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(1)ß.  July 15, 2005.                                

(2)   

(3)                   n  n ≥ 6    . 1.

(4) 1.

(5)  .  . f (x1 , x2 , . . . , xn ). x1 , x2 , . . . , xn. .    1.1.1. P P . . . P  n  n  (P P . . . P )   i, j  i ≡ j (mod n)   P  P 

(6)   P P P   n   (i)      1.1.2.  Monge   P P P P P  1.1. 0. 1. n−1. 0. i. 0. 1. j. 1. 2. n−1. i−1. 3. i. i+1. 4. (P0 P1 P2 )(P0 P3 P4 ) + (P0 P2 P3 )(P0 P1 P4 ) = (P0 P1 P3 )(P0 P2 P4 )..    I  P P , P P , P P , P P , T , T , T , T  0. 1. 0. 2. P4. I. 0. 3. 0. 4. 1. 2. 3. 4. P3. T4 P2. T3 T2. P0. T1 P1.  (P T T ) = p, (P T T ) = q, (P T T ) = r 0 1 2. 0 2 3. 0 3 4. pr + q(p + q + r) = (p + q)(q + r),.  (P0 T1 T2 )(P0 T3 T4 ) + (P0 T2 T3 )(P0 T1 T4 ) = (P0 T1 T3 )(P0 T2 T4 ). (1).    1 ≤ i, j ≤ 4, i = j   (P0 Pi Pj ) P0 Pi × P0 Pj . = (P0 Ti Tj ) P0 Ti × P0 Tj 2. (2).

(7)  α=.  (1) (2)  . P0 P1 × P0 P2 × P0 P3 × P0 P4 , P0 T1 × P0 T2 × P0 T3 × P0 T4. (P0 P1 P2 )(P0 P3 P4 ) + (P0 P2 P3 )(P0 P1 P4 ) − (P0 P1 P3 )(P0 P2 P4 )   = α (P0 T1 T2 )(P0 T3 T4 ) + (P0 T2 T3 )(P0 T1 T4 ) − (P0 T1 T3 )(P0 T2 T4 ) = 0,.    P P P P P  0. 1. 2. 3. 4. c1 = (0) + (1) + (2) + (3) + (4),. . (3). c2 = (0)(1) + (1)(2) + (2)(3) + (3)(4) + (4)(0)..  c , c  

(8)     1.1.3.    P P P P P  A  Monge. 1. 2. 0. 1. 2. 3. 4. A2 − c1 A + c2 = 0..   (P0 P1 P2 ) = (1), (P0 P3 P4 ) = (4), (P0 P1 P4 ) = (0), (P0 P2 P3 ) = A − (1) − (4), (P0 P1 P3 ) = A − (2) − (4), (P0 P2 P4 ) = A − (1) − (3),.  Monge  .      (1)(4) + (0) A − (1) − (4) = A − (2) − (4) A − (1) − (3) .. . A2 − c1 A + c2 = 0.. 3. (4).

(9)     P P P P P  A  0. 1. 2. 3. 4. x2 − c1 x + c2 = 0. (5).           P P P P P  A  P P P P P   P P P P P  A  P P P P P P P P P P   0. 0. 2. 4. 1. 1. 2. 3. . 4. . 3. 0.  3. 1.  4.  0. 2.  0. 3.  1. 4.  1.  2.  3.  4.  2. A = (P0 P3 P2 ) + (P1 P4 P3 ) + (P2 P0 P4 ) + (P3 P1 P0 ) + (P4 P2 P1 ) + A . P0. P3. P1. P2 O. P4. P4 P1. P0 P2. P3.

(10)  1.1.4.  x − c x + c = 0   A  A − (A + A )   P P P P P 

(11)  O  (OP P ) = t , 0 ≤ i, j ≤ 4  2.  0. 1.  1.  2. . 2.  3.  4. . i j. ij. A = t01 + t12 + t23 + t34 + t40 , A + A = t02 + t24 + t41 + t13 + t30 ,.  (i) = ti−1 i + ti. . c1 =. 4 . i+1. − ti−1. i+1 ,. 0 ≤ i ≤ 4.. (i) = 2A − (A + A ).. i=0.   A        A − (A + A ) . . 4.

(12)      (3)  (4)     a , a , a , a , a      A     

(13) (a , a , a , a , a , A)   1.2. 1. 1. 2. 3. 4. 2. 3. 4. 5. 5. f (x1 , x2 , x3 , x4 , x5 , x) = x2 −(x1 +x2 +x3 +x4 +x5 )x+(x1 x2 +x2 x3 +x3 x4 +x4 x5 +x5 x1 ) = 0.       

(14) (a , a , a , a , a , A)      (a , a , a , a , a , A)     

(15)      1.1.3 

(16)  1.1.4   

(17) 

(18)  1.2.1.  A   P P P P P  C , C  1. 1. 2. 3. 4. 2. 3. 4. 5. 5. 0. C1 = C2 =. 4  i=0 4 . 1. 2. 3. 4. 1. 2. (Pi Pi+1 Pi+2 Pi+3 ), (Pi Pi+1 Pi+2 Pi+3 )(Pi+1 Pi+2 Pi+3 Pi+4 ),. i=0.  (A, C , C )      1. 2. (Pi Pi+1 Pi+2 Pi+3 ) = A − (Pi+3 Pi+4 Pi ),. 0 ≤ i ≤ 4,.  (3)  C1 = 5A − c1 , C2 = 5A2 − 2c1 A + c2 ..  1.1.3  A2 − C1 A + C2 = A2 − (5A − c1 )A + (5A2 − 2c1 A + c2 ) = 0..  5.

(19)

(20)  1.2.2.  A   P P P P P 

(21)  1.1.4  0. 1. 2. 3. 4. (A + A , c1 , c2 ).  . c1 = 2A − (A + A ),   c2 = A A − (A + A ) ..   2   (A + A )2 − c21 + 4c2 = (A + A )2 − 2A − (A + A ) + 4A A − (A + A ) = 0,.  

(22)   

(23)           

(24) 

(25) .   1.2.3.  ABC  a  AB, BC, CA  F, D, E   AD, BE, CF   ABE  b  BCF  c  CAD  d  (a, b, c, d)   A E F. B. D. C.     . AF : F B = 1 : s, BD : DC = 1 : t.. CE : EA = st : 1, 1 c s d t b = , = , = . a 1 + st a 1+s a 1+t 6.

(26)  s=.  . c d ,t = c−a d−a. a3 − (b + c + d)a2 + (bc + cd + db)a − 2bcd = 0,.   1.2.4.    a  ABCD  a , a , a    (a, a , a , a )   1. 1. 2. 2. 3. 3. A. M. D a2. a1 N a3 B. C.   MD = s, NB = t  a1 (a − a1 ) = a1 (a2 + a3 + st) = a1 a2 + a1 a3 + a2 a3 ,.   

(27)             . 7.

(28)       . 2.   

(29)     2.1.1. C , C , . . . , C    f (x , x , . . . , x , y)  x ,x ,...,x ,y   t ∈ C , i = 1, 2, . . . , n  2.1. 1. 1. 2. 2. n. 1. n. i. 2. n. i. f (t1 , t2 , . . . , tn , y) = 0.    f (x , x , . . . , x , y)   2.1.1  

(30)            2.1.2. ! P, Q  a  (P QR) = a  R   L 

(31)    L   

(32)  2.1.3.  a , a , a , a , a   (a , a , a , a , a )      a < a < a < a  a , a , a , a     a  P P P   P P  L  P  L  ∠P P P + ∠P P P > π 

(33)   P P " L .  P P # L  a > a = (P P P )  L   P P   P  P P    P , P  L 

(34)  1. 2. n. a P,Q. 1. 2. 3. 1. a P,Q. 4. 5. 1. 2. 3. 5. 1. 1. a2 P1 ,P2. 0. 0. 0. 1. 1. 1. 2. 0. 3. 2. a5 P0 ,P1. 0. 2. 3. 1. 1. 5. 2. 1. 2. 0. 1. 2. 1. 1. 2. LaP50 ,P1 P3 P0. P1. P2. 8. LaP21 ,P2. 3. 4. 5. 5. a2 P1 ,P2. 2. 2. 1. 2. 3. a5 P0 ,P1 a5 P0 ,P1. 3. a3 P2 ,P3.

(35)  L , L $  P P  P P   h P   P P # P   P P "

(36)  P  L "  L  L !  h 

(37)    P P P P P      P P P   (P P P ) 

(38)    f (x , x , x , x , x ) 

(39)  (a , a , a , a , a ) f (a , a , a , a , a )      a < a < a < a   x      f (a , a , a , x , a ) = 0          2.1.1  f (x , x , x , x , x )  a5 P1 ,P0. 1. a3 P2 ,P3. 0. 4. 4. 2. 4. 1. 3. 3. 2. 4. a2 P1 ,P2. a5 P1 ,P0. a3 P2 ,P3. 0. 0. 1. 1. 2. 3. 4. 2. 3. 2. 

(40)   2.2.1.  0 < t , t , t. ≤. 3. 4. 3. 0. 3. 4. 4. 3. 5. 4. 1. 1. 1. 2. 4. 2. 4. 5. 4. 1. 3. 1. 2. 2. 3. 3. 4. 5. 5. 5. 5. 2.2. 1. 2. 3. 1 4. . ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩.   x ,x ,x    1. 2. x1 x1 +1 x2 x2 +1 x3 x3 +1. × × ×. 1 x2 +1 1 x3 +1 1 x1 +1. = t1 (6). = t2 = t3. 3. (t3 − t1 t3 )x2 + (t1 + t2 + t3 − 2t1 t3 − 1)x − t1 t3 + t1 = 0..  0 < t ,t ,t 1. 2. 3. ≤. 1 4. (7). . D = (t1 + t2 + t3 − 2t1 t3 − 1)2 − 4t3 (1 − t1 )(−t1 t3 + t1 ) = (t1 + t2 + t3 − 1)2 − 4t1 t2 t3 ≥ 0,.  (7)       −.  0 < t , t , t  1. 2. 3. ≤. 1 4. t1 + t2 + t3 − 2t1 t3 − 1 , t3 (1 − t1 ).  1   (7)  x2 =. x1 − 1. t1 (x1 + 1) 9. 1 2.  x  1.

(41)  x > >    1 2. 1. 1 3. x1 x1 +1. >. 1 4.  0 < t x3 =.  (1 − t. 1. − t2 )x1 >. 1 2. ×. 1 2. =. 1 4. ≤. 1. ≥ t2. 1 4. x. 2. =. x1 t1 (x1 +1). x2 − 1. t2 (x2 + 1).  (1 − t )x 1. 1. + t1 > t2 x1. −1 >0. .  x  2. (1 − t1 )x1 + t1 x2 = > t2 , x2 + 1 x1.  x = − 1 > 0  % x2 t2 (x2 +1). 3. 1 x3 × = t3 , x3 + 1 x1 + 1.  x , x , x   (6)  x , x   1. 2. 3. 2. 3. 1 (1 − t2 )x2 − t2 (1 − t1 − t2 )x1 − t1 x3 1 × = = × . x3 + 1 x1 + 1 x2 x1 + 1 (1 − t1 )x21 + (1 − 2t1 )x1 − t1.  x  1. (t3 − t1 t3 )x2 + (t1 + t2 + t3 − 2t1 t3 − 1)x − t1 t3 + t1 = 0. . .   (1 − t1 − t2 )x1 − t1 = t3 (1 − t1 )x21 + (1 − 2t1 )x1 − t1 ,.  × = t   2.2.2.  a, b, c, d   a  ABC  AB, BC, CA  F, D, E  (AF E) = b, (F BD) = c, (EDC) = d   (a, b, c, d)  x3 x3 +1. 1 x1 +1. 3. A b. E. F d. c B. D. 10. C.

(42)  2.2.3.  a, b, c, d   a ≥ 4max{b, c, d}  (a, b, c, d)    a  ABC   0 < t ,t ,t 1. 2. 3. b c d t1 = , t2 = , t3 = , a a a. ≤ 1/4.  2.2.1   ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩. . x1 x1 +1 x2 x2 +1 x3 x3 +1. × × ×. 1 x2 +1 1 x3 +1 1 x1 +1. = t1 = t2 = t3.  x ,x ,x   AB  F  AF : F B = 1 : x

(43)  BC  BD : DC = 1 : x

(44)  CA  E  CE : EA = 1 : x   1. 2. 3. 1. 2. D. . 3. x1 1 (F BD) b = × = t1 = , (ABC) x1 + 1 x2 + 1 a.  (F BD) = b

(45)  (AF E) = c, (EDC) = d  (a, b, c, d)   2.2.4.  (a, b, c, d)      f (x , x , x , x )  (a, b, c, d) f (a, b, c, d)   2.2.3   a, b, c, d  a ≥ 4max{b, c, d}  (a, b, c, d)      b, c, d f (x , b, c, d)    2.1.1  f (x , x , x , x )  1. 2. 3. 4. 1. 1. 2. 3. 4. 11.

(46) 3.    . &          &

(47)       &   x   & P P P P P P  P = (0, 0), P = (1, 0), P = (1, 1), P = (0, 1 + x), P = (− , 1 + x), P = (− , 1) . & + + 2  2.1.1    3.1. 0. 2. 3. 4. 1 2. x 2. 1. 2. 1 x. 3. 4. 5. 5. 0. 1. 1 x. 1 2x. P4. P3 P2. P5. P0. P1.  & &    3.1.1.  && &   P4. P3. P5. P2. P0. P1.   & P P P P P P  0. 1. 2. 3. 4. 5. P0 P1 = P2 P3 = P4 P5 ,. . P5 P0 = P1 P2 ,. ∠P4 P5 P0 = ∠P5 P0 P1 = ∠P0 P1 P2 = ∠P1 P2 P3 = 12. 2π , 3.

(48) & P P P P P P   &  P P = P P ∠P P P = ∠P P P  P P P P   P , P , P , P  

(49)

(50)  P , P , P , P   P , P , P , P   & P P P P P P & 

(51)   P P  P P $  P P P P P P  & ∠P P P = ∠P P P =  P P P P   ∠P P P =   P P ! P P  P P  P P $  P P P P P P  & P P ! P P  P P P P   4. 0. 1. 0. 1. 5. 0. 2. 4. 4. 0. 1. 3. 5. 5. 4. 5. 1. 1. 2. 3. 0. 5. 4. 0. 2. 2. 0. 1. 0. 4. 1. 2. 0. 5. 1. 0. 2. 1. 3. 5. 1. 4. 4. 5. 1. 2. 5. 0. 1. 3. 3. 3. 4. 5. 2π 3. 2. 2. 3. 1. 2. 4. 4. 3. 0. 5. 0. 1. 2. 3. 4. 5. 1. 4. 5. 0. 1. 2. 3. 4. 1. π 3. 2. 5. 4. P0 P1 = P2 P3 = P4 P5 , P5 P0 = P1 P2 = P3 P4 ,. &  &  x   P0 P1 = x, P1 P2 =. 1 , x.    P P  P P $  P  P P P   P P P  x  & P P P P P P    &

(52)   2.1.1    √. 3 4. 0. 0. 1. 3. 1 4 √ 3 2 4. 0. 0. 1. 2. 3. 1. 4. 5.   

(53)       n ≥ 6  n      3.2.1.  n  3   a > a > · · · > a  n − 2  a ,a ,...,a  n  P P . . . P   3.2. 1. 1. 2. n−2. 0. 1. (Pi−1 Pi Pi+1 ) = ai ,. 2. n−2. n−1. i = 1, . . . , n − 2,. (8). ∠Pn−1 P0 P1 + ∠Pn−2 Pn−1 P0 < π..    2n  A , A , . . . , A  (A A A ) =   P A $ a  P = A , P = A , P = A  a > a   L 0. 1. 0. 0. 1. 1. 2. 2. 1. 13. 2. 1. 2n−1. a2 P1 ,P2. 0. 2. 1. 3. 2.

(54) .   P 

(55)   a > a > · · · > a    P A   (P P P ) = a , i = 3, 4, . . . , n − 2   P P . . . P    A A . . . A   P , P , . . . ,  '  ∠P P P + ∠P P P < π  n  P P . . . P   3. Pi+1. i−1. Pn−1. 2. i. i+1. 0. 1. n−2. i. i+1. i. n−1. 0. n−1. 0. 3. 1. 0. 1. n−2. 1. n−1. n−1. 0. 1. 0. n−1.  3.2.2.  n  5   a , a , . . . , a  n  n    (a , a , . . . , a )     a > a > · · · > a > a > a  n−1  a , a , . . . , a ,  a , a  3.2.1  n − 2  P P . . . P 1. 1. 2. n. n. n−1. n. 1. n−1. 2. 2. n−3. n. 0. 1. 1. 2. n−3. n−3. i = 1, . . . , n − 4,. (Pi−1 Pi Pi+1 ) = ai ,. ∠Pn−3 P0 P1 + ∠Pn−4 Pn−3 P0 < π..  n  5  P , P , . . . , P ! P , P , P , P    P  P P !  P T P   L   PT  PP   0. 0. 0. 0. n−3. 1. n−3. 0. n−4. 0. 1. n−4. n−3. an P0 ,P1. n−1. n−3. T Pn−1. P0. Pn−3 LaPn0 ,P1 Pn−4. P1.  P P  L  (P P P ) = a > a  P .  P P   L   P P    P , P  L    P P " L  a > a  P P ...P.  P , P  L   P L L $ h  P  P P    P   P P # P   P P "

(56)  P    L L !  h 

(57)    0. an−1 P0 ,Pn−1. 0 n−1 an−1 P0 ,Pn−1. n−2. n−2. an−1 P0 ,Pn−1. n−1. n−3. 1 n−3 an−1 P0 ,Pn−1. 0 n−1 an−1 P0 ,Pn−1. n−4 an−3 Pn−3 ,Pn−4. 0. 1. n−1 0. n−4. n−2. n−2 an−3 Pn−3 ,Pn−4. n−4. 14. n−1. 1. n−1. an−3 Pn−3 ,Pn−4. an−3 Pn−3 ,Pn−4. n. n−3. 0. n−3. 0. 0. 1. n−3. n−1. 1. n−3.

(58)  P P . . . P P  n   P   (P P P ) 

(59)    2.1.1  (a , a , . . . , a )   n. 0. 1. n−2. n−1. n−3. n−2. 1. 2. n−3 Pn−2 Pn−1. . n−1. n.  [1].    19 . [2] Dragutin Svrtan, Darko Veljan, Vladimir Volenec, Geometry of pentagons : from Gauss to Robbins, arXiv : math. MG/0403503 (2004), 1 - 10. [3] G. Hasel, J. M. Rabin, Polygons Whose Vertex Triangles Have Equal Area, Amer. Math. Monthly, 110 (2003), 606 - 616. [4] D. P. Robbins, Areas of polygons inscribed in a circle, Amer. Math. Monthly, 102 (1995), 523 - 530.. 15.

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