

9 76 Jian Liu where u 1 = a 3 (b + c)a a(b + c 6bc) + (b + c)(b c) ; u = b 3 (c + a)b b(c + a 6ca) + (c + a)(c a) ; u 3 = c 3 (a + b)c c(a + b 6ab) + (a + b)(a b) ; v 1 = b 3 + (c a)b b(a c) + (a c)(a + c) ; v = c 3 + (a b)c c(b a) + (b a)(b + a) ; v 3 = a 3 + (b c)a a(b c) + (c b)(c + b) ; w 1 = c 3 + (b a)c c(a b) + (a b)(a + b) ; w = a 3 + (c b)a a(b c) + (b c)(b + c) ; w 3 = b 3 + (a c)b b(a c) + (c a)(c + a) : According to Lemma.5, we know that the equalities of (36) and its equivalent form (33) hold if and only if xu 1 v 1 w 1 = yu v w = zu 3 v 3 w 3. Further, the equality in (3) holds if and only if x a u 1v 1 w 1 = y b u v w = z c u 3v 3 w 3 : Then by Lemma.3, the equality in (7) holds if and only if P coincide with the point I 0 whose barycentric coordinates is a b u 1 v 1 w 1 ; u v w ; c u 3 v 3 w 3. This barycentric coordinates is much complicated. In fact it can be expressed in a simple way. The author nds that the following continued equality: (5) u 1 v 1 w 1 k 1 = u v w k = u 3 v 3 w 3 k 3 holds. It can easily be checked by using the mathematics software. Therefore, the barycentric coordinates of I 0 can also be expressed by (ak 1 ; bk ; ck 3 ). Now, we will prove that I 0 is the symmetrical point I 0 of the incentre I with respect to the circumcenter O of 4ABC. We rst prove three points I; O; I 0 are collinear. As it is well known, the barycentric coordinates of the incentre I and the circumentre O are (a; b; c) and a (b + c a ); b (c + a b ); c (a + b c ) respectively. Hence according to Lemma.4, to prove I; O; I 0 to be collinear, we need to prove that a b c a (b + c a ) b (c + a b ) c (a + b c ) ak 1 bk ck 3 = 0: Again, by the properties of the determinant, the proof can be turned into k 1 k k 3 a(b + c a ) b(c + a b ) c(a + b c ) = 0; Expending gives (53) k 1 b(c + a b ) c(a + b c ) +k c(a + b c ) a(b + c a ) +k 3 a(b + c a ) b(c + a b ) = 0:

10 Two inequalities for a point in the plane of a triangle 77 Putting (46)-(48) into the left of (53), one can easily check that (53) is true. Hence we proved that I; O; I 0 are collinear. Successively, we will prove that I 0 is the the symmetrical point of I with respect to the circumenter O. In Lemma.1, for P = O, then we get the known formula: (54) OM = R yza + zxb + xyc (x + y + z) : Letting M = I 0 in the above, we obtain that (55) OI 0 = R abc(ak k 3 + bk 3 k 1 + ck 1 k ) (ak 1 + bk + ck 3 ) : By (46), (47) and (43), it is easy to get two identities: (56) (57) so that ak 1 + bk + ck 3 = (a + b + c)(b + c a)(c + a b)(a + b c); ak k 3 + bk 3 k 1 + ck 1 k = (a + b + c)(b + c a) (c + a b) (a + b c) ; (58) OI 0 = R abc a + b + c : Since a + b + c = s and abc = 4Rrs (r is the inradius of 4ABC), we further get OI 0 = R Rr which is the same as the famous Euler formula OI = R Rr (see e.g. [, page 79]). Thus we have OI 0 = OI. Since also we have proved that I 0 ; O; I are collinear before, the point I 0 therefore must be the symmetrical point of the incentre I with respect to the circumcenter O. Case. One of k 1 ; k ; k 3 equals zero. There is no harm in assuming that k 1 = 0. In this case, 4p p 3 q1 = 0 follows from (40). Thus by Lemma.5, we know that the equality in (36) holds if and only if x = 0; p y = q 1 z. Further, the equality in (3) holds only when y x = 0; p b = q z 1 c ; i.e. the barycentric coordinates of P is (0; bq 1 ; p c). Indeed, this point just is the point with barycentric coordinates (0; bk ; ck 3 ). Due to the homogeneity of barycentric coordinates (cf. [6]) it will be su cient to prove that q 1 : k = p : k 3, i.e. (59) p k q 1 k 3 = 0: It is easy to check that p k q 1 k 3 = ab(a + b c)[a 3 + (3c b)a a(b + c) + (b c)(3c + b)(b + c)]k 1 ; which shows obviously that (59) holds true when k 1 = 0. Again, from the proof of Case 1, we see that P is still the symmetrical point I 0 of the incentre I with respect to the circumcenter O if k 1 = 0. Therefore, the equality in (7) under the second case holds only when P = I 0. The point I 0 does not lie on the boundary of 4ABC under Case 1 (since k 1 k k 3 6= 0). The point I 0 lies the boundary (except the vertices) of 4ABC under Case. Combing with the arguments of Case 1 and Case, we know

11 78 Jian Liu that the statements for the equality in Theorem 1.1 under these two cases are right. The proof of Theorem 1.1 is completed... Proof of Theorem 1.. Lemma.6. Let I be the incentre of 4ABC and let O be its circumcenter. The symmetrical point of I with respect to O is I 0. Then the distance between I 0 and the vertices A is given by (60) I 0 A = R p 1 sin B sin C; where R is the circumradius of 4ABC and B; C are the angles of 4ABC. Proof. In the proof of Theorem 1.1, we have known that the barycentric coordinates of I 0 is (ak 1 ; bk ; ck 3 ) (where the values of k 1 ; k ; k 3 are the same as in (46), (47), (48)). Hence by Lemma. we have that (61) I 0 A = bc(k c + k 3 b) ak 1 + bk + ck 3 abc(ak k 3 + bk 3 k 1 + ck 1 k ) (ak 1 + bk + ck 3 ) : Putting k ; k 3 into (61) and using (56), (57), we further get (6) I 0 A = bc(4bca + a 4 + b 4 + c 4 b c c a a b ) b c + c a + a b a 4 b 4 c 4 : Using the equivalent form of Heron s formula: (63) 16S = b c + c a + a b a 4 b 4 c 4 ; it follows that I 0 A = bc(16asr 16S ) 16S = (ar S)bc S = bc(ar ah a) S = abc(r h a) = R(R h a ) = R(R R sin B sin C) S = 4R (1 sin B sin C): Hence I 0 A = R p 1 sin B sin C is valid. This completes the proof of Lemma.6. As a straightforward important consequence of Lemma.1, we have that Lemma.7. For any point P in the plane of 4ABC and all real numbers x; y; z, (64) (x + y + z)(xp A + yp B + zp C ) yza + zxb + xyc ; with equality if and only if x : y : z = ~ S 4P BC : ~ S 4P CA : ~ S 4P AB. The inequality (64) is called The polar moment of the inertia inequality of Klamkin. This is one of the most important results for triangle geometric inequalities. A number of triangle inequalities can be derived from it (see e.g. [15], [16], []) We now prove Theorem 1.. Proof. Clearly, we can assume that the orientation of 4ABC is counterclockwise (see gure 1), then ~ S = S and we have to prove (65) ~r 1 + ~r + ~r 3 + R ~ S p S R:

12 Two inequalities for a point in the plane of a triangle 79 By Lemma.6, Lemma.7 and the sine rule, we obviously get the following weighted trigonometric inequality: (66) (x + y + z) [x(1 sin B sin C) + y(1 sin C sin A) + z(1 sin A sin B)] yz sin A + zx sin B + xy sin C; with equality if and only if x : y : z = ak 1 : bk : ck 3 (k 1 ; k ; k 3 are the same as in (46), (47), (48), respectively). That is (x + y + z) (x + y + z)(x sin B sin C + y sin C sin A + z sin A sin B) + yz sin A + zx sin B + xy sin C: Making substitutions x! x sin A; y! y sin B; z! z sin C, we get (x sin A + y sin B + z sin C) sin A sin B sin C[(x + y + z)(x sin A +y sin B + z sin C) + yz sin A + zx sin B + xy sin C]: Multiplying both sides by 4R and using the law of sines and the known formula S = R sin A sin B sin C, we obtain the equivalent inequality: (67) (xa + yb + zc) S [(x + y + z)(xa + yb + zc) + yza + zxb + xyc]; R with equality if and only if x : y : z = k 1 : k : k 3. If we put x = ~r 1 ; y = ~r ; z = ~r 3 in (67), then using the identities a~r 1 + b~r + c~r 3 = S (by (6) and hypothesis) and (68) we get Hence a~r ~r 3 + b~r 3 ~r 1 + c~r 1 ~r = 4R ~ S p ; 4S S R [S(~r 1 + ~r + ~r 3 ) + 4R ~ S p ]: SR S(~r 1 + ~r + ~r 3 ) + R ~ S p ; and a division by S produces inequality (65). By virtue of the equality condition of (67), it is seen that the equality in (65) holds if and only if the the barycentric coordinates of P is (ak 1 ; bk ; ck 3 ), i.e. P coincides with point I 0. The proof of Theorem 1. is complete. 3. Some remarks Remark 3.1. We have the following inequality similar to (8): (69) R + R 3 r r 3 r b r c + R 3 + R 1 r 3 r 1 r c r a + R 1 + R r 1 r 3 r a r b ; which is equivalent with (70) R + R 3 r r3 s a + R 3 + R 1 r3 r1 s b + R 1 + R r1 r 3 s c s: The equality in (69) or (70) is the same as in (7). In fact, by using Lemma. and Lemma.3 we can prove the following geometric identity (we omit details here): (71) X R 1 r1 = 1 X R + R3 r r3 ; h a r a r b r c

13 80 Jian Liu where P denote the circle sum. Therefore, (69) can be obtained by (8) and (71). Remark 3.. The point I 0 in Theorem 1.1 or Theorem 1. may be either in the interior (including the boundaries, except the vertexes) of 4ABC or outside the triangle. We have found some properties about point I 0. For example, four points I 0 ; N; H; I form a parallelogram, where N is the Nagel point, H the orthocenter and I the incenter of 4ABC respectively (see Figure 3). Figure 3 Remark 3.3. Actually, the inequality (7) of Theorem 1.1 is a generalization of Heron s formula. In fact, by using (31) we can prove the following equality: (7) R 1 r 1 = s bc holds when P = I 0. Thus by Theorem 1.1 we get the identity: a(s a)(s bc) + b(s b)(s ca) + c(s c)(s ab) = S : Further, it is easy to obtain the Heron s formula: (73) S = p s(s a)(s b)(s c): In addition, when P = I 0 we have the following equality similar to (7): (74) R + R 3 r r 3 = (s a) which evidently shows the equality condition in (70) is right. Remark 3.4. The inequality (9) of Theorem 1. actually is equivalent to the weighted inequality (67) and the later can also be proved by Lemma.5. In addition, by inequality (9) and the known inequality (see [17]): (75) S p r p S R ; where r p is the inradious of the pedal triangle DEF of interior point P with respect to triangle ABC and S p is its area, it is seen that the beautiful linear inequality (76) r 1 + r + r 3 + r p R holds for any interior point P of 4ABC.

14 Two inequalities for a point in the plane of a triangle 81 Remark 3.5. From inequality (9) and the known inequality used in [19] recently: (77) RS p S r 1r r 3 r ; we can get the following inequality (for interior point P ): (78) r 1 + r + r 3 + r 1r r 3 r R; which does not discriminate strength or weakness with (76). Remark 3.6. If we apply geometrical transformations to Theorem 1.1 or Theorem 1. or their consequences, one can obtain some new geometric inequalities. For example, applying the isogonal transformation (see e.g.[], [3]) to inequality (8), we get (79) a(s a)(r 1r 1 r r 3)+b(s b)(r r r 3r 1)+c(s c)(r 3r 3 r 1r ) 8R S p; which holds for any point P in the plane. Applying inequality (10) and other geometrical transformations, we can obtain the following inequalities (holds for any interior point P of triangle ABC): (80) 1 R R + 1 R R p R 1R R 3 r 1 r r 3 R ; (81) 1 R a + 1 R b + 1 R c + 1 R S RS p ; (8) R a + R b + R c + R R p R 1R R 3 r 1 r r 3 ; where R a ; R b ; R c are the circumradius of 4P BC; 4P CA; 4P AB respectively. 4. Two conjectures In [14], we have posed some conjectures for the Erdös-Mordell inequality. Here, we present two related new conjecture again. Conjecture 4.1. For any interior point of 4ABC, we have (83) R 1 + R + R 3 r 1 + r + r 3 R r p : Conjecture 4.. For any interior point of 4ABC, we have (84) R 1 + R + R 3 r 1 + r + r 3 R + R p r + r p : In passing, we have known that there is no comparison between (83) and (84).

15 8 Jian Liu References [1] Avez, A., A short proof of a theorem of Erdös and Mordell, Amer. Math. Monthly, 100(1993), [] Alsina, C. and Nelsen, R.B., A visual proof of the Erdös-Mordell inequality, Forum Geom., 7(007), [3] Bbanko, L., An elementary proof of the Erdös CMordell theorem, Amer. Math. Monthly, 65(1958), 51. [4] Bombardelli, M. and Wu, S.H., Reverse Inequalities of Erdös-Mordell type, Math.Inequal.Appl., 1()(009), [5] Bottema, O., On the Area of a Triangle in Barycentric Coordinates, Crux. Math., 8(198), [6] Bµandilµa, V., O generalizare a unei relatii a lui Leibniz si aplicarea ei la calculul distantlor dintre unele puncte remarcabile ale unui triunghi, Gaz. Mat (Bucharest), 90(1985), [7] Coxeter, H.S.M., Barycentric Coordinates, Introduction to Geometry, nd ed, New York: Wiley, 16 1, [8] Dergiades, N., Signed distances and the Erdös-Mordell inequality, Forum Geom., 4(004), [9] Dar, S. and Gueron, S., A weighted Erdös CMordell inequality, Amer. Math. Monthly, 108(001), [10] Erdös, P., Problem 3740, Amer. Math. Monthly, 4(1935), 396. [11] Janous, W., Further inequalities of Erdös CMordell type, Forum Geom., 4(004), [1] Komornik, V., A short proof of the Erdös-Mordell theorem, Amer. Math. Monthly, 104(1997), [13] Lee, H., Another proof of the Erdös-Mordell theorem, Forum Geom., 1(001), 7 8. [14] Liu, J., A new proof of the Erdös-Mordell inequality, Int. Electron. J. Geom., 4()(011), [15] Liu, J., A new geometric inequality and its applications, J.Inequal.Pure Appl.Math., 9()(008), art.58. [16] Liu, J., A weighted geometric inequality and its applications, Journal of science and arts, 14(1)(011), 5 1. [17] Liu, J., Some new inequalities for an interior point of a triangle, J. Math. Inequal., 6()(01), [18] Liu, J., Several new inequalities for the triangle, Mathematics Competition (in Chinese), 15(199), [19] Liu, J., A pedal triangle inequality with the exponents, Int. J. Open Problems Compt. Math., 6(4)(01), [0] Leon, S. J., Linear Algebra with Applications, Prentice Hall. New Jersey, 005. [1] Mordell, L. J. and Barrow, D. F., Solution of Problem 3740, Amer. Math. Monthly, 44(1937), [] Mitrinović, D. S., Peµcarić, J. E. and Volenec, V., Recent Advances in Geometric Inequalities, Kluwer Academic Publishers, Dordrecht-Boston-London [3] Oppenheim, A., The Erdös-Mordell Inequality and other inequalities for a triangle, Amer. Math. Monthly, 68(1961), [4] Pambuccian, V., The Erdös-Mordell inequality is equivalent to non-positive curvature, J. Geom., 88(008), [5] Satnoianu, R. A., Erdös - Mordell type inequality in a triangle, Amer. Math. Monthly, 110(003), [6] Yiu, P., The uses of homogeneous barycentric coordinates in plane euclidean geometry, Int. J. Math. Educ. Sci. Technol., 31(000), EAST CHINA JIAOTONG UNIVERSITY JIANGXI PROVINCE NANCHANG CITY, , CHINA address: China99jian@163.com