دروس مادة الرياضيات للفصل الثاني شعب علمية سنة ثانية ثانوي

ur uuur uuur uuur uuur Uuu=urCB + 2AB + 3AD : ‫ﻭ ﻤﻨﻪ‬ ( )urCB =uuuDr A :u‫ﺎ‬u‫ﻴﻨ‬u‫ﺩ‬r‫ﻀﻼﻉ ﻴﻜﻭﻥ ﻟ‬u‫ﺃ‬u‫ﻱ‬ur‫ ﻤﺘﻭﺍﺯ‬uAuBurCD ‫ﺒﻤﺎ ﺃﻥ‬ U = 2AD + 2AB = 2 AD + AB : ‫ﻭﻤﻨﻪ‬ uuur uuur uuur AD + ABur= ACuuur: ‫ﻭﻟﺩﻴﻨﺎ ﻜﺫﻟﻙ‬ U = 2AC : ‫ﻭﻤﻨﻪ‬ : 2 ‫ﺘﻁﺒﻴﻕ‬[ ] [ ]‫ ﻤﻨﺘﺼﻑ‬C′ ، AC ‫ ﻤﻨﺘﺼﻑ‬B′ BC ‫ ﻤﻨﺘﺼﻑ‬A′ ‫ ﻨﺴﻤﻲ‬. ‫ ﻤﺜﻠﺙ‬ABC uuur uuur uuur ur . [AB] AB′ + BB′ + CC′ = O : ‫ ﺒﺭﻫﻥ ﺃﻥ‬-uuuur uuur uuur uuur uuur uuur uuur uuur uuu:ur‫ﺤل‬( ) ( ) ( )AA′+BB′+CC′ = AB +BA′ + BC+CB′ + CA+AC′ uuur uuur uuur uuur ur AB + BC + CA = AA = O : ‫ﻟﻜﻥ‬uuur uuur uuur uuur uuuur uuurBA′ = 1 BC ; CB′ = 1 CA ; AC′ = 1 CA : ‫ﻭﺒﻤﺎ ﺃﻥ‬ 2 2 2 uuuur uuur BB′ ( )AA′ + uuur = 1 uuur uuur uuur + CC′ 2 BC + CA + AB : ‫ﻴﻜﻭﻥ‬ uuuur + uuur + uuur = 1 r = r : ‫ﺇﺫﻥ‬ AA′ BB′ CC′ 2 0 0 ( ‫ ) ﺇﻨﺸﺎﺀ ﻨﻘﻁﺔ ﻤﻌﺭﻓﺔ ﺒﻤﺴﺎﻭﺍﺓ ﺸﻌﺎﻋﻴﺔ‬: 3 ‫ﺘﻁﺒﻴﻕ‬ . ‫ ﻤﺜﻠﺙ‬ABC ‫ﻟﻴﻜﻥ‬ : ‫ﺘﺤﻘﻕ‬u‫ﻱ‬u‫ﻭ‬u‫ﺘ‬u‫ﺴ‬r‫ﻤﻥ ﺍﻟﻤ‬uMuur‫ﻁﺔ ﻭﺤﻴﺩﺓ‬u‫ﻘ‬u‫ﻨ‬u‫ﺩ‬ur‫ﺃﻨﻪ ﺘﻭﺠ‬u‫ﻥ‬u‫ﻫ‬ur‫ﺒﺭ‬ 2AM - 3 AB + 4 MC = BC . M ‫ ﺃﻨﺸﺊ ﺍﻟﻨﻘﻁﺔ‬- uuuur uuur uuuur uuur : ‫ﺤل‬ : ‫ ﺘﻜﺎﻓﺊ‬2 AM - 3 AB + 4 MC = BC : ‫ﻟﺩﻴﻨﺎ‬

‫‪uuuur uuuur uuur uuur‬‬‫‪2AM + 4 MC =uuBuurC + 3 AuuBuur uuur uuur uuur‬‬ ‫ﺃﻱ ‪:‬‬ ‫‪( )2AM + 4 MC + AC = BC + 3AB‬‬ ‫‪uuuur uuur uuur uuur‬‬ ‫‪−2uAuuMur =uBuurC +u3uuAr B - u4uAurC uuur‬‬ ‫‪−2AuMuuur= BAuu+ur ACuu+ur3 AB - 4 AC‬‬ ‫‪−2uAuuMur = u2uBuuAur - 3uAuuCuur‬‬ ‫‪AM = AM1 + AM2‬‬ ‫‪uuuuur uuur‬‬ ‫‪uuuuur‬‬ ‫‪uuur‬‬ ‫ﻭﻤﻨﻪ ‪:‬‬ ‫‪AM1 = - AB :‬‬ ‫= ‪AM2‬‬ ‫‪AC‬‬‫‪M1‬‬ ‫‪3‬‬ ‫ﺤﻴﺙ ‪:‬‬ ‫‪2‬‬ ‫ﺘﻭﺠﺩ ﺇﺫﻥ ﻨﻘﻁﺔ ﻭﺤﻴﺩﺓ ‪M‬‬ ‫ﻭﻀﻌﻴﺘﻬﺎ ﻤﺤﺩﺩﺓ ﻓﻲ ﺍﻟﺸﻜل ﺍﻟﻤﻘﺎﺒل‬‫‪AM‬‬‫‪uurC uur‬‬ ‫‪ (4‬ﻤﻨﺘﺼﻑ ﻗﻁﻌﺔ ﻤﺴﺘﻘﻴﻡ ‪:‬‬‫‪[ ]B‬‬‫=‪AMI 2‬‬ ‫‪IB‬‬ ‫‪:‬‬ ‫ﻜﺎﻥ‬ ‫‪ AB‬ﺇﺫﺍ ﻭﺍﻓﻕ ﺇﺫﺍ‬ ‫ﺍﻟﻘﻁﻌﺔ‬ ‫‪ I‬ﻤﻨﺘﺼﻑ‬ ‫ﺘﻌﺭﻴﻑ ‪ :‬ﺍﻟﻨﻘﻁﺔ‬ ‫‪uur uur ur‬‬ ‫ﺃﻭ‬ ‫‪IA + IB = O‬‬ ‫‪uur‬‬ ‫‪1‬‬ ‫‪uuur‬‬ ‫ﺃﻭ‬ ‫= ‪AI‬‬ ‫‪2‬‬ ‫‪AB‬‬‫‪ur ur‬‬ ‫‪ (5‬ﺘﻭﺍﺯ‪r‬ﻱ‪u‬ﺸﻌﺎﻋ‪r‬ﻴ‪u‬ﻥ ‪:‬‬‫ﺘﻌﺭﻴﻑ ‪ V ; U :‬ﺸﻌﺎﻋﻴﻥ ﻏﻴﺭ ﻤﻌﺩﻭﻤﻴ‪ur‬ﻥ ‪.‬ﻴﻜﻭﻥ ﺍﻟﺸﻌ‪r‬ﺎ‪u‬ﻋﻴﻥ ‪ U‬و ‪ V‬ﻤﺘﻭﺍﺯﻴﻴﻥ ﺇﺫﺍ ﻭﻓﻘﻁ ﺇﺫﺍ‬ ‫ﻭﺠﺩ ﻋﺩﺩ ﺤﻘﻴﻕ ﻏﻴ‪ur‬ﺭ ﻤﻌﺩﻭﻡ ‪ k‬ﺒﺤﻴﺙ ‪U = k . V :‬‬ ‫ﺍﻟﺸﻌﺎﻉ ﺍﻟﻤﻌﺩﻭﻡ ‪ O‬ﻴﻭﺍﺯﻱ ﻜل ﺸﻌﺎﻉ ‪.‬‬ ‫‪ (6‬ﺇﺴﺘﻘﺎﻤﻴﺔ ﺜﻼﺙ ﻨﻘﻁ ‪:‬‬‫ﺘﻜ‪ur‬ﻭ‪u‬ﻥ‪u‬ﺍﻟﻨﻘﻁ ‪ A , B u,uCur‬ﻋﻠﻰ ﺍﺴﺘﻘﺎﻤﺔ ﻭﺍﺤﺩﺓ ) ﻓﻲ ﺍﺴﺘﻘﺎﻤﻴﺔ (ﺇﺫﺍ ﻭﻓﻘﻁ ﺇﺫﺍ ﻜﺎﻥ ‪:‬‬ ‫‪ AB‬و ‪ AC‬ﻤﺘﻭﺍﺯﻴﺎﻥ‪.‬‬ ‫‪( )r r‬‬ ‫‪ (7‬ﻤﺭﻜﻴﺒﺘﺎ ﺸﻌﺎﻉ ‪:‬‬‫‪uuur r‬‬ ‫ﺍﻟﻤﺴﺘﻭﻱ ﻤﻨﺴﻭﺏ ﺇﻟﻰ ﺍﻟﻤﻌﻠﻡ ‪r O ; i , j‬‬‫ﺘﻌﺭﻴﻑ ‪ ( xA ; yA ) :‬ﺇﺤﺩﺍﺜﻴﺎ ﺍﻟﻨﻘﻁﺔ ‪ A‬ﻤﻌﻨﺎﻩ ‪OA = xA i + yA j :‬‬

: ‫ ﻓﺈﻥ‬B ( xB ; yB ) ; A ( xA ; yA ) : ‫( ﺇﺫﺍ ﻜﺎﻥ‬1 uuur  xB - xA  AB    yB - yA  ur  x′  ‫و‬ ur x  ‫ﻜﺎﻥ‬ ‫ﺇﺫﺍ‬ (2 V  y′  U ury  ur ( y = y′ ‫ ﻭ‬x = x′) ‫ ﺘﻜﺎﻓﺊ‬U = V : ‫ﻓﺈﻥ‬ x + x′  : ‫ﻫﻤﺎ‬ ur uur ‫•ﻤﺭﻜﺒﺘﺎ ﺍﻟﺸﻌﺎﻉ‬  +  U + U′  y y′  k∈ ¡ ,  kx  ur uur  kyur : ‫ ﻫﻤﺎ‬k.U ‫•ﻤﺭﻜﺒﺘﺎ‬  ur uur U′ = k.U : ‫ ﺘﻜﺎﻓﺊ‬U // U ′ • uur ur U′  x′  ‫و‬ U  x  : ‫ﺨﺎﺼﻴﺔ‬  y′   y ur uur ur uur( ‫ ﻤﺭﺘﺒﻁﺎﻥ ﺨﻁﻴﺎ‬U , U′ ‫ ) ﺃﻱ‬xy′ - yx′ = 0 ‫ ﺘﻜﺎﻓﺊ‬U // U ′ : ‫ﺘﻁﺒﻴﻘﺎﺕ‬ : ‫ﺍﻟﺒﺭﻫﺎﻥ ﻋﻠﻰ ﺃﻥ ﺸﻌﺎﻋﻴﻥ ﻤﺘﻭﺍﺯﻴﻴﻥ‬ ur ur : 1‫ﻤﺜﺎل‬ur : ‫ ﺸﻌﺎﻋﺎﻥ ﺤﻴﺙ‬V ; U . ‫ ﻤﺭﺒﻊ‬ABCD ‫ﻟﻴﻜﻥ‬V uuur uuur ur uuur uuur = DB + 1 BC ; U = 2 AB + DA 2 ur ur .‫ ﻤﺘﻭﺍﺯﻴﺎﻥ‬V ‫ و‬U : ‫ﺒﺭﻫﻥ ﺃﻥ‬ uuur uuur : ‫ﺤل‬ ur uDuuAr = CuuBur : ‫ ﻤﺭﺒﻊ ﻓﺈﻥ‬ABCD ‫ﺒﻤﺎ ﺃﻥ‬ U = 2 AB + CB : ‫ﻭﻤﻨﻪ‬

ur = uuur uuur 1 uuur V AD + AB + 2 BC = uuur uuur + 1 uuur CB + AB 2 BC ur uuur 1 uuur V= AB + 2 CB ur ur U = 2V ur ur : ‫ﺇﺫﻥ‬ ‫ ﻤﺘﻭﺍﺯﻴﺎﻥ‬V ‫ و‬U : ‫ﻭﻤﻨﻪ‬ : 2‫ﻤﺜﺎل‬ : ‫ ﺒـ‬F‫ ﻭ‬E ‫ ﻨﻌﺭﻑ ﺍﻟﻨﻘﻁﺘﻴﻥ‬ABCD ‫ﻓﻲ ﻤﺘﻭﺍﺯﻱ ﺃﻀﻼﻉ‬ uuur uuur uuur uuur AF = 1 DA ; AE = 1 AB 3 4 .‫ ﻋﻠﻰ ﺍﺴﺘﻘﺎﻤﺔ ﻭﺍﺤﺩﺓ‬C , E , F ‫ﺒﺭﻫﻥ ﺃﻥ ﺍﻟﻨﻘﻁ‬ : ‫ﺤل‬uuur uuur uuur uuur uuur uuurFE =uu31ur AD + 1 AB : ‫ ﺇﺫﻥ‬FE = FA + AE 4 uuur uuur uuur FE = FA + AD + DO uuur uuur uuur : ‫ﻭﺒﺎﻟﻤﺜل‬ AD + AD + AB = 1 3uuur uuur uuur u31uuruAuDur u+uur41 uuurFC = 4 AD + AB =4 AB  3  uFuurC =u4uurFE : ‫ﻭﺒﻤﺎ ﺃﻥ‬ . ‫ ﻤﺘﻭﺍﺯﻴﺎﻥ‬FE ‫ ﻭ‬FC ‫ﻓﺈﻥ ﺍﻟﺸﻌﺎﻋﻴﻥ‬ . ‫ ﻋﻠﻰ ﺍﺴﺘﻘﺎﻤﺔ ﻭﺍﺤﺩﺓ‬C , E , F ‫ﻭﻤﻨﻪ ﺍﻟﻨﻘﻁ‬ ( )ur ( u‫ﺕ‬r‫ ) ﺍﺴﺘﻌﻤﺎل ﺍﻹﺤﺩﺍﺜﻴﺎ‬: 3‫ﻤﺜﺎل‬ : ‫ ﻨﻌﺘﺒﺭ ﺍﻟﻨﻘﻁ‬O ; i , j ‫ﻓﻲ ﻤﺴﺘﻭ ﻤﻨﺴﻭﺏ ﺇﻟﻰ ﻤﻌﻠﻡ‬ D (-1 ; 9) , C (1 ; -1) , B (-2 ; 2) , A (3 , 1)

uuur: ‫ﺘﺤﻘﻕ‬uMuu‫ﺓ‬r‫ﻁﺔ ﻭﺤﻴﺩ‬u‫ﻘ‬u‫ ﻨ‬u‫ﺩ‬u‫ﺠ‬r‫ﻥ ﺃﻨﻪ ﺘﻭ‬u‫ﻫ‬u‫ﺭ‬u‫ﺒ‬r(‫ﺃ‬ 2MA + 3MB - 4 MC = 2 AC . M ‫ﻋﻴﻥ ﺇﺤﺩﺍﺜﻴﻲ ﺍﻟﻨﻘﻁﺔ‬ ‫ ﻋﻠﻰ ﺍﺴﺘﻘﺎﻤﺔ ﻭﺍﺤﺩﺓ ؟‬D ; B ; M ‫ﺏ( ﻫل ﺍﻟﻨﻘﻁ‬ : ‫ﺤل‬uuuur : ‫ ﺇﻥ ﻭﺠﺩﺕ‬M (x ; y) ‫ﺃ( ﻟﺘﻜﻥ‬MC uuur uMuuAr   1 - x  ; MB  -2 - x  ;  3 - x : ‫ﻟﺩﻴﻨﺎ‬  1 - y   2 - y  1 -      y  uuur  -6 - 3x  ; uuur  6 - 2x  : ‫ﻭﻤﻨﻪ‬ 3MB   2 MA    6 - 3y   2 - 2y  uuur  -2  ; uuuur  -4 + 4x  AC   -4 MC    -2   4 + 4y  uuur uuur uuuur u: u‫ﻪ‬u‫ﻨ‬r‫ﻭﻤ‬ : ‫ ﺘﻜﺎﻓﺊ‬2MA + 3MB - 4 MC = 2 AC 6 - 2x - 6 - 3x - 4 + 4x = -4 2 - 2y + 6 - 3y + 4 + 4y = -4 x = 0 − x - 4 = -4  : ‫ﺃﻱ‬ -y + 12 = -4 : ‫ﺃﻱ‬  y = 16 M (0 ; 16) : ‫ ﻤﻭﺠﻭﺩﺓ ﻭﻭﺤﻴﺩﺓ ﺤﻴﺙ‬M ‫ﺍﻟﻨﻘﻁﺔ‬ uuur uBuMuuruuur124 BD  1  ;  (‫ﺏ‬  7  uuuur  uuur uuuur    BD ‫ ﻭ‬BM ‫ ﻓﺈﻥ ﺍﻟﺸﻌﺎﻋﻴﻥ‬BM = 2BD : ‫ﻭﺒﻤﺎ ﺃﻥ‬ D , B , M : ‫ﻤﺘﻭﺍﺯﻴﺎﻥ ﻭﻤﻨﻪ ﺍﻟﻨﻘﻁ‬ . ‫ﻋﻠﻰ ﺍﺴﺘﻘﺎﻤﺔ ﻭﺍﺤﺩﺓ‬ [ ]BM ‫ ﻫﻭ ﻤﻨﺘﺼﻑ‬D ‫ﺒﺎﻹﻀﺎﻓﺔ ﻓﺈﻥ‬ : ‫ ﻤﺭﺠﺢ ﻨﻘﻁﺘﻴﻥ‬- II

‫‪ (1‬ﻤﺒﺭﻫﻨﺔ ﻭﺘﻌﺭﻴﻑ ‪:‬‬‫‪ A‬ﻭ‪ B‬ﻨﻘﻁﺘﺎﻥ ﻤﻥ ﺍﻟﻤﺴﺘﻭﻱ‪ α .‬ﻭ ‪ β‬ﻋﺩﺩﺍﻥ ﺤﻘﻴﻘﻴﺎﻥ ﺤﻴﺙ ‪ 0 ≠ β + α :‬؛ ﺘﻭﺠﺩ ﻨﻘﻁﺔ‬ ‫ﻭ‪r‬ﺤ‪u‬ﻴﺩﺓ ‪rG‬ﺘ‪u‬ﺤﻘﻕ ‪ur :‬‬ ‫‪αG A + βG B = O‬‬ ‫ﺍﻟﻨﻘﻁﺔ ‪ G‬ﺘﺴﻤﻰ ﻤﺭﺠﺢ ﺍﻟﻨﻘﻁﺘﻴﻥ ‪ A‬ﻭ‪ B‬ﺍﻟﻤﺭﻓﻘﺘﻴﻥ ﺒﺎﻟﻌﺎﻤﻠﻴﻥ ‪ α‬و ‪ β‬ﻋﻠﻰ ﺍﻟﺘﺭﺘﻴﺏ ‪.‬‬ ‫ﻤﻼﺤﻅﺔ ‪:‬‬ ‫ﻨﻘﻭل ﻜﺫﻟﻙ ﺃﻥ ‪ G‬ﻫﻲ ﻤﺭﺠﺢ ﺍﻟﺠﻤﻠﺔ ) ‪(B , β ) ; (A , α‬‬ ‫• ﻤﺭﺠﺢ ﺍﻟﺠﻤﻠﺔ )‪ (A , 1‬ﻭ)‪ (B , 1‬ﻫﻭ ﻤﻨﺘﺼﻑ ‪[ ]AB‬‬ ‫• ﻤﺭﺠﺢ ﺍﻟﻨﻘﻁﺘﻴﻥ ﺍﻟﻤﺨﺘﻠﻔﺘﻴﻥ ‪ A‬ﻭ‪ B‬ﻴﻨﺘﻤﻲ ﺇﻟﻰ ﺍﻟﻤﺴﺘﻘﻴﻡ )‪(AB‬‬ ‫• ﺇﺫﺍ ﻜﺎﻥ ‪ α‬ﻭ ‪ β‬ﻤﻥ ﻨﻔﺱ ﺍﻹﺸﺎﺭﺓ ﻓﺈﻥ ‪ G‬ﻴﻨﺘﻤﻲ ﺇﻟﻰ ‪[ ]. AB‬‬ ‫• ﺇﺫﺍ ﻜﺎﻥ ‪ α‬ﻭ ‪ β‬ﻤﺨﺘﻠﻔﺎﻥ ﻓﻲ ﺍﻹﺸﺎﺭﺓ ﻓﺈﻥ ‪ G‬ﺨﺎﺭﺝ ‪[ ]. AB‬‬ ‫‪ (2‬ﺍﻟﺨﺎﺼﻴﺔ ﺍﻷﺴﺎﺴﻴﺔ ﻟﻠﻤﺭﺠﺢ ‪:‬‬‫ﺇﺫﺍ ﻜﺎﻥ ‪ G‬ﻤﺭﺠﺢ ﺍﻟﺠﻤﻠﺔ ) ‪ (B , β ) ; (A , α‬ﺤﻴﺙ ‪ . α + β ≠ 0‬ﻓﺈﻥ ﻤﻥ ﺃﺠل‬ ‫ﻜل‪ur‬ﻨ‪u‬ﻘ‪u‬ﻁ‪u‬ﺔ ‪ M‬ﻤﻥ ﺍﻟﻤﺴﺘﻭﻯ ‪uuur uuur :‬‬ ‫‪αMA + βMB = ( α + β ) MG‬‬‫‪( )r r‬‬ ‫‪ -‬ﺇﺤﺩﺍﺜﻴﺎﺕ ﺍﻟﻤﺭﺠﺢ ‪:‬‬‫ﺍﻟﻤﺴﺘﻭﻱ ﻤﻨﺴﻭﺏ ﺇﻟﻰ ﺍﻟﻤﻌﻠﻡ ‪O ; i , j‬‬‫ﻟﺘﻜﻥ ) ‪ G (xG ; yG ) . B (xB ; yB ) , A (xA , yA‬ﻤﺭﺠﺢ ﺍﻟﺠﻤﻠﺔ‬ ‫) ‪. (B , β ) ; (A , α‬‬‫ﺒﺘﻌﻭﻴﺽ ‪M‬ﺒـ ‪ O‬ﻓﻲ ﺍﻟﺨﺎﺼﺔ ﺍﻷﺴﺎﺴﻴﺔ ﻨﺠﺩ ‪:‬‬‫‪‬‬ ‫‪xG‬‬ ‫=‬ ‫‪αx A‬‬ ‫‪+ βxB‬‬‫‪‬‬ ‫‪α‬‬ ‫‪+β‬‬‫‪‬‬‫‪‬‬ ‫‪αyA + βYB‬‬‫‪‬‬ ‫‪yG‬‬ ‫=‬ ‫‪α +β‬‬ ‫ﻤﺜﺎل ‪ : 1‬ﺇﻨﺸﺎﺀ ﻤﺭﺠﺢ ﻨﻘﻁﺘﻴﻥ‬‫ﺃ( ﺃﻨﺸﺊ ﺍﻟﻤﺭﺠﺢ ‪ G‬ﻟﻠﺠﻤﻠﺔ )‪(B , 1) ; (A , 2‬‬‫ﺏ( ﻫل ﻴﻤﻜﻥ ﺇﻨﺸﺎﺀ ﻤﺭﺠﺢ ﺍﻟﺠﻤﻠﺔ )‪ (B , -3) ; (A , 3‬؟‬

‫ﺤل ‪:‬‬ ‫ﺃ( ﺍﻟﻤﺭﺠﺢ ﻤﻭﺠﻭﺩ ﻷﻥ ‪uu1ur+ 2uu≠ur0‬ﻭ ﻤﻥ ﺃﺠ‪r‬ل‪uu‬ﻜ‪u‬ل‪u‬ﻨﻘﻁﺔ ‪ M‬ﻤﻥ‬ ‫‪3MG = 2MA + MB‬‬ ‫‪uuur‬‬ ‫‪uuur‬‬ ‫ﺍﻟﻤﺴﺘﻭﻱ ﻟﺩﻴﻨﺎ ‪:‬‬ ‫= ‪AG‬‬ ‫‪AB‬‬ ‫‪1‬‬ ‫ﺒﻭﻀﻊ ‪ M = A :‬ﻨﺠﺩ ‪:‬‬ ‫‪3‬‬‫‪AG‬‬ ‫ﺏ( ﺍﻟﻤﺭﺠﺢ‪ur‬ﻏﻴﺭ ﻤﻭﺠ‪ur‬ﻭ‪u‬ﺩ‪u‬ﻷﻥ ‪uuurB3 + u(-u3u)r= 0‬‬ ‫ﺒﺎﻟﻔﻌل ‪3MA - 3MB = 3BA ≠ O :‬‬ ‫ﻤﺜﺎل ‪ ) : 2‬ﺍﺴﺘﻌﻤﺎل ﺍﻟﻤﺭﺠﺢ ﻓﻲ ﺍﻟﺤﺴﺎﺏ ﺍﻟﺸﻌﺎﻋﻲ (‬ ‫ﻟﺘﻜﻥ ﺜﻼﺙ ﻨﻘﻁ ‪ A , B , C‬ﻤﻥ ﺍﻟ‪r‬ﻤ‪uu‬ﺴﺘ‪u‬ﻭﻱ‪uuur uuur .‬‬ ‫ﺒﺭﻫﻥ ﺃﻨﻪ ﺘﻭﺠﺩ ﻨﻘﻁﺔ ﻭﺤﻴﺩﺓ ﺘﺤﻘﻕ ‪MA - 3MB = AC :‬‬ ‫ﺃﻨﺸﺊ ‪. M‬‬ ‫ﺤل ‪:‬‬ ‫ﻟﺘﻜﻥ ‪ G‬ﻤﺭﺠﺢ‪ ur‬ﺍ‪u‬ﻟ‪u‬ﺠﻤ‪u‬ﻠﺔ )‪uuur(B , -3u) u; u(Ar , 1‬‬ ‫ﺒﻤﺎ ﺃﻥ ‪uuuur uuur MAu-u3urMBu=uu-r2MuGuur :‬‬‫ﻓﺈﻥ ‪ MA - 3MB = AC :‬ﺘﻜﺎﻓﺊ ‪−2MG = AC :‬‬‫‪uuur‬‬ ‫‪uuur‬‬ ‫‪uuuur‬‬ ‫‪uuur‬‬‫= ‪AG‬‬ ‫‪3‬‬ ‫‪AB‬‬ ‫ﻨﻨﺸﺊ ﺃﻭﻻ ‪ G‬ﺤﻴﺙ ‪:‬‬ ‫= ‪. GM‬‬ ‫‪1‬‬ ‫‪AC‬‬ ‫ﺃﻱ ‪:‬‬ ‫‪2‬‬ ‫‪2‬‬ ‫‪uuuur‬‬ ‫‪1‬‬ ‫‪uuur‬‬ ‫= ‪GM‬‬ ‫‪2‬‬ ‫‪AC‬‬ ‫ﺜﻡ ﺍﻟﻨﻘﻁﺔ ﺍﻟﻭﺤﻴﺩﺓ ‪ M‬ﺍﻟﺘﻲ ﺘﺤﻘﻕ‬ ‫‪C‬‬ ‫‪M‬‬‫‪A‬‬ ‫‪BG‬‬ ‫‪ (3‬ﻤﺭﺠﺢ ﺜﻼﺙ ﻨﻘﻁ ‪:‬‬ ‫ﻤﺒﺭﻫﻨﺔ ﻭﺘﻌﺭﻴﻑ ‪:‬‬ ‫‪ A , B , C‬ﺜﻼﺙ ﻨﻘﻁ ﻤﻥ ﺍﻟﻤﺴﺘﻭﻱ‪.‬‬ ‫‪ α , β , γ‬ﺃﻋﺩﺍﺩ ﺤﻘﻴﻘﻴﺔ ﺤﻴﺙ ‪α + β + γ ≠ 0 :‬‬

‫‪r-1‬ﺘﻭﺠﺩ ﻨ‪r‬ﻘ‪u‬ﻁ‪u‬ﺔ‪u‬ﻭﺤﻴﺩﺓ ‪ uGr‬ﺘ‪u‬ﺤ‪u‬ﻘﻕ ‪uuur :‬‬ ‫‪αGA + βGB + γGC = 0‬‬ ‫‪ -2‬ﺍﻟﻨﻘﻁﺔ ‪ G‬ﺘﺴﻤﻰ ﻤﺭﺠﺢ ﺍﻟﺠﻤﻠﺔ ‪:‬‬ ‫)‪(A , α) ; (B , β) ; (C , γ‬‬ ‫ﺨﻭﺍﺹ ‪:‬‬ ‫‪ -‬ﻻ ﻴﺘﻐﻴﺭ ﺍﻟﻤﺭﺠﺢ ﺇﺫﺍ ﻀﺭﺒﻨﺎ ﺍﻟﻤﻌﺎﻤﻼﺕ ﻓﻲ ﻨﻔﺱ ﺍﻟﻌﺩﺩ ﻏﻴﺭ ﺍﻟﻤﻌﺩﻭﻡ‪.‬‬‫‪ -‬ﺇﺫﺍ ﻜﺎﻥ ‪ G‬ﻤﺭﺠﺢ ﺍﻟﺠﻤﻠﺔ ) ‪ (A , α) ; (B , β) ; (C , γ‬ﺤﻴﺙ‬ ‫‪ α + β + γ ≠ 0‬ﻓﺈﻥ ‪:‬‬ ‫ﻤ‪r‬ﻥ‪uu‬ﺃﺠ‪u‬ل‪ u‬ﻜل ﻨﻘﻁﺔ ‪ M‬ﻤﻥ ﺍﻟﻤﺴﺘﻭﻱ‪uuuur uuur uuu:r‬‬ ‫‪αMA + βMB + γMC = (α + β + γ) MG‬‬ ‫‪( )r r‬‬ ‫‪ (4‬ﺇﺤﺩﺍﺜﻴﺎﺕ ﺍﻟﻤﺭﺠﺢ ‪:‬‬ ‫ﺍﻟﻤﺴﺘﻭﻱ ﻤﻨﺴﻭﺏ ﺇﻟﻰ ﺍﻟﻤﻌﻠﻡ ‪O ; i , j‬‬‫ﻟﺘﻜﻥ ‪A (xA ; yA ) , B (xB ; yB ) , C (xC ; yC ) :‬‬‫) ‪ G (xG ; yG‬ﻤﺭﺠﺢ ﺍﻟﺠﻤﻠﺔ ‪(A , α ) ; (B , β ) ; (C , γ ) :‬‬‫‪‬‬ ‫‪xG‬‬ ‫=‬ ‫‪αxA + βxB + γxC‬‬‫‪‬‬ ‫‪α +β + γ‬‬‫‪‬‬‫‪‬‬ ‫‪αyA + βyB + γyC‬‬‫‪‬‬ ‫‪yG‬‬ ‫=‬ ‫‪α +β + γ‬‬‫‪uuur uuur uuur ur‬‬ ‫ﺨﺎﺼﻴﺔ ‪:‬‬‫ﻤﺭﻜﺯ ﺜﻘل ﺍﻟﻤﺜﻠﺙ ‪ ABC‬ﻫﻭ ﺍﻟﻨﻘﻁﺔ ‪ G‬ﺍﻟﺘﻲ ﺘﺤﻘﻕ ‪GA + GB + GC = O :‬‬ ‫)*( ﺨﺎﺼﻴﺔ ﺍﻟﺘﺠﻤﻴﻊ ‪:‬‬ ‫‪ A , B , C‬ﺜﻼﺙ ﻨﻘﻁ ﻤﻥ ﺍﻟﻤﺴﺘﻭﻱ‬ ‫‪ α , β , γ‬ﺃﻋﺩﺍﺩ ﺤﻘﻴﻘﻴﺔ ﺤﻴﺙ ‪α + β + γ ≠ 0 :‬‬‫ﺇﺫﺍ ﻜﺎﻥ ‪ α + β ≠ 0 :‬ﻓﺈﻥ ﺍﻟﻤﺭﺠﺢ ‪ G‬ﻟﻠﺠﻤﻠﺔ ) ‪ (A , α) ; (B , β) ; (C , γ‬ﻫﻭ‬‫ﻜﺫﺍﻟﻙ ﻤﺭﺠﺢ ﺍﻟﺠﻤﻠﺔ )‪ (C , α) ; (H , α + β‬ﺤﻴﺙ ‪ H‬ﻤﺭﺠﺢ ﺍﻟﺠﻤﻠﺔ‬ ‫)‪(B , β) ; (A , α‬‬ ‫ﻤﺜﺎل ‪ ) : 1‬ﺇﻨﺸﺎﺀ ﻤﺭﺠﺢ ﺜﻼﺙ ﻨﻘﻁ (‬

‫‪ ABC‬ﻤﺜﻠﺙ ‪ .‬ﺃﻨﺸﺊ ﺒﻁﺭﻴﻘﺘﻴﻥ ﻤﺨﺘﻠﻔﺘﻴﻥ ﻤﺭﺠﺢ ﺍﻟﺠﻤﻠﺔ‬ ‫)‪(A , 1) ; (B , 1) ; (C , 2‬‬ ‫ﺤل ‪:‬‬ ‫ﺍﻟﻤﺭﺠﺢ ‪ G‬ﻤﻭﺠﻭﺩ ﻷﻥ ‪11 + 1 + 2 ≠ 0‬‬ ‫• ﺒﺎﺴﺘﻌﻤﺎل ﺍﻟﺨﺎﺼﺔ ﺍﻷﺴﺎ‪ur‬ﺴﻴ‪u‬ﺔ‪uuuur uuur uuur u:u‬‬ ‫ﻤﻥ ﺃﺠل ﻜل ﻨﻘﻁﺔ ‪MA + MB + 2MC = 4MG : M‬‬ ‫‪uuur‬‬ ‫‪uuur‬‬ ‫‪uuur‬‬ ‫‪CG‬‬ ‫=‬ ‫‪1‬‬ ‫‪(CA‬‬ ‫‪+‬‬ ‫)‪CB‬‬ ‫ﺒﺘﻌﻭﻴﺽ ‪ M‬ﺒـ ‪ C :‬ﻨﺠﺩ ‪:‬‬ ‫‪4‬‬‫‪A‬‬ ‫• )*( ﺒﺎﺴﺘﻌﻤﺎل ﺨﺎﺼﻴﺔ ﺍﻟﺘﺠﻤﻴﻊ ‪:‬‬ ‫ﻟﻴﻜﻥ ‪ I‬ﻤﻨﺘﺼﻑ ‪[ ]. AB‬‬‫‪ G‬ﻫﻭ ﻤﺭﺠﺢ ‪I (C , 2) ; (I , 2) :‬‬ ‫‪G‬‬ ‫‪C‬‬ ‫ﺇﺫﻥ ‪ G :‬ﻤﻨﺘﺼﻑ ‪[ ]IC‬‬‫‪B‬‬ ‫ﻤﺜﺎل ‪: 2‬‬‫ﻓﻲ‪ ur‬ﺍ‪u‬ﻟﻤ‪u‬ﺜﻠﺙ ‪ .uuAurBC‬ﻨﺴﻤﻰ ‪ I‬ﻤﻨﺘﺼﻑ ‪ AB‬؛ ‪ J‬ﻤﻨﺘﺼﻑ ‪ CI‬؛ ﻭ ‪ K‬ﻨﻘﻁﺔ ﺤﻴﺙ ‪[ ] [ ]:‬‬ ‫‪3BK = 2BC‬‬ ‫ﺒﺭﻫﻥ ﺃﻥ ﺍﻟﻨﻘﻁ ‪ K ; J ; A‬ﻋﻠﻰ ﺍﺴﺘﻘﺎﻤﺔ ﻭﺍﺤﺩﺓ‪.‬‬ ‫ﺤل ‪uuur uuur uuur :‬‬ ‫ﻟﺩﻴﻨﺎ ‪uuur 3BuuKur = 2ur(BK + KC) :‬‬ ‫‪−BK + 2KC =uuuOr suuu ur‬‬ ‫ﺃﻱ ‪KB + 2KC = O :‬‬ ‫ﻭﻤﻨﻪ ‪ K :‬ﻫﻭ ﻤﺭﺠﺢ ﺍﻟﺠﻤﻠﺔ )‪(C , 2) ; (B , 1‬‬ ‫ﻟﺩﻴﻨﺎ ‪ J‬ﻤﻨﺘﺼﻑ ‪ CI‬ﻭﻤﻨﻪ ‪ J‬ﻫﻭ ﻤﺭﺠﺢ )‪[ )(I , 2) ; (C , 2‬‬ ‫ﻭﻟﻜﻥ ‪ I‬ﻤﻨﺘﺼﻑ ‪ AB‬ﻓﻬﻭ ﻤﺭﺠﺢ )‪[ ](B , 1) ; (A , 1‬‬ ‫ﺒﺘﻁﺒﻴﻕ ﺨﺎﺼﻴﺔ ﺍﻟﺘﺠﻤﻴﻊ ﻨﺠﺩ ﺃﻥ ‪:‬‬ ‫‪ J‬ﻤﺭﺠﺢ )‪(B , 1) ; (A , 1) ; (C , 2‬‬ ‫ﻭ ﺒﻤﺎ ﺃﻥ ‪ K‬ﻤﺭﺠﺢ )‪(C , 2) ; (B , 1‬‬ ‫ﻓﺈﻥ ﺨﺎﺼﻴﺔ ﺍﻟﺘﺠﻤﻴﻊ ﺘﺒﻴﻥ ﺃﻥ ‪ J‬ﻤﺭﺠﺢ )‪(K , 3) ; (A , 1‬‬

‫ﻨﺴﺘﻨﺘﺞ ﺃﻥ ‪ J‬ﻴﻨﺘﻤﻲ ﺇﻟﻰ ﺍﻟﻤﺴﺘﻘﻴﻡ)‪(AK‬‬ ‫ﺇﺫﻥ ‪ A ; J ; K :‬ﻋﻠﻰ ﺍﺴﺘﻘﺎﻤﺔ ﻭﺍﺤﺩﺓ‪.‬‬ ‫ﻤﺜﺎل ‪: 3‬‬ ‫‪ ABCD‬ﺭﺒﺎﻋﻲ ‪ I , J , K , L .‬ﻤﻨﺘﺼﻔﺎﺕ ﻋﻠﻰ ﺍﻟﺘﺭﺘﻴﺏ ﻟﻸﻀﻼﻉ‬ ‫]‪. [AB] ; [BC] ; [CD] ; [DA‬‬ ‫‪ M‬ﻭ‪ N‬ﻤﻨﺘﺼﻔﺎ ﺍﻟﻘﻁﺭﻴﻥ ‪ AC‬و ‪ BD‬ﻋﻠﻰ ﺍﻟﺘﺭﺘﻴﺏ ‪[ ] [ ].‬‬ ‫ﺒﺭﻫﻥ ﺃﻥ ﺍﻟﻤﺴﺘﻘﻴﻤﺎﺕ )‪ (IK) ; (JL) ; (MN‬ﺘﺘﻘﺎﻁﻊ ﻓﻲ ﻨﻘﻁﺔ ‪.‬‬ ‫ﺤل ‪:‬‬‫ﻟﺘﻜﻥ ‪ O‬ﻤﺭﺠﺢ ﺍﻟﺠﻤﻠﺔ )‪ (A , 1) ; (B , 1) ; (C , 1) ; (D , 1‬ﺒﺘﺠﻤﻴﻊ ‪ A‬ﻤﻊ ‪B‬‬ ‫ﻭ ‪ C‬ﻤﻊ‪ D‬ﻨﺠﺩ ﺃﻥ ‪ O‬ﻫﻭ ﻤﺭﺠﺢ ‪:‬‬ ‫)‪(K , 2) ; (I , 2‬‬‫ﺒﺘﺠﻤﻴﻊ ‪ C‬ﻤﻊ ‪ A‬ﻭ ‪ B‬ﻤﻊ ‪ D‬ﻨﺠﺩ ‪ O :‬ﻤﺭﺠﺢ ‪ (N , 2) ; (M , 2) :‬ﺒﺘﺠﻤﻴﻊ ‪ A‬ﻤﻊ ‪ D‬ﻭ ‪ C‬ﻤﻊ‬ ‫‪ B‬ﻨﺠﺩ ‪ O‬ﻤﺭﺠﺢ ‪(J , 2) ; (L , 2) :‬‬ ‫ﻭﻤﻨﻪ ‪ O :‬ﻴﻨﺘﻤﻲ ﺇﻟﻰ ﺍﻟﻤﺴﺘﻘﻴﻤﺎﺕ ‪(JL) ; (MN) ; (IK) :‬‬ ‫ﺇﺫﻥ ‪ :‬ﻫﺫﻩ ﺍﻟﻤﺴﺘﻘﻴﻤﺎﺕ ﺘﺘﻼﻗﻰ ﻓﻲ ﺍﻟﻨﻘﻁﺔ ‪. O‬‬

‫ﺕﻤـﺎریـﻦ و ﻡﺸﻜﻼت‬ ‫ﺍﻟﺘﻤﺭﻴﻥ‪ . 1‬ﺼﺤﻴﺢ ﺃﻡ ﺨﻁﺄ ‪:‬‬‫‪ur uuur uuur‬‬ ‫‪ (1‬ﻤﻥ ﺃﺠل ﺜﻼ‪ur‬ﺙ‪ u‬ﻨ‪u‬ﻘﻁ ﻜﻴﻔﻴﺔ‪ur‬ﻤ‪u‬ﻥ‪u‬ﺍﻟﻤﺴﺘﻭ‪ur‬ﻱ‪u: u‬‬‫‪V = BA - 2AC‬‬ ‫ﻟﺩﻴﻨﺎ ‪2uAr B +uuuBr C =u2uuAr C :‬‬ ‫‪ (2‬ﺍﻟﺸﻌﺎﻋﺎﻥ ‪ U = AB + 2AC :‬ﻭ‬ ‫ﻤﺭﺘﺒﻁﺎﻥ ﺨﻁﻴﺎ‪.‬‬‫‪ (3‬ﻤﺭﺠﺢ ﺍﻟﺠﻤﻠ‪r‬ﺔ )‪ ( A ; -2u)u;ur(B ;u-u5ur‬ﺃﻗﺭﺏ ﻤﻥ ‪{ }B‬‬‫‪ (4‬ﺇﺫﺍ ﻜﺎﻥ‪ AB + 2AC = 0 :‬ﻓﺈﻥ‪ B :‬ﻫﻭ ﻤﺭﺠﺢ ﺍﻟﺠﻤﻠﺔ‬ ‫})‪{( A ; 1) ; (C ; 2‬‬ ‫‪ (5‬ﻤﺭﺠﺢ )‪ C (-1 ; 2) ; B (3 ; 4) ; A (1 ; 2‬ﺍﻟﻤﻔﺭﻗﺔ ﺒﻨﻔﺱ‬ ‫ﺍﻟﻤﻌﺎﻤل ﻓﺎﺼﻠﺘﻪ ‪ 1‬ﻓﻲ ﻤﺴﺘﻭ ﻤﻨﺴﻭﺏ ﺇﻟﻰ ﻤﻌﻠﻡ ‪.‬‬ ‫‪ (6‬ﻤﺭﺠﺢ ﺍﻟﺠﻤﻠﺔ )‪ (C ; 4) ; (B ; 2) ; (A ; 2‬ﻫﻭ ﺃﻴﻀﺎ ﻤﻨﺘﺼﻑ‬ ‫]‪ [IC‬ﺤﻴﺙ ﻤﻨﺘﺼﻑ ]‪[AB‬‬ ‫‪ (7‬ﺇﺫﺍ ﻜﺎﻥ ‪ G‬ﻤﺭﺠﺢ ﺍﻟﺠﻤﻠﺔ )‪ (C ; 2) ; (B ; 2) ; (A ; 1‬ﻓﺈﻥ ‪G‬‬‫ﻫﻭ ﺃﻴﻀﺎ ﻤﺭﺠﺢ ﺍﻟﺠﻤﻠﺔ ‪ I ; 47 ; I ; 3‬ﺤﻴﺙ ‪ I‬ﻤﺭﺠﻊ) ( ) (} {‬ ‫})‪ {(B ; 2) ; ( A ; 1‬ﻭ ‪ J‬ﻤﻨﺘﺼﻑ ]‪[BC‬‬ ‫‪ (8‬ﺇﺫﺍ ﻜﺎﻥ ‪ F‬ﻤﺭﺠﺢ ﺍﻟﺠﻤﻠﺔ ‪:‬‬‫})‪{(D ; 1) ; (G ; 4) ; (B ; 2) ; (A ; 1‬‬ ‫ﻓﺈﻥ ﺍﻟﻨﻘﻁ ‪ F‬ﻭ ﻤﻨﺘﺼﻑ ‪ AD‬ﻭ ﻤﺭﺠﺢ ﺍﻟﺠﻤﻠﺔ ‪[ ]:‬‬‫)‪ (C ; 2) ; (B ; 1‬ﻋﻠﻰ ﺍﺴﺘﻘﺎﻤﺔ ﻭﺍﺤﺩﺓ ‪{ }.‬‬‫ﺃﻋﻁ ﺍﻷﺠﻭﺒﺔ ﺍﻟﺼﺤﻴﺤﺔ ﻭ ﺍﻟﺨﺎﻁﺌﺔ ﻟﻜل ﺍﻟﺘﻤﺎﺭﻴﻥ ﺍﻟﺘﺎﻟﻴﺔ ﻤﻥ ‪ 2‬ﺇﻟﻰ ‪: 7‬‬‫‪ur uuur ur uuur‬‬ ‫ﺍﻟﺘﻤﺭﻴﻥ‪. 2‬‬‫‪ ABCD‬ﻤﺴﺘﻁﻴل ﻨﻀﻊ ‪V = AD ; U = AB :‬‬

‫‪ur ur ur ur‬‬ ‫‪U + V = U - V (1‬‬ ‫‪ur‬‬ ‫‪2‬‬ ‫‪+‬‬ ‫‪ur‬‬ ‫‪2‬‬ ‫=‬ ‫‪ur‬‬ ‫‪-‬‬ ‫‪ur‬‬ ‫‪2‬‬ ‫‪U‬‬ ‫‪V‬‬ ‫‪U‬‬ ‫‪V‬‬ ‫‪ur ur ur ur ur‬‬ ‫‪(2‬‬ ‫‪U + V + U - V = 2 U (3‬‬ ‫ﺍﻟﺘﻤﺭﻴﻥ‪. 3‬‬‫‪ G‬ﻤﺭﺠﺢ ﺍﻟﺠﻤﻠﺔ ) ‪ ( A ; α) ; (B ; β) ; (C ; γ‬ﺤﻴﺙ ‪ γ , β , α‬ﻤﻥ} {‬ ‫ﻨﻔﺱ ﺍﻹﺸﺎﺭﺓ ‪.‬‬‫‪ (1‬ﺘﻭﺠﺩ ﻤﻌﺎﻤﻼﺕ ﺤﻘﻴﻘﻴﺔ ‪ γ′ , β′ , α′ :‬ﺒﺤﻴﺙ ‪γ′ + β′ + α′ = 1 :‬‬‫ﻭ ‪ G‬ﻫﻭ ﻤﺭﺠﺢ ﺍﻟﺠﻤﻠﺔ ‪. (C , γ ) ; (B ; β) ; ( A ; α) :‬‬‫‪uuur‬‬ ‫‪uuur‬‬ ‫‪uuur‬‬‫= ‪CG‬‬ ‫‪α‬‬ ‫‪CA +‬‬ ‫‪α‬‬ ‫‪CB‬‬ ‫‪ (2‬ﺍﻟﻨﻘﻁﺔ ‪ G‬ﺘﺤﻘﻕ ‪:‬‬ ‫‪α+β+γ‬‬ ‫‪α+β+γ‬‬ ‫‪ (3‬ﺍﻟﻨﻘﻁﺔ ‪ G‬ﺘﻘﻊ ﺩﺍﺨل ﺍﻟﻤﺜﻠﺙ ‪. ABC‬‬ ‫ﺍﻟﺘﻤﺭﻴﻥ‪. 4‬‬ ‫‪ ABC‬ﻤﺜﻠﺙ ﻤﺘﺴﺎﻭﻱ ﺍﻟﺴﺎﻗﻴﻥ ﻓﻲ ‪ AH . A‬ﺍﺭﺘﻔﺎﻉ ‪[ ].‬‬ ‫‪‬‬ ‫‪A‬‬ ‫;‬ ‫‪1‬‬ ‫‪,‬‬ ‫‪‬‬ ‫‪H‬‬ ‫;‬ ‫‪2 ‬‬ ‫‪ G‬ﻤﺭﺠﺢ ﺍﻟﺠﻤﻠﺔ ‪:‬‬ ‫‪‬‬ ‫‪3 ‬‬ ‫‪‬‬ ‫‪‬‬ ‫‪‬‬ ‫‪3‬‬ ‫‪‬‬ ‫‪uur‬‬ ‫=‬ ‫‪1‬‬ ‫‪uuur‬‬ ‫‪(1‬‬ ‫‪CI‬‬ ‫‪3‬‬ ‫)‪ (B G‬ﻴﻘﻁﻊ )‪ (AC‬ﻓﻲ ‪ I‬ﺒﺤﻴﺙ ‪CA :‬‬ ‫‪uuur uuur uuur r‬‬ ‫‪GA + GB + GC = 0 (2‬‬ ‫‪ A (3‬ﻤﺭﺠﻊ ﺍﻟﺠﻤﻠﺔ )‪{ }(G ; 3) ; (B ; 1) ; (C ; 1‬‬ ‫ﺍﻟﺘﻤﺭﻴﻥ‪. 5‬‬‫‪{ }(A‬‬‫;‬‫)‪-7‬‬‫‪,‬‬ ‫‪(B‬‬ ‫;‬ ‫)‪3‬‬ ‫‪,‬‬ ‫‪(C‬‬ ‫;‬ ‫)‪5‬‬ ‫‪ ABC‬ﻤﺜﻠﺙ ‪ G ،‬ﻤﺭﺠﺢ ﺍﻟﺠﻤﻠﺔ‬ ‫)‬ ‫‪ G (1‬ﻴﻘﻊ ﺩﺍﺨل ﺍﻟﻘﻁﺎﻉ ﺍﻟﺯﺍﻭﻱ ﺍﻟﻤﻌﺭﻑ ﺒـ ‪BAC‬‬ ‫ﻭﺨﺎﺭﺝ ﺍﻟﻤﺜﻠﺙ ‪.ABC‬‬ ‫‪ (2‬ﺍﻟﻤﺴﺘﻘﻴﻡ)‪ (BG‬ﻴﻘﻁﻊ ﺍﻟﻤﺴﺘﻘﻴﻡ )‪ (AC‬ﻓﻲ ﻤﺭﺠﺢ ﺍﻟﺠﻤﻠﺔ‬

‫} )‪. {(A ; -7) , (C ; 5‬‬‫‪ B (3‬ﻫﻭ ﻤﺭﺠﺢ ﺍﻟﺠﻤﻠﺔ } ) ‪. {(A ; -7 ) , (C ; 5 ) , ( G ; 1‬‬ ‫ﺍﻟﺘﻤﺭﻴﻥ‪. 6‬‬‫‪ ABC‬ﻤﺜﻠﺙ ‪ G ،‬ﻤﺭﺠﺢ )‪{ }(C ; 3uu)r , (Buu;ur2) , ( A ; 1‬‬ ‫‪ (AG) (1‬ﻴﻘﻁﻊ )‪ (BC‬ﻓﻲ ﺍﻟﻨﻘﻁﺔ ‪ I‬ﺤﻴﺙ ‪5BI = 2BC :‬‬ ‫‪ (2‬ﻨﻘﻁ‪r‬ﺔ ﺘﻘﺎﻁ‪r‬ﻊ‪ (BG)uu‬ﻭ‪ (AuCuu)r‬ﻫﻲ ﺍﻟﻨﻘﻁﺔ ‪ J‬ﺤﻴﺙ ‪:‬‬ ‫‪GB + 2GJ = 0‬‬ ‫‪ (CG) (3‬ﻴﻘﻁﻊ )‪ (AB‬ﻓﻲ ﻤﻨﺘﺼﻑ ‪[ ]. AB‬‬ ‫‪uur uuur‬‬ ‫ﺍﻟﺘﻤﺭﻴﻥ‪. 7‬‬‫‪3AJ = 2AC‬‬ ‫‪ uAuBurC‬ﻤﺜﻠﺙ ‪ K , Juu, uIr.‬ﻨﻘﻁ ﺤ‪r‬ﻴﺙ ‪uur uur :‬‬ ‫‪; -3IB + 4IC = 0 ; 2uKurA =u3uKur B‬‬ ‫‪. IC = 3BC (1‬‬ ‫‪ J (2‬ﻫﻭ ﻤﺭﺠﺢ ﺍﻟﺠﻤﻠﺔ )‪{ }. ( A ; 3) ; (C ; 2‬‬ ‫‪ (3‬ﺍﻟﻤﺴﺘﻘﻴﻤﺎﺕ )‪ (C k) ; (BJ) ; (AI‬ﺘﺘﻘﺎﻁﻊ ﻓﻲ ﻤﺭﺠﺢ ﺍﻟﺠﻤﻠﺔ‬ ‫})‪. {( A ; 2) ; (B ; -3) ; (C ; 4‬‬ ‫ﺍﻟﺘﻤﺭﻴﻥ‪. 8‬‬ ‫‪ A‬ﻭ‪ B‬ﻨﻘﻁﺘﺎﻥ ﺤﻴﺙ ‪AB = 5 cm :‬‬ ‫ﺃﻨﺸﺊ ﺍﻟﻨﻘﻁﺔ ‪ G‬ﻤﺭﺠﺢ ﺍﻟﺠﻤﻠﺔ ‪{ }( ) ( )B ; -2 ; A ; 7‬‬ ‫ﺍﻟﺘﻤﺭﻴﻥ‪. 9‬‬ ‫‪ABC‬ﻤﺜﻠﺙ ﻜﻴﻔﻲ ‪.‬‬‫ﺃﻨﺸﺊ ﺍﻟﻨﻘﻁﺔ ‪ G‬ﻤﺭﺠﺢ ﺍﻟﺠﻤﻠﺔ )‪{ }. ( A ; 2) ; (B ; -1) ; (C ; -2‬‬ ‫ﺍﻟﺘﻤﺭﻴﻥ‪. 10‬‬ ‫‪ ABC‬ﺸﺒﻪ ﻤﺤﺭﻑ ﻗﺎﺌﻡ ﺤﻴﺙ ‪:‬‬ ‫‪CD = 2 cm ; AD = 3 cm ; AB = 6 cm‬‬

‫‪( )r‬‬‫‪1‬‬‫‪uuur‬‬‫و‬ ‫‪r‬‬ ‫=‬ ‫‪1‬‬ ‫‪uuur‬‬ ‫ﻨﻀﻊ ‪:‬‬‫‪3‬‬ ‫‪AD‬‬ ‫‪i‬‬ ‫‪r6‬‬ ‫‪AB‬‬‫=‪j‬‬ ‫‪r‬‬ ‫‪ (1‬ﺒﻴﻥ ﺃﻥ ‪A ; i , j‬‬ ‫ﻤﻌﻠﻡ ﻟﻠﻤﺴﺘﻭﻱ‬ ‫‪ (2‬ﺍﺤﺴﺏ ﺇﺤﺩﺍﺜﻴﻲ ﺍﻟﻨﻘﻁﺔ ‪ G‬ﻤﺭﺠﺢ ﺍﻟﺠﻤﻠﺔ ‪:‬‬ ‫)‪(D ; 3) ; (C ; -2) ; (B ; 5) ; (A ; 4‬‬ ‫ﺍﻟﺘﻤﺭﻴﻥ‪. 11‬‬ ‫‪ A ; B ; C ; D‬ﺃﺭﺒﻊ ﻨﻘﻁ ﺤﻴﺙ ‪:‬‬ ‫‪DA = 6 cm ; CD = 7 cm ; AB = 5 cm‬‬ ‫‪ G‬ﻤﺭﺠﺢ ﺍﻟﺠﻤﻠﺔ )‪(A ; 2) ; (B ; 3) ; (C ; 3) ; (D ; 4‬‬‫ﺃ( ﺃﻨﺸﺊ ﺍﻟﻨﻘﻁﺔ ‪ I‬ﻤﺭﺠﺢ ﺍﻟﺠﻤﻠﺔ ‪{ }( ) ( )A ; 2 ; B ; 3‬‬‫ﺏ( ﺃﻨﺸﺊ ﺍﻟﻨﻘﻁﺔ ‪ J‬ﻤﺭﺠﺢ ﺍﻟﺠﻤﻠﺔ ‪{ }( ) ( )C ; 3 ; D ; 4‬‬‫ﺟـ ( ﺍﺴﺘﻨﺘﺞ ﺃﻥ ﺍﻟﻨﻘﻁﺔ ‪ G‬ﺘﻨﺘﻤﻲ ﺇﻟﻰ ﺍﻟﻤﺴﺘﻘﻴﻡ )‪. (IJ‬‬ ‫ﺍﻟﺘﻤﺭﻴﻥ‪. 12‬‬‫‪ A‬ﻭ‪ B‬ﻨﻘﻁﺘﺎﻥ ﺤﻴﺙ ‪ AB = 6 cm :‬ﻭ ‪ M‬ﻨﻘﻁﺔ ﻜﻴﻔﻴﺔ ﻤﻥ ﺍﻟﻤﺴﺘﻭﻱ‪.‬‬‫‪ (1‬ﻋﻴ‪r‬ﻥ‪ u‬ﺍ‪u‬ﻟﻤ‪u‬ﺭﺠﺢ ‪urI‬ﻟ‪u‬ﻠ‪u‬ﺠﻤ‪u‬ﻠﺔ )‪ . (A ; 1u)u;ur(B ; 1‬ﻭﻋﺒﺭ ﻋﻥ ﺍﻟﻤﺠﻤﻭﻉ ‪:‬‬ ‫‪ MA + MB‬ﺒﺩﻻﻟﺔ ‪. MI‬‬ ‫‪ (2‬ﻋﻴ‪r‬ﻥ‪ u‬ﺍ‪u‬ﻟﻤ‪u‬ﺭﺠﺢ ‪rG‬ﻟ‪u‬ﻠ‪u‬ﺠ‪u‬ﻤﻠ‪u‬ﺔ )‪ (A ;u1u)u;ur(B ; 2‬ﻭﻋﺒﺭ ﻋﻥ ﺍﻟﻤﺠﻤﻭﻉ ‪:‬‬ ‫‪ MA + 2MB‬ﺒﺩﻻﻟﺔ ‪MG‬‬ ‫ﺍﻟﺘﻤﺭﻴﻥ‪. 13‬‬ ‫ﻟﻴﻜﻥ ﺍﻟﻘﻁﻌﺔ ‪ AB‬ﻁﻭﻟﻬﺎ ‪[ ]. 6 cm‬‬ ‫‪ - 1‬ﻤﺎ ﻫﻲ ﻤﺠﻤﻭﻋﺔ ﺍﻟﻨﻘﻁ ‪ M‬ﻤﻥ ﺍﻟﻤﺴﺘﻭﻱ ﺒﺤﻴﺙ ‪ AM = 6 cm‬؟‬‫‪ -2‬ﻟﻴﻜﻥ ‪ I‬ﻤﻨﺘﺼﻑ ‪[ ]uuur uuuur uu.urAB‬‬‫ﺃ ( ﻋﺒﺭ ﻋﻥ ﺍﻟﻤﺠﻤﻭ‪ur‬ﻉ‪ MuAuu+r MB u: u‬ﺒ‪ur‬ﺩ‪u‬ﻻﻟ‪u‬ﺔ‪u‬ﺍﻟﺸﻌﺎﻉ ‪. MI‬‬ ‫ﺏ( ﺒﺭﻫﻥ ﺃﻨﻪ ‪MA + MB = 2 MI :‬‬ ‫ﺟـ( ﻋﻴﻥ ﻤﺠﻤﻭﻋﺔ ﺍﻟﻨﻘﻁ ‪ M‬ﻤﻥ ﺍﻟﻤﺴﺘﻭﻱ ﺒﺤﻴﺙ ‪:‬‬

‫‪uuuur uuur‬‬‫‪MA + MB = 6 cm‬‬ ‫ﺍﻟﺘﻤﺭﻴﻥ‪. 14‬‬ ‫‪ A‬ﻭ‪ B‬ﻨﻘﻁﺘﺎﻥ ﻤﻥ ﺍﻟﻤﺴﺘﻭﻱ ‪.‬‬‫‪ f‬ﺘﻁﺒﻴﻕ ﻴﺭﻓﻕ ﺒﻜل ﻨﻘﻁﺔ ‪ M‬ﻤﻥ ﺍﻟﻤﺴﺘﻭﻱ ﺍﻟﻨﻘ‪ur‬ﻁﺔ‪ Muu′uu‬ﺤﻴ‪r‬ﺙ‪uuuur uu:u‬‬‫‪MA + MuBuur= MuMuur′‬‬ ‫‪ (1‬ﻟﻴﻜﻥ ‪ I‬ﻤﻨﺘﺼﻑ ‪ . AB‬ﺒﺭﻫﻥ ﺃﻥ ‪[ ]. IM′ = -IM :‬‬ ‫‪ (2‬ﺤﺩﺩ ﻁﺒﻴﻌﺔ ﺍﻟﺘﻁﺒﻴﻕ ‪ f‬ﻭﻋﻨﺎﺼﺭﻩ ﺍﻟﻤﻤﻴﺯﺓ‪.‬‬ ‫)*( ﺍﻟﺘﻤﺭﻴﻥ‪. 15‬‬ ‫‪ ABC‬ﻤﺜﻠﺙ ﻭ ‪ G‬ﻤﺭﺠﺢ ﺍﻟﺠﻤﻠﺔ ‪:‬‬ ‫)‪(C ; -1) , (B ; 2) , (A ; 1‬‬‫‪uuuur‬‬ ‫‪ (1‬ﺃﻨﺸﺊ ﺍﻟﻨﻘﻁﺔ ‪uuuur uuur uuur . G‬‬‫‪ (2‬ﻋﺒﺭ ﻋﻥ ﺍﻟﻤﺠﻤﻭﻉ ‪uuMurA +uu2urMB - MuuuCur :‬ﺒﺩﻻﻟﺔ ‪uMuuGur‬‬‫ﺜﻡ ﺒﺭﻫﻥ ﺃﻥ ‪MA + 2MB - MC = 2 MG :‬‬ ‫‪ (3‬ﻋﻴﻥ ﻤﺠﻤﻭﻋ‪r‬ﺔ‪u‬ﺍﻟ‪u‬ﻨﻘ‪u‬ﻁ ‪rM‬ﻤ‪u‬ﻥ‪ u‬ﺍ‪u‬ﻟﻤﺴﺘﻭﻱ‪r‬ﺒ‪uu‬ﺤﻴ‪u‬ﺙ‪: u‬‬ ‫‪MA + 2MB - MC = 5‬‬ ‫ﺍﻟﺘﻤﺭﻴﻥ‪. 16‬‬‫‪uAuuBurC‬ﻤﺜ‪u‬ﻠﺙ ‪rf .‬ﺘ‪u‬ﻁ‪u‬ﺒﻴ‪u‬ﻕ ﻴﺭﻓ‪r‬ﻕ‪u‬ﺒﻜ‪u‬ل‪ u‬ﻨﻘﻁﺔ‪uMuur‬ﻤ‪u‬ﻥ ﺍﻟﻤﺴﺘﻭﻱ ﺍﻟﻨﻘﻁﺔ ‪ M′‬ﺒﺤﻴﺙ ‪:‬‬ ‫‪MA + MB + MC = MM′‬‬ ‫‪ (1‬ﻟﻴﻜﻥ ‪ G‬ﻤﺭﻜﺯ‪ur‬ﺜ‪u‬ﻘل‪ u‬ﺍ‪u‬ﻟﻤﺜﻠﺙ ‪uu.uAurBC‬‬ ‫ﺒﺭﻫﻥ ﺃﻥ ‪GM′ = -2GM :‬‬ ‫‪ (2‬ﺤﺩﺩ ﻁﺒﻴﻌﺔ ﺍﻟﺘﻁﺒﻴﻕ ‪ f‬ﻭﻋﻨﺎﺼﺭﻩ ﺍﻟﻤﻤﻴﺯﺓ ‪.‬‬ ‫‪( )r r‬‬ ‫ﺍﻟﺘﻤﺭﻴﻥ‪. 17‬‬ ‫ﺍﻟﻤﺴﺘﻭﻱ ﻤﻨﺴﻭﺏ ﺇﻟﻰ ﺍﻟﻤﻌﻠﻡ ‪ O ; i , j‬ﻟﺘﻜﻥ ﺍﻟﻨﻘﻁﺘﺎﻥ‬ ‫)‪B (3 ; 6) ; A (6 ; 3‬‬ ‫‪ (1‬ﺍﺤﺴﺏ ﺇﺤﺩﺍﺜﻴﻲ ﻤﺭﻜﺯ ﺜﻘل ﺍﻟﻤﺜﻠﺙ ‪.OAB‬‬ ‫‪ (2‬ﺍﺤﺴﺏ ﺇﺤﺩﺍﺜﻴﺎﺕ ﻤﺭﺍﻜﺯ ﺍﻟﺜﻘل ‪ J , I , H‬ﻟﻠﻤﺜﻠﺜﺎﺕ‪.‬‬

‫‪AGB ; OGB ; AGA‬‬‫‪ (3‬ﺒﺭﻫﻥ ﺃﻥ ﺍﻟﻤﺜﻠﺜﻴﻥ ‪ OAB‬ﻭ ‪ HIJ‬ﻟﻬﻤﺎ ﻨﻔﺱ ﻤﺭﻜﺯ ﺍﻟﺜﻘل ‪.‬‬‫‪( )r r‬‬ ‫ﺍﻟﺘﻤﺭﻴﻥ‪. 18‬‬‫ﺍﻟﻤﺴﺘﻭﻱ ﻤﻨﺴﻭﺏ ﺇﻟﻰ ﻤﻌﻠﻡ ﻤﺘﻌﺎﻤﺩ ﻭ ﻤﺘﺠﺎﻨﺱ ‪O ; i , j‬‬ ‫ﻟﺘﻜﻥ ﺍﻟﻨﻘﻁ )‪C (5 ; 2) ; B (-2 ; 3) ; A ( 1 ; 4‬‬‫)‪C′(2 ; 5) ; B′(-2 ; 3) ; A′(4 ; 1‬‬ ‫‪ (D) (1‬ﻫﻭ ﺍﻟﻤﺴﺘﻘﻴﻡ ﺫﻭ ﺍﻟﻤﻌﺎﺩﻟﺔ ‪. y = x‬‬‫ﺒﺭﻫﻥ ﺃﻥ ﺍﻟﻤﺴﺘﻘﻴﻡ )‪ (D‬ﻫﻭ ﻤﺤﻭﺭ ﺍﻟﻘﻁﻊ ﺍﻟﻤﺴﺘﻘﻴﻤﺔ‬ ‫]‪[CC′] ; [BB′] ; [AA′‬‬ ‫‪ (2‬ﻟﻴﻜﻥ ‪ G‬ﻤﺭﺠﺢ ﺍﻟﺠﻤﻠﺔ )‪(A ; -3) ; (B ; 2) ; (C ; 4‬‬ ‫ﺍﺤﺴﺏ ﺇﺤﺩﺍﺜﻴﻲ ‪.G‬‬ ‫‪ (3‬ﻟﻴﻜﻥ ‪ G′‬ﻤﺭﺠﺢ ﺍﻟﺠﻤﻠﺔ ‪:‬‬‫)‪ . (C′ ; 4) , (B′ ; 2) , ( A′ ; -3‬ﺍﺤﺴﺏ ﺇﺤﺩﺍﺜﻴﺘﻲ ‪. G′‬‬‫‪ (4‬ﻫل ﺍﻟﻤﺴﺘﻘﻴﻡ )‪ (D‬ﻤﺤﻭﺭ ﻟﻠﻘﻁﻌﺔ ‪ GG′‬؟] [‬ ‫)*( ﺍﻟﺘﻤﺭﻴﻥ‪. 19‬‬ ‫‪ ABC‬ﻤﺜﻠﺙ ﻤﺘﻘﺎﻴﺱ ﺍﻷﻀﻼﻉ ﻁﻭل ﻀﻠﻌﻪ ‪. 5cm‬‬‫‪ -1‬ﺃﻨﺸﺊ ﺍﻟﻨﻘﻁﺔ ‪ G‬ﻤﺭﺠﺢ ﺍﻟﺠﻤﻠﺔ )‪(C . 1) , (B ; -1) , ( A ; 1‬‬ ‫ﺜﻡ ﺒﻴﻥ ﺃﻥ ‪ ABCG‬ﻤﺘﻭﺍﺯﻱ ﺃﻀﻼﻉ ‪.‬‬‫‪ -2‬ﺃ( ﺍﺴﺘﻌﻤل ﺍﻟﺴﺅﺍل ﺍﻟﺴﺎﺒﻕ ﻟﺘﻌﻴﻴﻥ ﺍﻟﻤﺠﻤﻭﻋﺔ )‪ (Γ‬ﻟﻠﻨﻘﻁ ‪ M‬ﻤﻥ‬‫‪uuuur uuur uuur‬‬ ‫=‬ ‫‪53‬‬ ‫ﺍﻟﻤﺴﺘﻭﻱ ﺤﻴﺙ ‪:‬‬‫‪MA - MB + MC‬‬ ‫‪2‬‬‫ﺏ( ﺘﺤﻘﻕ ﺃﻥ ﻤﻨﺘﺼﻑ ‪ AC‬ﻴﻨﺘﻤﻲ ﺇﻟﻰ )‪ . (Γ‬ﺃﻨﺸﺊ )‪[ ]. (Γ‬‬ ‫)*( ﺍﻟﺘﻤﺭﻴﻥ‪. 20‬‬ ‫ﻟﻴﻜﻥ ‪ ABC‬ﻤﺜﻠﺙ ﻏﻴﺭ ﻗﺎﺌﻡ ‪ A′ , B′ , C′‬ﻤﻨﺘﺼﻔﺎﺕ ﺍﻷﻀﻠﻊ‬‫‪ AB , AC , BC‬ﻋﻠﻰ ﺍﻟﺘﺭﺘﻴﺏ ﻭﻟﻴﻜﻥ ‪ O‬ﻤﺭﻜﺯ ﺍﻟﺩﺍﺌﺭﺓ ﺍﻟﻤﺤﻴﻁﺔ ﺒﺎﻟﻤﺜﻠﺙ ‪[ ] [ ] [ ].‬‬

‫‪A‬‬ ‫‪ -1‬ﻟﻴﻜﻥ‪uHuur‬ﻨﻘﻁﺔ‪r‬ﻤ‪u‬ﻥ‪u‬ﺍﻟ‪u‬ﻤﺴﺘﻭ‪r‬ﻱ‪u‬ﺍ‪u‬ﻟﻤ‪u‬ﻌﺭﻑ ﺒ‪r‬ـ‪uu: u‬‬ ‫'‪H B‬‬ ‫‪uuuOr Hu=uurOA +uuOuurB + OC‬‬ ‫ﺃ( ﺒﻴﻥ ﺃﻥ ‪. OB + OCuu=uur2OAu′ uur:‬‬ ‫‪C' G‬‬ ‫‪O‬‬ ‫ﺏ( ﺍﺴﺘﻨﺘﺞ ‪ AH‬ﺒﺩﻻﻟﺔ ‪. OA′‬‬‫'‪B A‬‬ ‫ﺟـ( ﺒﺭﻫﻥ ﺃﻥ ﺍﻟﻤﺴﺘﻘﻴﻡ )‪ (AH‬ﻋﻤﻭﺩﻱ ﻋﻠﻰ )‪. (BC‬‬ ‫ﺩ( ﻟﻤﺎﺫﺍ ﺍﻟﻤﺴﺘﻘﻴﻡ )‪ (BH‬ﻋﻤﻭﺩﻱ ﻋﻠﻰ )‪ (AC‬؟‬ ‫ﺍﺴﺘﻨﺘﺞ ﻁﺒﻴﻌﺔ ﺍﻟﻨﻘﻁﺔ ‪ H‬ﺒﺎﻟﻨﺴﺒﺔ ﻟﻠﻤﺜﻠﺙ ‪. ABC‬‬ ‫‪ -2‬ﻟﻴﻜﻥ ‪ G‬ﻤﺭﻜﺯ‪r‬ﺜ‪u‬ﻘل‪ u‬ﺍ‪u‬ﻟﻤﺜﻠﺙ ‪uuAurBC‬‬ ‫ﺃ( ﺒﺭﻫﻥ ﺃﻥ ‪. OH = 3 OG :‬‬ ‫ﺏ( ﻓﻲ ﺃﻱ ﺤﺎﻟﺔ ﻴﻜﻭﻥ ‪:‬‬ ‫‪O=G=H‬؟‬ ‫ﺟـ( ﺒﺎﺴﺘﺜﻨﺎﺀ ﺍﻟﺤﺎﻟﺔ ﺍﻟﺴﺎﺒﻘﺔ ﺒﻴﻥ‬ ‫ﺃﻥ ﺍﻟﻨﻘﻁ ‪C H , G , O‬‬ ‫ﺘﻨﺘﻤﻲ ﺇﻟﻰ ﻤﺴﺘﻘﻴﻡ‬ ‫) ﻴﺴﻤﻰ ﻫﺫﺍ ﺍﻟﻤﺴﺘﻘﻴﻡ ﻤﺴﺘﻘﻴﻡ \"ﺃﻭﻟﻴﺭ\" (‬‫ﻡﺴﺘﻘﻴﻢ أوﻟﻴﺮ‬ ‫ﺩ( ﺒﺭﻫﻥ ﺃﻥ ﺍﻟﻨﺘﻴﺠﺔ ﺍﻟﺴﺎﺒﻘﺔ ﺘﺒﻘﻰ ﺼﺤﻴﺤﺔ ﻓﻲ ﺤﺎﻟﺔ ﻤﺜﻠﺙ ﻗﺎﺌﻡ‬ ‫ﺍﻟﺘﻤﺭﻴﻥ‪. 21‬‬ ‫‪ A , B‬ﻨﻘﻁﺘﺎﻥ ﻤﺘﻤﺎﻴﺯﺘﺎﻥ ﻤﻥ ﺍﻟﻤﺴﺘﻭﻱ ‪.‬‬ ‫‪ I‬ﻨﻘﻁﺔ ﻤﻥ ﺍﻟﻤﺴﺘﻘﻴﻡ ‪( ). AB‬‬ ‫‪ -‬ﺒﺭﻫﻥ ﺃﻨﻪ ﻴﻭﺠﺩ ﻋﺩﺩﺍﻥ ﺤﻘﻴﻘﻴﺎﻥ ‪ a‬ﻭ ‪ b‬ﺒﺤﻴﺙ ‪:‬‬ ‫ﻴﻜﻭﻥ ‪ I‬ﻤﺭﺠﺢ ﺍﻟﻨﻘﻁﺘﻴﻥ ‪( ) ( )A ; a , B ; b‬‬ ‫ﺍﻟﺘﻤﺭﻴﻥ‪. 22‬‬‫ﻟﻴﻜﻥ ‪ ABC‬ﻤﺜﻠﺙ ‪ .‬ﻋﻴﻥ ﻭﺃﻨﺸﺊ ﺍﻟﻤﺠﻤﻭﻋﺔ )‪ (E‬ﻟﻠﻨﻘﻁ ‪ M‬ﻤﻥ ﺍﻟﻤﺴﺘﻭﻱ ﺒﺤﻴﺙ ﻴﻜﻭﻥ ‪:‬‬ ‫‪uuur uuuur uuur uuur‬‬ ‫) ‪ ( MA + 2 MB + MC‬ﻭ ‪ AC‬ﻤﺭﺘﺒﻁﺎﻥ ﺨﻁﻴﺎ ‪.‬‬

‫‪uuur uuur‬‬ ‫ﺍﻟﺘﻤﺭﻴﻥ‪. 23‬‬‫ﻓﻲ ﻤﺜﻠﺙ ‪ ABC‬ﻟﺘﻜﻥ ‪ E‬ﻨﻘﻁﺔ ﺤﻴﺙ ‪EB = - 2 EA :‬‬‫ﻨﺴﻤﻲ ‪ A′‬و ‪ A1‬ﻤﻨﺘﺼﻔﺎ ‪ BC‬و ‪ AA′‬ﻋﻠﻰ ﺍﻟﺘﺭﺘﻴﺏ] [ ] [‬‫‪ -‬ﺒﺭﻫﻥ ﺃﻥ ﺍﻟﻨﻘﻁ ‪ A1 , E , C‬ﻋﻠﻰ ﺍﺴﺘﻘﺎﻤﺔ ﻭﺍﺤﺩﺓ ‪.‬‬ ‫ﺍﻟﺘﻤﺭﻴﻥ‪. 24‬‬‫ﻟﺘﻜﻥ ‪ A , B , C , D‬ﺃﺭﺒﻊ ﻨﻘﻁ ﻤﻥ ﺍﻟﻤﺴﺘﻭﻱ ﻭ ‪ G‬ﻤﺭﺠﺢ ﻟﻠﺠﻤﻠﺔ‬‫)‪. (A ; 1) , (B ; 2) , (C ; 2) , (D ; 1‬‬‫ﻨﺴﻤﻲ ‪ M‬ﻭ ‪ N‬ﻤﻨﺘﺼﻔﺎ ‪ AD‬و ‪ BC‬ﻋﻠﻰ ﺍﻟﺘﺭﺘﻴﺏ ‪[ ] [ ].‬‬‫‪ -‬ﺒﺭﻫﻥ ﺃﻥ ﺍﻟﻨﻘﻁ ‪ G , N , M‬ﻋﻠﻰ ﺍﺴﺘﻘﺎﻤﺔ ﻭﺍﺤﺩﺓ ‪.‬‬

‫اﻟﺤـﻠــــــﻮل‬ ‫ﺍﻟﺘﻤﺭﻴﻥ‪. 1‬‬ ‫‪ (3‬ﺼﺤﻴﺢ‬ ‫‪ (2‬ﺼﺤﻴﺢ‬ ‫‪ (1‬ﺨﻁﺄ‬ ‫‪ (6‬ﺼﺤﻴﺢ‬ ‫‪ (5‬ﺼﺤﻴﺢ‬ ‫‪ (4‬ﺨﻁﺄ‬ ‫‪ (8‬ﺼﺤﻴﺢ ‪.‬‬ ‫‪ (7‬ﺨﻁﺄ‬ ‫ﺍﻟﺘﻤﺭﻴﻥ‪. 2‬‬ ‫‪ (3‬ﺨﻁﺄ‬ ‫‪ (2‬ﺼﺤﻴﺢ‬ ‫‪ (1‬ﺼﺤﻴﺢ‬ ‫ﺍﻟﺘﻤﺭﻴﻥ‪. 3‬‬ ‫‪ (3‬ﺼﺤﻴﺢ‬ ‫‪ (2‬ﺼﺤﻴﺢ‬ ‫‪ (1‬ﺼﺤﻴﺢ‬ ‫ﺍﻟﺘﻤﺭﻴﻥ‪. 4‬‬ ‫‪ (3‬ﺨﻁﺄ‬ ‫‪ (2‬ﺼﺤﻴﺢ‬ ‫‪ (1‬ﺨﻁﺄ‬ ‫ﺍﻟﺘﻤﺭﻴﻥ‪. 5‬‬ ‫‪ (3‬ﺨﻁﺎ‬ ‫‪ (2‬ﺼﺤﻴﺢ‬ ‫‪ (1‬ﺼﺤﻴﺢ‬ ‫ﺍﻟﺘﻤﺭﻴﻥ‪. 6‬‬ ‫‪ (3‬ﺨﻁﺄ‬ ‫‪ (2‬ﺼﺤﻴﺢ‬ ‫‪ (1‬ﺨﻁﺎ‬ ‫ﺍﻟﺘﻤﺭﻴﻥ‪. 7‬‬ ‫‪ (3‬ﺼﺤﻴﺢ‬ ‫‪ (2‬ﺨﻁﺄ‬ ‫‪ (1‬ﺨﻁﺄ‬ ‫ﺍﻟﺘﻤﺭﻴﻥ‪. 8‬‬ ‫ﻁﺭﻴﻘﺔ ‪:‬‬ ‫ﺃ ( ﺍﺤﺴﺏ ﻤﺠﻤﻭﻉ‪r‬ﺍ‪u‬ﻟﻤ‪u‬ﻌ‪u‬ﺎﻤﻼﺕ ﻭﺘﺤﻘﻕ‪ ur‬ﺃ‪u‬ﻨﻬ‪u‬ﺎ ﻏﻴﺭ ﻤﻌﺩﻭﻤﺔ‬ ‫ﺏ ( ﻋﺒﺭ ﻋﻥ ‪ AG‬ﺒﺩﻻﻟﺔ ‪. AB‬‬ ‫ﺟـ ( ﺍﺭﺴﻡ ﺍﻟﻨﻘﻁﺔ ‪. G‬‬ ‫ﺤـل ‪:‬‬ ‫ﺃ ( ﻤﺠﻤﻭﻉ ﺍﻟﻤﻌﺎﻤﻼﺕ ‪( )7 + - 2 = 5 ≠ 0 :‬‬ ‫‪uuur‬‬ ‫‪uuur‬‬ ‫‪AG = -‬‬ ‫‪2‬‬ ‫‪AB‬‬ ‫ﺏ(‬ ‫‪5‬‬‫‪GA‬‬ ‫‪B‬‬

‫ﺍﻟﺘﻤﺭﻴﻥ‪. 9‬‬‫ﺃ( ﻤﺠﻤﻭﻋﺔ ﺍﻟﻤﻌﺎﻤﻼﺕ ‪2 + (- 1) + (- 2) + - 1 ≠ 0 :‬‬ ‫‪uuur‬‬ ‫‪uuur‬‬ ‫‪--u21uuruAuCur‬‬ ‫= ‪AG‬‬ ‫‪-1‬‬ ‫‪AB‬‬ ‫‪+‬‬ ‫ﺏ(‬ ‫‪-1‬‬ ‫‪uuur‬‬ ‫‪uuur‬‬ ‫‪AG = AB + 2AC‬‬‫‪AC‬‬ ‫ﺟـ( ﺍﻟﺭﺴﻡ ‪:‬‬‫‪B‬‬ ‫ﺍﻟﺘﻤﺭﻴﻥ‪G . 10‬‬ ‫‪ (1‬ﺍﻟﻤﺴﺘﻘﻴﻤﺎﻥ‪r ABuuur‬و‪ ADuuu‬ﻤﺘﻌﺎﻤﺩﺍﻥ) ( ) (‬ ‫ﻭﻤﻨﻪ ‪ AB :‬و‪ rAD r‬ﻏﻴﺭ ﻤﺘﻭﺍﺯﻴﻴﻥ ) ﻤﺭﺘﺒﻁﻴﻥ ﺨﻁﻴﺎ (‬ ‫ﻨﺴﺘﻨﺘﺞ ﺇﺫﻥ‪r‬ﺃﻥ ‪ I r:‬و ‪ J‬ﻏﻴﺭ ﻤﺘﻭﺍﺯﻴﺎﻥ ‪( ).‬‬ ‫ﻭﻤﻨﻪ ‪ A ; I , J :‬ﻤﻌﻠﻡ ﻟﻠﻤﺴﺘﻭﻱ ‪.‬‬ ‫‪ (2‬ﺍﻟﻨﻘﻁﺔ ‪ A‬ﻫﻲ ﻤﺒﺩﺃ ﺍﻟﻤﻌﻠﻡ ﻭﻤﻨﻪ )‪A (0 ; 0‬‬ ‫‪uuur r‬‬ ‫‪r‬‬ ‫‪uuur‬‬‫ﻨﺴﺘﻨﺘﺞ ‪ AB = 6 i :‬ﻭﻤﻨﻪ ‪B ( 6 ; 0) :‬‬ ‫=‪i‬‬ ‫‪1‬‬ ‫‪AB‬‬ ‫ﻤﻥ ‪:‬‬ ‫‪6‬‬ ‫‪uuur r‬‬ ‫‪r‬‬ ‫‪1‬‬ ‫‪uuur‬‬‫ﻨﺴﺘﻨﺘﺞ ‪ AD = 3 j :‬ﻭﻤﻨﻪ‪D ( 0 ; 3 ) :‬‬ ‫=‪j‬‬ ‫‪3‬‬ ‫ﻤﻥ ‪AD‬‬ ‫‪ -‬ﺍﻟﻤﺴﺘ‪ur‬ﻘﻴ‪u‬ﻤﺎ‪u‬ﻥ )‪ (DC‬ﻭ‪ (AB) r‬ﻤﺘﻭﺍﺯﻴﺎﻥ‬‫ﺍﻟﺸﻌﺎﻋﺎﻥ ‪ DCr :‬و ‪ i‬ﻟﻬﻤﺎ ﻨﻔﺱ ﺍﻻ‪r‬ﺘﺠﺎﻩ ‪uuur‬‬‫ﺇﺫ‪ur‬ﻥ‪DC = 2 i u:u‬‬ ‫‪ uuri‬؛‪u‬‬ ‫ﻭ‪=1‬‬ ‫‪DC = 2‬‬ ‫‪uuur‬‬ ‫ﻭﻤﻨﻪ‪ur AC r= ArD + rDC r :‬‬‫ﺇﺫﻥ ‪ AC = 3 j + 2 i = 2 i + 3j :‬ﻭﻤﻨﻪ ‪C (2 ; 3) :‬‬ ‫‪ -‬ﻟﻴﻜﻥ ‪( )G xG ; yG :‬‬

‫‪‬‬ ‫‪xG‬‬ ‫=‬ ‫‪4×0‬‬ ‫‪+ 5×6 +‬‬ ‫‪(- 2)× 2 +‬‬ ‫‪3×0‬‬ ‫‪‬‬ ‫‪4+5+‬‬ ‫‪(- 2) + 3‬‬ ‫‪‬‬ ‫‪‬‬ ‫ﻟﺩﻴﻨﺎ ‪:‬‬ ‫‪4×0‬‬ ‫‪+‬‬ ‫‪5× 0 + (- 2)× 3‬‬ ‫‪+‬‬ ‫‪3×3‬‬ ‫‪‬‬ ‫‪yG‬‬ ‫=‬ ‫‪4 + 5 + (- 2) + 3‬‬ ‫‪‬‬ ‫ﺃﻱ‪ xG = 2,6 ; yG = 0,3 :‬ﻭ ﻤﻨﻪ ‪G ( 2,5 ; 0,3 ) :‬‬ ‫ﺍﻟﺘﻤﺭﻴﻥ‪. 11‬‬ ‫ﺃ ( ﻤﺠﻤﻭﻉ ﺍﻟﻤﻌﺎﻤﻼﺕ ‪2 + 3 = 5 ≠ 0‬‬ ‫‪uur‬‬ ‫‪uuur‬‬ ‫‪AI‬‬ ‫=‬ ‫‪3‬‬ ‫‪AB‬‬ ‫ﻭﻤﻨﻪ‪:‬‬ ‫‪5‬‬ ‫ﺏ ( ﻤﺠﻤﻭﻉ ﺍﻟﻤﻌﺎﻤﻼﺕ ‪3 + 4 = 7 ≠ 0 :‬‬ ‫‪uuur‬‬ ‫‪uuur‬‬ ‫= ‪AC‬‬ ‫‪4‬‬ ‫‪CD‬‬ ‫ﻭﻤﻨﻪ‪:‬‬ ‫‪7‬‬ ‫ﺟـ ( ﺍﻟﻨﻘﻁﺔ ‪ G‬ﻫﻲ ﻤﺭﺠﺢ ﺍﻟﺠﻤﻠﺔ )‪(B , 3) ; ( A , 2‬‬ ‫)‪ (D , 4) , (C , 3‬ﻓﻬﻲ ّﺇﺫﻥ ﻜﺫﻟﻙ ﻤﺭﺠﺢ ﺍﻟﺠﻤﻠﺔ ‪:‬‬ ‫)‪ ( J , 7) , ( I , 5‬ﺇﺫﻥ ﺍﻟﻨﻘﻁﺔ ‪ G‬ﺘﻨﺘﻤﻲ ﺇﻟﻰ ﺍﻟﻤﺴﺘﻘﻴﻡ ) ‪( IJ‬‬ ‫‪uur‬‬ ‫‪uur‬‬ ‫‪D‬‬ ‫‪GJ‬‬ ‫=‬ ‫‪5‬‬ ‫‪IJ‬‬ ‫ﻭ ﺘﺤﻘﻕ ‪:‬‬ ‫‪12‬‬ ‫‪G‬‬ ‫‪J‬‬‫‪A‬‬ ‫‪C‬‬ ‫‪IB‬‬ ‫ﺍﻟﺘﻤﺭﻴﻥ‪. 12‬‬ ‫‪ – 1‬ﺃ ( ‪urI‬ﻫ‪u‬ﻭ‪ u‬ﻤﻨﺘﺼﻑ ‪[ ]uuuur AuuBur‬‬ ‫‪MA + MB = 2 MI‬‬ ‫‪uuur‬‬ ‫‪uuur‬‬ ‫ﺏ(‬ ‫‪uuuur‬‬ ‫= ‪AG‬‬ ‫‪2‬‬ ‫‪AB‬‬ ‫‪–2‬ﺃ(‬ ‫‪uuur‬‬ ‫‪3‬‬ ‫‪uuuur‬‬ ‫ﺏ ( ‪MA + 2 MB = 3 MG‬‬

. 13‫ﺍﻟﺘﻤﺭﻴﻥ‬ 6cm u‫ﻫﺎ‬u‫ﺭ‬u‫ﻁ‬ur‫ﻨﺼﻑ ﻗ‬u‫ﻭ‬uAur ‫ﻲ ﻤﺭﻜﺯﻫﺎ‬u‫ﻟﺘ‬u‫ﺍ‬ur‫ ﺍﻟﺩﺍﺌﺭﺓ‬- 1 uuuMur A u+uuMr B = 2uMuurI ( ‫ – ﺃ‬2 MA + MB = 2 MI ( ‫ﺏ‬ uuuur uuur uuur MA + MB = 2 MIuuuur uuur uuurMA + MB = 6 ⇔ MI = 6 ( ‫ﺟـ‬‫ ﻫﻲ ﺍﻟﺩﺍﺌﺭﺓ ﺫﺍﺕ‬M ‫ ﺇﺫﻥ ﻤﺠﻤﻭﻋﺔ ﺍﻟﻨﻘﻁ‬. MI = 6 : ‫ﺃﻱ‬ . 3cm‫ ﻭﻨﺼﻑ ﺍﻟﻘﻁﺭ‬I ‫ﺍﻟﻤﺭﻜﺯ‬uuur uuuuur uuuur uuur u.u1ur4 ‫ﺍﻟﺘﻤﺭﻴﻥ‬2 MI = uMuurM′ : ‫ﻥ‬u‫ﺃ‬u‫ﺞ‬ur‫ ﻨﺴﺘﻨﺘ‬MAuu+ur MBuuu=r 2 MuuuIr ‫( ﻤﻥ‬1IM′ = - IM :‫ ﺃﻱ‬2 MI = MI + IM′ : ‫ﺇﺫﻥ‬ . I ‫ ﻫﻭ ﺍﻟﺘﻨﺎﻅﺭ ﺍﻟﺫﻱ ﻤﺭﻜﺯﻩ‬f ‫( ﺍﻟﺘﻁﺒﻴﻕ‬2 . 15‫)*( ﺍﻟﺘﻤﺭﻴﻥ‬ uuur uuur uuurC AG = AB - 1 AC (1 2ABuuuur G uuur uuur uuuurMA + 2 MB u-uMuurC = 2uuMurG uuur uuuur(2 MA + 2 MB - MC = 2 MGuuuur uuur uuur uuuurMA + 2 MB - MC = 2 . MGuuuur uuur uuur uuuurMA + 2 MB - MC = 2 MGuuuur uuur uuur uuuurMA + 2 MB - MC = 5 ⇔ 2 MG = 5 (3

‫⇔‬ ‫‪uuuur‬‬ ‫=‬ ‫‪5‬‬ ‫‪uuur‬‬ ‫‪MG‬‬ ‫‪2‬‬‫‪uuuur‬‬ ‫‪uuur‬‬‫ﻤﺠﻤﻭﻋﺔ ﺍﻟﻨﻘﻁ ‪ M‬ﺒﺤﻴﺙ ‪MA + 2 MB - MC = 5 :‬‬ ‫‪5‬‬ ‫ﻫﻲ ﺍﻟﺩﺍﺌﺭﺓ ﺫﺍﺕ ﺍﻟﻤﺭﻜﺯ ‪ G‬ﻭﻨﺼﻑ ﺍﻟﻘﻁﺭ ‪2‬‬ ‫ﺍﻟﺘﻤﺭﻴﻥ‪uuuur uuur uuur uu.u1ur6‬‬ ‫‪ (1‬ﻤﻥ ‪uuuur uuuurMA + MuBuuu+r MCuu=uu3ur MG :‬‬ ‫ﻨﺴﺘﻨﺘﺞ ﺃﻥ ‪ 3 MG = MM′ :‬ﺇﺫﻥ ‪−2 GM = GM′ :‬‬ ‫‪ f (2‬ﻫﻭ ﺍﻟﺘﺤﺎﻜﻲ ﺍﻟﺫﻱ ﻤﺭﻜﺯﻩ ‪ G‬ﻭﻨﺴﺒﺘﻪ ‪. -2‬‬ ‫ﺍﻟﺘﻤﺭﻴﻥ‪. 17‬‬ ‫‪G ( 3 ; 3 ) (1‬‬ ‫‪ H ( 3 ; 2 ) (2‬؛ ) ‪ I ( 2 ; 3‬؛ ) ‪J ( 4 ; 4‬‬ ‫‪ (3‬ﺇﺤﺩﺍﺜﻴﺘﺎ ﻤﺭﻜﺯ ﺜﻘل ﺍﻟﻤﺜﻠﺙ ‪ HIJ‬ﻫﻤﺎ ) ‪( 3 ; 3‬‬ ‫ﺍﻟﺘﻤﺭﻴﻥ‪. 18‬‬ ‫‪ (1‬ﺍﻟﻨﻘﻁﺘﺎﻥ ‪ O ( 0 ; 0 ) :‬ﻭ ) ‪ M ( 4 ; 4‬ﺘﻨﺘﻤﻴﺎﻥ ﺇﻟﻰ ) ‪( D‬‬‫‪MA′ = 3 ; MA = 3 ; OA′ = 17 ; OA = 17‬‬ ‫ﺇﺫﻥ ‪ ( D ) :‬ﻫﻭ ﻤﺤﻭﺭ ‪[ ]AA′‬‬‫‪MB′ = 37 ; MB = 37 ; OB′ = 13 ; OB = 13‬‬ ‫ﺇﺫﻥ ‪ ( D ) :‬ﻤﺤﻭﺭ ‪[ ]BB′‬‬‫‪MC′ = 5 ; MC = 5 ; OC′ = 29 ; OC = 29‬‬ ‫ﺇﺫﻥ ‪ ( D ) :‬ﻤﺤﻭﺭ ‪[ ]CC′‬‬ ‫‪.‬‬ ‫‪G‬‬ ‫‪‬‬ ‫‪23‬‬ ‫;‬ ‫‪-8 ‬‬ ‫‪(2‬‬ ‫‪‬‬ ‫‪3‬‬ ‫‪3 ‬‬ ‫= ‪. OG′‬‬ ‫‪593‬‬ ‫= ‪; OG‬‬ ‫‪593‬‬ ‫‪3‬‬ ‫‪3 (3‬‬

‫= ‪MG′‬‬ ‫‪521‬‬ ‫;‬ ‫= ‪MG‬‬ ‫‪521‬‬ ‫‪3‬‬ ‫‪3‬‬ ‫ﺇﺫﻥ ‪ ( D ) :‬ﻤﺤﻭﺭ ‪[ ]. GG′‬‬ ‫)*( ﺍﻟﺘﻤﺭﻴﻥ‪. 19‬‬ ‫‪ (1‬ﻟﻴﻜﻥ ‪ J‬ﻤﻨﺘﺼﻑ ‪[ ]AC‬‬ ‫‪ J‬ﻫﻭ ﻤﺭﺠﺢ )‪(C , 1) ; ( A , 1‬‬ ‫‪ G‬ﻫﻭ ﻤﺭ‪r‬ﺠ‪u‬ﺢ‪ u‬ﺍﻟﻜﻠﻤﺔ ‪(B , - 1) ; ( )J u,u2ur‬‬ ‫ﻭﻤﻨﻪ ‪ JG = - JB :‬ﺇﺫﻥ ‪ J‬ﻫﻭ ﻤﻨﺘﺼﻑ ‪[ ]BG‬‬ ‫)‪ ( AC‬ﻭ ) ‪ ( BG‬ﻨﻔﺱ ﺍﻟﻤﻨﺘﺼﻑ ‪J‬‬ ‫ﺇﺫﻥ ‪ ABCG‬ﻤﺘﻭﺍﺯﻱ ﺃﻀﻼﻉ‬ ‫ﻤﻼﺤﻅﺔ‪:‬‬ ‫ﺒﻤﺎ ﺃﻥ ‪ BA = BC :‬ﻓﺈﻥ ‪ ABCG‬ﻤﻌﻴﻥ‬ ‫‪ u–ur2‬ﺃ‪ u( u‬ﺒﺎﺨﺘﺼﺎ‪r‬ﺭ‪u‬ﺍﻟ‪u‬ﺠ‪u‬ﻤﻊ ﺍﻟﺸ‪r‬ﻌﺎ‪u‬ﻋ‪u‬ﻲ‪ u‬ﻨﺠﺩ‪uuuu:r‬‬ ‫‪ MA - MB + MC = MG‬ﻭﻤﻨﻪ ‪:‬‬ ‫‪uuuur uuuur uuur‬‬ ‫‪MA - MB + MC‬‬ ‫=‬ ‫‪53‬‬ ‫)‪ M ∈ (Γ‬ﺃﻱ ‪:‬‬ ‫‪2‬‬ ‫‪uuuur‬‬ ‫‪MG‬‬ ‫=‬ ‫‪53‬‬ ‫ﻭﻤﻨﻪ ‪:‬‬ ‫‪2‬‬‫‪r‬‬ ‫=‬ ‫‪53‬‬ ‫)‪ (Γ‬ﻫﻲ ﺩﺍﺌﺭﺓ ﺫﺍﺕ ﺍﻟﻤﺭﻜﺯ ‪ G‬ﻭﻨﺼﻑ ﺍﻟﻘﻁﺭ‬ ‫ﺇﺫﻥ ‪:‬‬ ‫‪2‬‬‫)‪(Γ‬‬ ‫‪C‬‬ ‫‪G‬‬ ‫‪J‬‬ ‫‪B‬‬ ‫‪A‬‬

‫ﺏ ( ‪ GJ‬ﺍﺭﺘﻔﺎﻉ ﻟﻠﻤﺜﻠﺙ ﺍﻟﻤﺘﻘﺎﻴﺱ ﺍﻷﻀﻼﻉ ‪ GAC‬ﻁﻭل ﻀﻠﻌﻪ ‪[ ]5‬‬‫)‪J ∈ (Γ‬‬ ‫ﻭﻤﻨﻪ ‪:‬‬ ‫= ‪GJ‬‬ ‫‪53‬‬ ‫ﺇﺫﻥ ‪2 :‬‬ ‫)*( ﺍﻟﺘﻤﺭﻴﻥ‪. 20‬‬ ‫‪ – 1‬ﺃ ( ﺒﻤﺎ ﺃﻥ ‪ A′‬ﻤ‪r‬ﻨ‪u‬ﺘ‪u‬ﺼ‪u‬ﻑ‪ur BC u‬ﻓﺈ‪u‬ﻥ‪ u‬ﻤﻥ ﺃﺠل‪ur‬ﻨﻘ‪u‬ﻁ‪u‬ﻪ ‪ M‬ﻤﻥ] [‬ ‫ﺍﻟﻤﺴﺘﻭﻱ ﻴﻜﻭﻥ ‪uuur uuurMB + uMur C = 2 MA′ :‬‬ ‫ﺒﺘﻌﻭﻴ‪ur‬ﺽ‪uu‬ﺍﻟﻨﻘﻁﺔ‪ uMuur‬ﺒﺎﻟﻨﻘﻁ‪ur‬ﺔ‪ Ouu‬ﻨﺠﺩ ‪OB + OC = 20 A′ :‬‬ ‫ﺏ ( ‪uuur uuur AuHuur= AuOuur+ OuHuur‬‬ ‫‪AuuHur = AuuOur + OuuAur + OBuu+r OC‬‬ ‫ﻭﻤﻨﻪ ‪AH = OB + OC = 20 A′ :‬‬ ‫ﺟـ ( ﻟﺩﻴﻨﺎ ‪ (OuurA′) :‬ﻤﺤﻭﺭ ‪ [BCuu]ur‬ﻭﻤﻨﻪ ‪(BC) ⊥ (OA′) :‬‬‫ﻤﻥ ﺍﻟﻤﺴﺎﻭﺍﺓ ‪ AH = 20 A′‬ﻨﺴﺘﻨﺘﺞ ﺃﻥ ‪( AH) // ( )OA′‬‬‫ﻭﻤﻨﻪ ‪uuur ( AH) uur⊥ (BC) :‬‬‫ﺩ ( ﻭﺒﺎﻟﻤﺜل ﻨﺒﺭﻫﻥ ﺃﻥ ‪BH = 20 B′ :‬‬ ‫ﻭﻤﻨﻪ ‪(BH) ⊥ ( AC) :‬‬‫ﺇﺫﻥ ‪ AH :‬ﻭ ‪ BH‬ﺍﺭﺘﻔﺎﻉ ﻟﻠﻤﺜﻠﺙ ‪[ ] [ ]. ABC‬‬‫ﺍﻟﻨﻘﻁﺔ ‪ H‬ﻤﺸﺘﺭﻜﺔ ﺒﻴﻨﻬﻤﺎ ﻓﻬﻲ ﺇﺫﻥ ﻨﻘﻁﺔ ﺘﻼﻗﻲ ﺍﻻﺭﺘﻔﺎﻋﺎﺕ‬‫‪ – 2‬ﺃ ( ‪uurG‬ﻫﻭ‪uu‬ﻤﺭﻜﺯ ﺜﻘل ﺍﻟﻤ‪r‬ﺜﻠ‪u‬ﺙ‪uuuur uuur ABCuu‬‬‫ﻭﻤﻨﻪ ‪MA + MB + MC = 3 MG :‬‬‫‪uuur uuur uuur‬‬ ‫ﻤﻥ ﺃﺠل ﻜل ﻨﻘﻁﺔ ‪ M‬ﻤﻥ ﺍﻟﻤ‪r‬ﺴ‪u‬ﺘﻭﻱ‬‫ﻤﻥ ﺃﺠل ‪ M = O‬ﻨﺠﺩ‪OA + OB + OCuu=ur30 Guuu:r‬‬‫ﻭﻤﻨﻪ ﻨﺴﺘﻨﺘﺞ ﺃﻥ ‪OH = 3 OG :‬‬‫ﺏ( ‪urO = G = H‬ﻓﻘﻁ ﻓﻲ ﺤﺎﻟﺔ ‪ AuBuCur‬ﻤﺘﻘﺎﻴﺱ ﺍ‪ur‬ﻷ‪u‬ﻀ‪u‬ﻼﻉ ‪uuur‬‬‫ﺟـ( ﻤﻥ ﺍﻟﻌﻼﻗﺔ ‪ OH = 30 G :‬ﻓﺈﻥ ‪ OH :‬ﻭ ‪OG‬‬ ‫ﻤﺭﺘﺒﻁﺎﻥ ﺨﻁﻴﺎ ‪.‬‬

‫ﺇﺫﻥ ‪ O , G , H :‬ﻋﻠﻰ ﺍﺴﺘﻘﺎﻤﺔ ﻭﺍﺤﺩﺓ ‪.‬‬ ‫ﺩ ( ﺇﺫﺍ ﻜﺎﻥ ﺍﻟﻤﺜﻠﺙ ﻗﺎﺌﻡ ﻓﻲ ‪ ) A‬ﻤﺜﻼ (‬‫ﻓﺈﻥ ‪uuuur uuuurO = A′uuu;r H =uuuAr :‬‬‫ﻭﻨﺠﺩ ‪ OH = 3 OG :‬ﻷﻥ ‪A′A = 3 A′G :‬‬ ‫ﻭﻤﺴﺘﻘﻴﻡ \" ﺃﻭﻟﺭ \" ﻫﻭ ‪( ). OA‬‬‫ﺍﻟﺘﻤﺭﻴﻥ‪uuur . 21‬‬‫ﺒﻤﺎ ﺃﻥ ‪ I :‬ﻴﻨﺘﻤﻲ ﺇﻟﻰ ﺍﻟﻤ‪ur‬ﺴﺘ‪u‬ﻘﻴ‪u‬ﻡ ‪ur AB‬ﺍﻟ‪u‬ﺫﻱ ﻴﻤﺭ ﻤﻥ ‪ A‬ﻭﺸﻌﺎﻉ ﺘﻭﺠﻴﻬﻪ ‪ AB‬ﻓﺈﻨﻪ ﻴﻭﺠﺩ) (‬ ‫ﻋﺩﺩ ﺤﻘﻴﻘﻲ ‪ k‬ﺘﺤ‪r‬ﻘ‪uu‬ﻕ ‪( )AuuIr = k ×uAurB‬‬ ‫‪−‬‬ ‫‪AI‬‬ ‫‪+‬‬ ‫‪k AI‬‬ ‫‪+‬‬ ‫= ‪IB‬‬ ‫‪0‬‬ ‫ﻭﻤﻨﻪ‪:‬‬ ‫‪uur‬‬ ‫‪uur‬‬ ‫‪uur‬‬ ‫‪ur‬‬ ‫ﺃﻱ‪IA + k × AuIr + k × uIuBr = uOr :‬‬ ‫‪(1 - R) IA + k × IB = O‬‬ ‫ﻟﻜﻥ ‪1 - k + k = 1 ≠ 0 :‬‬ ‫ﻭﻤﻨﻪ‪ I :‬ﻤﺭﺠﺢ ﺍﻟﺠﻤﻠﺔ ) ‪{ }( A ; 1 - k ) ; (B ; k‬‬‫‪1+2 +1 ≠ 0‬‬ ‫‪ C ; 1‬ﻤﻭﺠﻭﺩ ﻷﻥ ‪( ):‬‬ ‫ﺍﻟﺘﻤﺭﻴﻥ‪. 22‬‬ ‫ﻟﻴﻜﻥ ‪ G‬ﻤﺭﺠﺢ ﺍﻟﺠﻤﻠﺔ ‪:‬‬ ‫)‪, (B ; 2) , (A ; 1‬‬ ‫ﻤﻥ ﺃﺠل‪r‬ﻜ‪u‬ل‪ u‬ﻨ‪u‬ﻘ‪u‬ﻁﺔ ‪ M‬ﻤﻥ‪r‬ﺍﻟ‪u‬ﻤ‪u‬ﺴ‪u‬ﺘﻭﻱ ‪uuuur uuur:‬‬ ‫ﻟﺩﻴﻨﺎ ‪( ) ( )uuMurA +u2uuMur B + MC = 4 MG :‬‬‫ﺇﺫﻥ‪ M ∈ E :‬ﻴﻜﺎﻓﺊ ‪ MG‬و ‪uur AC‬ﻤ‪u‬ﺭﺘﺒﻁﺎﻥ ﺨﻁﻴﺎ) (‬ ‫ﺃﻱ ‪ M :‬ﻴﺭﺴﻡ ﺍﻟﻤﺴﺘﻘﻴﻡ ﺍﻟﺫﻱ ﻴﻤﺭ ﻤﻥ ‪ G‬ﻭﺸﻌﺎﻉ ﺘﻭﺠﻴﻬﻪ ‪AC‬‬ ‫ﻹﻨﺸﺎﺀ ‪ G‬ﻨﺭﺴﻡ ﺃﻭﻻ ﺍﻟﻤﻨﺘﺼﻑ ‪ I‬ﻟـ ‪[ ]AC‬‬ ‫ﻭﺒﺎﺴﺘﻌﻤﺎل ﺨﺎﺼﻴﺔ ﺍﻟﺘﺠﻤﻊ ﻟﻠﻤﺭﺠﺢ ‪.‬‬‫‪ G‬ﻫﻭ ﻤﺭﺠﺢ )‪ (I ; 2‬و )‪ (B ; 2‬ﺃﻱ‪ G :‬ﻤﻨﺘﺼﻑ ]‪[IB‬‬

‫‪A‬‬ ‫‪(E) I‬‬ ‫‪G‬‬ ‫‪BC‬‬ ‫ﺍﻟﺘﻤﺭﻴﻥ‪. 23‬‬ ‫ﻟﺒﺭﻫﺎﻥ ﻋﻠﻰ ﺇﺴﺘﻘﺎﻤﻴﺔ ﺍﻟﻨﻘﻁ ‪A1 , E , C‬‬‫ﻴﻜﻔﻲ ﺃﻥ ﻨﺒ‪ur‬ﺭ‪u‬ﻫ‪u‬ﻥ ﺃﻥ ‪r A1‬ﻫ‪u‬ﻭ‪uu‬ﻤﺭﺠﺢ ﺍﻟﻨﻘﻁﺘﻴﻥ ‪rC‬ﻭ ‪ E‬ﺍ‪ur‬ﻟﻤ‪u‬ﺭ‪u‬ﻓﻘﺘﻴﻥ ﺒﻤﻌﺎ‪r‬ﻤ‪u‬ﻠﻴ‪u‬ﻥ‪ u‬ﻴﻁﻠﺏ ﺘﻌﻴﻴﻨﻬﻤﺎ ‪.‬‬ ‫ﺒﻤﺎ ﺃﻥ‪ EB = - 2 EA :‬ﻴﻜﻭﻥ ﻟﺩﻴﻨﺎ ‪EB + 2 EA = 0 :‬‬ ‫ﻭﻤﻨﻪ‪ E :‬ﻫﻭ ﻤﺭﺠﻊ )‪(B ; 1) , ( A ; 2‬‬ ‫‪ A1‬ﻤﻨﺘﺼﻑ ‪ [ ]AA′‬ﻓﻬﻭ ﻤﺭﺠﺢ )‪( A′ ; 2) , ( A ; 2‬‬ ‫ﻭﻟﻜﻥ ‪ A′‬ﻫﻭ ﻤﻨﺘﺼﻑ ‪ BC‬ﻓﻬﻭ ﻤﺭﺠﺢ ‪[ ](C ; 1) , (B ; )1‬‬ ‫ﺒﺎﺴﺘﻌﻤﺎل ﺨﺎﺼﻴﺔ ﺍﻟﺘﺠﻤﻴﻊ ﻨﺴﺘﻨﺘﺞ ﺃﻥ ‪ A1‬ﻫﻭ ﻤﺭﺠﺢ ‪( ) ( )C ; 1 , E ; 3‬‬ ‫ﺇﺫﻥ ‪ C , E , A:‬ﻋﻠﻰ ﺇﺴﻘﺎﻤﺔ ﻭﺍﺤﺩﺓ ‪.‬‬ ‫ﺍﻟﺘﻤﺭﻴﻥ‪. 24‬‬ ‫ﻟﺩﻴﻨﺎ ‪ G‬ﻤﺭﺠﺢ ﺍﻟﺠﻤﻠﺔ ‪( ) ( )D ; 1 , A ; 1‬‬ ‫)‪(C ; 2) , (B ; 2‬‬ ‫‪ M‬ﻤﻨﺘﺼﻑ ‪ AD‬ﻭ ‪ N‬ﻤﻨﺘﺼﻑ ‪[ ] [ ]BC‬‬ ‫ﺇﺫﻥ ﺒﺎﺴﺘﻌﻤﺎل ﺨﺎﺼﻴﺔ ﺍﻟﺘﺠﻤﻴﻊ ﻓﺈﻥ ‪ G‬ﻫﻭ ﻜﺫﻟﻙ ﻤﺭﺠﺢ ﺍﻟﺠﻤﻠﺔ‪:‬‬ ‫)‪(N ; 4) , (M ; 2‬‬ ‫ﺇﺫﻥ ‪ N , M , G :‬ﻋﻠﻰ ﺍﺴﺘﻘﺎﻤﺔ ﻭﺍﺤﺩﺓ‬

‫ﺍﻟﺠـﺩﺍﺀ ﺍﻟﺴﻠﻤـﻲ‬ ‫ﺍﻟﻜﻔﺎﺀﺍﺕ ﺍﻟﻤﺴﺘﻬﺩﻓﺔ‬‫‪ -1‬ﺤﺴﺎﺏ ﺍﻟﺠﺩﺍﺀ ﺍﻟﺴﻠﻤﻲ ﻟﺸﻌﺎﻋﻴﻥ ﻓﻲ ﺍﻟﻤﺴﺘﻭﻱ ‪.‬‬ ‫‪ -2‬ﺇﺜﺒﺎﺕ ﻋﻼﻗﺎﺕ ﺍﻟﺘﻌﺎﻤﺩ ‪.‬‬ ‫‪ -3‬ﺤﺴﺎﺏ ﻤﺴﺎﻓﺎﺕ ‪.‬‬ ‫‪ -4‬ﺤﺴﺎﺏ ﺃﻗﻴﺎﺱ ﺯﻭﺍﻴﺎ ‪.‬‬ ‫‪ -5‬ﺘﻌﻴﻴﻥ ﻤﻌﺎﺩﻟﺔ ﻤﺴﺘﻘﻴﻡ ﻋﻠﻡ ﻨﺎﻅﻡ ﻟﻪ‪.‬‬ ‫‪ -6‬ﻜﺘﺎﺒﺔ ﻤﻌﺎﺩﻟﺔ ﻜﺭﺓ‪.‬‬ ‫‪ -7‬ﺩﺭﺍﺴﺔ ﻤﺠﻤﻭﻋﺎﺕ ﻨﻘﻁ ‪.‬‬ ‫ﺘﺼﻤﻴﻡ ﺍﻟﺩﺭﺱ‬ ‫ﺘﻌﺎﺭﻴﻑ‬ ‫ﺗﻤـﺎرﻳـﻦ و ﻡﺸﻜﻼت‬ ‫اﻟﺤﻠﻮل‬

‫ﺘﻌﺎﺭﻴﻑ‬‫=‪p‬‬ ‫‪r‬‬ ‫‪r‬‬ ‫ﺤﻴﺙ ‪:‬‬ ‫ﺍﻟ‪r‬ﺤﻘﻴﻘﻲ‪p r‬‬ ‫ﻫﻭ ﺍﻟﻌﺩﺩ‬ ‫ﺍ•ﻟﺠﺇ‪1‬ﺩﺫﺍﺍ‪-‬ﺀﻜﺍﺎﺘﻟﻌﻥﺴﺭﻠﻴﻤ‪ur‬ﻑﻲ ‪:‬ﻟ‪r‬وﺸﻌﺎ‪vr‬ﻋﻴ‪r‬ﻏﻥﻴﺭ‪r‬ﻤ‪u‬ﻌﺩ‪r,‬ﻭﻤﻴ‪vr‬ﻥ‬ ‫×‪u‬‬ ‫‪v‬‬ ‫‪(u , v‬‬ ‫‪× cos‬‬ ‫ﻓﺈ‪r‬ﻥ‪) :‬‬ ‫• ﺇﺫﺍ ﻜﺎﻥ ‪ u = 0 :‬ﺃﻭ ‪ rv =r0‬ﻓﺈﻥ ‪. p = 0 :‬‬ ‫ﻭ ﻨﺭﻤﺯ ﻟﻠﺠﺩﺍﺀ ﺍﻟﺴ‪r‬ﻠﻤﻲ ﺒﺎ‪r‬ﻟﺭﻤﺯ ‪r ru . v‬‬ ‫‪rr‬‬ ‫ﺇﺫﻥ‪u . v = u × v × cos (u , v) :‬‬ ‫• ﺇﺫﻥ ﺍﻟﺠﺩﺍﺀ ﺍﻟﺴﻠﻤﻲ ﻟﺸﻌﺎﻋﻴﻥ ﻫﻭ ﻋﺩﺩ ﺤﻘﻴﻘﻲ ﻭﻟﻴﺱ ﺸﻌﺎﻋﺎ ‪.‬‬ ‫‪r‬‬ ‫‪r‬‬ ‫‪rr‬‬‫‪(u‬‬ ‫‪,‬‬ ‫=)‪v‬‬ ‫‪π‬‬ ‫‪+‬‬ ‫‪2kπ‬‬ ‫;‬ ‫‪k∈¢‬‬ ‫• ﺇﺫﺍ ﻜﺎﻥ ‪ u ⊥ v :‬ﻓﺈﻥ ‪:‬‬ ‫‪2‬‬ ‫‪r‬‬ ‫‪r‬‬ ‫‪r‬‬ ‫‪rr‬‬ ‫ﻭﻋﻠﻴﻪ ‪ cos r(u ,rv ) = 0‬ﻭﻤ‪r‬ﻨﻪ ‪r u .r v = r0 :‬‬ ‫ﻭﺇﺫﺍ ﻜﺎﻥ ‪ u . v = 0 :‬ﻭﻜﺎﻥ ‪ u ≠ 0‬ﻭ ‪ v ≠ 0‬ﻓﺈﻥ ‪:‬‬ ‫‪r‬‬ ‫‪r‬‬ ‫‪rr‬‬‫‪(u ,‬‬ ‫= )‪v‬‬ ‫‪π‬‬ ‫‪+ 2kπ ; k ∈ ¢‬‬ ‫ﻭﻤﻨﻪ ‪:‬‬ ‫‪cos (u , v ) = 0‬‬ ‫‪2‬‬ ‫‪rr‬‬ ‫ﺇﺫﻥ‪r r . u ⊥ rv :r‬‬ ‫‪rr‬‬ ‫• )‪u . u = u × u × cos (u , u‬‬‫‪rr‬‬ ‫‪r‬‬ ‫‪2‬‬ ‫‪rr‬‬ ‫‪r‬‬ ‫‪2 × cos‬‬ ‫‪0‬‬ ‫=‬ ‫‪r‬‬ ‫‪2‬‬ ‫‪1‬‬‫=‪u.u‬‬ ‫‪u‬‬ ‫=‪u.u‬‬ ‫‪u‬‬ ‫‪u‬‬ ‫‪:‬‬ ‫ﻭﻤﻨﻪ‬ ‫×‬ ‫ﻭ ﻴﺴﻤﻰ‪r‬ﺍﻟ‪u‬ﻤﺭﺒﻊ‪ r‬ﺍﻟﺴﻠ‪r‬ﻤﻲ‪.‬‬ ‫ﻤﺜﺎل ‪ u , v , w :‬ﺜﻼﺙ ﺃﺸﻌﺔ ﺤﻴﺙ ‪:‬‬‫‪( )r r‬‬ ‫‪ur r‬‬ ‫‪r‬‬ ‫=‬ ‫‪π‬‬ ‫‪+ 2kπ‬‬ ‫;‬ ‫‪w =2 ; v = 2 ; u =4‬‬ ‫‪u,v‬‬ ‫‪4‬‬ ‫‪( )r ur‬‬ ‫=‬ ‫‪π‬‬ ‫‪+ 2kπ‬‬ ‫‪; k∈¢‬‬ ‫‪2‬‬ ‫‪r‬‬ ‫‪r ur‬‬ ‫‪u,w‬‬ ‫‪u‬‬ ‫‪2‬‬ ‫‪u.w‬‬ ‫;‬ ‫‪rr‬‬ ‫;‬ ‫‪ -‬ﺍﺤﺴﺏ ﺍﻟﺠﺩﺍﺀﺍﺕ ﺍﻟﺴﻼﻤﻴﺔ ﺍﻵﺘﻴﺔ ‪u . v :‬‬

rr r r rr π : ‫ﺤـل‬u . v = u × v cos (u , v) = 4 × 2 cos 4 : ‫* ﻟﺩﻴﻨﺎ‬ rr 2 2 =4 : ‫ﻭ ﻤﻨﻪ‬ u. v=4 × 2 r ur r ur cos r ur = 4 × 2 cos π : ‫* ﻟﺩﻴﻨﺎ‬ u.w= u× w r (u , w) 2 ur u . w = 4 × 2 × 0 = 0 : ‫ﻭ ﻤﻨﻪ‬ r r u 2 = u 2 = (4)2 = 16 : ‫* ﻟﺩﻴﻨﺎ‬ : ‫ ﻋﺒﺎﺭﺓ ﺃﺨﺭﻯ ﻟﻠﺠﺩﺍﺀ ﺍﻟﺴﻠﻤﻲ‬-2‫ ﺍﻟﻤﺴﻘﻁ ﺍﻟﻌﻤﻭﺩﻱ‬H ‫ ﻭﻜﺎﻨﺕ‬A ≠ B : ‫ ﺜﻼﺙ ﻨﻘﻁ ﻤﻥ ﺍﻟﻤﺴﺘﻭﻱ ﺤﻴﺙ‬A , B , C ‫ﺇﺫﺍ ﻜﺎﻨﺕ‬ C uuur uuur u(uAurB) ‫ﻰ‬u‫ﻋﻠ‬uuCr ‫ﻟﻠﻨﻘﻁﺔ‬ B AB . AC u=uuArB .uuAurH : ‫ﻓﺈﻥ‬ H uuur uuur: ‫ ﻤﻥ ﻨﻔﺱ ﺍﻻﺘﺠﺎﻩ ﻓﺈﻥ‬AH ‫ و‬AB ‫ﻓﺈﺫﺍ ﻜﺎﻥ‬A AB . AC = AB . AH uuur uuur uuur uuu:r‫ ﻤﺨﺘﻠﻔﻴﻥ ﻓﻲ ﺍﻻﺘﺠﺎﻩ ﻓﺈﻥ‬AH ‫ و‬AB ‫ﻭﺇﺫﺍ ﻜﺎﻥ‬ AB . AC = -AB . AH : ‫ﻤﺜﺎل‬ uuur uuur . ‫ ﻤﺴﺘﻁﻴل‬ABCD AB = 4 : ‫ ﻋﻠﻤﺎ ﺃﻥ‬AB . AC : ‫ ﺍﺤﺴﺏ ﺍﻟﺠﺩﺍﺀ ﺍﻟﺴﻠﻤﻲ‬- uuur uuur : ‫ﺤـل‬D C AB . AC = AB u.uAurB u=uur4 × 4 : ‫ﻟﺩﻴﻨﺎ‬ AB . AC = 16 : ‫ﻭ ﻤﻨﻪ‬AB u:r‫ﻭﺍﺹ‬r‫ ﺨ‬-r3 .‫ ﻋﺩﺩ ﺤﻘﻴﻘﻲ‬λ . ‫ﻲ ﺍﻟﻤﺴﺘﻭﻱ‬r‫ﺸﻌﺔ ﻓ‬r‫ ﺃ‬w r, v r, u r r u . v =rv .ru (1 (λ u ) . v = λ . (u . v ) (2

rr rr u .ur(λ v) = rλ (u . v) r r r ur (3 ( )r u . v + w = u . v + u . w (4 ( )r r 2 = ur 2 r r + vr 2 v + 2u . v u+ (5 ( )r r 2 = r rr . r 2 v u - 2u . v v u- (6 ( ) ( )r r rr = ur 2 - r 2 (7 u-v v u+v r r r r r r u + v 2 = u 2 + v 2 2.u. v + (8 r - r 2 = r 2 + r 2 − 2. r r u v u v u. v rr (9 λu = λ . u (10( )r r r r : ‫ﻤﺜﺎل‬ v= u u,v = π + 2kπ ; 2, =4 rr 4 : ‫ ﺸﻌﺎﻋﺎﻥ ﺤﻴﺙ‬v , u r r 2 . ‫ ﻋﺩﺩ ﺤﻘﻴﻘﻲ‬λ ‫ﻟﻴﻜﻥ‬ ( ) ( )r r r λu + v .v ; λu + v : ‫ ﺍﺤﺴﺏ ﻤﺎ ﻴﻠﻲ‬- ( )r r 2 rr u - 3v ; u + v ( )r r r r . r + r 2 : ‫ﺤـل‬ .v = λu v v : ‫ﻟﺩﻴﻨﺎ‬ λu + v rr rr r2 = λ . u × v cos (u , v ) + v = λ×4× 2× cos π + ( 2 )2 4 = 4λ 2 × 2 +2 2

( )r r 2 λ 2 r 2 = 4λ + 2 r. 2 u rr v λu + v = : ‫ﻟﺩﻴﻨﺎ‬ + 2λu . v + = λ2( 4 )2 + 2 (4λ) + ( 2 )2 r = 16λ2 + 8rλ + 2 . r 3v = r r r( u - )2 u 2 - 6u . v+9 v 2 : ‫ﻟﺩﻴﻨﺎ‬ = (4)2 - 6 . 4 + 9 ( 2 )2: ‫ﻟﺩﻴﻨﺎ‬ = 16 - 24 + 18 = 10 . r r r r r r v u + v 2 = u 2 + v 2 + 2u . = (4)2 + ( 2 )2 + 2 × 4 = 16 + 2 + 8 = 26 . : ‫ ﺍﻟﻌﺒﺎﺭﺍﺕ ﺍﻟﺘﺤﻠﻴﻠﻴﺔ ﻟﻠﺠﺩﺍﺀ ﺍﻟﺴﻠﻤﻲ‬-4rr r ru . v = x x′ + y y′ : ‫ﻓﺈﻥ‬ u  x , v  x′  : ‫ﺇﺫﺍ ﻜﺎﻥ‬  y  r  y′ r x x′ + y y′ = 0 : ‫ﺫﺍ ﻭﻓﻘﻁ ﺇﺫﺍ ﻜﺎﻥ‬r‫ ﺇ‬v ‫ ﻋﻤﻭﺩﻴﺎ ﻋﻠﻰ‬u ‫• ﻴﻜﻭﻥ‬ u = x2 + y2 • rr rr ru . vr : ‫• ﻟﺩﻴﻨﺎ‬ cos (u , v ) = u.v rr x x′ + y y′ : ‫ﻭﻋﻠﻴﻪ‬cos ( u , v ) = x2 + y2 . x′2 + y′2 : ‫ ﻨﻘﻁﺘﺎﻥ ﺤﻴﺙ‬M2 , M1 ‫• ﺇﺫﺍ ﻜﺎﻨﺕ‬ M2 (x2 ; y2 ) , M1 (x1 ; y1 ) M1 M2 = (x2 - x1 )2 + (y2 - y1 )2 : ‫ﻤﺜﺎل‬

rr: ‫ ( ﺤﻴﺙ‬O ; i , j ) ‫ ﺜﻼﺙ ﻨﻘﻁ ﻓﻲ ﺍﻟﻤﺴﺘﻭﻱ ﺍﻟﻤﺯﻭﺩ ﺒﻤﻌﻠﻡ ﻤﺘﻌﺎﻤﺩ ﻭﻤﺘﺠﺎﻨﺱ‬A , B , C . A (2 ; -1) , B (-3 ; 4)uuu,r Cu(u-2ur; -1) uuu.r AB .uuAurC ‫( ﺍﺤﺴﺏ‬1 AB , AC . uuur uuur ‫( ﺍﺤﺴﺏ‬2 . cos(AB , AC) ‫( ﺍﺤﺴﺏ‬3 . C‫ ﻭ‬B ‫( ﺍﺤﺴﺏ ﺍﻟﻤﺴﺎﻓﺔ ﺒﻴﻥ‬4 uuur uuur : ‫ﺤـل‬ AB AC : ‫ﻭ ﻤﻨﻪ‬  -5 ;  -4  : ‫( ﻟﺩﻴﻨﺎ‬1    uuur uuur  5   0 AB . AuCuur= (-5) (-4) + 5 (0) = 20 AB = (-5)2 + (5)2 = 50 = 5 2uuur (2AC = (-4)2 + 02 = 16 = 4 uuur uuur uuur uuur :uucuors(AuuBur , AC) ‫( ﺤﺴﺎﺏ‬3 cos (AB , AC) = uAuurB . AuuCur AB × AC : ‫ﻟﺩﻴﻨﺎ‬ uuur uuur 20 = 1 : ‫ﻭﻤﻨﻪ‬ cos(AB , AC) = 2 : ‫ﻭﻋﻠﻴﻪ‬ 5 2×4 uuur uuur cos(AB , AC) = 2 2 : C‫ ﻭ‬B ‫( ﺤﺴﺎﺏ ﺍﻟﻤﺴﺎﻓﺔ ﺒﻴﻥ‬4BC = ( xC - xB )2 + ( yC - yB )2 = (-2 + 3)2 + (-1 - 4)2 = 1 + 25 = 26 : ‫ ﻤﻌﺎﺩﻟﺔ ﻤﺴﺘﻘﻴﻡ ﻋﻠﻡ ﺸﻌﺎﻉ ﻨﺎﻅﻤﻲ ﻟﻪ ﻭﻨﻘﻁﺔ ﻤﻨﻪ‬-5

‫‪r‬‬ ‫‪α‬‬ ‫ﺃﻱ ﻴﻌﺎﻤﺩﻩ ‪( ) ( ).‬‬‫∆‬ ‫‪u‬‬ ‫∆‬ ‫ﺸﻌﺎﻉ ﻨﺎﻅﻤﻲ ﻟﻠﻤﺴﺘﻘﻴﻡ‬ ‫‪‬‬ ‫‪β‬‬ ‫‪‬‬ ‫ﻤﺴﺘﻘﻴﻡ‬ ‫ﻟﻴﻜﻥ‬ ‫‪‬‬ ‫‪‬‬‫) ‪ A (x0 ; y0‬ﻨﻘﻁﺔ ﻤﻥ ‪ . ∆ r‬ﻟﺘ‪ur‬ﻜ‪u‬ﻥ‪ M (x ; yu)u‬ﻨﻘﻁﺔ ﻜﻴﻔﻴﺔ ﻤﻥ ﺍﻟﻤﺴﺘﻭﻱ ‪ .‬ﺘﻜﻭﻥ ‪ M‬ﻨﻘﻁﺔ) (‬ ‫ﻤﻥ ∆ ﺇﺫﺍ ﻭﻓﻘﻁ ﺇﺫﺍ ﻜﺎﻥ ‪( )AM ⊥ u‬‬ ‫‪r‬‬ ‫‪α‬‬ ‫;‬ ‫‪uuuur‬‬ ‫‪‬‬ ‫‪x‬‬ ‫‪-‬‬ ‫‪x0 ‬‬ ‫‪u‬‬ ‫‪‬‬ ‫‪‬‬ ‫‪AM‬‬ ‫‪‬‬ ‫‪y‬‬ ‫‪-‬‬ ‫‪‬‬ ‫ﻭ ﻟﺩﻴﻨﺎ ‪:‬‬ ‫‪‬‬ ‫‪β‬‬ ‫‪‬‬ ‫‪‬‬ ‫‪y0‬‬ ‫‪‬‬ ‫ﻭﻤﻨﻪ ‪α (x - x0 ) + β (y - y0 ) = 0 :‬‬ ‫ﺇﺫﻥ ﻤﻌﺎﺩﻟﺔ ∆ ﻫﻲ ‪αx + βy + (-αx0 - βy0 ) = 0 ( ):‬‬ ‫ﻭﻫﻲ ﻤﻥ ﺍﻟﺸﻜل ‪α x + β x + δ = 0 :‬‬ ‫∆) (‬ ‫ﻫﻭ ﺍﻟﺸﻌﺎﻉ ﺍﻟﻨﻅﺎﻤﻲ ﻟﻠﻤﺴﺘﻘﻴﻡ‬ ‫‪r‬‬ ‫‪α‬‬ ‫‪:‬‬ ‫ﺤﻴﺙ‬ ‫‪u‬‬ ‫‪ β ‬‬ ‫‪ -6‬ﺍﻟﻤﺴﺎﻓﺔ ﺒﻴﻥ ﻨﻘﻁﺔ ﻭ ﻤﺴﺘﻘﻴﻡ ‪:‬‬ ‫∆ ﻤﺴﺘﻘﻴﻡ ﻤﻌﺎﺩﻟﺘﻪ ‪( )α x + β y + δ = 0 :‬‬ ‫ﻟﺘﻜﻥ ) ‪ M0 (x0 ; y0‬ﻨﻘﻁﺔ ﻤﻥ ﺍﻟﻤﺴﺘﻭﻱ ‪ .‬ﺘﻌﻁﻰ ﺍﻟﻤﺴﺎﻓﺔ ﺒﻴﻥ ‪M0‬‬ ‫‪αx0 + β y0 + δ‬‬ ‫ﻭ ∆ ﺒﺎﻟﻌﺒﺎﺭﺓ ‪( ):‬‬ ‫‪α2 + β2‬‬‫‪r‬‬ ‫ﺘﻁﺒﻴﻕ ‪:‬‬‫‪u‬‬ ‫‪ −1‬‬‫ﻨﺎﻅﻤﻲ) (‬ ‫‪ 4‬‬ ‫ﻭ ﺸﻌﺎﻉ ‪‬‬ ‫‪M0‬‬ ‫ﺍﻟﺫﻱ ﻴﺸﻤل ﺍﻟﻨﻘﻁﺔ )‪(2 ; 3‬‬ ‫∆‬ ‫ﻋﻴﻥ ﻤﻌﺎﺩﻟﺔ ﺍﻟﻤﺴﺘﻘﻴﻡ‬ ‫‪‬‬ ‫ﻟﻪ ‪.‬‬ ‫‪ -‬ﺍﺤﺴﺏ ﺍﻟﻤﺴﺎﻓﺔ ﺒﻴﻥ ∆ ﻭ ﺍﻟﻨﻘﻁﺔ )‪( ). H (4 ; 2‬‬ ‫ﺤل ‪:‬‬ ‫‪ -‬ﻤﻌﺎﺩﻟﺔ ∆ ﻤﻥ ﺍﻟﺸﻜل ‪ − x + 4y + δ = 0 :‬ﻭﺒﻤﺎ ﺃﻥ ‪ M0‬ﻨﻘﻁﺔ) (‬ ‫ﻤﻥ ) ∆ ( ﻓﺈﻥ‪ -2 + 4(3) + δ = 0 :‬ﻭﻤﻨﻪ ‪δ = -10 :‬‬

‫ﺇﺫﻥ ﻤﻌﺎﺩﻟﺔ ∆ ﻫﻲ ‪( )-x + 4y – 10 = 0 :‬‬ ‫‪ -‬ﺍﻟﻤﺴﺎﻓﺔ ﺒﻴﻥ ‪ H‬ﻭ ∆ ﻫﻲ ‪( ):‬‬‫= ‪-4 + 4 (2) - 10‬‬ ‫= ‪-6‬‬ ‫‪6‬‬ ‫=‬ ‫‪6 17‬‬ ‫‪( −1 )2 + (4)2‬‬ ‫‪17‬‬ ‫‪17‬‬ ‫‪17‬‬ ‫‪ -7‬ﻤﻌﺎﺩﻟﺔ ﺍﻟﺩﺍﺌﺭﺓ ‪r r :‬‬‫ﺍﻟﻤﺴﺘﻭﻱ ﻤﻨﺴﻭﺏ ﺇﻟﻰ ﻤﻌﻠﻡ ﻤﺘﻌﺎﻤﺩ ﻭﻤﺘﺠﺎﻨﺱ )‪(O ; i , j‬‬ ‫• ﻤﻌﺎﺩﻟﺔ ﺩﺍﺌﺭﺓ ُﻋﻠﻡ ﻤﺭﻜﺯﻫﺎ ﻭ ﻨﺼﻑ ﻗﻁﺭﻫﺎ ‪:‬‬ ‫)‪ (C‬ﺩﺍﺌﺭﺓ ﻤﺭﻜﺯﻫﺎ ) ‪ ω(x0 ; y0‬ﻭ ﻨﺼﻑ ﻗﻁﺭﻫﺎ ‪ α‬ﺤﻴﺙ ‪. α > 0‬‬‫)‪ M(x ; y‬ﻨﻘﻁﺔ ﻤﻥ ﺍﻟﻤﺴﺘﻭﻯ ؛ ﺘﻜﻭﻥ ‪ M‬ﻤﻥ ﺍﻟﺩﺍﺌﺭﺓ )‪ (C‬ﺇﺫﺍ ﻭﻓﻘﻁ ﺇﺫﺍ ﻜﺎﻥ ‪:‬‬ ‫‪ ωM = α‬ﺃﻱ ‪ωM2 = α2 :‬‬‫ﺇﺫﻥ ‪ ( x - x0 )2 + (y - y0 )2 = α2 :‬ﻭﻫﻲ ﻤﻌﺎﺩﻟﺔ ﺍﻟﺩﺍﺌﺭﺓ )‪. (C‬‬ ‫ﻭﺒﻨﺸﺭ ﻫﺫﻩ ﺍﻟﻌﺒﺎﺭﺓ ﻨﺠﺩ ‪:‬‬‫‪x2 + y2 - 2x0 x - 2y0 y + x02 - α2 = 0‬‬ ‫ﻭﻫﻲ ﻤﻥ ﺍﻟﺸﻜل ‪:‬‬ ‫‪x2 + y2 + α x + β y + δ = 0‬‬ ‫ﻤﺜﺎل ‪:‬‬‫ﺍﻜﺘﺏ ﻤﻌﺎﺩﻟﺔ ﺍﻟﺩﺍﺌﺭﺓ ﺫﺍﺕ ﺍﻟﻤﺭﻜﺯ )‪ ω(1 ; -3‬ﻭﻨﺼﻑ ﺍﻟﻘﻁﺭ ‪r = 5‬‬ ‫ﺤـل ‪:‬‬‫ﻤﻌﺎﺩﻟﺔ ﺍﻟﺩﺍﺌﺭﺓ ﻫﻲ ‪( x - 1)2 + (y + 3)2 = 25 :‬‬ ‫ﺃﻱ ‪x2 + y2 - 2x + 6y - 15 = 0 :‬‬ ‫• ﻤﻌﺎﺩﻟﺔ ﺩﺍﺌﺭﺓ ُﻋﻠﻡ ﻗﻁﺭﻫﺎ ‪:‬‬‫)‪ (C‬ﺩﺍﺌﺭﺓ ﻗﻁﺭﻫﺎ ‪ AB‬ﺤﻴﺙ ) ‪[ ]B (x1 ; y1 ) , A (x0 ; y0‬‬‫ﻟﺘﻜﻥ )‪ Mu(uxu,ry‬ﻨﻘ‪ur‬ﻁﺔ‪uu‬ﻤﻥ ﺍﻟﻤﺴﺘﻭﻱ ‪ .‬ﺘﻜﻭﻥ ﺍﻟﻨﻘﻁﺔ ‪ M‬ﻤﻥ )‪ (C‬ﺇﺫﺍ ﻭﻓﻘﻁ ﺇﺫﺍ ﻜﺎﻥ ‪:‬‬ ‫‪MA . MB = 0‬‬‫‪uuur‬‬ ‫‪ x1‬‬ ‫‪- x‬‬ ‫‪,‬‬ ‫‪uuur‬‬ ‫‪‬‬ ‫‪x0‬‬ ‫‪- x‬‬‫‪MB‬‬ ‫‪‬‬ ‫‪‬‬ ‫‪MA‬‬ ‫‪‬‬ ‫‪y0‬‬ ‫‪‬‬ ‫ﻟﻜﻥ‬ ‫‪‬‬ ‫‪y1‬‬ ‫‪-‬‬ ‫‪y‬‬ ‫‪‬‬ ‫‪‬‬ ‫‪-‬‬ ‫‪y‬‬ ‫‪‬‬

‫ﻭﻋﻠﻴﻪ ‪( x0 - x) (x1 - x) + (y0 - y) + (y1 - y) = 0 :‬‬‫ﺇﺫﻥ‪x0 x1 - x0 x - x1 x + x2 + y0 y1 - y0 y + y1 y + y2 = 0 :‬‬ ‫ﻭﻤﻨﻪ ﻤﻌﺎﺩﻟﺔ )‪ (C‬ﻫﻲ ‪:‬‬‫‪x2 + y2 - (x0 + x1 ) x - (y0 + y1 ) y + x0 x1 + y0 y1 = 0‬‬ ‫ﻤﺜﺎل ‪:‬‬ ‫ﺍﻜﺘﺏ ﻤﻌﺎﺩﻟﺔ ﺍﻟﺩﺍﺌﺭﺓ )‪ (C‬ﺫﺍﺕ ﺍﻟﻘﻁﺭ ‪ AB‬ﺤﻴﺙ ‪[ ]:‬‬ ‫)‪A (-1 ; 4) ; B (4 ; 2‬‬ ‫‪uuuur uuur‬‬ ‫ﺤـل ‪:‬‬ ‫‪MA . MB = 0‬‬ ‫)‪ M (x ; y‬ﻨﻘﻁﺔ ﻤﻥ ﺍﻟﺩﺍﺌﺭﺓ )‪ (C‬ﺃﻱ ‪:‬‬ ‫‪uuur‬‬ ‫‪uuuur‬‬ ‫‪MB‬‬ ‫‪4‬‬ ‫‪-‬‬ ‫‪x‬‬ ‫;‬ ‫‪MA‬‬ ‫‪ -1 - x ‬‬ ‫ﻟﺩﻴﻨﺎ ‪:‬‬ ‫‪‬‬ ‫‪-‬‬ ‫‪‬‬ ‫‪‬‬ ‫‪‬‬ ‫‪‬‬ ‫‪2‬‬ ‫‪y‬‬ ‫‪‬‬ ‫‪‬‬ ‫‪4‬‬ ‫‪-‬‬ ‫‪y‬‬ ‫‪‬‬ ‫ﻭﻤﻨﻪ ‪(4 – x) (-1 – x) + (2 – y) + (4 – y) = 0 :‬‬ ‫ﺃﻱ ‪- 4 - 4x + x + x2 + 8 - 2y + 4y + 4y + y2 = 0 :‬‬ ‫ﺇﺫﻥ ﻤﻌﺎﺩﻟﺔ )‪ (C‬ﻫﻲ ‪x2 + y2 - 3x + 2y + 4 = 0 :‬‬ ‫• ﻤﺠﻤﻭﻋﺔ ﺍﻟﻨﻘﻁ )‪ M (x ; y‬ﺤﻴﺙ ‪:‬‬‫‪x2 + y2 + α x + βy + δ = 0‬‬‫‪‬‬ ‫‪x‬‬ ‫‪+‬‬ ‫‪α 2‬‬ ‫‪-‬‬ ‫‪α2‬‬ ‫‪+‬‬ ‫‪‬‬ ‫‪x‬‬ ‫‪+‬‬ ‫‪β 2‬‬ ‫‪-‬‬ ‫‪β2‬‬ ‫‪+δ =0‬‬ ‫ﻭﻋﻠﻴﻪ ‪:‬‬‫‪‬‬ ‫‪2 ‬‬ ‫‪4‬‬ ‫‪‬‬ ‫‪2 ‬‬ ‫‪4‬‬‫‪‬‬ ‫‪x‬‬ ‫‪+‬‬ ‫‪α‬‬ ‫‪2‬‬ ‫‪+‬‬ ‫‪‬‬ ‫‪x‬‬ ‫‪+‬‬ ‫‪β 2‬‬ ‫=‬ ‫‪α2‬‬ ‫‪+ β2‬‬ ‫‪- 4δ‬‬ ‫ﺇﺫﻥ ‪:‬‬‫‪‬‬ ‫‪2‬‬ ‫‪‬‬ ‫‪‬‬ ‫‪2 ‬‬ ‫‪4‬‬ ‫ﺒﻭﻀﻊ ‪ α2 + β2 - 4δ = λ‬ﻨﺠﺩ ‪:‬‬ ‫• ﻟﻤﺎ ‪ λ < 0 :‬ﻤﺠﻤﻭﻋﺔ ﺍﻟﻨﻘﻁ ﺨﺎﻟﻴﺔ ‪.‬‬ ‫‪.‬‬ ‫‪ω‬‬ ‫‪‬‬ ‫‪-α‬‬ ‫;‬ ‫‪-β ‬‬ ‫ﻟﻤﺎ ‪ λ = 0 :‬ﻤﺠﻤﻭﻋﺔ ﺍﻟﻨﻘﻁ ﻫﻲ ﺍﻟﻨﻘﻁﺔ‬ ‫•‬ ‫‪‬‬ ‫‪2‬‬ ‫‪2 ‬‬ ‫• ﻟﻤﺎ ‪ : λ > 0 :‬ﻤﺠﻤﻭﻋﺔ ﺍﻟﻨﻘﻁ ﻫﻲ ﺍﻟﺩﺍﺌﺭﺓ ﺫﺍﺕ ﺍﻟﻤﺭﻜﺯ‬

‫‪.‬‬ ‫‪λ‬‬ ‫‪:‬‬ ‫ﺍﻟﻘﻁﺭ‬ ‫ﻭﻨﺼﻑ‬ ‫‪ω‬‬ ‫‪‬‬ ‫‪-α‬‬ ‫;‬ ‫‪-β ‬‬ ‫‪2‬‬ ‫‪‬‬ ‫‪2‬‬ ‫‪2 ‬‬ ‫ﺃﻤﺜﻠﺔ ‪:‬‬ ‫ﻋﻴﻥ ﻤﺠﻤﻭﻋﺔ ﺍﻟﻨﻘﻁ )‪ M (x ; y‬ﻓﻲ ﻜل ﺤﺎﻟﺔ ﻤﻤﺎ ﻴﻠﻲ ‪:‬‬ ‫‪x2 + y2 - 4x + 3y + 4 = 0 (1‬‬ ‫‪2x 2‬‬ ‫‪+ 2y2 +‬‬ ‫‪2x - 4y +‬‬ ‫‪5‬‬ ‫‪=0‬‬ ‫‪(2‬‬ ‫‪2‬‬ ‫‪x2 + y2 + 8x + 4y + 40 = 0 (3‬‬ ‫ﺤـل ‪:‬‬ ‫‪ (1‬ﻟﺩﻴﻨﺎ ‪x2 + y2 - 4x + 3y + 4 = 0 :‬‬ ‫ﻭ ﻤﻨﻪ ‪x2 - 4x + y2 + 3y + 4 = 0 :‬‬‫‪( x - 2)2 −‬‬ ‫‪(2)2 +‬‬ ‫‪‬‬ ‫‪y‬‬ ‫‪+‬‬ ‫‪3‬‬ ‫‪2‬‬ ‫‪−‬‬ ‫‪‬‬ ‫‪3‬‬ ‫‪2‬‬ ‫‪+‬‬ ‫‪4=0‬‬ ‫ﺃﻱ ‪:‬‬ ‫‪‬‬ ‫‪2‬‬ ‫‪‬‬ ‫‪‬‬ ‫‪2‬‬ ‫‪‬‬ ‫‪( x - 2)2 +‬‬ ‫‪‬‬ ‫‪y‬‬ ‫‪+‬‬ ‫‪3‬‬ ‫‪2‬‬ ‫=‬ ‫‪4+‬‬ ‫‪9‬‬ ‫‪-4‬‬ ‫ﺃﻱ ‪:‬‬ ‫‪‬‬ ‫‪2‬‬ ‫‪‬‬ ‫‪4‬‬ ‫‪( x - 2)2 +‬‬ ‫‪‬‬ ‫‪y+‬‬ ‫‪3 2‬‬ ‫=‬ ‫‪9‬‬ ‫ﺇﺫﻥ ‪:‬‬ ‫‪‬‬ ‫‪2 ‬‬ ‫‪4‬‬‫‪ ω 2‬ﻭ ﻨﺼﻑ‬ ‫;‬ ‫‪-3 ‬‬ ‫ﻭﻋﻠﻴﻪ ﻤﺠﻤﻭﻋﺔ ﺍﻟﻨﻘﻁ ‪ M‬ﻫﻲ ﺩﺍﺌﺭﺓ ﻤﺭﻜﺯﻫﺎ‬ ‫‪2 ‬‬ ‫‪.‬‬ ‫‪R‬‬ ‫=‬ ‫‪3‬‬ ‫ﻗﻁﺭﻫﺎ‬ ‫‪2‬‬ ‫‪2x2 + 2 y2 + 2x - 4y + 5 = 0‬‬ ‫‪2‬‬ ‫‪ (2‬ﻟﺩﻴﻨﺎ ‪:‬‬ ‫‪2‬‬ ‫‪‬‬ ‫‪x‬‬ ‫‪2‬‬ ‫‪+‬‬ ‫‪y2 +‬‬ ‫‪x - 2y +‬‬ ‫‪5‬‬ ‫‪=0‬‬ ‫ﻭ ﻤﻨﻪ ‪:‬‬ ‫‪‬‬ ‫‪4 ‬‬

‫‪x2 +‬‬ ‫‪y2‬‬ ‫‪+ x - 2y +‬‬ ‫‪5‬‬ ‫‪=0‬‬ ‫ﺇﺫﻥ ‪:‬‬ ‫‪4‬‬ ‫ﻭ ﻤﻨﻪ ‪:‬‬ ‫‪5‬‬ ‫‪x2 +‬‬ ‫‪x + y2‬‬ ‫‪- 2y +‬‬ ‫‪4‬‬ ‫‪=0‬‬‫‪‬‬ ‫‪x‬‬ ‫‪+‬‬ ‫‪1 2‬‬ ‫‪-‬‬ ‫‪‬‬ ‫‪1 2‬‬ ‫‪+‬‬ ‫‪( y - 1)2 - 1 +‬‬ ‫‪5‬‬ ‫‪=0‬‬ ‫ﻭ ﺒﺎﻟﺘﺎﻟﻲ ‪:‬‬‫‪‬‬ ‫‪2 ‬‬ ‫‪‬‬ ‫‪2 ‬‬ ‫‪4‬‬ ‫‪‬‬ ‫‪x‬‬ ‫‪+‬‬ ‫‪1‬‬ ‫‪2‬‬ ‫‪+‬‬ ‫‪( y - 1)2‬‬ ‫‪-‬‬ ‫‪1‬‬ ‫‪-1‬‬ ‫‪+‬‬ ‫‪5‬‬ ‫‪=0‬‬ ‫ﺃﻱ ‪:‬‬ ‫‪‬‬ ‫‪2‬‬ ‫‪‬‬ ‫‪4‬‬ ‫‪4‬‬ ‫‪‬‬ ‫‪x‬‬ ‫‪+‬‬ ‫‪1‬‬ ‫‪2‬‬ ‫‪+‬‬ ‫‪( y - 1)2‬‬ ‫‪=0‬‬ ‫ﺃﻱ ‪:‬‬ ‫‪‬‬ ‫‪2‬‬ ‫‪‬‬ ‫‪.‬‬ ‫‪ω ‬‬ ‫‪-1‬‬ ‫;‬ ‫‪1‬‬ ‫ﻫﻲ ﺍﻟﻨﻘﻁﺔ‬ ‫ﺇﺫﻥ ﻤﺠﻤﻭﻋﺔ ﺍﻟﻨﻘﻁ ‪M‬‬ ‫‪2‬‬ ‫‪ (3‬ﻟﺩﻴﻨﺎ ‪x2 + y2 + 8x + 4y + 40 = 0 :‬‬ ‫‪x2 + 8x + y2 + 4y + 40 = 0‬‬ ‫ﻭ ﻤﻨﻪ ‪:‬‬‫‪( x + 4)2 − (4)2 + (y + 2)2 − (2)2 + 40 = 0‬‬‫‪( x + 4)2 − 16 + (y + 2)2 − 4 + 40 = 0‬‬‫‪( x + 4)2 + (y + 2)2 = -20‬‬ ‫ﺇﺫﻥ ﻤﺠﻤﻭﻋﺔ ﺍﻟﻨﻘﻁ ﺨﺎﻟﻴﺔ ‪.‬‬ ‫‪ -8‬ﺍﻟﻌﻼﻗﺎﺕ ﺍﻟﻤﺘﺭﻴﺔ ﻓﻲ ﺍﻟﻤﺜﻠﺙ ‪:‬‬ ‫‪ ABC‬ﻤﺜﻠﺙ ﻜﻴﻔﻲ ‪ I .‬ﻤﻨﺘﺼﻑ ‪ BC‬ﻟﺩﻴﻨﺎ ﺍﻟﻌﻼﻗﺎﺕ ﺍﻟﺘﺎﻟﻴﺔ ‪[ ]:‬‬ ‫‪BC2 = AB2 + AC2 - 2AB . AC .cos Aµ (1‬‬ ‫ﻭﻜﺫﻟﻙ ‪AB2 = CA2 + CB2 - 2CA . CB .cos Cµ :‬‬ ‫)ﻭﻫﻲ ﻨﻅﺭﻴﺔ ﺍﻟﻜﺎﺘﻲ(‬ ‫‪AB2 + AC2 = 2AI2 + 2IB2 (2‬‬

BC = AC = AB = 2R (3 sin Aµ sinBµ sinCµ . ABC ‫ﺎﻟﻤﺜﻠﺙ‬u‫ﺒ‬u‫ﺔ‬u‫ﻁ‬r‫ﺍﻟﻤﺤﻴ‬u‫ﺓ‬u‫ﺭ‬u‫ﺌ‬r‫ ﻨﺼﻑ ﻗﻁﺭ ﺍﻟﺩﺍ‬R ‫ﺤﻴﺙ‬ AB . AC = AI2 - IB2 (4 : ‫ ﺤﻴﺙ‬S ‫ ﻫﻲ‬ABC ‫( ﻤﺴﺎﺤﺔ ﺍﻟﻤﺜﻠﺙ‬5S = 1 BA . BC . sin Bµ ‫ﺃﻭ‬ S = 1 AB . AC . sin Aµ 2 2 1 S = 2 CA . CB . sin Cµ ‫ﺃﻭ‬ BC = AC = AB = AB . AC . BC sin Aµ sin Bµ sin Cµ 2S : ‫ﻤﺜﺎل‬ [ ]: ‫ ﺤﻴﺙ‬BC ‫ ﻤﻨﺘﺼﻑ‬I . ‫ ﻤﺜﻠﺙ‬ABC (cm ‫ )ﻭﺤﺩﺓ ﺍﻟﻁﻭل ﻫﻲ‬BC = 6 ; AC = 4 ; AB = 3 . cos Aµ ‫( ﺍﺤﺴﺏ‬1 . AI ‫( ﺍﺤﺴﺏ‬2 ABC ‫ﺩﺍﺌﺭﺓ ﺍﻟﻤﺤﻴﻁﺔ ﺒﺎﻟﻤﺜﻠﺙ‬u‫ﺍﻟ‬u‫ﺭ‬ur‫ﻑ ﻗﻁ‬u‫ﺼ‬u‫ﻨ‬urR ‫( ﺍﺤﺴﺏ‬3 AB . AC ‫( ﺍﺤﺴﺏ‬4 ABC ‫( ﺍﺤﺴﺏ ﻤﺴﺎﺤﺔ ﺍﻟﻤﺜﻠﺙ‬5 : ‫ﺤـل‬ : cos Aµ ‫( ﺤﺴﺎﺏ‬1 BC 2 = AB2 + AC2 − 2AB . AC .cos Aµ : ‫ﻟﺩﻴﻨﺎ‬ ( 6 )2 = (3)2 + (4)2 − 2 . 3 . 4 cos Aµ : ‫ﻭﻤﻨﻪ‬ 11 = -24 cos Aµ : ‫ ﺃﻱ‬36 = 25 - 24 cos Aµ : ‫ﻭﻤﻨﻪ‬ . cos Aµ = -11 : ‫ﻭ ﺒﺎﻟﺘﺎﻟﻲ‬ 24

: AI ‫( ﺤﺴﺎﺏ‬2: ‫ ﻭ ﻤﻨﻪ‬AB2 + AC2 = 2AI2 + + 2(3)2 : ‫ﻟﺩﻴﻨﺎ‬25 = 2AI2 + 18 : ‫ ( ﺃﻱ‬3 )2 + (4)2 = 2AI2 + 2 (3)2 AI 2 = 7 : ‫ﻭﻋﻠﻴﻪ‬ 2 AI 2 = 7 : ‫ﻭﻤﻨﻪ‬ 2 . AI = 14 : ‫ﺇﺫﻥ‬ AI = 7 2 2 : ‫ﻭﻤﻨﻪ‬ : R ‫( ﺤﺴﺎﺏ‬3 6 Aµ = 2R : ‫ﻭﻤﻨﻪ‬ BC = 2R : ‫ﻟﺩﻴﻨﺎ‬ sin sin Aµcos2 Aµ + sin2 Aµ = 1 ‫ﻭ‬ cos Aµ = -11 : ‫ﻟﻜﻥ‬ 24 : ‫ ﻭ ﺒﺎﻟﺘﺎﻟﻲ‬sin2 Aµ = 1- cos2 Aµ : ‫ﻭﻋﻠﻴﻪ‬ sin2 Aµ = 455 ‫ﺃﻱ‬ sin2 Aµ = 1 -  -11 2 (24)2  24 6× 24 = 2R : ‫ﻭﻤﻨﻪ‬ sin Aµ = 455 455 24 : ‫ﺇﺫﻥ‬. R = 72 455 : ‫ﺇﺫﻥ‬ R = 72 : ‫ﻭﻋﻠﻴﻪ‬ 455 uuur 45u5uur uuur uuur : AB . AC ‫( ﺤﺴﺎﺏ‬4 AB . AC = AI2 − IB2 : ‫ﻟﺩﻴﻨﺎ‬uuur uuur =  14 2 −  3 2AB . AC    2  : ‫ﻭﻤﻨﻪ‬ 2   = 14 - 9 = 5 4 4 4

uuur . uuur = 5 : ‫ﺇﺫﻥ‬ AB AC 4 : ABC ‫( ﻤﺴﺎﺤﺔ ﺍﻟﻤﺜﻠﺙ‬5S = 1 AB . AC . sin Aµ 2S= 1 × 3× 4× 455 2 24 . S= 455 cm2 : ‫ﻭ ﻤﻨﻪ‬ 4

‫ﺗﻤـﺎرﻳـﻦ و ﻡﺸﻜﻼت‬ ‫‪rr‬‬‫ﻓﻲ ﻜل ﻤﻤﺎ ﻴﻠﻲ ﺍﻟﻤﺴﺘﻭﻱ ﻤﻨﺴﻭﺏ ﺇﻟﻰ ﻤﻌﻠﻡ ﻤﺘﻌﺎﻤﺩ ﻤﺘﺠﺎﻨﺱ ) ‪(O ; i , j‬‬ ‫ﺍﻟﺘﻤﺭﻴﻥ‪. 1‬‬‫‪ur‬‬ ‫‪r‬‬ ‫‪ x‬ﻋﺩﺩ ﺤﻘﻴﻘﻲ‬‫‪v‬‬ ‫‪u‬‬ ‫‪rr‬‬ ‫‪‬‬ ‫‪−1‬‬ ‫‪‬‬ ‫;‬ ‫‪ x + 3‬‬ ‫‪ u , v‬ﺸﻌﺎﻋﺎﻥ ﺤﻴﺙ ‪:‬‬ ‫‪‬‬ ‫‪x+‬‬ ‫‪1‬‬ ‫‪‬‬ ‫‪‬‬ ‫‪‬‬ ‫‪‬‬ ‫‪x‬‬ ‫‪+‬‬ ‫‪1‬‬ ‫‪‬‬ ‫‪rr‬‬ ‫‪ (1‬ﺃﺤﺴﺏ ‪r r u . v‬‬ ‫‪ (2‬ﻋﻴﻥ ‪ x‬ﺒﺤﻴ‪r‬ﺙ ﻴﻜﻭﻥ‪ vr‬و ‪ u‬ﻤﺘﻌﺎﻤﺩﻴﻥ‬ ‫‪ (3‬ﺍﺤﺴﺏ ‪. v ; u‬‬ ‫‪rr‬‬ ‫‪ (4‬ﻋﻴﻥ ‪ x‬ﺒﺤﻴﺙ ‪u = v‬‬ ‫‪( )r r‬‬ ‫‪ (5‬ﻨﻔﺭﺽ ‪ x = 3‬ﻋﻴﻥ ﻗﻴﻤﺔ ﻤﻘﺭﺒﺔ ﻟـ ‪u , v‬‬ ‫ﺍﻟﺘﻤﺭﻴﻥ‪r r . 2‬‬ ‫ﻨﻌﺘﺒﺭ ﺍﻟﺸﻌﺎﻋﺎﻥ ‪ u , v‬ﺤﻴﺙ ‪:‬‬ ‫‪r‬‬ ‫‪r‬‬ ‫‪r‬‬ ‫‪r‬‬ ‫‪r‬‬ ‫‪r‬‬ ‫‪v‬‬ ‫=‬ ‫‪3i +‬‬ ‫‪6‬‬ ‫‪j‬‬ ‫;‬ ‫‪u‬‬ ‫=‬ ‫‪1‬‬ ‫‪i-‬‬ ‫‪5‬‬ ‫‪j‬‬ ‫‪5 r 2r 4‬‬ ‫‪v‬‬ ‫;‬ ‫‪u‬‬ ‫‪.‬‬ ‫‪r‬‬ ‫‪r‬‬ ‫‪ (1‬ﺍﺤﺴﺏ‬ ‫‪ (2‬ﺍﺤﺴﺏ ‪ u , v‬ﻤﺎﺫﺍ ﺘﺴﺘﻨﺘﺞ ؟‬ ‫ﺍﻟﺘﻤﺭﻴﻥ‪. 3‬‬ ‫ﻨﻌﺘﺒﺭ ﺍﻟﻨﻘﻁ ‪:‬‬‫) ‪A(2 ; 2 ) , B( 5 ; -2 ) , C(7 ; -1 ) , D(4 ; 3‬‬ ‫‪ (1‬ﺍﺤﺴﺏ ﻜل ﻤﻥ‪. AD , BC , AC , AB :‬‬

‫‪ (2‬ﻤﺎ ﻫﻲ ﻁﺒﻴﻌﺔ ﺍﻟﺭﺒﺎﻋﻲ ‪ ABCD‬؟‬ ‫ﺍﻟﺘﻤﺭﻴﻥ‪. 4‬‬ ‫ﻋﻴﻥ ﺍﻟﻌﺩﺩ ﺍﻟﺤﻘﻴﻘﻲ ‪ λ‬ﺒﺤﻴﺙ ﻴﻜﻭﻥ ﺍﻟﺸﻌﺎﻋﺎﻥ‬ ‫‪r‬‬ ‫‪1‬‬ ‫‪,‬‬ ‫‪r‬‬ ‫‪ 3‬‬ ‫ﻴﻌﻁﻰ ﺍﻟﺸﻌﺎﻋﺎﻥ ‪:‬‬ ‫‪u‬‬ ‫‪ 2‬‬ ‫‪v‬‬ ‫‪ 4 ‬‬ ‫‪r‬‬ ‫‪rr‬‬ ‫‪ λu - v :‬و ‪ λu + v‬ﻤﺘﻌﺎﻤﺩﻴﻥ ‪( ) ( ).‬‬‫‪r‬‬ ‫ﺍﻟﺘﻤﺭﻴﻥ‪. 5‬‬ ‫‪ β , α‬ﻋﺩﺩﺍﻥ ﺤﻘﻴﻘﻴﺎﻥ ‪ .‬ﻋﻴﻥ‬ ‫‪r‬‬ ‫‪ -1 ‬‬ ‫‪,‬‬ ‫‪r‬‬ ‫‪α‬‬ ‫‪u‬‬ ‫‪‬‬ ‫‪‬‬ ‫‪v‬‬ ‫ﺤﻴﺙ‬ ‫‪‬‬ ‫‪2‬‬ ‫‪‬‬ ‫‪‬‬ ‫‪β‬‬ ‫‪‬‬ ‫ﻴﻌﻁﻰ ﺍﻟﺸﻌﺎﻋﺎﻥ ‪:‬‬ ‫‪‬‬ ‫‪‬‬ ‫‪ β , α‬ﻓﻲ‪r‬ﻜل ﺤﺎﻟﺔ ﻤﻤﺎ ﻴﻠ‪r‬ﻲ ‪r :‬‬ ‫‪ v = 1 (1‬و ‪. u // v‬‬ ‫‪rr r‬‬ ‫‪ v = 2 (2‬ﻭ ‪. u ⊥ v‬‬ ‫‪rr r‬‬ ‫‪ v = 12 (3‬و ‪u . v = 3‬‬ ‫ﺍﻟﺘﻤﺭﻴﻥ‪r . r6‬‬ ‫ﺍﺤﺴﺏ )‪ (v , u‬ﻓﻲ ﻜل ﺤﺎﻟﺔ ﻤﻤﺎ ﻴﻠﻲ ‪:‬‬‫‪r‬‬ ‫‪‬‬ ‫‪-3‬‬ ‫‪‬‬ ‫‪(4‬‬ ‫;‬ ‫‪r‬‬ ‫‪2‬‬ ‫‪‬‬ ‫‪(3‬‬ ‫;‬ ‫‪r‬‬ ‫‪ -5‬‬ ‫‪(2‬‬ ‫;‬ ‫‪r‬‬ ‫‪‬‬ ‫‪0‬‬ ‫‪‬‬ ‫‪(1‬‬‫‪u‬‬ ‫‪‬‬ ‫‪4‬‬ ‫‪‬‬ ‫‪u‬‬ ‫‪‬‬ ‫‪‬‬ ‫‪u‬‬ ‫‪ 0 ‬‬ ‫‪u‬‬ ‫‪‬‬ ‫‪3‬‬ ‫‪‬‬ ‫‪‬‬ ‫‪‬‬ ‫‪‬‬ ‫‪5‬‬ ‫‪‬‬ ‫‪-2‬‬ ‫‪3‬‬ ‫‪4‬‬ ‫ﺍﻟﺘﻤﺭﻴﻥ‪. 7‬‬ ‫‪uuuur uuur‬‬ ‫ﻨﻌﺘﺒﺭ ﺍﻟﻨﻘﻁﺘﺎﻥ ‪C (-1 ; 2) ; B (3 ; -4) :‬‬‫ﻋﻴﻥ ﻤﺠﻤﻭﻋﺔ ﺍﻟﻨﻘﻁ ‪ M‬ﻤﻥ ﺍﻟﻤﺴﺘﻭﻱ ﺒﺤﻴﺙ ‪ BM . BC = 0 :‬ﺜﻡ ﺃﻨﺸﺌﻬﺎ ﻭ ﺃﻨﺸﺊ ﺍﻟﻤﺴﺘﻘﻴﻡ‬ ‫)‪ (BC‬ﻤﺎﺫﺍ ﺘﻼﺤﻅ ؟‬ ‫ﺍﻟﺘﻤﺭﻴﻥ‪. 8‬‬ ‫ﻨﻌﺘﺒﺭ ﺍﻟﻨﻘﻁﺘﺎﻥ )‪B (-2 ; -3) , A (1 ; -4‬‬ ‫ﻋﻴﻥ ﻤﺠﻤﻭﻋﺔ ﺍﻟﻨﻘﻁ ‪ M‬ﻤﻥ ﺍﻟﻤﺴﺘﻭﻱ ﺤﻴﺙ ‪:‬‬

‫‪uuuur uuur‬‬ ‫‪ AM . AB = -4‬ﺜﻡ ﺃﻨﺸﺌﻬﺎ‬ ‫ﺍﻟﺘﻤﺭﻴﻥ‪. 9‬‬ ‫‪ ABC‬ﻤﺜ‪ur‬ﻠ‪u‬ﺙ‪uuur u‬‬ ‫ﺍﺤﺴﺏ ‪ AC . BC‬ﺇﺫﺍ ﻋﻠﻤﺕ ﺃﻥ ‪:‬‬‫‪AB = 4 ; AC = 5 ; BC = 10‬‬ ‫‪r‬‬ ‫ﺍ‪r‬ﻟﺘﻤﺭﻴ‪r‬ﻥ‪. 10‬‬ ‫; ‪v =5‬‬ ‫‪ rv , u‬ﺸﻌﺎ‪r‬ﻋﺎﻥ ﺤﻴﺙ ‪r :‬‬ ‫‪( ) ( )r r r r‬‬ ‫‪u =1 ; v.u=5‬‬ ‫‪rr‬‬ ‫ﺍﺤﺴﺏ ﻤﺎ ﻴﻠﻲ ‪u + v u - v ; 3u.(-v) :‬‬ ‫‪( ) ( )r r r r r r‬‬ ‫‪u + v ; u + 2v u - 3v‬‬ ‫ﺍ‪r‬ﻟﺘﻤﺭﻴ‪r‬ﻥ‪. 11‬‬ ‫‪rv , u‬ﺸﻌﺎﻋﺎﻥ ﺤﻴﺙ ‪r :‬‬ ‫‪rr‬‬ ‫‪α , u . v = 12‬‬ ‫و ‪v =5‬‬ ‫‪u =2‬‬‫ﻋﺩﺩ ﺤﻘﻴﻘﻲ‬ ‫‪r rr‬‬ ‫ﻭ‪r‬‬‫‪ -1‬ﻋﻴﻥ ‪ α‬ﺒﺤﻴﺙ ﻴﻜﻭﻥ ﺍﻟﺸﻌﺎﻋﺎﻥ ‪ u - rv‬ﻭ‪ u + αvr‬ﻤﺘﻌﺎﻤﺩﺍﻥ‬ ‫‪ – 2‬ﻋﻴﻥ ‪ α‬ﺒﺤﻴﺙ ﻴﻜﻭﻥ ‪u + αv = 5 :‬‬ ‫‪rr‬‬ ‫ﺍ‪r‬ﻟﺘﻤﺭﻴ‪r‬ﻥ‪. 12‬‬ ‫‪u‬‬ ‫=‬ ‫‪v‬‬ ‫‪ v , u‬ﺸ‪r‬ﻌﺎﻋﺎﻥ ﻤ‪r‬ﻥ ﺍﻟﻤﺴﺘﻭﻱ ﺤﻴﺙ‪:r‬‬ ‫‪r‬‬ ‫ﺃﺜﺒﺕ ﺃﻥ ‪ u + v :‬ﻋﻤﻭﺩﻱ ﻋﻠﻰ ‪u - v‬‬ ‫ﺍ‪r‬ﻟﺘﻤﺭﻴ‪r‬ﻥ‪. 13‬‬ ‫‪r v , u‬ﺸﻌﺎﻋﺎﻥ ﻤﻥ ﺍﻟﻤﺴﺘﻭ‪r‬ﻱ ﺤﻴ‪r‬ﺙ ‪:‬‬ ‫‪ u =2‬و ‪u. v=4‬‬‫‪uuur r uuur r‬‬‫‪ Ar , B , D‬ﺜﻼ‪r‬ﺙ ﻨﻘﻁ‪r‬ﺤﻴﺙ ‪ AB = u :‬و ‪AC = v‬‬ ‫‪ -1‬ﺃﺜﺒﺕ ﺃﻥ ‪ u‬ﻭ ‪ v - u‬ﻤﺘﻌﺎﻤﺩﻴﻥ ﺜﻡ ﺍﺴﺘﻨﺘﺞ ﻨﻭﻉ ﺍﻟﻤﺜﻠﺙ ‪. ABC‬‬

‫‪( )r r‬‬ ‫=‬ ‫‪π‬‬ ‫; ‪+ 2k π‬‬ ‫‪k∈¢‬‬ ‫‪ _2‬ﺃﻨﺸﺊ ‪ C‬ﺇﺫﺍ ﻋﻠﻤﺕ ﺃﻥ ‪:‬‬ ‫‪3‬‬ ‫‪v,u‬‬ ‫ﻤﻊ ﺤﺴﺎﺏ ‪.BC , AC‬‬‫‪A‬‬ ‫‪B‬‬ ‫ﺍﻟﺘﻤﺭﻴﻥ‪. 14‬‬ ‫‪ ABCD‬ﺸﺒﻪ ﻤﻨﺤﺭﻑ ﺤﻴﺙ ‪:‬‬ ‫‪AD = 2 , DC = 5‬‬ ‫‪ AB = 4‬ﻜﻤﺎ ﻓﻲ ﺍﻟﺸﻜل ‪.‬‬‫‪uAuDurD.‬‬ ‫‪uuur‬‬ ‫‪,‬‬ ‫‪uuur uuur‬‬ ‫‪,‬‬ ‫‪DuuCCur‬‬ ‫‪.‬‬ ‫‪uuur‬‬ ‫ﺍﺤﺴﺏ ﻤﺎ‪ur‬ﻴﻠ‪u‬ﻲ‪uuur :u‬‬ ‫‪BC‬‬ ‫‪CA . CHB‬‬ ‫‪DB‬‬ ‫‪, AB . BC‬‬ ‫ﺍﻟﺘﻤﺭﻴﻥ‪. 15‬‬ ‫‪ ABC‬ﻤﺜﻠﺙ‬ ‫‪ (1‬ﺒﺭﻫﻥ ﺃﻨ‪r‬ﻪ‪uu‬ﻤ‪u‬ﻥ ﺃﺠل‪r‬ﻜ‪u‬ل‪u‬ﻨ‪u‬ﻘﻁﺔ ‪urM‬ﻤ‪u‬ﻥ‪ u‬ﺍﻟﻤﺴ‪r‬ﺘ‪u‬ﻭ‪u‬ﻱ‪u‬ﻓﺈﻥ ‪uuur uuur:‬‬ ‫‪MA . BC + MB . CA + MC . AB = 0‬‬ ‫‪ (2‬ﺒﺭﻫﻥ ﺃﻥ ﺍﻻﺭﺘﻔﺎﻋﺎﺕ ﺍﻟﺜﻼﺜﺔ ﻓﻲ ﺍﻟﻤﺜﻠﺙ ﺘﺘﻘﺎﻁﻊ ﻓﻲ ﻨﻘﻁﺔ ﻭﺍﺤﺩﺓ‪.‬‬ ‫‪( )r r‬‬ ‫ﺍ‪r‬ﻟﺘﻤﺭﻴ‪r‬ﻥ‪. 16‬‬ ‫‪ v , u‬ﺸﻌﺎﻋﺎﻥ ﻏﻴﺭ ﻤﻌﺩﻭﻤﻴﻥ ‪ α .‬ﻗﻴﺱ ﺒﺎﻟﺭﺍﺩﻴﺎﻥ ﻟﻠﺯﺍﻭﻴﺔ ‪ u , v‬ﺤﻴﺙ ‪:‬‬ ‫‪0<α<π‬‬ ‫‪rr‬‬ ‫ﺃﺤﺴﺏ ‪ α‬ﻓ‪r‬ﻲ ﻜل ﻤﻥ ﺍﻟﺤﺎﻻﺕ ﺍﻵﺘﻴﺔ‪:r‬‬ ‫‪u . v = -3 3 ; v = 3 ; u = 2 (1‬‬ ‫‪rr r r‬‬ ‫‪u . v = +3 3 ; v = 3 ; u = 6 (2‬‬ ‫‪rr r r‬‬ ‫‪u . v = 10 ; v = 4 ; u = 5 (3‬‬ ‫ﺍ‪r‬ﻟﺘﻤﺭﻴ‪r‬ﻥ‪. 17‬‬ ‫‪ v , u‬ﺸﻌﺎﻋﺎﻥ ﺤﻴﺙ ‪:‬‬‫‪( )r r‬‬ ‫‪r‬‬ ‫‪r‬‬ ‫=‬ ‫‪π2ur+‬‬ ‫‪2kπ‬‬ ‫‪; k∈¢‬‬ ‫‪; v =2‬‬ ‫‪; u =3‬‬ ‫‪u,v‬‬ ‫‪r‬‬ ‫‪rr‬‬ ‫‪rr‬‬ ‫‪ur r‬‬ ‫ﻭﻟﻴﻜﻥ ﺍﻟﺸﻌﺎﻋﺎﻥ ‪ x , y‬ﺤﻴﺙ ‪x = 2u - 3v ; y = u + v :‬‬

‫‪r ur r r‬‬‫‪y‬‬ ‫‪, x,v‬‬ ‫‪,‬‬ ‫‪u‬‬‫‪xr 2‬‬ ‫‪r‬‬ ‫‪r‬‬ ‫ﺃﻨﺸﺊ ﻜل ﻤ‪r‬ﻥ‬ ‫‪(1‬‬ ‫ﺜﻡ‬ ‫‪y‬‬ ‫‪2‬‬ ‫ﺜﻡ‬ ‫‪u‬‬ ‫‪h‬ﺤﺴﺏ ‪.rv‬‬ ‫‪(2‬‬ ‫‪r‬‬ ‫‪ (3‬ﺍﺴﺘﻨﺘﺞ ‪y , x‬‬ ‫ﺍﻟﺘﻤﺭﻴﻥ‪. 18‬‬ ‫‪ ABCD‬ﻤﺴﺘﻁﻴل ﺤﻴﺙ ‪ AB = 4 :‬ﻭ ‪BC = 8‬‬‫‪ M‬ﻤﻨﺘﺼﻑ ‪ CD‬؛ ‪ K‬ﻤﺭﺠﺢ ﺍﻟﺠﻤﻠﺔ )‪[ ]. (D , 1) , (A , 3‬‬‫‪ I‬ﺍﻟﻤﺴﻘﻁ ﺍﻟﻌﻤﻭﺩﻱ ﻟﻠﻨﻘﻁﺔ ‪ K‬ﻋﻠﻰ )‪. (BM‬‬‫‪uuur uuur‬‬ ‫‪ (1‬ﺃﻨﺸﺊ ﺍﻟﺸ‪r‬ﻜ‪u‬ل‪uuur u.u‬‬‫‪ (2‬ﺍﺤﺴﺏ ‪. BA . BK ; uBuCur . BuuMur‬‬‫‪ (3‬ﺍﺤﺴﺏ ‪( )uuuur uuur BK . BM‬‬ ‫‪ (4‬ﺍﺴﺘﻨﺘﺞ ‪ BI‬ﺜﻡ ﻗﻴﻡ ﻤﻘﺭﺒﺔ ﻟـ ‪BM , BK‬‬ ‫ﺍﻟﺘﻤﺭﻴﻥ‪. 19‬‬‫‪ ABCD‬ﻤﺴﺘﻁﻴل ﻤﺭﻜﺯﻩ ‪ O‬ﺒﺭﻫﻥ ﺃﻨﻪ ﻤﻥ ﺃﺠل ﻜل ﻨﻘﻁﺔ ‪ M‬ﻤﻥ ﺍﻟﻤﺴﺘﻭﻱ ﻓﺈﻥ ‪:‬‬ ‫‪MA2 + MC2 = MB2 + MD2‬‬ ‫ﺍﻟﺘﻤﺭﻴﻥ‪. 20‬‬ ‫‪ ABC‬ﻤﺜﻠﺙ ‪.‬‬ ‫ﺍﺤﺴﺏ ﻜل ﻤﻥ ‪ BC ; AC ; Cµ :‬ﻋﻠﻤﺎ ﺃﻥ ‪:‬‬ ‫‪AB = 16 , Aµ = 68o , Bµ = 25o‬‬ ‫ﺘﻌﻁﻰ ﺍﻟﻤﺴﺎﻓﺎﺕ ﺒﺘﻘﺭﻴﺏ ‪ 0,01‬ﻭ ﺍﻟﺯﻭﺍﻴﺎ ﺒﺘﻘﺭﻴﺏ ‪. 0,1‬‬ ‫ﺍﻟﺘﻤﺭﻴﻥ‪. 21‬‬ ‫‪ ABC‬ﻤﺜﻠﺙ ‪.‬‬ ‫ﺍﺤﺴﺏ ﻜل ﻤﻥ ‪ BC‬ﻭ ‪ Bµ‬ﻭ ‪ Cµ‬ﻋﻠﻤﺎ ﺃﻥ ‪:‬‬ ‫‪Aµ = 100o , AC = 30 , AB = 20‬‬ ‫ﺜﻡ ﺍﺴﺘﻨﺘﺞ ‪ S‬ﻤﺴﺎﺤﺔ ﺍﻟﻤﺜﻠﺙ ‪.‬‬ ‫ﺍﻟﺘﻤﺭﻴﻥ‪. 22‬‬

‫‪ ABC‬ﻤﺜﻠﺙ ﻜﻴﻔﻲ ‪.‬‬ ‫‪ (1‬ﺒﻴﻥ ﺃﻥ ‪:‬‬‫‪sin2 Aµ = sin2 Bµ + sin2 Cµ - 2 sin Bµ . sin Cµ . cos Aµ‬‬ ‫‪ (2‬ﺒﻴﻥ ﺃﻥ ﺍﻟﻤﺜﻠﺙ ‪ ABC‬ﻴﻜﻭﻥ ﻗﺎﺌﻤﺎ ﻓﻲ ‪ A‬ﺇﺫﺍ ﻭﻓﻘﻁ ﺇﺫﺍ ﻜﺎﻥ ‪:‬‬‫‪sin2 Aµ = sin2 Bµ + sin2Cµ‬‬ ‫‪ (3‬ﺒﻴﻥ ﺃﻥ ﺍﻟﻤﺜﻠﺙ ‪ ABC‬ﻴﻜﻭﻥ ﻗﺎﺌﻤﺎ ﺇﺫﺍ ﻭﻓﻘﻁ ﺇﺫﺍ ﻜﺎﻥ ‪:‬‬‫‪sin2 Aµ + sin2Bµ + sin2Cµ = 2‬‬ ‫)*( ﺍﻟﺘﻤﺭﻴﻥ‪. 23‬‬‫‪uuur uuuur‬‬ ‫‪ A‬ﻭ‪ B‬ﻨﻘﻁﺘﺎﻥ ﺤﻴﺙ ‪AB = 6 :‬‬‫‪ (1‬ﻋﻴﻥ ﻭ ﺃﻨﺸﺊ ‪ ∆1‬ﻤﺠﻤﻭﻋﺔ ﺍﻟﻨﻘﻁ ‪ M‬ﻤﻥ ﺍﻟﻤﺴﺘﻭﻱ ﺒﺤﻴﺙ ‪( )uuAurB .uuAuuMr = 18 :‬‬‫‪ (2‬ﻋﻴﻥ ﻭ ﺃﻨﺸﺊ ‪ ∆ 2‬ﻤﺠﻤﻭﻋﺔ ﺍﻟﻨﻘﻁ ‪ M‬ﻤﻥ ﺍﻟﻤﺴﺘﻭﻱ ﺒﺤﻴﺙ ‪( )ABuuu.rAMuuu=ur -12 :‬‬‫‪ (3‬ﻋﻴﻥ ﻭ ﺃﻨﺸﺊ ‪ π1‬ﻤﺠﻤﻭﻋﺔ ﺍﻟﻨﻘﻁ ﻤﻥ ﺍﻟﻤﺴﺘﻭﻱ ﺒﺤﻴﺙ ‪( )AB . AM ≥ 30 :‬‬ ‫)*( ﺍﻟﺘﻤﺭﻴﻥ‪. 24‬‬ ‫‪ AB‬ﻗﻁﻌﺔ ﻤﺴﺘﻘﻴﻤﺔ ﺤﻴﺙ ‪[ ]AB = 4 :‬‬‫‪ (1‬ﻋﻴﻥ ﻤﺠﻤﻭﻋﺔ ﺍﻟﻨﻘﻁ ‪ M‬ﻤﻥ ﺍﻟﻤﺴﺘﻭﻱ ﺒﺤﻴﺙ ‪MA2 - MB2 = 12 :‬‬‫‪ (2‬ﻋﻴﻥ ﻤﺠﻤﻭﻋﺔ ﺍﻟﻨﻘﻁ ‪ M‬ﻤﻥ ﺍﻟﻤﺴﺘﻭﻱ ﺒﺤﻴﺙ ‪MAu2u+urMuBu2ur= 40 :‬‬ ‫‪ (3‬ﻋﻴﻥ ﻤﺠﻤﻭﻋﺔ ﺍﻟﻨﻘﻁ ‪ M‬ﻤﻥ ﺍﻟﻤﺴﺘﻭﻱ ﺒﺤﻴﺙ ‪MA - MB = λ :‬‬ ‫)*( ﺍﻟﺘﻤﺭﻴﻥ‪. 25‬‬ ‫‪ ABC‬ﻤﺜﻠﺙ ﻜﻴﻔﻲ ﻨﻀﻊ ‪BC = a , AC = b , AB = c :‬‬ ‫= ‪1 + cosAµ‬‬ ‫)‪(a + b +c) (b + c - a‬‬ ‫‪ (1‬ﺒﻴﻥ ﺃﻥ ‪:‬‬ ‫‪2bc‬‬ ‫‪(a‬‬ ‫‪+‬‬ ‫‪c‬‬ ‫‪-b) (a‬‬ ‫‪+‬‬ ‫‪b‬‬ ‫‪-‬‬ ‫)‪c‬‬ ‫‪1‬‬ ‫‪-‬‬ ‫= ‪cosAµ‬‬ ‫‪2bc‬‬ ‫‪ (2‬ﻨﻀﻊ ‪ 2p) a + b + c = 2p :‬ﻤﺤﻴﻁ ﺍﻟﻤﺜﻠﺙ(‬‫= ‪sin2 Aµ‬‬ ‫)‪4p (p - a) (p - b) (p - c‬‬ ‫ﺍﺴﺘﻨﺘﺞ ﻤﻥ )‪ (1‬ﺃﻥ ‪:‬‬ ‫‪b2 c2‬‬ ‫‪ (3‬ﻟﺘﻜﻥ ‪ S‬ﻤﺴﺎﺤﺔ ﺍﻟﻤﺜﻠﺙ ‪ ABC‬ﺒﻴﻥ ﺃﻥ ‪:‬‬

‫=‪S‬‬ ‫)‪p (p - a) (p - b) (p - c‬‬ ‫)ﻋﺒﺎﺭﺓ ﻫﻴﺭﻭﻥ(‬ ‫)ﺍﻟﻭﺤﺩﺓ ﻫﻲ ‪(cm‬‬ ‫‪ (4‬ﺍﺤﺴﺏ ﻤﺴﺎﺤﺔ ﺍﻟﻤﺜﻠﺙ ‪ ABC‬ﺤﻴﺙ ‪:‬‬ ‫‪BC = 10 ; AC = 15 ; AB = 9‬‬ ‫ﺍﻟﺘﻤﺭﻴﻥ‪. 26‬‬ ‫‪ -‬ﻨﻌﺘﺒﺭ ﺍﻟﻤﺴﺘﻘﻴﻡ ∆ ﺍﻟﺫﻱ ﻤﻌﺎﺩﻟﺘﻪ ‪( )3x + 3y – 5 = 0 :‬‬ ‫‪ -1‬ﺍﻜﺘﺏ ﻤﻌﺎﺩﻟﺔ ﺍﻟﻤﺴﺘﻘﻴﻡ ‪ ∆1‬ﺍﻟﺫﻱ ﻴﺸﻤل )‪ A (-1 ; 2‬ﻭﻴﻌﺎﻤﺩ ∆ ‪( ) ( ).‬‬ ‫‪ -2‬ﻋﻴﻥ ﻨﻘﻁﺔ ﺘﻘﺎﻁﻊ ∆ ﻭ ‪ ∆1‬ﻭﻟﺘﻜﻥ ‪( ) ( ). B‬‬ ‫‪ -3‬ﺍﻜﺘﺏ ﻤﻌﺎﺩﻟﺔ ﺍﻟﺩﺍﺌﺭﺓ ﺫﺍﺕ ﺍﻟﻘﻁﺭ ‪[ ]. AO‬‬ ‫ﺍﻟﺘﻤﺭﻴﻥ‪. 27‬‬ ‫ﻨﻌﺘﺒﺭ ﺍﻟﻨﻘﻁ )‪C (1 ; 2) , B (1 ; -4) , A (-2 ; 2‬‬ ‫‪ -1‬ﻋﻴﻥ ﻤﺭﻜﺯ ﺜﻘل ﺍﻟﻤﺜﻠﺙ ‪.ABC‬‬ ‫‪ -2‬ﻋﻴﻥ ﻤﻌﺎﺩﻻﺕ ﺍﻟﻤﺘﻭﺴﻁﺎﺕ ﺍﻟﺜﻼﺜﺔ ﻟﻠﻤﺜﻠﺙ ‪. ABC‬‬ ‫‪ -3‬ﻋﻴﻥ ﻨﻘﻁﺔ ﺘﻘﺎﻁﻊ ﺍﻟﻤﺘﻭﺴﻁﺎﺕ ﺍﻟﺜﻼﺜﺔ ‪ .‬ﻤﺎﺫﺍ ﺘﻼﺤﻅ؟‬ ‫ﺍﻟﺘﻤﺭﻴﻥ‪. 28‬‬ ‫ﻨﻌﺘﺒﺭ ﺍﻟﻨﻘﻁ ‪C (0 ; 3) , B (-4 ; 0) , A (1 ; 2) :‬‬ ‫‪ -1‬ﺍﻜﺘﺏ ﻤﻌﺎﺩﻻﺕ ﺃﻋﻤﺩﺓ ﺍﻟﻤﺜﻠﺙ ‪. ABC‬‬ ‫‪ -2‬ﻋﻴﻥ ﻨﻘﻁﺔ ﺘﻘﺎﻁﻊ ﺍﻷﻋﻤﺩﺓ ‪.‬‬ ‫‪ -3‬ﻋﻴﻥ ﻤﺭﻜﺯ ﻭ ﻨﺼﻑ ﻗﻁﺭ ﺍﻟﺩﺍﺌﺭﺓ ﺍﻟﻤﺤﻴﻁﺔ ﺒﺎﻟﻤﺜﻠﺙ ‪.ABC‬‬ ‫ﺍﻟﺘﻤﺭﻴﻥ‪. 29‬‬ ‫‪ (1‬ﺍﻜﺘﺏ ﻤﻌﺎﺩﻟﺔ ﺍﻟﺩﺍﺌﺭﺓ ‪ Γ1‬ﺫﺍﺕ ﺍﻟﻤﺭﻜﺯ )‪ ω (1 ; -3‬ﻭ ﺘﺸﻤل) (‬ ‫ﺍﻟﻨﻘﻁﺔ )‪. A (5 ; 2‬‬ ‫‪ (2‬ﺍﻜﺘﺏ ﻤﻌﺩﻟﺔ ﺍﻟﺩﺍﺌﺭﺓ ﺫﺍﺕ ﺍﻟﻘﻁﺭ ‪[ ]. ωA‬‬ ‫‪ (3‬ﺍﻜﺘﺏ ﻤﻌﺎﺩﻟﺔ ﺍﻟﺩﺍﺌﺭﺓ ﺍﻟﻤﺤﻴﻁﻴﺔ ﺒﺎﻟﻤﺜﻠﺙ ‪ ABω‬ﺤﻴﺙ ‪B (1 ; -2) :‬‬ ‫‪ (4‬ﺍﻜﺘﺏ ﻤﻌﺩﻟﺔ ﺍﻟﺩﺍﺌﺭﺓ ﺫﺍﺕ ﺍﻟﻤﺭﻜﺯ ‪ ω‬ﻭﺘﻘﺒل ﺍﻟﻤﺴﺘﻘﻴﻡ ∆ ﺍﻟﺫﻱ) (‬ ‫ﻤﻌﺎﺩﻟﺘﻪ ‪ x – y + 2 = 0 :‬ﻤﻤﺎﺴﺎ ﻟﻬﺎ ‪.‬‬ ‫ﺍﻟﺘﻤﺭﻴﻥ‪. 30‬‬ ‫ﻋﻴﻥ ﻤﺠﻤﻭﻋﺔ ﺍﻟﻨﻘﻁ )‪ M (x ; y‬ﻤﻥ ﺍﻟﻤﺴﺘﻭﻱ ﻓﻲ ﻜل ﺤﺎﻟﺔ ﻤﻤﺎ ﻴﻠﻲ ‪:‬‬

‫‪. x2 + y2 - 2x + 4y - 11 = 0 (1‬‬ ‫‪. 2x2 + 2y2 - 4x + 8y + 10 = 0 (2‬‬ ‫‪. - x2 - y2 + 6x + 10y - 60 = 0 (3‬‬ ‫‪. y2 + x2 - 4xy + 5y2 = 0 (4‬‬ ‫ﺍﻟﺘﻤﺭﻴﻥ‪. 31‬‬ ‫ﻨﺎﻗﺵ ﺤﺴﺏ ﻗﻴﻡ ﺍﻟﻭﺴﻴﻁ ﺍﻟﺤﻘﻴﻘﻲ ‪ m‬ﻤﺠﻤﻭﻋﺔ ﺍﻟﻨﻘﻁ )‪ M (x ; y‬ﺤﻴﺙ ‪:‬‬‫‪x2 + y2 - 2m x - 2 (1 + m) y + 6m + 1 = 0‬‬ ‫ﺍﻟﺘﻤﺭﻴﻥ‪. 32‬‬ ‫ﻟﺘﻜﻥ ‪ Cm‬ﻤﺠﻤﻭﻋﺔ ﺍﻟﻨﻘﻁ )‪ M (x ; y‬ﺤﻴﺙ ‪( ):‬‬‫‪x2‬‬ ‫‪+‬‬ ‫‪y2‬‬ ‫‪-2‬‬ ‫‪(1‬‬ ‫‪-‬‬ ‫‪m) x‬‬ ‫‪+‬‬ ‫‪(1‬‬ ‫‪+ 6m) y‬‬ ‫‪+ 5m‬‬ ‫‪-‬‬ ‫‪5‬‬ ‫=‬ ‫‪0‬‬ ‫‪4‬‬ ‫‪ M‬ﻭﺴﻴﻁ ﺤﻘﻴﻘﻲ ‪.‬‬ ‫‪ (1‬ﺃﺜﺒﺕ ﺃﻥ ‪ Cm‬ﺩﺍﺌﺭﺓ‪( ).‬‬ ‫‪ (2‬ﺃﻨﺸﺊ ﻓﻲ ﻨﻔﺱ ﺍﻟﻤﻌﻠﻡ ‪( ) ( )C1 , C0‬‬ ‫‪ (3‬ﻟﺘﻜﻥ ‪ ωm‬ﻤﺭﻜﺯ ﺍﻟﺩﺍﺌﺭﺓ ‪ . Cm‬ﻤﺎ ﻫﻲ ﻤﺠﻤﻭﻋﺔ ﺍﻟﻨﻘﻁ ‪ ωm‬ﻋﻨﺩﻤﺎ) (‬ ‫ﻴﺘﻐﻴﺭ ‪ m‬ﻓﻲ ¡ ‪.‬‬ ‫‪ (4‬ﺃﺜﺒﺕ ﺃﻥ ‪ Cm‬ﺘﺸﻤل ﻨﻘﻁﺘﺎﻥ ﺜﺎﺒﺘﺘﺎﻥ ‪ B , A‬ﻴﻁﻠﺏ ﺘﻌﻴﻴﻨﻬﻤﺎ‪( ).‬‬ ‫ﺍﻟﺘﻤﺭﻴﻥ‪. 33‬‬‫ﺃﺩﺭﺱ ﻭﻀﻌﻴﺔ ﺍﻟﺩﺍﺌﺭﺓ )‪ (C‬ﺍﻟﺘﻲ ﻤﻌﺎﺩﻟﺘﻬﺎ ‪ x2 + y + m = 0‬ﻭﺍﻟﻤﺴﺘﻘﻴﻡ ‪ ∆m‬ﺍﻟﺫﻱ) (‬ ‫ﻤﻌﺎﺩﻟﺘﻪ ‪ m , x – y + m = 0 :‬ﻭﺴﻴﻁ ﺤﻘﻴﻘﻲ‪.‬‬ ‫ﺍﻟﺘﻤﺭﻴﻥ‪. 34‬‬ ‫‪ (1‬ﻋﻴﻥ ﺍﻟﻤﺠﻤﻭﻋﺔ ‪( )x2 + y2 + 2x - y = 5 : C1‬‬ ‫‪(2‬ﺃﻜﺘﺏ ﻤﻌﺎﺩﻟﺔ ﺍﻟﺩﺍﺌﺭﺓ ‪ C2‬ﺫﺍﺕ ﺍﻟﻤﺭﻜﺯ )‪ ω2(4 ; 3‬ﻭﻨﺼﻑ ﻗﻁﺭﻫﺎ ‪( ).5‬‬ ‫‪ (3‬ﺃﻨﺸﺊ ‪ C1‬و ‪ C2‬ﻤﻌﻴﻨﺎ ﻨﻘﻁﺘﻲ ﺘﻘﺎﻁﻌﻬﻤﺎ ‪ F‬ﻭ‪ H‬ﺤﺴﺎﺒﻴﺎ ‪( ) ( ).‬‬ ‫‪ (4‬ﺍﻜﺘﺏ ﻤﻌﺎﺩﻟﺔ ﺍﻟﻤﻤﺎﺱ ﻋﻨﺩ ‪ F‬ﻟﻜل ﻤﻥ ‪ C1‬و ‪ C2‬ﻭ ﺒﻴﻥ ﺃﻨﻬﻤﺎ) ( ) (‬

‫ﻤﺘﻌﺎﻤﺩﺍﻥ ؛ ﻨﻔﺱ ﺍﻟﺴﺅﺍل ﻋﻨﺩ ‪.H‬‬