الألمبياد .pdf



Nom original: الألمبياد.pdf
Titre: International Mathematical Olympiad
Auteur: Omran Kouba

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This book is downloaded from this site:
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This book is downloaded from this site:
www.syCourses.com




2010
‫ﺔ ﻟﻸﻭﻟﻤﺒﻴﺎﺩ ﺍﻟﻌﻠﻤﻲ ﺍﻟﺴﻮﺭﻱ‬‫ﺍﻟﻬﻴﺌﺔ ﺍﻟﻮﻃﻨﻴ‬

This book is downloaded from this site:
www.syCourses.com

This book is downloaded from this site:
www.syCourses.com

This book is downloaded from this site:
www.syCourses.com

‫اﻷوﳌﺒﻴﺎد اﻟﻌﺎﳌﻲ‬
‫ﻟﻠﺮايﺿ ّﻴﺎت‬





2010

This book is downloaded from this site:
www.syCourses.com

‫ﺣﻘﻮﻕ ﺍﻟﺘﺄﻟﻴﻒ ﻭﺍﻟﻄﺒﺎﻋﺔ ﻭﺍﻟﻨﺸﺮ ﳏﻔﻮﻇﺔ‬

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‫‪This book is downloaded from this site:‬‬
‫‪www.syCourses.com‬‬

‫ ‬
‫ﻻ ﺑ‪‬ﺪ‪ ‬ﺃﻥ ﻓﻜﺮﺓ ﻭﺿﻊ ﻃﻼﹼ ﺏﹴ ﰲ ﻣﻮﺍﺟﻬﺔ ﻣﺴﺎﺋﻞ ﰲ ﺍﻟﺮﻳﺎﺿﻴ‪‬ﺎﺕ ﻗﺪﳝﺔﹲ‪ .‬ﻓﻨﺮﻯ ﻣﺴﺎﺑﻘﺎﺕ‬
‫ﺍﻟﺮﻳﺎﺿﻴ‪‬ﺎﺕ ﰲ ﺗﻘﺎﻟﻴﺪ ﺍﻟﻌﺪﻳﺪ ﻣﻦ ﺍﻟﺒﻠﺪﺍﻥ ﻭﻋﻠﻰ ﻣﺪﻯ ﺍﻟﻌﺼﻮﺭ‪ .‬ﺇﺫ ﻛﺎﻥ ﺍﻹﻏﺮﻳﻖ ﻳﺘﺒﺎﺭﻭﻥ ﰲ ﺣﻞﹼ‬
‫ﻣﺴﺎﺋﻞ ﺍﳍﻨﺪﺳﺔ ﺍﻹﻗﻠﻴﺪﻳ‪‬ﺔ‪ .‬ﻭﰲ ﺍﻟﻘﺮﻥ ﺍﻟﺴﺎﺩﺱ ﻋﺸﺮ ﻛﺎﻥ ﺍﻹﻳﻄﺎﻟﻴ‪‬ﻮﻥ ﻳﺘﺒﺎﺭﻭﻥ ﰲ ﺣﻞﹼ ﺍﳌﻌﺎﺩﻻﺕ ﻣﻦ‬
‫ﺍﻟﺪﺭﺟﺔ ﺍﻟﺜﺎﻟﺜﺔ‪ .‬ﻭﺃﻋﻠﻦ ﺍﻟﻔﺮﻧﺴﻴ‪‬ﻮﻥ ﻋﻦ ﻣﺴﺎﺑﻘﺎﺕ ﰲ ﺍﻟﺮﻳﺎﺿﻴ‪‬ﺎﺕ ﰲ ﺍﻟﻘﺮﻥ ﺍﻟﺜﺎﻣﻦ ﻋﺸﺮ‪ .‬ﻭﻟﻌﻞﹼ‬
‫ﻣﺴﺎﺑﻘﺎﺕ ‪ Eötvös‬ﺍﻟﱵ ﻧﻈﹼﻤﺘﻬﺎ ﻫﻨﻐﺎﺭﻳﺎ ﻋﺎﻡ ‪ 1894‬ﺃﻗﺮﺏ ﺳﻠﻒ ﳌﺴﺎﺑﻘﺎﺕ ﺍﻷﻭﳌﺒﻴﺎﺩ ﺍﳌﹸﻌﺎﺻﺮﺓ‪.‬‬
‫ﻟﻘﺪ ﺟﺮﻯ ﺗﻨﻈﻴﻢ ﺃﻭ‪‬ﻝ ﺃﻭﳌﺒﻴﺎﺩ ﻟﻠﺮﻳﺎﺿﻴ‪‬ﺎﺕ ﰲ ﻣﺪﻳﻨﺔ ﻟﻴﻨﻨﻐﺮﺍﺩ ) ﺳﺎﻥ ﺑﻴﺘﺮﺳﺒﻮﻍ ﺣﺎﻟﻴﺎﹰ ( ﻋﺎﻡ ‪، 1934‬‬
‫ﻣﻦ ﻗﺒﻞ ﺍﻟﺮﻳﺎﺿﻴ‪‬ﻴ‪‬ﻦ ‪ Delone‬ﻭ ‪ . Frijtengolts‬ﻭﺃﻗﺎﻣﺖ ﺭﻭﻣﺎﻧﻴﺎ ﺃﻭ‪‬ﻝ ﺃﻭﳌﺒﻴﺎﺩ ﻋﺎﳌﻲ ﻟﻠﺮﻳﺎﺿﻴﺎﺕ‬
‫ﻋﺎﻡ ‪ ، 1959‬ﻭﻛﺎﻥ ﻟﻪ ﺻﻴﻐﺔ ﻣﺴﺎﺑﻘﺔ ﻣﻮ ﺟ‪‬ﻬﺔ ﻟﺪﻭﻝ ﺃﻭﺭﺑﺎ ﺍﻟﺸﺮﻗﻴ‪‬ﺔ ﻭﺷﺎﺭﻛﺖ ﻓﻴﻬﺎ ﺳﺒﻊ ﺩﻭﻝ‪.‬‬
‫ﻭﺻﺎﺭﺕ ﺍﳌﺴﺎﺑﻘﺔ‪ ،‬ﻣﻨﺬ ﺫﻟﻚ ﺍﳊﲔ‪ ،‬ﺗ‪‬ﺠﺮﻯ ﻛﻞ ﻋﺎﻡ ﺑﺎﺳﺘﺜﻨﺎﺀ ﻋﺎﻡ ‪ . 1980‬ﺑﺪﺍﻳﺔﹰ‪ ،‬ﻛﺎﻧﺖ ﺍﳌﺸﺎﺭﻛﺔ‬
‫ﺑﺎﳌﺴﺎﺑﻘﺔ ﳏﺼﻮﺭﺓ ﺑﺎﻟﺪﻭﻝ ﻧﻔﺴﻬﺎ‪ ،‬ﺇﻻﹼ ﺃﻧ‪‬ﻬﺎ ﺃﺧﺬﺕ ﺑﺎﻻﺗﺴﺎﻉ ﻟﺘﺸﻤﻞ ﺣﺎﻟﻴ‪‬ﺎﹰ ﺃﻛﺜﺮ ﻣﻦ ﺗﺴﻌﲔ ﺩﻭﻟﺔ‪ ،‬ﻣﻦ‬
‫ﺍﻟﻘﺎﺭﺍﺕ ﺍﳋﻤﺲ‪ .‬ﻳﺘﻐﻴ‪‬ﺮ ﺍﻟﺒﻠﺪ‪ ‬ﺍ ﳌﹸﻀﻴﻒ ﺳﻨﻮ‪‬ﻳﺎﹰ‪ ،‬ﻭﺇﻥ ﻛﺎﻥ ﻫﻨﺎﻙ ﺍﻟﻌﺪﻳﺪ ﻣﻦ ﺍﻟﺒﻠﺪﺍﻥ ﺍﻟﱵ ﺍﺳﺘﻀﺎﻓﺖ‬
‫ﺍﻷﻭﳌﺒﻴﺎﺩ ﺃﻛﺜﺮ ﻣﻦ ﻣﺮ‪‬ﺓ‪ .‬ﻳﺮﺍﻓﻖ ﺍﳌﺴﺎﺑﻘﺔ ﻋﺎﺩﺓ ﺑﺮﻧﺎﻣﺞ ﺛﻘﺎﰲ ﻭﺳﻴﺎﺣﻲ ﰲ ﺍﻟﺒﻠﺪ ﺍ ﳌﹸﻀﻴﻒ ﻳﻬﺪﻑ ﺇﱃ‬
‫ﺗﻌﻤﻴﻖ ﺍﻟﺘﻔﺎﻫﻢ ﺍﳌﺘﺒﺎﺩﻝ ﺑﲔ ﺍﻟﻔ‪‬ﺮ‪‬ﻕ ﺍﳌﺸﺎﺭﻛﺔ ﻭﺗﻨﻤﻴ‪‬ﺔ ﻋﻼﻗﺎﺕ ﺍﻟﺼﺪﺍﻗﺔ ﻭﺍﻟﻮﺩ‪ ‬ﺑﲔ ﺍﳌﺘﺴﺎﺑﻘﲔ ﻭﺍﳌﺸﺮﻓﲔ‪.‬‬
‫ﰲ ﺍﻟﺒﺪﺍﻳﺔ ﻛﺎﻧﺖ ﻛﻞﱡ ﺩﻭﻟﺔ ﺗﺸﺎﺭﻙ ﺑﺜﻤﺎﻧﻴﺔ ﻣﺘﺴﺎﺑﻘﲔ ﺃﻭ ﻣﺘﺴﺎﺑﻘﺎﺕ ﻋﻠﻰ ﺍﻷﻛﺜﺮ‪ ،‬ﺧ‪‬ﻔﱢﺾ‪ ‬ﻫﺬﺍ ﺍﻟﻌﺪﺩ‪‬‬
‫ﺇﱃ ﺃﺭﺑﻌﺔ ﻣﺘﺴﺎﺑﻘﲔ ﻋﺎﻡ ‪ ، 1982‬ﺛﹸﻢ‪ ‬ﺭ‪‬ﻓ‪‬ﻊ‪ ‬ﻣﻦ ﺟﺪﻳﺪ ﺇﱃ ﺳﺘﺔ ﻣﺸﺎﺭﻛﲔ‪ ،‬ﻭﺑﻘﻲ ﻋﻠﻰ ﻫﺬﻩ ﺍﳊﺎﻝ ﺣﺘ‪‬ﻰ‬
‫ﻭﻗﺘﻨﺎ ﻫﺬﺍ‪.‬‬
‫ﳚﺐ ﺃ ﻻﹼ ﻳﺘﺠﺎﻭﺯ ﻋﻤﺮ ﺍﳌﺘﺴﺎﺑﻖ ﻋﺸﺮﻳﻦ ﻋﺎﻣﺎﹰ ﻭﺃ ﻻﹼ ﻳﻜﻮ ﻥﹶ ﻗﺪ ﺗﻠﻘﹼﻰ ﺗﻌﻠﻴﻤﺎﹰ ﻣﺎ ﺑﻌﺪ ﺍﻟﺜﺎﻧﻮﻱ‪.‬‬
‫ﻭﳝﻜﻦ ﻟﻠﻤﺘﺴﺎﺑﻖ ﺃﻥ ﻳﺸﺎﺭﻙ ﺃﻛﺜﺮ ﻣﻦ ﻣﺮ‪‬ﺓ ﺇﺫﺍ ﺣﻘﹼﻖ ﺍﻟﺸﺮﻃﲔ ﺍﻟﺴﺎﺑﻘﲔ‪ .‬ﻭﻣﻊ ﺃﻥﹼ ﺍﳌﺘﺴﺎﺑﻘﲔ ﳝﺜﻠﹼﻮﻥ‬
‫ﺑﻠﺪﺍ‪‬ﻢ ﰲ ﺍﳌﺴﺎﺑﻘﺔ ﺇ ﻻﹼ ﺃ ﻥﹼ ﻋﻤﻠﻴ‪‬ﺔ ﺗﻘﻴﻴﻢ ﺃﺩﺍﺋﻬﻢ ﻋﻤﻠﻴ‪‬ﺔ ﻓﺮﺩ ﻳ‪‬ﺔ‪ ،‬ﺇ ﺫﹾ ﻟﻴﺲ ﻫﻨﺎﻙ ﺗﻘﻴﻴﻢ ﲨﺎﻋﻲ ﻟﻠﻔﺮﻕ‬
‫ﺍﳌﹸﺸﺎﺭﻛﺔ‪.‬‬

‫‪vThis book is downloaded from this site:‬‬
‫‪www.syCourses.com‬‬
‫ﺍﻷﻭﳌﺒﻴﺎﺩ ﺍﻟﻌﺎﳌﻲ ﻟﻠﺮﻳﺎﺿﻴ‪‬ﺎﺕ‬

‫‪ii‬‬

‫ﻳﺘﺒﻊ ﻛﻞﱡ ﺑﻠﺪ‪ ‬ﻃﺮﻳﻘﺘﻪ ﺍﳋﺎﺻ‪‬ﺔ ﰲ ﺍﺧﺘﻴﺎﺭ ﻣﺮﺷ‪‬ﺤﻴﻪ ﻟﻠﻤﺸﺎﺭﻛﺔ ﰲ ﻫﺬﻩ ﺍﳌﺴﺎﺑﻘﺔ‪ ،‬ﺇﻻﹼ ﺃﻧ‪‬ﻬﺎ ﲨﻴﻌﹰﺎ‬
‫ﺗﺘﻄﻠﹼﺐ ﻣﻦ ﳑﺜﹼﻠﻴﻬﺎ ﻣﻬﺎﺭﺓ ﻋﺎﻟﻴﺔ ﰲ ﳎﺎﻝ ﺍﻟﺮﻳﺎﺿﻴ‪‬ﺎﺕ‪ .‬ﻛﻤﺎ ﲤﻨﺢ ﺍﳌﺴﺎﺑﻘﺔ ﺍ ﳌﹸﺸﺎﺭﻛﲔ ﻓﹸﺮﺻﺔ ﻋﺮﺽ‬
‫ﻣﻬﺎﺭﺍ‪‬ﻢ ﻭﺣﻨﻜﺘﻬﻢ ﻭﺳﺮﻋﺔ ﺑﺪﻳﻬﺘﻬﻢ ﰲ ﻣﻮﺍﺟﻬﺔ ﻣﺴﺎﺋﻞ ﰲ ﺍﻟﺮﻳﺎﺿﻴ‪‬ﺎﺕ‪ .‬ﳝﺘﺪ‪ ‬ﺍﳊﺪﺙ ﻋﻠﻰ ﻣﺪﻯ‬
‫ﺃﺳﺒﻮﻋﲔ ﲢﺘﻞﹼ ﻓﻴﻬﻤﺎ ﺍﳌﺴﺎﺑﻘﺔ ﻓﻌﻠﻴﺎﹰ ﻳﻮﻣﲔ ﻣﺘﺘﺎﻟﻴﲔ ﻓﻘﻂ‪ ،‬ﺇﺫﹾ ﻳ‪‬ﻤﻀﻲ ﺍﳌﺘﺴﺎﺑﻘﻮﻥ ﺑﻘﻴﺔ ﺍﻟﻮﻗﺖ ﰲ ﺯﻳﺎﺭﺓ‬
‫ﺍﻟﺒﻠﺪ ﺍ ﳌﹸﻀﻴﻒ ﻭﺍﻻﻃﻼﻉ ﻋﻠﻰ ﺛﻘﺎﻓﺘﻪ‪ ،‬ﻭﺯﻳﺎﺭﺓ ﺍﳌﻮﺍﻗﻊ ﺍﻟﺴﻴﺎﺣﻴ‪‬ﺔ ﺍﳌﻬﻤ‪‬ﺔ ﻓﻴﻪ‪ ،‬ﻭﰲ ﺗﻨﻤﻴ‪‬ﺔ ﺍﻟﺼﻼﺕ‬
‫ﺍﻻﺟﺘﻤﺎﻋﻴ‪‬ﺔ ﻓﻴﻤﺎ ﺑﻴﻨﻬﻢ‪ .‬ﻭﺑﺎﻟﻨﺴﺒﺔ ﺇﱃ ﺍﻟﻌﺪﻳﺪ ﻣﻦ ﺍﳌﺸﺎﺭﻛﲔ ﲢﺘﻞﹼ ﺍﻟﺬﻛﺮﻳﺎﺕ ﻭﺍﻟﺼﺪﺍﻗﺎﺕ ﺍﻟﱵ ﺗ‪‬ﻨﺴﺞ‬
‫ﰲ ﻫﺬﻩ ﺍﳌﻨﺎﺳﺒﺔ ﻣﺮﻛﺰﺍﹰ ﺃ ﻛﺜﺮ ﺃﳘﻴ‪‬ﺔ ﰲ ﻧﻔﻮﺳﻬﻢ ﻣﻦ ﺍﻟﺪﺭﺟﺎﺕ ﺍﻟﱵ ﻳ‪‬ﺤﺼ‪‬ﻠﻮ‪‬ﺎ ﺃﻭ ﺍﳌﻴﺪﺍﻟﻴ‪‬ﺎﺕ ﺍﻟﱵ‬
‫ﳛﺼﻠﻮﻥ ﻋﻠﻴﻬﺎ‪ .‬ﻛﻤﺎ ﺣﻘﹼﻖ ﺍﻟﻌﺪﻳﺪ ﻣﻨﻬﻢ ﻻﺣﻘﺎﹰ ﺇﳒﺎﺯﺍﺕ ﻣﻬﻤ‪‬ﺔ ﰲ ﺍﻟﺮﻳﺎﺿﻴﺎﺕ‪.‬‬
‫ﻳﺘﻜﹼﻮﻥ ﻛﻞﹼ ﻓﺮﻳﻖ ﻣﺸﺎﺭ ﻙ‪ ‬ﻣﻦ ﺳﺘﺔ ﻣﺘﺴﺎﺑﻘﲔ ﻋﻠﻰ ﺍﻷﻛﺜﺮ‪ ،‬ﻭﻣﻦ ﻗﺎﺋﺪ‪‬ﻳﻦ ﺍﺛﻨﲔ ﻟﻠﻔﺮﻳﻖ ﻋﻠﻰ‬
‫ﺍﻷﻛﺜﺮ‪ .‬ﻳﻜﺘﺐ ﻛﻞﱡ ﻣﺘﺴﺎﺑﻖﹴ ﻭﺭﻗﺘﲔ ﰲ ﻳﻮ ﻣ‪‬ﻲ ﺍﳌﺴﺎﺑﻘﺔ ﺍﳌﺘﺘﺎﻟﻴﲔ‪ ،‬ﲢﻮﻱ ﻛﻞﱡ ﻭﺭﻗﺔ ﺣﻠﻮﻟﹶﻪ ﻟﻠﻤﺴﺎﺋﻞ‬
‫ﺍﻟﺜﻼﺙ ﺍﻟﱵ ﻃﹸﺮﺣﺖ ﻋﻠﻴﻪ ﰲ ﺫﻟﻚ ﺍﻟﻴﻮﻡ‪ .‬ﻭﺗ‪‬ﺼﺤ‪‬ﺢ‪ ‬ﻛﻞﱡ ﻣﺴﺄﻟﺔ ﻭﺗﻌﻄﻰ ﺍﻟﻨﺘﻴﺠﺔ ﺑﺼﻴﻐﺔ ﻋﺪﺩ‪ ‬ﺻﺤﻴﺢﹴ‬
‫ﺑﲔ ‪ 0‬ﻭ ‪ ) . 7‬ﻓﻘﻂ ﰲ ﺃﻭﳌﺒﻴﺎﺩ ﻋﺎﻡ ‪ 1962‬ﻃﹸﺮﹺﺣﺖ ﺳﺒﻊ ﻣﺴﺎﺋﻞ ﺑﺪﻻﹰ ﻣﻦ ﺳﺖ‪ ‬ﻭﻫﺬﺍ ﻫﻮ ﺍﻻﺳﺘﺜﻨﺎﺀ‬
‫ﺍﻟﻮﺣﻴﺪ ( ‪.‬‬

‫ﺣﻘﺎﺋﻖ ﻋﺎ ّﻣﺔ ﻋﻦ اﻷوﳌﺒﻴﺎد اﻟﻌﺎﳌﻲ ﻟﻠﺮايﺿ ّﻴﺎت‬
‫‪ å‬ﺃﻭﺭﺍﻕ ﺍﻷﺳﺌﻠﺔ‪ .‬ﻗﺒﻞ ﺃﺭﺑﻌﺔ ﺃﺷﻬﺮ ﻋﻠﻰ ﺍﻷﻗﻞ ﻣﻦ ﻣﻮﻋﺪ ﺍﳌﺴﺎﺑﻘﺔ‪ ،‬ﳝﻜﻦ ﻟﻜﻞﹼ ﺩﻭﻟﺔ‪ ‬ﺿﻴﻒ ﺃﻥ‬
‫ﺗﻘﺘﺮﺡ ﺳﺘﺔ ﺃﺳﺌﻠﺔ ﻋﻠﻰ ﺍﻷﻛﺜﺮ ‪‬ﺪﻑ ﺩﺭﺍﺳﺘﻬﺎ ﻟﺘﻜﻮﻥ ﺑﲔ ﺍﻷﺳﺌﻠﺔ ﺍﳌﹸﺨﺘﺎﺭﺓ ﻟﻠﻤﺴﺎﺑﻘﺔ‪ .‬ﲡﺮﻱ ﺩﺭﺍﺳﺔ‬
‫ﻫﺬﻩ ﺍ ﳌﹸﻘﺘﺮﺣﺎﺕ ﰲ ﺍﻟﺒﻠﺪ ﺍ ﳌﹸﻀﻴﻒ ﻣﻦ ﻗ‪‬ﺒ‪‬ﻞ ﳉﻨﺔ ﺍﳌﺴﺎﺑﻘﺔ ﺍﻟﱵ ﲣﺘﺎﺭ ﻣﻦ ﳎﻤﻮﻉ ﻫﺬﻩ ﺍﻷﺳﺌﻠﺔ ﻗﺎﺋﻤﺔ‬
‫ﻗﺼﲑﺓ ﻣﻜﻮ‪‬ﻧﺔ ﻣﻦ ﺣﻮﺍﱄ ﺛﻼﺛﲔ ﺳﺆﺍﻻﹰ‪ .‬ﻭﻟﻘﺪ ﺭﺃﻳﻨﺎ ﰲ ﺍﻟﺴﻨﻮﺍﺕ ﺍﻷﺧﲑﺓ ﻗﺎﺋﻤﺔ ﺃﻗﺼﺮ‪ ،‬ﻣﻜﻮ‪‬ﻧﺔ ﻣﻦ ﺍﺛﲏ‬
‫ﻋﺸﺮ ﺳﺆﺍﻻﹰ‪ ،‬ﲤﺜﹼﻞ ﺍﻷﺳﺌﻠﺔ ﺍﳌﻔﻀ‪‬ﻠﺔ ﻟﺪﻯ ﺍﻟﻠﺠﻨﺔ‪ .‬ﺃﻣ‪‬ﺎ ﺍﻻﺧﺘﻴﺎﺭ ﺍﻟﻨﻬﺎﺋﻲ ﻟﻸﺳﺌﻠﺔ ﺍﻟﱵ ﺳﺘﻄﺮﺡ ﰲ ﺍﳌﺴﺎﺑﻘﺔ‬
‫ﻓﺘ‪‬ﺠﺮﻳﻪ ﳉﻨﺔ ﺍﳊﻜﹼﺎﻡ ﺍﻟﺪﻭﻟﻴ‪‬ﺔ‪ ،‬ﻭﻫﻲ ﻣﻜﻮ‪‬ﻧﺔ ﻣﻦ ﻗﺎﺋﺪ ﻛﻞﱢ ﻓﺮﻳﻖ ﻣﺸﺎﺭﻙ‪ ،‬ﺑﺎﻹﺿﺎﻓﺔ ﺇﱃ ﺃﺭﺑﻌﺔ ﺣﻜﹼﺎﻡ‬
‫ﺗﻨﻔﻴﺬ‪‬ﻳﲔ ﳜﺘﺎﺭﻫﻢ ﺍﻟﺒﻠﺪ ﺍﳌﹸﻀﻴﻒ‪ ،‬ﻭﺗ‪‬ﺘ‪‬ﺨ‪‬ﺬ ﺍﻟﻘﺮﺍﺭﺍﺕ ﺑﺎﻷﻛﺜﺮﻳ‪‬ﺔ‪ .‬ﲡﺘﻤﻊ ﳉﻨﺔ ﺍﳊﻜﹼﺎﻡ ﺍﻟﺪﻭﻟﻴ‪‬ﺔ ﻗﺒﻞ ﺃﻳ‪‬ﺎﻡﹴ ﻣﻦ‬
‫ﻣﻮﻋﺪ ﺍﳌﺴﺎﺑﻘﺔ ﰲ ﻣﻜﺎﻥ ﻣﻌﺰﻭﻝﹴ ﻭﺳﺮ‪‬ﻱ‪ ‬ﻭﳚﺮﻱ ﺍﻻﺧﺘﻴﺎﺭ ﺍﻟﻨﻬﺎﺋﻲ ﻷﺳﺌﻠﺔ ﺍﳌﺴﺎﺑﻘﺔ‪ .‬ﺍﻟﻠﻐﺎﺕ ﺍﻟﺮﲰﻴ‪‬ﺔ ﰲ‬
‫ﺍﳌﺴﺎﺑﻘﺔ ﻫﻲ ﺍﻻﳒﻠﻴﺰ ﻳ‪‬ﺔ ﻭﺍﻟﻔﺮﻧﺴﻴ‪‬ﺔ ﻭﺍﻷﳌﺎﻧﻴ‪‬ﺔ ﻭﺍﻟﺮﻭﺳﻴ‪‬ﺔ ﻭﻫﻨﺎﻙ ﺑﻌﺾ ﺍﻟﻠﻐﺎﺕ » ﻏﲑ ﺍﻟﺮ‪‬ﲰﻴ‪‬ﺔ «‬

‫ﻛﺎﻹﺳﺒﺎﻧﻴ‪‬ﺔ ﻭﺍﻟﻌﺮﺑﻴ‪‬ﺔ‪ .‬ﺃﻣ‪‬ﺎ ﺍﻟﻄﻠﺒﺔ ﺍﻟﻌﺮﺏ ﻓﺘﻮﺯ‪‬ﻉ ﻋﻠﻴﻬﻢ ﺍﻷﺳﺌﻠﺔ ﺑﺎﻟﻠﻐﺎﺕ ﺍﻟﻌﺮﺑﻴ‪‬ﺔ ﻭﺍﻹﻧﻜﻠﻴﺰﻳ‪‬ﺔ ﻭﺍﻟﻔﺮﻧﺴﻴ‪‬ﺔ‪.‬‬

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‫ﺍﻟﻜﺘﺎﺏ ﺍﻟﺬﻫﱯ‬

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‫‪ å‬ﳉﻨﺔ ﺍﳊﻜﹼﺎﻡ ﺍﻟﺪﻭﻟﻴ‪‬ﺔ‪ .‬ﻳﺘﺴﻠﹼﻢ ﺃﻋﻀﺎﺀ ﺍﻟﻠﺠﻨﺔ ﺍﻟﻘﺎﺋﻤﺔ ﺍﻟﻘﺼﲑﺓ ﻋﻨﺪ ﻭﺻﻮﳍﻢ ﺇﱃ ﺍﳌﻜﺎﻥ ﺍﳌﻌﺰﻭﻝ‬
‫ﻭﻳ‪‬ﻤﻨﺤﻮﻥ ﺑﻌﺾ ﺍﻟﻮﻗﺖ ﻟﻠﻨﻈﺮ ﰲ ﺍﻷﺳﺌﻠﺔ ﻗﺒﻞ ﺍﻻﺟﺘﻤﺎﻉ ﳌﻨﺎﻗﺸﺔ ﺍﳌﺴﺎﺋﻞ ﺍﻟﱵ ﺳﻴﺠﺮﻱ ﺍﺧﺘﻴﺎﺭﻫﺎ‪.‬‬
‫ﻳﻨﺒﻐﻲ ﻋﻠﻰ ﺍﻟﻠﺠﻨﺔ ﺃﻥ ﺗﺴﺘﺜﲏ ﺃﻱ ﺳﺆﺍﻝ ﻣﻄﺮﻭﺡ ﺳﺎﺑﻘﺎﹰ‪ ،‬ﺃﻭ ﻣﻨﺼﻮﺹ ﻋﻨﻪ ﰲ ﺃﺣﺪ ﺍﻟﻜﺘﺐ‪ ،‬ﺃﻭ‬
‫ﻣﺴﺘﺨﺪﻡ ﰲ ﻓﺘﺮﺓ ﺍﻟﺘﺪﺭﻳﺐ‪ .‬ﲢﺬﻑ ﺑﻌﺾ ﺍﳌﺴﺎﺋﻞ ﻣﺒﺎﺷﺮﺓ ﺇﺫﺍ ﻭ‪‬ﺟﺪﺕ ﺻﻌﺒﺔ ﺟﺪ‪‬ﺍﹰ ﺃﻭ ﺳﻬﻠﺔ ﺟﺪ‪‬ﺍﹰ‪.‬‬
‫ﻭ ﺑﻌﺪ ﺟﺪﺍﻝ ﻗﺪ ﻳﺪﻭﻡ ﻃﻮﻳﻼﹰ ﳚﺮﻱ ﺍﺧﺘﻴﺎﺭ ﺍﳌﺴﺎﺋﻞ ﺍﻟﺼﻌﺒﺔ ﺍﻟﱵ ﲢﻤﻞ ﺍﻷﺭﻗﺎﻡ ‪ 3‬ﻭ ‪ ، 6‬ﺛﱡﻢ ﺍﳌﺴﺎﺋﻞ‬
‫ﺍﻷﻗﻞ ﺻﻌﻮﺑﺔ ﻭﺍﻟﱵ ﲢﻤﻞ ﺍﻷﺭﻗﺎﻡ ‪ 1‬ﻭ ‪ 2‬ﻭ ‪ 4‬ﻭ ‪ ، 5‬ﻭﳚﺮﻱ ﺍﻟﺘﺼﻮﻳﺖ ﻋﻠﻰ ﻛﻞﱢ ﻭﺍﺣﺪﺓ ﻣﻨﻬﺎ‪ .‬ﻭﺃﺧﲑﺍﹰ‬
‫ﻳ‪‬ﺘﺮ ﺟﹺﻢ‪ ‬ﻗﺎﺩﺓ ﺍﻟﻔ‪‬ﺮ‪‬ﻕ‪ ،‬ﺍﻟﱵ ﻳﺘﻄﻠﹼﺐ ﻃﻼﹼ‪‬ﺎ ﻟﻐﺎﺕ ﺃﺧﺮﻯ‪ ،‬ﺍﻷﺳﺌﻠﺔ ﺇﱃ ﻟﻐﺎ‪‬ﻢ‪ ،‬ﻭﳚﺮﻱ ﺗﺪﻗﻴﻖ ﲨﻴﻊ‬
‫ﺍﻟﺘﺮﲨﺎﺕ ﻣﻦ ﻗ‪‬ﺒ‪‬ﻞ ﲨﻴﻊ ﺃﻋﻀﺎﺀ ﺍﻟﻠﺠﻨﺔ‪ ،‬ﻟﻠﺘﻴﻘﹼﻦ ﻣﻦ ﺣ‪‬ﺴﻦ ﺍﻟﺘﺮﲨﺔ‪.‬‬
‫‪ å‬ﺍﳌﺴﺎﺑﻘﺔ‪ .‬ﻳﺼﻞ ﺍﳌﺘﺴﺎﺑﻘﻮﻥ ﺇﱃ ﺍﻟﺒﻠﺪ ﺍﳌﹸﻀﻴﻒ ﻗﺒﻞ ﻣﻮﻋﺪ ﺍﳌﺴﺎﺑﻘﺔ ﺑﻌﺪ‪‬ﺓ ﺃﻳﺎﻡ‪ ،‬ﳌﻨﺤﻬﻢ ﺍﻟﻮﻗﺖ‬
‫ﺍﻟﻼﺯﻡ ﻟﻠﺘﺄﻗﻠﻢ‪ .‬ﻭﺗﺘﻜﻮ‪‬ﻥ ﺍﳌﺴﺎﺑﻘﺔ ﻣﻦ ﻭﺭﻗﺘﲔ‪ ،‬ﺗﺸﻤﻞ ﻛﻞﹼ ﻭﺍﺣﺪﺓ ﻣﻨﻬﻤﺎ ﺛﻼﺛﺔ ﺃﺳﺌﻠﺔ‪ ،‬ﳚﺐ ﺍﻹﺟﺎﺑﺔ‬
‫ﻋﻨﻬﺎ ﰲ ﺃﺭﺑﻊ ﺳﺎﻋﺎﺕ ﻭﻧﺼﻒ ﺍﻟﺴﺎﻋﺔ‪ .‬ﺍﻷﻭﱃ ﰲ ﺍﻟﻴﻮﻡ ﺍﻷﻭ‪‬ﻝ ﻭﺍﻟﺜﺎﻧﻴﺔ ﰲ ﺍﻟﻴﻮﻡ ﺍﻟﺬﻱ ﻳﻠﻴﻪ‪ .‬ﺗﻘﻠﻴﺪﻳﺎﹰ‪،‬‬
‫ﺍﻟﺴﺆﺍﻝ ﺭﻗﻢ ‪ 1‬ﻫﻮ ﺍﻷﺳﻬﻞ‪ ،‬ﻭﺍﻟﺴﺆﺍﻝ ﺭﻗﻢ ‪ 6‬ﻫﻮ ﺍﻷﺻﻌﺐ‪ .‬ﻭﰲ ﺍﻟﻮﻗﺖ ﺍﻟﺬﻱ ﳚﺮﻱ ﻓﻴﻪ ﺗﺼﺤﻴﺢ‬
‫ﺍﻷﻭﺭﺍﻕ‪ ،‬ﳜﻀﻊ ﺍﳌﺘﺴﺎﺑﻘﻮﻥ ﻟﱪﻧﺎﻣﺞ ﺗﺮﻓﻴﻬﻲ ﻭﺛﻘﺎﰲ ﻭﺳﻴﺎﺣﻲ ﻳﻬﺪﻑ ﺇﱃ ﺗﻌﺮ‪‬ﻑ ﺍﻟﺒﻠﺪ ﺍﳌﹸﻀﻴﻒ‪.‬‬
‫‪ å‬ﺗﺼﺤﻴﺢ ﺍﻷﻭﺭﺍﻕ‪ .‬ﻧﻈﺮﺍﹰ ﺇﱃ ﺗﻌﺪ‪‬ﺩ ﻟﻐﺎﺕ ﺍﻟﺒﻠﺪﺍﻥ ﺍﳌﺸﺎﺭﻛﺔ‪ ،‬ﻳﺼﺤ‪‬ﺢ ﺑﺪﺍﻳﺔﹰ ﻗﺎﺩ ﺓﹸ ﻛﻞﱢ ﻓﺮﻳﻖ‬
‫ﺃﻭﺭﺍﻕ‪ ‬ﻓﺮﻳﻘﻬﻢ‪ ،‬ﻭﻟﻜﻦ ﺩﻭﻥ ﺃﻥ ﻳﻀﻌﻮﺍ ﺃﻱ‪ ‬ﺩﺭﺟﺎﺕ ﻋﻠﻰ ﺃﻭﺭﺍﻕ ﺍﻹﺟﺎﺑﺔ‪ .‬ﺛﹸﻢ‪ ‬ﻳﺴﻠﹼﻤﻮﻥ ﺍﻷﻭﺭﺍﻕ‪ ،‬ﺑﻌﺪ‬
‫ﺗﺮﲨﺔ ﺑﻌﺾ ﺍﻷﺟﺰﺍﺀ ﺍﻟﱵ ﺗﺘﻄﻠﹼﺐ ﺍﻟﺘﺮﲨﺔ ﻣﻨﻬﺎ‪ ،‬ﺇﱃ ﻓﺮﻳﻖ ﻭﺍﺿﻌﻲ ﺍﻟﺪﺭﺟﺎﺕ‪ ،‬ﺍﳌﺴﻤ‪‬ﻰ ﻓﺮﻳﻖ ﺍﳌﹸﻨﺴ‪‬ﻘﲔ‪،‬‬
‫ﺍﻟﺬﻳﻦ ﻳ‪‬ﻌﻴ‪‬ﻨﻬﻢ ﺍﻟﺒﻠﺪ ﺍﳌﹸﻀﻴﻒ‪ .‬ﻭﰲ ﺍﻟﻨﻬﺎﻳﺔ ﳚﺐ ﺃﻥ ﻳﺘﻔﹼﻖ ﻗﺎﺩﺓ ﺍﻟﻔ‪‬ﺮ‪‬ﻕ ﻣﻊ ﺍﳌﻨﺴ‪‬ﻘﲔ ﻋﻠﻰ ﺍﻟﺪﺭﺟﺎﺕ ﺍﻟﱵ‬
‫ﺗ‪‬ﻤﻨﺢ ﻟﻜﻞﹼ ﻭﺭﻗﺔ ﺇﺟﺎﺑﺔ‪ ،‬ﻓﺘﺴﺠ‪‬ﻞ ﻫﺬﻩ ﺍﻟﺪﺭﺟﺎﺕ ﰲ ﺍﻟﺘﻘﺮﻳﺮ ﺍﻟﺮﲰﻲ ﺍﻟﺬﻱ ﻳﻮﻗﹼﻌﻪ ﺍﻟﻔﺮﻳﻘﺎﻥ‪ .‬ﻭﺇﺫﺍ ﻭﻗﻊ‬
‫ﺧﻼ ﻑ‪ ‬ﻳ‪‬ﺤﺎﻭﻝ ﻗﺎﺋﺪ ﻓﺮﻳﻖ ﺍﳌﻨﺴ‪‬ﻘﲔ ﺍﻟﺘﻮﺳ‪‬ﻂ ﻟﻠﺘﻮﺻ‪‬ﻞ ﺇﱃ ﺍﺗﻔﺎﻕ‪ ،‬ﻭﺇﺫﺍ ﺑﻘﻲ ﺍﳋﻼﻑ ﻗﺎﺋﻤﺎﹰ‪ ،‬ﺗ‪‬ﻌﺮﺽ‬
‫ﺍﳌﺸﻜﻠﺔ ﻋﻠﻰ ﻛﺎﻣﻞ ﳉﻨﺔ ﺍﻟﺘﺤﻜﻴﻢ ﺍﻟﺪﻭﻟﻴ‪‬ﺔ ﻟﻴ‪‬ﺘ‪‬ﺨﺬ ﺍﻟﻘﺮﺍﺭ ﻓﻴﻬﺎ ﺑﺎﻷﻛﺜﺮﻳﺔ‪.‬‬
‫‪ å‬ﺍﻟﻨﺘﺎﺋﺞ‪ .‬ﺇﻥﹼ ﺍﻷﻭﳌﺒﻴﺎﺩ ﺍﻟﻌﺎﳌﻲ ﻟﻠﺮﻳﺎﺿﻴ‪‬ﺎﺕ ﻣﺴﺎﺑﻘﺔ ﻓﺮﺩﻳ‪‬ﺔ‪ ،‬ﻳﺘﺒﺎﺭﻯ ﻓﻴﻬﺎ ﺍﻷﻓﺮﺍﺩ ﻭﻟﻴﺲ ﺍﻟﻔ‪‬ﺮ‪‬ﻕ‪.‬‬
‫ﻳ‪‬ﻤﻨﺢ ﻓﻴﻪ ﻧﺼﻒ ﺍﳌﺘﺴﺎﺑﻘﲔ ﺗﻘﺮﻳﺒﺎﹰ ﻣﻴﺪﺍﻟﻴ‪‬ﺎﺕ ﺫﻫﺒﻴ‪‬ﺔ ﻭﻓﻀﻴ‪‬ﺔ ﻭﺑﺮﻭﻧﺰ ﻳ‪‬ﺔ ﺗﺘﻨﺎﺳﺐ ﺃﻋﺪﺍﺩﻫﺎ ﻣﻊ ‪ 1‬ﻭ ‪2‬‬

‫ﻭ ‪ ، 3‬ﻋﻠﻰ ﺃﻻﹼ ﺗﺰﻳﺪ ﻧﺴﺒﺔ ﺍﳌﺘﺴﺎﺑﻘﲔ ﺍﻟﺬﻳﻦ ﳛﺼﻠﻮﻥ ﻋﻠﻰ ﻣﻴﺪﺍﻟﻴ‪‬ﺎﺕ ﺫﻫﺒﻴ‪‬ﺔ ﻋﻠﻰ ‪ ، 121‬ﻭﺃﻻﹼ ﺗﺰﻳﺪ ﻧﺴﺒﺔ‬
‫ﺍﳌﺘﺴﺎﺑﻘﲔ ﺍﻟﺬﻳﻦ ﳛﺼﻠﻮﻥ ﻋﻠﻰ ﻣﻴﺪﺍﻟﻴ‪‬ﺎﺕ ﺫﻫﺒﻴ‪‬ﺔ ﺃﻭ ﻓﻀﻴ‪‬ﺔ ﻋﻠﻰ ‪ ، 14‬ﻭﺃﻻﹼ ﺗﺰﻳﺪ ﻧﺴﺒﺔ ﺍﳌﺘﺴﺎﺑﻘﲔ ﺍﻟﺬﻳﻦ‬
‫ﳛﺼﻠﻮﻥ ﻋﻠﻰ ﻣﻴﺪﺍﻟﻴ‪‬ﺔ ﻣﺎ ﻋﻠﻰ ‪ ، 12‬ﻭ ﻟﺘﺸﺠﻴﻊ ﺍﳌﺘﺴﺎﺑﻘﲔ ﻋﻠﻰ ﺍﳌﺜﺎﺑﺮﺓ ﻭﺇﳒﺎﺯ ﺍﳊﻠﻮﻝ ﺗ‪‬ﻤﻨﺢ ﺷﻬﺎﺩﺓ‬
‫ﺗﻘﺪﻳﺮ ﺇﱃ ﺍﳌﺘﺴﺎﺑﻘﲔ ﺍﻟﺬﻳﻦ ﳛﻠﹼﻮﻥ ﻭﺍﺣﺪﺓ ﻣﻦ ﺍﳌﺴﺎﺋﻞ ﺣﻼﹰ ﻛﺎﻣﻼﹰ ﺩﻭﻥ ﺃﻥ ﳛﺼﻠﻮﺍ ﻋﻠﻰ ﻣﻴﺪﺍﻟﻴﺔ‪.‬‬

‫ﺍﻷﻭﳌﺒﻴﺎﺩ ﺍﻟﻌﺎﳌﻲ ﻟﻠﺮﻳﺎﺿﻴ‪‬ﺎﺕ‬

‫‪iv‬‬



‫ﺗ‪‬ﺸﺠ‪‬ﻊ ﺍﳌﺸﺎﺭﻛﺔ ﰲ ﻫﺬﻩ ﺍﻟﺘﻈﺎﻫﺮﺓ ﺍﻟﺒ‪‬ﻠﺪﺍﻥ ﻋﻠﻰ ﻭﺿﻊ ﺁﻟﻴ‪‬ﺎﺕ ﻟﺘﺪﺭﻳﺐ ﺍﻟﻔ‪‬ﺮ‪‬ﻕ‪ ،‬ﻭﺍﺧﺘﻴﺎﺭ ﻫﺎ‬
‫ﺑﻄﺮﺍﺋﻖ ﺗﻨﺎﻓﺴﻴ‪‬ﺔ ﲢﻔﹼﺰ ﺍﻟﻄﻼﺏ‪ ،‬ﻷﻥﹼ ﺫﻟﻚ ﻳﻨﻌﻜﺲ ﺇﳚﺎﺑﺎﹰ‪ ،‬ﻟﻴﺲ ﻓﻘﻂ ﻋﻠﻰ ﺍﻟﻔﺌﺔ ﺍﳌﻌﻨﻴ‪‬ﺔ ﺑﺎﳌﺸﺎﺭﻛﺔ ﰲ‬
‫ﺍﳌﺴﺎﺑﻘﺔ ﻓﺤﺴﺐ‪ ،‬ﺑﻞ ﻋﻠﻰ ﻛﺎﻣﻞ ﺃﺳﺎﻟﻴﺐ ﺗﺪﺭﻳﺲ ﺍﻟﺮﻳﺎﺿﻴ‪‬ﺎﺕ ﺧﺼﻮﺻﺎﹰ ﻭﺍﻟﻌﻠﻮﻡ ﻋﻤﻮﻣﺎﹰ‪.‬‬
‫ﻟﺬﻟﻚ ﺭﺃﻳﺖ‪ ‬ﺃﻥ ﺃﲨﻊ ﰲ ﻛﺘﺎ ﺏﹴ ﻭﺍﺣﺪ‪ ‬ﻭﺑﺎﻟﻠﻐﺔ ﺍﻟﻌﺮﺑﻴ‪‬ﺔ‪ ،‬ﲨﻴﻊ ﺍﳌﺴﺎﺋﻞ ﺍﻟﱵ ﻃﹸﺮﹺ ﺣﺖ‪ ‬ﰲ‬
‫ﺍﻷﻭﳌﺒﻴﺎﺩ ﺍﻟﻌﺎﳌﻲ ﻟﻠﺮﻳﺎﺿﻴ‪‬ﺎﺕ‪ ،‬ﻣﻨﺬ ﺍﻟﺒﺪﺍﻳﺔ ﻭﺣﱴ ﺗﺎﺭﻳﺦ ﺻﺪﻭﺭ ﻫﺬﺍ ﺍﻟﻜﺘﺎﺏ‪ ،‬ﻭﺃﻥ ﺃﻋﺮﺽ‪ ‬ﺣﻠﻮﻻﹰ ﳍﺬﻩ‬
‫ﺍﳌﺴﺎﺋﻞ‪ .‬ﺭﺍﺟﻴﺎﹰ ﺃﻥ ﳚﺪ ﺍﻟﻘﺎﺭﺉ ﺍﳌﻬﺘﻢ‪ ‬ﻭﺍﶈﺐ‪ ‬ﻟﻠﺮﻳﺎﺿﻴﺎﺕ ﻓﺎﺋﺪﺓ ﰲ ﺫﻟﻚ‪.‬‬
‫ﻛﻤﺎ ﳋﹼﺼﺖ‪ ‬ﰲ ﻣ‪‬ﻠﺤﻖﹴ‪ ،‬ﳎﻤﻮﻋﺔﹰ ﻣﻦ ﺍﳌﻌﺎﺭﻑ ﻭﺍﳋﻮﺍﺹ ﺍﻟﱵ ﺍﻓﺘﺮﺿﺖ‪ ‬ﰲ ﺍﻟﻘﺎﺭﺉ ﻣﻌﺮﻓﺘﻬﺎ‪.‬‬
‫ﻭﺃﺧﲑﺍﹰ‪ ،‬ﺑﻐﻴﺔ ﺍﻛﺘﻤﺎﻝ ﺍﻟﻔﺎﺋﺪﺓ ﻣﻦ ﻫﺬﺍ ﺍﻟﻜﺘﺎﺏ‪ ،‬ﺃﻭﺻﻲ ﺍﻟﻘﺎﺭﺉ ﺍﻟﻜﺮﱘ ﺑﻌﺪﻡ ﺍﻟﺘﺴﺮ‪‬ﻉ ﰲ ﻗﺮﺍﺀﺓ ﺍﳊﻠﻮﻝ‬
‫ﺍﳌﻘﺘﺮﺣﺔ ﻟﻠﻤﺴﺎﺋﻞ ﺑﻞ ﻣ‪‬ﻘﺎﺭﻋﺘﻬﺎ ﺗﻔﻜﲑﺍﹰ ﻭﲝﺜﺎﹰ‪ .‬ﻭﻋﻨﺪﻫﺎ ﺳﻴﺠﺪ ﺍﻟﻘﺎﺭﺉ ﻣﺘﻌﺔ ﰲ ﺩﺭﺍﺳﺔ ﻫﺬﺍ ﺍﻟﻜﺘﺎﺏ‬
‫ﻭﰲ ﺇﳚﺎﺩ ﺣﻠﻮﻝ ﺃﺧﺮﻯ ﻏﲑ ﺍﻟﱵ ﺍﺧﺘﺮﺗ‪‬ﻬﺎ‪ ،‬ﻭﰲ ﺫﻟﻚ ﺟ‪‬ﻞﹼ ﺍﻟﻔﺎﺋﺪﺓ ﻭﺍﳌﺘﻌﺔ‪.‬‬
‫ﻭﺃﺧﲑﺍﹰ ﺃﺧﺘﻢ ﻫﺬﻩ ﺍﳌﻘﺪ‪‬ﻣﺔ ﺑﺘﻘﺪﱘ ﺍﻟﺸﻜﺮﺍﻟﻌﻤﻴﻖ ﺇﱃ ﺍﻟﺴﻴ‪‬ﺪ ﺍﻟﺮﺋﻴﺲ ﺑﺸﺎﺭ ﺍﻷﺳﺪ ﻭﺇﱃ ﺍﻟﺴﻴ‪‬ﺪﺓ‬
‫ﻋﻘﻴﻠﺘﻪ ﻟﻼﻫﺘﻤﺎﻡ ﺍﻟﺒﺎﻟﻎ ﺍﻟﺬﻱ ﺃﻭﻟﻴﺎﻩ ﻟﻌﻤﻠﻴ‪‬ﺔ ﺇﻋﺪﺍﺩ ﺍﻟﻔﺮﻳﻖ ﺍﻟﻮﻃﲏ ﰲ ﺍﳉﻤﻬﻮﺭﻳ‪‬ﺔ ﺍﻟﻌﺮﺑﻴ‪‬ﺔ ﺍﻟﺴﻮﺭﻳ‪‬ﺔ‪ ،‬ﻭﺍﻟﺬﻱ‬
‫ﺗﺘﻮ‪‬ﺝ ﺑﺈﻧﺸﺎﺀ ﺍﳍﻴﺌﺔ ﺍﻟﻮﻃﻨﻴﺔ ﻟﻸﳌﺒﻴﺎﺩ ﺍﻟﻌﻠﻤﻲ ﺍﻟﺴﻮﺭﻱ‪ ،‬ﻭﻟﺘﺸﺠﻴﻌﻬﻤﺎ ﺍﳊﺜﻴﺚ ﻋﻠﻰ ﺍﳌﺸﺎﺭﻛﺔ ﰲ ﻫﺬﻩ‬
‫ﺍﳌﺴﺎﺑﻘﺔ ﺍﻟﻌﺎﳌﻴ‪‬ﺔ‪.‬‬
‫ﺩﻣﺸﻖ ﰲ ‪ 7‬ﺃﻳﻠﻮﻝ ‪.2010‬‬

‫ﺍﻟﺪﻛﺘﻮﺭ ﻋﻤﺮﺍﻥ ﻗﻮﺑﺎ‬

‫ﺍﻟﻜﺘﺎﺏ ﺍﻟﺬﻫﱯ‬

‫‪v‬‬

‫ ‬
‫ﺗﻘﺪﱘ‪i .................................................................................................................................................‬‬
‫ﺍﶈﺘﻮﻯ ‪v ..............................................................................................................................................‬‬

‫ﺍﻷﻭﳌﺒﻴﺎﺩ ﺍﻷﻭ‪‬ﻝ‬
‫ﺍﻷﻭﳌﺒﻴﺎﺩ ﺍﻟﺜﺎﱐ‬
‫ﺍﻷﻭﳌﺒﻴﺎﺩ ﺍﻟﺜﺎﻟﺚ‬
‫ﺍﻷﻭﳌﺒﻴﺎﺩ ﺍﻟﺮﺍﺑﻊ‬
‫ﺍﻷﻭﳌﺒﻴﺎﺩ ﺍﳋﺎﻣﺲ‬
‫ﺍﻷﻭﳌﺒﻴﺎﺩ ﺍﻟﺴﺎﺩﺱ‬
‫ﺍﻷﻭﳌﺒﻴﺎﺩ ﺍﻟﺴﺎﺑﻊ‬
‫ﺍﻷﻭﳌﺒﻴﺎﺩ ﺍﻟﺜﺎﻣﻦ‬
‫ﺍﻷﻭﳌﺒﻴﺎﺩ ﺍﻟﺘﺎﺳﻊ‬
‫ﺍﻷﻭﳌﺒﻴﺎﺩ ﺍﻟﻌﺎﺷﺮ‬
‫ﺍﻷﻭﳌﺒﻴﺎﺩ ﺍﳊﺎﺩﻱ ﻋﺸﺮ‬
‫ﺍﻷﻭﳌﺒﻴﺎﺩ ﺍﻟﺜﺎﱐ ﻋﺸﺮ‬
‫ﺍﻷﻭﳌﺒﻴﺎﺩ ﺍﻟﺜﺎﻟﺚ ﻋﺸﺮ‬
‫ﺍﻷﻭﳌﺒﻴﺎﺩ ﺍﻟﺮﺍﺑﻊ ﻋﺸﺮ‬
‫ﺍﻷﻭﳌﺒﻴﺎﺩ ﺍﳋﺎﻣﺲ ﻋﺸﺮ‬
‫ﺍﻷﻭﳌﺒﻴﺎﺩ ﺍﻟﺴﺎﺩﺱ ﻋﺸﺮ‬
‫ﺍﻷﻭﳌﺒﻴﺎﺩ ﺍﻟﺴﺎﺑﻊ ﻋﺸﺮ‬
‫ﺍﻷﻭﳌﺒﻴﺎﺩ ﺍﻟﺜﺎﻣﻦ ﻋﺸﺮ‬
‫ﺍﻷﻭﳌﺒﻴﺎﺩ ﺍﻟﺘﺎﺳﻊ ﻋﺸﺮ‬
‫ﺍﻷﻭﳌﺒﻴﺎﺩ ﺍﻟﻌﺸﺮﻭﻥ‬
‫ﺍﻷﻭﳌﺒﻴﺎﺩ ﺍﳊﺎﺩﻱ ﻭﺍﻟﻌﺸﺮﻭﻥ‬
‫ﺍﻷﻭﳌﺒﻴﺎﺩ ﺍﻟﺜﺎﱐ ﻭﺍﻟﻌﺸﺮﻭﻥ‬
‫ﺍﻷﻭﳌﺒﻴﺎﺩ ﺍﻟﺜﺎﻟﺚ ﻭﺍﻟﻌﺸﺮﻭﻥ‬
‫ﺍﻷﻭﳌﺒﻴﺎﺩ ﺍﻟﺮﺍﺑﻊ ﻭﺍﻟﻌﺸﺮﻭﻥ‬

‫‪1 ................................................................................... 1959‬‬
‫‪9 .................................................................................. 1960‬‬
‫‪17 ................................................................................. 1961‬‬
‫‪25 ................................................................................. 1962‬‬
‫‪37 ................................................................................. 1963‬‬
‫‪47 ................................................................................. 1964‬‬
‫‪55 ................................................................................. 1965‬‬
‫‪65 ................................................................................. 1966‬‬
‫‪71 ................................................................................. 1967‬‬
‫‪79 ................................................................................. 1968‬‬
‫‪85 ................................................................................. 1969‬‬
‫‪95 ................................................................................. 1970‬‬
‫‪101 ............................................................................... 1971‬‬
‫‪113 ............................................................................... 1972‬‬
‫‪121 ............................................................................... 1973‬‬
‫‪129 ............................................................................... 1974‬‬
‫‪137 ............................................................................... 1975‬‬
‫‪145 ............................................................................... 1976‬‬
‫‪153 ............................................................................... 1977‬‬
‫‪163 ............................................................................... 1978‬‬
‫‪177 ............................................................................... 1979‬‬
‫‪187 ............................................................................... 1981‬‬
‫‪197 ............................................................................... 1982‬‬
‫‪207 ............................................................................... 1983‬‬

‫‪This book is downloaded from this site:‬‬
‫‪www.syCourses.com‬‬

‫‪vi‬‬

‫ﺍﻷﻭﳌﺒﻴﺎﺩ ﺍﻟﻌﺎﳌﻲ ﻟﻠﺮﻳﺎﺿﻴ‪‬ﺎﺕ‬

‫ﺍﻷﻭﳌﺒﻴﺎﺩ ﺍﳋﺎﻣﺲ ﻭﺍﻟﻌﺸﺮﻭﻥ‬
‫ﺍﻷﻭﳌﺒﻴﺎﺩ ﺍﻟﺴﺎﺩﺱ ﻭﺍﻟﻌﺸﺮﻭﻥ ‪225 ............................................................................... 1985‬‬
‫ﺍﻷﻭﳌﺒﻴﺎﺩ ﺍﻟﺴﺎﺑﻊ ﻭﺍﻟﻌﺸﺮﻭﻥ ‪235 ............................................................................... 1986‬‬
‫ﺍﻷﻭﳌﺒﻴﺎﺩ ﺍﻟﺜﺎﻣﻦ ﻭﺍﻟﻌﺸﺮﻭﻥ‬
‫‪243 ............................................................................... 1987‬‬
‫ﺍﻷﻭﳌﺒﻴﺎﺩ ﺍﻟﺘﺎﺳﻊ ﻭﺍﻟﻌﺸﺮﻭﻥ ‪251 ............................................................................... 1988‬‬
‫ﺍﻷﻭﳌﺒﻴﺎﺩ ﺍﻟﺜﻼﺛﻮﻥ‬
‫‪265 ............................................................................... 1989‬‬
‫ﺍﻷﻭﳌﺒﻴﺎﺩ ﺍﳊﺎﺩﻱ ﻭﺍﻟﺜﻼﺛﻮﻥ ‪275 ............................................................................... 1990‬‬
‫ﺍﻷﻭﳌﺒﻴﺎﺩ ﺍﻟﺜﺎﱐ ﻭﺍﻟﻌﺸﺮﻭﻥ‬
‫‪291 ............................................................................... 1991‬‬
‫ﺍﻷﻭﳌﺒﻴﺎﺩ ﺍﻟﺜﺎﻟﺚ ﻭﺍﻟﺜﻼﺛﻮﻥ‬
‫‪301 ............................................................................... 1992‬‬
‫‪217 ............................................................................... 1984‬‬

‫ﺍﻷﻭﳌﺒﻴﺎﺩ ﺍﻟﺮﺍﺑﻊ ﻭﺍﻟﺜﻼﺛﻮﻥ‬
‫ﺍﻷﻭﳌﺒﻴﺎﺩ ﺍﳋﺎﻣﺲ ﻭﺍﻟﺜﻼﺛﻮﻥ‬
‫ﺍﻷﻭﳌﺒﻴﺎﺩ ﺍﻟﺴﺎﺩﺱ ﻭﺍﻟﺜﻼﺛﻮﻥ‬
‫ﺍﻷﻭﳌﺒﻴﺎﺩ ﺍﻟﺴﺎﺑﻊ ﻭﺍﻟﺜﻼﺛﻮﻥ‬
‫ﺍﻷﻭﳌﺒﻴﺎﺩ ﺍﻟﺜﺎﻣﻦ ﻭﺍﻟﺜﻼﺛﻮﻥ‬
‫ﺍﻷﻭﳌﺒﻴﺎﺩ ﺍﻟﺘﺎﺳﻊ ﻭﺍﻟﺜﻼﺛﻮﻥ‬
‫ﺍﻷﻭﳌﺒﻴﺎﺩ ﺍﻷﺭﺑﻌﻮﻥ‬
‫ﺍﻷﻭﳌﺒﻴﺎﺩ ﺍﳊﺎﺩﻱ ﻭﺍﻷﺭﺑﻌﻮﻥ‬
‫ﺍﻷﻭﳌﺒﻴﺎﺩ ﺍﻟﺜﺎﱐ ﻭﺍﻷﺭﺑﻌﻮﻥ‬
‫ﺍﻷﻭﳌﺒﻴﺎﺩ ﺍﻟﺜﺎﻟﺚ ﻭﺍﻷﺭﺑﻌﻮﻥ‬
‫ﺍﻷﻭﳌﺒﻴﺎﺩ ﺍﻟﺮﺍﺑﻊ ﻭﺍﻷﺭﺑﻌﻮﻥ‬
‫ﺍﻷﻭﳌﺒﻴﺎﺩ ﺍﳋﺎﻣﺲ ﻭﺍﻷﺭﺑﻌﻮﻥ‬
‫ﺍﻷﻭﳌﺒﻴﺎﺩ ﺍﻟﺴﺎﺩﺱ ﻭﺍﻷﺭﺑﻌﻮﻥ‬
‫ﺍﻷﻭﳌﺒﻴﺎﺩ ﺍﻟﺴﺎﺑﻊ ﻭﺍﻷﺭﺑﻌﻮﻥ‬
‫ﺍﻷﻭﳌﺒﻴﺎﺩ ﺍﻟﺜﺎﻣﻦ ﻭﺍﻷﺭﺑﻌﻮﻥ‬
‫ﺍﻷﻭﳌﺒﻴﺎﺩ ﺍﻟﺘﺎﺳﻊ ﻭﺍﻷﺭﺑﻌﻮﻥ‬
‫ﺍﻷﻭﳌﺒﻴﺎﺩ ﺍﳋﻤﺴﻮﻥ‬
‫ﺍﻷﻭﳌﺒﻴﺎﺩ ﺍﳊﺎﺩﻱ ﻭﺍﳋﻤﺴﻮﻥ‬

‫‪315 ............................................................................... 1993‬‬
‫‪327 ............................................................................... 1994‬‬
‫‪335 ............................................................................... 1995‬‬
‫‪345 ............................................................................... 1996‬‬
‫‪355 ............................................................................... 1997‬‬
‫‪369 ............................................................................... 1998‬‬
‫‪381 ............................................................................... 1999‬‬
‫‪391 ............................................................................... 2000‬‬
‫‪407 ............................................................................... 2001‬‬
‫‪417 ............................................................................... 2002‬‬
‫‪431 ............................................................................... 2003‬‬
‫‪441 ............................................................................... 2004‬‬
‫‪455 ............................................................................... 2005‬‬
‫‪465 ............................................................................... 2006‬‬
‫‪477 ............................................................................... 2007‬‬
‫‪489 ............................................................................... 2008‬‬
‫‪503 ............................................................................... 2009‬‬
‫‪513 ............................................................................... 2010‬‬

‫ﻣﻠﺤﻖ ﺃﻭ‪‬ﻝ‪ .‬ﺑﻌﺾ ﺍﳌﻌﺎﺭﻑ ﻭﺍﳌﺮﺍﺟﻊ ﺍﳌﻔﻴﺪﺓ‪........................................................... .‬‬

‫‪525‬‬

‫ﻣﻠﺤﻖ ﺛﺎﻥ‪ .‬ﺑﻌﺾ ﺍﻟﺮﻣﻮﺯ ﺍﳌﺴﺘﺨﺪﻣﺔ‪...................................................................‬‬

‫‪531‬‬

‫ﻣﻠﺤﻖ ﺛﺎﻟﺚ‪ .‬ﺃﻳﻦ ﻭﻣﱴ ﻭﻧﻮﻉ ﺍﳌﺴﺎﺋﻞ ﺍﳌﻄﺮﻭﺣﺔ‪....................................................... .‬‬

‫‪533‬‬

‫ّ‬
‫ّ‬
‫ﺃﻭﳌﺒﻴﺎﺩ ﺍﻟﺮﻳﺎﺿﻴﺎﺕ ﺍﻷﻭﻝ‬
‫‪21n + 4‬‬
‫ ﺃﺛﺒﺖ ﺃﻥﹼ ﺍﻟﻜﺴﺮ‬
‫‪14n + 3‬‬

‫ﻻ ﻳﻘﺒﻞ ﺍﻻﺧﺘﺰﺍﻝ ﻣﻬﻤﺎ ﻛﺎﻧﺖ ﻗﻴﻤﺔ ﺍﻟﻌﺪﺩ ﺍﻟﺼﺤﻴﺢ ‪. n‬‬

‫ ﰲ ﺍﳊﻘﻴﻘﺔ‪ ،‬ﳚﺐ ﺇﺛﺒﺎﺕ ﺃ ﻥﹼ ﺍﻟﻌﺪﺩﻳﻦ ‪ 21n + 4‬ﻭ ‪ 14n + 3‬ﺃ ﻭ‪‬ﻟﻴ‪‬ﺎﻥ ﻓﻴﻤﺎ ﺑﻴﻨﻬﻤﺎ‪ ،‬ﻭﺫﻟﻚ ﻣﻬﻤﺎ‬
‫ﻛﺎﻧﺖ ﻗﻴﻤﺔ ﺍﻟﻌﺪﺩ ﺍﻟﺼﺤﻴﺢ ‪ . n‬ﻭﻫﺬﺍ ﻳﻨﺘﺞ ﻭﺿﻮﺣﺎﹰ ﻣﻦ ﺍﳌﺘﻄﺎﺑﻘﺔ ﺍﻟﺘﺎﻟﻴﺔ ‪:‬‬
‫‪3 ( 14n + 3 ) − 2 ( 21n + 4 ) = 1‬‬

‫‪ù‬‬

‫ ‬
‫ ﺣﻞﹼ ﰲ ‪ ℝ‬ﺍﳌﻌﺎﺩﻟﺔ ‪ ، x + 2x − 1 + x − 2x − 1 = A :‬ﺇﺫ ﻧﻔﺘﺮﺽ ﺃ ﻥﹼ ﲨﻴﻊ‬

‫ﺍﳉﺬﻭﺭ ﺍﻟﺘﺮﺑﻴﻌﻴ‪‬ﺔ ﻫﻲ ﺟﺬﻭﺭ ﺗﺮﺑﻴﻌﻴ‪‬ﺔ ﻷﻋﺪﺍﺩ ﻣﻮﺟﺒﺔ‪ .‬ﻭﺫﻟﻚ ﰲ ﺍﳊﺎﻻﺕ ﺍﻟﺘﺎﻟﻴﺔ ‪:‬‬
‫ ‬

‫‪.A = 2‬‬

‫ ‬

‫‪.A = 1‬‬

‫ ‬

‫‪.A = 2‬‬

‫ ﰲ ﺍﳊﻘﻴﻘﺔ‪ ،‬ﻟﻨﻼﺣﻆ ﺃﻥﹼ‬
‫‪1‬‬
‫‪2‬‬
‫) ‪x + 2x − 1 = ( 1 + 2x − 1‬‬
‫‪2‬‬
‫‪1‬‬
‫‪2‬‬
‫) ‪x − 2x − 1 = ( 1 − 2x − 1‬‬
‫‪2‬‬

‫ﻭﻋﻠﻴﻪ ﺗﻜﻮﻥ ﲨﻴﻊ ﺍﳉﺬﻭﺭ ﺍﻟﺘﺮﺑﻴﻌﻴ‪‬ﺔ ﰲ ﺍﳌﻌﺎﺩﻟﺔ ﺍﳌﻌﻄﺎﺓ ﺟﺬﻭﺭﺍﹰ ﺗﺮﺑﻴﻌﻴ‪‬ﺔ ﻷﻋﺪﺍﺩ ﻣﻮﺟﺒﺔ ﺇﺫﺍ ﻭﻓﻘﻂ ﺇﺫﺍ‬
‫‪1‬‬
‫ﲢﻘﹼﻘﺖ ﺍﳌﺘﺮﺍﺟﺤﺔ‬
‫‪2‬‬

‫≥ ‪ . x‬ﻭﻋﻨﺪﺋﺬ ﻳﻜﻮﻥ ﻟﺪﻳﻨﺎ‬

‫‪2‬‬
‫‪2‬‬
‫) ‪( 1 + 2x − 1 ) + ( 1 − 2x − 1‬‬

‫= ‪x + 2x − 1 + x − 2x − 1‬‬

‫‪2‬‬

‫ﺃﻭ‬

‫‪1 + 2x − 1 + 1 − 2x − 1‬‬
‫‪2‬‬

‫= ‪x + 2x − 1 + x − 2x − 1‬‬

‫ﻭﻣﻦ ﺛﹶﻢ‪‬‬

‫) ‪2 max ( 1, 2x − 1‬‬

‫= ‪x + 2x − 1 + x − 2x − 1‬‬

‫ﻭﻟﻜﻦ‪ ‬ﺍﻟﺘﺎﺑﻊ ‪ x ֏ 2x − 1‬ﻣﺘﺰﺍﻳﺪ‪ ‬ﲤﺎﻣﺎﹰ ﻋﻠﻰ ﺍ‪‬ﺎﻝ [ ∞‪ [ 21 , +‬ﻭﻳﺄﺧﺬ ﺍﻟﻘﻴﻤﺔ ‪ 1‬ﻋﻨﺪ ‪. 1‬‬
‫‪1‬‬

‫‪This book is downloaded from this site:‬‬
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‫ﺍﻷﻭﳌﺒﻴﺎﺩ ﺍﻟﻌﺎﳌﻲ ﻟﻠﺮﻳﺎﺿﻴ‪‬ﺎﺕ‬

‫‪2‬‬

‫ﻧﺴﺘﻨﺘﺞ ﳑﺎ ﺳﺒﻖ ﺃﻥﹼ‬
‫] ‪: x ∈ [ 21 ,1‬‬

‫‪ 2‬‬
‫‪x + 2x − 1 + x − 2x − 1 = ‬‬
‫[ ∞‪ 4x − 2 : x ∈ ]1, +‬‬
‫ ﺇﺫﻥ ﰲ ﺣﺎﻟﺔ ‪ ، A = 2‬ﺗﺘﺄﻟﹼﻒ ﳎﻤﻮﻋﺔ ﺣﻠﻮﻝ ﺍﳌﻌﺎﺩﻟﺔ ﻣﻦ ﺍ‪‬ﺎﻝ ] ‪. [ 21 ,1‬‬

‫ ﻭﰲ ﺣﺎﻟﺔ ‪ ، A = 1‬ﺗﻜﻮﻥ ﳎﻤﻮﻋﺔ ﺍﳊﻠﻮﻝ ﳎﻤﻮﻋﺔ ﺧﺎﻟﻴﺔ‪.‬‬
‫ ﻭﺃﺧﲑﺍﹰ ﰲ ﺣﺎﻟﺔ ‪ A = 2‬ﺗﻜﻮﻥ ﳎﻤﻮﻋﺔ ﺍﳊﻠﻮﻝ ﻫﻲ } ‪. { 3/2‬‬

‫‪ù‬‬

‫ ‬
‫ ﻟﺘﻜﻦ ‪ a‬ﻭ ‪ b‬ﻭ ‪ c‬ﺃﻋﺪﺍﺩﺍﹰ ﺣﻘﻴﻘﻴ‪‬ﺔ‪ .‬ﻧﺘﺄﻣ‪‬ﻞ ﺍﳌﻌﺎﺩﻟﺔ ‪a cos x + b cos x + c = 0 :‬‬
‫‪2‬‬

‫) ‪(E‬‬

‫ﺑﺎ‪‬ﻬﻮﻝ ‪ . cos x‬ﺃﻭﺟﺪ ﻣﻌﺎﺩﻟﺔ ) ‪ ( Eɶ‬ﻣﻦ ﺍﻟﺪﺭﺟﺔ ﺍﻟﺜﺎﻧﻴﺔ ﺑﺎ‪‬ﻬﻮﻝ ‪ cos 2x‬ﺗﻜﻮﻥ ﲨﻴﻊ ﻗﻴﻢ ‪x‬‬
‫ﺍ ﻟﱵ ﲢﻞﹼ ﺍ ﳌﻌﺎ ﺩ ﻟﺔ ) ‪ ( E‬ﺣﻠﻮ ﻻﹰ ﳍﺎ‪ .‬ﺛﹸﻢ‪ ‬ﻗﺎ ﺭ ﻥ ﻫﺎ ﺗﲔ ﺍ ﳌﻌﺎ ﺩ ﻟﺘﲔ ﰲ ﺣﺎ ﻟﺔ ‪ a = 4‬ﻭ ‪b = 2‬‬

‫ﻭ ‪. c = −1‬‬
‫ ﰲ ﺍﳊﻘﻴﻘﺔ‪ ،‬ﻧﺴﺘﻔﻴﺪ ﻣﻦ ‪ . 1 + cos 2x = 2 cos2 x‬ﻓﻨﺴﺘﻨﺘﺞ ﻣﻦ ) ‪ ( E‬ﺃﻥﹼ‬
‫ﻭﺑﺎﻟﺘﺮﺑﻴﻊ ﻭﺍﻹﺻﻼﺡ ﳒﺪ‬

‫‪a cos 2x + a + 2c = −2b cos x‬‬

‫) ‪a 2 cos2 2x + ( a + 2c )2 + 2a ( a + 2c ) cos 2x = 2b 2 ( cos 2x + 1‬‬

‫ﺃﻭ‬

‫‪( Eɶ ) a cos 2x + 2 (a + 2ca − b ) cos 2x + (a + 2c ) − 2b = 0‬‬
‫ﻭﰲ ﺣﺎﻟﺔ ‪ a = 4‬ﻭ ‪ b = 2‬ﻭ ‪ ، c = −1‬ﺗ‪‬ﺼﺒﺢ ﻫﺬﻩ ﺍﳌﻌﺎﺩﻟﺔ ﺑﻌﺪ ﺍﻻﺧﺘﺼﺎﺭ ﻋﻠﻰ ‪ 4‬ﺑﺎﻟﺸﻜﻞ‬
‫‪2‬‬

‫‪2‬‬

‫ﻭﻫﻲ ﺍﳌﻌﺎﺩﻟﺔ ) ‪ ( E‬ﻧﻔﺴﻬﺎ‪.‬‬

‫‪2‬‬

‫‪2‬‬

‫‪2‬‬

‫‪2‬‬

‫‪4 cos2 2x + 2 cos 2x − 1 = 0‬‬

‫‪ù‬‬

‫ﻣﻼﺣﻈﺔ ‪ :‬ﺗ‪‬ﻜﺎﻓﺊ ﺍﳌﻌﺎﺩﻟﺔ ) ‪ ( E‬ﰲ ﺣﺎﻟﺔ ‪ a = 4‬ﻭ ‪ b = 2‬ﻭ ‪ c = −1‬ﻣﺎ ﻳﻠﻲ ‪:‬‬
‫‪2 cos 2x + 2 cos x + 1 = 0‬‬
‫ﻭﻫﺬﻩ ﺗ‪‬ﻜﺎﻓﺊ ‪ ، e −2 i x + e − i x + 1 + e i x + e 2 i x = 0‬ﺃﻭ‬
‫‪1 + e i x + e2 i x + e 3 i x + e 4 i x = 0‬‬
‫ﻭﺃﺧﲑﺍﹰ ﻷﻥﹼ ‪ ، e i x ≠ 1‬ﻓﺈﻥﹼ ﺍﳌﻌﺎﺩﻟﺔ ﺍﻟﺴﺎﺑﻘﺔ ﺗ‪‬ﻜﺎﻓﺊ ‪ e 5 i x = 1‬ﻭ ‪≠ 1‬‬
‫‪2πk‬‬
‫= ‪ x‬ﺣﻴﺚ ‪ k‬ﻋﻨﺼﺮ‪ ‬ﻣﻦ ‪ℤ\5ℤ‬‬
‫‪5‬‬
‫ ‬

‫‪ . e i x‬ﻭﻣﻦ ﺛﹶﻢ‪‬‬

‫ﻋﺎﻡ ‪1959‬‬

‫‪3‬‬

‫‪π‬‬
‫ ﻧ‪‬ﻌﻄﻰ ﺍﻟﻄﻮﻝ ‪ ، b = AC‬ﺃﻧﺸﺊ ﻣﺜﻠﹼﺜﺎﹰ ‪ ABC‬ﻳﻜﻮﻥ ﻓﻴﻪ‬
‫‪2‬‬
‫] ‪ [ BM‬ﺍﳌﺴﺎﻭﺍﺓ ‪. BM 2 = AB ⋅ BC‬‬

‫= ‬
‫‪ ، ABC‬ﻭﻳ‪‬ﺤﻘﹼﻖ ﺍﳌﺘﻮﺳ‪‬ﻂ‬

‫ ﺍﻟﺘﺤﻠﻴﻞ ‪ :‬ﻟﻨﻔﺘﺮﺽ ﺃﻥﹼ ‪ ABC‬ﻣﺜﻠﹼﺚﹲ ﻳ‪‬ﺤﻘﹼﻖ ﺍﳋﻮﺍﺹ‪ ‬ﺍﳌﺸﺎﺭ ﺇﻟﻴﻬﺎ‪ ،‬ﻋﻨﺪﺋﺬ ﻧﺮﻯ ﻣﺒﺎﺷﺮﺓ ﻣﺎ ﻳﻠﻲ ‪:‬‬
‫ ﺍﻟﻨﻘﻄﺔ ‪ B‬ﺗﻘﻊ ﻋﻠﻰ ﺍﻟﺪﺍﺋﺮﺓ ‪ C‬ﺍﻟﱵ ﻗﻄﺮﻫﺎ ‪. AC‬‬
‫‪b‬‬
‫‪2‬‬

‫ ﻃﻮﻝ ﺍﳌﺘﻮﺳ‪‬ﻂ ] ‪ [ BM‬ﻳﺴﺎﻭﻱ ﻧﺼﻒ ﻗﻄﺮ ﺍﻟﺪﺍﺋﺮﺓ ‪ C‬ﺃﻱ ‪.‬‬
‫ ﺿﻌﻔﺎ ﻣﺴﺎﺣﺔ ﺍﳌﺜﻠﹼﺚ ‪ ABC‬ﻳﺴﺎﻭﻱ ﻣﻦ ﺟﻬﺔ ﺃﻭﱃ ﺟﺪﺍﺀ ﺿﺮﺏ ﺍﻟﻀﻠﻌﲔ ﺍﻟﻘﺎﺋﻤﲔ‪ ،‬ﺃﻱ‬
‫‪b2‬‬
‫‪4‬‬

‫ﻭﺫﻟﻚ ﻋﻤﻼﹰ ﺑﺎﻟﻔﺮ‪‬ﺽ‪ ،‬ﻭﻫﻮ ﻳﺴﺎﻭﻱ ﺃﻳﻀﺎﹰ ﺟﺪﺍﺀ ﺿﺮﺏ ﻃﻮﻝ ﺍﻟﻀﻠﻊ ] ‪ [ AC‬ﺃﻱ ‪b‬‬

‫ﺑﻄﻮﻝ ﺍﻻﺭﺗﻔﺎﻉ ﺍﻟﻨﺎﺯﻝ ﻣﻦ ‪ B‬ﻭﻟﻴﻜﻦ ‪ . h‬ﻭﻋﻠﻴﻪ ﻧﺮﻯ ﺃﻥﹼ‬
‫‪b2‬‬
‫‪4‬‬

‫= ‪bh‬‬

‫‪1‬‬
‫‪4‬‬

‫ﺃﻭ ‪. h = b‬‬
‫‪1‬‬
‫‪4‬‬

‫ ﻧﺴﺘﻨﺘﺞ ﺇﺫﻥ ﺃﻥﹼ ‪ B‬ﺗﺒﻌﺪ ﻋﻦ ﺍﳌﺴﺘﻘﻴﻢ ) ‪ ( AC‬ﻣﺴﺎﻓﺔ ﺗﺴﺎﻭﻱ ‪ . b‬ﺃﻭ ﺃﻥﹼ ‪ B‬ﺗﻘﻊ ﻋﻠﻰ‬
‫‪1‬‬
‫‪4‬‬

‫ﺃﺣﺪ ﺍﳌﺴﺘﻘﻴﻤﲔ ﺍﳌﻮﺍﺯﻳﲔ ﻟﻠﻤﺴﺘﻘﻴﻢ ) ‪ ( AC‬ﻭﻳﺒﻌﺪ ﻛﻞﱞ ﻣﻨﻬﻤﺎ ﻋﻦ ﻫﺬﺍ ﺍﻷﺧﲑ ﻣﺴﺎﻓﺔ ‪. b‬‬
‫ﺍﻹﻧﺸﺎﺀ ‪:‬‬
‫ ﺍﺭﺳﻢ ﺍﻟﺪﺍﺋﺮﺓ ‪ C‬ﺍﻟﱵ ﻗﻄﺮﻫﺎ ﻗﻄﻌﺔ ﻣﺴﺘﻘﻴﻤﺔ ] ‪ [ AC‬ﻃﻮﳍﺎ ‪ . b‬ﻭﻟﻴﻜﻦ ‪ M‬ﻣﺮﻛﺰﻫﺎ‪.‬‬
‫ ﺃﻧﺸﺊ ﻧﺼﻒ ﺍﻟﻘﻄﺮ ] ‪ [ MA′‬ﰲ ﺍﻟﺪﺍﺋﺮﺓ ‪ C‬ﺍﻟﻌﻤﻮﺩﻱ ﻋﻠﻰ ] ‪. [ AC‬‬
‫ ﺍﺧﺘﺮ ‪ B‬ﻧﻘﻄﺔ ﻣﻦ ﻧﻘﻄﱵ ﺗﻘﺎﻃﻊ ﳏﻮﺭ ﺍﻟﻘﻄﻌﺔ ] ‪ [ MA′‬ﻣﻊ ﺍﻟﺪﺍﺋﺮﺓ ‪. C‬‬
‫ﻓﻨﺤﺼﻞ ﺑﺬﻟﻚ ﻋﻠﻰ ﺍﳌﺜﻠﹼﺚ ‪ ABC‬ﺍﳌﻄﻠﻮﺏ‪.‬‬
‫‪A′‬‬
‫‪C‬‬

‫‪B‬‬

‫‪h‬‬
‫‪C‬‬

‫‪M‬‬

‫ ‬

‫‪A‬‬

‫‪ù‬‬

‫ﺍﻷﻭﳌﺒﻴﺎﺩ ﺍﻟﻌﺎﳌﻲ ﻟﻠﺮﻳﺎﺿﻴ‪‬ﺎﺕ‬

‫‪4‬‬

‫ ﺍﻟﻨﻘﻄﺔ ‪ M‬ﻧﻘﻄﺔﹲ ﻣﺎ ﻣﻦ ﺍﻟﻘﻄﻌﺔ ﺍﳌﺴﺘﻘﻴﻤﺔ ] ‪ . [ AB‬ﻧﻨﺸﺊ ﺍﳌﺮﺑ‪‬ﻌﲔ ‪ AMCD‬ﻭ ‪ MBEF‬ﻣﻦ‬
‫ﺟﻬﺔ ﻭﺍﺣﺪﺓ ﺑﺎﻟﻨﺴﺒﺔ ﺇﱃ ﺍﳌﺴﺘﻘﻴﻢ ) ‪ . ( AB‬ﺗﺘﻘﺎﻃﻊ ﺍﻟﺪﺍﺋﺮﺗﺎﻥ ‪ C‬ﻭ ‪ C ′‬ﺍﳌﺮﺳﻮﻣﺘﺎﻥ ﻋﻠﻰ ﻫﺬﻳﻦ‬
‫ﺍﳌﺮﺑ‪‬ﻌﲔ ﺑﺎﻟﺘﺮﺗﻴﺐ‪ ،‬ﻋﻨﺪ ﺍﻟﻨﻘﻄﺘﲔ ‪ M‬ﻭ ‪. N‬‬
‫‪ .1‬ﺃﺛﺒﺖ ﺃﻥﹼ ﺍﳌﺴﺘﻘﻴﻤﲔ ) ‪ ( AF‬ﻭ ) ‪ ( BC‬ﻳﺘﻘﺎﻃﻌﺎﻥ ﰲ ‪. N‬‬
‫‪ .2‬ﺃﺛﺒﺖ ﺃﻥﹼ ﺍﳌﺴﺘﻘﻴﻤﺎﺕ ) ‪ ( MN‬ﲤﺮ‪ ‬ﺑﻨﻘﻄﺔ ﺛﺎﺑﺘﺔ ‪ S‬ﻋﻨﺪﻣﺎ ﺗﺘﺤﻮ‪‬ﻝ ﺍﻟﻨﻘﻄﺔ ‪ M‬ﻋﻠﻰ ] ‪. [ AB‬‬
‫‪ .3‬ﻧﻔﺘﺮﺽ ﺃﻥﹼ ‪ P‬ﻭ ‪ Q‬ﳘﺎ ﻣﺮﻛﺰﺍ ﺍﻟﺪﺍﺋﺮﺗﲔ ‪ C‬ﻭ ‪ C ′‬ﺑﺎﻟﺘﺮﺗﻴﺐ‪ .‬ﻋﻴ‪‬ﻦ ﺍﶈﻞﹼ ﺍﳍﻨﺪﺳﻲ ﳌﻨﺘﺼﻔﺎﺕ‬
‫ﺍﻟﻘﻄﻊ ﺍﳌﺴﺘﻘﻴﻤﺔ ] ‪ [ PQ‬ﻋﻨﺪﻣﺎ ﺗﺘﺤﻮ‪‬ﻝ ﺍﻟﻨﻘﻄﺔ ‪ M‬ﻋﻠﻰ ] ‪. [ AB‬‬
‫= ‬
‫‪ ، ANC‬ﻭﻫﺬﺍ ﻳﱪﻫﻦ ﻋﻠﻰ ﺃﻥﹼ‬

‫‪π‬‬
‫ ‪ .1‬ﺇﻥﹼ ] ‪ [ AC‬ﻫﻮ ﻗﻄﺮ‪ ‬ﰲ ﺍﻟﺪﺍﺋﺮﺓ ‪ ، C‬ﻭﻋﻠﻴﻪ ﻧﺮﻯ ﺃﻥﹼ‬
‫‪2‬‬
‫)‪(1‬‬
‫) ‪( AN ) ⊥ ( NC‬‬
‫‪ = 1 APM‬‬
‫ ﻭﻣﻦ ﺟﻬﺔ ﺃﺧﺮﻯ‪ ،‬ﲟﻼﺣﻈﺔ ﺍﻟﺪﺍﺋﺮﺓ ‪ C‬ﳒﺪ ‪ = π‬‬‫‪ ، ANM‬ﻭﻛﺬﻟﻚ ﲟﻼﺣﻈﺔ‬
‫‪2‬‬
‫‪4‬‬
‫‪ = 1 MQB‬‬
‫ﺍﻟﺪﺍﺋﺮﺓ ‪ C ′‬ﳒﺪ ‪ = π‬‬
‫‪ . MNB‬ﻭﻋﻠﻴﻪ ﻓﺈﻥﹼ‬
‫‪2‬‬
‫‪4‬‬
‫‪π‬‬
‫‪π‬‬
‫‪π‬‬
‫‪ = ANM‬‬
‫‪ + MNB‬‬
‫= ‪ = +‬‬
‫‪ANB‬‬
‫‪4 4‬‬
‫‪2‬‬

‫ﻭﻫﺬﺍ ﻳﱪﻫﻦ ﻋﻠﻰ ﺃﻥﹼ‬

‫) ‪( AN ) ⊥ ( NB‬‬

‫ﻭﻣﻦ ) ‪ ( 1‬ﻭ ) ‪ ( 2‬ﻧﺴﺘﻨﺘﺞ ﺃﻥﹼ ﺍﻟﻨﻘﻄﺔ ‪ N‬ﻫﻲ‬
‫ﺍﳌﺴﻘﻂ ﺍﻟﻘﺎﺋﻢ ﻟﻜﻞﱟ ﻣﻦ ﺍﻟﻨﻘﻄﺘﲔ ‪ B‬ﻭ ‪ C‬ﻋﻠﻰ‬
‫ﺍﳌﺴﺘﻘﻴﻢ ) ‪ ( AN‬ﻓﺎﻟﻨﻘﺎﻁ ‪ B‬ﻭ ‪ C‬ﻭ ‪ N‬ﺗﻘﻊ ﻋﻠﻰ‬
‫ﺍﺳﺘﻘﺎﻣﺔ ﻭﺍﺣﺪﺓ ﻭ ) ‪ ( BC‬ﻋﻤﻮﺩﻱ ﻋﻠﻰ ) ‪. ( AN‬‬

‫‪E‬‬

‫‪F‬‬
‫‪N‬‬

‫‪ ،‬ﻭﻫﺬﺍ‬

‫‪D‬‬

‫‪C‬‬

‫‪C′‬‬

‫ ﻭﻛﺬﻟﻚ‪ ،‬ﻧﺮﻯ ﺃﻥﹼ ﺍﻟﺰﺍﻭﻳﺔ ‬‫‪ BNF‬ﺗﻘﺎﺑﻞ ﺍﻟﻘﻄﺮ‬

‫‪π‬‬
‫] ‪ [ FB‬ﰲ ﺍﻟﺪﺍﺋﺮﺓ ‪ C ′‬ﻓﻬﻲ ﺇﺫﻥ ﺗﺴﺎﻭﻱ‬
‫‪2‬‬

‫)‪(2‬‬

‫‪Q‬‬

‫‪P‬‬
‫‪B‬‬

‫‪M‬‬

‫‪C‬‬
‫‪A‬‬

‫ﻳﱪﻫﻦ ﻋﻠﻰ ﺃﻥﹼ‬

‫) ‪( FN ) ⊥ ( NB‬‬

‫)‪(3‬‬

‫ﻭﻣﻦ ) ‪ ( 2‬ﻭ ) ‪ ( 3‬ﻧﺴﺘﻨﺘﺞ ﺃﻥﹼ ﺍﻟﻨﻘﻄﺔ ‪ N‬ﻫﻲ ﺍﳌﺴﻘﻂ ﺍﻟﻘﺎﺋﻢ ﻟﻜﻞﱟ ﻣﻦ ﺍﻟﻨﻘﻄﺘﲔ ‪ F‬ﻭ ‪ A‬ﻋﻠﻰ‬
‫ﺍﳌﺴﺘﻘﻴﻢ ) ‪ ( NB ) = ( BC‬ﻓﺎﻟﻨﻘﺎﻁ ‪ F‬ﻭ ‪ A‬ﻭ ‪ N‬ﺗﻘﻊ ﻋﻠﻰ ﺍﺳﺘﻘﺎﻣﺔ ﻭﺍﺣﺪﺓ ﻭﺍﳌﺴﺘﻘﻴﻢ‬
‫) ‪ ( BC‬ﻋﻤﻮﺩﻱ ﻋﻠﻰ ) ‪ . ( AF‬ﻭﺍﳌﺴﺘﻘﻴﻤﺎﻥ ﺍﳌﺘﻌﺎﻣﺪﺍﻥ ) ‪ ( BC‬ﻭ ) ‪ ( AF‬ﻳﺘﻘﺎﻃﻌﺎﻥ ﰲ ‪. N‬‬

‫ﻋﺎﻡ ‪1959‬‬

‫‪5‬‬

‫‪ .2‬ﻟﻘﺪ ﺭﺃﻳﻨﺎ ﺃﻥﹼ ‪ = π‬‬
‫‪ ، ANB‬ﻭﻫﺬﺍ ﻳ‪‬ﺜﺒﺖ‪ ‬ﺃﻥﹼ ﺍﻟﻨﻘﻄﺔ ‪ N‬ﺗﺘﺤﻮ‪‬ﻝ ﻋﻠﻰ ﻧﺼﻒ ﺍﻟﺪﺍﺋﺮﺓ ‪ C‬ﺍﻟﱵ‬
‫‪2‬‬

‫ﻗﻄﺮﻫﺎ ﺍﻟﻘﻄﻌﺔ ﺍﳌﺴﺘﻘﻴﻤﺔ ] ‪ ، [ AB‬ﻭﻟﻴﻜﻦ ‪ M ′‬ﻣﺮﻛﺰﻫﺎ‪.‬‬
‫ﺍﳌﺴﺘﻘﻴﻢ ) ‪ ( MN‬ﻳﻘﻄﻊ ﺍﻟﺪﺍﺋﺮﺓ ‪ C‬ﰲ ﻧﻘﻄﺔ ﺛﺎﻧﻴﺔ‬
‫ﻏﲑ ‪ N‬ﻭﻟﺘﻜﻦ ‪ . S‬ﻭﻟﻜﻦ‪ ‬ﺍﻟﺰﺍﻭﻳﺔ ﺍﳌﺮﻛﺰﻳ‪‬ﺔ‬
‫ ‬
‫‪′S‬‬
‫‪ AM‬ﺗﺴﺎﻭﻱ ﺿﻌﻔﻲ ﻗﻴﻤﺔ ﺍﻟﺰﺍﻭﻳﺔ ﺍﶈﻴﻄﻴ‪‬ﺔ‬

‫‪E‬‬

‫‪F‬‬

‫‪N‬‬

‫ ‪C‬‬

‫‪C′‬‬

‫‪C‬‬

‫‪Q‬‬

‫‪ = ANM‬‬
‫= ‬
‫‪ . ANS‬ﻭﻋﻠﻴﻪ ﻻ ﺑ‪‬ﺪ‪ ‬ﺃﻥ ﻳﻜﻮﻥ‬

‫‪π‬‬
‫‪4‬‬
‫‪π‬‬
‫ ‬
‫= ‪. AM ′S‬‬
‫‪2‬‬

‫‪D‬‬

‫‪C‬‬

‫‪P‬‬
‫‪M′‬‬

‫‪B‬‬

‫‪A‬‬

‫‪M‬‬

‫ﻭﻫﺬﺍ ﻳﱪﻫﻦ ﺃﻥﹼ ﻧﻘﻄﺔ ﺗﻘﺎﻃﻊ ﺍﳌﺴﺘﻘﻴﻢ ) ‪ ( MN‬ﻣﻊ‬
‫ﺍﻟﺪﺍﺋﺮﺓ ‪ C‬ﺍﳌﺨﺘﻠﻔﺔ ﻋﻦ ﺍﻟﻨﻘﻄﺔ ‪ N‬ﻫﻲ ﻧﻘﻄﺔ ﺛﺎﺑﺘﺔ‬
‫‪ S‬ﻻﺗﺘﻌﻠﹼﻖ ﲟﻮﻗﻊ ‪ M‬ﻋﻠﻰ ﺍﻟﻘﻄﻌﺔ ﺍﳌﺴﺘﻘﻴﻤﺔ ] ‪. [ AB‬‬
‫‪ .3‬ﻟﺘﻜﻦ ‪ S ′‬ﻧﻘﻄﺔ ﺗﻘﺎﻃﻊ ﺍﳌﺴﺘﻘﻴﻤﲔ ) ‪( BQ‬‬
‫ﻭ ) ‪ . ( AP‬ﳌﹼﺎ ﻛﺎﻥ‬
‫‪ ′ = MBQ‬‬
‫‪ =π‬‬
‫‪ABS‬‬
‫‪4‬‬
‫‪ ′ = MAP‬‬
‫‪ = π‬‬
‫‪BAS‬‬
‫ﻭ‬
‫‪4‬‬
‫‪π‬‬
‫ ‬
‫= ‪′B‬‬
‫‪ ، AS‬ﻓﺎﻟﻨﻘﻄﺔ ‪ S ′‬ﻫﻲ‬
‫ﺍﺳﺘﻨﺘﺠﻨﺎ ﺃﻥﹼ‬
‫‪2‬‬
‫ﺍﻟﻨﻘﻄﺔ ﺍﳌﹸﻘﺎﺑﻠﺔ ﻗﻄﺮ‪‬ﻳﺎﹰ ﻟﻠﻨﻘﻄﺔ ‪ S‬ﰲ ﺍﻟﺪﺍﺋﺮﺓ ‪. C‬‬

‫‪S‬‬

‫‪F‬‬

‫‪E‬‬

‫‪S′‬‬

‫ ‪C‬‬

‫‪Q‬‬

‫‪C‬‬

‫ ‬
‫ ‬

‫‪K‬‬
‫‪B‬‬

‫‪D‬‬

‫‪M′‬‬
‫‪M‬‬

‫‪P‬‬
‫‪A‬‬

‫‪S‬‬

‫ﻭﳌﹼﺎ ﻛﺎﻥ ) ‪ ( MQ ) ⊥ ( BQ‬ﻭﻛﺬﻟﻚ ) ‪ ( MP ) ⊥ ( AP‬ﺍﺳﺘﻨﺘﺠﻨﺎ ﺃﻥﹼ‬
‫‪ ′ = MPS‬‬
‫‪ ′ = π‬‬
‫‪MQS‬‬
‫‪2‬‬
‫ﻭﻫﻜﺬﺍ ﻧﻜﻮﻥ ﻗﺪ ﺃﺛﺒﺘﻨﺎ ﺃ ﹼﻥ ﺍﻟﺮﺑﺎﻋﻲ ‪ MQS ′P‬ﻣﺴﺘﻄﻴﻞ‪ ،‬ﻭﻋﻠﻴﻪ ﻓﺈ ﻥﹼ ‪ K‬ﻣﻨﺘﺼﻒ ] ‪ [ PQ‬ﻫﻮ‬
‫ﻧﻔﺴﻪ ﻣﻨﺘﺼﻒ ] ‪ . [ MS ′‬ﺃﻱ ﺇ ﻥﹼ ‪ K‬ﻫﻲ ﺻﻮﺭﺓ ‪ M‬ﻭﻓﻖ ﺍﻟﺘﺤﺎﻛﻲ ‪ HS ′, 1‬ﺍﻟﺬﻱ ﻣﺮﻛﺰﻩ ‪S ′‬‬
‫‪2‬‬

‫‪1‬‬
‫‪2‬‬

‫ﻭﻧﺴﺒﺘﻪ ‪ ،‬ﻭﻋﻨﺪﻣﺎ ﺗﺘﺤﻮ‪‬ﻝ ‪ M‬ﻋﻠﻰ ] ‪ [ AB‬ﺗﺘﺤﻮ‪‬ﻝ ‪ K‬ﻋﻠﻰ ﺻﻮﺭﺓ ﻫﺬﻩ ﺍﻟﻘﻄﻌﺔ ﺍﳌﺴﺘﻘﻴﻤﺔ ﻭﻓﻖ‬
‫‪ HS ′, 12‬ﺃﻱ ﻋﻠﻰ ﺍﻟﻘﻄﻌﺔ ﺍﳌﺴﺘﻘﻴﻤﺔ ﺍﻟﻮﺍﺻﻠﺔ ﺑﲔ ﻣﻨﺘﺼﻔﻲ ﺍﻟﻘﻄﻌﺘﲔ ] ‪ [ S ′A‬ﻭ ] ‪. [ S ′B‬‬
‫ ‬
‫‪This book is downloaded from this site:‬‬
‫‪www.syCourses.com‬‬

‫‪ù‬‬

‫ﺍﻷﻭﳌﺒﻴﺎﺩ ﺍﻟﻌﺎﳌﻲ ﻟﻠﺮﻳﺎﺿﻴ‪‬ﺎﺕ‬

‫‪6‬‬


ﻧﺘﺄﻣ‪‬ﻞ ﻣﺴﺘﻮﻳﲔ ﻏﲑ ﻣﺘﻮﺍﺯﻳﲔ ‪ P‬ﻭ ‪ .Q‬ﺍﻟﻨﻘﻄﺔ ‪ A‬ﺗﻨﺘﻤﻲ ﺇﱃ ‪ P‬ﻭﻻ ﺗﻨﺘﻤﻲ ﺇﱃ ‪ ، Q‬ﻭﺍﻟﻨﻘﻄﺔ‬
‫‪ C‬ﺗﻨﺘﻤﻲ ﺇﱃ ‪ Q‬ﻭﻻ ﺗﻨﺘﻤﻲ ﺇﱃ ‪ . P‬ﺃﻧﺸﺊ ﺍﻟﻨﻘﻄﺘﲔ ‪ B‬ﰲ ‪ P‬ﻭ ‪ D‬ﰲ ‪ Q‬ﺍﻟﻠﺘﲔ ﲡﻌﻼﻥ‬
‫ﺍﻟﺮﺑﺎﻋﻲ ‪ ABCD‬ﻳ‪‬ﺤﻘﹼﻖ ﺍﻟﺸﺮﻭﻁ ﺍﻟﺘﺎﻟﻴﺔ ‪:‬‬
‫‪ .1‬ﺍﻟﺮﺑﺎﻋﻲ ‪ ABCD‬ﻳﻘﻊ ﰲ ﻣﺴﺘﻮﹴ ﻭﺍﺣﺪ‪.‬‬
‫‪ .2‬ﺍﳌﺴﺘﻘﻴﻤﺎﻥ ) ‪ ( AB‬ﻭ ) ‪ (CD‬ﻣﺘﻮﺍﺯﻳﺎﻥ‪.‬‬
‫‪. AD = BC .3‬‬
‫‪ .4‬ﳝﻜﻦ ﺭﺳﻢ ﺩﺍﺋﺮﺓ ﲤﺲ ﺃﺿﻼﻉ ﺍﻟﺮﺑﺎﻋﻲ ‪ ABCD‬ﺩﺍﺧﻼﹰ‪.‬‬
‫ ﺍﻟﺘﺤﻠﻴﻞ ‪ :‬ﻟﻨﻔﺘﺮﺽ ﺃﻥﹼ ﺍﻹﻧﺸﺎﺀ ﺍﳌﻄﻠﻮﺏ ﻣ‪‬ﻨﺠﺰ‪ .‬ﻭﻟﻨﻌﺮ‪‬ﻑ ‪ d‬ﺍﻟﻔﺼﻞ ﺍﳌﺸﺘﺮﻙ ﻟﻠﻤﺴﺘﻮﻳﲔ ‪ P‬ﻭ ‪،Q‬‬
‫ﻛﻤﺎ ﻟﻨﻌﺮ‪‬ﻑ ‪ R‬ﻣﺴﺘﻮﻱ ﺍﻟﺮﺑﺎﻋﻲ ‪ . ABCD‬ﻭﻟﻨﻼﺣﻆ ﻣﺎ ﻳﻠﻲ ‪:‬‬
‫ ﺍﳌﺴﺘﻘﻴﻢ ) ‪ dA = ( AB‬ﻫﻮ ﺍﻟﻔﺼﻞ ﺍﳌﺸﺘﺮﻙ ﻟﻠﻤﺴﺘﻮﻳﲔ ‪ R‬ﻭ ‪. P‬‬
‫ ﺍﳌﺴﺘﻘﻴﻢ ) ‪ dC = (CD‬ﻫﻮ ﺍﻟﻔﺼﻞ ﺍﳌﺸﺘﺮﻙ ﻟﻠﻤﺴﺘﻮﻳﲔ ‪ R‬ﻭ ‪.Q‬‬
‫ ﻭﳌﹼﺎ ﻛﺎﻥ ‪ dA dC‬ﺍﺳﺘﻨﺘﺠﻨﺎ ﺃﻥﹼ ‪ dA d‬ﻭ ‪. dC d‬‬
‫ ﰲ ﺍ ﳌﺴﺘﻮ ﻱ ‪ R‬ﺍ ﻟﺮ ﺑﺎ ﻋﻲ ‪ ABCD‬ﻫﻮ ﺇ ﺫ ﻥ ﺷﺒﻪ ﻣﻨﺤﺮﻑ‬
‫ﻣﺘﺴﺎﻭﻱ ﺍﻟﺴﺎﻗﲔ‪ ،‬ﳝﻜﻦ ﺭﺳﻢ ﺩﺍﺋﺮﺓ ‪ C‬ﲤﺲ‪ ‬ﺃﺿﻼﻋﻪ ﺩﺍﺧﻼﹰ‪.‬‬

‫‪A‬‬

‫‪S‬‬
‫ ‬

‫‪B‬‬

‫‪H‬‬
‫ ‬

‫‪T‬‬
‫‪C‬‬

‫‪D‬‬

‫ ‬
‫‪S′‬‬

‫‪C‬‬

‫ﻟﻨﻌﺮ‪‬ﻑ ﺇﺫﻥ ﺍﻟﻨﻘﻄﺔ ‪ H‬ﺑﺄ ﻧ‪‬ﻬﺎ ﺍﳌﺴﻘﻂ ﺍﻟﻘﺎﺋﻢ ﻟﻠﻨﻘﻄﺔ ‪ C‬ﻋﻠﻰ ﺍﳌﺴﺘﻘﻴﻢ ‪ ، dA‬ﻭ ﻟﻨﺮﻣﺰ ﺑﺎﻟﺮﻣﻮ ﺯ ‪S‬‬

‫ﻭ ‪ S ′‬ﻭ ‪ T‬ﺇﱃ ﻧﻘﺎﻁ ﲤﺎﺱ ﺍﻟﺪﺍﺋﺮﺓ ‪ C‬ﻣﻊ ﺍﻷﺿﻼﻉ ] ‪ [ AB‬ﻭ ] ‪ [CD‬ﻭ ] ‪ [ DA‬ﺑﺎﻟﺘﺮﺗﻴﺐ‪.‬‬
‫ ﺇﻥﹼ ﺍﳌﺴﺘﻘﻴﻢ ) ‪ ( SS ′‬ﳏﻮﺭ ﺗﻨﺎﻇﺮ ﻟﺸﺒﻪ ﺍﳌﻨﺤﺮﻑ ‪ ، ABCD‬ﻭﻣﻦ ﺛﹶﻢ‪‬‬
‫‪HS = CS ′ = S ′D‬‬
‫ﻭﻷﻥﹼ ﺍﻟﺪﺍﺋﺮﺓ ‪ C‬ﲤﺲ ﺃﺿﻼﻉ ﺷﺒﻪ ﺍﳌﻨﺤﺮﻑ ‪ ABCD‬ﺍﺳﺘﻨﺘﺠﻨﺎ ﺃ ﹼﻥ ‪S ′D = DT‬‬

‫ﻭﻛﺬﻟﻚ ﺃﻥﹼ ‪ . SA = TA‬ﻭﻋﻠﻴﻪ ﻧﺮﻯ ﻣﺎ ﻳﻠﻲ ‪:‬‬
‫‪HA = HS + SA = DT + TA = DA‬‬
‫ﺇﺫﻥ ‪ . HA = DA = BC‬ﺇﺫﻥ ﺗﻘﻊ ‪ D‬ﻋﻠﻰ ﺍﻟﺪﺍﺋﺮﺓ ﺍﻟﱵ ﻣﺮﻛﺰﻫﺎ ‪ A‬ﻭﻧﺼﻒ ﻗﻄﺮﻫﺎ‬

‫‪ ، AH‬ﻭﺗﻘﻊ ‪ B‬ﻋﻠﻰ ﺍﻟﺪﺍﺋﺮﺓ ﺍﻟﱵ ﻣﺮﻛﺰﻫﺎ ‪ C‬ﻭﻧﺼﻒ ﻗﻄﺮﻫﺎ ‪. AH‬‬
‫ ﻭﺃﺧﲑﺍﹰ ﳌﹼﺎ ﻛﺎﻥ ‪ BC ≤ CH‬ﺍﺳﺘﻨﺘﺠﻨﺎ ﺃﻧ‪‬ﻪ ﰲ ﺣﺎﻝ ﻭﺟﻮﺩ ﺣﻞﱟ ﻟﺪﻳﻨﺎ ‪. AH ≤ CH‬‬
‫‪This book is downloaded from this site:‬‬
‫‪www.syCourses.com‬‬

‫ﻋﺎﻡ ‪1959‬‬

‫‪7‬‬

‫ﺍﻹﻧﺸﺎﺀ ‪:‬‬
‫ ﺃﻧﺸﺊ ﺍﳌﺴﺘﻘﻴﻢ ‪ dA‬ﺍﳌﺎﺭ ﺑﺎﻟﻨﻘﻄﺔ ‪ A‬ﻣﻮﺍﺯﻳﺎﹰ ﺍﻟﻔﺼﻞ ﺍﳌﺸﺘﺮﻙ ‪ d‬ﻟﻠﻤﺴﺘﻮﻳﲔ ‪ P‬ﻭ ‪.Q‬‬
‫ ﺃﻧﺸﺊ ﻛﺬﻟﻚ ﺍﳌﺴﺘﻘﻴﻢ ‪ dC‬ﺍﳌﺎﺭ ﺑﺎﻟﻨﻘﻄﺔ ‪ C‬ﻣﻮﺍﺯﻳﺎﹰ ﺍﻟﻔﺼﻞ ﺍﳌﺸﺘﺮﻙ ‪. d‬‬
‫ ﰲ ﺍﳌﺴﺘﻮﻱ ‪ R‬ﺍﳌﻌﻴ‪‬ﻦ ﺑﺎﳌﺴﺘﻘﻴﻤﲔ ﺍﳌﺘﻮﺍﺯﻳﲔ ‪ dA‬ﻭ ‪ ، dC‬ﻋﻴ‪‬ﻦ ﺍﻟﻨﻘﻄﺔ ‪ H‬ﺍﳌﺴﻘﻂ ﺍﻟﻘﺎﺋﻢ‬
‫ﻟﻠﻨﻘﻄﺔ ‪ C‬ﻋﻠﻰ ﺍﳌﺴﺘﻘﻴﻢ ‪. dA‬‬
‫ ﰲ ﺣﺎﻟﺔ ‪ AH > CH‬ﻳﻜﻮﻥ ﺍﻹﻧﺸﺎﺀ ﻣﺴﺘﺤﻴﻼﹰ‪ .‬ﺃ ﻣ‪‬ﺎ ﰲ ﺣﺎﻟﺔ ‪ ، AH ≤ CH‬ﻓﻨﻌﻴ‪‬ﻦ‬
‫‪ D‬ﰲ ﺍﳌﺴﺘﻮﻱ ‪ R‬ﻣﻦ ﻛﻮ‪‬ﺎ ﻧﻘﻄﺔ ﺗﻘﺎﻃﻊ ﺍﳌﺴﺘﻘﻴﻢ ‪ dC‬ﻣﻊ ﺍﻟﺪﺍﺋﺮﺓ ﺍﻟﱵ ﻣﺮﻛﺰﻫﺎ ‪A‬‬
‫ﻭﻧﺼﻒ ﻗﻄﺮﻫﺎ ‪ . AH‬ﻭﻛﺬﻟﻚ ﻧﻌﻴ‪‬ﻦ ‪ B‬ﰲ ﺍﳌﺴﺘﻮﻱ ‪ R‬ﻣﻦ ﻛﻮ‪‬ﺎ ﻧﻘﻄﺔ ﺗﻘﺎﻃﻊ ﺍﳌﺴﺘﻘﻴﻢ‬
‫‪ dA‬ﻣﻊ ﺍﻟﺪﺍﺋﺮﺓ ﺍﻟﱵ ﻣﺮﻛﺰﻫﺎ ‪ C‬ﻭﻧﺼﻒ ﻗﻄﺮﻫﺎ ‪ AH‬ﺃﻳﻀﺎﹰ‪ .‬ﻭﻧﺮﻯ ﺃﻧ‪‬ﻪ‪ ،‬ﺑﻮﺟﻪ ﻋﺎﻡ‪ ،‬ﻫﻨﺎﻙ‬
‫ﺣﻼﹼﻥ ﻟﻠﻤﺴﺄﻟﺔ ﺍﳌﻄﺮﻭﺣﺔ‪.‬‬

‫‪dA‬‬

‫‪A‬‬

‫‪B‬‬
‫‪H‬‬

‫‪B′‬‬
‫‪d‬‬

‫‪dC‬‬

‫‪D‬‬

‫‪D′‬‬

‫‪C‬‬

‫ﻓﻔﻲ ﺍﻟﺸﻜﻞ ﺃﻋﻼﻩ ﻛﻞﱞ ﻣﻦ ‪ ABCD‬ﻭ ‪ AB ′CD ′‬ﻫﻮ ﺣﻞﱞ ﻟﻠﻤﺴﺄﻟﺔ ﺍﳌﻄﺮﻭﺣﺔ‪.‬‬
‫ ‬

‫‪ù‬‬

‫ﺎﺕ‬‫ﺍﻷﻭﳌﺒﻴﺎﺩ ﺍﻟﻌﺎﳌﻲ ﻟﻠﺮﻳﺎﺿﻴ‬

8

QWE
AgD
ZXC

This book is downloaded from this site:
www.syCourses.com

‫ّ‬
‫ﺃﻭﳌﺒﻴﺎﺩ ﺍﻟﺮﻳﺎﺿﻴﺎﺕ ﺍﻟﺜﺎﻧﻲ‬
‫ ﺃﻭﺟﺪ ﲨﻴﻊ ﺍﻷﻋﺪﺍﺩ ‪ n‬ﺍﻟﱵ ﺗ‪‬ﻜﺘﺐ ﺑﺜﻼﺙ ﺧﺎﻧﺎﺕ ﻋﺸﺮ ﻳ‪‬ﺔ‪ ،‬ﻭﺗﻘﺒﻞ ﺍﻟﻘﺴﻤﺔ ﻋﻠﻰ ﺍﻟﻌﺪﺩ ‪11‬‬

‫ﻭﺧﺎﺭﺝ ﻗﺴﻤﺘﻬﺎ ﻋﻠﻰ ﺍﻟﻌﺪﺩ ‪ 11‬ﻳﺴﺎﻭﻱ ﳎﻤﻮﻉ ﻣﺮﺑ‪‬ﻌﺎﺕ ﺧﺎﻧﺎ‪‬ﺎ ﺍﻟﺜﻼﺙ‪.‬‬
‫ ﺃﺣﺪ ﺍﻟﻄﺮﺍﺋﻖ ﺍﳌﻤﻜﻨﺔ ﻫﻲ ﲡﺮﻳﺐ ﲨﻴﻊ ﻣﻀﺎﻋﻔﺎﺕ ﺍﻟﻌﺪﺩ ‪ 11‬ﺍﶈﺼﻮﺭﺓ ﺑﲔ ‪ 110‬ﻭ ‪ 990‬ﻭﻋﺪﺩﻫﺎ‬
‫‪ 81‬ﻋﺪﺩﺍﹰ‪ .‬ﻭﻟﻜﻦ ﻟﻴﺴﺖ ﻫﺬﻩ ﺍﻟﻄﺮﻳﻘﺔ ﺃﻧﻴﻘﺔ‪.‬‬
‫ ﻟﻨﻔﺘﺮﺽ ﺃ ﻥﹼ ﺁ ﺣﺎﺩ ﺍﻟﻌﺪﺩ ‪ n‬ﻫﻲ ‪ a‬ﻭﻋﺸﺮﺍﺗﻪ ﻫﻲ ‪ b‬ﻭﻣﺌﺎﺗﻪ ﻫﻲ ‪ c‬ﺣﻴﺚ ‪ . c ≥ 1‬ﻋﻨﺪﺋﺬ‬
‫ﻳﻜﻮﻥ ﻟﺪﻳﻨﺎ ﻣﻦ ﺟﻬﺔ ﺃﻭﱃ‬
‫‪n = a + 10b + 100c‬‬
‫ ﻭﻷﻥﹼ ‪ n‬ﻳﻘﺒﻞ ﺍﻟﻘﺴﻤﺔ ﻋﻠﻰ ﺍﻟﻌﺪﺩ ‪ 11‬ﺍﺳﺘﻨﺘﺠﻨﺎ ﺃﻥﹼ ﺍﻟﻌﺪﺩ ‪ a + c − b‬ﻣﻦ ﻣﻀﺎﻋﻔﺎﺕ ﺍﻟﻌﺪﺩ‬
‫) ‪(1‬‬

‫‪ ، 11‬ﻭﻟﻜﻦ‪ ‬ﻫﺬﺍ ﺍﻷﺧﲑ ﻳﻨﺘﻤﻲ ﺇﱃ ﺍ‪‬ﻤﻮﻋﺔ } ‪ ، { −8, −7, …, 0,1, …,18‬ﺇﺫﻥ‬
‫} ‪a + c − b ∈ { 0,11‬‬

‫ ﺃﻣ‪‬ﺎ ﺍﻟﻔﺮ‪‬ﺽ ﺍﻟﺜﺎﱐ ﻓﻴ‪‬ﻜﺘﺐ ﺑﺎﻟﺸﻜﻞ‬
‫) ‪n = 11 ( a 2 + b 2 + c 2‬‬

‫)‪(2‬‬

‫ ﻟﻨﻨﺎﻗﺶ ﺇﺫﻥ ﺣﺎﻟﺘﲔ ‪:‬‬
‫ ﺣﺎﻟﺔ ‪ . a + c − b = 0‬ﰲ ﻫﺬﻩ ﺍﳊﺎﻟﺔ ﺗ‪‬ﻜﺘﺐ ) ‪ ( 2‬ﺑﺎﻟﺸﻜﻞ‬
‫‪11 ( a 2 + ( a + c )2 + c 2 ) = a + 10 ( a + c ) + 100c‬‬

‫ﻭﻫﺬﺍ ﻳ‪‬ﻜﺎﻓﺊ‬
‫) ‪2a 2 + ( 2c − 1 )a = 2c ( 5 − c‬‬
‫ﻓﺎﻟﻌﺪﺩ ‪ a‬ﻋﺪﺩ‪ ‬ﺯﻭﺟﻲ‪ ،‬ﺃﻱ ‪ ،a = 2a ′‬ﻭﻣﻨﻪ‬
‫) ‪4a ′2 + ( 2c − 1 )a ′ = c ( 5 − c‬‬
‫ﻓﺈﺫﺍ ﺍﻓﺘﺮﺿﻨﺎ ﺃﻥﹼ ‪ a ′ ≥ 1‬ﻧﺘﺞ ﻣﻦ ﺫﻟﻚ ﺃﻥﹼ‬
‫‪c ( 5 − c ) ≥ 4 + ( 2c − 1 ) = 2c + 3‬‬

‫ﻭﻧﺼﻞ ﻣﻦ ﺛﹶﻢ‪ ‬ﺇﱃ ﺍﻟﺘﻨﺎﻗﺾ ﺍﻟﺘﺎﱄ ‪:‬‬
‫‪3 2 3‬‬
‫‪+ = c 2 − 3c + 3 ≤ 0‬‬
‫‪2‬‬
‫‪4‬‬
‫ﻫﺬﺍ ﻳ‪‬ﺜﺒﺖ‪ ‬ﺃﻥﹼ ‪ ، a ′ = 0‬ﻭﻣﻦ ﺛﹶﻢ‪ c = 5 ‬ﻭ ‪ a = 0‬ﻭ ‪ b = 5‬ﺃﻱ ‪. n = 550‬‬

‫)‬

‫‪9‬‬

‫‪c−‬‬

‫(‬

‫ﺍﻷﻭﳌﺒﻴﺎﺩ ﺍﻟﻌﺎﳌﻲ ﻟﻠﺮﻳﺎﺿﻴ‪‬ﺎﺕ‬

‫‪10‬‬

‫ ﺣﺎﻟﺔ ‪ . a + c − b = 11‬ﰲ ﻫﺬﻩ ﺍﳊﺎﻟﺔ ﻳ‪‬ﻜﺘﺐ ﺍﻟﺸﺮﻁ ) ‪ ( 2‬ﺑﺎﻟﺸﻜﻞ‬
‫‪11 ( ( b − c + 11 )2 + b 2 + c 2 ) = b − c + 11 + 10b + 100c‬‬

‫ﻭﻫﺬﺍ ﻳ‪‬ﻜﺎﻓﺊ‬
‫‪( b − c + 11 )2 + b 2 + c 2 = 1 + b + 9c‬‬

‫ﺃﻭ‬
‫‪2b 2 + ( 21 − 2c ) b + 2c 2 − 31c + 120 = 0‬‬

‫ﻭﺃﺧﲑﺍﹰ‬
‫‪2b 2 + ( 21 − 2c )b + ( 2c − 15 )( c − 8 ) = 0‬‬
‫ﻓﺈﺫﺍ ﺍﻓﺘﺮﺿﻨﺎ ﺃﻥﹼ ‪ b ≥ 1‬ﻧﺘﺞ ﻣﻦ ﻛﻮﻥ ‪ 21 − 2c > 0‬ﺃﻥﹼ‬
‫‪( 2c‬‬

‫‪− 15 )( c − 8 ) < 0‬‬
‫ﻭﻫﺬﺍ ﻳﻘﻮﺩ ﺇﱃ ﺍﻟﺘﻨﺎﻗﺾ ‪ 7.5 < c < 8‬ﻷﻥﹼ ‪ c‬ﻋﺪﺩ‪ ‬ﻃﺒﻴﻌﻲ‪ .‬ﺇﺫﻥ ﳚﺐ ﺃﻥ ﻳﻜﻮﻥ‬

‫‪ ، b = 0‬ﻭﻫﺬﺍ ﻳﻘﺘﻀﻲ ﺃﻥﹼ ‪ ، c = 8‬ﻭﺃﺧﲑﺍﹰ ﺃﻥﹼ ‪ a = 3‬ﺃﻱ ‪. n = 803‬‬
‫ﻭﺑﺎﻟﻌﻜﺲ ﻧﺘﻴﻘﻦ ﺑﺎﳊﺴﺎﺏ ﺍﳌﺒﺎﺷﺮ ﺃﻥﹼ ﺍﻟﻌﺪﺩﻳﻦ ‪ 550‬ﻭ ‪ 803‬ﳘﺎ ﺣﻼﹼﻥ ﻟﻠﻤﺴﺄﻟﺔ ﺍﳌﻄﺮﻭﺣﺔ‪.‬‬

‫‪ù‬‬

‫ ‬

‫ ﻋﻴ‪‬ﻦ ﳎﻤﻮﻋﺔ ﺍﻷﻋﺪﺍﺩ ﺍﳊﻘﻴﻘﻴ‪‬ﺔ ‪ x‬ﺍﻟﱵ ﺗ‪‬ﺤﻘﹼﻖ ﺍﳌﺘﺮﺍﺟﺤﺔ‬
‫‪4x 2‬‬
‫‪2 < 2x + 9‬‬
‫) ‪( 1 − 2x + 1‬‬

‫ ﻟﻨﻼﺣﻆ ﺃ ﻥﹼ ﺍﻟﻜﺴﺮ ﰲ ﺍﻟﻄﺮﻑ ﺍﻷﳝﻦ ﻣﻦ ﺍﳌﺘﺮﺍﺟﺤﺔ ﺍﻟﺴﺎﺑﻘﺔ ﻣﻌﺮ‪ ‬ﻑ‪ ‬ﺇﺫﺍ ﻭﻓﻘﻂ ﺇﺫﺍ ﺍﻧﺘﻤﺖ ‪ x‬ﺇﱃ‬
‫ﺍ‪‬ﻤﻮﻋﺔ [∞‪ . D = [ − 12 , 0 [ ∪ ]0, +‬ﻭﰲ ﺣﺎﻟﺔ ‪ x‬ﻣﻦ ‪ D‬ﻟﺪﻳﻨﺎ‬
‫‪( 2x + 1 − 1 )( 2x + 1 + 1 ) = 2x‬‬

‫ﻭﻣﻦ ﺛﹶﻢ‪‬‬
‫‪4x 2‬‬
‫‪2‬‬
‫) ‪2 = ( 2x + 1 + 1‬‬
‫) ‪( 1 − 2x + 1‬‬

‫‪∀x ∈ D,‬‬

‫‪= 2x + 2 + 2 2x + 1‬‬

‫ﻭﻫﻨﺎ ﻧﻼﺣﻆ ﺃﻧ‪‬ﻪ ﳝﻜﻦ ﲤﺪﻳﺪ ﺍﻟﻜﺴﺮ ﰲ ﺍﻟﻄﺮﻑ ﺍﻷﻳﺴﺮ ﻣﻦ ﺍﳌﺘﺮﺍﺟﺤﺔ ﺑﺎﻻﺳﺘﻤﺮﺍﺭ ﻋﻨﺪ ‪ 0‬ﻭﺫﻟﻚ‬
‫ﺑﺈﻋﻄﺎﺋﻪ ﺍﻟﻘﻴﻤﺔ ‪ 4‬ﻋﻨﺪ ﺗﻠﻚ ﺍﻟﻨﻘﻄﺔ‪.‬‬

‫ﻋﺎﻡ ‪1960‬‬

‫‪11‬‬

‫ﻭﻋﻠﻴﻪ‪ ،‬ﰲ ﺣﺎﻟﺔ ‪ x‬ﻣﻦ [∞‪ ، [ − 12 , +‬ﺗ‪‬ﻜﺎﻓﺊ ﺍﳌﺘﺮﺍﺟﺤﺔﹸ ﺍﳌﻌﻄﺎﺓﹸ ﺍﳌﺘﺮﺍﺟﺤﺔﹶ‬
‫‪2x + 2 + 2 2x + 1 < 2x + 9‬‬

‫ﺃﻭ‬
‫‪7‬‬
‫‪2‬‬

‫< ‪2x + 1‬‬

‫ﻭﺃﺧﲑﺍﹰ ‪ . x < 458‬ﻓﻤﺠﻤﻮﻋﺔ ﺣﻠﻮﻝ ﺍﳌﺘﺮﺍﺟﺤﺔ ﺍﳌﻌﻄﺎﺓ ﻫﻲ [ ‪ ، [ − 12 , 458‬ﻭﺫﻟﻚ ﺷﺮﻁ ﲤﺪﻳﺪ ﻃﺮﻓﻬﺎ‬
‫‪ù‬‬
‫ﺍﻷﳝﻦ ﺑﺎﻻﺳﺘﻤﺮﺍﺭ ﻋﻨﺪ ‪ ، 0‬ﺃﻭ ﻳ‪‬ﺤﺬﻑ ﺍﻟﻌﺪﺩ ‪ 0‬ﻣﻦ ﻫﺬﻩ ﺍ‪‬ﻤﻮﻋﺔ‪.‬‬
‫ ‬
‫ ﻧﺘﺄﻣ‪‬ﻞ ﻣﺜﻠﹼﺜﺎﹰ ‪ ABC‬ﻗﺎﺋﻢ ﺍﻟﺰﺍﻭﻳﺔ ﰲ ‪ ، A‬ﻃﻮﻝ ﻭﺗﺮﻩ ] ‪ [ BC‬ﻳﺴﺎﻭﻱ ‪ . a‬ﻳ‪‬ﻘﺴ‪‬ﻢ ﺍﻟﻮﺗﺮ ﺇﱃ ‪n‬‬

‫ﻗﻄﻌﺔ ﻣﺴﺘﻘﻴﻤﺔ ﻣﺘﺴﺎﻭﻳﺔ ﺍﻟﻄﻮﻝ‪ ،‬ﻭ ‪ n‬ﻋﺪﺩ‪ ‬ﻓﺮﺩﻱ‪ .‬ﺗ‪‬ﺮﻯ ﺍﻟﻘﻄﻌﺔ ﺍﻟﻮﺍﻗﻌﺔ ﰲ ﺍﳌﻨﺘﺼﻒ ﲢﺖ ﺯﺍﻭﻳﺔ‬
‫ﻗﺪﺭﻫﺎ ‪ α‬ﻣﻦ ﺍﻟﺮﺃﺱ ‪ ، A‬ﻭﺇﺫﺍ ﻋﻠﻤﺖ‪ ‬ﺃﻥﹼ ﻃﻮﻝ ﺍﻻﺭﺗﻔﺎﻉ ﺍﻟﻨﺎﺯﻝ ﻣﻦ ‪ A‬ﻳﺴﺎﻭﻱ ‪ ، h‬ﻓﺄﺛﺒﺖ ﺃﻥﹼ‬
‫‪4hn‬‬
‫)‪a (n2 − 1‬‬

‫= ‪tan α‬‬

‫ ﻟﻌﻞﹼ ﺃﺑﺴﻂ ﻃﺮﻳﻘﺔ‪ ،‬ﻫﻲ ﺍﻟﻄﺮﻳﻘﺔ ﺍﻟﺘﺤﻠﻴﻠﻴ‪‬ﺔ‪ .‬ﻟﻨﻨﺴﺐ ﺍﳌﺜﻠﹼﺚ ﺇﱃ ﲨﻠﺔ ﻣﺘﻌﺎﻣﺪﺓ‬
‫ﻧﻈﺎﻣﻴ‪‬ﺔ ﻣﺒﺪﺅﻫﺎ ‪ ، A‬ﻭﻓﻴﻬﺎ ﺇﺣﺪﺍﺛﻴ‪‬ﺎﺕ ﺍﻟﻨﻘﻄﺔ ‪ B‬ﻫﻲ ) ‪ ( β, 0‬ﻭﺇﺣﺪﺍﺛﻴ‪‬ﺎﺕ‬
‫ﺍﻟﻨﻘﻄﺔ ‪ C‬ﻫﻲ ) ‪ . ( 0, γ‬ﻋﻨﺪﺋﺬ ﺗﻜﻮﻥ ﻧﻘﺎﻁ ﺗﻘﺴﻴﻢ ﺍﻟﻮﺗﺮ ﻫﻲ ﺍﻟﻨﻘﺎﻁ‬
‫‪ ( Dk )0≤k ≤n‬ﺍﳌﻌﺮ‪‬ﻓﺔ ﻛﻤﺎ ﻳﻠﻲ ‪:‬‬

‫‪C‬‬

‫‪α‬‬

‫‪D1‬‬

‫ ‬
‫ ‪k‬‬
‫‪BD‬‬
‫=‬
‫‪BC‬‬
‫‪k‬‬
‫‪B‬‬
‫‪n‬‬
‫ﺃﻣ‪‬ﺎ ﺍﻟﻘﻄﻌﺔ ﺍﳌﻮﺟﻮﺩﺓ ﰲ ﺍﳌﻨﺘﺼﻒ ﻓﻬﻲ ‪ .  Dn −2 1 , Dn +2 1 ‬ﺇﺫﻥ‬
‫ ‬
‫ ‬
‫‪n −1‬‬
‫‪ADn −2 1 = AB + BDn −2 1 = ( β, 0 ) +‬‬
‫) ‪( −β, γ‬‬
‫‪2n‬‬
‫‪n +1 n −1‬‬
‫=‬
‫‪β,‬‬
‫‪γ‬‬
‫‪2n‬‬
‫‪2n‬‬

‫)‬

‫ﻭ‬

‫(‬

‫ ‬
‫ ‬
‫‪n +1‬‬
‫‪ADn +1 = AB + BDn +1 = ( β, 0 ) +‬‬
‫) ‪( −β, γ‬‬
‫‪2‬‬
‫‪2‬‬
‫‪2n‬‬
‫‪n −1 n +1‬‬
‫=‬
‫‪β,‬‬
‫‪γ‬‬
‫‪2n‬‬
‫‪2n‬‬

‫)‬

‫(‬

‫‪A‬‬

‫ﺍﻷﻭﳌﺒﻴﺎﺩ ﺍﻟﻌﺎﳌﻲ ﻟﻠﺮﻳﺎﺿﻴ‪‬ﺎﺕ‬

‫‪12‬‬

‫ﻟﻨﻌﺮ‪‬ﻑ ﺇﺫﻥ ﺍﻟﻄﻮﻟﲔ‬
‫ﻋﻨﺪﺋﺬ‬

‫ﻭ‬

‫ ‬
‫‪ L = ADn −2 1‬ﻭ‬

‫ ‬
‫‪L ′ = ADn +1‬‬
‫‪2‬‬

‫ ‬
‫‪ADn −2 1 ⋅ ADn +1‬‬
‫‪β2 + γ2‬‬
‫‪2‬‬
‫= ‪cos α‬‬
‫‪= (n 2 − 1 ) 2‬‬
‫‪LL ′‬‬
‫‪4n LL ′‬‬
‫‪2‬‬
‫‪a‬‬
‫‪= (n2 − 1) 2‬‬
‫‪4n LL ′‬‬
‫ ‬
‫) ‪det ( ADn −2 1 , ADn +1‬‬
‫‪βγ‬‬
‫‪2‬‬
‫= ‪sin α‬‬
‫‪= ( ( n + 1 )2 − ( n − 1 )2 ) 2‬‬
‫‪LL ′‬‬
‫‪4n LL ′‬‬
‫‪βγ‬‬
‫‪ah‬‬
‫‪= 4n 2‬‬
‫‪= 4n 2‬‬
‫‪4n LL ′‬‬
‫‪4n LL ′‬‬

‫ﻭﻋﻠﻴﻪ ﻓﺈﻥﹼ‬

‫‪4n‬‬
‫‪h‬‬
‫×‬
‫‪n −1 a‬‬
‫‪2‬‬

‫‪ù‬‬

‫= ‪tan α‬‬

‫ ‬
‫ ﺃﻧﺸﺊ ﻣﺜﻠﹼﺜﺎﹰ ‪ ABC‬ﲟﻌﺮﻓﺔ ﻃﻮﻟﹶﻲ ﺍﻻﺭﺗﻔﺎﻋﲔ ‪ hA‬ﻭ ‪ hB‬ﺍﻟﻨﺎﺯﻟﲔ ﻣﻦ ﺍﻟﺮﺃﺳﲔ ‪ A‬ﻭ ‪ ، B‬ﻭﻃﻮﻝ‬

‫ﺍﳌﺘﻮﺳ‪‬ﻂ ‪ m‬ﺍﳌﺮﺳﻮﻡ ﻣﻦ ﺍﻟﺮﺃﺱ ‪. A‬‬
‫ ﺍﻟﺘﺤﻠﻴﻞ ‪ :‬ﻟﻨﻔﺘﺮﺽ ﺃ ﻥﹼ ﻫﻨﺎﻙ ﻣﺜﻠﹼﺜﺎﹰ ﻳ‪‬ﺤﻘﹼﻖ ﺍﳋﻮﺍﺹ ﺍﳌﻄﻠﻮﺑﺔ ﻭﻟﺘﻜﻦ ‪ A′‬ﻭ ‪ B ′‬ﺍﳌﺴﻘﻄﲔ ﺍﻟﻘﺎﺋﻤﲔ‬
‫ﻟﻠﻨﻘﻄﺘﲔ ‪ A‬ﻭ ‪ B‬ﻋﻠﻰ ﺍﳌﺴﺘﻘﻴﻤﲔ ) ‪ ( BC‬ﻭ ) ‪ ( AC‬ﺑﺎﻟﺘﺮﺗﻴﺐ‪ .‬ﻭﺃﺧﲑﺍﹰ ﻟﺘﻜﻦ ‪ M‬ﻣﻨﺘﺼﻒ‬
‫ﺍﻟﻀﻠﻊ ] ‪ . [ BC‬ﺍﺳﺘﻨﺎﺩﺍﹰ ﺇﱃ ﺍﻟﻔﺮ‪‬ﺽ ﻟﺪﻳﻨﺎ ‪ ، hA = AA′‬ﻭ ‪ ، hB = BB ′‬ﻭ ‪. m = AM‬‬
‫‪A‬‬
‫ﻟﻨﺘﺄﻣ‪‬ﻞ ﺍﻟﺪﺍﺋﺮﺓ ‪ C‬ﺍﻟﱵ ﻗﻄﺮﻫﺎ ] ‪. [ AM‬‬
‫ ‬
‫ ﳌﹼﺎ ﻛﺎ ﻧﺖ ‪′ M = π2‬‬
‫‪ AA‬ﺍﺳﺘﻨﺘﺠﻨﺎ ﺃ ﻥﹼ ‪ A′ ∈ C‬ﺇ ﺫﻥ‬
‫‪ A′‬ﻫﻲ ﻧﻘﻄﺔ ﺗﻘﺎﻃﻊﹴ ﻟﻠﺪﺍﺋﺮﺗﲔ ‪ C‬ﻭﺍﻟﺪﺍﺋﺮﺓ ) ‪C ( A, hA‬‬
‫ﺍﻟﱵ ﻣﺮﻛﺰﻫﺎ ‪ A‬ﻭﻧﺼﻒ ﻗﻄﺮﻫﺎ ‪ ، hA‬ﻭﳒﺪ ‪. hA ≤ m‬‬

‫‪B′‬‬
‫‪M′‬‬
‫‪C‬‬

‫‪C‬‬
‫‪m hA‬‬
‫‪hB‬‬

‫‪A′‬‬

‫‪M‬‬

‫‪B‬‬

‫ ﻟﻨﺘﺄ ﻣ‪‬ﻞ ﺃﻳﻀﺎﹰ ‪ ،‬ﺍﻟﻨﻘﻄﺔ ‪ ، M ′‬ﺍﻟﻨﻘﻄﺔ ﺍﳌﺨﺘﻠﻔﺔ ﻋﻦ ‪ A‬ﺍﻟﱵ ﺗﺘﻘﺎﻃﻊ ﻋﻨﺪﻫﺎ ﺍﻟﺪﺍﺋﺮﺓ ‪ C‬ﻣﻊ‬
‫ﺍﳌﺴﺘﻘﻴﻢ ) ‪ . ( AC‬ﳌﹼﺎ ﻛﺎﻧﺖ‬

‫‪π‬‬
‫‪2‬‬

‫ ‬
‫= ‪′A‬‬
‫‪ MM‬ﺍﺳﺘﻨﺘﺠﻨﺎ ﺃﻥﹼ ) ‪. ( MM ′ ) ( BB ′‬‬

‫ﻋﺎﻡ ‪1960‬‬

‫‪13‬‬

‫ ﻳﻨﺘﺞ ﻣﻦ ﺫﻟﻚ ﺃ ﻥﹼ ﺍﳌﺜﻠﹼﺜﲔ ‪ CM ′M‬ﻭ ‪ CB ′B‬ﻣﺘﺸﺎ‪‬ﺎﻥ ‪ ،‬ﻭ ﻷ ﻥﹼ ‪ M‬ﻣﻨﺘﺼﻒ‬
‫ﺍﻗﺘﻀﻰ ﻫﺬﺍ ﺃﻥﹼ‬

‫] ‪[ BC‬‬

‫‪1‬‬
‫‪h‬‬
‫‪BB ′ = B‬‬
‫‪2‬‬
‫‪2‬‬
‫ﺇﺫﻥ ‪ M ′‬ﻫﻲ ﻧﻘﻄﺔ ﺗﻘﺎﻃﻊﹴ ﻟﻠﺪﺍﺋﺮﺗﲔ ‪ C‬ﻭ ) ‪ C ( M , 12 hB‬ﺍﻟﱵ ﻣﺮﻛﺰﻫﺎ ‪ ، M‬ﻭﻧﺼﻒ‬
‫= ‪MM ′‬‬

‫ﻗﻄﺮﻫﺎ ‪ ، 12 hB‬ﻭﻣﻨﻪ ﺍﻟﺸﺮﻁ ﺍﻟﻼﺯﻡ ﺍﻟﺜﺎﱐ ‪. hB < 2m‬‬
‫ﺍﻹﻧﺸﺎﺀ ‪:‬‬

‫ﻧﻔﺘﺮﺽ ﲢﻘﱡﻖ ﺍﻟﺸﺮﻃﲔ ‪ hA ≤ m‬ﻭ ‪. hB < 2m‬‬
‫ ﻧﻨﺸﺊ ﺍﻟﺪﺍﺋﺮﺓ ‪ C‬ﺍﻟﱵ ﻗﻄﺮﻫﺎ ﺍﻟﻘﻄﻌﺔ ﺍﳌﺴﺘﻘﻴﻤﺔ ] ‪ [ AM‬ﺍﻟﱵ ﻃﻮﳍﺎ ‪. m‬‬
‫ ﻧﻨﺸﺊ ‪ A′‬ﻧﻘﻄﺔ ﺗﻘﺎﻃﻊﹴ ﻟﻠﺪﺍﺋﺮﺓ ‪ C‬ﻣﻊ ﺍﻟﺪﺍﺋﺮﺓ ) ‪.C ( A, hA‬‬
‫ ﻧﻨﺸﺊ ‪ ) M ′‬ﻋﻠﻰ ﺍﻟﺘﺮﺗﻴﺐ ‪ ′‬‬
‫‪ (M‬ﻧﻘﻄﺔ ﺗﻘﺎﻃﻊ ﺍﻟﺪﺍﺋﺮﺓ ‪ C‬ﻣﻊ ﺍﻟﺪﺍﺋﺮﺓ ) ‪.C ( M , 12 hB‬‬
‫ ﻧﻌﻴ‪‬ﻦ ‪ C‬ﻧﻘﻄﺔ ﺗﻘﺎﻃﻊ ﺍﳌﺴﺘﻘﻴﻤﲔ ) ‪ ( MA′‬ﻭ ) ‪ ، ( AM ′‬ﻭﳝﻜﻦ ﺃﻥ ﻧﻌﻴ‪‬ﻦ ‪ C‬ﻧﻘﻄﺔ ﺗﻘﺎﻃﻊ‬
‫ﺍﳌﺴﺘﻘﻴﻤﲔ ) ‪ ( MA′‬ﻭ )‪ ′‬‬
‫‪. (AM‬‬

ﻧﻌﻴ‪‬ﻦ ‪ B‬ﻧﻈﲑﺓ ‪ C‬ﺑﺎﻟﻨﺴﺒﺔ ﺇﱃ ‪ ، M‬ﻭﳝﻜﻦ ﺃﻥ ﻧﻌﻴ‪‬ﻦ ‬
‫‪ B‬ﻧﻈﲑﺓ ‪ C‬ﺑﺎﻟﻨﺴﺒﺔ ﺇﱃ ‪. M‬‬
‫‪A‬‬

‫‪hB‬‬

‫ ‬
‫‪B‬‬

‫‪hA‬‬

‫‪M′‬‬
‫‪C‬‬

‫‪hB‬‬

‫‪A′‬‬

‫‪C‬‬
‫ ‬
‫‪M′‬‬
‫‪hB‬‬
‫‪2‬‬

‫‪M‬‬
‫‪B‬‬

‫ﻭﳓﺼﻞ ﺑﻮﺟﻪ ﻋﺎﻡ‪ ‬ﻋﻠﻰ ﺣﻠﹼﲔ ‪ ABC‬ﻭ ‬
‫‪ ABC‬ﻟﻠﻤﺴﺄﻟﺔ ﺍﳌﻄﺮﻭﺣﺔ‪.‬‬

‫ ‬
‫‪C‬‬

‫‪ù‬‬

‫ ‬
‫ ﻧﺘﺄ ﻣ‪‬ﻞ ﻣﻜﻌ‪‬ﺒﺎﹰ ‪ ، ABCDA′ B ′C ′D ′‬ﻣﻊ ‪ A‬ﻓﻮﻕ ‪ A′‬ﻭ ‪ B‬ﻓﻮﻕ ‪ B ′‬ﻭﻫﻜﺬﺍ‪ ، ...‬ﻭﻧﺘﺄ ﻣ‪‬ﻞ‬

‫ﻧﻘﻄﺔ ﻣﺎ ‪ X‬ﻣﻦ ﺍﻟﻘﻄﻌﺔ ﺍﳌﺴﺘﻘﻴﻤﺔ ] ‪ ، [ AC‬ﻭﻧﻘﻄﺔ ﻣﺎ ‪ Y‬ﻣﻦ ] ‪. [ B ′D ′‬‬
‫‪ .1‬ﺃﻭﺟﺪ ﺍﶈﻞﹼ ﺍﳍﻨﺪﺳﻲ ﳌﻨﺘﺼﻒ ﺍﻟﻘﻄﻌﺔ ] ‪. [ XY‬‬
‫‪ .2‬ﺃﻭﺟﺪ ﺍﶈﻞﹼ ﺍﳍﻨﺪﺳﻲ ﻟﻠﻨﻘﺎﻁ ‪ Z‬ﻣﺮﻛﺰ ﺍﻷﺑﻌﺎﺩ ﺍﳌﺘﻨﺎﺳﺒﺔ ﻟﻠﻨﻘﻄﺘﲔ ) ‪ ( X ;2‬ﻭ ) ‪. (Y ;1‬‬

‫ﺍﻷﻭﳌﺒﻴﺎﺩ ﺍﻟﻌﺎﳌﻲ ﻟﻠﺮﻳﺎﺿﻴ‪‬ﺎﺕ‬

‫‪14‬‬

‫ ﺳﻨﺒﺤﺚ ﻋﻦ ﺣﻞﱟ ﲢﻠﻴﻠﻲ ﳍﺬﻩ ﺍﳌﺴﺄﻟﺔ‪ ،‬ﻓﻨﺘﺄﻣ‪‬ﻞ ﲨﻠﺔ ﻣﺘﻌﺎﻣﺪﺓ ﻧﻈﺎﻣﻴ‪‬ﺔ ﻓﻴﻬﺎ‬
‫) ‪D ( 0,1, 0‬‬

‫‪B ( 1, 0, 0 ), C ( 1,1, 0 ),‬‬

‫‪A ( 0, 0, 0 ),‬‬

‫) ‪A′ ( 0, 0,1 ), B ′ ( 1, 0,1 ), C ′ ( 1,1,1 ), D ′ ( 0,1,1‬‬

‫ﻓﺘﻜﻮﻥ ) ‪ X ( x , x , 0‬ﻣﻊ ‪ ، 0 ≤ x ≤ 1‬ﻭﺗﻜﻮﻥ ) ‪ Y ( y,1 − y,1‬ﻣﻊ ‪. 0 ≤ y ≤ 1‬‬
‫‪ .1‬ﻭﻋﻠﻰ ﻫﺬﺍ ﻳﻜﻮﻥ ﺍﶈﻞﹼ ﺍﳍﻨﺪﺳﻲ ﳌﻨﺘﺼﻒ ﺍﻟﻘﻄﻌﺔ ] ‪ [ XY‬ﻫﻮ ﳎﻤﻮﻋﺔ ﺍﻟﻨﻘﺎﻁ‬

‫} ] ‪{( x +2 y , x + 21 − y , 21 ) : ( x, y ) ∈ [ 0,1‬‬
‫‪2‬‬

‫=‪L‬‬

‫ﻋﻠﻴﻨﺎ ﺇﺫﻥ ﺗﻌﻴﲔ ﺻﻮﺭﺓ ﺍﻟﺘﺎﺑﻊ‬
‫‪‬‬
‫‪‬‬
‫‪ x + y x + 1 − y ‬‬
‫‪‬‬
‫‪Φ : [ 0,1 ] → ℝ , Φ ( x , y ) = ‬‬
‫‪,‬‬
‫ ‪‬‬
‫‪2‬‬
‫‪‬‬


‫‪ 2u‬‬
‫‪‬‬
‫‪v‬‬
‫‪2‬‬

‫ﻧﺴﺘﻨﺘﺞ ﻣﻦ ﺍﳌﺴﺎﻭﺍﺗﲔ‬

‫‪x +y‬‬
‫‪2‬‬

‫=‪u‬ﻭ‬

‫‪x +1−y‬‬
‫‪2‬‬

‫‪2‬‬

‫= ‪ v‬ﺃﻥﹼ‬

‫‪u + v − 12 = x‬‬
‫‪u − v + 12 = y‬‬

‫ﻭﻋﻠﻴﻪ ‪ ( u, v ) ∈ Im Φ‬ﺇﺫﺍ ﻭﻓﻘﻂ ﺇﺫﺍ ﲢﻘﹼﻘﺖ ﺍﳌﺘﺮﺍﺟﺤﺘﺎﻥ‬
‫‪ 0 ≤ u + v − 12 ≤ 1‬ﻭ ‪0 ≤ u − v + 12 ≤ 1‬‬
‫ﺃﻭ‬
‫‪3‬‬
‫‪2‬‬

‫≤ ‪≤u +v‬‬

‫‪1‬‬

‫‪1‬‬
‫‪2‬‬

‫ﻭ‬
‫ﺇﺫﻥ ‪ Im Φ‬ﻫﻲ ﺍﳌﺮﺑ‪‬ﻊ ﺍﻟﺬﻱ ﺭﺅﻭﺳﻪ ﺍﻟﻨﻘﺎﻁ ) ‪ ( 12 , 0‬ﻭ ) ‪( 1, 21‬‬
‫ﻭ ) ‪ ( 12 ,1‬ﻭ ) ‪ ( 0, 12‬ﻛﻤﺎ ﻳﺒﻴ‪‬ﻦ ﺍﻟﺸﻜﻞ ﺍ‪‬ﺎﻭﺭ‪.‬‬
‫‪1‬‬
‫‪2‬‬

‫‪v‬‬

‫‪1‬‬
‫‪2‬‬

‫≤ ‪− 21 ≤ u − v‬‬

‫‪1‬‬
‫‪2‬‬

‫‪1 u‬‬

‫‪D′‬‬

‫ﻭﻋﻠﻰ ﻫﺬﺍ ﻓﺈﻥﹼ ﺍﶈﻞﹼ ﺍﳍﻨﺪﺳﻲ ‪ L‬ﺍﳌﻄﻠﻮﺏ ﻫﻮ ﺍﳌﺮﺑ‪‬ﻊ ﺍﻟﺬﻱ ﺭﺅﻭﺳﻪ‬
‫ﻣﺮﺍﻛﺰ ﺍﻟﻮﺟﻮﻩ ﺍﳉﺎﻧﺒﻴ‪‬ﺔ ‪ AA′ B ′B‬ﻭ ‪ BB ′C ′C‬ﻭ ‪CC ′D ′D‬‬
‫ﻭ ‪ . DD ′A′ A‬ﻛﻤﺎ ﻫﻮ ﻣﺒﻴ‪‬ﻦ ﰲ ﺍﻟﺸﻜﻞ ﺍ‪‬ﺎﻭﺭ‪.‬‬

‫‪0‬‬

‫‪A′‬‬
‫‪C′‬‬

‫‪Y‬‬

‫‪B′‬‬

‫‪A‬‬

‫‪D‬‬

‫‪X‬‬
‫‪C‬‬

‫‪B‬‬

‫ﻋﺎﻡ ‪1960‬‬

‫‪15‬‬

‫‪ .2‬ﻭﻣﻦ ﺟﻬﺔ ﺃﺧﺮﻯ‪ ،‬ﺍﶈﻞ ﺍﳍﻨﺪﺳﻲ ﻟﻠﻨﻘﺎﻁ ‪ Z‬ﻫﻮ ﳎﻤﻮﻋﺔ ﺍﻟﻨﻘﺎﻁ‬

‫} ] ‪{( 2x 3+ y , 2x +31 − y , 13 ) : ( x, y ) ∈ [ 0,1‬‬
‫‪2‬‬

‫= ‪L′‬‬

‫ﻋﻠﻴﻨﺎ ﺇﺫﻥ ﺗﻌﻴﲔ ﺻﻮﺭﺓ ﺍﻟﺘﺎﺑﻊ‬
‫‪‬‬
‫‪‬‬
‫‪ 2x + y 2x + 1 − y ‬‬
‫‪‬‬
‫‪Ψ : [ 0,1 ] → ℝ , Ψ ( x , y ) = ‬‬
‫‪,‬‬

‪‬‬
‫‪3‬‬
‫‪ ‬‬

‪ 3u‬‬
‫‪v‬‬
‫‪2‬‬

‫‪2‬‬

‫ﻧﺴﺘﻨﺘﺞ ﻣﻦ ﺍﳌﺴﺎﻭﺍﺗﲔ ‪ u = 2x3+y‬ﻭ ‪ v = 2x +31−y‬ﺃﻥﹼ‬
‫‪ u + v − 13 = 43 x‬ﻭ ‪u − v + 13 = 23 y‬‬
‫ﻭﻋﻠﻴﻪ ‪ ( u, v ) ∈ Im Ψ‬ﺇﺫﺍ ﻭﻓﻘﻂ ﺇﺫﺍ ﲢﻘﹼﻘﺖ ﺍﳌﺘﺮﺍﺟﺤﺘﺎﻥ‬
‫‪ 0 ≤ u + v − 13 ≤ 43‬ﻭ ‪0 ≤ u − v + 13 ≤ 23‬‬
‫ﺃﻭ‬
‫‪ 13 ≤ u + v ≤ 53‬ﻭ ‪− 13 ≤ u − v ≤ 13‬‬

‫‪v‬‬

‫‪1‬‬
‫‪2‬‬
‫‪3‬‬

‫ﺇﺫﻥ ‪ Im Φ‬ﻫﻲ ﺍﳌﺴﺘﻄﻴﻞ ﺍﻟﺬﻱ ﺭﺅﻭﺳﻪ ﺍﻟﻨﻘﺎﻁ ) ‪( 13 , 0‬‬

‫‪1‬‬
‫‪3‬‬

‫ﻭ ) ‪ ( 1, 23‬ﻭ ) ‪ ( 23 ,1‬ﻭ ) ‪ ( 0, 23‬ﻛﻤﺎ ﻳﺒﻴ‪‬ﻦ ﺍﻟﺸﻜﻞ ﺍ‪‬ﺎﻭﺭ‪.‬‬

‫‪1‬‬
‫‪3‬‬

‫‪1 u‬‬

‫‪D′‬‬

‫ﻭﻋﻠﻰ ﻫﺬﺍ ﻓﺈﻥﹼ ﺍﶈﻞﹼ ﺍﳍﻨﺪﺳﻲ ‪ L′‬ﺍﳌﻄﻠﻮﺏ ﻫﻮ ﺍﳌﺴﺘﻄﻴﻞ ﺍﻟﺬﻱ‬
‫ﺭﺅﻭﺳﻪ ﺍﻟﻨﻘﺎﻁ ‪:‬‬
‫) ‪ ( 13 , 0, 13‬ﻭ ) ‪ ( 1, 23 , 13‬ﻭ ) ‪ ( 23 ,1, 13‬ﻭ ) ‪( 0, 23 , 13‬‬
‫ﻛﻤﺎ ﻫﻮ ﻣﺒﻴ‪‬ﻦ ﰲ ﺍﻟﺸﻜﻞ ﺍ‪‬ﺎﻭﺭ‪.‬‬

‫‪A′‬‬
‫‪C′‬‬

‫‪D‬‬

‫‪X‬‬
‫‪C‬‬

‫ﻭﺑﺬﺍ ﻳﺘﻢ ﺍﻹﺛﺒﺎﺕ‪.‬‬

‫‪0‬‬

‫‪Y‬‬
‫‪B′‬‬

‫‪A‬‬
‫‪B‬‬

‫‪ù‬‬
‫ ‬

‫ﺲ ﺳﻄﺤﻪ ﺍﳉﺎﻧﱯ ﻭﻗﺎﻋﺪﺗﻪ‪ .‬ﻭﻧﺘﺄ ﻣ‪‬ﻞ ﺃﺳﻄﻮﺍﻧﺔ ‪C‬‬
‫ ﻧﺘﺄ ﻣ‪‬ﻞ ﳐﺮﻭﻃﺎﹰ ﺩﻭﺭ ﺍﻧﻴ‪‬ﺎﹰ ‪ Q‬ﻓﻴﻪ ﻛﺮﺓ ‪ S‬ﲤ ‪‬‬

‫ﲢﻮﻱ ﺍﻟﻜﺮﺓ ‪ S‬ﻧﻔﺴﻬﺎ‪ ،‬ﻭﲤﺲ‪ ‬ﻫﺬﻩ ﺍﻟﻜﺮﺓ ﻗﺎﻋﺪﰐ ﺍﻷﺳﻄﻮﺍﻧﺔ ﻭﺳﻄﺤﻬﺎ ﺍﳉﺎﻧﱯ‪ .‬ﻟﻴﻜﻦ ‪ V1‬ﺣﺠﻢ‬
‫ﺍﳌﺨﺮﻭﻁ ‪ Q‬ﻭﻟﻴﻜﻦ ‪ V2‬ﺣﺠﻢ ﺍﻷﺳﻄﻮﺍﻧﺔ ‪ .C‬ﺃﺛﺒﺖ ﺃﻥﹼ ‪ .V1 ≠ V2‬ﻭﻋﻴ‪‬ﻦ ﺃﺻﻐﺮ ﻗﻴﻤﺔ ﺗﺄﺧﺬﻫﺎ‬
‫ﺍﻟﻨﺴﺒﺔ ‪ ،V1 /V2‬ﻭﺃﻧﺸﺊ ﻧﺼﻒ ﺯﺍﻭﻳﺔ ﺭﺃﺱ ﺍﳌﺨﺮﻭﻁ ﰲ ﻫﺬﻩ ﺍﳊﺎﻟﺔ‪.‬‬

‫ﺍﻷﻭﳌﺒﻴﺎﺩ ﺍﻟﻌﺎﳌﻲ ﻟﻠﺮﻳﺎﺿﻴ‪‬ﺎﺕ‬

‫‪16‬‬

‫ ﻟﻨﻔﺘﺮﺽ ﺃ ﻥﹼ ‪ a‬ﻫﻮ ﻧﺼﻒ ﻗﻄﺮ ﻗﺎﻋﺪﺓ ﺍﳌﺨﺮﻭﻁ ‪ ، Q‬ﻭﺃ ﻥﹼ ‪ r‬ﻫﻮ‬
‫ﻧﺼﻒ ﻗﻄﺮ ﺍ ﻟﻜﺮ ﺓ ‪ . S‬ﻭ ﻟﻴﻜﻦ ‪ h‬ﺍ ﺭ ﺗﻔﺎ ﻉ ﺍ ﳌﺨﺮ ﻭ ﻁ‪ .‬ﻋﻨﺪ ﺋﺬ‬
‫‪r‬‬
‫‪h‬‬
‫‪ = tan 2θ‬ﻣﻊ ‪= tan θ‬‬
‫‪a‬‬
‫‪a‬‬
‫ﺍﺳﺘﻨﺘﺠﻨﺎ ﺃﻥﹼ ‪. 0 < r < a‬‬

‫‪h‬‬
‫‪r‬‬

‫‪ ،‬ﻭﻷ ﻧ‪‬ﻪ ﻟﺪﻳﻨﺎ [ ‪2θ ∈ ] 0, π2‬‬
‫‪θ‬‬
‫‪θ‬‬
‫|‬

‫ﻭﻋﻠﻴﻪ‬

‫‪2 tan θ‬‬
‫‪2a 2r‬‬
‫=‬
‫‪1 − tan2 θ‬‬
‫‪a2 − r 2‬‬

‫‪a‬‬

‫|‬

‫‪h =a‬‬

‫ﺇﺫﻥ ﻳ‪‬ﻌﻄﻰ ﺣﺠﻢ ﺍﳌﺨﺮﻭﻁ ‪ Q‬ﺑﺎﻟﺼﻴﻐﺔ ﺍﻟﺘﺎﻟﻴﺔ ‪:‬‬

‫‪1 2‬‬
‫‪2π‬‬
‫‪a 4r‬‬
‫= ‪πa h‬‬
‫‪⋅ 2‬‬
‫‪3‬‬
‫‪3 a − r2‬‬
‫ﺃﻣ‪‬ﺎ ﺍﻷﺳﻄﻮﺍﻧﺔ ‪ C‬ﻓﻨﺼﻒ ﻗﻄﺮ ﻗﺎﻋﺪ‪‬ﺎ ‪ ، r‬ﻭﺍﺭﺗﻔﺎﻋﻬﺎ ‪ ، 2r‬ﻭﻳ‪‬ﻌﻄﻰ ﺣﺠﻢ ﺍﻷﺳﻄﻮﺍﻧﺔ ‪ C‬ﺑﺎﻟﺼﻴﻐﺔ‪:‬‬
‫= ‪V1‬‬

‫‪V2 = 2πr 3‬‬

‫‪1‬‬
‫‪3‬‬

‫ﻓﻤﻦ ﺟﻬﺔ ﺃﻭﱃ ﺗ‪‬ﻜﺎﻓﺊ ﺍﳌﺴﺎﻭﺍﺓﹸ ‪ V1 = V2‬ﺍﳌﺴﺎﻭﺍﺓﹶ ‪ r 4 − r 2a 2 + a 4 = 0‬ﻭﻫﻲ ﺑﺪﻭﺭﻫﺎ‬
‫‪2‬‬

‫ﺗ‪‬ﻜﺎﻓﺊ ﺍﳌﺴﺎﻭﺍﺓ ﺍﳌﺴﺘﺤﻴﻠﺔ ‪ . ( r 2 − a2 ) + a12 = 0‬ﺇﺫﻥ ‪.V1 ≠ V2‬‬
‫‪4‬‬

‫‪2‬‬

‫ﻭﻣﻦ ﺟﻬﺔ ﺃﺧﺮﻯ‪،‬‬
‫‪4‬‬

‫‪V1‬‬
‫‪1‬‬
‫‪a‬‬
‫‪1‬‬
‫‪= ⋅ 2‬‬
‫=‬
‫‪2‬‬
‫‪2‬‬
‫‪2‬‬
‫‪V2‬‬
‫‪3 (a − r )r‬‬
‫) ‪3 f ( ar 2‬‬
‫‪1‬‬
‫ﻣﻊ ) ‪ . f ( x ) = x ( 1 − x‬ﻭﻧﻌﻠﻢ ﺃﻥﹼ ﺍﳊﺪ‪ ‬ﺍﻷﻋﻠﻰ ﻟﻠﺘﺎﺑﻊ ‪ f‬ﻋﻠﻰ ﺍ‪‬ﺎﻝ [‪ ] 0,1‬ﻳﺴﺎﻭﻱ‬
‫‪4‬‬
‫‪a‬‬
‫‪4‬‬
‫‪V‬‬
‫‪1‬‬
‫= ‪ . r‬ﺃﻭ‬
‫ﻳﻮﺍﻓﻖ = ‪ . x‬ﺇﺫﻥ ﺃﺻﻐﺮ ﻗﻴﻤﺔ ﻟﻠﻨﺴﺒﺔ ‪ 1‬ﻫﻲ ‪ ،‬ﻭﻫﻲ ﺗﻮﺍﻓﻖ‬
‫‪3‬‬
‫‪V2‬‬
‫‪2‬‬
‫‪2‬‬
‫‪1‬‬
‫= ‪tan θ‬‬
‫‪2‬‬

‫ﻭﻫﻮ‬

‫ﻭﻣﻦ ﺛﹶﻢ‪‬‬

‫‪1‬‬
‫‪3‬‬

‫=‬

‫‪1‬‬
‫‪2‬‬
‫‪1‬‬
‫‪2‬‬

‫‪2‬‬

‫‪1−‬‬
‫‪1 − tan θ‬‬
‫=‬
‫‪2‬‬
‫‪1+‬‬
‫‪1 + tan θ‬‬

‫= ‪cos 2θ‬‬

‫ﺇﺫﻥ ﻧﺼﻒ ﺯﺍﻭﻳﺔ ﺭﺃﺱ ﺍﳌﺨﺮﻭﻁ ﺗﺴﺎﻭﻱ ) ‪. arcsin ( 13‬‬
‫ ‬
‫‪This book is downloaded from this site:‬‬
‫‪www.syCourses.com‬‬

‫‪ù‬‬

‫ّ‬
‫ﺃﻭﳌﺒﻴﺎﺩ ﺍﻟﺮﻳﺎﺿﻴﺎﺕ ﺍﻟﺜﺎﻟﺚ‬
‫ ﺣﻞﹼ ﲨﻠﺔ ﺍﳌﻌﺎﺩﻻﺕ ﺍﻟﺘﺎﻟﻴﺔ ‪:‬‬
‫‪ x + y + z = a‬ﻭ ‪ x 2 + y 2 + z 2 = b 2‬ﻭ ‪xy = z 2‬‬

‫ﺑﺎﻟﻨﺴﺒﺔ ﺇﱃ ﺍ‪‬ﺎﻫﻴﻞ ‪ x‬ﻭ ‪ y‬ﻭ ‪ . z‬ﻭﻋﻴ‪‬ﻦ ﺍﻟﺸﺮﻭﻁ ﻋﻠﻰ ) ‪ (a,b‬ﺣﺘ‪‬ﻰ ﺗﻜﻮﻥ ﺍﳊﻠﻮﻝ ﺃﻋﺪﺍﺩﺍﹰ‬
‫ﻣﺘﺒﺎﻳﻨﺔ ﻣﻮﺟﺒﺔ ﲤﺎﻣﺎﹰ‪.‬‬
‫ ‪ ä‬ﻟﻨﺘﺄ ﻣ‪‬ﻞ ﺣﺎﻟﺔ ‪ ، a = 0‬ﻭ ﻟﻴﻜﻦ ) ‪ ( x , y, z‬ﺣﻼﹰ ﻟﻠﺠﻤﻠﺔ ﺍﳌﻌﻄﺎﺓ‪ .‬ﻋﻨﺪﺋﺬ ﻧﺴﺘﻨﺘﺞ ﻣﻦ ﺍﳌﻌﺎﺩﻟﺔ‬
‫‪ x + y = −z‬ﺃ ﻥﹼ ‪ ، x 2 + y 2 + 2xy − z 2 = 0‬ﻭ ﺑﺎ ﻻﺳﺘﻔﺎ ﺩ ﺓ ﻣﻦ ‪ xy = z 2‬ﳒﺪ‬
‫‪ ، x 2 + y 2 + z 2 = 0‬ﻓﻼ ﺑ‪‬ﺪ‪ ‬ﺃﻥ ﻳﻜﻮﻥ ‪ . b = 0‬ﻭﻋﻨﺪﺋﺬ ﺃ ‪‬ﻳﺎﹰ ﻛﺎﻧﺖ ‪ z‬ﻛﺎﻥ ‪ x‬ﻭ ‪ y‬ﳘﺎ‬
‫ﺣﻼﹼ ﺍﳌﻌﺎﺩﻟﺔ ‪ T 2 + zT + z 2 = 0‬ﺑﺎﻟﻨﺴﺒﺔ ﺇﱃ ﺍ‪‬ﻬﻮﻝ ‪ ،T‬ﻭﻣﻨﻪ ﳎﻤﻮﻋﺔ ﺍﳊﻠﻮﻝ‬

‫} ‪{( j z, j2 z, z ), z ∈ ℂ } ∪ {( j2 z, j z, z ), z ∈ ℂ‬‬
‫ﻣﻊ ‪ . j = − 21 + i 23‬ﻭﻻ ﳝﻜﻦ ﰲ ﻫﺬﻩ ﺍﳊﺎﻟﺔ ﺃﻥ ﺗﻜﻮﻥ ﺍﳊﻠﻮﻝ ﺃﻋﺪﺍﺩﺍﹰ ﻣﺘﺒﺎﻳﻨﺔ ﻣﻮﺟﺒﺔ ﲤﺎﻣﺎﹰ‪.‬‬
‫‪ ä‬ﺣﺎﻟﺔ ‪ . a ≠ 0‬ﻟﻴﻜﻦ ) ‪ ( x , y, z‬ﺣﻼﹰ ﻟﻠﺠﻤﻠﺔ ﺍﳌﻌﻄﺎﺓ‪ .‬ﺑﺘﺮﺑﻴﻊ ﺍﳌﻌﺎﺩﻟﺔ ﺍﻷﻭﱃ ﳒﺪ‬
‫‪2‬‬
‫ ‪a 2 = x‬‬
‫‪+ y 2 + z 2 + 2(xy‬‬
‫) ‪ + yz + zx‬‬

‫‪b2‬‬

‫‪z2‬‬

‫ ‪= b 2 + 2z (z‬‬
‫‪+ y + x ) = b 2 + 2az‬‬
‫‪a‬‬

‫‪a 2 − b2‬‬
‫ﻭﻣﻨﻪ ﻧﺴﺘﻨﺘﺞ ﺃﻥﹼ‬
‫‪2a‬‬

‫= ‪ . z‬ﺇﺫﻥ‬

‫‪ a 2 − b 2 2‬‬
‫‪a 2 + b2‬‬
‫‪xy = ‬‬
‫ﻭ‬
‫‪x‬‬
‫‪+‬‬
‫‪y‬‬
‫=‬
‫‪a‬‬
‫‪−‬‬
‫‪z‬‬
‫=‬
‫‪‬‬
‫‪2a‬‬
‫‪ 2a ‬‬

‫ﻭﻋﻠﻴﻪ ﻓﺈﻥﹼ ‪ x‬ﻭ ‪ y‬ﳘﺎ ﺣﻼﹼ ﺍﳌﻌﺎﺩﻟﺔ‬
‫‪ a 2 − b 2 2‬‬
‫‪a 2 + b2‬‬
‫‪T −‬‬
‫‪T + ‬‬
‫‪=0‬‬
‫‪ 2a ‬‬
‫‪2a‬‬
‫‪2‬‬

‫)‬

‫∗‬

‫(‬

‫ﺑﺎﻟﻨﺴﺒﺔ ﺇﱃ ﺍ‪‬ﻬﻮﻝ ‪ ،T‬ﻭﻫﺬﻩ ﺍﳌﻌﺎﺩﻟﺔ ﺗ‪‬ﻜﺎﻓﺊ‬
‫‪2‬‬
‫‪2 2‬‬
‫‪2 2‬‬
‫‪2 2‬‬
‫) ‪3a 2 − b 2 )( 3b 2 − a 2‬‬
‫‪‬‬
‫‪ 2‬‬
‫‪ 2‬‬
‫( = ‪T − a + b  =  a + b  −  2a − 2b ‬‬
‫‪4a ‬‬
‫‪4a‬‬
‫‪‬‬
‫‪ 4a ‬‬
‫‪‬‬
‫‪‬‬
‫‪16a 2‬‬

‫ﻭﻫﻜﺬﺍ ﳒﺪ ﺣﻠﻮﻝ ﺍﳉﻤﻠﺔ ﺍﳌﻌﻄﺎﺓ ﰲ ‪. ℂ‬‬
‫‪17‬‬

‫ﺍﻷﻭﳌﺒﻴﺎﺩ ﺍﻟﻌﺎﳌﻲ ﻟﻠﺮﻳﺎﺿﻴ‪‬ﺎﺕ‬

‫‪18‬‬

‫‪ ä‬ﻟﻨﻔﺘﺮﺽ ﺍﻵﻥ ﺃﻥﹼ ﺍﳊﻠﻮﻝ ) ‪ ( x , y, z‬ﺃﻋﺪﺍﺩ‪ ‬ﻣﻮﺟﺒﺔ ﲤﺎﻣﺎﹰ ﻭﻣﺘﺒﺎﻳﻨﺔ‪ ،‬ﻋﻨﺪﺋﺬ ﳚﺐ ﺃﻥ ﻳﻜﻮﻥ‬
‫‪ a ∈ ℝ∗+‬ﻷﻥﹼ ‪ ، x + y + z = a‬ﻭﳚﺐ ﺃﻥ ﻳﻜﻮﻥ ∗‪ . b ∈ ℝ‬ﻭﻧﺴﺘﻨﺘﺞ ﻣﻦ ﺍﳌﺴﺎﻭﺍﺓ‬
‫‪a 2 − b2‬‬
‫‪2a‬‬

‫= ‪ z‬ﺃﻥﹼ ‪ . b < a‬ﻭﺣﺘ‪‬ﻰ ﻳﻜﻮﻥ ‪ x‬ﻭ ‪ y‬ﻣﻮﺟﺒﲔ ﲤﺎﻣﺎﹰ ﻭﳐﺘﻠﻔﲔ ﻳﻜﻔﻲ ﺃﻥ ﻳﻜﻮﻥ‬

‫ﺟﺬﺭﺍ ﺍﳌﻌﺎﺩﻟﺔ ) ∗ ( ﺣﻘﻴﻘﻴ‪‬ﺎﻥ ﻭﳐﺘﻠﻔﺎﻥ‪ ،‬ﻭﻫﺬﺍ ‪‬ﻳﻜﺎﻓﺊ‬

‫‪( 3a 2 − b 2 )( 3b 2 − a 2 ) > 0‬‬
‫ﻭﻷﻥﹼ ‪ b < a‬ﺍﺳﺘﻨﺘﺠﻨﺎ ﺃﻥﹼ ‪ 3a 2 − b 2 > 0‬ﻭﻣﻦ ﺛﹶﻢ‪ 3b 2 > a 2 ‬ﺃﻭ ‪. 3 b > a‬‬
‫ﻟﻘﺪ ﺃﺛﺒﺘﻨﺎ ﺇﺫﻥ ﺃﻥﹼ ﺍﻟﺸﺮﻁ ‪ b < a < 3 b‬ﺷﺮﻁﹲ ﻻﺯﻡ‪ ‬ﻟﺘﻜﻮﻥ ﺣﻠﻮﻝ ﺍﳉﻤﻠﺔ ﺍﳌﺪﺭﻭﺳﺔ ﺃﻋﺪﺍﺩﺍﹰ‬
‫ﺣﻘﻴﻘﻴ‪‬ﺔ ﻣﻮﺟﺒﺔ ﻭﻣﺘﺒﺎﻳﻨﺔ‪.‬‬
‫ﻭﺑﺎﻟﻌﻜﺲ‪ ،‬ﰲ ﺣﺎﻟﺔ ﻋﺪﺩﻳﻦ ﺣﻘﻴﻘﻴ‪‬ﲔ ‪ a‬ﻭ ‪ b‬ﻳ‪‬ﺤﻘﹼﻘﺎﻥ ‪ b < a < 3 b‬ﺗﻜﻮﻥ ﺣﻠﻮﻝ ﺍﳉﻤﻠﺔ‬
‫ﺍﳌﻌﻄﺎﺓ ﺃﻋﺪﺍﺩﺍﹰ ﺣﻘﻴﻘﻴ‪‬ﺔ ﻣﻮﺟﺒﺔ ﲤﺎﻣﺎﹰ ﻭﻣﺘﺒﺎﻳﻨﺔ‪ ،‬ﻭﻫﻲ‬

‫) ‪( 3a 2 − b 2 )( 3b 2 − a 2‬‬

‫‪a 2 + b2 +‬‬

‫‪4a‬‬
‫‪2‬‬

‫) ‪( 3a − b 2 )( 3b 2 − a 2‬‬

‫‪2‬‬

‫‪2‬‬

‫‪2‬‬

‫‪2‬‬

‫‪a +b −‬‬

‫‪4a‬‬
‫‪a −b‬‬
‫‪2a‬‬

‫= ‪x‬‬
‫= ‪y‬‬
‫= ‪z‬‬

‫ﺃﻭ‬

‫) ‪( 3a 2 − b 2 )( 3b 2 − a 2‬‬

‫‪a 2 + b2 −‬‬

‫‪4a‬‬

‫) ‪( 3a 2 − b 2 )( 3b 2 − a 2‬‬

‫‪a 2 + b2 +‬‬

‫‪4a‬‬
‫‪2‬‬

‫= ‪x‬‬
‫= ‪y‬‬

‫‪2‬‬

‫‪a −b‬‬
‫‪2a‬‬
‫ﻓﺎﻟﺸﺮﻁ ﺍﳌﻄﻠﻮﺏ ﻫﻮ ﺃﻥ ﻳﻜﻮﻥ ‪ a‬ﻭ ‪ b‬ﻋﺪﺩﻳﻦ ﺣﻘﻴﻘﻴ‪‬ﲔ ﻳ‪‬ﺤﻘﹼﻘﺎﻥ ‪3 b‬‬

‫= ‪z‬‬

‫< ‪.b <a‬‬

‫ ‬

‫ ﺍﻷﻋﺪﺍﺩ ‪ a‬ﻭ ‪ b‬ﻭ ‪ c‬ﻫﻲ ﺃﻃﻮﺍﻝ ﺃﺿﻼﻉ ﻣﺜﻠﹼﺚ ﻣﺴﺎﺣﺘﻪ ‪ A‬ﺃﺛﺒﺖ ﺻﺤ‪‬ﺔ ﺍﳌﺘﺮﺍﺟﺤﺔ‬
‫‪a 2 + b 2 + c 2 ≥ 4 3A‬‬

‫ﻭﺑﻴ‪‬ﻦ ﻣﱴ ﺗﺘﺤﻘﹼﻖ ﺍﳌﺴﺎﻭﺍﺓ ﻓﻴﻬﺎ‪.‬‬

‫‪ù‬‬

‫ﻋﺎﻡ ‪1961‬‬

‫‪19‬‬

‫ ﻃﺮﻳﻘﺔ ﺃﻭﱃ ‪ :‬ﳝﻜﻦ ﺍﻻﻧﻄﻼﻕ ﻣﻦ ﻋﻼﻗﺔ ‪ Heron‬ﺍﻟﱵ ﺗ‪‬ﻌﻄﻲ ﻣﺴﺎﺣﺔ ﺍﳌﺜﻠﹼﺚ ﺑﺪﻻﻟﺔ ﺃﻃﻮﺍﻝ ﺃﺿﻼﻋﻪ‬
‫‪a +b +c‬‬
‫‪2‬‬

‫=‪p‬‬

‫ ‬

‫=‪A‬‬

‫) ‪p ( p − a )( p − b )( p − c‬‬

‫ﻭﻣﻨﻪ‬
‫‪1‬‬
‫) ‪(a + b + c )( c + b − a )( a + c − b )( a + b − c‬‬
‫‪4‬‬
‫‪1‬‬
‫=‬
‫) ‪( (a + b )2 − c 2 )(c 2 − (b − a )2‬‬
‫‪4‬‬
‫‪1‬‬
‫=‬
‫‪2 (a 2b 2 + b 2c 2 + c 2a 2 ) − a 4 − b 4 − c 4‬‬
‫‪4‬‬

‫=‪A‬‬

‫ﻭﻋﻠﻰ ﻫﺬﺍ ﺇﺫﺍ ﻋﺮ‪‬ﻓﻨﺎ‬
‫‪2‬‬

‫‪2‬‬

‫) ‪∆ = ( a 2 + b 2 + c 2 ) − ( 4 3A‬‬

‫ﻛﺎﻥ ﻟﺪﻳﻨﺎ‬
‫‪2 2‬‬

‫) ‪) − 3 ( 2 (a 2b 2 + b 2c 2 + c 2a 2 ) − a 4 − b 4 − c 4‬‬

‫‪∆ = (a 2 + b 2 + c‬‬

‫) ) ‪= 4 (a 4 + b 4 + c 4 − ( a 2b 2 + b 2c 2 + c 2a 2‬‬
‫) ) ‪= 2 ( ( a 4 + b 4 − 2a 2b 2 ) + (b 4 + c 4 − 2b 2c 2 ) + (c 4 + a 4 − 2a 2c 2‬‬

‫)‬

‫‪2‬‬

‫‪2‬‬

‫‪2‬‬

‫(‬

‫) ‪= 2 ( a 2 − b 2 ) + (b 2 − c 2 ) + ( a 2 − a 2‬‬

‫ﺇﺫﻥ ‪ ∆ ≥ 0‬ﻣﻊ ﻣﺴﺎﻭﺍﺓ ﺇﺫﺍ ﻭﻓﻘﻂ ﺇﺫﺍ ﻛﺎﻥ ‪. a = b = c‬‬
‫ﻃﺮﻳﻘﺔ ﺛﺎﻧﻴﺔ ‪ :‬ﻟﻨﺮﻣﺰ ﺑﺎﻟﺮﻣﺰ ‪ h‬ﺇﱃ ﻃﻮﻝ ﺍﻻﺭﺗﻔﺎﻉ ﺍﻟﻨﺎﺯﻝ ﻣﻦ ﺃﺣﺪ‬
‫ﺭﺅﻭﺱ ﺍﳌﺜﻠﹼﺚ‪ ،‬ﳔﺘﺎﺭ ﺍﻟﺮﺃﺱ ﺍﳌﻮﺍﻓﻖ ﻟﻠﺰﺍﻭﻳﺔ ﺍﳌﻨﻔﺮﺟﺔ ﰲ ﺣﺎﻝ ﻛﻮﻥ‬
‫ﺍﳌﺜﻠﹼﺚ ﻣﻨﻔﺮﺝ ﺍﻟﺰﺍﻭﻳﺔ‪ .‬ﻳﻘﺴﻢ ﻣﻮﻗﻊ ﻫﺬﺍ ﺍﻻﺭﺗﻔﺎﻉ ﺍﻟﻀﻠﻊ ﺍﳌﻘﺎﺑﻠﺔ ﺇﱃ‬
‫ﻗﻄﻌﺘﲔ ﻃﻮﻟﻴﻬﻤﺎ ‪ s‬ﻭ ‪ . t‬ﻋﻨﺪﺋﺬ ﻳﻜﻮﻥ‬

‫‪h‬‬
‫‪t‬‬

‫‪a 2 + b 2 + c 2 = h 2 + s 2 + h 2 + t 2 + ( s + t )2‬‬

‫) ‪= 2 ( h 2 + s 2 + t 2 + st‬‬

‫ﻭ‬
‫‪2A = ( s + t ) h‬‬

‫ﻓﺈﺫﺍ ﻋﺮ‪‬ﻓﻨﺎ‬
‫‪δ = ( a 2 + b 2 + c 2 ) − 4 3A‬‬

‫‪s‬‬

‫ﺍﻷﻭﳌﺒﻴﺎﺩ ﺍﻟﻌﺎﳌﻲ ﻟﻠﺮﻳﺎﺿﻴ‪‬ﺎﺕ‬

‫‪20‬‬

‫ﻛﺎﻥ ﻟﺪﻳﻨﺎ‬
‫‪δ = 2 ( h 2 + s 2 + t 2 + st ) − 2 3 ( s + t ) h‬‬

‫)‬

‫‪2‬‬

‫‪+ t ) ) + s 2 + t 2 + st − 34 ( s + t )2‬‬

‫(‪3‬‬
‫‪2 s‬‬

‫‪+ t )) +‬‬

‫(‪3‬‬
‫‪2 s‬‬

‫)‬

‫‪− t )2‬‬

‫‪1 (s‬‬
‫‪4‬‬

‫‪2‬‬

‫(‬
‫‪= 2 (( h −‬‬
‫‪= 2 (h −‬‬

‫ﻭﻫﺬﺍ ﻳ‪‬ﺜﺒﺖ ﺃ ﻥﹼ ‪ δ ≥ 0‬ﻣﻊ ﻣﺴﺎﻭﺍﺓ ﺇﺫﺍ ﻭﻓﻘﻂ ﺇﺫﺍ ﻛﺎﻥ ‪ s = t‬ﻭ ) ‪+ t‬‬

‫(‪3‬‬
‫‪2 s‬‬

‫‪‬ﻳﻜﺎﻓﺊ ﻛﻮﻥ ﺍﳌﺜﻠﹼﺚ ﻣﺘﺴﺎﻭﻱ ﺍﻷﺿﻼﻉ‪.‬‬

‫= ‪ ، h‬ﻭﻫﺬﺍ‬
‫‪ù‬‬

‫ ‬
‫ ﺣﻞﱠ ﰲ ‪ ℝ‬ﺍﳌﻌﺎﺩﻟﺔ ‪ cosn x − sinn x = 1‬ﻭ ‪ n‬ﻋﺪﺩ‪ ‬ﻃﺒﻴﻌﻲ ﻣﻦ ‪. ℕ‬‬
‫∗‬

‫ ‬

‫ ‬

‫ﺣﺎﻟﺔ ‪ . n = 1‬ﰲ ﻫﺬﻩ ﺍﳊﺎﻟﺔ‪ ،‬ﺗ‪‬ﻜﺎﻓﺊ ﺍﳌﻌﺎﺩﻟﺔﹸ ‪ cos x − sin x = 1‬ﺍﳌﻌﺎﺩﻟﺔﹶ‬
‫) ‪cos ( x + π4 ) = cos ( π4‬‬

‫ﻭﺣﻠﻮﻝ ﻫﺬﻩ ﺍﻷﺧﲑﺓ ﻫﻲ ) ‪. 2πℤ ∪ ( − π2 + 2πℤ‬‬
‫ ﺣﺎﻟﺔ ‪ . n = 2‬ﰲ ﻫﺬﻩ ﺍﳊﺎﻟﺔ‪ ،‬ﺗ‪‬ﻜﺎﻓﺊ ﺍﳌﻌﺎﺩﻟﺔ ‪ cos2 x − sin2 x = 1‬ﺍﳌﹸﻌﺎﺩﻟﺔﹶ‬
‫‪cos 2x = 1‬‬

‫ﻭﺣﻠﻮﻝ ﻫﺬﻩ ﺍﻷﺧﲑﺓ ﻫﻲ ‪. πℤ‬‬
‫ ﺣﺎﻟﺔ ‪ . n > 2‬ﻟﻴﻜﻦ ‪ x‬ﺣﻼﹼ ﻟﻠﻤﻌﺎﺩﻟﺔ ﺍﳌﻄﺮﻭﺣﺔ‪ ،‬ﻋﻨﺪﺋﺬ‪‬‬
‫‪≤ cos2 x + sin2 x = 1‬‬

‫‪n‬‬

‫‪+ sin x‬‬

‫‪n‬‬

‫‪1 = cosn x − sinn x ≤ cos x‬‬

‫ﺍﳌﺴﺎﻭﺍﺓ ﰲ ﺍﳌﺘﺮﺍﺟﺤﺔ ﺍﻟﺴﺎﺑﻘﺔ ﺗﻘﺘﻀﻲ ﺃﻥﹼ‬
‫‪= cos2 x‬‬

‫‪n‬‬

‫‪ cosn x = cos x‬ﻭ ‪= sin2 x‬‬

‫‪n‬‬

‫‪− sinn x = sin x‬‬

‫ ﺇﺫﺍ ﻛﺎﻥ ‪ n‬ﺯﻭﺟﻴ‪‬ﺎﹰ‪ ،‬ﺍﺳﺘﻨﺘﺠﻨﺎ ﳑﺎ ﺳﺒﻖ ﺃﻥﹼ ‪ sin x = 0‬ﻭ ‪ cos2 x = 1‬ﻓﻤﺠﻤﻮﻋﺔ‬
‫ﺍﳊﻠﻮﻝ ﳏﺘﻮﺍﺓ ﰲ ‪ . πℤ‬ﻭﺑﺎﻟﻌﻜﺲ‪ ،‬ﻧﺮﻯ ﻣﺒﺎﺷﺮﺓ ﺃﻥﹼ ﲨﻴﻊ ﻋﻨﺎﺻﺮ ‪ πℤ‬ﻫﻲ ﺣﻠﻮﻝ ﻟﻠﻤﻌﺎﺩﻟﺔ‬
‫ﺍﳌﺪﺭﻭﺳﺔ ﰲ ﻫﺬﻩ ﺍﳊﺎﻟﺔ‪.‬‬
‫ ﺇﺫﺍ ﻛﺎﻥ ‪ n‬ﻓﺮﺩ‪‬ﻳﺎﹰ‪ ،‬ﺍﺳﺘﻨﺘﺠﻨﺎ ﳑﺎ ﺳﺒﻖ ﺃﻧ‪‬ﻪ ﺇﻣ‪‬ﺎ ‪ sin x = −1‬ﺃﻭ ‪ cos x = 1‬ﻓﻤﺠﻤﻮﻋﺔ‬
‫ﺍﳊﻠﻮﻝ ﳏﺘﻮﺍﺓ ﰲ ) ‪ ، 2πℤ ∪ ( − π2 + 2πℤ‬ﻭﺑﺎﻟﻌﻜﺲ‪ ،‬ﻧﺮﻯ ﻣﺒﺎﺷﺮﺓ ﺃﻥﹼ ﲨﻴﻊ ﻋﻨﺎﺻﺮ‬
‫ﻫﺬﻩ ﺍ‪‬ﻤﻮﻋﺔ ﻫﻲ ﺣﻠﻮﻝ ﻟﻠﻤﻌﺎﺩﻟﺔ ﺍﳌﺪﺭﻭﺳﺔ ﰲ ﻫﺬﻩ ﺍﳊﺎﻟﺔ‪.‬‬
‫‪ù‬‬
‫ﻭﺑﺬﺍ ﻧﻜﻮﻥ ﻗﺪ ﺃﻭﺟﺪﻧﺎ ﺣﻠﻮﻝ ﺍﳌﻌﺎﺩﻟﺔ ﺍﳌﻌﻄﺎﺓ ﰲ ﲨﻴﻊ ﺍﻷﺣﻮﺍﻝ‪.‬‬
‫ ‬
‫‪This book is downloaded from this site:‬‬
‫‪www.syCourses.com‬‬

‫ﻋﺎﻡ ‪1961‬‬

‫‪21‬‬

‫ ﺍﻟﻨﻘﻄﺔ ‪ P‬ﻫﻲ ﻧﻘﻄﺔ ﺩﺍﺧﻞ ﻣﺜﻠﹼﺚ ‪ . ABC‬ﻳﺘﻘﺎﻃﻊ ) ‪ ( PA‬ﻣﻊ ) ‪ ( BC‬ﰲ ‪ ، A′‬ﻭﻳﺘﻘﺎﻃﻊ‬
‫) ‪ ( PB‬ﻣﻊ ) ‪ ( AC‬ﰲ ‪ ، B ′‬ﻭﻳﺘﻘﺎﻃﻊ ) ‪ ( PC‬ﻣﻊ ) ‪ ( AB‬ﰲ ‪ .C ′‬ﺃﺛﺒﺖ ﺃﻥﹼ‬
‫‪AP BP CP‬‬
‫‪,‬‬
‫‪,‬‬
‫‪≤2‬‬
‫‪( PA‬‬
‫) ‪′ PB ′ PC ′‬‬

‫‪ min‬ﻭ‬

‫‪AP BP CP‬‬
‫‪,‬‬
‫‪,‬‬
‫‪≥2‬‬
‫‪( PA‬‬
‫) ‪′ PB ′ PC ′‬‬

‫‪max‬‬

‫ ﳌﹼﺎ ﻛﺎﻧﺖ ﺍﻟﻨﻘﻄﺔ ‪ P‬ﺗﻘﻊ ﺩﺍﺧﻞ ﺍﳌﺜﻠﹼﺚ ‪ ABC‬ﺍﺳﺘﻨﺘﺠﻨﺎ ﺃ ﻧ‪‬ﻬﺎ ﻣﺮﻛﺰ ﺍﻷﺑﻌﺎﺩ ﺍﳌﺘﻨﺎﺳﺒﺔ ﻟﻠﻨﻘﺎﻁ‬
‫) ‪ ( A, α‬ﻭ ) ‪ ( B, β‬ﻭ ) ‪ (C , γ‬ﻣﻊ ) ‪ ( α, β, γ‬ﻣﻦ ‪ ( ℝ∗+ )3‬ﺗ‪‬ﺤﻘﹼﻖ ‪. α + β + γ = 1‬‬
‫ ﻟﺘﻜﻦ ‪ X‬ﻣﺮﻛﺰ ﺍﻷﺑﻌﺎﺩ ﺍﳌﺘﻨﺎﺳﺒﺔ ﻟﻠﻨﻘﻄﺘﲔ ) ‪ ( B, β‬ﻭ ) ‪ ، (C , γ‬ﻋﻨﺪﺋﺬ ﺗﻨﺘﻤﻲ ‪ X‬ﺇﱃ‬
‫ﺍﳌﺴﺘﻘﻴﻢ ) ‪ ( BC‬ﻛﻤﺎ ﺇ‪‬ﺎ ﺗﻨﺘﻤﻲ ﺇﱃ ﺍﳌﺴﺘﻘﻴﻢ ) ‪ ( AP‬ﻷ ﻥﹼ ‪ P‬ﻣﺮﻛﺰ ﺍﻷﺑﻌﺎﺩ ﺍﳌﺘﻨﺎﺳﺒﺔ‬
‫ﻟﻠﻨﻘﻄﺘﲔ ) ‪ ( X , β + γ‬ﻭ ) ‪ . ( A, α‬ﺇﺫﻥ ‪ . X = A′‬ﻭﻋﻠﻴﻪ‬








‫‪αPA + ( 1 − α ) PA′ = 0‬‬

‫ﺇﺫﻥ‬
‫‪AP‬‬
‫‪1‬‬
‫‪= −1‬‬
‫‪α‬‬
‫‪PA′‬‬

‫ ﻭﳒﺪ ﺑﺄﺳﻠﻮﺏ ﳑﺎﺛﻞ ﺃﻥﹼ‬
‫‪CP‬‬
‫‪1‬‬
‫‪BP‬‬
‫‪1‬‬
‫ﻭ ‪= −1‬‬
‫‪= −1‬‬
‫‪γ‬‬
‫‪β‬‬
‫‪PC ′‬‬
‫‪PB ′‬‬

‫ ﻭﻫﻜﺬﺍ ﻧﺮﻯ ﺃﻥﹼ‬
‫‪AP BP CP‬‬
‫‪1 1 1‬‬
‫‪1‬‬
‫‪,‬‬
‫‪,‬‬
‫= ‪= min  , ,  − 1‬‬
‫‪−1‬‬
‫‪( PA‬‬
‫)‬
‫‪′ PB ′ PC ′‬‬
‫‪α β γ ‬‬
‫) ‪max ( α, β, γ‬‬
‫‪1 1 1‬‬
‫‪AP BP CP‬‬
‫‪1‬‬
‫( ‪max‬‬
‫‪,‬‬
‫‪,‬‬
‫= ‪= max  , ,  − 1‬‬
‫‪−1‬‬
‫)‬
‫‪α β γ ‬‬
‫) ‪min ( α, β, γ‬‬
‫‪PA′ PB ′ PC ′‬‬
‫‪min‬‬

‫ﻭﻟﻜﻦ ﻣﻊ ﺍﻟﺸﺮﻁ ‪ α + β + γ = 1‬ﻧﺮﻯ ﻣﺒﺎﺷﺮﺓ ﺃﻥﹼ‬
‫‪1‬‬
‫‪1‬‬
‫≤ ) ‪ min ( α, β, γ‬ﻭ‬
‫‪3‬‬
‫‪3‬‬

‫≥ ) ‪max ( α, β, γ‬‬

‫ﻭﻫﺬﺍ ﻳﻘﺘﻀﻲ ﺃﻥﹼ‬
‫‪AP BP CP‬‬
‫‪,‬‬
‫‪,‬‬
‫‪≤2‬‬
‫‪( PA‬‬
‫) ‪′ PB ′ PC ′‬‬

‫‪ min‬ﻭ‬

‫‪AP BP CP‬‬
‫‪,‬‬
‫‪,‬‬
‫‪≥2‬‬
‫‪( PA‬‬
‫) ‪′ PB ′ PC ′‬‬

‫ﻭﳘﺎ ﺍﳌﺘﺮﺍﺟﺤﺘﺎﻥ ﺍﳌﻄﻠﻮﺑﺘﺎﻥ‪.‬‬

‫‪max‬‬

‫‪ù‬‬
‫ ‬

‫ﺍﻷﻭﳌﺒﻴﺎﺩ ﺍﻟﻌﺎﳌﻲ ﻟﻠﺮﻳﺎﺿﻴ‪‬ﺎﺕ‬

‫‪22‬‬

‫ ﺃﻧﺸﺊ ﻣﺜﻠﹼﺜﺎﹰ ‪ ، ABC‬ﻓﻴﻪ ‪ AC = b‬ﻭ ‪ AB = c‬ﻭ ‪ = α‬‬
‫‪ AMB‬ﺣﻴﺚ ‪ M‬ﻫﻲ ﻣﻨﺘﺼﻒ‬

‫ﺍﻟﻀﻠﻊ ] ‪ . [ BC‬ﺃﺛﺒﺖ ﺃ ﻥﹼ ﻫﺬﺍ ﺍﻹﻧﺸﺎﺀ ﳑﻜﻦ ﺇﺫﺍ ﻭﻓﻘﻂ ﺇﺫﺍ ﻛﺎﻥ ‪، b tan ( α2 ) ≤ c < b‬‬
‫ﻭﺗﻔﺤ‪‬ﺺ ﺣﺎﻟﺔ ﺍﳌﺴﺎﻭﺍﺓ‪.‬‬
‫ ﺍﻟﺘﺤﻠﻴﻞ ‪ :‬ﻟﻨﻔﺘﺮﺽ ﺃ ﻥﹼ ﺍﻹﻧﺸﺎﺀ ﻣ‪‬ﻨﺠﺰ‪ ، ‬ﻛﻤﺎ ﰲ ﺍﻟﺸﻜﻞ ﺍ‪‬ﺎﻭﺭ‪ .‬ﻭﻟﻨﺘﺄ ﻣ‪‬ﻞ ﺍﻟﻨﻘﻄﺔ ‪ N‬ﻧﻈﲑﺓ ‪B‬‬

‫ﺑﺎﻟﻨﺴﺒﺔ ﺇﱃ ‪. A‬‬
‫ ﳌﹼﺎ ﻛﺎ ﻥ ) ‪ ( AM ) ( NC‬ﺍﺳﺘﻨﺘﺠﻨﺎ ﺃ ﻥﹼ ﺍﻟﺰ ﺍﻭ ﻳﺘﲔ‬
‫‪ NCB‬ﻭ ‬
‫ ‬
‫‪ AMB‬ﻣﺘﺴﺎﻭﻳﺘﺎﻥ‪ ،‬ﻭﻣﻦ ﺛﹶﻢ‪‬‬
‫ ‬

‫‪ =α‬‬
‫‪NCB‬‬
‫ﻭﻋﻠﻰ ﻫﺬﺍ‪ ،‬ﺗﻨﺘﻤﻲ ﺍﻟﻨﻘﻄﺔ ‪ C‬ﺇﱃ ﺍﻟﻘﻮﺱ ‪ C‬ﻣﻦ ﺍﻟﺪﺍﺋﺮﺓ‬

‫‪X‬‬

‫‪C‬‬

‫ ‪C‬‬

‫‪α‬‬

‫‪C‬‬
‫‪b‬‬

‫‪M‬‬

‫‪α‬‬

‫ﺍﻟﱵ ﲤﺮ ﺑﺎﻟﻨﻘﻄﺘﲔ ‪ N‬ﻭ ‪ ، B‬ﻭﺗ‪‬ﺮﻯ ﻣﻦ ﻧﻘﺎﻃﻬﺎ ﺍﻟﻘﻄﻌﺔ‬
‫] ‪ [ NB‬ﺑﺰﺍﻭﻳﺔ ﻗﺪﺭﻫﺎ ‪. α‬‬
‫ ﻭﻣﻦ ﺟﻬﺔ ﺃﺧﺮﻯ ‪ ،‬ﺗﻨﺘﻤﻲ ‪ C‬ﺇﱃ ﺍﻟﺪﺍﺋﺮﺓ ) ‪ C = C ( A, b‬ﺍﻟﱵ ﻣﺮﻛﺰﻫﺎ ‪ A‬ﻭﻧﺼﻒ‬
‫ﻗﻄﺮﻫﺎ ‪. b‬‬
‫ ﺇ ﻥﹼ ﺷﺮﻁ ﺗﻘﺎﻃﻊ ﺍﻟﺪﺍﺋﺮﺓ ‪ C‬ﻭﺍﻟﻘﻮﺱ ‪ C‬ﻫﻮ ﺃﻥ ﻳﻜﻮﻥ ‪ ، c < b ≤ AX‬ﺣﻴﺚ ﺭﻣﺰﻧﺎ‬
‫‪B‬‬

‫‪c‬‬

‫‪A‬‬

‫‪c‬‬

‫‪N‬‬

‫ﺑﺎﻟﺮﻣﺰ ‪ X‬ﺇﱃ ﻧﻘﻄﺔ ﺗﻘﺎﻃﻊ ﳏﻮﺭ ﺍﻟﻘﻄﻌﺔ ] ‪ [ NB‬ﻣﻊ ﺍﻟﻘﻮﺱ ‪ . C‬ﻭﳌﹼﺎ ﻛﺎﻥ ‪ = α‬‬
‫‪AXB‬‬
‫‪2‬‬
‫‪c‬‬
‫= ‪ ، AX‬ﻭﻋﻠﻴﻪ ﻳ‪‬ﺼﺒﺢ ﺷﺮﻁ ﺇﻣﻜﺎﻥ ﺍﻹﻧﺸﺎﺀ ﺑﺎﻟﺸﻜﻞ‬
‫ﺍﺳﺘﻨﺘﺠﻨﺎ ﺃﻥﹼ‬
‫) ‪tan ( α /2‬‬
‫‪b tan ( α2 ) ≤ c < b‬‬

‫ﺃﻣ‪‬ﺎ ﺣﺎﻟﺔ ) ‪ c = b tan ( α2‬ﻓﺘﻮﺍﻓﻖ ﺍﳊﺎﻟﺔ ﺍﻟﱵ ﻳﻜﻮﻥ ﻓﻴﻬﺎ ﺍﳌﺜﻠﹼﺚ ‪ ABC‬ﻗﺎﺋﻤﺎﹰ ﰲ ‪. A‬‬
‫ﺍﻹﻧﺸﺎﺀ ‪:‬‬
‫ ﻧﻨﺸﺊ ﺍﻟﻘﻄﻌﺔ ﺍﳌﺴﺘﻘﻴﻤﺔ ] ‪ [ NB‬ﺑﻄﻮﻝ ﻗﺪﺭﻩ ‪ ، 2c‬ﻭﻧﻌﻴ‪‬ﻦ ﻣﻨﺘﺼﻔﻬﺎ ‪. A‬‬

ﻧﻨﺸﺊ ﺍﻟﻘﻮﺱ ‪ C‬ﺍﻟﺬﻱ ﳝﺜﹼﻞ ﳎﻤﻮﻋﺔ ﺍﻟﻨﻘﺎﻁ ﺍﻟﱵ ﺗﺮﻯ ﻣﻨﻬﺎ ﺍﻟﻘﻄﻌﺔ ﺍﳌﺴﺘﻘﻴﻤﺔ ] ‪ [ NB‬ﺑﺰﺍﻭﻳﺔ‬
‫ﻗﺪﺭﻫﺎ ‪ ) . α‬ﻟﺘﺤﻘﻴﻖ ﺫﻟﻚ ﺃﻧﺸﺊ ﺍﳌﺜﻠﹼﺚ ‪ NBB ′‬ﺍﻟﻘﺎﺋﻢ ﰲ ‪ B‬ﻭﻓﻴﻪ ‪ ′ = π − α‬‬
‫‪، BNB‬‬
‫‪2‬‬
‫ﻓﻴﻜﻮﻥ ﻣﻨﺘﺼﻒ ﺍﻟﻮﺗﺮ ] ‪ [ BB ′‬ﻫﻮ ﻣﺮﻛﺰ ﺍﻟﺪﺍﺋﺮﺓ ‪( . C‬‬
‫ ﻋﻴ‪‬ﻦ ‪ C‬ﻧﻘﻄﺔ ﺗﻘﺎﻃﻊ ﺍﻟﻘﻮﺱ ‪ C‬ﻣﻊ ﺍﻟﺪﺍﺋﺮﺓ ) ‪ .C ( A, b‬ﻭﻻﺣﻆ ﺃ ﻥﹼ ﻫﻨﺎﻙ ﺣﻠﻴ‪‬ﻦ ﻟﻠﻤﺴﺄﻟﺔ‬
‫ﺑﻮﺟﻪ ﻋﺎﻡ‪.‬‬
‫‪ù‬‬

‫ ‬
‫‪This book is downloaded from this site:‬‬
‫‪www.syCourses.com‬‬

‫ﻋﺎﻡ ‪1961‬‬

‫‪23‬‬

‫ ﻧ‪‬ﻌﻄﻰ ﺛﻼﺙ ﻧﻘﺎﻁ ﻟﻴﺴﺖ ﻋﻠﻰ ﺍﺳﺘﻘﺎﻣﺔ ﻭﺍﺣﺪﺓ‪ ،‬ﻭﻣﺴﺘﻮﻳﺎﹰ ‪ P‬ﻻ ﻳﻮﺍﺯﻱ ) ‪ ( ABC‬ﻭﲝﻴﺚ ﺗﻘﻊ‬
‫ﺍﻟﻨﻘﺎﻁ ‪ A‬ﻭ ‪ B‬ﻭ ‪ C‬ﰲ ﺟﻬﺔ ﻭﺍﺣﺪﺓ ﻣﻦ ‪ . P‬ﳔﺘﺎﺭ ﺛﻼﺙ ﻧﻘﺎ ﻁ‪ A′ ‬ﻭ ‪ B ′‬ﻭ ‪ C ′‬ﻧﻘﺎﻃﺎﹰ ﻻ‬
‫ﻋﻠﻰ ﺍﻟﺘﻌﻴﲔ ﻣﻦ ‪ ، P‬ﻭﻟﺘﻜﻦ ‪ A′′‬ﻣﻨﺘﺼﻒ ] ‪ ، [ AA′‬ﻭ ‪ B ′′‬ﻣﻨﺘﺼﻒ ] ‪ [ BB ′‬ﻭ ‪ C ′′‬ﻣﻨﺘﺼﻒ‬
‫] ‪ ، [CC ′‬ﻭﺃﺧﲑﺍﹰ ﻟﺘﻜﻦ ‪ O‬ﻣﺮﻛﺰ ﺛﻘﻞ ﺍﳌﺜﻠﹼﺚ ‪ . A′′ B ′′C ′′‬ﺃﻭﺟﺪ ﺍﶈﻞﹼ ﺍﳍﻨﺪﺳﻲ ﻟﻠﻨﻘﻄﺔ ‪O‬‬
‫ﻋﻨﺪﻣﺎ ﺗﺘﺤﻮ‪‬ﻝ ﺍﻟﻨﻘﺎﻁ ‪ A′‬ﻭ ‪ B ′‬ﻭ ‪.C ′‬‬
‫ ﻟﻨﻌﺮ‪‬ﻑ ‪ G‬ﻣﺮﻛﺰ ﻧﻘﻞ ﺍﳌﺜﻠﹼﺚ ‪ ، ABC‬ﻭﻟﻨﻌﺮ‪‬ﻑ ‪ G ′‬ﻣﺮﻛﺰ ﺛﻘﻞ ﺍﳌﺜﻠﹼﺚ ‪ . A′ B ′C ′‬ﳌﹼﺎ ﻛﺎﻥ‬

‫)‬
‫ﻭﻛﺎﻥ‬

















‪1‬‬
‫‪GO = GA′′ + GB ′′ + GC ′′‬‬
‫‪3‬‬

‫(‬












‪1‬‬
‫‪GA′′ = GA + GA′‬‬
‫‪2‬‬











‪1‬‬
‫‪GB ′′ = GB + GB ′‬‬
‫‪2‬‬












‪1‬‬
‫‪GC ′′ = GC + GC ′‬‬
‫‪2‬‬

‫)‬
‫)‬
‫)‬

‫ﻭ‬
‫ﻭ‬

‫‪A′‬‬

‫(‬
‫(‬
‫(‬

‫‪B‬‬

‫‪C′‬‬

‫‪O B ′′‬‬

‫‪C ′′‬‬
‫‪A′′‬‬

‫‪G‬‬

‫‪C‬‬

‫‪B′‬‬
‫‪A‬‬

‫‪P′‬‬

‫‪P‬‬

‫ﺍﺳﺘﻨﺘﺠﻨﺎ ﺃﻥﹼ‬









‪1 


















 1‬‬
‫‪′ + GB‬‬
‫‪′ + GC ′  = GG ′‬‬
‫‪+ GB + GC‬‬
‫‪GO = GA‬‬
‫‪ + GA‬‬
‫ ‬





‫ ‬
‫‪‬‬
‫ ‬
‫ ‪6 ‬‬
‫‪0‬‬
‫‪ 2‬‬
‫‪3GG ′‬‬

‫ﻭﻋﻠﻴﻪ ﺇﺫﺍ ﻛﺎﻥ ‪ HG , 12‬ﻫﻮ ﺍﻟﺘﺤﺎﻛﻲ ﺍﻟﺬﻱ ﻣﺮﻛﺰﻩ ‪ G‬ﻭﻧﺴﺒﺘﻪ‬

‫‪1‬‬
‫‪2‬‬

‫ﻛﺎﻧﺖ ‪ O‬ﺻﻮﺭﺓ ‪ G ′‬ﻭﻓﻖ‬

‫ﺍﻟﺘﺤﺎﻛﻲ ‪ . HG , 12‬ﻭﻟﻜﻦ ﻋﻨﺪﻣﺎ ﺗﺘﺤﻮ‪‬ﻝ ‪ A′‬ﻭ ‪ B ′‬ﻭ ‪ C ′‬ﰲ ‪ P‬ﺗﺮﺳﻢ ‪ G ′‬ﻛﺎﻣﻞ ﺍﳌﺴﺘﻮﻱ ‪، P‬‬
‫ﻓﺎﶈﻞ ﺍﳍﻨﺪﺳﻲ ﻟﻠﻨﻘﻄﺔ ‪ O‬ﻫﻮ ) ‪ ، P ′ = HG , 1 ( P‬ﻓﻬﻮ ﺇﺫﻥ ﺍﳌﺴﺘﻮﻱ ﺍﳌﻮﺍﺯﻱ ﻟﻠﻤﺴﺘﻮﻱ ‪P‬‬
‫‪2‬‬

‫ﻭﻳﺒﻌﺪ ﻋﻦ ‪ G‬ﻧﺼﻒ ﺑ‪‬ﻌﺪ ‪ G‬ﻋﻦ ‪. P‬‬

‫‪ù‬‬
‫ ‬

‫ﺎﺕ‬‫ﺍﻷﻭﳌﺒﻴﺎﺩ ﺍﻟﻌﺎﳌﻲ ﻟﻠﺮﻳﺎﺿﻴ‬

24

QWE
AgD
ZXC

This book is downloaded from this site:
www.syCourses.com

‫ّ‬
‫ﺃﻭﳌﺒﻴﺎﺩ ﺍﻟﺮﻳﺎﺿﻴﺎﺕ ﺍﻟﺮﺍﺑﻊ‬
‫ ﺃﻭﺟﺪ ﺃﺻﻐﺮ ﻋﺪﺩ ﻃﺒﻴﻌﻲ ‪ n‬ﺁﺣﺎﺩﻩ ﰲ ﺍﻟﻜﺘﺎﺑﺔ ﺍﻟﻌﺸﺮ ﻳ‪‬ﺔ ﺗﺴﺎﻭﻱ ‪ ، 6‬ﻭﻋﻨﺪ ﻧﻘﻞ ﻫﺬﻩ ﺍﳋﺎﻧﺔ‬
‫ﻟﺘﻮﺿﻊ ﰲ ﺍﻟﻨﻬﺎﻳﺔ ﳓﺼﻞ ﻋﻠﻰ ﻋﺪﺩ‪ ‬ﻳﺴﺎﻭﻱ ﺃﺭﺑﻌﺔ ﺃﺿﻌﺎﻑ ﺍﻟﻌﺪﺩ ‪. n‬‬
‫‪n = ( abc ⋯d 6 )10 ⇝ ( 6abc ⋯d )10‬‬
‫ ﻟﻨﻔﺘﺮﺽ ﺍﻟﻌﺪﺩ ﺍﳌﻄﻠﻮﺏ ‪ n‬ﻳ‪‬ﻜﺘﺐ ﺑﻌﺪﺩ ‪ m + 1‬ﻣﻦ ﺍﳋﺎﻧﺎﺕ ﺍﻟﻌﺸﺮﻳ‪‬ﺔ ﻭ ‪ . m ≥ 1‬ﻭﻟﻨﻀﻊ ‪p‬‬

‫ﺑﺎﻗﻲ ﻗﺴﻤﺔ ‪ n‬ﻋﻠﻰ ‪ ، 10‬ﻓﻴﻜﻮﻥ ‪ n = 10p + 6‬ﻣﻊ ‪ . 0 ≤ p < 10m‬ﻋﻨﺪﺋﺬ ﻳ‪‬ﻌﺒ‪‬ﺮ ﻋﻦ‬
‫ﺍﻟﺸﺮﻁ ﺍﳌﻄﻠﻮﺏ ﺑﺎﻟﺼﻴﻐﺔ‬
‫‪4 ( 10p + 6 ) = 6 ⋅ 10m + p‬‬

‫ﺃﻭ ‪ ، 13p + 8 = 2 ⋅ 10m‬ﻭﻫﺬﺍ ﻳﻘﺘﻀﻲ ﺃﻥﹼ ‪ ، p = 2q‬ﻭﻣﻨﻪ‬
‫‪13q + 4 = 10m‬‬

‫ﻭﺑﻮﺟﻪ ﺧﺎﺹ ﻳﻨﺒﻐﻲ ﺃﻥ ﻳﻜﻮﻥ ‪ ، 10m = 4 mod13‬ﻭﻟﻜﻦ ﻟﻨﺘﺄﻣ‪‬ﻞ ﺍﳉﺪﻭﻝ ﺍﻟﺘﺎﱄ ‪:‬‬
‫‪4 5 6‬‬

‫‪3‬‬

‫‪1‬‬

‫‪2‬‬

‫‪m‬‬

‫‪10m mod13 10 9 12 3 4 1‬‬

‫ﻭﻋﻠﻰ ﻫﺬﺍ ﻧﺮﻯ ﺃﻥﹼ ﺭﺗﺒﺔ ‪ 10‬ﰲ ﺍﻟﺰﻣﺮﺓ )×‪ ( ( ℤ /13ℤ )∗ ,‬ﺗﺴﺎﻭﻱ ‪ 6‬ﻭﺃﻥﹼ ﳎﻤﻮﻋﺔ ﺣﻠﻮﻝ‬
‫ﺍﳌﻌﺎﺩﻟﺔ ‪ 10m = 4 mod13‬ﻫﻲ ‪ ، m = 5 mod 6‬ﻓﺄﺻﻐﺮ ﻗﻴﻤﺔ ﺃﻛﱪ ﻣﻦ ﺍﻟﻮﺍﺣﺪ ﻟﻠﻌﺪﺩ ‪m‬‬
‫‪m‬‬
‫ﺗ‪‬ﺤﻘﹼﻖ ﺍﳌﻄﻠﻮﺏ ﻫﻲ ‪ ، m = 5‬ﻭﻋﻨﺪﻫﺎ ﻳﻜﻮﻥ ‪ ، q = 1013−4 = 7692‬ﻭﻣﻦ ﺛﹶﻢ ﳒﺪ‬
‫‪ ، p = 15384‬ﻭﺃﺧﲑﺍﹰ ‪. n = 153846‬‬
‫‪ù‬‬
‫ ‬

‫ ﺃﻭﺟﺪ ﲨﻴﻊ ﺍﻷﻋﺪﺍﺩ ﺍﳊﻘﻴﻘﻴ‪‬ﺔ ‪ x‬ﺍﻟﱵ ﺗ‪‬ﺤﻘﹼﻖ‬
‫‪1‬‬
‫‪2‬‬
‫ ﻟﻨﻼ ﺣﻆ ﺃ ﻭ‪ ‬ﻻﹰ ﺃ ﹼﻥ ﺍ ﻟﺘﺎ ﺑﻊ ‪3 − x − 1 + x‬‬

‫> ‪3−x − x +1‬‬

‫= ) ‪ x ֏ f ( x‬ﺗﺎ ﺑﻊ‪ ‬ﻣﻌﺮ‪ ‬ﻑ‪ ‬ﻋﻠﻰ ﺍ ‪‬ﺎﻝ‬
‫] ‪ . I = [ −1, 3‬ﻭﻫﻮ ﻣﺴﺘﻤﺮ‪ ‬ﻭﻣﺘﻨﺎﻗﺺ‪ ‬ﲤﺎﻣﺎﹰ ﻋﻠﻰ ﻫﺬﺍ ﺍ‪‬ﺎﻝ ﻷﻧ‪‬ﻪ ﳎﻤﻮﻉ ﺗﺎﺑﻌﲔ ﻣﺘﻨﺎﻗﺼﲔ ﲤﺎﻣﺎﹰ‬
‫ﻋﻠﻴﻪ‪ .‬ﻛﻤﺎ ﻧﻼﺣﻆ ﺃ ﻥﹼ ‪ f ( −1 ) = 2‬ﻭ ‪ ، f ( 1 ) = 0‬ﻓﻴﻮﺟﺪ ﻋﺪ ﺩ‪ ‬ﻭﺣﻴﺪ‪ x 0 ‬ﻣﻦ ] ‪[ −1,1‬‬
‫ﻳ‪‬ﺤﻘﹼﻖ ‪ f ( x 0 ) = 12‬ﻭﺗﻜﻮﻥ ﳎﻤﻮﻋﺔ ﺣﻠﻮﻝ ﺍﳌﺘﺮﺍﺟﺤﺔ ﻫﻲ ﺍ‪‬ﺎﻝ [ ‪. [ −1,x 0‬‬
‫‪25‬‬

‫ﺍﻷﻭﳌﺒﻴﺎﺩ ﺍﻟﻌﺎﳌﻲ ﻟﻠﺮﻳﺎﺿﻴ‪‬ﺎﺕ‬

‫‪26‬‬

‫ﻟﺘﻌﻴﲔ ‪ x 0‬ﳓﻞﹼ ﺍﳌﻌﺎﺩﻟﺔ‬

‫‪1‬‬
‫‪2‬‬

‫= ) ‪ . f ( x‬ﺑﺎﻟﺘﺮﺑﻴﻊ ﳒﺪ‬
‫= ‪3 + 2x − x 2‬‬

‫‪4 − ( 1 − x )2‬‬

‫ﻭﺑﺘﺮﺑﻴﻊ ﺛﺎﻥ‪ ‬ﳒﺪ‬
‫‪225‬‬
‫‪31‬‬
‫=‬
‫‪64‬‬
‫‪64‬‬

‫ﻭﻣﻨﻪ‬
‫ﻭﻷﻥﹼ ‪ x 0 < 1‬ﻧﺴﺘﻨﺘﺞ ﺃﻥﹼ‬

‫‪15‬‬
‫=‬
‫‪8‬‬

‫‪( 1 − x )2 = 4 −‬‬

‫‪‬‬
‫‪31‬‬
‫‪31 ‬‬
‫‪x ∈ 1 −‬‬
‫‪,1 +‬‬
‫‪‬‬
‫‪8‬‬
‫‪8 ‬‬
‫‪‬‬
‫‪31‬‬
‫‪8‬‬

‫‪ x 0 = 1 −‬ﻓﻤﺠﻤﻮﻋﺔ ﺍﳊﻠﻮﻝ ﻫﻲ ‪‬‬
‫‪‬‬

‫‪31‬‬
‫‪8‬‬

‫‪.  −1,1 −‬‬

‫‪ù‬‬

‫ ‬

‫ ﻧﺘﺄ ﻣ‪‬ﻞ ﺍﳌﻜﻌﺐ ‪ ABCDA′ B ′C ′D ′‬ﺍﻟﺬﻱ ﻭﺟﻬﻪ ﺍﻟﻌﻠﻮﻱ ‪ ABCD‬ﻭﻭﺟﻬﻪ ﺍﻟﺴﻔﻠﻲ ﻫﻮ‬
‫‪ A′ B ′C ′D ′‬ﻣﻊ ‪ A′‬ﻓﻮﻕ ‪ A‬ﻣﺒﺎﺷﺮﺓ‪ ،‬ﻭﻫﻜﺬﺍ‪ ....‬ﺗﺘﺤﺮ‪‬ﻙ ﻧﻘﻄﺔ ‪ X‬ﺑﺴﺮﻋﺔ ﺧﻄﻴ‪‬ﺔ ﺛﺎﺑﺘﺔ ﻋﻠﻰ‬
‫ﳏﻴﻂ ﺍﳌﺮ ﺑ‪‬ﻊ ‪ ABCD‬ﻭﺗﺘﺤﺮ‪‬ﻙ ﻧﻘﻄﺔ ‪ Y‬ﺑﺎﻟﺴﺮﻋﺔ ﺍﳋﻄﻴ‪‬ﺔ ﺍﻟﺜﺎﺑﺘﺔ ﻧﻔﺴﻬﺎ ﻋﻠﻰ ﳏﻴﻂ ﺍﳌﺮ ﺑ‪‬ﻊ‬
‫‪ . B ′C ′CB‬ﺗ‪‬ﻐﺎﺩﺭ ‪ X‬ﺍﻟﻨﻘﻄﺔ ‪ A‬ﺑﺎﲡﺎﻩ ‪ B‬ﰲ ﺍﻟﻠﺤﻈﺔ ﻧﻔﺴﻬﺎ ﺍﻟﱵ ﺗ‪‬ﻐﺎﺩﺭ ﻓﻴﻬﺎ ‪ Y‬ﺍﻟﻨﻘﻄﺔ ‪B ′‬‬

‫ﺑﺎﲡﺎﻩ ‪ .C ′‬ﻋﻴ‪‬ﻦ ﺍﶈﻞﹼ ﺍﳍﻨﺪﺳﻲ ﻟﻠﻨﻘﻄﺔ ‪ Z‬ﻣﻨﺘﺼﻒ ﺍﻟﻘﻄﻌﺔ ] ‪. [ XY‬‬
‫ ‬

‫ ‪ .1‬ﳝﻜﻦ ﺃﻥ ﻧﻔﺘﺮﺽ ﺷﻌﺎﻉ ﺳﺮﻋﺔ ‪ X‬ﻋﻠﻰ ] ‪ [ AB‬ﻫﻮ ‪ ، AB‬ﻭﺃ ﻥﹼ ﺷﻌﺎﻉ ﺳﺮﻋﺔ ‪ Y‬ﻋﻠﻰ‬
‫ ‬

‫ ‬

‫ ‬

‫] ‪ [ B ′C ′‬ﻫﻮ ‪ ، B ′C ′‬ﻋﻨﺪﺋﺬ‪ ،‬ﰲ ﺍﻟﻠﺤﻈﺔ ‪ t‬ﻣﻦ ] ‪ ، [ 0,1‬ﻳﻜﻮﻥ ‪ ، AX = tAB‬ﻭﻳﻜﻮﻥ‬
‫ ‬

‫ ‬

‫‪ . B ′Y = tB ′C ′‬ﻭﻋﻠﻴﻪ ﺇﺫﺍ ﻋﺮ‪‬ﻓﻨﺎ ‪ T‬ﻣﻨﺘﺼﻒ ] ‪ [ AB ′‬ﻛﺎﻥ ﻟﺪﻳﻨﺎ ﰲ ﺍﻟﻠﺤﻈﺎﺕ ‪ t‬ﻣﻦ ﺍ‪‬ﺎﻝ‬
‫] ‪ [ 0,1‬ﻣﺎ ﻳﻠﻲ ‪:‬‬

‫ ‪ 1‬‬
‫) ‪TZ = (TX + TY‬‬
‫‪2‬‬
‫ ‬
‫ ‪1‬‬
‫‪ TA + TB ′ = 0‬‬
‫‪= TA + AX + TB ′ + B ′Y‬‬
‫‪2‬‬
‫ ‬
‫ ‪1‬‬
‫ ‪t‬‬
‫) ‪= tAB + tB ′C ′ = ( AB + BC‬‬
‫‪2‬‬
‫‪2‬‬
‫ ‬
‫ ‪t‬‬
‫‪= AC = tTU‬‬
‫‪2‬‬
‫ﻭﻗﺪ ﻋﺮ‪‬ﻓﻨﺎ ‪ U‬ﻣﻨﺘﺼﻒ ﺍﻟﻘﻄﻌﺔ ﺍﳌﺴﺘﻘﻴﻤﺔ ] ‪ . [ BC ′‬ﻓﻌﻨﺪﻣﺎ ﺗﺘﺤﺮ‪‬ﻙ ‪ X‬ﻋﻠﻰ ] ‪ ، [ AB‬ﻭ ‪ Y‬ﻋﻠﻰ‬

‫)‬

‫)‬

‫] ‪ [ B ′C ′‬ﺗﺘﺤﺮ‪‬ﻙ ‪ Z‬ﻋﻠﻰ ﺍﻟﻘﻄﻌﺔ ] ‪ [TU‬ﻣﻦ ‪ T‬ﺇﱃ ‪.U‬‬

‫(‬
‫(‬

‫ﻋﺎﻡ ‪1962‬‬

‫‪27‬‬

‫ ‬
‫‪ .2‬ﻭ ﻛﺬ ﻟﻚ ‪ ،‬ﻳﻜﻮ ﻥ ﺷﻌﺎ ﻉ ﺳﺮ ﻋﺔ ‪ X‬ﻋﻠﻰ ] ‪ [ BC‬ﻫﻮ ‪ ، BC‬ﻭﺷﻌﺎ ﻉ ﺳﺮ ﻋﺔ ‪ Y‬ﻋﻠﻰ‬
‫ ‬
‫ ‬
‫ ‬
‫] ‪ [C ′C‬ﻫﻮ ‪ ، C ′C‬ﻋﻨﺪ ﺋﺬ ‪ ،‬ﰲ ﺍ ﻟﻠﺤﻈﺔ ‪ t‬ﻣﻦ ] ‪ ، [ 1, 2‬ﻳﻜﻮ ﻥ ‪، BX = ( t − 1 ) BC‬‬
‫ ‬
‫ ‬
‫ﻭﻳﻜﻮﻥ ‪ .C ′Y = ( t − 1 )C ′C‬ﻭﻋﻠﻴﻪ ﰲ ﺍﻟﻠﺤﻈﺎﺕ ‪ t‬ﻣﻦ ﺍ‪‬ﺎﻝ ] ‪ [ 1, 2‬ﻟﺪﻳﻨﺎ ‪:‬‬
‫ ‪ 1‬‬
‫) ‪UZ = (UX + UY‬‬
‫‪2‬‬
‫ ‬
‫ ‪1‬‬
‫‪ UB + UC ′ = 0‬‬
‫‪= UB + BX + UC ′ + C ′Y‬‬
‫‪2‬‬
‫ ‬
‫ ‬
‫ ‪t − 1‬‬
‫‪1‬‬
‫= ‪= ( t − 1 ) BC + ( t − 1 )C ′C‬‬
‫‪B ′C ′ + C ′C‬‬
‫‪2‬‬
‫‪2‬‬
‫ ‬
‫ ‪t − 1‬‬
‫=‬
‫‪B ′C = ( t − 1 )UC‬‬
‫‪2‬‬
‫ﻓﻌﻨﺪﻣﺎ ﺗﺘﺤﺮ‪‬ﻙ ‪ X‬ﻋﻠﻰ ] ‪ ، [ BC‬ﻭ ‪ Y‬ﻋﻠﻰ ] ‪ [C ′C‬ﺗﺘﺤﺮ‪‬ﻙ ‪ Z‬ﻋﻠﻰ ﺍﻟﻘﻄﻌﺔ ] ‪ [UC‬ﻣﻦ ‪U‬‬

‫)‬

‫)‬
‫)‬

‫(‬

‫(‬
‫(‬

‫ﺇﱃ ‪.C‬‬
‫ ‬
‫‪ .3‬ﻭﻣﻦ ﺛﹶﻢ‪ ،‬ﻳﻜﻮﻥ ﺷﻌﺎﻉ ﺳﺮﻋﺔ ‪ X‬ﻋﻠﻰ ] ‪ [CD‬ﻫﻮ ‪ ، CD‬ﻭﺷﻌﺎﻉ ﺳﺮﻋﺔ ‪ Y‬ﻋﻠﻰ ] ‪[CB‬‬
‫ ‬
‫ ‬
‫ ‬
‫ﻫﻮ ‪ ، CB‬ﻋﻨﺪﺋﺬ‪ ،‬ﰲ ﺍﻟﻠﺤﻈﺔ ‪ t‬ﻣﻦ ] ‪ ، [ 2, 3‬ﻳﻜﻮﻥ ﻣﻦ ﺟﻬﺔ ﺃﻭﱃ ‪،CX = ( t − 2 )CD‬‬
‫ ‬
‫ ‬
‫ﻭﻳﻜﻮﻥ ‪ .CY = ( t − 2 )CB‬ﻭﻋﻠﻴﻪ ﰲ ﺍﻟﻠﺤﻈﺎﺕ ‪ t‬ﻣﻦ ﺍ‪‬ﺎﻝ ] ‪ [ 2, 3‬ﻟﺪﻳﻨﺎ ‪:‬‬
‫ ‪ 1‬‬
‫ ‬
‫ ‬
‫‪1‬‬
‫) ‪CZ = (CX + CY ) = ( ( t − 2 )CD + ( t − 2 )CB‬‬
‫‪2‬‬
‫‪2‬‬
‫ ‬
‫ ‪t − 2‬‬
‫ ‪t − 2‬‬
‫=‬
‫= ) ‪CD + CB‬‬
‫‪CA = ( t − 2 )CV‬‬
‫(‬
‫‪2‬‬
‫‪2‬‬

‫ﻭﻗﺪ ﻋﺮ‪‬ﻓﻨﺎ ‪ V‬ﻣﻨﺘﺼﻒ ﺍﻟﻘﻄﻌﺔ ] ‪ . [CA‬ﻓﻌﻨﺪﻣﺎ ﺗﺘﺤﺮ‪‬ﻙ ‪ X‬ﻋﻠﻰ ] ‪ ، [CD‬ﻭ ‪ Y‬ﻋﻠﻰ‬
‫ﺗﺘﺤﺮ‪‬ﻙ ‪ Z‬ﻋﻠﻰ ﺍﻟﻘﻄﻌﺔ ] ‪ [CV‬ﻣﻦ ‪ C‬ﺇﱃ ‪.V‬‬

‫] ‪[CB‬‬

‫ ‬

‫‪ .4‬ﻭﺃﺧﲑﺍﹰ‪ ،‬ﻳﻜﻮﻥ ﺷﻌﺎﻉ ﺳﺮﻋﺔ ‪ X‬ﻋﻠﻰ ] ‪ [ DA‬ﻫﻮ ‪ ، DA‬ﻭﺷﻌﺎﻉ ﺳﺮﻋﺔ ‪ Y‬ﻋﻠﻰ ] ‪[ BB ′‬‬
‫ ‬

‫ ‬

‫ ‬

‫ﻫﻮ ‪ ، BB ′‬ﻋﻨﺪﺋﺬ‪ ،‬ﰲ ﺍﻟﻠﺤﻈﺔ ‪ t‬ﻣﻦ ] ‪ ، [ 3, 4‬ﻳﻜﻮﻥ ﻣﻦ ﺟﻬﺔ ﺃﻭﱃ ‪، DX = ( t − 3 ) DA‬‬
‫ ‬

‫ ‬

‫ﻭﻳﻜﻮﻥ ‪ . BY = ( t − 3 ) BB ′‬ﻭﻋﻠﻴﻪ‪ ،‬ﳒﺪ ﺑﺄﺳﻠﻮﺏ ﳑﺎﺛﻞﹴ ﻟ‪‬ﻤﺎ ﺳﺒﻖ ‪:‬‬
‫ ‬
‫ ‬
‫‪VZ = ( t − 3 )VT‬‬

‫‪∀t ∈ [ 3, 4 ],‬‬

‫ﻭﻗﺪ ﻋﺮ‪‬ﻓﻨﺎ ‪ V‬ﻣﻨﺘﺼﻒ ﺍﻟﻘﻄﻌﺔ ] ‪ . [CA‬ﻓﻌﻨﺪﻣﺎ ﺗﺘﺤﺮ‪‬ﻙ ‪ X‬ﻋﻠﻰ ] ‪ ، [CD‬ﻭ ‪ Y‬ﻋﻠﻰ‬
‫ﺗﺘﺤﺮ‪‬ﻙ ‪ Z‬ﻋﻠﻰ ﺍﻟﻘﻄﻌﺔ ] ‪ [CV‬ﻣﻦ ‪ C‬ﺇﱃ ‪.V‬‬

‫] ‪[CB‬‬

‫ﺍﻷﻭﳌﺒﻴﺎﺩ ﺍﻟﻌﺎﳌﻲ ﻟﻠﺮﻳﺎﺿﻴ‪‬ﺎﺕ‬

‫‪28‬‬

‫ﻭﻫﻜﺬﺍ‪ ،‬ﺗﺘﺤﺮ‪‬ﻙ ‪ Z‬ﻋﻠﻰ ﻣﺘﻮﺍﺯﻱ ﺍﻷﺿﻼﻉ ‪ .TUCV‬ﻛﻤﺎ ﻫﻮ ﻣﻮﺿ‪‬ﺢ ﺑﺎﻟﺸﻜﻞ ﺍﻟﺘﺎﱄ‬
‫‪D‬‬
‫‪C‬‬

‫‪V‬‬
‫‪X‬‬

‫‪B‬‬
‫‪C′‬‬

‫‪U‬‬

‫‪A‬‬

‫‪T Z‬‬

‫‪Y‬‬

‫‪A′‬‬

‫‪ù‬‬

‫‪B′‬‬

‫ ‬

‫ ﺃﻭﺟﺪ ﲨﻴﻊ ﺍﳊﻠﻮﻝ ﺍﳊﻘﻴﻘﻴ‪‬ﺔ ﻟﻠﻤﻌﺎﺩﻟﺔ‬
‫‪cos2 x + cos2 2x + cos2 3x = 1‬‬

‫ ﺑﺎﺳﺘﺨﺪﺍﻡ ‪ ، 2 cos2 θ = 1 + cos 2θ‬ﺗ‪‬ﻜﺎﻓﺊ ﺍﳌﻌﺎﺩﻟﺔﹸ ﺍﳌﻌﻄﺎﺓ ﺍﳌﻌﺎﺩﻟﺔﹶ ﺍﻟﺘﺎﻟﻴﺔ ‪:‬‬
‫‪cos 2x + 2 cos2 2x + cos 6x = 0‬‬
‫ﻭﺑﺎﺳﺘﺨﺪﺍﻡ ‪ ، cos 3θ = 4 cos3 θ − 3 cos θ‬ﺗ‪‬ﻜﺎﻓﺊ ﺍﳌﻌﺎﺩﻟﺔﹸ ﺍﻟﺴﺎﺑﻘﺔ ﺍﳌﻌﺎﺩﻟﺔﹶ ﺍﻟﺘﺎﻟﻴﺔ ‪:‬‬
‫‪2 cos3 2x + cos2 2x − cos 2x = 0‬‬

‫ﺃﻭ‬

‫‪cos 2x ( cos 2x + 1 )( 2 cos 2x − 1 ) = 0‬‬

‫ﻓﻤﺠﻤﻮﻋﺔ ﺍﳊﻠﻮﻝ ﻫﻲ ‪:‬‬

‫) ‪( π4 + π2 ℤ ) ∪ ( π2 + πℤ ) ∪ ( π6 + πℤ ) ∪ ( − π6 + πℤ‬‬

‫‪ù‬‬

‫ ‬

‫ ﻧﺘﺄﻣ‪‬ﻞ ﺛﻼﺙ ﻧﻘﺎﻁ ﻣﻌﻄﺎﺓ ‪ A‬ﻭ ‪ B‬ﻭ ‪ C‬ﻣﻦ ﺩﺍﺋﺮﺓ ‪ . C‬ﻳ‪‬ﻄﻠﹶﺐ‪ ‬ﺇﻧﺸﺎﺀ ﻧﻘﻄﺔ ‪ D‬ﻣﻦ ‪ ، C‬ﲝﻴﺚ‬
‫ﻳﻜﻮﻥ ﻣﻦ ﺍﳌﻤﻜﻦ ﺭﺳﻢ ﺩﺍﺋﺮﺓ ﲤﺲ‪ ‬ﺃﺿﻼﻉ ﺍﻟﺮﺑﺎﻋﻲ ‪ ABCD‬ﺩﺍﺧﻼﹰ‪.‬‬
‫ ﺍﻟﺘﺤﻠﻴﻞ ‪ :‬ﻟﻨﻔﺘﺮﺽ ﺃﻥﹼ ﺍﻹﻧﺸﺎﺀ ﻣ‪‬ﻨﺠ‪‬ﺰ‪ .‬ﻭﻟﻴﻜﻦ ‪ O‬ﻣﺮﻛﺰ ﺍﻟﺪﺍﺋﺮﺓ ﺍﳌﺎﺳ‪‬ﺔ‬
‫ﻷﺿﻼﻉ ﺍﻟﺮﺑﺎﻋﻲ ‪ ABCD‬ﺩﺍﺧﻼﹰ‪ .‬ﻣﻦ ﺍﻟﻮﺍﺿﺢ ﺃ ﻥﹼ ‪ O‬ﻳﻘﻊ ﻋﻠﻰ‬
‫ﺍﳌﻨﺼ‪‬ﻒ ‪ d‬ﻟﻠﺰﺍﻭﻳﺔ ‬
‫‪ . ABC‬ﻭ ﳌﹼﺎ ﻛﺎﻥ‬

‫‪D‬‬
‫‪C‬‬

‫ ‬
‫ ‬
‫ ‬
‫ ‬
‫ ‬
‫ ‬
‫ ‬
‫ ‬
‫ ‬
‫‪(BO, BA) + (AB, AO ) + (OA,OB ) = π‬‬

‫ﻭ‬

‫ ‬
‫ ‬
‫ ‬
‫ ‬
‫ ‬
‫ ‬
‫ ‬
‫ ‬
‫ ‬
‫‪(BC , BO ) + (OB,OC ) + (CO,CB ) = π‬‬

‫‪A‬‬

‫‪O‬‬

‫‪C‬‬

‫‪B‬‬

‫ﻋﺎﻡ ‪1962‬‬

‫‪29‬‬

‫ﺍﺳﺘﻨﺘﺠﻨﺎ ﲜﻤﻊ ﺍﳌﺴﺎﻭﺍﺗﲔ ﺍﻟﺴﺎﺑﻘﺘﲔ ﻃﺮﻓﺎﹰ ﻣﻊ ﻃﺮﻑ ﺃﻥﹼ ‪:‬‬

‫ ‬
‫ ‬
‫ ‬
‫ ‬
‫ ‬
‫ ‬
‫ ‬
‫ ‬
‫ ‬
‫ ‬
‫ ‬
‫ ‬
‫‪(BC , BA) + (OA,OC ) + (AB, AO ) + (CO,CB ) = 2π‬‬
‫ﻭﻟﻜﻦ ) ‪ (OA‬ﻫﻮ ﻣﻨﺼ‪‬ﻒ ﺍﻟﺰﺍﻭﻳﺔ ‬
‫‪ ، BAD‬ﻭ ) ‪ (OC‬ﻫﻮ ﻣﻨﺼ‪‬ﻒ ﺍﻟﺰﺍﻭﻳﺔ ‬
‫‪ ، BCD‬ﺇﺫﻥ‬
‫ ‬
‫ ‬
‫ ‬
‫ ‬
‫ ‬
‫ ‬
‫ ‪1‬‬
‫ ‪1‬‬
‫ ‬
‫ ‬
‫ ‬
‫ ‬
‫) ‪ (AB, AO ) = (AB, AD‬ﻭ ) ‪(CO,CB ) = (CD,CB‬‬
‫‪2‬‬
‫‪2‬‬
‫ﻓﺈﺫﺍ ﺍﺳﺘﻔﺪﻧﺎ ﻣﻦ ﻛﻮﻥ ﺍﻟﺮﺑﺎﻋﻲ ‪ ABCD‬ﺭﺑﺎﻋﻴ‪‬ﹰﺎ ﺩﺍﺋﺮﻳﺎﹰ ﺍﺳﺘﻨﺘﺠﻨﺎ ﺃﻥﹼ‬
‫ ‬
‫ ‬
‫ ‬
‫ ‬
‫ ‬
‫ ‬
‫ ‬
‫ ‪1‬‬
‫‪π‬‬
‫ ‬
‫ ‬
‫ ‬
‫ ‬
‫= ) ‪(CO,CB ) + (AB, AO ) = (AB, AD ) + (CD,CB‬‬
‫‪2‬‬
‫‪2‬‬
‫ ‬
‫ ‬
‫ ‬
‫ ‬
‫ ‬
‫ ‬
‫ ‬
‫ ‬
‫ ‬
‫ ‬
‫ ‬
‫ ‬
‫ﻭﻋﻠﻰ ﻫﺬﺍ ‪ ، (BC , BA) + (OA,OC ) = 2π − π2‬ﻭﻷﻥﹼ )‪2π − (OA,OC ) = (OC ,OA‬‬

‫)‬

‫ﺍﺳﺘﻨﺘﺠﻨﺎ‬

‫(‬

‫ ‬
‫ ‬
‫ ‬
‫ ‬
‫‪π‬‬
‫ ‬
‫ ‬
‫)‪(OC ,OA) = + (BC , BA‬‬
‫‪2‬‬
‫ ‬
‫ ‬
‫ ‬
‫‪ ، (BC‬ﻧﻀﻊ ‪ G‬ﻣﺮﻛﺰ ﺍﻟﺪﺍﺋﺮﺓ ‪ . C‬ﻭﻧﻌﺮ‪‬ﻑ ‪ O ′‬ﻧﻘﻄﺔ ﺗﻘﺎﻃﻊ‬
‫≠ )‪, BA‬‬

‫ ﰲ ﺣﺎﻟﺔ‬
‫ﺍﳌﻤﺎﺳﲔ ﻟﻠﺪﺍﺋﺮﺓ ‪ C‬ﺍﳌﺮﺳﻮﻣﲔ ﻣﻦ ‪ C‬ﻭ ‪ ) ، A‬ﻋﻨﺪﺋﺬ ‪. (O ′A = O ′C‬‬
‫‪π‬‬
‫‪2‬‬

‫ ‬

‫ ‬

‫ ‬
‫ ‬
‫‪ ، (GC‬ﻭﻣﻦ ﺛﹶﻢ‪‬‬
‫ﻧﻼﺣﻆ ﺃﻥﹼ )‪,GA) = 2(BC , BA‬‬

‫ ‬
‫ ‬
‫ ‬
‫ ‬
‫ ‬
‫ ‬
‫ ‬
‫ ‬
‫ ‬
‫)‪(O ′A,O ′C ) = π − (GC ,GA) = π − 2(BC , BA‬‬

‫ﻭﺃﺧﲑﺍﹰ‬

‫ ‬
‫ ‬
‫ ‬
‫ ‬
‫ ‬
‫ ‬
‫ ‬
‫ ‬
‫ ‬
‫)‪(O ′C ,O ′A) = π + 2(BC , BA) = 2(OC ,OA‬‬
‫ﻓﺈﺫﺍ ﺭﲰﻨﺎ ﺍﻟﺪﺍﺋﺮﺓ ‪ C ′‬ﺍﻟﱵ ﻣﺮﻛﺰﻫﺎ ‪ O ′‬ﻭﲤﺮ ﺑﺎﻟﻨﻘﻄﺔ ‪ ، A‬ﺍﺳﺘﻨﺘﺠﻨﺎ ﻣﻦ ﺍﳌﺴﺎﻭﺍﺓ ﺍﻟﺴﺎﺑﻘﺔ ﺃ ﹼﻥ‬

‫‪ O‬ﺗﻘﻊ ﻋﻠﻰ ﺍﻟﻘﻮﺱ ﻣﻦ ‪ C ′‬ﺍﶈﺘﻮﻯ ﺩﺍﺧﻞ ‪ ، C‬ﻷﻥﹼ ﻫﺬﺍ ﺍﻟﻘﻮﺱ ﻫﻮ ﳎﻤﻮﻋﺔ ﺍﻟﻨﻘﺎﻁ ﺍﻟﱵ‬

‫ ‬
‫ ‬
‫ ‬
‫ ‬
‫ ‬
‫ ‬
‫‪ (OC‬ﺛﺎﺑﺘﺔ ﻭﺗﺴﺎﻭﻱ )‪. 21 (O ′C ,O ′A‬‬
‫ﺗ‪‬ﺮﻯ ﻣﻨﻬﺎ ﺍﻟﻘﻄﻌﺔ ﺍﳌﺴﺘﻘﻴﻤﺔ ] ‪ [CA‬ﺑﺰﺍﻭﻳﺔ )‪,OA‬‬
‫‪A‬‬

‫‪B‬‬

‫‪O′‬‬

‫‪O‬‬

‫‪C′‬‬

‫ ﺃﻣ‪‬ﺎ ﰲ ﺣﺎﻟﺔ‬

‫‪π‬‬
‫‪2‬‬

‫ ‬

‫‪G‬‬

‫‪C‬‬
‫‪C‬‬

‫ ‬
‫‪ ، (BC‬ﻓﻨﺮﻯ ﺃﻥﹼ ‪ O‬ﺗﻘﻊ ﻋﻠﻰ ﺍﻟﻘﻄﻌﺔ ﺍﳌﺴﺘﻘﻴﻤﺔ ] ‪. [CA‬‬
‫= )‪, BA‬‬

‫ﺍﻷﻭﳌﺒﻴﺎﺩ ﺍﻟﻌﺎﳌﻲ ﻟﻠﺮﻳﺎﺿﻴ‪‬ﺎﺕ‬

‫‪30‬‬

‫ﺗ‪‬ﺜﺒﺖ ﺍﳌﻨﺎﻗﺸﺔ ﺍﻟﺴﺎﺑﻘﺔ ﻭﺣﺪﺍﻧﻴ‪‬ﺔ ﺍﳊﻞﱢ ﰲ ﺣﺎﻝ ﻭﺟﻮﺩﻩ‪ ،‬ﺇﺫ ﺇﻥﹼ ﻣﺮﻛﺰ ﺍﻟﺪﺍﺋﺮﺓ ﺍﳌﻤﺎﺳ‪‬ﺔ ﺩﺍﺧﻼﹰ ﻟﻠﺮﺑﺎﻋﻲ‬
‫‪ ABCD‬ﳏﺪ‪‬ﺩ‪ ‬ﲤﺎﻣﺎﹰ ﻛﺘﻘﺎﻃﻊ ﻣﻨﺼ‪‬ﻒ ﺍﻟﺰﺍﻭﻳﺔ ‬
‫‪ ABC‬ﻣﻊ ﺍﻟﻘﻮﺱ ﻣﻦ ﺍﻟﺪﺍﺋﺮﺓ ‪ C ′‬ﺍﶈﺘﻮﻯ ﰲ ‪، C‬‬

‫ﺃﻭ ﻣﻊ ﺍﻟﻘﻄﻌﺔ ﺍﳌﺴﺘﻘﻴﻤﺔ ] ‪ [CA‬ﰲ ﺣﺎﻟﺔ‬

‫‪π‬‬
‫‪2‬‬

‫= ‬
‫‪. ABC‬‬

‫ﻹﺛﺒﺎﺕ ﻭﺟﻮﺩ ﺣﻞﱟ ﻟﻠﻤﺴﺄﻟﺔ ﺍﳌﻄﺮﻭﺣﺔ ﺳﻨﺴﺘﻔﻴﺪ ﻣﻦ ﺍﳋﺎﺻ‪‬ﺔ ﺍﻟﺘﺎﻟﻴﺔ ﺍﻟﱵ ﺳﻨﺜﺒﺘ‪‬ﻬﺎ ﻻﺣﻘﺎﹰ‪.‬‬
‫ﻣﱪﻫﻨﺔ ‪ :‬ﳝﻜﻦ ﺭﺳﻢ ﺩﺍﺋﺮﺓ ﲤﺲ‪ ‬ﺃﺿﻼﻉ ﺍﻟﺮﺑﺎﻋﻲ ﺍﶈﺪ‪‬ﺏ ‪ ABCD‬ﺇﺫﺍ ﻭﻓﻘﻂ ﺇﺫﺍ ﻛﺎﻥ‬
‫‪AB + CD = AD + CB‬‬

‫ﻟﻨﺘﺄﻣ‪‬ﻞ ﺇﺫﻥ ﺛﻼﺙ ﻧﻘﺎﻁ ‪ ABC‬ﻣﻦ ﺍﻟﺪﺍﺋﺮﺓ ‪ . C‬ﻭﻟﻴﻜﻦ ‬
‫‪ CA‬ﺍﻟﻘﻮﺱ ﺍﳌﻔﺘﻮﺡ ﻣﻦ ‪ C‬ﺍﻟﺬﻱ ﻃﺮﻓﺎﻩ‬
‫ﺍﻟﻨﻘﻄﺘﺎﻥ ‪ A‬ﻭ ‪ ،C‬ﻭﻻ ﺗﻨﺘﻤﻲ ﺇﻟﻴﻪ ﺍﻟﻨﻘﻄﺔ ‪ . B‬ﻭﻟﻨﺘﺄﻣ‪‬ﻞ ﺍﻟﺘﺎﺑﻊ ﺍﳌﺴﺘﻤﺮ‪‬‬
‫‪ → ℝ, f ( M ) = AB + CM − AM − CB‬‬
‫‪f : CA‬‬

‫ﻧﻼﺣﻆ ﺃﻥﹼ‬
‫‪f ( M ) = AB + CA − CB > 0‬‬

‫‪f ( M )>0‬‬
‫‪ր‬‬

‫‪lim‬‬

‫ ‬
‫‪M →A,M ∈CA‬‬

‫ﻭ‬

‫‪A‬‬

‫‪M‬‬
‫‪D‬‬

‫‪B‬‬

‫‪M′‬‬

‫‪f ( M ) = AB − AC − CB < 0‬‬

‫‪lim‬‬

‫ ‬
‫‪M →C ,M ∈CA‬‬

‫‪ց‬‬
‫‪f (M′)<0‬‬

‫‪C‬‬

‫ﺇﺫﻥ ﺍﻋﺘﻤﺎﺩﺍﹰ ﻋﻠﻰ ﻣﱪﻫﻨﺔ ﺍﻟﻘﻴﻤﺔ ﺍﻟﻮﺳﻄﻰ ﻧﺴﺘﻨﺘﺞ ﺃﻧ‪‬ﻪ ﺗﻮﺟﺪ ‪ D‬ﺗﻨﺘﻤﻲ ﺇﱃ ﺍﻟﻘﻮﺱ ‬
‫‪ CA‬ﺗ‪‬ﺤﻘﹼﻖ‬
‫‪ . f ( D ) = 0‬ﻭﻋﻤﻼﹰ ﺑﺎﻟﺘﻮﻃﺌﺔ ﺍﳌﺸﺎﺭ ﺇﻟﻴﻬﺎ ﺃﻋﻼﻩ‪ ،‬ﻧﺴﺘﻨﺘﺞ ﻭﺟﻮﺩ ﺣﻞﱟ ﻟﻠﻤﺴﺄﻟﺔ ﺍﳌﻄﺮﻭﺣﺔ‪.‬‬

‫ﺍﻹﻧﺸﺎﺀ ‪:‬‬
‫ ﻧﻨﺸﺊ ﺍﳌﻨﺼ‪‬ﻒ ‪ d‬ﻟﻠﺰﺍﻭﻳﺔ ‬
‫‪. ABC‬‬
‫ ﰲ ﺣﺎﻟﺔ ‪ = π‬‬
‫‪ ABC‬ﻛﺎﻧﺖ ‪ O‬ﻫﻲ ﻧﻘﻄﺔ ﺗﻘﺎﻃﻊ ‪ d‬ﻣﻊ ] ‪. [ AC‬‬
‫‪2‬‬
‫‪π‬‬
‫ ‬

ﺃ ﻣ‪‬ﺎ ﰲ ﺍ ﳊﺎ ﻟﺔ ﺍ ﻟﻌﺎ ﻣ‪‬ﺔ ‪ ، ABC ≠ 2‬ﻓﻨﻨﺸﺊ ‪ O ′‬ﻧﻘﻄﺔ ﺗﻘﺎ ﻃﻊ ﺍ ﳌﻤﺎﺳﲔ ﻟﻠﺪ ﺍ ﺋﺮ ﺓ ‪C‬‬
‫ﺍﳌﺮﺳﻮﻣﲔ ﻣﻦ ‪ A‬ﻭ ‪ ،C‬ﺛﹸﻢ‪ ‬ﻧ‪‬ﻨﺸﺊ ﺍﻟﻘﻮﺱ ﻣﻦ ﺍﻟﺪﺍﺋﺮﺓ ﺍﻟﱵ ﻣﺮﻛﺰﻫﺎ ‪ O ′‬ﻭﲤﺮ ﺑﺎﻟﻨﻘﻄﺘﲔ ‪A‬‬
‫ﻭ ‪ C‬ﻭﺍﶈﺘﻮﻯ ﺩﺍﺧﻞ ‪ ، C‬ﻓﺘﺘﻘﺎﻃﻊ ﻫﺬﻩ ﺍﻟﻘﻮﺱ ﻣﻊ ‪ d‬ﰲ ‪.O‬‬
‫ ﻧ‪‬ﻨﺸﺊ ﺍﻟﺪﺍﺋﺮﺓ ﺍﻟﱵ ﻣﺮﻛﺰﻫﺎ ‪ O‬ﻭﲤﺲ‪ ، ( AB ) ‬ﻓﻴﺘﻘﺎﻃﻊ ﺍﳌﻤﺎﺱ ﺍﻟﺜﺎﱐ‪ ،‬ﺍﳌﺮﺳﻮﻡ ﻣﻦ ‪، A‬‬
‫ﳍﺬﻩ ﺍﻟﺪﺍﺋﺮﺓ ﻣﻊ ﺍﻟﺪﺍﺋﺮﺓ ‪ C‬ﰲ ﺍﻟﻨﻘﻄﺔ ‪ D‬ﺍﳌﻄﻠﻮﺑﺔ‪.‬‬

‫ﻋﺎﻡ ‪1962‬‬

‫‪31‬‬

‫ﻧﻮﺿ‪‬ﺢ ﰲ ﺍﻟﺸﻜﻞ ﺍﻟﺘﺎﱄ ﺧﻄﻮﺍﺕ ﻫﺬﺍ ﺍﻹﻧﺸﺎﺀ‪.‬‬
‫‪D‬‬
‫‪ւ‬‬

‫‪C′‬‬
‫‪A‬‬

‫‪O′‬‬

‫‪B‬‬
‫‪O‬‬

‫‪C‬‬

‫‪G‬‬

‫‪C‬‬

‫ﻧﺄﰐ ﺍﻵﻥ ﺇﱃ ﺇﺛﺒﺎﺕ ﺍﳌﱪﻫﻨﺔ‪.‬‬
‫ﺍﻟﺸﺮﻁﹸ ﻻﺯ ‪‬ﻡ ‪ :‬ﻟﻨﻔﺘﺮﺽ ﺃﻧ‪‬ﻪ ﺗﻮﺟﺪ ﺩﺍﺋﺮﺓ ﲤﺲ‪ ‬ﺃﺿﻼﻉ ﺍﻟﺮﺑﺎﻋﻲ‬

‫‪D‬‬
‫‪Z‬‬

‫ﺍﶈﺪ‪‬ﺏ ‪ . ABCD‬ﻭﻟﻨﻔﺘﺮﺽ ﺃﻥﹼ ﻧﻘﺎﻁ ﺍﻟﺘﻤﺎﺱ ﻫﻲ ‪ X‬ﻭ ‪Y‬‬

‫ﻭ ‪ Z‬ﻭ ‪ T‬ﻛﻤﺎ ﰲ ﺍﻟﺸﻜﻞ ﺍ‪‬ﺎﻭﺭ‪.‬‬
‫ﻋﻨﺪﺋﺬ‪ ،‬ﻧﺴﺘﻨﺘﺞ ﻣﻦ ﺗﺴﺎﻭﻱ ﻃﻮﻟﹶﻲ‪ ‬ﺍﳌﻤﺎﺳﲔ ﻟﺪﺍﺋﺮﺓ ﺍﳌﻨﺒﻌﺜﲔ ﻣﻦ‬
‫ﺍﻟﻨﻘﻄﺔ ﻧﻔﺴﻬﺎ‪ ،‬ﻣﺎ ﻳﻠﻲ ‪:‬‬

‫‪T‬‬

‫‪A‬‬

‫‪X‬‬

‫‪C‬‬

‫‪Y‬‬
‫‪B‬‬

‫‪AD + BC = AT + TD + BY + YC‬‬

‫‪= AX + ZD + XB + CZ‬‬
‫‪= AX + XB + ZD + CZ = AB + CD‬‬
‫ﺍﻟﺸﺮﻁ ﻛﺎﻑ‪ : ‬ﻟﻨﺘﺄﻣ‪‬ﻞ ﺭﺑﺎﻋﻴ‪‬ﺎﹰ ﳏﺪ‪‬ﺑﺎﹰ ‪ ABCD‬ﻳ‪‬ﺤﻘﹼﻖ ﺍﻟﺸﺮﻁ ‪:‬‬
‫‪AB + CD = AD + BC‬‬
‫ﻭﻟﺘﻜﻦ ‪ C‬ﺍﻟﺪﺍﺋﺮﺓ ﺍﻟﱵ ﲤﺲ ﺍﻷﺿﻼﻉ ] ‪ [ AD‬ﻭ ] ‪[ AB‬‬

‫ﻭ ] ‪ . [ BC‬ﺍﳌﻤﺎﺱ ﺍﻟﺜﺎﱐ ﻟﻠﺪﺍﺋﺮﺓ ‪ C‬ﺍﳌﺮﺳﻮﻡ ﻣﻦ ﺍﻟﻨﻘﻄﺔ ‪ C‬ﻳﻘﻄﻊ‬
‫ﺍﳌﺴﺘﻘﻴﻢ ) ‪ ( AD‬ﰲ ‪ . D ′‬ﳌﹼﺎ ﻛﺎﻧﺖ ‪ C‬ﲤﺲ‪ ‬ﺃﺿﻼﻉ ﺍﻟﺮﺑﺎﻋﻲ‬
‫‪ ABCD ′‬ﺍﺳﺘﻨﺘﺠﻨﺎ ﻣﻦ ﻟﺰﻭﻡ ﺍﻟﺸﺮﻁ ﺃﻥﹼ‬

‫‪D′ D‬‬

‫‪C‬‬

‫‪A‬‬

‫‪C‬‬
‫‪B‬‬

‫‪AB + CD ′ = AD ′ + BC‬‬
‫ﺇﺫﻥ ‪ .CD − CD ′ = AD − AD ′‬ﻭﻟﻜﻦ ‪ AD − AD ′‬ﻳﺴﺎﻭﻱ ‪ DD ′‬ﺃﻭ ‪، −DD ′‬‬
‫ﻭﻛﻞﱞ ﻣﻦ ﺍﳌﺴﺎﻭﺍﺗﲔ ‪ CD = CD ′ + D ′D‬ﺃﻭ ‪ CD ′ = CD + DD ′‬ﺗﻔﺘﻀﻲ ﺍﻧﺘﻤﺎﺀ ‪D ′‬‬

‫ﺇﱃ ) ‪ ، (CD‬ﺇﺫﻥ ‪ D ′‬ﻫﻲ ﻧﻘﻄﺔ ﺗﻘﺎ ﻃﻊ ﺍﳌﺴﺘﻘﻴﻤﲔ ) ‪ ( AD‬ﻭ ) ‪ ، (CD‬ﻭ ﻣﻨﻪ ‪، D = D ′‬‬
‫ﻭﺍﻟﺪﺍﺋﺮﺓ ‪ C‬ﲤﺲ ﺃﺿﻼﻉ ﺍﻟﺮﺑﺎﻋﻲ ‪. ABCD‬‬
‫‪ù‬‬
‫ ‬

‫ﺍﻷﻭﳌﺒﻴﺎﺩ ﺍﻟﻌﺎﳌﻲ ﻟﻠﺮﻳﺎﺿﻴ‪‬ﺎﺕ‬

‫‪32‬‬

‫ ﻧﺼﻒ ﻗﻄﺮ ﺍﻟﺪﺍﺋﺮﺓ ﺍﳌﺎ ﺭ‪‬ﺓ ﺑﺮﺅﻭﺱ ﻣﺜﻠﹼﺚ ﻣﺘﺴﺎﻭﻱ ﺍﻟﺴﺎﻗﲔ ﻳﺴﺎﻭﻱ ‪ ، R‬ﻭﻧﺼﻒ ﻗﻄﺮ ﺍﻟﺪﺍﺋﺮﺓ‬
‫ﺍﳌﻤﺎﺳﺔ ﻷﺿﻼﻋﻪ ﺩﺍﺧﻼﹰ ﻫﻮ ‪ . r‬ﺃﺛﺒﺖ ﺃﻥﹼ ﺍﳌﺴﺎﻓﺔ ﺑﲔ ﻣﺮﻛﺰ‪‬ﻱ ﺍﻟﺪﺍﺋﺮﺗﲔ ﺗﺴﺎﻭﻱ‬
‫) ‪R ( R − 2r‬‬

‫ ﻟﻨﺘﺄ ﻣ‪‬ﻞ ﻣﺜﻠﹼﺜﺎﹰ ‪ ABC‬ﻣﺘﺴﺎ ﻭ ﻱ ﺍ ﻟﺴﺎ ﻗﲔ ﻓﻴﻪ ‪ . AB = AC‬ﻭ ﻟﻨﻌﺮ‪‬ﻑ‬
‫ﺍ ﳌﻘﺪ ﺍ ﺭ ﻳﻦ ‪ a‬ﻣﻦ ‪ ℝ ∗+‬ﻭ ‪ θ‬ﻣﻦ [ ‪ ] 0, π2‬ﺑﺎ ﻟﻌﻼ ﻗﺘﲔ ‪BC = 2a‬‬
‫ﻭ‪ = θ‬‬
‫‪ . ABC‬ﻭﻟﺘﻜﻦ ‪ P‬ﻣﻨﺘﺼﻒ ] ‪ . [ BC‬ﻭﻟﻨﺮﻣﺰ ﺑﺎﻟﺮﻣﺰ ‪ O‬ﺇﱃ‬
‫ﻣﺮﻛﺰ ﺍﻟﺪﺍﺋﺮﺓ ﺍﳌﺎﺭ‪‬ﺓ ﺑﺮﺅﻭﺱ ﺍﳌﺜﻠﹼﺚ ‪ ، ABC‬ﻭﺑﺎﻟﺮﻣﺰ ‪ I‬ﺇﱃ ﻣﺮﻛﺰ ﺍﻟﺪﺍﺋﺮﺓ‬
‫ﺍﳌﻤﺎﺳ‪‬ﺔ ﻷﺿﻼﻋﻪ ﺩﺍﺧﻼﹰ‪ .‬ﺗﻘﻊ ﺍﻟﻨﻘﻄﺘﺎﻥ ‪ O‬ﻭ ‪ I‬ﻋﻠﻰ ﺍﳌﺴﺘﻘﻴﻢ ) ‪ ( AP‬ﳏﻮﺭ‬
‫ﺍﻟﻘﻄﻌﺔ ﺍﳌﺴﺘﻘﻴﻤﺔ ] ‪. [ BC‬‬

‫‪A‬‬

‫‪O‬‬
‫‪I‬‬
‫‪θ‬‬

‫‪C‬‬

‫‪P‬‬

‫‪a‬‬

‫‪B‬‬

‫‪a‬‬
‫‪BC‬‬
‫‪2a‬‬
‫ ﻣﻦ ﺟﻬﺔ ﺃﻭﱃ‪،‬‬
‫= ‪ 2R‬ﺇﺫﻥ‬
‫=‬
‫ ‬
‫‪sin 2θ‬‬
‫‪sin 2θ‬‬
‫‪sin A‬‬
‫ ﻭﻣﻦ ﺟﻬﺔ ﺃﺧﺮﻯ‪ ،‬ﻷﻥﹼ ) ‪ ( BI‬ﻣﻨﺼ‪‬ﻒ ﺍﻟﺰﺍﻭﻳﺔ ‬
‫‪ ABC‬ﺍﺳﺘﻨﺘﺠﻨﺎ ﺃﻥﹼ ) ‪. r = a tan ( θ2‬‬

‫= ‪. AO = R‬‬

‫ ﻭﺃﺧﲑﺍﹰ‪ ،‬ﳌﹼﺎ ﻛﺎﻥ ‪ I‬ﻳﻨﺘﻤﻲ ﺇﱃ ﺍﻟﻘﻄﻌﺔ ﺍﳌﺴﺘﻘﻴﻤﺔ ] ‪ [ AP‬ﺍﺳﺘﻨﺘﺠﻨﺎ ﺃﻥﹼ‬
‫‪AI = AP − PI = AP − r‬‬

‫) ‪= a tan θ − a tan ( θ2‬‬
‫) ‪sin ( θ2‬‬
‫) ‪cos θ cos ( 2θ‬‬

‫‪=a‬‬

‫ﻭﳌﹼﺎ ﻛﺎﻧﺖ ﺍﻟﻨﻘﻄﺘﺎﻥ ‪ O‬ﻭ ‪ I‬ﺗﻘﻌﺎﻥ ﻋﻠﻰ ﺍﳌﺴﺘﻘﻴﻢ ) ‪ ( AP‬ﰲ ﺟﻬﺔ ﻭﺍﺣﺪﺓ ﺑﺎﻟﻨﺴﺒﺔ ﺇﱃ ﺍﻟﻨﻘﻄﺔ ‪، A‬‬
‫ﺍﺳﺘﻨﺘﺠﻨﺎ ﺃﻥﹼ‬
‫) ‪sin ( θ2‬‬
‫‪1‬‬
‫‪d = OI = AO − AI = a‬‬
‫‪−‬‬
‫) ‪sin 2θ cos θ cos ( θ2‬‬
‫ﻭﻟﻜﻦ‬
‫) ‪sin 2θ = 4 cos θ cos ( θ2 ) sin ( θ2‬‬

‫ﺇﺫﻥ‬
‫) ‪sin ( θ2‬‬
‫‪a‬‬
‫‪1 − 2 cos θ‬‬
‫=‬
‫‪1 − 4 sin2 ( 2θ ) = a‬‬
‫‪θ‬‬
‫‪sin 2θ‬‬
‫‪sin 2θ‬‬
‫) ‪cos θ cos ( 2‬‬

‫‪d = a 1−‬‬

‫ﻋﺎﻡ ‪1962‬‬

‫‪33‬‬

‫ﻭﻣﻦ ﺟﻬﺔ ﺃﺧﺮﻯ‪،‬‬
‫‪a‬‬
‫‪a‬‬
‫) ‪− 2a tan ( θ2‬‬
‫‪sin 2θ sin 2θ‬‬
‫‪a2‬‬
‫=‬
‫) ) ‪( 1 − 2 sin 2θ tan ( 2θ‬‬
‫‪sin2 2θ‬‬

‫)‬

‫ﻭﻟﻜﻦ‬

‫(‬

‫= ) ‪R ( R − 2r‬‬

‫) ‪1 − 2 sin 2θ tan ( θ2 ) = 1 − 8 cos θ sin2 ( θ2‬‬
‫‪= 1 − 4 cos θ ( 1 − cos θ ) = ( 1 − 2 cos θ )2‬‬

‫ﺇﺫﻥ‬
‫‪( 1 − 2 cos θ )2‬‬

‫‪R ( R − 2r ) = a 2‬‬

‫‪sin2 2θ‬‬
‫ﻭﻫﺬﺍ ﻳﱪﻫﻦ ﺃﻥﹼ ) ‪ ، d 2 = R ( R − 2r‬ﻭﻳﺘﻢ‪ ‬ﺍﻹﺛﺒﺎﺕ‪.‬‬

‫‪ù‬‬

‫ ‬

‫ ﺃﺛﺒﺖ ﺃﻥﹼ ﺍﳌﺴﺘﻘﻴﻤﺎﺕ ﺍﻟﺴﺘﺔ ﺍﻟﱵ ﲢﻤﻞ ﺃﺿﻼﻉ ﺭﺑﺎﻋﻲ ﻭﺟﻮﻩ ﻣﻨﺘﻈﻢ ﲤﺲ ﲬﺲ ﻛﹸﺮﺍﺕ‪ .‬ﻭﺑﺮﻫﻦ‬
‫ﺑﺎﻟﻌﻜﺲ‪ ،‬ﺃﻧ‪‬ﻪ ﺇﺫﺍ ﻭﺟﺪﺕ ﲬﺲ ﻛﺮﺍﺕ ﲤﺲ ﺍﳌﺴﺘﻘﻴﻤﺎﺕ ﺍﻟﺴﺘﺔ ﺍﻟﱵ ﲢﻤﻞ ﺃﺿﻼﻉ ﺭ‪‬ﺑﺎﻋﻲ ﻭﺟﻮﻩ‪‬‬
‫ﻛﺎﻥ ﺭﺑﺎﻋﻲ ﺍﻟﻮﺟﻮﻩ ﻫﺬﺍ ﻣﻨﺘﻈﻤﺎﹰ‪.‬‬

‫‪A‬‬

‫ ﻣﻘﺪ‪‬ﻣﺔ ‪ :‬ﻟﻴﻜﻦ ﺭﺑﺎﻋﻲ ﺍﻟﻮﺟﻮﻩ ﺍﳌﻨﺘﻈﻢ ‪ ، ABCD‬ﺍﻟﺬﻱ ﻧﻔﺘﺮﺽ‬
‫ﺃﻥ ﻃﻮﻝ ﺿﻠﻌﻪ ﻳﺴﺎﻭﻱ ‪ . a‬ﻭﻟﻴﻜﻦ ‪ O‬ﻣﺮﻛﺰ ﺛﻘﻠﻪ ‪ ،‬ﺍﳌﻌﺮ‪‬ﻑ‬
‫ﺑﺎﻟﺼﻴﻐﺔ ‪:‬‬
‫ ‬
‫‪OA + OB + OC + OD = 0‬‬

‫)‪(1‬‬

‫‪O‬‬

‫‪D‬‬

‫‪B‬‬
‫‪A′‬‬

‫‪P‬‬

‫ ‬

‫ ‬

‫‪C‬‬

‫ﻟﻴﻜﻦ ‪ A′‬ﻣﺮﻛﺰ ﺛﻘﻞ ﺍﻟﻮﺟﻪ ‪ . BCD‬ﻧﺴﺘﻨﺘﺞ ﻣﻦ ) ‪ ( 1‬ﺃﻥﹼ ‪ OA + 3OA′ = 0‬ﺇﺫﻥ‬
‫ ‬
‫ ‬
‫‪AO = 43 AA′‬‬

‫)‪(2‬‬

‫ﻭ ﳌﹼﺎ ﻛﺎﻥ ﺍﳌﺜﻠﹼﺚ ‪ BCD‬ﻣﺘﺴﺎﻭﻱ ﺍﻷﺿﻼﻉ ﺍﺳﺘﻨﺘﺠﻨﺎ ﺃ ﻥﹼ ‪ A′‬ﻣﺘﺴﺎﻭﻳﺔ ﺍﻟﺒ‪‬ﻌﺪ ﻋﻦ ﺭﺅﻭﺱ ﻫﺬﺍ‬
‫ﺍﳌﺜﻠﹼﺚ‪ . A′ B = A′ C = A′ D = a3 ،‬ﻭ ﳌﹼﺎ ﻛﺎﻧﺖ ‪ A‬ﺃﻳﻀﺎﹰ ﻣﺘﺴﺎﻭﻳﺔ ﺍﻟﺒ‪‬ﻌﺪ ﻋﻦ ﺭﺅﻭﺱ‬
‫ﺍﳌﺜﻠﹼﺚ ‪ BCD‬ﻧﻔﺴﻪ‪ ،‬ﺍﺳﺘﻨﺘﺠﻨﺎ ﺃﻥﹼ ﲨﻴﻊ ﻧﻘﺎﻁ ﺍﳌﺴﺘﻘﻴﻢ ) ‪ ( AA′‬ﻣﺘﺴﺎﻭﻳﺔ ﺍﻟﺒ‪‬ﻌﺪ ﻋﻦ ﺭﺅﻭﺱ ﺍﳌﺜﻠﹼﺚ‬
‫‪ ، BCD‬ﻭﺃﻧ‪‬ﻪ ﻋﻤﻮﺩﻱ ﻋﻠﻰ ﺍﳌﺴﺘﻮﻱ ) ‪ . ( BCD‬ﻭﻣﻨﻪ‬
‫‪2‬‬

‫‪2‬‬

‫) ) ‪AA′2 = AP 2 − A′ P 2 = AP 2 − ( 31 DP ) = ( 98 ) ( a sin ( π3‬‬

‫ﺍﻷﻭﳌﺒﻴﺎﺩ ﺍﻟﻌﺎﳌﻲ ﻟﻠﺮﻳﺎﺿﻴ‪‬ﺎﺕ‬

‫‪34‬‬

‫ﺇﺫﻥ ‪ ، AA′ = a 36‬ﻭﺑﺎﻻﺳﺘﻔﺎﺩﺓ ﻣﻦ ) ‪ ( 2‬ﻧﺴﺘﻨﺘﺞ ﺃ ﻥﹼ‬
‫ﺍﳌﺴﺄﻟﺔ ﻧﺴﺘﻨﺘﺞ ﺃﻥﹼ‬
‫‪6‬‬
‫‪4‬‬

‫‪6‬‬
‫‪4‬‬

‫‪ ، OA = a‬ﻭﺍﺳﺘﻨﺎﺩﺍﹰ ﺇﱃ ﺗﻨﺎﻇﺮ‬

‫‪OA = OB = OC = OD = a‬‬

‫ﺇﺛﺒﺎﺕ ﺍﳋﺎﺻ‪‬ﺔ ‪ :‬ﻧﺴﺘﻨﺘﺞ ﳑﺎ ﺳﺒﻖ ﺗﻄﺎﺑﻖ ﺍﳌﺜﻠﹼﺜﺎﺕ ﺍﻟﺘﺎﻟﻴﺔ ﺍﳌﺘﺴﺎﻭﻳ‪‬ﺔ ﺍﻟﺴﺎﻗﲔ ﻭﺍﳌﺸﺘﺮﹺﻛﹶﺔ ﺑﺎﻟﺮﺃﺱ ‪: O‬‬
‫‪ OAB‬ﻭ ‪ OAC‬ﻭ ‪ OAD‬ﻭ ‪ OBC‬ﻭ ‪ OBD‬ﻭ ‪OCD‬‬

‫ﻭﻫﺬﺍ ﻳﻘﺘﻀﻲ‪ ،‬ﻣﻦ ﺛﹶﻢ‪ ،‬ﺗﻄﺎﺑﻖ ﺍﺭﺗﻔﺎﻋﺎ‪‬ﺎ ﺍﻟﻨﺎﺯﻟﺔ ﻣﻦ ‪ ،O‬ﺃﻱ ﺇﻥﹼ ﺍﻟﻨﻘﻄﺔ ‪ O‬ﺗﺒﻌﺪ ﺃﺑﻌﺎﺩﺍﹰ ﻣﺘﺴﺎﻭﻳﺔ ﻋﻦ‬
‫ﺍﳌﺴﺘﻘﻴﻤﺎﺕ ) ‪ ( AB‬ﻭ ) ‪ ( AC‬ﻭ ) ‪ ( AD‬ﻭ ) ‪ ( BC‬ﻭ ) ‪ ( BD‬ﻭ ) ‪ . (CD‬ﻭﻫﺬﺍ ﺍﻟﺒ‪‬ﻌﺪ ﻳﺴﺎﻭﻱ‬
‫‪a‬‬

‫‪2‬‬

‫= ) ‪r = OA2 − ( 12 AB‬‬

‫‪2 2‬‬
‫ﺇﺫﻥ ﺍﻟﻜﺮﺓ ) ‪ S = S (O, 2 a 2‬ﲤﺲ‪ ‬ﲨﻴﻊ ﺃﺿﻼﻉ ﺭﺑﺎﻋﻲ ﺍﻟﻮﺟﻮﻩ ‪. ABCD‬‬

‫ ﺗﺘﻘﺎﻃﻊ ﺍﻟﻜﺮﺓ ‪ S‬ﻣﻊ ﺍﳌﺴﺘﻮﻱ ) ‪ ( ABC‬ﺑﺪﺍﺋﺮﺓ ‪ C‬ﲤ ‪‬‬
‫ﺲ‬
‫ﺃﺿﻼ ﻉ ﺍ ﳌﺜﻠﹼﺚ ‪ ABC‬ﺩ ﺍ ﺧﻼﹰ ‪ ،‬ﻭ ﻷ ﻥﹼ ﻫﺬ ﺍ ﺍ ﳌﺜﻠﹼﺚ ﻣﺘﺴﺎ ﻭﻱ‬
‫ﺍﻷﺿﻼﻉ‪ ،‬ﺍﺳﺘﻨﺘﺠﻨﺎ ﺃﻥﹼ ‪ ، D ′‬ﻣﺮﻛﺰ ‪ ، C‬ﻫﻮ ﻣﺮﻛﺰ ﺛﻘﻠﻪ‪ .‬ﻭﺇﺫﺍ ﻛﺎﻥ‬

‫‪A‬‬
‫‪K‬‬

‫ ‬
‫ ‬
‫‪ AP‬ﳏﻮ ﺭ ﺍ ﻟﻀﻠﻊ ‪ BC‬ﻛﺎ ﻥ ‪ . AD ′ = 23 AP‬ﻟﺘﻜﻦ ‪P ′‬‬

‫‪P′‬‬

‫‪L‬‬

‫‪D′‬‬

‫‪C‬‬

‫ﺍ ﻟﻨﻘﻄﺔ ﻣﻦ ‪ C‬ﺍ ﳌﹸﻘﺎ ﺑﻠﺔ ﻗﻄﺮ ‪‬ﻳﺎﹰ ﻟﻠﻨﻘﻄﺔ ‪ ، P‬ﻳﻘﻄﻊ ﺍ ﳌﺴﺘﻘﻴﻢ‪ ‬ﺍ ﳌﺎ ‪‬ﺭ‬
‫ﺑﺎﻟﻨﻘﻄﺔ ‪ P ′‬ﻣﻮﺍﺯﻳﺎﹰ ) ‪ ( BC‬ﺍﻟﻀﻠﻌﲔ ] ‪ [ AB‬ﻭ ] ‪ [ AC‬ﰲ ﺍﻟﻨﻘﻄﺘﲔ ‪ K‬ﻭ ‪ L‬ﻋﻠﻰ ﺍﻟﺘﺮﺗﻴﺐ‪.‬‬
‫‪B‬‬

‫ ‬

‫ ‬

‫ ‬

‫‪P‬‬

‫‪C‬‬

‫ ‬

‫ﻭﻧﻼﺣﻆ ﺃﻥﹼ ‪ AK = 31 AB‬ﻭ ‪. AL = 13 AC‬‬
‫ﻟﻴﻜﻦ ) ‪ S(OD ′‬ﺍﻟﺘﻨﺎﻇﺮ ﺍﻟﻘﺎﺋﻢ ﺑﺎﻟﻨﺴﺒﺔ ﺇﱃ ﺍﳌﺴﺘﻘﻴﻢ ) ‪ . (OD ′‬ﻋﻨﺪﺋﺬ ﻧﻼﺣﻆ ﺃﻥﹼ‬

‫) ‪ S(OD ′‬ﻳ‪‬ﺤﺎﻓﻆ‬

‫ﻋﻠﻰ ﻛﻞﱟ ﻣﻦ ﺍﻟﻜﺮﺓ ‪ S‬ﻭﺍﳌﺴﺘﻮﻱ ) ‪ ، ( ABC‬ﻭﻫﻮ ﻣﻦ ﺛﹶﻢ‪ ،‬ﻳ‪‬ﺤﺎﻓﻆ ﻋﻠﻰ ﺍﻟﺪﺍﺋﺮﺓ ‪ . C‬ﻭﻣﻨﻪ ﻧﺴﺘﻨﺘﺞ‬
‫ﺃﻥﹼ ﺍﻟﻜﺮﺓ ‪ S‬ﲤﺲ ﺍﳌﺴﺘﻘﻴﻢ ) ‪ ( KL‬ﻷﻥﹼ ) ‪. S(OD ′ )(( BC )) = ( KL‬‬
‫ ‬

‫ ‬

‫ ﻭﺇﺫﺍ ﻋﺮ‪‬ﻓﻨﺎ ﺍﻟﻨﻘﻄﺔ ‪ M‬ﺑﺎﻟﻌﻼﻗﺔ ‪ ، AM = 31 AD‬ﺍﺳﺘﻨﺘﺠﻨﺎ‪ ،‬ﺑﺄﺳﻠﻮﺏ ﳑﺎﺛﻞﹴ ﻟ‪‬ﻤﺎ ﺳﺒﻖ‪ ،‬ﺃﻥﹼ‬
‫ﺍﻟﻜﺮﺓ ‪ S‬ﲤﺲ ﺃﻳﻀﺎﹰ ﺍﳌﺴﺘﻘﻴﻤﲔ ) ‪ ( LM‬ﻭ ) ‪. ( MK‬‬
‫ ﻟﻴﻜﻦ ﺇﺫﻥ ‪ HA,3‬ﺍﻟﺘﺤﺎﻛﻲ ﺍﻟﺬﻱ ﻣﺮﻛﺰﻩ ‪ A‬ﻭﻧﺴﺒﺘﻪ ‪ . 3‬ﻭﻟﻨﻌﺮ‪‬ﻑ ﺍﻟﻜﺮﺓ ‪ SA‬ﺑﺄ‪‬ﺎ ﺻﻮﺭﺓ‬
‫‪ S‬ﻭﻓﻖ ﺍﻟﺘﺤﺎﻛﻲ ‪ HA,3‬ﺃﻱ ) ‪ . SA = HA,3 ( S‬ﻋﻨﺪﺋﺬ ﲤﺲ‪ ‬ﺍﻟﻜﺮﺓ ‪ SA‬ﻛﻼﹰ ﻣﻦ ﺍﳌﺴﺘﻘﻴﻤﺎﺕ‬
‫) ‪ ( AB‬ﻭ ) ‪ ( AC‬ﻭ ) ‪ ( AD‬ﻷ ﻥﹼ ﺍ ﻟﺘﺤﺎ ﻛﻲ ‪ HA,3‬ﻳ‪‬ﺤﺎ ﻓﻆ ﻋﻠﻴﻬﺎ ‪ ،‬ﻭ ﻫﻲ ﲤﺲ‪ ‬ﺍ ﳌﺴﺘﻘﻴﻤﺎﺕ‬
‫) ‪ (CB‬ﻭ ) ‪ ( DC‬ﻭ ) ‪ ( BD‬ﻷ‪‬ﺎ ﺻﻮﺭ ) ‪ ( KL‬ﻭ ) ‪ ( LM‬ﻭ ) ‪ ( MK‬ﻭﻓﻖ ‪. HA,3‬‬

‫ﻋﺎﻡ ‪1962‬‬

‫‪35‬‬

‫ ﻭﺃﺧﲑﺍﹰ ﺇﺫﺍ ﻋﺮ‪‬ﻓﻨﺎ ﺑﺎ ﳌﹸﻤﺎﺛﻠﺔ ﺍﻟﺘﺤﺎﻛﻴﺎﺕ ‪ HB,3‬ﻭ ‪ HC ,3‬ﻭ ‪ HD,3‬ﻛﺎﻧﺖ ﺃﻳﻀﺎﹰ ﺍﻟﻜﺮﺍﺕ‬
‫) ‪ SB = HB,3 ( S‬ﻭ ) ‪ SC = HC ,3 ( S‬ﻭ ) ‪ SD = HD,3 ( S‬ﳑﺎﺳ‪‬ﺔ ﳉﻤﻴﻊ ﺍ ﳌﺴﺘﻘﻴﻤﺎﺕ‬
‫ﺍﳊﺎﻣﻠﺔ ﻷﺿﻼﻉ ﺭﺑﺎﻋﻲ ﺍﻟﻮﺟﻮﻩ ‪ . ABCD‬ﻭﻫﻜﺬﺍ ﺗﻜﻮﻥ ﺍﻟﻜﺮﺍﺕ ﺍﳋﻤﺲ ‪ S‬ﻭ ‪ SA‬ﻭ ‪ SB‬ﻭ ‪SC‬‬
‫ﻭ ‪ SD‬ﳑﺎﺳ‪‬ﺔ ﳉﻤﻴﻊ ﺍﳌﺴﺘﻘﻴﻤﺎﺕ ﺍﳊﺎﻣﻠﺔ ﻷﺿﻼﻉ ﺭﺑﺎﻋﻲ ﺍﻟﻮﺟﻮﻩ ‪. ABCD‬‬
‫ﺇﺛﺒﺎﺕ ﺍﻟﻌﻜﺲ ‪ :‬ﻟﻨﺮﻣﺰ ﺇﱃ ﺭﺅﻭﺱ ﺭﺑﺎﻋﻲ ﺍﻟﻮﺟﻮﻩ ﺑﺎﻟﺮﻣﻮﺯ ‪ . ( Ai )1≤i ≤4‬ﻭﻟﺘﻜﻦ ﺇﺫﻥ ‪ S‬ﺍﻟﻜﺮﺓ‬
‫ﺍﳌﻤﺎﺳ‪‬ﺔ ﺩﺍﺧﻼﹰ ﻷﺿﻼﻉ ﺭﺑﺎﻋﻲ ﺍﻟﻮﺟﻮﻩ‪ .‬ﻭﻟﻨﺮﻣﺰ ﺑﺎﻟﺮﻣﺰ ‪ Si‬ﺇﱃ ﺍﻟﻜﺮﺓ ﺍﳌﻤﺎﺳ‪‬ﺔ ﺧﺎﺭﺟﺎﹰ ﳌﻤﺪ‪‬ﺩﺍﺕ‬
‫ﺃﺿﻼﻉ ﺭﺑﺎﻋﻲ ﺍﻟﻮﺟﻮﻩ ﻭﺍﻟﱵ ﺗ‪‬ﻘﺎﺑﻞ ﺍﻟﺮﺃﺱ ‪. Ai‬‬

‫‪S1‬‬

‫‪X3‬‬

‫‪ℓ4‬‬

‫‪X4‬‬

‫‪X2‬‬

‫‪ւ‬‬

‫‪ℓ 2 A2‬‬

‫‪A3‬‬
‫‪ℓ3‬‬

‫‪A1 ℓ 1‬‬

‫ ﺍﻟﻔﻜﺮﺓ ﺍﻷﺳﺎﺳﻴ‪‬ﺔ ﻫﻲ ﰲ ﻛﻮﻥ ﺃﻃﻮﺍﻝ ﺍﳌﻤﺎﺳﺎﺕ ﻟﻜﺮﺓ‪ ،‬ﻭﺍﳌﻨﺒﻌﺜﺔ ﻣﻦ ﻧﻘﻄﺔ ﺧﺎﺭﺟﻬﺎ‪ ،‬ﻣﺘﺴﺎﻭﻳﺔ‪.‬‬
‫ﻷﻥﹼ ﺍﳌﺴﺘﻮﻱ ﺍﻟﺬﻱ ﻳﻌﻴ‪‬ﻨﻪ ﺃﻱ ﳑﺎﺳﲔ ﻟﻜﺮﺓ ﻳﺘﻘﺎﻃﻊ ﻣﻌﻬﺎ ﻭﻓﻖ ﺩﺍﺋﺮﺓ ﻳﻜﻮﻧﺎﻥ ﳑﺎﺳﲔ ﳍﺎ‪.‬‬
‫ﻟﻨﻌﺮ‪‬ﻑ ﺇﺫﻥ ﺍﻟﻄﻮﻝ ‪ ℓ i‬ﺑﺄﻧ‪‬ﻪ ﻃﻮﻝ ﻗﻄﻌﺔ ﺍﳌﻤﺎﺱ ﺍﳌﺮﺳﻮﻡ ﻣﻦ ﺍﻟﺮﺃﺱ ‪ Ai‬ﻟﻠﻜﺮﺓ ‪ ، S‬ﻭﺫﻟﻚ ﰲ ﺣﺎﻟﺔ‬
‫‪ . 1 ≤ i ≤ 4‬ﻓﻴﻜﻮﻥ ﻃﻮﻝ ﺍﻟﻀﻠﻊ ] ‪ [ Ai Aj‬ﻣﺴﺎﻭﻳﺎﹰ ‪. Ai Aj = ℓ i + ℓ j‬‬
‫ ﻟﻨﺮﻣﺰ ﺇﱃ ﻧﻘﻄﺔ ﲤﺎﺱ ﺍﳌﺴﺘﻘﻴﻢ ) ‪ ( A1Aj‬ﻣﻊ ﺍﻟﻜﺮﺓ ‪ S1‬ﺑﺎﻟﺮﻣﺰ ‪ X j‬ﰲ ﺣﺎﻟﺔ ‪ j‬ﻣﻦ‬
‫ﺍ‪‬ﻤﻮﻋﺔ } ‪ . { 2, 3, 4‬ﻭﻟﻨﻼﺣﻆ ﺃ ﻥﹼ ‪ S1‬ﺗﺘﻘﺎﻃﻊ ﻣﻊ ﺍﳌﺴﺘﻮﻱ ) ‪ ( A2A3A4‬ﰲ ﺍﻟﺪﺍﺋﺮﺓ ﺍﳌﻤﺎﺳ‪‬ﺔ‬
‫ﺩﺍﺧﻼﹰ ﻷﺿﻼﻉ ﺍﳌﺜﻠﹼﺚ ‪ ، A2A3A4‬ﻭﻫﻲ ﻧﻔﺴﻬﺎ ﺍﻟﺪﺍﺋﺮﺓ ﺍﻟﱵ ﺗﺘﻘﺎﻃﻊ ﻓﻴﻬﺎ ﺍﻟﻜﺮﺓ ‪ S‬ﻣﻊ ﺍﳌﺴﺘﻮﻱ‬
‫) ‪ . ( A2A3A4‬ﺇﺫﻥ ﺃﻃﻮﺍﻝ ﻗﻄﻊ ﺍﳌﻤﺎﺳﺎﺕ ﺍﳌﻨﺒﻌﺜﺔ ﻣﻦ ‪ Aj‬ﻟﻠﻜﺮﺓ ‪ S‬ﺗﺴﺎﻭﻱ ﺃﻃﻮﺍﻝ ﻗﻄﻊ ﺍﳌﻤﺎﺳﺎﺕ‬
‫ﺍﳌﻨﺒﻌﺜﺔ ﻣﻦ ﺍﻟﻨﻘﻄﺔ ﻧﻔﺴﻬﺎ ﻟﻠﻜﺮﺓ ‪ ، S1‬ﻭﻋﻠﻴﻪ ﻳﻜﻮﻥ ‪ Aj X j = ℓ j‬ﻭﺫﻟﻚ ﰲ ﺣﺎﻟﺔ ‪ j‬ﻣﻦ ﺍ‪‬ﻤﻮﻋﺔ‬
‫} ‪ . { 2, 3, 4‬ﻧﺴﺘﻨﺘﺞ ﺇﺫﻥ ﻣﻦ ﻛﻮﻥ ‪ A1X 2 = A1X 3 = A1X 4‬ﺃﻥﹼ‬
‫‪ℓ 1 + 2ℓ 2 = ℓ 1 + 2 ℓ 3 = ℓ 1 + 2 ℓ 4‬‬

‫ﻭﻫﺬﺍ ﻳﱪﻫﻦ ﻋﻠﻰ ﺃﻥﹼ ‪. ℓ 2 = ℓ 3 = ℓ 4‬‬

‫ﺍﻷﻭﳌﺒﻴﺎﺩ ﺍﻟﻌﺎﳌﻲ ﻟﻠﺮﻳﺎﺿﻴ‪‬ﺎﺕ‬

‫‪36‬‬

‫ ﻭﺑﺘﻄﺒﻴﻖ ﺍﻟﺪﺭﺍﺳﺔ ﺍﻟﺴﺎ ﺑﻘﺔ ﺍﻧﻄﻼﻗﺎﹰ ﻣﻦ ﺍﻟﺮﺃﺱ ‪ A2‬ﻭﺍﻟﻜﺮﺓ ‪ S2‬ﻧﺴﺘﻨﺘﺞ ﺑﺄﺳﻠﻮﺏ ﳑﺎﺛﻞﹴ ﺃ ﹼﻥ‬
‫‪ . ℓ 1 = ℓ 3 = ℓ 4‬ﻭﺑﻨﺎ ﺀً ﻋﻠﻰ ﻫﺬﺍ ﻧﺮﻯ ﺃ ﻥﹼ ﺍﻷﻃﻮﺍﻝ ‪ ( Ai Aj )1≤i < j ≤4‬ﻣﺘﺴﺎﻭﻳﺔ‪ ،‬ﻓﺮﺑﺎﻋﻲ‬
‫ﺍﻟﻮﺟﻮﻩ ‪ A1A2A3A4‬ﻣﻨﺘﻈﻢ‪ .‬ﻭﻳﺘﻢ‪ ‬ﺍﻹﺛﺒﺎﺕ‪.‬‬

‫‪ù‬‬
‫ ‬

‫‪QWE‬‬
‫‪AgD‬‬
‫‪ZXC‬‬

‫ّ‬
‫ﺃﻭﳌﺒﻴﺎﺩ ﺍﻟﺮﻳﺎﺿﻴﺎﺕ ﺍﳋﺎﻣﺲ‬
‫ ﻋﻴ‪‬ﻦ ﳎﻤﻮﻋﺔ ﻗﻴﻢ ‪ p‬ﻣﻦ ‪ ℝ‬ﺍﻟﱵ ﺗﻘﺒﻞ ﻋﻨﺪﻫﺎ ﺍﳌﻌﺎﺩﻟﺔ‬
‫‪x2 − p + 2 x2 − 1 = x‬‬

‫ﺣﻠﻮﻻﹰ ﺣﻘﻴﻘﻴ‪‬ﺔ‪ .‬ﻭﺃﻭﺟﺪ ﻋﻨﺪﺋﺬ ﻫﺬﻩ ﺍﳊﻠﻮﻝ‪.‬‬
‫ ﻟﻨﻔﺘﺮﺽ ﺃﻧ‪‬ﻪ ﻳﻮﺟﺪ ﻋﺪﺩ‪ ‬ﺣﻘﻴﻘﻲ ‪ x‬ﻳﻜﻮﻥ ﺣﻼﹰ ﻟﻠﻤﻌﺎﺩﻟﺔ ﺍﳌﻌﻄﺎﺓ‪.‬‬
‫ ﻻ ﺑ‪‬ﺪ‪ ‬ﺃﻥ ﻳﻜﻮﻥ ‪ x‬ﻣﻮﺟﺒﺎﹰ ﻷﻥﹼ ﺍﻟﻄﺮﻑ ﺍﻷﻳﺴﺮ ﻣﻮﺟﺐ‪ ،‬ﻭﻻ ﺑ‪‬ﺪ‪ ‬ﺃﻥ ﻳﻜﻮﻥ ﺍﳉﺬﺭﺍﻥ ﻣﻌﺮ‪‬ﻓﲔ‪.‬‬
‫ﻭﻫﺬﺍ ﻳﻘﺘﻀﻲ ﲢﻘﹼﻖ ﺍﳌﺘﺮﺍﺟﺤﺔ ‪. x ≥ max ( 1, p ) ≥ 1 :‬‬
‫ ﻭﺗ‪‬ﻜﺎﻓﺊ ﺍﳌﺘﺮﺍﺟﺤﺔ ‪ x 2 − p ≤ x‬ﺃﻥ ﻳﻜﻮﻥ ‪. p ≥ 0‬‬
‫ ﻛﻤﺎ ﺗﻜﺎ ﻓﺊ ﺍ ﳌﺘﺮ ﺍ ﺟﺤﺔ ‪ 2 x 2 − 1 ≤ x‬ﺃ ﻥ ﻳﻜﻮ ﻥ‬
‫‪ p ≤ x 2‬ﺍﺳﺘﻨﺘﺠﻨﺎ ﺃﻥﹼ ‪. p ≤ 43‬‬
‫ﻭﻫﻜﺬﺍ ﻧﺮﻯ ﺃﻥﹼ ﺍﻧﺘﻤﺎﺀ ‪ p‬ﺇﱃ ﺍ‪‬ﺎﻝ ] ‪ [ 0, 43‬ﻫﻮ ﺷﺮﻁﹲ ﻻﺯﻡ ﻟﻮﺟﻮﺩ ﺣﻠﻮﻝ ﺣﻘﻴﻘﻴ‪‬ﺔ ﻟﻠﻤﻌﺎﺩﻟﺔ‪.‬‬
‫ﻭﺑﺎﻟﻌﻜﺲ‪ ،‬ﻟﻨﻔﺘﺮﺽ ﺃﻥﹼ ‪ p‬ﻋﻨﺼﺮ‪ ‬ﻣﻦ ﺍ‪‬ﺎﻝ ] ‪ . [ 0, 43‬ﺇﻥﹼ ﻛﻞﱠ ﺣﻞﱟ ﻟﻠﻤﻌﺎﺩﻟﺔ ﺍﳌﻌﻄﺎﺓ ﻳ‪‬ﺤﻘﹼﻖ ﺑﺎﻟﺘﺮﺑﻴﻊ‬
‫‪4‬‬
‫‪3‬‬

‫≤ ‪ . x 2‬ﻭ ﺇ ﺫ ﺍ ﺗﺬ ﻛﹼﺮ ﻧﺎ ﺃﻥﹼ‬

‫‪p‬‬
‫‪4‬‬

‫‪( x 2 − 1 )( x 2 − p ) = 1 + − x 2‬‬
‫ﻭﳒﺪ ﺑﺎﻟﺘﺮﺑﻴﻊ ﻣﻦ ﺟﺪﻳﺪ ﻭﺍﻹﺻﻼﺡ‬

‫‪( 4 − p )2‬‬
‫) ‪8 (2 − p‬‬

‫= ‪x2‬‬

‫ﻭﳌﹼﺎ ﻛﺎﻥ ﺍﳊﻞﱡ ﺍﳌﺮﺟﻮ‪ ‬ﻣﻮﺟﺒﺎﹰ ﺍﺳﺘﻨﺘﺠﻨﺎ ﺃﻥﹼ‬
‫‪4−p‬‬
‫‪2 4 − 2p‬‬

‫= ‪x = xp‬‬

‫ﻭﰲ ﻫﺬﻩ ﺍﳊﺎﻟﺔ ﻧﺘﻴ‪‬ﻘﹼﻦ ﻣﺒﺎﺷﺮﺓ ﺃﻥﹼ‬
‫‪( 4 − 3p )2‬‬
‫‪4 − 3p‬‬
‫=‬
‫) ‪8 (2 − p‬‬
‫‪2 4 − 2p‬‬

‫‪( 4 − p )2‬‬
‫= ‪−p‬‬
‫) ‪8 (2 − p‬‬

‫= ‪x p2 − p‬‬

‫‪p2‬‬
‫‪p‬‬
‫=‬
‫) ‪8(2 − p‬‬
‫‪2 4 − 2p‬‬

‫‪( 4 − p )2‬‬
‫= ‪−1‬‬
‫) ‪8 (2 − p‬‬

‫= ‪x p2 − 1‬‬

‫ﻭﻣﻦ ﺛﹶﻢ‪ . x p2 − p + 2 x p2 − 1 = x p ‬ﻭ ‪ x p‬ﻫﻲ ﺍﳊﻞﹼ ﺍﻟﻮﺣﻴﺪ ﻟﻠﻤﻌﺎﺩﻟﺔ ﰲ ﻫﺬﻩ ﺍﳊﺎﻟﺔ‪.‬‬
‫ﻓﺎﻟﺸﺮﻁ ﺍﻟﻼﺯﻡ ﻭﺍﻟﻜﺎﰲ ﻟﻮﺟﻮﺩ ﺣﻞﹼ ﺣﻘﻴﻘﻲ ﳍﺬﻩ ﺍﳌﻌﺎﺩﻟﺔ ﻫﻮ ﺍﻧﺘﻤﺎﺀ ‪ p‬ﺇﱃ ] ‪. [ 0, 43‬‬
‫‪ù‬‬
‫‪37‬‬

‫ﺍﻷﻭﳌﺒﻴﺎﺩ ﺍﻟﻌﺎﳌﻲ ﻟﻠﺮﻳﺎﺿﻴ‪‬ﺎﺕ‬

‫‪38‬‬

‫ ﻧ‪‬ﻌﻄﻰ ﻧﻘﻄﺔ ‪ A‬ﻭﻗﻄﻌﺔ ﻣﺴﺘﻘﻴﻤﺔ ] ‪ . [ BC‬ﻋﻴ‪‬ﻦ ﺍﶈﻞﹼ ﺍﳍﻨﺪﺳﻲ ﻟﻠﻨﻘﺎﻁ ‪ P‬ﻣﻦ ﺍﻟﻔﺮﺍﻍ ﺍﻟﱵ‬
‫ﺗﻮﺟﺪ ﻋﻨﺪ ﻛﻞﱟ ﻣﻨﻬﺎ ﻧﻘﻄﺔ ‪ X‬ﻣﻦ ﺍﻟﻘﻄﻌﺔ ﺍﳌﺴﺘﻘﻴﻤﺔ ] ‪ [ BC‬ﺗ‪‬ﺤﻘﹼﻖ‬

‫‪π‬‬
‫‪2‬‬

‫= ‬
‫‪. APX‬‬

‫ ﰲ ﺍﳊﻘﻴﻘﺔ‪ ،‬ﻟﻨﻼﺣﻆ ﺃﻥﹼ ‪ X‬ﺗﻨﺘﻤﻲ ﺇﱃ ] ‪ [ BC‬ﺇﺫﺍ ﻭﻓﻘﻂ ﺇﺫﺍ ﻛﺎﻧﺖ ‪ X‬ﻣﺮﻛﺰ ﺍﻷﺑﻌﺎﺩ ﺍﳌﺘﻨﺎﺳﺒﺔ‬
‫ﻟﻠﻨﻘﻄﺘﲔ ) ‪ ( B,1 − t‬ﻭ ) ‪ (C , t‬ﰲ ﺣﺎﻟﺔ ] ‪ . t ∈ [ 0,1‬ﻟﻨﺮﻣﺰ ﺑﺎﻟﺮﻣﺰ ‪ L‬ﺇﱃ ﺍﶈﻞﹼ ﺍﳍﻨﺪﺳﻲ‬
‫ﺍﳌﻄﻠﻮﺏ‪ .‬ﻋﻨﺪﺋﺬ‬
‫ ‬
‫‪P ∈ L ⇔ ∃X ∈ [ BC ], AP ⋅ PX = 0‬‬
‫ ‬
‫ ‬
‫ ‬
‫‪⇔ ∃t ∈ [ 0,1 ], AP ⋅ ( ( 1 − t ) PB + tPC ) = 0‬‬
‫ ‬
‫ ‬
‫‪⇔ ∃t ∈ [ 0,1 ], ( 1 − t ) AP ⋅ PB + tAP ⋅ PC = 0‬‬
‫ ‬
‫‪⇔ ( AP ⋅ PB )( AP ⋅ PC ) ≤ 0‬‬

‫ﻭﻟﻜﻦ‪ ،‬ﺇﺫﺍ ﻋﺮ‪‬ﻓﻨﺎ ‪ T ′‬ﻣﻨﺘﺼﻒ ﺍﻟﻘﻄﻌﺔ ﺍﳌﺴﺘﻘﻴﻤﺔ ] ‪ [ AT‬ﺍﺳﺘﻨﺘﺠﻨﺎ ﺃﻥﹼ‬
‫ ‬
‫ ‬
‫ ‬
‫ ‬
‫‪AP ⋅ PT = AT ′ + T ′P ⋅ PT ′ + T ′T‬‬
‫ ‬
‫‪= AT ′ − PT ′ PT ′ + AT ′ = AT ′2 − PT ′2‬‬

‫)‬
‫)‬

‫()‬
‫()‬

‫(‬
‫(‬

‫ﻓﺈﺫﺍ ﻛﺎﻧﺖ ‪ B ′‬ﻣﻨﺘﺼﻒ ] ‪ ، [ AB‬ﻭ ‪ C ′‬ﻣﻨﺘﺼﻒ ] ‪ [ AC‬ﺍﺳﺘﻨﺘﺠﻨﺎ ﳑﺎ ﺳﺒﻖ ﺃﻥﹼ‬
‫‪P ∈ L ⇔ ( AB ′2 − PB ′2 )( AC ′2 − PC ′2 ) ≤ 0‬‬

‫‪⇔ ( AB ′ − PB ′ )( AC ′ − PC ′ ) ≤ 0‬‬

‫ﻧﻌﺮ‪‬ﻑ‪ ،‬ﰲ ﺣﺎﻟﺔ ﻧﻘﻄﺔ ‪ M‬ﻭﻋﺪﺩ‪ ‬ﻣﻮﺟﺐﹴ ‪ ، r‬ﺍﻟﻜﺮﺓ ﺍﳌﻔﺘﻮﺣﺔ ﺍﳌﻠﻴﺌﺔ ) ‪ ، B ( M , r‬ﻭﺍﻟﻜﺮﺓ ﺍﳌﻐﻠﻘﺔ‬
‫ﺍﳌﻠﻴﺌﺔ ) ‪ B ( M , r‬ﺍﻟﱵ ﻣﺮﻛﺰﻫﺎ ‪ M‬ﻭﻧﺼﻒ ﻗﻄﺮﻫﺎ ‪ ، r‬ﻛﻤﺎ ﻳﻠﻲ ‪:‬‬
‫} ‪ B ( M , r ) = { P : PM < r‬ﻭ } ‪B ( M , r ) = { P : PM ≤ r‬‬

‫ﻋﻨﺪﺋﺬ ﻧﺮﻯ ﻣﺒﺎﺷﺮﺓ ﺃﻥﹼ ‪ P ∈ L‬ﻳ‪‬ﻜﺎﻓﺊ ﺍﻟﺸﺮﻁ‬
‫) ) ‪( ( PB ′ ≤ AB ′ ) ∧ ( AC ′ ≤ PC ′ ) ) ∨ ( ( AB ′ ≤ PB ′ ) ∧ ( PC ′ ≤ AC ′‬‬
‫ﺃﻭ‬
‫) ) ‪L = ( B ( B ′, AB ′ )\B (C ′, AC ′ ) ) ∪ ( B (C ′, AC ′ )\B ( B ′, AB ′‬‬

‫ﻭﻋﻠﻴﻪ ﺇﺫﺍ ﻋﺮ‪‬ﻓﻨﺎ ﺑﺎﻟﺘﺮﺗﻴﺐ ‪ BB‬ﻭ ‪ B B‬ﺑﺄ‪‬ﻤﺎ ﺍﻟﻜﺮﺗﺎﻥ ﺍﳌﻠﻴﺌﺘﺎﻥ ﺍﳌﻔﺘﻮﺣﺔ ﻭﺍﳌﻐﻠﻘﺔ ﺍﻟﻠﺘﺎﻥ ﺗﻘﺒﻼﻥ‬
‫] ‪ [ AB‬ﻗﻄﺮﺍﹰ‪ ،‬ﻭﻋﺮ‪‬ﻓﻨﺎ ﺑﺄﺳﻠﻮﺏ ﳑﺎﺛﻞﹴ ‪ BC‬ﻭ ‪ BC‬ﺑﺄ‪‬ﻤﺎ ﺍﻟﻜﺮﺗﺎﻥ ﺍﳌﻠﻴﺌﺘﺎﻥ ﺍﳌﻔﺘﻮﺣﺔ ﻭﺍﳌﻐﻠﻘﺔ ﺍﻟﻠﺘﺎﻥ‬
‫ﺗﻘﺒﻼﻥ ] ‪ [ AC‬ﻗﻄﺮﺍﹰ‪ ،‬ﻛﺎﻥ ) ‪. L = ( B B \BC ) ∪ ( BC \BB‬‬
‫‪ù‬‬

‫ﻋﺎﻡ ‪1963‬‬

‫‪39‬‬

‫ ﻧﺘﺄ ﻣ‪‬ﻞ ﻣﻀﻠﹼﻌﺎﹰ ﻟﻪ ‪ n‬ﺿﻠﻌﺎﹰ‪ .‬ﻧﻔﺘﺮﺽ ﺃ ﻥﹼ ﲨﻴﻊ ﺯﻭﺍﻳﺎﻩ ﻣﺘﺴﺎﻭ ﻳ‪‬ﺔ‪ ،‬ﻭﺃ ﻥﹼ ﺃﻃﻮﺍﻝ ﺃﺿﻼﻋﻪ ﺍﳌﺘﺘﺎﻟﻴﺔ‬
‫‪ (ai )0≤i <n‬ﺗ‪‬ﺤﻘﹼﻖ ﺍﳌﺘﺮﺍﺟﺤﺔ ‪ . a 0 ≥ a1 ≥ ⋯ ≥ an −1‬ﺃﺛﺒﺖ ﺃﻧ‪‬ﻪ ﻣﻀﻠﹼﻊ ﻣﻨﺘﻈﻢ‪.‬‬
‫ ﳚﺐ ﺃﻥ ﻧﻔﺘﺮﺽ ﺃﻥﹼ ﺭﺅﻭﺱ ﻫﺬﺍ ﺍﳌﻀﻠﹼﻊ ﻣﺘﺒﺎﻳﻨﺔ‪ .‬ﻟﻨﺮﻣﺰ ﺇﱃ ﻫﺬﻩ ﺍﻟﺮﺅﻭﺱ ﺑﺎﻟﺮﻣﻮﺯ ‪. ( Ai )0≤i <n‬‬
‫ﻟﻨﻤﺪ‪ ‬ﺩ ﺍ ﳌﺘﺘﺎ ﻟﻴﺘﲔ ‪ (ai )0≤i <n‬ﻭ ‪ ( Ai )0≤i <n‬ﺑﻮﺿﻊ ‪ ak = ak mod n‬ﻭ ‪. Ak = Ak mod n‬‬
‫ﻟﻨﺘﻤﻜﹼﻦ ﻣﻦ ﺍﻋﺘﺒﺎﺭ ﺍﻟﻀﻠﻊ ﺍﻷﻭ‪‬ﻝ ﺍﻟﻀﻠﻊ‪ ‬ﺍﻟﺬﻱ ﻳﻠﻲ ﺍﻷﺧﲑ‪ .‬ﻧﻔﺘﺮﺽ ﺇﺫﻥ ﺃﻥﹼ‬
‫‪∀k ∈ { 0,1, …, n − 1 }, Ak Ak +1 = ak‬‬

‫ﳝﻜﻦ ﺃﻥ ﻧﻨﺴﺐ ﺍﳌﺴﺘﻮﻱ ﺇﱃ ﲨﻠﺔ ﻣﺘﻌﺎﻣﺪﺓ ﻧﻈﺎﻣﻴ‪‬ﺔ ﻣﺒﺪ ﺅﻫﺎ ‪ ، A0‬ﻭﳏﻮﺭ ﻓﻮﺍﺻﻠﻬﺎ ﻣﻮ ﺟ‪‬ﻪ ﺑﺎﻟﺸﻌﺎﻉ‬
‫ ‬
‫‪ ، A0A1‬ﻭﳔﺘﺎﺭ ﳏﻮﺭ ﺗﺮﺍﺗﻴﺒﻬﺎ ﻟﻴﻜﻮﻥ ﺗﺮﺗﻴﺐ ‪ A2‬ﻣﻮﺟﺒﺎﹰ‪ .‬ﺇﺫﻥ ﻳﻮﺟﺪ ‪ θ‬ﻣﻦ [ ‪ ] 0, π‬ﻳ‪‬ﺤﻘﹼﻖ‬
‫ ‬
‫ ‬
‫ ‬
‫‪(Ak −1Ak , Ak Ak +1 ) = θ‬‬

‫‪∀k ∈ {1, …, n },‬‬

‫ﺑﺎﻻﺳﺘﻔﺎﺩﺓ ﻣﻦ ﻋﻼﻗﺔ ﺷﺎﻝ‪ ،‬ﰲ ﺣﺎﻟﺔ ‪ k‬ﻣﻦ } ‪ ، {1, …, n‬ﻟﺪﻳﻨﺎ‬
‫ ‬
‫ ‬
‫ ‬

‫‪k‬‬

‫‪∑ (Aj −1Aj , Aj Aj +1) = kθ‬‬

‫ ‬
‫ ‬
‫ ‬
‫= ) ‪(A0A1, Ak Ak +1‬‬

‫‪j =1‬‬

‫ﻭﻧﺴﺘﻨﺘﺞ ﻣﻦ ﺫﻟﻚ ﰲ ﺣﺎﻟﺔ ‪ k = n‬ﺃﻥﹼ‬
‫‪n θ = 0 mod 2π‬‬

‫)‪(1‬‬

‫ﻭﺃﺧﲑﺍﹰ‪ ،‬ﺳﻨﻄﺎﺑﻖ ﺍﳌﺴﺘﻮﻱ ﻣﻨﺴﻮﺑﹰﺎ ﺇﱃ ﺍﳉﻤﻠﺔ ﺍﻟﺴﺎﺑﻘﺔ ﻭﺍﳌﺴﺘﻮﻱ ﺍﻟﻌﻘﺪﻱ ‪ . ℂ‬ﻓﻴﻜﻮﻥ ‪، A0 = 0‬‬
‫ﻭ ‪ A1 = a0‬ﻭ ‪ ، A2 = a 0 + a1e i θ‬ﻭﺑﻮﺟﻪ ﻋﺎﻡ ﻧﺴﺘﻨﺘﺞ ﳑﺎ ﺳﺒﻖ ﺃﻥﹼ‬
‫‪Ak +1 = Ak + ake i k θ‬‬

‫ﰲ ﺣﺎﻟﺔ ‪ k‬ﻣﻦ } ‪ . { 0,1, …, n − 1‬ﻭﻫﺬﺍ ﻳﱪﻫﻦ ﺑﺎﻟﺘﺪﺭﻳﺞ ﺃﻥﹼ‬
‫‪k‬‬

‫‪∑ a je i j θ‬‬

‫= ‪Ak +1‬‬

‫‪∀k ∈ { 0, …, n − 1 },‬‬

‫‪j =0‬‬

‫ﻭﺑﻮﺟﻪ ﺧﺎﺹ‪ ،‬ﰲ ﺣﺎﻟﺔ ‪ k = n − 1‬ﻟﺪﻳﻨﺎ ‪ ، An = A0‬ﺇﺫﻥ‬
‫‪n −1‬‬

‫‪=0‬‬

‫‪∑ a je i j θ‬‬
‫‪j =0‬‬

‫ﻟﻨﻌﺮ‪‬ﻑ ) ‪cos j θ‬‬

‫(‪1‬‬
‫‪2 1−‬‬

‫= ) ‪ λj = sin2 ( j2θ‬ﰲ ﺣﺎﻟﺔ ‪ ، j ∈ ℤ‬ﻓﻨﺠﺪ‬

‫) ‪λj +1 − λj = 12 ( cos j θ − cos ( j + 1 ) θ ) = sin ( θ2 ) sin ( ( j + 12 ) θ‬‬

‫)‪(2‬‬

‫ﺍﻷﻭﳌﺒﻴﺎﺩ ﺍﻟﻌﺎﳌﻲ ﻟﻠﺮﻳﺎﺿﻴ‪‬ﺎﺕ‬

‫‪40‬‬

‫ﻧﺴﺘﻨﺘﺞ ﻣﻦ ) ‪ ( 2‬ﺃﻥﹼ‬
‫‪n −1‬‬

‫‪=0‬‬

‫‪∑ a j sin ( θ2 )e i( j +1/2 )θ‬‬
‫‪j =0‬‬

‫ﻭﺇﺫﺍ ﺗﺄﻣ‪‬ﻠﻨﺎ ﺍﳉﺰﺀ ﺍﻟﺘﺨﻴ‪‬ﻠﻲ ﻭﺟﺪﻧﺎ‬
‫‪n −1‬‬

‫‪∑ a j ( λj +1 − λj ) = 0‬‬
‫‪j =0‬‬

‫ﻭﻟﻜﻦ‬
‫‪n −1‬‬

‫‪n −1‬‬

‫‪n‬‬

‫‪∑ a j −1λj − ∑ a j λj‬‬
‫‪j =0‬‬

‫=‬

‫‪j =1‬‬

‫‪λ0 = λn = 0‬‬

‫‪n −1‬‬

‫‪n −1‬‬

‫‪∑ a j ( λj +1 − λj ) = ∑ a j λj +1 − ∑ a j λj‬‬
‫‪j =0‬‬

‫‪j =0‬‬

‫‪j =0‬‬

‫‪n −1‬‬

‫‪¤‬‬

‫‪∑ (a j −1 − a j )λj‬‬

‫=‬

‫‪j =1‬‬

‫ﺇﺫ ﺇﻥﹼ ‪ λn = 0‬ﺑﻨﺎﺀً ﻋﻠﻰ ) ‪ . ( 1‬ﻭﻣﻨﻪ‬
‫‪n −1‬‬

‫‪=0‬‬

‫‪∑ (a j −1 − a j )λj‬‬
‫‪j =1‬‬

‫ﻭﻷﻥ ﲨﻴﻊ ﺣﺪﻭﺩ ﺍ‪‬ﻤﻮﻉ ﺍﻟﺴﺎﺑﻖ ﺃﻛﱪ ﺃﻭ ﺗﺴﺎﻭﻱ ‪ ، 0‬ﻧﺴﺘﻨﺘﺞ‬
‫‪(a j −1 − a j )λj = 0‬‬

‫‪∀j ∈ {1,2, …, n − 1 },‬‬

‫)‪(3‬‬

‫ﻟﻨﻌﺮ‪‬ﻑ‬
‫} ‪j0 = min { j ≥ 1 : λj = 0‬‬

‫ﻷﻥﹼ ‪ ، 0 < θ < π‬ﻧﺮﻯ ﻭﺿﻮﺣﺎﹰ ﺃﻥﹼ ‪ 3 ≤ j 0 ≤ n‬ﻓﺈﺫﺍ ﺍﻓﺘﺮﺿﻨﺎ ﺃﻥﹼ ‪ j0 < n‬ﺍﺳﺘﻨﺘﺠﻨﺎ ﳑﺎ‬
‫ﺳﺒﻖ ﺃﻥﹼ ‪ . a 0 = a2 = ⋯ = a j0 −1‬ﻭﻣﻦ ﺛﹶﻢ‪‬‬
‫‪e i j 0θ − 1‬‬
‫‪=0‬‬
‫‪eiθ − 1‬‬

‫‪j0 −1‬‬

‫‪= a0 ∑ e i j θ = a0‬‬

‫‪i jθ‬‬

‫‪j =0‬‬

‫‪j0 −1‬‬

‫‪∑ a je‬‬

‫= ‪Aj0‬‬

‫‪j =0‬‬

‫ﺇﺫ ﺍﺳﺘﻔﺪﻧﺎ ﻣﻦ ﺃﻥﹼ ﺍﻟﺸﺮﻁ ‪ λj0 = 0‬ﻳﻘﺘﻀﻲ ‪ . e i j0θ = 1‬ﻭﻫﺬﺍ ﻳﺘﻨﺎﻗﺾ ﻣﻊ ﺍﻓﺘﺮﺍﺿﻨﺎ ﺍﻟﻨﻘﺎﻁ‬
‫‪ ( Aj )0≤ j <n‬ﻣﺘﺒﺎﻳﻨﺔ‪ .‬ﺇﺫﻥ ﻻ ﺑ‪‬ﺪ‪ ‬ﺃﻥ ﻳﻜﻮﻥ ‪ ، j0 = n‬ﻭﻣﻦ ﺛﹶﻢ‪‬‬
‫‪a 0 = a2 = ⋯ = an −1‬‬

‫ﻭﺍﳌﻀﻠﹼﻊ ﻣﻨﺘﻈﻢ‪.‬‬




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