ﺩﻭﺭﻩ ﻯ ﺷﺎﻧﺰﺩﻫﻢ
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ﺯﻣﺴﺘﺎﻥ 1389
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ﺷﻤﺎﺭﻩ ﻯ 2
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48ﺻﻔﺤﻪ
56 ﺭﻳﺎﺿﯽ ﺩﻭﺭﻩﻯ ﺭﺍﻫﻨﻤﺎﻳﻰ ﺗﺤﺼﻴﻠﻰ
ﻭﺯﺍﺭﺕ ﺁﻣﻮﺯﺵ ﻭ ﭘﺮﻭﺭﺵ ﺳﺎﺯﻣﺎﻥ ﭘﮋﻭﻫﺶ ﻭ ﺑﺮﻧﺎﻣﻪﺭﻳﺰﻯ ﺁﻣﻮﺯﺷﻰ ﺩﻓﺘﺮ ﺍﻧﺘﺸﺎﺭﺍﺕ ﻛﻤﻚﺁﻣﻮﺯﺷﻰ
ﻫﻤﺖ ﻣﻀﺎﻋﻒ، ﻛﺎﺭ ﻣﻀﺎﻋﻒ ﻣﺪﻳﺮ ﻣﺴﺌﻮﻝ :ﻣﺤﻤﺪ ﻧﺎﺻﺮﻯ ﺳﺮﺩﺑﻴﺮ :ﺣﻤﻴﺪﺭﺿﺎ ﺍﻣﻴﺮﻯ ﻣﺪﻳﺮ ﺩﺍﺧﻠﻰ :ﺣﺴﻴﻦ ﻧﺎﻣﻰ ﺳﺎﻋﻰ ﺍﻋﻀﺎﻯ ﻫﻴﺌﺖ ﺗﺤﺮﻳﺮﻳﻪ :ﺣﺴﻦ ﺍﺣﻤﺪﻯ ،ﺣﻤﻴﺪﺭﺿﺎ ﺍﻣﻴﺮﻯ ،ﺯﻫﺮﻩ ﭘﻨﺪﻯ، ﺳﭙﻴﺪﻩ ﭼﻤﻦﺁﺭﺍ ،ﺧﺴﺮﻭ ﺩﺍﻭﺩﻯ ،ﻣﻴﺮﺷﻬﺮﺍﻡ ﺻﺪﺭ ،ﺣﺴﻴﻦ ﻧﺎﻣﻰ ﺳﺎﻋﻰ ،ﺳﻴﺪ ﻣﺤﻤﺪﺭﺿﺎ ﻫﺎﺷﻤﻰ ﻣﻮﺳﻮﻯ ﻭﻳﺮﺍﺳﺘﺎﺭ ﺍﺩﺑﻰ :ﻣﺮﺗﻀﻰ ﺣﺎﺟﻌﻠﻰﻓﺮﺩ ﻃﺮﺍﺡ ﮔﺮﺍﻓﻴﻚ :ﻋﻠﻰ ﺩﺍﻧﺸﻮﺭ ﺗﺼﻮﻳﺮﮔﺮ :ﺳﺎﻡ ﺳﻠﻤﺎﺳﻰ ﻧﺸﺎﻧﻰ ﺩﻓﺘﺮ ﻣﺠﻠﻪ :ﺗﻬﺮﺍﻥ ،ﺍﻳﺮﺍﻧﺸﻬﺮ ﺷﻤﺎﻟﻰ ،ﭘﻼﻙ ،266ﺻﻨﺪﻭﻕ ﭘﺴﺘﻰ 6585ـ 15875 ﻧﻤﺎﺑﺮ 88301478 : ﺗﻠﻔﻦ 9 :ـ 8 8831161ـ 021ﺩﺍﺧﻠﻰ374: ﺭﺍﻳﺎﻧﺎﻣﻪ
[email protected] : ﻭﺑﮕﺎﻩ www.roshdmag.ir : ﺗﻠﻔﻦ ﭘﻴﺎﻡﮔﻴﺮ ﻧﺸﺮﻳﺎﺕ ﺭﺷﺪ 88301482: ﻛﺪ ﻣﺪﻳﺮ ﻣﺴﺌﻮﻝ 102:ﻛﺪ ﺩﻓﺘﺮ ﻣﺠﻠﻪ 113 :ﻛﺪ ﻣﺸﺘﺮﻛﻴﻦ 102 : ﻧﺸﺎﻧﻰ ﺍﻣﻮﺭ ﻣﺸﺘﺮﻛﻴﻦ :ﺗﻬﺮﺍﻥ ،ﺻﻨﺪﻭﻕ ﭘﺴﺘﻰ16595 / 111: ﺗﻠﻔﻦ ﺍﻣﻮﺭ ﻣﺸﺘﺮﻛﻴﻦ 77336656 : ﭼﺎپ :ﺷﺮﻛﺖ ﺍﻓﺴﺖ )ﺳﻬﺎﻣﻰ ﻋﺎﻡ( ﺷﻤﺎﺭﮔﺎﻥ 22000:ﻧﺴﺨﻪ
ﻓﻬﺮﺳﺖ ﺣﺮﻑ ﺍﻭﻝ ﺭﻳﺎﺿﻴﺎﺕ ﻭ ﻣﺴﺎﺋﻞ ﺭﻭﺯﻣﺮﻩ /ﺣﻤﻴﺪﺭﺿﺎ ﺍﻣﻴﺮﻯ2/
ﻫﻤﺮﺍﻩ ﺑﺎ ﻛﺘﺎﺏ
ﻧﻤﺎﻳﺶ ﺍﻋﺪﺍﺩ ﺩﺭ ﻣﺒﻨﺎﻫﺎﻱ ﻣﺘﻔﺎﻭﺕ ) ﻗﺴﻤﺖ
ﺩﻭﻡ( /ﺳﭙﻴﺪﻩ ﭼﻤﻦﺁﺭﺍ 3 /
ﺩﺍﻧﺶﺍﻓﺰﺍﻳﻲ
ﻭﺍژﻩﻫﺎﻱ ﺭﻳﺎﺿﻲ /ﺷﺎﺩﻱ ﺑﻬﺎﺭﻱ 17 /
ﺭﺍﺑﻄﻪﻱ ﻓﻴﺜﺎﻏﻮﺭﺱ ،ﺍﺛﺒﺎﺕ ﻭ ﻛﺎﺭﺑﺮﺩﻫﺎﻱ ﺁﻥ /ﻣﻮﺳﻲ ﻫﺎﺷﻤﻠﻮ19 / ﻧﻘﺎﻁ ﺍﻣﻦ ) ﻗﺴﻤﺖ ﺩﻭﻡ( /ﺣﺴﻦ ﺍﺣﻤﺪﻱ28 / ﻣﺮﻳﻢ ﺳﻌﻴﺪﻱ37 /
ﺣﺪﺱ ﺑﺰﻧﻴﺪ /
ﻓﺮﻣﻮﻟﻲ ﺑﺮﺍﻱ ﻳﻚ ﺍﻟﮕﻮ /ﺯﻫﺮﻩ ﭘﻨﺪﻱ38 /
ﭼﻨﺪ ﻗﺪﻡ ﺑﺎ ﺭﻳﺎﺿﻲﺩﺍﻧﺎﻥ /ﺳﺎﻳﻪ ﻣﻬﺮﺑﺎﻥ42 / ﮔﻔﺖ ﻭ ﮔﻮ ﻫﻤﻪ ﺍﺳﺘﻌﺪﺍﺩ ﺭﻳﺎﺿﻲ ﺩﺍﺭﻧﺪ /ﺁﺯﺍﺩﻩ ﺷﺎﻛﺮﻱ 8 /
ﺭﻳﺎﺿﻲ ﻭ ﻛﺎﺭﺑﺮﺩ ﺁﻥ
ﭼﻪ ﻛﺴﻲ ﻣﺮﺩﻡﺩﺍﺭ ﺍﺳﺖ؟ /ﺗﺮﺟﻤﻪﻱ ﺣﺴﻦ ﻳﺎﻭﺭﺗﺒﺎﺭ 25 / ﻛﻤﻲ ﻓﻜﺮ ﻛﻨﻴﺪ ﺗﻌﺠﺐ ﻧﻜﻨﻴﺪ ،ﻓﻜﺮ ﻛﻨﻴﺪ! /ﺧﺴﺮﻭ ﺩﺍﻭﺩﻱ23 /
ﻣﺤﺎﺳﺒﻪﻱ ﺍﻭﻗﺎﺕ ﺷﺮﻋﻲ ﻭ ﺳﺮﻋﺖ ﭼﺮﺧﺶ
ﺯﻣﻴﻦ /ﻛﺎﻇﻢ ﺍﻳﺮﺍﻧﻲ6 /
ﻣﻨﻄﻖ ﻭ ﺭﻳﺎﺿﻲ ﻧﺠﺎﺕ ﻣﻨﻄﻘﻲ ﻋﺒﺪﺍﻟﻜﺎﻇﻢ /ﺣﻤﻴﺮﺍ ﻇﻔﺮﻗﻨﺪ 15/
ﺑﺪﻭﻥ ﺩﺭﺻﺪ ﺍﻣﻜﺎﻥ ﻧﺪﺍﺭﺩ! /ﺳﻜﻴﻨﻪ
ﺑﻤﺎﻧﻴﺎﻥ 12 /
ﺟﺪﻭﻝ ﻧﺎﻣﻪﻫﺎ ﻳﻚ ﻧﺎﻣﻪ ﻳﻚ ﺩﻭﺳﺖ30 / ﺭﻳﺎﺿﻲ ﻭ ﺑﺎﺯﻱ ﺑﺎﺯﻱﻫﺎﻱ ﭼﻨﺪ ﻧﻔﺮﻩ /ﺯﻫﺮﻩ ﭘﻨﺪﻱ32 / ﺗﺎﺭﻳﺦ ﺭﻳﺎﺿﻴﺎﺕ ﺍﺯ ﺧﻴﺎﻡ ﻧﻴﺸﺎﺑﻮﺭﻱ ﺗﺎ ﭘﺮﻭﻓﺴﻮﺭ ﻫﺸﺘﺮﻭﺩﻱ ﻣﺤﻤﺪ ﻋﺰﻳﺰﻱﭘﻮﺭ 26 /
/ﺳﻴﺮﻭﺱ ﻏﻔﺎﺭﻳﺎﻥ34 / ﻣﻌﻤﺎ ﻭ ﺳﺮﮔﺮﻣﻲ ﺳﺮﮔﺮﻣﻲﻫﺎﻱ ﺭﻳﺎﺿﻲ /ﻣﺤﻤﺪ ﻋﺰﻳﺰﻱﭘﻮﺭ40 /
ﺳﺆﺍﻝﻫﺎﻯ ﻣﺴﺎﺑﻘﻪﺍﻯ
/ﻣﺴﺎﺑﻘﻪﻯ ﺭﻳﺎﺿﻰ ﺍﺳﺘﺮﺍﻟﻴﺎ
) / (2008ﺗﺮﺟﻤﻪﻯ ﺳﭙﻴﺪﻩ ﭼﻤﻦﺁﺭﺍ45 /
ﻣﻌﺮﻓﻰ ﻛﺘﺎﺏ
/ﻟﺬﺕ ﺍﻧﺪﻳﺸﻪﻭﺭﺯﻱ /ﺟﻌﻔﺮ ﺭﺑﺎﻧﻰ 48 /
ﺍﻧﺪﻳﺸﻪﻭﺭﺯﻱ ﺍﺗﻮﻣﺒﻴﻞ ﺑﺎﺯﻳﭽﻪ /ﺗﺮﺟﻤﻪﻱ ﺣﺴﻦ ﻧﺼﻴﺮﻧﻴﺎ 14 /
ﻗﺎﺑﻞ ﺗﻮﺟﻪ ﻧﻮﻳﺴﻨﺪﮔﺎﻥ ﻭ ﻣﺘﺮﺟﻤﺎﻥ: ﻼ ﺩﺭ ﺟﺎﻯ ﺩﻳﮕﺮﻯ ﭼﺎپ ﻧﺸﺪﻩ ﺑﺎﺷﺪ .ـ ﻣﻘﺎﻟﻪﻫﺎﻯ ﺗﺮﺟﻤﻪ ﺷﺪﻩ ﺑﺎﻳﺪ ﺑﺎ ﻣﺘﻦ ﺍﺻﻠﻰ ﻫﻤﺨﻮﺍﻧﻰ ﺩﺍﺷﺘﻪ ﺑﺎﺷﺪ ﻭ ﻣﺘﻦ ﺍﺻﻠﻰ ﻧﻴﺰ ﻫﻤﺮﺍﻩ ﺁﻥ ـ ﻣﻘﺎﻟﻪﻫﺎﻳﻰ ﻛﻪ ﺑﺮﺍﻯ ﺩﺭﺝ ﺩﺭ ﻣﺠﻠﻪ ﻣﻰﻓﺮﺳﺘﻴﺪ ،ﺑﺎﻳﺪ ﺑﺎ ﺍﻫﺪﺍﻑ ﻭ ﺳﺎﺧﺘﺎﺭ ﺍﻳﻦ ﻣﺠﻠﻪ ﻣﺮﺗﺒﻂ ﺑﺎﺷﺪ ﻭ ﻗﺒ ً ﺑﺎﺷﺪ .ﭼﻨﺎﻥﭼﻪ ﻣﻘﺎﻟﻪ ﺭ ﺍ ﺧﻼﺻﻪ ﻣﻰﻛﻨﻴﺪ ،ﺍﻳﻦ ﻣﻮﺿﻮﻉ ﺭﺍ ﻗﻴﺪ ﺑﻔﺮﻣﺎﻳﻴﺪ .ـ ﻣﻘﺎﻟﻪ ﻳﻚ ﺧﻂ ﺩﺭ ﻣﻴﺎﻥ ،ﺩﺭ ﻳﻚ ﺭﻭﻯ ﻛﺎﻏﺬ ﻭ ﺑﺎ ﺧﻂ ﺧﻮﺍﻧﺎ ﻧﻮﺷﺘﻪ ﻳﺎ ﺗﺎﻳﭗ ﺷﻮﺩ .ﻣﻘﺎﻟﻪﻫﺎ ﻣﻰﺗﻮﺍﻧﻨﺪ ﺑﺎ ﻧﺮﻡﺍﻓﺰﺍﺭ wordﻭ ﺑﺮ ﺭﻭﻯ CDﻳﺎ ﻓﻼﭘﻰ ﻭ ﻳﺎ ﺍﺯ ﻃﺮﻳﻖ ﺭﺍﻳﺎﻧﺎﻣﻪ ﻣﺠﻠﻪ ﺍﺭﺳﺎﻝ ﺷﻮﻧﺪ .ـ ﻧﺜﺮ ﻣﻘﺎﻟﻪ ﺑﺎﻳﺪ ﺭﻭﺍﻥ ﻭ ﺍﺯ ﻧﻈﺮ ﺩﺳﺘﻮﺭ ﺯﺑﺎﻥ ﻓﺎﺭﺳﻰ ﺩﺭﺳﺖ ﺑﺎﺷﺪ ﻭ ﺩﺭ ﺍﻧﺘﺨﺎﺏ ﻭﺍژﻩﻫﺎﻯ ﻋﻠﻤﻰ ﻭ ﻓﻨﻰ ﻭﻗﺖ ﻻﺯﻡ ﻣﺒﺬﻭﻝ ﺷﻮﺩ .ـ ﻣﺤﻞ ﻗﺮﺍﺭ ﺩﺍﺩﻥ ﺟﺪﻭﻝﻫﺎ ،ﺷﻜﻞﻫﺎ ﻭ ﻋﻜﺲﻫﺎ ﺩﺭ ﻣﺘﻦ ﻣﺸﺨﺺ ﺷﻮﺩ. ـ ﻣﻘﺎﻟﻪ ﺑﺎﻳﺪ ﺩﺍﺭﺍﻯ ﭼﻜﻴﺪﻩ ﺑﺎﺷﺪ ﻭ ﺩﺭ ﺁﻥ ﻫﺪﻑﻫﺎ ﻭ ﭘﻴﺎﻡ ﻧﻮﺷﺘﺎﺭ ﺩﺭ ﭼﻨﺪ ﺳﻄﺮ ﺗﻨﻈﻴﻢ ﺷﻮﺩ .ـ ﻛﻠﻤﺎﺕ ﺣﺎﻭﻯ ﻣﻔﺎﻫﻴﻢ ﻧﻤﺎﻳﻪ )ﻛﻠﻰﺩﻭﺍژﻩﻫﺎ( ﺍﺯ ﻣﺘﻦ ﺍﺳﺘﺨﺮﺍﺝ ﻭ ﺭﻭﻯ ﺻﻔﺤﻪﺍﻯ ﺟﺪﺍﮔﺎﻧﻪ ﻧﻮﺷﺘﻪ ﺷﻮﻧﺪ .ـ ﻣﻘﺎﻟﻪ ﺑﺎﻳﺪ ﺩﺍﺭﺍﻯ ﺗﻴﺘﺮ ﺍﺻﻠﻰ ،ﺗﻴﺘﺮﻫﺎﻯ ﻓﺮﻋﻰ ﺩﺭ ﻣﺘﻦ ﻭ ﺳﻮﺗﻴﺘﺮ ﺑﺎﺷﺪ .ـ ﻣﻌﺮﻓﻰﻧﺎﻣﻪﻯ ﻛﻮﺗﺎﻫﻰ ﺍﺯ ﻧﻮﻳﺴﻨﺪﻩ ﻳﺎ ﻣﺘﺮﺟﻢ ﻫﻤﺮﺍﻩ ﻳﻚ ﻗﻄﻌﻪ ﻋﻜﺲ ،ﻋﻨﺎﻭﻳﻦ ﻭ ﺁﺛﺎﺭ ﻭﻯ ﭘﻴﻮﺳﺖ ﺷﻮﺩ .ـ ﻣﺠﻠﻪ ﺩﺭ ﺭﺩ ،ﻗﺒﻮﻝ ،ﻭﻳﺮﺍﻳﺶ ﻭ ﺗﻠﺨﻴﺺ ﻣﻘﺎﻟﻪﻫﺎﻯ ﺭﺳﻴﺪﻩ ﻣﺨﺘﺎﺭ ﺍﺳﺖ .ـ ﻣﻘﺎﻻﺕ ﺩﺭﻳﺎﻓﺘﻰ ﺑﺎﺯﮔﺮﺩﺍﻧﺪﻩ ﻧﻤﻰﺷﻮﺩ .ـ ﺁﺭﺍﻯ ﻣﻨﺪﺭﺝ ﺩﺭ ﻣﻘﺎﻟﻪ ﺿﺮﻭﺭﺗﺎً ﻣﺒﻴﻦ ﺭﺃﻯ ﻭ ﻧﻈﺮ ﻣﺴﺌﻮﻻﻥ ﻣﺠﻠﻪ ﻧﻴﺴﺖ.
ﺍﻭﻝ ﺣﺮﻑ ﺍﻭﻝ ﺣﺮﻑ
ر︀︲﹫︀ت و ﹞︧︀﹏ روز﹞︣ه ﺯﻧﺪﮔـﻲ ﭘﺮ ﺍﺯ ﻣﺴـﺌﻠﻪﻫﺎﻱ ﮔﻮﻧﺎﮔـﻮﻥ ﺑﺎ ﺭﺍﻩﺣﻞﻫـﺎﻱ ﻣﺘﻔﺎﻭﺕ ﺧﺎﺹ ﻫﺮ ﺷـﺨﺺ ،ﻛﻪ ﺗﻘﺮﻳﺒ ًﺎ ﻫﻴﭻ ﺍﻟﮕﻮﻱ ﺍﺳﺖ؛ ﻣﺴـﺌﻠﻪﻫﺎﻱ ِ ﺛﺎﺑﺘﻲ ﺑﺮﺍﻱ ﺣﻞ ﺁﻥﻫﺎ ﻭﺟﻮﺩ ﻧﺪﺍﺭﺩ. ﺷـﻤﺎ ﺩﺭ ﺯﻧﺪﮔﻲ ﺭﻭﺯﻣﺮﻩﻱ ﺧﻮﺩ ﺩﺭ ﺑﺮﺍﺑﺮ ﻣﺴﺎﺋﻞ ﺑﺴﻴﺎﺭﻱ ﻗﺮﺍﺭ ﻣﻲﮔﻴﺮﻳﺪ ﺑﺮﺍﻱ ﺣﻞ ﺁﻥﻫﺎ ﺑﻪ ﻣﺸﻮﺭﺕ ﻭ ﻫﻤﻜﺎﺭﻱ ﺍﻓﺮﺍﺩ ﻣﻄﻤﺌﻦ ﻭ ﻛﺎﺭﺩﺍﻥ ﺍﺣﺘﻴﺎﺝ ﺩﺍﺭﻳﺪ ﺍﻣﺎ ﺩﺭ ﻧﻬﺎﻳﺖ ﺷﻤﺎ ﺧﻮﺩﺗﺎﻥ ﺑﺎﻳﺪ ﺗﺼﻤﻴﻢ ﺑﮕﻴﺮﻳﺪ ﻭ ﻣﺴﺌﻠﻪ ﺭﺍ ﺣﻞ ﻛﻨﻴﺪ. ﺣـﺎﻻ ﺑﻴﺎﻳﻴﺪ ﻣﺴـﺌﻠﻪﻫﺎﻱ ﺭﻭﺯﻣـﺮﻩ ﺭﺍ ﺑﺎ ﻳﻚ ﻣﺴـﺌﻠﻪ ﺭﻳﺎﺿﻲ ﻣﻘﺎﻳﺴﻪ ﻛﻨﻴﻢ ،ﺑﺮﺍﻱ ﺣﻞ ﺻﺤﻴﺢ ﻭ ﻛﻮﺗﺎﻩ ﻳﻚ ﻣﺴﺌﻠﻪﻱ ﺭﻳﺎﺿﻲ ﭼﻪ ﻋﺎﻣﻞ ﻳﺎ ﻋﻮﺍﻣﻠﻲ ﺑﻪ ﺷﻤﺎ ﻛﻤﻚ ﻣﻲﻛﻨﻨﺪ؟ ﺩﺭﺳﺖ ﻣﻲﮔﻮﻳﻴﺪ: ﺍﻭﻟﻴﻦ ﻋﺎﻣﻞ ﻣﻬﻢ ﺩﺭ ﺣﻞ ﻳﻚ ﻣﺴﺌﻠﻪﻱ ﺭﻳﺎﺿﻲ »ﺁﮔﺎﻫﻲ ﺩﻗﻴﻖ ﺍﺯ ﻣﻔﺮﻭﺿﺎﺕ ﻳﺎ ﺩﺍﺩﻩﻫﺎﻱ ﻣﺴﺌﻠﻪ« ﺍﺳﺖ؛ ﻳﻌﻨﻲ ﺁﮔﺎﻫﻲ ﺍﺯ ﺁﻧﭽﻪ ﺩﺍﺭﻳﻢ ﻭ ﺁﻧﭽـﻪ ﻣﻲﺧﻮﺍﻫﻴﻢ؛ ﺑﻪ ﻋﺒﺎﺭﺕ ﺩﻳﮕﺮ :ﻓﺮﺽ ﻣﺴـﺌﻠﻪ ﻭ ﺣﻜﻢ ﻣﺴﺌﻠﻪ .ﻋﺎﻣﻞ ﻣﻬﻢ ﺩﻳﮕﺮ ﺍﻧﺘﺨﺎﺏ ﺭﻭﺵ ﻳﺎ ﺭﻭﺵﻫﺎﻱ ﻗﺎﺑﻞ ﻗﺒﻮﻝ ﻭ ﺍﺳﺘﺪﻻﻝﻫﺎﻱ ﺻﺤﻴﺢ ﺍﺳﺖ .ﺍﺯ ﻋﻮﺍﻣﻞ ﻣﻬﻢ ﺩﻳﮕﺮ ،ﺩﻗﺖ ﺩﺭ ﻋﻤﻠﻴﺎﺕ ،ﻣﺤﺎﺳﺒﺎﺕ ،ﺍﺳﺘﻔﺎﺩﻩﻱ ﺻﺤﻴﺢ ﻭ ﺩﻗﻴﻖ ﺍﺯ ﺍﺑﺰﺍﺭﻫﺎ، ﻗﻀﻴﻪﻫﺎ ،ﺗﻌﺎﺭﻳﻒ ﻭ ﺍﺻﻮﻝ ﺍﺳﺖ. ﺣـﺎﻻ ،ﺑـﻪ ﻧﻈـﺮ ﺷـﻤﺎ ﺩﺭ ﺣـﻞ ﻣﺴـﺎﺋﻞ ﺭﻭﺯﻣـﺮﻩ ﺯﻧﺪﮔﻲ ﭼﻪ ﻣﺸـﺎﺑﻬﺖﻫﺎﻳﻲ ﺑﺎ ﺁﻧﭽﻪ ﺩﺭﺑﺎﺭﻩﻱ ﺣﻞ ﻣﺴـﺎﺋﻞ ﺭﻳﺎﺿﻲ ﮔﻔﺘﻴﻢ، ﻭﺟﻮﺩ ﺩﺍﺭﺩ؟ ﺯﻧﺪﮔﻲ ﻳﻚ ﺳـﺎﻟﻦ ﺑﺰﺭگ ﺍﻣﺘﺤﺎﻥ ﺍﺳـﺖ ﻛﻪ ﺩﺭ ﻫﺮ ﮔﻮﺷـﻪﻱ ﺁﻥ ،ﻳﻚ ﻧﻔﺮ ﻧﺸﺴـﺘﻪ ﺍﺳـﺖ ﻭ ﺭﻭﻱ ﻣﺴﺌﻠﻪﺍﻱ ﻓﻜﺮ ﻣﻲﻛﻨﺪ .ﻣﺎ ﺑﺎﻳﺪ ﺑﺘﻮﺍﻧﻴﻢ ﺑﻪ ﺧﻮﺑﻲ ﻭ ﺷﺎﻳﺴـﺘﮕﻲ ﺍﺯ ﻋﻬﺪﻩﻱ ﺣﻞ ﺁﻥ ﻣﺴﺎﺋﻞ ﺑﺮﺁﻳﻴﻢ ،ﺍﻳﻦ ﺍﻣﺘﺤﺎﻥ ﮔﺮﻓﺘﻦ ﺳـﻨّﺖ ﺍﻟﻬﻲ ﺍﺳـﺖ ﻭ ﻣﺎ ﺑﺎﻳﺪ ﺑﺮﺍﻱ ﺍﻳـﻦ ﺍﻣﺘﺤﺎﻧﺎﺕ ﺁﻣﺎﺩﻩ ﺑﺎﺷـﻴﻢ ﺑﺎﻳﺪ ﺻﻮﺭﺕ ﻣﺴـﺌﻠﻪﻫﺎ ﺭﺍ ﺧﻮﺏ ﺑﺸﻨﺎﺳـﻴﻢ ،ﺑﺎﻳﺪ ﺑـﺮﺍﻱ ﺍﻳﻦ ﭼﺎﻟـﺶ ﻣﻬ ّﻴﺎ ﺑﻮﺩ ،ﺑﺎﻳﺪ ﺑﻪ ﺳـﻼﺡ ﻣﻌﺮﻓﺖ ﻭ ﺁﮔﺎﻫﻲ ﻣﺴـﻠﺢ ﺑﺎﺷﻴﻢ ﻭ ﺭﻭﺵﻫﺎﻱ ﺑﺮﺧﻮﺭﺩ ﺑﺎ ﻣﺴﺎﺋﻞ ﺭﺍ ﺍﺯ ﺑﺰﺭﮔﺎﻥ ،ﺍﻭﻟﻴﺎ ﻭ ﻣﻌﺼﻮﻣﻴﻦ )ﻉ( ﺑﻴﺎﻣﻮﺯﻳﻢ ﻭ ﺑﻪﻛﺎﺭ ﺑﮕﻴﺮﻳﻢ. 2
ﺭﺍﻫﻨﻤﺎﻳﻲ
ﺩﻭﺭﻩﻯ ﺷﺎﻧﺰﺩﻫﻢ ،ﺷﻤﺎﺭﻩﻯ ،2ﺯﻣﺴﺘﺎﻥ 1389
ﺍﮔﺮ ﺩﺭ ﺣﻞ ﻳﻚ ﻣﺴـﺌﻠﻪ ﻧﺎﻛﺎﻡ ﺷـﺪﻳﻢ ﻧﺒﺎﻳﺪ ﻧﺎﺍﻣﻴﺪ ﺷﻮﻳﻢ ﺑﺎﻳﺪ ﻫﺮ ﺷﻜﺴـﺘﻲ ﺭﺍ ﻣﻘﺪﻣﻪﺍﻱ ﺑﺮﺍﻱ ﭘﻴﺮﻭﺯﻱ ﺑﻌـﺪﻱ ﺑﺪﺍﻧﻴﻢ .ﻧﺒﺎﻳﺪ ﺍﻳﻤﺎﻥ ﺭﺍﺳـﺦ ﺑﻪ ﻣﻴـﺪﺍﻥ ﺭﺍ ﺧﺎﻟﻲ ﻛﻨﻴﻢ .ﻫﻤﻮﺍﺭﻩ ﺑﺎﻳﺪ ﺑﺎ ﺍﻣﻴﺪ ﻭ ِ ﺣﺮﻛﺖ ﺭﻭ ﺑﻪ ﺟﻠﻮﻱ ﺧﻮﺩ ﺍﺩﺍﻣﻪ ﺩﻫﻴﻢ. ﺭﻳﺎﺿـﻲ ﺧﻮﺩ ﺭﺍ ﺷـﻤﺎ ﺍﮔـﺮ ﺑﺨﻮﺍﻫﻴﺪ ﻣﻲﺗﻮﺍﻧﻴـﺪ ﺫﻫﻦ ﺧﻼﻕ ﻭ ِ ﭘﺮﻭﺭﺵ ﺩﻫﻴﺪ ﻭ ﺁﻥ ﺭﺍ ﺗﺮﺑﻴﺖ ﻛﻨﻴﺪ ﻭ ﺍﻳﻦ ﻧﻴﺮﻭﻳﻲ ﺭﺍ ﻛﻪ ﺧﺪﺍﻭﻧﺪ ﺑﻪ ﺻﻮﺭﺕ ﻧﻬﻔﺘﻪ ﺩﺭ ﻭﺟﻮﺩ ﻫﻤﻪﻱ ﺷﻤﺎ ﻧﻬﺎﺩﻩ ﺍﺳﺖ ﻇﺎﻫﺮ ﻛﺮﺩﻩ ﻭ ﻛﺎﻣﻞ ﻛﻨﻴﺪ ،ﺑﺮﺍﻱ ﺭﺷـﺪ ﻭ ﺗﻜﺎﻣﻞ ﺫﻫﻦ ﺭﻳﺎﺿﻲ ﻭ ﺧﻼﻕ ﺧﻮﺩ ﭼﻨﺪ ﺷﺮﻁ ﺍﺳﺎﺳﻲ ﻭﺟﻮﺩ ﺩﺍﺭﺩ. .1ﺑﺎ ﺭﻳﺎﺿﻴﺎﺕ ﺍُﻧﺲ ﺑﮕﻴﺮﻳﺪ ﻭ ﺑﻪ ﺁﻥ ﻋﻼﻗﻪﻣﻨﺪ ﺑﺎﺷﻴﺪ ﻭ ﻫﻤﻮﺍﺭﻩ ﺑﺨﺸﻲ ﺍﺯ ﻓﺮﺻﺖﻫﺎﻱ ﺧﻮﺩ ﺭﺍ ﺻﺮﻑ ﺷﻨﺎﺧﺖ ﺑﻴﺸﺘﺮ ﻧﺴﺒﺖ ﺑﻪ ﺭﻳﺎﺿﻴﺎﺕ ﻛﺮﺩﻩ ﻭ ﺩﺭ ﺍﻳﻦ ﺭﺍﻩ ﺗﻼﺵ ﻛﻨﻴﺪ. .2ﺑﻪ ﺍﻳﻦ ﺣﻘﻴﻘﺖ ﻭ ﺑﺎﻭﺭ ﺑﺮﺳﻴﺪ ﻛﻪ ﻛﻤﺘﺮ ﺍﺯ ﺣﺎﻓﻈﻪ ﻭ ﺑﻴﺸﺘﺮ ﺍﺯ ﻓﻜﺮ ﻭ ﺍﻧﺪﻳﺸﻪﻱ ﻣﻨﻄﻘﻲ ﺍﺳﺘﻔﺎﺩﻩ ﻛﻨﻴﺪ. .3ﻣﻔﺎﻫﻴﻢ ﺭﻳﺎﺿﻲ ﺭﺍ ﺑﻪ ﺧﺎﻃﺮ ﺯﻳﺒﺎﻳﻲﻫﺎﻱ ﺩﺭﻭﻧﻲ ﻭ ﻛﺎﺭﺑﺮﺩﻫﺎﻱ ﺑﻴﺮﻭﻧـﻲ ﺁﻥ ﻳﺎﺩ ﺑﮕﻴﺮﻳﺪ ﻧﻪ ﺑﻪ ﺧﺎﻃـﺮ ﺗﻜﺎﻟﻴﻔﻲ ﻛﻪ ﺑﻪ ﻋﻬﺪﻩﺗﺎﻥ ِ ﮔﺬﺍﺷـﺘﻪ ﺷﺪﻩ ﺍﺳﺖ ﻭ ﻳﺎ ﺑﻪ ﺩﻟﻴﻞ ﻧﻤﺮﻩﺍﻱ ﻛﻪ ﺑﻪ ﺁﻥ ﻧﻴﺎﺯ ﺩﺍﺭﻳﺪ ﺍﻟﺒﺘﻪ ﺍﮔﺮ ﺷـﻤﺎ ﻣﻔﻬﻮﻣﻲ ﺭﺍ ﺧﻮﺏ ﻳﺎﺩ ﺑﮕﻴﺮﻳﺪ ،ﻧﻤﺮﻩﻱ ﺧﻮﺏ ﻫﻢ ﺧﻮﺍﻫﻴﺪ ﮔﺮﻓﺖ! ﺗﻜﺎﻣـﻞ ﺫﻫﻦ ﺧﻼﻕ ﻭ .4ﺍﺯ ﺩﻳﮕﺮ ﺭﺍﻩﻫﺎﻱ ﺭﺳـﻴﺪﻥ ﺑﻪ ﺭﺷـﺪ ﻭ ِ ﻛﺎﻭﺷـﮕ ِﺮ ﺭﻳﺎﺿـﻲ ،ﺗﺠﺰﻳـﻪ ﻭ ﺗﺤﻠﻴﻞ ﻣﺴـﺌﻠﻪ ﻭ ﻛﻨـﺪﻭﻛﺎﻭ ﺩﺭ ﭘﻴﺮﺍﻣﻮﻥ ﺁﻥ ﺍﺳﺖ. ﻣﻔﺎﻫﻴﻢ ﻭ ﻣﺴﺌﻠﻪﻫﺎﻱ ِ ﺳـﺎﺯﺍﻥ ﺍﻳﺮﺍﻥ ﻋﺰﻳـﺰﺍﻥ ﺩﺍﻧﺶﺁﻣﻮﺯ ﺁﻳﻨﺪﻩ ﺑﻪ ﻫﺮ ﺻﻮﺭﺕ ﺷـﻤﺎ ِ ِ ﺍﺳـﻼﻣﻲ ﺧﻮﺍﻫﻴﺪ ﺑـﻮﺩ ﻭ ﺑﺎﻳﺪ ﺍﺯ ﻫﻤﻴـﻦ ﺍﻻﻥ ﺧﻮﺩﺗﺎﻥ ﺭﺍ ﺑﺮﺍﻱ ﻗﺒﻮﻝ ﻣﺴﺌﻮﻟﻴﺖﻫﺎﻱ ﺍﺟﺘﻤﺎﻋﻲ ﻛﻮﭼﻚ ﻭ ﺑﺰﺭگ ﺁﻣﺎﺩﻩ ﻛﻨﻴﺪ .ﻭ ﺧﻼﻕ ﺧﻮﺩ ﺣﻔﻆ ﻛﻨﻴﺪ. ﺗﻜﺎﻣﻞ ﺫﻫﻦ ﺍﻳﻦ ﺁﻣﺎﺩﮔﻲ ﺭﺍ ﺑﺎ ﺭﺷﺪ ﻭ ِ ِ
ﺍﻥﺷﺎءﺍ...
ﺳﺮﺩﺑﻴﺮ
)﹇︧﹝️ دوم( ﺳﭙﻴﺪﻩ ﭼﻤﻦﺁﺭﺍ
ﻛﻠﻴﺪﻭﺍژﻩﻫﺎ :ﻣﺒﻨﺎ ،ﺍﺭﺯﺵ ﻣﻜﺎﻧﻰ ،ﺗﻮﺍﻥﻫﺎﻱ ﺍﻋﺪﺍﺩ. ﺩﺭ ﺍﻳﻦ ﺷـﻤﺎﺭﻩ ﺍﺯ ﻣﺠﻠﻪ ﻧﻴﺰ ﺑﻪ ﻃﺮﺡ ﭼﻨﺪ ﭘﺮﺳـﺶ ﺩﺭﺑﺎﺭﻩﻯ ﻧﻤﺎﻳﺶ ﺍﻋﺪﺍﺩ ﺩﺭ ﻣﺒﻨﺎﻫﺎﻯ ﻣﺨﺘﻠـﻒ ﻣﻰﭘﺮﺩﺍﺯﻳﻢ ﻭ ﺗﻼﺵ ﻣﻰﻛﻴﻨﻢ ﺑﺎ ﺍﺳﺘﻔﺎﺩﻩ ﺍﺯ ﺍﺭﺯﺵ ﻣﻜﺎﻧﻰ ﺍﺭﻗﺎﻡ ﺩﺭ ﻫﺮ ﻣﺒﻨﺎ ،ﺑﺎ ﺳﺮﻋﺖ ﺑﻴﺶﺗﺮﻯ ﺑﻪ ﺁﻥ ﭘﺮﺳﺶﻫﺎ ﭘﺎﺳﺦ ﺩﻫﻴﻢ. ﭘﻴﺶ ﺍﺯ ﻃﺮﺡ ﭘﺮﺳﺶﻫﺎ ﺑﺪ ﻧﻴﺴﺖ ﻳﺎﺩﺁﻭﺭﻯ ﻛﻨﻢ ﻛﻪ ﺍﺭﺯﺵ ﻣﻜﺎﻧﻰ ﺩﺭ ﻣﺒﻨﺎﻯ ،10ﺗﻮﺍﻥﻫﺎﻯ 10ﺍﺳﺖ: ﻳﻜﺎﻥ=100
ﺩﻫﮕﺎﻥ=101
...
ﺻﺪﮔﺎﻥ= 102ﻳﻜﺎﻥ ﻫﺰﺍﺭ= 103ﺩﻫﮕﺎﻥ ﻫﺰﺍﺭ=104
)ﺟﺪﻭﻝ ﺍﺭﺯﺵ ﻣﻜﺎﻧﻰ ﺩﺭ ﻣﺒﻨﺎﻯ (10
ﻭ ﻧﻴﺰ ﻳﺎﺩﺁﻭﺭ ﺷــﻮﻳﻢ ﻛﻪ ﺩﺭ ﻫﺮ ﻣﺒﻨﺎﻯ ﻋﺪﺩ ﻃﺒﻴﻌﻰ ) ، (n 〉1ﺍﺭﺯﺵ ﻣﻜﺎﻧﻰ ﺍﺭﻗﺎﻡ ﻳﻚ ﻋﺪﺩ ﺩﺭ ﺁﻥ ﻣﺒﻨﺎ ،ﺍﺯ ﺭﺍﺳــﺖ ﺑﻪ ﭼﭗ ،ﺗﻮﺍﻥﻫﺎﻯ nﺍﺳــﺖ ﻛﻪ ﺍﻓﺰﺍﻳﺶ ﻣﻰﻳﺎﺑﺪ: ﻳﻜﻰﻫﺎ=n0
n2ﺗﺎﻳﻰ
n1ﺗﺎﻳﻰ
n3ﺗﺎﻳﻰ
n4ﺗﺎﻳﻰ
...
n5ﺗﺎﻳﻰ
)ﺟﺪﻭﻝ ﺍﺭﺯﺵ ﻣﻜﺎﻧﻰ ﺩﺭ ﻣﺒﻨﺎﻯ (n
ﺍﻳﻨﻚ ﭘﺮﺳﺶﻫﺎ ﺭﺍ ﻣﻄﺮﺡ ﻣﻰﻛﻨﻴﻢ: ﭘﺮﺳــﺶ :1ﻧﻤﺎﻳــﺶ ﻋــﺪﺩ 81ﺩﺭ ﻣﺒﻨــﺎﻯ 3ﭼﻴﺴــﺖ؟ ﻳﻌﻨــﻰ 81=(?)3 ﭘﺎﺳﺦ :ﻃﺒﻴﻌﻰ ﺍﺳﺖ ﻛﻪ ﺑﻪ ﺳﺮﺍﻍ ﺗﻘﺴﻴﻢﻫﺎﻯ ﻣﺘﻮﺍﻟﻰ ﺑﺮﻭﻳﻢ ﺗﺎ ﺑﻔﻬﻤﻴﻢ 81ﺭﺍ ﺩﺭ ﻣﺒﻨــﺎﻯ 3ﭼﮕﻮﻧﻪ ﻧﻤﺎﻳﺶ ﻣﻰﺩﻫﻨﺪ .ﻭﻟﻰ ﺍﮔﺮ ﺗﻮﺍﻥﻫﺎﻯ 3ﺭﺍ ﺣﻔﻆ ﺑﺎﺷﻴﻢ ،ﻳﺎﺩﻣﺎﻥ ﻣﻰﺁﻳﺪ ﻛﻪ ،81=34ﭘﺲ ﻓﻮﺭﻯ ﺩﺭ ﺟﺪﻭﻝ ﺍﺭﺯﺵ ﻣﻜﺎﻧﻰ ﺩﺭ ﻣﺒﻨﺎﻯ 3ﻣﻰﻧﻮﻳﺴﻴﻢ:
ﭘﺮﺳﺶ ﺑﻌﺪﻯ ﻧﻴﺰ ﻣﺸﺎﺑﻪ ﭘﺮﺳﺶ 1ﺍﺳﺖ. ﭘﺮﺳﺶ :2ﻧﻤﺎﻳﺶ 128ﺩﺭ ﻣﺒﻨﺎﻯ 2ﭼﻴﺴﺖ؟ ﭘﺎﺳــﺦ :ﺑﺪﻭﻥ ﺍﻳﻦﻛﻪ ﺑﻪ ﺳﺮﺍﻍ ﺗﻘﺴﻴﻢﻫﺎﻯ ﻣﺘﻮﺍﻟﻰ ﺑﺮﻳﻢ ،ﻳﺎﺩﻣﺎﻥ ﺑﺎﺷﺪ ﻛﻪ ،128=27ﭘﺲ ﺩﺭ ﺟﺪﻭﻝ ﺍﺭﺯﺵ ﻣﻜﺎﻧﻰ 2ﺩﺍﺭﻳﻢ: 0
2
21
22
23
24
25
26
27
0
0
0
0
0
0
0
1
3
31
32
33
34
ﻳﻌﻨﻰ:
0
0
0
0
1
128=(10000000)2 ﺁﻳــﺎ ﺍﺯ ﭘﺎﺳــﺦ ﺍﻳــﻦ ﺩﻭ ﭘﺮﺳــﺶ ،ﺭﺍﺑﻄــﻪﻯ ﺟﺪﻳــﺪ ﺩﻳﮕﺮﻯ ﻛﺸــﻒ ﻧﻜﺮﺩﻩﺍﻳﺪ؟
0
ﻳﻌﻨﻰ 81=(10000)3
ﺩﻭﺭﻩﻯ ﺷﺎﻧﺰﺩﻫﻢ ،ﺷﻤﺎﺭﻩ ﻯ ،2ﺯﻣﺴﺘﺎﻥ 1389
ﺭﺍﻫﻨﻤﺎﻳﻲ
3
ﻫﻤﺮﺍﻩ ﺑﺎ ﻛﺘﺎﺏ
﹡﹝︀︩ ا︻︡اد در ﹞︊﹠︀﹨︀ی ﹞︐﹀︀وت
ﺑﻠﻪ ﺩﺭﺳــﺖ ﺍﺳــﺖ .ﺩﺭ ﭘﺮﺳــﺶ 1ﻛﻪ 81=34ﺑﻮﺩ ،ﻧﻤﺎﻳﺶ ﺁﻥ ﺷﺪ ﻛﻮﭼﻚﺗﺮﻳﻦ ﻋﺪﺩ ﭘﻨﺞ ﺭﻗﻤﻰ ﺩﺭ ﻣﺒﻨﺎﻯ ،3ﻳﻌﻨﻰ ﺯﺍﻭﻳﻪ 1ﺑﺎ 4ﺗﺎ ﺻﻔﺮ ﺟﻠﻮﻯ ﺁﻥ ،81=(10000)3 :ﭼﻮﻥ ﺍﺭﺯﺵ ﻣﻜﺎﻧﻰ ﺍﺯ 30ﺷﺮﻭﻉ ﻣﻰﺷﻮﺩ، ﻟﺬﺍ 34ﺩﺭ ﻣﻜﺎﻥ ﭘﻨﺠﻢ ﺟﺪﻭﻝ ﻗﺮﺍﺭ ﻣﻰﮔﻴﺮﺩ .ﺑﻪ ﺩﻟﻴﻞ ﻣﺸﺎﺑﻪ ،ﻧﻤﺎﻳﺶ 27=128ﺷﺪ ﻳﻚ 1ﺑﺎ 7ﺗﺎ ﺻﻔﺮ ﺟﻠﻮ ﺁﻥ ،ﻳﻌﻨﻰ ﻛﻮﭼﻚﺗﺮﻳﻦ ﻋﺪﺩ 8 ﺭﻗﻤﻰ ﺩﺭ ﻣﺒﻨﺎﻯ :2 128=(10000000)2 ﺣﺎﻝ ﻛﻪ ﺍﻳﻦ ﺭﺍﺑﻄﻪ ﺭﺍ ﻣﺘﻮﺟﻪ ﺷﺪﻳﺪ ،ﺳﺮﻳﻊ ﺑﮕﻮﻳﻴﺪ. ﭘﺮﺳﺶ :3 625=( ?)5 ﭘﺎﺳﺦ ،625=54 :ﭘﺲ 625=(10000)5 ﭘﺲ ﺑﺪ ﻧﻴﺴﺖ ﻗﺒﻞ ﺍﺯ ﺍﺩﺍﻣﻪﻯ ﺍﻳﻦ ﻣﻘﺎﻟﻪ ،ﻳﻚ ﺟﺪﻭﻝ ﺗﻮﺍﻥ ﺑﺮﺍﻯ ﺍﻋﺪﺍﺩ 1ﺗﺎ 10ﺑﺮﺍﻯ ﺧﻮﺩﺗﺎﻥ ﺩﺭﺳﺖ ﻛﻨﻴﺪ ﺗﺎ ﺑﺎ ﻛﻤﻚ ﺁﻥ ،ﺳﺮﻳﻊﺗﺮ ﺑﺘﻮﺍﻧﻴﺪ ﺑﻪ ﺍﻳﻦ ﭘﺮﺳــﺶﻫﺎ ﭘﺎﺳﺦ ﺩﻫﻴﺪ .ﺍﮔﺮ ﻗﺴﻤﺘﻰ ﺍﺯ ﺁﻥ ﺭﺍ ﺣﻔﻆ ﻛﻨﻴﺪ ﻛﻪ ﺧﻴﻠﻰ ﺑﻬﺘﺮ ﺍﺳﺖ! ﭘﺮﺳﺶ :4ﻧﻤﺎﻳﺶ ﻋﺪﺩ 80ﺩﺭ ﻣﺒﻨﺎﻯ 3ﭼﻴﺴﺖ؟ ﭘﺎﺳــﺦ :ﮔﺮﭼﻪ ،80ﻫﻴﭻ ﺗﻮﺍﻧﻰ ﺍﺯ 3ﻧﻴﺴــﺖ ،ﻭﻟﻰ ﺍﺯ ،81ﻳﻜﻰ ﻛﻢﺗﺮ
ﺍﺳﺖ .ﭘﺲ ﭼﻮﻥ ،(10000)3=81ﻳﻌﻨﻰ ﻛﻮﭼﻚﺗﺮﻳﻦ ﻋﺪﺩ ﭘﻨﺞ ﺭﻗﻤﻰ ﺩﺭ ﻣﺒﻨﺎﻯ 3ﺍﺳــﺖ 80 ،ﺑﺎﻳﺪ ﺑﺰﺭگﺗﺮﻳﻦ ﻋﺪﺩ ﭼﻬﺎﺭ ﺭﻗﻤﻰ ﺩﺭ ﻣﺒﻨﺎﻯ 3ﺑﺎﺷﺪ ،ﻳﻌﻨﻰ: 80=(2222)3 )ﻳﺎﺩﺗﺎﻥ ﺑﺎﺷﺪ ﺑﺰﺭگﺗﺮﻳﻦ ﺭﻗﻢ ﺩﺭ ﻣﺒﻨﺎﻯ ،3ﺭﻗﻢ 2ﺑﻮﺩ!( ﭘﺮﺳﺶ :5ﻧﻤﺎﻳﺶ 127ﺩﺭ ﻣﺒﻨﺎﻯ 2ﭼﻴﺴﺖ؟ 127=(1111111)2 128=27=(10000000)2 ﭘﺮﺳﺶ :6ﻧﻤﺎﻳﺶ 63ﺩﺭ ﻣﺒﻨﺎﻯ 4ﭼﻴﺴﺖ؟ ﭼﻮﻥ ،64=43ﭘﺲ 64=(1000)4؛ ﻟﺬﺍ 63=(333)4 ﺍﺣﺘﻤﺎﻻً ﺳــﻪ ﭘﺮﺳــﺶ ﺍﺧﻴﺮ ﺭﺍ ﺩﺭ ﻣﺪﺕِ ﺧﻴﻠﻰ ﻛﻮﺗﺎﻫﻰ ﭘﺎﺳﺦ ﺩﺍﺩﻳﺪ، ﻭﻟﻰ ﺍﮔﺮ ﻣﻰﺧﻮﺍﺳــﺘﻴﺪ ﺑﺎ ﺗﻘﺴﻴﻢﻫﺎﻯ ﻣﺘﻮﺍﻟﻰ ﺑﻪ ﭘﺎﺳﺦ ﺑﺮﺳﻴﺪ ،ﺧﻴﻠﻰ ﺧﻴﻠﻰ ﺑﻴﺶﺗﺮ ﻃﻮﻝ ﻣﻰﻛﺸﻴﺪ! ﻓﻘﻂ ﺑﺎﻳﺪ ﺟﺪﻭﻝ ﺗﻮﺍﻥ ﺭﺍ ﺩﻡ ﺩﺳﺖ ﻳﺎ ﺩﺭ ﺫﻫﻦ ﺩﺍﺷﺘﻪ ﺑﺎﺷﻴﺪ! ﭘﺮﺳــﺶ :7ﺣﺎﺻﻞ ﺟﻤــﻊ 128 +64ﺩﺭ ﻣﺒﻨﺎﻯ 2ﭼﮕﻮﻧﻪ ﻧﻤﺎﻳﺶ ﺩﺍﺩﻩ ﻣﻰﺷﻮﺩ؟
ﺟﺪﻭﻝ ﺗﻮﺍﻥﻫﺎﻯ ﺍﻋﺪﺍﺩ 1ﺗﺎ ) 10ﺗﺎ ﺗﻮﺍﻥ ﺩﻫﻢ( )ﺧﻮﺩﺗﺎﻥ ﺍﻳﻦ ﺟﺪﻭﻝ ﺭﺍ ﺑﺎ ﻛﻤﻚ ﻣﺎﺷﻴﻦ ﺣﺴﺎﺏ ،ﻛﺎﻣﻞ ﻛﻨﻴﺪ!(
ﺗﻮﺍﻥ
10
9
8
7
6
5
4
3
2
.......
......
28=...
27=128
26=64
25=32
24=16
23=8
22=4
2
.......
........
34=81
33=27
32=9
3
........
44=256
43=64
42=16
4
ﭘﺎﻳﻪ
5 6 7 8 9 10
4
ﺭﺍﻫﻨﻤﺎﻳﻲ
ﺩﻭﺭﻩﻯ ﺷﺎﻧﺰﺩﻫﻢ ،ﺷﻤﺎﺭﻩﻯ ،2ﺯﻣﺴﺘﺎﻥ 1389
ﭘﺎﺳﺦ :ﺍﺣﺘﻤﺎﻻً ﻣﺘﻮﺟﻪ ﺷﺪﻩﺍﻳﺪ ﻛﻪ ﻫﺮ ﺩﻭ ﻋﺪﺩ 64ﻭ ،128ﺗﻮﺍﻥﻫﺎﻳﻰ ﺍﺯ 2ﻫﺴﺘﻨﺪ .ﭘﺲ ﺑﺪﻭﻥ ﺍﻳﻦﻛﻪ ﺑﻪ ﺳﺮﺍﻍ ﺟﻤﻊ ﺯﺩﻥ ﺁﻥﻫﺎ ﻭ ﺳﭙﺲ ﺗﻘﺴﻴﻢ ﻛﺮﺩﻥﻫﺎﻯ ﻣﺘﻮﺍﻟﻰ ﺑﺮﻭﻳﻢ ،ﺍﺯ ﺟﺪﻭﻝ ﺍﺭﺯﺵ ﻣﻜﺎﻧﻰ 2ﺍﺳﺘﻔﺎﺩﻩ ﻣﻰﻛﻨﻴﻢ:
=128+64
2
20
21
22
23
24
25
26
27
0 0
0 0
0 0
0 0
0 0
0 0
0
1
)0
0
0
0
0
0
1
1
+
⎫ ⎪ 64=26 ⇒ ⎬ ⎭⎪ 128=27
(1
ﺟﻤﻊ ﺍﻧﺠﺎﻡ ﺷﺪ! ﭘﺮﺳﺶ :8ﻋﺪﺩ 66ﺩﺭ ﻣﺒﻨﺎﻯ 4ﭼﮕﻮﻧﻪ ﻧﻮﺷﺘﻪ ﻣﻰﺷﻮﺩ؟ ﭘﺎﺳﺦ :ﺑﺎﺯ ﻫﻢ ﻳﺎﺩ ﺗﻮﺍﻥﻫﺎﻯ 4ﻣﻰﺍﻓﺘﻴﻢ 66 .ﺗﻮﺍﻧﻰ ﺍﺯ 4ﻧﻴﺴﺖ ﻭﻟﻰ ﺍﺯ 2 ،43=64ﻭﺍﺣﺪ ﺑﻴﺶﺗﺮ ﺍﺳﺖ .ﭘﺲ 40
41
42
43
0
0
0
1
0
(1
2 2)4 =66
ﺣﺎﻝ ﺷــﻤﺎ ﻳﻚ ﭘﺮﺳﺶ ﻣﺸــﺎﺑﻪ ﻣﻄﺮﺡ ﻛﻨﻴﺪ ﻛﻪ ﺑﺎ ﺍﺳﺘﻔﺎﺩﻩ ﺍﺯ ﺍﺭﺯﺵ ﻣﻜﺎﻧﻰ ،ﺗﻮﺍﻥﻫﺎﻯ ﺍﻋﺪﺍﺩ ﻭ ﺍﺭﺗﺒﺎﻁ ﺁﻥﻫﺎ ﺑﺎ ﻫﻢ ،ﺑﺘﻮﺍﻥ ﺑﻪ ﺳﺮﻋﺖ ﺑﻪ ﺁﻥ ﭘﺎﺳﺦ ﺩﺍﺩ: ﭘﺮﺳﺶ ...................... :12 ﻣﻮﻓﻖ ﺑﺎﺷﻴﺪ.
+ 0
)ﺗﻮﺟﻪ ﻛﻨﻴﺪ ﻛﻪ ،2ﺭﻗﻤﻰ ﻣﺠﺎﺯ ﺩﺭ ﻣﺒﻨﺎﻯ 4ﺍﺳﺖ ﻭ .(2=(2)4 ﭘﺮﺳــﺶ :9ﻋــﺪﺩ ﺑﻌــﺪ ﺍﺯ (333)4ﺩﺭ ﻣﺒﻨــﺎﻯ 2ﭼﮕﻮﻧﻪ ﻧﻮﺷــﺘﻪ ﻣﻰﺷﻮﺩ؟ ﭘﺎﺳــﺦ :ﺗﻮﺟﻪ ﻛﻨﻴــﺪ ﻛــﻪ ،(333)4ﺑﺰﺭگﺗﺮﻳﻦ ﻋﺪﺩ ﺳــﻪ ﺭﻗﻤﻰ ﺩﺭ ﺍﻳﻦ ﻣﺒﻨﺎﺳــﺖ .ﭘﺲ ﻋﺪﺩ ﺑﻌــﺪﻯ ﺁﻥ ،ﻛﻮﭼﻚﺗﺮﻳﻦ ﻋﺪﺩ ﭼﻬﺎﺭ ﺭﻗﻤﻰ، ﻳﻌﻨــﻰ (1000)4ﺍﺳــﺖ .ﺍﻳﻦ ﻫــﻢ ﻳﻌﻨﻰ ،43ﻳﺎ ﺑﻪ ﻋﺒﺎﺭﺗــﻰ ) 26ﺯﻳﺮﺍ .(43=(22)3=26ﭘﺲ ﻧﻤﺎﻳﺶ ﺁﻥ ﺩﺭ ﻣﺒﻨﺎﻯ ،2ﭼﻨﻴﻦ ﺍﺳﺖ: .(1000000)2ﻳﻌﻨﻰ ﺑﻪ ﻃﻮﺭ ﺧﻼﺻﻪ: (333)4+1=(1000000)2 ﭘﺮﺳــﺶ :10ﺍﺧﺘﻼﻑ 150ﻭ 125ﺩﺭ ﻣﺒﻨﺎﻯ 5ﭼﮕﻮﻧﻪ ﻧﻮﺷــﺘﻪ ﻣﻰﺷﻮﺩ؟ ﭘﺎﺳﺦ: 150-125=25 2 25=5 ﭘﺲ 150-125=(100)5 ﭘﺮﺳﺶ .11ﺍﺧﺘﻼﻑ 127ﻭ 129ﺩﺭ ﻣﺒﻨﺎﻯ 5ﺍﺳﺖ ،ﭘﺲ 129-127=2 ﻭ ﭼﻮﻥ ،2ﺭﻗﻤﻰ ﻣﺠﺎﺯ ﺩﺭ ﻣﺒﻨﺎﻯ 5ﺍﺳﺖ ،ﭘﺲ 129-127=(2)5 ﺩﻭﺭﻩﻯ ﺷﺎﻧﺰﺩﻫﻢ ،ﺷﻤﺎﺭﻩ ﻯ ،2ﺯﻣﺴﺘﺎﻥ 1389
ﺭﺍﻫﻨﻤﺎﻳﻲ
5
ﺩﺍﻧﺶﺍﻓﺰﺍﻳﻲ
﹞︀︨︊﹤ی او﹇︀ت ︫︣︻ ﹩و ︨︣︻️ ︩︣︠ ز﹞﹫﹟ ﻛﺎﻇﻢ ﺍﻳﺮﺍﻧﻲ
ﻛﻠﻴﺪﻭﺍژﻩﻫﺎ :ﺗﻮﺍﻥ ،ﭘﺎﻳﻪ ،ﻧﻤﺎ. ﻣﺤﺎﺳﺒﻪﻯ ﻇﻬﺮ ﻣﺤﻠﻰ ﻭ ﺍﻭﻗﺎﺕ ﺷﺮﻋﻰ ﺍﺯ ﺟﻤﻠﻪ ﻣﺴﺎﺋﻠﻰ ﺍﺳﺖ ﻛﻪ ﻫﻤﻪﺭﻭﺯﻩ ﺑﺎ ﺁﻥ ﻣﻮﺍﺟﻪ ﻫﺴﺘﻴﻢ .ﺑﺎ ﻳﻚ ﻣﺤﺎﺳﺒﻪﻯ ﺑﺴﻴﺎﺭ ﺳﺎﺩﻩﻯ ﺭﻳﺎﺿﻰ ﻧﻪ ﺗﻨﻬﺎ ﻣﻰﺗﻮﺍﻥ ﺳﺮﻋﺖ ﺯﺍﻭﻳﻪﺍﻯ ﻭ ﺳﺮﻋﺖ ﻣﺴﺎﻓﺘﻰ ﺯﻣﻴﻦ ﺑﻪ ﺩﻭﺭ ﺧﻮﺩ ﺭﺍ ﺑﻪ ﺩﺳﺖ ﺁﻭﺭﺩ ،ﺑﻠﻜﻪ ﻣﻰﺗﻮﺍﻥ ﺍﻭﻗﺎﺕ ﺷﺮﻋﻰ ﻫﺮ ﻧﻘﻄﻪ ﺍﺯ ﺟﻬﺎﻥ ﺭﺍ ﺑﺎ ﺍﺳﺘﻔﺎﺩﻩ ﺍﺯ ﺍﻭﻗﺎﺕ ﺷﺮﻋﻰ ﻳﻚ ﻧﻘﻄﻪﻯ ﻣﺒﺪﺃ ﻭ ﻃﻮﻝ ﺟﻐﺮﺍﻓﻴﺎﻳﻰ ﻣﻜﺎﻥ ﻣﻮﺭﺩ ﻧﻈﺮ ﻭ ﻣﺒﺪﺃ ﺑﻪ ﺩﺳﺖ ﺁﻭﺭﺩ .ﺍﺑﺘﺪﺍ ﻣﻰﺧﻮﺍﻫﻴﻢ ﺳﺮﻋﺖ ﺯﺍﻭﻳﻪﺍﻯ ﻭ ﺳﺮﻋﺖ ﻣﺴﺎﻓﺘﻰ ﺯﻣﻴﻦ ﺭﺍ ﻣﺤﺎﺳﺒﻪ ﻛﻨﻴﻢ ﻭ ﺳﭙﺲ ﻣﺤﺎﺳﺒﻪﻯ ﺍﻭﻗﺎﺕ ﺷﺮﻋﻰ ﺭﺍ ﺗﻮﺿﻴﺢ ﺩﻫﻴﻢ. ﻣﻰﺩﺍﻧﻴﻢ ﻛﻪ ﻳﻚ ﺩﻭﺭ ﭼﺮﺧﺶ ﺯﻣﻴﻦ ﺑﻪ ﺩﻭﺭ ﺧﻮﺩ ﺩﺭ ﻳﻚ ﺭﻭﺯ ﺭﺍ ﺣﺮﻛﺖ ﻭﺿﻌﻰ ﺯﻣﻴﻦ ﻣﻰﻧﺎﻣﻨﺪ .ﺯﻣﻴﻦ ﺩﺭ 24ﺳﺎﻋﺖ 360ﺩﺭﺟﻪ ﺑﻪ ﺩﻭﺭ ﺧﻮﺩ ﻣﻰﭼﺮﺧﺪ .ﺑﺎ ﺗﻘﺴﻴﻢ 360ﺑﺮ 24ﻋﺪﺩ 15ﺑﻪ ﺩﺳﺖ ﻣﻰﺁﻳﺪ ﻭ ﺍﻳﻦ ﺳﺮﻋﺖ ﺯﺍﻭﻳﻪﺍﻯ ﺯﻣﻴﻦ ﺍﺳﺖ ،ﻳﻌﻨﻰ ﺯﻣﻴﻦ ﺩﺭ ﻳﻚ ﺳﺎﻋﺖ 15 ﺩﺭﺟﻪ ﺑﻪ ﺩﻭﺭ ﺧﻮﺩ ﻣﻰﭼﺮﺧﺪ .ﺍﺯ ﻃﺮﻓﻰ ﭼﻮﻥ ﻫﺮ ﺩﺭﺟﻪ ﺑﺮﺍﺑﺮ ﺑﺎ 60 ﺩﻗﻴﻘﻪﻯ ﺯﺍﻭﻳﻪﺍﻯ ﻭ ﻫﺮ ﺩﻗﻴﻘﻪ ﺑﺮﺍﺑﺮ ﺑﺎ 60ﺛﺎﻧﻴﻪﻯ ﺯﺍﻭﻳﻪﺍﻯ ﻭ ﻫﺮ ﺳﺎﻋﺖ ﻣﻌﺎﺩﻝ 60ﺩﻗﻴﻘﻪﻯ ﺯﻣﺎﻧﻰ ﻭ ﻫﺮ ﺩﻗﻴﻘﻪ ﻣﻌﺎﺩﻝ 60ﺛﺎﻧﻴﻪﻯ ﺯﻣﺎﻧﻰ ﺍﺳﺖ ،ﭘﺲ ﺳﺮﻋﺖ ﺯﻣﻴﻦ ﺩﺭ ﻳﻚ ﺩﻗﻴﻘﻪﻯ ﺯﻣﺎﻧﻰ ﺑﺮﺍﺑﺮ 15ﺩﻗﻴﻘﻪﻯ ﻣﻜﺎﻧﻰ ﻭ ﺩﺭ ﻳﻚ ﺛﺎﻧﻴﻪﻯ ﺯﻣﺎﻧﻰ ﺑﺮﺍﺑﺮ ﺑﺎ 15ﺛﺎﻧﻴﻪﻯ ﻣﻜﺎﻧﻰ ﺍﺳﺖ. ﺧﻼﺻﻪﻯ ﺍﻳﻦ ﻣﺤﺎﺳﺒﺎﺕ ﺩﺭ ﺟﺪﻭﻝ ﺯﻳﺮ ﺁﻣﺪﻩ ﺍﺳﺖ.
6
ﺭﺍﻫﻨﻤﺎﻳﻲ
ﻣﻜﺎﻧﻰ )ﺯﺍﻭﻳﻪﺍﻯ(
ﺯﻣﺎﻧﻰ
15º
1ﺳﺎﻋﺖ
َ15
1ﺩﻗﻴﻘﻪ
ً15
1ﺛﺎﻧﻴﻪ
ﺩﻭﺭﻩﻯ ﺷﺎﻧﺰﺩﻫﻢ ،ﺷﻤﺎﺭﻩﻯ ،2ﺯﻣﺴﺘﺎﻥ 1389
ﺑﺮﺍﻯ ﻣﺤﺎﺳﺒﻪﻯ ﺳﺮﻋﺖ ﻣﺴﺎﻓﺘﻰ ،ﺷﻜﻞ ﺯﻣﻴﻦ ﺭﺍ ﺑﻪ ﺻﻮﺭﺕ ﻛﺮﻩ )ﺷﻜﻞ ﻭﺍﻗﻌﻰ ﻧﺰﺩﻳﻚ ﺑﻪ ﺑﻴﻀﻰﮔﻮﻥ ﺍﺳﺖ( ﺩﺭ ﻧﻈﺮ ﻣﻰﮔﻴﺮﻳﻢ .ﺷﻌﺎﻉ ﻛﺮﻩﻯ ﺯﻣﻴﻦ ﺗﻘﺮﻳﺒﺎً ﻣﻌﺎﺩﻝ 6.37 × 106 mﺍﺳﺖ ﻛﻪ ﺑﺎ ﺟﺎﻯﮔﺰﺍﺭﻯ ﺩﺭ ﻓﺮﻣﻮﻝ ﻣﺤﻴﻂ ﺩﺍﻳﺮﻩ ) (I = 2πrﻃﻮﻝ ﻣﺪﺍﺭ ﺯﻣﻴﻦ )ﻣﺪﺍﺭ ﺍﺳﺘﻮﺍ( ﺑﺮﺍﺑﺮ ﺑﺎ 4.00036 × 107ﻣﺘﺮ ﺧﻮﺍﻫﺪ ﺷﺪ ﻛﻪ ﺍﮔﺮ ﺁﻥ ﺭﺍ ﺑﻪ 24ﺗﻘﺴﻴﻢ ﻛﻨﻴﻢ، ﺳﺮﻋﺖ ﻣﺴﺎﻓﺘﻰ ﺑﺮﺍﺑﺮ ﺑﺎ 1.6668 × 106ﻣﺘﺮ ﺑﺮ ﺳﺎﻋﺖ ﻳﻌﻨﻰ 1666 .8 ﻛﻴﻠﻮﻣﺘﺮ ﺑﺮ ﺳﺎﻋﺖ ﺍﺳﺖ ﻛﻪ ﺍﺯ ﺳﺮﻳﻊﺗﺮﻳﻦ ﻫﻮﺍﭘﻴﻤﺎﻫﺎ ﻧﻴﺰ ﺳﺮﻳﻊﺗﺮ ﺣﺮﻛﺖ ﻣﻰﻛﻨﺪ. ﺣﺎﻝ ﺑﺎ ﺍﺳﺘﻔﺎﺩﻩ ﺍﺯ ﺍﻳﻦ ﺍﻃﻼﻋﺎﺕ ﺑﻪ ﻧﺤﻮﻩﻯ ﻣﺤﺎﺳﺒﻪﻯ ﺍﻭﻗﺎﺕ ﺷﺮﻋﻰ ﻣﻨﻄﻘﻪﻯ ﺩﻟﺨﻮﺍﻩ ﺍﺯ ﺭﻭﻯ ﻃﻮﻝ ﺟﻐﺮﺍﻓﻴﺎﻳﻰ ﺷﻬﺮ ﻣﻮﺭﺩ ﻧﻈﺮ ﻭ ﻳﻚ ﺷﻬﺮ ﻣﺒﺪﺃ ﻭ ﺍﻭﻗﺎﺕ ﺷﺮﻋﻰ ﺷﻬﺮ ﻣﺒﺪﺃ ﻣﻰﭘﺮﺩﺍﺯﻳﻢ. ﺍﮔﺮ ﻃﻮﻝ ﺟﻐﺮﺍﻓﻴﺎﻳﻰ ﺷﻬﺮ ﻣﺒﺪﺃ ﺭﺍ a0.b0ﺩﺭ ﻧﻈﺮ ﺑﮕﻴﺮﻳﻢ )ﻛﻪ ﺩﺭ ﺁﻥ a0ﺑﻪ ﺩﺭﺟﻪ ﻭ b0ﺑﻪ ﺩﻗﻴﻘﻪ ﺍﺳﺖ( ﻭ ﻃﻮﻝ ﺟﻐﺮﺍﻓﻴﺎﻳﻰ ﺷﻬﺮ ﻣﻮﺭﺩ ﻧﻈﺮ ﺭﺍ a.bﺑﮕﻴﺮﻳﻢ ،ﺗﻔﺎﺿﻞ ﻃﻮﻝ ﺟﻐﺮﺍﻓﻴﺎﻳﻰ ﺷﻬﺮ ﻣﻮﺭﺩ ﻧﻈﺮ ﺭﺍ
ﺍﺯ ﺷﻬﺮ ﻣﺒﺪﺃ ﺑﻪ ﺻﻮﺭﺕ ﺯﻳﺮ ﭘﻴﺪﺍ ﺧﻮﺍﻫﻴﻢ ﻛﺮﺩ. )a1.b1 = a0.b0− a.b = (a0 − a).(b0− b
ﻛﻪ ﺩﺭ ﺁﻥ ﺑﺮﺍﻯ ﺳﺎﺩﮔﻰ ﻣﺤﺎﺳﺒﺎﺕ ﺩﺭﺟﻪ ﻭ ﺩﻗﻴﻘﻪ ﺭﺍ ﻫﻢ ﻋﻼﻣﺖ ﻣﻰﮔﻴﺮﻳﻢ ﻳﻌﻨﻰ ﻫﺮ ﺩﻭ ﻣﺜﺒﺖ ﻳﺎ ﻫﺮ ﺩﻭ ﻣﻨﻔﻰ .ﺑﻪ ﻋﺒﺎﺭﺕ ﺩﻳﮕﺮ ،ﻫﺮﮔﺎﻩ a0〉 aﻋﺒﺎﺭﺕ a1.b1 = a0.b0− a.bﺭﺍ ﻣﺤﺎﺳﺒﻪ ﻣﻰﻛﻨﻴﻢ a .b − a.bﺭﺍ ﻣﺤﺎﺳﺒﻪ ﻣﻰﻛﻨﻴﻢ ،ﺳﭙﺲ ﻭ ﺍﮔﺮ a 〉 a 0ﺑﺎﺷﺪ ،ﺍﺑﺘﺪﺍ 0 0 ﻗﺮﻳﻨﻪﻯ ﺁﻥ ﺭﺍ ﺑﻪ ﺟﺎﻯ ﻋﺒﺎﺭﺕ a0.b 0− a.bﻗﺮﺍﺭ ﻣﻰﺩﻫﻴﻢ ،ﻳﻌﻨﻰ ) . a1.b1 = −(a.b − a0 .b0ﺩﺭ ﺿﻤﻦ ﻫﺮﮔﺎﻩ b0〈 bﺑﺎﺷﺪ ،ﻳﻚ ﻭﺍﺣﺪ ﺍﺯ a0ﻛﻢ ﻭ 60ﻭﺍﺣﺪ ﺑﻪ b0ﺍﺿﺎﻓﻪ ﻣﻰﻛﻨﻴﻢ. ﺍﺧﺘﻼﻑ ﺯﻣﺎﻧﻰ ﺍﻭﻗﺎﺕ ﺷﺮﻋﻰ ﺍﺯ ﻓﺮﻣﻮﻝ ﻣﻰﺷﻮﺩ ﻛﻪ ﺩﺭ ﺁﻥ xﺑﻪ ﺩﻗﻴﻘﻪ ﺍﺳﺖ ﻭ ﺍﮔﺮ
b1
x = 4a1 +ﻣﺤﺎﺳﺒﻪ
15 b1
15
ﺍﻋﺸﺎﺭﻯ ﺑﻮﺩ ﺁﻥ ﺭﺍ
ﺩﺭ 60ﺿﺮﺏ ﻣﻰﻛﻨﻴﻢ ﻭ ﺁﻥ ﺭﺍ ﺛﺎﻧﻴﻪﻯ ﺯﻣﺎﻧﻰ ﺩﺭ ﻧﻈﺮ ﻣﻰﮔﻴﺮﻳﻢ .ﺩﺭ
ﻭﻗﺖ ﺷﺮﻋﻰ ﻣﺒﺪﺃ ﺟﻤﻊ ﻣﻰﻛﻨﻴﻢ ﺗﺎ ﻭﻗﺖ ﺷﺮﻋﻰ ﻣﻜﺎﻥ ﻣﻮﺭﺩ ﻧﻈﺮ ﺑﻪ ﺩﺳﺖ ﺁﻳﺪ .ﺩﺭ ﺿﻤﻦ ،ﺩﺭ ﻋﻤﻞ ﺍﺯ ﻣﺤﺎﺳﺒﻪﻯ ﺛﺎﻧﻴﻪ ﺻﺮﻑﻧﻈﺮ ﻣﻰﻛﻨﻴﻢ ﻭ ﺍﮔﺮ ﻭﻗﺖ ﺷﺮﻋﻰ ﺩﺍﺭﺍﻯ ﺛﺎﻧﻴﻪ ﺑﻮﺩ ،ﺁﻥ ﺭﺍ ﺣﺬﻑ ﻭ ﻳﻚ ﻭﺍﺣﺪ ﺑﻪ ﺩﻗﻴﻘﻪ ﺍﺿﺎﻓﻪ ﻣﻰﻛﻨﻴﻢ. ﺑﺮﺍﻯ ﺟﻤﻊ ﻳﺎ ﺗﻔﺮﻳﻖ ﺯﻣﺎﻥ ﻫﻢ ،ﻛﺎﺭﻯ ﻣﺘﺸﺎﺑﻪ ﺁﻧﭽﻪ ﮔﻔﺘﻪ ﺷﺪ ﺍﻧﺠﺎﻡ ﻣﻰﺩﻫﻴﻢ ،ﺑﺪﻳﻦ ﻣﻌﻨﺎ ﺳﺎﻋﺖ ﺭﺍ ﺑﺎ ﺳﺎﻋﺖ ﻭ ﺩﻗﻴﻘﻪ ﺭﺍ ﺑﺎ ﺩﻗﻴﻘﻪ ﻭ ﺛﺎﻧﻴﻪ ﺭﺍ ﺑﺎ ﺛﺎﻧﻴﻪ ﺟﻤﻊ ﻳﺎ ﻣﻨﻬﺎ ﻣﻰﻛﻨﻴﻢ .ﺩﺭ ﺗﻔﺎﺿﻞ ﻫﺮﮔﺎﻩ ﻋﺪﺩ ﻣﻨﻔﻰ ﺑﻮﺩ ،ﻳﻚ ﻭﺍﺣﺪ ﺍﺯ ﺑﺰﺭگﺗﺮﻯ ﻛﻢ ﻭ ﺷﺼﺖ ﻭﺍﺣﺪ ﺑﻪ ﺁﻥ ﺍﺿﺎﻓﻪ ﻣﻰﻛﻨﻴﻢ. ﺩﺭ ﺟﺪﻭﻝ ﺯﻳﺮ ﻃﻮﻝ ﺟﻐﺮﺍﻓﻴﺎﻳﻰ ﭼﻨﺪ ﺷﻬﺮ ﺩﺍﺩﻩ ﺷﺪﻩ ﻛﻪ ﺩﺭ ﺁﻥ ﺗﻬﺮﺍﻥ ﺑﻪ ﻋﻨﻮﺍﻥ ﺷﻬﺮ ﻣﺒﺪﺃ ﺩﺭ ﻧﻈﺮ ﮔﺮﻓﺘﻪ ﺷﺪﻩ ﺍﺳﺖ .ﺑﺮﺍﻯ ﻣﺜﺎﻝ ،ﺍﺫﺍﻥ ﻣﻐﺮﺏ ﺷﻬﺮﻫﺎﻯ ﺍﺭﻭﻣﻴﻪ ﻭ ﻣﺸﻬﺪ ﺭﺍ ﻣﺤﺎﺳﺒﻪ ﻣﻰﻛﻨﻴﻢ. ﻃﻮﻝ ﺟﻐﺮﺍﻓﻴﺎﻳﻰ
a1.b1
6 .6 ′
44 .58′
4 .38′
46 .26′
0
51 .4′
−3 . 0
−5 .22′
ﭘﺎﻳﺎﻥ ،ﻋﺪﺩ xﺑﻪ ﺩﺳﺖ ﺁﻣﺪﻩ ﺭﺍ ﻛﻪ ﻋﺪﺩﻯ ﻣﺜﺒﺖ ﻳﺎ ﻣﻨﻔﻰ ﺍﺳﺖ ﺑﺎ
−8 .33′
ﺷﻬﺮ ﺍﺭﻭﻣﻴﻪ ﺗﺒﺮﻳﺰ ﺗﻬﺮﺍﻥ
54 .4′
ﻳﺰﺩ
56 .26′
ﺑﻨﺪﺭﻋﺒﺎﺱ
59 .37′
ﻣﺸﻬﺪ
ﺑـــــــــﺮﺍﻯ ﺷــــــﻬــــــــﺮ ﺍﺭﻭﻣــــــﻴـــﻪ x = 4 × 6 + 6 = (24.4)′ = 24′.24′′ﻭ ﺍﮔﺮ ﺛﺎﻧﻴﻪ 15
ﺭﺍ ﺣﺬﻑ ﻭ ﻳﻚ ﻭﺍﺣﺪ ﺑﻪ ﺩﻗﻴﻘﻪ ﺍﺿﺎﻓﻪ ﻛﻨﻴﻢ ،ﺩﺍﺭﻳﻢ: ً x=25ﻳﻌﻨﻰ ﺍﮔﺮ ﻭﻗﺖ ﺷﺮﻋﻰ ﺗﻬﺮﺍﻥ ﺭﺍ ﺑﺎ 25ﺩﻗﻴﻘﻪ ﺟﻤﻊ ﻛﻨﻴﻢ ،ﻭﻗﺖ ﺷﺮﻋﻰ ﺍﺭﻭﻣﻴﻪ ﺑﻪ ﺩﺳﺖ ﻣﻰﺁﻳﺪ .ﺍﺫﺍﻥ ﻣﻐﺮﺏ ﺗﻬﺮﺍﻥ ﺩﺭ ﭘﻨﺠﻢ ﺑﻬﻤﻦ ﻣﺎﻩ ﺳﺎﻋﺖ َ 18. 11ﺍﺳﺖ ،ﺑﻨﺎﺑﺮﺍﻳﻦ ﺍﺫﺍﻥ ﻣﻐﺮﺏ ﺍﺭﻭﻣﻴـــﻪ َ18 .36 ﺧﻮﺍﻫﺪ ﺑﻮﺩ. ⎞ 33 ⎛ ﺑﺮﺍﻯ ﺷﻬﺮ ﻣﺸﻬﺪ ⎟ = ( −34.2)′ = −34′.12′′ ⎠ 15 ⎝ ﻭ ﺍﮔﺮ ﺛﺎﻧﻴﻪ ﺭﺍ ﺣﺬﻑ ﻛﻨﻴﻢ ،ﺩﺍﺭﻳﻢ ،x=-34َ :ﻳﻌﻨﻰ ﺍﮔﺮ ﻭﻗﺖ ﺷﺮﻋﻰ x = −⎜4× 8 +
ﺗﻬﺮﺍﻥ ﺭﺍ ﺑﺎ -34ﺩﻗﻴﻘﻪ ﺟﻤﻊ ﻛﻨﻴﻢ ،ﻭﻗﺖ ﺷﺮﻋﻰ ﻣﺸﻬﺪ ﺭﺍ ﺑﻪ ﺩﺳﺖ ﻣﻰﺁﻳﺪ .ﺍﺫﺍﻥ ﻣﻐﺮﺏ ﺗﻬﺮﺍﻥ ﺩﺭ ﭘﻨﺠﻢ ﺑﻬﻤﻦ ﻣﺎﻩ ﺳﺎﻋﺖ َ 18 .11ﻫﺴﺖ ﻭ ﺑﻨﺎﺑﺮﺍﻳﻦ ﺍﺫﺍﻥ ﻣﻐﺮﺏ ﻣﺸﻬﺪ 17 .37ﺧﻮﺍﻫﺪ ﺑﻮﺩ.
ﺩﻭﺭﻩﻯ ﺷﺎﻧﺰﺩﻫﻢ ،ﺷﻤﺎﺭﻩ ﻯ ،2ﺯﻣﺴﺘﺎﻥ 1389
ﺭﺍﻫﻨﻤﺎﻳﻲ
7
ﮔﻘﺖ ﻭ ﮔﻮ
﹎﹀️ و ﹎︡﹫︍︨ ︀︋ ﹢ه ﹝﹟آرا ﹜﹚︺﹞ ،ر︀︲ ﹩دوره را﹨﹠﹝︀﹩
﹨﹝﹤ ا︨︐︺︡اد ر︀︲ ﹩دار﹡︡ ﮔﻔﺖ ﻭ
ﮔﻮ :ﺁﺯﺍﺩﻩ ﺷﺎﻛﺮﻯ
ﻛﻠﻴﺪﻭﺍژﻩﻫﺎ :ﻣﻌﻠﻢ ﺭﻳﺎﺿﻰ ،ﺩﺭﺱ ﺭﻳﺎﺿﻲ ،ﺩﺍﻧﺶﺁﻣﻮﺯﺍﻥ . ﺍﺷﺎﺭﻩ:
ـ ﺧﺎﻧﻢ ﺍﺟﺎﺯﻩ! ﭼﺮﺍ ﺭﻳﺎﺿﻰ ﺍﻳﻨﻘﺪﺭ ﺳﺨﺘﻪ؟ ـ ﺑﺒﺨﺸـﻴﺪ ﺁﻗﺎ! ﻣﻦ ﺍﺻ ً ﻼ ﺍﻳﻦ ﺩﺭﺳﻮ ﻧﻔﻬﻤﻴﺪﻡ ،ﻣﻰﺷﻪ ﺩﻭﺑﺎﺭﻩ ﺗﻮﺿﻴﺢ ﺑﺪﻳﺪ؟ ـ ﺁﻗﺎ ﺍﺟﺎﺯﻩ! ﭼﻪ ﻛﺘﺎﺑﻰ ﺑﺨﻮﻧﻴﻢ ﻛﻪ ﺭﻳﺎﺿﻴﻤﻮﻥ ﻗﻮﻯ ﺑﺸﻪ؟ ـ ﺧﺎﻧﻢ! ﺷﻤﺎ ﻛﻪ ﻫﻢﺳﻦ ﻣﺎ ﺑﻮﺩﻳﺪ ﭼﻄﻮﺭﻯ ﺭﻳﺎﺿﻰ ﻣﻰﺧﻮﻧﺪﻳﺪ؟ ـ ﺍﺟﺎﺯﻩ! ﻣﻰﺷﻪ ﺍﻣﺮﻭﺯ ﺑﻬﻤﻮﻥ ﺗﻜﻠﻴﻒ ﺭﻳﺎﺿﻰ ﻧﺪﻳﺪ؟ ﺍﻳﻨـﻬــﺎ ﻭ ﺳـﺆﺍﻻﺗﻰ ﺩﻳﮕﺮ ﺭﺍ ﺧﻴﻠﻰ ﺍﺯ ﺑﭽﻪﻫﺎ ،ﺳـﺮ ﻛﻼﺱ ﺭﻳﺎﺿﻰ ﺍﺯ ﻣﻌﻠﻢﻫﺎﻯ ﺭﻳﺎﺿﻰﺷﺎﻥ ﻣﻰﭘﺮﺳﻨﺪ. ﺧﺎﻧﻢ ﺳـﭙﻴﺪﻩ ﭼﻤﻦﺁﺭﺍ ﻳﻜـﻰ ﺍﺯ ﺍﻳﻦ ﻣﻌﻠـﻢ ﺭﻳﺎﺿﻰﻫﺎﺳـﺖ .ﺍﻭ ﺩﺍﺭﺍﻯ ﻓﻮﻕ ﻟﻴﺴـﺎﻧﺲ ﺭﻳﺎﺿﻰ ﻭ ﻓﻮﻕ ﻟﻴﺴـﺎﻧﺲ ﺁﻣﻮﺯﺵ ﺭﻳﺎﺿﻰ ﻭ ﺩﺭ ﺣﺎﻝ ﺣﺎﺿﺮ، ﺩﺍﻧﺸـﺠﻮﻯ ﺩﻛﺘﺮﻯ ﺁﻣﻮﺯﺵ ﺭﻳﺎﺿﻰ ﺍﺳـﺖ 21 .ﺳﺎﻝ ﺍﺳـﺖ ،ﺭﻳﺎﺿﻰ ﺩﺭﺱ ﻣﻰﺩﻫـﺪ ،ﺍﻣﺎ ﺍﻧـﮕﺎﺭ ﻫﻨﻮﺯ ﺍﺯ ﺣﺎﻝ ﻭ ﻫـﻮﺍﻯ ﻧﻮﺟﻮﺍﻧﻰ ﻭ ﻳـﺎﺩ ﻭ ﺧﺎﻃﺮﺍﺕ ﺁﻥ ﺭﻭﺯﻫﺎ ﺧﻴﻠﻰ ﺩﻭﺭ ﻧﺸـﺪﻩ ﺍﺳـﺖ؛ ﭼﻮﻥ ﻭﻗﺘﻰ ﺍﺯ ﺧﺎﻃﺮﺍﺕ ﻣﻌﻠﻢ ﺭﻳﺎﺿﻰﻫﺎﻯ ﺭﺍﻫﻨﻤﺎﻳﻰﺍﺵ ﻣﻰﭘﺮﺳـﻴﻢ ،ﺍﺷﻚ ﺩﺭ ﭼﺸﻤﺎﻧﺶ ﺣﻠﻘﻪ ﻣﻰﺯﻧﺪ ﻭ ﺑﻪ ﻳﺎﺩ ﻣﻌﻠﻢ ﺳﺎﻝﻫﺎﻯ ﺩﻭﺭﺵ ﺑﻐﺾ ﻣﻰﻛﻨﺪ. ﭘﺎﻯ ﺻﺤﺒﺖ ﺧﺎﻧﻢ ﭼﻤﻦﺁﺭﺍ ﻧﺸﺴﺘﻴﻢ ﻭ ﭘﺮﺳﺶﻫﺎﻯ ﺷﻤﺎ ﺑﭽﻪﻫﺎ ﺭﺍ ﺍﺯ ﺍﻭ ﭘﺮﺳﻴﺪﻳﻢ ،ﺍﻳﻦ ﮔﻔﺖﻭ ﮔﻮ ﺭﺍ ﺩﺭ ﺍﺩﺍﻣﻪ ﻣﻰﺧﻮﺍﻧﻴﺪ.
ﭼﻄﻮﺭ ﺑﻪ ﺭﻳﺎﺿﻰ ﻋﻼﻗﻤﻨﺪ ﺷﺪﻳﺪ؟ ﻣﻦ ﺍﺻﻮﻻً ﺩﺭﺱ ﺧﻮﺍﻧﺪﻥ ﻭ ﻳﺎﺩ ﮔﺮﻓﺘﻦ ﭼﻴﺰﻫﺎﻯ ﺑﻴﺸــﺘﺮ ﺭﺍ ﺧﻴﻠﻰ ﺩﻭﺳﺖ ﺩﺍﺷــﺘﻢ .ﺗﺎ ﺣﺪﻭﺩﻯ ﻫﻢ ﺗﺤﺖ ﺗﺄﺛﻴﺮ ﺑﺮﺍﺩﺭ ﻛﻮﭼﻜﻢ ﺑﻪ ﺭﻳﺎﺿﻰ ﻋﻼﻗﻤﻨــﺪ ﺷــﺪﻡ .ﻣﺎ ﺑﺎ ﻫﻢ ﺧﻴﻠــﻰ ﺻﻤﻴﻤﻰ ﺑﻮﺩﻳــﻢ ﻭ ﺭﻭﻯ ﻫﻢ ﺗﺄﺛﻴﺮ ﻣﻰﮔﺬﺍﺷــﺘﻴﻢ .ﺑﺮﺍﺩﺭﻡ ﺩﺭ ﻳﻚ ﺩﻭﺭﻩﺍﻯ ،ﻣﻌﻠﻤﺎﻧﻰ ﺩﺍﺷﺖ ﻛﻪ ﺭﻳﺎﺿﻰ ﺭﺍ ﺑــﻪ ﺷــﻜﻞ ﻣﺘﻔﺎﻭﺗﻰ ﺑﺎ ﺍﻭ ﻛﺎﺭ ﻣﻰﻛﺮﺩﻧﺪ ﻭ ﺑــﻪ ﻫﻤﻴﻦ ﺩﻟﻴﻞ ﺑﻪ ﺭﻳﺎﺿﻰ ﺧﻴﻠﻰ ﻋﻼﻗﻤﻨﺪ ﺷــﺪﻩ ﺑﻮﺩ ﻭ ﺭﻳﺎﺿﻰ ﺭﺍ ﻛﺎﺭﺑﺮﺩﻯ ﻭ ﻋﻴﻨﻰ ﻣﻰﺩﻳﺪ ﻭ ﺍﺯ ﺍﻳــﻦ ﺩﺭﺱ ﻟﺬﺕ ﻣﻰﺑﺮﺩ ﻭ ﺍﻳﻦ ﺣﺲ ﺭﺍ ﺑﻪ ﺧﺎﻧﻪ ﻫﻢ ﻣﻨﺘﻘﻞ ﻣﻰﻛﺮﺩ ﻭ ﻣﻦ ﻫﻢ ﺑﻪ ﺭﻳﺎﺿﻰ ﻋﻼﻗﻤﻨﺪ ﺷــﺪﻡ ﻭ ﺭﺷﺘﻪﻯ ﺭﻳﺎﺿﻰ ﺭﺍ ﺩﺭ ﺩﺑﻴﺮﺳﺘﺎﻥ ﺍﻧﺘﺨﺎﺏ ﻛﺮﺩﻡ .ﺑﺮﺍﻯ ﺍﺩﺍﻣﻪ ﺗﺤﺼﻴﻞ ﺩﺭ ﺩﺍﻧﺸﮕﺎﻩ ﻫﻢ ﺑﺎ ﺍﻳﻨﻜﻪ ﻣﻰﺗﻮﺍﻧﺴﺘﻢ ﺭﺷــﺘﻪﻫﺎﻯ ﻣﻬﻨﺪﺳــﻰ ﺭﺍ ﺍﻧﺘﺨﺎﺏ ﻛﻨﻢ ﺗﺮﺟﻴﺢ ﺩﺍﺩﻡ ﺭﻳﺎﺿﻰ ﺭﺍ ﺩﻧﺒﺎﻝ ﻛﻨــﻢ .ﭼﻮﻥ ﺭﻳﺎﺿــﻰ ،ﺩﻗﺘﺶ ﻭ ﺍﺭﺗﺒﺎﻃﻰ ﻛﻪ ﺑﻴــﻦ ﻣﻮﺿﻮﻋﺎﺕ ﺭﻳﺎﺿﻰ 8
ﺭﺍﻫﻨﻤﺎﻳﻲ
ﺩﻭﺭﻩﻯ ﺷﺎﻧﺰﺩﻫﻢ ،ﺷﻤﺎﺭﻩﻯ ،2ﺯﻣﺴﺘﺎﻥ 1389
ﺑﺮﻗﺮﺍﺭ ﻣﻰﺷﺪ ﺭﺍ ﺩﻭﺳﺖ ﺩﺍﺷﺘﻢ. ﭼﺮﺍ ﺑﻌﻀﻰ ﺍﺯ ﺑﭽﻪﻫﺎ ﻓﻜﺮ ﻣﻰﻛﻨﻨﺪ ،ﺭﻳﺎﺿﻰ ﺩﺭﺱ ﺳـﺨﺘﻰ ﺍﺳﺖ؟ ﺍﮔــﺮ ﺧﻮﺩ ﺑﭽﻪﻫﺎ ﻫﻢ ﺍﻳﻦﻃﻮﺭ ﻓﻜﺮ ﻧﻜﻨﻨــﺪ ،ﭘﺪﺭ ﻭ ﻣﺎﺩﺭﻫﺎ ﻭ ﺣﺘﻰ ﻣﻌﻠﻢﻫﺎﻯ ﺭﻳﺎﺿﻰ ﺍﻳﻦ ﺭﺍ ﺑﻪ ﺁﻧﻬﺎ ﺍﻟﻘﺎ ﻣﻰﻛﻨﻨﺪ .ﺑﻌﻀﻰ ﻭﻗﺘﻬﺎ ﻣﻌﻠﻢﻫﺎﻯ ﺭﻳﺎﺿﻰ ﺟــﻮﺭﻯ ﺑﺮﺧﻮﺭﺩ ﻣﻰﻛﻨﻨﺪ ﻛﻪ ﻧﺸــﺎﻥ ﺩﻫﻨﺪ ﺩﺭﺱ ﻣﻦ ﺧﻴﻠﻰ ﺳــﺨﺖ ﻭ ﻣﻬﻢ ﺍﺳﺖ ﻭ ﺑﺎﻳﺪ ﺧﻴﻠﻰ ﺩﺭﺱ ﺑﺨﻮﺍﻧﻴﺪ ﻭ ﻛﺎﺭ ﻛﻨﻴﺪ ﺗﺎ ﻣﻮﻓﻖ ﺷﻮﻳﺪ .ﻳﺎ ﭘﺪﺭ ﻭ ﻣﺎﺩﺭ ﺧﻴﻠﻰ ﺭﻭﻯ ﺩﺭﺱ ﺭﻳﺎﺿﻰ ﺳﺨﺖﮔﻴﺮﻯ ﻣﻰﻛﻨﻨﺪ ﻭ ﺑــﻪ ﺩﺭﺱﻫﺎﻯ ﺩﻳﮕﺮ ﺍﻫﻤﻴﺖ ﻧﻤﻰﺩﻫﻨﺪ ﻭ ﺑﭽﻪﻫﺎ ﻓﻜﺮ ﻣﻰﻛﻨﻨﺪ ﺣﺘﻤﺎً ﺭﻳﺎﺿﻰ ﺩﺭﺱ ﺳﺨﺖ ﻭ ﭘﻴﭽﻴﺪﻩﺍﻯ ﺍﺳﺖ ﻛﻪ ﺍﻳﻨﻘﺪﺭ ﻫﻤﻪ ﺑﻪ ﺁﻥ ﺍﻫﻤﻴﺖ ﻣﻰﺩﻫﻨــﺪ .ﺑﻪ ﻧﻈﺮ ﻣﻦ ﺍﮔــﺮ ﺍﻃﺮﺍﻓﻴﺎﻥ ﺣﺴﺎﺳﻴﺘﺸــﺎﻥ ﺭﺍ ﻛﻤﺘﺮ ﻛﻨﻨﺪ، ﺑﭽﻪﻫﺎ ﻫــﻢ ﺭﺍﺣﺖﺗﺮ ﺑﺎ ﺩﺭﺱ ﺭﻳﺎﺿﻰ ﺑﺮﺧــﻮﺭﺩ ﻣﻰﻛﻨﻨﺪ .ﺑﻪ ﻧﻈﺮ ﻣﻦ ﺩﺭﺱﻫﺎﻳﻰ ﻣﺜﻞ ﻋﻠﻮﻡ ،ﺍﺟﺘﻤﺎﻋﻰ ،ﺟﻐﺮﺍﻓﻴﺎ ﻭ ﺗﺎﺭﻳﺦ ﻫﻢ ﺳﺨﺖ ﻫﺴﺘﻨﺪ ﺍﻣﺎ ﻛﺴﻰ ﺭﺍﺟﻊﺑﻪ ﺳﺨﺘﻰ ﺍﻳﻦ ﺩﺭﺱﻫﺎ ﺻﺤﺒﺖ ﻧﻤﻰﻛﻨﺪ. ﺑﻌﻀﻰ ﺑﭽﻪﻫﺎ ﭘﻴﺶ ﺍﺯ ﺣﻞ ﻣﺴﺎﺋﻞ ﺭﻳﺎﺿﻰ ،ﺍﺣﺴﺎﺱ ﻧﮕﺮﺍﻧﻰ ﻭ ﺩﻟﻬﺮﻩ ﻣﻰﻛﻨﻨﺪ ﻛﻪ ﻣﺒﺎﺩﺍ ﻧﺘﻮﺍﻧﻨﺪ ﻣﺴـﺌﻠﻪ ﺭﺍ ﺣﻞ ﻛﻨﻨﺪ ،ﺷـﻤﺎ ﭼﻪ ﺗﻮﺻﻴﻪﺍﻯ ﺑﺮﺍﻯ ﭼﻨﻴﻦ ﺑﭽﻪﻫﺎﻳﻰ ﺩﺍﺭﻳﺪ؟ ﺭﻳﺎﺿﻰ ﺍﻳﻦ ﻣﺎﻫﻴﺖ ﺭﺍ ﺩﺍﺭﺩ ﻛﻪ ﺍﺯ ﻣﺎ ﻣﻰﺧﻮﺍﻫﻨﺪ ﻭﻗﺘﻰ ﺑﺎ ﻣﻮﺿﻮﻋﻰ ﺁﺷــﻨﺎ ﻣﻰﺷــﻮﻳﻢ ،ﺗﻌﺮﻳﻒ ﻭ ﻳﺎ ﺭﺍﺑﻄﻪﺍﻯ ﺭﺍ ﻣﻰﺧﻮﺍﻧﻴﻢ ﺍﺯ ﺁﻥ ﺍﺳــﺘﻔﺎﺩﻩ ﻛﻨﻴﻢ ﻭ ﻣﺴــﺌﻠﻪ ﺣﻞ ﻛﻨﻴﻢ .ﺩﺭ ﻋﻠﻮﻡ ﻫﻢ ﻛﻢ ﻭ ﺑﻴﺶ ﭼﻨﻴﻦ ﺣﺎﻟﺘﻰ ﺭﺍ ﺩﺍﺭﻳﻢ ،ﺍﻣﺎ ﻧﻪ ﺑﻪ ﺍﻳﻦ ﺷﺪﺕ. ﻣــﻦ ﻓﻜﺮ ﻣﻰﻛﻨﻢ ﻧﻮﻉ ﺗﺪﺭﻳﺲ ﻣــﺎ ،ﻛﻪ ﺍﻧﺘﻈﺎﺭ ﺩﺍﺭﻳﻢ ﺣﺘﻤﺎً ﺑﭽﻪﻫﺎ ﻳﻚ ﻧﻮﻉ ﻣﺴﺎﺋﻞ ﺧﺎﺻﻰ ﺭﺍ ﺣﻞ ﻛﻨﻨﺪ ،ﺑﺎﻋﺚ ﺷﺪﻩ ﺍﺳﺖ ﻛﻪ ﺭﻳﺎﺿﻰ ﺑﺮﺍﻯ ﺑﭽﻪﻫﺎ ﺳﺨﺖ ﺷﻮﺩ .ﭼﻮﻥ ﻣﺎ ﺍﺯ ﻫﻤﻪﻯ ﺗﻮﺍﻥ ﺑﭽﻪﻫﺎ ﺍﺳﺘﻔﺎﺩﻩ ﻧﻤﻰﻛﻨﻴﻢ. ﺧﻴﻠﻰ ﻭﻗﺖﻫﺎ ﻣﻰﺑﻴﻨﻴﻢ ﻛﻪ ﻳﻜﻰ ﺍﺯ ﺑﭽﻪﻫﺎ ،ﻳﻚ ﻣﺴﺌﻠﻪﻯ ﻫﻨﺪﺳﻪ ﺭﺍ ﺑﺎ ﻳﻚ ﺷﻜﻞ ﺟﺎﻟﺐ ﻭ ﺍﺑﺘﻜﺎﺭﻯ ﺣﻞ ﻣﻰﻛﻨﺪ ﻭﻟﻰ ﻣﻌﻠﻢ ﻣﻰﮔﻮﻳﺪ :ﻧﻪ! ﺑﺎﻳﺪ
ﻣﺴــﺌﻠﻪ ﺭﺍ ﻫﻤﺎﻥ ﺟﻮﺭﻯ ﺣﻞ ﻛﻨﻰ ﻛﻪ ﻣﻦ ﺍﺯ ﺗﻮ ﺧﻮﺍﺳــﺘﻪﺍﻡ .ﻭ ﺩﺭ ﺍﻳﻦ ﺣﺎﻟﺖ ﺍﺳﺖ ﻛﻪ ﺑﭽﻪ ﻓﻜﺮ ﻣﻰﻛﻨﺪ ﻛﻪ ﺭﻳﺎﺿﻰ ﺳﺨﺖ ﺍﺳﺖ ﻭ ﻧﻤﻰﺗﻮﺍﻧﺪ ﺁﻥ ﺭﺍ ﻳــﺎﺩ ﺑﮕﻴﺮﺩ .ﮔﺎﻫﻰ ﺍﻭﻗﺎﺕ ﺍﺯ ﺑﭽﻪﻫﺎ ﻣﻰﺧﻮﺍﻫﻴﻢ ﻣﻌﺎﺩﻟﻪﺍﻯ ﺭﺍ ﺣﻞ ﻛﻨﻨﺪ ﻭ ﻣﻰﮔﻮﻳﻴﻢ ﻛﻪ ﺣﺘﻤﺎً ﺑﺎﻳﺪ ﺁﻥ ﺭﺍ ،ﺁﻥﻃﻮﺭ ﻛﻪ ﻣﻦ ﻣﻰﺧﻮﺍﻫﻢ ﺣﻞ ﻛﻨﻰ ﻭﻟﻰ ﺍﻭ ﺑﺎ ﻓﺮﺍﻳﻨﺪﻫﺎﻯ ﺫﻫﻨﻰ ﺧﻮﺩﺵ ﻣﺴﺌﻠﻪ ﺭﺍ ﺗﺤﻠﻴﻞ ﻣﻰﻛﻨﺪ ﻭ ﻣﺘﻮﺟﻪ ﻣﻰﺷﻮﺩ ﻣﺠﻬﻮﻝ ﻣﻌﺎﺩﻟﻪ ﭼﻴﺴﺖ .ﺩﺭ ﻭﺍﻗﻊ ﺑﻪ ﺧﺎﻃﺮ ﻣﺤﺪﻭﺩﻳﺖ ﺭﻭﺵﻫﺎﻳﻰ ﻛﻪ ﺑﻪ ﺑﭽﻪﻫﺎ ﺍﺭﺍﺋﻪ ﻣﻰﻛﻨﻴﻢ ﻳﺎ ﻓﻀﺎﻳﻰ ﻛﻪ ﺩﺭ ﻛﻼﺱ ﺍﻳﺠﺎﺩ ﻣﻰﻛﻨﻴﻢ ،ﺍﻳﻦ ﻧﮕﺮﺍﻧﻰ ﺩﺭ ﺑﭽﻪﻫﺎ ﺍﻳﺠﺎﺩ ﻣﻰﺷﻮﺩ .ﻣﻦ ﺍﻋﺘﻘﺎﺩ ﺩﺍﺭﻡ ﻫﻤﻪﻯ ﺑﭽﻪﻫﺎ ﺩﺭ ﺭﻳﺎﺿﻰ ﺍﺳﺘﻌﺪﺍﺩ ﺩﺍﺭﻧﺪ ،ﻭﻟﻰ ﻫﺮ ﻛﺪﺍﻡ ﺩﺭ ﻳﻚ ﻗﺴﻤﺘﻰ .ﻧﻮﻉ ﺍﺳــﺘﻌﺪﺍﺩﻫﺎ ﻣﺘﻔﺎﻭﺕ ﺍﺳــﺖ ،ﻭﻟﻰ ﻫﻤﻪﻯ ﺑﭽﻪﻫﺎ ﻣﻰﺗﻮﺍﻧﻨﺪ ﺭﻳﺎﺿﻰ ﺭﺍ ﺑﻔﻬﻤﻨﺪ ﻭ ﺁﻧﻄﻮﺭ ﻛﻪ ﺧﻮﺩﺷﺎﻥ ﻣﻰﺗﻮﺍﻧﻨﺪ ﺍﺯ ﺁﻥ ﺍﺳﺘﻔﺎﺩﻩ ﻛﻨﻨﺪ ﻭﻟﻰ ﻧﻮﻉ ﺁﻣــﻮﺯﺵ ﻣﺎ ﭼﻮﻥ ﺧﻂ ﺧﺎﺻﻰ ﺭﺍ ﺩﻧﺒﺎﻝ ﻣﻰﻛﻨﺪ ﺑﻪ ﻫﻤﻪ ﺍﻳﻦ ﺍﻣﻜﺎﻥ ﺭﺍ ﻧﻤﻰﺩﻫﺪ ﻛﻪ ﺍﺯ ﺗﻮﺍﻧﺎﻳﻰ ﻣﺘﻔﺎﻭﺕﺷﺎﻥ ﺍﺳﺘﻔﺎﺩﻩ ﻛﻨﻨﺪ. ﺑﭽﻪﻫﺎ ﺑﺎﻳﺪ ﺍﻋﺘﻤﺎﺩ ﺑﻪ ﻧﻔﺲ ﺩﺍﺷﺘﻪ ﺑﺎﺷﻨﺪ ﻭ ﺍﺟﺎﺯﻩ ﻧﺪﻫﻨﺪ ﺍﺿﻄﺮﺍﺏ ﺑﻴﺮﻭﻧﻰ ﺁﻧﻬﺎ ﺭﺍ ﺍﺫﻳﺖ ﻛﻨﺪ .ﺁﻧﻬﺎ ﺑﺎﻳﺪ ﻫﺮ ﻣﻮﻓﻘﻴﺖ ﻛﻮﭼﻜﻰ ﺭﺍ ﺩﺭ ﺭﻳﺎﺿﻰ ﺟﺪﻯ ﺑﮕﻴﺮﻧﺪ .ﺍﻟﺒﺘﻪ ﻣﻌﻠﻢﻫﺎ ﻫﻢ ﺑﺎﻳﺪ ﺑﻪ ﺑﭽﻪﻫﺎ ﻛﻤﻚ ﻛﻨﻨﺪ .ﺧﻮﺩ ﻣﻦ ﻣﻰﺑﻴﻨﻢ ﻛﻪ ﺑﻌﻀﻰ ﺍﺯ ﺑﭽﻪﻫﺎ ﺩﺭ ﺍﻣﺘﺤﺎﻧﺎﺕ ﻧﻤﺮﺍﺕ ﺧﻮﺑﻰ ﻧﻤﻰﮔﻴﺮﻧﺪ ﺍﻣﺎ ﺩﺭ ﺑﺤﺚﻫﺎﻯ ﻛﻼﺳﻰ ﺍﻳﺪﻩﻫﺎﻯ ﺧﻴﻠﻰ ﺧﻮﺑﻰ ﻣﻰﺩﻫﻨﺪ ﻭ ﻣﻨﻄﻘﻰ ﺑﺤﺚ ﻣﻰﻛﻨﻨﺪ .ﭼﻮﻥ ﺑﺤﺚﻫﺎﻯ ﻛﻼﺳــﻰ ﺑﺎﺯﺗﺮ ﺍﺳــﺖ ﻭ ﺍﺯ ﻛﻠﻴﺸﻪﻫﺎ ﺧﺎﺭﺝ ﺍﺳﺖ .ﺍﻳﻦ ﻧﺸﺎﻥ ﻣﻰﺩﻫﺪ ﻛﻪ ﺍﻳﻦ ﺑﭽﻪﻫﺎ ﺗﻮﺍﻥ ﺭﻳﺎﺿﻰ ﺩﺍﺭﻧﺪ ،ﭼﻮﻥ ﻧﻘﺪ ﻛــﺮﺩﻥ ﻭ ﺩﻧﺒﺎﻝ ﻛﺮﺩﻥ ﻣﻨﻄﻘﻰ ﻣﻮﺿﻮﻉ ،ﺗﻮﺍﻧﺎﻳﻰ ﻭﺍﻗﻌﻰ ﺭﻳﺎﺿﻰ ﺍﺳــﺖ ﻭ ﻧﻤﺮﻩ ﺧﻮﺏ ﻧﮕﺮﻓﺘﻦ ﺩﺭ ﺍﻣﺘﺤﺎﻥ ﺑﻪ ﺍﻳﻦ ﻣﻌﻨﺎ ﻧﻴﺴــﺖ ﻛﻪ ﺑﭽﻪﻫﺎ ﺗﻮﺍﻥ ﺭﻳﺎﺿــﻰ ﻧﺪﺍﺭﻧﺪ .ﻣﻦ ﺍﻳﻦ ﻣﻮﻓﻘﻴﺖﻫﺎ ﺭﺍ ﺑﻪ ﺑﭽﻪﻫﺎ ﻣﻰﮔﻮﻳﻢ ﺗﺎ ﺗﺸــﻮﻳﻖ ﺷــﻮﻧﺪ ﻭ ﺑﺮﺍﻯ ﺍﻣﺘﺤــﺎﻥ ﺑﻬﺘﺮ ﺩﺍﺩﻥ ﻫﻢ ﺗﻼﺵ ﻛﻨﻨﺪ .ﻣﻦ ﺷــﺎﮔﺮﺩﺍﻧﻰ ﺩﺍﺷــﺘﻪﺍﻡ ﻛــﻪ ﻛﻼﺱ ﺍﻭﻝ ﺭﺍﻫﻨﻤﺎﻳــﻰ ﺑﻮﺩﻩﺍﻧــﺪ ﻭ ﺧﻴﻠــﻰ ﺍﺯ ﺭﻳﺎﺿﻰ ﻣﻰﺗﺮﺳــﻴﺪﻧﺪ .ﻓﻜﺮ ﻣﻰﻛﺮﺩﻧﺪ ﻧﻤﻰﺗﻮﺍﻧﻨﺪ ﺩﺭ ﺭﻳﺎﺿﻰ ﻣﻮﻓﻖ ﺷﻮﻧﺪ ،ﺍﻣﺎ ﺭﺳــﻢﻫﺎﻯ ﺯﻳﺒﺎﻳﻰ ﻣﻰﻛﺸﻴﺪﻧﺪ ﻛﻪ ﻧﺸﺎﻥ ﻣﻰﺩﺍﺩ ﺩﻗﺖ ﻭ ﺗﻤﺮﻛﺰ ﺧﻮﺑﻰ ﺩﺍﺭﻧﺪ .ﻳﺎ ﺩﺭ ﺑﺤﺚﻫﺎﻯ ﻛﻼﺳﻰ ﻭ ﺩﺭ ﻛﺎﺭﻫﺎﻯ ﮔﺮﻭﻫﻰ ،ﺍﻳﺪﻩﻫﺎﻯ ﺧﻮﺑﻰ ﻣﻰﺩﺍﺩﻧﺪ ﻭ ﺑﺎﻋﺚ ﻣﻮﻓﻘﻴﺖ ﮔﺮﻭﻩ ﻣﻰﺷﺪﻧﺪ. ﺩﺭ ﻳﺎﺩﮔﻴـﺮﻯ ﺭﻳﺎﺿـﻰ ﻫﻮﺵ ﻭ ﺍﺳـﺘﻌﺪﺍﺩ ﻣﻬﻢﺗﺮ ﺍﺳـﺖ ﻳﺎ ﺗﻼﺵ ﻭ ﭘﺸﺘﻜﺎﺭ؟ ﻫﺮﺩﻭﻯ ﺍﻳﻦ ﻣﻮﺍﺭﺩ ﻣﻬﻢ ﺍﺳــﺖ .ﻫﻤﺎﻥﻃﻮﺭ ﻛﻪ ﮔﻔﺘﻢ ﻫﺮﻛﺲ ﻳﻚ ﻧﻮﻉ ﻫﻮﺵ ﻭ ﺍﺳﺘﻌﺪﺍﺩ ﺩﺍﺭﺩ .ﻣﻨﺘﻬﻰ ﺩﺭ ﺁﻣﻮﺯﺵ ﻣﺎ ﭼﻮﻥ ﻣﺎ ﻣﻰﺧﻮﺍﻫﻴﻢ ﺍﺯ ﭘﺲ ﺍﻣﺘﺤﺎﻧﺎﺕ ﺑﺮ ﺑﻴﺎﻳﻴﻢ ،ﺗﻼﺵ ﻭ ﭘﺸــﺘﻜﺎﺭ ﻣﻬﻢﺗﺮ ﺍﺳﺖ .ﭼﻮﻥ ﺑﺎﻳﺪ ﺳﺒﻚﻫﺎﻯ ﺧﺎﺻﻰ ﺭﺍ ﻛﺎﺭ ﻛﻨﻴﻢ ،ﺗﻤﺮﻳﻦ ﻛﻨﻴﻢ ﻭ ﺳﺮﻋﺖ ﻭ ﻣﻬﺎﺭﺕ ﻻﺯﻡ ﺭﺍ ﭘﻴﺪﺍ ﻛﻨﻴﻢ .ﺍﻣﺎ ﺩﺭ ﻛﺎﺭ ﺍﻳﻦ ﺩﻭ ﻋﺎﻣﻞ ﻋﻼﻗﻪ ﺷﺨﺼﻰ ﻫﻢ ﻣﻬﻢ ﺍﺳﺖ. ﻣﺜ ً ﻼ ﺍﮔﺮ ﻛﺴــﻰ ﺑﻪ ﺩﻻﻳﻠﻰ ﺑﻪ ﺗﺎﺭﻳﺦ ﻛﺸﻮﺭﺵ ﻋﻼﻗﻤﻨﺪ ﺑﺎﺷﺪ ،ﻫﻤﻴﺸﻪ ﻧﻤﺮﻩﻫﺎﻯ ﺧﻮﺑﻰ ﺩﺭ ﺍﻳﻦ ﺩﺭﺱ ﻣﻰﮔﻴﺮﺩ .ﭼﻮﻥ ﺍﻧﮕﻴﺰﻩ ﺷــﺨﺼﻰ ﺩﺍﺭﺩ. ﺍﮔﺮ ﻣﺎ ﻣﻌﻠﻢﻫــﺎ ﻫﻢ ﺑﺘﻮﺍﻧﻴﻢ ﺩﺭ ﺑﭽﻪﻫﺎ ﺍﻧﮕﻴﺰﻩ ﺷــﺨﺼﻰ ﺑﺮﺍﻯ ﺭﻳﺎﺿﻰ ﺩﻭﺭﻩﻯ ﺷﺎﻧﺰﺩﻫﻢ ،ﺷﻤﺎﺭﻩ ﻯ ،2ﺯﻣﺴﺘﺎﻥ 1389
ﺭﺍﻫﻨﻤﺎﻳﻲ
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ﺍﻳﺠﺎﺩ ﻛﻨﻴﻢ ﺑﻪ ﻣﻮﻓﻘﻴﺖﺷﺎﻥ ﺩﺭ ﺍﻳﻦ ﺩﺭﺱ ﻛﻤﻚ ﻛﺮﺩﻩﺍﻳﻢ.
ﺭﻳﺎﺿﻰ ﻭ ﺍﺳﺘﻌﺪﺍﺩﻫﺎﻯ ﻣﺨﺘﻠﻒ ﭼﻪ ﺗﻮﺻﻴﻪﺍﻯ ﺑﺮﺍﻯ ﺑﭽﻪﻫﺎﻳﻰ ﺩﺍﺭﻳﺪ ﻛﻪ ﻧﺴـﺒﺖ ﺑﻪ ﺑﻘﻴﻪ ﺍﺯ ﺍﺳﺘﻌﺪﺍﺩ ﺭﻳﺎﺿﻰ ﻛﻤﺘﺮﻯ ﺑﺮﺧﻮﺭﺩﺍﺭﻧﺪ؟ ﻫﻤﻪ ﺩﺭ ﺭﻳﺎﺿﻰ ﺍﺳــﺘﻌﺪﺍﺩ ﺩﺍﺭﻧﺪ ﺍﻣﺎ ﺑﺎﻳﺪ ﺑﻪ ﻛﻤﻚ ﻣﻌﻠﻢﻫﺎﻳﺸــﺎﻥ ﺑﻔﻬﻤﻨﺪ ﻛﻪ ﺩﺭ ﭼﻪ ﻗﺴــﻤﺘﻰ ﺍﺯ ﺭﻳﺎﺿﻰ ﺍﺳﺘﻌﺪﺍﺩ ﺑﻴﺸﺘﺮﻯ ﺩﺍﺭﻧﺪ).ﭼﻮﻥ ﻣﻌﻠــﻢ ﺭﻳﺎﺿﻰ ﻣﻰﺗﻮﺍﻧﺪ ﺷــﺎﮔﺮﺩﺍﻧﺶ ﺭﺍ ﺧﻮﺏ ﺑﺸﻨﺎﺳــﺪ ﻭ ﺑﻔﻬﻤﺪ ﻛﻪ ﻫﺮﻛﺲ ﺩﺭ ﻛﺪﺍﻡ ﻗﺴــﻤﺖ ﺍﺳــﺘﻌﺪﺍﺩ ﺩﺍﺭﺩ (.ﻭ ﺍﺯ ﺁﻥ ﺍﺳــﺘﻌﺪﺍﺩ ﻛﻤﻚ ﺑﮕﻴﺮﻧﺪ ﺗﺎ ﺩﺭ ﺯﻣﻴﻨﻪﻫﺎﻯ ﺩﻳﮕﺮ ﻫﻢ ﭘﻴﺸــﺮﻓﺖ ﻛﻨﻨﺪ .ﻣﻦ ﻣﻌﺘﻘﺪ ﻧﻴﺴﺘﻢ ﻛﻪ ﺑﭽﻪﻫﺎ ﺩﺭ ﺭﻳﺎﺿﻰ ﺍﺳﺘﻌﺪﺍﺩ ﻛﻤﻰ ﺩﺍﺭﻧﺪ .ﺍﮔﺮ ﺑﭽﻪﺍﻯ ﻣﺴﺘﻌﺪ ﺍﺳﺖ ﻭ ﺍﺩﺑﻴﺎﺕ ،ﻋﻠﻮﻡ ﻭ ﺟﻐﺮﺍﻓﻰﺍﺵ ﺧﻮﺏ ﺍﺳﺖ ،ﺣﺘﻤﺎً ﻣﻰﺗﻮﺍﻧﺪ ﺭﻳﺎﺿﻰ ﺭﺍ ﻫﻢ ﺧﻮﺏ ﻳﺎﺩ ﺑﮕﻴﺮﺩ .ﺍﻣﺎ ﺍﺣﺘﻤﺎﻻً ﻳﺎ ﺍﺯ ﺭﻳﺎﺿﻰ ﻣﻰﺗﺮﺳﺪ ﻳﺎ ﺩﻭﺳﺘﺶ ﻧﺪﺍﺭﺩ، ﻳﺎ ﺩﺭﺱ ﺭﺍ ﭘﺸــﺖ ﮔﻮﺵ ﻣﻰﺍﻧﺪﺍﺯﺩ .ﻫﺮ ﻣﻄﻠﺒﻰ ﺑﻪ ﻣﺮﻭﺭ ﺧﻮﺍﻧﺪﻩ ﻧﺸﻮﺩ، ﺗﻠﻨﺒﺎﺭ ﻭ ﻳﺎﺩﮔﻴﺮﻯﺍﺵ ﺳﺨﺖ ﻣﻰﺷﻮﺩ. ﺭﻳﺎﺿﻰ ﺩﺭ ﭼﻪ ﻋﻠﻮﻣﻰ ﻧﻘﺶ ﺩﺍﺭﺩ؟ ﺩﺭ ﺗﻤــﺎﻡ ﻋﻠــﻮﻡ ﻣﺎﻧﻨــﺪ ﻛﺎﻣﭙﻴﻮﺗﺮ ،ﻋﻠــﻮﻡ ﺍﺭﺗﺒﺎﻃــﺎﺕ ،ﺍﻗﺘﺼﺎﺩ ﻭ ﺑﺎﻧﻜﺪﺍﺭﻯ ،ﺭﻳﺎﺿﻰ ﻧﻘﺶ ﻋﻤﺪﻩﺍﻯ ﺩﺍﺭﺩ .ﺧﻴﻠﻰ ﺍﺯ ﻛﺴــﺎﻧﻰ ﻛﻪ ﺩﻛﺘﺮﺍﻯ ﺭﻳﺎﺿﻰ ﺩﺍﺭﻧﺪ ﺩﺭ ﺑﺎﻧﻚﻫﺎ ﻣﺸﻐﻮﻝ ﺑﻪ ﻛﺎﺭﻧﺪ .ﺑﺮﺍﻯ ﺍﻳﻦﻛﻪ ﺑﺨﺶﻫﺎﻳﻰ ﺍﺯ ﺑﺎﻧﻜﺪﺍﺭﻯ ﻧﻴﺎﺯﻣﻨﺪ ﻣﺤﺎﺳﺒﺎﺗﻰ ﺍﺳﺖ ﻛﻪ ﻣﺘﺨﺼﺺ ﺭﻳﺎﺿﻰ ﺑﺎﻳﺪ ﺁﻥﻫﺎ ﺭﺍ ﺍﻧﺠﺎﻡ ﺩﻫﺪ ،ﻧﻪ ﻳﻚ ﻣﺘﺨﺼﺺ ﺍﻗﺘﺼﺎﺩ. ﺩﺭ ﺭﺷــﺘﻪﺍﻯ ﻣﺜــﻞ ﻧﺠــﻮﻡ ﻫﻢ ﺭﻳﺎﺿــﻰ ﻛﺎﺭﺑــﺮﺩ ﺩﺍﺭﺩ .ﺩﺭ ﻗﺪﻳﻢ ﺭﻳﺎﺿﻰﺩﺍﻥﻫــﺎ ﻣﻨﺠﻢ ﻫﻢ ﺑﻮﺩﻩﺍﻧﺪ .ﺍﻻﻥ ﻫﻢ ﺧﻴﻠﻰ ﺍﺯ ﻓﻴﺰﻳﻚﺩﺍﻥﻫﺎ ﻛﻪ ﻣﺪﺍﺭﺝ ﺑﺎﻻﺗﺮ ﺭﺍ ﺩﻧﺒﺎﻝ ﻣﻰﻛﻨﻨﺪ ﻭ ﻳﺎ ﻧﻈﺮﻳﻪﻫﺎﻯ ﺧﻴﻠﻰ ﻣﻬﻤﻰ ﺩﺭﻣﻮﺭﺩ ﺟﻬــﺎﻥ ﻣﻰﺩﻫﻨﺪ ،ﺭﻳﺎﺿــﻰ ﻣﻄﺎﻟﻌﻪ ﻣﻰﻛﻨﻨﺪ .ﺷــﺎﻳﺪ ﻛﻢﺗﺮﻳﻦ ﻛﺎﺭﺑﺮﺩ ﺭﻳﺎﺿﻰ ﺩﺭ ﺷــﻴﻤﻰ ﻭ ﺯﻳﺴﺖﺷﻨﺎﺳﻰ ﺑﺎﺷﺪ ﻛﻪ ﺩﺭ ﺣﺪ ﻣﻄﺎﻟﻌﺎﺕ ﺁﻣﺎﺭﻯ ﺍﺯ ﺁﻥ ﺍﺳــﺘﻔﺎﺩﻩ ﻣﻰﺷﻮﺩ ﺍﻣﺎ ﺩﺭ ﻫﺮ ﺣﺎﻝ ﺁﻥ ﻫﻢ ﺭﻳﺎﺿﻰ ﺍﺳﺖ .ﺑﻰﺧﻮﺩ ﻧﻴﺴﺖ ﻛﻪ ﺑﻪ ﺭﻳﺎﺿﻰ »ﻣﺎﺩﺭ ﻋﻠﻮﻡ« ﻣﻰﮔﻮﻳﻨﺪ.
ﺧﻮﺏ ﺧﻮﺍﻧﺪﻥ ﻣﻬﻢﺗﺮ ﺍﺯ ﺯﻳﺎﺩ ﺧﻮﺍﻧﺪﻥ ﺁﻳﺎ ﺍﺳـﺘﻔﺎﺩﻩ ﺍﺯ ﻛﺘﺎﺏﻫـﺎﻯ ﻛﻤﻚ ﺁﻣﻮﺯﺷـﻰ ﻣﻰﺗﻮﺍﻧﺪ ﺑﻪ ﺑﭽﻪﻫﺎ ﻛﻤﻚ ﻛﻨﺪ؟ ﻣﻦ ﺗﺄﻛﻴﺪﻯ ﺭﻭﻯ ﺍﺳــﺘﻔﺎﺩﻩﻯ ﺑﭽﻪﻫﺎ ﺍﺯ ﻛﺘﺎﺏﻫﺎﻯ ﻛﻤﻚ ﺩﺭﺳﻰ ﻧﺪﺍﺭﻡ .ﭼﻮﻥ ﺧﻴﻠﻰ ﻭﻗﺖﻫﺎ ﺷــﺎﮔﺮﺩﻯ ﻛــﻪ ﺩﺭ ﺗﻤﺎﻡ ﺩﻭﺭﻩﻯ ﺭﺍﻫﻨﻤﺎﻳﻰ ﻧﻤﺮﺍﺕ ﺭﻳﺎﺿﻰﺍﺵ 20ﺍﺳــﺖ ﻧﻤﻰﺗﻮﺍﻧﺪ ﻳﻚ ﻣﻌﻤﺎﻯ ﺳﺎﺩﻩﻯ ﺭﻳﺎﺿﻰ ﺭﺍ ﺣﻞ ﻛﻨﺪ ،ﻭﻟﻰ ﺷﺎﮔﺮﺩﻯ ﻛﻪ ﻧﻤﺮﺍﺕ ﺭﻳﺎﺿﻰ ﺧﻮﺑﻰ ﻧﻤﻰﮔﺮﻓﺘﻪ ،ﺑﺎ ﻛﻤﻰ ﻓﻜﺮ ﻛﺮﺩﻥ ﺁﻥ ﺭﺍ ﺣﻞ ﻣﻰﻛﻨﺪ .ﺑﻪ ﻧﻈﺮ ﻣﻦ ﺧﻮﺏ ﺧﻮﺍﻧﺪﻥ ﻳﻚ ﺩﺭﺱ، ﭼــﻪ ﺭﻳﺎﺿﻰ ﻭ ﭼــﻪ ﺗﺎﺭﻳﺦ ،ﺟﻐﺮﺍﻓﻴﺎ ﻭ ﻫــﺮ ﺩﺭﺱ ﺩﻳﮕﺮﻯ ﻳﻌﻨﻰ ﺍﻳﻨﻜﻪ ﻧﮕــﺮﺵ ﺁﻥ ﺩﺭﺱ ﺭﺍ ﭘﻴــﺪﺍ ﻛﻨﻴﻢ .ﺑﻔﻬﻤﻴﻢ ﻛﻪ ﺑﺎﻳــﺪ ﺩﺭ ﺁﻥ ﺩﺭﺱ ﭼﻪ 10
ﺭﺍﻫﻨﻤﺎﻳﻲ
ﺩﻭﺭﻩﻯ ﺷﺎﻧﺰﺩﻫﻢ ،ﺷﻤﺎﺭﻩﻯ ،2ﺯﻣﺴﺘﺎﻥ 1389
ﻣﻄﺎﻟﻌﻪﻯ ﺩﺍﺳﺘﺎﻥﻫﺎﻯ ﺭﻳﺎﺿﻰ ﻳﺎ ﻣﻌﻤﺎ ﺁﻥ ﺣﺎﻟﺖ ﺧﺸﻜﻰ ﺭﻳﺎﺿﻰ ﺭﺍ ﻛﻪ ﺩﺭ ﺑﻌﻀﻰ ﺍﺯ ﻛﻼﺱﻫﺎ ﻭﺟﻮﺩ ﺩﺍﺭﺩ ،ﺍﺯ ﺑﻴﻦ ﻣﻰﺑﺮﺩ ﻭ ﺑﻪ ﺑﭽﻪﻫﺎ ﻛﻤﻚ ﻣﻰﻛﻨﺪ ﺑﻪ ﺭﻳﺎﺿﻰ ﻭﺍﻗﻌﻰ ﻧﺰﺩﻳﻚ ﺷﻮﻧﺪ
ﺗﻮﺍﻧﺎﻳﻰﻫﺎﻳﻰ ﺭﺍ ﻛﺴــﺐ ﻛﻨﻴﻢ .ﺩﺭ ﺭﻳﺎﺿﻰ ﻣﻬﻢﺗﺮﻳﻦ ﺗﻮﺍﻧﺎﻳﻰ ﻓﻜﺮﻛﺮﺩﻥ ﺍﺳﺖ .ﺩﺍﻧﺶﺁﻣﻮﺯ ﺑﺎﻳﺪ ﺑﺘﻮﺍﻧﺪ ﺧﻮﺏ ﻓﻜﺮ ﻛﻨﺪ ،ﺧﻮﺏ ﺍﺭﺗﺒﺎﻁ ﺑﺮﻗﺮﺍﺭ ﻛﻨﺪ ﻭ ﺑﺘﻮﺍﻧﺪ ﺍﺯ ﭘﺲ ﻣﺴــﺌﻠﻪ ﺑﺮﺑﻴﺎﻳﺪ .ﻛﻪ ﺍﻳﻦ ﻣﺴﺌﻠﻪ ﻣﻰﺗﻮﺍﻧﺪ ﻳﻚ ﺍﻣﺘﺤﺎﻥ ﺭﻳﺎﺿــﻰ ،ﻳﻚ ﻣﻌﻤﺎﻯ ﺗﻔﺮﻳﺤﻰ ﻣﺜﻞ ﻳــﻚ ﻣﻌﻤﺎﻯ ﭼﻮﺏ ﻛﺒﺮﻳﺘﻰ ﻭ ﻳﺎ ﻳﻚ ﻣﺴﺌﻠﻪﻯ ﺷﻄﺮﻧﺞ ﺑﺎﺷﺪ. ﺍﻻﻥ ﭼﻴــﺰﻯ ﻛﻪ ﺧﻴﻠﻰ ﺑﺎﺏ ﺷــﺪﻩ ﺍﻳﻦ ﺍﺳــﺖ ﻛــﻪ ﺑﭽﻪﻫﺎﻳﻰ ﺩﺭ ﺭﻳﺎﺿــﻰ ﻣﻮﻓﻖﺗﺮ ﻫﺴــﺘﻨﺪ ﻛﻪ ﻛﺘﺎﺏﻫﺎﻯ ﻛﻤﻚﺁﻣﻮﺯﺷــﻰ ﺑﻴﺸــﺘﺮﻯ ﺑﺨﻮﺍﻧﻨﺪ .ﻣﻦ ﺧﻴﻠﻰ ﺑﻪ ﺍﻳﻦ ﻣﻮﺿﻮﻉ ﺍﻋﺘﻘﺎﺩ ﻧﺪﺍﺭﻡ ﻛﻪ ﺍﮔﺮ ﺩﺍﻧﺶﺁﻣﻮﺯﻯ ﺩﺭ ﻳﻚ ﻣﻮﺿﻮﻉ ﻳﺎ ﻣﻬﺎﺭﺕ ﺿﻌﻴﻒ ﺍﺳــﺖ ،ﻣﻤﻜﻦ ﺍﺳــﺖ ﺑﻪ ﺍﻳﻦ ﻋﻠﺖ ﺑﺎﺷﺪ ﻛﻪ ﺁﻥ ﻣﻮﺿﻮﻉ ﺭﺍ ﻧﻤﻰﻓﻬﻤﺪ ﻳﺎ ﺑﺎ ﺗﻤﺮﻳﻦ ﺑﻴﺸﺘﺮ ﻣﻬﺎﺭﺗﺶ ﺑﻴﺸﺘﺮ ﻣﻰﺷﻮﺩ. ﺑﻴﺸــﺘﺮ ﺣﻞ ﻛﺮﺩﻥ ﺑﻪ ﻣﻌﻨﻰ ﺑﻴﺸــﺘﺮ ﻳﺎﺩ ﮔﺮﻓﺘﻦ ﻧﻴﺴــﺖ .ﻣﻦ ﺑﻪ ﺧﺼﻮﺹ ﺍﻳﻦ ﻣﻄﺎﻟﺐ ﺭﺍ ﺑﺮﺍﻯ ﺧﺎﻧﻮﺍﺩﻩﻫﺎ ﻣﻰﮔﻮﻳﻢ ،ﭼﻮﻥ ﺧﻴﻠﻰ ﻭﻗﺖﻫﺎ ﻣﺎﺩﺭﻫﺎ ﺍﻳﻦ ﺭﺍ ﻣﻰﭘﺮﺳــﻨﺪ ﻛﻪ ﺑﭽﻪﻯ ﻣﻦ ﺑﺮﺍﻯ ﺗﺎﺑﺴﺘﺎﻥ ﭼﻪ ﻛﺘﺎﺑﻰ ﺭﺍ ﺑﺨﻮﺍﻧﺪ ﻭ ﻳﺎ ﺍﻻﻥ ﻛﻪ ﻣﺪﺍﺭﺱ ﺷﺮﻭﻉ ﺷﺪﻩ ﭼﻪ ﻛﺘﺎﺑﻰ ﺭﺍ ﻳﺮﺍﻯ ﺍﻭ ﺑﮕﻴﺮﻡ ﺗﺎ ﺩﺭ ﺭﻳﺎﺿﻰ ﻗﻮﻯﺗﺮ ﺷﻮﺩ .ﺗﻮﺻﻴﻪﻯ ﻣﻦ ﺍﻳﻦ ﺍﺳﺖ ﻛﻪ ﺍﺯ ﺑﭽﻪﻫﺎ ﺑﺨﻮﺍﻫﻨﺪ ﺳــﺮ ﻛﻼﺱ ﺑﻪ ﺑﺤﺚﻫﺎ ﻭ ﮔﻔﺖﻭ ﮔﻮﻫﺎ ﮔﻮﺵ ﺑﺪﻫﻨﺪ .ﺗﻜﻠﻴﻒ ﺭﺍ ﺑﺎ ﺩﻗﺖ ﺍﻧﺠﺎﻡ ﺩﻫﻨﺪ ﻭ ﺍﮔﺮ ﻭﻗﺘﻰ ﺗﻤﺮﻳﻦ ﺳــﺮ ﻛﻼﺱ ﺣﻞ ﻣﻰﺷــﻮﺩ ،ﺍﺷــﻜﺎﻟﻰ ﺩﺍﺭﻧﺪ ﺑﺒﻴﻨﻨﺪ ﺍﺷﻜﺎﻝﺷﺎﻥ ﻛﺠﺎ ﺑﻮﺩﻩ ﺍﺳﺖ. ﺑــﻪ ﻧﻈﺮ ﻣﻦ ﻛﻼﺱ ﺍﺿﺎﻓﻪ ﺭﻓﺘﻦ ﻭ ﻛﺘــﺎﺏ ﺧﻮﺍﻧﺪﻥ ﻫﻴﭻ ﺿﻤﺎﻧﺘﻰ ﺑﺮﺍﻯ ﻳﺎﺩ ﮔﺮﻓﺘﻦ ﺑﻬﺘﺮ ﺭﻳﺎﺿﻰ ﻧﺪﺍﺭﺩ .ﻣﻤﻜﻦ ﺍﺳــﺖ ﺩﺍﻧﺶﺁﻣﻮﺯ ﻧﻤﺮﻩﻯ ﺑﻬﺘﺮﻯ ﺑﮕﻴﺮﺩ ﭼﻮﻥ ﺑﻌﻀﻰ ﻣﺴــﺎﺋﻞ ﺭﺍ ﺁﻧﻘــﺪﺭ ﺗﻜﺮﺍﺭ ﻛﺮﺩﻩ ﻛﻪ ﻣﻠﻜﻪﻯ ﺫﻫﻨﺶ ﺷــﺪﻩ ﺍﺳﺖ ﺍﻣﺎ ﻳﻚ ﻣﺎﻩ ﺑﻌﺪ ﺍﺯ ﺍﻣﺘﺤﺎﻥ ﭼﻴﺰﻯ ﺍﺯ ﺩﺭﺱ ﻳﺎﺩﺵ ﻧﻤﻰﺁﻳﺪ .ﻣﻬﻢ ﺍﻳﻦ ﺍﺳــﺖ ﻛﻪ ﺑﭽﻪﻫﺎ ﺗﻔﻜﺮ ﺭﻳﺎﺿﻰ ﺭﺍ ﭘﻴﺪﺍ ﻛﻨﻨﺪ .ﺷﺎﻳﺪ ﻛﺘﺎﺑﻬــﺎﻯ ﻣﻌﻤﺎ ،ﺑﺎﺯﻯﮔﻮﻧﻪ ﻭ ﻛﺘﺎﺏ ﻗﺼﻪﻫﺎﻳﻰ ﻛﻪ ﭼﻴﺰﻫﺎﻳﻰ ﺍﺯ ﺭﻳﺎﺿﻰ ﺩﺭ ﺁﻧﻬﺎ ﮔﻔﺘﻪ ﺷــﺪﻩ )ﻛﻪ ﺑﺮﺍﻯ ﺑﭽﻪﻫﺎ ﺟﺎﻟﺐ ﺍﺳﺖ( ﺑﻴﺸﺘﺮ ﻛﻤﻚﺷﺎﻥ ﻛﻨﺪ ﺗﺎ ﺭﻳﺎﺿﻰ ﺭﺍ ﺑﻬﺘﺮ ﻳﺎﺩ ﺑﮕﻴﺮﻧﺪ ﺗﺎ ﺍﻳﻨﻜﻪ ﻓﻘﻂ ﺗﻤﺮﻳﻨﺎﺕ ﻛﻠﻴﺸــﻪﺍﻯ ﺭﻳﺎﺿــﻰ ﺭﺍ ﺣﻞ ﻛﻨﻨــﺪ .ﻭﺍﻟﺪﻳﻦ ﻓﻜﺮ ﻧﻜﻨﻨﺪ ﺑﺎ ﺧﺮﻳﺪ ﻛﺘﺎﺑﻬﺎﻯ ﺧﺎﺹ ﻭ ﻳــﺎ ﺑﺎ ﻛﻼﺱﻫﺎﻯ ﺧــﺎﺹ ﺭﻳﺎﺿﻰ ﺑﭽﻪﻫﺎ ﺑﻬﺘﺮ ﻣﻰﺷــﻮﺩ .ﻧﻪ! ﺍﻳﻦ ﻭﺍﻗﻌﺎً ﺿﺮﻭﺭﺗﻰ ﻧﺪﺍﺭﺩ. ﻣﻄﺎﻟﻌﻪﻯ ﺩﺍﺳﺘﺎﻥﻫﺎﻯ ﺭﻳﺎﺿﻰ ﻳﺎ ﻣﻌﻤﺎ ﺁﻥ ﺣﺎﻟﺖ ﺧﺸﻜﻰ ﺭﻳﺎﺿﻰ
ﺭﺍ ﻛــﻪ ﺩﺭ ﺑﻌﻀﻰ ﺍﺯ ﻛﻼﺱﻫﺎ ﻭﺟــﻮﺩ ﺩﺍﺭﺩ ،ﺍﺯ ﺑﻴﻦ ﻣﻰﺑﺮﺩ ﻭ ﺑﻪ ﺑﭽﻪﻫﺎ ﻛﻤﻚ ﻣﻰﻛﻨﺪ ﺑﻪ ﺭﻳﺎﺿﻰ ﻭﺍﻗﻌﻰ ﻧﺰﺩﻳﻚ ﺷﻮﻧﺪ.
ﺗﻤﺮﻛﺰ ﺩﺭ ﻛﻼﺱ ﺩﺭﺱ ﻣﻬﻢ ﺍﺳﺖ ﺗﻤﺮﻛـﺰ ﺩﺭ ﻛﻼﺱ ﺩﺭﺱ ،ﭼﻘﺪﺭ ﺩﺭ ﻳﺎﺩﮔﻴﺮﻯ ﺩﺭﺱ ﺭﻳﺎﺿﻰ ﻣﻬﻢ ﺍﺳﺖ؟ ﺧﻴﻠﻰ ﻣﻬﻢ ﺍﺳــﺖ .ﺗﻤﺮﻛﺰ ﺑﺎﻋﺚ ﻣﻰﺷــﻮﺩ ﻛﻪ ﺑﭽﻪﻫــﺎ ﺍﺭﺗﺒﺎﻃﺎﺕ ﺭﻳﺎﺿﻰ ﺭﺍ ﻛﻪ ﺩﺭ ﻛﻼﺱ ﻣﻄﺮﺡ ﻣﻰﺷــﻮﺩ )ﻭ ﺍﺣﻴﺎﻧﺎً ﺑﻪ ﺑﺤﺚ ﮔﺬﺍﺷــﺘﻪ ﻣﻰﺷــﻮﻧﺪ( ﺭﺍ ﺩﺭﻙ ﻛﻨﻨﺪ .ﺑﭽﻪﻫﺎ ﺍﮔﺮ ﺗﻤﺮﻛﺰ ﺩﺍﺷﺘﻪ ﺑﺎﺷﻨﺪ ﺑﺎ ﺷﻨﻴﺪﻥ ﺣﺮﻑﻫﺎﻯ ﻳﻜﺪﻳﮕﺮ ﻫﻢ ﺍﻳﺪﻩﻫﺎﻯ ﺯﻳﺎﺩﻯ ﻣﻰﮔﻴﺮﻧﺪ ﻭ ﺩﺭ ﻫﻤﺎﻥ ﺣﺮﻑﻫﺎ ﻫﻢ ﭼﻴﺰﻫــﺎﻯ ﺟﺪﻳﺪﻯ ﻳﺎﺩ ﻣﻰﮔﻴﺮﻧﺪ ﻭ ﺑــﺪﻭﻥ ﺍﻳﻨﻜﻪ ﻣﻄﺎﻟﻌﻪ ﺍﺿﺎﻓﻰ ﺩﺍﺷﺘﻪ ﺑﺎﺷــﻨﺪ ﺍﺯ ﺑﺤﺚ ﻛﺘﺎﺏ ﻓﺮﺍﺗﺮ ﻣﻰﺭﻭﻧﺪ .ﺣﺘﻰ ﺍﮔﺮ ﻣﺤﻴﻂ ﻛﻼﺱ ﻫﻢ ﻣﺤﻴﻂ ﺑﺤﺚ ﻧﺒﺎﺷــﺪ ﻭ ﻳﻚ ﻣﺤﻴﻂ ﺳــﻨﺘﻰ ﺑﺎﺷــﺪ ﻛﻪ ﻓﻘﻂ ﻣﻌﻠﻢ ﺻﺤﺒﺖ ﻣﻰﻛﻨﺪ ،ﺗﻤﺮﻛﺰ ﺑﺎﻋﺚ ﻣﻰﺷــﻮﺩ ﻛﻪ ﺩﺍﻧﺶﺁﻣﻮﺯ ﺩﺭﺻﺪ ﺯﻳﺎﺩﻯ ﺍﺯ ﻣﻄﺎﻟــﺐ ﺭﺍ ﻫﻤﺎﻧﺠﺎ ﻳﺎﺩ ﺑﮕﻴﺮﺩ .ﻳﻌﻨﻰ ﺯﺣﻤﺖ ﺧﻮﺩﺵ ﻛﻢ ﻣﻰﺷــﻮﺩ. ﺍﻧﺘﻈﺎﺭﺍﺕ ﻣﻌﻠﻢ ﺭﺍ ﺑﻬﺘﺮ ﻣﻰﻓﻬﻤﺪ ،ﻣﺘﻮﺟﻪ ﻣﻰﺷــﻮﺩ ﻣﻌﻠﻢ ﺑﻴﺸﺘﺮ ﺭﻭﻯ ﭼــﻪ ﻣﻄﺎﻟﺒــﻰ ﺗﺄﻛﻴﺪ ﺩﺍﺭﺩ .ﭼــﻪ ﭼﻴﺰﻫﺎﻳﻰ ﺑﺮﺍﻯ ﻣﻌﻠﻢ ﻣﻬﻢ ﺍﺳــﺖ ﻭ ﻣﻰﺗﻮﺍﻧﺪ ﺭﻭﻯ ﺁﻧﻬﺎ ﺗﻤﺮﻳﻦ ﺑﻴﺸﺘﺮﻯ ﺍﻧﺠﺎﻡ ﺩﻫﺪ ﻭ ﻣﻮﻓﻖﺗﺮ ﺷﻮﺩ. ﺩﺭﺱ ﺧﻮﺍﻧـﺪﻥ ﮔﺮﻭﻫﻰ ﺑﻪ ﻳﺎﺩﮔﻴـﺮﻯ ﺩﺭﺱ ﺭﻳﺎﺿﻰ ﻛﻤﻚ ﻣﻰﻛﻨﺪ؟ ﺍﻻﻥ ﻧﻈﺮﻳــﻪﺍﻯ ﻣﻄﺮﺡ ﺷــﺪﻩ ﺍﺳــﺖ ﻛــﻪ ﺑﭽﻪﻫــﺎ ﺩﺭ ﮔﺮﻭﻩﻫﺎﻯ ﻫﻢﺳﺎﻝ ﺧﻮﺩﺷﺎﻥ ﺧﻴﻠﻰ ﭼﻴﺰﻫﺎ ﻳﺎﺩ ﻣﻰﮔﻴﺮﻧﺪ ﻭﻟﻰ ﺍﮔﺮ ﺍﻳﻦ ﻫﻢﺳﺎﻻﻥ ﺗﻮﺍﻧﺎﻳﻰﻫﺎﻯ ﻣﺘﻔﺎﻭﺕ ﺩﺍﺷﺘﻪ ﺑﺎﺷﻨﺪ ﺑﻪ ﻫﻢ ﺑﻴﺸﺘﺮ ﻛﻤﻚ ﻣﻰﻛﻨﻨﺪ. ﺑﻌﻀﻰﻫــﺎ ﺩﺭ ﮔﺮﻭﻩ ﺗﻤﺮﻛﺰ ﺑﻬﺘﺮﻯ ﺩﺍﺭﻧﺪ .ﭼﻮﻥ ﻭﻗﺘﻰ ﺣﻮﺍﺳﺸــﺎﻥ ﭘﺮﺕ ﻣﻰﺷــﻮﺩ ،ﻫﻢﮔﺮﻭﻫﻰ ﺑﻪ ﺁﻧﻬﺎ ﺍﺷﺎﺭﻩﺍﻯ ﻣﻰﻛﻨﺪ ﻭ ﻣﺜ ً ﻼ ﻣﻰﮔﻮﻳﺪ ﺍﻳﻦ ﻣﻄﻠﺐ ﻣﻬﻢ ﺍﺳﺖ ﻭ ﺑﻪ ﺁﻥ ﺗﻮﺟﻪ ﻛﻦ .ﮔﺮﻭﻩ ﻭ ﻫﻢﺳﺎﻝ ﺑﻪ ﺍﻭ ﻛﻤﻚ ﻣﻰﻛﻨﺪ ﺧﻂ ﻣﺸﺨﺼﻰ ﺭﺍ ﺩﻧﺒﺎﻝ ﻛﻨﺪ ﻛﻪ ﺧﻮﺩﺵ ﺑﻪ ﺗﻨﻬﺎﻳﻰ ﻧﻤﻰﺗﻮﺍﻧﺪ ﺍﻳﻦ ﻛﺎﺭ ﺭﺍ ﺑﻜﻨﺪ .ﺑﺴــﺘﮕﻰ ﺑﻪ ﺁﺩﻣﺶ ﺩﺍﺭﺩ .ﺑﺮﺍﻯ ﻫﻤﻪ ﻧﻤﻰﺷــﻮﺩ ﻳﻚ ﻧﺴــﺨﻪ ﭘﻴﭽﻴﺪ ﻛﻪ ﮔﺮﻭﻫﻰ ﺩﺭﺱ ﺧﻮﺍﻧﺪﻥ ﺧﻮﺏ ﺍﺳﺖ ﻳﺎ ﺑﺪ .ﺑﻌﻀﻰﻫﺎ ﻧﻈﻢ ﺫﻫﻨﻰ ﺧﻮﺩﺷــﺎﻥ ﺭﺍ ﻧﻤﻰﺗﻮﺍﻧﻨﺪ ﺩﺭ ﮔﺮﻭﻩ ﺩﺍﺷــﺘﻪ ﺑﺎﺷﻨﺪ ﻭ ﻣﻤﻜﻦ ﺍﺳﺖ ﮔﺮﻭﻩ ﺭﺍ ﻫﻢ ﺍﺫﻳﺖ ﻛﻨﻨﺪ.
ﻫﺮ ﺗﻜﻠﻴﻒ ﺣﺘﻤ ًﺎ ﻫﺪﻓﻰ ﺩﺍﺭﺩ ﺗﻜﻠﻴﻒ ﻣﻨﺰﻝ ﻛﻪ ﻣﻌﻠﻢ ﺑﻪ ﺩﺍﻧﺶﺁﻣﻮﺯﺍﻥ ﻣﻰﺩﻫﺪ ﭼﻪ ﺗﺄﺛﻴﺮﻯ ﺑﺮ ﻣﻴﺰﺍﻥ ﻭ ﺳﺮﻋﺖ ﻳﺎﺩﮔﻴﺮﻯ ﺑﭽﻪﻫﺎ ﺩﺍﺭﺩ؟ ﻫــﺮ ﺗﻜﻠﻴﻔﻰ ﺣﺘﻤﺎً ﻫﺪﻓﻰ ﺩﺍﺭﺩ .ﺍﮔﺮ ﻫﺪﻑ ﺗﻜﻠﻴﻒ ﺍﻳﻦ ﺑﺎﺷــﺪ ﻛﻪ ﺭﻭﻯ ﻣﻮﺿﻮﻋــﻰ ﻓﻜﺮ ﻛﻨﻴﻢ ﻭ ﺩﺭ ﺟﻠﺴــﻪﻯ ﺑﻌﺪ ،ﺑــﺮﺍﻯ ﻛﻼﺱ ﺁﻣﺎﺩﻩ ﺑﺎﺷــﻴﻢ؛ ﺍﮔﺮ ﺩﺍﻧﺶﺁﻣﻮﺯ ﺑﺪﻭﻥ ﺁﻣﺎﺩﮔﻰ ﺳــﺮ ﻛﻼﺱ ﺑﻴﺎﻳﺪ ،ﺍﺯ ﺩﻭﺳﺘﺎﻥ ﺩﻳﮕﺮﺵ ﻋﻘﺐ ﻣﻰﻣﺎﻧﺪ ﻭ ﻧﻤﻰﺗﻮﺍﻧﺪ ﻣﻮﺿﻮﻉ ﺭﺍ ﺧﻮﺏ ﺷﺮﻭﻉ ﻛﻨﺪ.
ﺍﮔﺮ ﺗﻜﻠﻴﻒ ﺑﺮﺍﻯ ﺍﻳﻦ ﺑﺎﺷــﺪ ﻛﻪ ﺑﻪ ﻣﻮﺿﻮﻋﻰ ﻛﻪ ﻗﺒ ً ﻼ ﺩﺭﺱ ﺩﺍﺩﻩ ﺷــﺪﻩ ﺍﺳﺖ ،ﻣﺴــﻠﻂ ﺷــﻮﻳﻢ ،ﺍﮔﺮ ﺁﻥ ﺭﺍ ﺍﻧﺠﺎﻡ ﺩﻫﻴﻢ ﻳﺎ ﺑﺎ ﺩﻗﺖ ﺍﻧﺠﺎﻡ ﻧﺪﻫﻴﻢ ،ﺍﺭﺗﺒﺎﻁ ﺑﺎ ﻣﻮﺿﻮﻉ ﺑﻌﺪﻯ ﺭﺍ ﺍﺯ ﺩﺳﺖ ﻣﻰﺩﻫﻴﻢ. ﺍﮔﺮ ﺗﻜﻠﻴﻒ ﻣﻬﺎﺭﺗﻰ ﺍﺳــﺖ ﻛﻪ ﺳــﺮﻋﺖﻣﺎﻥ ﺭﺍ ﺩﺭ ﻣﻮﺿﻮﻉ ﺧﺎﺻﻰ ﺑﻴﺸﺘﺮ ﻣﻰﻛﻨﺪ ،ﺑﺎﺯ ﻫﻢ ﻧﻘﺶ ﺧﻮﺩﺵ ﺭﺍ ﺩﺍﺭﺩ. ﺩﺍﻧﺶﺁﻣـﻮﺯﺍﻥ ﺑﺎﻳﺪ ﺍﺯ ﻣﻌﻠﻤﺎﻥﺷـﺎﻥ ﭼﻪ ﺗﻮﻗﻌﺎﺗﻰ ﺩﺍﺷـﺘﻪ ﺑﺎﺷﻨﺪ؟ ﻣﻬﻤﺘﺮﻳــﻦ ﺗﻮﻗــﻊ ﺩﺍﻧﺶﺁﻣــﻮﺯﺍﻥ ﺍﺯ ﻣﻌﻠــﻢ ﺑﺎﻳــﺪ ﺍﻳﻦ ﺑﺎﺷــﺪ ﻛﻪ ﺗﻮﺍﻧﺎﻳﻰﻫﺎﻯ ﻣﺘﻔﺎﻭﺕﺷــﺎﻥ ﺩﺭ ﻛﻼﺱ ﺭﻳﺎﺿﻰ ﺩﻳﺪﻩ ﺷــﻮﺩ .ﺑﻪ ﺁﻧﻬﺎ ﺑﻬﺎ ﺩﺍﺩﻩ ﺷــﻮﺩ ﻭ ﺍﻳﻨﻜﻪ ﻣﻌﻠﻢ ﺑﻪ ﺩﺍﻧﺶﺁﻣــﻮﺯﺍﻥ ﻧﻘﺶ ﻣﻬﻤﻰ ﺩﺭ ﻳﺎﺩﮔﻴﺮﻯ ﺭﻳﺎﺿﻰ ﺑﺪﻫﺪ .ﺍﮔﺮ ﺑﭽﻪﻫﺎ ﺍﺣﺴﺎﺱ ﻛﻨﻨﺪ ﺧﻮﺩﺷﺎﻥ ﺩﺭ ﻳﺎﺩﮔﻴﺮﻯ ﺭﻳﺎﺿﻰ ﻧﻘﺶ ﺩﺍﺭﻧﺪ ،ﺧﻴﻠﻰ ﺑﻴﺸــﺘﺮ ﺁﻥ ﺭﺍ ﺩﻭﺳﺖ ﺧﻮﺍﻫﻨﺪ ﺩﺍﺷﺖ ﻭ ﻗﺪﺭﺵ ﺭﺍ ﺑﻴﺸــﺘﺮ ﻣﻰﺩﺍﻧﻨﺪ .ﺍﻣﺎ ﺍﮔﺮ ﻣﻌﻠﻢ ﻫﻤﻪﻯ ﺯﺣﻤﺖﻫﺎ ﺭﺍ ﺑﻜﺸﺪ ﻭ ﺑﭽﻪﻫﺎ ﺑﻪ ﺭﻳﺎﺿﻰ ﺑﻪ ﻋﻨﻮﺍﻥ ﻣﻮﺿﻮﻋﻰ ﻧﮕﺎﻩ ﻛﻨﻨﺪ ﻛﻪ ﻛﺎﺭﺑﺮﺩ ﻧﺪﺍﺭﺩ ،ﺳﺨﺖ ﺍﺳﺖ، ﻛﺎﺭﺵ ﺯﻳﺎﺩ ﺍﺳﺖ ﻭ ﻣﺴﺌﻠﻪﺍﺵ ﻫﻢ ﻗﺎﺑﻞ ﺣﻞ ﻧﻴﺴﺖ؛ ﺗﻼﺷﻰ ﻫﻢ ﺑﺮﺍﻯ ﺁﻥ ﻧﻤﻰﻛﻨﻨﺪ .ﻳﻌﻨﻰ ﺍﺻ ً ﻼ ﺍﻧﮕﻴﺰﻩﺍﻯ ﻫﻢ ﺑﺮﺍﻯ ﺗﻼﺵ ﺑﻴﺸﺘﺮ ﻧﺪﺍﺭﻧﺪ. ﺗﺠﺮﺑﻪﻯ ﺧﻮﺩﺗﺎﻥ ﺭﺍ ﺩﺭ ﻳﺎﺩﮔﻴﺮﻯ ﺭﻳﺎﺿﻰ ﺩﺭ ﺩﻭﺭﻩ ﺭﺍﻫﻨﻤﺎﻳﻰ ﺑﻪ ﻳﺎﺩ ﺩﺍﺭﻳﺪ؟ ﭼﻄﻮﺭ ﺩﺭﺱ ﻣﻰﺧﻮﺍﻧﺪﻳﺪ؟ ﻧﻪ .ﭼﻮﻥ ﻣﺮﺑﻮﻁ ﺑﻪ ﺧﻴﻠﻰ ﻭﻗﺖ ﭘﻴﺶ ﺍﺳــﺖ ،ﺯﻳﺎﺩ ﻳﺎﺩﻡ ﻧﻤﻰﺁﻳﺪ، ﻓﻘﻂ ﻳﺎﺩﻡ ﻣﻰﺁﻳﺪ ﻛﻪ ﻣﻌﻠﻢ ﺧﻴﻠﻰ ﺧﻮﺑﻰ ﺩﺍﺷــﺘﻢ .ﻣﻦ ﺍﺻﻮﻻً ﺍﺯ ﻫﻤﺎﻥ ﺍﻭﻝ ﺩﺭﺱﺧــﻮﺍﻥ ﺑﻮﺩﻡ .ﻭ ﻛﺎﺭﻯ ﺑﻪ ﺟﺰ ﺩﺭﺱ ﺧﻮﺍﻧﺪﻥ ﻧﺪﺍﺷــﺘﻢ .ﻭﻗﺘﻰ ﺑﻪ ﺧﺎﻧﻪ ﻣﻰﺭﺳﻴﺪﻡ ﻛﻴﻒ ﻭ ﻛﺘﺎﺑﻢ ﺭﺍ ﻣﻰﮔﺬﺍﺷﺘﻢ .ﻛﺘﺎﺏﻫﺎ ﺭﺍ ﺭﻭﻯ ﻫﻢ ﻣﻰﭼﻴﺪﻡ .ﺑﺮﺟﻰ ﺍﺯ ﻛﺘﺎﺏ ﻭ ﺩﻓﺘﺮ ﺩﺭﺳــﺖ ﻣﻰﺷﺪ ﻭ ﻣﻦ ﻣﻰﻧﺸﺴﺘﻢ ﻭ ﻣﺸــﻖﻫﺎﻯ ﻓﺎﺭﺳﻰ ،ﺭﻳﺎﺿﻰ ﻭ ﻋﻠﻮﻡ ﺭﺍ ﻣﻰﻧﻮﺷﺘﻢ .ﺑﺮﺍﻯ ﺟﻐﺮﺍﻓﻰ ﻧﻘﺸﻪ ﻣﻰﻛﺸــﻴﺪﻡ .ﻛﻠﻰ ﻛﺎﺭ ﺑﺮﺍﻯ ﺍﻧﺠﺎﻡ ﺩﺍﺩﻥ ﺩﺍﺷﺘﻢ .ﺗﻜﻠﻴﻒ ﻫﺮ ﺩﺭﺳﻰ ﺭﺍ ﻫﻢ ﻫﻤﺎﻥ ﺭﻭﺯﻯ ﻛﻪ ﺑﻪ ﻣﻦ ﻣﻰﺩﺍﺩﻧﺪ ﺍﻧﺠﺎﻡ ﻣﻰﺩﺍﺩﻡ .ﻣﻦ ﻓﻜﺮ ﻣﻰﻛﻨﻢ ﺧﻴﻠﻰ ﻣﻬﻢ ﺍﺳــﺖ ﻛﻪ ﺗﻜﻠﻴﻒ ﺩﺭﺱ ﺩﺭ ﻫﻤﺎﻥ ﺭﻭﺯ ﺍﻧﺠﺎﻡ ﺷــﻮﺩ .ﭼﻮﻥ ﺍﮔﺮ ﻫﻤﺎﻥ ﺭﻭﺯ ﻧﮕﺎﻫﻰ ﺑﻪ ﺩﺭﺱ ﺍﻧﺪﺍﺧﺘﻪ ﺷــﻮﺩ ﻭ ﺗﻜﻠﻴﻒ ﺍﻧﺠﺎﻡ ﺷــﻮﺩ، ﻣﻄﻠــﺐ ﺑﻬﺘﺮ ﺩﺭ ﺫﻫﻦ ﺑﺎﻗﻰ ﻣﻰﻣﺎﻧﺪ .ﭼﻮﻥ ﺩﺍﻧﺶﺁﻣﻮﺯ ﺍﺯ ﺩﺭﺱ ﻓﺎﺻﻠﻪ ﻧﮕﺮﻓﺘﻪ ،ﺣﺮﻑﻫﺎﻯ ﻣﻄﺮﺡ ﺷــﺪﻩ ﺩﺭ ﻛﻼﺱ ﺭﺍ ﻫﻨﻮﺯ ﻳﺎﺩﺵ ﺍﺳﺖ ﻭ ﺁﻥ ﺍﺭﺗﺒــﺎﻁ ﻻﺯﻡ ﺑﻬﺘﺮ ﺩﺭ ﺫﻫﻨﺶ ﺑﺮﻗﺮﺍﺭ ﻣﻰﺷــﻮﺩ ﻭﻟﻰ ﺍﮔﺮ ﺩﺭﺱ ﺭﺍ ﺑﺮﺍﻯ ﺩﻭ ﺳــﻪ ﺭﻭﺯ ﺑﻌﺪ ﺑﮕﺬﺍﺭﺩ ﻣﻤﻜﻦ ﺍﺳﺖ ﺷــﺒﻰ ﻛﻪ ﻣﻰﺧﻮﺍﻫﺪ ﺗﻜﺎﻟﻴﻔﺶ ﺭﺍ ﺍﻧﺠﺎﻡ ﺩﻫﺪ ،ﺑﻌﻀﻰ ﻣﻄﺎﻟﺐ ﺭﺍ ﻓﺮﺍﻣﻮﺵ ﻛﺮﺩﻩ ﺑﺎﺷــﺪ ﭼﻮﻥ ﺣﺎﻓﻈﻪﻯ ﻫﺮ ﻛﺴــﻰ ﮔﻨﺠﺎﻳــﺶ ﺧﺎﺹ ﺧــﻮﺩﺵ ﺭﺍ ﺩﺍﺭﺩ ﻭ ﺩﺭﺱ ﺭﺍ ﻛﻪ ﺭﻭﺯ ﺍﻭﻝ ﺑﻪ ﺭﺍﺣﺘﻰ ﻣﻰﺷــﺪ ﻳــﺎﺩ ﮔﺮﻓﺖ ،ﺩﻳﮕﺮ ﻳﺎﺩ ﻧﻤﻰﮔﻴﺮﺩ ﻳﺎ ﻣﺠﺒﻮﺭ ﺍﺳــﺖ ﺯﺣﻤﺖ ﺑﻴﺸــﺘﺮﻯ ﺑﺮﺍﻳﺶ ﺑﻜﺸــﺪ ﻭ ﻣﻤﻜﻦ ﺍﺳــﺖ ﺍﻳﻦ ﻣﺸﻜﻞ ﺑﺮﺍﻳﺶ ﺑﺎﻗﻰ ﺑﻤﺎﻧﺪ. ﺷــﺎﻳﺪ ﻣﻬﻤﺘﺮﻳﻦ ﻋﺎﻣﻞ ﻣﻮﻓﻘﻴﺖ ﻣــﻦ ﺩﺭ ﺩﺭﺱﻫﺎﻳﻢ ﺑﻌﺪ ﺍﺯ ﻟﻄﻒ ﺧﺪﺍﻭﻧﺪ ﺍﻳﻦ ﺑﻮﺩ ﻛﻪ ﻫﺮ ﺗﻜﻠﻴﻔﻰ ﺭﺍ ﺩﺭ ﻫﻤﺎﻥ ﺭﻭﺯ ﺍﻧﺠﺎﻡ ﻣﻰﺩﺍﺩﻡ. ﺩﻭﺭﻩﻯ ﺷﺎﻧﺰﺩﻫﻢ ،ﺷﻤﺎﺭﻩ ﻯ ،2ﺯﻣﺴﺘﺎﻥ 1389
ﺭﺍﻫﻨﻤﺎﻳﻲ
11
ﺭﻳﺎﺿﻲ ﻭ ﻛﺎﺭﺑﺮﺩ ﺁﻥ
ﺳﻜﻴﻨﻪ ﺑﻤﺎﻧﻴﺎﻥ
ﻛﻠﻴﺪﻭﺍژﻩﻫﺎ :ﺩﺭﺻﺪ ﻋﻨﺎﺻﺮ ،ﻛﺎﺭﺑﺮﺩ ﺭﻳﺎﺿﻴﺎﺕ ،ﻓﻮﻻﺩ . ﻫﻤﻪﻯ ﺷﻤﺎ ﺑﺎ ﻧﺎﻡ ﻓﻮﻻﺩ ﺁﺷﻨﺎ ﻫﺴﺘﻴﺪ ﻭ ﺗﻘﺮﻳﺒﺎً ﺩﺭﺑﺎﺭﻩﻯ ﻛﺎﺭﺑﺮﺩ ﺁﻥ ﺍﻃﻼﻋﺎﺗﻰ ﺩﺍﺭﻳﺪ .ﺁﻳﺎ ﻫﻤــﻪﻯ ﻓﻮﻻﺩﻫﺎﻯ ﻛﺎﺭﺑﺮﺩﻯ ﺩﺭ ﺟﺎﻫﺎﻯ ﻣﺨﺘﻠﻒ ﻳﻚ ﻧﻮﻉ ﺳﺎﺧﺘﺎﺭ ﻣﻮﻟﻜﻮﻟﻰ ﺩﺍﺭﺩ؟ ﺑﻪ ﻧﻈﺮ ﺷﻤﺎ ﺁﻳﺎ ﺭﺍﺑﻄﻪﺍﻯ ﺑﻴﻦ ﺩﺭﺻﺪ ﻭ ﻋﻨﺎﺻﺮ ﺗﺸﻜﻴﻞﺩﻫﻨﺪﻩﻯ ﻓﻮﻻﺩ ﻭﺟﻮﺩ ﺩﺍﺭﺩ؟ ﺩﺭ ﺯﻣﺎﻥ ﺍﻧﻘﻼﺏ ﺻﻨﻌﺘﻰ ﻗﺮﻥ ﻧﻮﺯﺩﻫﻢ ﺍﺯ ﻓﻮﻻﺩ ﺩﺭ ﻣﻘﻴﺎﺱ ﻭﺳــﻴﻊ ﺑﺮﺍﻯ ﺳﺎﺧﺘﻦ ﭘُﻞ ،ﺁﺳــﻤﺎﻥﺧﺮﺍﺵ ،ﻫﻮﺍﭘﻴﻤﺎ ،ﺧﻮﺩﺭﻭ ،ﻛﺸﺘﻰ ﻭ ﻗﻮﻃﻰ ﻛﻨﺴــﺮﻭ ﺍﺳﺘﻔﺎﺩﻩ ﺷﺪﻩ ﺍﺳــﺖ .ﺍ ّﻣﺎ ﺩﺭ 15ﺳﺎﻝ ﮔﺬﺷﺘﻪ ﺗﻮﺳﻌﻪﻯ ﻣﻮﺍﺩ ﺟﺪﻳﺪﻯ ﻛﻪ ﺩﺭ ﺁﻥﻫﺎ ﺳﺒﻜﻰ ﻭﺯﻥ ﻭ ﺍﺳﺘﺤﻜﺎﻡ ﺑﺎ ﻫﻢ ﺩﺭﺁﻣﻴﺨﺘﻪ ﺍﺳﺖ، ﺳــﺎﺯﻧﺪﮔﺎﻥ ﻓﻮﻻﺩ ﺭﺍ ﻧﺎﭼﺎﺭ ﻛﺮﺩﻩ ﺍﺳــﺖ ﺗﺎ ﻓﻮﻻﺩ ﺗﻮﻟﻴــﺪﻯ ﺧﻮﺩ ﺭﺍ ﺑﻪ ﮔﻮﻧﻪﺍﻯ ﻃﺮﺍﺣﻰ ﻛﻨﻨﺪ ﻛﻪ ﺧﻮﺍﺹ ﻣﻄﻠﻮﺑﻰ ﺑﺮﺍﻯ ﺍﻳﻦ ﻗﺒﻴﻞ ﻛﺎﺭﺑﺮﺩﻫﺎﻯ ﺻﻨﻌﺘﻰ ﺩﺍﺷﺘﻪ ﺑﺎﺷﺪ .ﺍﺯ ﺍﻳﻦ ﺭﻭ ﺣﺪﻭﺩ ٪70ﻓﻮﻻﺩﻯ ﻛﻪ ﺍﻣﺮﻭﺯﻩ ﻭﺟﻮﺩ ﺩﺍﺭﺩ ،ﺣﺘﻰ ﺩﺭ 10ﺳﺎﻝ ﮔﺬﺷﺘﻪ ﻧﻴﺰ ﻭﺟﻮﺩ ﻧﺪﺍﺷﺘﻪ ﺍﺳﺖ.
ﺗﺮﻛﻴﺐ ﻓﻮﻻﺩ ﺍﺻﻞ ﻓــﻮﻻﺩ ﻣﺨﻠﻮﻃﻰ )ﺁﻟﻴﺎژﻯ( ﺍﺯ ﺁﻫﻦ ﻭ ﻛﺮﺑﻦ ﺍﺳــﺖ ﻛﻪ ﺍﻟﺒﺘﻪ ﻣﻘﺎﺩﻳﺮ ﻛﻤﻰ ﺍﺯ ﭼﻨﺪ ﻋﻨﺼﺮ ﺩﻳﮕﺮ ﭼﻮﻥ P، S، Mn، Siﻧﻴﺰ ﺑﻪ ﻫﻤﺮﺍﻩ ﺩﺍﺭﺩ .ﺧﻮﺍﺹ ﻓﻴﺰﻳﻜﻰ ﻓﻮﻻﺩ ،ﻣﺎﻧﻨﺪ ﺳــﺨﺘﻰ ،ﻗﺎﺑﻠﻴﺖ ﻣﻔﺘﻮﻝ ﺷــﺪﻥ ﻭ ﺍﺳــﺘﺤﻜﺎﻡ ﺑﻪ ﻧﺴــﺒﺖ ﻣﻘﺪﺍﺭ ﻛﺮﺑﻦ ﻣﻮﺟﻮﺩ ﺩﺭ ﺁﻥ ﺗﻌﻴﻴﻦ ﻣﻰﺷﻮﺩ .ﺩﺭ ﺩﻣﺎﻫﺎﻯ ﺑﺎﻻ ﻛﻪ ﺑﺮﺍﻯ ﺗﻮﻟﻴﺪ ﻓﻮﻻﺩ ﻣﺎﻳﻊ ﺿﺮﻭﺭﻯ ﺍﺳﺖ ،ﻛﺮﺑﻦ ﺩﺭ ﻓﻮﻻﺩ ﻣﺎﻳﻊ ﺣﻞ ﻣﻰﺷــﻮﺩ ﻳﺎ ﺑﺎ ﺍﺗﻢﻫﺎﻯ ﺁﻫﻦ ﺗﺮﻛﻴﺐ ﻣﻰﺷــﻮﺩ ﻭ ﺁﻫﻦ ﻛﺮﺑﻴﺪ ﺗﺸﻜﻴﻞ ﻣﻰﺩﻫﺪ )ﻛﺮﺑﻴﺪﻫﺎ ﻣﻮﺍﺩ ﺑﺴﻴﺎﺭ ﺳﺨﺘﻰ ﻫﺴﺘﻨﺪ(. ﻫﺮﮔﻮﻧــﻪ ﺍﺧﻼﻟــﻰ ﺩﺭ ﺁﺭﺍﻳــﺶ ﻣﻨﻈــﻢ ﺍﺗﻢﻫﺎ ،ﺑﻪ ﻃــﻮﺭﻯ ﻛﻪ ﺩﺭ ﺷــﻜﻞ ﺭﻭﺑﻪﺭﻭ ﻧﺸــﺎﻥ ﺩﺍﺩﻩ ﺷﺪﻩ ﺍﺳــﺖ ،ﻣﺎﻧﻊ ﺍﺯ ﻟﻐﺰﺵ ﺍﺗﻢﻫﺎﻯ ﺁﻫﻦ ﺍﺯ ﻛﻨــﺎﺭ ﻳﻜﺪﻳﮕﺮ ﻣﻰﺷــﻮﺩ ﻭ ﺑﻪ ﺍﻳﻦ ﺗﺮﺗﻴﺐ ﺍﺳــﺘﺤﻜﺎﻡ ﻓﻮﻻﺩ ﺍﻓﺰﺍﻳﺶ ﻣﻰﻳﺎﺑﺪ .ﻣﻮﺍﺭﺩ ﺯﻳﺮ ﺗﻮﺟﻪ ﺷﻤﺎ ﺭﺍ ﺑﻪ ﻛﺎﺭﺑﺮﺩ ﺩﺭﺻﺪ ﺩﺭﺍﻧﻮﺍﻉ ﻓﻮﻻﺩ ﺟﻠﺐ ﻣﻰﻛﻨﺪ. ـ ﻓــﻮﻻﺩ ﻛــﻢ ﻛﺮﺑﻦ ﻛﻪ ٪15ﺗــﺎ ٪30ﺩﺭﺻﺪ ﻛﺮﺑــﻦ ﺩﺍﺭﺩ؛ ﺍﻳﻦ 12
ﺭﺍﻫﻨﻤﺎﻳﻲ
ﺩﻭﺭﻩﻯ ﺷﺎﻧﺰﺩﻫﻢ ،ﺷﻤﺎﺭﻩﻯ ،2ﺯﻣﺴﺘﺎﻥ 1389
ﻓــﻮﻻﺩ ﻣﺤﻜﻢ ﺍﺳــﺖ ﻭ ﻗﺎﺑﻠﻴﺖ ﻣﻔﺘــﻮﻝ ﺷــﺪﻥ ﺭﺍ ﺩﺍﺭﺩ ﻭ ﺩﺭ ﻛﺎﺭﻫﺎﻯ ﺳــﺎﺧﺘﻤﺎﻧﻰ ﻣﺎﻧﻨﺪ ﺁﺳﻤﺎﻥﺧﺮﺍﺵ ،ﺳﻮﻟﻪﻯ ﻓﺮﻭﺷﮕﺎﻩﻫﺎ ﻭ ﻛﺸﺘﻰﺳﺎﺯﻯ ﺑﻪ ﻛﺎﺭ ﻣﻰﺭﻭﺩ ،ﺯﻳﺮﺍ ﭼﻨﻴﻦ ﻓﻮﻻﺩﻯ ﺭﺍ ﻣﻰﺗﻮﺍﻥ ﺑﻪ ﺁﺳﺎﻧﻰ ﺟﻮﺵ ﺩﺍﺩ. ـ ﻓــﻮﻻﺩ ﻧﻴﻤﭽﻪ ﻛﺮﺑﻦ ﻛﻪ ٪30ﺗﺎ ٪60ﺩﺭﺻﺪ ﻛﺮﺑﻦ ﺩﺍﺭﺩ؛ ﻛﻤﻰ ﺳﺨﺖﺗﺮ ﻭ ﻣﺤﻜﻢﺗﺮ ﺍﺳﺖ ﻭ ﺩﺭ ﺳﺎﺧﺘﻦ ﺭﻳﻞ ﺭﺍﻩﺁﻫﻦ ﻭ ﺩﻧﺪﻩﻫﺎ ﻭ ﻣﺤﻮﺭ ﺧﻮﺩﺭﻭﻫﺎ ﻛﺎﺭﺑﺮﺩ ﺩﺍﺭﺩ. ـ ﻓــﻮﻻﺩ ﭘُﺮ ﻛﺮﺑﻦ ﻛــﻪ ٪60ﺗﺎ ٪90ﺩﺭﺻﺪ ﻛﺮﺑﻦ ﺩﺍﺭﺩ؛ ﺑﺴــﻴﺎﺭ ﺳــﺨﺖ ﻭ ﻣﺤﻜﻢ ﺍﺳــﺖ ﻭ ﺍﺯ ﺁﻥ ﺑﺮﺍﻯ ﺳــﺎﺧﺘﻦ ﺍﺑﺰﺍﺭﻯ ﻣﺎﻧﻨﺪ ﭼﻜﺶ، ﭘﻴﭻﮔﺸﺎ )ﭘﻴﭻﮔﻮﺷﺘﻰ( ﻭ ﺑﻌﻀﻰ ﺍﺯ ﺗﻴﻐﻪﻫﺎ ﺍﺳﺘﻔﺎﺩﻩ ﻣﻰﺷﻮﺩ. ـ ﻓــﻮﻻﺩﻯ ﻛﻪ ﻣﻘﺪﺍﺭ ﻛﺮﺑﻦ ﺁﻥ ٪90ﺗﺎ 1/15ﺩﺭﺻﺪ ﺍﺳــﺖ؛ ﺍﻳﻦ ﻓﻮﻻﺩ ﺑﺮﺍﻯ ﻣﻮﺍﺭﺩ ﻭﻳﮋﻩﺍﻯ ﻛﻪ ﺑﻪ ﺳــﺨﺘﻰ ﺯﻳﺎﺩ ﻧﻴﺎﺯ ﺍﺳﺖ ﻣﺎﻧﻨﺪ ﺑﺪﻧﻪﻯ ﻣ ّﺘﻪ ﺑﻪ ﻛﺎﺭ ﻣﻰﺭﻭﺩ. ﻣﻌﻴﻦ ﺩﺭ ﻓــﻮﻻﺩ ﺑﻪ ﺍﻳﻦ ﺁﻟﻴﺎژ، ﻋﻨﺼﺮﻫﺎﻯ ﻣﺨﺘﻠــﻒ ﺑﺎ ﺍﻧﺪﺍﺯﻩﻫﺎﻯ ّ ﺧﻮﺍﺹ ﻣﺘﻔﺎﻭﺗﻰ ﻣﻰﺑﺨﺸــﻨﺪ .ﻛﺎﺭ ﻓﻮﻻﺩﺳــﺎﺯﻯ ﺩﺭ ﻭﺍﻗﻊ ﺍﺿﺎﻓﻪ ﻛﺮﺩﻥ ﻋﻨﺼﺮﻫﺎﻳــﻰ ﻣﺎﻧﻨــﺪ Ni ،Cr ،Mn ،Si، Cﺑﺎ ﺩﺭﺻﺪﻫﺎﻯ ﻣﺨﺘﻠﻒ ﺑﻪ ﺧﻮﺍﺹ ﻓﻴﺰﻳﻜﻰ ﻣﺎﻧﻨﺪ ﺍﺳﺘﺤﻜﺎﻡ ﺯﻳﺎﺩ ،ﺳﺨﺘﻰ ﻓﻮﻻﺩ ﺍﺳــﺖ ﺗﺎ ﺑﻬﺘﺮﻳﻦ ّ ﺯﻳﺎﺩ ،ﻳﺎ ﻣﻘﺎﻭﻣﺖ ﺩﺭ ﺑﺮﺍﺑﺮ ﺧﻮﺭﺩﮔﻰ ﺑﻪ ﺩﺳﺖ ﺁﻳﺪ. ـ ﺑﺎ ﺍﺿﺎﻓﻪ ﻛﺮﺩﻥ Mnﺑﻪ ﻓﻮﻻﺩ ،ﺳﺨﺘﻰ ﻣﺤﺼﻮﻝ ﺍﻓﺰﺍﻳﺶ ﻣﻰﻳﺎﺑﺪ. ﺯﻳــﺮﺍ Mnﺑــﺎ ،Cﻛﺮﺑﻴﺪ ﭘﺎﻳــﺪﺍﺭﻯ ﻣﻰﺩﻫﺪ Mn(Mn3c) .ﺑﻪ ﻃﻮﺭ ﺩﺭﻭﻥ ﺷﺒﻜﻪﺍﻯ ﺩﺭ ﺁﻫﻦ ﺣﻞ ﻣﻰﺷﻮﺩ ﻭ ﺩﺭ ﻧﺘﻴﺠﻪ ﻓﻮﻻﺩﻯ ﺑﺎ ﺍﺳﺘﺤﻜﺎﻡ ﺑﻴﺶﺗــﺮ ﺑﻪ ﺩﺳــﺖ ﻣﻰﺁﻳﺪ .ﺍﻓﺰﻭﻥ ﺑﺮ ﺍﻳﻦ Mn ،ﺑــﻪ ﺭﺍﺣﺘﻰ ﺑﺎ ﮔﻮﮔﺮﺩ ﻣﻮﺟــﻮﺩ ﺩﺭ ﺁﻫﻦ ﻣﺬﺍﺏ ﺗﺮﻛﻴﺐ ﻣﻰﺷــﻮﺩ ﻭ ﻣﻨﮕﺰ ﺳــﻮﻟﻔﻴﺪ )(MnS ﻣﺨﺮﺑﻰ ﺩﺍﺭﺩ ،ﺯﻳﺮﺍ ﺗﻮﻟﻴﺪ ﺁﻫﻦ ﺳﻮﻟﻔﻴﺪ ﻣﻰﺩﻫﺪ .ﮔﻮﮔﺮﺩ ﺩﺭ ﻓﻮﻻﺩ ﻧﻘﺶ ّ ﻣﻰﻛﻨﺪ ﻛﻪ ﺑﻪ ﺻﻮﺭﺕ ﻻﻳﻪﻯ ﻧﺎﺯﻛــﻰ ﺍﻃﺮﺍﻑ ﺑﻠﻮﺭﻫﺎﻯ ﻓﻮﻻﺩ ﺟﺎﻣﺪ ﺭﺍ
ﻫﻤﺮﺍﻩ ﺑﺎ ﻛﺘﺎﺏ
︋︡ون در︮ــــ︡ ا﹞﹊︀ن ﹡︡ارد!
ﻣﻰﮔﻴﺮﺩ ﻭ ﺳﺒﺐ ﻣﻰﺷــﻮﺩ ﻓﻮﻻﺩ ﻫﻨﮕﺎﻡ ﻗﺎﻟﺐﮔﻴﺮﻯ ﺗﺮﻙﻫﺎﻯ ﻣﻮﻳﻰ ﺑﺮﺩﺍﺭﺩ .ﺑﺮﻋﻜﺲ ﻭﺟﻮﺩ MnSﺳﺒﺐ ﻣﻰﺷﻮﺩ ﻛﻪ ﻣﺎﺷﻴﻦ ﻛﺎﺭﻯ ﻓﻮﻻﺩ ﺑﻪ ﺭﺍﺣﺘﻰ ﺍﻧﺠﺎﻡ ﺷﻮﺩ. ـ ﺑــﺎ ﺍﺿﺎﻓــﻪ ﻛﺮﺩﻥ ،Crﻓــﻮﻻﺩ ﺯﻧﮓ ﻧﺰﻥ)ﺿﺪ ﺯﻧﮓ( ﺑﻪ ﺩﺳــﺖ ﺟﺮﺍﺣﻰ ﺍﺳــﺘﻔﺎﺩﻩ ﻣﻰﺁﻳﺪ ﻛﻪ ﺩﺭ ﺳــﺎﺧﺘﻦ ﻛﺎﺭﺩ ﻭ ﭼﻨﮕﺎﻝ ﻭ ﻭﺳــﺎﻳﻞ ّ ﻣﻰﺷــﻮﺩ .ﻛﺮﻭﻡ ﺑﺎ ﺗﺸﻜﻴﻞ ﻳﻚ ﻻﻳﻪﻯ ﺑﺴــﻴﺎﺭ ﻧﺰﻙ ﻭ ﭘﺎﻳﺪﺍﺭ ﺍﻛﺴﻴﺪ، ﻣﺎﻧــﻊ ﺧﻮﺭﺩﮔﻰ ﻣﻰﺷــﻮﺩ .ﺩﺭ ﻓﻮﻻﺩ ﺯﻧﮓ ﻧﺰﻥ ﻣﻘــﺪﺍﺭ ﻛﺮﻭﻡ ﻣﻰﺗﻮﺍﻧﺪ ﺗﺎ ﻣﺮﺯ ٪20ﺑﺮﺳــﺪ .ﻛﺮﻭﻡ ﻫﻢﭼﻨﻴﻦ ﺑﺎ ﻛﺮﺑﻦ ﻛﺮﺑﻴﺪ ﭘﺎﻳﺪﺍﺭ )(Cr3C2 ﺗﻮﻟﻴﺪ ﻣﻰﻛﻨﺪ ﻭ ﻓﻮﻻﺩ ﺯﻧﮓ ﻧﺰﻥ ﺑﺴﻴﺎﺭ ﺳﺨﺘﻰ ﺑﻪ ﻭﺟﻮﺩ ﻣﻰﺁﻭﺭﺩ. ـ ﺑــﺎ ﺍﺿﺎﻓــﻪ ﻛــﺮﺩﻥ Niﺑﻪ ﻓﻮﻻﺩ ﻛــﻢ ﻛﺮﺑﻦ ﻳﺎ ﻧﻴﻤﭽــﻪ ﻛﺮﺑﻦ، ﺍﺳﺘﺤﻜﺎﻡ ﻛﺸﺸــﻰ ﻓﻮﻻﺩ ﺍﻓﺰﺍﻳﺶ ﻣﻰﻳﺎﺑﺪ ،ﺩﺭ ﺣﺎﻟﻰﻛﻪ ﻓﻮﻻﺩ ﻗﺎﺑﻠﻴﺖ ﻣﻔﺘﻮﻝ ﺷــﺪﻥ ﻧﺴــﺒﻰ ﺧﻮﺩ ﺭﺍ ﺣﻔﻆ ﻣﻰﻛﻨﺪ .ﻧﻴــﻜﻞ ﺭﺍ ﻫﻢﭼﻨﻴﻦ ﺑﻪ ﻓــﻮﻻﺩ ﺯﻧﮓ ﻧﺰﻥ ،ﺍﺿﺎﻓﻪ ﻣﻰﻛﻨﺪ ،ﺯﻳﺮﺍ ﺳــﺒﺐ ﻣﻰﺷــﻮﺩ ﻛﻪ ﻓﻮﻻﺩ ﺩﺭ ﺑﺮﺍﺑﺮ ﺣﻤﻠﻪﻯ ﺍﺳــﻴﺪﻫﺎﻳﻰ ﭼﻮﻥ ﻫﻴﺪﺭﻭﻛﻠﺮﻳﻚ ،ﺳﻮﻟﻔﻮﺭﻳﻚ ﻭ ﻧﻴﺘﺮﻳﻚ ﻣﻘﺎﻭﻡ ﺑﺎﺷﺪ. ـ ﺍﺿﺎﻓﻪ ﻛﺮﺩﻥ Pﺳــﺒﺐ ﻣﻰﺷﻮﺩ ﻛﻪ ﺣﺘﻰ ﻓﻮﻻﺩ ﺑﺎ ﺗﺸﻜﻴﻞ ﺁﻫﻦ ﻓﺴــﻔﻴﺪ ) (FcP2ﺯﻳﺎﺩ ﺷــﻮﺩ ،ﺍ ّﻣﺎ ﭼﻨﺎﻥﭼﻪ ﺑﻴﺶ ﺍﺯ ٪05ﻓﺴﻔﺮ ﺍﺿﺎﻓﻪ ﺷــﻮﺩ ،ﺷــﻜﻨﻨﺪﮔﻰ ﻓﻮﻻﺩ ﺑﻪ ﺣﺪﻯ ﻣﻰﺭﺳــﺪ ﻛﻪ ﺩﻳﮕﺮ ﻗﺎﺑﻞ ﺍﺳﻔﺎﺩﻩ ﻧﻴﺴﺖ. ـ ﺍﺿﺎﻓــﻪ ﻛــﺮﺩﻥ ALﻭ Siﺑﻪ ﻓــﻮﻻﺩ ﻣﺎﻳﻊ ،ﺍﻛﺴــﻴﮋﻧﻰ ﺭﺍ ﻛﻪ ﺑﻪ ﺻﻮﺭﺕ ﺍﻛﺴــﻴﺪﻫﺎﻯ ﻓﻠــﺰﻯ ﺑﺎﻗﻰ ﻣﺎﻧﺪﻩ ﺍﺳــﺖ ﺣــﺬﻑ ﻣﻰﻛﻨﺪ .ﺍﻳﻦ
ﺍﻛﺴﻴﺪﻫﺎ ﻣﻰﺗﻮﺍﻧﻨﺪ ﻣﻮﺟﺐ ﻧﻘﺺ ﺩﺭ ﻓﺮﺍﻧﺪﻫﺎﻯ ﻧﻬﺎﻳﻰ ﺷﻮﻧﺪ. ﻣﻼﺣﻈــﻪ ﻛﺮﺩﻳﺪ ﻛﻪ ﺩﺭ ﻣﻮﺍﺭﺩ ﺫﻛﺮ ﺷــﺪﻩ ،ﺍﺿﺎﻓــﻪ ﻛﺮﺩﻥ ﻋﻨﺎﺻﺮ ﺧﻮﺍﺹ ﻣﻜﺎﻧﻴﻜﻰ ﻭ ﻣﻮﺍﺭﺩ ﻣﻌﻴﻦ ﺩﺭ ّ ﮔﻮﻧﺎﮔــﻮﻥ ﺑﺎ ﺍﻧﺪﺍﺯﻩﻫﺎ ﻭ ﺩﺭﺻﺪﻫﺎﻯ ّ ﻛﺎﺭﺑﺮﺩ ﻓﻮﻻﺩ ،ﭼﻪ ﺍﺛﺮﺍﺕ ﺣﻴﺎﺗﻰ ﺩﺍﺭﺩ!
ﭘﻴﺸﺮﻓﺖ ﺟﺪﻳﺪ ﺍﮔﺮﭼﻪ ﺗﻮﻟﻴﺪ ﻓﻮﻻﺩ ﺩﺭ ﻣﻘﺎﻳﺴﻪ ﺑﺎ ﺁﻟﻮﻣﻴﻨﻴﻮﻡ ﻭ ﺑﻌﻀﻰ ﺍﺯ ﺑﺴﻰﭘﺎﺭﻫﺎ ﺧﻴﻠﻰ ﺍﺭﺯﺍﻧﺘﺮ ﺍﺳــﺖ ،ﺍ ّﻣﺎ ﺩﺭ ﺑﻌﻀﻰ ﻣﻮﺍﺭﺩ ﺳــﺒﻚ ﺑﻮﺩﻥ ﺁﻥﻫﺎ ﻣﻤﻜﻦ ﺍﺳــﺖ ﺑــﺎﻻ ﺑﻮﺩﻥ ﻗﻴﻤﺖ ﺭﺍ ﺗﺤﺖﺍﻟﺸــﻌﺎﻉ ﻗﺮﺍﺭ ﺩﻫﺪ .ﺑــﺎ ﺗﻮﺟﻪ ﺑﻪ ﺍﻳﻦ ﻧﻜﺘﻪ ﺩﺭ ﭘﺎﺳﺦ ﺑﻪ ﺗﻼﺵ ﺟﻬﺎﻧﻰ ﺻﻨﻌﺖ ﺧﻮﺩﺭﻭﺳﺎﺯﻯ ﺑﺮﺍﻯ ﻛﺎﺳﺘﻦ ﺍﺯ ﻭﺯﻥ ﺧﻮﺩﺭﻭ ،ﺻﺎﺣﺒﺎﻥ ﺻﻨﺎﻳﻊ ﻓﻮﻻﺩ ،ﺑﺪﻧﻪﻯ ﻓﻮﻻﺩﻯ ﺑﺴــﻴﺎﺭ ﺳﺒﻜﻰ ﺭﺍ ﺑﺮﺍﻯ ﺧﻮﺩﺭﻭ ﺳــﺎﺧﺘﻪﺍﻧﺪ ﻛﻪ ﻭﺯﻥ ﻳﻚ ﺧﻮﺩﺭﻭ ﺍﺳﺘﺎﻧﺪﺍﺭﺩ ﺭﺍ ﺑﻪ ﺍﻧﺪﺍﺯﻩﻯ ٪36ﻛﺎﻫﺶ ﻣﻰﺩﻫﺪ ،ﺩﺭ ﻋﻴﻦ ﺣﺎﻝ ﻛﻪ ﺍﺳــﺘﺤﻜﺎﻡ ﺁﻥ ﺣﻔﻆ ﻣﻰﺷﻮﺩ ﻭ ﺍﻳــﻦ ﺍﻣﺮ ﺑﻪ ﻛﺎﻫﺶ ﻣﺼﺮﻑ ﺳــﻮﺧﺖ ﻛﻤﻚ ﻣﻰﻛﻨــﺪ .ﺍﻳﻦ ﻛﺎﺭ ﺑﺎ ﺑﻪ ﻛﺎﺭ ﺑــﺮﺩﻥ ﻓﻮﻻﺩﻫﺎﻳﻰ ﺑﺎ ﺍﺳــﺘﺤﻜﺎﻡ ﺯﻳﺎﺩ ﻭ ﻓﻮﻻﺩ ﺳــﺎﻧﺪﻭﻳﭽﻰ )ﻳﻌﻨﻰ ﺑﺴــﻰﭘﺎﺭﮔﺮﻣﺎﻧﺮﻣﻰ ﻛﻪ ﺑﻴﻦ ﺩﻭ ﻭﺭﻗﻪﻯ ﻓﻮﻻﺩﻯ ﺳﺎﻧﺪﻭﻳﭻ ﺷﺪﻩ ﺍﺳﺖ( ﺍﻣﻜﺎﻥﭘﺬﻳﺮ ﺷﺪﻩ ﺍﺳﺖ. ﻳﻚ ﻣﺜﺎﻝ ﺩﻳﮕﺮ ﺍﺯ ﻓﻮﻻﺩﻫﺎﻯ ﻭﻳﮋﻩ ،ﻛﺎﺑﻞﻫﺎﻳﻰ ﺑﺎ ﺍﺳﺘﺤﻜﺎﻡ ﻛﺸﺸﻰ ﺯﻳﺎﺩ ﺍﺳــﺖ ﻛﻪ ﺩﺭ ﺳﺎﺧﺘﻦ ﭘﻞﻫﺎﻯ ﻣﻌﻠّﻖ ﺑﻪ ﻛﺎﺭ ﻣﻰﺭﻭﺩ .ﺩﺭ ﺍﻳﻦ ﻣﻮﺭﺩ ﺩﺍﻧﺸﻤﻨﺪﺍﻥ ﻳﻚ ﻧﻮﻉ ﻓﻮﻻﺩ ﺟﺪﻳﺪ ﻛﻢ ﻛﺮﺑﻦ ﺗﻮﻟﻴﺪ ﻛﺮﺩﻩﺍﻧﺪ ﻭ ﺍﺳﺘﺤﻜﺎﻡ ﻛﺸﺸﻰ ﺁﻥ ﺭﺍ ﺑﺎ ﺍﺿﺎﻓﻪ ﻛﺮﺩﻥ ﺳﻴﻠﻴﺴﻴﻢ ﺑﺎﻻ ﺑﺮﺩﻩﺍﻧﺪ.
ﺩﻭﺭﻩﻯ ﺷﺎﻧﺰﺩﻫﻢ ،ﺷﻤﺎﺭﻩ ﻯ ،2ﺯﻣﺴﺘﺎﻥ 1389
ﺭﺍﻫﻨﻤﺎﻳﻲ
13
ﺍﻧﺪﻳﺸﻪﻭﺭﺯﻱ
ا︑︀︋ ﹏﹫︊﹞﹢ز︙﹤ ﻧﻮﻳﺴﻨﺪ ﻩ :ﻣﺎﻳﻜﻞ ﺳﺘﻴﻮﺑﻦ ﺗﺮﺟﻤﻪ ﻯ ﺣﺴﻦ ﻧﺼﻴﺮﻧﻴﺎ
ﻛﻠﻴﺪﻭﺍژﻩﻫﺎ :ﺍﺗﻮﻣﺒﻴﻞ ،ﺟﻬﺖ ﺣﺮﻛﺖ ،ﺣﺮﻛﺖ ﺟﻬﺸﻰ. ﺑﻬــﺮﻭﺯ ﻭ ﻛﺎﻭﻩ ﻳــﻚ ﺍﺗﻮﻣﺒﻴــﻞ ﺍﺑﺘﻜﺎﺭﻯ ﺳــﺎﺧﺘﻪﺍﻧﺪ ﻛﻪ ﺑﺮ ﺧﻼﻑ ﺍﺗﻮﻣﺒﻴﻞﻫﺎﻯ ﻣﻌﻤﻮﻟﻰ ﺩﻭ ﺭﺍﻧﻨﺪﻩ ﺩﺍﺭﺩ ﻭ ﺑﻪ ﺟﺎﻯ ﺣﺮﻛﺖ ﺭﻭﻯ ﭼﺮﺥﻫﺎ ﺑﺎ »ﺣﺮﻛﺖﻫﺎﻯ ﺟﻬﺸــﻰ« ﭘﻴﺶ ﻣﻰﺭﻭﺩ .ﻧﺨﺴﺖ ﺑﻬﺮﻭﺯ ﺍﺗﻮﻣﺒﻴﻞ ﺭﺍ ﺩﺭ ﺭﺍﺳﺘﺎﻯ ﻳﻜﻰ ﺍﺯ ﺟﻬﺖﻫﺎﻯ ﻫﺸﺖﮔﺎﻧﻪ )ﺷﻤﺎﻝ ،ﺷﻤﺎﻝ ﺷﺮﻗﻰ ،ﻣﺸﺮﻕ، ﺟﻨﻮﺏ ﺷــﺮﻗﻰ ،ﺟﻨــﻮﺏ ،ﺟﻨﻮﺏ ﻏﺮﺑﻰ ،ﻣﻐﺮﺏ ﻭ ﺷــﻤﺎﻝ ﻏﺮﺑﻰ( ﻗﺮﺍﺭ ﻣﻰﺩﻫﺪ .ﺳــﭙﺲ ﻛﺎﻭﻩ ﺍﺗﻮﻣﺒﻴﻞ ﺭﺍ ﺭﻭﺷــﻦ ﻣﻰﻛﻨــﺪ ﻭ ﺑﺮﺍﻯ ﺟﻬﺶ ﺑﻪ ﺳــﻮﻯ ﺟﻠﻮ ﻳﺎ ﻋﻘﺐ ﺁﻥ ﺭﺍ ﺩﺭ ﺩﻧﺪﻩﻯ ﻣﻨﺎﺳــﺐ ﻣﻰﮔﺬﺍﺭﺩ .ﺍﮔﺮ ﺑﻬﺮﻭﺯ ﺍﺗﻮﻣﺒﻴﻞ ﺭﺍ ﺩﺭ ﺭﺍﺳــﺘﺎﻯ ﺷﻤﺎﻝ ﺷــﺮﻗﻰ ﻗﺮﺍﺭ ﺩﻫﺪ ،ﻛﺎﻭﻩ ﻓﻘﻂ ﻣﻰﺗﻮﺍﻧﺪ ﺍﺗﻮﻣﺒﻴﻞ ﺭﺍ ﻳﺎ ﺩﺭ ﻫﻤﺎﻥ ﺭﺍﺳﺘﺎ ﻳﻌﻨﻰ ﺷﻤﺎﻝ ﺷﺮﻗﻰ ﻳﺎ ﺩﺭ ﺟﻬﺖ ﻣﻌﻜﻮﺱ ﻳﻌﻨﻰ ﺟﻨﻮﺏ ﻏﺮﺑﻰ ﺑﺮﺍﻧﺪ.
ﺍﺗﻮﻣﺒﻴــﻞ ﺭﺍ ﻣﻰﺗﻮﺍﻥ ﺩﺭ ﻫﺮ ﻫﺸــﺖ ﺟﻬﺖ ﻣﺬﻛﻮﺭ ﺣﺮﻛﺖ ﺩﺍﺩ ،ﺑﻪ ﻃﻮﺭﻯ ﻛﻪ ﻧﻘﺎﻁ ﻣﺨﺘﻠﻒ ﺍﺳــﺘﻘﺮﺍﺭ ﺁﻥ ﻳﻚ ﺷــﺒﻜﻪﻯ ﻣﺮﺑﻊ ﺷــﻜﻞ ﺭﺍ ﺗﺸــﻜﻴﻞ ﺑﺪﻫﻨﺪ .ﻃﻮﻝ ﻫﺮ ﺣﺮﻛﺖ ﺟﻬﺸــﻰ ﺍﺗﻮﻣﺒﻴﻞ ﺑﻪ ﺳﻤﺖ ﺷﻤﺎﻝ ﻳﻚ ﻣﺘﺮ ﺍﺳﺖ.
ﺣﺎﻝ ﺑﺎ ﺩﺭ ﻧﻈﺮ ﮔﺮﻓﺘﻦ ﻧﻜﺎﺕ ﮔﻔﺘﻪ ﺷــﺪﻩ ﺑﻪ ﺩﻭ ﻣﻌﻤﺎﻯ ﺯﻳﺮ ﭘﺎﺳﺦ ﺩﻫﻴﺪ: 14
ﺭﺍﻫﻨﻤﺎﻳﻲ
ﺩﻭﺭﻩﻯ ﺷﺎﻧﺰﺩﻫﻢ ،ﺷﻤﺎﺭﻩﻯ ،2ﺯﻣﺴﺘﺎﻥ 1389
-1ﺑﺎ ﺗﻮﺟﻪ ﺑﻪ ﺍﻳﻦﻛﻪ ﺍﺗﻮﻣﺒﻴﻞ ﺑﺮﺍﻯ ﺳــﻔﺮ ﺳــﺎﺧﺘﻪ ﻧﺸﺪﻩ ﺍﺳﺖ، ﺑﻬﺮﻭﺯ ﻭ ﻛﺎﻭﻩ ﺍﺯ ﺁﻥ ﺑﻪ ﻋﻨﻮﺍﻥ ﻳﻚ ﻭﺳــﻴﻠﻪﻯ ﺑﺴــﻴﺎﺭ ﻣﻨﺎﺳــﺐ ﺗﻔﻨﻨﻰ ﺩﺭ ﻳﻚ ﺑﺎﺯﻯ ﺳــﺎﺩﻩ ﺍﺳــﺘﻔﺎﺩﻩ ﻣﻰﻛﻨﻨﺪ .ﺁﻧﺎﻥ ﻣﻴﺪﺍﻥ ﺑﺎﺯﻯ ﺭﺍ ﺑﻪ ﺷﻜﻞ ﻣﺮﺑﻊ ﺭﻭﻯ ﺯﻣﻴﻦ ﻋﻼﻣﺖﮔﺬﺍﺭﻯ ﻭ ﻣﺸــﺨﺺ ﻣﻰﻛﻨﻨﺪ ﻭ ﺳﻮﺍﺭ ﺍﺗﻮﻣﺒﻴﻞ ﻣﻰﺷﻮﻧﺪ .ﺣﺎﻝ ﻛﺎﻭﻩ ﻧﻘﻄﻪﺍﻯ ﺭﺍ ﺑﻪ ﻋﻨﻮﺍﻥ ﺁﻏﺎﺯﮔﺮ ﺑﺎﺯﻯ ﺩﺭ ﻣﺤﺪﻭﺩﻩﻯ ﻣﺮﺑﻊ ﺗﺮﺳﻴﻢﺷــﺪﻩ ﺍﻧﺘﺨﺎﺏ ﻣﻰﻛﻨﺪ ﻭ ﺑﺎﺯﻯ ﺷــﺮﻭﻉ ﻣﻰﺷﻮﺩ .ﺑﻬﺮﻭﺯ ﺩﺭ ﻫﺮ ﺑﺎﺭ ِ ﻧﻮﺑﺖ ﺧﻮﺩ ﻣﻰﻛﻮﺷــﺪ ﺍﺗﻮﻣﺒﻴــﻞ ﺭﺍ ﺍﺯ ﻣﺤﺪﻭﺩﻩﻯ ﻣﻴﺪﺍﻥ ﺑﺎﺯﻯ ﺧﺎﺭﺝ ﻛﻨﺪ ،ﺩﺭ ﺣﺎﻟﻰ ﻛﻪ ﻛﺎﻭﻩ ﺗﻼﺵ ﻣﻰﻛﻨﺪ ﺗﺎ ﻋﻤﻞ ﺍﻭ ﺭﺍ ﺧﻨﺜﻰ ﻛﻨﺪ ﻭ ﻧﮕﺬﺍﺭﺩ ﺍﺗﻮﻣﺒﻴﻞ ﺍﺯ ﭼﺎﺭﭼﻮﺏ ﻣﻮﺭﺩ ﻧﻈﺮ ﻓﺮﺍﺗﺮ ﺭﻭﺩ. ﺁﻳﺎ ﺑﻪ ﻧﻈﺮ ﺷــﻤﺎ ﻣﻰﺗــﻮﺍﻥ ﻣﻴﺪﺍﻥ ﺑﺎﺯﻯ ﺭﺍ ﭼﻨــﺎﻥ ﺑﺰﺭگ ﺍﻧﺘﺨﺎﺏ ﻛﺮﺩ ﻛــﻪ ﺑﻬﺮﻭﺯ ﻫﺮﮔﺰ ﻧﺘﻮﺍﻧﺪ ﺑﺮﻧﺪﻩﻯ ﺑﺎﺯﻯ ﺷــﻮﺩ؟ ﺍﮔﺮ ﭼﻨﻴﻦ ﻭﺿﻌﻰ ﺍﻣﻜﺎﻥﭘﺬﻳﺮ ﺑﺎﺷﺪ ،ﺣﺪﺍﻗﻞ ﺑﺰﺭﮔﻰ ﻣﻴﺪﺍﻥ ﺑﺎﺯﻯ ﭼﻪﻗﺪﺭ ﺑﺎﻳﺪ ﺑﺎﺷﺪ؟ -2ﺑﻬﺮﻭﺯ ﻭ ﻛﺎﻭﻩ ﻛﻪ ﺍﺯ ﺍﻳﻦ ﺑﺎﺯﻯ ﺧﺴــﺘﻪ ﺷﺪﻩﺍﻧﺪ ،ﺑﻪ ﻛﺎﺭﮔﺎﻩ ﺧﻮﺩ ﺑﺎﺯﻣﻰﮔﺮﺩﻧــﺪ ﻭ ﺍﺗﻮﻣﺒﻴﻞ ﺩﻳﮕﺮﻯ ﻣﻰﺳــﺎﺯﻧﺪ ﻛــﻪ ﻣﻰﺗﻮﺍﻧﺪ ﺩﺭ ﻫﻤﻪﻯ ﻣﻌﺎﺩﻝ ﻳــﻚ ﻣﺘﺮ ﺑﻪ ﭘﻴﺶ ﺑﺠﻬﺪ ،ﺍﻳــﻦ ﺍﺗﻮﻣﺒﻴﻞ ﻣﻰﺗﻮﺍﻧﺪ ﺍﺯ ﺟﻬﺖﻫــﺎ ِ ﻣﺤﻞ ﺍﺳﺘﻘﺮﺍﺭ ﺑﻪ ﺗﻤﺎﻡ ﻧﻘﺎﻁِ ﮔﺮﺩﺍﮔﺮ ِﺩ ﺧﻮﺩ ﺑﺠﻬﺪ ،ﺑﻪ ﻃﻮﺭﻯ ﻛﻪ ﻧﻘﺎﻁ ﺟﺪﻳﺪ ﺍﺳــﺘﻘﺮﺍﺭ ﻳﻚ ﺩﺍﻳﺮﻩ ﺗﺸﻜﻴﻞ ﺑﺪﻫﻨﺪ .ﺍﻳﻦ ﺍﺗﻮﻣﺒﻴﻞ ﻧﻴﺰ ﺩﻭ ﺭﺍﻧﻨﺪﻩ ﻣﻰﺧﻮﺍﻫــﺪ .ﺑﺎﺯ ﺑﻬﺮﻭﺯ ﺍﺗﻮﻣﺒﻴﻞ ﺭﺍ ﺩﺭ ﺭﺍﺳــﺘﺎﻯ ﻣــﻮﺭﺩ ﻧﻈﺮ ﺧﻮﺩ ﻗﺮﺍﺭ ﻣﻰﺩﻫــﺪ ﻭ ﻛﺎﻭﻩ ﺩﺭ ﻧﻮﺑﺖ ﺑﺎﺯﻯ ﺧﻮﺩ ﺗﺼﻤﻴﻢ ﻣﻰﮔﻴﺮﺩ ﻛﻪ ﺍﺗﻮﻣﺒﻴﻞ ﺭﺍ ﺑــﻪ ﺟﻠﻮ ﻳﺎ ﻋﻘﺐ ﺑﺮﺍﻧﺪ .ﺣﺎﻻ ﻛﻪ ﺑﻬﺮﻭﺯ ﻭ ﻛﺎﻭﻩ ﻣﻮﻓﻖ ﺑﻪ ﺳــﺎﺧﺘﻦ ﻳﻚ ﻭﺳــﻴﻠﻪﻯ ﺗﻔﻨﻨﻰ ﭘﻴﺸﺮﻓﺘﻪ ﻭ ﻧﻮ ﺷــﺪﻩﺍﻧﺪ ،ﺁﻳﺎ ﻣﻰﺗﻮﺍﻥ ﻣﻴﺪﺍﻥ ﺑﺎﺯﻯ ﺭﺍ ﭼﻨﺎﻥ ﺑﺰﺭگ ﺍﻧﺘﺨﺎﺏ ﻛﺮﺩ ﻛﻪ ﺑﻬﺮﻭﺯ ﻫﺮﮔﺰ ﻧﺘﻮﺍﻧﺪ ﺑﺮ ﺣﺮﻳﻒ ﭼﻴﺮﻩ ﺷﻮﺩ؟ )ﺩﺭ ﺍﻳﻦ ﺑﺎﺯﻯ ﺑﺎﺯ ﻫﺪﻑ ﺑﻬﺮﻭﺯ ﺁﻥ ﺍﺳﺖ ﻛﻪ ﺍﺗﻮﻣﺒﻴﻞ ﺑــﻪ ﺧﺎﺭﺝ ﺍﺯ ﻣﺤﻴﻂ ﺩﺍﻳﺮﻩ ﺭﺍﻧﺪﻩ ﺷــﻮﺩ( ﺍﻣﻜﺎﻥ ﭘﺰﻳﺮ ﺍﺳﺖ ﻛﻮﭼﻚ ﺗﺮﻳﻦ ﻣﻘﺪﺍﺭ ﻣﺴــﺎﺣﺖ ﭼﻨﻴﻦ ﻣﻴﺪﺍﻧﻰ ﭼﻪﻗﺪﺭ ﺑﺎﺷﺪ ﭘﺎﺳﺦ؛ ﺻﻔﺤﻪﻱ 31
ﻣﻨﻄﻖ ﺭﻳﺎﺿﯽ
﹡︖︀ت ﹞﹠︴﹆︡︊︻ ﹩ا﹛﹊︀︸﹜ ﺣﻤﻴﺮﺍ ﻇﻔﺮﻗﻨﺪ
ﻛﻠﻴﺪ ﻭﺍژﻩﻫﺎ :ﺭﻳﺎﺿﻴﺎﺕ ،ﻣﻨﻄﻖ ،ﻣﺴﺌﻠﻪﻫﺎﻯ ﻓﻜﺮﻯ ِ ﺑــﺰﺭگ ﺍﻣﻴﺮ ﺍﺑــﻦ ﺍﻟﺤﻮﺽ ،ﻳﻜــﻰ ﺍﺯ ﺭﻗﻴﺒﺎﻥ ﺧــﻮﺩ ﺑﻪ ﻧﺎﻡ ﻭﺯﻳــﺮ ﻋﺒﺪﺍﻟﻜﺎﻇــﻢ ﺭﺍ ﺑﻪ ﺍﺗﻬﺎﻡ ﺁﻥﻛﻪ ﻗﺮﺻﻰ ﺳــﻤﻰ ﻓﺮﻭﺧﺘــﻪ ﺑﻮﺩ ،ﺑﻪ ﺍﻋﺪﺍﻡ ﻣﺤﻜﻮﻡ ﻛﺮﺩ .ﺩﻳﺮ ﺯﻣﺎﻧﻰ ﺑﻮﺩ ﻛﻪ ﻭﺯﻳﺮ ﻣﻰﻛﻮﺷــﻴﺪ ﺗﺎ ﻋﺒﺪﺍﻟﻜﺎﻇﻢ ﺭﺍ ﺍﺯ ﺳــﺮ ﺭﺍﻩ ﺧﻮﺩ ﺑﺮﺩﺍﺭﺩ ،ﺍﻣﺎ ﻳﻚ ﻣﻮﺿﻮﻉ ﺑﺮﺍﻯ ﺍﻭ ﻧﮕﺮﺍﻥﻛﻨﻨﺪﻩ ﺑﻮﺩ .ﺑﻨﺎ ﺑﺮ ﻗﺎﻧﻮﻥ ،ﻫﺮ ﻣﺤﻜﻮﻡ ﺣﻖ ﺩﺍﺷــﺖ ﺩﺭ ﺑﺎﻣــﺪﺍﺩ ﺭﻭﺯ ﺍﺟﺮﺍﻯ ﺣﻜﻢ ﺍﺯ ﺍﻣﻴﺮ ﺗﻘﺎﺿــﺎﻯ ﻋﻔﻮ ﻛﻨﺪ .ﻭﺯﻳﺮ ﻣﻰﺩﺍﻧﺴــﺖ ﻛﻪ ﺍﻣﻴــﺮ ﻋﺒﺪﺍﻟﻜﺎﻇﻢ ﺭﺍ ﺧﻮﺍﻫﺪ ﺑﺨﺸــﻴﺪ .ﺍﺯ ﺍﻳﻦ ﺭﻭ ﻧﻘﺸﻪﺍﻯ ﻣﺎﻫﺮﺍﻧﻪ ﻃﺮﺡﺭﻳﺰﻯ ﻛﺮﺩ .ﻋﺒﺪﺍﻟﻜﺎﻇﻢ ﺭﺍ ﺑﻪ ﻳﻜﻰ ﺍﺯ ﻧﮕﻬﺒﺎﻧﺎﻥ ﻃﺮﻑ ﺍﻋﺘﻤﺎﺩ ﺧﻮﺩ ﺳــﭙﺮﺩ ﻭ ﺁﻥﮔﻮﻧﻪ ﻛﻪ ﻋﺒﺪﺍﻟﻜﺎﻇﻢ ﺑﺸــﻨﻮﺩ ،ﻫﻔﺘــﻪﺍﻯ ﺭﺍ ﻣﻌﻴــﻦ ﻛﺮﺩ ﻭ ﺑــﻪ ﻧﮕﻬﺒﺎﻥ ﮔﻔــﺖ» :ﺩﺭ ﻳﻜﻰ ﺍﺯ ﺭﻭﺯﻫــﺎﻯ ﺍﻳﻦ ﻫﻔﺘﻪ ﻋﺒﺪﺍﻟﻜﺎﻇﻢ ﺭﺍ ﺍﻋــﺪﺍﻡ ﻣﻰﻛﻨﻰ .ﺑﺮﺍﻯ ﺍﻧﺘﺨﺎﺏ ﺭﻭﺯ ﺍﻋــﺪﺍﻡ ﺧﻮﺩﺕ ﺗﺼﻤﻴﻢ ﻣﻰﮔﻴﺮﻯ ،ﺍﻣﺎ ﺍﻳــﻦ ﺭﻭﺯ ﺭﺍ ﺑﺎﻳﺪ ﭼﻨﺎﻥ ﺍﻧﺘﺨﺎﺏ ﻛﻨﻰ ﻛﻪ ﻋﺒﺪﺍﻟﻜﺎﻇﻢ ﻧﺘﻮﺍﻧﺪ ﺁﻥ ﺭﺍ ﭘﻴﺶﺑﻴﻨﻰ ﻛﻨﺪ .ﺑﺎﻳﺪ ﻛﺎﻣ ً ﻼ ﻏﺎﻓﻞﮔﻴﺮ ﺷــﻮﺩ .ﻫﺮ ﺑﺎﻣﺪﺍﺩ ﺧﻮﺩﺕ ﻧﺎﺷــﺘﺎﻳﻰ ﻋﺒﺪﺍﻟﻜﺎﻇﻢ ﺭﺍ ﺑﺮﺍﻳﺶ ﻣﻰﺑﺮﻯ .ﺍﮔﺮ ﺩﺭ ﺁﻥ ﻫﻨــﮕﺎﻡ ﺑﺎ ﺩﻟﻴﻞﻫﺎﻯ ﻣﻨﻄﻘﻰ ﺛﺎﺑﺖ ﻛﺮﺩ ﻛــﻪ ﺩﺭ ﺁﻥ ﺭﻭﺯ ﺍﻋﺪﺍﻡ ﻣﻰﺷﻮﺩ ،ﺩﺭ ﺁﻥ ﺻﻮﺭﺕ ﻭﻇﻴﻔﻪ ﺩﺍﺭﻯ ﺍﺟﺎﺯﻩ ﺩﻫﻰ ﺍﻭ ﺍﺯ ﺍﻣﻴﺮ ﺩﺭﺧﻮﺍﺳﺖ ﺑﺨﺸــﺎﻳﺶ ﻛﻨﺪ ،ﺍﻣﺎ ﺍﮔﺮ ﺁﻥ ﻫﻮﺷﻤﻨﺪﻯ ﺭﺍ ﻧﺪﺍﺷﺖ ﻛﻪ ﺩﺭ ﺭﻭﺯ ﺍﻧﺘﺨﺎﺏ ﺷــﺪﻩ ﺑﻪ ﻫﻨﮕﺎﻡ ﺩﺭﻳﺎﻓﺖ ﻧﺎﺷﺘﺎﻳﻰ ﺍﺯ ﺍﻋﺪﺍﻡ ﺧﻮﺩ ﺩﺭ ﺁﻥ ﺭﻭﺯ ﺧﺒﺮ ﺩﻫﺪ، ﺣﻖ ﻧﺨﻮﺍﻫﺪ ﺩﺍﺷﺖ ﺩﺭﺧﻮﺍﺳﺖ ﺑﺨﺸﺎﻳﺶ ﻛﻨﺪ ﻭ ﺑﻌﺪ ﺍﺯ ﻧﻴﻢﺭﻭﺯ ﻫﻤﺎﻥ ﺭﻭﺯ ﺍﻋﺪﺍﻣﺶ ﻣﻰﻛﻨﻰ«. ﻧﮕﻬﺒــﺎﻥ ،ﻋﺒﺪﺍﻟﻜﺎﻇﻢ ﺭﺍ ﺑﺎ ﺧﻮﺩ ﺑﺮﺩ .ﺍﻭ ﺍﻧﺘﺨﺎﺏ ﺭﻭﺯ ﺍﻋﺪﺍﻡ ﺭﺍ ﻛﺎﺭﻯ ﺳﺎﺩﻩ ﻣﻰﺍﻧﮕﺎﺷﺖ ،ﺍﻣﺎ ﻛﻤﻰ ﺑﻴﺶﺗﺮ ﻛﻪ ﺩﺭﺑﺎﺭﻩﻯ ﺁﻥ ﻓﻜﺮ ﻛﺮﺩ ،ﺩﺍﻧﺴﺖ ﭼﻨﺪﺍﻥ ﺳﺎﺩﻩ ﻧﻴﺴــﺖ .ﻋﺒﺪﺍﻟﻜﺎﻇﻢ ﻫﻢ ﺩﺭ ﺍﻳﻦ ﺍﻧﺪﻳﺸﻪ ﺑﻮﺩ ﻛﻪ ﭼﮕﻮﻧﻪ ﺑﻔﻬﻤﺪ ﺩﺭ ﻛﺪﺍﻡ ﺭﻭﺯ ﺍﺯ ﺁﻥ ﻫﻔﺘﻪ ﺍﺣﺘﻤﺎﻝ ﺍﻋﺪﺍﻣﺶ ﺑﻴﺶﺗﺮ ﺍﺳــﺖ .ﻫﻢ ﻧﮕﻬﺒﺎﻥ ﻭ ﻫﻢ ﻋﺒﺪﺍﻟﻜﺎﻇﻢ ﺑﺎ ﻳﻚ ﻣﺴــﺌﻠﻪﻯ ﻓﻜﺮﻯ ﺭﻭﺑﻪﺭﻭ ﺷﺪﻩ ﺑﻮﺩﻧﺪ. ﺑﺮﺍﻯ ﻧﮕﻬﺒﺎﻥ ﻣﺴــﺌﻠﻪ ﭘﻴﭽﻴﺪﻩﺗﺮ ﺑﻮﺩ؛ ﺯﻳﺮﺍ ﻣﻰﺑﺎﻳﺴــﺖ ﺭﻭﺯ ﺍﻋﺪﺍﻡ ﺭﺍ ﺍﺯ ﭘﻴﺶ ﻣﻌﻠﻮﻡ ﻣﻰﻛﺮﺩ ﻭ ﻣﻘﺪﻣﺎﺕ ﻛﺎﺭ ﺭﺍ ﻓﺮﺍﻫﻢ ﻣﻰﺁﻭﺭﺩ.
ﻣﺤﻜﻮﻡ ﺷﺪﻥ ﻋﺒﺪﺍﻟﻜﺎﻇﻢ ،ﻣﺮﺩﻡ ﺷﻬﺮ ﺭﺍ ﻣﺘﺄﺛﺮ ﻛﺮﺩﻩ ﺑﻮﺩ .ﺍﻣﻴﺮﺯﺍﺩﻩ ﻫﻢ ﻛﻪ ﺍﺯ ﺍﻳﻦ ﭘﻴﺸــﺎﻣﺪ ﻧﺎﺭﺍﺣﺖ ﺷــﺪﻩ ﺑﻮﺩ ،ﺍﺯ ﺍﻣﻴﺮ ﺧﻮﺍﺳﺖ ﺗﺎ ﻓﺮﻣﺎﻥ ﻟﻐﻮ ﺣﻜﻢ ﺷــﮕﻔﺖﺁﻭﺭ ﻭﺯﻳﺮ ﺭﺍ ﺻﺎﺩﺭ ﻛﻨﺪ .ﺍﻣﻴﺮ ﺩﺭ ﭘﺎﺳﺦ ﺑﻪ ﺩﺭﺧﻮﺍﺳﺖ ﭘﺴﺮﺵ ﮔﻔﺖ ﻛﻪ ﺣﻜﻢ ﻭﺯﻳﺮ ﻧﻘﻄﻪﺿﻌﻔﻰ ﺩﺍﺭﺩ ﻭ ﻫﻤﺎﻥ ﺍﺯ ﺍﺟﺮﺍﻯ ﺣﻜﻢ ﻧﮕﻬﺒﺎﻥ ﺭﺍ ﻓﺮﺍ ﺧﻮﺍﻧﺪ ﺟﻠﻮﮔﻴﺮﻯ ﺧﻮﺍﻫﺪ ﻛــﺮﺩ .ﭘﺲ ﺍﺯ ﺁﻥ ،ﻋﺒﺪﺍﻟﻜﺎﻇﻢ ِ ﻭ ﺑﻪ ﺍﻭ ﮔﻔﺖ: ـ ﺍﺯ ﺩﺳــﺘﻮﺭﻯ ﻛﻪ ﻭﺯﻳﺮ ﺑﻪ ﺗﻮ ﺩﺍﺩﻩ ﺍﺳــﺖ ،ﺁﮔﺎﻫﻰ ﺩﺍﺭﻡ .ﺁﻳﺎ ﺑﺮﺍﻯ ﺍﺟﺮﺍﻯ ﺁﻥ ﺗﻮﺍﻧﺴﺘﻪﺍﻯ ﺗﺼﻤﻴﻢ ﺑﮕﻴﺮﻯ؟ ﻧﮕﻬﺒﺎﻥ ﭘﺎﺳــﺦ ﺩﺍﺩ :ﻫﻨﻮﺯ ﻓﻜﺮﻡ ﺑﻪ ﺟﺎﻳﻰ ﻧﺮﺳــﻴﺪﻩ ﺍﺳــﺖ .ﺑﻨﺎﺑﺮ ﺩﺳﺘﻮﺭ ﻭﺯﻳﺮ ﺑﺎﻳﺪ ﺭﻭﺯﻯ ﺍﺯ ﺁﻥ ﻫﻔﺘﻪ ﺭﺍ ﺑﺮﮔﺰﻳﻨﻴﻢ ﻛﻪ ﻋﺒﺪﺍﻟﻜﺎﻇﻢ ﻧﺘﻮﺍﻧﺪ ﺁﻥ ﺭﺍ ﺗﺸــﺨﻴﺺ ﺩﻫﺪ .ﺩﻟﻢ ﻣﻰﺧﻮﺍﻫــﺪ ﺑﺘﻮﺍﻧﻢ ﺑﺮﺍﻯ ﻋﺒﺪﺍﻟﻜﺎﻇﻢ ﻣﻔﻴﺪ ﻭﺍﻗﻊ ﺷﻮﻡ ،ﺍﻣﺎ ﺑﺎﻳﺪ ﺩﺳﺘﻮﺭ ﺭﺍ ﺍﺟﺮﺍ ﻛﻨﻢ. ـ ﺗﺤﺼﻴــﻼﺕ ﺗﻮ ﺗﺎ ﭼﻪ ﺣﺪ ﺍﺳــﺖ؟ ﺁﻳﺎ ﺭﻳﺎﺿﻴــﺎﺕ ﺗﻮ ﺧﻮﺏ ﺑﻮﺩﻩ ﺍﺳﺖ؟ ـ ﺩﺑﻴﺮﺳﺘﺎﻥ ﺭﺍ ﮔﺬﺭﺍﻧﺪﻩﺍﻡ .ﺭﻳﺎﺿﻴﺎﺕ ﻭ ﻣﻨﻄﻖ ﺭﺍ ﺩﻭﺳﺖ ﺩﺍﺷﺘﻪﺍﻡ. ﻫﻨﻮﺯ ﻫﻢ ﻫﺮﮔﺎﻩ ﻛﻪ ﭘﻴﺶ ﺁﻳﺪ ،ﺧﻮﺩﻡ ﺭﺍ ﺑﺎ ﻣﺴﺌﻠﻪﻫﺎﻯ ﻓﻜﺮﻯ ﻣﺸﻐﻮﻝ ﻣﻰﻛﻨﻢ. ـ ﺍﻛﻨــﻮﻥ ﻫﻢ ﻛــﻪ ﺑﻪ ﻃﻮﺭ ﻋﻤﻠﻰ ﺑﺎ ﻳﻚ ﻣﺴــﺌﻠﻪﻯ ﻓﻜﺮﻯ ﺩﺭﮔﻴﺮ ﺷــﺪﻩﺍﻯ .ﺩﺭ ﺁﻥ ﻫﻔﺘﻪﺍﻯ ﻛﻪ ﻭﺯﻳﺮ ﻣﻌﻠﻮﻡ ﻛﺮﺩﻩ ﺍﺳﺖ ﺑﺎﻳﺪ ﻋﺒﺪﺍﻟﻜﺎﻇﻢ ﺭﺍ ﺍﻋــﺪﺍﻡ ﻛﻨــﻰ .ﺧﻮﺍﻫﺶ ﻣﻦ ﺍﻳﻦ ﺍﺳــﺖ ﻛﻪ ﺍﮔﺮ ﻣﻰﺗﻮﺍﻧــﻰ ﺁﺩﻳﻨﻪ ﺭﺍ ﺑﺮﮔﺰﻳﻨﻰ. ﻧﮕﻬﺒﺎﻥ ﻓﻜﺮﻯ ﻛﺮﺩ ﻭ ﺑﺎ ﻗﺎﻃﻌﻴﺖ ﭘﺎﺳﺦ ﺩﺍﺩ: ـ ﻧــﻪ ،ﺍﻳــﻦ ﻣﻤﻜﻦ ﻧﻴﺴــﺖ ،ﻋﺒﺪﺍﻟﻜﺎﻇﻢ ﻣﺮﺩﻯ ﻛﻮﺩﻥ ﻧﻴﺴــﺖ ﻭ ﻣﻰﺩﺍﻧﺪ ﻛﻪ ﺩﺳﺘﻮﺭ ﻭﺯﻳﺮ ﭼﻪ ﺷﺮﻁﻫﺎﻳﻰ ﺩﺍﺭﺩ .ﻭﻗﺘﻰ ﭘﻨﺠﺸﻨﺒﻪ ﺑﮕﺬﺭﺩ ﻭ ﺯﻧﺪﻩ ﺑﻤﺎﻧﺪ .ﺑﺎﻣﺪﺍﺩ ﺭﻭﺯ ﺁﺩﻳﻨﻪ ﻛﻪ ﻧﺎﺷﺘﺎﻳﻰ ﺍﻭ ﺭﺍ ﺑﺮﺍﻳﺶ ﺑﺒﺮﻡ ،ﺧﻮﺍﻫﺪ ﮔﻔﺖ» :ﻧﮕﻬﺒﺎﻥ ،ﻣﻰﺩﺍﻧﻢ ﻛﻪ ﺗﺼﻤﻴﻢ ﺩﺍﺭﻯ ﺍﻣﺮﻭﺯ ﻣﺮﺍ ﺍﻋﺪﺍﻡ ﻛﻨﻰ ،ﺯﻳﺮﺍ ﺍﻣﺮﻭﺯ ﻭﺍﭘﺴﻴﻦ ﺭﻭﺯﻯ ﺍﺳﺖ ﻛﻪ ﻣﻰﺗﻮﺍﻧﻰ ﺣﻜﻢ ﺭﺍ ﺍﺟﺮﺍ ﻛﻨﻰ!« ﺩﻭﺭﻩﻯ ﺷﺎﻧﺰﺩﻫﻢ ،ﺷﻤﺎﺭﻩ ﻯ ،2ﺯﻣﺴﺘﺎﻥ 1389
ﺭﺍﻫﻨﻤﺎﻳﻲ
15
ـ ﺑﻨﺎﺑﺮﺍﻳــﻦ ،ﻫﻢ ﺗﻮ ﻭ ﻫﻢ ﻋﺒﺪﺍﻟﻜﺎﻇــﻢ ﻣﻰﺩﺍﻧﻴﺪ ﻛﻪ ﺍﺟﺮﺍﻯ ﺣﻜﻢ ﺑــﻪ ﺭﻭﺯ ﺁﺩﻳﻨﻪ ﻧﺨﻮﺍﻫﺪ ﺍﻓﺘــﺎﺩ .ﺍﻳﻦ ﻛﺎﺭ ﺩﺭ ﻳﻜﻰ ﺍﺯ ﺭﻭﺯﻫﺎﻯ ﺷــﻨﺒﻪ ﺗﺎ ﭘﻨﺠﺸــﻨﺒﻪ ﻋﻤﻠﻰ ﺧﻮﺍﻫﺪ ﺑﻮﺩ ،ﭘﺲ ﺍﮔﺮ ﻣﻤﻜﻦ ﺍﺳــﺖ ،ﭘﻨﺠﺸــﻨﺒﻪ ﺭﺍ ﺍﻧﺘﺨﺎﺏ ﻛﻦ. ﻧﮕﻬﺒﺎﻥ ﺑﺎﺯ ﺑﻪ ﻓﻜﺮ ﻓﺮﻭ ﺭﻓﺖ ﻭ ﺑﻌﺪ ﮔﻔﺖ: ـ ﻧﻪ ،ﺟﻨﺎﺏ ﺍﻣﻴﺮ ،ﭘﻨﺠﺸــﻨﺒﻪ ﻫﻢ ﻧﻤﻰﺗﻮﺍﻧــﻢ ،ﺯﻳﺮﺍ ﻫﻢ ﻣﻦ ﻭ ﻫﻢ ﻋﺒﺪﺍﻟﻜﺎﻇﻢ ﻣﻰﺩﺍﻧﻴﻢ ﻛﻪ ﭘﻨﺠﺸــﻨﺒﻪ ﺁﺧﺮﻳﻦ ﺭﻭﺯﻯ ﺍﺳﺖ ﻛﻪ ﻣﻰﺷﻮﺩ ﺣﻜﻢ ﺭﺍ ﺍﺟﺮﺍ ﻛﺮﺩ ﻭ ﻣﻦ ﻧﺒﺎﻳﺪ ﺭﻭﺯﻯ ﺭﺍ ﺑﺮﮔﺰﻳﻨﻢ ﻛﻪ ﻋﺒﺪﺍﻟﻜﺎﻇﻢ ﺑﺘﻮﺍﻧﺪ ﺁﻥ ﺭﺍ ﺗﺸﺨﻴﺺ ﺩﻫﺪ. ـ ﭘﺲ ﭘﻨﺠﺸﻨﺒﻪ ﺭﺍ ﻫﻢ ﺑﺎﻳﺪ ﻛﻨﺎﺭ ﺑﮕﺬﺍﺭﻯ .ﻣﻰﻣﺎﻧﺪ ﻳﻜﻰ ﺍﺯ ﺭﻭﺯﻫﺎﻯ ﺷﻨﺒﻪ ﺗﺎ ﭼﻬﺎﺭﺷﻨﺒﻪ ،ﺍﻳﻦﻃﻮﺭ ﻧﻴﺴﺖ؟ ـ ﺩﻗﻴﻘﺎً ﻫﻤﻴﻦﻃﻮﺭ ﺍﺳﺖ ﺟﻨﺎﺏ ﺍﻣﻴﺮ. ـ ﺍﺳﺘﺪﻻﻝ ﺭﺍ ﻛﻪ ﺍﺩﺍﻣﻪ ﺩﻫﻴﻢ ،ﻧﺘﻴﺠﻪ ﺧﻮﺍﻫﻰ ﮔﺮﻓﺖ ﻛﻪ ﭼﻬﺎﺭﺷﻨﺒﻪ ﺭﺍ ﻫﻢ ﺑﺎﻳﺪ ﻛﻨﺎﺭ ﺑﮕﺬﺍﺭﻯ. ﻧﮕﻬﺒﺎﻥ ﺑﺎ ﺑﻬﺖ ﻭ ﺗﻌﺠﺐ ﮔﻔﺖ: ـ ﻛﺎﻣ ً ﻼ ﺻﺤﻴﺢ ﺍﺳــﺖ .ﭼﻬﺎﺭﺷﻨﺒﻪ ﺭﺍ ﺑﺎﻳﺪ ﻛﻨﺎﺭ ﺑﮕﺬﺍﺭﻡ ﻭ ﺳﻪﺷﻨﺒﻪ ﺁﺧﺮﻳﻦ ﺭﻭﺯ ﺧﻮﺍﻫﺪ ﺑﻮﺩ ،ﺍﻣﺎ ﺑﻪ ﻫﻤﺎﻥ ﺩﻟﻴﻞ ،ﺳﻪﺷﻨﺒﻪ ﺭﺍ ﻫﻢ ﺑﺎﻳﺪ ﻛﻨﺎﺭ
16
ﺭﺍﻫﻨﻤﺎﻳﻲ
ﺩﻭﺭﻩﻯ ﺷﺎﻧﺰﺩﻫﻢ ،ﺷﻤﺎﺭﻩﻯ ،2ﺯﻣﺴﺘﺎﻥ 1389
ﺑﮕﺬﺍﺭﻡ .ﺩﻭﺷــﻨﺒﻪ ﺁﺧﺮﻳﻦ ﺭﻭﺯ ﻣﻰﺷــﻮﺩ ﻭ ﺁﻥ ﺭﺍ ﻫﻢ ﺑﺎﻳﺪ ﻛﻨﺎﺭ ﺑﮕﺬﺍﺭﻡ. ﺑﺎﺯ ﺑﻪ ﻫﻤﻴﻦ ﺩﻟﻴﻞ ،ﺩﻭﺷــﻨﺒﻪ ،ﻫﻢﭼﻨﻴﻦ ﻳﻜﺸﻨﺒﻪ ﻭ ﺳﺮﺍﻧﺠﺎﻡ ﺷﻨﺒﻪ ﺭﺍ ﻫﻢ ﺑﺎﻳﺪ ﻛﻨﺎﺭ ﺑﮕﺬﺍﺭﻡ. ـ ﺑﻨﺎﺑﺮﺍﻳﻦ ،ﺍﺟﺮﺍﻯ ﺣﻜﻢ ،ﺁﻥﮔﻮﻧﻪ ﻛﻪ ﻭﺯﻳﺮ ﺩﺳــﺘﻮﺭ ﺩﺍﺩﻩ ﻧﺎﻣﻤﻜﻦ ﺍﺳــﺖ .ﺑﺮﻭ ﻭ ﺣﻘﻴﻘﺖ ﺍﻣﺮ ﺭﺍ ﺑﻪ ﻭﺯﻳﺮ ﮔﺰﺍﺭﺵ ﻛﻦ ﻭ ﺑﻪ ﺍﻭ ﺑﮕﻮ ﺗﺎ ﺩﺳﺘﻮﺭ ﺁﺯﺍﺩﻯ ﻋﺒﺪﺍﻟﻜﺎﻇﻢ ﺭﺍ ﺑﺪﻫﺪ. ﺍﻣﻴﺮﺯﺍﺩﻩ ﻛﻪ ﮔﻔﺖ ﻭ ﮔﻮ ﺭﺍ ﺷﻨﻴﺪﻩ ﺑﻮﺩ ،ﺍﺯ ﺍﻣﻴﺮ ﭘﺮﺳﻴﺪ: ـ ﺁﻳﺎ ﺍﻳﻦ ﺍﺳﺘﺪﻻﻝ ﻫﻢ ﺍﺳﺘﻘﺮﺍﻯ ﺭﻳﺎﺿﻰ 1ﺍﺳﺖ؟ ـ ﮔﻮﻧﻪﺍﻯ ﺍﺯ ﺍﺳــﺘﺪﻻﻝ ﺑﺎ ﺭﻭﺵ ﺍﺳــﺘﻘﺮﺍﻯ ﺭﻳﺎﺿﻰ ﺍﺳــﺖ ﻛﻪ ﺑﻪ ﺁﻥ ﺍﺳــﺘﺪﻻﻝ ﺑﺎ ﺭﻭﺵ ﺑﺎﺯﮔﺸﺘﻰ ﻣﻰﮔﻮﻳﻨﺪ .ﺧﻮﺷﺤﺎﻟﻢ ﻛﻪ ﻋﺒﺪﺍﻟﻜﺎﻇﻢ ﻭ ﻫــﻢ ﻧﮕﻬﺒﺎﻥ ﻭﻯ ﺁﻥ ﺍﻧﺪﺍﺯﻩ ﻫﻮﺷــﻤﻨﺪ ﺑﻮﺩﻧﺪ ﻛــﻪ ﺑﺘﻮﺍﻧﻨﺪ ﺍﻳﻦﮔﻮﻧﻪ ﺍﺳﺘﺪﻻﻝ ﺭﺍ ﺩﺭﻙ ﻛﻨﻨﺪ ﻭﮔﺮﻧﻪ ﻛﺎﺭ ﺑﻪ ﺍﻳﻦ ﺁﺳﺎﻧﻰ ﭘﻴﺶ ﻧﻤﻰﺭﻓﺖ. ﭘﻰﻧﻮﺷﺖ ﺍﺳــﺘﺪﺍﻻﻝ ﺍﺳﺘﻘﺮﺍﻳﻰ ﺑﺎ ﺭﻭﺵ ﺑﺎﺯﮔﺸﺘﻰ ﺑﺮ ﺍﻳﻦ ﺍﺻﻞ ﺍﺳﺘﻮﺍﺭ ﺍﺳﺖ :ﺩﺭ ﺣﻮﺯﻩﻯ ﻋﺪﺩﻫﺎﻯ ﻃﺒﻴﻌــﻰ ﻣﻮﻛﻮﻝ ﺑﻪ ﺩﺭﺳــﺘﻰ ﺁﻥ ﺣﻜــﻢ ﺑﻪ ﺍﺯﺍﻯ ﻋﺪﺩ ﻃﺒﻴﻌﻰ k-1ﺑﺎﺷــﺪ ﻭ ﺍﮔﺮ ﺁﻥ ﺣﻜﻢ ﺑﻪ ﺍﺯﺍﻯ ﻳﻚ )ﻳﺎ ﺑﻪ ﺍﺯﺍﻯ ﻛﻮﭼﻚﺗﺮﻳﻦ ﻋﺪﺩ ﻣﻤﻜﻦ( ﺩﺭﺳــﺖ ﻧﺒﺎﺷﺪ ،ﺁﻥ ﺣﻜﻢ ﺑﻪ ﺍﺯﺍﻯ ﻫﻴﭻ ﻋﺪﺩ ﻃﺒﻴﻌﻰ ﺩﺭﺳﺖ ﻧﺨﻮﺍﻫﺪ ﺑﻮﺩ.
ﺩﺍﻧﺶﺍﻓﺰﺍﻳﯽ
واژه﹨︀ی ر︀︲﹩ ﺷﺎﺩﻯ ﺑﻬﺎﺭﻯ
ﻛﻠﻴﺪ ﻭﺍژﻩﻫﺎ :ﻭﺗﺮ ،ﻣﺜﻠﺚ ﻗﺎﺋﻢﺍﻟﺰﺍﻭﻳﻪ ،ﭘﺎﺭﻩﺧﻂ. ﺩﺭ ﺍﻳﻦ ﺷــﻤﺎﺭﻩ ﺍﺯ ﻣﺠﻠﻪ ،ﺩﺭ ﺳﺘﻮﻥ ﻭﺍژﻩﻫﺎﻯ ﺭﻳﺎﺿﻰ ،ﻭﺍژﻩﻯ ﻭﺗﺮ ﺭﺍ ﻣﻌﺮﻓﻰ ﻣﻰﻛﻨﻴﻢ. ﺍﻳﻦ ﻛﻠﻤﻪ ﺭﺍ ﻛﺠﺎ ﺷﻨﻴﺪﻩﺍﻳﺪ؟ ﺑﻠﻪ ﺩﺭﺳــﺖ ﺍﺳــﺖ ،ﺩﺭ ﻣﺜﻠــﺚ ﻗﺎﺋﻢﺍﻟﺰﺍﻭﻳﻪ ،ﺑــﻪ ﺑﺰﺭگﺗﺮﻳﻦ ﺿﻠﻊ ﻛﻪ ﺭﻭﺑﻪﺭﻭﻯ ﺯﺍﻭﻳﻪﻯ 90°ﻫﺴﺖ ،ﻭﺗﺮ ﻣﺜﻠﺚ ﻣﻰﮔﻮﻳﻨﺪ.
ﻣﺜ ً ﻼ ﺩﺭ ﺷﻜﻞ ،1ﺩﺭ ﻣﺜﻠﺚ ،ABCﺯﺍﻭﻳﻪﻯ ˆ ، 90 = Aﭘﺲ ﺿﻠﻊ ،BC ﻭﺗﺮ ﻣﺜﻠﺚ ﻗﺎﺋﻢﺍﻟﺰﺍﻭﻳﻪﻯ ABCﺍﺳﺖ .ﺍ ّﻣﺎ ﺩﺍﻧﺶﺁﻣﻮﺯﺍﻥ ﺍﻭﻝ ﺭﺍﻫﻨﻤﺎﻳﻰ ﺍﻛﻨــﻮﻥ ﻣﻰﺩﺍﻧﻨﺪ ﻛﻪ ﺩﺭ ﺩﺍﻳــﺮﻩ ﻧﻴﺰ ﻛﻠﻤﻪﻯ ﻭﺗﺮ ﺭﺍ ﺑــﻪ ﻛﺎﺭ ﻣﻰﺑﺮﻳﻢ. ﻳﺎﺩﺗﺎﻥ ﻫﺴﺖ ﻭﺗﺮ ﺩﺍﻳﺮﻩ ،ﭼﻪ ﺑﻮﺩ؟ ﺑﻠﻪ ،ﺑﻪ ﭘﺎﺭﻩﺧﻄﻰ ﻛﻪ ﺩﻭ ﺳــﺮ ﺁﻥ ﺭﻭﻯ ﺩﺍﻳﺮﻩ ﺑﺎﺷــﺪ ،ﻭﺗﺮ ﺩﺍﻳﺮﻩ ﮔﻮﻳﻨﺪ. ﻣﺜ ً ﻼ ﺑﻪ ﺷﻜﻞ 2ﻧﮕﺎﻩ ﻛﻨﻴﺪ. G
C
C A
B
ﻭﺗﺮ
.
O E B )ﺷﻜﻞ .1ﻣﺜﻠﺚ ﻗﺎﺋﻢﺍﻟﺰﺍﻭﻳﻪ(
A )ﺷﻜﻞ .2ﺩﺍﻳﺮﻩ ﻭ ﻭﺗﺮﻫﺎﻯ ﺁﻥ(
D F
ﺩﻭﺭﻩﻯ ﺷﺎﻧﺰﺩﻫﻢ ،ﺷﻤﺎﺭﻩ ﻯ ،2ﺯﻣﺴﺘﺎﻥ 1389
ﺭﺍﻫﻨﻤﺎﻳﻲ
17
ﭘﺎﺳﺦ ﻣﻌﻤﺎﻯ ﭼﻪ ﻛﺴﻰ ﻣﺮﺩﻡﺩﺍﺭ ﺍﺳﺖ؟ ﺳــﻌﻰ ﻛﻨﻴﺪ ﺗﻤﺎﻡ ﻭﺗﺮﻫﺎﻳــﻰ ﺍﺯ ﺩﺍﻳﺮﻩ ﺭﺍ ﻛﻪ ﺩﺭ ﺷــﻜﻞ ﻣﻰﺑﻴﻨﻴﺪ ،ﻧﺎﻡ ﺑﺒﺮﻳﺪ. ﻛﺎﻣ ً ﻼ ﺩﺭﺳــﺖ ﺍﺳــﺖ AB ،ﻭ CDﻭ DEﻭ ،GFﭼﻬــﺎﺭ ﻭﺗﺮ ﺩﺍﻳﺮﻩ ﻫﺴــﺘﻨﺪ .ﺩﺭ ﺑﻴﻦ ﺁﻥﻫﺎ ،GF ،ﻗﻄﺮ ﺩﺍﻳﺮﻩ ﻧﻴﺰ ﻫﺴﺖ ،ﺯﻳﺮﺍ ﺍﺯ ﻣﺮﻛﺰ ﺁﻥ، ﻳﻌﻨﻰ ﻧﻘﻄﻪﻯ Oﮔﺬﺷﺘﻪ ﺍﺳﺖ .ﻭﻟﻰ OEﻭﺗﺮ ﺩﺍﻳﺮﻩ ﻧﻴﺴﺖ ،ﺯﻳﺮﺍ ﻳﻚ ﺳﺮ ﺁﻥ Oﺍﺳﺖ ﻛﻪ ﺭﻭﻯ ﺩﺍﻳﺮﻩ ﻗﺮﺍﺭ ﻧﺪﺍﺭﺩ. ﺣﺎﻝ ﺑﻪ ﺟﻤﻠﻪﻯ ﺯﻳﺮ ﻛﻪ ﺩﺭﺑﺎﺭﻩﻯ ﺷﻜﻞ 3ﺍﺳﺖ ،ﺩﻗﺖ ﻛﻨﻴﺪ ﻭ ﻣﺸﺨﺺ ﻛﻨﻴﺪ ﻫﺮ ﻛﻠﻤﻪﻯ ﻭﺗﺮ ﻛﻪ ﺩﺭ ﺟﻤﻠﻪ ﺑﻪ ﻛﺎﺭ ﺭﻓﺘﻪ ،ﺑﻪ ﻛﺪﺍﻡ ﻣﻌﻨﺎﺳﺖ؟ »ﻗﻄﺮ ،ABﻭﺗﺮﻯ ﺍﺯ ﺩﺍﻳﺮﻩ ﺍﺳــﺖ ﻛــﻪ ﺍﺯ ﻭﺗﺮﻫﺎﻯ ﺩﻳﮕﺮ ﺁﻥ ،ﺑﺰﺭگﺗﺮ ﺍﺳــﺖ .ﻭﺗﺮﻫﺎﻯ BCﻭ CAﻭ ،ABﻳﻚ ﻣﺜﻠﺚ ﻗﺎﺋﻢﺍﻟﺰﺍﻭﻳﻪ ﺗﺸــﻜﻴﻞ ﺩﺍﺩﻩﺍﻧﺪ ﻛﻪ ،ABﻭﺗﺮ ﺁﻥ ﺍﺳﺖ«. ﺷﺎﻳﺪ ﺑﺎ ﺩﻳﺪﻥ ﺍﻳﻦ ﺟﻤﻠﻪ ،ﺩﺍﻧﺶﺁﻣﻮﺯﺍﻥ ﺳﻮﻡ ﺭﺍﻫﻨﻤﺎﻳﻰ ﻣﺘﻮﺟﻪ ﺑﺸﻮﻧﺪ ﻛﻪ ﭼﺮﺍ ﻭﺍژﻩﻯ »ﻭﺗﺮ« ﺍﻳﻦ ﺩﻭ ﻣﻌﻨﺎ ﺭﺍ ﺩﺍﺭﺩ؟
B
.
O
C
A
) ﺷﻜﻞ( 3
ﭘﺎﺳﺦﻫﺎ: ﻭﺗﺮ ﺩﺍﻳﺮﻩ ،ﻭﺗﺮ ﺩﺍﻳﺮﻩ ،ﻭﺗﺮ ﺩﺍﻳﺮﻩ ،ﻭﺗﺮ ﻣﺜﻠﺚ. ﻫﻤﻴﺸــﻪ ،ﺯﺍﻭﻳﻪﻯ ﻣﺤﺎﻃﻰ ﺭﻭﺑﻪﺭﻭ ﻗﻄﺮ 90° ،ﺍﺳــﺖ ﻭ ﻟﺬﺍ ﻳﻚ ﻣﺜﻠﺚ ﻗﺎﺋﻢﺍﻟﺰﺍﻭﻳﻪ ﺗﺸــﻜﻴﻞ ﻣﻰﺷــﻮﺩ ﻛﻪ ﻗﻄﺮ ﺩﺍﻳﺮﻩ ،ﻭﺗﺮ ﻣﺜﻠﺚ ﻧﻴﺰ ﻫﺴﺖ! 18
ﺭﺍﻫﻨﻤﺎﻳﻲ
ﺩﻭﺭﻩﻯ ﺷﺎﻧﺰﺩﻫﻢ ،ﺷﻤﺎﺭﻩﻯ ،2ﺯﻣﺴﺘﺎﻥ 1389
ﻣﻰﺩﺍﻧﻴﻢ ﻛﻪ ﻫﺮ ﻣﺮﺩ ،ﺳﻪ ﻭﻳﮋﮔﻰ ﺩﺍﺭﺩ .ﺑﻨﺎﺑﺮﺍﻳﻦ ﺍﺯ ﺟﻤﻠﻪﻫﺎﻯ 1ﻭ 2ﺩﺭﻣﻰﻳﺎﺑﻴﻢ ﻛﻪ ﺍﺣﻤﺪ ﺩﺍﺭﺍﻯ ﻳﻜﻰ ﺍﺯ ﻣﺠﻤﻮﻋﻪ ﻭﻳﮋﮔﻰﻫﺎﻯ ﺯﻳﺮ ﺍﺳﺖ: ـ ﺑﺬﻟﻪﮔﻮ ،ﺧﻮﺵﻗﻴﺎﻓﻪ ،ﻧﻴﺮﻭﻣﻨﺪ ـ ﺑﺬﻟﻪﮔﻮ ،ﺧﻮﺵﻗﻴﺎﻓﻪ ،ﻣﺮﺩﻡﺩﺍﺭ ـ ﺧﻮﺵﻗﻴﺎﻓﻪ ،ﻧﻴﺮﻭﻣﻨﺪ ،ﻣﺮﺩﻡﺩﺍﺭ ـ ﻧﻴﺮﻭﻣﻨﺪ ،ﺑﺎﻫﻮﺵ ،ﻣﺮﺩﻡﺩﺍﺭ ﺑــﻪ ﻫﻤﻴﻦ ﺗﺮﺗﻴﺐ ﺍﺯ ﺟﻤﻠﻪﻫﺎﻯ 1ﻭ 3ﻧﻴﺰ ﻧﺘﻴﺠﻪ ﻣﻰﮔﻴﺮﻳﻢ ﻛﻪ ﺑﻬﺮﻭﺯ ﻳﻜﻰ ﺍﺯ ﻣﺠﻤﻮﻋﻪ ﻭﻳﮋﮔﻰﻫﺎﻯ ﺯﻳﺮ ﺭﺍ ﺩﺍﺭﺩ: ـ ﺑﺬﻟﻪﮔﻮ ،ﺑﺎﻫﻮﺵ ،ﺧﻮﺵﻗﻴﺎﻓﻪ ـ ﺑﺎﻫﻮﺵ ،ﺧﻮﺵﻗﻴﺎﻓﻪ ،ﻧﻴﺮﻭﻣﻨﺪ ـ ﺑﺎﻫﻮﺵ ،ﺧﻮﺵﻗﻴﺎﻓﻪ ،ﻣﺮﺩﻡﺩﺍﺭ ـ ﺧﻮﺵﻗﻴﺎﻓﻪ ،ﻧﻴﺮﻭﻣﻨﺪ ،ﻣﺮﺩﻡﺩﺍﺭ ﺍﺯ ﺟﻤﻠﻪﻫﺎﻯ 1ﻭ 4ﻧﻴﺰ ﺩﺭﻣﻰﻳﺎﺑﻴﻢ ﻛﻪ ﺳــﺎﻣﺎﻥ ﺩﺍﺭﺍﻯ ﻳﻜﻰ ﺍﺯ ﻣﺠﻤﻮﻋﻪ ﻭﻳﮋﮔﻰﻫﺎﻯ ﺯﻳﺮ ﺍﺳﺖ: ـ ﺧﻮﺵﻗﻴﺎﻓﻪ ،ﻧﻴﺮﻭﻣﻨﺪ ،ﺑﺎﻫﻮﺵ ـ ﺧﻮﺵﻗﻴﺎﻓﻪ ،ﻧﻴﺮﻭﻣﻨﺪ ،ﻣﺮﺩﻡﺩﺍﺭ ـ ﻧﻴﺮﻭﻣﻨﺪ ،ﺑﺬﻟﻪﮔﻮ ،ﻣﺮﺩﻡﺩﺍﺭ ـ ﺑﺎﻫﻮﺵ ،ﺑﺬﻟﻪﮔﻮ ،ﻣﺮﺩﻡﺩﺍﺭ ﺍﺯ ﻣﺠﻤﻮﻋــﻪ ﻭﻳﮋﮔﻰﻫﺎﻯ ﺑﺎﻻ ﻭ ﺍﺯ ﺟﻤﻠــﻪﻯ 1ﭘﻰ ﻣﻰﺑﺮﻳﻢ ﻛﻪ ﺍﮔﺮ ﺍﺣﻤﺪ ﻣﺮﺩﻡﺩﺍﺭ ﺑﺎﺷــﺪ ،ﺑﻬﺮﻭﺯ ﻭ ﺳــﺎﻣﺎﻥ ﻫﺮ ﺩﻭ ﺑﺎﻫﻮﺵ ﻭ ﺧﻮﺵﻗﻴﺎﻓﻪﺍﻧــﺪ .ﺍﺯ ﺍﻳﻦ ﺭﻭ ،ﺍﺣﻤﺪ ﻧﺒﺎﻳــﺪ ﺑﺎﻫﻮﺵ ﻳﺎ ﺧﻮﺵﻗﻴﺎﻓﻪ ﺑﺎﺷﺪ .ﺍﺯ ﺁﻥﺟﺎ ﻛﻪ ﺍﻳﻦ ﻭﺿﻊ ﻧﺎﻣﻤﻜﻦ ﺍﺳﺖ ،ﺍﺣﻤﺪ ﻧﺒﺎﻳﺪ ﻣﺮﺩﻡﺩﺍﺭ ﺑﺎﺷﺪ. ﺍﺯ ﻣﺠﻤﻮﻋــﻪ ﻭﻳﮋﮔﻰﻫﺎﻯ ﺑﺎﻻ ﻭ ﺍﺯ ﺟﻤﻠﻪﻯ 1ﻣﻰﻓﻬﻤﻴﻢ ﻛﻪ ﺍﮔﺮ ﺑﻬﺮﻭﺯ ﻣﺮﺩﻡﺩﺍﺭ ﺑﺎﺷﺪ ،ﺍﺣﻤﺪ ﻭ ﺳﺎﻣﺎﻥ ﻫﺮ ﺩﻭ ﺧﻮﺵﻗﻴﺎﻓﻪﺍﻧﺪ. ﺑﻨﺎﺑﺮﺍﻳﻦ ،ﺑﻬﺮﻭﺯ ﻧﺒﺎﻳﺪ ﻣﺮﺩﻡﺩﺍﺭ ﺑﺎﺷﺪ. ﺑﻨﺎﺑﺮﺍﻳﻦ ،ﺳﺎﻣﺎﻥ ﺑﺎﻳﺪ ﻣﺮﺩﻡﺩﺍﺭ ﺑﺎﺷﺪ. ﺣﺎﻝ ،ﻣﻰﺗﻮﺍﻥ ﺳﻪ ﻭﻳﮋﮔﻰ ﻓﻘﻂ ﻳﻜﻰ ﺍﺯ ﻣﺮﺩﺍﻥ ﻭ ﺩﻭ ﻭﻳﮋﮔﻰ ﻫﺮﻳــﻚ ﺍﺯ ﻣــﺮﺩﺍﻥ ﺩﻳﮕﺮ ﺭﺍ ﺗﻌﻴﻴــﻦ ﻛﺮﺩ .ﺍﺯ ﺁﻥﺟﺎ ﻛﻪ ﺳــﺎﻣﺎﻥ ﻣﺮﺩﻡﺩﺍﺭ ﺍﺳــﺖ ،ﭘﺲ ﺍﺣﻤﺪ ﺑﺬﻟﻪﮔــﻮ ،ﺧﻮﺵﻗﻴﺎﻓﻪ ﻭ ﻧﻴﺮﻭﻣﻨﺪ ﻭ ﺑﻬﺮﻭﺯ ﻫﻢ ﺧﻮﺵﻗﻴﺎﻓﻪ ﻭ ﺑﺎﻫﻮﺵ ﺍﺳﺖ.ﺳــﺮﺍﻧﺠﺎﻡ ،ﭼﻮﻥ ﺳــﺎﻣﺎﻥ ﻧﻤﻰﺗﻮﺍﻧﺪ ﺧﻮﺵﻗﻴﺎﻓﻪ ﺑﺎﺷﺪ ،ﺑﺎﻫﻮﺵ ﻭ ﻣﺮﺩﻡﺩﺍﺭ ﺍﺳﺖ.
ﺩﺍﻧﺶﺍﻓﺰﺍﻳﯽ
را︋︴﹤ی ﹁﹫︓︀︾﹢رس ،ا︔︊︀ت و﹋︀ر︋︣د﹨︀ی آن ﻣﻮﺳﻲ ﻫﺎﺷﻤﻠﻮ
ﻛﻠﻴﺪ ﻭﺍژﻩﻫﺎ :ﻓﻴﺜﺎﻏﻮﺭﺱ ،ﻣﺜﻠﺚ ﻗﺎﺋﻢﺍﻟﺰﺍﻭﻳﻪ ،ﻋﺪﺩ ﮔﻨﮓ ،ﻋﺪﺩ ﮔﻮﻳﺎ ،ﻣﺴﺎﺣﺖ ﻣﺜﻠﺚ ،ﺣﺠﻢ ﻣﻜﻌﺐ ،ﻫﺮﻡ، ﺗﺸﺎﺑﻪ . ﺍﺷـﺎﺭﻩ :ﺍﺑﺘـﺪﺍ ﺑـﻪ ﺗﺎﺭﻳﺨﭽـﻪﺍﻯ ﻣﺨﺘﺼﺮ ﺩﺭﺑـﺎﺭﻩﻯ ﭘﻴﺪﺍﻳـﺶ ﻣﻜﺘـﺐ ﻭ ﺭﺍﺯ ﻭ ﺭﻣـﺰ
ﻣﻰﺳــﺎﺧﺘﻨﺪ .ﺁﻧﺎﻥ ﻣﻰﺩﺍﻧﺴــﺘﻨﺪ ﻛﻪ ﻣﺜﻠﺜﻰ ﺑﺎ ﺍﻳﻦ ﺍﺑﻌــﺎﺩ ﺩﺍﺭﺍﻯ ﻳﻚ ﺯﺍﻭﻳﻪﻯ ﻗﺎﺋﻤﻪ )ﻳﻌﻨﻰ ﺩﺍﺭﺍﻯ ﺩﻭ ﭘﺎﺭﻩﺧﻂ ﺭﺍﺳــﺖ ﻋﻤﻮﺩ ﺑﺮ ﻫﻢ( ﺍﺳــﺖ. ﺑﻨﺎﺑﺮﺍﻳﻦ ﺑﺎ ﺍﻳﻦ ﺭﻭﺵ ﺑﻪ ﺳﺎﺩﮔﻰ ،ﻗﻄﻌﻪ ﺯﻣﻴﻦﻫﺎﻳﻰ ﻣﺎﻧﻨﺪ ﺷﻜﻞ ﺯﻳﺮ ﺭﺍ ﻛﻪ ﺩﺍﺭﺍﻯ ﮔﻮﺷﻪ ﻳﺎ ﮔﻮﺷﻪﻫﺎﻳﻰ ﻗﺎﺋﻢ ﺑﻮﺩﻧﺪ ،ﺗﻘﺴﻴﻢ ﻣﻰﻛﺮﺩﻧﺪ.
ﻓﻴﺜﺎﻏﻮﺭﺳـﻴﺎﻥ ﺍﺷـﺎﺭﻩ ﻣﻰﻛﻨﻴﻢ ﻭ ﺳﭙﺲ ﺑﻪ ﺭﻭﺵﻫﺎﻳﻰ ﻛﻪ ﺑﻪ ﺍﺛﺒﺎﺕ ﺭﺍﺑﻄﻪﻯ ﻓﻴﺜﺎﻏﻮﺭﺱ ﻣﻰﺍﻧﺠﺎﻣﻨـﺪ ﻣﻰﭘﺮﺩﺍﺯﻳﻢ ﻭ ﺩﺭ ﺁﺧﺮ ﺑﺮﺧﻰ ﺍﺯ ﻛﺎﺭﺑﺮﺩﻫﺎﻯ ﺁﻥ ﺭﺍ ﺑﺮﺭﺳﻰ ﻣﻰﻛﻨﻴﻢ.
ﺗﺎﺭﻳﺨﭽﻪ ﺗﻘﺴــﻴﻢﺑﻨﺪﻯ ﺯﻣﻴﻦﻫﺎﻯ ﺍﻃﺮﺍﻑ ﻧﻴﻞ ،ﻛﻪ ﻫﺮ ﺳﺎﻝ ﺑﺮ ﺍﺛﺮ ﻃﻐﻴﺎﻥ ﺁﺏ ﺍﺯ ﺑﻴﻦ ﻣﻰﺭﻓﺖ ﺳﺒﺐ ﻣﻰﺷﺪ ﻛﻪ ﻫﺮﭼﻨﺪ ﻳﻚ ﺑﺎﺭ ﺍﻳﻦ ﺗﻘﺴﻴﻢﺑﻨﺪﻯ ﺗﺠﺪﻳﺪ ﺷﻮﺩ. ﺷﻴﻮﻩﻯ ﺗﻘﺴﻴﻢﺑﻨﺪﻯ ﺑﻪ ﺍﻳﻦ ﺻﻮﺭﺕ ﺑﻮﺩ ﻛﻪ ﻣﺜﻠﺜﻰ ﺑــﺎ ﺍﺑﻌــﺎﺩ 3ﻭ 4ﻭ ) 5ﻭﺍﺣﺪ ﻃﻮﻝ( ﺑﺎ ﺳــﻨﮓﭼﻴﻦ ﻳﺎ ﻧﻰ ﻳــﺎ ﭼﻮﺏ ﻳﺎ ﻣــﻮﺍﺩﻯ ﺩﻳﮕﺮ
ﻓﻴﺜﺎﻏــﻮﺭﺱ ﺭﺍﺑﻄﻪﻯ ﺑﻴﻦ ﺿﻠﻊﻫﺎﻯ ﻣﺜﻠﺚ ﻣﺼﺮﻯ ﺭﺍ ﭘﻴﺪﺍ ﻛﺮﺩ ﻛﻪ ﺑﺎ ﺭﺍﺑﻄﻪﻯ 32+42=52ﺑﻴﺎﻥ ﻣﻰﺷــﻮﺩ .ﻓﻴﺜﺎﻏﻮﺭﺱ ﻧﺸــﺎﻥ ﺩﺍﺩ ﻛﻪ ﺍﻳﻦ ﺭﺍﺑﻄﻪ ﺑﺮﺍﻯ ﻫﺮ ﻣﺜﻠﺚ ﻗﺎﺋﻢﺍﻟﺰﺍﻭﻳﻪ ﺑﺎ ﺿﻠﻊﻫﺎﻯ b، aﻭ cﺑﻪ ﺻﻮﺭﺕ ﺯﻳﺮ a2+b2=c2ﺩﺭﺳﺖ ﺍﺳﺖ ﻭ ﺭﺍﺑﻄﻪﻫﺎﻳﻰ ﺑﺮﺍﻯ ﺍﻳﻦ ﺿﻠﻊﻫﺎ ﭘﻴﺪﺍ ﻛﺮﺩ. ﺍﻣﺮﻭﺯﻩ ﻣﺜﻠﺚﻫﺎﻯ ﻗﺎﺋﻢﺍﻟﺰﺍﻭﻳــﻪﺍﻯ ﻛﻪ ﺿﻠﻊﻫﺎﻯ ﺁﻥﻫﺎ ﺑﺎ ﻋﺪﺩﻫﺎﻯ ﻃﺒﻴﻌﻰ ﺑﻴﺎﻥ ﺷﻮﻧﺪ ،ﻣﺜﻠﺚﻫﺎﻯ ﻓﻴﺜﺎﻏﻮﺭﺳﻰ ﻧﺎﻣﻴﺪﻩ ﻣﻰﺷﻮﻧﺪ. ﺍﻳﻦ ﻣﻄﻠــﺐ ﺭﺍ ﻣﺘﻌﻠﻖ ﺑﻪ ﻓﻴﺜﺎﻏﻮﺭﺱ ﻣﻰﺩﺍﻧﻨــﺪ ﻛﻪ ﺩﺭ ﻫﺮ ﻣﺜﻠﺚ ﻗﺎﺋﻢﺍﻟﺰﺍﻭﻳﻪ ،ﻣﺮﺑﻌﻰ ﻛﻪ ﺭﻭﻯ ﻭﺗﺮ ﺳــﺎﺧﺘﻪ ﺷــﻮﺩ ﺑﺮﺍﺑﺮ ﺍﺳﺖ ﺑﺎ ﻣﺠﻤﻮﻉ ﻣﺮﺑﻊﻫﺎﻳﻰ ﻛﻪ ﺭﻭﻯ ﺩﻭ ﺿﻠﻊ ﺩﻳﮕﺮ ﺳﺎﺧﺘﻪ ﻣﻰﺷﻮﺩ. ﺭﺍﺯ ﻣﻜﺘﺐ ﻓﻴﺜﺎﻏﻮﺭﺳﻰ ﺍﺑﺘﺪﺍ ﮔﻤﺎﻥ ﻣﻰﻛﺮﺩﻧﺪ ﻛﻪ ﺿﻠﻊﻫﺎﻯ ﻫﺮ ﻣﺜﻠﺚ ﻗﺎﺋﻢﺍﻟﺰﺍﻭﻳﻪ ﺭﺍ ﻣﻰﺗﻮﺍﻥ ﺑﺎ ﻋﺪﺩﻫﺎﻯ ﻃﺒﻴﻌﻰ ﺑﻴﺎﻥ ﻛﺮﺩ ،ﻭﻟﻰ ﺑﺮﺭﺳﻰﻫﺎﻯ ﺭﻳﺎﺿﻰﺩﺍﻥﻫﺎﻯ ﻣﻜﺘﺐ ﻓﻴﺜﺎﻏﻮﺭﺳﻰ ﻧﺸﺎﻥ ﺩﺍﺩ ﻛﻪ ﺍﻳﻦ ﺗﺼﻮﺭ ﺩﺭﺳﺖ ﻧﻴﺴﺖ ،ﺑﺮﺍﻯ ﻣﺜﺎﻝ ،ﻣﺜﻠﺚ ﻗﺎﺋﻢﺍﻟﺰﺍﻭﻳﻪﻯ ﻣﺘﺴﺎﻭﻯﺍﻟﺴﺎﻗﻴﻦ ﺷﻤﺎﺭﻩ ﻯ ،2ﺯﻣﺴﺘﺎﻥ 1389 ﺩﻭﺭﻩﻯ ﺷﺎﻧﺰﺩﻫﻢ ،ﺷﻤﺎ
ﺭﺍﻫﻨﻤﺎﻳﻲ
19
ﻳﻚ ﻣﺜﻠﺚ ﻓﻴﺜﺎﻏﻮﺭﺳــﻰ ﻧﻴﺴﺖ ،ﺯﻳﺮﺍ ﻧﻤﻰﺗﻮﺍﻥ ﻋﺪﺩﻫﺎﻳﻰ ﺻﺤﻴﺢ ﭘﻴﺪﺍ ﻛﺮﺩ ﻛﻪ ﺩﺭ ﺭﺍﺑﻄﻪﻯ ﺯﻳﺮ ﺻﺪﻕ ﻛﻨﻨﺪ: 2 2 2 2 2 ﻳﺎ a = b: a +a =c 2a = c )ﺍﻣﺮﻭﺯﻩ ﻣﻰﺩﺍﻧﻴﻢ ﻛﻪ ﺗﺴﺎﻭﻯ ﺍﺧﻴﺮ ،ﻣﻌﺎﺩﻝ ﺗﺴﺎﻭﻯ ﺯﻳﺮ ﺍﺳﺖ: = 2
c a
ﻳﺎ
c=a 2
)ﻣﻰﺗــﻮﺍﻥ ﺛﺎﺑﺖ ﻛﺮﺩ ﻛﻪ 2ﺭﺍ ﻧﻤﻰﺗﻮﺍﻥ ﺑﻪ ﺻﻮﺭﺕ ﻳﻚ ﻛﺴــﺮ ﻧﻮﺷﺖ ،ﻳﻌﻨﻰ ﻋﺪﺩﻯ ﮔﻮﻳﺎ ﻣﺴﺎﻭﻯ 2ﻭﺟﻮﺩ ﻧﺪﺍﺭﺩ(. ﺍﻳﻦ ﻛﺸﻒ ﺑﺮﺍﻯ ﻓﻴﺜﺎﻏﻮﺭﺳــﻰﻫﺎ ﻣﺼﻴﺒﺖﺑﺎﺭ ﺑﻮﺩ ،ﺯﻳﺮﺍ ﺍﻳﻦ ﺍﻋﺘﻘﺎﺩ ﺁﻥﻫﺎ ﺭﺍ ﻛﻪ ﻫﻤﻪﻯ ﭘﺪﻳﺪﻩﻫﺎ ﺑﺎ ﻋﺪﺩﻫﺎﻯ ﻃﺒﻴﻌﻰ ﻗﺎﺑﻞ ﺑﻴﺎﻥ ﻫﺴــﺘﻨﺪ، ﺩﭼﺎﺭ ﺷﻜﺴــﺖ ﻛﺮﺩ .ﻛﺸــﻒ ﺍﻳﻦ ﻣﻄﻠﺐ ﻛﻪ ﺩﻧﻴــﺎﻯ ﻋﺪﺩﻫﺎ ﺑﺎ ﺩﻧﻴﺎﻯ ﺳــﺎﺧﺘﻤﺎﻥﻫﺎﻯ ﻫﻨﺪﺳــﻰ ﻣﺘﻔﺎﻭﺕ ﺍﺳــﺖ ،ﭼﻨﺎﻥ ﺍﺛﺮ ﺑﺰﺭﮔﻰ ﺩﺍﺷــﺖ ﻛﻪ ﺩﺍﻧﺸــﻤﻨﺪﺍﻥ ﻓﻴﺜﺎﻏﻮﺭﺳــﻰ ﺁﻥ ﺭﺍ ﺑﻪ ﻋﻨﻮﺍﻥ ﺭﺍﺯﻯ ﻣﺨﻔﻰ ﻛﺮﺩﻧﺪ ﻭ ﺑﺮﺭﺳﻰﻫﺎﻯ ﻫﻨﺪﺳﻰ ﺭﺍ ﺑﻪ ﻃﻮﺭ ﻛﻠﻰ ﺍﺯ ﺣﺴﺎﺏ ﺟﺪﺍ ﻛﺮﺩﻧﺪ. ﺍﻳﻦ ﻣﻄﻠﺐ ﺑﻪ ﻃﻮﺭ ﺟﺪﻯ ﻣﺎﻧﻊ ﭘﻴﺸﺮﻓﺖ ﺣﺴﺎﺏ ﺩﺭ ﻳﻮﻧﺎﻥ ﺷﺪ ،ﺩﺭ ﺣﺎﻟﻰﻛﻪ ﻫﻨﺪﺳﻪ ﺭﺍ ﺑﻪ ﻧﺤﻮ ﺳﺮﻳﻌﻰ ﺗﻜﺎﻣﻞ ﺩﺍﺩ. ﺍﻓﺘﺨﺎﺭ ﺗﻔﻜﺮ ﺭﻳﺎﺿﻰ ﻓﻴﺜﺎﻏﻮﺭﺳﻰ ﻣﺸﻬﻮﺭﺗﺮﻳﻦ ﻛﺸﻒ ﻓﻴﺜﺎﻏﻮﺭﺳﻰ ﺍﻳﻦ ﺍﺳﺖ: »ﻣﺮﺑﻌﻰ ﻛﻪ ﺭﻭﻯ ﻭﺗﺮ ﻣﺜﻠﺚ ﻗﺎﺋﻢﺍﻟﺰﺍﻭﻳﻪ ﺳﺎﺧﺘﻪ ﺷﻮﺩ ﺑﺮﺍﺑﺮ ﺍﺳﺖ ﺑﺎ ﻣﺠﻤﻮﻉ ﻣﺮﺑﻊﻫﺎﻳﻰ ﻛﻪ ﺭﻭﻯ ﺿﻠﻊﻫﺎﻯ ﻣﺠﺎﻭﺭ ﺑﻪ ﺯﺍﻭﻳﻪﻯ ﻗﺎﺋﻤﻪ ﺳﺎﺧﺘﻪ ﻣﻰﺷﻮﻧﺪ «.ﻋﻜﺲ ﺍﻳﻦ ﻣﻄﻠﺐ ﻫﻢ ﺻﺤﻴﺢ ﺍﺳﺖ( a 2 + b2 = c2 ) : ﺍﮔﺮ ﺿﻠﻊﻫﺎﻯ b، aﻭ cﺍﺯ ﻣﺜﻠﺜﻰ ﺩﺭ ﺷــﺮﺍﻳﻂ ﻓﻴﺜﺎﻏﻮﺭﺳــﻰ:ﺻﺪﻕ ﻛﻨﻨــﺪ ،ﺩﺭ ﺍﻳﻦ ﺻﻮﺭﺕ ﻣﺜﻠــﺚ ﻣﻔﺮﻭﺽ ﻗﺎﺋﻢﺍﻟﺰﺍﻭﻳﻪ ﺍﺳــﺖ ﻭ ﺯﺍﻭﻳﻪﻯ ﻗﺎﺋﻤﻪﻯ ﺁﻥ ﺭﻭﺑﻪﺭﻭﻯ ﺿﻠﻊ ) cﻭﺗﺮ ﻣﺜﻠﺚ( ﺍﺳﺖ )ﻣﻄﺎﺑﻖ ﺷﻜﻞ ﺯﻳﺮ(.
ﺑﻪ ﻭﻳﮋﻩ ﻣﺜﻠﺜﻰ ﺟﺎﻟﺐ ﺍﺳــﺖ ﻛﻪ ﺳﻪ ﺿﻠﻊ ﺁﻥ ﺑﺎ ﻋﺪﺩﻫﺎﻯ ﺻﺤﻴﺢ ﺑﻴﺎﻥ ﺷﻮﺩ ﻭ ﺷﺮﻁ ﻓﻴﺜﺎﻏﻮﺭﺳﻰ ﺩﺭ ﻣﻮﺭﺩ ﺁﻥﻫﺎ ﺑﺮﻗﺮﺍﺭ ﺑﺎﺷﺪ. ﺑﺮﺍﻯ ﻣﺜﺎﻝ ،ﻣﺜﻠﺚ ﺑﺎ ﺿﻠﻊﻫﺎﻯ 4 ،3ﻭ 5ﺷــﺮﻁ ﻓﻴﺜﺎﻏﻮﺭﺳــﻰ ﺭﺍ ﻗﺒﻮﻝ ﺩﺍﺭﺩ: 2 2 2 3 +4 =5 ﻭ ﺍﻳﻦ ﺳﺎﺩﻩﺗﺮﻳﻦ ﻭ ﻣﻌﺮﻭﻑﺗﺮﻳﻦ ﻣﺜﻠﺚ ﻓﻴﺜﺎﻏﻮﺭﺳﻰ ﺍﺳﺖ. ﺩﺭ ﺍﻳﻦ ﺟﺎ ﭼﻨﺪ ﻣﺜﻠﺚ ﻓﻴﺜﺎﻏﻮﺭﺳﻰ ﺁﻭﺭﺩﻩﺍﻳﻢ: a=3 b=4 c=5 a=5 b=12 c=13 a=15 b=8 c=17 a=7 b=24 c=25 c=29 a=21 b=20 c=41 a=9 b=40 b=24 c=26 a=10 ﺑــﻪ ﺳــﺎﺩﮔﻰ ﺩﻳــﺪﻩ ﻣﻰﺷــﻮﺩ ﻛﻪ ﻫﻤــﻪﻯ ﺍﻳــﻦ ﻣﺜﻠﺚﻫــﺎ ﺩﺭ ﺷــﺮﻁ ﻓﻴﺜﺎﻏﻮﺭﺳــﻰ a 2 + b2 = c2ﺻــﺪﻕ ﻣﻰﻛﻨﻨــﺪ ﻭ ﺑﻨﺎﺑﺮﺍﻳــﻦ ﻗﺎﺋﻢﺍﻟﺰﺍﻭﻳﻪﺍﻧﺪ. ﺩﺭ ﻣﺼﺮ ﻗﺪﻳﻢ ﻭ ﺳﺎﻳﺮ ﻛﺸﻮﺭﻫﺎﻯ ﺷﺮﻕ ﺁﺳﻴﺎ ﺍﺯ ﻣﺜﻠﺜﻰ ﻛﻪ ﺿﻠﻊﻫﺎﻯ ﺁﻥ 4، 3ﻭ 5ﺑﺎﺷــﺪ ،ﺑﺮﺍﻯ ﺳــﺎﺧﺘﻦ ﺯﺍﻭﻳﻪﻯ ﻗﺎﺋﻤﻪ )ﻳﻌﻨﻰ ﺑﺮﺍﻯ ﺭﺳﻢ ﺩﻭ ﺧﻂ ﺭﺍﺳــﺖ ﻋﻤﻮﺩ ﺑﺮ ﻫﻢ( ﺩﺭ ﻋﻤﻞ ﺍﺳﺘﻔﺎﺩﻩ ﻣﻰﻛﺮﺩﻩﺍﻧﺪ .ﺗﺼﺎﺩﻓﻰ ﻧﻴﺴــﺖ ﻛﻪ ﺑﺎﺳﺘﺎﻥﺷﻨﺎﺳــﺎﻥ ﭼﻨﻴــﻦ ﻧﺴــﺒﺖﻫﺎﻳﻰ ﺭﺍ ﺩﺭ ﺍﻧﺪﺍﺯﻩﻫﺎﻯ ﺳــﻨﮓﻫﺎﻯ ﺗﺮﺍﺷﻴﺪﻩ ﺷﺪﻩﻯ ﻫﺮﻡ ﺧﻔﺮﻭﻥ ﭘﻴﺪﺍ ﻛﺮﺩﻩﺍﻧﺪ .ﺍﻳﻦ ﺣﻘﻴﻘﺖ ﺑﺴﻴﺎﺭ ﺟﺎﻟﺐ ﺍﺳﺖ ﻛﻪ ﺍﺗﺎﻕ ﻓﺮﻋﻮﻥ ﺩﺭ ﻫﺮﻡ ﻣﺸﻬﻮﺭ ﺧﺌﻮﭘﺲ ﺍﻧﺪﺍﺯﻩﻫﺎﻳﻰ ﺩﺍﺭﺩ ﻛــﻪ ﻛﺎﻣ ً ﻼ ﺑﻪ ﻋﺪﺩﻫــﺎﻯ 4 ،3ﻭ 5ﻣﺮﺑﻮﻁﺍﻧﺪ .ﺍﮔﺮ ﻗﻄﺮ ﺗﻤﺎﻡ ﺍﺗﺎﻕ ﺭﺍ 5ﻭﺍﺣﺪ ﺑﮕﻴﺮﻳﻢ ،ﺑﺰﺭگﺗﺮﻳﻦ ﺩﻳﻮﺍﺭ ﺁﻥ 4ﻭ ﻗﻄﺮ ﻛﻮﭼﻚﺗﺮﻳﻦ ﺩﻳﻮﺍﺭ ﺁﻥ ﻣﺴﺎﻭﻯ 3ﻭﺍﺣﺪ ﺍﺳﺖ. ﻣﻄﺎﺑﻖ ﺷﻜﻞ ﺯﻳﺮ:
5 A c B
b a
C
3
4 )ﺍﺗﺎﻕ ﻓﺮﻋﻮﻥ ﺩﺭ ﻫﺮﻡ ﺧﺌﻮﭘﺲ( ﺩﺭ ﺩﻭﺭﺍﻥ ﺑﺎﺳــﺘﺎﻥ ،ﻣﺜﻠﺜﻰ ﺭﺍ ﻛﻪ ﺿﻠﻊﻫﺎﻯ ﺁﻥ 4 ،3ﻭ 5ﺑﺎﺷــﺪ، ﺷــﻜﻠﻰ ﺍﺳــﺮﺍﺭﺁﻣﻴﺰ ﻭ ﺟﺎﺩﻭﻳﻰ ﺑﻪ ﺣﺴــﺎﺏ ﻣﻰﺁﻭﺭﺩﻧﺪ .ﭼﻨﻴﻦ ﻣﺜﻠﺜﻰ ﺧﺎﺻﻴﺖﻫــﺎﻯ ﺟﺎﻟــﺐ ﺩﻳﮕﺮﻯ ﻫﻢ ﺩﺍﺭﺩ .ﻣﺤﻴــﻂ ﺁﻥ ﺑﺎ ﻋﺪﺩ 12ﺑﻴﺎﻥ ّ ﻣﻰﺷــﻮﺩ ﻭ ﻣﺴﺎﺣﺖ ﺁﻥ ﺑﺮﺍﺑﺮ ﺑﺎ 6ﺍﺳﺖ ،ﻳﻌﻨﻰ ﻋﺪﺩﻯ ﻛﻪ ﺩﺭﺳﺖ ﺑﻌﺪ
20
ﺭﺍﻫﻨﻤﺎﻳﻲ
ﺩﻭﺭﻩﻯ ﺷﺎﻧﺰﺩﻫﻢ ،ﺷﻤﺎﺭﻩﻯ ،2ﺯﻣﺴﺘﺎﻥ 1389
ﺍﺯ ﺳــﻪ ﻋﺪﺩ ﺿﻠﻊﻫﺎ ﻗﺮﺍﺭ ﮔﺮﻓﺘﻪ ﺍﺳﺖ؛ ﺑﺎﻻﺗﺮ ﺍﺯ ﻫﻤﻪ ﻣﻰﺗﻮﺍﻥ ﺳﻪ ﻋﺪﺩ ﭘﻴــﺪﺍ ﻛﺮﺩ ﻛﻪ ﻣﺠﻤــﻮﻉ ﻣﻜﻌﺐﻫﺎﻯ ﺁﻥﻫﺎ ﺧﻮﺩ ﻣﻜﻌﺐ ﻛﺎﻣﻠﻰ ﺑﺎﺷــﺪ، ﺑﺮﺍﻯ ﻣﺜﺎﻝ: 3 3 3 3 )ﺭﺍﺑﻄﻪﻯ ﻣﻨﺘﺴﺐ ﺑﻪ ﺍﻓﻼﻃﻮﻥ( 3 +4 +5 =6 ﺍﻳﻦ ﺭﺍﺑﻄﻪ ﺑﻪ ﻣﻌﻨﺎﻯ ﺁﻥ ﺍﺳــﺖ ﻛﻪ ﺣﺠﻢ ﻣﻜﻌﺒﻰ ﺑﻪ ﺿﻠﻊ 6ﻭﺍﺣﺪ ﺑﺮﺍﺑﺮ ﺑﺎ ﻣﺠﻤﻮﻉ ﺣﺠﻢﻫﺎﻯ ﺳــﻪ ﻣﻜﻌﺐ ﺑﺎ ﺿﻠﻊﻫﺎﻯ ﻣﺴﺎﻭﻯ 4 ،3ﻭ 5 ﻭﺍﺣﺪ ﺍﺳﺖ )ﺍﻳﻦ ﻗﻀﻴﻪ ﻣﻨﺘﺴﺐ ﺑﻪ ﺍﻓﻼﻃﻮﻥ ﺍﺳﺖ(. ﻋﺪﺩﻫﺎﻯ ﻓﻴﺜﺎﻏﻮﺭﺳــﻰ ﺧﺎﺻﻴﺖﻫﺎﻯ ﺟﺎﻟﺐ ﺩﻳﮕﺮﻯ ﻫﻢ ﺩﺍﺭﻧﺪ ﻛﻪ ﺩﺭ ﺍﻳﻦﺟﺎ ﺑﺪﻭﻥ ﺍﺛﺒﺎﺕ ﺁﻥﻫﺎ ﺭﺍ ﺫﻛﺮ ﻣﻰﻛﻨﻴﻢ: (1ﻳﻜــﻰ ﺍﺯ »ﺿﻠﻊﻫــﺎﻯ ﻣﺠﺎﻭﺭ ﺑﻪ ﺯﺍﻭﻳﻪﻯ ﻗﺎﺋﻤــﻪ« ﻣﻀﺮﺑﻰ ﺍﺯ 3 ﺍﺳﺖ. (2ﻳﻜــﻰ ﺍﺯ »ﺿﻠﻊﻫــﺎﻯ ﻣﺠﺎﻭﺭ ﺑﻪ ﺯﺍﻭﻳﻪﻯ ﻗﺎﺋﻤــﻪ« ﻣﻀﺮﺑﻰ ﺍﺯ 4 ﺍﺳﺖ. (3ﻳﻜﻰ ﺍﺯ ﺳﻪ ﻋﺪﺩ ﻓﻴﺜﺎﻏﻮﺭﺳﻰ ﻣﻀﺮﺑﻰ ﺍﺯ 5ﺍﺳﺖ. ﺑﺪﻭﻥ ﺗﺮﺩﻳﺪ ،ﻫﻨﻮﺯ ﻫﻢ ﻧﺠﺎﺭﻫﺎﻯ ﺭﻭﺳــﺘﺎﻫﺎ ﻣﻮﻗﻊ ﺳﺎﺧﺘﻦ ﺧﺎﻧﻪﻫﺎ ﻳﺎ ﺍﻧﺒﺎﺭﻫﺎﻯ ﭼﻮﺑﻰ ﺑﺮﺍﻯ ﺍﻳﻦﻛﻪ ﺯﺍﻭﻳﻪﻯ ﻗﺎﺋﻤﻪ ﺑﻪ ﺩﺳﺖ ﺁﻭﺭﻧﺪ ﺍﺯ ﻣﺜﻠﺜﻰ ﺑﻪ ﺿﻠﻊﻫﺎﻯ 4 ،3ﻭ 5ﺍﺳﺘﻔﺎﺩﻩ ﻣﻰﻛﻨﻨﺪ؛ ﻭ ﺍﻳﻦ ﺩﺭﺳﺖ ﻫﻤﺎﻥ ﺷﻴﻮﻩﺍﻯ ﺍﺳــﺖ ﻛﻪ ﺩﺭ ﻫﺰﺍﺭﺍﻥ ﺳــﺎﻝ ﻗﺒﻞ ﺑﺮﺍﻯ ﺳﺎﺧﺘﻤﺎﻥ ﻣﻌﺒﺪﻫﺎﻯ ﺑﺰﺭگ ﺩﺭ ﻣﺼﺮ ،ﺑﺎﺑﻞ ،ﭼﻴﻦ ﻭ ﺍﺣﺘﻤﺎﻻً ﺩﺭ ﻣﻜﺰﻳﻚ ﺑﻪ ﻛﺎﺭ ﻣﻰﺭﻓﺘﻪ ﺍﺳﺖ. ﺑﻨﺎﺑﺮﺍﻳــﻦ ،ﻓﻴﺜﺎﻏــﻮﺭﺱ ﺍﻳﻦ ﺧﺎﺻﻴﺖ ﻣﺜﻠﺚ ﻗﺎﺋﻢﺍﻟﺰﺍﻭﻳﻪ ﺭﺍ ﻛﺸــﻒ ﻧﻜﺮﺩ ،ﺑﻠﻜﻪ ﺍﻭ ﺑﺮﺍﻯ ﻧﺨﺴــﺘﻴﻦ ﺑﺎﺭ ﺗﻮﺍﻧﺴــﺖ ﺍﻳــﻦ ﺧﺎﺻﻴﺖ ﺭﺍ ﺗﻌﻤﻴﻢ ﺩﻫﺪ ،ﺁﻥ ﺭﺍ ﺛﺎﺑﺖ ﻛﻨﺪ ﻭ ﺍﺯ ﻧﻈﺮ ﻋﻠﻤﻰ ﺑﻪ ﺟﻨﺒﻪﻫﺎﻯ ﻋﻤﻠﻰ ﺁﻥ ﺑﺮﺳﺪ. ﺭﻭﺵﻫﺎﻳﻰ ﺑﺮﺍﻯ ﺍﺛﺒﺎﺕ ﺭﺍﺑﻄﻪﻯ ﻓﻴﺜﺎﻏﻮﺭﺱ ﻣﺜﻠــﺚ ﻗﺎﺋﻢﺍﻟﺰﺍﻭﻳﻪﻯ ﻣﺘﺴﺎﻭﻯﺍﻟﺴــﺎﻗﻴﻦ ﺍﻭﻟﻴﻦ ﺣﺎﻟﺖ ﺧﺎﺹ ﻣﻮﺭﺩ ﻧﻈــﺮ ﻓﻴﺜﺎﻏﻮﺭﺱ ﺑﻮﺩﻩ ﺍﺳــﺖ ﻛﻪ ﺑﺎ ﺗﻮﺟﻪ ﺑﻪ ﺷــﻜﻞ ﺯﻳﺮ ،ﺑﻪ ﺳــﺎﺩﮔﻰ ﻣﻰﺗﻮﺍﻥ ﻧﺘﻴﺠﻪﻯ ﻣﻮﺭﺩ ﻧﻈﺮ ﺭﺍ ﺩﺭ ﺁﻥ ﺑﻪ ﺩﺳﺖ ﺁﻭﺭﺩ: 2 2 2 a +b =c
a=bﻭ
2
2
a
C a B
a c
c
A
b I
c2
b
ш
п
c
b2
b
IV
b
c ) ﺷﻜﻞ(1
c b
ш
b I
a
a2
a
п
IV
2
a +a =c
ﻭ ﻳﺎ
a
ﺑﺪﻳﻬﻰ ﺍﺳــﺖ ﻛﻪ ﺳــﻄﺢ ﻣﺮﺑﻊﻫﺎﻯ ﺩﻭ ﺿﻠﻊ ﻗﺎﺋﻤﻪ ﻛﻪ ﻫﺮ ﻳﻚ ﺑﻪ ﺩﻭ ﻣﺜﻠﺚ ﻣﺴــﺎﻭﻯ ﺗﻘﺴﻴﻢ ﺷــﺪﻩﺍﻧﺪ ،ﻣﺮﺑﻊ ﻭﺗﺮ ﻛﻪ ﺑﻪ ﭼﻬﺎﺭ ﻣﺜﻠﺚ ﻣﺴﺎﻭﻯ ﺗﻘﺴﻴﻢ ﺷﺪﻩ ﺍﺳﺖ ﺭﺍ ﻣﻰﭘﻮﺷﺎﻧﻨﺪ. ﺗﻘﺮﻳﺒﺎً ﻫﺮ ﻗﺮﻧﻰ ﻛﻪ ﻣﻰﮔﺬﺭﺩ ﺭﻭﺵﻫﺎﻯ ﺟﺪﻳﺪﻯ ﺑﺮﺍﻯ ﺍﺛﺒﺎﺕ ﺍﻳﻦ ﻗﻀﻴﻪ ﺍﺭﺍﺋﻪ ﻣﻰﺷــﻮﺩ ﻳﺎ ﻻﺍﻗﻞ ﻓﻜﺮ ﺭﻭﺵﻫــﺎﻯ ﺟﺪﻳﺪ ﺍﺛﺒﺎﺕ ﺑﻪ ﻭﺟﻮﺩ ﻣﻰﺁﻳﺪ .ﻫﻨﻮﺯ ﻫﻢ ﺍﻓﺰﺍﻳﺶ ﺗﻌﺪﺍﺩ ﺍﻳﻦ ﺍﺛﺒﺎﺕﻫﺎ ﺑﻪ ﭘﺎﻳﺎﻥ ﻧﺮﺳــﻴﺪﻩ ﺍﺳﺖ. ﺍﻗﻠﻴﺪﺱ ﺩﺭ ﻛﺘﺎﺏ ﻣﻘﺪﻣﺎﺕ ﺧﻮﺩ ﺑﻪ 8ﻃﺮﻳﻖ ﺍﻳﻦ ﺭﺍﺑﻄﻪ ﺭﺍ ﺛﺎﺑﺖ ﻛﺮﺩﻩ ﺍﺳﺖ .ﺧﻮﺍﺟﻪ ﻧﺼﻴﺮﺍﻟﺪﻳﻦ ﻃﻮﺳﻰ ﻧﻴﺰ ﺁﻥ ﺭﺍ ﺩﺭ ﺳﺎﻝ 1594ﺛﺎﺑﺖ ﻛﺮﺩ. ﺩﺭ ﺍﻳﻦ ﺟﺎ ﺑﻪ ﺑﻴﺎﻥ ﭼﻨﺪ ﺭﻭﺵ ﺟﺎﻟﺐ ﻛﻪ ﺑﻪ ﻫﻢﺍﺭﺯﻯ ﺷﻜﻞﻫﺎ ﻭ ﺗﺴﺎﻭﻯ ﻣﺴﺎﺣﺖﻫﺎ ﻣﺮﺑﻮﻁ ﻣﻰﺷﻮﺩ ،ﻣﻰﭘﺮﺩﺍﺯﻳﻢ. ﺭﻭﺵ ﺍﻭﻝ: ﺍﻳﻦ ﺭﻭﺵ ،ﺍﺛﺒﺎﺕ ﺍﺣﺘﻤﺎﻟﻰ ﺧﻮﺩ ﻓﻴﺜﺎﻏﻮﺭﺱ ﺍﺳﺖ:
2 2 2a = c
b
) ﺷﻜﻞ(2
c
b
c
ﻣﺮﺑﻌﻰ ﻣﻰﺳــﺎﺯﻳﻢ ﻛﻪ ﺿﻠﻊ ﺁﻥ ﻣﺴــﺎﻭﻯ ﻣﺠﻤــﻮﻉ bﻭ cﻳﻌﻨﻰ ﺿﻠﻊﻫــﺎﻯ ﻣﺠﺎﻭﺭ ﺑــﻪ ﺯﺍﻭﻳﻪﻯ ﻗﺎﺋﻤﻪ ﺍﺯ ﻣﺜﻠــﺚ ﻗﺎﺋﻢﺍﻟﺰﺍﻭﻳﻪﻯ ﻣﻔﺮﻭﺽ ﺑﺎﺷــﺪ )ﺷــﻜﻞ ) .((1ﺍﻳﻦ ﻣﺮﺑﻊ ﺭﺍ ﺑﻪ ﺩﻭ ﻣﺮﺑﻊ c2 ، b2ﻭ ﺩﻭ ﻣﺴﺘﻄﻴﻞ ﻣﺴﺎﻭﻯ ﺑﻪ ﺿﻠﻊﻫﺎﻯ bﻭ cﺗﻘﺴﻴﻢ ﻣﻰﻛﻨﻴﻢ. ﺍﻳﻦ ﻣﺴــﺘﻄﻴﻞﻫﺎ ﺭﺍ ﺑﻪ ﻧﻮﺑﻪﻯ ﺧﻮﺩ ﺑــﻪ ﭼﻬﺎﺭ ﻣﺜﻠﺚ ﻗﺎﺋﻢﺍﻟﺰﺍﻭﻳﻪﻯ ﻣﺴــﺎﻭﻯ ΙΙΙ ، ΙΙ ، Ιﻭ ، IVﺗﻘﺴــﻴﻢ ﻣﻰﻛﻨﻴــﻢ .ﺍﻳﻦ ﻣﺜﻠﺚﻫﺎ ﺭﺍ ﺁﻥﻃﻮﺭ ﻛﻪ ﺩﺭ ﺷﻜﻞ ) (2ﺩﻳﺪﻩ ﻣﻰﺷﻮﺩ ﻗﺮﺍﺭ ﻣﻰﺩﻫﻴﻢ .ﺩﺭ ﺍﻳﻦ ﺻﻮﺭﺕ ﺩﻭﺭﻩﻯ ﺷﺎﻧﺰﺩﻫﻢ ،ﺷﻤﺎﺭﻩ ﻯ ،2ﺯﻣﺴﺘﺎﻥ 1389
ﺭﺍﻫﻨﻤﺎﻳﻲ
21
ﺑﻼﻓﺎﺻﻠــﻪ ﻣﺮﺑﻊ a2ﺑﻪ ﺩﺳــﺖ ﻣﻰﺁﻳﺪ ﻭ ﭼﻮﻥ ﻣﺮﺑــﻊ ) (1ﺑﺎ ﻣﺮﺑﻊ )(2 ﻣﻌﺎﺩﻝ ﻭ ﺑﺮﺍﺑﺮ ﺍﺳﺖ؛ ﻣﻰﺗﻮﺍﻥ ﻧﻮﺷﺖ: 2 2 ﺷﻜﻞ ) b + c + 4bc = a 2 + 4bc (2ﺷﻜﻞ )(1 ﺍﺯ ﺣﺬﻑ 4bcﺍﺯ ﺩﻭ ﻃﺮﻑ ﺗﺴﺎﻭﻯ ﺑﺎﻻ:
ﻣﺜﻠﺜــﻰ ﺑﺎ ﻫﻤﺎﻥ ﺍﺑﻌﺎﺩ ﺑﺎ ﻗﺎﻋﺪﻩﻯ bﺭﺍ ﺩﺭ ﻛﻨﺎﺭ ﺁﻥ ،ﻣﺎﻧﻨﺪ ﺷــﻜﻞ ﺯﻳﺮ ﻗﺮﺍﺭ ﻣﻰﺩﻫﻴﻢ: E
2 2 2 b +c =a
ﻧﺘﻴﺠــﻪ :ﺑﺎ ﺗﻮﺟﻪ ﺑﻪ ﺷــﻜﻞﻫﺎﻯ ) (1ﻭ ) ،(2ﺍﮔــﺮ ﺍﺯ ﻣﺮﺑﻊ ﺑﻪ ﺿﻠﻊ b+cﻣﻘــﺪﺍﺭ 2bcﺭﺍ ﻛﻢ ﻛﻨﻴــﻢ ،ﺍﺯ ﻳﻚ ﻃﺮﻑ b2+c2ﻭ ﺍﺯ ﻃﺮﻑ ﺩﻳﮕﺮ a2ﺭﺍ ﻣﻰﺩﻫﺪ ،ﻳﻌﻨﻰ: a2+b2=c2
ﺭﻭﺵ ﺩﻭﻡ: ﺍﺛﺒﺎﺕ ﺑﻬﺎ ﺳﻜﺎﺭﺍ )ﻣﺆﻟﻒ ﻣﻌﺮﻭﻑ ﻗﺮﻥ ﺩﻭﺍﺯﺩﻫﻢ(. ﺭﻳﺎﺿــﻰﺩﺍﻥ ﺑﺰﺭگ ﻫﻨﺪ ،ﺯﻳﺮ ﺍﻳﻦ ﺷــﻜﻞ ﺗﻨﻬﺎ ﻳﻚ ﻛﻠﻤﻪ ﻧﻮﺷــﺘﻪ ﺍﺳﺖ :ﻧﮕﺎﻩ ﻛﻨﻴﺪ. a
1 bc 2
a
b
2
b
c 2
D
C
3
1
B
a
C
ﻧﻘﻄﻪﻯ Eﺭﺍ ﺑﺎ ﺧﻂ ﺭﺍﺳــﺘﻰ ﺑﻪ Aﻭﺻــﻞ ﻣﻰﻛﻨﻴﻢ ﺗﺎ ﺫﻭﺯﻧﻘﻪﻯ ABDEﺣﺎﺻﻞ ﺷﻮﺩ .ﺑﺪﻳﻬﻰ ﺍﺳﺖ ﻛﻪ ∠c1 + c2 = 90ﻭ ﺩﺭ ﻧﺘﻴﺠﻪ ﺩﺍﺭﻳﻢ: ∠c = 90
1 bc 2 a
1
(c - b)2
1 bc 2
ﺍﺯ ﻃﺮﻓﻰ ﻣﻰﺩﺍﻧﻴﻢ ﻛﻪ ﻣﺴﺎﺣﺖ ﺫﻭﺯﻧﻘﻪ ﭼﻨﻴﻦ ﺍﺳﺖ: 1 )(a + c)(a + c 2
a
= ﻣﺠﻤﻮﻉ ﺩﻭ ﻗﺎﻋﺪﻩ × ﺍﺭﺗﻔﺎﻉ × = 1ﻣﺴﺎﺣﺖ ABCD 2
2 )(a + c
1 bc 2 a
1 2
=
ﺍﺯ ﻃﺮﻓﻰ ﺩﻳﮕﺮ ،ﻣﺴﺎﺣﺖ ﺫﻭﺯﻧﻘﻪ ﺑﺮﺍﺑﺮ ﻣﺠﻤﻮﻉ ﻣﺴﺎﺣﺖ ﺳﻪ ﻣﺜﻠﺚ ﻗﺎﺋﻢﺍﻟﺰﺍﻭﻳﻪ ﺍﺳﺖ:
ﺑﺎ ﺗﻮﺟﻪ ﺑﻪ ﺷﻜﻞ ﺩﺍﺭﻳﻢ: 1 2 2 )a = (c − b) + 4( bc 2
ﻳﺎ: 2 2 2 a = b +c
ﺭﻭﺵ ﺳﻮﻡ ﻣﺜﻠﺚ ﻗﺎﺋﻢﺍﻟﺰﺍﻭﻳﻪﻯ ﺯﻳﺮ ﺭﺍ ﺩﺭ ﻧﻈﺮ ﻣﻰﮔﻴﺮﻳﻢ:
2
2
2
1 2 = ac + b 2
ﻳﺎ:
2 2 2 a = c + b − 2bc + 2bc
1
1
1
= a.c + a.c + b.bﻣﺴﺎﺣﺖ ﺫﻭﺯﻧﻘﻪﻯ ABCD
ﺑﻨﺎﺑﺮﺍﻳﻦ ﺩﺍﺭﻳﻢ: = 1 (a + c )2 = ac + 1 b2ﻣﺴﺎﺣﺖ ﺫﻭﺯﻧﻘﻪﻯ ABCD 2
ﻭ ﺩﺭ ﻧﺘﻴﺠﻪ ﺧﻮﺍﻫﻴﻢ ﺩﺍﺷﺖ:
2
1 2 = ac + b 2
ﻃﺮﻓﻴﻦ ﺗﺴﺎﻭﻯ ﺍﺧﻴﺮ ﺭﺍ ﺩﺭ 2ﺿﺮﺏ ﻣﻰﻛﻨﻴﻢ: A
b
C 22
ﺭﺍﻫﻨﻤﺎﻳﻰ
2
c ﻭ ﻳﺎ ﭘﺲ ﺍﺯ ﺍﺧﺘﺼﺎﺭ ﻻﺯﻡ ﺩﺍﺭﻳﻢ: a
2
) (a + c
B
ﺩﻭﺭﻩﻯ ﺷﺎﻧﺰﺩﻫﻢ ،ﺷﻤﺎﺭﻩﻯ ،2ﺯﻣﺴﺘﺎﻥ 1389
= 2ac + b
1 2
2
) (a + c
2 2 2 a + c + 2ac = 2ac + b 2
2 2 a +c = b
ﻛﻤﻲ ﻓﻜﺮ ﻛﻨﻴﺪ
︑︺︖︉ ﹡﹊﹠﹫︡! ︡﹫﹠﹋ ︣﹊﹁ ، ﺧﺴﺮﻭ ﺩﺍﻭﺩﻱ
ﺭﻳﺎﺿﻰ ﺩﺍﻧﺎﻥ ﻣﺸﻬﻮﺭ
ﺟﻤﻊ ﺳﺎﻋﺖ
10ﻋﺒﺎﺭﺕ ﺯﻳﺮ ﺗﻮﺻﻴﻔﻰ ﺍﺯ ﺷــﺶ ﺭﻳﺎﺿﻰ ﺩﺍﻥ ﻣﺸــﻬﻮﺭﻯ ﺍﺳــﺖ ﻛﻪ 1400ﺳــﺎﻝ ﻗﺒﻞ ﺍﺯ ﻣﻴﻼﺩ ،ﻳﻌﻨﻰ ﺣﺪﻭﺩ ﺳــﺎﻝﻫﺎﻯ،450 ،400، ،1800 ،1100، 800ﺑﻌﺪ ﺍﺯ ﻣﻴﻼﺩ ﻣﺴﻴﺢ ﺯﻧﺪﮔﻰ ﻣﻰﻛﺮﺩﻧﺪ . ﺑﻪ ﻛﻤﻚ ﺍﻳﻦ ﺳــﺮ ﻧﺦﻫﺎ ﺁﻳﺎ ﻣﻰﺗﻮﺍﻧﻴﺪ ﺩﺭﻳﺎﺑﻴﺪ ﻛﻪ ﻫﺮ ﻛﺴــﻰ ﺩﺭ ﻛﺠــﺎ ﻭ ﺩﺭ ﭼﻪ ﺯﻣﺎﻧــﻰ ﺯﻧﺪﮔﻰ ﻣﻰﻛﺮﺩﻩ ﻭ ﺑﻪ ﭼﻪ ﻋﻨﻮﺍﻧﻰ ﻣﻌﺮﻭﻑ ﺑﻮﺩﻩ ﺍﺳﺖ ؟ .1ﭼﻮﺍﻧﮓ – ﭼﻰ ﻣﺪﺕ ﺯﻣﺎﻥ ﺯﻳﺎﺩﻯ ﺭﺍ ﺑﺮﺍﻯ ﻣﺤﺎﺳﺒﻪﻯ ﻣﻘﺪﺍﺭ ﻋﺪﺩ лﺻﺮﻑ ﻛﺮﺩ . .2ﺭﻳﺎﺿﻰ ﺩﺍﻧﻰ ﻛﻪ ﻗﺒﻞ ﺍﺯ ﻣﻴﻼﺩ ﻣﺴﻴﺢ ﺩﺭ ﻣﺼﺮ ﺯﻧﺪﮔﻰ ﻣﻰ ﻛﺮﺩ .ﺍﻭ ﻫﻰ ﭘﺎﺗﻴﺎ ﻧﺒﻮﺩ . .3ﺭﻳﺎﺿــﻰ ﺩﺍﻥ ﻫﻨﺪﻯ ﺑﻪﻧﺎﻡ ﺑﺎﺳــﻜﺎﺭﺍ ﻛﻪ ﺑﻌﺪ ﺍﺯ ﭘﺪﺭ ﺟﺒﺮ ﺯﻧﺪﮔﻰ ﻣﻰ ﻛﺮﺩ . .4ﺭﻳﺎﺿﻰ ﺩﺍﻥ ﺯﻥ ﻛﻪ ﺣﻮﺍﻟﻰ ﺳﺎﻝ 400ﺑﻌﺪ ﺍﺯ ﻣﻴﻼﺩ ﻣﺴﻴﺢ ﺯﻧﺪﮔﻰ ﻣﻰ ﻛﺮﺩ ﻭ ﺩﺭ ﻋﻠﻢ ﻫﻨﺪﺳﻪ ﻣﻄﺎﻟﻌﻪ ﺩﺍﺷﺖ . .5ﺭﻳﺎﺿــﻰ ﺩﺍﻥ ﺁﻟﻤﺎﻧــﻰ ﻛﻪ ﺩﺭﺑﺎﺭﻩ ﻯ ﺗﺌﻮﺭﻯ ﺍﻋﺪﺍﺩ ﻧﻮﺷــﺘﻪ ﺑﻮﺩ ﺍﻭ ﺗﻰ ﺳــﻮ ﭼﻮ ﺁﻧﮓ ﻧﺒﻮﺩ .ﺍﻭ ﺩﺭ ﺳــﺎﻝ 450ﺑﻌﺪ ﺍﺯ ﻣﻴﻼﺩ ﺯﻧﺪﮔﻰ ﻣﻰ ﻛﺮﺩ . .6ﺍﻟﻜــﻮ ﺁﺭﻳﺴــﻢ ﺩﺭ ﺳــﺎﻝ 830ﺑﻌﺪ ﺍﺯ ﻣﻴﻼﺩ ﺯﻧﺪﮔــﻰ ﻣﻰ ﻛﺮﺩ .ﺍﻭ ﺭﻳﺎﺿﻰ ﺩﺍﻥ ﻳﻮﻧﺎﻧﻰ ﻧﺒﻮﺩ . .7ﺍﺣﻤﺲ ﺑﻴﺶ ﺍﺯ 1000ﺳــﺎﻝ ﻗﺒﻞ ﺍﺯ ﺭﻳﺎﺿﻰ ﺩﺍﻥ ﭼﻴﻨﻰ ﺯﻧﺪﮔﻰ ﻣﻰ ﻛﺮﺩ . . 8ﺭﻳﺎﺿﻰ ﺩﺍﻥ ﻋﺮﺑﻰ ﻣﻌﺎﺩﻻﺕ ﺩﺭﺟﻪ ﻯ ﺩﻭﻡ ﺭﺍ ﻣﻄﺎﻟﻌﻪ ﻧﻜﺮﺩ . .9ﻣﺮﺩﻯ ﻛﻪ ﺑﻪ ﺣﻞ ﻣﺴﺎﻟﻪ ﻣﻰ ﺍﻧﺪﻳﺸﻴﺪ ﻭ ﺑﻴﺶ ﺍﺯ 3000ﺳﺎﻝ ﻗﺒﻞ ﺍﺯ ﺳﻮﻓﻴﻪ ﺟﺮﻣﻴﻦ ﺯﻧﺪﮔﻰ ﻣﻰ ﻛﺮﺩ . .10ﻣﺮگ ﻫﻰ ﭘﺎﺗﻴﺎ ﺩﺭ ﺷــﻬﺮ ﺍﻟﻜﺴــﺎﻧﺪﺭﻳﺎﻯ ﻣﺼﺮ ﻧﺸﺎﻧﻪﺍﻯ ﺍﺯ ﭘﺎﻳﺎﻥ ﺩﻭﺭﻩﻯ ﺭﻳﺎﺿﻴﺎﺕ ﻳﻮﻧﺎﻧﻰ ﺑﻮﺩ .
ﺁﻳﺎ ﻣﻰ ﺗﻮﺍﻧﻴﺪ ﻳﻚ ﺻﻔﺤﻪﻯ ﺳﺎﻋﺖ ﺭﺍ ﺑﻪ 5ﺧﻂ ﻣﺴﺘﻘﻴﻢ ﺗﻘﺴﻴﻢ ﻛﻨﻴــﺪ ﺑﻄﻮﺭﻯﻛﻪ ﺍﻳﻦ ﺧﻄﻮﻁ ﻫﻤﺪﻳﮕــﺮ ﺭﺍ ﻗﻄﻊ ﻧﻜﻨﻨﺪ ﻭ ﺍﻋﺪﺍﺩ ﺟﻔﺖ ﺟﻤﻊ ﻫﻤﺎﻥ ﺗﻌﺪﺍﺩ ﻛﻞ ﺑﺎﺷﻨﺪ ؟
ﭼﻨﺪ ﻃﺎﻟﺒﻰ ؟ - 1ﺧﺎﻧﻤﻰ ﺑﻪ ﻣﻐﺎﺯﻩﻯ ﻣﻴﻮﻩ ﻓﺮﻭﺷــﻰ ﺭﻓﺖ ﺗﺎ ﭼﻨﺪ ﻃﺎﻟﺒﻰ ﺑﺨﺮﺩ ﻭ ﺩﺭ ﻛﻴﺴــﻪﺍﻯ ﺣﻤﻞ ﻛﻨﺪ .ﺍﻭ ﺩﺭﻳﺎﻓﺖ ﻛﻪ ﺍﮔﺮ ﻃﺎﻟﺒﻰﻫﺎ ﺭﺍ ﺩﻭ ﺗﺎ ﺩﻭﺗﺎ ﺷــﻤﺎﺭﺵ ﻛﻨﺪ ﻳﻚ ﻃﺎﻟﺒﻰ ﺑﺎﻗﻰ ﺧﻮﺍﻫﺪ ﻣﺎﻧﺪ .ﺍﻣﺎ ﺍﮔﺮ ﺳــﻪ ﺗﺎ ﺳــﻪ ﺗﺎ ﺷﻤﺎﺭﺵ ﻛﻨﺪ ،ﺩﻭ ﻃﺎﻟﺒﻰ ﺑﺎﻗﻰ ﺧﻮﺍﻫﺪ ﻣﺎﻧﺪ .ﺗﻌﺪﺍﺩ ﻛﻞ ﺑﻴﻦ 60ﻭ 70 ﺑﻮﺩ .ﭼﻨﺪ ﻃﺎﻟﺒﻰ ﺩﺭ ﻛﻴﺴﻪﻯ ﺍﻭ ﺑﻮﺩ ؟ ﺩﻭﺭﻩ ﻯ ﺷﺎﻧﺰﺩﻫﻢ ،ﺷﻤﺎﺭﻩ ﻯ ،2ﺯﻣﺴﺘﺎﻥ 1389
ﺭﺍﻫﻨﻤﺎﻳﻰ
23
ﭼﻪ ﺭﻭﺯﻯ ﺍﺳﺖ ؟ ﺍﮔﺮ ﻓﺮﺩﺍﻯ ﺩﻳﺮﻭﺯ ﭘﻨﺠﺸــﻨﺒﻪ ﺑﺎﺷــﺪ ،ﺭﻭﺯ ﺑﻌﺪ ﺍﺯ ﺩﻳﺮﻭﺯ ﻓﺮﺩﺍ ﭼﻪ ﺭﻭﺯﻯ ﺧﻮﺍﻫﺪ ﺑﻮﺩ ؟
︎︀︨ ︨︣﹎︣﹞ ﹩و ︗︡ول
ﺟﻤﻊ ﺩﻳﮕﺮ ﺳﺎﻋﺖ؟ ﺁﻳــﺎ ﻣﻰ ﺗﻮﺍﻧﻴــﺪ ﻳﻚ ﺻﻔﺤﻪﻯ ﺳــﺎﻋﺖ ﺭﺍ ﺑﻪ ﻭﺳــﻴﻠﻪﻯ ﺩﻭ ﺧﻂ ﻣﺴــﺘﻘﻴﻢ ﺗﻘﺴــﻴﻢ ﻛﻨﻴﺪ ﺑﻪ ﻃﻮﺭﻯ ﻛﻪ ﺍﻳﻦ ﺩﻭ ﺧــﻂ ﻳﻜﺪﻳﮕﺮ ﺭﺍ ﻗﻄﻊ ﻧﻜﻨﻨﺪ ﻭ ﺟﻤﻊ ﻫﺮ ﺑﺨﺶ ﺑﺎ ﻣﻘﺪﺍﺭ ﻛﻞ ﻳﻜﻰ ﺑﺎﺷﺪ ؟
-١ 8
ﺑﺰﺭگﺗﺮ ﭼﻪ ﻛﻠﻤﺎﺗﻰ ؟ ﻙ :ﭼﻪ ﻛﻠﻤﻪ ﺍﻯ ﺍﺯ ﭘﻨﺞ ﺣﺮﻑ ﺗﺸــﻜﻴﻞ ﺷــﺪﻩ ﺍﺳﺖ ﻛﻪ ﺍﮔﺮ ﺩﻭ ﺣﺮﻑ ﺁﻥ ﺭﺍ ﺑﺮﺩﺍﺭﻳﻢ ﻛﻠﻤﻪ ﺍﻯ ﺑﻪ ﻣﻌﻨﺎﻯ ﻋﺪﺩ ﻳﻚ ﺑﺎﻗﻰ ﻣﻰ ﻣﺎﻧﺪ ؟ ) oneﺑﻪ ﻣﻌﻨﺎﻯ ﻳﻚ ﺍﺳﺖ( . ﺍﻟﻒ Stone :
)، 9ﺧﻮﺭﺩﻥ 7، 8 ( ate
ﺁﻥﻫﺎ ﺩﺭ ﭼﻪ ﺭﻭﺯﻯ ﻣﺘﻮﻟﺪ ﺷﺪﻧﺪ ؟ ﺁﻥ ﻓﺮﺍﻧﻚ ﺩﺭ ﺩﻭﺍﺯﺩﻫﻢ ﻣﺎﻩ ﺟﻮﻥ ﺳﺎﻝ 1929ﻣﺘﻮﻟﺪ ﺷﺪ .ﻣﺎﺭﺗﻴﻦ ﻟﻮﺗﺮ ﺩﺭ ﭘﺎﻧﺰﺩﻫﻢ ژﺍﻧﻮﻳﻪ ﺳــﺎﻝ 1929ﻣﺘﻮﻟﺪ ﺷــﺪ .ﺟﻴﻦ ﺁﺳــﺘﻦ ﺩﺭ ﺷــﺎﻧﺰﺩﻫﻢ ﺩﺳــﺎﻣﺒﺮ 1775ﻣﺘﻮﻟﺪ ﺷــﺪ .ﺍﻳﻨﺪﻳﺮﺍ ﮔﺎﻧﺪﻯ ﺩﺭ ﻧﻮﺯﺩﻫﻢ ﻧﻮﺍﻣﺒﺮ 1917ﻣﺘﻮﻟﺪ ﺷــﺪ .ﻭﻳﻠﻴﺎﻡ ﺷﻜﺴﭙﻴﺮ ﺩﺭ ﺑﻴﺴﺖ ﻭ ﺳﻮﻡ ﺁﻭﺭﻳﻞ 1564ﻣﺘﻮﻟﺪ ﺷﺪ . ﺁﻳﺎ ﻣــﻰ ﺗﻮﺍﻧﻴﺪ ﭘﻴﺪﺍ ﻛﻨﻴﺪ ﻛﻪ ﺍﻓﺮﺍﺩ ﻧﺎﻡ ﺑــﺮﺩﻩ ﺩﺭ ﻛﺪﺍﻡ ﺭﻭﺯ ﻫﻔﺘﻪ ﻣﺘﻮﻟﺪ ﺷﺪﻧﺪ ؟
24
ﺭﺍﻫﻨﻤﺎﻳﻲ
ﺩﻭﺭﻩﻯ ﺷﺎﻧﺰﺩﻫﻢ ،ﺷﻤﺎﺭﻩﻯ ،2ﺯﻣﺴﺘﺎﻥ 1389
9
5
11
10
12
2 7
1
-٢ ٧
ﺗﺮﺱ : ﻙ :ﭼﺮﺍ 6ﺍﺯ 7ﻣﻰ ﺗﺮﺳﺪ ؟ ﺍﻟﻒ :ﺯﻳﺮﺍ 9 ،7ﺭﺍ ﻣﻰ ﺧﻮﺭﺩ
4
3
ﻙ :ﭼﻪ ﻛﺴﻰ ﺑﺰﺭگﺗﺮ ﺍﺳﺖ ؟ ﺁﻗﺎﻯ ﺭﺿﺎﻳﻰ ﻳﺎ ﺑﭽﻪﻯ ﺩﺧﺘﺮ ﺍﻭ ؟ ﺍﻟﻒ :ﺑﭽﻪﻯ ﺩﺧﺘﺮ ﺍﻭ .ﺍﻭ ﻛﻤﻰ ﺑﺰﺭگﺗﺮ ﺍﺳﺖ .
6
٨ ٦
٥
١ ٣
٤ ٧
-٣
٩ ٢
ﺳﺮ 14 + 8 = 22
ﭘﺎﺳﺦ 14 :ﺍﺳﺐ ﻭ 8ﻣﺮﺩ ﺩﺭ ﺁﻥ ﻣﺤﻞ ﻫﺴﺘﻨﺪ. ﭘﺎ ( 8 × 2) + ( 14 × 4) = 72
-٤ ٤٣ ٦١ ٧ ١ ٣٧ ٧٣
-٥
٦٧ ١٣ ٣١
ﻣﻨﻄﻖ ﻭ ﺭﻳﺎﺿﻲ
﹤ ﹋︧︣﹞ ﹩دمدار ا︨️؟ ﻧﻮﻳﺴ ﻨﺪﻩ :ﺟﻮﺭﺝ ﺳ ِﺎﻣﺮﺯ ﺗﺮﺟﻤﻪ ﻯ ﺣﺴﻦ ﻳﺎﻭﺭﺗﺒﺎﺭ
ﻛﻠﻴﺪﻭﺍژﻩﻫﺎ :ﻣﻨﻄﻖ ،ﻣﻌﻤﺎ ،ﺍﺳﺘﺪﻻﻝ . ﺍﺣﻤﺪ ،ﺑﻬﺮﻭﺯ ﻭ ﺳــﺎﻣﺎﻥ ،ﻫﺮ ﻳﻚ ﺳــﻪ ﻭﻳﮋﮔﻰ ﺑﺮﺟﺴــﺘﻪ ﺩﺍﺭﻧﺪ. ﺍﻃﻼﻋﺎﺕ ﻣﺎ ﺩﺭﺑﺎﺭﻩﻯ ﺁﻧﺎﻥ ﺑﻪ ﺷﺮﺡ ﺯﻳﺮ ﺍﺳﺖ: -1ﺩﻭ ﻧﻔﺮ ﺍﺯ ﺁﻧﺎﻥ ﺑﺎﻫﻮﺵ ،ﺩﻭ ﻧﻔﺮ ﺧﻮﺵﻗﻴﺎﻓﻪ ،ﺩﻭ ﻧﻔﺮ ﻧﻴﺮﻭﻣﻨﺪ ﻭ ﺩﻭ ﻧﻔﺮ ﺑﺬﻟﻪﮔﻮ ﻫﺴﺘﻨﺪ .ﻳﻜﻰ ﺍﺯ ﺁﻥﻫﺎ ﻫﻢ ﻣﺮﺩﻡﺩﺍﺭ ﺍﺳﺖ. -2ﺩﺭﺑﺎﺭﻩﻯ ﺍﺣﻤﺪ ﺑﺎﻳﺪ ﮔﻔﺖ: ﺍﻟﻒ( ﺍﮔﺮ ﺍﻭ ﺑﺬﻟﻪﮔﻮ ﺑﺎﺷﺪ ،ﺧﻮﺵﻗﻴﺎﻓﻪ ﺍﺳﺖ. ﺏ( ﺍﮔﺮ ﺍﻭ ﺧﻮﺵﻗﻴﺎﻓﻪ ﺑﺎﺷﺪ ،ﺑﺎﻫﻮﺵ ﻧﻴﺴﺖ. -3ﺩﺭﺑﺎﺭﻩﻯ ﺑﻬﺮﻭﺯ ﺑﺎﻳﺪ ﮔﻔﺖ: ﺍﻟﻒ( ﺍﮔﺮ ﺍﻭ ﺑﺬﻟﻪﮔﻮ ﺑﺎﺷﺪ ،ﺑﺎﻫﻮﺵ ﺍﺳﺖ. ﺏ( ﺍﮔﺮ ﺍﻭ ﺑﺎﻫﻮﺵ ﺑﺎﺷﺪ ،ﺧﻮﺵﻗﻴﺎﻓﻪ ﺍﺳﺖ. -4ﺩﺭﺑﺎﺭﻩﻯ ﺳﺎﻣﺎﻥ ﺑﺎﻳﺪ ﮔﻔﺖ: ﺍﻟﻒ( ﺍﮔﺮ ﺍﻭ ﺧﻮﺵﻗﻴﺎﻓﻪ ﺑﺎﺷﺪ ،ﻧﻴﺮﻭﻣﻨﺪ ﺍﺳﺖ.
ﺏ( ﺍﮔﺮ ﺍﻭ ﻧﻴﺮﻭﻣﻨﺪ ﺑﺎﺷــﺪ ،ﺑﺬﻟﻪﮔﻮ ﻧﻴﺴﺖ. ﺑﻪ ﻧﻈﺮ ﺷــﻤﺎ ،ﻛﺪﺍﻡ ﻳﻚ ﻣﺮﺩﻡﺩﺍﺭ ﺍﺳﺖ؟ )ﺭﺍﻫﻨﻤﺎﻳﻰ :ﺑﺮﺍﻯ ﺁﺳﺎﻥ ﺷﺪﻥ ّ ﺣﻞ ﻣﻌﻤﺎ ،ﻧﺨﺴﺖ ﻣﺠﻤﻮﻋﻪﻯ ﻭﻳﮋﮔﻰﻫﺎﻳﻰ ﺭﺍ ﻛﻪ ﻣﻰﺗﻮﺍﻥ ﺑﻪ ﻫﺮ ﻛﺪﺍﻡ ﻧﺴﺒﺖ ﺩﺍﺩ، ﻣﺸــﺨﺺ ﻛﻨﻴﺪ .ﺳــﭙﺲ ﻓﺮﺽ ﻛﻨﻴﺪ ﻳﻜﻰ ﺍﺯ ﺍﻳﻦ ﺳــﻪ ﻧﻔﺮ ﻣﺮﺩﻡﺩﺍﺭ ﺍﺳــﺖ. ﻳﺎﺩﺗﺎﻥ ﺑﺎﺷــﺪ ﻛﻪ ﻓﻘﻂ ﺩﺭ ﻳﻜﻰ ﺍﺯ ﺍﻳﻦ ﻣﺠﻤﻮﻋــﻪ ﻭﻳﮋﮔﻰﻫــﺎ ،ﻫﻴــﭻ ﺗﻨﺎﻗﻀﻰ ﻭﺟﻮﺩ ﻧﺪﺍﺭﺩ(. ﺩﻭﺭﻩﻯ ﺷﺎﻧﺰﺩﻫﻢ ،ﺷﻤﺎﺭﻩ ﻯ ،2ﺯﻣﺴﺘﺎﻥ 1389
ﺭﺍﻫﻨﻤﺎﻳﻲ
25
ﺟﺪﻭﻝ
︗︡ول ٤٦
ا﹁﹆﹩
ﻣﺤﻤﺪ ﻋﺰﻳﺰﻱﭘﻮﺭ
6
5
4
2
3
8
1 7
10
11
9
13
12 14
17
15
16 20
19
18
23
22 25
21 24
.1ﺷــﺎﻧﺰﺩﻩ ﺑﺮﺍﺑــﺮ 4ﻋﻤــﻮﺩﻯ 17 .2ﻋﻤــﻮﺩﻯ ﻣﻨﻬــﺎﻯ 1933 5 .7ﻋﻤﻮﺩﻯ ﻣﻨﻬﺎﻯ ﭼﻬﻞ ﻭ ﺩﻭ 21 .8ﺍﻓﻘﻰ ﻣﻨﻬﺎﻯ ﺑﻴﺴــﺖ ﻭ ﻧﻪ 14 .9ﺍﻓﻘﻰ ﻣﻨﻬﺎﻯ 6 .12 8906ﻋﻤﻮﺩﻯ ﺗﻘﺴﻴﻢ ﺑﺮ ﻫﻔﺪﻩ .13 12ﺍﻓﻘــﻰ ﺑﻪﻋﻼﻭﻩﻯ ﺑﻴﺴــﺖ ﻭ ﻫﺸــﺖ 25 .14ﺍﻓﻘﻰ ﺿﺮﺏ ﺩﺭ 2 ﻋﻤﻮﺩﻯ 3 .15ﻋﻤﻮﺩﻯ ﺿﺮﺏ ﺩﺭ ﺩﻭ 15 .16ﺍﻓﻘﻰ ﺑﻪﻋﻼﻭﻩﻯ ﻧﻮﺩ ﻭ ﻫﻔﺖ .18ﻳﺎﺯﺩﻩ ﺑﺮﺍﺑﺮ 24ﺍﻓﻘﻰ 12 .21ﺍﻓﻘﻰ ﺗﻘﺴﻴﻢ ﺑﺮ ﻫﻔﺖ 2 .23ﻋﻤــﻮﺩﻯ ﻣﻨﻬــﺎﻯ ﻳﻚ 2 .24ﻋﻤﻮﺩﻯ ﺿﺮﺏ ﺩﺭ 19ﻋﻤﻮﺩﻯ .25ﺗﻌــﺪﺍﺩ ﺳــﺎﻧﺘﻰﻣﺘﺮﻫﺎ ﺩﺭ ﺩﻭﺍﺯﺩﻩ ﻣﺘﺮ 2 .19ﻋﻤﻮﺩﻯ ﺿﺮﺏ ﺩﺭ ﭘﻨﺞ 21 .20ﺍﻓﻘﻰ ﺿﺮﺏ ﺩﺭ ﻧ ُﻪ
︻﹝﹢دی
24 .1ﺍﻓﻘﻰ ﻣﻨﻬﺎﻯ .2 161ﻳﻚ ﻋﺪﺩ ﺍ ّﻭﻝ .3ﺗﻌﺪﺍﺩ ﺳــﺎﻧﺘﻰﻣﺘﺮ ﺩﺭ ﭼﻬﺎﺭ ﻣﺘﺮ .4ﻣﺮﺑﻊ ﻳﻚ ﻋﺪﺩ 13 .5ﺍﻓﻘﻰ ﺗﻘﺴﻴﻢ ﺑﺮ ﻫﻔﺖ .6 17ﻋﻤﻮﺩﻯ ﺑﻪﻋﻼﻭﻩﻯ ﺑﻴﺴــﺖ ﻭ ﻳﻚ 9 .9ﺍﻓﻘﻰ ﺑﻪﻋﻼﻭﻩﻯ 3115 1 .10ﺍﻓﻘــﻰ ﺑﻪﻋــﻼﻭﻩﻯ 9 .11 13800ﻋﻤﻮﺩﻯ ﺑﻪﻋﻼﻭﻩﻯ 676 1 .15ﺍﻓﻘــﻰ ﻣﻨﻬﺎﻯ ﺷــﺶ .17ﺗﻌﺪﺍﺩ ﻣﺘﺮﻫــﺎ ﺩﺭ ﻫﻔﺖ ﻛﻴﻠﻮﻣﺘﺮ 2 .19ﻋﻤــﻮﺩﻯ ﺿﺮﺏ ﺩﺭ ﭘﻨــﺞ 21 .20ﺍﻓﻘﻰ ﺿﺮﺏ ﺩﺭ ﻧ ُﻪ .22 23ﻋﻤــﻮﺩﻯ ﺑﻪ ﻋﻼﻭﻩﻯ ﭘﻨﺠﺎﻩ ﻭ ﭼﻬــﺎﺭ .23ﺗﻌﺪﺍﺩ ﻣﻴﻠﻰﻣﺘﺮﻫﺎ ﺩﺭ ﭼﻬﺎﺭ ﺳﺎﻧﺘﻰﻣﺘﺮ
︎︀︨ ︗︡ول ︫﹝︀رۀ ٤٥ ﺍﻓﻘﻲ 8 .1ﻋﻤــﻮﺩﻱ ﻣﻨﻬﺎﻱ .5 17897ﺗﻌﺪﺍﺩ ﺳــﺎﻧﺘﻴﻤﺘﺮ ﺩﺭ ﻫﻔﺖ ﻣﺘﺮ 5 .6ﺍﻓﻘﻲ ﻣﻨﻬﺎﻱ ﻫﺸــﺖ 1 .8ﻋﻤﻮﺩﻱ ﺑﻪﻋﻼﻭﻩﻱ ﺑﻴﺴــﺖ ﻭ ﻳﻚ .9ﺗﻌﺪﺍﺩ ﻣﻴﻠﻴﻤﺘﺮ ﺩﺭ ﺩﻭﺍﺯﺩﻩ ﺳﺎﻧﺘﻴﻤﺘﺮ 4 .11ﻋﻤﻮﺩﻱ ﻣﻨﻬﺎﻱ ﻫﻔــﺪﻩ 7 .13ﻋﻤــﻮﺩﻱ ﻣﻨﻬــﺎﻱ 14 .15 122ﻋﻤﻮﺩﻱ ﻣﻨﻬﺎﻱ ﭼﻬــﻞ .16ﺗﻌﺪﺍﺩ ﻣﻴﻠﻴﻤﺘﺮ ﺩﺭ 10ﻣﺘــﺮ 20 .17ﻋﻤﻮﺩﻱ ﻣﻨﻬﺎﻱ ﻧــﻪ 5 .19ﻋﻤﻮﺩﻱ ﻣﻨﻬــﺎﻱ 26 .21 112 ﻋﻤﻮﺩﻱ ﻣﻨﻬﺎﻱ ﺳﻲ ﻭ ﭘﻨﺞ 15 .22ﺍﻓﻘﻲ ﻣﻨﻬﺎﻱ 27 .24 172ﺍﻓﻘﻲ ﺗﻘﺴــﻴﻢ ﺑﺮ ﻫﺸــﺖ 22 .25ﻋﻤﻮﺩﻱ ﺑﻪﻋﻼﻭﻩﻱ ﭼﻬــﻞ 25 .27ﺍﻓﻘــﻲ ﺑﻪﻋﻼﻭﻩﻱ ﺳــﻲ ﻭ ﺷــﺶ 8 .29ﻋﻤﻮﺩﻱ ﺑﻪﻋﻼﻭﻩ 9008 26
ﺭﺍﻫﻨﻤﺎﻳﻲ
ﺩﻭﺭﻩﻯ ﺷﺎﻧﺰﺩﻫﻢ ،ﺷﻤﺎﺭﻩﻯ ،2ﺯﻣﺴﺘﺎﻥ 1389
ﻋﻤﻮﺩﻱ 22 .1ﻋﻤــﻮﺩﻱ ﺗﻘﺴــﻴﻢ ﺑﺮ ﭘﻨﺞ 15 .2ﻋﻤــﻮﺩﻱ ﻣﻨﻬﺎﻱ 104 9 .3ﺍﻓﻘﻲ ﺿﺮﺏ ﺩﺭ ﻫﺸــﺖ 18 .4ﻋﻤﻮﺩﻱ ﺗﻘﺴــﻴﻢ ﺑﺮ ﻫﻔﺖ .5 ﻧﻪ ﺑﺮﺍﺑﺮ 28ﻋﻤﻮﺩﻱ 28 .7ﻋﻤﻮﺩﻱ ﺿﺮﺏ ﺩﺭ ﺳــﻪ .8ﺳــﻪ ﺑﺮﺍﺑﺮ 12ﻋﻤــﻮﺩﻱ 16 .10ﺍﻓﻘــﻲ ﺿــﺮﺏ ﺩﺭ ﺩﻭ 28 .12ﻋﻤﻮﺩﻱ ﺿﺮﺏ ﺩﺭ 18ﻋﻤــﻮﺩﻱ .14ﻳﺎﺯﺩﻩ ﺑﺮﺍﺑــﺮ 24ﺍﻓﻘﻲ .15ﭘﻨﺞ ﺑﺮﺍﺑﺮ 8ﺍﻓﻘﻲ .18ﻣﻜﻌــﺐ ﻳﻚ ﻋــﺪﺩ 3 .20ﻋﻤﻮﺩﻱ ﻣﻨﻬﺎﻱ ﭼﻬﺎﺭﺩﻩ .22ﺗﻌﺪﺍﺩ ﺳــﺎﻧﺘﻴﻤﺘﺮ ﺩﺭ ﺳــﻪ ﻣﺘﺮ 26 .23ﻋﻤﻮﺩﻱ ﺿﺮﺏ ﺩﺭ ﻳﺎﺯﺩﻩ 18 .26 ﻋﻤﻮﺩﻱ ﺗﻘﺴﻴﻢ ﺑﺮ ﻫﻔﺖ .28ﻳﻚ ﻋﺪﺩ ﺍ ّﻭﻝ 4
3
7 12
2
9
︗︡ول ︫﹝︀رۀ ٤٥
15
ﻣﺠﻠﻪﻫﺎﻱ ﺭﺷـﺪ ﺗﻮﺳـﻂ ﺩﻓﺘﺮ ﺍﻧﺘﺸـﺎﺭﺍﺕ ﻛﻤﻚﺁﻣﻮﺯﺷﻲ ﺳـﺎﺯﻣﺎﻥ ﭘﮋﻭﻫـﺶ ﻭ ﺑﺮﻧﺎﻣﻪﺭﻳﺰﻱ ﺁﻣﻮﺯﺷـﻲ ﻭﺍﺑﺴـﺘﻪ ﺑﻪ ﻭﺯﺍﺭﺕ ﺁﻣـﻮﺯﺵ ﻭ ﭘـﺮﻭﺭﺵ ﺗﻬﻴـﻪ ﻭ ﻣﻨﺘﺸـﺮ ﻣﻲﺷـﻮﻧﺪ:
1
6
11
ﺑﺎ ﻣﺠﻠﻪﻫﺎﻱ ﺭﺷﺪﺁﺷﻨﺎ ﺷﻮﻳﺪ
ﻣﺠﻠﻪ ﻫﺎﻱ ﺩﺍﻧﺶ ﺁﻣﻮﺯﻱ ) ﺑﻪ ﺻﻮﺭﺕ ﻣﺎﻫﻨﺎﻣﻪ ﻭ 8ﺷﻤﺎﺭﻩ ﺩﺭ ﻫﺮ ﺳﺎﻝ ﺗﺤﺼﻴﻠﻲ ﻣﻨﺘﺸﺮ ﻣﻲﺷﻮﻧﺪ ( :
5 10
ﺩﻓﺘﺮ ﺍﻧﺘﺸﺎﺭﺍﺕ ﻛﻤﻚ ﺁﻣﻮﺯﺷﻲ
14
8
)ﺑﺮﺍﻱ ﺩﺍﻧﺶﺁﻣﻮﺯﺍﻥ ﺁﻣﺎﺩﮔﻲ ﻭ ﭘﺎﻳﻪﻱ ﺍﻭﻝ ﺩﻭﺭﻩﻱ ﺩﺑﺴﺘﺎﻥ(
13
)ﺑﺮﺍﻱ ﺩﺍﻧﺶﺁﻣﻮﺯﺍﻥ ﭘﺎﻳﻪﻫﺎﻱ ﺩﻭﻡ ﻭ ﺳﻮﻡ ﺩﻭﺭﻩﻱ ﺩﺑﺴﺘﺎﻥ( )ﺑﺮﺍﻱ ﺩﺍﻧﺶﺁﻣﻮﺯﺍﻥ ﭘﺎﻳﻪﻫﺎﻱ ﭼﻬﺎﺭﻡ ﻭ ﭘﻨﺠﻢ ﺩﻭﺭﻩﻱ ﺩﺑﺴﺘﺎﻥ(
16 19
20
18
)ﺑﺮﺍﻱ ﺩﺍﻧﺶﺁﻣﻮﺯﺍﻥ ﺩﻭﺭﻩﻱ ﺭﺍﻫﻨﻤﺎﻳﻲ ﺗﺤﺼﻴﻠﻲ(
17
)ﺑﺮﺍﻱ ﺩﺍﻧﺶﺁﻣﻮﺯﺍﻥ ﺩﻭﺭﻩﻱ ﻣﺘﻮﺳﻄﻪﻭﭘﻴﺶﺩﺍﻧﺸﮕﺎﻫﻲ(
23
24 28
21
22 26
27
ﻣﺠﻠﻪ ﻫﺎﻱ ﺑﺰﺭﮔﺴﺎﻝ ﻋﻤﻮﻣﯽ
25
)ﺑﻪ ﺻﻮﺭﺕ ﻣﺎﻫﻨﺎﻣﻪ ﻭ 8ﺷﻤﺎﺭﻩ ﺩﺭ ﻫﺮ ﺳﺎﻝ ﺗﺤﺼﻴﻠﻲ ﻣﻨﺘﺸﺮ ﻣﻲﺷﻮﻧﺪ(:
29
ﺭﺷﺪ ﺁﻣـﻮﺯﺵ ﺍﺑﺘــﺪﺍﻳﻲ ﺁﻣﻮﺯﺷﻲ
2
︎︀︨ ︗︡ول ︫﹝︀رۀ ٤٥
2
3
7
7
o
4
9
9
6 o
4 o
9
9
7
4 6
3
2
3
6
o
o
1 5
o o
5
o
o
o
5 7
3
9
9
o
2
3 o
4
o
9
ﺭﺷﺪ ﻣﺪﺭﺳﻪ ﻓﺮﺩﺍ
ﺭﺷﺪ ﻣﺪﻳﺮﻳﺖ ﻣﺪﺭﺳﻪ
ﺭﺷﺪ ﻣﻌﻠﻢ
ﻣﺠﻠﻪ ﻫﺎﻱ ﺑﺰﺭﮔﺴﺎﻝ ﻭ ﺩﺍﻧﺶﺁﻣﻮﺯﻱ ﺍﺧﺘﺼﺎﺻﻲ )ﺑﻪ ﺻﻮﺭﺕ ﻓﺼﻠﻨﺎﻣﻪ ﻭ 4ﺷﻤﺎﺭﻩ ﺩﺭ ﻫﺮ ﺳﺎﻝ ﺗﺤﺼﻴﻠﻲ ﻣﻨﺘﺸﺮ ﻣﻲﺷﻮﻧﺪ(:
7
ﺭﺷﺪ ﺑﺮﻫﺎﻥ ﺭﺍﻫﻨﻤﺎﻳﻲ )ﻣﺠﻠﻪ ﺭﻳﺎﺿﻲ ﺑﺮﺍﻱ ﺩﺍﻧﺶﺁﻣﻮﺯﺍﻥ ﺩﻭﺭﻩﻱ ﺭﺍﻫﻨﻤﺎﻳﻲ ﺗﺤﺼﻴﻠﻲ(
1
8
1
1
1 7
ﺭﺷﺪ ﺁﻣﻮﺯﺵ ﺭﺍﻫﻨﻤـﺎﻳﻲ ﺗﺤﺼﻴﻠﻲ
ﺭﺷﺪ ﺗﻜﻨﻮﻟﻮﮊﻱ
2 3
9
4
1
3
ﺭﺷﺪ ﺑﺮﻫﺎﻥ ﻣﺘﻮﺳﻄﻪ )ﻣﺠﻠﻪ ﺭﻳﺎﺿﻲ ﺑﺮﺍﻱ ﺩﺍﻧﺶﺁﻣﻮﺯﺍﻥ ﺩﻭﺭﻩﻱ ﻣﺘﻮﺳﻄﻪ(
ﺭﺷﺪ ﺁﻣﻮﺯﺵ
ﻗﺮﺁﻥ ﺭﺷﺪ ﺁﻣﻮﺯﺵ ﻣﻌﺎﺭﻑ ﺍﺳﻼﻣﻲ ﺭﺷﺪ ﺁﻣﻮﺯﺵ ﺯﺑﺎﻥ ﻭ ﺍﺩﺏ ﻓﺎﺭﺳﻲ ﺭﺷﺪ ﺁﻣﻮﺯﺵ ﻫﻨﺮ ﺭﺷـﺪ ﻣﺸـﺎﻭﺭ ﻣﺪﺭﺳﻪ ﺭﺷﺪ ﺁﻣﻮﺯﺵ ﺗﺮﺑﻴﺖﺑﺪﻧﻲ ﺭﺷﺪ ﺁﻣﻮﺯﺵ ﻋﻠﻮﻡ ﺍﺟﺘﻤﺎﻋﻲ ﺭﺷـﺪ ﺁﻣﻮﺯﺵ ﺗﺎﺭﻳﺦ ﺭﻳﺎﺿﻲ
ﺭﺷـﺪ ﺁﻣﻮﺯﺵ ﺟﻐﺮﺍﻓﻴﺎ ﺭﺷـﺪ ﺁﻣﻮﺯﺵ ﺯﺑﺎﻥ ﺭﺷﺪ ﺁﻣﻮﺯﺵ
ﺭﺷـﺪ ﺁﻣﻮﺯﺵ ﻓﻴﺰﻳﻚ
ﺭﺷﺪ ﺁﻣﻮﺯﺵ ﺷﻴﻤﻲ
ﺭﺷﺪ ﺁﻣﻮﺯﺵ ﺯﻳﺴﺖﺷﻨﺎﺳﻲ
ﺭﺷﺪ ﺁﻣﻮﺯﺵ ﺯﻣﻴﻦﺷﻨﺎﺳﻲ ﺭﺷﺪ ﺁﻣﻮﺯﺵ ﻓﻨﻲﻭﺣﺮﻓﻪﺍﻱ ﺭﺷﺪ ﺁﻣﻮﺯﺵ ﭘﻴﺶ ﺩﺑﺴﺘﺎﻧﻲ
ﻣﺠﻠﻪﻫﺎﻱ ﺭﺷــﺪ ﻋﻤﻮﻣــﻲ ﻭ ﺍﺧﺘﺼﺎﺻﻲ ﺑــﺮﺍﻱ ﺁﻣﻮﺯﮔﺎﺭﺍﻥ ،ﻣﻌﻠﻤــﺎﻥ ،ﻣﺪﻳﺮﺍﻥ ﻭ ﻛﺎﺭﻛﻨــﺎﻥ ﺍﺟﺮﺍﻳــﻲ ﻣــﺪﺍﺭﺱ ،ﺩﺍﻧﺶﺟﻮﻳﺎﻥ ﻣﺮﺍﻛــﺰ ﺗﺮﺑﻴﺖﻣﻌﻠﻢ ﻭ ﺭﺷــﺘﻪﻫﺎﻱ ﺩﺑﻴﺮﻱ ﺩﺍﻧﺸــﮕﺎﻩﻫﺎ ﻭ ﻛﺎﺭﺷﻨﺎﺳــﺎﻥ ﺗﻌﻠﻴﻢ ﻭ ﺗﺮﺑﻴﺖ ﺗﻬﻴﻪ ﻭ ﻣﻨﺘﺸــﺮ ﻣﻲﺷﻮﻧﺪ.
ﻧﺸـﺎﻧﻲ :ﺗﻬــﺮﺍﻥ ،ﺧﻴﺎﺑــﺎﻥ ﺍﻳﺮﺍﻧﺸــﻬﺮ ﺷﻤﺎﻟﻲ،ﺳــﺎﺧﺘﻤﺎﻥ ﺷــﻤﺎﺭﻩﻱ4 ﺁﻣﻮﺯﺵﻭﭘﺮﻭﺭﺵ ،ﭘﻼﻙ ،266ﺩﻓﺘﺮ ﺍﻧﺘﺸﺎﺭﺍﺕ ﻛﻤﻚﺁﻣﻮﺯﺷﻲ. ﺗﻠﻔﻦ ﻭ ﻧﻤﺎﺑﺮ 88301478 :ـ 021
ﺩﻭﺭﻩﻯ ﺷﺎﻧﺰﺩﻫﻢ ،ﺷﻤﺎﺭﻩ ﻯ ،2ﺯﻣﺴﺘﺎﻥ 1389
ﺭﺍﻫﻨﻤﺎﻳﻲ
27
ﺑﺮگ ﺍﺷﺘﺮﺍﻙ ﻣﺠﻠﻪﻫﺎﻱ ﺭﺷﺪ
ﺩﺍﻧﺶﺍﻓﺰﺍﻳﻲ
ﻫﻤّﺖ ﻣﻀﺎﻋﻒ،ﻛﺎﺭﻣﻀﺎﻋﻒ
﹡﹆︀ط ا﹞﹟ ﹇︧﹝️ دوم ﺣﺴﻦ ﺍﺣﻤﺪﻯ
ﺷﺮﺍﻳﻂ: .1ﭘﺮﺩﺍﺧﺖ ﻣﺒﻠﻎ 70/000ﺭﻳﺎﻝ ﺑﻪ ﺍﺯﺍﻱ ﻳﻚ ﺩﻭﺭﻩ ﻳﻚ ﺳﺎﻟﻪ ﻣﺠﻠﻪﻱ ﺩﺭﺧﻮﺍﺳﺘﻲ، ﺑﻪ ﺻﻮﺭﺕ ﻋﻠﻲﺍﻟﺤﺴﺎﺏ ﺑﻪ ﺣﺴﺎﺏ ﺷﻤﺎﺭﻩﻱ 39662000ﺑﺎﻧﻚ ﺗﺠﺎﺭﺕ ﺷﻌﺒﻪ ﻱ ﺳﻪ ﺭﺍﻩ ﺁﺯﻣﺎﻳﺶ )ﺳﺮﺧﻪﺣﺼﺎﺭ( ﻛﺪ 395ﺩﺭ ﻭﺟﻪ ﺷﺮﻛﺖ ﺍﻓﺴﺖ. .2ﺍﺭﺳﺎﻝ ﺍﺻﻞ ﻓﻴﺶ ﺑﺎﻧﻜﻲ ﺑﻪ ﻫﻤﺮﺍﻩ ﺑﺮگ ﺗﻜﻤﻴﻞ ﺷﺪﻩﻱ ﺍﺷﺘﺮﺍﻙ ﺑﺎﭘﺴﺖﺳﻔﺎﺭﺷﻲ) .ﻛﭙﻲﻓﻴﺶﺭﺍﻧﺰﺩﺧﻮﺩﻧﮕﻪ ﺩﺍﺭﻳﺪ(.
ﻧﺎﻡ ﻣﺠﻠﻪ ﻫﺎﻱﺩﺭﺧﻮﺍﺳﺘﻲ: ...................................................................................... ......................................................................................
ﻛﻠﻴﺪﻭﺍژﻩﻫﺎ :ﻓﺎﺻﻠﻪﻯ ﻋﻤﻮﺩﻯ ،ﻛﻮﺗﺎﻩﺗﺮﻳﻦ ﻓﺎﺻﻠﻪ . ﺍﺷﺎﺭﻩ: ﺩﺭ ﺷﻤﺎﺭﻩﻯ ﭘﻴﺸﻴﻦ ﺑﺎ ﺍﺳﺘﻔﺎﺩﻩ ﺍﺯ ﻣﻔﻬﻮﻡ ﻓﺎﺻﻠﻪﻯ ﻋﻤﻮﺩﻯ )ﻛﻮﺗﺎﻩﺗﺮﻳﻦ ﻓﺎﺻﻠﻪ( ،ﺗﻮﺍﻧﺴـﺘﻴﻢ ﺍﻣﻦﺗﺮﻳﻦ ﻧﻘﻄﻪ ﻧﺴﺒﺖ ﺑﻪ ﻳﻚ ِ ﺧﻂ ﻧﺎﺍﻣﻦ ﺭﺍ ﺑﻴﺎﺑﻴﻢ .ﺩﺭ ﺍﻳﻦ ﺷﻤﺎﺭﻩ ﺣﺎﻟﺖﻫﺎﻯ ﺩﻳﮕﺮﻯ ﺍﺯ ﻧﺎﺍﻣﻨﻰ ﺭﺍ ﺑﺮﺭﺳﻰ ﺧﻮﺍﻫﻴﻢ ﻛﺮﺩ.
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ﻧﺎﻡ ﻭ
ﻧﺎﻡﺧﺎﻧﻮﺍﺩﮔﻲ. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .:
ﺗﺎﺭﻳﺦ
ﺗﻮﻟﺪ. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .:
ﻣﻴﺰﺍﻥ ﺗﺤﺼﻴﻼﺕ:
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ﺗﻠﻔﻦ. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .:
ﻧﺸﺎﻧﻲ ﻛﺎﻣﻞ ﭘﺴﺘﻲ:
ﺍﺳﺘﺎﻥ . . . . . . . . . . . . . . . . . . . . . . . . . . . . :ﺷﻬﺮﺳﺘﺎﻥ: ﺧﻴﺎﺑﺎﻥ: ﻛﺪﺍﺷﺘﺮﺍﻙ. . . . . . . . . . . . . . . . . . . . . . . . : ﭘﻼﻙ . . . . . . . . . . . . . . . . . . . :ﺷﻤﺎﺭﻩﻱ ﭘﺴﺘﻲ. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . : .............................
ﺝ( ﻓﺮﺽ ﻛﻨﻴﺪ ﺩﻭ ﺭﺷﺘﻪ ﻛﻮﻩ )ﺧﻂ( ﻧﺎﺍﻣﻦ L1ﻭ L2ﺩﺍﺭﻳﻢ ﻛﻪ ﺑﻪ ﺻــﻮﺭﺕ ﻣﻮﺍﺯﻯ ﺑﺎ ﻫﻢ ﻗﺮﺍﺭ ﮔﺮﻓﺘﻪﺍﻧﺪ .ﻣﻰﺧﻮﺍﻫﻴﻢ ﺑﻴﻦ ﺍﻳﻦ ﺩﻭ ﺭﺷــﺘﻪ ﻛﻮﻩ ،ﻣﺴﻴﺮﻯ ﺗﻌﻴﻴﻦ ﻛﻨﻴﻢ ﻛﻪ ﺷﻬﺮﻫﺎﻯ )ﻧﻘﺎﻁ( Aﻭ Bﺭﺍ ﺑﻪ ﻳﻜﺪﻳﮕﺮ ﻣﺘﺼﻞ ﻛﻨﺪ .ﺑﻪ ﻧﻈﺮ ﺷــﻤﺎ ﭼﮕﻮﻧﻪ ﻣﻰﺗﻮﺍﻥ ﺍﻣﻦﺗﺮﻳﻦ ﻣﺴﻴﺮ ﺭﺍ ﺍﻧﺘﺨﺎﺏ ﻛﺮﺩ؟ )ﻓﺮﺽ ﻛﻨﻴﺪ ﺷﺪﺕ ﻧﺎﺍﻣﻨﻰ ﺩﺭ ﻫﺮ ﺩﻭ ﺭﺷﺘﻪ ﺑﺮﺍﺑﺮ ﺍﺳﺖ(.
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ﺩﺭ ﺻﻮﺭﺗﻲ ﻛﻪ ﻗﺒ ًﻼ ﻣﺸﺘﺮﻙ ﻣﺠﻠﻪ ﺑﻮﺩﻩﺍﻳﺪ ،ﺷﻤﺎﺭﻩﻱ ﺍﺷﺘﺮﺍﻙ ﺧﻮﺩ ﺭﺍ ﺑﻨﻮﻳﺴﻴﺪ:
ﺍﻣﻀﺎ:
ﻃﺒﻴﻌﻰ ﺍﺳﺖ ﻛﻪ ﺑﺎﻳﺪ ﻧﻘﺎﻁ ﺍﻣﻦ ﺩﺭ ﺩﻭﺭﺗﺮﻳﻦ ﻣﺤﻞ ﺍﺯ ﺧﻄﻮﻁ ﻗﺮﺍﺭ ﮔﻴﺮﻧﺪ ،ﻭﻟﻰ ﻣﺸــﻜﻞ ﺍﻳﻦﺟﺎﺳــﺖ ﻛﻪ ﻭﻗﺘﻰ ﺍﺯ ﻳﻚ ﺧﻂ ﺩﻭﺭ ﻣﻰﺷﻮﻳﻢ ﺑﻪ ﺧﻂ ﺩﻳﮕﺮ ﻧﺰﺩﻳﻚ ﺧﻮﺍﻫﻴﻢ ﺷــﺪ .ﺑﺮﺍﻯ ﺭﻓﻊ ﺍﻳﻦ ﻣﺸــﻜﻞ ،ﻳﻚ ﺍﻳﺪﻩ ﻣﻰﺗﻮﺍﻧﺪ ﺍﻳﻦ ﺑﺎﺷﺪ ﻛﻪ ﺍﺯ ﻳﻚ ﺧﻂ ﺑﻪ ﺳﻮﻯ ﺧﻂ ﺩﻳﮕﺮ ﺣﺮﻛﺖ ﻛﻨﻴﻢ ﻭ A
15875/6567 ﺻﻨـﺪﻭﻕ ﭘﺴﺘﻲ ﻣﺮﻛﺰﺑﺮﺭﺳﻲﺁﺛﺎﺭ: 16595/111 ﺻﻨـﺪﻭﻕ ﭘﺴﺘﻲ ﺍﻣﻮﺭﻣﺸﺘﺮﻛﻴﻦ: www.roshdmag.ir ﻧﺸﺎﻧﻲ ﺍﻳﻨﺘﺮﻧﺘﻲ: 77335110ـ 77336656ـ021 ﺍﻣﻮﺭ ﻣﺸﺘﺮﻛﻴﻦ: 88301482ـ021 ﭘﻴﺎﻡﮔﻴﺮ ﻣﺠﻠﻪ ﻫﺎﻱ ﺭﺷﺪ:
a b
ﻳﺎﺩﺁﻭﺭﻱ:
ﻫﺰﻳﻨﻪﻱ ﺑﺮﮔﺸﺖ ﻣﺠﻠﻪ ﺩﺭ ﺻﻮﺭﺕ ﺧﻮﺍﻧﺎ ﻭ ﻛﺎﻣﻞ ﻧﺒﻮﺩﻥ ﻧﺸﺎﻧﻲ ﻭ ﻋﺪﻡ ﺣﻀﻮﺭ ﮔﻴﺮﻧﺪﻩ ،ﺑﺮﻋﻬﺪﻩﻱ ﻣﺸﺘﺮﻙ ﺍﺳﺖ. ﻣﺒﻨﺎﻱ ﺷﺮﻭﻉ ﺍﺷﺘﺮﺍﻙ ﻣﺠﻠﻪ ﺍﺯ ﺯﻣﺎﻥ ﺩﺭﻳﺎﻓﺖ ﺑﺮگ ﺍﺷﺘﺮﺍﻙ ﺧﻮﺍﻫﺪ ﺑﻮﺩ.
B 28
ﺭﺍﻫﻨﻤﺎﻳﻲ
ﺩﻭﺭﻩﻯ ﺷﺎﻧﺰﺩﻫﻢ ،ﺷﻤﺎﺭﻩﻯ ،2ﺯﻣﺴﺘﺎﻥ 1389
L
ﺩﺭ ﻣﺴﻴﺮ ﺣﺮﻛﺖ ،ﻓﺎﺻﻠﻪﻣﺎﻥ ﺭﺍ ﺍﺯ ﻫﺮ ﺩﻭ ﺧﻂ ﺍﻧﺪﺍﺯﻩ ﺑﮕﻴﺮﻳﻢ ﻭ ﻣﻘﺎﻳﺴﻪ ﻛﻨﻴﻢ. ﻫﺮﭼــﻪ ﺍﺯ L1ﺩﻭﺭ ﺷــﻮﻳﻢ ،ﻓﺎﺻﻠــﻪﻯ aﺑﺰﺭگﺗــﺮ ﻭ bﻛﻮﭼﻚﺗﺮ ﻣﻰﺷــﻮﺩ .ﺍﻟﺒﺘﻪ ﭼﻮﻥ ﺩﺭ ﻫﺮﻳﻚ ﺍﺯ ﺍﻳﻦ ﺧﻄﻮﻁ ﻣﻤﻜﻦ ﺍﺳﺖ ﺯﻟﺰﻟﻪ ﺭﺥ ﺩﻫﺪ ،ﻣﺠﺒﻮﺭﻳﻢ ﻓﺎﺻﻠﻪﻯ ) aﻛﻢﺗﺮ( ﺭﺍ ﻓﺎﺻﻠﻪﻯ ﺍﻳﻤﻨﻰ ﺩﺭ ﻧﻈﺮ ﺑﮕﻴﺮﻳﻢ. ﺩﺭ ﻭﺍﻗﻊ ﻣﺎ ﺑﺎﻳﺪ ﺳــﻌﻰ ﻛﻨﻴﻢ ﺍﻳﻦ ﻓﺎﺻﻠﻪﻯ ﻛﻢﺗﺮ ﺭﺍ ﺑﻪ ﺣﺪﺍﻛﺜﺮ ﻣﻤﻜﻦ ﺑﺮﺳــﺎﻧﻴﻢ .ﭘﺲ ﺣﺮﻛﺘﻤﺎﻥ ﺑﻪ ﺳﻮﻯ L2ﺭﺍ ﺗﺎ ﺁﻥﺟﺎ ﺍﺩﺍﻣﻪ ﻣﻰﺩﻫﻴﻢ ﻛﻪ aﺑﺎ bﺑﺮﺍﺑﺮ ﺷــﻮﻧﺪ .ﺁﺷﻜﺎﺭ ﺍﺳــﺖ ﻛﻪ ﺍﮔﺮ ﺑﻪ ﺍﻳﻦ ﺣﺮﻛﺖ ﺍﺩﺍﻣﻪ ﺩﻫﻴﻢ، bﻛﻮﭼﻚﺗﺮ ﺧﻮﺍﻫﺪ ﺷــﺪ ﻭ ﻓﺎﺻﻠﻪﻯ ﺍﻳﻤﻨﻰ ﺩﻭﺑﺎﺭﻩ ﻛﻢ ﻣﻰﺷــﻮﺩ .ﭘﺲ ﺍﻣﻦﺗﺮﻳﻦ ﺣﺎﻟﺖ ﺯﻣﺎﻧﻰ ﺭﺥ ﻣﻰﺩﻫﺪ ﻛﻪ aﻭ bﺑﺮﺍﺑﺮ ﺑﺎﺷﻨﺪ. ﺑﻪ ﺭﺍﺣﺘﻰ ﻣﻰﺗﻮﺍﻥ ﻧﺸﺎﻥ ﺩﺍﺩ ﻧﻘﺎﻃﻰ ﻛﻪ ﺍﺯ L1ﻭ L2ﺑﻪ ﻳﻚ ﻓﺎﺻﻠﻪ ﺑﺎﺷﻨﺪ ،ﺭﻭﻯ ﺧﻄﻰ ﻗﺮﺍﺭ ﺩﺍﺭﻧﺪ ﻛﻪ ﻭﺳﻂ ﺍﻳﻦ ﺩﻭ ﺧﻂ ﻭ ﻣﻮﺍﺯﻯ ﺑﺎ ﺁﻥﻫﺎ ﻛﺸﻴﺪﻩ ﺷﺪﻩ ﺑﺎﺷﺪ. ﺩ( ﻓﺮﺽ ﻛﻨﻴﺪ ﺩﻭ ﺭﺷﺘﻪ ﻛﻮﻩ )ﺧﻂ( ﻧﺎ ﺍﻣﻦ L1ﻭ L2ﺩﺍﺭﻳﻢ ﻛﻪ ﺑﻪ ﺻﻮﺭﺕ ﻧﺎﻣﻮﺍﺯﻯ ﺑﺎ ﻫﻢ ﻗﺮﺍﺭ ﮔﺮﻓﺘﻪﺍﻧﺪ .ﻣﻰﺧﻮﺍﻫﻴﻢ ﺑﻴﻦ ﺍﻳﻦ ﺩﻭ ﺭﺷــﺘﻪ ﻛﻮﻩ ،ﻣﺴﻴﺮﻯ ﺗﻌﻴﻴﻦ ﻛﻨﻴﻢ ﻛﻪ ﺷﻬﺮﻫﺎﻯ )ﻧﻘﺎﻁ( Aﻭ Bﺭﺍ ﺑﻪ ﻳﻜﺪﻳﮕﺮ ﻣﺘﺼﻞ ﻛﻨﺪ .ﺑﻪ ﻧﻈﺮ ﺷــﻤﺎ ﭼﮕﻮﻧﻪ ﻣﻰﺗﻮﺍﻥ ﺍﻣﻦﺗﺮﻳﻦ ﻣﺴﻴﺮ ﺭﺍ ﺍﻧﺘﺨﺎﺏ
ﻛﺮﺩ؟ )ﻓﺮﺽ ﻛﻨﻴﺪ ﺷﺪﺕ ﻧﺎﺍﻣﻨﻰ ﺩﺭ ﻫﺮ ﺩﻭ ﺭﺷﺘﻪ ﺑﺮﺍﺑﺮ ﺍﺳﺖ(. A
L
B ﺑــﻪ ﺍﻳﻦ ﺣﺎﻟﺖ ﺧﻮﺏ ﻓﻜﺮ ﻛﻨﻴﺪ ﻭ ﭘﺎﺳــﺦﻫﺎﻯ ﺧﻮﺩ ﺭﺍ ﺑﺮﺍﻯ ﻣﺠﻠﻪ ﺑﻔﺮﺳــﺘﻴﺪ .ﺩﺭ ﺷــﻤﺎﺭﻩﻯ ﺁﻳﻨــﺪﻩ ﺣﺎﻟﺖﻫﺎﻯ ﭘﻴﭽﻴﺪﻩﺗﺮﻯ ﺭﺍ ﺑﺮﺭﺳــﻰ ﺧﻮﺍﻫﻴﻢ ﻛﺮﺩ.
ﺩﻭﺭﻩﻯ ﺷﺎﻧﺰﺩﻫﻢ ،ﺷﻤﺎﺭﻩ ﻯ ،2ﺯﻣﺴﺘﺎﻥ 1389
ﺭﺍﻫﻨﻤﺎﻳﻲ ﻲ ﺭﺍﻫﻨﻤﺎﻳ
29
ﻧﺎﻣﻪﻫﺎ
﹈ ﹡︀﹞﹤ ،﹈ دو︨️ ﺩﺭ ﺣﺮﻑ ﺍﻭﻝ ﺷﻤﺎﺭﻩﻱ 54ﺑﺮﻫﺎﻥ ﺭﺍﻫﻨﻤﺎﻳﻲ ﺳﺮﺩﺑﻴﺮ ﻣﺠﻠﻪ ﺳﺆﺍﻻﺗﻲ ﺭﺍ ﻣﻄﺮﺡ ﻧﻤﻮﺩﻩ ﺑﻮﺩﻧﺪ ﻛﻪ ﺁﻗﺎﻱ ﻣﻬﺪﻱ ﻧﺼﻴﺮﻱ ﺩﺍﻧﺶﺁﻣﻮﺯ ﺳﺎﻋﻲ ﺳﺎﻝ ﺳﻮﻡ ﻣﺪﺭﺳﻪ ﺭﺍﻫﻨﻤﺎﻳﻲ ﺭﺍﺯﻱ ﺍﺯ ﻣﻨﻄﻘﻪﻱ 16ﺗﻬــﺮﺍﻥ ﺑﻪ ﺁﻥ ﺳﺆﺍﻻﺕ ﭘﺎﺳﺦﻫﺎﻱ ﺩﻗﻴﻖ ﻭ ﺧﻮﺍﻧﺪﻧﻲﺍﻱ ﺩﺍﺩﻩﺍﻧﺪ ﻛﻪ ﺑﺎ ﻫﻢ ﻣﻲﺧﻮﺍﻧﻴﻢ:
ﺷﻴﺮﻳﻦ ﻭ ﺩﻭﺳﺖﺩﺍﺷﺘﻨﻰ ﺍﺳﺖ. ﺁﻳﺎ ﺑـﺮﺍﻯ ﻣﻠﻤﻮﺱ ﻛـﺮﺩﻥ ﺭﻳﺎﺿﻴـﺎﺕ ﻭ ﻋﻼﻗﻪﻣﻨﺪ ﻛﺮﺩﻥ ﺩﻭﺳﺘﺎﻥ ﺷﻤﺎ ﺑﻪ ﺁﻥ ،ﺭﺍﻫﻰ ﭘﻴﺸﻨﻬﺎﺩ ﻣﻰﻛﻨﻴﺪ؟ ﺁﻣﻮﺯﺵ ﻣﻔﺎﻫﻴﻢ ﺭﻳﺎﺿﻰ ﺑﻪﻭﺳــﻴﻠﻪﻯ ﻓﻴﻠﻢ ﻭ ﺍﻧﻴﻤﻴﺸــﻦ ،ﺑﺮﮔﺰﺍﺭﻯ ﻣﺴــﺎﺑﻘﺎﺕ ﺭﻳﺎﺿﻰ ﺍﺳــﺘﺎﻧﻰ ﻭ ﻛﺸﻮﺭﻯ؛ ﺗﺸــﻜﻴﻞ ﺍﻧﺠﻤﻦ ﺩﻭﺳﺖﺩﺍﺭﺍﻥ
ﺷﻤﺎ ﺗﺎ ﭼﻪ ﺣﺪ ﺑﻪ ﺭﻳﺎﺿﻴﺎﺕ ﻋﻼﻗﻪ ﺩﺍﺭﻳﺪ؟ ﻣﻦ ﺭﻳﺎﺿﻴﺎﺕ ﺭﺍ ﺑﺎ ﺩﻝ ﻭ ﺟﺎﻥ ﺩﻭﺳﺖ ﺩﺍﺭﻡ ﻭ ﻋﺎﺷﻖ ﺁﻥ ﻫﺴﺘﻢ. ﺁﻳـﺎ ﺍﺯ ﻭﻳﮋﮔﻰﻫـﺎﻯ ﺭﻳﺎﺿﻴـﺎﺕ ﻭ ﺭﻳﺎﺿﻰﺩﺍﻥﻫـﺎ ﻳـﺎ ﺭﻳﺎﺿﻰﺧﻮﺍﻥﻫﺎ ﺍﻃﻼﻋﻰ ﺩﺍﺭﻳﺪ؟ ﺗﺎ ﺁﻥﺟﺎ ﻛﻪ ﺯﻧﺪﮔﻰﻧﺎﻣﻪﻫﺎﻯ ﺁﻥﻫﺎ ﺭﺍ ﺧﻮﺍﻧﺪﻩﺍﻡ ،ﻣﻌﻤﻮﻻً ﺍﻧﺴﺎﻥﻫﺎﻯ ﻓﻮﻕﺍﻟﻌﺎﺩﻩ ﻣﻨﻈﻢ ،ﻣﺮﺗﺐ ﻭ ﻓﻜﻮﺭ ﻭ ﺣﻼﻝ ﻣﺴﺌﻠﻪﺍﻧﺪ ﻭ ﺯﻧﺪﮔﻰ ﻓﻮﻕﺍﻟﻌﺎﺩﻩﺍﻯ
ﺭﻳﺎﺿــﻰ ﺭﺍﻫﻨﻤﺎﻳﻰ؛ ﺁﻣﻮﺯﺵ ﺩﺍﺩﻥ ﻣﻌﻠﻤﺎﻥ ﺭﻳﺎﺿــﻰ ﺗﺎ ﺑﺘﻮﺍﻧﻨﺪ ﺧﻮﺏ ﻭ ﺟﺬﺍﺏ ،ﻫﻤﺮﺍﻩ ﺑﺎ ﻭﺳﺎﻳﻞ ﺩﺭﺱ ﺑﺪﻫﻨﺪ. ﺑﻪ ﻧﻈﺮ ﺷـﻤﺎ ﺭﻳﺎﺿﻴـﺎﺕ ﭼﻪﻗﺪﺭ ﺑﺎ ﻓﻄﺮﺕ ﺁﺩﻣﻰ ﺳـﺎﺯﮔﺎﺭ ﺍﺳﺖ؟ ﺑﻪ ﻧﻈﺮ ﻣﻦ ،ﻓﻄﺮﺕ ﺍﻭﻟﻴﻪﻯ ﺑﺸﺮ ﺑﺮ ﺍﺳﺎﺱ ﻣﺤﺎﺳﺒﺎﺕ ﺍﻭﻟﻴﻪ ﻭ ﭘﺎﻳﻪﻯ ﺭﻳﺎﺿﻰ ﺑﻨﺎ ﺷﺪﻩ ﻭ ﺭﺍﺯ ﺁﻓﺮﻳﻨﺶ ﺩﺭ ﺭﻳﺎﺿﻴﺎﺕ ﻧﻬﻔﺘﻪ ﺍﺳﺖ. ﺁﻳﺎ ﺑﻪ ﺗﺎﺭﻳﺦ ﺭﻳﺎﺿﻴﺎﺕ ﻋﻼﻗﻪ ﺩﺍﺭﻳﺪ؟
ﺩﺍﺭﻧﺪ. ﻓﻜﺮ ﻣﻰﻛﻨﻴﺪ ﺭﻳﺎﺿﻴـﺎﺕ ﺩﺭ ﺩﺍﻧﺶﻫﺎﻯ ﺩﻳﮕﺮ ﻭ ﻋﻠﻮﻡ ﻏﻴﺮ ﺭﻳﺎﺿﻰ ﻣﺎﻧﻨﺪ ﻓﻴﺰﻳﻚ ،ﻣﻜﺎﻧﻴﻚ ،ﺷﻴﻤﻰ ،ﺯﻳﺴﺖﺷﻨﺎﺳﻰ ،ﻧﺠﻮﻡ، ﺍﻗﺘﺼﺎﺩ ﻭ ﺟﺎﻣﻌﻪﺷﻨﺎﺳﻰ ﭼﻪ ﻧﻘﺸﻰ ﺩﺍﺭﺩ؟ ﺭﻳﺎﺿﻴﺎﺕ ﭘﺎﻳﻪﻯ ﺍﺳﺎﺳﻰ ﺑﺴﻴﺎﺭﻯ ﺍﺯ ﻋﻠﻮﻡ ﻭ ﺩﺍﻧﺶ ﺑﺸﺮ ﺍﺳﺖ ﻭ ﻣﺎﺩﺭ ﻫﻤﻪﻯ ﻋﻠﻮﻡ ﻭ ﻓﻨﻮﻥ ﻭﺩﺍﻧﺶﻫﺎﻯ ﺍﻧﺴﺎﻥ ﺑﻪ ﺷﻤﺎﺭ ﻣﻰﺁﻳﺪ. ﺭﺍﺳـﺘﻰ ،ﺭﻳﺎﺿﻴﺎﺕ ﭼﻪ ﻧﻘﺸﻰ ﺩﺭ ﺯﻧﺪﮔﻰ ﺭﺭﻭﺯﻣﺮﻩﻯ ﺷﻤﺎ
ﺧﻴﻠﻰ ﺯﻳﺎﺩ .ﺍﻣﺎ ﻛﺘﺎﺏ ﻣﻨﺎﺳﺒﻰ ﺩﺭ ﺣﺪ ﺳﻦ ﺧﻮﺩﻡ ﻛﻪ ﺑﺘﻮﺍﻧﻢ ﺧﻮﺏ ﺁﻥ ﺭﺍ ﺑﻔﻬﻤﻢ ،ﭘﻴﺪﺍ ﻧﻜﺮﺩﻩﺍﻡ. ﺍﺯ ﺗﺎﺭﻳـﺦ ﺭﻳﺎﺿﻴـﺎﺕ ،ﺍﻳﺮﺍﻧﻰ ﻭ ﺍﺳـﻼﻣﻰ ،ﭼﻪﻗﺪﺭ ﺁﮔﺎﻫﻰ ﺩﺍﺭﻳﺪ؟ ﻣﺜ ً ﻼ ﺍﺑﻮﺭﻳﺤﺎﻥ ﺑﻴﺮﻭﻧﻰ ،ﺧﻴﺎﻡ ﻭ ﻛﺎﺷﺎﻧﻰ ﺭﺍ ﺧﻮﺏ ﻣﻰﺷﻨﺎﺳﻢ. ﺑـﻪ ﻧﻈﺮ ﺷـﻤﺎ ﻣﻌﻠـﻢ ﺭﻳﺎﺿـﻰ ﺧـﻮﺏ ﻭ ﻣﻮﻓـﻖ ﺑﺎﻳﺪ ﭼﻪ ﺧﺼﻮﺻﻴﺎﺗﻰ ﺩﺍﺷﺘﻪ ﺑﺎﺷﺪ؟
ﺩﺍﺭﺩ؟ ﺩﺭ ﺯﻧﺪﮔــﻰ ﻣﻦ ﻭ ﺩﻭﺳــﺘﺎﻧﻢ ﺭﻳﺎﺿﻴﺎﺕ ﻧﻘﺶ ﻣﻬﻤــﻰ ﺩﺍﺭﺩ؛ ﻣﺎﻧﻨﺪ:
ﺑﺎﻳﺪ ﺩﺭ ﺍﺑﺘﺪﺍ ﺑﺎ ﺑﭽﻪﻫﺎ ﺩﻭﺳﺖ ﻭ ﺻﻤﻴﻤﻰ ﺑﺎﺷﺪ ،ﺧﻮﺩﺵ ﺑﻪ ﺭﻳﺎﺿﻰ
ﺑﺮﻧﺎﻣﻪﺭﻳــﺰﻯ ﺯﻧﺪﮔﻰ ،ﺗﺤﺼﻴﻠﻰ ﻭ ...؛ ﭼﻴﻨﺶ ﺍﺗﺎﻕ؛ ﭘﻴﺶﺑﻴﻨﻰ ﻭﺳــﺎﻳﻞ
ﻋﻼﻗﻪﻣﻨﺪ ﺑﺎﺷــﺪ ،ﺭﻳﺎﺿﻰ ﺭﺍ ﺧﻮﺏ ﻳﺎﺩ ﺑﺪﻫﺪ ﻭ ﺳــﻌﻰ ﻛﻨﺪ ﻣﺴﺌﻠﻪﻫﺎﻯ
ﻣــﻮﺭﺩ ﻧﻴﺎﺯ؛ ﺑﺮﺁﻭﺭﺩ ﻧﻴﺎﺭﻫﺎﻯ ﺭﻭﺯﺍﻧﻪ؛ ﺧﺮﻳﺪ ﻭﺳــﺎﻳﻞ ﻣــﻮﺭﺩ ﻧﻴﺎﺯ ﻣﻨﺰﻝ ﻭ
ﺳﺎﺩﻩ ﺭﺍ ﺑﺎ ﻋﻼﻗﻪ ﺑﻪ ﺑﭽﻪﻫﺎ ﺍﺭﺍﺋﻪ ﺩﻫﺪ.
ﺧﻴﻠﻰ ﭼﻴﺰﻫﺎﻯ ﺩﻳﮕﺮ. ﺍﮔﺮ ﺭﻳﺎﺿﻴﺎﺕ ﺑﺮﺍﻯ ﺷـﻤﺎ ﺩﺭﺳـﻰ ﺳـﺨﺖ ،ﺑـﺪﻭﻥ ﺭﻭﺡ ﻭ ﺧﺸﻚ ﺑﻪ ﻧﻈﺮ ﻣﻰﺭﺳﺪ ،ﺩﻟﻴﻞ ﺁﻥ ﺭﺍ ﺩﺭ ﭼﻪ ﻣﻰﺩﺍﻧﻴﺪ؟ ﺑﺮﺍﻯ ﻣﻦ ﺑﺮﻋﻜﺲ ،ﺭﻳﺎﺿﻰ ﺩﺭﺱ ﺧﺸــﻜﻰ ﻧﻴﺴﺖ ،ﺑﻠﻜﻪ ﻓﻮﻕﺍﻟﻌﺎﺩﻩ 30
ﺭﺍﻫﻨﻤﺎﻳﻲ
ﺩﻭﺭﻩﻯ ﺷﺎﻧﺰﺩﻫﻢ ،ﺷﻤﺎﺭﻩﻯ ،2ﺯﻣﺴﺘﺎﻥ 1389
ﺭﻳﺎﺿﻴﺎﺕ ﺑﺎ ﻧﻈﻢ ﭼﻪ ﺍﺭﺗﺒﺎﻃﻰ ﺩﺍﺭﺩ؟ ﺑﺴﻴﺎﺭﻯ ﺍﺯ ﻧﻈﻢﻫﺎﻯ ﺑﺸﺮ ﺑﺮ ﺍﺳﺎﺱ ﺭﻳﺎﺿﻴﺎﺕ ﺍﺳﺖ .ﻛﺴﻰ ﻛﻪ ﺧﻮﺏ ﺭﻳﺎﺿﻰ ﻣﻰﺩﺍﻧﺪ ﻭ ﻣﻰﻓﻬﻤﺪ ،ﻣﻌﻤﻮﻻً ﺍﻧﺴــﺎﻥ ﻣﻨﻈﻢ ﻭ ﻣﺮﺗﺒﻰ ﺍﺳﺖ ﻭ ﺩﺭ ﺧﻴﻠﻰ ﻛﺎﺭﻫﺎ ﻣﻮﻓﻖ ﻭ ﻣﻤﺘﺎﺯ ﻣﻰﺷﻮﺩ.
︎︀︨ ا︑︀︋ ﹏﹫︊﹞﹢ز︙﹤ _ 1ﺑﻠــﻰ ﺍﻣﻜﺎﻥ ﺍﻧﺘﺨﺎﺏ ﭼﻨﻴﻦ ﻣﻴﺪﺍﻥ ﺑﺰﺭﮔﻰ ﻣﻴﺴــﺮ ﺍﺳــﺖ ﻭ ﻛﻢﺗﺮﻳﻦ ﻣﻘﺪﺍﺭ ﻣﺴــﺎﺣﺖ ﻣﻴﺪﺍﻥ ﺑﺎﺯﻯ ،ﻣﺮﺑﻌﻰ ﺑﻪ ﺿﻠﻊ ﺳﻪ ﻣﺘﺮ ﺍﺳﺖ. ﺍﮔﺮ ﻛﺎﻭﻩ ﺑﺎﺯﻯ ﺭﺍ ﺍﺯ ﻧﻘﻄﻪﻯ ﻣﺸﺨﺺ ﺷﺪﻩ ﺩﺭ ﺍﻳﻦ ﺗﺼﻮﻳﺮ ﺁﻏﺎﺯ ﻛﻨﺪ:
ﻣﻌﻴﻦ ﺷﺪﻩ ﺩﺭ ﺑﻬﺮﻭﺯ ﺧﻮﺍﻫﺪ ﺗﻮﺍﻧﺴﺖ ﺍﺗﻮﻣﺒﻴﻞ ﺭﺍ ﺑﻪ ﻳﻜﻰ ﺍﺯ ﻧﻘﺎﻁ ﻣﻌ ﻦ ﻴﻦ ﺷﻜﻞ ﺯﻳﺮ ﺑﺮﺍﻧﺪ.
ﺒﻴﻞ ﺍﺗﻮﻣﺒﻴﻞ ﻞ ﻮﻣﺒﻴ ﺸﻮﺩ ﺍﺗﺍﺗﻮﻣ ﻧﺸﻮﺩ ﻣﻮﻓﻖ ﻧ ﻮ ﺣﺮﻳﻒ ﻣﻮ ﻖ ﻓﻖ ﻛﻨﺪ ﻛﻪ ﻒ ﺑﺎﺯﻯ ﻛﻨﺪ ﺎﺯﻯ ﻯ ﭼﻨﺎﻥ ﺑﺑﺎﺯ ﺗﻮﺍﻧﺪ ﭼﻨﺎﻥ ﻣﻰﺗﻮﺍﺍﻧﺪ ﺍﻣﺎ ﻛﻛﺎﻭﻩ ﻣﻰ ﺍﻣﺎ ﺯﻳﺮﺍ ﺑﺑﺎﺎ ﺑﺮﺳــﺎﻧﺪ ،ﺯﺯﻳﺮﺍ ﻧﺪ، ـﺎﻧﺪ ﺳــﺎ ﺑﺮﺳـ ﻣﺮﺑﻊ ﺑﺑﺮ ﺮﺑﻊ ﻊ ﻫﺎﻯ ﻣﻣﺮﺑ ﮔﻮﺷــﻪﻫﺎﻯ ﺷــﻪ ـﻪ ﮔﻮﺷـ ﻭﺍﻗﻊ ﺩﺩﺭ ﮔﻮ ﻧﻘﻄﻪﻯ ﻭﺍﻗﻊ ﭼﻬﺎﺭ ﻧﻘﻄﻪ ﻳﻜﻰ ﺍﺯﺍﺯ ﭼﻬﺎﺭ ﺍﺭﺍ ﺑﻪ ﻳﻜﻰ ﻥ ﭼﻬﺎﺭ ﻧﻧﻘﻄﻪ ،ﺑ ُﺮﺩ ﺍﺯ ﺁﺁﻥ ﻥ ﺎﺭ ﻧﻘﻄ ﻚ ﺯﺍﺯ ﭼﻬﺎ ﻄﻪ ﺁﺁﻥ ﻧﻘﻄﻪ ﭼﻬﺎﺭ ﺳﺘﻘﺮﺍﺭ ﺍﺗﻮﻣﺒﻴﻞ ﺩﺭ ﻫﺮ ﻳﻳﻚ ﺍﺳﺘﻘﺮﺍﺭ ﺎﻧﻰ ﻰ ﺗﻨﻬﺎﺎ ﻣﻜﺎﺎﻧﻧ ﮔﻮﺷﻪﺍﻯ ﺗ ﻬ ﻨﻬ ﻮﺷﻪ ﻚ ﻧﻘﻄﻪ ﮔﮔﻮﺷ ﻣﻜﺎﻧﻰ ﻬﺮﻭﺯ ﺧﻮﺍﻫﺪ ﺑﻮﺩ ﺳﺒﺐ ﺁﻥ ﺍﺳﺖ ﻛﻪ ﻳﻳﻚ ﺑﻬﺮﻭﺯ ِ ﻮﺭ ِﺩ ﻧﻈ ﺟﻬﺖﻫﺎﻯ ﻣﻮ ﻣﺨﺎﻟﻒ ﺟﻬﺖ ﺟﻬﺖ ﻣﺨﺎﻟﻒ ﺳــﺖ ﻛﻪ ﻣﻰﺗﻮﺍﻥ ﺍﺗﻮﻣﺒﻴﻞ ﺭﺍ ﺩﺭ ﻈﺮﺮ ﻣﻮﺭ ﺍﺳــﺖ ﻧﻈ ِ ﺝ ﺣﺮﻳﻒ ﺑﻪ ﺣﺮﻛﺖ ﺩﺭ ﺁﻭﺭﺩ ﻭ ﺁﻥ ﺭﺍ ﺍﺯ ﻣﺤﺪﻭﺩﻩﻯ ﻣﻴﺪﺍﻥ ﺑﺎﺯﻯ ﺍﺭﺍ ﺧﺎﺎﺭ ﺧﺎﺭﺝ ﻛﺮﺩ. _ 2ﺩﺭ ﺍﻳﻦ ﻣﻮﺭﺩ ﭘﺎﺳــﺦ ﻣﻨﻔﻰ ﺍﺳــﺖ ﻭ ﺑﻬــﺮﻭﺯ ﻫﻤﻮﺍﺭﻩ ﺑﺎﺯﻯ ﺍﺭﺍ ﻯ ﺁﻏﺎﺯ ﺗﺎ ﻫﻫﺮ ﻣ ﺎﻓﺘ ﻧﻘﻄﻪﻯ ﺑــﺮﺩ .ﺍﻭ ﻣﻣﻰﺗﻮﺍﻧ ﺪ ﻣﻣﻰﺑ ﺩ ﻣﺴــﺎﻓﺘﻰ ﻛﻪ ﺗﻮﺍﻧــﺪ ﺍﺗﻮﻣﺒﻴﻞ ﺍﺭﺍ ﺍﺯ ﻧﻘﻄﻪ ﺑﺨﻮﺍﻫﺪ ﺩﻭﺭ ﻛﻨﺪ .ﺻﺮﻑ ﻧﻈﺮ ﺍﺯ ﺁﻥ ﻛﻪ ﺩﺭ ﻛﺪﺍﻡ ﻣﺤﻞ ﺍﺯ ﺑﺎﺯﻯ ﻣﺴﺘﻘﺮ ﺑﺎﺷــﺪ ،ﺑﻬﺮﻭﺯ ﺍﺗﻮﻣﺒﻴﻞ ﺑﻪ ﺩﻭﺭﺗﺮﻳﻦ ﻧﻘﻄﻪﻯ ﻣﻴﺪﺍﻥ ﺑﺎﺯﻯ ﺍﻧﺘﻘﺎﻝ ﺩﻫﺪ. ﺑﻬﺮﻭﺯ ﺩﺭ ﻫﻤﻪ ﺣﺎﻝ ﺑﺎﻳﺪ ﻣﺴﻴﺮ ﺣﺮﻛﺖ ﺭﺍ ﭼﻨﺎﻥ ﺑﺮﮔﺰﻳﺪﻧﺪ ﻛﻪ ﺑﺮ ﺧﻂ ﻣﻴﺎﻥ ﻧﻘﻄــﻪﻯ ﺁﻏﺎﺯ ﺑﺎﺯﻯ ﻭ ﻧﻘﻄﻪﻯ ﺍﺳــﺘﻘﺮﺍﺭ ﻛﻨﻮﻧﻰ ﺍﺗﻮﻣﺒﻴﻞ ﺭﺍﺑﻄــ ِﻪ ِ ﻋﻤﻮﺩ ﺑﺎﺷــﺪ .ﺑﺎ ﭘﻴﺮﻭﺯﻯ ﺍﺯ ﺍﻳــﻦ ﺗﺪﺑﻴﺮ ،ﺍﻭ ﻫﻤــﻮﺍﺭﻩ ﻓﺎﺻﻠﻪﻯ ﺧﻮﺩ ﺭﺍ ﺍﺯ ﻧﻘﻄــﻪﻯ ﺁﻏﺎﺯ ﺑﺎﺯﻯ ﺍﻓﺰﺍﻳﺶ ﻣﻰﺩﻫــﺪ ﻭ ﻣﻮﺿﻮﻋﻰ ﻛﻪ ﻛﺎﻭﻩ ﺍﻧﺘﺨﺎﺏ ﻣﻰﻛﻨﺪ ﺑﺮ ﺍﻭ ﺗﺄﺛﻴﺮﻯ ﻧﺪﺍﺭﺩ.
ﺩﺭﺳﺘﻰ ﻳﺎ ﻧﺎﺩﺭﺳﺘﻰ ﺍﻳﻦ ﺷﻴﻮﻩ ﺭﺍ ﻣﻰﺗﻮﺍﻥ ﺍﺯ ﻣﻌﺎﺩﻟﻪﻯ ﻓﻴﺜﺎﻏﻮﺭﺱ ﻣﻌﻴــﻦ ﻛﺮﺩ .ﺑﺮﺍﻯ ﺭﻭﺷــﻦﺗﺮ ﺷــﺪﻥ ﻣﻮﺿــﻮﻉ ﻣﻰﺗﻮﺍﻥ ﻳــﻚ ﻣﺜﻠﺚ ﻟﺰﺍﺍﻭﻳﻪ ﺋﻢﺍﺍﺍﻟﺰ ﻗﺎﺋﻢ ﺍﻟﺰﺍﻭﻳﻪ ﺑﻪ ﺍﺿﻼﻉ ﺳــﻪ ﻣﺴــﺎﻓﺘﻰ ﻛﻪ ﺩﺭ ﭘﻰ ﺫﻛﺮ ﻣﻰﺷﻮﺩ ،ﺗﺮﺳﻴﻢ ﻗﺎﺋﻢ ﻛــﺮﺩ» :ﻓﻓﺎﺻﻠــﻪ ﻛــﺮﺩ: »ﻓﺎﺻﻠــﻪﻯ ﻣﺤﻞ ﺟﺪﻳﺪ ﺍﺳــﺘﻘﺮﺍﺭ ﺍﺗﻮﻣﺒﻴﻞ ﺗﺎ ﻧﻘﻄــﻪﻯ ﺁﻏﺎﺯ«، »ﻓﺎﺻﻠﻪﻯ ﻣﺤﻞ ﻣﺤﻞ ﺍﺳﺘﻘﺮﺍﺭ ﭘﻴﺸﻴﻦ ﺍﺗﻮﻣﺒﻴﻞ ﺗﺎ ﻧﻘﻄﻪﻯ ﺁﻏﺎﺯ« ﻭ »ﻣﺴﺎﻓﺖ ﻴﺪﺍﺳ ﺍﺳ « .ﭘﭘﻴﺪ ﻴﺪ ﺪﻩ«. ﺩﻩ ﺷﺪﺪﻩ ﭘﻴﻤﻮﺩﻩﻩ ﭘﻴ ﻮ ﻤﻮ ﺍﺳﺖ ﻛﻪ ﻣﺴﺎﻓﺖ ﭘﻴﻤﻮﺩﻩ ﺷﺪﻩ ﺩﺭ ﺁﺧﺮﻳﻦ ﺩﻓﻌﻪﻯ ﭘﻴﺪﺍﺳﺖ ﺷﺪﻩ«. ﻟﺰﻭﺍﻭﻳﻪ ﻗﺎﺋﻢﺍ ﺰ ﻣﺜﻠﺚ ﻗﺎﺋﻢ ﺜﻠﺚ ﺚ ﻭﺗﺮ ﻣﻣﺜﻠ ﺜﻠ ﻫﻤﺎﻥ ﻭﺗﺮ ﺍﺗﻮﻣﺒﻴﻞ ،ﻫﻫﻤﺎﻥ ﻴﻞ،، ﻣﺒﻴﻞ ﻴﻞ ﺗﻮﻣﺒ ﺣﺮﻛﺖ ﺍﺍﺗﻮ ﺣﺮﻛﺖ ﺍﻟﺰﺍﻭﻳﻪ ﺍﺳــﺖ ﻛﻪ ﻫﻤﻮﺍﺭﻩ ﻳﻚ ﭘﻴﻤﻮﺩﻩ ﺷﺪ ﻮﺩﻩﻩ ﻴﻤﻮﺩ ﺟﺪﻳﺪ ﭘﭘﻴﻤ ﻴﻤ ﻣﺴﺎﻓﺖ ﺟﺪﻳﺪ ﻣﺮﺑﻊ ﻣﺴﻓﺎﻓﺎﻓﺖ ﺖ ﺮﺑﻊ ﻊ ﺑﻨﺎﺑﺮﺍﻳﻦ ﻣﻣﺮﺑ ﺮﺍ ﻦ ﻳﻦ ﻣﺘﺮ ﻃﻮﻝ ﺩﺍﺭﺩ .ﺑﻨﺎﺑ ﺍ ﺷﺪﻩ ﺑﺮﺍﺑﺮ ﺍﺳﺖ ﺍﻋﺪﺍ ِ ﺪﺍﺩ ﺭﺷﺘﻪﻯ ﺍﺍﻋﺪﺍ ﻳﻚ .ﺭﺷﺘﻪ ﻋﻼﻭﻩﻯ ﻚ ﻳﻚ ﻼﻭﻩﻩ ﺷﺪﻩ ﺑﻪ ﻋﻋﻼﻭ ﭘﻴﻤﻮﺩﻩ ﺷﺪﻩ ﻴﻤﻮﻮﺩﻩ ﻣﺴﺎﻓﺖ ﭘﭘﻴﻤ ﺴﺎﻓﺖ ﻓﺖ ﻣﺮﺑﻊ ﻣﻣﺴﺎ ﺴﺎ ﻣﺮﺑﻊ ﺑﺎﺑﺎ ﺮ ﺍﺩﺩ ﻧﻤﺎﻳﺎﻧﮕ ِﺮ ﺒﺎﺭﺕ ﺑﺎﺯﻯ ﻋﻋﺒﺎ ﺩﺭ ﺑﺎﺯﻯ ﺷﺪﻩ ﺩﺭ ﭘﻴﻤﻮﺩﻩ ﺷﺪﻩ ﻤﻮﺩﻩ ﺩﻩ ﻣﻣﺴﺎﻓﺖﻫﺎﻯ ﭘﻴﻴﻤﻮ ﻋﺒﺎﺭﺕﺍﻧﺪ ﺍﺯ: 5ﻭ 2ﻭ 3ﻭ 2ﻭ 1 ...ﻭ 3ﻭ 2
2ﻭ 7ﻭ6
ﺍﻋﺪﺍﺩ ﻫﻤﻮﺍﺭﻩ ﺭﻭ ﺪﺍﺩ ﺭﺷــﺘﻪ ﻋﺍﻋﺪﺍ ـﺘﻪ ﺭﺷــﺘ ﺷـ ﺍﻳﻦ ﺭﺷ ﻣﻘﺪﺍﺭ ﺍ ﻦ ﻳﻦ ﻘﺪﺍﺭﺍﺭ ﺷــﻮﺩ ،ﻣﻣﻘﺪ ﻘﺪ ـﻮﺩ، ﺩ، ﺷــﻮ ـﻮ ﺷـ ﻣﻰﺷ ﺩﻳﺪﻩ ﻣﻰ ﺩﻳﺪﻩ ﻛﻪ ﻳ ﻃﻮﺭﻯ ﻛﻪ ﺑﻪ ﻃ ﺭﻯ ﺑﻪ ﻴﺴــﺖ. ﻣﺘﺼﻮﺭ ﻧﻴﻴﺴ ﺘﺼﻮﻮﺭ ﺘﺼﻮﺭ ﺁﻥﻫﺎ ﻣﻣﺘﺼ ﺣﺪﻯ ﺑﺮ ﺁﻥ ﻫﻴﭻ ﺣﺪﻯ ﻫﻴﭻ ﺭﻭﺩ ﻭ ﻴ ﻣﻰ ﻭ ﺭﻭ ﻓﺰﻭﻧﻰ ﻣﻰ ﺰﻭﻧﻰ ﻧﻰ ﺑﻪ ﻓﻓﺰﻭ ﺰﻭ ﻧﻴﺴــﺖ .ﺍﺯ ﺍﻳﻦ ﺭﻭ ﺩﺭ ﺻﻮﺭﺗﻰﻛــﻪ ﺭﺗﻰ ﺻﻮﺭﺗ ﺻﻮ ﺻ ﻛــﻪ ﺑﻬﺮﻭﺯ ﻣﺠﺎﻝ ﻛﺎﻓﻰ ﺑﺮﺍﻯ ﻧﻮﺑﺖﻫﺎﻯ ﻣﺘﻌﺪﺩ ﺩﺭ ﺑﺎﺯﻯ ﺑﻴﺎﺑﺪ، ﻣﻰﺗﻮﺍﻧﺪ ﺗﺎ ﻫﺮ ﻧﻘﻄﻪﻯ ﺩﻟﺨﻮﺍﻩ ﺍﺯ ﻧﻘﻄﻪﻯ ﺁﺁﻏﺎﺯ ﺑﺎﺯﻯ ﻓﺎﺻﻠﻪ ﺑﮕﻴﺮﺩ.
ﺩﻭﺭﻩﻯ ﺷﺎﻧﺰﺩﻫﻢ ،ﺷﻤﺎﺭﻩ ﻯ ،2ﺯﻣﺴﺘﺎﻥ 1389
ﺭﺍﻫﻨﻤﺎﻳﻲ
31
ﺭﻳﺎﺿﯽ ﻭ ﺑﺎﺯﯼ
︋︀زی﹨︀ی ﹠︡ ﹡﹀︣ه ﺯﻫﺮﻩ ﭘﻨﺪﻯ
ﻛﻠﻴﺪﻭﺍژﻩﻫﺎ :ﺑﺎﺯﻯ ،ﺯﻭﺝ ،ﻓﺮﺩ ،ﺣﺎﺻﻞ ﺟﻤﻊ ﻭ ﺗﻔﺮﻳﻖ ،ﺏ.ﻡ.ﻡ ،ﺭﻭﺵ ﺑﺮﺩ ،ﺑﺮﺭﺳﻰ ﺑﺎﺯﻯ .
ﺑﺎﺯﯼ ١
ﻭﺳﺎﻳﻞ ﻻﺯﻡ :ﺳﻪ ﺩﺳﺘﻪ ﻟﻮﺑﻴﺎ )ﻳﺎ ﻫﺮ ﻧﻮﻉ ﻣﻬﺮﻩﻯ ﺩﻳﮕﺮ( ﺑﻪ ﺗﺮﺗﻴﺐ ﺷﺎﻣﻞ 16 ،11ﻭ 21ﻟﻮﺑﻴﺎ ﺷﺮﺡ ﺑﺎﺯﻯ: ﺍﺑﺘﺪﺍ ﺑﺎ ﻗﺮﻋﻪﻛﺸــﻰ ،ﺑﺎﺯﻳﻜﻦ ﺍﻭﻝ ﺭﺍ ﺍﻧﺘﺨﺎﺏ ﻛﻨﻴﺪ .ﺑﻪ ﺗﺮﺗﻴﺐ ﺑﺎﺯﻯ ﻛﻨﻴﺪ ﻭ ﻫﺮ ﻛﺪﺍﻡ ﺩﺭ ﻧﻮﺑﺖ ﺧﻮﺩ ﻳﻜﻰ ﺍﺯ ﺩﺳﺘﻪ ﻟﻮﺑﻴﺎﻫﺎ ﺭﺍ ﺑﺮﮔﺰﻳﻨﻴﺪ ﻭ ﺁﻥ ﺭﺍ ﺑﻪ ﺩﻭ ﺩﺳــﺘﻪﻯ ﻛﻮﭼﻚﺗﺮ ﺗﻘﺴﻴﻢ ﻛﻨﻴﺪ .ﺑﺮﻧﺪﻩﻯ ﺑﺎﺯﻯ ﻛﺴﻰ ﺍﺳﺖ ﻛــﻪ ﺁﺧﺮﻳﻦ ﺣﺮﻛﺖ ﺭﺍ ﺍﻧﺠﺎﻡ ﻣﻰﺩﻫﺪ ﻭ ﭘﺲ ﺍﺯ ﺁﻥ ﺣﺮﻛﺖ ﺩﺳــﺘﻪﺍﻯ ﺑﺮﺍﻯ ﺗﻘﺴﻴﻢ ﺷﺪﻥ ﺑﺎﻗﻰ ﻧﻤﻰﻣﺎﻧﺪ. ﺁﻳﺎ ﭘﻴﺶ ﺍﺯ ﺑﺎﺯﻯ ﻣﻰﺗﻮﺍﻧﻴﺪ ﺣﺪﺱ ﺑﺰﻧﻴﺪ ﭼﻪ ﻛﺴــﻰ ﺩﺭ ﺍﻳﻦ ﺑﺎﺯﻯ ﺑﺮﻧﺪﻩ ﻣﻰﺷﻮﺩ؟
ﺑﺎﺯﯼ ٢
ﻭﺳـﺎﻳﻞ ﻻﺯﻡ 20 :ﻛﺎﺭﺕ ﻛﻪ ﺭﻭﻯ ﺁﻥﻫﺎ ﺍﻋﺪﺍﺩ 1ﺗﺎ 20ﻧﻮﺷــﺘﻪ ﺷﺪﻩ ﺍﺳﺖ ﻭ ﻳﻚ ﻛﺎﻏﺬ ﻭ ﺧﻮﺩﻛﺎﺭ ﺷﺮﺡ ﺑﺎﺯﻯ: ﺍﺑﺘﺪﺍ ﺑﺎ ﻗﺮﻋﻪﻛﺸــﻰ ﺑﺎﺯﻳﻜﻦ ﺍﻭﻝ ﺭﺍ ﺍﻧﺘﺨﺎﺏ ﻛﻨﻴﺪ .ﺑﺎﺯﻳﻜﻦ ﺍﻭﻝ ﺩﺭ ﺍﻭﻟﻴــﻦ ﺣﺮﻛﺖ ﺑﺎﻳﺪ ﻳﻜﻰ ﺍﺯ ﻛﺎﺭﺕﻫﺎ ﺭﺍ ﺍﻧﺘﺨﺎﺏ ﻛﻨﺪ ﻭ ﺩﺭ ﺍﺑﺘﺪﺍﻯ ﻳﻚ ﺳــﻄﺮ ﻛﺎﻏﺬ ﻗﺮﺍﺭ ﺩﻫﺪ ﻭ ﺟﻠﻮ ﺁﻥ ﺭﻭﻯ ﻛﺎﻏﺬ ﻳﻚ ﻋﻼﻣﺖ ﺟﻤﻊ ) (+ﻳﺎ ﺗﻔﺮﻳﻖ ) (-ﺑﮕﺬﺍﺭﺩ .ﺑﺎﺯﻳﻜﻦ ﺩﻭﻡ ﺑﺎﻳﺪ ﻳﻜﻰ ﺩﻳﮕﺮ ﺍﺯ ﻛﺎﺭﺕﻫﺎ ﺭﺍ ﺍﻧﺘﺨﺎﺏ ﻛﻨــﺪ ﻭ ﺟﻠﻮ ﻋﺒــﺎﺭﺕ ﻗﺒﻠﻰ ﻗﺮﺍﺭ ﺩﻫﺪ ﻭ ﺟﻠــﻮ ﻛﺎﺭﺕ ﺧﻮﺩ ﻳﻚ ﻋﻼﻣﺖ ﺟﻤﻊ ) (+ﻳﺎ ﺗﻔﺮﻳﻖ ) (-ﺑﻨﻮﻳﺴﺪ .ﺑﺎﺯﻯ ﺑﻪ ﻫﻤﻴﻦ ﺗﺮﺗﻴﺐ ﺗﺎ ﺗﻤﺎﻡ ﺷﺪﻥ ﻛﺎﺭﺕﻫــﺎ ﺑﻪ ﻧﻮﺑﺖ ﺍﺩﺍﻣﻪ ﻣﻰﻳﺎﺑﺪ .ﺩﺭ ﺍﻳﻦ ﻣﺮﺣﻠﻪ ﺑﺎﺯﻳﻜﻨﺎﻥ ﺑﺎﻳﺪ ﻣﻘﺪﺍﺭ ﻋﺒﺎﺭﺕ ﺭﺍ ﻣﺤﺎﺳﺒﻪ ﻛﻨﻨﺪ .ﺍﮔﺮ ﺣﺎﺻﻞ ﺑﻪﺩﺳﺖﺁﻣﺪﻩ ﺯﻭﺝ ﺑﺎﺷﺪ ،ﺑﺎﺯﻳﻜﻦ 32
ﺍﻭﻝ ﻭ ﺩﺭ ﻏﻴﺮ ﺍﻳﻦ ﺻﻮﺭﺕ ﺑﺎﺯﻳﻜﻦ ﺩﻭﻡ ﺑﺮﻧﺪﻩ ﺍﺳــﺖ) .ﺩﺭ ﺍﻋﺪﺍﺩ ﻣﻨﻔﻰ ﺑــﺮﺍﻯ ﺗﻌﻴﻴﻦ ﺍﻳﻦ ﻛﻪ ﻋﺪﺩ ﺯﻭﺝ ﺍﺳــﺖ ﻳﺎ ﻓﺮﺩ ،ﺑــﻪ ﻋﺪﺩ ﺑﺪﻭﻥ ﻋﻼﻣﺖ ﺗﻮﺟﻪ ﻣﻰﻛﻨﻨﺪ (.ﺁﻳﺎ ﭘﻴﺶ ﺍﺯ ﺑﺎﺯﻯ ﻣﻰﺗﻮﺍﻧﻴﺪ ﺣﺪﺱ ﺑﺰﻧﻴﺪ ﭼﻪ ﻛﺴــﻰ ﺩﺭ ﺍﻳﻦ ﺑﺎﺯﻯ ﺑﺮﻧﺪﻩ ﻣﻰﺷﻮﺩ؟
ﺑﺎﺯﯼ ٣
ﻭﺳﺎﻳﻞ ﻻﺯﻡ :ﻛﺎﻏﺬ ﻭ ﺧﻮﺩﻛﺎﺭ ﺷﺮﺡ ﺑﺎﺯﻯ: ﺍﺑﺘﺪﺍ ﺑﺎ ﻗﺮﻋﻪﻛﺸــﻰ ﺑﺎﺯﻳﻜﻦ ﺍﻭﻝ ﺭﺍ ﺍﻧﺘﺨﺎﺏ ﻛﻨﻴﺪ .ﻋﺪﺩﻫﺎﻯ 14ﻭ 34ﺭﺍ ﺭﻭﻯ ﻛﺎﻏﺬ ﺑﻨﻮﻳﺴــﻴﺪ .ﺑﺎﺯﻳﻜــﻦ ﺍﻭﻝ ﺑﺎﻳﺪ ﺣﺎﺻﻞ ﺗﻔﺮﻳﻖ ﺍﻳﻦ ﺩﻭ ﻋﺪﺩ ﺭﺍ ﺭﻭﻯ ﻛﺎﻏﺬ ﺑﻨﻮﻳﺴﺪ .ﺑﻪ ﻫﻤﻴﻦ ﺗﺮﺗﻴﺐ ﺑﺎﺯﻯ ﺍﺩﺍﻣﻪ ﻣﻰﻳﺎﺑﺪ ﻭ ﻫﺮ ﺑﺎﺯﻳﻜﻦ ﺩﺭ ﻧﻮﺑﺖ ﺧﻮﺩ ،ﺩﻭ ﻋﺪﺩ ﺍﺯ ﻣﻴﺎﻥ ﺍﻋﺪﺍﺩ ﻧﻮﺷــﺘﻪ ﺷﺪﻩ ﺭﺍ ﺍﻧﺘﺨﺎﺏ ﻣﻰﻛﻨﺪ ﻭ ﺣﺎﺻﻞ ﺗﻔﺮﻳﻖ ﺁﻥﻫﺎ ﺭﺍ ﻣﻰﻧﻮﻳﺴــﺪ ﺑﻪ ﺷــﺮﻃﻰ ﻛﻪ ﺍﻳﻦ ﻋﺪﺩ ﻗﺒﻼ ﻧﻮﺷــﺘﻪ ﻧﺸﺪﻩ ﺑﺎﺷــﺪ .ﺑﺮﻧﺪﻩﻯ ﺑﺎﺯﻯ ﻛﺴﻰ ﺍﺳﺖ ﻛﻪ ﺁﺧﺮﻳﻦ ﻋﺪﺩ ﺭﺍ ﺑﻨﻮﻳﺴﺪ. ﺁﻳﺎ ﭘﻴﺶ ﺍﺯ ﺑﺎﺯﻯ ﻣﻰﺗﻮﺍﻧﻴﺪ ﺣﺪﺱ ﺑﺰﻧﻴﺪ ﭼﻪ ﻛﺴــﻰ ﺩﺭ ﺍﻳﻦ ﺑﺎﺯﻯ ﺑﺮﻧﺪﻩ ﻣﻰﺷﻮﺩ؟ ﺑــﺎﺯﻯ ﻛﺮﺩﻳﺪ؟ ﻟﺬﺕ ﺑﺮﺩﻳﺪ؟ ﺍﻳﻦ ﺑﺎﺭ ﺳــﻪ ﺑﺎﺯﻯ ﺑﻪ ﺷــﻤﺎ ﻣﻌﺮﻓﻰ ﻛﺮﺩﻳــﻢ ﻛﻪ ﺑﻴﺶﺗﺮ ﺷــﺒﻪ ﺑﺎﺯﻯﺍﻧﺪ ﺗﺎ ﺑﺎﺯﻯ! ﺩﺭ ﻫﺮﺳــﻪﻯ ﺍﻳﻦ ﺑﺎﺯﻯﻫﺎ ﺑﻴــﺶ ﺍﺯ ﺁﻏﺎﺯ ﺑﺎﺯﻯ ﻣﻰﺗﻮﺍﻥ ﻓﻬﻤﻴﺪ ﻛﻪ ﺩﺭ ﻫﺮ ﺣﺎﻝ ﺑﺎﺯﻳﻜﻦ ﺍﻭﻝ ﺑﺮﻧﺪﻩ ﺍﺳﺖ! ﺗﻌﺠﺐ ﻛﺮﺩﻳﺪ؟ ﺑﻴﺎﻳﻴﺪ ﻳﻚ ﺑﺎﺭ ﺑﺎﺯﻯﻫﺎ ﺭﺍ ﺑﺎ ﻫﻢ ﻣﺮﻭﺭ ﻛﻨﻴﻢ. ﭘﺎﺳﺦ؛ ﺻﻔﺤﻪﻱ 33
ﺭﺍﻫﻨﻤﺎﻳﻲ
ﺩﻭﺭﻩﻯ ﺷﺎﻧﺰﺩﻫﻢ ،ﺷﻤﺎﺭﻩﻯ ،2ﺯﻣﺴﺘﺎﻥ 1389
︎︀︨﹨︀ ﺑﺎﺯﯼ ١ ﺩﺭ ﺁﻏﺎﺯ ﺍﻳﻦ ﺑﺎﺯﻯ 3ﺩﺳــﺘﻪ ﻟﻮﺑﻴﺎ ﺩﺍﺭﻳﻢ .ﺑﺎﺯﻳﻜﻦ ﺍﻭﻝ ﻫﺮ ﺩﺳﺘﻪﺍﻯ ﺭﺍ ﻛــﻪ ﺍﻧﺘﺨﺎﺏ ﻛﻨﺪ ﻭ ﻫﺮ ﺗﺼﻤﻴﻤــﻰ ﻛﻪ ﺑﮕﻴﺮﺩ ﺑﻪ ﻫﺮ ﺣﺎﻝ ﺗﻌﺪﺍﺩ ﺍﻳﻦ ﺩﺳــﺘﻪﻫﺎ ﺭﺍ ﺑﻪ 4ﺗﺒﺪﻳﻞ ﻣﻰﻛﻨﺪ .ﺳــﭙﺲ ﺑﺎﺯﻳﻜﻦ ﺩﻭﻡ ﺗﻌﺪﺍﺩ ﺩﺳﺘﻪﻫﺎ ﺭﺍ ﺑﻪ 5ﻣﻰﺭﺳــﺎﻧﺪ ﻭ ﺑﺎﺯﻯ ﺍﺩﺍﻣﻪ ﻣﻰﻳﺎﺑﺪ ﻭ ﺑﺎﺯﻳﻜﻦ ﺍﻭﻝ ﺩﺭ ﻫﺮ ﺣﺮﻛﺖ ﺗﻌﺪﺍﺩ ﺩﺳﺘﻪﻫﺎ ﺭﺍ ﺑﻪ ﻳﻚ ﻋﺪﺩ ﺯﻭﺝ ﻭ ﺑﺎﺯﻳﻜﻦ ﺩﻭﻡ ﺩﺭ ﻫﺮ ﺣﺮﻛﺖ ،ﺗﻌﺪﺍﺩ ﺩﺳــﺘﻪﻫﺎ ﺭﺍ ﺑﻪ ﻳﻚ ﻋﺪﺩ ﻓﺮﺩ ﺗﺒﺪﻳﻞ ﻣﻰﻛﻨﺪ .ﺩﺭ ﺍﻧﺘﻬﺎﻯ ﺑﺎﺯﻯ ﻫﻤﻪﻯ ﻟﻮﺑﻴﺎﻫﺎ ﺍﺯ ﻫﻢ ﺟﺪﺍ ﻣﻰﺷــﻮﻧﺪ ﻭ 48=21+16+11ﺩﺳــﺘﻪ ﻟﻮﺑﻴﺎ ﻛﻪ ﻫﺮ ﻳﻚ ﺷﺎﻣﻞ ﻳﻚ ﻟﻮﺑﻴﺎﺳﺖ ﺑﻪ ﻭﺟﻮﺩ ﻣﻰﺁﻳﺪ 48 .ﻋﺪﺩﻯ ﺯﻭﺝ ﺍﺳﺖ ،ﭘﺲ ﺑﺎﺯﻳﻜﻦ ﺍﻭﻝ ﺑﺮﻧﺪﻩﻯ ﺑﺎﺯﻯ ﺍﺳﺖ! ! ! ﺗﻌﺪﺍﺩ ﻟﻮﺑﻴﺎﻫﺎ ﺩﺭ ﻳﻜﻰ ﺍﺯ ﺩﺳــﺘﻪﻫﺎ ﺭﺍ ﻃﻮﺭﻯ ﺗﻐﻴﻴﺮ ﺩﻫﻴﺪ ﻛﻪ ﻧﻔﺮ ﺩﻭﻡ ﺑﺮﻧﺪﻩ ﺷﻮﺩ.
ﺑﺎﺯﯼ ٣ ﺑﺮﺭﺳــﻰ ﺍﻳﻦ ﺑﺎﺯﻯ ﺍﺯ ﺩﻭ ﺑﺎﺯﻯ ﻗﺒﻠﻰ ﺳﺨﺖﺗﺮ ﺍﺳﺖ .ﭘﺲ ﺑﻴﺶﺗﺮ ﺗﻮﺟﻪ ﻛﻨﻴﺪ! ﺩﺭ ﺧﻼﻝ ﺍﻧﺠﺎﻡ ﺑﺎﺯﻯ ﺩﻳﺮ ﻳﺎ ﺯﻭﺩ ﺏ.ﻡ.ﻡ ﺩﻭ ﻋﺪﺩ ﺑﻪ ﺻﻮﺭﺕ ﺣﺎﺻﻞ ﺗﻔﺮﻳﻖ ﺩﻭ ﻋﺪﺩ ﺍﻧﺘﺨﺎﺏ ﺷــﺪﻩ ﺑﻪ ﺩﺳــﺖ ﻣﻰﺁﻳﺪ ﻭ ﻧﻮﺷــﺘﻪ ﻣﻰﺷﻮﺩ. ﻫﻤــﻪﻯ ﻣﻀﺮﺏﻫﺎﻯ ﺍﻳﻦ ﺏ.ﻡ.ﻡ ﺗﺎ ﺟﺎﻳﻰ ﻛﻪ ﺍﺯ ﺑﺰﺭگﺗﺮﻳﻦ ﻋﺪﺩ ﺍﻭﻟﻴﻪ ﺑﺰﺭگﺗﺮ ﻧﺒﺎﺷﻨﺪ ،ﺩﺭ ﻣﻴﺎﻥ ﺍﻋﺪﺍﺩ ﺧﻮﺍﻫﻨﺪ ﺑﻮﺩ. ﺩﺭﺳﺘﻰ ﺍﻳﻦ ﻣﻄﻠﺐ ﺭﺍ ﺑﺎ ﺭﻭﺵ ﺯﻳﺮ ﺑﺮﺭﺳﻰ ﻛﻨﻴﺪ: ﺑــﺎﺯﻯ ﺭﺍ ﺑﺎ ﻫﺮﻳﻚ ﺍﺯ ﺟﻔــﺖ ﻋﺪﺩﻫــﺎﻯ ) 25ﻭ 18) ،(5ﻭ ،(12 ) 13ﻭ (26ﻭ ) 13ﻭ (1ﺑﺎﺯﻯ ﻛﻨﻴﺪ ﻭ ﺑﺒﻴﻨﻴﺪ ﺩﺭ ﻫﺮ ﻣﻮﺭﺩ ﭼﻪ ﺍﻋﺪﺍﺩﻯ ﻧﻮﺷﺘﻪ ﻣﻰﺷﻮﻧﺪ. ﺭﻭﺵ ﺗﻘﺴﻴﻢﻫﺎﻯ ﻣﺘﻮﺍﻟﻰ ﺑﺮﺍﻯ ﻳﺎﻓﺘﻦ ﺏ.ﻡ.ﻡ ﭼﻪ ﺍﺭﺗﺒﺎﻃﻰ ﺑﺎ ﺍﻳﻦ ﺑﺎﺯﻯ ﺩﺍﺭﺩ؟
ﺑﺎﺯﯼ ٢ ﺩﺭ ﺍﻳــﻦ ﺑﺎﺯﻯ ﻓﺮﺩ ﺑﻮﺩﻥ ﻳــﺎ ﺯﻭﺝ ﺑﻮﺩﻥ ﺣﺎﺻﻞ ﻋﺒــﺎﺭﺕ ﺑﻪ ﻫﻴﭻ ﻣﻮﺭﺩﻯ ﺑﻪ ﺟﺰ ﺗﻌﺪﺍﺩ ﻋﺪﺩﻫﺎﻯ ﻓﺮﺩ ﻣﻮﺟﻮﺩ ﺩﺭ ﻋﺒﺎﺭﺕ ﺑﺴــﺘﮕﻰ ﻧﺪﺍﺭﺩ. ﺩﺭ ﻣﻴﺎﻥ ﺍﻋﺪﺍﺩ 1ﺗﺎ ،20ﺩﻩ ﻋﺪﺩ ﻓﺮﺩ ﻭﺟﻮﺩ ﺩﺍﺭﺩ ﻛﻪ ﭼﻮﻥ ﺗﻌﺪﺍﺩﺷﺎﻥ ﺯﻭﺝ ﺍﺳــﺖ ،ﺣﺎﺻﻞ ﺟﻤﻊ ﻳﺎ ﺗﻔﺮﻳﻖ ﺁﻥﻫﺎ ﺑﻪ ﻫــﺮ ﺗﺮﺗﻴﺐ ﻋﺪﺩﻯ ﺯﻭﺝ ﺍﺳﺖ .ﭘﺲ ﺑﺎﺯﻳﻜﻦ ﺍﻭﻝ ﻫﻤﻴﺸﻪ ﺑﺮﻧﺪﻩ ﺍﺳﺖ! ! ! ﻳﻜــﻰ ﺍﺯ ﻛﺎﺭﺕﻫﺎ ﺭﺍ ﺍﺯ ﺑﺎﺯﻯ ﺣﺬﻑ ﻛﻨﻴﺪ ﺑــﻪ ﻃﻮﺭﻯ ﻛﻪ ﻧﻔﺮ ﺩﻭﻡ ﺑﺮﻧﺪﻩﻯ ﺑﺎﺯﻯ ﺷﻮﺩ.
ﺩﺭ ﺑــﺎﺯﻯ ،3ﺏ.ﻡ.ﻡ 14ﻭ 34ﺑﺮﺍﺑــﺮ 2ﺍﺳــﺖ ﻭ ﺑﺎ ﻫــﺮ ﺗﺮﺗﻴﺒﻰ ﺍﻋﺪﺍﺩ ﺑﻌﺪﻯ ﺭﺍ ﺑﻨﻮﻳﺴﻴﻢ ،ﺩﺭ ﺍﻧﺘﻬﺎﻯ ﺑﺎﺯﻯ ﺍﻋﺪﺍﺩ ،ﻳﻚ ﺯﻭﺝ ﻛﻮﭼﻜﺘﺮ ﻭ ﻣﺴﺎﻭﻯ 34ﺭﺍ ﺧﻮﺍﻫﻴﻢ ﺩﺍﺷﺖ .ﭼﻮﻥ ﺗﻌﺪﺍﺩ ﺍﻳﻦ ﻋﺪﺩﻫﺎ ﻓﺮﺩ ﺍﺳﺖ ،ﭘﺲ ﺑﺎﺯﻳﻜﻦ ﺍﻭﻝ ﺩﺭ ﻫﺮ ﺣﺎﻝ ﺑﺮﻧﺪﻩﻯ ﺑﺎﺯﻯ ﺍﺳﺖ! ! ! ﻳﻜﻰ ﺍﺯ ﺩﻭ ﻋﺪﺩ ﺍﻭﻟﻴﻪ ﺭﺍ ﻃﻮﺭﻯ ﺗﻐﻴﻴﺮ ﺩﻫﻴﺪ ﻛﻪ ﺑﺎﺯﻛﻦ ﺩﻭﻡ ﺑﺮﻧﺪﻩ ﺷﻮﺩ! ﺁﻳﺎ ﭘﻴﺶ ﺍﺯ ﺑﺎﺯﻯ ﻣﻰﺗﻮﺍﻧﻴﺪ ﺣﺪﺱ ﺑﺰﻧﻴﺪ ﭼﻪ ﻛﺴﻰ ﺩﺭ ﺍﻳﻦ ﺑﺎﺯﻯ ﺑﺮﻧﺪﻩ ﻣﻰﺷﻮﺩ؟ ﻣﻨﺒﻊ :ﻛﺘﺎﺏ ﻣﺤﺎﻓﻞ ﺭﻳﺎﺿﻰ )ﺗﺠﺮﺑﻪﻯ ﺭﻭﺱﻫﺎ( ،ﻣﺆﺳﺴﻪﻯ ﺍﻧﺘﺸﺎﺭﺍﺕ ﻓﺎﻃﻤﻰ، .1386
ﺩﻭﺭﻩﻯ ﺷﺎﻧﺰﺩﻫﻢ ،ﺷﻤﺎﺭﻩ ﻯ ،2ﺯﻣﺴﺘﺎﻥ 1389
ﺭﺍﻫﻨﻤﺎﻳﻲ
33 3
ﺗﺎﺭﻳﺦ ﺭﻳﺎﺿﻴﺎﺕ
﹡﹍︀﹨ ﹤︋ ﹩ز﹡︡﹎ ﹩ر︀︲﹩دا﹡︀ن ︮︀︉ ︻﹚﹢م و ﹁﹠﹢ن
از ︠﹫︀م ﹡﹫︪︀︋﹢ری ︑︀ ︎︣و﹁︧﹢ر ﹨︪︐︣ودی ﺳﻴﺮﻭﺱ ﻏﻔﺎﺭﻳﺎﻥ
ﻛﻠﻴﺪﻭﺍژﻩﻫﺎ :ﺧﻴﺎﻡ ﻧﻴﺸﺎﺑﻮﺭﻯ ،ﭘﺮﻭﻓﺴﻮﺭ ﻫﺸﺘﺮﻭﺩﻯ ،ﺭﻳﺎﺿﻰﺩﺍﻥ ،ﺷﺎﻋﺮ . ﺩﺭ ﺗﺎﺭﻳﺦ ﻋﻠﻢ ﺩﺭ ﺍﻳﺮﺍﻥ ﺍﺳــﻼﻣﻰ ﺑﻪ ﻧﺎﻡ ﺍﻓﺮﺍﺩﻯ ﺑﺮﺧﻮﺭﺩ ﻣﻰﻛﻨﻴﻢ ﻛــﻪ ﺑﻪ ﮔﻔﺘــﻪﻯ ﺍﺭﻭﭘﺎﻳﻴــﺎﻥ ﺍﺯ ﻭﻳﮋﮔﻰ ﭘﻠﻰ ﺗﻜﻨﻴــﻚ ) ﺑﺮﺧﻮﺭﺩﺍﺭﻯ ﺍﺯ ﻫﻤﻪﻯ ﻓﻨﻮﻥ( ﺑﻬﺮﻩﻣﻨﺪ ﺑﻮﺩﻧﺪ ﻭ ﺑﻪ ﻧﺤﻮ ﺑﺎﺭﺯﺗﺮﻯ ﺟﺎﻣﻊ ﺟﻤﻴﻊ ﻋﻠﻮﻡ ﻭ ﻓ ّﻨﺎﻭﺭﻯ ﺯﻣﺎﻥ ﺧﻮﺩ ﺑﻮﺩﻧﺪ .ﻣﺎ ﺳﻌﻰ ﺩﺍﺭﻳﻢ ﺍﺯ ﻫﻤﻪﻯ ﺁﻥﻫﺎ ﻧﺎﻡ ﻧﺒﺮﻳﻢ .ﺍﺯ ﻣﻴﺎﻥ ﺳــﺘﺎﺭﮔﺎﻥ ﻗﺪﺭ ﺍﻭﻝ ﺁﺳﻤﺎﻥ ﻋﻠﻮﻡ ﻭ ﻓﻦ ﺑﻪ ﻋﻨﻮﺍﻥ ﻧﻤﻮﻧﻪ ﻣﻰﺗﻮﺍﻥ ﺍﺯ ﻗــﺮﻥ ﭼﻬﺎﺭﻡ ﻭ ﭘﻨﺠﻢ ﻫﺠﺮﻯ ﻗﻤﺮﻯ ﺑﻪ ﺑﻌــﺪ ﺍﺯ ﺍﻓﺮﺍﺩ ﺯﻳﺮ ﻧﺎﻡ ﺑﺒﺮﻳﻢ ﻭ ﺳــﭙﺲ ﺑﻪ ﺯﻳﺴﺖﻧﮕﺎﺭﻯ ) ﺑﻴﻮﮔﺮﺍﻓﻰ( ﻣﺮﺣﻮﻡ ﻫﺸﺘﺮﻭﺩﻯ ﺑﭙﺮﺩﺍﺯﻳﻢ ﻭ ﻣﺸﺘﺮﻛﺎﺕ ﺍﻭﺭﺍ ﺑﺎ ﺧﻴﺎﻡ ﺑﺮﺭﺳﻰ ﻛﻨﻴﻢ. ﺍﺑﻦﺳــﻴﻨﺎ ،ﺭﻭﺳﺘﺎﺯﺍﺩﻩ ﺍﻓﺸــﻨﻪﺍﻯ ﺑﺨﺎﺭﺍﻳﻰ ،ﺭﻳﺎﺿﻰﺩﺍﻥ ،ﻓﻴﻠﺴﻮﻑ، ﭘﺰﺷــﻚ ،ﺩﺍﺭﻭﺷﻨﺎﺱ ﻭ ﺣﻜﻴﻢ ﺍﻟﻬﻰ ﺑﻮﺩ ﻭ ﺩﺍﻧﺸﻤﻨﺪﺍﻥ ﻣﻌﺎﺻــــــﺮ ﺍﻭ » ﺍﺑﻮﺭﻳﺤﺎﻥ ﺑﻴﺮﻭﻧﻰ ﺧﻮﺍﺭﺯﻣﻰ« ،ﺭﻳﺎﺿﻰﺩﺍﻥ ﻣﻨﺠﻢ ،ﻣﻮﺭﺥ ،ﺟﻐﺮﺍﻓﻰﺩﺍﻥ، ﺧﻴﺎﻡ ﻧﻴﺸــﺎﺑﻮﺭﻯ ﻭ ﺍﺑﻮﺍﻟﻮﻓﺎ ﺑﻮﺯﺟﺎﻧﻰ ﺭﻳﺎﺿﻰﺩﺍﻥ ﻭ ﻣﻨﺠﻢ ﺑﻮﺩﻧﺪ ﻛﻪ ﺩﺭ ﺑﻴﻦ ﺍﻳﻦ ﺩﻭ ﺧﻴﺎﻡ ﺷﺎﻋﺮ ﻭ ﻓﻴﻠﺴﻮﻑ ﻧﻴﺰ ﺑﻪ ﺣﺴﺎﺏ ﻣﻰﺁﻣﺪ ،ﺯﻳﺮﺍ ﻫﺮﻳﻚ ﺍﺯ ﺭﺑﺎﻋﻴﺎﺕ ﺍﻭ ﺩﺭﺑﺮﺩﺍﺭﻧﺪﻩﻯ ﻓﻠﺴﻔﻪ ﻭ ﺟﻬﺎﻥﺑﻴﻨﻰ ﺧﺎﺹ ﺍﻭ ﺑﻮﺩ. ﻏﻴﺎﺙﺍﻟﺪﻳﻦ ﺟﻤﺸﻴﺪ ﻛﺎﺷﺎﻧﻰ ،ﺭﻳﺎﺿﻰﺩﺍﻥ ،ﻣﻨﺠﻢ ﻭ ﺍﻳﺠﺎﺩﻛﻨﻨﺪﻩﻯ ﺑﺰﺭگﺗﺮﻳــﻦ ﺭﺻﺪﺧﺎﻧــﻪ ﺩﺭ ﻗــﺮﻥ ﻫﺸــﺘﻢ ﻭ ﻗﻄﺐﺍﻟﺪﻳﻦ ﺷــﻴﺮﺍﺯﻯ، ﺭﻳﺎﺿﻰﺩﺍﻥ ،ﻃﺒﻴﺐ ،ﻣﻨﺠﻢ ﻭ ﻓﻴﻠﺴﻮﻑ ﺑﻮﺩ. ﺩﺭ ﻋﺼــﺮ ﻣﺎ ﺍﺯ ﺯﻣﺎﻥ ﺗﺄﺳــﻴﺲ ﺩﺍﺭﺍﻟﻔﻨﻮﻥ ﺗﻬــﺮﺍﻥ ﻭ ﺗﺒﺮﻳﺰ ،ﻳﻌﻨﻰ ﺍﺯ ﺩﻭﺭﻩﻯ ﺍﻣﻴﺮﻛﺒﻴــﺮ ﻭ ﻋﺼــﺮ ﻧﺎﺻﺮﻯ ﺑﻪ ﺑﻌﺪ ،ﻣﻴــﺮﺯﺍ ﻋﺒﺪﺍﻟﻐﻔﺎﺭ ﺧﺎﻥ ﻧﺠﻢﺍﻟﺪﻭﻟــﻪ ﻧﻮﺭﻯ ،ﺭﻳﺎﺿﻰﺩﺍﻥ ،ﻣﻨﺠﻢ ﻭ ﻣﻬﻨﺪﺱ ﻭ ﺍﺳــﺘﺎﺩ ﺩﺍﺭﺍﻟﻔﻨﻮﻥ ﺑﻮﺩ .ﻭﻯ ﺩﺭ ﻫﺸــﺘﺎﺩ ﺳــﺎﻝ ﺍﺧﻴــﺮ ﺍﺳــﺘﺎﺩ ﺫﻭﺍﻟﻔﻨــﻮﻥ ،ﺭﻳﺎﺿﻰﺩﺍﻥ، ﺗﻘﻮﻳﻢﺷﻨﺎﺱ ،ﻣﻨﺠﻢ ﻭ ﻋﺎﻟﻢ ﺍﻟﻬﻰ ﺑﻮﺩ ﻛﻪ ﭘﻴﺶﺑﻴﻨﻰ ﺧﺴﻮﻑ ﻭ ﻛﺴﻮﻑ ﺩﺭ ﺍﻳﺮﺍﻥ ﺗﺎ ﺁﺧﺮ ﺳــﺎﻝ ) 1331ﺳــﺎﻟﻰ ﻛﻪ ﺳــﺎﻟﻨﺎﻣﻪﻯ ﭘﺎﺭﺱ ﺭﺍ ﻛﻪ ﻣﺮﺣﻮﻡ ﺍﻣﻴﺮ ﺟﺎﻫﺪ ﻣﻨﺘﺸــﺮ ﻣﻰﻛﺮﺩ( ﺑﻪ ﻋﻬﺪﻩﻯ ﺍﻭ ﺑﻮﺩ ﻭ ﺗﺎ ﺁﺧﺮ ﻋﻤﺮ ﺗﻘﻮﻳﻢ ﺗﻄﺒﻴﻘﻰ ﺳﺎﻝﻫﺎﻯ ﻫﺠﺮﻯ ﻗﻤﺮﻯ ،ﻫﺠﺮﻯ ﺷﻤﺴﻰ ﻭ ﻣﻴﻼﺩﻯ ﺭﺍ ﺍﺯ 1305ﺧﻮﺭﺷﻴﺪﻯ ﺑﻪ ﺑﻌﺪ ﺩﺭ ﺳﺎﻟﻨﺎﻣﻪﻯ ﭘﺎﺭﺱ ﻣﻨﺘﺸﺮ ﻣﻰﻛﺮﺩ. 34
ﺭﺍﻫﻨﻤﺎﻳﻲ
ﺩﻭﺭﻩﻯ ﺷﺎﻧﺰﺩﻫﻢ ،ﺷﻤﺎﺭﻩﻯ ،2ﺯﻣﺴﺘﺎﻥ 1389
ﺍﺣﻤــﺪ ﺑﻴﺮﺷــﻚ ،ﺭﻳﺎﺿﻰﺩﺍﻧﻰ ﻛــﻪ ﺩﻭﺭﻩﻯ ﻛﺘﺎﺏﻫــﺎﻯ ﺭﻳﺎﺿﻰ ﺩﺑﻴﺮﺳﺘﺎﻧﻰ ﺭﺍ ﺑﺎ ﻋﺪﻩﺍﻯ ﺍﺯ ﺩﺑﻴﺮﺍﻥ ﺑﺎ ﺗﺠﺮﺑﻪ ﺗﺤﺖ ﻋﻨﻮﺍﻥ »ﻣﺠﻤﻮﻋﻪﻯ ﺧﺮﺩ« ﻣﻨﺘﺸﺮ ﻛﺮﺩ ﻛﻪ ﻣﻮﺭﺩ ﺍﺳﺘﻔﺎﺩﻩﻯ ﺩﺍﻧﺶﺁﻣﻮﺯﺍﻥ ﻛﺸﻮﺭ ﻗﺮﺍﺭ ﮔﺮﻓﺖ ﻭ ﻋﻼﻭﻩ ﺑﺮ ﺁﻥ ﺍﻳﻦ ﺷــﺨﺺ ﺗﺎ ﺁﺧﺮ ﻋﻤﺮ ﺩﺭ ﺩﺍﻧﺸﻨﺎﻣﻪﻯ ﺑﺰﺭگ ﻓﺎﺭﺳﻰ ﻛﺎﺭ ﻣﻰﻛﺮﺩ ﻭ ﮔﺎﻩﺷﻤﺎﺭﻯ ﺗﻄﺒﻴﻘﻰ ﺳﻪ ﻫﺰﺍﺭ ﺳﺎﻟﻪ ﺭﺍ ﺑﻪ ﺭﺷﺘﻪﻯ ﺗﺤﺮﻳﺮ ﻛﺸــﻴﺪ ﻭ ﻋﻤﺮ ﭘﺮ ﺑﺮﻛﺖ ﺧﻮﺩ ﺭﺍ ﺩﺭ ﺭﺍﻩ ﺗﺄﻟﻴﻒ ﺗﻘﻮﻳﻢ ﺳــﻪ ﻫﺰﺍﺭ ﺳﺎﻟﻪ، ﻛﺘﺐ ﺭﻳﺎﺿﻰ ﻭ ﺷﺮﺡ ﺣﺎﻝ ﺩﺍﻧﺸﻤﻨﺪﺍﻥ ﺟﻬﺎﻥ ﮔﺬﺭﺍﻧﺪ. ﺭﻳﺎﺿﻰ ﻭ ﻮﺭﺥ ﻣﺘﺮﺟﻢ ﻛﺘﺐ ﺭ ﻰ ﺩﺍﻥ ،ﺮ ﻢ ﺭﻳﺎﺿﻰ ﻥ ﺁﺭﺍﻡ ،ﺭ ﻰ ﻣﺮﺣــﻮﻡ ﺍﺣﻤﺪ ﺭ ﻡ ﺮ ﻮﻡ ﻣﻮﺭﺥ ﺑﻪ ﻧﻧﺤﻮﻯ ﻛﻪ ﺑﺨﺸــﻰ ﺍﺯ ﻛﺘﺎﺏﻫﺎﻯ ﺗﺎﺭﻳﺦ ﺗﻤﺪﻥ ﻧﻮﺷــﺘﻪﻯ ﻭﻳﻞ ﺩﻭﺭﺍﻧﺖ ﺍﺯ ﺟﻤﻠﻪ ﻣﺸــﺮﻕ ﺯﻣﻴﻦ ،ﮔﺎﻫﻮﺍﺭﻩﻯ ﺗﻤﺪﻥ ﺭﺍ ﺗﺮﺟﻤﻪ ﻛﺮﺩ .ﺍﻭ ﻫﻢﭼﻨﻴﻦ ﻓﻓﻴﻠﺴــﻮﻑ ﻭ ﻣﻌﻠﻢ ﻓﻠﺴــﻔﻪ ﻧﻴﺰ ﺑﻮﺩ ﺑﻪﻃﻮﺭﻯ ﻛﻪ ﺩﺭ ﺯﻣﺎﻥ ﺍﻭﺝ ﺍﺷﺘﻬﺎﺭ ﻛﻛﺘﺎﺏ ﻓﻠﺴﻔﻪ ﺳــﻴﺮ ﺣﻜﻤﺖ ﺩﺭ ﺍﺭﻭﭘﺎ ﻧﻮﺷﺘﻪﻯ ﻣﺤﻤﺪﻋﻠﻰ ﻓﺮﻭﻏﻰ ﻛﻪ ﺍﻭﻟﻴﻦ ﺗﺮﺟﻤﻪﻯ ﻓﻠﺴــﻔﻪﻯ ﺍﻭﭘﺎﻳﻰ ﺑﻮﺩ ،ﺩﺭ 1326ﺧﻮﺭﺷــﻴﺪﻯ ﻛﺘﺎﺏ ﻣﻣﻘﺪﻣﻪﺍﻯ ﺑﺮ ﻓﻠﺴــﻔﻪ ﻧﻮﺷــﺘﻪﻯ ﺍﻭﺯﻭﺍﻟﺪﻛﻮﺍﭘﻪ ﺭﺍ ﺗﺮﺟﻤﻪ ﻭ ﺍﻧﺘﺸــﺎﺭﺍﺕ ﻫﺎﻯ ﻠ ﺷــﺨﺼﻴﺖ ﺎ ﺯﻣﺎﻥ ﺷ ﺨ ﺍﻣﻴﺮﻛﺒﻴــﺮ ﺁﺁﻥ ﺍﺭﺍ ﻣﻨﺘﺸــﺮ ﻛﻛﺮﺩ .ﺩﺭ ﺁﺁﻥ ﺎ ﺍ ﻛ ﻋﻠﻤﻰ ﺍﺍﺯ ﺟﻤﻠــﻪ ﺩﻛﺘﺮ ﻋﻠﻰﺍﻛﺒﺮ ﺳﻴﺎﺳــﻰ ﻛﻪ »ﻣﺒﺎﻧﻰ ﻓﻠﺴــﻔﻪ ﻭ ﺭﻭﺍﻧﺸﻨﺎﺳــﻰ
ﺟﺪﻳﺪ« ﺭﺍ ﺗﺪﺭﻳﺲ ﻣﻰﻛﺮﺩ ،ﺍﺯ ﻛﺎﺭ ﺍﻭ ﺗﺠﻠﻴﻞ ﻛﺮﺩ .ﻭ ﺳﺮﺍﻧﺠﺎﻡ ﭘﺮﻭﻓﺴﻮﺭ ﻣﺤﺴــﻦ ﻫﺸــﺘﺮﻭﺩﻯ ﻛﻪ ﻋﻼﻭﻩ ﺑــﺮ ﺭﻳﺎﺿﻰ ﺩﺭ ﺯﻣﻴﻨﻪﻫــﺎﻯ ﻣﻜﺎﻧﻴﻚ ﺁﺳــﻤﺎﻧﻰ ،ﻫﻴﺌﺖ ﻭ ﻧﺠﻮﻡ ،ﻓﻠﺴﻔﻪ ﻭ ﺍﺩﺑﻴﺎﺕ ﺻﺎﺣﺐﻧﻈﺮ ﺑﻮﺩ ﻭ ﻣﻰﺗﻮﺍﻥ ﺍﺯ ﺍﻭ ﺑﻪ ﻋﻨﻮﺍﻥ ﺧﻴﺎﻡ ﺩﻳﮕﺮﻯ ﺩﺭ ﺩﻫﻪﻯ ﺷﺸﻢ ﻗﺮﻥ ﺑﻴﺴﺘﻢ ﻧﺎﻡ ﺑﺮﺩ .ﺑﻰ ﻣﻨﺎﺳﺒﺖ ﻧﻴﺴﺖ ﻛﻪ ﺍﺑﺘﺪﺍ ﺷﺮﺣﻰ ﺍﺯ ﺍﻗﺪﺍﻣﺎﺕ ﻋﻠﻤﻰ ﻋﻤﺮ ﺧﻴﺎﻡ ﺭﺍ ﭘﻴﺶ ﺭﻭﻯ ﺧﻮﺍﻧﻨﺪﮔﺎﻥ ﻣﺤﺘﺮﻡ ﻗﺮﺍﺭ ﺩﻫﻴﻢ ﻭ ﺳــﭙﺲ ﻭﺟﻮﻩ ﻣﺸﺘﺮﻙ ﺯﻧﺪﮔﻰ ﭘﺮﻭﻓﺴﻮﺭ ﻫﺸﺘﺮﻭﺩﻯ ﻭ ﺧﻴﺎﻡ ﺭﺍ ﻣﺘﺬﻛﺮ ﺷﻮﻳﻢ. ﻧﮕﺎﻫﻰ ﺑﻪ ﺯﻧﺪﮔﻰ ﺧﻴﺎﻡ ﻗﺮﻥ ﻳﺎﺯﺩﻫــﻢ ﻣﻴﻼﺩﻯ ) ﻗﺮﻥ ﭘﻨﺠﻢ ﻫﺠــﺮﻯ ﻗﻤﺮﻯ( ﺭﺍ ﻣﻰﺗﻮﺍﻥ ﻋﺼﺮ ﺧﻴﺎﻡ ﺩﺍﻧﺴــﺖ )ﺹ 168ﺯﻧﺪﮔﻰﻧﺎﻣﻪ ﺍﺳــﺘﺎﺩ ﭘﺮﻭﻳﺰ ﺷﻬﺮﻳﺎﺭﻯ(، ﺯﻳﺮﺍ ﺍﻭ ﻃﺒﻘﻪﺑﻨﺪﻯ ﺷﺎﻳﺴــﺘﻪﺍﻯ ﺍﺯ ﻣﻌــﺎﺩﻻﺕ ﺭﺍ ﻋﺮﺿﻪ ﻛﺮﺩ .ﺍﺯ ﺟﻤﻠﻪ ﺳــﻴﺰﺩﻩ ﺻﻮﺭﺕ ﺍﺯ ﻣﻌﺎﺩﻻﺕ ﺩﺭﺟﻪﻯ ﺳــﻮﻡ ﺭﺍ ﻃﺒﻘﻪﺑﻨــﺪﻯ ﻛﺮﺩ ﻭ ﺍﻭ ﻛﻮﺷﻴﺪ ﻛﻪ ﻫﻤﻪﻯ ﺁﻥﻫﺎ ﺭﺍ ﺣﻞ ﻛﻨﺪ ﻭ ﺑﺮﺍﻯ ﺗﻌﺪﺍﺩﻯ ﺍﺯ ﺁﻥﻫﺎ ﺍﺯ ﻃﺮﻳﻖ ﻣﻘﺎﻃﻊ ﻣﺨﺮﻭﻃــﻰ ﺭﺍﻩﺣﻠﻰ ﭘﻴﺪﺍ ﻛﻨﺪ .ﺍﻭ ﻣﻌﺎﺩﻟﻪﻯ x 3 + a = cx 2ﺭﺍ ﺣــﻞ ﻛﺮﺩ .ﺧﻴــﺎﻡ ﻣﻰﮔﻮﻳﺪ ﻛﻪ ﺍﻳــﻦ ﻣﻌﺎﺩﻟﻪ ﺭﺍ »ﻣﺎﻫﺎﻧــﻰ« ﺍﺭﺍﺋﻪ ﺩﺍﺩ، ﺍ ّﻣــﺎ ﻗﺎﺩﺭ ﺑــﻪ ﺣﻞ ﺁﻥ ﻧﺒﻮﺩ ﺗــﺎ ﺁﻥﻛﻪ ﺍﻭ ﻭ ﺍﺑﻮﺟﻌﻔــﺮ ﺧﺎﺯﻧﻰ ﺑﻪ ﻛﻤﻚ ﻣﻘﺎﻃﻊ ﻣﺨﺮﻭﻃﻰ ﺁﻥ ﺭﺍ ﺣﻞ ﻛﺮﺩﻧﺪ ﻭ ﺭﺍﻩﺣﻞ ﻫﻨﺪﺳــﻰ ﻳﮕﺎﻧﻪ ﺭﺍﻩﺣﻞ ﺍﻳﻦﮔﻮﻧﻪ ﻣﻌﺎﺩﻻﺕ ﺍﺳــﺖ .ﺑﺰﺭگﺗﺮﻳﻦ ﺧﺪﻣــﺖ ﺧﻴﺎﻡ ﺩﺭ ﻋﻠﻢ ﺭﻳﺎﺿﻰ،
ﺍﻳﺠﺎﺩ ﮔﺎﻩﺷــﻤﺎﺭﻯ )ﺗﻘﻮﻳﻢ( ﺟﻼﻟﻰ )ﻣﻠﻜﺸﺎﻫﻰ( ﺍﺳﺖ ﻛﻪ ﻃﺒﻖ ﺩﻋﻮﺕ ﺧﻮﺍﺟﻪ ﻧﻈﺎﻡ ﺍﻟﻤﻠﻚ ،ﻭﺯﻳﺮ ﺩﺍﻧﺸﻤﻨﺪ ﺟﻼﻝﺍﻟﺪﻳﻦ ﻣﻠﻜﺸﺎﻩ ﺳﻠﺠﻮﻗﻰ ،ﺑﻪ ﺍﻳﻦ ﺍﻣﺮ ﺍﻗﺪﺍﻡ ﺷﺪ .ﺑﻪ ﮔﻔﺘﻪﻯ ﺍﺳﺘﺎﺩ ﭘﺮﻭﻳﺰ ﺷﻬﺮﻳﺎﺭﻯ ،ﺩﺍﻧﺸﻤﻨﺪ ﺭﻳﺎﺿﻰ ﻣﻌﺎﺻﺮ ﺩﺭ ﺹ 168ﻛﺘﺎﺏ ﺯﻧﺪﮔﻰﻧﺎﻣﻪ ﺧﻮﺩﺵ ﺩﺭﺑﺎﺭﻩﻯ ﺭﻳﺎﺿﻰﺩﺍﻧﺎﻥ ﮔﺬﺷــﺘﻪ »ﺧﻴﺎﻡ ﻧﻴﺸﺎﺑﻮﺭﻯ ﮔﺎﻩﺷــﻤﺎﺭﻯ ﺗﺎﺯﻩﺍﻯ ﺭﺍ ﺑﻨﻴﺎﻥ ﮔﺬﺍﺷﺖ ﻛﻪ ﺩﻗﺖ ﺑﻰﺍﻧﺪﺍﺯﻩﺍﻯ ﺩﺍﺷﺖ«. ﺧﻴﺎﻡ ﺩﺭ 467ﺑﺮﺍﻯ ﺍﺻﻼﺡ ﮔﺎﻩﺷــﻤﺎﺭﻯ )ﺗﻘﻮﻳﻢ( ﺑﻪ ﺍﺻﻔﻬﺎﻥ ﺭﻓﺖ ﻭ ﺑﺎ ﻋﺪﻩﺍﻯ ﺍﺯ ﺩﺍﻧﺸــﻤﻨﺪﺍﻥ ﻣﺎﻧﻨﺪ »ﻟﺴــﻔﺰﺍﺭﻯ« ﻭ »ﻟﻮﻛﺮﻯ« ﺩﺳﺖ ﺑﻪ ﭼﻨﻴﻦ ﺍﻗﺪﺍﻣﻰ ﺯﺩ .ﺩﺭ ﺍﻳﺮﺍﻥ ﻗﺪﻳﻢ ﺍﺯ ﮔﺎﻩﺷــﻤﺎﺭﻯ ﺧﻮﺭﺷﻴﺪﻯ ﺍﺳﺘﻔﺎﺩﻩ ﻣﻰﻛﺮﺩ ﻛﻪ ﺁﺧﺮﻳﻦ ﺁﻥﻫﺎ ﺩﺭ ﺩﻭﺭﻩﻯ ﺳﺎﺳﺎﻧﻰ ﺑﻪ ﻧﺎﻡ ﺗﻘﻮﻳﻢ ﺧﻮﺭﺷﻴﺪﻯ ﻳﺰﺩﮔﺮﺩﻯ ﺑﻮﺩ .ﺑﺎ ﻓﺮﻭﭘﺎﺷﻰ ﺳﺎﺳﺎﻧﻴﺎﻥ ،ﮔﺎﻩﺷﻤﺎﺭﻯ ﻗﻤﺮﻯ ﻣﻌﻤﻮﻝ ﺷﺪ. ﺍﻳﻦ ﮔﺎﻩﺷﻤﺎﺭﻯ )ﮔﺎﻩﺷﻜﺎﺭﻯ ﺧﻮﺭﺷــﻴﺪﻯ( ﺑﺮﺍﻯ ﺍﻗﺘﺼﺎﺩ ﻛﺸﺎﻭﺭﺯﻯ ﻭ ﺁﮔﺎﻫﻰ ﺍﺯ ﻓﺼﻮﻝ ﺳﺎﻝ ﺿﺮﻭﺭﺕ ﺩﺍﺷﺘﻨﺪ ،ﻛﺸﺎﻭﺭﺯﺍﻥ ﺑﺎﻳﺪ ﺍﺯ ﺯﻣﺎﻥ ﻛﺸﺖ ﻭ ﺁﺑﻴﺎﺭﻯ ﻭ ﺑﺮﺩﺍﺷﺖ ﺁﮔﺎﻫﻰ ﭘﻴﺪﺍ ﻣﻰﻛﺮﺩﻧﺪ ﻭ ﺍﻳﻦ ﺍﻣﺮ ﻣﻤﻜﻦ ﻧﻤﻰﺷﺪ ﻣﮕﺮ ﺑﺎ ﮔﺎﻩﺷــﻤﺎﺭﻯ ﺧﻮﺭﺷــﻴﺪﻯ .ﺩﻭﻟﺖ ﻫﻢ ﺑــﺮﺍﻯ ﮔﺮﻓﺘﻦ ﻣﺎﻟﻴﺎﺕ ﺍﺯ ﺩﻫﻘﺎﻧﺎﻥ ﻭ ﻣﺎﻟﻜﺎﻥ ﺑﻪ ﺩﺷﻮﺍﺭﻯ ﺑﺮﺧﻮﺭﺩ ﻛﺮﺩﻩ ﺑﻮﺩ ،ﭼﺮﺍ ﻛﻪ ﺭﺍﻫﻰ ﺑﺮﺍﻯ ﺗﺸــﺨﻴﺺ ﺯﻣﺎﻥ ﻣﺎﻟﻴﺎﺕ ﻭﺟﻮﺩ ﻧﺪﺍﺷﺖ .ﺍﻟﺒﺘﻪ ﺍﻛﺜﺮ ﻣﺮﺩﻡ ،ﻧﻮﺭﻭﺯ ﺭﺍ ﻛﻪ ﺛﺎﺑﺖ ﺑﻮﺩ ﭘﺎﻳﻪ ﻗﺮﺍﺭ ﻣﻰﺩﺍﺩﻧﺪ )ﺍﻭﻝ ﺍﻋﺘﺪﺍﻝ ﺑﻬﺎﺭﻯ( ،ﺩﺭ ﺻﻮﺭﺗﻰ ﻛﻪ ﺍﮔﺮ ﻣﻰﺧﻮﺍﺳــﺘﻨﺪ ﻣﺎﻟﻴﺎﺕ ﺭﺍ ﺑﻪ ﻣﺎﻩﻫﺎﻯ ﻗﻤﺮﻯ ﺑﮕﻴﺮﻧﺪ ﻳﻚ ﺳــﺎﻝ ﺑﻌﺪ ﺍﺯ ﺑﺮﺩﺍﺷﺖ ﻣﺤﺼﻮﻝ ﺑﻪ ﻓﺮﻭﺭﺩﻳﻦ ﺑﺮﺧﻮﺭﺩ ﻣﻰﻛﺮﺩ ﻭ ﺯﻣﺎﻧﻪ ﺑﺮﺍﻯ ﻛﺎﺷﺖ ﻣﻬﺮﻣﺎﻩ ﻧﺒﻮﺩ ﻭ ﺳــﺎﻝ ﺩﺭ ﺣــﺎﻝ ﮔﺮﺩﺵ ﺑﻮﺩ ﻭ ﺑﺮﺍﻯ ﻣﺎﻟﻚ ﻭ ﺯﺍﺭﻉ ﻛﺎﺭ ﺭﺍ ﻣﺸــﻜﻞ ﻣﻰﻛــﺮﺩ .ﺫﻫﻦ ﺧﻮﺍﺟﻪ ﻧﻈــﺎﻡ ﺍﻟﻤﻠﻚ ﻛﻪ ﺑﻪ ﻭﻳﮋﻩ ﺩﺭ ﺟﻬﺖ ﺍﺳﺘﻮﺍﺭ ﺳــﺎﺧﺘﻦ ﻧﻈﺎﻡ ﻓﺌﻮﺩﺍﻟﻰ ﻭ ﻧﻈﻢ ﺩﺍﺩﻥ ﺑــﻪ ﻗﺎﻧﻮﻥ ﺩﺭ ﻛﺎﺭﻫﺎﻯ ﺩﻭﻟﺘﻰ ﻭ ﺍﺯ ﺟﻤﻠﻪ ﻭﺻﻮﻝ ﻣﺎﻟﻴﺎﺕﻫﺎ ﻛﺎﺭ ﻣﻰﻛــﺮﺩ ﻭﺍﺟﺐ ﺷــﺪ ﺗﺎ ﺑﺎ ﺣﻤﺎﻳــﺖ ﺍﺯ ﺧﻴﺎﻡ ﻭ ﺩﻳﮕﺮ ﺩﺍﻧﺸﻤﻨﺪﺍﻥ ﺑﻪ ﺍﻳﻦ ﻛﺎﺭ ﺍﻗﺪﺍﻡ ﻛﻨﺪ .ﻟﺬﺍ ﮔﺎﻩﺷﻤﺎﺭﻯ ﺟﻼﻟﻰ ) ﻣﻠﻜﻰ( ﺭﺍ ﻣﻌﻤﻮﻝ ﻛﺮﺩ ﻛﻪ ﺍﺑﺘﺪﺍﻯ ﻓﺮﻭﺭﺩﻳﻦ ﺳــﺎﻝ 458 ﺧﻮﺭﺷﻴﺪﻯ ﺍﺑﺘﺪﺍﻯ ﺁﻥ ﺑﻮﺩ ﻭ ﺍﻳﻦ ﮔﺎﻩﺷﻤﺎﺭﻯ ﺑﺴﻴﺎﺭ ﺩﻗﻴﻖﺗﺮ ﺍﺯ ﺗﻘﻮﻳــﻢ ﮔﺮﻳﮕﻮﺭﻯ ﺑﻮﺩ ﻛﻪ ﺍﻭﻝ ﺳــﺎﻝ ﺁﻥﻫــﺎ ﺍﻭﻝ ژﺍﻧﻮﻳﻪ ﺑﻪ ﺣﺴﺎﺏ ﻣﻰﺁﻣﺪ. ﺧﻴــﺎﻡ ﺭﻭﻯ ﻫــﻢ ﺭﻓﺘﻪ ﻫﺠﺪﻩ ﺳــﺎﻝ ﺩﺭ ﺍﺻﻔﻬﺎﻥ ﺑﻮﺩ ﺗﺎ ﺁﻥﻛﻪ ﻛﺎﺭﻫﺎﻯ ﻋﻠﻤﻰ ﺧﻮﺩ ﺭﺍ ﺑﻪ ﺳﺎﻣﺎﻥ ﺭﺳﺎﻧﻴﺪ ﻭ ﺑﻪ ﻧﻴﺸﺎﺑﻮﺭ ﺑﺎﺯﮔﺸــﺖ ﻭ ﺩﺭ 546ﻫﺠﺮﻯ ﻗﻤﺮﻯ )1131ﻣﻴﻼﺩﻯ( ﺩﺭ ﺳﻦ 83ﺳﺎﻟﮕﻰ ﺩﺭﮔﺬﺷــﺖ .ﺑﻌﺪﻫﺎ ﻛﺎﺭﻫﺎﻯ ﺧﻴﺎﻡ ﺑﺎ ﻧﻮﺷﺘﻪﻫﺎﻯ ﺧﻮﺍﺟــﻪ ﻧﺼﻴﺮﺍﻟﺪﻳﻦ ﻃﻮﺳــﻰ ﺑﻪ ﺍﻭﭘﺎ ﺭﺍﻩ ﻳﺎﻓﺖ .ﺑﺴــﻴﺎﺭﻯ ﺑﺮ ﺍﻳﻦ ﻋﻘﻴﺪﻩ ﺑﻮﺩﻧﺪ ﻛﻪ ﻣﺜﻠﺚ ﺣﺴــﺎﺑﻰ ﭘﺎﺳﻜﺎﻝ ﺭﺍ ﺑﺎﻳﺪ »ﻣﺜﻠﺚ ﺣﺴــﺎﺑﻰ ﺧﻴﺎﻡ ﻧﺎﻣﻴﺪ ﻭ ﺑﺮﺧــﻰ ﭘﺎ ﺭﺍ ﺍﺯ ﺁﻥ ﻓﺮﺍﺗﺮ ﮔﺬﺍﺷــﺘﻪ ﻭ ﻣﻌﺘﻘﺪﻧﺪ ﻛﻪ ﺩﻭ ﺟﻤﻠــﻪﺍﻯ ﻧﻴﻮﺗﻮﻥ ﺭﺍ ﺑﺎﻳــﺪ »ﺩﻭﺟﻤﻠﻪﺍﻯ ﺧﻴــﺎﻡ« ﻧﺎﻣﻴﺪ .ﺩﻭ ﺩﻭﺭﻩﻯ ﺷﺎﻧﺰﺩﻫﻢ ،ﺷﻤﺎﺭﻩ ﻯ ،2ﺯﻣﺴﺘﺎﻥ 1389
ﺭﺍﻫﻨﻤﺎﻳﻲ
35
ﻣﻌﻤﺎ ﻭ ﺳﺮﮔﺮﻣﯽ
ﺟﻤﻠﻪﺍﻯ ﻧﻴﻮﺗﻮﻥ ﺩﺭﺑﺎﺭﻩﻯ (a + b) nﻭﻗﺘﻰ ﻗﺎﺑﻞ ﺑﺴﻂ ﺍﺳﺖ ﻛﻪ ﻋﺪﺩ ﺩﺭﺳﺖ ﻭ ﻣﺜﺒﺖ ﺑﺎﺷﺪ .ﺩﺭ ﺍﻳﻦ ﺩﻭ ﺟﻤﻠﻪﺍﻯ ﺍﺯ ﺗﻮﺍﻥ )ﻗﻮﻩ( ﺻﻔﺮ ﺗﺎ ﺗﻮﺍﻥ ) (3ﺭﺍ ﻧﺸﺎﻥ ﻣﻰﺩﻫﻴﻢ. 0 )(a + b) = 1(1
)(1
1 )(a + b) = a + b(1, 1
)(1, 1
2 2 2 )(a + b) = a + 2ab + b (1, 2, 1
)(1, 2, 1
3 3 2 2 )(a + b) = a + 3 a + 3 ab + b(1, 3, 3, 1
)(1, 3, 3, 1
ﻋﺪﺩﻫﺎﻯ ﺩﺍﺧﻞ ﭘﺮﺍﻧﺘﺰﻫﺎ ﻣﻌــﺮﻑ ﺿﺮﻳﺐﻫﺎﻯ ﻋﺪﺩﻯ ﺟﻤﻠﻪﻫﺎ ﺩﺭ ﺑﺴﻂ ﺩﻭﺟﻤﻠﻪﺍﻯ ﺍﺳﺖ .ﭘﺎﺳﻜﺎﻝ ﻓﺮﺍﻧﺴﻮﻯ ﻛﻪ ﺑﺎ ﻧﻴﻮﺗﻮﻥ ﻫﻢﺯﻣﺎﻥ ﺑﻮﺩﻩ، ﺍﻳﻦ ﻣﺜﻞ ﺭﺍ ﺍﺯ ﺭﻭﻯ ﻣﻄﺎﻟﻌﺎﺕ ﺧﻴﺎﻡ ﺑﻪ ﺻﻮﺭﺕ ﺯﻳﺮ ﻧﻮﺷﺖ:
1 1
6
1 5
1 2 1 3 3 1 3 6 4 5 1 0 10
1 1 1 1 1 1
6 15
1
20 15
ﺍﻭ ﻫﻤــﻪ ﻓﻦ ﺣﺮﻳــﻒ ﺑﻮﺩ ﻭ ﺭﺑﺎﻋﻴــﺎﺕ ﻧﻴﺰ ﻣﻰﮔﻔــﺖ .ﻧﻤﻮﻧﻪﺍﻯ ﺍﺯ ﺭﺑﺎﻋﻴﺎﺕ ﺍﻭ » ﭼﻮﻥ ﺍﺑﺮ ﺑﻪ ﻧﻮﺭﻭﺯ ﺭﺥ ﻻﻟﻪ ﺑﺸﺴﺖ ﺑﺮﺧﻴﺰ ﻭ ﺑﻪ ﺟﺎﻡ ﺑﺎﺩﻩ ﻛﻦ ﻋﻬﺪ ﺩﺭﺳﺖ« »ﻛﺎﻳﻦ ﺳﺒﺰﻩ ﻛﻪ ﺍﻣﺮﻭﺯ ﺗﻤﺎﺷﺎﮔﻪ ﺗﻮﺳﺖ ﻓﺮﺩﺍ ﻫﻤﻪ ﺍﺯ ﺧﺎﻙ ﻭ ﺑﺮﺕ ﺧﻮﺍﻫﺪ ُﺭﺳﺖ« ﺧﻴﺎﻡ ﺩﺭﺑﺎﺭﻩﻯ ﺣﺎﻟﺖ ﺍﻧﺪﻳﺸﻤﻨﺪﺍﻥ ﻭ ﺩﺍﻧﺸﻤﻨﺪﺍﻥ ﻣﻰﮔﻮﻳﺪ: »ﻭﺭ ﻋﻠﻢ ﺑُﺪﻯ ﺑﻪ ﻛﺎﺭﻫﺎ ﺩﺭ ﮔﺮﺩﻭﻥ ﻛﻰ ﺧﺎﻃﺮ ﺍﻫﻞ ﻋﻠﻢ ﺁﺯﺭﺩﻩ ﺑُﺪﻯ«
ﺁﻳﺎ ﭘﺮﻭﻓﺴﻮﺭ ﻫﺸﺘﺮﻭﺩﻯ ﻣﺎﻧﻨﺪ ﺧﻴﺎﻡ ﺑﻮﺩ؟ ﺩﺭ 884ﺳــﺎﻝ ﭘﺲ ﺍﺯ ﺧﻴﺎﻡ ،ﻳﻌﻨﻰ ﺣﺪﻭﺩ 9ﻗﺮﻥ ﺑﻌﺪ ،ﺩﺭ 1960 ﻣﻴﻼﺩﻯ ) 1339ﺧﻮﺭﺷﻴﺪﻯ( ﻛﺴﻰ ﻫﻤﭽﻮﻥ ﭘﺮﻭﻓﺴﻮﺭ ﻫﺸﺘﺮﻭﺩﻯ ﺩﺭ ﺍﻳﺮﺍﻥ ﻣﻰﺯﻳﺴﺖ ﻛﻪ ﺍﺷﺘﻬﺎﺭ ﺟﻬﺎﻧﻰ ﺩﺍﺷﺖ .ﺍﻭ ﺭﻳﺎﺿﻰﺩﺍﻥ ﺷﺎﻋﺮ ،ﺍﺩﻳﺐ، ﺩﺍﻧﺸﻤﻨﺪ ﻋﻠﻮﻡ ﻓﻀﺎﺋﻰ ﻭ ﻣﺘﺨﺼﺺ ﻣﻜﺎﻧﻴﻚ ﺁﺳﻤﺎﻧﻰ ﺑﻮﺩ ﻭ ﻓﻠﺴﻔﻪ ﻫﻢ ﻣﻰﺩﺍﻧﺴﺖ .ﺯﻣﺎﻧﻰ ﻛﺎﺭﻝ ﻳﺎﺳﭙﺮﺱ ،ﻓﻴﻠﺴﻮﻑ ﭘﻴﺮﻭ »ﺍﮔﺰﻳﺴﺘﺎﻧﺴﻴﺎﻟﻴﺴﺖ«، ﮔﻔﺘﻪ ﺑﻮﺩ ﻛﻪ »ﻓﻠﺴــﻔﻴﺪﻥ ﻳﻌﻨﻰ ﺩﺭ ﺭﺍﻩ ﺑﻮﺩﻥ« .ﭘﺮﻭﻓﺴــﻮﺭ ﻣﺤﺴــﻦ ﻫﺸــﺘﺮﻭﺩﻯ ،ﻫﻤﻮﺍﺭﻩ ﺩﺭ ﺭﺍﻩ ﻭ ﺍﻧﺪﻳﺸﻤﻨﺪﻯ »ﺩﻳﻨﺎﻣﻴﻚ« )ﭘﻮﻳﺎ( ﺑﻮﺩ .ﺍﻭ ﺧــﻮﺩ ﺭﺍ ﻓﻘﻂ ﺩﺭ ﺗﺨﺼﺺ ﺧﻮﺩ ،ﻳﻌﻨﻰ ﺭﻳﺎﺿﻴﺎﺕ ﻭ ﻣﻜﺎﻧﻴﻚ ﺁﺳــﻤﺎﻧﻰ 36
ﺭﺍﻫﻨﻤﺎﻳﻲ
ﺩﻭﺭﻩﻯ ﺷﺎﻧﺰﺩﻫﻢ ،ﺷﻤﺎﺭﻩﻯ ،2ﺯﻣﺴﺘﺎﻥ 1389
ﻭ ﻫﻴﺌــﺖ ﻣﺤﺼﻮﺭ ﻧﻜــﺮﺩ ﻭ ﺣﺎﻟﺖ »ﺍﺳــﺘﺎﺗﻴﻚ« )ﺍﻳﺴــﺘﺎﻳﻰ( ﺭﺍ ﻛﻨﺎﺭ ﮔﺬﺍﺷــﺖ .ﺍﻭ ﻋﻼﻭﻩ ﺑﺮ ﻣﻌﺎﺷﺮﺕ ﺑﺎ ﺩﺍﻧﺸﻤﻨﺪﺍﻥ ﻋﻠﻮﻡ ﻋﻤﻠﻰ )ﭘﻮﺯﻳﻮﺗﻴﻮ( ﻣﺎﻧﻨــﺪ ﻓﻴﺰﻳﻚﺩﺍﻧﺎﻥ ،ﻣﺘﺨﺼﺼﺎﻥ »ﺁﺳــﺘﺮﻭﻧﻮﻣﻰ« )ﺳﺘﺎﺭﻩﺷﻨﺎﺳــﻰ( ) (Astramomyﺑﺎ ﻓﻴﻠﺴﻮﻓﺎﻥ ،ﺷــﺎﻋﺮﺍﻥ ،ﻫﻨﺮﻣﻨﺪﺍﻥ ﻭ ﻧﻘﺎﺷﺎﻥ ﻧﻴﺰ ﻣﻌﺎﺷــﺮﺕ ﺩﺍﺷــﺖ .ﺍﻭ ﻓﺮﺩﻯ ﺟﺎﻣﻊﺍﻻﻃﺮﺍﻑ ﺑﻮﺩ ﻭ ﻋــﺪﻩﺍﻯ ﺍﺯ ﻗﻮﻝ ﺍﻭ ﮔﻔﺘﻪﺍﻧﺪ ﻛﻪ ﺍﻓﺮﺍﺩ ﺻﺮﻓﺎً ﻣﺘﺨﺼﺺ ﺑﻪ ﻣﻨﺰﻟﻪﻯ ﻛﺴــﺎﻧﻰ ﻫﺴﺘﻨﺪ ﻛﻪ ﺩﺭ ﺷــﺐ ﺗﺎﺭﻳﻚ ﻓﻘﻂ ﺷــﻤﻌﻰ ﺩﺭ ﺩﺳــﺖ ﺩﺍﺭﻧﺪ ﻭ ﻓﻘﻂ ﺟﻠﻮ ﭘﺎﻯ ﺧﻮﺩ ﺭﺍ ﻣﻰﺑﻴﻨﻨﺪ ﻧﻪ ﺑﻴﺸــﺘﺮ ،ﻭﻟﻰ ﻓﺮﺩ ﻣﺘﺨﺼﺺ ﺟﺎﻣﻊﺍﻻﻃﺮﺍﻑ ،ﺩﻳﺪﻯ ﻭﺳﻴﻊ ﻧﺴﺒﺖ ﺑﻪ ﻣﺴــﺎﺋﻞ ﺟﻬﺎﻥ ﺩﺍﺭﺩ ﻭ ﺩﺭ ﭘﻴﭻ ﻣﺒﺤﺜﻰ ﻭﺍﻣﺎﻧﺪﻩ ﻧﻤﻰﺷﻮﺩ .ﺍﻭ ﺩﺭ ﺯﻧﺪﮔﻰ ﺧﻮﺩ ﺑﺎ ﻛﺴــﺎﻧﻰ ﻣﺎﻧﻨﺪ ﺟــﻼﻝ ﺁﻝ ﺍﺣﻤﺪ ،ﺩﻛﺘﺮ ﻣﺤﻤﺪ ﺑﺎﻗﺮ ﻫﻮﺷﻤﻨﺪ ،ﺍﺳﺘﺎﺩ ﺳﻌﻴﺪ ﻧﻔﻴﺴﻰ ،ﺍﺑﻮﺍﻟﺤﺴﻦ ﻓﺮﻭﻏﻰ ،ﺣﺴﻴﻦ ﻋﻠﻰ ﺭﺍﺷﺪ، ﻣﺤﻤﺪ ﻋﻠﻰ ﺑﺎﻣﺪﺍﺩ ،ﺩﻛﺘﺮ ﻣﻬﺪﻯ ﻣﺤﻘﻖ ،ﺩﻛﺘﺮ ﺟﻌﻔﺮ ﺷﻬﻴﺪﻯ ،ﻣﺤﻤﺪ ﻋﻠﻰ ﻣﺠﺘﻬﺪﻯ ،ﺍﺣﻤﺪ ﺑﻴﺮﺷــﻚ ،ﻏﻼﻣﺤﺴــﻴﻦ ﺭﻫﻨﻤــﺎ ،ﺩﻛﺘﺮ ﺍﺣﻤﺪ ﻓﺮﺩﻳﺪ ،ﻧﺎﺻﺮ ﻓﺮﺑﺪ ،ﺍﺣﻤﺪ ﻋﻠﻰ ﻣﻮﺭﺥ ﺍﻟﺪﻭﻟﻪ ﺳــﭙﻬﺮ ،ﺩﻛﺘﺮ ﭘﺮﻭﻳﺰ ﻧﺎﺗﻞ ﺧﺎﻧﻠﺮﻯ ،ﺣﺴﻴﻨﻘﻠﻰ ﻣﺴﺘﻌﺎﻥ ،ﻣﺤﻤﻮﺩ ﺣﺴﺎﺑﻰ ،ﻣﺤﻤﺪ ﻋﻠﻰ ﺟﻤﺎﻟﺰﺍﺩﻩ ﻭ ﺻﺪﻫﺎ ﻓﻴﻠﺴﻮﻑ ،ﺷﺎﻋﺮ ﻭ ﻧﻮﻳﺴﻨﺪﻩ ﻭ ﺩﺍﻧﺸﻤﻨﺪ ﺩﻳﮕﺮ ﻣﻌﺎﺷﺮﺕ ﺩﺍﺷﺖ ﻭ ﻋﻼﻭﻩ ﺑﺮ ﺑﺤﺚﻫﺎﻯ ﻋﻠﻤﻰ ،ﺍﺷﻌﺎﺭ ﺧﻮﺩ ﺭﺍ ﺑﺮﺍﻯ ﺁﻥﻫﺎ ﻗﺮﺍﺋﺖ ﻣﻰﻛﺮﺩ. ﺍﺯ ﻗﺪﻣﺎ ﺑﻪ ﺧﻴﺎﻡ ﺑﻪ ﺳﺒﺐ ﺭﺑﺎﻋﻴﺎﺕ ﻭ ﺭﻳﺎﺿﻴﺎﺕ ﺍﻭ ﺑﺴﻴﺎﺭ ﻋﻼﻗﻪﻣﻨﺪ ﺑﻮﺩ ﻭ ﺳﭙﺲ ﺑﻪ ﺍﺑﻮﺭﻳﺤﺎﻥ ﻭ ﻏﻴﺎﺙﺍﻟﺪﻳﻦ ﺟﻤﺸﻴﺪ ﻛﺎﺷﺎﻧﻰ ﻋﺸﻖ ﻣﻰﻭﺭﺯﻳﺪ. ﺑﻰ ﺟﻬﺖ ﻧﻴﺴــﺖ ﻛــﻪ ﺍﻭ ﺭﺍ ﻣﻰﺗﻮﺍﻥ ﺣﺘﻰﺍﻟﻤﻘﺪﻭﺭ ﻫــﻢ ﺭﺩﻳﻒ ﺧﻴﺎﻡ ﺩﺍﻧﺴــﺖ ،ﺯﻳﺮﺍ ﺩﺭ ﻋﻠﻢ ﻧﺠﻮﻡ ﻭ ﺭﻳﺎﺿﻰ ﻋﺼﺮ ﺧﻮﺩ ﺩﺭ ﺍﻳﺮﺍﻥ ﺳﺮﺁﻣﺪ ﺑﻮﺩ ﻭ ﻳﻜﻰ ﺍﺯ ﺩﻭﻳﺴــﺖ ﺍﺳﺘﺎﺩ ﺑﻴﻦﺍﻟﻤﻠﻠﻰ ﺭﻳﺎﺿﻴﺎﺕ ﺑﻪ ﺣﺴﺎﺏ ﻣﻰﺁﻣﺪ ﺯﻳﺮﺍ ﺩﺭ ﺩﺍﻧﺸﮕﺎﻩﻫﺎﻯ ﻛﺸﻮﺭ ﺗﺪﺭﻳﺲ ﻣﻜﺎﻧﻴﻚ ﺍﺳﺘﺪﻻﻟﻰ ،ﻫﻨﺪﺳﻪ ،ﻫﻴﺌﺖ ﻭ ﻧﺠﻮﻡ ،ﺁﻣﺎﺭ ،ﺁﻧﺎﻟﻴﺰ ﻭ ﺭﻳﺎﺿﻰ ﻋﻤﻮﻣﻰ ﺭﺍ ﺑﺮ ﻋﻬﺪﻩ ﺩﺍﺷﺖ ﻭ ﻋﻼﻭﻩ ﺑﺮ ﺁﻥ ﭘﺲ ﺍﺯ ﭘﺮﺗﺎﺏ ﺍﻭﻟﻴﻦ ﻗﻤﺮ ﻣﺼﻨﻮﻋﻰ ﺭﻭﺱﻫﺎ ﺑﻪ ﻧﺎﻡ »ﺍﺳــﭙﻮﺗﻨﻴﻚ« ﺩﺭ ﺍﻭﺍﺧﺮ ﺩﻫﻪﻯ 1950ﺑﻪ ﺩﻋﻮﺕ ﺍﻧﺠﻤﻦ ﻓﻀﺎﻧﻮﺭﺩﺍﻥ ﺷــﻮﺭﻭﻯ ﺑﻪ ﻣﺴﻜﻮ ﺩﻋﻮﺕ ﺷﺪ ﻭ ﺩﺭﺑﺎﺭﻩﻯ ﺗﺴــﺨﻴﺮ ﻓﻀﺎ ﺳﺨﻨﺮﺍﻧﻰ ﻛﺮﺩ .ﺯﻣﺎﻧﻰ ﻫﻤﺴﺮﺵ ﮔﻔﺘﻪ ﺑﻮﺩ ﻛﻪ ﭘﺮﻭﻓﺴﻮﺭ ﺣﺪﺍﻗﻞ ﭼﻬﺎﺭﺻﺪ ﺳﺎﻝ ﺯﻭﺩﺗﺮ ﺑﻪ ﺩﻧﻴﺎ ﺁﻣﺪﻩ ﺑﻮﺩ، ﭼــﻮﻥ ﺍﻳﻦ ﺩﻧﻴﺎ ﺑﺮﺍﻳﺶ ﻛﻮﭼﻚ ﺑﻮﺩ ،ﺍﺷــﻌﺎﺭﻯ ﻛﻪ ﺩﺭ ﭘﻨﺞ ﺳــﺎﻟﮕﻰ ﺍﺯ ﺣﻔﻆ ﻛﺮﺩﻩ ﺑﻮﺩ ﺗﺎ ﻟﺤﻈﻪﻯ ﻣﺮگ ﺑﻪ ﻳﺎﺩ ﺩﺍﺷﺖ. ﻣﻨﺎﺑﻊ
.1ﺯﻧﺪﮔﻰﻧﺎﻣﻪ ﻭ ﺧﺪﻣﺎﺕ ﻋﻠﻤﻰ ﻭ ﻓﺮﻫﻨﮕﻰ ﻣﺮﺣﻮﻡ ﭘﺮﻭﻓﺴﻮﺭ ﻣﺤﺴﻦ ﻫﺸﺘﺮﻭﺩﻯ، ﺍﻧﺠﻤﻦ ﻣﻔﺎﺧﺮ ،ﺗﻬﺮﺍﻥ.1380 ، .2ﺯﻧﺪﮔﻰﻧﺎﻣــﻪ ﻭ ﺧﺪﻣﺎﺕ ﻋﻠﻤﻰ ﻭ ﻓﺮﻫﻨﮕﻰ ﺍﺳــﺘﺎﺩ ﭘﺮﻭﻳﺰ ﺷــﻬﺮﻳﺎﺭﻯ ،ﺍﻧﺠﻤﻦ ﻣﻔﺎﺧﺮ ،ﺗﻬﺮﺍﻥ.1385 ، » .3ﺧﻴــﺎﻡ ﻧﻴﺸــﺎﺑﻮﺭﻯ« ﺩﺍﻳﺮﻩ ﺍﻟﻤﻌﺎﺭﻑ ﻣﺼﺎﺣــﺐ ﻭ ﻫﻢﭼﻨﻴﻦ ﻣﻘﺎﻻﺕ ﻋﺒﺎﺱ ﺍﻗﺒﺎﻝ ﺩﺭﺑﺎﺭﻩﻯ ﺧﻴﺎﻡ ﺑﻪ ﻛﻮﺷﺶ ﺍﺳﺘﺎﺩ ﺩﺑﻴﺮ ﺳﻴﺎﻗﻰ. .4ﻣﻘﺪﻣﻪﺍﻯ ﺑﺮ ﻓﻠﺴــﻔﻪ ،ﻧﻮﺷــﺘﻪﻯ ﺍﻭﺯﻭﺍﻟﺪﻛﻮﺍﭘﻪ ،ﺗﺮﺟﻤﻪﻯ ﺍﺣﻤﺪ ﺁﺭﺍﻡ ،ﺗﻬﺮﺍﻥ ﺍﻧﺘﺸﺎﺭﺍﺕ ﻋﻠﻤﻰ.1326 ، . 5ﺯﻧﺪﮔﻰﻧﺎﻣﻪﻯ ﺭﻳﺎﺿﻰﺩﺍﻧﺎﻥ ﺍﺳﻼﻣﻰ ،ﺍﺑﻮﺍﻟﻘﺎﺳﻢ ﻗﺮﺑﺎﻧﻰ ،ﻣﺮﻛﺰ ﻧﺸﺮ ﺩﺍﻧﺸﮕﺎﻫﻰ، .1368
﹇︧﹝️ دوم
ﺩﺍﻧﺶﺍﻓﺰﺍﻳﯽ
︡س ︋︤﹡﹫︡ ﻣﺮﻳﻢ ﺳﻌﻴﺪﻯ
ﻛﻠﻴﺪﻭﺍژﻩﻫﺎ :ﻣﻨﺤﻨﻰ ،ﻣﺴﻴﺮ ،ﺯﺍﻭﻳﻪ . ﺷﻜﻞ ﺯﻳﺮ ﻳﻚ ﻣﺴﺌﻠﻪﻯ ﺗﻘﺮﻳﺒﺎً ﻣﻌﺮﻭﻑ ﺍﺳﺖ ﻛﻪ ﻣﻰﺗﻮﺍﻧﻴﺪ ﺟﻮﺍﺏ ﺭﺍ ﺩﻗﻴﻘﺎً ﺑﻪ ﺩﺳﺖ ﺁﻭﺭﻳﺪ. ﺍﻣﺎ ﺍﺑﺘﺪﺍ ﺟﻮﺍﺏ ﺭﺍ ﺣﺪﺱ ﺑﺰﻥ ﻭ ﺑﻌﺪ ﺑﺎ ﺟﻮﺍﺏ ﺩﻗﻴﻘﻰ ﻛﻪ ﺑﻪ ﺩﺳﺖ ﻣﻰﺁﻭﺭﻯ ﻣﻘﺎﻳﺴﻪ ﻛﻦ.
︎︣وژۀ ١ ﺑﻪ ﺩﻭ ﺷﻜﻞ ﺯﻳﺮ ﺩﻗﺖ ﻛﻨﻴﺪ.
ﺍﻟﻒ ﺏ )ﺷﻜﻞ (3
)ﺷﻜﻞ (1
︎︣وژۀ ٢ ﺍﻟﻒ ﺏ
)ﺷﻜﻞ (2 ﺣﺪﺱ ﻣﻰﺯﻧﻴﺪ ﺩﺭ ﻫﺮ ﺷﻜﻞ ﻛﺪﺍﻡ ﻣﺴﻴﺮ ﻃﻮﻻﻧﻰﺗﺮ ﺍﺳﺖ؟ »ﺍﻟﻒ« ﻳﺎ »ﺏ« ﺷــﻜﻞﻫﺎﻯ ﺩﻳﮕﺮﻯ ﺑﻪ ﻫﻤﻴﻦ ﺻﻮﺭﺕ ﺭﺳــﻢ ﻛﻨﻴــﺪ ﻭ ﻳﺎ ﻧﻘﻄﻪﻯ ﺧﻴﺎﺑﺎﻥﻫــﺎﻯ ﺗﻬﺮﺍﻥ ﺭﺍ ﺭﻭﻯ ﺩﻳﻮﺍﺭ ﺍﺗﺎﻗﺘﺎﻥ ﺑﭽﺴــﺒﺎﻧﻴﺪ ﻭ ﺑﺮﺍﻯ ﺧﻮﺩﺗﺎﻥ ﻣﺒــﺪﺃ ﻭ ﻣﻘﺼﺪ ﺍﻧﺘﺨﺎﺏ ﻛﻨﻴﺪ .ﺍﺯ ﻣﺴــﻴﺮﻫﺎﻯ ﻣﺘﻔﺎﻭﺕ ﺍﺯ ﻣﺒﺪﺃ ﺣﺮﻛﺖ ﻛﻨﻴﺪ ﺗﺎ ﺑﻪ ﻣﻘﺼﺪ ﺑﺮﺳﻴﺪ ﻭ ﺑﺎ ﺩﻗﺖ ﺑﻪ ﻣﺴﻴﺮﻫﺎﻯ ﺍﻧﺘﺨﺎﺏ ﺷﺪﻩ ،ﺣﺪﺱ ﺧﻮﺩ ﺭﺍ ﺍﻣﺘﺤﺎﻥ ﻛﻨﻴﺪ .ﻛﺪﺍﻡ ﻣﺴﻴﺮ ﻃﻮﻻﻧﻰﺗﺮ ﺍﺳﺖ؟
ﻳﻚ ﺑﺴﻜﺘﺒﺎﻟﻴﺴــﺖ ﺩﺭ ﻳﻚ ﻓﺎﺻﻠﻪﻯ ﻣﺸــﺨﺺ ﺍﺯ ﭘﺎﻳﻪﻯ ﺳــﺒﺪ ﺑﺴــﺘﻜﺒﺎﻝ ﺍﻳﺴﺘﺎﺩﻩ ﻭ ﺍﺯ ﺁﻥﺟﺎ ﻣﻰﺧﻮﺍﻫﺪ ﺷﺎﻧﺲ ﭘﺮﺗﺎﺏ ﻣﻮﻓﻖ ﺧﻮﺩ ﺭﺍ ﺩﺭ ﺯﺍﻭﻳﻪﻫــﺎﻯ ﻣﺘﻔﺎﻭﺕ ﺍﻣﺘﺤﺎﻥ ﻛﻨﺪ .ﭘﺲ ﺷــﺮﻭﻉ ﻣﻰﻛﻨﺪ 10 :ﭘﺮﺗﺎﺏ ﺑــﺎ ﺯﺍﻭﻳﻪﻯ ﺗﻘﺮﻳﺒﺎً 10 ،30°ﭘﺮﺗﺎﺏ ﺑﺎ ﺯﺍﻭﻳﻪﻯ ﺗﻘﺮﻳﺒﺎً ،45°ﻭ ﺑﻪ ﻫﻤﻴﻦ ﺗﺮﺗﻴﺐ ﺑﺎ ﺯﺍﻭﻳﻪﻫﺎﻯ 60°ﻭ .80° ﺣﺎﻝ ،ﻣﻨﺤﻨﻰ ﻣﻴﺰﺍﻥ ﻣﻮﻓﻘﻴﺖ ﺭﺍ ﻧﺴﺒﺖ ﺑﻪ ﺯﺍﻭﻳﻪ ﺭﺳﻢ ﻣﻰﻛﻨﺪ. ـ ﺣﺪﺱ ﻣﻰﺯﻧﻴﺪ ﺩﺭ ﻛﺪﺍﻡ ﺯﺍﻭﻳﻪ ،ﻣﻴﺰﺍﻥ ﻣﻮﻗﻌﻴﺖ ﺑﻴﺶﺗﺮ ﺍﺳﺖ؟ ـ ﺁﻳﺎ ﻣﻰﺗﻮﺍﻧﻴﺪ ﺭﺍﺑﻄﻪﺍﻯ ﺭﻳﺎﺿﻰ ﺑﻴﻦ ﻧﺮﺥ ﻣﻮﻓﻘﻴﺖ )ﺩﺭﺻﺪ ﻧﺴﺒﺖ ﭘﺮﺗﺎﺏﻫﺎﻯ ﻣﻮﻓﻖ ﺑﻪ ﻛﻞ ﭘﺮﺗﺎﺏﻫﺎ( ﻭ ﺯﺍﻭﻳﻪﻯ ﭘﺮﺗﺎﺏ ﺭﺍ ﭘﻴﺪﺍ ﻛﻨﻴﺪ. ـ ﻣﻰﺗﻮﺍﻧﻴــﺪ ﭘﻴﺶﺗﺮ ﺑﺮﻭﻳﺪ ﻭ ﺩﺭ ﻓﺎﺻﻠﻪﻫﺎﻯ ﻣﺘﻔﺎﻭﺕ ﻧﺴــﺒﺖ ﺑﻪ ﺳﺒﺪ ،ﺍﻳﻦ ﺗﻐﻴﻴﺮﺍﺕ ﺭﺍ ﺑﺴــﻨﺠﻴﺪ .ﺩﺭ ﻫﻤﻪﻯ ﻣﺮﺍﺣﻞ ﻛﺸﻴﺪﻥ ﻣﻨﺤﻨﻰ ﺗﻐﻴﻴﺮﺍﺕ ﺭﺍ ﻓﺮﺍﻣﻮﺵ ﻧﻜﻨﻴﺪ. ـ ﭘﺲ ﭼﺮﺍ ﻣﻌﻄﻠﻴﺪ؟ ﺍﻣﺘﺤﺎﻥ ﻛﻨﻴﺪ .ﺍﻳﻦ ﻳﻚ ﭘﺮﻭژﻩﻯ ﺧﻮﺏ ﺑﺮﺍﻯ ﺩﺭﺱ ﺭﻳﺎﺿﻰﺗﺎﻥ ﺧﻮﺍﻫﺪ ﺑﻮﺩ. ﺩﻭﺭﻩﻯ ﺷﺎﻧﺰﺩﻫﻢ ،ﺷﻤﺎﺭﻩ ﻯ ،2ﺯﻣﺴﺘﺎﻥ 1389
ﺭﺍﻫﻨﻤﺎﻳﻲ
37
ﺩﺍﻧﺶﺍﻓﺰﺍﻳﻲ
﹁︣﹞︣︋ ﹩﹛﹢ای ﹈ ا﹛﹍﹢ ﺯﻫﺮﻩ ﭘﻨﺪﻯ
ﻛﻠﻴﺪﻭﺍژﻩﻫﺎ :ﻣﺴﺌﻠﻪ ،ﭼﻮﺏ ﻛﺒﺮﻳﺖ ،ﻣﺮﺑﻊ ،ﻣﺴﺘﻄﻴﻞ . ﻣﺮﺍﺟﻌﻪ ﻣﺴﻴﺮ...... ﻣﻘﺎﻟﻪﻯ ﺁﺩﺭﺱ ﻣﻘﺎﻟﻪﻣﻰﺗﻮﺍﻥ ﺣﺘﻰ ﭘﺮﺍﻛﻨﺪﻩ ﻛﺎﺭ ﺻــﻮﺭﺕ ﺍﻟﮕﻮﻣﺠﻠﻪ ﺷــﻤﺎﺭﻩﻯ ﭘﻴﺪﺍﻫﺎ ﺭﺍ ﻣﺴــﺌﻠﻪ ﺍﻳﻦ ﺭﺳﻴﺪﻥ ﺑﻪ ﺑــﻪﻗﺪﻡ، ﻛﻪﭼﻨﺪ ﺩﺍﺩ ﺩﺭ ﺍﻳﻢ ﺗﺎ ﺳـﻌﻰ ﻛﺮﺩﻩ ﻛﺮﺩ.ﺍﻳﻦ ﺭﺳـﺪ .ﺩﺭ ﻣﺸـﻜﻞ ﻣﻰ ﮔﺎﻫﻰﺑﻪﺑﻪ ﻧﻈﺮ ﻫﻤﻴﻦﺑﺮﺍﻯ ﻳﻚ ﺗﻮﺍﻥﻯﺩﺭﻋﻤﻮﻣﻰ ﻣﻰﺟﻤﻠﻪ ﻛﺮﺩﻥ ﭼﻜﻴﺪﻩ: ﻛﻨﻴﺪ.ﻯ ﻋﻤﻮﻣﻰ ﺭﺍ ﺩﺭ ﺍﻟﮕﻮﻫﺎﻯ ﺳﺎﺩﻩ ﺗﺮﺳﻴﻢ ﻛﻨﻴﻢ. ﺟﻤﻠﻪ ﺩﺭ ﺷﻜﻞ ﺯﻳﺮ ﻳﻚ ﺷﻜﻞ Lﻣﺎﻧﻨﺪ 4ﺩﺭ 4ﺭﺍ ﻣﻼﺣﻈﻪ ﻣﻰﻛﻨﻴﺪ:
ﺑﺮﺍﻯ ﺳــﺎﺧﺘﻦ ﺍﻳﻦ ﺷــﻜﻞ ﺍﺯ 22ﭼﻮﺏ ﻛﺒﺮﻳﺖ ﺍﺳــﺘﻔﺎﺩﻩ ﺷﺪﻩ ﺍﺳــﺖ .ﻣﺤﻴﻄﻰ ﺑﺮﺍﺑﺮ ﺑﺎ ﻃﻮﻝ 16ﭼﻮﺏ ﻛﺒﺮﻳﺖ ﺩﺍﺭﺩ ﻭ ﺩﺭ ﺁﻥ 7ﻣﺮﺑﻊ ﺩﻳﺪﻩ ﻣﻰﺷﻮﺩ. ﺑﺮﺍﻯ ﺳﺎﺧﺘﻦ ﻳﻚ ﺷﻜﻞ Lﻣﺎﻧﺪ 25ﭼﻨﺪ ﭼﻮﺏ ﻛﺒﺮﻳﺖ ﻻﺯﻡ ﺍﺳﺖ؟ ﻣﺤﻴﻂ ﺁﻥ ﭼﻪﻗﺪﺭ ﺍﺳﺖ؟ ﭼﻨﺪ ﻣﺮﺑﻊ ﺩﺭ ﺁﻥ ﺩﻳﺪﻩ ﻣﻰﺷﻮﺩ؟ ﻳﻚ ﺷﻜﻞ 100ﺩﺭ 100ﭼﻪﻃﻮﺭ؟ ﻳﻚ ﺷﻜﻞ nﺩﺭ nﭼﻪﻃﻮﺭ؟
ﺷﻜﻞ ﺯﻳﺮ ﺍﻟﮕﻮﻳﻰ ﺷﺎﻣﻞ ﺩﻭ ﻣﺮﺑﻊ ﺭﺍ ﻧﺸﺎﻥ ﻣﻰﺩﻫﺪ ﻛﻪ ﭼﻬﺎﺭ ﻧﻘﻄﻪﻯ ﺳﻴﺎﻩ ﺩﺍﺭﺩ. ﺩﺭ ﺍﻳﻦ ﺷﻜﻞ 27ﻧﻘﻄﻪﻯ ﺳﻔﻴﺪ ﺩﻳﺪﻩ ﻣﻰﺷﻮﺩ ﻭ ﺑﺮﺍﻯ ﺳﺎﺧﺘﻦ ﺁﻥ ﺍﺯ 48ﭘﺎﺭﻩﺧﻂ ﻛﻮﭼﻚ ﺍﺳﺘﻔﺎﺩﻩ ﺷﺪﻩ ﺍﺳﺖ.
ﺑﺮﺍﻯ ﺳــﺎﺧﺘﻦ ﺷــﻜﻠﻰ ﺑﺎ ﻫﻤﻴﻦ ﺍﻟﮕﻮ ﺍﻣﺎ ﺑﺎ 25ﻧﻘﻄﻪﻯ ﺳــﻴﺎﻩ ﭼﻨﺪ ﭘﺎﺭﻩﺧﻂ ﻛﻮﭼﻚ ﻻﺯﻡ ﺍﺳــﺖ؟ ﺩﺭ ﺁﻥ ﺷــﻜﻞ ﭼﻨﺪ ﻧﻘﻄﻪﻯ ﺳﻔﻴﺪ ﺩﻳﺪﻩ ﻣﻰﺷﻮﺩ؟ ﻳﻚ ﺷﻜﻞ ﺑﺎ 100ﻧﻘﻄﻪﻯ ﺳﻴﺎﻩ ﭼﻪﻃﻮﺭ؟ ﻳﻚ ﺷﻜﻞ ﺑﺎ nﻧﻘﻄﻪﻯ ﺳﻴﺎﻩ ﭼﻪﻃﻮﺭ؟
38
ﺭﺍﻫﻨﻤﺎﻳﻲ
ﺷﻤﺎﺭﻩﻯ ،2ﺯﻣﺴﺘﺎﻥ 1389 ﺩﻭﺭﻩﻯ ﺷﺎﻧﺰﺩﻫﻢ ،ﺷﻤﺎﺭﻩﻯ ﺩﻭﺭﻩﻯ
ﺷــﻜﻞ ﺯﻳﺮ ﻣﺮﺑﻌﻰ ﺑﻪ ﺿﻠﻊ 5ﺭﺍ ﻧﺸــﺎﻥ ﻣﻰﺩﻫﺪ ﻛﻪ ﻣﺤﻴﻂ ﺁﻥ ﺍﺯ 16ﻣﺮﺑﻊ ﻛﻮﭼﻚ ﺗﺸــﻜﻴﻞ ﺷﺪﻩ ﺍﺳــﺖ .ﺍﻳﻦ ﺍﻟﮕﻮ ﺑﺎ 48ﭘﺎﺭﻩﺧﻂ ﻛﻮﭼﻚ ﺳﺎﺧﺘﻪ ﺷﺪﻩ ﺍﺳﺖ.
ﻣﺮﺑﻌﻰ ﺑﺎ ﻫﻤﻴﻦ ﺍﻟﮕﻮ ﻭ ﺑﻪ ﺿﻠﻊ 25ﺍﺯ ﭼﻨﺪ ﻣﺮﺑﻊ ﻛﻮﭼﻚ ﺗﺸﻜﻴﻞ ﻣﻰﺷﻮﺩ؟ ﭼﻨﺪ ﭘﺎﺭﻩﺧﻂ ﻛﻮﭼﻚ ﺑﺮﺍﻯ ﺳﺎﺧﺘﻦ ﺁﻥ ﻻﺯﻡ ﺩﺍﺭﻳﻢ؟ ﻣﺮﺑﻌﻰ ﺑﻪ ﺿﻠﻊ 100ﭼﻪﻃﻮﺭ؟ ﻣﺮﺑﻌﻰ ﺑﻪ ﺿﻠﻊ nﭼﻪﻃﻮﺭ؟
ﺷــﻜﻞ ﺯﻳﺮ ﻣﺮﺑﻌﻰ ﺑﻪ ﺿﻠﻊ 3ﺭﺍ ﻧﺸــﺎﻥ ﻣﻰﺩﻫﺪ ﻛﻪ ﺑﺎ 24ﭘﺎﺭﻩﺧﻂ ﻛﻮﭼﻚ ﺳــﺎﺧﺘﻪ ﺷــﺪﻩ ﺍﺳــﺖ .ﺩﺭ ﺍﻳﻦ ﺷــﻜﻞ 25ﻧﻘﻄﻪﻯ ﺳﻔﻴﺪ ﺩﻳﺪﻩ ﻣﻰﺷﻮﺩ.
ﺑﺮﺍﻯ ﺳﺎﺧﺘﻦ ﻣﺮﺑﻌﻰ ﺑﺎ ﻫﻤﻴﻦ ﺍﻟﮕﻮ ﻭ ﺑﻪ ﺿﻠﻊ 25ﭼﻨﺪ ﭘﺎﺭﻩﺧﻂ ﻛﻮﭼﻚ ﻻﺯﻡ ﺩﺍﺭﻳﻢ؟ ﭼﻨﺪ ﻧﻘﻄﻪﻯ ﺳﻔﻴﺪ ﺩﺭ ﺁﻥ ﺩﻳﺪﻩ ﻣﻰﺷﻮﺩ؟ ﻣﺮﺑﻌﻰ ﺑﻪ ﺿﻠﻊ 100ﭼﻪﻃﻮﺭ؟ ﻣﺮﺑﻌﻰ ﺑﻪ ﺿﻠﻊ nﭼﻪﻃﻮﺭ؟
ﻣﺴﺘﻄﻴﻞ ﺯﻳﺮ ﺍﺯ ﻛﻨﺎﺭ ﻫﻢ ﻗﺮﺍﺭ ﮔﺮﻓﺘﻦ ﺩﻭ ﻣﺮﺑﻊ ﻣﺴﺎﻭﻯ ﺑﻪ ﺿﻠﻊ 3ﺩﺭﺳﺖ ﺷﺪﻩ ﺍﺳﺖ .ﺑﺮﺍﻯ ﺳﺎﺧﺘﻦ ﺍﻳﻦ ﻣﺴﺘﻄﻴﻞ ﺍﺯ 21ﭘﺎﺭﻩﺧﻂ ﻛﻮﭼﻚ ﻣﺴﺎﻭﻯ ﺍﺳﺘﻔﺎﺩﻩ ﺷﺪﻩ ﺍﺳﺖ ﻭ ﺩﺭ ﺁﻥ 28ﻧﻘﻄﻪﻯ ﺳﻔﻴﺪ ﺩﻳﺪﻩ ﻣﻰﺷﻮﺩ.
ﺑﺮﺍﻯ ﺳﺎﺧﺘﻦ ﻣﺴﺘﻄﻴﻠﻰ ﺑﺎ ﻫﻤﻴﻦ ﺍﻟﮕﻮ ﻛﻪ ﺍﺯ ﻛﻨﺎﺭ ﻫﻢ ﻗﺮﺍﺭ ﮔﺮﻓﺘﻦ ﺩﻭ ﻣﺮﺑﻊ ﻣﺴﺎﻭﻯ ﺑﻪ ﺿﻠﻊ 15ﺩﺭﺳﺖ ﺷﺪﻩ ﺍﺳﺖ ،ﭼﻨﺪ ﭘﺎﺭﻩﺧﻂ ﻛﻮﭼﻚ ﻻﺯﻡ ﺩﺍﺭﻳﻢ؟ ﭼﻨﺪ ﻧﻘﻄﻪﻯ ﺳﻔﻴﺪ ﺩﺭ ﺁﻥ ﺩﻳﺪﻩ ﻣﻰﺷﻮﺩ؟ ﻣﺴﺘﻄﻴﻠﻰ ﻛﻪ ﺍﺯ ﻛﻨﺎﺭ ﻫﻢ ﻗﺮﺍﺭ ﮔﺮﻓﺘﻦ ﺩﻭ ﻣﺮﺑﻊ ﺑﻪ ﺿﻠﻊ 100ﺩﺭﺳﺖ ﺷﺪﻩ ﺍﺳﺖ ،ﭼﻪﻃﻮﺭ؟ ﺩﻭ ﻣﺮﺑﻊ ﺑﻪ ﺿﻠﻊ nﭼﻪﻃﻮﺭ؟
ﺩﻭﺭﻩﻯ ﺷﺎﻧﺰﺩﻫﻢ ،ﺷﻤﺎﺭﻩ ﻯ ،2ﺯﻣﺴﺘﺎﻥ 1389
ﺭﺍﻫﻨﻤﺎﻳﻲ
39
ﻣﻌﻤﺎ ﻭ ﺳﺮﮔﺮﻣﯽ
︨︣﹎︣﹞︀﹨﹩ی ر︀︲﹩ ﻣﺤﻤﺪ ﻋﺰﻳﺰﻯﭘﻮﺭ
ﻛﻠﻴﺪﻭﺍژﻩﻫﺎ :ﺍﻋﺪﺍﺩ ،ﻣﻌﻤﺎ ،ﺟﺪﻭﻝ . -١ﺣﻠﻘﻪﻫــﺎﯼ ﺍﻟﻤﭙﻴــﻚ :ﺍﻋــﺪﺍﺩ 1ﺗﺎ 12ﺭﺍ ﻃــﻮﺭﻯ ﺩﺭ ﺧﺎﻧﻪﻫﺎ ﻗﺮﺍﺭ ﺩﻫﻴﺪ ﻛﻪ ﺣﺎﺻﻞ ﺟﻤﻊ ﺍﻋﺪﺍﺩ ﺧﺎﻧﻪﻫﺎﻯ ﻫﺮ ﺩﺍﻳﺮﻩ 28ﺷــﻮﺩ .ﺑﺮﺍﻯ ﺁﺳﺎﻥﺗﺮ ﺷﺪﻥ ّ ﺣﻞ ﻣﻌﻤﺎ ،ﭼﻬﺎﺭ ﻋﺪﺩ ﺭﺍ ﺩﺭ ﺧﺎﻧﻪﻫﺎ ﻗﺮﺍﺭ ﺩﺍﺩﻳﻢ.
8 9 10
-٢ﻫﻔﺖ ﺧﻮﺵ ﻳُﻤﻦ :ﺩﺭ ﻫﺮ ﺧﺎﻧﻪ ،ﻳﻜﻰ ﺍﺯ ﺍﺭﻗﺎﻡ 1ﺗﺎ 9ﺭﺍ ﻃﻮﺭﻯ ﺑﮕﺬﺍﺭﻳﺪ ﺗﺎ ﺗﺴﺎﻭﻯ ﺑﺮﻗﺮﺍﺭ ﺷﻮﺩ.
٧ 40
ﺭﺍﻫﻨﻤﺎﻳﻲ
ﺩﻭﺭﻩﻯ ﺷﺎﻧﺰﺩﻫﻢ ،ﺷﻤﺎﺭﻩﻯ ،2ﺯﻣﺴﺘﺎﻥ 1389
11
-٣ﺩﺭ ﻃﻮﻳﻠﻪﺍﻯ ﭼﻨﺪ ﻣﺮﺩ ﻭ ﭼﻨﺪ ﺍﺳــﺐ ﻭﺟﻮﺩ ﺩﺍﺭﺩ .ﺁﻥﻫﺎ ﺩﺭ ﻣﺠﻤﻮﻉ 22ﺳــﺮ ﻭ 72ﭘﺎ ﺩﺍﺭﻧﺪ .ﺣﺴــﺎﺏ ﻛﻨﻴﺪ ﻛﻪ ﭼﻨﺪ ﻣﺮﺩ ﻭ ﭼﻨﺪ ﺍﺳﺐ ﺩﺭ ﺁﻧﺠﺎ ﻫﺴﺘﻨﺪ؟
-٤ﺟﺪﻭﻝ ﺟﺎﺩﻭﻳﻲ :ﭼﮕﻮﻧﻪ ﻣﻰﺗﻮﺍﻧﻴﻢ ﺍﻳﻦ ﻣﺮﺑﻊ ﺭﺍ ﺑﺎ ﺍﻋﺪﺍﺩ ﺩﻳﮕﺮ ﺍﻭﻝ ﻛﻮﭼﻚﺗﺮ ﺍﺯ 100ﻛﺎﻣﻞ ﻛﻨﻴﻢ ﺗﺎ ﻣﺮﺑﻊ ﺟﺎﺩﻭﻳﻰ ﺷﻮﺩ. ﺑﻪ ﺍﻳﻦ ﻣﻌﻨﺎ ﻛﻪ ﻣﺠﻤﻮﻉ ﺍﻋﺪﺍﺩ ﻧﻮﺷﺘﻪ ﺩﺭ ﻫﺮ ﺳﻄﺮ ،ﻫﺮ ﺳﺘﻮﻥ ﻭ ﻫﺮ ﻗﻄﺮ ﺑﺎ ﻫﻢ ﻣﺴﺎﻭﻯ ﺑﺎﺷﺪ. ﺗﻮﺟﻪ :ﺍﻋﺪﺍﺩ ﺍﻭﻝ ﻣﺎﻧﻨﺪ 7 ،13 ،2ﺟﺰ ﺑﺮ ﺧﻮﺩﺷﺎﻥ ﻭ ﻋﺪﺩ 1ﺑﺮ ﻋﺪﺩ ﺩﻳﮕﺮﻯ ﺑﺨﺶﭘﺬﻳﺮ ﻧﻴﺴﺘﻨﺪ.
١
٧
ﭘﺎﺳﺦ؛ ﺻﻔﺤﻪﻱ 24
١٣ -٥ﭼﮕﻮﻧﻪ ﻣﻰﺗﻮﺍﻥ ﺷﺶﺿﻠﻌﻰ ﻣﻨﺘﻈﻢ ﺯﻳﺮ ﺭﺍ ﺑﻪ ﻫﺸﺖ ﻗﺴﻤﺖ ﻳﻜﺴﺎﻥ ﺗﻘﺴﻴﻢ ﻛﺮﺩ. ﺑﺮﺍﻯ ﻳﺎﻓﺘﻦ ﺩﻭ ﭘﺎﺳﺦ ﮔﻮﻧﺎﮔﻮﻥ ﺗﻼﺵ ﻛﻨﻴﺪ.
ﺩﻭﺭﻩﻯ ﺷﺎﻧﺰﺩﻫﻢ ،ﺷﻤﺎﺭﻩ ﻯ ،2ﺯﻣﺴﺘﺎﻥ 1389
ﺭﺍﻫﻨﻤﺎﻳﻲ
41
ﺩﺍﻧﺶﺍﻓﺰﺍﻳﯽ
ﺳﺎﻳﻪ ﻣﻬﺮﺑﺎﻥ
ﻛﻠﻴﺪﻭﺍژﻩﻫﺎ :ﭼــﻮﺏ ﻛﺒﺮﻳﺖ، ﭼﻴﻨﺶ ،ﻗﺪﻡ ﺍﻭﻝ ،ﻗﺪﻡ nﺍﻡ .
ﭼﻴﻨﺶ ﻣﺎﻧﯽ
ﻗﺪﻡ ﻫﻔﺘﻢ
ﺑﻪ ﺷﻜﻞﻫﺎﻯ ﺯﻳﺮ ﻧﮕﺎﻩ ﻛﻨﻴﺪ .ﺳﻪ ﺩﺍﻧﺶﺁﻣﻮﺯ ﭼﻮﺏ ﻛﺒﺮﻳﺖﻫﺎ ﺭﺍ ﺑﺮﺍﻯ ﺭﺳﻴﺪﻥ ﺑﻪ ﺍﻟﮕﻮﻯ ﺑﺎﻻ ﺑﺎ ﺳﻪ ﺗﺮﺗﻴﺐ ﻣﺨﺘﻠﻒ ﭼﻴﺪﻩ ﻭ ﺷﻤﺮﺩﻩﺍﻧﺪ!
ﭼﻴﻨﺶ ﻋﻠﯽ
3ﭼﻮﺏ ﻛﺒﺮﻳﺖ
2×3ﭼﻮﺏ ﻛﺒﺮﻳﺖ 7ﭼﻮﺏ ﻛﺒﺮﻳﺖ
2×7ﭼﻮﺏ ﻛﺒﺮﻳﺖ
) 2×7 + (2+7ﭼﻮﺏ ﻛﺒﺮﻳﺖ 42
ﺭﺍﻫﻨﻤﺎﻳﻲ
ﺩﻭﺭﻩﻯ ﺷﺎﻧﺰﺩﻫﻢ ،ﺷﻤﺎﺭﻩﻯ ،2ﺯﻣﺴﺘﺎﻥ 1389
3×7ﭼﻮﺏ ﻛﺒﺮﻳﺖ
(3×7 ) +1ﭼﻮﺏ ﻛﺒﺮﻳﺖ
ﺩﺍﻧﺶﺍﻓﺰﺍﻳﯽ
﹠︡ ﹇︡م ︋︀ ر︀︲﹩دا﹡︀ن
ﻫﺮ ﻳــﻚ ﺍﺯ ﺍﻳﻦ ﺩﺍﻧﺶﺁﻣــﻮﺯﺍﻥ ﺑﺮﺍﻯ ﭼﻴﺪﻥ ﭼــﻮﺏ ﻛﺒﺮﻳﺖﻫﺎ ﺍﺯ ﺗﺮﺗﻴــﺐ ﺧﺎﺻﻰ ﭘﻴﺮﻭﻯ ﻛﺮﺩﻩﺍﻧﺪ .ﻫﺮﻳﻚ ﭼﻨــﺪ ﭼﻮﺏ ﻛﺒﺮﻳﺖ ﺭﺍ ﺑﺮﺍﻯ ﺳﺎﺧﺘﻦ 7ﻣﺮﺑﻊ ﺍﺳﺘﻔﺎﺩﻩ ﻛﺮﺩﻩﺍﻧﺪ؟
ﭼﻴﻨﺶ ﺍﻣﻴﺮ 4ﭼﻮﺏ ﻛﺒﺮﻳﺖ
ﻗﺪﻡ ﺑﻴﺴﺖ ﻭ ﭘﻨﺠﻢ
3+4ﭼﻮﺏ ﻛﺒﺮﻳﺖ
) 4 + (3×2ﭼﻮﺏ ﻛﺒﺮﻳﺖ
ﺣــﺎﻻ ﻓﻜﺮ ﻛﻨﻴﺪ ﻛﻪ ﻫﺮﻳﻚ ﺍﺯ ﺁﻥﻫﺎ ﻗﺮﺍﺭ ﺍﺳــﺖ ﺑﺎ ﺗﻌﺪﺍﺩﻯ ﭼﻮﺏ ﻛﺒﺮﻳﺖ 25 ،ﻣﺮﺑﻊ ﺑﻪ ﻫﻤﺎﻥ ﺗﺮﺗﻴﺐ ﺑﺎﻻ ﻛﻨﺎﺭ ﻫﻢ ﺑﭽﻴﻨﻨﺪ. ﺑﻪ ﭼﻴﻨﺶ ﻋﻠﻰ ﻧﮕﺎﻩ ﻛﻨﻴﺪ ﻭ ﺗﻮﺿﻴﺢ ﺩﻫﻴﺪ ﻛﻪ ﺍﻭ ﺑﺮﺍﻯ ﭼﻴﺪﻥ 25 ﻣﺮﺑﻊ ﺑﻪ ﺗﺮﺗﻴﺐ ﭼﻪ ﻛﺎﺭ ﺧﻮﺍﻫﺪ ﻛﺮﺩ؟ ﺍﻭ ﭼﻨﺪ ﭼﻮﺏ ﻛﺒﺮﻳﺖ ﺭﺍ ﺑﻪ ﺻﻮﺭﺕ ﺍﻓﻘﻰ ﻗﺮﺍﺭ ﻣﻰﺩﻫﺪ؟ ﭼﻨﺪ ﭼــﻮﺏ ﻛﺒﺮﻳﺖ ﺭﺍ ﺑﻪ ﺻﻮﺭﺕ ﻋﻤﻮﺩﻯ ﻗﺮﺍﺭ ﻣﻰﺩﻫﺪ ﺗﺎ ﺍﻟﮕﻮ ﺭﺍ ﻛﺎﻣﻞ ﻛﻨﺪ؟ ﻋﻠــﻰ ﺑﺮﺍﻯ ﺳــﺎﺧﺘﻦ 25ﻣﺮﺑــﻊ ﺍﺯ ﭼﻨﺪ ﭼﻮﺏ ﻛﺒﺮﻳﺖ ﺍﺳــﺘﻔﺎﺩﻩ ﻣﻰﻛﻨﺪ؟ ﺑــﻪ ﭼﻴﻨﺶ ﻣﺎﻧﻰ ﻧــﮕﺎﻩ ﻛﻨﻴﺪ ﻭ ﺗﻮﺿﻴﺢ ﺩﻫﻴــﺪ ﺍﻭ ﺑﻪ ﭼﻪ ﺗﺮﺗﻴﺒﻰ ﻋﻤﻞ ﻣﻰﻛﻨﺪ ﻭ ﺍﺯ ﭼﻨﺪ ﭼﻮﺏ ﻛﺒﺮﻳﺖ ﺑﺮﺍﻯ ﺳﺎﺧﺘﻦ 25ﻣﺮﺑﻊ ﺍﺳﺘﻔﺎﺩﻩ ﻣﻰﻛﻨﺪ؟ ﺍﻣﻴــﺮ ﭼﮕﻮﻧــﻪ ﻋﻤﻞ ﻣﻰﻛﻨــﺪ ﻭ ﺍﺯ ﭼﻨﺪ ﭼﻮﺏ ﻛﺒﺮﻳﺖ ﺍﺳــﺘﻔﺎﺩﻩ ﻣﻰﻛﻨﺪ؟
ﻗﺪﻡ ﺻﺪﻡ ﻫﺮﻳﻚ ﺍﺯ ﺍﻳﻦ ﺳــﻪ ﻧﻔﺮ ﺑﺮﺍﻯ ﺳــﺎﺧﺘﻦ 100ﻣﺮﺑﻊ ﺑﺎ ﻫﻤﺎﻥ ﺍﻟﮕﻮﻯ ﺑﺎﻻ ﺍﺯ ﭼﻨﺪ ﭼﻮﺏ ﻛﺒﺮﻳﺖ ﺍﺳﺘﻔﺎﺩﻩ ﺧﻮﺍﻫﻨﺪ ﻛﺮﺩ؟ ﺑﺮﺍﻯ ﺳﺎﺧﺘﻦ ﻳﻚ ﻣﻴﻠﻴﻮﻥ ﻣﺮﺑﻊ ﭼﻪﻃﻮﺭ؟
ﻗﺪﻡ nﺍﻡ ) + (3×6ﭼﻮﺏ ﻛﺒﺮﻳﺖ
ﺩﺭ ﺍﻳــﻦ ﻗﺪﻡ ﺑﻪ ﺩﻧﺒــﺎﻝ ﻋﺒﺎﺭﺗﻰ ﻣﻰﮔﺮﺩﻳﻢ ﻛﻪ ﺑﺎ ﺍﺳــﺘﻔﺎﺩﻩ ﺍﺯ ﺁﻥ ﺑﺘﻮﺍﻧﻴﻢ ﺗﻌﺪﺍﺩ ﭼﻮﺏ ﻛﺒﺮﻳﺖﻫﺎﻳﻰ ﺭﺍ ﻛﻪ ﺑﺮﺍﻯ ﺳﺎﺧﺘﻦ nﻣﺮﺑﻊ ﺑﺎ ﺍﻟﮕﻮﻯ ﺑﺎﻻ ﺍﺳــﺘﻔﺎﺩﻩ ﻣﻰﺷــﻮﺩ ﭘﻴﺪﺍ ﻛﻨﻴﻢ n .ﻣﺘﻐﻴﺮ ﺍﺳﺖ ﻭ ﻣﻤﻜﻦ ﺍﺳﺖ ﻫﺮ
ﺩﻭﺭﻩﻯ ﺷﺎﻧﺰﺩﻫﻢ ،ﺷﻤﺎﺭﻩ ﻯ ،2ﺯﻣﺴﺘﺎﻥ 1389
ﺭﺍﻫﻨﻤﺎﻳﻲ
43
ﻳﻚ ﺍﺯ ﺍﻋﺪﺍﺩ ﻃﺒﻴﻌﻰ 4 ،3 ،2 ،1ﻭ ...ﺑﺎﺷﺪ .ﺑﺎﻳﺪ ﻋﺒﺎﺭﺕ ﻣﻮﺭﺩ ﻧﻈﺮ ﺭﺍ ﻃــﻮﺭﻯ ﭘﻴﺪﺍ ﻛﻨﻴﻢ ﻛﻪ ﺑﺎ ﻗﺮﺍﺭ ﺩﺍﺩﻥ ﻫــﺮ ﻳﻚ ﺍﺯ ﺍﻳﻦ ﺍﻋﺪﺍﺩ ﺑﻪ ﺟﺎﻯ n ﺗﻌﺪﺍﺩ ﭼﻮﺏ ﻛﺒﺮﻳﺖﻫﺎ ﺭﺍ ﺑﻪ ﺩﺭﺳﺘﻰ ﻧﺸﺎﻥ ﺩﻫﺪ. ﻳﻚ ﺑــﺎﺭ ﺩﻳﮕﺮ ﺑﻪ ﭼﻴﻨﺶ ﻋﻠــﻰ ﻧﮕﺎﻩ ﻛﻨﻴﺪ :ﻋﻠﻰ ﺑﺮﺍﻯ ﺳــﺎﺧﺘﻦ nﻣﺮﺑــﻊ ﺍﺑﺘﺪﺍ nﭼــﻮﺏ ﻛﺒﺮﻳﺖ ﺭﺍ ﺑــﻪ ﺻﻮﺭﺕ ﺍﻓﻘﻰ ﻗــﺮﺍﺭ ﻣﻰﺩﻫﺪ، ﺳــﭙﺲ nﭼﻮﺏ ﻛﺒﺮﻳﺖ ﺩﻳﮕﺮ ﺭﺍ ﺯﻳــﺮ ﺁﻥﻫﺎ ﻣﻰﮔﺬﺍﺭﺩ ﻭ ﺩﺭ ﺁﺧﺮ n+1 ﭼــﻮﺏ ﻛﺒﺮﻳﺖ ﺭﺍ ﺑﻪ ﺻﻮﺭﺕ ﻋﻤﻮﺩﻯ ﻗــﺮﺍﺭ ﻣﻰﺩﻫﺪ .ﭘﺲ ﺭﻭﻯ ﻫﻢ ﺍﺯ ) n + n + (n + 1ﭼﻮﺏ ﻛﺒﺮﻳﺖ ﺍﺳﺘﻔﺎﺩﻩ ﻣﻰﻛﻨﺪ. ﺣــﺎﻻ ﭼﻴﻨﺶ ﻣﺎﻧﻰ ﺭﺍ ﻣﺮﻭﺭ ﻛﻨﻴﺪ .ﻣﺎﻧﻰ ﺍﺑﺘﺪﺍ ﺳــﻪ ﭼﻮﺏ ﻛﺒﺮﻳﺖ ﻣﺮﺑــﻊ ﺍﻭﻝ ﻗﺮﺍﺭ ﻣﻰﺩﻫﺪ ﻭ ﺍﻳﻦ ﻛﺎﺭ ﺭﺍ ﺑــﺮﺍﻯ ﺑﻘﻴﻪﻯ ﻣﺮﺑﻊﻫﺎ ﻫﻢ ﺗﻜﺮﺍﺭ ﻣﻰﻛﻨﺪ ﻭ ﺩﺭ ﺁﺧﺮ ،ﺁﺧﺮﻳــﻦ ﭼﻮﺏ ﻛﺒﺮﻳﺖ ﺭﺍ ﻗﺮﺍﺭ ﻣﻰﺩﻫﺪ .ﭘﺲ ﺭﻭﻯ ﻫﻢ ﺍﺯ 3n+1ﭼﻮﺏ ﻛﺒﺮﻳﺖ ﺍﺳﺘﻔﺎﺩﻩ ﻣﻰﻛﻨﺪ. ﭼﻴﻨــﺶ ﺍﻣﻴﺮ ﺭﺍ ﻫﻢ ﻳﻚ ﺑﺎﺭ ﺩﻳﮕﺮ ﺑﺮﺭﺳــﻰ ﻛﻨﻴــﺪ :ﺍﻣﻴﺮ ﺍﺑﺘﺪﺍ 4 ﭼﻮﺏ ﻛﺒﺮﻳﺖ ﻣﺮﺑﻊ ﺍﻭﻝ ﺭﺍ ﻗﺮﺍﺭ ﻣﻰﺩﻫﺪ .ﺳﭙﺲ ﺑﺮﺍﻯ ﻫﺮ ﻳﻚ ﺍﺯ n-1 ﻣﺮﺑﻊ ﺑﻌﺪﻯ 3 ،ﭼﻮﺏ ﻛﺒﺮﻳﺖ ﺍﺳــﺘﻔﺎﺩﻩ ﻣﻰﻛﻨﺪ ﺗﺎ ﺍﻟﮕﻮ ﻛﺎﻣﻞ ﺷــﻮﺩ، ﻳﻌﻨﻰ ﺭﻭﻯ ﻫﻢ ﺍﺯ ) 4 + 3(n − 1ﭼﻮﺏ ﻛﺒﺮﻳﺖ ﺑﺮﺍﻯ ﺳﺎﺧﺘﻦ nﻣﺮﺑﻊ ﺍﺳﺘﻔﺎﺩﻩ ﻣﻰﻛﻨﺪ. ﺁﻳﺎ ﺳﻪ ﻋﺒﺎﺭﺕ ﺑﻪ ﺩﺳﺖ ﺁﻣﺪﻩ ﺑﺎ ﻫﻢ ﺑﺮﺍﺑﺮﻧﺪ؟
44
ﺭﺍﻫﻨﻤﺎﻳﻲ
ﺩﻭﺭﻩﻯ ﺷﺎﻧﺰﺩﻫﻢ ،ﺷﻤﺎﺭﻩﻯ ،2ﺯﻣﺴﺘﺎﻥ 1389
ﻗﺪﻡﻫﺎﯼ ﺑﻌﺪﯼ ﺑﻪ ﻫﻤﻴﻦ ﺗﺮﺗﻴﺐ ﻣﻰﺗﻮﺍﻧﻴﺪ ﺑﺮﺍﻯ ﺍﻟﮕﻮﻫﺎﻯ ﺩﻳﮕﺮ ﻫﻢ ﻋﺒﺎﺭﺗﻰ ﺑﻴﺎﺑﻴﺪ ﻛﻪ ﺑﺎ ﺍﺳﺘﻔﺎﺩﻩ ﺍﺯ ﺁﻥ ﺑﺘﻮﺍﻥ ﺑﺎ ﺗﻐﻴﻴﺮ nﺑﻪ ﭘﺎﺳﺦﻫﺎﻯ ﻣﻮﺭﺩ ﻧﻈﺮ ﺭﺳﻴﺪ. ﺑﺮﺍﻯ ﻣﺜﺎﻝ ﺑﻪ ﺷﻜﻞ ﺯﻳﺮ ﻧﮕﺎﻩ ﻛﻨﻴﺪ:
ﺑﺮﺍﻯ ﺳــﺎﺧﺘﻦ ﻳﻚ ﻣﺴﺘﻄﻴﻞ 2×3ﺍﺯ 17ﭼﻮﺏ ﻛﺒﺮﻳﺖ ﺍﺳﺘﻔﺎﺩﻩ ﺷــﺪﻩ ﺍﺳﺖ .ﺍﻳﻦ ﻣﺴــﺘﻄﻴﻞ ﻣﺤﻴﻄﻰ ﺑﺮﺍﺑﺮ ﺑﺎ ﻃﻮﻝ 10ﭼﻮﺏ ﻛﺒﺮﻳﺖ ﺩﺍﺭﺩ .ﺩﺭ ﺍﻳﻦ ﺷﻜﻞ 12ﻧﻘﻄﻪﻯ ﺗﻘﺎﻃﻊ ﺩﻳﺪﻩ ﻣﻰﺷﻮﺩ. ﺗﺮﺗﻴﺒﻰ ﺑــﺮﺍﻯ ﭼﻴﻨﺶ ﭼــﻮﺏ ﻛﺒﺮﻳﺖﻫﺎ ﺩﺭ ﻧﻈــﺮ ﺑﮕﻴﺮﻳﺪ .ﺑﺮﺍﻯ ﺳﺎﺧﺘﻦ ﻳﻚ ﻣﺴــﺘﻄﻴﻞ 2×25ﭼﻨﺪ ﻛﺒﺮﻳﺖ ﻻﺯﻡ ﺍﺳﺖ؟ ﻣﺤﻴﻂ ﺍﻳﻦ ﻣﺴﺘﻄﻴﻞ ﭼﻪ ﻗﺪﺭ ﺍﺳﺖ؟ ﭼﻨﺪ ﻧﻘﻄﻪﻯ ﺗﻘﺎﻃﻊ ﺩﺭ ﺁﻥ ﺩﻳﺪﻩ ﻣﻰﺷﻮﺩ؟ ﻳﻚ ﻣﺴﺘﻄﻴﻞ 2×100ﭼﻪ ﻃﻮﺭ؟ ﻳﻚ ﻣﺴﺘﻄﻴﻞ 2×nﭼﻪ ﻃﻮﺭ؟ ﺍﮔﺮ ﭘﺎﺳــﺦ 5 n +2ﺭﺍ ﺑﺮﺍﻯ ﺗﻌﺪﺍﺩ ﭼــﻮﺏ ﻛﺒﺮﻳﺖﻫﺎ 2n +4 ،ﺭﺍ ﺑﺮﺍﻯ ﻣﺤﻴﻂ ﻣﺴﺘﻄﻴﻞ ﻭ 3 n +3ﺭﺍ ﺑﺮﺍﻯ ﺑﺮﺍﻯ ﺗﻌﺪﺍﺩ ﻧﻘﺎﻁ ﺗﻘﺎﻃﻊ ﺑﻪ ﺩﺳﺖ ﺁﻭﺭﺩﻩﺍﻳﺪ ﺑﺪﺍﻧﻴﺪ ﻛﻪ ﺩﺭﺳﺖ ﻋﻤﻞ ﻛﺮﺩﻩﺍﻳﺪ!
ﭘﺎﯾﻪﻫﺎی ﺩﻭﻡ ﻭ ﺳﻮﻡ ﺭﺍﻫﻨﻤﺎﯾﯽ
ﺗﺮﺟﻤﻪ
ﻯ ﺳﭙﻴﺪﻩ ﭼﻤﻦﺁﺭﺍ
ﻛﻠﻴﺪﻭﺍژﻩﻫﺎ :ﻣﺴﺎﺑﻘﻪﻯ ﺭﻳﺎﺿﻰ ،ﺯﺍﻭﻳﻪ ،ﻣﺮﺑﻊ ،ﻣﺤﻴﻂ ،ﺍﺳﺘﺮﺍﻟﻴﺎ . ﺳـﺆﺍﻝﻫﺎﻯ 1ﺗـﺎ ،10ﻫﺮﻳـﻚ 3ﺍﻣﺘﻴـﺎﺯ ﺩﺍﺭﺩ. .1ﺣﺎﺻﻞ ﻋﺒﺎﺭﺕ 2008+8002ﭼﻴﺴﺖ؟ ﺍﻟﻒ( 1010 ﺏ( 4004 پ( 10008 ﺕ( 8910 ﺙ( 10010
.3ﻣﺤﻴﻂ ﺷــﻜﻞ ﻣﻘﺎﺑﻞ ﺑﺮ ﺣﺴﺐ ﺳﺎﻧﺘﻰﻣﺘﺮ ﭼﻴﺴﺖ؟ 2CM
2CM
4CM
4CM
.2ﺩﺭ ﻣﻴﺎﻥ ﺍﻋﺪﺍﺩ ﺯﻳﺮ ،ﺑﺰﺭگﺗﺮﻳﻦ ﻋﺪﺩ ﻛﺪﺍﻡ ﺍﺳﺖ؟ ﺏ( 2/2 ﺍﻟﻒ( 2/15 ﺕ( 2/1 پ( 2/08 ﺙ( 2/185
ﺏ( 10 ﺕ(16
ﺍﻟﻒ( 8 پ(12 ﺙ( 20 ِ ﻧﺼﻒ .4 ﺍﻟﻒ( 1 2 1 پ( 99 4 3 ﺙ( 99 4
1 2
95
199
ﭼﻴﺴﺖ؟ ﺏ( ﺕ(
3 4 1 99 2
ﺷﻤﺎﺭﻩ ﻯ ،2ﺯﻣﺴﺘﺎﻥ 1389 ﻧﺰﺩﻫﻢ ،ﺷ ﻫﻢ، ﺷﺎﻧﺰﺩﻫﻢ، ﺰﺩﻫﻢ ﻧﺰﺩﺩ ﺷﺎﻧﺰ ﺩﻭﺭﻩﻯ ﺷ ﻭﺭﻩ ﺩﻭﺭ
95
ﺭﺍﻫﻨﻤﺎﻳﻲ ﺭﺍﻫﻨﻤﺎﻳﻰ
45
ﺳﺆﺍﻝﻫﺎﯼ ﻣﺴﺎﺑﻘﻪﺍﯼ
) 31ژﻭﻻﻯ (2008
.5ﺩﺭ ﺷﻜﻞ ﺯﻳﺮ ،ﻣﻘﺪﺍﺭ xﺑﺮﺍﺑﺮ ﺍﺳﺖ ﺑﺎ
˚135
˚X
ﺏ( 95° ﺕ( 55°
ﺍﻟﻒ( 135° پ( 35° ﺙ( 45°
.7ﺩﺭ ﺷــﻜﻞ ﺯﻳﺮ ،ﭼﻨﺪ ﻣﺮﺑــﻊ )ﺑﺎ ﺍﻧﺪﺍﺯﻩﻫﺎﻯ ﻣﺨﺘﻠﻒ( ﻣﻰﺑﻴﻨﻴﺪ؟ 2 1 1 1
1
1
1 1
ﺍﻟﻒ( 9 پ( 12 ﺙ( 16
1
ﺏ( 11 ﺕ( 14
.8ﻗﻄــﺎﺭﻯ ﺩﺭ ﺳــﺎﻋﺖ 8:58ﺻﺒﺢ ،ﺷــﻬﺮ ﺯﻧﺠــﺎﻥ ﺭﺍ ﺗﺮﻙ ﻛﺮﺩ ﻭ ﺳــﺎﻋﺖ 9:34ﺑﻪ ﺗﺒﺮﻳﺰ ﺭﺳــﻴﺪ .ﭼﻨﺪ ﺩﻗﻴﻘــﻪ ﺩﺭ ﺭﺍﻩ ﺑﻮﺩﻩ ﺍﺳﺖ؟ ﺏ( 22 ﺍﻟﻒ( 82 ﺕ( 38 پ( 36 ﺙ( 78 .9ﺑﺎ ﺍﺳــﺘﻔﺎﺩﻩ ﺍﺯ ﺭﻗﻢﻫــﺎﻯ ،9 ،8 ،7 ،6 ،5 ﻋﺪﺩﻫﺎﻯ ﭘﻨﺞ ﺭﻗﻤﻰ ﺯﻭﺝ ﺳﺎﺧﺘﻴﻢ .ﺭﻗﻢ
46
ﺭﺍﻫﻨﻤﺎﻳﻲ
Q
S
˚20
L P
M ˚X
.6ﺣﺎﺻﻞ ﻋﺒﺎﺭﺕِ ، 200×8ﺑﺮﺍﺑﺮ ﺍﺳﺖ ﺑﺎ 200 ÷8 ﺍﻟﻒ( 1 پ( 16 ﺏ(8 ﺕ(64 ﺙ(200
2
ﺩﻫــﮕﺎﻥ ﺑﺰﺭگﺗﺮﻳﻦ ﻋﺪﺩﻯ ﻛﻪ ﺳــﺎﺧﺘﻪﺍﻳﻢ ﭼﻴﺴﺖ؟ ﺏ( 6 ﺍﻟﻒ( 5 ﺕ(8 پ( 7 ﺙ( 9
ﺍﻟﻒ( 70 پ( 90 ﺙ( 110
ﺏ( 80 ﺕ( 100
PQRS .10ﻳــﻚ ﻣﺮﺑــﻊ ﺍﺳــﺖ ﻭ ﻧﻘﺎﻁ E ﻭ Fﺩﺭ ﺧــﺎﺭﺝ ﺍﺯ ﻣﺮﺑــﻊ ﻗــﺮﺍﺭ ﺩﺍﺭﻧﺪ ﺑﻪ ﻃﻮﺭﻯ ﻛﻪ ﻣﺜﻠﺚﻫــﺎﻯ PQEﻭ QRF ﻣﺘﺴــﺎﻭﻯﺍﻻﺿﻼﻉ ﻫﺴــﺘﻨﺪ .ﺍﻧــﺪﺍﺯﻩﻯ ﺯﺍﻭﻳﻪﻯ ∠EQFﭼﻨﺪ ﺩﺭﺟﻪ ﺍﺳﺖ؟ ﺍﻟﻒ( 60 ﺏ(90 پ( 120 ﺕ( 150 ﺙ( 180 ﺳـﺆﺍﻝﻫﺎﻯ 11ﺗـﺎ ،20ﻫﺮﻳـﻚ 4ﺍﻣﺘﻴﺎﺯ ﺩﺍﺭﻧﺪ. .11ﻣﺴــﺘﻄﻴﻠﻰ ﻛــﻪ ﻃــﻮﻝ ﺁﻥ ،ﺩﻭ ﺑﺮﺍﺑــﺮ ﻋﺮﺿﺶ ﺍﺳﺖ ،ﻣﺴﺎﺣﺘﺶ 72ﺳﺎﻧﺘﻰﻣﺘﺮ ﻣﺮﺑﻊ ﺍﺳــﺖ .ﻣﺤﻴﻂ ﺍﻳﻦ ﻣﺴﺘﻄﻴﻞ ﭼﻨﺪ ﺳﺎﻧﺘﺘﻰﻣﺘﺮ ﺍﺳﺖ؟ ﺏ( 36 ﺍﻟﻒ( 34 ﺕ( 48 پ( 42 ﺙ( 54 .12ﺩﺭ ﻳﻚ ﻛﺎﺳــﻪ ﺗﻌﺪﺍﺩﻯ ﺁﺏﻧﺒﺎﺕ ﺑﺎ ﺳــﻪ ﺭﻧﮓ ﻣﺘﻔــﺎﻭﺕ ﺭﻳﺨﺘﻪﺍﻳﻢ 2 .ﺁﺏﻧﺒﺎﺕﻫﺎ 5 ﻗﺮﻣــﺰ 1 ،ﺁﻥﻫﺎ ﺳــﺒﺰ ﻭ 12ﺗﺎﻯ ﺑﻘﻴﻪ 3 ﺯﺭﺩﻧــﺪ .ﺗﻌﺪﺍﺩ ﻫﻤــﻪﻯ ﺁﺏﻧﺒﺎﺕﻫﺎ ﺭﻭﻯ ﻫﻢ ﭼﻪﻗﺪﺭ ﺍﺳﺖ؟ ﺍﻟﻒ( 30 پ( 54 ﺙ( 90
ﺏ( 45 ﺕ( 60
.13ﺩﺭ ﺷــﻜﻞ ،ﻣﺜﻠـــﺚﻫــﺎﻯ PQRﻭ LMNﻫــﺮ ﺩﻭ ﻣﺘﺴــﺎﻭﻯﺍﻻﺿﻼﻉﺍﻧﺪ ﻭ . ∠QSM = 20ﻣﻘــﺪﺍﺭ Xﭼﻪﻗــﺪﺭ ﺍﺳﺖ؟
ﺩﻭﺭﻩﻯ ﺷﺎﻧﺰﺩﻫﻢ ،ﺷﻤﺎﺭﻩﻯ ،2ﺯﻣﺴﺘﺎﻥ 1389
R
N
.14ﺩﺭ ﻳﻚ ﻧﻴﻤﻪﻯ ﺑﺎﺯﻯ ﻓﻮﺗﺒﺎﻝ ﺑﻴﻦ ﺭﺍﻩﺁﻫﻦ ﺍﺻﻔﻬﺎﻥ ﻭ ﺗﺮﺍﻛﺘﻮﺭﺳﺎﺯﻯ ﺗﺒﺮﻳﺰ ،ﺍﻣﺘﻴﺎﺯﻫﺎ 1ﺑﺮ 0ﺑﻪ ﻧﻔــﻊ ﺭﺍﻩﺁﻫﻦ ﺍﺻﻔﻬﺎﻥ ﺑﻮﺩ .ﺩﺭ ﻧﻴﻤــﻪﻯ ﺩﻭﻡ ﺍﻳــﻦ ﺑﺎﺯﻯ ،ﺳــﻪ ﮔﻞ ﺯﺩﻩ ﺷــﺪ .ﻛﺪﺍﻡ ﻳﻚ ﺍﺯ ﻣــﻮﺍﺭﺩ ﺯﻳﺮ ﻧﻤﻰﺗﻮﺍﻧﺪ ﻧﺘﻴﺠﻪﻯ ﺍﻳﻦ ﺑﺎﺯﻯ ﺑﺎﺷﺪ. ﺍﻟﻒ( ﺑﺎﺯﻯ ﺑﻪ ﺗﺴﺎﻭﻯ ﺍﻧﺠﺎﻣﻴﺪ. ﺏ( ﺭﺍﻩﺁﻫــﻦ ﺍﺻﻔﻬﺎﻥ ﺑﺎ 2ﮔﻞ ﺑﻴﺶﺗﺮ ﺑﺮﻧﺪﻩ ﺷﺪ. پ( ﺗﺮﺍﻛﺘﻮﺭﺳــﺎﺯﻯ ﺗﺒﺮﻳﺰ ﺑــﺎ 2ﮔﻞ ﺑﻴﺶﺗﺮ ﺑﺮﻧﺪﻩ ﺷﺪ. ﺕ( ﺭﺍﻩﺁﻫﻦ ﺍﺻﻔﻬﺎﻥ ﺑﺎ ﻳﻚ ﮔﻞ ﺑﻴﺶﺗﺮ ﺑﺮﻧﺪﻩ ﺷﺪ. ﺙ( ﺭﺍﻩﺁﻫــﻦ ﺍﺻﻔﻬﺎﻥ ﺑﺎ 4ﮔﻞ ﺑﻴﺶﺗﺮ ﺑﺮﻧﺪﻩ ﺷﺪ. .15ﺑــﻪ ﭼﻨﺪ ﻃﺮﻳﻖ ﻣﺨﺘﻠــﻒ ﻣﻰﺗﻮﺍﻥ ﻋﺪﺩ 12ﺭﺍ ﺑﻪ ﺻــﻮﺭﺕ ﻣﺠﻤﻮﻉ ﺩﻭ ﻳﺎ ﺑﻴﺶ ﺍﺯ ﺩﻭ ﻋﺪﺩ ﻃﺒﻴﻌﻰ ﻣﺨﺘﻠﻒ ﻧﻮﺷﺖ؟ )ﺗﻐﻴﻴﺮ ﺗﺮﺗﻴﺐ ﺍﻋــﺪﺍﺩ ﺩﺭ ﺟﻤﻊ ،ﺣﺎﻟﺖ ﺟﺪﻳﺪ ﺑﻪ ﺣﺴﺎﺏ ﻧﻤﻰﺁﻳﺪ(. ﺏ( 13 ﺍﻟﻒ( 12 ﺕ( 15 پ( 14 ﺙ( 16 .16ﭼﻨــﺪ ﻋﺪﺩ ﺻﺤﻴﺢ ﻣﺜﺒﺖ ﻣﺨﺘﻠﻒ ﺩﺍﺭﻳﻢ ﺣﺎﺻﻞ ﺿﺮﺏ ﺩﻭ ﻋﺪﺩ ﻓﺮﺩ ﻳﻚ ﺭﻗﻤﻰ ﻛﻪ ِ ﺑﺎﺷﻨﺪ؟ ﺏ( 15 ﺍﻟﻒ( 25 ﺕ( 13 پ( 14 ﺙ( 11
.17ﻣﺤﻴﻂ ﻣﺴــﺘﻄﻴﻞ ﺑﺰﺭگﺗــﺮ 700ﻣﺘﺮ ﺍﺳﺖ .ﺁﻥ ﺭﺍ ﺑﻪ ﺷﺶ ﻣﺴﺘﻄﻴﻞ ﻛﻮﭼﻚﺗﺮ ﻭ ﻳﻜﺴﺎﻥ ﻣﻄﺎﺑﻖ ﺷﻜﻞ ،ﺗﻘﺴﻴﻢ ﻛﺮﺩﻩﺍﻳﻢ. ﻣﺤﻴــﻂ ﻫﺮﻳــﻚ ﺍﺯ ﺷــﺶ ﻣﺴــﺘﻄﻴﻞ ﻛﻮﭼﻚﺗﺮ ،ﭼﻨﺪ ﻣﺘﺮ ﺍﺳﺖ؟
1 ﺍﻟﻒ( 3 پ(200
ﺏ( 300 ﺕ(150
116
ﺙ( 600 .18ﻛﻤﺪﻫــﺎﻯ )ﻗﻔﺴــﻪﻫﺎﻯ( ﺩﺍﻧﺶﺁﻣﻮﺯﺍﻥ ﻣﺪﺭﺳــﻪﺍﻯ ،ﺍﺯ ﺷــﻤﺎﺭﻩﻯ 1ﺗــﺎ 500 ﺷﻤﺎﺭﻩﮔﺰﺍﺭﻯ ﺷــﺪﻩ ﺍﺳﺖ .ﺍﻳﻦ ﺷﻤﺎﺭﻩﻫﺎ ﺑﺎ ﺍﺭﻗﺎﻡ ﭘﻼﺳــﺘﻴﻜﻰ ﻛﻪ ﻫﺮ ﻳﻚ 5ﺭﻳﺎﻝ ﻗﻴﻤﺖ ﺩﺍﺭﺩ ،ﺭﻭﻯ ﻛﻤﺪﻫﺎ ﭼﺴﺒﺎﻧﺪﻩ ﺷﺪﻩ ﺍﺳــﺖ .ﻗﻴﻤﺖ ﻫﻤﻪﻯ ﺭﻗﻢﻫﺎﻯ ﭼﺴﺒﺎﻧﺪﻩ ﺷﺪﻩ ﺭﻭﻯ ﻛﻤﺪﻫﺎ ﺑﺮﺍﺑﺮ ﺍﺳﺖ ﺑﺎ: ﺍﻟﻒ( 250ﺗﻮﻣﺎﻥ ﺏ( 63/25ﺗﻮﻣﺎﻥ پ( 69/50ﺗﻮﻣﺎﻥ ﺕ( 69/60ﺗﻮﻣﺎﻥ ﺙ( 85ﺗﻮﻣﺎﻥ .19ﺧﺎﻧﻪﻫﺎﻯ ﺟــﺪﻭﻝ ﺯﻳﺮ ﺭﺍ ﺑﺎ ﺍﺭﻗﺎﻡ 1ﻭ 2 ﻭ 3ﻭ 4ﭘﺮﻛﺮﺩﻩﺍﻳــﻢ ﺑﻪ ﻃﻮﺭﻯ ﻛﻪ ﺩﺭ ﻫﺮ ﺭﺩﻳﻒ ﻭ ﻫﺮ ﺳــﺘﻮﻥ ﻭ ﻫﺮ ﻗﻄﺮ ،ﻫﺮ ﻋﺪﺩ ﻓﻘﻂ ﻳﻚ ﺑﺎﺭ ﻇﺎﻫﺮ ﻣﻰﺷﻮﺩ .ﺑﺰﺭگﺗﺮﻳﻦ ﻣﻘﺪﺍﺭ ﻣﻤﻜﻦ ﺑﺮﺍﻯ Yﻭ Xﭼﻴﺴﺖ؟ ﺏ( 5 ﺍﻟﻒ( 4 ﺕ( 7 پ( 6 ﺙ( 8 1
2 X Y
3
.20ﻣﻴﺎﻧﮕﻴﻦ ﭼﻨﺪ ﻋﺪﺩ 4 ،ﺍﺳــﺖ .ﻣﻴﺎﻧﮕﻴﻦ ﮔــﺮﻭﻩ ﺩﻳﮕﺮﻯ ﺍﺯ ﺍﻋﺪﺍﺩ ﺑﻪ ﺗﻌﺪﺍﺩ ﺩﻭ ﺑﺮﺍﺑﺮ ﺍﻋﺪﺍﺩ ﺩﺳــﺘﻪﻯ ﺍﻭﻝ 10 ،ﺍﺳــﺖ .ﺍﮔﺮ ﺩﻭ ﮔــﺮﻭﻩ ﻋﺪﺩ ﺭﺍ ﺭﻭﻯ ﻫﻢ ﺑﺮﻳﺰﻳﻢ ،ﻣﻴﺎﻧﮕﻴﻦ ﺁﻥﻫﺎ ﭼﻨﺪ ﻣﻰﺷﻮﺩ؟ ﺏ( 6 ﺍﻟﻒ( 5 ﺕ( 8 پ( 7 ﺙ( 9 ﺳـﺆﺍﻝﻫﺎﻯ 21ﺗـﺎ 25ﻫﺮﻳـﻚ 5ﺍﻣﺘﻴﺎﺯ ﺩﺍﺭﻧﺪ. .21ﻣﻜﻌﺒﻰ ﺑﻪ ﺿﻠﻊ 2ﻣﺘﺮ ﺭﺍ ﺑﻪ ﻣﻜﻌﺐﻫﺎﻳﻰ ﺑﻪ ﺿﻠــﻊ 5ﺳــﺎﻧﺘﻰﻣﺘﺮ ﺑﺮﻳﺪﻩﺍﻳــﻢ .ﺍﮔﺮ ﻫﻤــﻪﻯ ﺍﻳــﻦ ﻣﻜﻌﺐﻫــﺎﻯ ﻛﻮﭼﻚﺗﺮ ﺭﺍ ﻳﻜﻰ ﻳﻜــﻰ ﺭﻭﻯ ﻫﻢ ﺑﭽﻴﻨﻴﻢ ﻭ ﻳﻚ ﺑﺮﺝ ﺑﺴﺎﺯﻳﻢ ،ﺍﺭﺗﻔﺎﻉ ﺍﻳﻦ ﺑﺮﺝ ﺑﺮﺍﺑﺮ ﺍﺳﺖ ﺑﺎ ﺍﻟﻒ( 32 km ﺏ( 160km ﺕ( 3/2 km پ( 1600km ﺙ( 320km .22ﺑﺰﺭگﺗﺮﻳــﻦ ﻋــﺪﺩﻯ ﺭﺍ ﺩﺭ ﻧﻈﺮ ﺑﮕﻴﺮﻳﺪ ﻛﻪ ﺍﺯ 2008ﻛﻮﭼﻚﺗﺮ ﺍﺳﺖ ،ﻓﺮﺩ ﺍﺳﺖ، ﺑﺎﻗﻰﻣﺎﻧﺪﻩﻯ ﺗﻘﺴــﻴﻢ ﺁﻥ ﺑــﺮ 3ﻋﺪﺩ 2 ﺍﺳــﺖ ﻭ ﺑﺎﻗﻰﻣﺎﻧﺪﻩﻯ ﺗﻘﺴــﻴﻢ ﺁﻥ ﺑﺮ ،5 ﻋﺪﺩ 4ﺍﺳﺖ .ﻣﺠﻤﻮﻉ ﺭﻗﻢﻫﺎﻯ ﺍﻳﻦ ﻋﺪﺩ ﭼﻴﺴﺖ؟ ﺏ( 25 ﺍﻟﻒ( 26 ﺕ( 23 پ( 24 ﺙ( 22 .23ﻳﻚ ﻣﺰﺭﻋــﻪﺩﺍﺭ ،ﺩﻭ ﺗﺎﻧﻜﺮ ﺁﺏ ﺩﺍﺭﺩ .ﺁﺑﻰ ﻛﻪ ﻫﻨﮕﺎﻡ ﺑﺎﺭﻧﺪﮔﻰ ﺍﺯ ﺳــﻘﻒ ﺧﺎﻧﻪﻯ ﺍﻭ ﺟﺎﺭﻯ ﻣﻰﺷﻮﺩ ﺩﺭ ﺗﺎﻧﻜﺮ 100ﻛﻴﻠﻮﻟﻴﺘﺮﻯ ﺟﻤﻊ ﻣﻰﺷﻮﺩ ﻭ ﺁﺑﻰ ﻛﻪ ﻫﻨﮕﺎﻡ ﺑﺎﺭﻧﺪﮔﻰ ﺍﺯ ﺳــﻘﻒ ﺍﻧﺒــﺎﺭ ﻣﺰﺭﻋﻪ ﺟﺎﺭﻯ ﻣﻰﺷــﻮﺩ ﺩﺭ ﺗﺎﻧﻜــﺮ 25ﻛﻴﻠﻮﻟﻴﺘــﺮﻯ ﺟﻤــﻊﺁﻭﺭﻯ ﻣﻰﺷــﻮﺩ .ﺳﻄﺢ ﺳﻘﻒ ﺧﺎﻧﻪﻯ ﻣﺰﺭﻋﻪﺩﺍﺭ 200ﻣﺘﺮ ﻣﺮﺑﻊ ﺍﺳﺖ ،ﺩﺭ ﺣﺎﻟﻰﻛﻪ ﺳﻘﻒ ﺍﻧﺒﺎﺭ ﻣﺰﺭﻋــﻪﻯ ﺍﻭ ،ﻓﻘــﻂ 80ﻣﺘﺮ ﻣﺮﺑﻊ ﻣﺴﺎﺣﺖ ﺩﺍﺭﺩ .ﻫﻢﺍﻛﻨﻮﻥ ﺩﺭ ﺗﺎﻧﻜﺮ ﻣﺮﺑﻮﻁ
ﺑﻪ ﺧﺎﻧﻪﻯ ﻣﺰﺭﻋﻪﺩﺍﺭ 35 ،ﻛﻴﻠﻮﻟﻴﺘﺮ ﻭ ﺩﺭ ﺗﺎﻧﻜﺮ ﻣﺮﺑﻮﻁ ﺑــﻪ ﺍﻧﺒﺎﺭ 13 ،ﻛﻴﻠﻮﻟﻴﺘﺮ ﺁﺏ ﻫﺴــﺖ .ﺑﺎﺭﻧﺪﮔﻰ ﺷــﺮﻭﻉ ﻣﻰﺷــﻮﺩ ﻭ ﺍﻭ ﻣﻰﺧﻮﺍﻫﺪ ﺑﻴﺶﺗﺮﻳﻦ ﺁﺏ ﻣﻤﻜﻦ ﺭﺍ ﺟﻤﻊ ﻛﻨﺪ .ﺍﻭ ﺑﺎﻳﺪ: ﺍﻟــﻒ( ﺁﺏ ﺩﺍﺧﻞ ﺗﺎﻧﻜﺮ ﺍﻧﺒﺎﺭ ﺭﺍ ﺩﺭ ﺗﺎﻧﻜﺮ ﺧﺎﻧﻪ ﺧﺎﻟﻰ ﻛﻨﺪ. ﺏ( ﺗﺎﻧﻜــﺮ ﻣﺰﺭﻋــﻪ ﺭﺍ ﺑﺎ ﺁﺏِ ﺗﺎﻧﻜــﺮ ﺧﺎﻧﻪ ﭘﺮ ﻛﻨﺪ. پ( 10ﻛﻴﻠﻮﻟﻴﺘﺮ ﺍﺯ ﺁﺏ ﺗﺎﻧﻜﺮ ﻣﺮﺑﻮﻁ ﺑﻪ ﺧﺎﻧﻪ ﺭﺍ ﺩﺭ ﺗﺎﻧﻜﺮ ﺍﻧﺒﺎﺭ ﺑﺮﻳﺰﺩ. ﺕ( 10ﻛﻴﻠﻮﻟﻴﺘــﺮ ﺍﺯ ﺁﺏ ﺗﺎﻧﻜــﺮ ﺍﻧﺒــﺎﺭ ﺭﺍ ﺩﺭ ﺗﺎﻧﻜﺮ ﺧﺎﻧﻪ ﺑﺮﻳﺰﺩ. ﺙ( ﻫﻴﭻ ﻛﺎﺭ ﻧﻜﻨﺪ. .24ﻳــﻚ ﺗﺎﻧﻜــﺮ ﺁﺏ ﺑﺎ ﻗﺎﻋﺪﻩﻯ ﻣﺴــﺘﻄﻴﻞ ﺷــﻜﻞ ﺑــﻪ ﺍﺑﻌــﺎﺩ 100 cmﺩﺭ cm 200ﻭ ﻋﻤــﻖ 100 cmﺗــﺎ ﺍﺭﺗﻔﺎﻉ 50 ﺳــﺎﻧﺘﻰﻣﺘﺮﻯ ﺍﺯ ﺁﺏ ﭘﺮ ﺷﺪﻩ ﺍﺳﺖ .ﻳﻚ ﻗﻄﻌــﻪ ﻣﻜﻌﺐ ﻣﺴــﺘﻄﻴﻞ ﻓﻠــﺰﻯ ﺗﻮ ﭘﺮ ﻛــﻪ ﺍﺑﻌــﺎﺩﺵ cm80ﺩﺭ 100 cmﺩﺭ 60 cmﺍﺳــﺖ ﺩﺭﻭﻥ ﺁﺏ ﺍﻳﻦ ﺗﺎﻧﻜﺮ ﻓﺮﻭ ﻣﻰﺑﺮﻳــﻢ ﺑﻪ ﻃﻮﺭﻯ ﻛﻪ ﻭﺟــ ِﻪ 80×100 ﺁﺏ ﺑﻪ ِ ﻛﻒ ﺗﺎﻧﻜﺮ ﺑﭽﺴﺒﺪ .ﺍﺭﺗﻔﺎﻋﻰ ﺍﺯ ﺁﺏ ﻛﻪ ﺑﺎﻻﻯ ﺍﻳﻦ ﻗﻄﻌﻪ ﻗﺮﺍﺭ ﻣﻰﮔﻴﺮﺩ ،ﭼﻨﺪ ﺳﺎﻧﺘﻰﻣﺘﺮ ﺍﺳﺖ؟ ﺏ( 14 ﺍﻟﻒ( 12 ﺕ( 18 پ( 16 ﺙ( 20
)ﭘﺎﺳـﺦ ﺳـﺆﺍﻻﺕ ﺭﺍ ﺑـﻪ ﻫﻤﺮﺍﻩ ﺣـﻞ ﺗﺸـﺮﻳﺤﻰ ﺗﻌـﺪﺍﺩﻯ ﺍﺯ ﺍﻳـﻦ ﻣﺴـﺎﺋﻞ ﻭ ﻧﻴﺰ ﭼﻨﺪ ﻣﺴﺌﻠﻪﻯ ﺩﻳﮕﺮ، ﺩﺭ ﺷـﻤﺎﺭﻩﻯ ﺁﻳﻨـﺪﻩﻯ ﺍﻳـﻦ ﻣﺠﻠﻪ ﺧﻮﺍﻫﻴﺪ ﺩﻳﺪ(.
ﺩﻭﺭﻩﻯ ﺷﺎﻧﺰﺩﻫﻢ ،ﺷﻤﺎﺭﻩ ﻯ ،2ﺯﻣﺴﺘﺎﻥ 1389
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47
ﻣﻌﺮﻓﯽ ﻛﺘﺎﺏ
ﺩﺍﻧﺶﺍﻓﺰﺍﻳﯽ
﹞︺︣﹁︀︐﹋ ﹩ب ﺟﻌﻔﺮ ﺭﺑﺎﻧﻰ
ﻟﺬﺕ ﺍﻧﺪﻳﺸﻪﯼ ﺭﻳﺎﺿﯽ ]ﭘﺮﺳﺶﻫﺎ، ﻣﻌﻤﺎﻫﺎ ﻭ ﺑﺎﺯﯼﻫﺎ ﺑﺮﺍﯼ ﻛﻮﺩﻛﺎﻥ ﻭ ﺑﺰﺭﮔﺴﺎﻻﻥ[ ﻧﻮﻳﺴﻨﺪﻩ :ﺍﺣﻤﺪ ﺳﻠﻴﻢ ﺳﻌﻴﺪﺍﻥ ﻣﺘﺮﺟﻢ :ﺳﺘﺎﺭ ﻋﻤﻮﺩﯼ ﻧﺎﺷﺮ :ﺷﺮﻛﺖ ﺍﻧﺘﺸﺎﺭﺍﺕ ﻋﻠﻤﯽ ﻭ ﻓﺮﻫﻨﮕﯽ، ﭼﺎﭖ ﺍﻭﻝ١٣٨٣ ، ﺍﺣﻤــﺪ ﺳــﻠﻴﻢ ﺳــﻌﻴﺪﺍﻥ )(1914-1991 ﺭﻳﺎﺿﻰﺩﺍﻥ ﻓﻠﺴــﻄﻴﻨﻰ ﺑﺮﺟﺴــﺘﻪﺍﻯ ﺑــﻮﺩ ﻛﻪ ﺩﺭ ﺟﻮﺍﻧﻰ ﭘﺲ ﺍﺯ ﺍﺷــﻐﺎﻝ ﺧﺎﻧﻪ ﻭ ﻛﺎﺷﺎﻧﻪﻯ ﭘﺪﺭﻯﺍﺵ ﺑﻪ ﺩﺳﺖ ﺍﺷﻐﺎﻟﮕﺮﺍﻥ ﺻﻬﻴﻮﻧﻴﺴﺖ ﺩﺭ ﺷﻬﺮ ﺁﻭﺍﺭﻩ ﺷﺪ ﻭ ﻧﺎﮔﺰﻳﺮ ﺟﻼﻯ ﻭﻃﻦ ﻛﺮﺩ .ﻧﺨﺴﺖ ﺑﻪ ﺳﻮﺩﺍﻥ ﺭﻓﺖ ﻭ ﭘﺲ ﺍﺯ ﭼﻨﺪ ﺳــﺎﻝ ﺑــﻪ ﺍﺭﺩﻥ ﺁﻣﺪ ﻭ ﺑﻪ ﺭﻳﺎﺳــﺖ ﺩﺍﻧﺸــﻜﺪﻩ ﻋﻠﻮﻡ ﻭ ﺳــﭙﺲ ﺩﺍﻧﺸــﮕﺎﻩ ﻗﺪﺱ ﺭﺳﻴﺪ. ﻭﻯ ﻣﺮﺩﻯ ﺷــﻴﻔﺘﻪ ﻋﻠﻢ ﻭ ﺩﺍﻧــﺶ ﺑﻮﺩ ﻭ ﺣﺪﻭﺩ 50 ﻛﺘﺎﺏ ﺩﺭﺳﻰ ﺑﺮﺍﻯ ﻣﺪﺍﺭﺱ ﺟﻬﺎﻥ ﻋﺮﺏ ﻧﻮﺷﺖ.ﻛﺘﺎﺏ ﻟﺬﺕ ﺍﻧﺪﻳﺸﻪﻯ ﺭﻳﺎﺿﻰ ،ﻫﻤﺎﻥﻃﻮﺭ ﻛﻪ ﺍﺯ ﺍﺳــﻤﺶ ﭘﻴﺪﺍﺳــﺖ ﺑﺎ ﻫﺪﻑ ﻋﻼﻗﻤﻨﺪ ﻛﺮﺩﻥ ﺩﺍﻧﺶﺁﻣﻮﺯﺍﻥ ﻭ ﻣﻌﻠﻤﺎﻥ ﺑﻪ ﺭﻳﺎﺿﻰ ﺗﺮﺟﻤﻪ ﺷﺪﻩ ﺍﺳﺖ. ﻛﺘﺎﺏ ﺷﺎﻣﻞ ﭼﻬﺎﺭ ﻓﺼﻞ ﺍﺳــﺖ .ﻧﻜﺘﻪﻫﺎﻯ ﻇﺮﻳﻒ ﺑﺎﺯﻯ ﺑﺎ ﺍﻋﺪﺍﺩ، ﭼﺸﻢﺑﻨﺪﻯﻫﺎﻯ ﺣﺴﺎﺑﻰ ﻣﺴﺎﺋﻞ ﻋﺮﺑﻰ. ﻋــﻼﻭﻩ ﺑﺮ ﺍﻳﻦ ﻳــﻚ ﻣﻘﺪﻣﻪ ﺩﺭ ﺷــﺮﻭﻉ ﻛﺎﺭﻫﺎﻯ ﻣﺆﻟــﻒ ﺩﺍﺭﺩ ﻛﻪ ﺧﻮﺍﻧﺪﻧــﻰ ﺍﺳــﺖ .ﺑﻪ ﻃﻮﺭ ﻛﻠﻰ ،ﺍﻳﻦ ﻛﺘﺎﺏ ﺑﻪ ﮔﻮﻧﻪﺍﻯ ﻧﻮﺷــﺘﻪ ﺷــﺪﻩ ﺍﺳــﺖ ﻛﻪ ﻣﻌﻠﻤﺎﻥ ﺭﻳﺎﺿﻰ ﻫﻢ ﻣﻰﺗﻮﺍﻧﺪ ﺑﺎ ﻃﺮﺡ ﭘﺎﺭﻩﺍﻯ ﺍﺯ ﻣﺴــﺎﺋﻞ ﺁﻥ 48
ﺭﺍﻫﻨﻤﺎﻳﻲ
ﺩﻭﺭﻩﻯ ﺷﺎﻧﺰﺩﻫﻢ ،ﺷﻤﺎﺭﻩﻯ ،2ﺯﻣﺴﺘﺎﻥ 1389
ﺩﺭ ﻛﻼﺱ ﺩﺍﻧﺶﺁﻣــﻮﺯﺍﻥ ﺧﻮﺩ ﺭﺍ ﺳﺮﮔﺮﻡ ﻛﻨﻨﺪ. ﻣﺴﺌﻠﻪﻯ ﺭﻭﻏﻦ 14ﻛﻴﻠــﻮ ﺭﻭﻏــﻦ ﺩﺍﺭﻡ ﻛﻪ ﻣﻰﺧﻮﺍﻫــﻢ ﺁﻥ ﺭﺍ ﺑــﻪ ﺩﻭ ﺑﺨﺶ ﻣﺴــﺎﻭﻯ ﺗﻘﺴــﻴﻢ ﻛﻨﻢ .ﻣﻦ ﺩﻭ ﻭﺯﻧﻪﻯ 9ﻛﻴﻠﻮﻳــﻰ ﻭ 5ﻛﻴﻠﻮﻳﻰ ﺩﺍﺭﻡ .ﭼﻪ ﻛﺎﺭ ﻛﻨﻢ؟ ﺯﻛﺎﺕ ﻣﺎﻝ ﻣﻰﺧﻮﺍﺳــﺘﻢ ﻣﺒﻠــﻎ ﺯﻛﺎﺕ ﺍﻣﻮﺍﻟــﻢ ﺭﺍ ﺑﻪ ﻃﻮﺭ ﻣﺴــﺎﻭﻯ ﺑﻴﻦ ﻓﻘﺮﺍ ﺗﻘﺴــﻴﻢ ﻛﻨــﻢ .ﻭﻗﺘﻰ ﺁﻥ ﺭﺍ ﺷــﻤﺮﺩﻡ ﺩﻳﺪﻡ ﺍﮔــﺮ ﺑﺨﻮﺍﻫﻢ ﺁﻥ ﺭﺍ ﻣﻴﺎﻥ 3ﻓﻘﺮ ﺗﻘﺴــﻴﻢ ﻛﻨﻢ ﻳﻚ ﺩﻳﻨﺎﺭ ﺑﺎﻗﻰ ﻣﻰﻣﺎﻧﺪ .ﺍﮔﺮ ﻣﻴﺎﻥ 5ﻧﻔﺮ ﺗﻘﺴــﻴﻢ ﻛﻨﻢ 4ﺩﻳﻨﺎﺭ ﻣﻰﻣﺎﻧﺪ ﻭ ﺍﮔﺮ ﻣﻴﺎﻥ 7ﻧﻔﺮ ﺗﻘﺴﻴﻢ ﻛﻨﻢ 3ﺩﻳﻨﺎﺭ ﻣﻰﻣﺎﻧﺪ .ﺩﺭ ﺍﻳﻦﺟﺎ ﻣﻦ ﻳﻚ ﺩﻳﻨﺎﺭ ﺑﻪ ﺍﻣﻮﺍﻝ ﺯﻛﺎﺕ ﺍﺿﺎﻓﻪ ﻛﺮﺩﻡ ﻭ ﺁﻥ ﺭﺍ ﻣﻴﺎﻥ 19ﻓﻘﻴﺮ ﺑﻪ ﻃﻮﺭ ﻣﺴــﺎﻭﻯ ﺗﻘﺴﻴﻢ ﻛﺮﺩﻡ .ﺯﻛﺎﺕ ﺍﻣﻮﺍﻟﻢ ﭼﻘﺪﺭ ﺑﻮﺩ؟ ﻣﻌﻠﻤﺎﻥ ﺭﻳﺎﺿﻰ ﺑﻪ ﺧﻮﺑﻰ ﻣﻰﺗﻮﺍﻧﻨﺪ ﺑﺎ ﭘﺎﺭﻩﺍﻯ ﺍﺯ ﻣﺴﺎﺋﻞ ﺍﻳﻦ ﻛﺘﺎﺏ ﺩﺍﻧﺶﺁﻣﻮﺯﺍﻥ ﺧﻮﺩ ﺭﺍ ﺳﺮﮔﺮﻡ ﻛﻨﻨﺪ.
ﺩﻭﺭﻩﻯ ﺷﺎﻧﺰﺩﻫﻢ ،ﺷﻤﺎﺭﻩ ﻯ ،2ﺯﻣﺴﺘﺎﻥ 1389
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