some geometric inequalities

Journal of Science and Arts Year 12, No 3(20), pp 291-296, 2012 ORIGINAL PAPER SOME GEOMETRIC INEQUALITIES OF RADON-ERDÖS-MORDELL TYPE DUMITRU M BĂTINEŢU-GIURGIU , NECULAI STANCIU _ Manuscript received: 11.07.2012; Accepted paper: 02.09.2012; Published online: 15.09.2012 Abstract Some Erdös-Mordell type inequalities for general convex poligons are presented.The main tool in the proofs is the Radon inequality Keywords: Erdös-Mordell type inequality, Radon inequality, convex polygon INTRODUCTION The purpose of this article is to establish some geometric inequalities (other than [2] ) on Erdös-Mordell - type , in convex polygons, used J Radon’s inequality (see for example[1]) n n k 1 k 1 Let a, b, c, d , xk , yk  R , k  1, n and X n   xk , Yn   yk *  Theorem (A generalization of J Radon’s inequality) If m, p, q, s  R , r  1,   such that cYns  d max yks , k  1, n , then: 1 k  n aX  bx  x  cY  dy  y p n n s n k 1 q m 1 r ( m 1) k k s m m k k an X  bX   cn  d  Y q  r m 1 n pr n q m s m ( s 1) n  n ( m 1)( q  r 1)  ms (1) Proof: We denoted  uk  aX  bx p n q k  x , v   cY r k s n k  dy s k n  y , k  1, n , V   v k n k 1 k and the LHS of (1) becomes: m 1 m 1 n  v u    Vn  k  k  k 1 Vn  v k   * * m 1 Since the function f : R  R , f ( x)  x is convex on R* , we use Jensen’s inequality and we obtain that: n  uk u km 1  v    k m k 1 v k k 1  vk n vk  u k  n f      k 1 Vn  v k  k 1 Therefore, n v u  n f  k  k     Vn v k  k 1  n    uk  u  f  k    k 1 m 1 Vn  Vn  m 1 v  k k 1 Vn n  uk   vk    m 1  n    uk    k 1 m 1 Vn m 1 “Matei Basarab” National College, Bucharest, Romania “George Emil Palade” General School, Buzau, Romania E-mail: stanciuneculai@yahoo.com ISSN: 1844 – 9581 Mathematics Section Some geometric inequalities … 292 Dumitru Batinetu-Giurgiu, Neculai Stanciu  n    uk   Vn   k 1 m 1 Vn u km 1  m k 1 v k n aX  bx  x  cY  dy  y q m 1 r ( m 1) k k s m m k k p n n s n k 1 m 1  n    uk    k 1 m  Vn m 1  m 1  n    aX np  bx kq x kr  k 1    m n     cYns  dy ks y k   k 1     m 1  m 1 m 1 n n n n  n        aX np x kr  bx kq  r   aX np  x kr  b x kq  r   aX np  x kr  b x kq  r  k 1 k 1 k 1 k 1      k   m m m n n n n    s   s 1    cYns y k  dy ks 1   cYn  y k  d  y ks 1   cYn  d  y ks 1  k 1 k 1 k 1  k 1       * r qr s1 Because the functions g , h, k : R  R , g ( x)  x , h( x)  x , k ( y )  y are convex     on R* , also by Jensen’s inequality we have: n X nr X nr 1 n  r ( )       x g x ng x n      k k k nr n r 1 , k 1 k 1  n k 1  n n X nq  r X nq  r 1 n  qr x k   h( x k )  nh   x k   n  q  r  q  r 1  n n k 1 k 1  n k 1  , s 1 s 1 n n n Y Y 1  y is 1   k ( y i )  nk    y i   n  ns 1  n s  n n i 1 i 1  n i 1  n Then, we deduce that: aX  bx  x  cY  dy  y p n n s n k 1 an X  bX   cn  d  Y q  r m 1 n pr n q s m m ( s 1) n q m 1 r ( m 1) k k s m m k k  n ms n ( m 1)( q  r 1)  X np  r X qr   a  r 1  b  q n r 1  n n   m  s 1 dYns 1   cYn  s  n   an X  bX   cn  d  Y q  r m 1 n pr n q s m 1 m m ( s 1) n   n ( m 1)( q  r 1)  ms , and we are done ■ Observation 1.1 If p  q  s  , then (1) becomes: a  b m1 xkr ( m 1)  c  d m y km k 1 n a  b m1 X nr ( m1)   ( m 1)( r 1) m m n c  d  Yn  x kr ( m 1) X nr ( m 1)   y km Ynm n ( m 1)( r 1) k 1 n (1) If we consider r = 1, then by (1’) we obtain: n x km 1 X nm 1  m  m Yn k 1 y k (R) i.e that is just the inequality of J Radon, with equality if and only if there exists t  R such *  that xk  tyk , k  1, n www.josa.ro Mathematics Section Some geometric inequalities … Dumitru Batinetu-Giurgiu, Neculai Stanciu 293 Observation 1.2 If m = 1, then (1) becomes: aX  cY  bx kq  x k2 r an q X np  r  bX nq  r    2( q  r 1)  s s s s s 1 cn  d Yn n k 1 n  dy k y k If we take p  q  s  0, r  , then by (1”) we obtain: n x k2 X n2   Yn k 1 y k n p n (1) (B) but that is just the inequality of H Bergström Next, we consider A1A2…An (n≥3), a convex polygon and for any point M from inside the polygon we use the notations: xk = MAk, yk the distance from M to the line AkAk+1, ak the length of the side [AkAk+1] of the polygon k  1, n , 2p is the perimeter of the polygon, S is   the area of the given polygon and An+1≡A1 and Theorem If A1A2…An (n≥3) is a convex polygon where we use the above notations a, b, c, d  R* m, p, q, s  R , r  1,   , such that cYns  d max y ks , k  1, n 1 k  n then: aX  bx  x  cY  dy  y p n n s n k 1 q m 1 r ( m 1) k k s m m k k an X  bX   cn  d  X q  r m 1 n pr n q m s m ( s 1) n ,     sec  n  m ( s 1)  n ( m 1)( q  r 1)  ms (2) Proof: By L.Fejes Tóth’s inequality (see e.g [2]) n n      y Y cos   x k   cos  X n  k n n  k 1 n ,   k 1 and by (1) we deduce what we have to show ■ Observation 2.1 If we put p =q, then by (2) we obtain: aX  bx  x  cY  dy  y n p n k 1 s n p m 1 r ( m 1) k k s m m k k an  b X  cn  d  X an  b X  cn  d  n m 1 p m s m 1 p s m ( p  r )( m 1) n m ( s 1) n     sec  n  m ( p  r  s 1)  p  r n ( m 1)( p  r 1)  ms m ( s 1)      sec  n  n ( m 1)( q  r 1)  ms  m ( s 1) (2’) If we consider p = s = 0, then by (2’) we deduce that: m x kr ( m 1)  X nm ( r 1)  r  m y k 1 k n    sec  n   ( m 1)( r 1) n (2”) If we take r = then by (2”) yields: m n x km 1    X n  sec   m n  k 1 y k (2”’) Remark 2.1 Putting m =1 in (2”’) we obtain the relation (18) from [2] ISSN: 1844 – 9581 Mathematics Section Some geometric inequalities … 294 Dumitru Batinetu-Giurgiu, Neculai Stanciu Theorem If we have the notations presented above, then n x km 1 X nm 1  , m  R  , p  N *  1  m 1 m m p p Y k 1 ( y k y k 1 y k  p 1 ) n (3) where yn  j  y j , j  0, p 1 Proof: By AM-GM inequality we have: p 1 p 1 p y   y k  j , k  1, n  k j p j 0 j 0 , which yields n  k 1 x km 1 x km 1 n  p 1  p     yk  j   j 1  m  p k 1  p 1    yk  j     j 0  m , then by (R) xkm 1 n  k 1 p  p 1    yk  j   j 1  m  p X nm 1 X nm 1  , p mYnm p m 1Ynm which completes the proof ■ Observation 3.1 By (3) and L Fejes Tóth’s inequality we deduce that: m n x km 1 X nm 1 Xn    sec     m m n p m 1   k 1 m 1 m  p 1  p X n  cos  p   2   yk  j    j 1  (3’) which is a Radon-Erdös-Mordel type inequality If we consider m = 1, then by (3’) we obtain that: n x k2   X n sec  p 1 n k 1 p  yk  j j 0 (3”) For the same convex polygon A1A2…An (n≥3), An+1≡A1 and for any point M of space which is not on the line AkAk+1, we denoted by yk(M) the distance from M to line AkAk+1, n sk ( M )  area  Ak MAk 1  , k  1, n and S ( M )   sk ( M ) k 1  Theorem If M and N are the points in space which is not on the line AkAk+1, k  1, n , then:   1  a k   m  y k ( N ) m k 1   yk (M ) n www.josa.ro   p m 1 S ( M ) m  S ( N ) m   S ( M ) S ( N )  m   (4) Mathematics Section Some geometric inequalities … Dumitru Batinetu-Giurgiu, Neculai Stanciu Proof: We have: n   n m 1  1 1    ak    U n   a k  m m  m   y k ( N )   k 1  a k y k ( M )  a k y k ( N ) m k 1   y k (M ) n a m 1  1     k m  m  m k 1  S k (M ) S k ( N )  , where we apply (R) and yields m 1 m 1    n   n     a  a    k k    k 1  k 1    Un  m   m m   m  n      Sk (N )     S k (M )   k 1      k 1 m 1 2 p     p m1 S ( M ) m  S ( N ) m  m  S ( M ) m S ( N ) n   S ( M ) S ( N ) m = , and this is what we had to prove  295      ■ Observation 4.1 If M is a point inside of the polygon A1A2…An and N is a point outside of the plane (A1A2…An), then the polygon is the base of the pyramid with the vertex N, so that (4) becomes: n   p m 1 S m  S lm 1   a   k m m      S  S l m y M y N ( ) ( ) k 1 k  k  (4’) where S is the area of the polygon with the perimeter 2p, and Sl is the lateral area of the pyramid with vertex N and base the polygon A1A2…An If M and N are inside of the polygon then S(M) = S(N) = S, and (4’) becomes: n   p m 1 S m p m 1 1    a   k m m  2m S Sm     ( ) ( ) y M y N k 1 k  k  (4”) If in addition M ≡ N, then we obtain that: n n ak ak p m 1 p m 1    2  m m Sm Sm k 1  y k ( M )  k 1  y k ( M )  (4”’)   Remark 4.1 Taking m =1, then by (4”’) we deduce that: n ak p2   S , k 1 y k ( M ) i.e the problem 10876 , proposed by D Buşneag in G.M.-B nr 1/1971, pp.35 For the convex polygon A1A2…An (n≥3), An+1≡A1 and M a point in space which is not on the line AkAk+1, k  1, n , we denote by mk    Ak MAk 1  , k  1, n , the measure in   radians of the angle Ak MAk 1 , k  1, n Theorem If A1A2…An (n≥3) is a convex polygon as above, and M is a point in space   which is not on the line AkAk+1, k  1, n , with mk   0,  , k  1, n , then:  2  ISSN: 1844 – 9581  Mathematics Section Some geometric inequalities … 296 Dumitru Batinetu-Giurgiu, Neculai Stanciu a k2 m  m2   mn  4n sin  2n k 1 x k x k 1 n (5) Proof: In the triangle AkMAk+1, by the Law of Cosines we have, a k2  x k2  x k21  x k x k 1 cosAk MAk 1   x k2  x k21  x k x k 1 cos mk , k  1, n , where we apply AM-GM inequality and we obtain that: m a k2 m a k2  x k x k 1 1  cos mk   x k x k 1 sin k , k  1, n   sin k , k  1, n x k x k 1 Hence: n a k2 m  sin k   k 1 x k x k 1 k 1 x     Since the function f :  0,   R , f ( x)  sin is convex on  0,  , by Jensen’s  2  2 inequality we have that: n m m  m2   mn sin k  n sin  2n k 1 n From the last two relations we get what must be demonstrated Remark 5.1 If n≥5, M  IntA1 A2 An , then n m k 1 k ■  2 and (5) becomes: a k2   4n sin  n, k 1 x k x k 1 i.e the problem11386, proposed by R.N Gologan in G.M.-B nr 8/1971, pp.487 n REFERENCES [1] [2] Bătineţu-Giurgiu, D.M., Gaz Mat.-B, 115(7-9), 359, 2010 Minculete, N., Gobej, A., Gaz Mat - A, 28(1-2), 1, 2010 www.josa.ro Mathematics Section .. .Some geometric inequalities … 292 Dumitru Batinetu-Giurgiu, Neculai Stanciu  n    uk   Vn   k 1... if there exists t  R such *  that xk  tyk , k  1, n www.josa.ro Mathematics Section Some geometric inequalities … Dumitru Batinetu-Giurgiu, Neculai Stanciu 293 Observation 1.2 If m = 1, then... m =1 in (2”’) we obtain the relation (18) from [2] ISSN: 1844 – 9581 Mathematics Section Some geometric inequalities … 294 Dumitru Batinetu-Giurgiu, Neculai Stanciu Theorem If we have the notations