ﺩﻭﺭﻩﻱ ﺷﺎﻧﺰﺩﻫﻢ
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ﺗﺎﺑﺴﺘﺎﻥ 1390
ﺷﻤﺎﺭﻩﻱ 4
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48ﺻﻔﺤﻪ
58
ﺭﻳﺎﺿﯽ ﺩﻭﺭﺓ ﺭﺍﻫﻨﻤﺎﻳﻰ ﺗﺤﺼﻴﻠﻰ
ﻭﺯﺍﺭﺕ ﺁﻣﻮﺯﺵ ﻭ ﭘﺮﻭﺭﺵ ﺳﺎﺯﻣﺎﻥ ﭘﮋﻭﻫﺶ ﻭ ﺑﺮﻧﺎﻣﻪﺭﻳﺰﻯ ﺁﻣﻮﺯﺷﻰ ﺩﻓﺘﺮ ﺍﻧﺘﺸﺎﺭﺍﺕ ﻛﻤﻚﺁﻣﻮﺯﺷﻰ
ﻣﺪﻳﺮ ﻣﺴﺆﻝ :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 :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] : ﭘﺎﻳﮕﺎﻩ ﺍﻃﻼﻉ ﺭﺳﺎﻧﻰ 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 19000:ﻧﺴﺨﻪ
ﻓﻬﺮﺳﺖ ﺣﺮﻑ ﺍﻭﻝ ﺯﻛﺎﺕ ﺩﺍﻧﺶ /ﺣﻤﻴﺪﺭﺿﺎ ﺍﻣﻴﺮﻱ2/
ﺩﺍﻧﺶﺍﻓﺰﺍﻳﻲ
ﮔﻔﺖ ﻭ ﮔﻮ
ﺭﻳﺎﺿﻴﺎﺕ ﺭﻗﺎﺑﺘﻲ /ﺍﺣﺴﺎﻥ ﻳﺎﺭﻣﺤﻤﺪﻱ3 /
ﻣﺴﺘﻄﻴﻞ ﻃﻮﻝ ﻛﺪﺍﻡ ﺍﺳﺖ؟ /ﺯﻳﻨﺐ ﻣﺮﺍﺩﺧﺎﻧﻲ7 /
ﺩﺭ
ﺑﺨﺶﭘﺬﻳﺮﻱ/
ﻣﺤﻤﻮﺩ ﺩﺍﻭﺭﺯﻧﻲ 8 /ﻧﻘﺎﻁ ﺍﻣﻦ /ﺣﺴﻦ ﺍﺣﻤﺪﻱ 19 /ﺧﻮﺍﻧﺪﻧﻲﻫﺎﻳﻲ ﺍﺯ ﺭﻳﺎﺿﻴﺎﺕ /ﺯﻳﻨﺐ ﮔﻠﺒﺮﺍﺭﻱ 22 /ﻣﺴﺌﻠﻪﻫﺎﻱ ﻭﺍﻗﻌﻲ /ﺍﻋﻈﻢ ﭘﻮﺭﭘﺮﻭﻳﻦ/ 25
ﻭﺍژﻩﻫﺎﻱ ﺭﻳﺎﺿﻲ »ﺳﺎﺩﻩﻛﺮﺩﻥ ﻋﺒﺎﺭﺕ«» ،ﺗﺴﺎﻭﻱ=« /ﺳﭙﻴﺪﻩ
ﭼﻤﻦﺁﺭﺍ30 /
ﺩﻭ ﻣﺴﺌﻠﻪﻱ ﺟﺎﻟﺐ /ﺷﺎﺩﻱ ﺑﻬﺎﺭﻱ 34 /ﺣﻞ ﻣﺴﺌﻠﻪ
ﻗﺪﻡ ﺑﻪ ﻗﺪﻡ /ﺳﺎﻳﻪ ﻣﻬﺮﺑﺎﻥ36 / ﺭﻳﺎﺿﻲ ﻭ ﺑﺎﺯﻱ ﺑﺎﺯﻱ ﺩﻭ ﻧﻔﺮﻩ /ﺯﻫﺮﻩ ﭘﻨﺪﻱ6 / ﺟﺪﻭﻝ ﻭ ﺳﺮﮔﺮﻣﻲ ﺟﺪﻭﻝ /ﻣﺤﻤّ ﺪ ﻋﺰﻳﺰﻱﭘﻮﺭ 11 /ﺟﺪﻭﻝ 1
ﻣﻲﺗﻮﺍﻥ ﻫﺮ ﻣﺴﺌﻠﻪﺍﻱ ﺭﺍ ﺣﻞ ﻛﺮﺩ! /ﺁﺯﺍﺩﻩ
ﺷﺎﻛﺮﻱ14 / ﻫﻤﺮﺍﻩ ﺑﺎ ﻛﺘﺎﺏ ﻧﮕﺎﻫﻲ ﻧﻮ ﺑﻪ ﻣﻘﺴﻮﻡﻋﻠﻴﻪ /ﻣﺠﻴﺪ ﻣﻨﺸﻮﺭﻱ/ 27
ﻣﻌﻤﺎ ﻭ ﺳﺮﮔﺮﻣﻲ
ﻣﻌﻤﺎﻫﺎﻳﻲ ﺑﻜﺮ ﺑﺮﺍﻱ ﺗﺎﺑﺴﺘﺎﻥ /ﻋﻠﻴﺮﺿﺎ
ﻳﻮﺳﻔﻲ38 / ﺳﺆﺍﻝﻫﺎﻯ ﻣﺴﺎﺑﻘﻪﺍﻯ ﻣﺴﺎﺑﻘﻪﻱ ﺭﻳﺎﺿﻲ ﺍﺳﺘﺮﺍﻟﻴﺎ )/(2010 ﺗﺮﺟﻤﻪﻱ ﺳﭙﻴﺪﻩ ﭼﻤﻦﺁﺭﺍ45 /
ﻣﻌﺮﻓﻰ ﻛﺘﺎﺏ
ﺭﻳﺎﺿﻴﺎﺕ ﺯﻳﺒﺎ ﻭ ﺩﻭﺳﺖﺩﺍﺷﺘﻨﻲ /ﺟﻌﻔﺮ
ﺭﺑﺎﻧﻲ48 /
ﺗﺎ /100ﻧﺴﺮﻳﻦ ﺷﺮﻳﻔﻴﺎﻥ13/
ﺍﻧﺪﻳﺸﻪﻭﺭﺯﻱ
ﻣﺨﺎﻃﺮﺍﺕ ﺳﻔﺮ ﺩﺭ ﺳﻴﺎﺭﻩﻱ ﻧﺎﻟﻮﻣﺮ /ﺗﺮﺟﻤﻪﻱ
ﺣﺴﻦ ﻳﺎﻭﺭﺗﺒﺎﺭ12 /
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 ﺩﻗﺖ 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 )ﻛﻠﻴﺪﻭﺍژﻩﻫﺎ( 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 ﭘﻴﻮﺳﺖ ﺷﻮﺩ .ـ ﻣﺠﻠﻪ ﺩﺭ ﺭﺩ ،ﻗﺒﻮﻝ ،ﻭﻳﺮﺍﻳﺶ ﻭ ﺗﻠﺨﻴﺺ ﻣﻘﺎﻟﻪﻫﺎﻯ ﺭﺳﻴﺪﻩ ﺁﺯﺍﺩ ﺍﺳﺖ .ـ ﻣﻘﺎﻻﺕ ﺩﺭﻳﺎﻓﺘﻰ ﺑﺎﺯﮔﺮﺩﺍﻧﺪﻩ ﻧﻤﻰﺷﻮﻧﺪ .ـ ﺁﺭﺍﻯ ﻣﻨﺪﺭﺝ ﺩﺭ ﻣﻘﺎﻟﻪ ﺿﺮﻭﺭﺗﺎً ﻣﺒﻴﻦ ﺭﺃﻯ ﻭ ﻧﻈﺮ ﻣﺴﺌﻮﻻﻥ ﻣﺠﻠﻪ ﻧﻴﺴﺖ.
ﺣﺮﻑ ﺍﻭﻝ
ز﹋︀ت دا﹡︩ ﺍﻟﻌﻠﻢ َﺗﻌﻠﻴﻤ ُﻪ َﻣﻦ ٰﻻ َﻳ ْﻌ َﻠ ُﻢ. ﮐـﺎ ُﺉ ْ ِ ْ ﻛﺴﻲ ﻛﻪ ﻧﻤﻲﺩﺍﻧﺪ. ﻗﺎﻝ ﺭﺳﻮﻝ ﺍﷲ )ﺹ(َ :ﺯ ٰ 1
ﺶ ،ﻳﺎﺩ ﺩﺍﺩﻥ ﺁﻥ ﺍﺳﺖ ﺑﻪ
ﺹ( ﻓﺮﻣﻮﺩﻧﺪ :ﺯﻛﺎﺕ 2ﺩﺍﻧ
ﭘﻴﺎﻣﺒﺮ )
ﺩﻫﻲ ﻭ ﻳﺎﺩﮔﻴﺮﻱ ﺭﻳﺎﺿﻲ،
ﻳﺎﺩﮔﻴﺮﻱ ،ﺑﻪﻭﻳـﮋﻩ ﻳﺎﺩ
ﻦ ﺭﻭﺵﻫﺎﻱ ﻳﺎﺩﺩﻫﻲ ﻭ ﺍﺯ ﺑﻬﺘﺮﻳـ ﺩﻩ ﺍﺯ ﺧﺮﺩﺟﻤﻌﻲ ﺍﺳﺖ. ﺵ ﻛﺎﺭ ﮔﺮﻭﻫﻲ ﻭ ﺍﺳﺘﻔﺎ ﺮﻳﻦ ﺭﺍﻩ ﺗﺜﺒﻴﺖ ﻭ ﺗﻌﻤﻴﻖ ﺭﻭ ِ ﺿﻴﺎﺕ ﻳﺎﺩ ﮔﺮﻓﺘﻪﺍﻳﺪ ،ﺑﻬﺘ ﺭﻳﺎ ﺩﺭ ﺭﺍ ﻲ ﺤﺜ ﻣﺒ ﻳﺎ ﻲ ﺩﺗﺎﻥ ﺍﺳﺖ ،ﺯﻳﺮﺍ ﻫﻨﮕﺎﻡ ﺍﮔﺮ ﻣﻄﻠﺒ ﺎﻥ ﻭ ﻫﻤﻜﻼﺳﻲﻫﺎﻱ ﺧﻮ ﻳﺎﺩ ﺩﺍﺩﻥ ﺁﻥ ﺑﻪ ﺩﻭﺳـﺘ ﻧﻘﺎﻁ ﻣﺒﻬﻢ ﺁﻥ ﻣﻲﺑﺮﻳﺪ ﺁﻥ ﻣﻮﺿﻮﻉ، ﮕﺮﻱ ،ﭘﻲ ﺑﻪ ﺍﺷﻜﺎﻻﺕ ﻭ ﻌﻠﻴﻢ ﻳﻚ ﻣﻮﺿﻮﻉ ﺑﻪ ﺩﻳ ﻓﺮﺍﻧﮕﺮﻓﺘﻪﺍﻳﺪ .ﻫﻢﭼﻨﻴﻦ ﺗﺪﺭﻳﺲ ﻭ ﺗ ﺍﺯ ﺁﻥ ﺭﺍ ﺑﻪﻃﻮﺭ ﻛﺎﻣـﻞ ﻲﺷـﻮﻳﺪ ﻛﻪ ﺑﺨﺶﻫﺎﻳﻲ ﻛﻤﻚ ﻛﻨﺪ ﺗﺎ ﺁﻥ ﻣﻄﻠﺐ ﻭ ﻣﺘﻮﺟﻪ ﻣ ﺕ ﻭ ﻧﻈﺮﺍﺕ ﺧﻮﺩ ﺑﻪ ﺷﻤﺎ ﺍﻻ ﺳﺆ ﻭﺳﺖ ﺷﻤﺎ ﻧﻴﺰ ﻣﻲﺗﻮﺍﻧﺪ ﺑﺎ ﻞ ﻣﺮﺑﻮﻁ ﺑﻪ ﺁﻥ ﻣﻮﺿﻮﻉ ﺩ ﺑﻪ ﻛﻤﻚ ﻳﻜﺪﻳﮕﺮ ﻣﺴﺎﺋ ﻛﺎﻣﻞ ﺟﺎ ﺑﻴﻔﺘﺪ ﻭ ﺑﺘﻮﺍﻧﻴﺪ ﺑﺮﺍﻱ ﺷﻤﺎ ﺭﺍ ﺣﻞ ﻛﻨﻴﺪ. ﺹ( ﻧﻘﻞ ﺷـﺪ ،ﺻﺤﺒﺖ ﺍﺯ ﻱ ﻛﻼﻡ ﺍﺯ ﭘﻴﺎﻣﺒﺮ ﺍﻛﺮﻡ ) ﺪﺍ ﺍﺑﺘ ﺩﺭ ﺣﺪﻳﺚ ﻧﺒﻮﻱ ﻛﻪ ﺩﺭ ﻌﻠﻴﻢ ﺑﻪ ﻛﺴﺎﻧﻲ ﺍﺳﺖ ﻛﻪ ﺍﻛﺮﻡ )ﺹ( ﺯﻛﺎﺕ ﻋﻠﻢ ،ﺗ ﺒﺮ ﻋﻠﻢ ﺍﺳـﺖ .ﺍﺯ ﻧﮕﺎﻩ ﭘﻴﺎﻣ ﻋﻠﻢ ﺧﻮﺩ ﺑﻪ ﺩﻳﮕﺮﺍﻥ ﺑﻪ ﺯﻛﺎ ِ ﺕ ﻤﺎ ﺑـﺎ ﺗﻌﻠﻴﻢ ﻭ ﻳﺎﺩﺩﻫﻲ ﻲ ﺑﻬﺮﻩﺍﻧﺪ .ﺑﻨﺎﺑﺮﺍﻳﻦ ﺷـ ﺍﺯ ﺁﻥ ﻋﻠﻢ ﺑ ﻲﺩﻫﻴﺪ ﻭ ﻫﻤﺎﻥﻃﻮﺭ ﻛﻪ ﻛﻪ ﺯﻛﺎﺕ ﻋﻠﻢ ﺧـﻮﺩ ﺭﺍ ﻣ ﺁﻥ ﻭﻝ ﺍ ﺪ: ﺩﻭ ﻣﻬﻢ ﺩﺳـﺖ ﻳﺎﻓﺘﻪﺍﻳ ﻠﻢ ﺷـﻤﺎ ﻧﻴﺰ ﺑﺎ ﻳﺎﺩﺩﻫﻲ ﻮﺩ ﺑﺎ ﺑﺮﻛﺖ ﺧﻮﺍﻫﺪ ﺑﻮﺩ ،ﻋ ﺷـ ﺖ ﻣﺎﻟﻲ ﻛﻪ ﺯﻛﺎﺗﺶ ﭘﺮﺩﺍﺧ ﺧﻮﺩﺗﺎﻥ ﻣﻮﺿﻮﻉ ﺭﺍ ﺑﻪﻃﻮﺭ ﻫﺮ ﺧﻮﺍﻫﺪ ﺷـﺪ .ﺩﻭﻡ ﺁﻥﻛﻪ ﺖ ﻭ ﺍ ِﻥﺷـﺎءﺍ ...ﻛﺎﺭﺑﺮﺩﻱ ﭘﺮﺑﺮﻛ ﻧﻴﺰ ﺑﻬﺮﻩﻣﻨﺪ ﻣﻲﺷﻮﻳﺪ. ﻴﺪ ﻭ ﺍﺯ ﻧﻈﺮﺍﺕ ﺩﻳﮕﺮﺍﻥ ﻛﺎﻣﻞ ﺩﺭﻙ ﻣﻲﻛﻨ ﺩﺭﺱ ﻭ ﺗﻤﺮﻳﻦ ﻭ ﺗﻼﺵ، ﺁﻳﻨﺪﻩ ﺑﺎ ﺗﻤﺮﻛﺰ ﺩﺭ ﻛﻼﺱ ﻲ ﺼﻴﻠ ﺗﺤ ﭘﺲ ﺑﻴﺎﻳﻴﺪ ﺩﺭ ﺳﺎﻝ ﺳﺘﺎﻥ ﻭ ﻫﻤﻜﻼﺳﻲﻫﺎﻱ ﻳﺪ ﻭ ﺳـﭙﺲ ﺁﻥ ﺭﺍ ﺑﻪ ﺩﻭ ﻴﺮ ﺑﮕ ﺐ ﺩﺭﺳـﻲ ﺭﺍ ﺧﻮﺏ ﻳﺎﺩ ﺯﻣﻴﻨﻪ ﺑﻪ ﻧﺘﻴﺠﻪﻱ ﺧﻮﺏ ﻣﻄﺎﻟ ﻳﺎﺩ ﺑﺪﻫﻴﺪ .ﺍﮔﺮ ﺩﺭ ﺍﻳﻦ ﺁﻥ ﺩﺭﺱ ﺿﻌﻒ ﺩﺍﺭﻧﺪ، ﺧﻮﺩ ﻛﻪ ﺩﺭ ﻭ ﺑﺮﺍﻱ ﻣﺎ ﺍﺭﺳﺎﻝ ﻛﻨﻴﺪ. ﻛﺮﺩﻳﺪ ﺁﻥ ﺭﺍ ﺑﻨﻮﻳﺴﻴﺪ ﺮﻩﻱ ﺟﺎﻟﺒﻲ ﺩﺳﺖ ﭘﻴﺪﺍ ﻳﺎ ﺧﺎﻃ ،2ﺣﺪﻳﺚ .10 ،2ﺻﻔﺤﻪﻱ 5 ﭘﻲﻧﻮﺷﺖ ﭼﺎپ ،(88ﺟﻠﺪ ﻣﻌﻨ ِﻲ ﭘﺎﻛﻴﺰﻩﺷ ﺠﻠﺴﻲ )ﺭﻩ() ، ﻭ ﻫﻤﭽﻨﻴﻦ ﺑﻪ ﻻﻧﻮﺍﺭ ،ﻋﻼﻣﻪ ﻣ ﻮ ﻭ ﻗﺪ ﻛﺸﻴﺪﻥ .1ﺑﺤﺎﺭﺍ ﺮﺁﻧﻲ ﺑﻪ ﻣﻔﻬﻮﻡ ﻧﻤ ّ ﺕ ﺩﺭ ﻓﺮﻫﻨﮓ ﻗ .2ﺯﻛﺎ
ﺪﻥ ﺁﻣﺪﻩ ﺍﺳﺖ.
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ﺭﺍﻫﻨﻤﺎﻳﻲ
ﺩﻭﺭﺓ ﺷﺎﻧﺰﺩﻫﻢ ،ﺷﻤﺎﺭﺓ ،4ﺗﺎﺑﺴﺘﺎﻥ 1390
ﺩﺍﻧﺶﺍﻓﺰﺍﻳﻲ
ر︀︲﹫︀ت ر﹇︀︋︐﹩ ﺍﺣﺴﺎﻥ ﻳﺎﺭﻣﺤﻤﺪﻱ
ﻛﻠﻴﺪﻭﺍژﻩﻫﺎ :ﺭﻗﺎﺑﺖ ،ﺗﻴﺰﻫﻮﺷﺎﻥ ،ﺁﺯﻣﻮﻥ ﻭﺭﻭﺩﻱ ،ﻣﻌﺎﺩﻟﻪﻱ ﺧﻄﻲ ،ﺩﺳﺘﮕﺎﻩ ﺩﻭ ﻣﻌﺎﺩﻟﻪ ﺩﻭ ﻣﺠﻬﻮﻝ.
ﻣﻘﺪﻣﻪ
ﺭﻳﺎﺿﻲﭘﮋﻭﻩ ﺑﺮﺍﻱ ﭘﻴﺮﻭﺯﻱ ﻭ ﻛﺎﻣﻴﺎﺑﻲ ﺩﺭ ﺍﻳﻦﮔﻮﻧﻪ ﺭﻗﺎﺑﺖﻫﺎﻱ ﺭﻳﺎﺿﻲ
ﺑـﺎ ﺗﻮﺟﻪ ﺑﻪ ﻋﻼﻗـﻪﻱ ﺭﻭﺯﺍﻓـﺰﻭﻥ ﻧﻮﺟﻮﺍﻧﺎﻥ ﻣﻤﻠﻜـﺖ ﻋﺰﻳﺰﻣﺎﻥ
ﺑﺎﻳﺪ ﺑﻪ ﺁﻥ ﻣﺠﻬﺰ ﺑﺎﺷﺪ .ﺩﺭ ﺍﻧﺘﻬﺎ ﻣﺘﺬﻛﺮ ﻣﻲﺷﻮﻡ ﺍﺯ ﺁﻥﺟﺎ ﻛﻪ ﻣﺴﺎﺋﻞ
ﺑﻪ ﺷـﺮﻛﺖ ﺩﺭ ﻣﺴـﺎﺑﻘﺎﺕ ﻭ ﺭﻗﺎﺑﺖﻫـﺎﻱ ﺭﻳﺎﺿﻲ ﻣﺎﻧﻨـﺪ ﺍﻟﻤﭙﻴﺎﺩﻫﺎﻱ
ﻣﻄﺮﺡ ﺷـﺪﻩ ﺩﺭ ﺍﻟﻤﭙﻴﺎﺩﻫﺎ ،ﻣﺴـﺎﺑﻘﺎﺕ ﻭ ﺁﺯﻣﻮﻥﻫﺎﻱ ﺭﻳﺎﺿﻲ ﻣﺨﺘﺺ
ﺭﻳﺎﺿـﻲ ،ﺁﺯﻣﻮﻥﻫﺎﻱ ﻭﺭﻭﺩﻱ ﻣﺪﺍﺭﺱ ﺗﻴﺰﻫﻮﺷـﺎﻥ ،ﻣـﺪﺍﺭﺱ ﻧﻤﻮﻧﻪ
ﻧﻮﺟﻮﺍﻧﺎﻥ ﻧﻴﺰ ﺷﺎﻣﻞ ﻣﺒﺎﺣﺜﻲ ﺩﺭ ﻫﻨﺪﺳﻪ ،ﻧﻈﺮﻳﻪﻱ ﻋﺪﺩﻫﺎ ،ﺟﺒﺮ ﻭ
ﻭ ...ﻭ ﻧﻴﺰ ﺍﺳـﺘﻘﺒﺎﻝ ﺁﻥﻫﺎ ﺍﺯ ﻣﺴـﺎﺋﻞ ﭼﺎﻟﺶﭘﺬﻳﺮ ﺑﻪ ﻣﻨﻈﻮﺭ ﻛﺴـﺐ
ﺣﺴﺎﺏ ﻭ ﻫﻮﺵ ﺍﺳﺖ ،ﺑﻪ ﺷﻤﺎ ﻧﻮﺟﻮﺍﻧﺎﻥ ﮔﺮﺍﻥﻗﺪﺭ ﭘﻴﺸﻨﻬﺎﺩ ﻣﻲﺷﻮﺩ
ﺁﮔﺎﻫـﻲ ﻭ ﺩﺍﻧـﺶ ﺑﻴﺶﺗـﺮ ﻭ ﺩﺭ ﻧﻬﺎﻳﺖ ﺍﺭﺗﻘﺎﻱ ﺳـﻄﺢ ﻛﻴﻔﻲ ﻭ ﻛﻤﻲ
ﻛﻪ ﺩﺭﺑﺎﺭﻩﻱ ﺍﺛﺒﺎﺕ ﻗﻀﺎﻳﺎ ،ﺭﺍﻩﺣﻞﻫﺎ ﻭ ﭘﺎﺳـﺦﻫﺎﻱ ﻣﺴﺎﺋﻞ ﺍﺭﺍﺋﻪ ﺷﺪﻩ،
ﺍﻃﻼﻋـﺎﺕ ﻭ ﻣﻬﺎﺭﺕﻫﺎﻱ ﭘﻴﺮﺍﻣﻮﻥ ﻣﻮﺿﻮﻋﺎﺕ ﮔﻮﻧﺎﮔﻮﻥ ﺭﻳﺎﺿﻲ ،ﺑﺮﺍﻱ
ﺍﺑﺘﺪﺍ ﺗﻔﻜﺮ ﻭ ﺗﻌﻤﻖ ﻣﻨﺎﺳـﺐ ﺭﺍ ﺩﺍﺷـﺘﻪ ﺑﺎﺷﻴﺪ ﻭ ﺑﻪ ﺻﻮﺭﺕ ﻣﺴﺘﻘﻴﻢ
ﺁﺷـﻨﺎﻳﻲ ﻫﺮﭼﻪ ﺑﻴﺶﺗـﺮ ﻧﻮﺟﻮﺍﻧﺎﻥ ﺍﻳﺮﺍﻥ ﻫﻤﻴﺸـﻪ ﺳـﺮﻓﺮﺍﺯ ﻛﻪ ﺩﺭ
ﺑﻪ ﺳـﺮﺍﻍ ﺑﺮﻫﺎﻥ ﻗﻀﺎﻳﺎ ،ﺭﺍﻩﺣﻞﻫﺎ ﻭ ﭘﺎﺳـﺦﻫﺎﻱ ﺍﺭﺍﻳﻪ ﺷﺪﻩ ﺍﺯ ﺳﻮﻱ
ﺩﻭﺭﻩﻱ ﺭﺍﻫﻨﻤﺎﻳﻲ )ﺳﻨﻴﻦ ﺩﻭﺍﺯﺩﻩ ،ﺳﻴﺰﺩﻩ ﻭ ﭼﻬﺎﺭﺩﻩ ﺳﺎﻟﮕﻲ( ﺑﻪ ﺳﺮ
ﻧﮕﺎﺭﻧـﺪﻩﻱ ﻣﻘﺎﻟﻪ ﻧﺮﻭﻳﺪ ﺗﺎ ﺑﺘﻮﺍﻧﻴﺪ ﺑﻪ ﻣﻬـﺎﺭﺕ ﻭ ﺗﻜﻨﻴﻚﻫﺎﻱ ﻻﺯﻡ ﻭ
ﻣﻲﺑﺮﻧﺪ ،ﺗﺼﻤﻴﻢ ﺑﺮ ﺁﻥ ﺷـﺪ ﻛﻪ ﺳﻠﺴـﻠﻪ ﻣﻘﺎﻻﺗﻲ ﭘﻴﺮﺍﻣﻮﻥ ﻣﻮﺿﻮﻉ
ﻛﺎﻓﻲ ﺩﺭ ﺍﻳﻦ ﺯﻣﻴﻨﻪ ﺑﺮﺳﻴﺪ) .ﺍﻟﺒﺘﻪ ﺑﻪ ﺍﻳﻦ ﻋﻠﺖ ﻛﻪ ﺍﺭﺍﻳﻪﻱ ﺭﺍﻩﺣﻞ ﺩﺭ
ﻳﺎﺩﺷـﺪﻩ ﺑﺎ ﻋﻨﺎﻭﻳﻦ ﻣﻔﻴﺪ ﻭ ﻣﻮﺛﺮ ﺑـﺮﺍﻱ ﺩﺍﻧﺶﺁﻣﻮﺯﺍﻥ ﻧﻮﺟﻮﺍﻥ ﺗﻬﻴﻪ
ﻣﻮﺿﻮﻋﺎﺕ ﻫﻨﺪﺳﻪ ﻭ ﻧﻈﺮﻳﻪﻱ ﻋﺪﺩﻫﺎ ﻧﻴﺎﺯ ﺑﻪ ﺩﺭﻙ ﺷﻬﻮﺩﻱ ﻣﺘﻌﺎﻟﻲ
ﺷﻮﺩ ﻛﻪ ﺍﻳﻦ ﺩﺍﻧﺶﺁﻣﻮﺯﺍﻥ ﺑﺎ ﺳﺒﻚ ﻭ ﺳﻴﺎﻕ ﻗﻀﺎﻳﺎ ،1ﻧﻜﺎﺕ ﻣﻮﺭﺩ ﻧﻴﺎﺯ،
ﻭ ﻗـﻮﻩﻱ ﺧﻼﻗﻴﺖ ﺑﺎﻻ ﺩﺍﺭﻧﺪ ﻭ ﺣﻞ ﻛﺮﺩﻥ ﻣﺴـﺎﺋﻞ ﺩﺭ ﺍﻳﻦ ﻋﻨﺎﻭﻳﻦ ﺍﺯ
ﻣﺴـﺎﺋﻠﻲ ﻛﻪ ﺩﺭ ﺍﻳﻦ ﺯﻣﻴﻨﻪ ﻣﻄﺮﺡ ﻣﻲﺷﻮﺩ :ﺑﻪ ﻫﻤﺮﺍﻩ ﺑﺮﻫﺎﻥ ،2ﭘﺎﺳﺦ
ﻳﻚ ﺍﺳـﻠﻮﺏ ﻭ ﺭﻭﺵ ﺧﺎﺹ ﭘﻴﺮﻭﻱ ﻧﻤﻲﻛﻨﻨـﺪ ﻭ ﺑﺮﺍﻱ ﺣﻞ ﻫﺮﻳﻚ ﺍﺯ
ﻭ ﺭﻭﺵ ﺣﻞ ﺁﻥﻫﺎ ﺁﺷﻨﺎ ﺷﻮﻧﺪ .ﺍﻟﺒﺘﻪ ﺑﺎ ﺑﺮﺭﺳﻲ ﺩﻗﻴﻖ ﻣﺴﺎﺋﻠﻲ ﻛﻪ ﺩﺭ
ﺁﻥﻫﺎ ﻧﻴﺎﺯ ﺑﻪ ﺩﺍﺷـﺘﻦ ﺧﻼﻗﻴﺘﻲ ﻣﺨﺘﺺ ﺑﻪ ﺧﻮﺩ ﺍﺳﺖ ،ﺩﺭ ﺍﻳﻦ ﻣﻮﺍﺭﺩ،
ﺍﻟﻤﭙﻴﺎﺩﻫﺎ ،ﻣﺴﺎﺑﻘﺎﺕ ﻭ ﺁﺯﻣﻮﻥﻫﺎﻱ ﺭﻳﺎﺿﻲ 3ﻛﻪ ﺑﺮﺍﻱ ﺳﻨﻴﻦ ﻧﻮﺟﻮﺍﻧﺎﻥ
ﻫﻤﺖ ﻭ ﻣﻤﺎﺭﺳﺖ ﺑﻴﺶﺗﺮﻱ ﺑﻪ ﺧﺮﺝ ﺩﻫﻴﺪ(.
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ﭼﻪ ﺩﺭ ﺩﺍﺧﻞ ﻛﺸـﻮﺭ ﻭ ﭼﻪ ﺩﺭ ﺧﺎﺭﺝ ﻛﺸﻮﺭ ﻃﺮﺍﺣﻲ ﻭ ﺍﺭﺍﺋﻪ ﻣﻲﺷﻮﺩ. ﺭﻭﺷـﻦ ﻭ ﻣﺸﺨﺺ ﻣﻲﺷـﻮﺩ ﻛﻪ ﻫﺮ ﺭﻳﺎﺿﻲﭘﮋﻭﻩ ﻧﻮﺟﻮﺍﻧﻲ ﻛﻪ ﺁﺭﺯﻭﻱ ﻣﻮﻓﻘﻴﺖ ﺩﺭ ﭼﻨﻴﻦ ﺭﻗﺎﺑﺖﻫﺎﻱ ﺭﻳﺎﺿﻲ ﺭﺍ ﺑﻪ ﺳـﺮ ﻣﻲﭘﺮﻭﺭﺍﻧﺪ ،ﺑﺎﻳﺪ ﺍﺯ ﺩﺍﻧﺸﻲ ﻓﺮﺍﺗﺮ ﺍﺯ ﻣﻮﻓﻘﻴﺖ ﺯﻣﺎﻧﻲ ﻭ ﻣﻜﺎﻧﻲ ﻛﻪ ﺳﻦ ﻭ ﺳﺎﻝ ﺍﻭ ﻣﻲﻃﻠﺒﺪ ﺑﺮﺧﻮﺭﺩﺍﺭ ﺑﺎﺷـﺪ .ﺑﻪ ﻫﻤﻴﻦ ﻋﻠﺖ ﺩﺭ ﺍﻳـﻦ ﻣﺠﻤﻮﻋﻪ ﻣﻘﺎﻻﺕ ﻋﻼﻭﻩ ﺑﺮ ﻣﻄﺮﺡ ﻛﺮﺩﻥ ﻋﻨﺎﻭﻳﻦ ﭘﻴﻜﺎﺭﺟﻮ ﺩﺭﺑﺎﺭﻩﻱ ﻣﻄﺎﻟﺐ ﺩﺭﺳﻲ ﻛﻪ ﻣﺘﻨﺎﺳﺐ ﺑـﺎ ﺩﻭﺭﻩﻱ ﺗﺤﺼﻴﻠـﻲ ﺍﻳﻦ ﻧﻮﺟﻮﺍﻧﺎﻥ ﺍﺳـﺖ ،ﺑﻪ ﻣﻨﻈﻮﺭ ﺍﻳﺠﺎﺩ ﺳـﻮﺍﺩ ﺭﻳﺎﺿـﻲ ﺍﻓﺰﻭﻥﺗﺮ ،ﻛﻪ ﺑﺎﻋﺚ ﻋﻤﻠﻲ ﺳـﺎﺧﺘﻦ ﺍﺳـﺘﻌﺪﺍﺩﻫﺎﻱ ﺑﺎﻟﻘﻮﻩﻱ ﺭﻳﺎﺿﻲﭘﮋﻭﻫﺎﻥ ﻧﻮﺑﺎﻭﻩ ﻣﻲﺷـﻮﺩ ،ﺩﺭ ﭘﺎﺭﻩﺍﻱ ﻣـﻮﺍﺭﺩ ﺑﻪ ﺍﺭﺍﺋﻪﻱ ﻗﻀﺎﻳﺎ ﻭ ﻧـﻜﺎﺕ ﺍﺭﺯﻧﺪﻩ ﻭ ﭘﻮﻳﺎ ﺩﺭ ﻛﻨﺎﺭ ﻣﺴـﺎﺋﻞ ﻣﺮﺗﺒﻂ ﺑﻪ ﺁﻥﻫﺎ ﻣﻲﭘﺮﺩﺍﺯﻳﻢ. ﻛﻪ ﺍﻳﻦ ﻣﻄﺎﻟﺐ ﺑﺎ ﻋﻨﺎﻭﻳﻦ ﻭ ﻣﻮﺍﺭﺩﻱ ﻣﻨﺎﺳـﺐ ﺍﺳـﺖ ﻛﻪ ﻫﺮ ﻧﻮﺑﺎﻭﻩﻱ
ﺩﻭﺭﺓ ﺷﺎﻧﺰﺩﻫﻢ ،ﺷﻤﺎﺭﺓ ،4ﺗﺎﺑﺴﺘﺎﻥ 1390
ﺭﺍﻫﻨﻤﺎﻳﻲ
3
ﺭﺍﻫﺒﺮﺩﻫـﺎﻱ ﺗﺸـﻜﻴﻞ ﻣﻌﺎﺩﻟـﻪﻱ ﺧﻄﻲ ﻭ ﺗﺸﻜﻴﻞ ﺩﺳﺘﮕﺎﻩ ﺩﻭ ﻣﻌﺎﺩﻟﻪ ﻭ ﺩﻭ ﻣﺠﻬﻮﻟﻲ ﺩﺭ ﺯﻳﺮ ﭘﺮﺳﺶﻫﺎﻱ ﭼﻬﺎﺭﮔﺰﻳﻨﻪﺍﻱ ﮔﻮﻧﺎﮔﻮﻥ ﻭ ﭘﻴﻜﺎﺭﺟﻮ ﺭﺍ ﺑﻪ ﻫﻤﺮﺍﻩ ﭘﺎﺳﺦﻫﺎﻱ ﺗﺸــﺮﻳﺤﻲ ﺁﻥﻫﺎ ﻛﻪ ﺟﻨﺒﻪﻱ ﺭﻗﺎﺑﺘﻲ ﺑﺮﺍﻱ ﺭﻳﺎﺿﻲﭘﮋﻭﻫﺎﻥ ﻧﻮﺟــﻮﺍﻥ ﺩﺍﺭﻧﺪ ﻭ ﺩﺭ ﺁﺷــﻨﺎﻳﻲ ﻭ ﺍﻋﺘــﻼﻱ ﺗﻮﺍﻥ ﺫﻫﻨﻲ ﺍﻳﺸــﺎﻥ ﺑﺮﺍﻱ ﻣﻮﻓﻘﻴﺖ ﺩﺭﺁﺯﻣﻮﻥﻫﺎﻱ ﺍﻟﻤﭙﻴﺎﺩﻫﺎﻱ ﺭﻳﺎﺿﻲ ﺩﺍﺧﻞ ﻭ ﺧﺎﺭﺝ ﺍﺯ ﻛﺸــﻮﺭ، ﺁﺯﻣﻮﻥﻫﺎﻱ ﻭﺭﻭﺩﻱ ﻣﺪﺍﺭﺱ ﺗﻴﺰﻫﻮﺷﺎﻥ ﻭ ﻣﺪﺍﺭﺱ ﻧﻤﻮﻧﻪ ﻭ ﻛﺎﺭﺑﺮﺩﻱ ﻭ ...ﻛﺎﺭﺑﺮﺩﻱ ﺍﺭﺯﻧﺪﻩ ﺩﺍﺭﻧﺪ ،ﺍﺭﺍﻳﻪ ﻣﻲﺩﻫﻴﻢ .ﺍﻟﺒﺘﻪ ﺩﺭ ﭘﺎﺭﻩﺍﻱ ﺍﺯ ﻣﻮﺍﺭﺩ ﻛﻪ ﺫﻛﺮ ﻧﻜﺎﺕ ﻳﺎ ﻣﻄﺎﻟﺒﻲ ﺑﺮﺍﻱ ﺩﺭﻙ ﺑﻬﺘﺮ ﺍﻳﻦ ﭘﺮﺳﺶﻫﺎﻱ ﭼﻬﺎﺭﮔﺰﻳﻨﻪﺍﻱ ﻻﺯﻡ ﺑﻪ ﻧﻈﺮ ﻣﻲﺭﺳﺪ ،ﺍﺯ ﺑﻴﺎﻥ ﻭ ﺍﺭﺍﻳﻪﻱ ﺁﻥ ﺩﺭﻳﻎ ﻧﻜﺮﺩﻩ ﻭ ﺁﻥ ﺭﺍ ﻫﻤﺮﺍﻩ ﺑﺎ ﭘﺎﺳﺦﻫﺎﻱ ﺗﺸﺮﻳﺤﻲ ﺁﻭﺭﺩﻩﺍﻳﻢ. .1ﺩﺭ ﻳﻚ ﻗﻠﻚ 625ﺭﻳﺎﻝ ﺍﺯ ﺳــﻜﻪﻫﺎﻱ 5ﺭﻳﺎﻟﻲ ﻭ 20ﺭﻳﺎﻟﻲ ﻭﺟﻮﺩ ﺩﺍﺭﺩ .ﺍﮔﺮ ﻣﺠﻤﻮﻋﺎ 35ﺳــﻜﻪ ﺩﺭ ﺍﻳﻦ ﻗﻠﻚ ﻭﺟﻮﺩ ﺩﺍﺷــﺘﻪ ﺑﺎﺷﺪ، ﭼﻨﺪ ﺳﻜﻪﻱ 20ﺭﻳﺎﻟﻲ ﺩﺭ ﺍﻳﻦ ﻗﻠﻚ ﻭﺟﻮﺩ ﺩﺍﺭﺩ؟ 15 (2 30 (1 5 (3 10 (4 .2ﻣﻘﺪﺍﺭ ﻛﺴــﺮﻱ ﺑﺮﺍﺑﺮ ﺑﺎ 2ﺍﺳــﺖ .ﺍﮔﺮ ﺍﺧﺘــﻼﻑ ﺻﻮﺭﺕ ﻭ ﻣﺨﺮﺝ 3 ﻛﺴــﺮ ﺑﺮﺍﺑﺮ ﺑﺎ 7ﺑﺎﺷﺪ ،ﻣﺠﻤﻮﻉ ﺻﻮﺭﺕ ﻭ ﻣﺨﺮﺝ ﻛﺴﺮ ﻛﺪﺍﻡﻳﻚ ﺍﺯ ﮔﺰﻳﻨﻪﻫﺎﻱ ﺯﻳﺮ ﺍﺳﺖ؟ 35 (2 30 (1 45 (3 40 (4 .3ﻣﺠﻤﻮﻉ ﺳﻪ ﻋﺪﺩ ﻓﺮﺩ ﻣﺘﻮﺍﻟﻲ ﺑﺮﺍﺑﺮ ﺑﺎ 87ﺍﺳﺖ ،ﻣﺠﻤﻮﻉ ﻳﻜﺎﻥﻫﺎﻱ ﺁﻥ ﺳﻪ ﻋﺪﺩ ﻛﺪﺍﻡ ﻳﻚ ﺍﺯ ﮔﺰﻳﻨﻪﻫﺎﻱ ﺯﻳﺮ ﺍﺳﺖ؟ 17 (2 7 (1 8 (3 18 (4
.4ﻧﻴﻤﺎ ﻭ ﺳﻴﻨﺎ ﺭﻭﻱ ﻫﻢ 10500ﺭﻳﺎﻝ ﭘﻮﻝ ﺩﺍﺭﻧﺪ .ﺑﻌﺪ ﺍﺯ ﺁﻥ ﻛﻪ ﻧﻴﻤﺎ 1ﭘﻮﻝ ﺧﻮﺩ ﻭ ﺳــﻴﻨﺎ 1ﭘﻮﻝ ﺧﻮﺩ ﺭﺍ ﺧﺮﺝ ﻛﺮﺩﻧﺪ ،ﻣﻘﺪﺍﺭ ﭘﻮﻝ 6 3 ﻧﻴﻤﺎ ﺩﻭ ﺑﺮﺍﺑﺮ ﻣﻘﺪﺍﺭ ﭘﻮﻝ ﺳــﻴﻨﺎ ﺷــﺪ .ﻣﻘﺪﺍﺭ ﭘﻮﻝ ﻫﺮﻳﻚ ﺍﺯ ﺁﻥﻫﺎ ﻛﺪﺍﻡ ﮔﺰﻳﻨﻪ ﺍﺳﺖ؟ (1ﻧﻴﻤﺎ 3000ﺭﻳﺎﻝ ﻭ ﺳﻴﻨﺎ 7500ﺭﻳﺎﻝ (2ﻧﻴﻤﺎ 7500ﺭﻳﺎﻝ ﻭ ﺳﻴﻨﺎ 3000ﺭﻳﺎﻝ (3ﻧﻴﻤﺎ 3500ﺭﻳﺎﻝ ﻭ ﺳﻴﻨﺎ 7000ﺭﻳﺎﻝ (4ﻧﻴﻤﺎ 7000ﺭﻳﺎﻝ ﻭ ﺳﻴﻨﺎ 3500ﺭﻳﺎﻝ .5ﺍﮔﺮ ﺑﻪ ﺻﻮﺭﺕ ﻭ ﻣﺨﺮﺝ ﻛﺴﺮﻱ 3ﻭﺍﺣﺪ ﺍﺿﺎﻓﻪ ﺷﻮﺩ ،ﺁﻥ ﻛﺴﺮ ﺑﺮﺍﺑﺮ 4 ﺑﺎ 5ﻭ ﺍﮔﺮ ﺍﺯ ﺻﻮﺭﺕ ﻭ ﻣﺨﺮﺝ ﺁﻥ ﻛﺴــﺮ 3ﻭﺍﺣﺪ ﻛﻢ ﺷــﻮﺩ ،ﺁﻥ 1 ﻛﺴــﺮ ﺑﺮﺍﺑﺮ ﺑﺎ ﺧﻮﺍﻫﺪ ﺷﺪ ،ﻣﺠﻤﻮﻉ ﺻﻮﺭﺕ ﻭ ﻣﺨﺮﺝ ﺁﻥ ﻛﺴﺮ ⎧ 2 ﻛﺪﺍﻡ ﻳﻚ ﺍﺯ ﮔﺰﻳﻨﻪﻫﺎﻱ ﺯﻳﺮ ﺍﺳﺖ؟ 34 (2 43 (1 12 (4 21 (3 .6ﺩﺭ ﺷــﻜﻞ ﺯﻳﺮ ﻣﺠﻤﻮﻉ ﻣﺴــﺎﺣﺖﻫﺎﻱ s2 ،s1ﻭ s3ﺑﺮﺍﺑﺮ ﺑﺎ 37ﻭ ﻣﺴﺎﺣﺖ ﻗﺴﻤﺖ ﻫﺎﺷﻮﺭﺯﺩﻩ ﺑﺮﺍﺑﺮ ﺑﺎ 6ﺍﺳﺖ ،ﻣﺤﻴﻂ ﻣﺴﺘﻄﻴﻞ s3 ﻛﺪﺍﻡ ﻳﻚ ﺍﺯ ﮔﺰﻳﻨﻪﻫﺎﻱ ﺯﻳﺮ ﺍﺳﺖ؟
S1 S2 21 (1 14 (3
S3 7 (2 12 (4
.7ﺳــﺎﻧﺎﺯ ﺍﺯ ﮔﻠﻨﺎﺭ ﭘﺮﺳﻴﺪ :ﭼﻨﺪ ﺳﺎﻝ ﺩﺍﺭﻱ؟ ﮔﻠﻨﺎﺭ ﺟﻮﺍﺏ ﺩﺍﺩ :ﻭﻗﺘﻲ ﺗﻮ ﺑﻪ ﺳﻦ ﺍﻣﺮﻭﺯ ﻣﻦ ﺑﺮﺳﻲ ،ﻣﻦ ﺩﻭ ﺑﺮﺍﺑﺮ ﺳﻦ ﺍﻵﻥ ﺗﻮ ﺳﻦ ﺧﻮﺍﻫﻢ ﺩﺍﺷﺖ .ﺍﮔﺮ ﻣﺠﻤﻮﻉ ﺳﻦ ﺩﻭ ﻧﻔﺮ ﺁﻥﻫﺎ 30ﺳﺎﻝ ﺑﺎﺷﺪ ،ﮔﻠﻨﺎﺭ ﭼﻨﺪ ﺳﺎﻝ ﺩﺍﺭﺩ؟ 14 (2 12 (1 18 (3 20 (4 .8ﻣﺪﺕ ﺯﻣﺎﻧﻲ ﻛﻪ ﺍﺯ ﺳــﺎﺧﺖ ﻳﻚ ﻛﺸــﺘﻲ ﻣﻲﮔــﺬﺭﺩ ،ﺑﺮﺍﺑﺮ ﻣﺪﺕ ﺯﻣﺎﻧﻲ ﺍﺳﺖ ﻛﻪ ﺍﺯ ﺳﺎﺧﺖ ﺩﻳﮓ ﺑﺨﺎﺭ ﺁﻥ ﺗﺎ ﺯﻣﺎﻧﻲ ﻛﻪ ﻋﻤﺮ ﻛﺸﺘﻲ ﺑــﻪ ﺍﻧﺪﺍﺯﻩﻱ ﻋﻤﺮ ﻓﻌﻠﻲ ﺩﻳﮓ ﺑﺨﺎﺭ ﺁﻥ ﺷــﻮﺩ ،ﻣﻲﮔﺬﺭﺩ .ﻣﺠﻤﻮﻉ 4
ﺭﺍﻫﻨﻤﺎﻳﻲ
ﺩﻭﺭﺓ ﺷﺎﻧﺰﺩﻫﻢ ،ﺷﻤﺎﺭﺓ ،4ﺗﺎﺑﺴﺘﺎﻥ 1390
ﻋﻤﺮ ﻓﻌﻠﻲ ﻛﺸــﺘﻲ ﻭ ﺩﻳﮓ ﺑﺨﺎﺭ ﺁﻥ 49ﺳــﺎﻝ ﺍﺳــﺖ .ﺍﺯ ﺳﺎﺧﺖ ﻛﺸﺘﻲ ﭼﻨﺪ ﺳﺎﻝ ﻣﻲﮔﺬﺭﺩ؟ 21 (2 25 (1 24 (4 28 (3 .9ﺍﮔﺮ ﺣﺴــﻦ 3ﺩﻓﺘﺮ ﻭ 5ﺧﻮﺩﻛﺎﺭ ﺑﺨﺮﺩ 10 ،ﺗﻮﻣﺎﻥ ﺍﺯ ﭘﻮﻟﺶ ﺑﺎﻗﻲ ﻣﻲﻣﺎﻧــﺪ .ﺍﮔﺮ ﺍﻭ 2ﺩﻓﺘﺮ ﻭ 8ﺧﻮﺩﻛﺎﺭ ﺑﺨــﺮﺩ ،ﭘﻮﻟﻲ ﺑﺮﺍﻱ ﺍﻭ ﺑﺎﻗﻲ ﻧﻤﻲﻣﺎﻧﺪ ﻭ ﻗﻴﻤﺖ 2ﺩﻓﺘﺮ ﺑﺎ ﻗﻴﻤﺖ 5ﺧﻮﺩﻛﺎﺭ ﺑﺮﺍﺑﺮ ﺍﺳــﺖ .ﻣﻘﺪﺍﺭ ﭘﻮﻝ ﺣﺴﻦ ﻛﺪﺍﻡ ﮔﺰﻳﻨﻪ ﺍﺳﺖ؟ 200 (1 240 (2 260 (3 280 (4 .10ﺩﻭ ﺷــﻤﻊ ﻫﻢﻃﻮﻝ ﺭﺍ ﺑﺎ ﻫﻢ ﺭﻭﺷــﻦ ﻣﻲﻛﻨﻴﻢ .ﺷــﻤﻊ ﺍﻭﻝ ﺩﺭ 4 ﺳــﺎﻋﺖ ﻭ ﺷﻤﻊ ﺩﻭﻡ ﺩﺭ 3ﺳﺎﻋﺖ ﻣﻲﺳــﻮﺯﺩ .ﺑﺎ ﻓﺮﺽ ﺁﻥﻛﻪ ﻫﺮ ﺷﻤﻊ ﺑﻪ ﻣﻴﺰﺍﻥ ﺛﺎﺑﺘﻲ ﺑﺴﻮﺯﺩ ،ﭘﺲ ﺍﺯ ﮔﺬﺷﺖ ﭼﻨﺪ ﺳﺎﻋﺖ ،ﻃﻮﻝ ﺷﻤﻊ ﺍﻭﻝ ﺩﻭ ﺑﺮﺍﺑﺮ ﺷﻤﻊ ﺩﻭﻡ ﻣﻲﺷﻮﺩ؟ 3 4
1/5 (2
2 (3
2/4 (4
(1
.11ﻣﻘﺪﺍﺭﻱ ﭘﻮﻝ ﺩﺭ ﺻﻨﺪﻭﻕ ﻭﺟﻮﺩ ﺩﺍﺭﺩ .ﻗﺮﺍﺭ ﺑﺮ ﺍﻳﻦ ﺍﺳــﺖ ﻛﻪ ﺳﻪ ﻧﻔــﺮ ﺑﻪ ﺗﺮﺗﻴﺐ ﻫﺮﻳﻚ ﺑﻪ ﺍﻧــﺪﺍﺯﻩﻱ ﻣﺒﻠﻐﻲ ﻛﻪ ﺩﺭ ﺻﻨﺪﻭﻕ ﻭﺟﻮﺩ ﺩﺍﺭﺩ ،ﭘــﻮﻝ ﺑﻪ ﺻﻨﺪﻭﻕ ﺍﺿﺎﻓﻪ ﻛﻨﻨﺪ ﻭ ﺳــﭙﺲ ﻣﺒﻠﻎ 40ﺗﻮﻣﺎﻥ ﺍﺯ ﺁﻥ ﺑﺮﺩﺍﺭﻧﺪ .ﺑﻌﺪ ﺍﺯ ﺍﻳﻦﻛﻪ ﻫﺮ ﺳﻪ ﻧﻔﺮ ﺍﻳﻦ ﻛﺎﺭ ﺭﺍ ﺍﻧﺠﺎﻡ ﺩﺍﺩﻧﺪ ،ﺩﺭ ﺻﻨــﺪﻭﻕ ﭘﻮﻟﻲ ﻧﻤﻲﻣﺎﻧﺪ .ﺩﺭ ﺍﺑﺘــﺪﺍ ﭼﻪﻗﺪﺭ ﭘﻮﻝ ﺩﺭ ﺻﻨﺪﻭﻕ ﺑﻮﺩﻩ ﺍﺳﺖ؟ 25 (2ﺗﻮﻣﺎﻥ 55 (1ﺗﻮﻣﺎﻥ 35 (4ﺗﻮﻣﺎﻥ 45 (3ﺗﻮﻣﺎﻥ
.12ﻃﻮﻝ ﻣﺴــﺘﻄﻴﻠﻲ ﺍﺯ 2ﺑﺮﺍﺑﺮ ﻋﺮﺽ ﺁﻥ 4ﻣﺘﺮ ﻛﻢﺗﺮ ﺍﺳﺖ .ﺍﮔﺮ 6 ﻣﺘﺮ ﺍﺯ ﻃﻮﻝ ﻛﻢ ﻛﻨﻴﻢ ﻭ 2ﻣﺘﺮ ﺑﻪ ﻋﺮﺽ ﺍﺿﺎﻓﻪ ﻛﻨﻴﻢ ،ﻣﺴــﺘﻄﻴﻞ ﺗﺒﺪﻳﻞ ﺑﻪ ﻣﺮﺑﻊ ﻣﻲﺷﻮﺩ .ﺍﺧﺘﻼﻑ ﻣﺴﺎﺣﺖ ﺍﻳﻦ ﻣﺴﺘﻄﻴﻞ ﺑﺎ ﻣﺮﺑﻊ ﺑﻪ ﻭﺟﻮﺩ ﺁﻣﺪﻩ ،ﻛﺪﺍﻡ ﻳﻚ ﺍﺯ ﮔﺰﻳﻨﻪﻫﺎﻱ ﺯﻳﺮ ﺍﺳﺖ؟ 12 (2 24 (1 44 (4 48 (3 .13ﺳﻪ ﭘﺴــﺮ ﺑﭽﻪ ﺗﻮﺍﻓﻖ ﻣﻲﻛﻨﻨﺪ ﻛﻪ ﺗﻌﺪﺍﺩﻱ ﻣﻬﺮﻩﻱ ﺩﺍﺧﻞ ﻳﻚ ﻛﻴﺴــﻪ ﺭﺍ ﺑﻪ ﺭﻭﺵ ﺯﻳﺮ ﺑﻴﻦ ﺧﻮﺩ ﺗﻘﺴﻴﻢ ﻛﻨﻨﺪ .ﺁﻥﭼﻪ ﭘﺴﺮﺑﭽﻪﻱ ﺍﻭﻝ ﺑﺮﻣﻲﺩﺍﺭﺩ ،ﻳﻚ ﻣﻬﺮﻩ ﺑﻴﺶﺗﺮ ﺍﺯ ﻧﺼﻒ ﻣﻬﺮﻩﻫﺎﺳﺖ .ﭘﺴﺮﺑﭽﻪﻱ 1 ﺩﻭﻡ 3 ،ﺑﺎﻗﻲﻣﺎﻧــﺪﻩ ﺭﺍ ﺑﺮﻣــﻲﺩﺍﺭﺩ ﻭ 4ﻣﻬــﺮﻩﻱ ﺑﺎﻗﻲﻣﺎﻧــﺪﻩ ﺭﺍ ﭘﺴﺮﺑﭽﻪﻱ ﺳــﻮﻡ ﺑﺮﻣﻲﺩﺍﺭﺩ .ﺗﻌﺪﺍﺩ ﻣﻬﺮﻩﻫﺎﻳﻲ ﻛﻪ ﺑﻪ ﭘﺴﺮﺑﭽﻪﻱ ﺩﻭﻡ ﺭﺳﻴﺪﻩ ،ﻛﺪﺍﻡ ﮔﺰﻳﻨﻪ ﺍﺳﺖ؟ 3 (2 2 (1 6 (4 4 (3 .14ﺩﺭ ﻳﻚ ﺳﺎﻟﻦ n ،ﺻﻨﺪﻟﻲ ﺑﻪ ﮔﻮﻧﻪﺍﻱ ﭼﻴﺪﻩ ﺷﺪﻩ ﺍﺳﺖ ﻛﻪ ﺗﻌﺪﺍﺩ ﺻﻨﺪﻟﻲﻫــﺎ ﺩﺭ ﻫــﺮ ﺭﺩﻳﻒ ﻭ ﻫﺮ ﺳــﺘﻮﻥ ﺑﺎ ﻫﻢ ﺑﺮﺍﺑﺮ ﺍﺳــﺖ ﻭ ﺍﮔﺮ ﺑﺨﻮﺍﻫﻴــﻢ ﺍﺯ ﻫﺮ ﺭﺩﻳﻒ 3ﺻﻨﺪﻟﻲ ﻛﻢ ﻛﻨﻴﻢ ،ﻳﻚ ﺳــﺘﻮﻥ ﺍﺿﺎﻓﻪ ﻣﻲﺷﻮﺩ ﻭ ﺍﮔﺮ ﺍﺯ ﻫﺮ ﺭﺩﻳﻒ 5ﺻﻨﺪﻟﻲ ﻛﻢ ﻛﻨﻴﻢ ،ﺩﻭ ﺳﺘﻮﻥ ﺍﺿﺎﻓﻪ ﻣﻲﺷﻮﺩ n.ﻛﺪﺍﻡ ﻳﻚ ﺍﺯ ﮔﺰﻳﻨﻪﻫﺎﻱ ﺯﻳﺮ ﺍﺳﺖ؟ 60 (1 40 (2 80 (3 50 (4 ﭘﻲﻧﻮﺷﺖ:
1. Tbcorens 2. Proof 3. Matbcmatics 4. Goomctryy 5. Numberr Thoory 6. Algcbra 7. Arithmetic etic 8. Intuitivee
ﺗﻮﺟﻪ :ﭘﺎﺳﺦ ﺍﻳﻦ ﭘﺮﺳﺶﻫﺎ ﺭﺍ ﺩﺭ ﺻﻔﺤـﺔ 42ﻣﺠﻠﻪ ﻣﻄﺎﻟﻌـﻪ ﻛﻨﻴــﺪ.
ﺩﻭﺭﺓ ﺷﺎﻧﺰﺩﻫﻢ ،ﺷﻤﺎﺭﺓ ،4ﺗﺎﺑﺴﺘﺎﻥ 1390
ﺭﺍﻫﻨﻤﺎﻳﻲ
5
ﺭﻳﺎﺿﻲ ﻭ ﺑﺎﺯﻱ
︋︀زی دو ﹡﹀︣ه ﺯﻫﺮﻩ ﭘﻨﺪﻱ
ﻓﻜﺮ ﻭ ﺑﻜﺮ ﺑﺎ ﻋﺪﺩﻫﺎ
ﻛﻠﻴﺪﻭﺍژﻩﻫﺎ :ﺑﺎﺯﻱ ﻭ ﺭﻳﺎﺿﻲ ،ﺣﺪﺱ ﻣﻨﻄﻘﻲ ،ﻋﺪﺩ ،ﺑﺮﻧﺪﻩ ،ﻛﺎﻏﺬ.
ﺍﻳﻦ ﺑﺎﺯﻱ ﻳﻚ ﺑﺎﺯﻱ ﺩﻭ ﻧﻔﺮﻩ ﺍﺳﺖ ﻛﻪ ﺑﺮﺍﻱ ﺍﻧﺠﺎﻡ ﺁﻥ ،ﻓﻘﻂ ﺑﻪ ﻛﺎﻏﺬ ﻭ ﻣﺪﺍﺩ ﺍﺣﺘﻴﺎﺝ ﺩﺍﺭﻳﺪ. • ﻫﺮ ﻛﺪﺍﻡ ﻳﻚ ﺑﺮﮔﻪﻱ ﻛﺎﻏﺬ ﺭﺍ ﺑﻪ ﺷﻜﻞ ﺯﻳﺮ ﺧﻂﻛﺸﻲ ﻛﻨﻴﺪ ﻭ ﺻﻔﺤﻪﻱ ﺑﺎﺯﻱ ﺧﻮﺩﺗﺎﻥ ﺭﺍ ﺑﺴﺎﺯﻳﺪ. ﻋﺪﺩ ﺣﺪﺱ ﺯﺩﻩ ﺷﺪﻩ
ﻗﻀﺎﻭﺕ
1 2 3 4 5 6 7
ﺷﺪﻩ ﺍﺳﺖ. ﻋﻼﻣﺖ × ﺑﺮﺍﻱ ﻫﺮ ﺭﻗﻤﻲ ﻛﻪ ﺩﺭﺳﺖ ﺍ ّﻣﺎ ﻧﺎﺑﻪﺟﺎ ﺣﺪﺱ ﺯﺩﻩ ﺷﺪﻩ ﺍﺳﺖ. ﺑﺮﺍﻱ ﻣﺜﺎﻝ ،ﺍﮔﺮ ﻋﺪﺩ ﺷﻤﺎ 274ﺍﺳﺖ ،ﺑﺮﺍﻱ ﺣﺪﺱ ،235ﻗﻀﺎﻭﺕ ،ﺑﺮﺍﻱ ﺣﺪﺱ 452ﻗﻀﺎﻭﺕ × × ﻭ ﺑﺮﺍﻱ ﺣﺪﺱ 247ﻗﻀﺎﻭﺕ × × ﺭﺍ ﺍﻧﺠﺎﻡ ﺩﻫﻴﺪ. ﺳﻌﻲ ﻛﻨﻴﺪ ﺑﺎ ﺗﻮﺟﻪ ﺑﻪ ﻗﻀﺎﻭﺕ ﺍﻧﺠﺎﻡ ﺷﺪﻩ ﺩﺭ ﺣﺪﺱﻫﺎﻱ ﻗﺒﻠﻲ ﺩﺭ ﻫﺮ ﻣﺮﺣﻠﻪ ﺣﺪﺱ ﺑﻬﺘﺮ ﻳﺎ ﻣﺆﺛﺮﺗﺮﻱ ﺑﺰﻧﻴﺪ! • ﺑﺮﻧﺪﻩﻱ ﺑﺎﺯﻱ ﻛﺴﻲ ﺍﺳﺖ ﻛﻪ ﺯﻭﺩﺗﺮ ﻋﺪﺩ ﺑﺎﺯﻳﻜﻦ ﻣﻘﺎﺑﻠﺶ ﺭﺍ ﺣﺪﺱ ﺑﺰﻧﺪ. ﺑﻪ ﺟﺪﻭﻝ ﭘﺮ ﺷﺪﻩﻱ ﺯﻳﺮ ﻧﮕﺎﻩ ﻛﻨﻴﺪ .ﺁﻳﺎ ﻣﻲﺗﻮﺍﻧﻴﺪ ﻋﺪﺩ ﺩﺭﺳﺖ ﺭﺍ ﺣﺪﺱ ﺑﺰﻧﻴﺪ؟
8
ﻗﻀﺎﻭﺕ
ﻋﺪﺩ ﺣﺪﺱ ﺯﺩﻩ ﺷﺪﻩ
9
××
762
1
×
862
2
516
3
942
4
10 11 12
• ﻫﺮﻳﻚ ،ﻋﺪﺩﻱ ﺳﻪ ﺭﻗﻤﻲ ﺍﻧﺘﺨﺎﺏ ﻛﻨﻴﺪ ﻭ ﻃﻮﺭﻱ ﻛﻪ ﺑﺎﺯﻳﻜﻦ ﻣﻘﺎﺑﻞ ﻧﺒﻴﻨﺪ ،ﺁﻥ ﺭﺍ ﭘﺸﺖ ﺻﻔﺤﻪﻱ ﺑﺎﺯﻱ ﺧﻮﺩﺗﺎﻥ ﺑﻨﻮﻳﺴﻴﺪ. • ﺩﺭ ﻃﻮﻝ ﺑﺎﺯﻱ ﻗﺮﺍﺭ ﺍﺳﺖ ﻫﺮﻳﻚ ﺍﺯ ﺷﻤﺎ ﺑﺎ ﺣﺪﺱﻫﺎﻱ ﻣﻨﻄﻘﻲ، ﻋﺪﺩ ﺍﻧﺘﺨﺎﺑﻲ ﺑﺎﺯﻳﻜﻦ ﺣﺮﻳﻒ ﺭﺍ ﺣﺪﺱ ﺑﺰﻧﻴﺪ .ﺑﺎﺯﻱ ﺭﺍ ﺑﺎ ﻳﻚ ﺣﺪﺱ ﺷﺮﻭﻉ ﻛﻨﻴﺪ ﻭ ﻫﺮ ﺩﻭ ﻳﻚ ﻋﺪﺩ ﺑﮕﻮﻳﻴﺪ. • ﻋﺪﺩﻱ ﺭﺍ ﻛﻪ ﻣﻲﮔﻮﻳﻴﺪ ،ﺩﺭ ﺳﺘﻮﻥ ﻋﺪﺩ ﺣﺪﺱ ﺯﺩﻩ ﺷﺪﻩ ﺑﻨﻮﻳﺴﻴﺪ ﻭ ﻗﻀﺎﻭﺕ ﺑﺎﺯﻳﻜﻦ ﻣﻘﺎﺑﻞ ﺭﺍ ﺭﻭﺑﻪﺭﻭﻱ ﺁﻥ ﺩﺭ ﺳﺘﻮﻥ ﻗﻀﺎﻭﺕ ﻭﺍﺭﺩ ﻛﻨﻴﺪ. • ﺧﻮﺩﺗﺎﻥ ﻧﻴﺰ ﻋﺪﺩ ﺣﺪﺱ ﺯﺩﻩ ﺷﺪﻩ ﺗﻮﺳﻂ ﺑﺎﺯﻳﻜﻦ ﻣﻘﺎﺑﻞ ﺭﺍ ﺑﺎ ﻋﺪﺩ ﺍﻧﺘﺨﺎﺑﻲ ﺧﻮﺩﺗﺎﻥ ﻣﻘﺎﻳﺴﻪ ﻛﻨﻴﺪ ﻭ ﺁﻥ ﺭﺍ ﺑﻪ ﺻﻮﺭﺕ ﺯﻳﺮ ﻗﻀﺎﻭﺕ ﻛﻨﻴﺪ. ﻋﻼﻣﺖ ﺑﺮﺍﻱ ﻫﺮ ﺭﻗﻤﻲ ﻛﻪ ﺩﺭﺳﺖ ﻭ ﺩﺭ ﺟﺎﻱ ﺧﻮﺩ ﺣﺪﺱ ﺯﺩﻩ 6
ﺭﺍﻫﻨﻤﺎﻳﻲ
ﺩﻭﺭﺓ ﺷﺎﻧﺰﺩﻫﻢ ،ﺷﻤﺎﺭﺓ ،4ﺗﺎﺑﺴﺘﺎﻥ 1390
ﺣﺪﺱ 1ﻭ 2ﺭﺍ ﺑﺎ ﻫﻢ ﻣﻘﺎﻳﺴﻪ ﻛﻨﻴﺪ .ﺁﻳﺎ ﻣﻲﺗﻮﺍﻥ ﻣﻄﻤﺌﻦ ﺑﻮﺩ ﻛﻪ ﺭﻗﻢ ﺻﺪﮔﺎﻥ 8ﺍﺳﺖ؟ ﺣﺎﻻ ﺑﻪ ﺣﺪﺱ 3ﻧﮕﺎﻩ ﻛﻨﻴﺪ 5 .ﻛﻪ ﺩﺭ ﺟﺎﻱ ﺧﻮﺩﺵ ﻧﻴﺴﺖ! ﻓﻜﺮ ﻣﻲﻛﻨﻴﺪ 1ﺩﺭ ﺟﺎﻱ ﺧﻮﺩﺵ ﺍﺳﺖ ﻳﺎ 6؟ ﺍﮔﺮ 1ﺩﺭ ﺟﺎﻱ ﺧﻮﺩﺵ ﺑﺎﺷﺪ، ﺩﺭ ﺣﺪﺱ ،2ﺭﻗﻢ 2ﺩﺭﺳﺖ ﺑﻪ ﻛﺎﺭ ﺭﻓﺘﻪ ﺍﺳﺖ ﻳﺎ 6؟ ﺍﮔﺮ 6ﺩﺭ ﺣﺪﺱ 3ﺩﺭ ﺟﺎﻱ ﺧﻮﺩﺵ ﺑﺎﺷﺪ ،ﭼﻪﻃﻮﺭ؟ ﺁﻳﺎ ﻣﻲﺗﻮﺍﻥ ﻣﻄﻤﺌﻦ ﺑﻮﺩ ﻛﻪ ﺭﻗﻢ ﻳﻜﺎﻥ 6ﺍﺳﺖ؟ ﺑﻪ ﺣﺪﺱ 4ﺗﻮﺟﻪ ﻛﻨﻴﺪ! ﻓﻜﺮ ﻣﻲﻛﻨﻴﺪ ﻛﺪﺍﻡ ﺭﻗﻢ ﺩﺭ ﺟﺎﻱ ﺧﻮﺩ ﻗﺮﺍﺭ ﺩﺍﺭﺩ؟ ﺁﻳﺎ ﻣﻲﺗﻮﺍﻥ ﻣﻄﻤﺌﻦ ﺑﻮﺩ ﻛﻪ ﻋﺪﺩ ﺍﻧﺘﺨﺎﺑﻲ 846 ﺑﻮﺩﻩ ﺍﺳﺖ؟ ﻓﻜﺮ ﻣﻲﻛﻨﻴﺪ ﻗﻀﺎﻭﺕﻫﺎﻱ ﺑﺮﺍﻱ ﺣﺪﺱﻫﺎﻱ ﺑﻬﺘﺮ ﻛﻤﻚ ﺑﻴﺶﺗﺮﻱ ﻣﻲﻛﻨﻨﺪ ﻳﺎ ﻗﻀﺎﻭﺕﻫﺎﻱ × ؟
؟
ﺩﺍﻧﺶﺍﻓﺰﺍﻳﻲ
در ﹞︧︐︴﹫﹏ ، ︵﹢ل ﹋︡ام ا︨️؟ ! ﺯﻳﻨﺐ ﻣﺮﺍﺩﺧﺎﻧﻰ
ﻛﻠﻴﺪﻭﺍژﻩﻫﺎ :ﻃﻮﻝ ،ﻋﺮﺽ ،ﻣﺴﺘﻄﻴﻞ ،ﺩﺳﺘﮕﺎﻩ ﻣﺨﺘﺼﺎﺕ. ﺩﺭ ﻣﺴﺘﻄﻴﻞﻫﺎﻯ ﻣﻘﺎﺑﻞ ﻃﻮﻝ ﻭ ﻋﺮﺽ ﺭﺍ ﻣﺸﺨﺺ ﻛﻨﻴﺪ :
ﺩﺭ ﺯﻧﺪﮔﻰ ﺭﻭﺯﻣﺮﻩ ﻣﺮﺳــﻮﻡ ﺍﺳﺖ ﻛﻪ ﺩﺭ ﻫﺮ ﻣﺴﺘﻄﻴﻞ ،ﺑﻠﻨﺪﺗﺮﻳﻦ ﺿﻠﻊ ﺭﺍ ﻃﻮﻝ ﻭ ﻛﻮﺗﺎﻩﺗﺮﻳﻦ ﺿﻠﻊ ﺭﺍ ﻋﺮﺽ ﻣﻰﮔﻴﺮﻧﺪ .ﺁﻳﺎ ﭼﻨﻴﻦ ﻣﻄﻠﺒﻰ ﻣﻰﺗﻮﺍﻧﺪ ﻭﺍﻗﻌﻴﺖ ﺩﺍﺷــﺘﻪ ﺑﺎﺷــﺪ؟ ﺑﺎ ﻣﺮﻭﺭﻯ ﺑﺮ ﻳــﻚ ﻛﺮﻩ ﺟﻐﺮﺍﻓﻴﺎﻳﻰ، ﻣﻰﺗــﻮﺍﻥ ﺑﻪ ﺍﻳﻦ ﻣﻄﻠﺐ ﺭﺳــﻴﺪ ﻛــﻪ ﻣﺪﺍﺭﻫﺎ ﺧﻂﻫــﺎﻯ ﺍﻓﻘﻰ ﻭ ﻧﺼﻒ ﺍﻟﻨﻬﺎﺭﻫــﺎ ﺧﻄﻬﺎﻯ ﻋﻤﻮﺩﻯ ﻫﺴــﺘﻨﺪ ،ﻛﻪ ﺍﻳﻦ ﻣﻮﺿــﻮﻉ ﺭﺍ ﻣﻰﺗﻮﺍﻥ ﺑﻪ ﺩﺳــﺘﮕﺎﻩ ﻣﺨﺘﺼﺎﺕ ﻧﺴﺒﺖ ﺩﺍﺩ ﻭ ﺁﻥ )ﺧﻂﻫﺎﻱ ﺍﻓﻘﻲ ﻭ ﻧﺼﻒﺍﻟﻨﻬﺎﺭﻫﺎ( ﺭﺍ ﻣﺤﻮﺭ ﻃﻮﻝ ﻭ ﻣﺤﻮﺭ ﻋﻤﻮﺩﻯ ﺭﺍ ﻣﺤﻮﺭ ﻋﺮﺽ ﻧﺎﻣﻴﺪ. ﺣﺎﻝ ﺍﮔﺮ ﺩﻭ ﻣﺴــﺘﻄﻴﻞ aﻭ bﺭﺍ ﺭﻭﻯ ﺩﺳﺘﮕﺎﻩ ﻣﺨﺘﺼﺎﺕ ﻧﻤﺎﻳﺶ ﺩﻫﻴﻢ ،ﺑﺎﻋﺚ ﺍﻳﻦ ﻛﺸــﻒ ﻣﻰ ﺷﻮﻳﻢ ﻛﻪ ﻃﻮﻝ ﻣﻰﺗﻮﺍﻧﺪ ﺍﺯ ﻋﺮﺽ ﻛﻢﺗﺮ ﻳﺎ ﺑﻴﺶﺗﺮ ﺑﺎﺷﺪ.
b
a
ﺩﻭﺭﺓ ﺷﺎﻧﺰﺩﻫﻢ ،ﺷﻤﺎﺭﺓ ،4ﺗﺎﺑﺴﺘﺎﻥ 1390
ﺭﺍﻫﻨﻤﺎﻳﻲ
7
ﺩﺍﻧﺶﺍﻓﺰﺍﻳﻲ
︋︩︎︢︣ی ﻣﺤﻤﻮﺩ ﺩﺍﻭﺭﺯﻧﻲ
ﻛﻠﻴﺪﻭﺍژﻩﻫﺎ :ﺑﺨﺶﭘﺬﻳﺮﻱ ،ﺗﻘﺴﻴﻢ ،ﺍﻋﺪﺍﺩ 1ﺗﺎ ،30ﺍﻋﺪﺍﺩ ﺯﻭﺝ ،ﺍﻋﺪﺍﺩ ﻓﺮﺩ ،ﻣﻀﺎﺭﺏ. ﺁﻳﺎ ﺗﺎ ﺑﻪ ﺣﺎﻝ ﺑﺎ ﺍﻳﻦ ﺳﺆﺍﻻﺕ ﺑﺮﺧﻮﺭﺩ ﻛﺮﺩﻩﺍﻳﺪ؟ ﺁﻳﺎ 11253ﺑﺮ 3ﺑﺨﺶﭘﺬﻳﺮ ﺍﺳﺖ؟ ﺁﻳﺎ 4591ﺑﺮ 7ﺑﺨﺶﭘﺬﻳﺮ ﺍﺳﺖ؟ ﺑﺮ 11ﭼﻪﻃﻮﺭ؟ ﺍﮔﺮ ﻣﺘﻦ ﺯﻳﺮ ﺭﺍ ﻣﻄﺎﻟﻌﻪ ﻛﻨﻴﺪ ،ﻣﻲﺗﻮﺍﻧﻴﺪ ﺑﺪﻭﻥ ﺍﻧﺠﺎﻡ ﻋﻤﻞ ﺗﻘﺴﻴﻢ، ﺑﺨﺶﭘﺬﻳﺮﻱ ﻫﺮ ﻋﺪﺩﻱ ﺭﺍ ﺑﺮ ﺑﻌﻀﻲ ﺍﺯ ﺍﻋﺪﺍﺩ 1ﺗﺎ 30ﺑﺮﺭﺳــﻲ ﻛﻨﻴﺪ. ﺑﻌﻀــﻲ ﺍﺯ ﻗﺴــﻤﺖﻫﺎ ﺑﻪ ﺗﻮﺿﻴﺢ ﺑﻴﺸــﺘﺮﻱ ﻧﻴﺎﺯ ﺩﺍﺭﻧﺪ ﻛــﻪ ﺁﻥﻫﺎ ﺭﺍ ﺑﺎ ﻋﻼﻣﺖ * ﻣﺸﺨﺺ ﻛﺮﺩﻩﺍﻳﻢ ﻭ ﺩﺭ ﭘﺎﻳﺎﻥ ﺑﻪ ﺁﻥﻫﺎ ﻣﻲﭘﺮﺩﺍﺯﻳﻢ. ﺑﺨﺶﭘﺬﻳﺮﻱ ﺑﺮ :1ﻫﻤﺔ ﺍﻋﺪﺍﺩ ﺑﺮ ﻋﺪﺩ 1ﺑﺨﺶﭘﺬﻳﺮﻧﺪ. ﺑﺨﺶﭘﺬﻳــﺮﻱ ﺑﺮ :2ﻫﻤﺔ ﺍﻋﺪﺍﺩ ﺯﻭﺝ ﺑــﺮ ﻋﺪﺩ 2ﺑﺨﺶﭘﺬﻳﺮﻧﺪ )ﺑــﻪ ﻋﺪﺩﻱ ﺯﻭﺝ ﻣﻲﮔﻮﻳﻴﻢ ﻛﻪ ﺭﻗﻢ ﻳﻜﺎﻥ ﺁﻥ ﻳﻜﻲ ﺍﺯ ﺍﻋﺪﺍﺩ ،8 ،6 ،4، 2 ،0ﺑﺎﺷﺪ( ﺑﺨﺶﭘﺬﻳــﺮﻱ ﺑــﺮ :3ﺍﮔــﺮ ﻣﺠﻤــﻮﻉ ﺍﺭﻗــﺎﻡ ﻳﻚ ﻋــﺪﺩ ﺑﺮ 3 ﺑﺨﺶﭘﺬﻳﺮ ﺑﺎﺷــﺪ ،ﺁﻥ ﻋﺪﺩ ﺑﺮ 3ﺑﺨﺶﭘﺬﻳﺮ ﺍﺳــﺖ. ﻣﺜ ً ﻼ 78ﺑﺮ 3ﺑﺨﺶﭘﺬﻳﺮ ﺍﺳــﺖ ،ﺯﻳــﺮﺍ 8+7 =15ﻭ 15ﺑﺮ 3ﺑﺨﺶﭘﺬﻳﺮ ﺍﺳﺖ .ﻭﻟﻲ ﻋﺪﺩ 259204ﺑﺮ 3 ﺑﺨﺶﭘﺬﻳﺮ ﻧﻴﺴــﺖ ،ﺯﻳﺮﺍ 2+5+9+2+0+4=22ﻭ 22 ﺑﺮ 3ﺑﺨﺶﭘﺬﻳﺮ ﻧﻴﺴﺖ. ﺑﺨﺶﭘﺬﻳﺮﻱ ﺑﺮ :4ﺍﮔﺮ ﺁﺧﺮﻳﻦ ﺩﻭ ﺭﻗﻢ ﻳﻚ ﻋﺪﺩ ﺑﺮ 4ﺑﺨﺶﭘﺬﻳﺮ ﺑﺎﺷــﻨﺪ ،ﺁﻥ ﻋﺪﺩ ﺑﺮ 4ﺑﺨﺶﭘﺬﻳﺮ ﺍﺳﺖ .ﻣﺜ ً ﻼ 716ﺑﺮ 4ﺑﺨﺶﭘﺬﻳﺮ ﺍﺳﺖ ،ﺯﻳﺮﺍ 16ﺑﺮ 4ﺑﺨﺶﭘﺬﻳﺮ ﺍﺳﺖ. ﺑﺨﺶﭘﺬﻳــﺮﻱ ﺑﺮ :5ﺍﮔــﺮ ﺭﻗﻢ ﻳﻜﺎﻥ ﻳﻚ ﻋﺪﺩ 0ﻳﺎ 5ﺑﺎﺷــﺪ، ﺁﻥ ﻋﺪﺩ ﺑﺮ 5ﺑﺨﺶﭘﺬﻳﺮ ﺍﺳــﺖ .ﻣﺜــ ً ﻼ 4995ﺑﺮ 5 ﺑﺨﺶﭘﺬﻳﺮ ﺍﺳﺖ. ﺑﺨﺶﭘﺬﻳــﺮﻱ ﺑﺮ :6ﺍﮔﺮ ﻳــﻚ ﻋﺪﺩ ﺯﻭﺝ ﺑﺎﺷــﺪ ﻭ ﻫﻢﭼﻨﻴﻦ ﺑﺮ 3ﻧﻴﺰ ﺑﺨﺶﭘﺬﻳﺮ ﺑﺎﺷــﺪ ،ﺣﺘﻤﺎً ﺑﺮ 6ﻧﻴﺰ ﺑﺨﺶﭘﺬﻳﺮ ﺍﺳــﺖ. ﻣﺜ ً ﻼ 73452ﺑﺮ 2ﻭ 3ﺑﺨﺶﭘﺬﻳﺮ ﺍﺳــﺖ ،ﭘﺲ ﺑﺮ 6 ﻧﻴﺰ ﺑﺨﺶﭘﺬﻳﺮ ﺍﺳﺖ. 8
ﺭﺍﻫﻨﻤﺎﻳﻲ
ﺩﻭﺭﺓ ﺷﺎﻧﺰﺩﻫﻢ ،ﺷﻤﺎﺭﺓ ،4ﺗﺎﺑﺴﺘﺎﻥ 1390
ﺑﺨﺶﭘﺬﻳﺮﻱ ﺑﺮ * :7
ﺑﺨﺶﭘﺬﻳــﺮﻱ ﺑــﺮ :8ﺍﮔﺮ ﺁﺧﺮﻳﻦ ﺳــﻪ ﺭﻗﻢ ﻳﻚ ﻋــﺪﺩ ﺑﺮ 8 ﺑﺨﺶﭘﺬﻳﺮ ﺑﺎﺷــﺪ ،ﺁﻥ ﻋﺪﺩ ﺑﺮ 8ﺑﺨﺶﭘﺬﻳﺮ ﺍﺳــﺖ. ﻣﺜ ً ﻼ 3128ﺑﺮ 8ﺑﺨﺶﭘﺬﻳﺮ ﺍﺳــﺖ ،ﺯﻳﺮﺍ 128ﺑﺮ 8 ﺑﺨﺶﭘﺬﻳﺮ ﺍﺳﺖ )(128=8×16 ﺑﺨﺶﭘﺬﻳــﺮﻱ ﺑــﺮ :9ﺍﮔﺮ ﻣﺠﻤﻮﻉ ﻳﻚ ﻋﺪﺩ ﺑــﺮ 9ﺑﺨﺶﭘﺬﻳﺮ ﺑﺎﺷﺪ ،ﺁﻥ ﻋﺪﺩ ﺑﺮ 9ﺑﺨﺶﭘﺬﻳﺮ ﺍﺳﺖ .ﻣﺜ ً ﻼ 5148ﺑﺮ 9ﺑﺨﺶﭘﺬﻳﺮ ﺍﺳــﺖ ،ﺯﻳﺮﺍ 5+1+4+8=18ﻭ 18ﺑﺮ 9 ﺑﺨﺶﭘﺬﻳﺮ ﺍﺳﺖ. ﺑﺨﺶﭘﺬﻳــﺮﻱ ﺑﺮ :10ﺍﮔﺮ ﻳﻚ ﻋﺪﺩ ﺑﻪ ﺻﻔﺮ ﺧﺘﻢ ﺷــﻮﺩ ،ﺁﻥ ﻋﺪﺩ ﺑﺮ 10ﺑﺨﺶﭘﺬﻳﺮ ﺍﺳــﺖ ،ﻣﺎﻧﻨــﺪ ﺍﻋﺪﺍﺩ 790ﻭ .23350 ﺑﺨﺶﭘﺬﻳــﺮﻱ ﺑﺮ :11ﺍﮔﺮ ﺍﺧﺘﻼﻑ ﺑﻴﻦ ﻣﺠﻤﻮﻉ ﺭﻗﻢﻫﺎﻳﻲ ﻛﻪ ﺩﺭ ﻣــﻜﺎﻥ ﺯﻭﺝ ﻗﺮﺍﺭ ﺩﺍﺭﻧﺪ ﺭﺍ ﺍﺯ ﻣﺠﻤﻮﻉ ﺭﻗﻢﻫﺎﻳﻲ ﻛﻪ ﺩﺭ ﻣــﻜﺎﻥ ﻓﺮﺩ ﻗﺮﺍﺭ ﺩﺍﺭﻧﺪ ،ﻣﺤﺎﺳــﺒﻪ ﻛﻨﻴﻢ ﻭ ﺣﺎﺻﻞ ﺻﻔﺮ ﺑﺎﺷــﺪ ﻭ ﻳــﺎ ﺑﺮ 11ﺑﺨﺶﭘﺬﻳﺮ ﺑﺎﺷــﺪ ،ﺁﻥ ﻋﺪﺩ ﺑﺮ 11ﺑﺨﺶﭘﺬﻳﺮ ﺍﺳــﺖ .ﻣﺜــ ً ﻼ ﻋﺪﺩ 54934ﺑﺮ 11 ﺑﺨﺶﭘﺬﻳﺮ ﺍﺳﺖ ،ﺯﻳﺮﺍ (4+9+5) - (3+4) =11ﻭ 11 ﺑﺮ 11ﺑﺨﺶﭘﺬﻳﺮ ﺍﺳﺖ. ﺑﺨﺶﭘﺬﻳــﺮﻱ ﺑــﺮ :12ﺍﮔﺮ ﻣﺠﻤــﻮﻉ ﺍﺭﻗﺎﻡ ﻳﻚ ﻋــﺪﺩ ﺑﺮ 3 ﺑﺨﺶﭘﺬﻳﺮ ﺑﺎﺷﺪ ﻭ ﺁﺧﺮﻳﻦ ﺩﻭ ﺭﻗﻢ ﺁﻥ ﻋﺪﺩ ﻧﻴﺰ ﺑﺮ 12 ﺑﺨﺶﭘﺬﻳﺮ ﺑﺎﺷــﺪ ،ﺁﻥ ﻋﺪﺩ ﺑﺮ 12ﺑﺨﺶﭘﺬﻳﺮ ﺍﺳﺖ. ﻣﺜ ً ﻼ ﻋﺪﺩ 62532ﺑﺮ 12ﺑﺨﺶﭘﺬﻳﺮ ﺍﺳﺖ ،ﺯﻳﺮﺍ =18 6+2+5+3+2ﻭ 18ﺑﺮ 3ﺑﺨﺶﭘﺬﻳﺮ ﺍﺳﺖ .ﻫﻢﭼﻨﻴﻦ، ﺁﺧﺮﻳﻦ ﺩﻭ ﺭﻗﻢ ﺁﻥ ﻳﻌﻨﻲ 32ﺑﺮ 4ﺑﺨﺶﭘﺬﻳﺮ ﺍﺳﺖ، ﭘﺲ ﺍﻳﻦ ﻋﺪﺩ ﺑﺮ 4ﺑﺨﺶﭘﺬﻳﺮ ﺍﺳﺖ. ﺑﺨﺶﭘﺬﻳﺮﻱ ﺑﺮ * :13
ﺑﺨﺶﭘﺬﻳﺮﻱ ﺑﺮ * :17 ﺑﺨﺶﭘﺬﻳﺮﻱ ﺑﺮ * :19
ﺑﺨﺶﭘﺬﻳــﺮﻱ ﺑﺮ :20ﺍﮔﺮ ﺭﻗﻢ ﺩﻫﮕﺎﻥ ﻳﻚ ﻋﺪﺩ ﺯﻭﺝ ﺑﺎﺷــﺪ ﻭ ﺭﻗــﻢ ﻳــﻜﺎﻥ ﺁﻥ ﻧﻴﺰ ﺻﻔﺮ ﺑﺎﺷــﺪ ،ﺁﻥ ﻋــﺪﺩ ﺑﺮ 20 ﺑﺨﺶﭘﺬﻳﺮ ﺍﺳﺖ ،ﻣﺎﻧﻨﺪ ﻋﺪﺩ .7960 ﺑﺨﺶﭘﺬﻳــﺮﻱ ﺑــﺮ :21ﺍﮔﺮ ﻳﻚ ﻋﺪﺩ ﺑﺮ 7ﺑﺨﺶﭘﺬﻳﺮ ﺑﺎﺷــﺪ ﻭ ﻣﺠﻤــﻮﻉ ﺍﺭﻗﺎﻡ ﺁﻥ ﻧﻴﺰ ﺑﺮ 3ﺑﺨﺶﭘﺬﻳﺮ ﺑﺎﺷــﺪ ،ﺁﻥ ﻋﺪﺩ ﺑﺮ 21ﺑﺨﺶﭘﺬﻳﺮ ﺍﺳــﺖ .ﻣﺜ ً ﻼ ﻋﺪﺩ 1638ﺑﺮ 7 ﺑﺨﺶﭘﺬﻳﺮ ﺍﺳﺖ ﻭ ﻫﻢﭼﻨﻴﻦ ﻣﺠﻤﻮﻉ ﺍﺭﻗﺎﻡ ﺁﻥ ﻳﻌﻨﻲ 1+6+3+8=18ﺑﺮ 3ﺑﺨﺶﭘﺬﻳﺮ ﺍﺳﺖ ،ﭘﺲ ﺍﻳﻦ ﻋﺪﺩ ﺑﺮ 21ﻧﻴﺰ ﺑﺨﺶﭘﺬﻳﺮ ﺍﺳﺖ. ﺑﺨﺶﭘﺬﻳﺮﻱ ﺑﻪ * :23 ﺑﺨﺶﭘﺬﻳﺮﻱ ﺑﺮ * :29 ﺑﺮﺍﻱ ﺑﺮﺭﺳــﻲ ﺑﺨﺶﭘﺬﻳﺮﻱﻫﺎﻳﻲ ﻛﻪ ﺩﺭ ﺑﺎﻻ ﺑﺎ ﻋﻼﻣﺖ * ﻣﺸﺨﺺ ﺷﺪﻩﺍﻧﺪ ،ﺭﻭﺵ ﺳﺎﺩﻩﺍﻱ ﺑﻪ ﻧﺎﻡ ﺭﻭﺵ ﻣﻀﺎﺭﺏ ﺑﺮﺭﺳﻲﻛﻨﻨﺪﺓ ﺑﺨﺶﭘﺬﻳﺮﻱ ﻭ ﻳــﺎ ﺑﻪ ﺍﺧﺘﺼﺎﺭ »ﻣﻀﺎﺭﺏ ﺑﺮﺭﺳــﻲﻛﻨﻨﺪﻩ« ﻭﺟــﻮﺩ ﺩﺍﺭﺩ ﻛﻪ ﺑﺮﺍﻱ ﻫﺮ ﺑﺨﺶﭘﺬﻳﺮﻱ ﺑﻪ ﻛﻤﻚ ﻋﺪﺩﻱ ﺧﺎﺹ ﺑﻪ ﻧﺎﻡ »ﻣﻀﺮﺏ ﺑﺮﺭﺳــﻲﻛﻨﻨﺪﻩ« ﺑﻪ ﺭﺍﺣﺘﻲ ﻣﻲﺗﻮﺍﻧﻴﻢ ﺑﺨﺶﭘﺬﻳﺮﻱ ﻫﺮ ﻋﺪﺩﻱ ﺍﺳﺘﻔﺎﺩﻩ ﻛﺮﺩ. ﺩﺭ ﺯﻳﺮ ﻧﺸﺎﻥ ﻣﻲﺩﻫﻴﻢ ﻛﻪ ﭼﻪﻃﻮﺭ ﺍﺯ »ﻣﻀﺮﺏ ﺑﺮﺭﺳﻲﻛﻨﻨﺪﻩ« ﺩﺭ ﺑﺨﺶﭘﺬﻳﺮﻱ ﻳﻚ ﻋﺪﺩ ﺍﺳــﺘﻔﺎﺩﻩ ﻣﻲﺷــﻮﺩ ﻭ ﺑﻌﺪ ﺍﺯ ﺭﻭﺵ ﭘﻴﺪﺍ ﻛﺮﺩﻥ ﺍﻳﻦ ﻣﻀﺮﺏ ﺭﺍ ﺗﻮﺿﻴﺢ ﻣﻲﺩﻫﻴﻢ.
ﺭﻭﺵ »ﻣﻀﺎﺭﺏ ﺑﺮﺭﺳﻲﻛﻨﻨﺪﻩ«
ﺑﺮﺍﻱ ﺑﺮﺭﺳــﻲ ﻛﺮﺩﻥ ﺍﻳﻦﻛﻪ ﻋﺪﺩﻱ ﺑﺮ 7ﺑﺨﺶﭘﺬﻳﺮ ﺍﺳﺖ ،ﻳﺎ ﻧﻪ، ﻋﺪﺩ 5ﺭﺍ ﺑﻪ ﻋﻨﻮﺍﻥ ﻣﻀﺮﺏ ﺑﺮﺭﺳــﻲﻛﻨﻨﺪﻩ ﺍﻧﺘﺨﺎﺏ ﻣﻲﻛﻨﻴﻢ 5 .ﺭﺍ ﺩﺭ ﺭﻗﻢ ﻳﻜﺎﻥ ﻋﺪﺩ ﺩﺍﺩﻩ ﺷﺪﻩ ﺿﺮﺏ ﻣﻲﻛﻨﻴﻢ ﻭ ﺣﺎﺻﻞ ﺭﺍ ﺑﻪ ﻋﺪﺩﻱ ﻛﻪ ﺍﺯ ﺣﺬﻑ ﺭﻗﻢ ﻳﻜﺎﻥ ﺑﻪ ﺩﺳــﺖ ﺁﻣﺪﻩ ﺍﺳﺖ ،ﻣﻲﺍﻓﺰﺍﻳﻴﻢ .ﺍﮔﺮ ﺍﻳﻦ ﻋﺪﺩ ﺑﺮ 7 ﺑﺨﺶﭘﺬﻳﺮ ﺑﺎﺷﺪ ،ﻋﺪﺩ ﺍﻭﻟﻴﻪ ﻧﻴﺰ ﺑﺮ 7ﺑﺨﺶﭘﺬﻳﺮ ﺍﺳﺖ ﻭ ﺍﮔﺮ ﺍﻳﻦ ﻋﺪﺩ ﺑﺮ 7ﺑﺨﺶﭘﺬﻳﺮ ﻧﺒﺎﺷﺪ ،ﻋﺪﺩ ﺍﻭﻟﻴﻪ ﻧﻴﺰ ﺑﺮ 7ﺑﺨﺶﭘﺬﻳﺮ ﻧﺨﻮﺍﻫﺪ ﺑﻮﺩ. ﺑــﺮﺍﻱ ﺍﻭﻟﻴﻦ ﻣﺜﺎﻝ ،ﻣﻲﺧﻮﺍﻫﻴﻢ ﺑﺮﺭﺳــﻲ ﻛﻨﻴﻢ ﻛــﻪ ﺁﻳﺎ 91ﺑﺮ 7 ﺑﺨﺶﭘﺬﻳﺮ ﺍﺳﺖ ﻳﺎ ﻧﻪ؟ ﺭﻗﻢ ﻳﻜﺎﻥ ﺍﻳﻦ ﻋﺪﺩ ﻳﻌﻨﻲ 1ﺭﺍ ﺩﺭ ﻣﻀﺮﺏ ﺑﺮﺭﺳــﻲﻛﻨﻨﺪﻩ ﻳﻌﻨﻲ 5 ﺿﺮﺏ ﻣﻲﻛﻨﻴﻢ .ﺟﻮﺍﺏ 5ﺍﺳــﺖ .ﺍﮔــﺮ 5ﺭﺍ ﺑﻪ 9ﺍﺿﺎﻓﻪ ﻛﻨﻴﻢ ،ﺣﺎﺻﻞ 14ﻣﻲﺷــﻮﺩ ﻛﻪ ﺑﺮ 7ﺑﺨﺶﭘﺬﻳﺮ ﺍﺳﺖ ،ﭘﺲ 91ﻧﻴﺰ ﺑﺮ 7ﺑﺨﺶﭘﺬﻳﺮ ﺍﺳﺖ. ﺩﺭ ﻣﺜﺎﻟﻲ ﺩﻳﮕﺮ ،ﻣﻲﺧﻮﺍﻫﻴﻢ ﺑﺒﻴﻨﻴﻢ ﻛﻪ ﺁﻳﺎ 123ﺑﺮ 7ﺑﺨﺶﭘﺬﻳﺮ ﺍﺳﺖ ﻳﺎ ﻧﻪ؟ ﺭﻗــﻢ ﻳﻜﺎﻥ ﺍﻳﻦ ﻋﺪﺩ ﻳﻌﻨــﻲ 3ﺭﺍ ﺩﺭ ) 5ﻣﻀﺮﺏ ﺑﺮﺭﺳــﻲﻛﻨﻨﺪﻩ( ﺿــﺮﺏ ﻣﻲﻛﻨﻴﻢ .ﺟﻮﺍﺏ ﺑﻪ ﺩﺳــﺖ ﺁﻣــﺪﻩ ﻳﻌﻨــﻲ 15ﺭﺍ ﺑﺎ 12ﺟﻤﻊ ﻣﻲﻛﻨﻴﻢ ﻛﻪ 27ﻣﻲﺷــﻮﺩ ﻭ ﭼﻮﻥ 27ﺑﺮ 7ﺑﺨﺶﭘﺬﻳﺮ ﻧﻴﺴــﺖ ،ﭘﺲ 123ﻧﻴﺰ ﺑﺮ 7ﺑﺨﺶﭘﺬﻳﺮ ﻧﻴﺴﺖ. ﺑﺮﺍﻱ ﺁﺧﺮﻳﻦ ﻣﺜﺎﻝ ،ﺁﻳﺎ 1638ﺑﺮ 7ﺑﺨﺶﭘﺬﻳﺮ ﺍﺳﺖ ﻳﺎ ﻧﻪ؟ ﺭﻗــﻢ ﻳﻜﺎﻥ ﺍﻳﻦ ﻋــﺪﺩ ﻳﻌﻨﻲ 8ﺭﺍ ﺩﺭ 5ﺿــﺮﺏ ﻣﻲﻛﻨﻴﻢ .ﺣﺎﺻﻞ 40ﻣﻲﺷــﻮﺩ .ﺍﮔﺮ 40ﺭﺍ ﺑــﻪ 163ﺍﺿﺎﻓﻪ ﻛﻨﻴﻢ ،ﺟﻮﺍﺏ 203ﺍﺳــﺖ. ﺑﺮﺍﻱ ﺍﻳﻦﻛﻪ ﺑﺒﻴﻨﻴﻢ 203ﺑﺮ 7ﺑﺨﺶﭘﺬﻳﺮ ﺍﺳــﺖ ﻳﺎ ﻧﻪ ،ﺩﻭﺑﺎﺭﻩ ﻫﻤﻴﻦ ﻗﺎﻋﺪﻩ ﺭﺍ ﺑﻪ ﻛﺎﺭ ﻣﻲﺑﺮﻳﻢ .ﺭﻗﻢ ﻳﻜﺎﻥ ﻋﺪﺩ 203ﻳﻌﻨﻲ 3ﺭﺍ ﺩﺭ 5ﺿﺮﺏ ﻣﻲﻛﻨﻴﻢ ،ﺣﺎﺻﻞ 15ﻣﻲﺷــﻮﺩ ،ﺣﺎﺻﻞ 15ﻣﻲﺷﻮﺩ .ﺍﮔﺮ 15ﺭﺍ ﺑﺎ 20 ﺟﻤﻊ ﻛﻨﻴﻢ ،ﺟﻮﺍﺏ 35ﻣﻲﺷﻮﺩ ﻛﻪ ﺑﺮ 7ﺑﺨﺶﭘﺬﻳﺮ ﺍﺳﺖ .ﭘﺲ 203 ﻭ ﺩﺭ ﻧﺘﻴﺠﻪ 1638ﻧﻴﺰ ﺑﺮ 7ﺑﺨﺶﭘﺬﻳﺮﻧﺪ. ﺳﺆﺍﻟﻲ ﻛﻪ ﻣﻤﻜﻦ ﺍﺳﺖ ﺍﻳﻦﺟﺎ ﻣﻄﺮﺡ ﺷﻮﺩ ،ﺍﻳﻦ ﺍﺳﺖ ﻛﻪ ﻣﻀﺮﺏ ﺑﺮﺭﺳﻲﻛﻨﻨﺪﻩ ﭼﻪﻃﻮﺭ ﻣﺸﺨﺺ ﻣﻲﺷﻮﺩ؟ ﺩﺭ ﺯﻳﺮ ﺑﻪ ﺟﻮﺍﺏ ﺍﻳﻦ ﺳﺆﺍﻝ ﻣﻲﭘﺮﺩﺍﺯﻳﻢ.
ﭼﻪﻃﻮﺭ ﻣﻀﺎﺭﺏ ﺑﺮﺭﺳﻲﻛﻨﻨﺪﻩ ﺭﺍ ﭘﻴﺪﺍ ﻛﻨﻴﻢ؟
ﺍﮔﺮ ﺑﺨﻮﺍﻫﻴــﻢ ﻣﻀﺮﺏ ﺑﺮﺭﺳــﻲﻛﻨﻨﺪﻩ ﺭﺍ ﺑــﺮﺍﻱ ﺑﺨﺶﭘﺬﻳﺮﻱ ﺑﺮ ﻋﺪﺩﻱ ﻣﺎﻧﻨﺪ aﭘﻴﺪﺍ ﻛﻨﻴﻢ a ،ﺭﺍ ﻋﺪﺩﻱ ﺿﺮﺏ ﻣﻲﻛﻨﻴﻢ ﻛﻪ ﺭﻗﻢ ﻳﻜﺎﻥ ﻋﺪﺩ ﺣﺎﺻﻞ 9ﺑﺎﺷــﺪ .ﺍﻛﻨــﻮﻥ 9ﺭﺍ ﺍﺯ ﺭﻗﻢ ﻳﻜﺎﻥ ﺣﺬﻑ ﻣﻲﻛﻨﻴﻢ ﻭ ﺑﻪ ﻋﺪﺩ ﺑﻪ ﺩﺳﺖ ﺁﻣﺪﻩ ﻳﻚ ﻭﺍﺣﺪ ﺍﺿﺎﻓﻪ ﻣﻲﻛﻨﻴﻢ .ﺍﻳﻦ ﻋﺪﺩ ﻫﻤﺎﻥ ﻣﻀﺮﺏ ﺑﺮﺭﺳﻲﻛﻨﻨﺪﻩ ﺍﺳﺖ. ﺩﻭﺭﺓ ﺷﺎﻧﺰﺩﻫﻢ ،ﺷﻤﺎﺭﺓ ،4ﺗﺎﺑﺴﺘﺎﻥ 1390
ﺭﺍﻫﻨﻤﺎﻳﻲ
9
ﺑﺮﺍﻱ ﻣﺜﺎﻝ ،ﺍﮔﺮ ﺑﺨﻮﺍﻫﻴﻢ ﻣﻀﺮﺏ ﺑﺮﺭﺳــﻲﻛﻨﻨﺪﻩ ﺭﺍ ﺑﺮ ﻋﺪﺩ 7ﭘﻴﺪﺍ ﻛﻨﻴﻢ ،ﻛﺎﻓﻲ ﺍﺳــﺖ 7ﺭﺍ ﺩﺭ 7ﺿﺮﺏ ﻛﻨﻴﻢ ﺗﺎ ﺭﻗﻢ ﻳﻜﺎﻥ ﻋﺪﺩ ﺣﺎﺻﻞ 9ﺷــﻮﺩ .ﺍﻛﻨــﻮﻥ ﺟﻮﺍﺏ ﺍﻳــﻦ ﺣﺎﺻﻞﺿﺮﺏ ﺭﺍ ﻛﻪ 49ﺍﺳــﺖ ﺩﺭ ﻧﻈﺮ ﻣﻲﮔﻴﺮﻳﻢ ﻭ 9ﺭﺍ ﺍﺯ ﺭﻗﻢ ﻳﻜﺎﻥ ﺣﺬﻑ ﻭ ﻳﻚ ﻭﺍﺣﺪ ﺑﻪ ﻗﻴﻤﺖ ﺑﺎﻗﻲﻣﺎﻧﺪﻩ، ﻳﻌﻨﻲ 4ﺍﺿﺎﻓﻪ ﻣﻲﻛﻨﻴﻢ .ﻋﺪﺩ ﺑﻪﺩﺳــﺖﺁﻣﺪﻩ ،ﻳﻌﻨﻲ ،5ﻫﻤﺎﻥ ﻣﻀﺮﺏ ﺑﺮﺭﺳــﻲﻛﻨﻨﺪﻩ ﺑﺮﺍﻱ ﺑﺨﺶﭘﺬﻳﺮﻱ ﺑﺮ ﻋﺪﺩ 7ﺍﺳﺖ .ﺩﺭ ﺑﺎﻻ ﺩﻳﺪﻳﻢ ﻛﻪ ﭼﻪﻃــﻮﺭ ﺑﻪ ﻛﻤــﻚ ﺍﻳﻦ ﻋﺪﺩ ﻣﻲﺗﻮﺍﻥ ﺑﺨﺶﭘﺬﻳــﺮﻱ ﻫﺮ ﻋﺪﺩ ﺑﺮ 7ﺭﺍ ﺑﺮﺭﺳﻲ ﻛﺮﺩ. ﺍﻛﻨﻮﻥ ﻣﻀﺎﺭﺏ ﺑﺮﺭﺳــﻲﻛﻨﻨﺪﻩ ﺭﺍ ﺑﺮﺍﻱ ﺑﺨﺶﭘﺬﻳﺮﻱ ﺑﺮ ﺍﻋﺪﺍﺩ ،13 23 ،19 ،17ﻭ 29ﭘﻴﺪﺍ ﻣﻲﻛﻨﻴﻢ. .1ﺑﺮﺍﻱ ﻋﺪﺩ 13ﺩﺍﺭﻳﻢ .13×3=39 :ﺭﻗﻢ ﻳﻜﺎﻥ 9ﺍﺳــﺖ ﻛﻪ ﺍﮔﺮ ﻳﻚ ﻭﺍﺣﺪ ﺑﻪ ﺩﻫﮕﺎﻥ ﻳﻌﻨﻲ ﻫﻤﺎﻥ ﻗﺴــﻤﺖ ﺑﺎﻗﻲﻣﺎﻧــﺪﻩ ﺍﺿﺎﻓﻪ ﻛﻨﻴﻢ، ﻋﺪﺩ 4ﺑﻪﺩﺳــﺖ ﻣﻲﺁﻳــﺪ ﻛﻪ ﺍﻳﻦ ﻋﺪﺩ ﻫﻤﺎﻥ ﻣﻀﺮﺏ ﺑﺮﺭﺳــﻲﻛﻨﻨﺪﺓ ﺑﺨﺶﭘﺬﻳﺮﻱ ﺑﺮ 13ﺍﺳﺖ. .2ﺑﺮﺍﻱ ﻋﺪﺩ 17ﺩﺍﺭﻳﻢ .17×7=119 :ﺍﮔﺮ ﻳﻚ ﻭﺍﺣﺪ ﺑﻪ 11ﺍﺿﺎﻓﻪ ﻛﻨﻴﻢ ،ﻋﺪﺩ 12ﺑﻪﺩﺳﺖ ﻣﻲﺁﻳﺪ ﻛﻪ ﻣﻀﺮﺏ ﺑﺮﺭﺳﻲﻛﻨﻨﺪﻩ ﺍﺳﺖ. .3ﺑــﺮﺍﻱ ﻋﺪﺩ 19ﺩﺍﺭﻳﻢ .19×1=19 :ﺍﮔــﺮ ﻳﻚ ﻭﺍﺣﺪ ﺑﻪ ﺩﻫﮕﺎﻥ ﺍﺿﺎﻓﻪ ﻛﻨﻴﻢ ،ﻋﺪﺩ 2ﺑﻪﺩﺳــﺖ ﻣﻲﺁﻳﺪ ﻛﻪ ﻫﻤﺎﻥ ﻣﻀﺮﺏ ﺑﺮﺭﺳﻲﻛﻨﻨﺪﻩ ﺍﺳﺖ.
10
ﺭﺍﻫﻨﻤﺎﻳﻲ
ﺩﻭﺭﺓ ﺷﺎﻧﺰﺩﻫﻢ ،ﺷﻤﺎﺭﺓ ،4ﺗﺎﺑﺴﺘﺎﻥ 1390
.4ﺑﺮﺍﻱ ﻋﺪﺩ 23ﺩﺍﺭﻳﻢ .23×3=69 :ﺑﺎ ﺍﺿﺎﻓﻪ ﻛﺮﺩﻥ ﻳﻚ ﻭﺍﺣﺪ ﺑﻪ ﺩﻫﮕﺎﻥ ،ﻋﺪﺩ 7ﺑﻪ ﻋﻨﻮﺍﻥ ﻣﻀﺮﺏ ﺑﺮﺭﺳﻲﻛﻨﻨﺪﻩ ﺑﻪﺩﺳﺖ ﻣﻲﺁﻳﺪ. .5ﺑﺮﺍﻱ ﻋﺪﺩ 29ﻧﻴﺰ ﺩﺍﺭﻳﻢ .29×1=29 :ﺍﮔﺮ ﻳﻚ ﻭﺍﺣﺪ ﺑﻪ ﺩﻫﮕﺎﻥ ﺍﺿﺎﻓﻪ ﺷﻮﺩ ،ﻋﺪﺩ 3ﺑﻪ ﻋﻨﻮﺍﻥ ﻣﻀﺮﺏ ﺑﺮﺭﺳﻲﻛﻨﻨﺪﻩ ﺑﻪ ﺩﺳﺖ ﻣﻲﺁﻳﺪ. ﺑــﻪ ﻋﻨــﻮﺍﻥ ﻣﺜﺎﻝ ،ﻣﻲﺧﻮﺍﻫﻴــﻢ ﺗﻘﺴــﻴﻢﭘﺬﻳﺮﻱ 351ﺑﺮ 132ﺭﺍ ﺑﺮﺭﺳــﻲ ﻛﻨﻴﻢ .ﻫﻤﺎﻥﻃﻮﺭ ﻛﻪ ﺩﺭ ﺑﺎﻻ ﺩﻳﺪﻳﻢ ،ﻣﻀﺮﺏ ﺑﺮﺭﺳــﻲﻛﻨﻨﺪﻩ ﺑﺮﺍﻱ ﺑﺨﺶﭘﺬﻳﺮﻱ ﺑﺮ ،13ﻋﺪﺩ 4ﺍﺳــﺖ .ﺍﮔﺮ 4ﺭﺍ ﺩﺭ ﺭﻗﻢ ﻳﻜﺎﻥ ﺍﻳﻦ ﻋﺪﺩ ﺿﺮﺏ ﻛﻨﻴﻢ 4×1=4 ،ﺑﻪ ﺩﺳــﺖ ﻣﻲﺁﻳﺪ ﻭ ﺍﮔﺮ ﺁﻥ ﺭﺍ ﺑﻪ ﺑﻘﻴﺔ ﺍﻳﻦ ﻋــﺪﺩ ﻳﻌﻨﻲ 39ﺍﺿﺎﻓﻪ ﻛﻨﻴﻢ 35+4=39 ،ﺣﺎﺻﻞ ﻣﻲﺷــﻮﺩ ﻛﻪ ﺑﺮ 13 ﺑﺨﺶﭘﺬﻳﺮ ﺍﺳﺖ .ﭘﺲ 351ﻧﻴﺰ ﺑﺮ 13ﺑﺨﺶﭘﺬﻳﺮ ﺍﺳﺖ. ﺩﺭ ﻣﺜﺎﻟﻲ ﺩﻳﮕﺮ ،ﻣﻲﺧﻮﺍﻫﻴﻢ ﺑﺪﺍﻧﻴﻢ ﻛﻪ ﺁﻳﺎ 578ﺑﺮ 17ﺑﺨﺶﭘﺬﻳﺮ ﺍﺳﺖ ﻳﺎ ﺧﻴﺮ. ﺭﻗﻢ ﻳﻜﺎﻥ ﺍﻳﻦ ﻋﺪﺩ ﻳﻌﻨﻲ 8ﺭﺍ ﺩﺭ ﻋﺪﺩ 12ﻛﻪ ﻣﻀﺮﺏ ﺑﺮﺭﺳﻲﻛﻨﻨﺪﺓ ﺑﺨﺶﭘﺬﻳﺮﻱ ﺑﺮ 17ﺍﺳﺖ ،ﺿﺮﺏ ﻣﻲﻛﻨﻴﻢ 8×12=96 :ﻭ ﺣﺎﺻﻞ ﺭﺍ ﺑﻪ ﺑﻘﻴﺔ ﻋﺪﺩ ﺍﺿﺎﻓﻪ ﻣﻲﻛﻨﻴﻢ .57+96=153 :ﺑﺮﺍﻱ ﺑﺨﺶﭘﺬﻳﺮﻱ 153ﺑﺮ ،17ﺩﻭﺑــﺎﺭﻩ ﻫﻤﻴﻦ ﺭﻭﺵ ﺭﺍ ﺑﻪ ﻛﺎﺭ ﻣﻲﺑﺮﻳﻢ .ﺭﻗﻢ ﻳﻜﺎﻥ ﻳﻌﻨﻲ 3ﺭﺍ ﺩﺭ 12ﺿــﺮﺏ ﻣﻲﻛﻨﻴﻢ 3×12=36 :ﻭ ﺣﺎﺻــﻞ ﺭﺍ ﺑﺎ ﺑﻘﻴﺔ ﻋﺪﺩ ﻳﻌﻨﻲ 15 ﺟﻤﻊ ﻣﻲﻛﻨﻴﻢ 36+15=51 :ﻭ ﭼﻮﻥ 51ﻣﻀﺮﺏ 17ﺍﺳﺖ ،ﭘﺲ 153 ﻭ ﺩﺭ ﻧﺘﻴﺠﻪ 578ﻧﻴﺰ ﺑﺮ 17ﺑﺨﺶﭘﺬﻳﺮﻧﺪ.
ﺟﺪﻭﻝ ﻭ ﺳﺮﮔﺮﻣﻲ
︗︡ول ﻣ ّ ﺤﻤﺪ ﻋﺰﻳﺰﻱﭘﻮﺭ
ﺍﻓﻘﻲ 29 .1ﺍﻓﻘــﻲ ﺑﻪﻋﻼﻭﻩﻱ 9 .5 14963ﺍﻓﻘﻲ ﺑﻪﻋﻼﻭﻩﻱ .6 152 ﻳــﺎﺯﺩﻩ ﺑﺮﺍﺑــﺮ 28ﻋﻤﻮﺩﻱ 15 .8ﺍﻓﻘﻲ ﺗﻘﺴــﻴﻢ ﺑﺮ ﻳﺎﺯﺩﻩ 14 .9 ﻋﻤــﻮﺩﻱ ﺑﻪﻋﻼﻭﻩﻱ ﻫﺸــﺘﺎﺩ ﻭ ﭘﻨﺞ 4 .11ﻋﻤــﻮﺩﻱ ﺑﻪﻋﻼﻭﻩﻱ ﺩﻭ .13ﻣﺮﺑــﻊ ﻳﻚ ﻋــﺪﺩ .15ﺗﻌﺪﺍﺩ ﺛﺎﻧﻴﻪﻫﺎ ﺩﺭ ﺷــﺶ ﺩﻗﻴﻘﻪ .16 10ﻋﻤــﻮﺩﻱ ﺑﻪﻋــﻼﻭﻩﻱ 15 .17 10019ﻋﻤﻮﺩﻱ ﻣﻨﻬﺎﻱ ﻧﻮﺩ ﻭ ﻧﻪ 22 .19ﻋﻤﻮﺩﻱ ﺑﻪﻋﻼﻭﻩﻱ 15 .21 118ﺍﻓﻘﻲ ﺗﻘﺴﻴﻢ ﺑﺮ ﺷــﺶ .22ﺳــﻪ ﺑﺮﺍﺑﺮ 18ﻋﻤﻮﺩﻱ 21 .24ﺍﻓﻘﻲ ﻣﻨﻬﺎﻱ ﺳﻲ ﻭ ﭘﻨﺞ 7 .25ﻋﻤﻮﺩﻱ ﺑﻪﻋﻼﻭﻩﻱ .27 243ﺗﻌﺪﺍﺩ ﺛﺎﻧﻴﻪﻫﺎ ﺩﺭ ﻧﻪ ﺩﻗﻴﻘﻪ 16 .29ﺍﻓﻘﻲ ﺑﻪﻋﻼﻭﻩﻱ 8984 4 7 12
2
3
1
6 10
11
︎︀︨ ︗︡ول ︫﹝︀رهی ﹇︊﹏
5 9
15
8 13
14
ﻋﻤﻮﺩﻱ .1ﻳﻚ ﻋﺪﺩ ﺍ ّﻭﻝ .2ﺗﻌﺪﺍﺩ ﻣﺎﻩﻫﺎ ﺩﺭ ﻳﺎﺯﺩﻩ ﺳــﺎﻝ 17 .3ﺍﻓﻘﻲ ﻣﻨﻬﺎﻱ ﻫﻔﺘﺎﺩ ﻭ ﭘﻨﺞ 5 .4ﻋﻤﻮﺩﻱ ﺗﻘﺴــﻴﻢ ﺑﺮ ﭼﻬﺎﺭ .5ﭼﻬﺎﺭ ﺑﺮﺍﺑــﺮ 1ﻋﻤــﻮﺩﻱ 6 .7ﺍﻓﻘﻲ ﺑﻪﻋﻼﻭﻩﻱ 10 .8 101ﻋﻤﻮﺩﻱ ﻣﻨﻬــﺎﻱ 13 .10 4026ﺍﻓﻘــﻲ ﺿــﺮﺏ ﺩﺭ 2ﻋﻤﻮﺩﻱ 25 .12 ﺍﻓﻘﻲ ﺑﻪﻋﻼﻭﻩﻱ 2 .14 9806ﻋﻤﻮﺩﻱ ﺑﻪﻋﻼﻭﻩﻱ ﺑﻴﺴﺖ ﻭ ﭼﻬﺎﺭ 8 .15ﺍﻓﻘﻲ ﺿﺮﺏ ﺩﺭ ﻫﻔﺖ .18ﺗﻌﺪﺍﺩ ﻣﺎﻩﻫﺎ ﺩﺭ ﻧﻪ ﺳﺎﻝ .20 ﺗﻌﺪﺍﺩ ﻣﺎﻩﻫﺎ ﺩﺭ ﺩﻩ ﺳﺎﻝ 26 .22ﻋﻤﻮﺩﻱ ﺿﺮﺏ ﺩﺭ ﻳﺎﺯﺩﻩ .23 11ﺍﻓﻘﻲ ﺿﺮﺏ ﺩﺭ ﭘﻨﺞ 22 .26ﺍﻓﻘﻲ ﺗﻘﺴــﻴﻢ ﺑﺮ ﻧﻪ 27 .28 ﺍﻓﻘﻲ ﺗﻘﺴﻴﻢ ﺑﺮ ﺩﻭﺍﺯﺩﻩ
3
20
18
19
2
23
24 28
27
3
21
22 26 29
25
7
9
16 17
O
7
O
8
6
9
O
4
2 8 6
4 9 7
3
1 2
6
1
8
O
9
6
5
O
3
O
8 6
5
2
4
O
8
5
4
3
1 4 1
5
O
O
4
5 1
6 O
4
6
4
5 3
O
ﺩﻭﺭﺓ ﺷﺎﻧﺰﺩﻫﻢ ،ﺷﻤﺎﺭﺓ ،4ﺗﺎﺑﺴﺘﺎﻥ 1390
6
ﺭﺍﻫﻨﻤﺎﻳﻲ
11
ﺍﻧﺪﻳﺸﻪﻭﺭﺯﻱ
﹞︀︵︣ات ︨﹀︣ در ︨﹫︀رهی ﹡︀﹛︣﹞﹢ ﺗﺮﺟﻤﻪﻱ
:ﺣﺴﻦ ﻳﺎﻭﺭ ﺗﺒﺎﺭ
ﻛﻠﻴﺪﻭﺍژﻩﻫﺎ :ﻣﻌﻤﺎ ،ﺁﺯﻣﻮﻥ ﻭ ﺧﻄﺎ. ﺍﺗﻮﺑﻴﺎ
ﺳﺎﻝ 2988ﻣﻴﻼﺩﻯ ﺍﺳﺖ ﻭ ﺳﺎﻛﻨﺎﻥ ﺳﻴﺎﺭﻩﻯ ))ﻧﺎﻟﻮﻣﺮ(( ﺩﺭ ﺑﺮﺍﺑﺮ ))ﺍﻣﭙﺮﺍﺗﻮﺭﻯ ﺟﺎﺑﺮﺍﻧﻪﻯ ﺯﻣﻴﻦ (( ﻋﻠﻢ ﻃﻐﻴﺎﻥ ﺑﺮﺍﻓﺮﺍﺷــﺘﻪﺍﻧﺪ .ﺷﻬﺮﺕ ﻭ ﺭﻳﻮ 1 ﺍﻋﺘﺒﺎﺭ ﺷــﻤﺎ ﺑﻪ ﻋﻨﻮﺍﻥ ﻫﻨﺮﻣﻨﺪﻯ ﭘﺮﻃﺮﻓﺪﺍﺭ ﺳﺒﺐ ﺷﺪﻩ ﺍﺳﺖ ﻛﻪ ﺑﺮﺍﻯ 3ﺯﻧﮕﺒﺎﺭ ﺳــﺮﮔﺮﻡ ﻛﺮﺩﻥ ﻧﻴﺮﻭﻫﺎﻯ ﻣﺤﺎﻓــﻆ ،ﺩﻭﺍﺯﺩﻩ ﭘﺎﻳﮕﺎﻩ ﻧﻈﺎﻣﻰ ﺯﻣﻴﻨﻴﺎﻥ ﺩﺭ ﺍﻳﻦ ﻛﺮﻩﻯ ﺟﻨﮓﺯﺩﻩ ،ﺑﻪ ﺁﻥﺟﺎ ﻓﺮﺍ ﺧﻮﺍﻧﺪﻩ ﺷﻮﻳﺪ .ﺍﻳﻦ ﭘﺎﻳﮕﺎﻩﻫﺎ ﻛﻪ ﺑﻪ ﻳﺎﻟﻲ 4 ﺳﻨﮕﺎﭘﻮﺭ o ﻓﺎﺻﻠــﻪﻯ ﻳﻚ ﻛﻴﻠﻮﻣﺘﺮﻯ ﺍﺯ ﻳﻜﺪﻳﮕﺮ ﺩﺭ ﻃﻮﻝ ﻳﻚ ﺑﺰﺭﮔﺮﺍﻩ ﻗﺮﺍﺭ ﮔﺮﻓﺘﻪ ﺍﻧﺪ ،ﭼﻨﺎﻥ ﺳــﺎﺧﺘﻪ ﺷــﺪﻩﺍﻧﺪ ﻛﻪ ﺭﻭﻯ ﻫﻢ ﻳﻚ ﻣﺴــﻴﺮ ﺑﺴﺘﻪﻯ ﺣﻠﻘﻪ 1 ﻣﺎﻧﻨﺪ ﺭﺍ ﺗﺸﻜﻴﻞ ﻣﻰﺩﻫﻨﺪ. ﻫﻨﮓ ُ oﺭﻡ ﺷــﻤﺎ ﻫﻤﺮﺍﻩ ﻳﻚ ﺍﺗﻮﺑﻮﺳــﻰ ﻣﺨﺼﻮﺹ ﺭﺍﻫﭙﻴﻤﺎﻳﻰ ﺩﺭ ﻧﺎﻟﻮﻣﺮ ،ﺩﺭ ﻛﻨﮓ o ﺭﻳﺠﺰ ﻳﻜﻰ ﺍﺯ ﭘﺎﻳﮕﺎﻫﻬﺎﻯ ﺩﻭﺍﺯﺩﻩﮔﺎﻧﻪ ﻳﺎﺩ ﺷﺪﻩ ﻓﺮﻭﺩ ﻣﻰﺁﻳﻴﺪ .ﻣﺎﻣﻮﺭﻳﺖ ﺷﻤﺎ ﻫﻠﻴﺲﺑﻴﻮﺩ o ﺁﻥ ﺍﺳــﺖ ﻛــﻪ ﺣﺮﻛﺖ ﺧــﻮﺩ ﺭﺍ ﺍﺯ ﻳﻚ ﭘﺎﻳــﮕﺎﻩ ﺁﻏﺎﺯ ﻛﻨﻴــﺪ ﻭ ﭘﺲ ﺍﺯ 1 ﮔﺬﺷﺘﻦ ﺍﺯ ﺳﺎﻳﺮ ﭘﺎﻳﮕﺎﻩﻫﺎ ﺑﻪ ﭘﺎﻳﮕﺎﻩ ﻣﺒﺪﺍ ﺑﺎﺯ ﮔﺮﺩﻳﺪ .ﺍﻳﻦ ﻳﻚ ﻣﺎﻣﻮﺭﻳﺖ ﺭﻭﻳﻴﻦ 1 ﺧﻄﺮﻧﺎﻙ ﺍﺳﺖ ،ﺯﻳﺮﺍ ﺩﺭ ﺻﻮﺭﺕ ﻭﻗﻮﻉ ﻫﺮﮔﻮﻧﻪ ﺍﺷﺘﺒﺎﻩ ﻳﺎ ﻟﻐﺰﺵ ﻣﻤﻜﻦ ﻧﻮﻭﺭ ﺍﺳﺖ ﺩﺭ ﻣﻴﺎﻧﻪﻯ ﺭﺍﻩ ﺑﻪ ﺩﺳﺖ ﺷﻮﺭﺷﻴﺎﻥ ﻧﺎﻟﻮﻣﺮ ﺩﺳﺘﮕﻴﺮ ﺷﻮﻳﺪ ﻭ ﻣﻮﺭﺩ ﺷﻜﻨﺠﻪﻯ ﺭﻭﺍﻧﻰ ﻗﺮﺍﺭ ﮔﻴﺮﻳﺪ. ﺍﻳﻦ ﻭﺳﻴﻠﻪﻯ ﻧﻘﻠﻴﻪﻯ ﺭﺍﺣﺖ ﺷﻤﺎ ﺩﺭ ﻫﺮ ﻛﻴﻠﻮﻣﺘﺮ ﻳﻚ ﻟﻴﺘﺮ ﺑﻨﺰﻳﻦ ﻣﺼــﺮﻑ ﻣﻰﻛﻨﺪ ﻭ ﺑﺎﻙ ﺑﻨﺰﻳﻦ ﺁﻥ ﻫﻨﮕﺎﻡ ﻓﺮﻭﺩ ﺑﺮ ﺳــﻴﺎﺭﻩ ﺑﺎﻳﺪ ﺧﺎﻟﻰ ﺣﺠﻢ ﺑﺎﻙ ﺑﻨﺰﻳﻦ ﺍﺗﻮﺑﻮﺱ ﺑﻪ ﺷــﻤﺎ ﺍﻣﻜﺎﻥ ﻣﻰﺩﻫﺪ ﻛﻪ ﺑﺘﻮﺍﻧﻴﺪ ﻫﻤﻪﻯ ﺑﻨﺰﻳﻦ ﻫﺮ ﻳﻚ ﺍﺯ ﭘﺎﻳﮕﺎﻩﻫﺎ ﺭﺍ ﺩﺭ ﺁﻥ ﺟﺎﻯ ﺩﻫﻴﺪ. ﺷﺪﻩ ﺑﺎﺷﺪ. ﺑﻪ ﺍﻳﻦ ﺗﺮﺗﻴﺐ ،ﺁﻳﺎ ﻣﻰﺗﻮﺍﻧﻴﺪ ﺑﻰ ﺁﻥﻛﻪ ﺑﻪ ﻋﻠﺖ ﺗﻤﺎﻡ ﺷﺪﻥ ﺳﻮﺧﺖ ﺧﻮﺷﺒﺨﺘﺎﻧﻪ ﺩﺭ ﺑﺮﺧﻰ ﺍﺯ ﭘﺎﻳﮕﺎﻩﻫﺎﻯ ﻧﻈﺎﻣﻰ ﻣﻘﺪﺍﺭﻯ ﺑﻨﺰﻳﻦ ﻭﺟﻮﺩ ﺩﺍﺭﺩ .ﺩﺭ ﻭﺍﻗﻊ ﻣﺠﻤﻮﻉ ﺑﻨﺰﻳﻦ ﻣﻮﺟﻮﺩ ﺩﺭ ﺗﻤﺎﻣﻰ ﭘﺎﻳﮕﺎﻩﻫﺎ ﺁﻥﻗﺪﺭ ﻫﺴﺖ ﺩﺭ ﻣﻴﺎﻥ ﺭﺍﻩ ﺑﻤﺎﻧﻴﺪ ،ﻳﻚ ﺩﻭﺭ ﻛﺎﻣﻞ ﻃﻮﻝ ﺑﺰﺭﮔﺮﺍﻩ ﺭﺍ ﺑﭙﻴﻤﺎﻳﻴﺪ؟ ﺑﺎ ﺍﻳﻦ ﻛﻪ ﺷــﻤﺎ ﺑﺘﻮﺍﻧﻴﺪ ﺑﺎ ﺍﺳــﺘﻔﺎﺩﻩ ﺍﺯ ﺁﻥ ،ﻳﻚ ﺩﻭﺭ ﻛﺎﻣﻞ ﻣﺴﻴﺮ ﺑﺰﺭﮔﺮﺍﻩ ﺭﺍ ﺣﺴﺎﺏ ﺑﺎﻳﺪ ﺩﺭ ﻛﺪﺍﻡ ﻳﻚ ﺍﺯ ﭘﺎﻳﮕﺎﻩﻫﺎ ﻓﺮﻭﺩ ﺁﻳﻴﺪ؟ .2ﻧﺨﺴﺘﻴﻦ ﻣﻌﻤﺎ ﺭﺍ ﻣﻰﺗﻮﺍﻥ ﺑﺎ ﺍﺳﺘﻔﺎﺩﻩ ﺍﺯ ﺭﻭﺵ ))ﺁﺯﻣﻮﻥ ﻭ ﺧﻄﺎ(( ﺑﭙﻴﻤﺎﻳﻴﺪ .ﺍﻣﺎ ﺷــﺮﺍﻳﻂ ﺟﻨﮕﻰ ﺍﻳﺠﺎﺏ ﻣﻰﻛﻨﺪ ﻛﻪ ﺩﺭ ﺑﻌﻀﻰ ﺍﺯ ﭘﺎﻳﮕﺎﻩﻫﺎ ﻫﻴﭻ ﺑﻨﺰﻳﻨﻰ ﻧﺒﺎﺷﺪ .ﺍﺯ ﺍﻳﻦ ﺭﻭ ،ﺑﻬﺘﺮ ﺍﺳﺖ ﺗﺮﺗﻴﺒﻰ ﺑﺪﻫﻴﺪ ﻛﻪ ﺑﻪ ﻫﻨﮕﺎﻡ ﺣﻞ ﻛﺮﺩ .ﺍﻣﺎ ﺍﮔﺮ ﻣﺠﺒﻮﺭ ﺷﻮﻳﺪ ﻳﻚ ﭼﻨﻴﻦ ﻣﺴﻴﺮﻯ ﺭﺍ ﺩﺭ ﻳﻚ ﺑﺰﺭﮔﺮﺍﻩ ﺭﺳــﻴﺪﻥ ﺑﻪ ﭘﺎﻳﮕﺎﻩﻫﺎﻯ ﻓﺎﻗﺪ ﺑﻨﺰﻳﻦ ،ﻣﻘﺪﺍﺭﻯ ﺳﻮﺧﺖ ﺩﺭ ﺑﺎﻙ ﺩﺍﺷﺘﻪ ﺣﻠﻘﻮﻯ ﺷﻜﻞ ﺩﺍﺭﺍﻯ ﺻﺪﻫﺎ ﭘﺎﻳﮕﺎﻩ ﻧﻈﺎﻣﻰ ﺑﭙﻴﻤﺎﻳﻴﺪ ﭼﻪ ﺧﻮﺍﻫﻴﺪ ﻛﺮﺩ؟ ﺩﺭ ﺍﻳﻦ ﺻﻮﺭﺕ ،ﺑﻰﺗﺮﺩﻳﺪ ﺑﺎﻳﺪ ﺷــﻴﻮﻩﺍﻯ ﻣﺼﻮﻥ ﺍﺯ ﺧﻄﺎ ﺑﺮﮔﺰﻳﻨﻴﺪ ﻛﻪ ﺑﺎﺷﻴﺪ .ﺣﺎﻝ ﺑﺎ ﺗﻮﺟﻪ ﺑﻪ ﺍﻳﻦ ﻧﻜﺎﺕ ﺑﻪ ﺩﻭ ﻣﻌﻤﺎﻯ ﺯﻳﺮ ﺗﻮﺟﻪ ﻛﻨﻴﺪ: .1ﻫﻤﺎﻥﮔﻮﻧﻪ ﻛﻪ ﺩﺭ ﺗﺼﻮﻳﺮ ﺩﻳﺪﻩ ﻣﻰﺷــﻮﺩ ،ﻣﺤﻞ ﻭ ﻧﺎﻡ ﭘﺎﻳﮕﺎﻩﻫﺎ ﺷــﻤﺎ ﺭﺍ ﺩﺭ ﺍﺗﺨﺎﺫ ﺗﺼﻤﻴﻢ ﺑﺮﺍﻯ ﻳﺎﻓﺘﻦ ﻧﻘﻄﻪﻯ ﺁﻏﺎﺯ ﺣﺮﻛﺖ )ﺍﮔﺮ ﻳﻚ ﻭ ﻣﻘــﺪﺍﺭ ﺑﻨﺰﻳﻦ ﻣﻮﺟﻮﺩ ﺩﺭ ﻫﺮ ﻳﻚ ﺍﺯ ﺁﻥﻫﺎ ﻣﺸــﺨﺺ ﺷــﺪﻩ ﺍﺳــﺖ .ﭼﻨﻴﻦ ﻧﻘﻄﻪﻯ ﺁﻏﺎﺯﻯ ﺍﻣﻜﺎﻥ ﭘﺬﻳﺮ ﺑﺎﺷــﺪ( ﻳﺎﺭﻯ ﻛﻨــﺪ .ﺁﻳﺎ ﻣﻰﺗﻮﺍﻧﻴﺪ )ﺷﻤﺎﺭﻩﻫﺎ ﻧﻤﺎﻳﺸﮕﺮ ﻣﻘﺪﺍﺭ ﺑﻨﺰﻳﻦ ﻣﻮﺟﻮﺩ ﺩﺭ ﻫﺮ ﭘﺎﻳﮕﺎﻩ ﺍﺳﺖ( ﺩﺭ ﺿﻤﻦ ﭼﻨﻴﻦ ﺷﻴﻮﻩﺍﻯ ﺭﺍ ﺑﻴﺎﺑﻴﺪ؟ 1
ﭘﺎﺳﺦ ﺩﺭ ﺻﻔﺤﻪﻱ 21
12
ﺭﺍﻫﻨﻤﺎﻳﻲ
ﺩﻭﺭﺓ ﺷﺎﻧﺰﺩﻫﻢ ،ﺷﻤﺎﺭﺓ ،4ﺗﺎﺑﺴﺘﺎﻥ 1390
o
ﻣﺮﺍﻛﺶ
ﺟﺪﻭﻝ ﻭ ﺳﺮﮔﺮﻣﻲ
︗︡ول ١٠٠ ︀︑ ١ ﻧﺴﺮﻳﻦ ﺷﺮﻳﻔﻴﺎﻥ
ﻛﻠﻴﺪﻭﺍژﻩﻫﺎ :ﺟﺪﻭﻝ ﺍﻋﺪﺍﺩ ،ﺍﻋﺪﺍﺩ 1ﺗﺎ ،100ﺣﺪﺱ.
ﺟﺪﻭﻝ ﺭﻭﺑﻪ ﺭﻭ ﺭﺍ ﺧﻴﻠﻰ ﺩﻳﺪﻩ ﺍﻳﺪ! ﺍﻳﻦ ﺑﺎﺭ ﻛﻤﻰ ﺩﻗﻴﻖﺗﺮ ﻧﮕﺎﻩ ﻛﻨﻴﺪ. ﻓﺮﺽ ﻛﻨﻴﺪ ﺩﻭ ﺗﺎ ﺍﺯ ﺍﻳﻦ ﺟﺪﻭﻝﻫﺎ ﺩﺭﺳﺖ ﻛﺮﺩﻩﺍﻳﻢ ،ﺩﻗﻴﻘﺎ ﻫﻢﺍﻧﺪﺍﺯﻩ ﻭ ﻫﻢﺷﻜﻞ. ﺩﻭ ﺗــﺎ ﺟــﺪﻭﻝ ﺭﺍ ﻃﻮﺭﻯ ﺭﻭﻯ ﻫــﻢ ﻗﺮﺍﺭ ﻣﻰﺩﻫﻴــﻢ ﻛﻪ ﻋﺪﺩﻫﺎ ﺭﻭﻯ ﻫﻢ ﻗﺮﺍﺭ ﺑﮕﻴﺮﻧﺪ! ﺑﻪ ﺟﻬﺖ ﻓﻠﺶﻫﺎ ﺩﺭ ﺷﻜﻞﻫﺎﻯ ﺯﻳﺮ ﺩﻗﺖ ﻛﻨﻴﺪ: ﺩﺭ ﻫــﺮ ﻳﻚ ﺍﺯ ﺣﺎﻟﺖ ﻫﺎﻯ ﺑﺎﻻ ،ﻛﺪﺍﻡ ﻋﺪﺩ ﺟــﺪﻭﻝ ﺯﻳﺮﻯ ،ﺯﻳﺮ ﻋــﺪﺩ 24ﺍﺯ ﺟﺪﻭﻝ ﺭﻭﻳﻰ ﻗﺮﺍﺭ ﮔﺮﻓﺘﻪ ﺍﺳﺖ؟ ﻣﻰﺗﻮﺍﻧﻴــﺪ ﺩﻭ ﺗﺎ ﺟﺪﻭﻝ ﺩﺭﺳــﺖ ﻛﻨﻴﺪ ﻭ ﺁﻥﻫــﺎ ﺭﺍ ﺩﺭ ﻫــﺮ ﭼﻬﺎﺭ ﺣﺎﻟــﺖ ﺭﻭﻯ ﻫﻢ ﻗﺮﺍﺭ ﺩﻫﻴﺪ .ﺳــﭙﺲ ﺍﺑﺘﺪﺍ ﺣﺪﺱ ﺑﺰﻧﻴﺪ ﺯﻳﺮ 24ﭼﻪ ﻋﺪﺩﻯ ﺍﺳــﺖ ،ﺳﭙﺲ ﺁﺭﺍﻡ ﺩﻭ ﺟﺪﻭﻝ ﺭﺍ ﺍﺯ ﻫﻢ ﺟﺪﺍ ﺳــﺎﺯﻳﺪ ﻭ ﺩﺭﺳﺘﻰ ﭘﺎﺳــﺨﺘﺎﻥ ﺭﺍ ﺑﺮﺭﺳﻰ ﻛﻨﻴﺪ.
ﭘﺸﺖ ﺟﺪﻭﻝ 1ﺗﺎ 100 ﺭﻭﻳﻲ
ﭘﺸﺖ ﺟﺪﻭﻝ 1ﺗﺎ 100 ﺭﻭﻳﻲ
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ﭘﺸﺖ ﺟﺪﻭﻝ 1ﺗﺎ 100 ﺭﻭﻳﻲ
ﭘﺸﺖ ﺟﺪﻭﻝ 1ﺗﺎ 100 ﺭﻭﻳﻲ
ﺩﻭﺭﺓ ﺷﺎﻧﺰﺩﻫﻢ ،ﺷﻤﺎﺭﺓ ،4ﺗﺎﺑﺴﺘﺎﻥ 1390
ﺭﺍﻫﻨﻤﺎﻳﻲ
13
ﮔﻔﺖ ﻭ ﮔﻮ
﹞﹢︑﹩ان ﹨︣ ﹞︧﹚﹤ای را ﹏ ﹋︣د! ﹎﹀️و ﹎ ︀︋ ﹢دا﹡︩آ﹞﹢زان و ﹞︺﹚﹜ ر︀︲﹩ ﹞︡ر︨﹤ی را﹨﹠﹝︀﹩︊︨︀﹝︵ ︡﹫︫ ﹩ ﺁﺯﺍﺩﻩ ﺷﺎﻛﺮﻱ
ﻛﻠﻴﺪﻭﺍژﻩﻫﺎ :ﮔﻔﺖﻭ ﮔﻮ ،ﻣﺪﺭﺳﻪﻱ ﺭﺍﻫﻨﻤﺎﻳﻲ ،ﺷﻬﻴﺪ ﻃﻬﻤﺎﺳﺒﻲ ،ﻣﻌﻠﻢ ﺭﻳﺎﺿﻲ ،ﺩﺭﺱ ﺭﻳﺎﺿﻲ.
ﺍﺷﺎﺭﻩ ﻭﻗﺘﻲ ﻛﺎﺭ ﺑﺎ ﺍﻋﺪﺍﺩ ﺭﺍ ﺷـﺮﻭﻉ ﻣﻲﻛﻨـﻲ ،ﺫﻫﻨﺖ ﻣﺜﻞ ﻳﺦ ﻣﻨﺠﻤﺪ ﺍﺳـﺖ ،ﻫﻤﻴﻦﻃﻮﺭ ﻛﻪ ﺑﺎ ﺍﻋﺪﺍﺩ ﺑﺎﺯﻱ ﻣﻲﻛﻨﻲ ،ﺍﻳﻦ ﻳﺦ ﺁﺏ ﻣﻲﺷـﻮﺩ، ﻳﻌﻨـﻲ ﺫﻫﻨـﺖ ﻧـﺮﻡ ﻣﻲﺷـﻮﺩ ﻭ ﺩﺭ ﺁﻥ ﻟﺤﻈـﻪ ﺗﻘﺮﻳﺒ ًﺎ ﻣﻲﺗـﻮﺍﻥ ﻫﺮ ﻣﺴﺌﻠﻪﺍﻱ ﺭﺍ ﺣﻞ ﻛﺮﺩ! ﺧﻴﻠﻲ ﺑﻪ ﺫﻫﻨﺘﺎﻥ ﻓﺸﺎﺭ ﻧﻴﺎﻭﺭﻳﺪ .ﺍﻳﻦ ﺟﻤﻼﺕ ،ﺣﺮﻑﻫﺎﻱ ﻫﻴﭻﻳﻚ ﺍﺯ ﻧﻮﺍﺑﻎ ﺭﻳﺎﺿﻲ ﻛﻪ ﺷـﻤﺎ ﻣﻲﺷﻨﺎﺳﻴﺪ ،ﻧﻴﺴـﺖ! ﺍﻳﻦ ﺣﺮﻑﻫﺎﻱ ﻳﻚ ﺷـﺎﮔﺮﺩ ﻛﻼﺱ ﺳﻮﻡ ﻣﺪﺭﺳـﻪﻱ ﺭﺍﻫﻨﻤﺎﻳﻲ ﺷـﻬﻴﺪ ﻃﻬﻤﺎﺳﺒﻲ ﺍﺳﺖ ﻛﻪ ﭘـﺎﻱ ﺻﺤﺒﺖ ﺍﻭ ﻭ ﺩﻭﺍﺯﺩﻩ ﻧﻔﺮ ﺍﺯ ﺩﻭﺳـﺘﺎﻥ ﻭ ﻣﻌﻠﻢ ﺭﻳﺎﺿﻲﺷـﺎﻥ ﻧﺸﺴـﺘﻴﻢ .ﺁﻥﻫﺎ ﺍﺯ ﺩﺭﺱ ﺭﻳﺎﺿﻲ ﺣـﺮﻑ ﺯﺩﻧﺪ ﻭ ﮔﻔﺘﻨﺪ ﻛﻪ ﺑﻌﻀﻲﻫﺎ ﻋﺎﺷﻘﺎﻧﻪ ﺩﻭﺳﺘﺶ ﺩﺍﺭﻧﺪ ﻭ ﻋﺪﻩﺍﻱ ﺍﺯ ﺗﻪ ﺩﻝ ﺍﺯ ﺁﻥ ﻣﺘﻨﻔﺮﻧﺪ. ﺧﺎﻧﻢ ﺍﻣﺎﻣﻲ ﻫﻢ ،ﻣﻌﻠﻢ ﺑﭽﻪﻫﺎ ﻭ ﻣﺪﺍﻓﻊ ﺳﺮﺳﺨﺖ ﺭﻳﺎﺿﻲ ﺍﺳﺖ ﻭ ﺭﺍﻩﺣﻞﻫﺎﻱ ﺧﻮﺑﻲ ﺑﺮﺍﻱ ﻣﻮﻓﻖ ﺑـﻮﺩﻥ ﺩﺭ ﺭﻳﺎﺿﻲ ﻣﻲﺩﺍﻧﺪ ﻛﻪ ﺑﻪ ﺩﺭﺩ ﻫﻤﻪﻱ ﺑﭽﻪﻫﺎﻱ ﺩﻭﺭﻩ ﺭﺍﻫﻨﻤﺎﻳﻲ ﻣﻲﺧﻮﺭﺩ. ﺁﻥﭼـﻪ ﺩﺭ ﺍﺩﺍﻣـﻪ ﻣﻲﺧﻮﺍﻧﻴﺪ ﺣﺮﻑﻫﺎﻱ ﺑﭽﻪﻫـﺎ ﻭ ﻣﻌﻠﻢ ﺭﻳﺎﺿﻲ ﻣﺪﺭﺳﻪﻱ ﺷﻬﻴﺪ ﻃﻬﻤﺎﺳﺒﻲ ﺍﺳﺖ.
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ﺭﺍﻫﻨﻤﺎﻳﻲ
ﺩﻭﺭﺓ ﺷﺎﻧﺰﺩﻫﻢ ،ﺷﻤﺎﺭﺓ ،4ﺗﺎﺑﺴﺘﺎﻥ 1390
ﺑﭽﻪﻫﺎ ﺩﻭ ﮔﺮﻭﻩ ﺷﺪﻧﺪ .ﻫﺮ ﺩﻭ ﮔﺮﻭﻩ ﻧﻤﺮﻩﻫﺎﻱ ﺭﻳﺎﺿﻲ ﺧﻮﺑﻲ ﮔﺮﻓﺘﻪﺍﻧﺪ .ﻳﻚ ﮔﺮﻭﻩ ﺑﺎ ﺷـﻮﺭ ﻭ ﻫﻴﺠﺎﻥ ﺍﺯ ﻋﻼﻗﻪﺷﺎﻥ ﺑﻪ ﺭﻳﺎﺿﻲ ﻣﻲﮔﻮﻳﻨﺪ ﻭ ﮔـﺮﻭﻩ ﺩﻳﮕﺮ ﺍﺯ ﺑﻲﻋﻼﻗﮕﻲﺷـﺎﻥ ﺑﻪ ﺩﺭﺱ ﺭﻳﺎﺿﻲ ﺣـﺮﻑ ﻣﻲﺯﻧﻨﺪ .ﺍﺯ ﺁﻥﻫﺎ ﻣﻲﺧﻮﺍﻫـﻢ ﻳﻜﻲ ﻳﻜﻲ ﺩﻻﻳﻞ ﻋﻼﻗﻪ ﻭ ﺑﻲﻋﻼﻗﮕﻲﺷﺎﻥ ﺭﺍ ﺑﻪ ﺍﻳﻦ ﺩﺭﺱ ﺑﮕﻮﻳﻨﺪ. ﺷـﺒﻨﻢ ﺍﺑﻮﺍﻟﻔﻀﻠﻲ ﻣﻲﮔﻮﻳﺪ :ﻣﻦ ﺑــﻪ ﺩﺭﺱ ﺭﻳﺎﺿﻲ ﻋﻼﻗﻪ ﺩﺍﺭﻡ، ﭼــﻮﻥ ﻓﻜﺮ ﻣﻲﻛﻨﻢ ﻳــﺎﺩ ﮔﺮﻓﺘﻦ ﺭﻳﺎﺿــﻲ ﺩﺭ ﺯﻧﺪﮔــﻲ ﺧﻴﻠﻲ ﻛﻤﻜﻢ ﻣﻲﻛﻨﺪ. ﻳﺎﺩ ﮔﺮﻓﺘﻦ ﺭﻳﺎﺿﻲ ﺩﺭ ﺁﻳﻨﺪﻩ ﻭ ﺍﻧﺘﺨﺎﺏ ﺷﻐﻞ ﻫﻢ ﻛﻤﻚ ﺯﻳﺎﺩﻱ ﺑﻪ ﻣﻦ ﻣﻲﻛﻨﺪ .ﻋﻼﻭﻩ ﺑﺮ ﺍﻳﻦﻫﺎ ﻣﻦ ﻓﻜﺮ ﻣﻲﻛﻨﻢ ﺭﻳﺎﺿﻲ ﺩﺭﺱ ﺷــﻨﻴﺪﻧﻲ ﺍﺳﺖ ﻛﻪ ﺩﺭ ﺑﻴﺶﺗﺮ ﺭﺷﺘﻪﻫﺎ ﺣﺘﻲ ﺍﺩﺑﻴﺎﺕ ﻧﻘﺶ ﺩﺍﺭﺩ. ﻧﺮﺟﺲ ﺳﻌﻴﺪﻱ ﺗﻮﺿﻴﺢ ﻣﻲﺩﻫﺪ :ﺭﻳﺎﺿﻲ ﺳﺨﺖ ﺍﺳﺖ .ﻣﻔﺎﻫﻴﻢ ﮔﻴﺞﻛﻨﻨﺪﻩﺍﻱ ﺩﺍﺭﺩ .ﺍﮔﺮ ﻣﻲﺷــﺪ ﺭﻳﺎﺿﻲ ﺭﺍ ﺑﺎ ﺷــﻜﻞ ﻳﺎ ﺭﻭﺷﻲ ﻏﻴﺮ ﺍﺯ ﺍﻳﻦ ﻳﺎﺩ ﺩﺍﺩ ،ﻫﻤﻪ ﺩﻭﺳﺘﺶ ﺩﺍﺷﺘﻨﺪ. ﺷـﺒﻨﻢ ﺍﺑﻮﺍﻟﻔﻀﻠﻲ ﻣﻌﺘﻘﺪ ﺍﺳﺖ ﻛﻪ ﺭﻳﺎﺿﻲ ﺳﺨﺖ ﺍﺳﺖ ،ﺍﻣﺎ ﺍﮔﺮ ﺁﺩﻡ ﺩﻧﺒﺎﻝ ﺩﺭﺱ ﺭﺍ ﺑﮕﻴﺮﺩ ﻳــﺎ ﺧﻮﺩﺵ ﺑﻴﻦ ﻋﻼﺋﻢ ﺭﻳﺎﺿﻲ ﺭﺍﺑﻄﻪﻫﺎﻳﻲ ﺑﺮﻗﺮﺍﺭ ﻛﻨﺪ ،ﺩﺭﻛﺶ ﺁﺳﺎﻥﺗﺮ ﻣﻲﺷﻮﺩ.
ﻧﺮﺟﺲ ﺳـﻌﻴﺪﻱ ﺍﻋﺘﻘﺎﺩ ﺩﻳﮕﺮﻱ ﺩﺍﺭﺩ :ﺍﻳﻦ ﺩﺭﺳــﺖ ﺍﺳــﺖ ﻛﻪ ﺩﺭﻙ ﺭﻳﺎﺿــﻲ ﺑﺎ ﺍﻳﻦ ﺭﻭﺵ ﺭﺍﺣﺖﺗﺮ ﻣﻲﺷــﻮﺩ ،ﺍﻣــﺎ ﻫﺮﻛﺲ ﻇﺮﻓﻴﺘﻲ ﺩﺍﺭﺩ ﻭ ﺭﺍﻩﺣﻠﻲ ﻛﻪ ﺑﻪ ﺫﻫﻦ ﻳﻚ ﻧﻔﺮ ﻣﻲﺭﺳﺪ ،ﺷﺎﻳﺪ ﺑﻪ ﺫﻫﻦ ﻧﻔﺮ ﺩﻳﮕﺮ ﻧﺮﺳﺪ. ﺑﻪ ﻧﺮﺟﺲ ﻣﻲﮔﻮﻳﻢ ﺗﻮ ﻧﻤــﺮﻩﻱ ﺭﻳﺎﺿﻲﺍﺕ ﺭﺍ 20ﮔﺮﻓﺘﻪﺍﻱ ،ﭘﺲ ﺑﻪ ﺧﺎﻃﺮ ﺳــﺨﺖ ﺑﻮﺩﻥ ﺭﻳﺎﺿﻲ ﻧﻴﺴــﺖ ﻛﻪ ﻧﺴﺒﺖ ﺑﻪ ﺁﻥ ﺑﻲﻋﻼﻗﻪﺍﻱ، ﺩﺭﺳﺖ ﺍﺳﺖ؟ ﻧﺮﺟﺲ ﺳﻌﻴﺪﻱ ﺟﻮﺍﺏ ﻣﻲﺩﻫﺪ :ﻣﻔﺎﻫﻴﻢ ﺭﻳﺎﺿﻲ ﺧﻴﻠﻲ ﺷﺒﻴﻪ ﺑﻪ ﻫﻢ ﻫﺴﺘﻨﺪ ﻭ ﺑﻪ ﻫﻤﻴﻦ ﺩﻟﻴﻞ ﮔﻴﺞﻛﻨﻨﺪﻩﺍﻧﺪ. ﺍﺯ ﺍﻭ ﻣﻲﭘﺮﺳﻢ :ﺗﻮ ﺍﻳﻦ ﻣﺸﻜﻞ ﺭﺍ ﭼﮕﻮﻧﻪ ﺣﻞ ﻛﺮﺩﻱ؟ ﻧﺮﺟﺲ ﺳﻌﻴﺪﻱ ﭘﺎﺳﺦ ﻣﻲﺩﻫﺪ :ﻣﻦ ﺁﻥﻗﺪﺭ ﺭﻳﺎﺿﻲ ﺭﺍ ﺧﻮﺍﻧﺪﻩﺍﻡ ﺗﺎ 20ﮔﺮﻓﺘﻢ ،ﺍﻣﺎ ﺑﻪ ﺁﻥ ﻋﻼﻗﻪﻣﻨﺪ ﻧﺸﺪﻡ. ﻓﺎﻃﻤـﻪ ﻧﺼﻴﺮﻱ ﻫــﻢ ﻭﺍﺭﺩ ﺑﺤﺚ ﻣﻲﺷــﻮﺩ ﻭ ﻣﻲﮔﻮﻳﺪ :ﻣﻦ ﻓﻜﺮ ﻣﻲﻛﻨﻢ ﻛﺴــﻲ ﻛﻪ ﺭﻳﺎﺿــﻲ ﺭﺍ 20ﻣﻲﮔﻴﺮﺩ ،ﻭﻟﻲ ﺑــﻪ ﺁﻥ ﻋﻼﻗﻪﻣﻨﺪ ﻧﻴﺴﺖ ،ﺩﭼﺎﺭ ﺩﻭﮔﺎﻧﮕﻲ ﺷﺪﻩ ﺍﺳﺖ. ﻣــﻦ ﺧﻮﺩﻡ ﻓﻜﺮ ﻣﻲﻛﻨﻢ ﺭﻳﺎﺿــﻲ ﺩﺭ ﺭﺃﺱ ﺯﻧﺪﮔﻲﺍﻡ ﻗﺮﺍﺭ ﺩﺍﺭﺩ .ﺑﺎ ﺭﻳﺎﺿﻲ ﺧﻮ ﮔﺮﻓﺘﻪﺍﻡ.ﺍﺣﺴــﺎﺱ ﻣﻲﻛﻨﻢ ﺩﺭ ﺯﻧﺪﮔﻲﺍﻡ ﺑﺎ ﺭﻳﺎﺿﻲ ﺧﻴﻠﻲ ﻛﺎﺭ ﺩﺍﺭﻡ .ﻳﻜﻲ ﺍﺯ ﺳﺎﺩﻩﺗﺮﻳﻦ ﻛﺎﺭﺑﺮﺩﻫﺎﻱ ﺭﻳﺎﺿﻲ ،ﺧﺮﻳﺪ ﻛﺮﺩﻥﻫﺎﻱ ﻫﺮ ﺭﻭﺯﻩﻱ ﻣﺎﺳﺖ .ﻛﺎﺭﺑﺮﺩ ﺩﻳﮕﺮ ،ﺣﻞ ﭼﻴﺴﺘﺎﻥﻫﺎﻱ ﺭﻳﺎﺿﻲ ﺍﺳﺖ.
1390 139 ﺎﻥ 0 ﻥ ﺗﺎﺑﺴﺘﺎﻥ ﺰﺩﻫﻢ ،ﺷﻤﺎﺭﺓ ،4ﺗﺎﺑﺴﺘ ﻢ، ﺷﺎﻧﺰﺩﻫﻢ، ﺷﺎﻧﺰﺩﻫ ﺩﻭﺭﺓ ﺷ ﻧ ﺩﺩﻭﺭﺓ
ﺭﺍﻫﻨﻤﺎﻳﻲ ﺎﻳﻲ ﻫﻨﻤﺎ ﺭﺍﻫﻨﻤ ﺭﺭﺍ
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ﺭﻳﺎﺿﻲ ،ﻛﺎﺭ ﻛﺮﺩﻥ ﻳﺎ ﺑﺎﺯﻱ ﻛﺮﺩﻥ ﺑﺎ ﺍﻋﺪﺍﺩ ﺍﺳﺖ .ﻣﻦ ﻭﻗﺘﻲ ﺣﺎﺻﻞ ﺟﻤﻴﻠﻪ ﺍﻣﺎﻣﻲ )ﻣﻌﻠﻢ ﺑﭽﻪﻫﺎ( ﭘﺎﺳــﺦ ﺑﺮﺧﻲ ﺍﺯ ﺣﺮﻑﻫﺎﻱ ﺑﭽﻪﻫﺎ ﻛﺎﺭ ﻛﺮﺩﻥ ﺑﺎ ﺍﻋﺪﺍﺩ ﺭﺍ ﻣﻲﺑﻴﻨﻢ ،ﺣﺲ ﻣﻲﻛﻨﻢ ﻛﺎﺭ ﺧﻮﺑﻲ ﺍﻧﺠﺎﻡ ﺩﺍﺩﻩﺍﻡ ﺭﺍ ﻣﻲﺩﻫــﺪ ﻭ ﻣﻲﮔﻮﻳــﺪ :ﺑﭽﻪﻫﺎ ﺷــﻤﺎ ﺑــﻪ ﺍﻳﻦ ﺩﻟﻴﻞ ﻓﻜــﺮ ﻣﻲﻛﻨﻴﺪ ﻭ ﻧﺘﻴﺠﻪﺍﺵ ﺭﺍ ﻣﺸﺎﻫﺪﻩ ﻣﻲﻛﻨﻢ. ﺑﻌﻀﻲ ﺍﺯ ﻣﻄﺎﻟﺐ ﺭﻳﺎﺿﻲ ﮔﻴﺞﻛﻨﻨﺪﻩ ﺍﺳــﺖ ﻛــﻪ ﺍﺛﺒﺎﺕ ﺍﻳﻦ ﻣﻄﺎﻟﺐ ﺭﺍ ﻣﻦ ﺍﺣﺴــﺎﺱ ﻣﻲﻛﻨﻢ ﻫﺮ ﻣﻌﺎﺩﻟﻪﺍﻱ ﺭﺍ ﻛﻪ ﺣﻞ ﻣﻲﻛﻨﻢ ،ﻧﺘﻴﺠﻪﻱ ﻳــﺎﺩ ﻧﮕﺮﻓﺘﻪﺍﻳــﺪ .ﺍﮔﺮ ﺑﻪ ﻣﺒﺎﺣﺚ ﺩﺭﺱﻫﺎﻱ ﺭﻳﺎﺿﻲﺗــﺎﻥ ﺩﺭ ﺍﺑﺘﺪﺍﻳﻲ ﻭ ﺗﻼﺷﻢ ﺭﺍ ﻣﻲﺑﻴﻨﻢ .ﺑﻪ ﻫﻤﻴﻦ ﺩﻟﻴﻞ ﻋﺎﺷﻖ ﺭﻳﺎﺿﻲ ﻫﺴﺘﻢ. ﺭﺍﻫﻨﻤﺎﻳــﻲ ﺩﻗﺖ ﻛﻨﻴــﺪ ،ﻣﻲﺑﻴﻨﻴﺪ ﻛﻪ ﺑﻪ ﺑﻌﻀﻲ ﺍﺯ ﻣﺴــﺎﺋﻞ ﺩﺭﻣﻘﻄﻊ ﺷـﻘﺎﻳﻖ ﺻﺎﺩﻗﻲ ﻫﻢ ﺣﺮﻑﻫﺎﻳﻲ ﺩﺍﺭﺩ :ﻣﻤﻜﻦ ﺍﺳــﺖ ﺭﻳﺎﺿﻲ ﺩﺭ ﺍﺑﺘﺪﺍﻳﻲ ﺍﺷــﺎﺭﻩ ﺷﺪﻩ ﺑﻮﺩ ﺗﺎ ﺫﻫﻦ ﺷــﻤﺎ ﺑﺎ ﺁﻥ ﺁﺷﻨﺎ ﺑﺎﺷﺪ ،ﻭﻟﻲ ﻣﻄﺎﻟﺐ ﺯﻧﺪﮔﻲ ﻛﺎﺭﺑﺮﺩ ﺯﻳﺎﺩﻱ ﺩﺍﺷــﺘﻪ ﺑﺎﺷــﺪ ،ﺍﻣﺎ ﻣﻦ ﺭﺍﺑﻄــﻪﻱ ﺑﻴﻦ ﺍﻋﺪﺍﺩ ﺭﺍ ﺗﻜﻤﻴﻠﻲﺗﺮ ﺭﺍ ﺑﻌﺪﻫﺎ ﺩﺭ ﻣﻘﻄﻊ ﺭﺍﻫﻨﻤﺎﻳﻲ ﺧﻮﺍﻧﺪﻳﺪ. ﺩﺭﻙ ﻧﻤﻲﻛﻨﻢ .ﺑﺮﺍﻱ ﻣﺜﺎﻝ ،ﭼﺮﺍ 2+2ﺑﺮﺍﺑﺮ ﺑﺎ 4ﺍﺳــﺖ .ﺭﻭﺍﺑﻄﻲ ﻣﺜﻞ ﺩﺭﺱﻫﺎﻱ ﺭﻳﺎﺿﻲ ﺭﺍﻫﻨﻤﺎﻳﻲ ﻫﻢ ﺑﻪ ﻫﻤﻴﻦ ﺷــﻜﻞ ﺍﺳــﺖ .ﺍﺛﺒﺎﺕ ﺍﻳﻦ ﺑﺎﻋﺚ ﺳــﺮﺩﺭﮔﻤﻲ ﻣﻦ ﻣﻲﺷﻮﺩ .ﻣﻦ ﻓﻜﺮ ﻣﻲﻛﻨﻢ ﺑﻌﻀﻲ ﺍﺯ ﺭﻭﺍﺑﻂ ﺑﺮﺧــﻲ ﺍﺯ ﻣﻔﺎﻫﻴﻢ ﺭﺍ ﺑﻌﺪﻫﺎ ﺩﺭ ﺩﺑﻴﺮﺳــﺘﺎﻥ ﻭ ﻣﻘﺎﻃــﻊ ﺑﺎﻻﺗﺮ ﺧﻮﺍﻫﻴﺪ ﺭﻳﺎﺿﻲ ﺩﻟﻴﻠﻲ ﻣﻨﻄﻘﻲ ﻧﺪﺍﺭﻧﺪ. ﺧﻮﺍﻧــﺪ .ﺍﻵﻥ ﺗﺪﺭﻳﺲ ﺍﻳــﻦ ﻣﻔﺎﻫﻴﻢ ﺑﺮﺍﻱ ﻓﺎﻃﻤﻪ ﺧﺪﺍﻳﻲ ﻣﻌﺘﻘﺪ ﺍﺳــﺖ :ﻣﻦ ﺷﻤﺎ ﻛﺎﺑﺮﺩﻱ ﻧﺪﺍﺭﺩ ﻭ ﺳﻨﮕﻴﻦ ﺍﺳﺖ. ﻓﻜﺮ ﻣﻲﻛﻨﻢ ﺍﻧﺴــﺎﻥ ﺍﮔــﺮ ﺧﻮﺩﺵ ﻫﻢ ﺷـﺒﻨﻢ ﺍﺑﻮﺍﻟﻔﻀﻠـﻲ ﺩﻟﻴــﻞ ﺩﻳﮕــﺮ ﻧﺨﻮﺍﻫﺪ ،ﺭﻳﺎﺿﻲ ﺭﺍ ﻳﺎﺩ ﻣﻲﮔﻴﺮﺩ .ﺷــﻤﺎ ﻋﻼﻗﻪﻣﻨﺪﻱﺍﺵ ﺑﻪ ﺍﻳــﻦ ﺩﺭﺱ ﺭﺍ ﻣﻄﺮﺡ ﺑﻪ ﺁﺩﻡﻫﺎﻱ ﺑﻲﺳﻮﺍﺩ ﺗﻮﺟﻪ ﻛﻨﻴﺪ .ﺁﻥﻫﺎ ﻣﻲﻛﻨﺪ :ﭘﻴﭽﻴﺪﮔﻲ ﺭﻳﺎﺿﻲ ﻳﻜﻲ ﺍﺯ ﺩﻻﻳﻞ ﻫﻢ ﺑﺎ ﺍﻳﻦﻛﻪ ﺩﺭﺳــﻲ ﺭﺍ ﺑﻪ ﻧﺎﻡ ﺭﻳﺎﺿﻲ ﻋﻼﻗﻪﻣﻨﺪﻱ ﻣﻦ ﺑﻪ ﺍﻳﻦ ﺩﺭﺱ ﺍﺳــﺖ .ﻣﻦ ﻧﺨﻮﺍﻧﺪﻩﺍﻧــﺪ ،ﺍﻣﺎ ﺭﻳﺎﺿــﻲ ﻣﻲﺩﺍﻧﻨﺪ ﻭ ﺍﺯ ﻋﺎﺷــﻖ ﭼﻴﺰﻫﺎﻱ ﻣﺠﻬﻮﻟﻢ .ﭼﻴﺰﻫﺎﻳﻲ ﻛﻪ ﺁﻥ ﺩﺭ ﺯﻧﺪﮔﻲﺷــﺎﻥ ﺍﺳﺘﻔﺎﺩﻩ ﻣﻲﻛﻨﻨﺪ. ﺧﻮﺩﻣﺎﻥ ﺑﺎﻳﺪ ﺑﮕﺮﺩﻳﻢ ﺗﺎ ﻭﺍﻗﻌﻴﺘﺶ ﺭﺍ ﭘﻴﺪﺍ ﺑــﺮﺍﻱ ﻣﺜــﺎﻝ ،ﻣﻴﻮﻩﻓﺮﻭﺷــﻲ ﺭﺍ ﺩﺭ ﻧﻈﺮ ﻛﻨﻴﻢ. ﺑﮕﻴﺮﻳﺪ ﻛﻪ ﺍﺻ ً ﻼ ﺭﻳﺎﺿﻲ ﻧﺨﻮﺍﻧﺪﻩ ﺍﺳﺖ. ﻓﺎﻃﻤﻪ ﻧﺼﻴﺮﻱ ﻣﻲﮔﻮﻳﺪ :ﻭﻗﺘﻲ ﻛﺎﺭ ﺑﺮﺍﻱ ﺣﺴﺎﺏ ﻭ ﻛﺘﺎﺏ ﻛﺎﺭ ﻭ ﺯﻧﺪﮔﻲﺍﺵ ﺑﺎ ﺍﻋﺪﺍﺩ ﺭﺍ ﺷــﺮﻭﻉ ﻣﻲﻛﻨﻲ ،ﺫﻫﻨﺖ ﻣﺜﻞ ﻣﺠﺒﻮﺭ ﺍﺳﺖ ﺍﺯ ﺭﻳﺎﺿﻲ ﺍﺳﺘﻔﺎﺩﻩ ﻛﻨﺪ ﻭ ﻳﺦ ﻣﻨﺠﻤﺪ ﺍﺳﺖ ﻭ ﻫﻤﻴﻦﻃﻮﺭ ﻛﻪ ﺑﺎ ﺍﻋﺪﺍﺩ ﺣﺘﻲ ﺑﺪﻭﻥ ﺍﺳﺘﻔﺎﺩﻩ ﺍﺯ ﻣﺎﺷﻴﻦ ﺣﺴﺎﺏ، ﺑﺎﺯﻱ ﻣﻲﻛﻨﻲ ،ﺍﻳﻦ ﻳﺦ ﺁﺏ ﻣﻲﺷﻮﺩ ،ﻳﻌﻨﻲ ﺫﻫﻨﺖ ﻧﺮﻡ ﻣﻲﺷﻮﺩ ﻭ ﺩﺭ ﺁﻥ ﻟﺤﻈﻪ ﺗﻘﺮﻳﺒﺎً ﻣﺤﺎﺳﺒﺎﺕ ﻛﺎﺭﻱﺍﺵ ﺭﺍ ﺍﻧﺠﺎﻡ ﻣﻲﺩﻫﺪ. ﻓﺎﻃﻤﻪ ﻧﺼﻴﺮﻱ ﻣﻲﮔﻮﻳﺪ :ﺭﻳﺎﺿﻲ ﻣﻲﺗﻮﺍﻥ ﻫﺮ ﻣﺴﺌﻠﻪﺍﻱ ﺭﺍ ﺣﻞ ﻛﺮﺩ! ﺩﺭ ﺗﻤﺎﻡ ﺭﺷﺘﻪﻫﺎ ﻛﺎﺭﺑﺮﺩ ﺩﺍﺭﺩ ﻭ ﺑﺎ ﺗﻤﺎﻡ ﻳﻜــﻲ ﺍﺯ ﺩﻻﻳﻠﻲ ﻛﻪ ﺑﺎﻋﺚ ﻣﻲﺷــﻮﺩ ﺁﻥﻫﺎ ﺩﺭ ﺍﺭﺗﺒﺎﻁ ﺍﺳﺖ .ﺩﻭﺳﺘﻢ ﻣﻲﮔﻔﺖ ﻛﻪ ﺑﻌﻀﻲ ﺑﭽﻪﻫﺎ ﺭﻳﺎﺿﻲ ﺭﺍ ﺩﻭﺳــﺖ ﻧﺪﺍﺷــﺘﻪ ﺑﺎﺷــﻨﺪ ﻳﺎ ﺁﻥ ﭘﻴﭽﻴﺪﮔﻲ ﺍﺯ ﻣﻔﺎﻫﻴﻢ ﺭﻳﺎﺿﻲ ﻣﻨﻄﻘﻲ ﻧﻴﺴﺘﻨﺪ ﻭ ﺍﻭ ﺍﻳﻤﺎﻥ ﻧﺪﺍﺭﺩ ﺭﺍ ﻳﺎﺩ ﻧﮕﻴﺮﻧﺪ ،ﺍﻳﻦ ﺍﺳــﺖ ﻛﻪ ﺩﺭﺱ ﺭﺍ ﺳــﺮ ﻛﻼﺱ ﻛﻪ ﺩﺭﺳــﺖ ﺑﺎﺷــﻨﺪ .ﻣﻦ ﻣﻲﮔﻮﻳﻢ ﻛﻪ ﺗﻮ ﻣﻲﺗﻮﺍﻧﻲ ﻳﺎﺩ ﻣﻲﮔﻴﺮﻧﺪ ﻭ ﺑﺎ ﺧﻮﺩﺷــﺎﻥ ﻣﻲﮔﻮﻳﻨﺪ ﺍﻳﻦ ﺩﺭﺱ ﺭﻳﺎﺿﻲ ﻳﻜﻲ ﺍﺯ ﺩﻻﻳﻞ ﺍﻳﻦ ﻣﻔﺎﻫﻴــﻢ ﺭﺍ ﺑﺮﺍﻱ ﺧﻮﺩﺕ ﺍﺛﺒﺎﺕ ﻛﻨﻲ ﻳﺎ ﺍﺯ ﻳﻚ ﭼﻪﻗﺪﺭ ﺁﺳــﺎﻥ ﺑــﻮﺩ! ﺍﻣﺎ ﺗﻮﺟﻪ ﻧﻤﻲﻛﻨﻨــﺪ ﻛﻪ ﺍﻳﻦ ﻋﻼﻗﻪﻣﻨﺪﻱ ﻣﻦ ﺑﻪ ﻣﺘﺨﺼﺺ ﺭﻳﺎﺿﻲ ﺑﺨﻮﺍﻫﻲ ﺁﻥ ﺭﺍ ﺑﺮﺍﻳﺖ ﺛﺎﺑﺖ ﻛﻨﺪ. ﺩﺭﺱ ﺩﺭ ﺣﺎﻓﻈــﻪﻱ ﻛﻮﺗﺎﻩﻣﺪﺕ ﺁﻥﻫﺎ ﺟﺎﻱ ﮔﺮﻓﺘﻪ ﺁﻥ ﺍﺳﺖ .ﻣﻦ ﻋﺎﺷﻖ ﻛﺴــﻲ ﻛﻪ ﺍﻳﻦ ﻣﻔﺎﻫﻴﻢ ﺭﺍ ﺑﻪ ﺧﻮﺑﻲ ﺑﺪﺍﻧﺪ ﻣﻲﺗﻮﺍﻧﺪ ﺍﺳــﺖ ﻭ ﺍﮔﺮ ﺗﻤﺮﻳــﻦ ﺑﻴﺶﺗﺮﻱ ﺭﻭﻱ ﺁﻥ ﻧﺪﺍﺷــﺘﻪ ﭼﻴﺰﻫﺎﻱ ﻣﺠﻬﻮﻟﻢ ﺑﺎ ﻳﻚ ﻧﻤﺎﻳﺶ ﺳــﺎﺩﻩ ﺁﻥ ﺭﺍ ﺍﺛﺒــﺎﺕ ﻛﻨﺪ .ﻣﻦ ﻓﻜﺮ ﺑﺎﺷــﻨﺪ ،ﺁﻥ ﺭﺍ ﻓﺮﺍﻣﻮﺵ ﻣﻲﻛﻨﻨﺪ .ﺍﮔﺮ ﺑﺨﻮﺍﻫﻴﻢ ﻳﻚ ﻣﻲﻛﻨﻢ ﺍﻓــﺮﺍﺩﻱ ﻛﻪ ﺭﻳﺎﺿﻲ ﺭﺍ ﺩﻭﺳــﺖ ﻧﺪﺍﺭﻧﺪ ﺑﻪ ﻣﺴﺌﻠﻪﻱ ﺭﻳﺎﺿﻲ ﻭﺍﺭﺩ ﺣﺎﻓﻈﻪﻱ ﺑﻠﻨﺪ ﻣﺪﺕ ﻣﺎ ﺷﻮﺩ، ﻛﻤﻚ ﺍﻓﺮﺍﺩ ﺑﺎﺗﺠﺮﺑﻪ ﻧﻴﺎﺯ ﺩﺍﺭﻧﺪ ﻛﻪ ﺑﻪ ﺁﻥﭼﻪ ﻣﻲﮔﻮﻳﻨﺪ ﺍﻳﻤﺎﻥ ﺩﺍﺷــﺘﻪ ﺑﺎﻳــﺪ ﺑﻴﺶﺗــﺮ ﺭﻭﻱ ﺁﻥ ﻛﺎﺭ ﻛﻨﻴﻢ ﻭ ﺩﺭ ﺍﻳﻦ ﺻﻮﺭﺕ ﺍﺳــﺖ ﻛﻪ ﺑﻪ ﺍﻳﻦ ﺑﺎﺷﻨﺪ ﻭ ﺑﺘﻮﺍﻧﻨﺪ ﺩﺭﺳــﺘﻲ ﻣﻄﺎﻟﺒﻲ ﺭﺍ ﻛﻪ ﺩﺭﺑﺎﺭﻩﺍﺵ ﺣﺮﻑ ﻣﻲﺯﻧﻨﺪ ﺑﻪ ﺩﺭﺱ ﻋﻼﻗﻪﻣﻨﺪ ﻣﻲﺷــﻮﻳﻢ ،ﻭﻟﻲ ﺍﮔﺮ ﺁﻥ ﺭﺍ ﻛﻨﺎﺭ ﺑﮕﺬﺍﺭﻳﻢ ،ﻧﺴــﺒﺖ ﺑﻪ ﺩﻳﮕﺮﺍﻥ ﺛﺎﺑﺖ ﻛﻨﻨﺪ. ﺁﻥ ﺳﺮﺩ ﻣﻲﺷﻮﻳﻢ ﻭ ﺩﺭﺱ ﺭﻳﺎﺿﻲ ﻛﻢﻛﻢ ﺍﺯ ﺯﻧﺪﮔﻲﻣﺎﻥ ﻛﻨﺎﺭ ﮔﺬﺍﺷﺘﻪ ﻧﺮﺟﺲ ﺳـﻌﻴﺪﻱ ﺑﺎﺯ ﻫﻢ ﮔﻼﻳﻪﺍﻱ ﺩﺍﺭﺩ :ﻣﻦ ﻳﻚ ﻣﺸــﻜﻞ ﺩﻳﮕﺮ ﻣﻲﺷﻮﺩ. ﻫﻢ ﺑﺎ ﺩﺭﺱ ﺭﻳﺎﺿــﻲ ﺩﺍﺭﻡ .ﺑﻪ ﻧﻈﺮ ﻣﻦ ﻓﻘﻂ 45ﺩﻗﻴﻘﻪﻱ ﺍﻭﻝ ﻛﻼﺱ ﻧﺮﺟﺲ ﺳـﻌﻴﺪﻱ ﺍﺯ ﻋﻘﺎﻳﺪ ﻗﺒﻠﻲﺍﺵ ﺩﻓــﺎﻉ ﻣﻲﻛﻨﺪ ﻭ ﻣﻲﮔﻮﻳﺪ: ﺭﻳﺎﺿﻲ ﺧﻮﺏ ﺍﺳﺖ. ﺣﺮﻑ ﻣﺎ ﺍﻳﻦ ﻧﻴﺴــﺖ ﻛﻪ ﺭﻳﺎﺿﻲ ﺍﺯ ﺯﻧﺪﮔﻲﻣﺎﻥ ﻛﻨﺎﺭ ﮔﺬﺍﺷﺘﻪ ﺷﻮﺩ ،ﻣﺎ 16
ﺭﺍﻫﻨﻤﺎﻳﻲ
ﺩﻭﺭﺓ ﺷﺎﻧﺰﺩﻫﻢ ،ﺷﻤﺎﺭﺓ ،4ﺗﺎﺑﺴﺘﺎﻥ 1390
ﻣﻲﮔﻮﻳﻴــﻢ ﺭﻳﺎﺿﻲ ﺭﺍ ﻓﻘﻂ ﺑﻪ ﺁﻥ ﺍﻧﺪﺍﺯﻩﺍﻱ ﺑﻪ ﻣﺎ ﺑﺪﻫﻨﺪ ﻛﻪ ﺩﺭ ﺯﻧﺪﮔﻲ ﻣﺎ ﻛﺎﺭﺑﺮﺩ ﺩﺍﺭﺩ. ﺟﻤﻴﻠﻪ ﺍﻣﺎﻣﻲ ﺍﺯ ﻛﺎﺭﺑﺮﺩﻫﺎﻱ ﺭﻳﺎﺿﻲ ﺻﺤﺒﺖ ﻣﻲﻛﻨﺪ: ﻣﻦ ﻫﻤﻴﺸــﻪ ﺑﻪ ﺷــﺎﮔﺮﺩﺍﻧﻢ ﮔﻔﺘﻪﺍﻡ ﻛﻪ ﺭﻳﺎﺿﻲ ﺩﺭ ﺗﻤﺎﻡ ﻣﺴــﺎﺋﻞ ﺯﻧﺪﮔﻲ ﻧﻘــﺶ ﺩﺍﺭﺩ .ﻛﺎﻓﻲ ﺍﺳــﺖ ﺩﻗﻴﻖﺗﺮ ﺑﻪ ﺍﻃﺮﺍﻓﻤــﺎﻥ ﻧﮕﺎﻩ ﻛﻨﻴﻢ. ﺗﻮﻧﻞﻫﺎ ،ﭘﻞﻫﺎ ﻭ ﺣﺘﻲ ﺳــﺎﺧﺘﻤﺎﻥﻫﺎﻳﻲ ﻛﻪ ﻣﺎ ﺩﺭ ﺁﻥ ﺯﻧﺪﮔﻲ ﻣﻲﻛﻨﻴﻢ ﻳﺎ ﻫﺮ ﺭﻭﺯ ﺍﺯ ﺁﻥﻫﺎ ﺍﺳــﺘﻔﺎﺩﻩ ﻣﻲﻛﻨﻴﻢ ﺑﺎ ﺍﺳﺘﻔﺎﺩﻩ ﺍﺯ ﻣﺤﺎﺳﺒﺎﺕ ﺭﻳﺎﺿﻲ ﺳــﺎﺧﺘﻪ ﺷــﺪﻩﺍﻧﺪ ،ﺍﺻ ً ﻼ ﻧﻤﻲﺷــﻮﺩ ﺗﺼﻮﺭ ﻛﺮﺩ ﻛﻪ ﺭﻳﺎﺿﻲ ﺭﺍ ﺍﺯ ﺳﺎﻳﺮ ﺭﺷﺘﻪﻫﺎ ﺣﺬﻑ ﻛﻨﻨﺪ. ﻳﻚ ﺳﺆﺍﻝ ﺣﺴﺎﺱ ﺍﺯ ﺑﭽﻪﻫﺎ ﻣﻲﭘﺮﺳﻢ: ﭼﻪﻗـﺪﺭ ﻣﻌﻠﻢﻫﺎﻱ ﺭﻳﺎﺿﻲ ﺩﺭ ﻋﻼﻗﻪﻣﻨﺪ ﻛـﺮﺩﻥ ﺷـﻤﺎ ﺑـﻪ ﺩﺭﺱ ﺭﻳﺎﺿـﻲ ﺗﺄﺛﻴـﺮ ﺩﺍﺷﺘﻪﺍﻧﺪ؟ ﺷـﻘﺎﻳﻖ ﺻﺎﺩﻗﻲ ﺩﺭ ﭘﺎﺳــﺦ ﺑﻪ ﺍﻳﻦ ﺳــﺆﺍﻝ ﻣﻲﮔﻮﻳﺪ :ﻣﻌﻠﻤﻲ ﻛﻪ ﺑﺸﺎﺵ ﻭ ﺧﻨﺪﻩﺭﻭ ﺑﺎﺷﺪ ﻭ ﺑﺎ ﺑﭽﻪﻫﺎ ﺻﻤﻴﻤﺎﻧﻪ ﺣﺮﻑ ﺑﺰﻧﺪ ﻭ ﺑﺨﻨﺪﺩ ﻭ ﺭﻳﺎﺿﻲ ﺭﺍ ﺑﺎ ﺷﻴﻮﻩﻫﺎﻳﻲ ﻣﺜﻞ ﻧﻤﺎﻳﺶ ﺩﺍﺩﻥ ﻳﺎ ﺗﻌﺮﻳﻒ ﻛﺮﺩﻥ ﺩﺍﺳــﺘﺎﻥ ﺩﺭﺱ ﺑﺪﻫﺪ ،ﻣﺴﻠﻤﺎً ﺩﺭ ﺗﺪﺭﻳﺲ ﺭﻳﺎﺿﻲ ﻣﻮﻓﻖﺗــﺮ ﺍﺳــﺖ ﻭ ﺑﭽﻪﻫــﺎ ﺩﺭﺱ ﺍﻭ ﺭﺍ ﺑﻬﺘــﺮ ﻳﺎﺩ ﻣﻲﮔﻴﺮﻧﺪ .ﻣﻦ ﺩﺭ ﺳــﺎﻝ ﮔﺬﺷــﺘﻪ ﻣﻌﻠﻤﻲ ﺩﺍﺷﺘﻢ ﻛﻪ ﻣﺮﺍ ﺑﻪ ﺩﺭﺱ ﺭﻳﺎﺿﻲ ﻋﻼﻗﻪﻣﻨﺪ ﻛﺮﺩ .ﺍﻳﻦ ﻣﻌﻠﻢ ﺩﺭﺱ ﻫﻨﺪﺳﻪ ﺭﺍ ﺑﺎ ﻧﻤﺎﻳﺶ ﻳﺎ ﺗﻌﺮﻳﻒ ﻳﻚ ﺩﺍﺳﺘﺎﻥ ﺑﻪ ﻣــﺎ ﺩﺭﺱ ﻣﻲﺩﺍﺩ .ﺍﻣﺎ ﺍﻳﻦﻫــﺎ ﻫﻢ ﻓﻘﻂ ﺩﺭ ﻳﺎﺩ ﮔﺮﻓﺘﻦ ﺩﺭﺱ ﺗﺄﺛﻴﺮ ﺩﺍﺷﺖ ﻧﻪ ﻋﻼﻗﻪﻣﻨﺪﻱ ﺑﻪ ﺁﻥ. ﻣﺮﻳـﻢ ﺟﺒﺎﺭﻱ ﺟــﻮﺍﺏ ﻣﻲﺩﻫــﺪ :ﻣﻦ ﻓﻜﺮ ﻣﻲﻛﻨــﻢ ﺭﻳﺎﺿــﻲ ﻫﺮﺟﻮﺭ ﻫﻢ ﻛﻪ ﺗﺪﺭﻳﺲ ﺷﻮﺩ ،ﺍﺻ ً ﻼ ﺟﺎﻟﺐ ﻧﻴﺴﺖ .ﺭﻳﺎﺿﻲ ﻣﺠﻤﻮﻋﻪﺍﻱ ﺍﺳﺖ ﻛﻪ ﺑﻪ ﺷﻤﺎ ﻣﻲﺩﻫﻨﺪ ﻭ ﻓﻘﻂ ﺍﺯ ﺷﻤﺎ ﺟﻮﺍﺏ ﻣﻲﺧﻮﺍﻫﻨﺪ .ﻓﻘﻂ ﻫﻢ ﺑﺎ ﺍﻋﺪﺍﺩ ﺳﺮ ﻭ ﻛﺎﺭ ﺩﺍﺭﻱ. ﻧﺮﺟﺲ ﺳـﻌﻴﺪﻱ ﻣﻲﮔﻮﻳــﺪ :ﻳﻜﻲ ﺍﺯ ﻣﻌﻠﻢﻫﺎﻱ ﻣﺎ ﺩﺭﺳــﻲ ﺭﺍ ﺑﺎ ﺷــﻜﻞ ﻳﺎﺩ ﻣﻲﺩﺍﺩ .ﻣﻦ ﻫﺮ ﭼﻴﺰﻱ ﺭﺍ ﻣﻤﻜﻦ ﺑﻮﺩ ﺳﺮ ﺍﻣﺘﺤﺎﻥ ﻓﺮﺍﻣﻮﺵ ﻛﻨﻢ ﺍﻣﺎ ﺁﻥ ﻣﺜﺎﻝ ﺭﺍ ﻛﻪ ﺷــﻜﻠﻲ ﺍﺯ ﻧﺎﻥ ﺑﺮﺑﺮﻱ ﺑﻮﺩ ﻫﻴﭻﻭﻗﺖ ﻓﺮﺍﻣﻮﺵ ﻧﻜﺮﺩﻡ. ﺯﻳﻨـﺐ ﺍﻃﻬﺮﻱ ﺟــﻮﺍﺏ ﻣﻲﺩﻫﺪ :ﻣــﻦ ﺑــﻪ ﺩﺭﺱ ﺭﻳﺎﺿﻲ ﺍﺻ ً ﻼ ﻋﻼﻗﻪﺍﻱ ﻧﺪﺍﺭﻡ .ﺭﻳﺎﺿﻲ ﺩﺭﺱ ﻛﺴﻞﻛﻨﻨﺪﻩﺍﻱ ﺍﺳﺖ .ﺩﻭ ﻋﺪﺩ ﺩﺍﺭﻳﻢ ﻛﻪ ﺑﻌﺪ ﺍﺯ ﺍﻧﺠﺎﻡ ﻣﺤﺎﺳﺒﺎﺗﻲ ﺭﻭﻱ ﺁﻥﻫﺎ ﻳﻚ ﻋﺪﺩ ﺑﻪ ﺩﺳﺖ ﻣﻲﺁﻭﺭﻳﻢ. ﺭﻋﻨﺎ ﺍﻣﺮﺍﻳﻲ ﻣﻌﺘﻘﺪ ﺍﺳــﺖ :ﺣﺠﻢ ﻣﻄﺎﻟﺒﻲ ﻛﻪ ﺩﺭ ﺩﺭﺱ ﺭﻳﺎﺿﻲ ﺑﺎﻳــﺪ ﻳﺎﺩ ﺑﮕﻴﺮﻳﻢ ﺧﻴﻠﻲ ﺯﻳﺎﺩ ﺍﺳــﺖ .ﺧﻴﻠﻲ ﺍﺯ ﺍﻳــﻦ ﻣﻄﺎﻟﺐ ﻫﻴﭻﻭﻗﺖ ﻛﺎﺭﺑــﺮﺩﻱ ﺩﺭ ﺯﻧﺪﮔﻲ ﻣﺎ ﻧﺪﺍﺭﻧــﺪ .ﻣﻐﺰ ،ﺭﺍ ﺁﻥﻗﺪﺭ ﭘﺮ ﻣﻲﻛﻨﻨﺪ ﻛﻪ ﺩﻳﮕﺮ ﻫﻴــﭻ ﻓﺎﻳﻞ ﺧﺎﻟﻲ ﺩﺭ ﺁﻥ ﺑﺎﻗﻲ ﻧﻤﻲﻣﺎﻧﺪ ﻭ ﭼﻴﺰ ﺟﺪﻳﺪﻱ ﺭﺍ ﻧﻤﻲﺗﻮﺍﻧﻴﻢ
ﻳﺎﺩ ﺑﮕﻴﺮﻳﻢ. ﺩﻟــﻢ ﻣﻲﺧﻮﺍﻫــﺪ ﺑﺪﺍﻧﻢ ﺍﻳﻦ ﺩﻭ ﮔــﺮﻭﻩ ﻋﻼﻗﻪﻣﻨــﺪ ﻭ ﺑﻲﻋﻼﻗﻪ ﺑﻪ ﺭﻳﺎﺿﻲ ﻓﻜﺮ ﻛﺮﺩﻩﺍﻧﺪ ﺩﺭ ﺁﻳﻨﺪﻩ ﭼﻪ ﺷﻐﻠﻲ ﺭﺍ ﺑﺮ ﻋﻬﺪﻩ ﺑﮕﻴﺮﻧﺪ .ﺍﺯ ﺑﭽﻪﻫﺎ ﻣﻲﭘﺮﺳــﻢ :ﺩﻭﺳــﺖ ﺩﺍﺭﻳﺪ ﺩﺭ ﺁﻳﻨﺪﻩ ﭼﻪﻛﺎﺭﻱ ﺷــﻮﻳﺪ ﻭ ﺁﻳﺎ ﻣﻲﺩﺍﻧﻴﺪ ﺷــﻐﻠﻲ ﺭﺍ ﻛﻪ ﻣﻲﺧﻮﺍﻫﻴﺪ ﺩﺭ ﺁﻳﻨﺪﻩ ﺑﺮ ﻋﻬﺪﻩ ﺑﮕﻴﺮﻳﺪ ﭼﻪﻗﺪﺭ ﺑﺎ ﺭﻳﺎﺿﻲ ﺳﺮﻭ ﻛﺎﺭ ﺩﺍﺭﺩ؟ ﺷـﻘﺎﻳﻖ ﺻﺎﺩﻗﻲ ﻣﻲﮔﻮﻳﺪ :ﻣﻦ ﺩﻭﺳﺖ ﺩﺍﺭﻡ ﺩﺍﺭﻭﺳﺎﺯ ﺷﻮﻡ .ﺍﻟﺒﺘﻪ ﺩﺍﺭﻭﺳﺎﺯﻱ ﺷﻐﻠﻲ ﺍﺳﺖ ﻛﻪ ﺑﻪ ﺭﻳﺎﺿﻴﺎﺕ ﺍﺣﺘﻴﺎﺝ ﺩﺍﺭﺩ ،ﺍﻣﺎ ﺑﻪ ﺭﻳﺎﺿﻴﺎﺕ ﭘﻴﺸﺮﻓﺘﻪﺍﻱ ﻧﻴﺎﺯ ﻧﺪﺍﺭﺩ. ﻣﺮﻳﻢ ﺟﺒﺎﺭﻱ ﺟﻮﺍﺏ ﻣﻲﺩﻫﺪ :ﻣﻦ ﻣﻲﺧﻮﺍﻫﻢ ﺩﻧﺪﺍﻥﭘﺰﺷــﻚ ﺷــﻮﻡ .ﺭﻳﺎﺿﻴﺎﺕ ﺗﺎ ﺣﺪﻱ ﺩﺭ ﺍﻳﻦ ﺷــﻐﻞ ﻣﻮﺭﺩ ﻧﻴﺎﺯ ﺍﺳــﺖ ﺍﻣﺎ ﺟﺰء ﺩﺭﺱﻫﺎﻱ ﺍﺻﻠﻲ ﺍﻳﻦ ﺭﺷــﺘﻪ ﻧﻴﺴﺖ .ﺩﻧﺪﺍﻥﭘﺰﺷــﻚ ﺑﻴﺶﺗﺮ ﺑﺎﻳﺪ ﺑﺎ ﻋﻠﻮﻡ ﺗﺠﺮﺑﻲ ﺁﺷﻨﺎ ﺑﺎﺷﺪ. ﺯﻳﻨﺐ ﺍﻃﻬﺮﻱ ﭘﺎﺳــﺦ ﻣﻲﺩﻫﺪ :ﻣﻦ ﺩﻭﺳــﺖ ﺩﺍﺭﻡ ﻣﻌﻠــﻢ ﺟﻐﺮﺍﻓﻲ ﺷــﻮﻡ .ﺩﻭﺳــﺖ ﺩﺍﺭﻡ ﺑﺪﺍﻧﻢ ﭘﺪﻳﺪﻩﻫﺎﻳﻲ ﻣﺜﻞ ﺟﻠﮕﻪ ﻭ ...ﭼﮕﻮﻧﻪ ﺍﻳﺠﺎﺩ ﻣﻲﺷﻮﻧﺪ. ﺍﻳﻦ ﺭﺷﺘﻪ ﺧﻴﻠﻲ ﺑﺎ ﺭﻳﺎﺿﻲ ﺳﺮ ﻭ ﻛﺎﺭ ﻧﺪﺍﺭﺩ. ﻧﺮﺟﺲ ﺳﻌﻴﺪﻱ ﺑﺮﻧﺎﻣﻪﻱ ﺩﻳﮕﺮﻱ ﺩﺍﺭﺩ :ﻣﻦ ﺩﻟــﻢ ﻣﻲﺧﻮﺍﻫﺪ ﻃﻠﺒﮕﻲ ﻭ ﻭﻛﺎﻟــﺖ ﺭﺍ ﺍﺩﺍﻣﻪ ﺩﻫﻢ. ﻫﻴﭽﻜﺪﺍﻡ ﺑﻪ ﺭﻳﺎﺿﻴﺎﺕ ﺍﺭﺗﺒﺎﻃﻲ ﻧﺪﺍﺭﻧﺪ. ﺭﻋﻨﺎ ﺍﻣﺮﺍﻳﻲ ﻫﻢ ﺍﺯ ﻋﻼﻗﻪﻣﻨﺪﻱﺍﺵ ﻣﻲﮔﻮﻳﺪ: ﻣﻦ ﺑﻪ ﺩﻭ ﺭﺷــﺘﻪﻱ ﺧﻴﻠﻲ ﻣﺘﻔــﺎﻭﺕ ﻋﻼﻗﻪﻣﻨﺪﻡ. ﻋﻜﺎﺳﻲ ﻭ ﺩﺍﺭﻭﺳــﺎﺯﻱ؛ ﻛﻪ ﺧﻴﻠﻲ ﻫﻢ ﺑﺎ ﺭﻳﺎﺿﻴﺎﺕ ﻛﺎﺭﻱ ﻧﺪﺍﺭﻧﺪ. ﺁﺭﺯﻭ ﻋﺎﺭﻓﻲ ﻣﻲﮔﻮﻳﺪ :ﻣﻦ ﻫﻢ ﺩﻭﺳــﺖ ﺩﺍﺭﻡ ﺷــﺎﻋﺮ ﻭ ﻧﻮﻳﺴﻨﺪﻩ ﺷﻮﻡ .ﺍﻟﺒﺘﻪ ﺍﮔﺮ ﻧﻮﻳﺴﻨﺪﻩ ﻧﺸﻮﻡ ،ﻧﻘﺎﺵ ﺧﻮﺍﻫﻢ ﺷﺪ. ﻓﺎﻃﻤﻪ ﺟﻴﺮﺍﻧﻲ ﺩﺭ ﭘﺎﺳﺦ ﺑﻪ ﺍﻳﻦ ﺳﺆﺍﻝ ﻣﻲﮔﻮﻳﺪ :ﻣﻦ ﻫﻢ ﺩﻭﺳﺖ ﺩﺍﺭﻡ ﺩﻛﺘﺮﺍﻱ ﺭﻳﺎﺿﻲ ﺑﮕﻴﺮﻡ ﻭ ﺍﺳﺘﺎﺩ ﺩﺍﻧﺸﮕﺎﻩ ﺷﻮﻡ. ﺷـﺒﻨﻢ ﺍﺑﻮﺍﻟﻔﻀﻠﻲ ﺭﻳﺎﺿﻲ ﺭﺍ ﺩﻭﺳﺖ ﺩﺍﺭﺩ ،ﺍﻣﺎ ﺷﻐﻠﺶ ﺭﺑﻄﻲ ﺑﻪ ﺍﻳﻦ ﺭﺷﺘﻪ ﻧﺪﺍﺭﺩ .ﺍﻭ ﺩﺭﺍﻳﻦﺑﺎﺭﻩ ﻣﻲﮔﻮﻳﺪ :ﻣﻦ ﺩﻟﻢ ﻣﻲﺧﻮﺍﻫﺪ ﻣﺎﻣﺎ ﺷﻮﻡ. ﺍﻟﺒﺘﻪ ﺍﻳﻦ ﺷﻐﻞ ﺭﺑﻄﻲ ﺑﻪ ﺭﻳﺎﺿﻲ ﻧﺪﺍﺭﺩ. ﺍﻗﻠﻴﻤﺎ ﺍﻟﺴـﺎﺩﺍﺕ ﺳﻴﺪﺯﺍﺩﻩ ﻣﻌﺘﻘﺪ ﺍﺳﺖ :ﺑﻪ ﻧﻈﺮ ﻣﻦ ﻛﻞ ﺟﻬﺎﻥ ﻫﺴــﺘﻲ ﺑﺮ ﭘﺎﻳﻪﻱ ﺭﻳﺎﺿﻲ ﺑﺮﭘﺎﺳﺖ ﻭ ﻫﻤﻪﻱ ﺭﺷﺘﻪﻫﺎ ﺑﺎ ﺭﻳﺎﺿﻲ ﺍﺭﺗﺒﺎﻁ ﺩﺍﺭﻧﺪ .ﺑﺮﺍﻱ ﻣﺜــﺎﻝ ،ﻣﻌﻠﻢ ﺟﻐﺮﺍﻓﻲ ﺑﺪﻭﻥ ﺩﺍﻧﺴــﺘﻦ ﺭﻳﺎﺿﻲ ﻧﻤﻲﺗﻮﺍﻧﺪ ﻣﻮﻓﻖ ﺑﺎﺷﺪ. ﻣﻦ ﻓﻜﺮ ﻣﻲﻛﻨﻢ ﺑﭽﻪﻫﺎﻳﻲ ﻛﻪ ﻣﻲﮔﻮﻳﻨﺪ ﺭﻳﺎﺿﻲ ﺭﺍ ﺑﺎﻳﺪ ﺑﺎ ﻧﻤﺎﻳﺶ ﻭ ﺩﺍﺳﺘﺎﻥ ﺗﺪﺭﻳﺲ ﻛﺮﺩ ،ﺩﺭ ﺍﺷﺘﺒﺎﻩﺍﻧﺪ .ﻓﺮﺽ ﻛﻨﻴﺪ ﻫﺮ ﺳﺎﻝ ﺭﻳﺎﺿﻲ ﺭﺍ ﺑﺎ ﻧﻤﺎﻳﺶ ﺑﻪ ﺷــﻤﺎ ﻳﺎﺩ ﺩﺍﺩﻧﺪ ﻭ ﺑﻪ ﺧﺎﻃﺮ ﺍﻳﻦ ﺭﻭﺵ ﺗﺪﺭﻳﺲ ﺑﻪ ﺭﻳﺎﺿﻲ ﺷﻤﺎﺭﺓ ،4ﺗﺎﺑﺴﺘﺎﻥ 1390 ﺷﻤﺎﺭﻩ ﻯ ﺷﺎﻧﺰﺩﻫﻢ، ﺷﺎﻧﺰﺩﻫﻢ، ﺩﻭﺭﺓ ﺩﻭﺭﻩﻯ
ﺭﺍﻫﻨﻤﺎﻳﻲ
17
ﻋﻼﻗﻪﻣﻨﺪ ﺷﺪﻳﺪ ﻭ ﺭﺷﺘﻪﻱ ﺭﻳﺎﺿﻲ ﺭﺍ ﺍﺩﺍﻣﻪ ﺩﺍﺩﻳﺪ .ﺁﻳﺎ ﺑﺎﻳﺪ ﺩﺭ ﺩﺍﻧﺸﮕﺎﻩ ﻛﻼﺳﻲ ﺑﺎﺷﻢ ﻗﺒﻞ ﺍﺯ ﺗﺪﺭﻳﺲ ﻫﺮ ﺩﺭﺳﻲ ﻓﻜﺮ ﻣﻲﻛﻨﻢ ﻭ ﺭﻭﺷﻲ ﺭﺍ ﺑﺮﺍﻱ ﻫﻢ ﺑﺮﺍﻱ ﺷــﻤﺎ ﺑﺎ ﻧﻤﺎﻳﺶ ﺩﺭﺱ ﺩﺍﺩﻩ ﺷــﻮﺩ؟ ﻣــﻦ ﻓﻜﺮ ﻣﻲﻛﻨﻢ ﺫﻫﻦ ﺗﺪﺭﻳﺲ ﺍﻧﺘﺨﺎﺏ ﻣﻲﻛﻨﻢ ﻛﻪ ﺑﭽﻪﻫﺎ ﺭﺍ ﻧﺴــﺒﺖ ﺑﻪ ﺩﺭﺱ ﻛﻨﺠﻜﺎﻭ ﻛﻨﺪ. ﺍﻧﺴﺎﻥ ﺧﻮﺍﻩﻧﺎﺧﻮﺍﻩ ﻛﻨﺠﻜﺎﻭ ﺍﺳــﺖ ﻭ ﺩﻭﺳﺖ ﺩﺍﺭﺩ ﭼﻴﺰﻫﺎﻱ ﺑﻴﺶﺗﺮﻱ ﺑــﺎ ﺩﺍﻧﺶﺁﻣﻮﺯﺍﻧﻢ ﺣﺮﻑ ﻣﻲﺯﻧﻢ ﻭ ﺣﺮﻑﻫﺎﻱ ﺁﻥﻫﺎ ﺭﺍ ﮔﻮﺵ ﻣﻲﻛﻨﻢ ﺗﺎ ﺑﺪﺍﻧﺪ .ﻣﻦ ﺩﻭﺳــﺖ ﺩﺍﺭﻡ ﺟﻮﺍﺏ ﻣﺴــﺌﻠﻪﺍﻱ ﺭﺍ ﻛﻪ ﻧﻤﻲﺩﺍﻧﻢ ﭘﻴﺪﺍ ﻛﻨﻢ ﺩﻟﻴﻞ ﺑﻲﻋﻼﻗﮕﻲﺷــﺎﻥ ﺭﺍ ﺑﻪ ﺩﺭﺱ ﺑﻔﻬﻤﻢ .ﺑﺎ ﺑﭽﻪﻫﺎ ﺩﻭﺳــﺘﺎﻧﻪ ﺣﺮﻑ ﻣﻦ ﺑﻪ ﺍﻳﻦ ﺩﻟﻴﻞ ﺑﻪ ﺭﻳﺎﺿﻲ ﻋﻼﻗﻪﻣﻨﺪ ﺷﺪﻡ .ﺭﻳﺎﺿﻲ ﺍﻳﻦ ﻓﺮﺻﺖ ﺭﺍ ﺑﻪ ﻣﻲﺯﻧﻢ ﻭ ﻛﺎﺭﻱ ﻣﻲﻛﻨﻢ ﺗﺎ ﺑﭽﻪﻫﺎ ﻋﺎﺷﻖ ﻣﻌﻠﻢ ﺭﻳﺎﺿﻲﺷﺎﻥ ﺷﻮﻧﺪ .ﺍﮔﺮ ﺁﺩﻡ ﻣﻲﺩﻫﺪ ﺗﺎ ﺩﺭﺑﺎﺭﻩﻱ ﺧﻮﺩﺵ ﻭ ﺟﻬﺎﻧﻲ ﻛﻪ ﺩﺭ ﺁﻥ ﺯﻧﺪﮔﻲ ﻣﻲﻛﻨﺪ ،ﺑﭽﻪﻫــﺎ ﺑﺎ ﺁﻣﺪﻥ ﻣﻌﻠﻢ ﺭﻳﺎﺿﻲ ﺑﻪ ﺳــﺮ ﻛﻼﺱ ﺑﮕﻮﻳﻨﺪ :ﻭﺍﻱ! ﺍﻳﻦ ﻣﻌﻠﻢ ﺩﻭﺑﺎﺭﻩ ﺁﻣﺪ! ﻳﺎ ﺍﺣﺴــﺎﺱ ﻛﻨﻨﺪ ﺍﻳﻦ ﻣﻌﻠﻢ ﻓﻘﻂ ﻗﺼﺪ ﺍﺫﻳﺖ ﻛﺮﺩﻧﺸﺎﻥ ﺭﺍ ﺍﻃﻼﻋﺎﺗﻲ ﻛﺴﺐ ﻛﻨﺪ. ﻣﻌﺼﻮﻣﻪ ﺳـﻴﺪ ﻣﺤﻤﺪﺯﺍﺩﻩ ﻫﻢ ﺍﺯ ﺁﺭﺯﻭﻫﺎﻳﺶ ﻣﻲﮔﻮﻳﺪ :ﻣﻦ ﺑﻪ ﺩﺍﺭﺩ ،ﻣﻄﻤﺌﻨﺎً ﺩﺭﺱ ﺭﺍ ﻳﺎﺩ ﻧﻤﻲﮔﻴﺮﻧﺪ. ﺍﻗﻠﻴﻤﺎ ﺍﻟﺴﺎﺩﺍﺕ ﺳﻴﺪﺯﺍﺩﻩ :ﻣﻦ ﺍﮔﺮ ﺑﺨﻮﺍﻫﻢ ﺑﭽﻪﻫﺎ ﺭﺍ ﺑﻪ ﺭﻳﺎﺿﻲ ﺩﻭ ﺷــﻐﻞ ﻣﻬﻨﺪﺳﻲ ﺳﺎﺧﺘﻤﺎﻥ ﻭ ﻧﻮﻳﺴــﻨﺪﮔﻲ ﻋﻼﻗﻪﻣﻨﺪﻡ .ﺭﻳﺎﺿﻴﺎﺕ ﻋﻼﻗﻪﻣﻨﺪ ﻛﻨﻢ ﺑﺎﻳﺪ ﺑﺎ ﺷﻮﺭ ﻭ ﺍﺷﺘﻴﺎﻕ ﭘﺎﻱ ﺗﺨﺘﻪ ﻛﺎﺭﺑﺮﺩ ﺯﻳﺎﺩﻱ ﺩﺭ ﻣﻬﻨﺪﺳﻲ ﺳﺎﺧﺘﻤﺎﻥ ﺩﺍﺭﺩ. ﺭﻳﺎﺿﻲ ﺍﻳﻦ ﻓﺮﺻﺖ ﺳــﻴﺎﻩ ﺑــﺮﻭﻡ ﻭ ﺩﺭﺱ ﺭﺍ ﺑﺎ ﺷــﻮﺧﻲﻫﺎﻱ ﺭﻳﺎﺿﻲ ﺍﻗﻠﻴﻤـﺎ ﺍﻟﺴـﺎﺩﺍﺕ ﺳـﻴﺪﺯﺍﺩﻩ ﻫــﻢ ﺷــﻐﻞ ﻫﻤﺮﺍﻩ ﻛﻨﻢ. ﺁﻳﻨﺪﻩﺍﺵ ﺭﺍ ﺍﻧﺘﺨﺎﺏ ﻛﺮﺩﻩ ﺍﺳﺖ :ﻣﻦ ﺑﻪ ﺍﺧﺘﺮﺷﻨﺎﺳﻲ ﺭﺍ ﺑﻪ ﺁﺩﻡ ﻣﻲﺩﻫﺪ ﺗﺎ ﺯﻫﺮﺍ ﻓﺘﺤﻲ ﭘﺎﺳـﺦ ﻣﻲﺩﻫﺪ :ﻣﻦ ﺩﻭﺳﺖ ﻋﻼﻗﻪﻣﻨﺪﻡ ﻭ ﺍﻳﻦ ﺭﺷﺘﻪ ﺍﺭﺗﺒﺎﻁ ﻧﺰﺩﻳﻜﻲ ﺑﺎ ﺭﻳﺎﺿﻴﺎﺕ ﺩﺭﺑﺎﺭﻩﻱ ﺧﻮﺩﺵ ﻭ ﺩﺍﺭﻡ ﺑﺎ ﺷﺎﮔﺮﺩﺍﻧﻢ ﺣﺮﻑ ﺑﺰﻧﻢ ﻭ ﺁﻥﻫﺎ ﺭﺍ ﻗﺎﻧﻊ ﻛﻨﻢ ﺩﺍﺭﺩ. ﺯﻧﺪﮔﻲ ﺁﻥ ﺩﺭ ﻛﻪ ﺟﻬﺎﻧﻲ ﻛﻪ ﺭﻳﺎﺿﻲ ﺩﺭﺱ ﺟﺎﻟﺐ ﻭ ﺷﻨﻴﺪﻧﻲ ﺍﺳﺖ. ﻓﺎﻃﻤﻪ ﻧﺼﻴﺮﻱ ﻣﻲﮔﻮﻳﺪ :ﻣﻦ ﺩﻭﺳﺖ ﺩﺍﺭﻡ ﺩﺭ ﺳـﻴﺪﻩ ﺯﻫـﺮﺍ ﺩﺭﻳﺎﺑﺎﺭﻱ ﺍﻳــﺪﻩﻱ ﺩﻳﮕﺮﻱ ﺁﻳﻨــﺪﻩ ﺩﻛﺘﺮ ﻣﺘﺨﺼﺺ ﺩﻳﺎﺑﺖ ﺷــﻮﻡ .ﭼﻮﻥ ﺧﻮﺩﻡ ﻣﻲﻛﻨﺪ ،ﺍﻃﻼﻋﺎﺗﻲ ﺩﺍﺭﺩ :ﻣﻦ ﻫﻢ ﺳــﻌﻲ ﻣﻲﻛﻨﻢ ﺑﭽﻪﻫﺎ ﺭﺍ ﺑﻪ ﺩﺭﺱ ﺩﻳﺎﺑﺖ ﺩﺍﺭﻡ ﺑﻴﻤﺎﺭﻡ ﺭﺍ ﻣﻲﺗﻮﺍﻧﻢ ﺩﺭﻙ ﻛﻨﻢ ﻭ ﺑﻔﻬﻤﻢ ﻛﺴﺐ ﻛﻨﺪ ﺟﺬﺏ ﻛﻨﻢ .ﺑﺮﺍﻱ ﻣﺜﺎﻝ ﺑﺎ ﺗﺸــﻮﻳﻖ ﻳﺎ ﻧﻤﺮﻩ ﺩﺍﺩﻥ ﺑﻴﻤــﺎﺭ ﻣﺒﺘﻼ ﺑــﻪ ﺩﻳﺎﺑﺖ ﺍﺯ ﺑﻴﻤــﺎﺭﻱﺍﺵ ﭼﻪ ﺭﻧﺠﻲ ﺑﺮﺍﻱ ﭘﺎﺳــﺦ ﺑﻪ ﻳﻚ ﺳــﺆﺍﻝ ﺧﺎﺹ .ﺑﺎ ﺍﻳﻦ ﺭﻭﺵ ﻣﻲﺑﺮﺩ. ﻣﻤﻜﻦ ﺍﺳﺖ ﺩﺭ ﺍﺑﺘﺪﺍ ﺩﺍﻧﺶﺁﻣﻮﺯﺍﻥ ﺑﺮﺍﻱ ﺗﺸﻮﻳﻖ ﻣﻦ ﻓﻜﺮ ﻣﻲﻛﻨﻢ ﻳﺎﺩ ﺩﺍﺩﻥ ﻭ ﻳﺎﺩ ﮔﺮﻓﺘﻦ ﺭﻳﺎﺿﻲ ﻫــﻢ ﺑﻪ ﺍﻳــﻦ ﺩﺭﻙ ﻣﺘﻘﺎﺑﻞ ﻧﻴﺎﺯ ﺩﺍﺭﺩ ،ﻳﻌﻨﻲ ﺗــﻮ ﺍﺯ ﻳﺎﺩ ﮔﺮﻓﺘﻦ ﺭﻳﺎﺿﻲ ﺷــﺪﻥ ﻳﺎ ﻧﻤﺮﻩ ﮔﺮﻓﺘﻦ ﺩﺭﺱ ﺑﺨﻮﺍﻧﻨــﺪ ،ﺍﻣﺎ ﻛﻢﻛﻢ ﺑﻪ ﺩﺭﺱ ﻋﻼﻗﻪﻣﻨﺪ ﻣﻲﺷﻮﻧﺪ. ﺑﻬﺮﻩﺍﻱ ﺑﺮﺩﻩ ﺑﺎﺷﻲ ﺑﺘﻮﺍﻧﻲ ﺁﻥ ﺭﺍ ﺑﻪ ﺩﻳﮕﺮﺍﻥ ﻧﻴﺰ ﻳﺎﺩ ﺑﺪﻫﻲ. ﺷﺒﻨﻢ ﺍﺑﻮﺍﻟﻔﻀﻠﻲ ﺗﺤﻠﻴﻞ ﺟﺎﻟﺒﻲ ﺩﺍﺭﺩ :ﺧﻴﻠﻲ ﺍﺯ ﻣﻌﻠﻢﻫﺎ ﺧﻮﺩﺷﺎﻥ ﻣﻦ ﻋﺎﺷﻖ ﻧﻮﻳﺴﻨﺪﮔﻲ ﻫﻢ ﻫﺴﺘﻢ. ﺟﻤﻴﻠـﻪ ﺍﻣﺎﻣﻲ )ﻣﻌﻠﻢ ﺑﭽﻪﻫــﺎ( ﺭﺍﻩﺣﻞﻫﺎﻳﻲ ﺍﺭﺍﺋــﻪ ﻣﻲﺩﻫﺪ ﻛﻪ ﻧﺴﺒﺖ ﺑﻪ ﺩﺭﺱ ﺭﻳﺎﺿﻲ ﻋﻼﻗﻪﺍﻱ ﻧﺪﺍﺭﻧﺪ .ﻣﻦ ﺍﮔﺮ ﺑﺨﻮﺍﻫﻢ ﺗﺪﺭﻳﺲ ﻛﻨﻢ ﺧﻴﻠﻲ ﺍﺯ ﻣﺸــﻜﻼﺕ ﺑﭽﻪﻫﺎ ﺩﺭ ﺩﺭﺱ ﺭﻳﺎﺿــﻲ ﺭﺍ ﺣﻞ ﻣﻲﻛﻨﺪ :ﻣﻦ ﺑﻪ ﺍﻭﻝ ﺩﺭ ﺧﻮﺩﻡ ﻋﻼﻗﻪ ﺍﻳﺠﺎﺩ ﻣﻲﻛﻨﻢ .ﺑﻌﺪ ﺍﺯ ﺁﻥ ﺩﺍﻧﺶﺁﻣﻮﺯ ﻫﻢ ﺩﺭﺱ ﺭﺍ ﺑﭽﻪﻫﺎ ﺗﻮﺻﻴﻪ ﻣﻲﻛﻨﻢ ﻛﻪ ﺩﺭﺱ ﺭﻳﺎﺿﻲ ﺭﺍ ﺩﺭ ﻫﻤﺎﻥ ﺭﻭﺯ ﺑﺨﻮﺍﻧﻨﺪ .ﺍﮔﺮ ﻳﺎﺩ ﻣﻲﮔﻴﺮﺩ ﻭ ﺑﻪ ﺁﻥ ﻋﻼﻗﻪﻣﻨﺪ ﻣﻲﺷﻮﺩ. ﻓﺎﻃﻤﻪ ﺧﺪﺍﻳﻲ ﻣﻲﮔﻮﻳﺪ :ﻣﻦ ﺍﮔﺮ ﻣﻌﻠﻢ ﺭﻳﺎﺿﻲ ﺷــﻮﻡ ﻫﻴﭻﻭﻗﺖ ﺑﭽﻪﻫﺎ ﻋﺼﺮ ﻫﻤــﺎﻥ ﺭﻭﺯﻱ ﻛﻪ ﻣﻌﻠﻢ ﻣﻄﻠﺐ ﺟﺪﻳــﺪﻱ ﺑﻪ ﺁﻥﻫﺎ ﮔﻔﺘﻪ ﺍﺳــﺖ ،ﺁﻥ ﺭﺍ ﻣﺮﻭﺭ ﻛﻨﻨﺪ ،ﺩﺭﺱ ﺭﺍ ﺑﻪ ﺳﺎﺩﮔﻲ ﻳﺎﺩ ﻣﻲﮔﻴﺮﻧﺪ ﻭ ﻓﺮﺍﻭﺵ ﻛﺴــﻲ ﺭﺍ ﺗﻨﺒﻴﻪ ﻧﻤﻲﻛﻨﻢ ،ﭼﻮﻥ ﺗﻨﺒﻴﻪ ﺩﺍﻧﺶﺁﻣــﻮﺯ ﺭﺍ ﺍﺯ ﺩﺭﺱ ﻣﺘﻨﻔﺮ ﻣﻲﻛﻨﺪ .ﻋﻼﻭﻩ ﺑﺮ ﺍﻳﻦ ،ﺑﺎ ﺷﺎﮔﺮﺩﺍﻧﻢ ﺻﺤﺒﺖ ﻣﻲﻛﻨﻢ ﻭ ﺑﻪ ﺁﻥﻫﺎ ﻛﻤﻚ ﻧﻤﻲﻛﻨﻨﺪ .ﺍﻳﻦ ﺑﺎﻋﺚ ﻣﻲﺷﻮﺩ ﺑﻪ ﺩﺭﺱ ﻋﻼﻗﻪﻣﻨﺪ ﺷﻮﻧﺪ. ﺑﻪ ﻧﻈــﺮ ﻣﻦ ،ﻫﻤﻪﻱ ﺍﻓﺮﺍﺩ ﻇﺮﻓﻴﺖ ﻳﺎﺩﮔﻴــﺮﻱ ﻫﻤﻪﭼﻴﺰ ﺭﺍ ﺩﺍﺭﻧﺪ .ﻣﻲﻛﻨﻢ ﺑﺮﺍﻱ ﺩﺭﺱ ﺧﻮﺍﻧﺪﻥﺷــﺎﻥ ﺑﺮﻧﺎﻣﻪ ﺩﺍﺷﺘﻪ ﺑﺎﺷﻨﺪ .ﺑﺮﺍﻱ ﺍﻳﻦ ﻛﺎﺭ ﺍﻳﻦﻛﻪ ﺑﻌﻀﻲﻫﺎ ﻣﻲﮔﻮﻳﻨﺪ ﻣﺎ ﺍﺳــﺘﻌﺪﺍﺩ ﻳﺎﺩﮔﻴــﺮﻱ ﺭﻳﺎﺿﻲ ﺭﺍ ﻧﺪﺍﺭﻳﻢ ،ﺑﻪ ﻫﺮﻳﻚ ﺍﺯ ﺷــﺎﮔﺮﺩﺍﻧﻢ ﺯﻣﺎﻥ ﻣﺸﺨﺼﻲ ﻣﻲﺩﻫﻢ ﺗﺎ ﺑﺘﻮﺍﻧﻢ ﺑﺎ ﻣﺸﻜﻼﺕ ﺣﺮﻑ ﺩﺭﺳﺘﻲ ﻧﻴﺴﺖ .ﺑﭽﻪﻫﺎ ﻛﻨﺠﻜﺎﻭ ﺑﺎﺷﻴﺪ ﻭ ﺳﻌﻲ ﻛﻨﻴﺪ ﺍﺳﺘﻌﺪﺍﺩﻫﺎﻱ ﻭ ﺩﻻﻳﻞ ﺑﻲﻋﻼﻗﮕﻲﺷﺎﻥ ﺑﻪ ﺩﺭﺱ ﺁﺷﻨﺎ ﺷﻮﻡ. ﺣــﺮﻑ ﭘﺎﻳﺎﻧﻲ ﺟﻤﻴﻠﻪ ﺍﻣﺎﻣﻲ ،ﻧﺼﻴﺤﺘﻲ ﺑﻪ ﺑﭽﻪﻫﺎﻳﻲ ﺍﺳــﺖ ﻛﻪ ﺍﺯ ﻣﺨﺘﻠﻒ ﺍﺯ ﺟﻤﻠﻪ ﺍﺳﺘﻌﺪﺍﺩ ﺭﻳﺎﺿﻲ ﺭﺍ ﺩﺭ ﻭﺟﻮﺩﺗﺎﻥ ﺷﻜﻮﻓﺎ ﻛﻨﻴﺪ. ﺍﺯ ﺑﭽﻪﻫﺎﻳﻲ ﻛﻪ ﺭﻳﺎﺿﻲ ﺭﺍ ﺩﻭﺳـﺖ ﺩﺍﺭﻧﺪ ﻣﻲﺧﻮﺍﻫﻢ ﺗﺼﻮﺭ ﻭﺿﻊ ﻛﻼﺱ ﻭ ﻧﺤﻮﻩﻱ ﺗﺪﺭﻳــﺲ ﺭﻳﺎﺿﻲ ﮔﻼﻳﻪ ﻣﻲﻛﻨﻨﺪ .ﺍﻭ ﻣﻲﮔﻮﻳﺪ: ﻛﻨﻨﺪ ﻣﻌﻠﻢ ﺷﺪﻩﺍﻧﺪ ﻭ ﺍﺯ ﺁﻥﻫﺎ ﺧﻮﺍﺳﺘﻪﺍﻧﺪ ﻛﻼﺳﻲ ﺭﺍ ﺍﺩﺍﺭﻩ ﻛﻨﻨﺪ ﻣﻴﺰﺍﻥ ﻓﻌﺎﻟﻴﺖ ﺩﺍﻧﺶﺁﻣﻮﺯﺍﻥ ﺩﺭ ﻛﻼﺱ ،ﻳﻜﻲ ﺍﺯ ﻋﻮﺍﻣﻠﻲ ﺍﺳﺖ ﻛﻪ ﺁﻥﻫﺎ ﻛﻪ ﺩﺍﻧﺶﺁﻣﻮﺯﺍﻧـﺶ ﻫﻴﭻ ﻋﻼﻗﻪﺍﻱ ﺑﻪ ﺩﺭﺱ ﺭﻳﺎﺿﻲ ﻧﺪﺍﺭﻧﺪ .ﺍﺯ ﺭﺍ ﺑﻪ ﺩﺭﺱ ﻋﻼﻗﻪﻣﻨﺪ ﻣﻲﻛﻨﺪ .ﺑﭽﻪﻫﺎ ﺑﺎﻳﺪ ﺧﻮﺩﺷــﺎﻥ ﺩﺭﺑﺎﺭﻩﻱ ﺷﻜﻞ ﺗﺪﺭﻳﺲ ﺭﻳﺎﺿﻲ ﻧﻈﺮ ﺑﺪﻫﻨﺪ .ﺩﺭ ﻛﻼﺱ ﻣﺸﺎﺭﻛﺖ ﺩﺍﺷﺘﻪ ﺑﺎﺷﻨﺪ ﻭ ﺳﻌﻲ ﺁﻥﻫﺎ ﻣﻲﭘﺮﺳﻢ :ﺍﻳﻦ ﻛﻼﺱ ﺭﺍ ﭼﻪﻃﻮﺭ ﺍﺩﺍﺭﻩ ﻣﻲﻛﻨﻴﺪ؟ ﻓﺎﻃﻤـﻪ ﻧﺼﻴﺮﻱ ﺟﻮﺍﺏ ﻣﻲﺩﻫﺪ :ﺑﻪ ﻧﻈﺮ ﻣﻦ ﻣﻌﻠﻢ ﻓﻘﻂ ﺗﺎ ﭘﻨﺞ ﻛﻨﻨﺪ ﺍﺯ ﺯﻣﺎﻧﺸﺎﻥ ﺑﻴﺶﺗﺮﻳﻦ ﺍﺳﺘﻔﺎﺩﻩ ﺭﺍ ﺑﻜﻨﻨﺪ. ﺩﺭﺻــﺪ ﺑﺮ ﻳﺎﺩﮔﻴــﺮﻱ ﺩﺍﻧﺶﺁﻣﻮﺯﺍﻥ ﺗﺄﺛﻴﺮ ﺩﺍﺭﺩ .ﺍﮔــﺮ ﻣﻦ ﻣﻌﻠﻢ ﭼﻨﻴﻦ 18
ﺭﺍﻫﻨﻤﺎﻳﻲ
ﺩﻭﺭﺓ ﺷﺎﻧﺰﺩﻫﻢ ،ﺷﻤﺎﺭﺓ ،4ﺗﺎﺑﺴﺘﺎﻥ 1390
ﺩﺍﻧﺶﺍﻓﺰﺍﻳﻲ
﹡﹆︀ط ا﹞﹟ ﹇︧﹝️ آ︠︣ ﺣﺴﻦ ﺍﺣﻤﺪﻱ
ﻛﻠﻴﺪﻭﺍژﻩﻫﺎ :ﻧﻘﺎﻁ ،ﺩﺍﻳﺮﻩ ،ﻣﺴﻴﺮ ﺍﻣﻦ ،ﺧﻂ ﺭﺍﺳﺖ. ﻓــﺮﺽ ﻛﻨﻴﺪ ﺩﻭ ﺭﺷــﺘﻪ ﻛﻮﻩ )ﺧﻂ( ﻧﺎﺍﻣــﻦ L1ﻭ L2ﺩﺍﺭﻳﻢ ﻛﻪ ﺑﻪ ﺻﻮﺭﺕ ﻣﻮﺍﺯﻱ ﺑﺎ ﻫﻢ ﻗﺮﺍﺭ ﮔﺮﻓﺘﻪﺍﻧﺪ .ﻣﻲﺧﻮﺍﻫﻴﻢ ﺑﻴﻦ ﺍﻳﻦ ﺩﻭ ﺭﺷــﺘﻪ ﻛﻮﻩ، ﻣﺴــﻴﺮﻱ ﺗﻌﻴﻴﻦ ﻛﻨﻴﻢ ﻛﻪ ﺷــﻬﺮﻫﺎﻱ )ﻧﻘﺎﻁ( Aﻭ Bﺭﺍ ﺑﻪ ﻳﻜﺪﻳﮕﺮ ﻣﺘﺼﻞ ﻛﻨﺪ .ﺑﻪ ﻧﻈﺮ ﺷــﻤﺎ ﭼﮕﻮﻧﻪ ﻣﻲﺗﻮﺍﻥ ﺍﻣﻦﺗﺮﻳﻦ ﻣﺴﻴﺮ ﺭﺍ ﺍﻧﺘﺨﺎﺏ ﻛﺮﺩ؟ )ﻓﺮﺽ ﻛﻨﻴﺪ ﺷﺪﺕ ﻧﺎﺍﻣﻨﻲ ﺩﺭ ﻫﺮ ﺩﻭ ﺭﺷﺘﻪ ﺑﺮﺍﺑﺮ ﺍﺳﺖ(.
L1
B
A
L2
ﺩﺭ ﺣﻞ ﺍﻳﻦ ﻣﺴﺌﻠﻪ ﻫﻢ ﺍﺯ ﺍﻳﺪﻩﻱ ﺩﺍﻳﺮﻩﻫﺎﻱ ﻣﻤﺎﺱ )ﺷﺒﻴﻪ ﻣﺴﺌﻠﻪﻱ ﻗﺒﻞ( ﺍﺳﺘﻔﺎﺩﻩ ﻣﻲﻛﻨﻴﻢ .ﺑﻪ ﺗﻌﺪﺍﺩ ﺩﻟﺨﻮﺍﻩ ﺩﺍﻳﺮﻩ ﺭﺳﻢ ﻣﻲﻛﻨﻴﻢ ﺑﻪﻃﻮﺭﻱ ﻛﻪ ﺑﺮ ﺍﻳﻦ ﺩﻭ ﻣﻨﺤﻨﻲ ﻣﻤﺎﺱ ﺑﺎﺷــﻨﺪ .ﺣﺎﻝ ﺍﮔﺮ ﻣﺮﻛﺰ ﺍﻳﻦ ﺩﺍﻳﺮﻩ ﺭﺍ ﺑﻪ ﻫﻢ ﻭﺻﻞ ﻛﻨﻴﻢ ﻣﺴــﻴﺮ ﺍﻣﻦ ﻣﻮﺭﺩ ﻧﻈﺮ ﺑﻪﺩﺳــﺖ ﺧﻮﺍﻫﺪ ﺁﻣﺪ .ﻫﺮ ﭼﻪ ﺗﻌﺪﺍﺩ ﺩﻭﺭﺓ ﺷﺎﻧﺰﺩﻫﻢ ،ﺷﻤﺎﺭﺓ ،4ﺗﺎﺑﺴﺘﺎﻥ 1390
ﺭﺍﻫﻨﻤﺎﻳﻲ
19
ﺩﺍﻳﺮﻩﻫﺎ ﺑﻴﺸﺘﺮ ﺑﺎﺷﺪ ﻣﺴﻴﺮ ﺩﻗﻴﻖﺗﺮﻱ ﺭﺍ ﻧﺸﺎﻥ ﺧﻮﺍﻫﺪ ﺩﺍﺩ .ﭼﺮﺍ؟ ) ﺑﺮﺍﻱ ﺭﺳﻢ ﻣﻲﺗﻮﺍﻧﻴﺪ ﺍﺯ ﭘﺮﮔﺎﺭ ﻳﺎ ﺷﺎﺑﻠﻮﻥ ﺍﺳﺘﻔﺎﺩﻩ ﻛﻨﻴﺪ(.
L1
B
A
L2
ﺑﻪ ﻧﻈﺮ ﺷﻤﺎ ﭼﺮﺍ ﻣﺴﻴﺮ ﺑﻪ ﺩﺳﺖ ﺁﻣﺪﻩ ﺍﻣﻦﺗﺮﻳﻦ ﻣﺴﻴﺮ ﺍﺳﺖ؟ ﺩﺭ ﺟﻬﺎﻥ ﻭﺍﻗﻌﻲ ﺁﻳﺎ ﻣﻲﺗﻮﺍﻥ ﺍﺯ ﺍﻳﻦ ﺭﻭﺵ ﺍﺳﺘﻔﺎﺩﻩ ﻛﺮﺩ؟ ﭼﻪ ﻣﻮﺍﻧﻊ ﻭﺟﻮﺩ ﺧﻮﺍﻫﺪ ﺩﺍﺷﺖ؟ ﻳﻜﻲ ﺍﺯ ﭘﻴﭽﻴﺪﮔﻲﻫﺎﻱ ﻣﺴــﺎﺋﻞ ﻭﺍﻗﻌﻲ ﺍﻳﻦ ﺍﺳــﺖ ﻛﻪ ﺷﺪﺕ ﻧﺎ ﺍﻣﻨﻲ ﺩﺭ ﺩﻭ ﺭﺷــﺘﻪ ﻛﻮﻩ ﺑﺮﺍﺑﺮ ﻧﺒﺎﺷﺪ .ﺑﺮﺍﻱ ﺣﻞ ﺍﻳﻦ ﻣﺸﻜﻞ ﻣﻲﺗﻮﺍﻧﻴﺪ ﺑﺎﺯ ﻫﻢ ﻣﺴــﺌﻠﻪ ﺭﺍ ﺩﺭ ﺣﺎﻟﺖﻫﺎﻱ ﺳــﺎﺩﻩ ﺷــﺪﻩ ) ﺩﺭ ﻧﻈﺮ ﮔﺮﻓﺘﻦ ﺧﻂ ﺭﺍﺳﺖ( ﺣﻞ ﻛﻨﻴﺪ ﻭ ﺑﻌﺪ ﺍﺯ ﺍﻳﻨﻜﻪ ﺍﻳﺪﻩﻫﺎﻱ ﺍﺻﻠﻲ ﺭﺍ ﭘﻴﺪﺍ ﻛﺮﺩﻳﺪ ،ﺁﻥﻫﺎ ﺭﺍ ﺩﺭ ﺣﺎﻟﺖ ﻛﻠﻲ ﻣﻄﺮﺡ ﻛﻨﻴﺪ. ﺑﺮﺍﻱ ﻣﺜﺎﻝ ﺭﻭﻱ ﺣﺎﻟﺘﻲ ﻛﺎﺭ ﻛﻨﻴﺪ ﻛﻪ ﺷــﺪﺕ ﻧﺎ ﺍﻣﻨﻲ ﺩﺭ ﻳﻜﻲ ﺍﺯ ﺭﺷــﺘﻪ ﻛﻮﻩﻫﺎ ﺩﻭ ﺑﺮﺍﺩﺭ ﺩﻳﮕﺮﻱ ﺑﺎﺷــﺪ .ﭘﺎﺳﺦﻫﺎﻱ ﺟﺎﻟﺐﺗﺎﻥ ﺭﺍ ﺑﺮﺍﻱ ﻣﺠﻠﻪ ﺍﺭﺳﺎﻝ ﻛﻨﻴﺪ.
20
ﺭﺍﻫﻨﻤﺎﻳﻲ
ﺩﻭﺭﺓ ﺷﺎﻧﺰﺩﻫﻢ ،ﺷﻤﺎﺭﺓ ،4ﺗﺎﺑﺴﺘﺎﻥ 1390
︎︀︨ ﹞︀︵︣ات ︨﹀︣ در ︨﹫︀رهی ﹡︀﹛︣﹞﹢
ﻣﻘﺪﺍﺭ ﺑﻨﺰﻳﻦ
ﺍﺗﻮﭘﻴﺎ
ﻣﺮﺍﻛﺶ ﺯﻧﮕﺒﺎﺭ ﺳﻨﮕﺎﭘﻮﺭ
ﻳﺎﻟﻲ ﺭﻳﻮ
ﻣﻘﺪﺍﺭ ﺑﻨﺰﻳﻦ
ﺭﻡ
ﺳﻨﮕﺎﭘﻮﺭ
ﻫﻠﻴﺲﺑﻴﻮﺩ ﺭﻭﻳﻴﻦ
ﻳﺎﻟﻲ ﻫﻨﮓﻛﻨﮓ ﺭﻳﺠﺰ ﻧﻮﻭﺭ
ﺯﻧﮕﺒﺎﺭ
ﻣﺮﺍﻛﺶ ﺍﺗﻮﭘﻴﺎ ﺭﻳﻮ
ﺯﻧﮕﺒﺎﺭ
ﺭﻡ ﺳﻨﮕﺎﭘﻮﺭ
ﺭﻭﻳﻴﻦ
ﻧﻮﻭﺭ ﻫﻠﻴﺲﺑﻴﻮﺩ
ﺳﻨﮕﺎﭘﻮﺭ
11 12 13 14 15 16
ﺭﻡ ﺭﻭﻳﻴﻦ ﻫﻠﻴﺲﺑﻴﻮﺩ ﻧﻮﻭﺭ ﺭﻳﺠﺰ ﻫﻨﮓﻛﻨﮓ
4 5 0 1 2 3
ﺯﻧﮕﺒﺎﺭ ﻣﺮﺍﻛﺶ ﺍﺗﻮﭘﻴﺎ ﺭﻳﻮ ﻳﺎﻟﻲ ﻫﻨﮓﻛﻨﮓ ﺭﻳﺠﺰ
ﺣﺪ ﻗﺮﺍﺭ ﺩﺍﺭﺩ .ﺣﺮﻛﺖ ﺍﺗﻮﺑﻮﺱ ﺑﺎ ﺑﺎﻙ ﺧﺎﻟﻰ ﻭ ﺍﻓﺰﺍﻳﺶﻫﺎ ﻭ ﻛﺎﻫﺶﻫﺎﻯ ﻣﻘﺪﺍﺭ ﺳﻮﺧﺖ ﺩﺭ ﻫﺮ ﻳﻚ ﺍﺯ ﭘﺎﻳﮕﺎﻩﻫﺎ ﺩﺭﺳﺖ ﺑﻪ ﻣﺜﺎﺑﻪ ﺣﺮﻛﺖ ﺍﺗﻮﺑﻮﺱ ﺩﺭ ﺍﻳﻦ ﺳــﻔﺮ ﻓﺮﺿﻰ ﺍﺳﺖ ﺑﻨﺎﺑﺮﺍﻳﻦ ،ﺗﻔﺎﻭﺕ ﻣﻴﺎﻥ ﺑﻨﺰﻳﻦ ﺑﺎﻙ ﺍﺗﻮﺑﻮﺱ ﺷــﻤﺎ )ﻧﻮﺳﺎﻥﻫﺎﻯ ﺣﺠﻢ ﺑﻨﺰﻳﻦ ﺩﺭ ﺑﺎﻙ( ﻭ ﻣﻘﺪﺍﺭ ﻧﺸﺎﻥ ﺩﺍﺩﻩ ﺷﺪﻩ ﺩﺭ ﻧﻤﻮﺩﺍﺭ ،ﺑﺎﻳﺪ ﻳﻜﺴــﺎﻥ ﺑﺎﺷــﺪ ﻭ ﺁﻥ ﻫﻤﻮﺍﺭﻩ ﺑﺮﺍﺑﺮ ﺍﺳــﺖ ﺑﺎ ﭘﺎﻳﻴﻦﺗﺮﻳﻦ ﻣﻘﺪﺍﺭ ﺑﻨﺰﻳــﻦ ﺩﺭ ﻧﻤﻮﺩﺍﺭ .ﭼﻮﻥ ﻧﻤﻮﺩﺍﺭ ﻫﺮﮔﺰ ﺑﻪ ﺯﻳﺮ ﭘﺎﻳﻴﻦﺗﺮﻳﻦ ﻧﻘﻄﻪ ﺧﻮﺩ ﻧﻤﻰﺁﻳﺪ ،ﻣﻘﺪﺍﺭ ﺑﻨﺰﻳﻦ ﺑﺎﻙ ﺷﻤﺎ ﻫﺮﮔﺰ ﺑﻪ ﺯﻳﺮ ﺻﻔﺮ ﻧﺨﻮﺍﻫﺪ ﺭﻓﺖ؛ ﺣﺘــﻰ ﺩﺭ ﻣﻮﺍﺭﺩﻯ ﻛــﻪ ﺩﺭ ﻧﻤﻮﺩﺍﺭ ﭼﻨﺪ ﻧﻘﻄﻪ ﺑــﺎ ﭘﺎﻳﻴﻦﺗﺮﻳﻦ ﻧﻘﻄﻪﻯ ﻧﻤﻮﺩﺍﺭ ﻫﻤﺴــﺎﻥ ﺑﺸﻮﻧﺪ .ﺍﻳﻦ ﺑﺪﺍﻥ ﻣﻌﻨﺎﺳــﺖ ﻛﻪ ﻭﺳﻴﻠﻪﻯ ﻧﻘﻠﻴﻪ ﺷﻤﺎ ﺑﻪ ﻫﻨﮕﺎﻡ ﺭﺳﻴﺪﻥ ﺑﻪ ﺁﻥ ﭘﺎﻳﮕﺎﻩﻫﺎ ﺑﺎﻙ ﺑﻨﺰﻳﻨﺶ ﺧﺎﻟﻰ ﻣﻰﺷﻮﺩ ﻭ ﺷﻤﺎ ﻣﻰﺗﻮﺍﻧﻴﺪ ﺩﺭ ﺁﻥ ﭘﺎﻳﮕﺎﻩ ﺑﺮﺍﻯ ﺍﺩﺍﻣﻪ ﺣﺮﻛﺖ ﺑﻨﺰﻳﻦ ﺑﮕﻴﺮﻳﺪ. ﻧﻤﻮﺩﺍﺭ 2ﻧﺸــﺎﻥ ﺩﻫﻨﺪﻩﻯ ﺳــﻄﺢ ﺑﻨﺰﻳﻦ ﺍﺗﻮﺑﻮﺱ ﺷــﻤﺎ ﺩﺭ ﻃﻰ ﺳﻔﺮﻯ ﺍﺳﺖ ﻛﻪ ﺍﺯ ﺯﻧﮕﺒﺎﺭ ﻭ ﺑﺎ ﺑﺎﻙ ﺧﺎﻟﻰ ﺁﻏﺎﺯ ﻣﻰﺷﻮﺩ )ﺗﻮﺟﻪ ﺩﺍﺷﺘﻪ ﺑﺎﺷﻴﺪ ﻛﻪ ﻫﺮ ﻧﻤﻮﺩﺍﺭ ﺩﺭ ﻭﺍﻗﻊ ﻳﻚ ﺧﻂ ﭘﻴﻮﺳﺘﻪ ﺭﺍ ﻧﺸﺎﻥ ﻣﻰﺩﻫﺪ ،ﺯﻳﺮﺍ ﻧﺨﺴﺘﻴﻦ ﻭ ﺁﺧﺮﻳﻦ ﻧﻘﻄﻪ ﺭﻭﻯ ﻧﻤﻮﺩﺍﺭ ﻳﻜﻰ ﻫﺴﺘﻨﺪ(.
ﻣﻘﺪﺍﺭ ﺑﻨﺰﻳﻦ
... .. . . . . .. . . . ..
1
. . . .. .. . . 3 . . .. . .. .. . 2 . . . . .. . ...
8 9 10 11 12 13 ﺳﻨﮕﺎﭘﻮﺭ
.١ﺍﮔــﺮ ﺩﺭ ﭘﺎﻳﮕﺎﻩ ﺯﻧﮕﺒﺎﺭ ﻓﺮﻭﺩ ﺁﻳﻴــﺪ ﻭ ﺍﺯ ﺁﻥ ﻧﻘﻄﻪ ﺩﺭ ﺟﻬﺖ ﺣﺮﻛﺖ ﻋﻘﺮﺑﻪﻫﺎﻯ ﺳﺎﻋﺖ ،ﻣﺎﻣﻮﺭﻳﺖ ﺧﻮﺩ ﺭﺍ ﺁﻏﺎﺯ ﻛﻨﻴﺪ ،ﺧﻮﺍﻫﻴﺪ ﺗﻮﺍﻧﺴﺖ ﻣﺴــﻴﺮ ﺑﺰﺭﮔﺮﺍﻩ ﺭﺍ ﻳﻚ ﺩﻭﺭ ﻛﺎﻣﻞ ﺑﭙﻴﻤﺎﻳﺪ .ﺍﻣﺎ ﭼﻨﺎﻥﭼﻪ ﺑﺨﻮﺍﻫﻴﺪ ﺩﺭ ﺟﻬﺖ ﻋﻜﺲ ﺣﺮﻛﺖ ﻋﻘﺮﺑﻪﻫﺎﻯ ﺳﺎﻋﺖ ﺭﺍﻧﻨﺪﮔﻰ ﻛﻨﻴﺪ ،ﺑﺎﻳﺪ ﺍﺯ ﻳﻜﻰ ﺍﺯ ﭘﺎﻳﮕﺎﻩﻫﺎﻯ ﻫﻨﮓﻛﻨﮓ ﻳﺎ ﺑﺎﻟﻰ ﺭﺍﻩ ﺑﻴﻔﺘﻴﺪ. .٢ﻓﺮﺽ ﻛﻨﻴﺪ ﺷــﻤﺎ ﺩﺭ ﻳﻚ ﭘﺎﻳﮕﺎﻩ ﺑﻪﺧﺼﻮﺹ ﻣﺎﻧﻨﺪ ﺳﻨﮕﺎﭘﻮﺭ ﻓﺮﻭﺩ ﺁﻣﺪﻩﺍﻳــﺪ ﻭ ﺑﺮﺍﻯ ﭘﻴﻤﻮﺩﻥ ﻃــﻮﻝ ﺭﺍﻩ ،ﺑﻨﺰﻳﻦ ﻛﺎﻓﻰ ﺩﺭ ﺑﺎﻙ ﺩﺍﺭﻳﺪ. ﺣﺎﻝ ﻧﻤﻮﺩﺍﺭﻯ ﺭﺳــﻢ ﻛﻨﻴﺪ ﻛﻪ ﻧﻤﺎﻳﺎﻧﮕﺮ ﻣﺤﺘﻮﺍﻯ ﺑﻨﺰﻳﻦ ﺑﺎﻙ ﺷﻤﺎ ﺩﺭ ﺗﻤﺎﻡ ﻣﺴــﻴﺮ ﺑﺎﺷﺪ ﻭ ﺩﺭ ﻋﻴﻦ ﺣﺎﻝ ﻧﺸــﺎﻥ ﺑﺪﻫﺪ ﻛﻪ ﺑﻪ ﻫﻨﮕﺎﻡ ﻋﺒــﻮﺭ ﺍﺯ ﭘﺎﻳﮕﺎﻩﻫﺎ ﺩﺭ ﻛﺪﺍﻡﻳﻚ ﺍﺯ ﺁﻥﻫﺎ ﺳــﻮﺧﺖﮔﻴﺮﻯ ﻣﻰﻛﻨﻴﺪ. ﺩﺭ ﺗﺼﻮﻳــﺮ 1ﻧﻤﻮﺩﺍﺭ ﻣﺮﺑﻮﻁ ﺑﻪ ﻣﻌﻤﺎﻯ 1ﺭﺍ ﻣﻰﺑﻴﻨﻴﺪ ﻛﻪ ﻣﺴــﻴﺮ ﺣﺮﻛﺖ ﺭﺍ ﻧﺸﺎﻥ ﻣﻰﺩﻫﺪ .ﺧﻂﻫﺎﻯ ﻋﻤﻮﺩﻯ ﻧﻤﺎﻳﺎﻧﮕﺮ ﺟﺎﻳﮕﺎﻩﻫﺎﻯ ﺑﻨﺰﻳﻦ ﻭ ﻋﺪﺩﻫﺎ ﻧﻤﺎﻳﻨﺪﻩ ﻣﻘﺪﺍﺭ ﺑﻨﺰﻳﻦ ﻣﻮﺟﻮﺩ ﺩﺭ ﺑﺎﻙ ﺍﺳﺖ. ﻣﻘﺪﺍﺭ ﺑﻨﺰﻳﻦ ﺍﺗﻮﺑﻮﺱ ﺩﺭ ﻧﻘﻄﻪﻯ ﺁﻏﺎﺯ ﺣﺮﻛﺖ ﻫﻤﻮﺍﺭﻩ ﺩﺭ ﭘﺎﻳﻴﻦﺗﺮﻳﻦ
ﺍﻣﺎ ﺗﻮﺟﻪ ﺩﺍﺷــﺘﻪ ﺑﺎﺷــﻴﺪ ﻣﻪ ﻧﻤﻮﺩﺍﺭ ﻓﻮﻕ ﻣﺴﻴﺮ ﺣﺮﻛﺘﻰ ﺭﺍ ﻧﺸﺎﻥ ﻣﻰﺩﻫﺪ ﻛﻪ ﺩﺭ ﺟﻬﺖ ﻋﻘﺮﺑﻪﻫﺎﻯ ﺳﺎﻋﺖ ﺻﻮﺭﺕ ﮔﺮﻓﺘﻪ ﺑﺎﺷﺪ. ﺑﺮﺍﻯ ﻧﺸﺎﻥ ﺩﺍﺩﻥ ﻳﻚ ﻣﺴﻴﺮ ﭘﻴﻤﻮﺩﻩ ﺷﺪﻩ ﺩﺭ ﺟﻬﺖ ﻋﻜﺲ ﺣﺮﻛﺖ ﻋﻘﺮﺑﻪﻫﺎﻯ ﺳــﺎﻋﺖ ،ﺑﺎﻳﺪ ﻧﻤﻮﺩﺍﺭ ﺟﺪﻳﺪﻯ ﺑﻜﺸــﻴﺪ .ﺷــﻜﻞ 3ﭼﻨﻴﻦ ﻧﻤﻮﺩﺍﺭﻯ ﺑﺮﺍﻯ ﻣﻌﻤﺎﻯ 1ﺭﺍ ﻧﺸــﺎﻥ ﻣﻰﺩﻫﺪ .ﻫﻤﺎﻥﻃﻮﺭ ﻛﻪ ﻣﻰﺑﻴﻨﻴﺪ، ﺩﺭ ﺍﻳﻦ ﻧﻤﻮﺩﺍﺭ ﺩﻭ ﻧﻘﻄﻪ ﺩﺭ ﭘﺎﻳﻴﻦﺗﺮﻳﻦ ﺳﻄﺢ ﻗﺮﺍﺭ ﺩﺍﺭﻧﺪ ﻛﻪ ﻣﻰﺗﻮﺍﻧﻴﺪ ﺳﻔﺮ ﺭﺍ ﺍﺯ ﻫﺮ ﻳﻚ ﺍﺯ ﺍﻳﻦ ﺩﻭ ﻧﻘﻄﻪ ﺁﻏﺎﺯ ﻛﻨﻴﺪ. ﭼــﻮﻥ ﺩﺭ ﻫﺮ ﻧﻤﻮﺩﺍﺭ ﺑﺎﻳﺪ ﺩﺳــﺖ ﻛﻢ ﻳﻚ ﻧﻘﻄــﻪ ﺩﺭ ﭘﺎﻳﻴﻦﺗﺮﻳﻦ ﺳــﻄﺢ ﻭﺍﻗﻊ ﺷﺪﻩ ﺑﺎﺷــﺪ ،ﻫﺮ ﻣﺴــﻴﺮ ﺣﺮﻛﺘﻰ ﻧﺎﮔﺰﻳﺮ ﻳﻚ ﺣﻠﻘﻪ )ﻳﻚ ﻣﺴﻴﺮ ﺣﻠﻘﻪ ﻣﺎﻧﻨﺪ ﻣﺴﺪﻭﺩ( ﺧﻮﺍﻫﺪ ﺑﻮﺩ.
ﺩﺍﻧﺶﺍﻓﺰﺍﻳﻲ
︠﹢ا﹡︡﹡︀﹨﹩ ﹩از ر︀︲﹫︀ت ﺯﻳﻨﺐ ﮔﻠﺒﺮﺍﺭﻯ
ﻛﻠﻴﺪﻭﺍژﻩﻫﺎ :ﺷﺎﺧﻪﻫﺎﻱ ﺭﻳﺎﺿﻲ ،ﺭﻳﺎﺿﻴﺎﺕ ﻣﺤﺾ ،ﻫﻨﺪﺳﻪ ،ﺍﻣﻜﺎﻥ
ﺭﻳﺎﺿﻴﺎﺕ ﻭ ﺷﺎﺧﻪﻫﺎﻱ ﺁﻥ
ﺭﻳﺎﺿﻴﺎﺕ ،ﻋﻠﻢ ﻧﻈﻢ ﺍﺳﺖ ﻭ ﻣﻮﺿﻮﻉ ﺁﻥ ﻳﺎﻓﺘﻦ ،ﺗﻮﺻﻴﻒ ﻭ ﺩﺭﻙ ﻧﻈﻤﻰ ﺍﺳـﺖ ﻛﻪ ﺩﺭ ﻭﺿﻌﻴﺖﻫـﺎﻯ ﺑﻪ ﻇﺎﻫﺮ ﭘﻴﭽﻴﺪﻩ ﻧﻬﻔﺘﻪ ﺍﺳـﺖ ﻭ ﺍﺑﺰﺍﺭﻫـﺎﻯ ﺍﺻﻮﻟﻰ ﺍﻳﻦ ﻋﻠﻢ ،ﻣﻔﺎﻫﻴﻤﻰ ﻫﺴـﺘﻨﺪ ﻛﻪ ﺑـﻪ ﻛﻤﻚ ﺁﻥﻫﺎ ﻣﻰﺗﻮﺍﻧﻴـﻢ ﺍﻳﻦ ﻧﻈﻢ ﺭﺍ ﺗﻮﺻﻴﻒ ﻛﻨﻴﻢ .ﻋﻠـﻢ ﺭﻳﺎﺿﻰ ﻗﺎﻧﻮﻧﻤﻨﺪ ﻛﺮﺩﻥ ﺗﺠﺮﺑﻴﺎﺕ ﻃﺒﻴﻌﻰ ﺍﺳـﺖ ﻛﻪ ﺩﺭ ﮔﻴﺎﻫﺎﻥ ﻭ ﺑﻘﻴﻪﻯ ﻣﺨﻠﻮﻗﺎﺕ ﻣﺸﺎﻫﺪﻩ ﻣﻰﻛﻨﻴـﻢ .ﻋﻠﻮﻡ ﺭﻳﺎﺿﻴﺎﺕ ﺍﻳﻦ ﺗﺠﺮﺑﻴﺎﺕ ﺭﺍ ﺩﺳـﺘﻪﺑﻨﺪﻯ ﻭ ﻗﺎﻧﻮﻧﻤﻨﺪ ﻣﻰﻛﻨﻨﺪ ﻭ ﺗﻮﺳﻌﻪ ﻣﻰﺩﻫﻨﺪ. ﺩﻛﺘﺮ ﺭﻳﺎﺿﻰ ،ﺍﺳـﺘﺎﺩ ﺭﻳﺎﺿﻰ ،ﻧﻴﺰ ﺩﺭ ﻣﻌﺮﻓﻰ ﺍﻳﻦ ﻋﻠﻢ ﻣﻰﮔﻮﻳﺪ: ))ﺭﻳﺎﺿﻴﺎﺕ ﻋﻠﻢ ﻣﺪﻝﺩﻫﻰ ﺑﻪ ﺳﺎﻳﺮ ﻋﻠﻮﻡ ﺍﺳﺖ ،ﻳﻌﻨﻰ ﺯﺑﺎﻥ ﻣﺸﺘﺮﻙ ﻧﻈﺮﻳﺎﺕ ﻋﻠﻤﻰ ﺳـﺎﻳﺮ ﻋﻠﻮﻡ ،ﻋﻠﻢ ﺭﻳﺎﺿﻰ ﺍﺳﺖ .ﺍﻣﺮﻭﺯﻩ ﺍﮔﺮ ﻋﻠﻤﻰ ﺭﺍ ﻧﺘﻮﺍﻥ ﺑﻪ ﺯﺑﺎﻥ ﺭﻳﺎﺿﻰ ﺑﻴﺎﻥ ﻛﺮﺩ ﻋﻠﻢ ﻧﻴﺴﺖ((.
ﻣﺎﻫﻴﺖ ﺭﻳﺎﺿﻴﺎﺕ ﺭﻳﺎﺿﻴــﺎﺕ ﺑﺮ ﺧــﻼﻑ ﺗﺼﻮﺭ ﺑﺮﺧــﻰ ﺍﻓﺮﺍﺩ ،ﻳﻚ ﺳــﺮﻯ ﻓﺮﻣﻮﻝ ﻭ ﻗﻮﺍﻋﺪ ﻧﻴﺴــﺖ ﻛﻪ ﻫﻤﻴﺸــﻪ ﻭ ﺩﺭ ﻫﻤﻪ ﺟﺎ ﺑﺘﻮﺍﻥ ﺍﺯ ﺁﻥ ﺍﺳﺘﻔﺎﺩﻩ ﻛﺮﺩ، ﺑﻠﻜﻪ ﺭﻳﺎﺿﻴﺎﺕ ﺩﺭﺳــﺖ ﻓﻬﻤﻴﺪﻥ ﺻﻮﺭﺕ ﻣﺴﺌﻠﻪ ﻭ ﺩﺭﺳﺖ ﻓﻜﺮ ﻛﺮﺩﻥ ﺑﺮﺍﻯ ﺭﺳــﻴﺪﻥ ﺑﻪ ﺟﻮﺍﺏ ﺍﺳــﺖ .ﺑﺮﺍﻯ ﺑﻪ ﺩﺳــﺖ ﺁﻭﺭﺩﻥ ﺍﻳﻦ ﺗﻮﺍﻧﺎﻳﻰ، ﺩﺍﻧﺸــﺠﻮ ﺑﺎﻳﺪ ﺻﺒﺮ ﻭ ﭘﺸــﺘﻜﺎﺭ ﻻﺯﻡ ﺭﺍ ﺩﺍﺷــﺘﻪ ﺑﺎﺷــﺪ ﺗﺎ ﺑﺘﻮﺍﻧﺪ ﺣﺘﻰ ﺑﻪ ﻣﺪﺕ ﭼﻨﺪﻳﻦ ﺳــﺎﻋﺖ ﺩﺭ ﺑﺎﺭﻩﻯ ﻳﻚ ﻣﺴــﺌﻠﻪﻯ ﺭﻳﺎﺿﻰ ﺑﻴﻨﺪﻳﺸﺪ ﻭ ﺩﺭ ﻧﻬﺎﻳــﺖ ﺑــﺎ ﺍﺑﺘﻜﺎﺭ ﻭ ﺧﻼﻗﻴﺖ ﺁﻥ ﺭﺍ ﺣﻞ ﻛﻨــﺪ .ﻓﺎﺭﻍ ﺍﻟﺘﺤﺼﻴﻼﻥ ﺍﻳــﻦ ﺭﺷــﺘﻪ ﻣﻰﺗﻮﺍﻧﻨﺪ ﭘﺲ ﺍﺯ ﭘﺎﻳــﺎﻥ ﺗﺤﺼﻴــﻼﺕ ،ﺩﺭ ﺍﺩﺍﺭﺍﺕ ﺩﻭﻟﺘﻰ ﺑﺮﺍﻯ ﻣﺴــﺌﻮﻟﻴﺖﻫﺎﻳﻰ ﻛﻪ ﺑﻪ ﻧﻮﻋﻰ ﺑﺎ ﺗﺠﺰﻳﻪ ﻭ ﺗﺤﻠﻴﻞ ﻣﺴﺎﺋﻞ ﺳﺮﻭﻛﺎﺭ ﺩﺍﺭﻧﺪ ،ﺩﺭ ﺑﺨﺶ ﺧﺼﻮﺻﻰ ﺩﺭ ﺍﻣﻮﺭﻯ ﻫﻤﺎﻧﻨﺪ ﻃﺮﺍﺣﻰ ﺳﻴﺴــﺘﻢﻫﺎ ﺩﺭ 22
ﺭﺍﻫﻨﻤﺎﻳﻲ
ﺩﻭﺭﺓ ﺷﺎﻧﺰﺩﻫﻢ ،ﺷﻤﺎﺭﺓ ،4ﺗﺎﺑﺴﺘﺎﻥ 1390
ﺍﻣﺮ ﺑﻬﻴﻨﻪﺳــﺎﺯﻯ ﻭ ﺑﻬﺮﻩﻭﺭﻯ ،ﺩﺭ ﺑﺨــﺶ ﺻﻨﻌﺖ ﺑﺮﺍﻯ ﺍﻣﻮﺭﻯ ﻫﻤﺎﻧﻨﺪ ﻣﺪﻝﺳــﺎﺯﻯﻫﺎﻯ ﺭﻳﺎﺿﻰ ﻭ ﺩﺭ ﺁﻣﻮﺯﺵ ﻭ ﭘﺮﻭﺭﺵ ﻭ ، ...ﻣﺴﺌﻮﻟﻴﺖﻫﺎﻯ ﻣﺘﻔﺎﻭﺗﻰ ﺭﺍ ﺑﺮ ﻋﻬﺪﻩ ﮔﻴﺮﻧﺪ .ﺭﺋﻴﺲ ﺍﺗﺤﺎﺩﻳﻪﻯ ﺑﻴﻦﺍﻟﻤﻠﻠﻰ ﺭﻳﺎﺿﻰﺩﺍﻧﺎﻥ ﺟﻬــﺎﻥ ﺩﺭ ﻳﺎﺯﺩﻫﻤﻴﻦ ﺍﺟﻼﺱ ﺁﻛﺎﺩﻣﻰ ﺟﻬﺎﻥ ﺳــﻮﻡ ﻛﻪ ﺑﻪ ﺗﺎﺯﮔﻰ ﺩﺭ ﺗﻬﺮﺍﻥ ﺑﺮﮔﺰﺍﺭ ﺷــﺪ ،ﻋﻨﻮﺍﻥ ﻛﺮﺩ ﻛﻪ ﺑﻬﺘﺮ ﺍﺳــﺖ ﺑﮕﻮﻳﻴﻢ ))ﺭﻳﺎﺿﻴﺎﺕ ﻭ ﻛﺎﺭﺑﺮﺩﻫﺎﻯ ﺁﻥ(( ﻧﻪ ﺍﻳﻦﻛﻪ ﺭﻳﺎﺿﻴﺎﺕ ﺭﺍ ﺑﻪ ﻣﺤﺾ ﻭ ﻛﺎﺭﺑﺮﺩﻯ ﺗﻔﻜﻴﻚ ﻛﻨﻴﻢ ،ﺯﻳﺮﺍ ﺑﻪ ﺍﻋﺘﻘﺎﺩ ﺭﻳﺎﺿﻰﺩﺍﻥﻫﺎ ﻫﻴﭻ ﻣﻘﻮﻟﻪﻯ ﺭﻳﺎﺿﻰ ﻧﻴﺴــﺖ ﻛﻪ ﺭﻭﺯﻯ ﻛﺎﺭﺑﺮﺩﻯ ﺑﺮﺍﻯ ﺁﻥ ﭘﻴﺪﺍ ﻧﺸﻮﺩ. ﺭﻳﺎﺿﻴﺎﺕ ﻣﺤﺾ ﺑﻴﺶﺗﺮ ﺑﻪ ﻗﻀﺎﻳﺎ ﻭ ﺍﺳــﺘﺪﻻﻝﻫﺎ ،ﻣﻨﻄﻖ ﻣﻮﺟﻮﺩ ﺩﺭ ﺁﻥﻫﺎ ﻭ ﭼﮕﻮﻧﮕﻰ ﺍﺛﺒﺎﺗﺸــﺎﻥ ﻣﻰﭘﺮﺩﺍﺯﺩ .ﺍﻣﺎ ﺩﺭ ﺭﻳﺎﺿﻴﺎﺕ ﻛﺎﺭﺑﺮﺩﻯ
ﭼﮕﻮﻧﻪ ﺍﺳﺘﻔﺎﺩﻩ ﻛﺮﺩﻥ ﻭ ﺑﻪ ﻛﺎﺭ ﮔﺮﻓﺘﻦ ﻗﻀﺎﻳﺎ ،ﺁﻣﻮﺯﺵ ﺩﺍﺩﻩ ﻣﻰﺷﻮﺩ. ﺑﻪ ﻋﺒــﺎﺭﺕ ﺩﻳﮕﺮ ،ﺩﺭ ﺍﻳﻦ ﺷــﺎﺧﻪ ،ﻛﺎﺭﺑﺮﺩ ﺭﻳﺎﺿﻴﺎﺕ ﺩﺭ ﻣﺴــﺎﺋﻞ ﻣﻮﺟــﻮﺩ ﺩﺭ ﺟﺎﻣﻌﻪ ﺑﻴﺎﻥ ﻣﻰﺷــﻮﺩ .ﻭﻗﺘﻰ ﺻﺤﺒــﺖ ﺍﺯ ﺭﻳﺎﺿﻰ ﻣﺤﺾ ﻣﻰﺷــﻮﺩ ،ﻧﺒﺎﻳﺪ ﺗﺼﻮﺭ ﻛﺮﺩ ﻛﻪ ﺗﻨﻬﺎ ﺑﺎﻳﺪ ﺩﺭ ﮔﻮﺷــﻪﺍﻯ ﻧﺸﺴــﺖ ﻭ ﺑﻪ ﺣﻞ ﻣﺴــﺎﺋﻞ ﺭﻳﺎﺿﻰ ﭘﺮﺩﺍﺧﺖ ﺑﻠﻜﻪ ﺍﻳــﻦ ﻋﻠﻢ ،ﺑﻪ ﻭﻳﮋﻩ ﺩﺭ ﻣﺪﺍﺭﺝ ﺑﺎﻻ، ﺍﺭﺗﺒــﺎﻁ ﻧﺰﺩﻳﻜﻰ ﺑــﺎ ﻃﺒﻴﻌﺖ ﺩﺍﺭﺩ ﺑﻪ ﻋﺒﺎﺭﺕ ﺩﻳﮕــﺮ ﺍﻳﺪﻩﻫﺎﻯ ﺭﻳﺎﺿﻰ ﺍﺯ ﺫﻫﻦ ﭘﮋﻭﻫﺸــﮕﺮﺍﻥ ﻧﻤﻰﺭﻭﻳﺪ ،ﺑﻠﻜﻪ ﺭﻳﺎﺿﻰﺩﺍﻥﻫﺎ ﺍﻏﻠﺐ ﺍﻟﻬﺎﻡ ﺧﻮﺩ ﺭﺍ ﺍﺯ ﻃﺒﻴﻌــﺖ ﻣﻰﮔﻴﺮﻧﺪ ﻭ ﺑﻪ ﻗــﻮﻝ ﺯﺍﻥ ﺑﺎﭘﺘﻴﺖ ﻓﻮﺭﻳــﻪ ،ﺭﻳﺎﺿﻰﺩﺍﻥ ﻣﺸﻬﻮﺭ ﻗﺮﻥ ﻧﻮﺯﺩﻫﻢ ﻓﺮﺍﻧﺴــﻪ ))ﺗﻌﻤﻖ ﺩﺭ ﻃﺒﻴﻌﺖ ،ﭘﺮﺑﺎﺭ ﺗﺮﻳﻦ ﻣﻨﺎﺑﻊ ﺍﻛﺘﺸــﺎﻓﺎﺕ ﺭﻳﺎﺿﻰ ﺍﺳﺖ(( .ﺭﻳﺎﺿﻴﺎﺕ ﻛﺎﺭﺑﺮﺩﻯ ﺑﻪ ﺷﺎﺧﻪﺍﻯ ﺍﺯ ﺭﻳﺎﺿﻰ ﮔﻔﺘﻪ ﻣﻰﺷــﻮﺩ ﻛﻪ ﻛﺎﺭﺑﺮﺩ ﻋﻠﻤﻰ ﻣﺸــﺨﺼﻰ ﺩﺍﺷﺘﻪ ﺑﺎﺷﺪ؛ ﺑﺮﺍﻯ ﻣﺜﺎﻝ ﺩﺭ ﺍﻗﺘﺼﺎﺩ ،ﻛﺎﻣﭙﻴﻮﺗﺮ ،ﻓﻴﺰﻳﻚ ﻳﺎ ﺁﻣﺎﺭ ﺍﺣﺘﻤﺎﻝ ﻛﺎﺭﺑﺮﺩ ﺩﺍﺷــﺘﻪ ﺑﺎﺷﺪ ﻭ ﺭﻳﺎﺿﻰ ﻣﺤﺾ ﻧﻴﺰ ﺷﺎﺧﻪﺍﻯ ﮔﻔﺘﻪ ﻣﻰﺷﻮﺩ ﻛﻪ ﺑﻪ ﻧﻈﺮﻳﻪﭘﺮﺩﺍﺯﻯ ﺭﻳﺎﺿﻰ ﻣﻰﭘﺮﺩﺍﺯﺩ ،ﺍﻣﺎ ﺑﺎﻳﺪ ﺗﻮﺟﻪ ﺩﺍﺷــﺖ ﻛﻪ ﺍﻣﺮﻭﺯﻩ ﺍﻳﻦ ﺩﻭ ﮔﺮﺍﻳﺶ ﺁﻥﭼﻨﺎﻥ ﺩﺭ ﻫﻢ ﺍﺩﻏﺎﻡ ﺷﺪﻩﺍﻧﺪ؛ ﻛﻪ ﻣﺮﺯﻯ ﺭﺍ ﻧﻤﻰﺗﻮﺍﻥ ﺑﻴﻦ ﺁﻥﻫﺎ ﻣﺸﺨﺺ ﻛﺮﺩ. ﮔﺎﻩ ﻳــﻚ ﺗﺌﻮﺭﻯ ﻛﺎﻣــ ً ﻼ ﻣﺤﺾ ﺑﺎ ﻭﺭﻭﺩ ﺑﻪ ﻣﺮﺣﻠــﻪﻯ ﻛﺎﺭﺑﺮﺩﻯ ﭼﻮﻥ ﺩﺭ ﻋﻤﻞ ﺑﺎ ﻣﺸــﻜﻞ ﺭﻭﺑﻪ ﺭﻭ ﻣﻰﺷــﻮﺩ ،ﺑﺎﺭ ﺩﻳﮕﺮ ﺑــﻪ ﺣﻮﺯﻩﻯ ﺗﺌﻮﺭﻯ ﺑﺮﻣﻰﮔــﺮﺩﺩ ﻭ ﺩﺭ ﻧﻬﺎﻳــﺖ ﭘﺲ ﺍﺯ ﺭﻓﻊ ﻧﻘﺎﻳﺺ ،ﺩﻭﺑــﺎﺭﻩ ﻭﺍﺭﺩ ﻣﺮﺣﻠﻪﻯ ﻛﺎﺭﺑﺮﺩﻯ ﺧﻮﺍﻫﺪ ﺷﺪ .ﻳﻌﻨﻰ ﻳﻚ ﺗﻌﺎﻣﻞ ﻭ ﺍﺭﺗﺒﺎﻁ ﺩﻭﺟﺎﻧﺒﻪ ﺑﻴﻦ ﺭﻳﺎﺿﻰ ﻛﺎﺭﺑﺮﺩﻯ ﻭ ﻣﺤﺾ ﻭﺟﻮﺩ ﺩﺍﺭﺩ. ﻛﺎﺭﺑﺮﺩ ﺭﻳﺎﺿﻰ ﺩﺭ ﻋﻠﻮﻡ ﻣﺨﺘﻠﻒ ﺍﻧﻜﺎﺭﻧﺎﭘﺬﻳﺮ ﺍﺳﺖ .ﺑﺮﺍﻯ ﻣﺜﺎﻝ ،ﺩﺭ ﺟﺎﻣﻌﻪﺷﻨﺎﺳﻰ ﻧﻈﺮﻳﻪﻯ ﺍﺣﺘﻤﺎﻝ ﻭ ﻧﻈﺮﻳﻪﻯ ﮔﺮﻭﻩﻫﺎ ﻧﻘﺶ ﺑﺴﻴﺎﺭ ﻣﻬﻤﻰ ﺍﻳﻔﺎ ﻣﻰﻛﻨﺪ .ﺩﺭ ﻛﻞ ﺑﺎﻳﺪ ﮔﻔﺖ ﻛﻪ ﻫﻤﻪﻯ ﺻﻨﺎﻳﻊ ﺯﻳﺮﺳــﺎﺧﺖ ﺭﻳﺎﺿﻰ ﺩﺍﺭﻧﺪ ﻭ ﺑﻪ ﻫﻤﻴﻦ ﺩﻟﻴــﻞ ﺩﺭ ﻫﻤﻪﻯ ﻣﺮﺍﻛﺰ ﺻﻨﻌﺘﻰ ﻭ ﺗﺤﻘﻴﻘﺎﺗﻰ ﺩﻧﻴﺎ، ﺭﻳﺎﺿﻰﺩﺍﻥﻫﺎ ﺩﺭ ﻛﻨﺎﺭ ﻣﻬﻨﺪﺳــﺎﻥ ﻭ ﺩﺍﻧﺸﻤﻨﺪﺍﻥ ﺳﺎﻳﺮ ﻋﻠﻮﻡ ﺣﻀﻮﺭﻯ ﻓﻌــﺎﻝ ﺩﺍﺭﻧﺪ ﻭ ﺁﻥ ﭼﻪ ﺩﺭ ﻧﻬﺎﻳﺖ ﺍﺭﺍﺋﻪ ﻣﻰﺷــﻮﺩ ﻧﺘﻴﺠﻪ ﻛﺎﺭ ﺗﻴﻤﻰ ﺁﻥ ﻫﺎﺳــﺖ .ﺍﮔﺮ ﺩﺭ ﺟﺎﻣﻌﻪﻯ ﻣﺎ ﻣﺸــﺎﻏﻞ ﺟﻨﺒﻪﻯ ﻋﻠﻤﻰ ﺩﺍﺷــﺘﻪ ﺑﺎﺷﻨﺪ، ﺑﻰﮔﻤﺎﻥ ﺑﻪ ﺗﻌﺪﺍﺩ ﻗﺎﺑﻞ ﺗﻮﺟﻬﻰ ﺭﻳﺎﺿﻰﺩﺍﻥ ﻧﻴﺎﺯ ﺧﻮﺍﻫﻴﻢ ﺩﺍﺷــﺖ ،ﺯﻳﺮﺍ ﻳﻚ ﺭﻳﺎﺿﻰﺩﺍﻥ ﻣﻲﺗﻮﺍﻧﺪ ﻣﺸــﻜﻼﺕ ﺭﺍ ﺑﻪ ﺭﻭﺵ ﻋﻠﻤﻰ ﺣﻞ ﻛﻨﺪ .ﺍﻟﺒﺘﻪ ﺍﻳﻦ ﺑﻪ ﺁﻥ ﻣﻌﻨﺎ ﻧﻴﺴــﺖ ﻛﻪ ﺩﺭ ﺣﺎﻝ ﺣﺎﺿﺮ ﻫﻴﭻ ﻓﺮﺻﺖ ﺷــﻐﻠﻰ ﺑﺮﺍﻯ ﻳﻚ ﺭﻳﺎﺿﻰﺩﺍﻥ ﻭﺟﻮﺩ ﻧــﺪﺍﺭﺩ ،ﺍﻣﺎ ﺑﺎﻳﺪ ﺣﻀﻮﺭ ﺭﻳﺎﺿﻰﺩﺍﻥﻫﺎ ﺩﺭ ﻣﺮﺍﻛﺰ ﺗﺤﻘﻴﻘﺎﺗﻰ ﻭ ﺻﻨﻌﺘﻰ ﭘﺮﺭﻧﮓﺗﺮ ﺑﺎﺷﺪ .ﻫﺮ ﻗﺪﺭ ﺷﻐﻞ ﻳﻚ ﻓﺮﺩ ﺗﺨﺼﺼﻰﺗﺮ ﺷﻮﺩ ،ﻣﻴﺰﺍﻥ ﺭﻳﺎﺿﻴﺎﺗﻰ ﻛﻪ ﻻﺯﻡ ﺩﺍﺭﺩ ،ﺑﻴﺶﺗﺮ ﻣﻰﺷﻮﺩ .ﺑﺮﺍﻯ ﻣﺜﺎﻝ ﻳﻚ ﻣﻬﻨﺪﺱ ،ﺍﻟﻜﺘﺮﻭﻧﻴﻚ ﺍﺯ ﺁﻧﺎﻟﻴﺰ ﺗﺎﺑﻌﻰ ﻭ ﻓﺮﺁﻳﻨﺪﻫﺎﻯ ﺗﺼﺎﺩﻓﻰ ﺍﺳــﺘﻔﺎﺩﻩ ﻣﻰﻛﻨﺪ ﻳﺎ ﻳﻚ ﺑﺮﻧﺎﻣﻪﺭﻳﺰ ﭘﺮﻭژﻩﻫﺎﻯ ﺍﻗﺘﺼﺎﺩﻯ ﺍﺯ ﻣﻄﺎﻟﺐ ﭘﻴﺸــﺮﻓﺘﻪﻯ ﺁﻣﺎﺭﻯ ﻣﺎﻧﻨﺪ ﺳــﺮﻯﻫﺎﻯ ﺯﻣﺎﻧﻰ ﺑﻪ ﻋﻨﻮﺍﻥ ﺍﺑﺰﺍﺭ ﻛﺎﺭ ﻳﺎﺭﻯ ﻣﻰﮔﻴﺮﺩ .ﺑﻪ ﻫﻤﻴﻦ ﺩﻟﻴﻞ ،ﺍﻣﺮﻭﺯﻩ ﺗﺮﺑﻴﺖ ﻣﺘﺨﺼﺼﺎﻥ ﻋﻠﻢ ﺭﻳﺎﺿﻰ ،ﻳﻌﻨﻰ ﺍﻓﺮﺍﺩﻯ ﻛﻪ ﺑﺘﻮﺍﻧﻨــﺪ ﺭﻳﺎﺿﻴﺎﺕ ﻣﻮﺭﺩ ﻧﻴﺎﺯ ﺭﺍ ﺁﻣﻮﺯﺵ ﺩﻫﻨﺪ ﻳﺎ ﺗﻮﻟﻴﺪ ﻛﻨﻨﺪ ،ﺍﻫﻤﻴﺖ ﺑﺴــﻴﺎﺭ ﺯﻳﺎﺩﻯ ﺩﺍﺭﺩ ،ﺯﻳﺮﺍ ﻻﺯﻣﻪﻯ ﭘﻴﺸــﺮﻓﺖ ﺩﺭ ﺗﻜﻨﻮﻟﻮژﻯ ،ﺗﻮﺟﻪ ﺑﻪ
ﺩﺍﻧﺶ ﺭﻳﺎﺿﻰ ﺍﺳﺖ .ﺩﺭﺳﺖ ﺍﺳﺖ ﻛﻪ ﺩﺭ ﺟﺎﻣﻌﻪﻯ ﻣﺎ ﻣﻜﺎﻥ ﻣﺸﺨﺼﻰ ﺑــﺮﺍﻯ ﺟﺬﺏ ﻓﺎﺭﻍ ﺍﻟﺘﺤﺼﻴﻼﻥ ﺭﻳﺎﺿﻰ ﻭﺟــﻮﺩ ﻧﺪﺍﺭﺩ ،ﺭﻳﺎﺿﻰ ﺑﻪ ﺩﻟﻴﻞ ﻧﻈﻢ ﻓﻜﺮﻯ ﻭ ﺑﻴﻨﺶ ﻋﻤﻴﻘﻰ ﻛﻪ ﺩﺭ ﺩﻭﺭﺍﻥ ﺗﺤﺼﻴﻞ ﺑﻪ ﺩﺳﺖ ﻣﻰﺁﻭﺭﺩ، ﻣﻰﺗﻮﺍﻧﺪ ﺑﺎ ﻣﻄﺎﻟﻌﻪ ﻭ ﺗﻼﺵ ﺷــﺨﺼﻰ ﺩﺭ ﺑﺴــﻴﺎﺭﻯ ﺍﺯ ﺷــﻐﻞﻫﺎ ﺣﺘﻰ ﺷﻐﻞﻫﺎﻳﻰ ﻛﻪ ﺩﺭ ﻇﺎﻫﺮ ﺍﺭﺗﺒﺎﻃﻰ ﺑﺎ ﺭﻳﺎﺿﻰ ﻧﺪﺍﺭﻧﺪ ،ﻣﻮﻓﻖ ﻣﻰﺷﻮﺩ .
ﺗﻮﺍﻧﺎﻳﻰﻫﺎﻯ ﻣﻮﺭﺩ ﻧﻴﺎﺯ ﻭ ﻗﺎﺑﻞ ﺗﻮﺻﻴﻪ ﺷــﺎﻳﺪ ﻣﻬﻢﺗﺮﻳﻦ ﺗﻮﺍﻧﺎﻳﻰ ﻋﻠﻤﻰ ﻳﻚ ﺩﺍﻧﺸﺠﻮﻯ ﺭﻳﺎﺿﻰ ،ﺗﺴﻠﻂ ﺑﺮ ﺩﺭﻭﺱ ﺭﻳﺎﺿﻰ ﺭﺍﻫﻨﻤﺎﻳﻰ ﻭ ﺩﺑﻴﺮﺳﺘﺎﻥ ﺑﺎﺷﺪ ﻛﻪ ﺍﻳﻦ ﺍﻣﺮ ﺗﻨﻬﺎ ﺯﺍﻳﻴﺪﻩﻯ ﻋﻼﻗﻪﻯ ﺷﺨﺼﻰ ﺑﻪ ﺍﻳﻦ ﺩﺭﺱ ﺍﺳﺖ .ﺍﻳﻦ ﺭﺷﺘﻪ ﻧﻴﺎﺯﻣﻨﺪ ﺩﺍﻧﺸﺠﻮﻳﺎﻧﻰ ﺍﺳﺖ ﻛﻪ ﺍﺯ ﻧﻈﺮ ﺫﻫﻨﻰ ﺁﻣﺎﺩﮔﻰ ﺟﺬﺏ ﺍﻳﺪﻩﻫﺎﻯ ﺟﺪﻳﺪ ﺭﺍ ﺩﺍﺷﺘﻪ ﺑﺎﺷﻨﺪ ﻭ ﺑﺘﻮﺍﻧﻨﺪ ﺍﻟﮕﻮﻫﺎ ﻭ ﻧﻈﻢ ﺭﺍ ﺗﺸــﺨﻴﺺ ﺩﻫﻨﺪ ﻭ ﻣﺴــﺎﺋﻞ ﻏﻴﺮ ﻣﺘﻌﺎﺭﻑ ﺭﺍ ﺣــﻞ ﻛﻨﻨﺪ .ﺑﻪ ﻋﺒﺎﺭﺕ ﺩﻳﮕﺮ ،ﺭﻭﺣﻴﻪﻯ ﻋﻠﻤﻰ ،ﺗﻔﻜﺮ ﺍﻧﺘﻘﺎﺩﻯ ﻭ ﺗﻮﺍﻧﺎﻳﻰ ﺗﺠﺰﻳﻪ ﻭ ﺗﺤﻠﻴﻞ ﺩﺍﺷﺘﻪ ﺑﺎﺷﻨﺪ. ﺍﺯ ﺁﻥﺟــﺎ ﻛﻪ ﺭﻳﺎﺿﻴﺎﺕ ﻭﺭﻭﺩ ﺑﻪ ﻋﺮﺻﻪﻫﺎﻯ ﻧﺎﺷــﻨﺎﺧﺘﻪ ﻭ ﻛﺸــﻒ ﻗﻮﺍﻧﻴﻦ ﺁﻥ ﺍﺳــﺖ ،ﻋﻼﻗﻪﻣﻨــﺪﻯ ﺑﻪ ﻣﺒﺎﺣﺚ ﺭﻳﺎﺿــﻰ ﺍﺯ ﻫﻤﺎﻥ ﺩﻭﺭﺍﻥ ﺗﺤﺼﻴﻞ ﺩﺭ ﻣﺪﺍﺭﺱ ﻣﺸــﺨﺺ ﻣﻰﺷﻮﺩ .ﻫﻤﻴﻦ ﻋﻼﻗﻪﻣﻨﺪﻯ ﺍﺳﺖ ﻛﻪ ﻣﻰﺗﻮﺍﻧﺪ ﺭﺍﻩﻫﺎﻯ ﺑﺴــﻴﺎﺭ ﺳﺨﺖ ﺭﺍ ﺑﺮﺍﻯ ﻋﻼﻗﻪﻣﻨﺪﺍﻥ ﺍﻳﻦ ﺭﺷﺘﻪ ﻫﻤﻮﺍﺭ ﺳــﺎﺯﺩ .ﻳﻚ ﺭﻳﺎﺿــﻰﺩﺍﻥ ﻗﺒﻞ ﺍﺯ ﻫﺮ ﭼﻴﺰ ﺑﺎﻳﺪ ﺟــﺮﺃﺕ ﻗﺪﻡﮔﺬﺍﺭﻯ ﺩﺭ ﻭﺍﺩﻯ ﻧﺎﺷﻨﺎﺧﺘﻪﻫﺎ ﺭﺍ ﺩﺍﺷﺘﻪ ﺑﺎﺷﺪ .ﺑﻪ ﻃﻮﺭ ﻛﻠﻰ ،ﺩﻗﺖ ،ﺗﺠﺰﻳﻪ ﻭ ﺗﺤﻠﻴﻞ ﺻﺤﻴﺢ ﻭ ﺻﺒﺮ ﻭ ﭘﺸﺘﻜﺎﺭ ،ﺳﻪ ﻋﺎﻣﻞ ﺍﺻﻠﻰ ﺩﺭ ﺗﻮﻓﻴﻖ ﺩﺍﻭﻃﻠﺐ ﺩﺭ ﺍﻳﻦ ﺭﺷــﺘﻪ ﺍﺳﺖ .ﺑﺎ ﺗﻮﺟﻪ ﺑﻪ ﻭﺿﻌﻴﺖ ﻧﻴﺎﺯ ﻛﺸــﻮﺭ ﺑﻪ ﺍﻳﻦ ﺭﺷﺘﻪ ،ﺩﺭ ﺣﺎﻝ ﺣﺎﺿﺮ ﻫﺮ ﻭﺯﺍﺭﺗﺨﺎﻧﻪ ﻳﺎ ﺷــﺮﻛﺖ ﻫﺎ ﻣﻰﺗﻮﺍﻧﻨﺪ ﻓﺎﺭﻍ ﺍﻟﺘﺤﺼﻴﻼﻥ ﺭﻳﺎﺿﻰ ﻣﺤﺾ ﻳﺎ ﻛﺎﺭﺑﺮﺩﻯ ﺭﺍ ﺟﺬﺏ ﻛﻨﻨﺪ .ﺭﺷﺘﻪﻫﺎﻯ ﻣﺨﺘﻠﻒ ﺭﻳﺎﺿﻰ ﺟﺎﻳﮕﺎﻩ ﻭﺳــﻴﻌﻰ ﺩﺭ ﺟﺎﻣﻌﻪ ﺩﺍﺭﻧﺪ ،ﺍﺯ ﺁﻥ ﺟﻤﻠﻪ :ﺗﻤﺎﻡ ﺭﺷــﺘﻪﻫﺎﻯ ﻣﻬﻨﺪﺳﻰ، ﺭﺷــﺘﻪﻫﺎﻯ ﻣﺨﺘﻠــﻒ ﻋﻠﻮﻡ ﭘﺎﻳﻪ )ﻓﻴﺰﻳﻚ ،ﺷــﻴﻤﻰ ،ﺯﻳﺴﺖﺷﻨﺎﺳــﻰ، ﺯﻣﻴﻦﺷﻨﺎﺳﻰ( ،ﭘﺰﺷﻜﻰ ،ﻋﻠﻮﻡ ﻛﺎﻣﭙﻴﻮﺗﺮ ،ﺍﻛﺘﺸﺎﻓﺎﺕ ﻓﻀﺎﻳﻰ ،ﺑﺎﺯﺭﮔﺎﻧﻰ، ﺑﺮﻧﺎﻣﻪﺭﻳﺰﻯﻫﺎﻯ ﺩﻭﻟﺘﻰ .ﺍﻏﻠﺐ ﺭﺷﺘﻪﻫﺎﻯ ﻭﺍﺑﺴﺘﻪ ﺑﻪ ﺻﻨﻌﺖ ،ﻣﺪﻳﺮﻳﺖ ﻭ ﺭﺷــﺘﻪﻫﺎﻯ ﻣﺨﺘﻠﻒ ﻛﺸﺎﻭﺭﺯﻯ ﺑﻪ ﺭﺷــﺘﻪﻯ ﺭﻳﺎﺿﻰ ﻭﺍﺑﺴﺘﻪ ﺍﻧﺪ ﻭ ﺍﺯ ﺁﻥ ﺑﻪ ﻃﻮﺭ ﻣﺴــﺘﻘﻴﻢ ﺍﺳــﺘﻔﺎﺩﻩ ﻣﻰﻛﻨﻨﺪ .ﻫﻢﭼﻨﻴﻦ ﺑﺨﺶ ﺑﺰﺭﮔﻰ ﺍﺯ ﻓﻌﺎﻟﻴﺖﻫﺎﻯ ﺍﻗﺘﺼﺎﺩﻯ ﻭ ﺗﻮﻟﻴﺪﻯ ﻛﺸــﻮﺭ ﺩﺭ ﻃﺮﺡﻫﺎﻯ ﻣﺨﺘﻠﻒ ،ﻣﺎﻧﻨﺪ ﻧﻔﺖ ،ﭘﺘﺮﻭﺷــﻴﻤﻰ ،ﺣﻤﻞ ﻭ ﻧﻘﻞ ﻭ ،...ﻣﺴــﺘﻘﻴﻢ ﻳﺎ ﻏﻴﺮ ﻣﺴــﺘﻘﻴﻢ ﺍﺯ ﺭﻳﺎﺿﻰ ﺍﺳﺘﻔﺎﺩﻩ ﻣﻰﻛﻨﻨﺪ.
ﺩﻭﺭﺓ ﺷﺎﻧﺰﺩﻫﻢ ،ﺷﻤﺎﺭﺓ ،4ﺗﺎﺑﺴﺘﺎﻥ 1390
ﺭﺍﻫﻨﻤﺎﻳﻲ
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ﻫﻨﺪﺳﻪ ﭼﻄﻮﺭ ﻭ ﭼﮕﻮﻧﻪ ﺁﻏﺎﺯ ﺷﺪ؟ ﭼﻪ ﻛﺴــﻰ ﺑﺮﺍﻯ ﻧﺨﺴــﺘﻴﻦ ﺑﺎﺭ ﺧﻂ ،ﺍﻧﺤﻨﺎ ﻭ ﺍﺷﻜﺎﻟﻰ ﺭﺍ ﻛﻪ ﻣﺎ ))ﺷﻜﻞﻫﺎﻯ ﻫﻨﺪﺳﻰ((ﻣﻰﻧﺎﻣﻴﻢ ﻛﺸﻒ ﻛﺮﺩ؟ ﺍﻳﻦ ﺷﻜﻞ ﺭﺍ ﺍﻧﺴﺎﻥﻫﺎﻯ ﺍﻭﻟﻴﻪﺍﻯ ﻛﺸــﻒ ﻛﺮﺩﻧﺪ ،ﺯﻳﺮﺍ ﺍﻳﻦ ﺷــﻜﻞ ﺩﺭ ﺟﺎﻯ ﺟــﺎﻯ ﻃﺒﻴﻌﺖ ،ﺍﻳﻦ ﻣﻮﺯﻩﻯ ﻫﻨﺮﻯ ﻋﻈﻴﻢ ﺧﺪﺍﺩﺍﺩﻯ ﻳﺎﻓﺖ ﻣﻰﺷﻮﻧﺪ ،ﺑﻴﺎﻳﻴﺪ ﺑﻪ ﺩﻩﻫﺎ ﻫﺰﺍﺭ ﺳﺎﻝ ﻗﺒﻞ ﺑﺮﮔﺮﺩﻳﻢ ﻭ ﺯﻣﺎﻧﻰ ﺭﺍ ﻣﺠﺴﻢ ﻛﻨﻴﻢ ﻛﻪ ﻧﺨﺴﺘﻴﻦ ﺍﻧﺴﺎﻥﻫﺎ ﺗﻚﺗﻚ ﻳﺎ ﺑﻪ ﺻﻮﺭﺕ ﮔﺮﻭﻫﻰ ﺭﻭﻯ ﺯﻣﻴــﻦ ﺁﺯﺍﺩ ﺑﻮﺩﻧﺪ .ﺗﻤﺎﻣﻰ ﺭﺍﺯﻫﺎﻯ ﺑﺰﺭگ ﻧﻬﻔﺘﻪ ﺑﺰﺭگ ﻭ ﻣﻨﺎﺑﻊ ﺍﻋﺠﺎﺏﺁﻭﺭ ﺳﺮ ﺑﺴﺘﻪ ﺑﻮﺩ ﻭ ﺑﻪ ﻛﺸﻒ ﻧﻴﺎﺯ ﺩﺍﺷﺖ. ﺍﻧﺴــﺎﻥ ﻫﺎﻯ ﺍﻭﻟﻴﻪ ﺍﺯ ﺗﺮﺱ ﺭﻋﺪ ﻭ ﺑﺮﻕ ،ﺧﻮﺩ ﺭﺍ ﭘﻨﻬﺎﻥ ﻣﻰﻛﺮﺩﻧﺪ، ﺍﺯ ﻧﻴﺮﻭﻫــﺎﻯ ﭘﺮ ﺭﻣــﺰ ﻭ ﺭﺍﺯ ﺟﻬﺎﻥ ﺁﻓﺮﻳﻨﺶ ﺩﺭ ﻫﺮﺍﺱ ﺑﻮﺩﻧﺪ ﻭ ﺑﺎ ﻛﻮﺗﺎﻩ ﺷﺪﻥ ﺭﻭﺯﻫﺎ ﻭ ﻏﺮﻭﺏ ﺧﻮﺭﺷــﻴﺪ ﻓﻜﺮ ﻣﻰﻛﺮﺩﻧﺪ ﻛﻪ ﺭﻭﺯ ﺑﺮﺍﻯ ﻫﻤﻴﺸﻪ ﺍﺯ ﺑﻴــﻦ ﻣﻰﺭﻭﺩ ﻭ ﺁﻧﺎﻥ ﺩﺭ ﺗﺎﺭﻳﻜﻰ ﺳــﺮﺩ ﻭ ﻣﻄﻠــﻖ ﺗﻨﻬﺎ ﺑﺎﻗﻰ ﺧﻮﺍﻫﻨﺪ ﻣﺎﻧــﺪ .ﺑــﻪ ﻫﻤﻴﻦ ﺩﻟﻴﻞ ﺩﺭ ﻛﻨــﺎﺭ ﺁﺗﺶ ﭘﺮﺍﺭﺯﺵ ﮔﺮﺩ ﻫــﻢ ﻣﻰﺁﻣﺪﻧﺪ. ﺁﺗﺶ ﻧﺨﺴــﺘﻴﻦ ﺭﺍﺯ ﺑﺰﺭﮔﻰ ﺑــﻮﺩ ﻛﻪ ﺍﺯ ﺩﻝ ﻃﺒﻴﻌﺖ ﺑﻴﺮﻭﻥ ﻛﺸــﻴﺪﻩ ﺷــﺪ .ﺍﻧﺴﺎﻥﻫﺎﻯ ﻣﺎﻗﺒﻞ ﺗﺎﺭﻳﺦ ﺁﺗﺶ ﺭﺍ ﺑﺮﺍﻯ ﺍﻭﻟﻴﻦ ﺑﺎﺭ ﺍﺯ ﺻﺎﻋﻘﻪﺍﻯ ﻛﻪ ﺩﺭﺧﺘﺎﻥ ﺭﺍ ﻣﻰﺳﻮﺯﺍﻧﺪ ﺑﻪ ﺩﺳﺖ ﺁﻭﺭﺩﻧﺪ ﻭ ﺳﭙﺲ ﺁﻣﻮﺧﺘﻨﺪ ﻛﻪ ﭼﮕﻮﻧﻪ ﺁﺗﺶ ﺭﺍ ﺗﻬﻴﻪ ﻛﻨﻨﺪ. ﺍﻣﺎ ﺍﻳﻦ ﻫﻢ ﻧﺘﻮﺍﻧﺴــﺖ ﺗﺮﺱ ﺍﺯ ﺩﺳﺖ ﺩﺍﺩﻥ ﺧﻮﺭﺷﻴﺪ ﺭﺍ ﺍﺯ ﺩﻟﺸﺎﻥ ﺑﺰﺩﺍﻳﺪ .ﺁﻥﻫﺎ ﺑﻪ ﺩﻟﻴﻞ ﺍﻳﻦ ﻛﻪ ﺗﺮﺳﺸــﺎﻥ ﺭﺍ ﺑﺎ ﻳﻜﺪﻳﮕﺮ ﺗﻘﺴﻴﻢ ﻛﻨﻨﺪ ﻭ ﻧﻴﺰ ﺑﺮﺍﻯ ﻛﻤﻚ ﺑﻪ ﺑﺎﺯﮔﺸﺖ ﺧﻮﺭﺷﻴﺪ ﺑﻪ ﻫﻨﮕﺎﻡ ﺍﺑﺮﻯ ﺷﺪﻥ ﻳﺎ ﺧﻮﺭﺷﻴﺪ ﮔﺮﻓﺘﮕــﻰ ﻳﺎ ﺩﻳﮕﺮ ﻭﻗﺎﻳﻊ ﻃﺒﻴﻌﻰ ،ﻣﺮﺍﺳــﻢ ﻭﻳﮋﻩﺍﻯ ﺑﺮﮔﺰﺍﺭ ﻣﻰﻛﺮﺩﻧﺪ ﻭ ﺑــﻪ ﺧﻮﺍﻧﺪﻥ ﺩﻋﺎ ﻭ ﻗﺮﺑﺎﻧﻰ ﺣﻴﻮﺍﻧﺎﺕ ﺑﺮﺍﻯ ﺁﻓﺮﻳﻨﻨﺪﻩﻯ ﺑﺰﺭگ ﻭ ﻣﻬﺮﺑﺎﻥ ﻣﻰﭘﺮﺩﺍﺧﺘﻨﺪ ﻭ ﺳــﭙﺲ ﺭﻓﺘﻪﺭﻓﺘﻪ ﺑﺎ ﺑﺎﺯﮔﺸــﺖ ﮔﺮﻣﺎ ﻭ ﻧﻮﺭ ،ﺭﻭﺣﻴﻪﻯ ﺧــﻮﺩ ﺭﺍ ﺑﺎﺯ ﻣﻰﻳﺎﻓﺘﻨــﺪ .ﺁﻳﺎ ﺗﺎ ﺑﻪ ﺣﺎﻝ ﺑﻠﻮﺭ ﻛﻮﺍﺭﺗــﺰ ﺭﺍ ﺩﻳﺪﻩ ﺍﻳﺪ؟ ﺍﻳﻦ
ﺑﻠﻮﺭﻫﺎ ﻣﻨﺸﻮﺭﻯ ﺷﺶ ﻭﺟﻬﻰ ﻫﺴﺘﻨﺪ ﻛﻪ ﻳﻚ ﻫﺮﻡ ﺷﺶ ﻭﺟﻬﻰ ﺭﻭﻯ ﺁﻥﻫﺎ ﻗﺮﺍﺭ ﮔﺮﻓﺘﻪ ﺍﺳﺖ. ﺁﻳﺎ ﻫﺮﮔــﺰ ﺍﻭﺍﻳﻞ ﺑﻬﺎﺭ ﺩﺭ ﺟﻨﮕﻞ ﻳﺎ ﺩﺷــﺖ ﺑﻮﺩﻩﺍﻳــﺪ؟ ﺩﺭﺧﺘﺎﻥ ﻭ ﮔﻴﺎﻫﺎﻥ ﺑﻪ ﻳﻜﺒﺎﺭﻩ ﺷــﻜﻮﻓﻪ ﻣﻰﻛﻨﻨﺪ ،ﺑﻌﻀﻰ ﺑﺎ ﺳﻪ ﮔﻠﺒﺮگ ﻭ ﺑﻌﻀﻰ ﺑﺎ ﭼﻬﺎﺭ ﮔﻠﺒﺮگ ﻭ ﺷﻜﻮﻓﻪﻫﺎﻯ ﺑﻌﻀﻰ ﭘﻨﺞ ﺿﻠﻌﻰ ﻫﺴﺘﻨﺪ ﺍﮔﺮ ﺧﻴﺎﺭﻯ ﺭﺍ ﺣﻠﻘﻪﺣﻠﻘﻪ ﻛﻨﻴﺪ ﺩﺍﻧﻪﻫﺎﻯ ﺁﻥ ﺭﺍ ﺩﺭ ﺳﻪ ﻗﺴﻤﺖ ﻭ ﺍﮔﺮ ﻓﻠﻔﻞ ﺳﺒﺰﻯ ﺭﺍ ﺍﺯ ﻫﻢ ﺑﮕﺸــﺎﻳﻴﺪ ﺩﺍﻧﻪﻫﺎﻯ ﺁﻥ ﺭﺍ ﺩﺭ ﭼﻬﺎﺭ ﻗﺴــﻤﺖ ﺧﻮﺍﻫﻴﺪ ﺩﻳﺪ .ﺣﺎﻝ ﭘﻴﺎﺯﻯ ﺭﺍ ﺣﻠﻘﻪﺣﻠﻘﻪ ﻛﻨﻴﺪ ،ﻣﺸــﺎﻫﺪﻩ ﺧﻮﺍﻫﻴﺪ ﻛﺮﺩ ﻛﻪ ﭘﻴﺎﺯ ﺑﻪ ﺻﻮﺭﺕ ﺩﻭﺍﻳﺮﻯ ﻣﻨﻈﻢ ﺍﺯﻫﻢ ﺟﺪﺍ ﻣﻰﺷﻮﺩ. ﺍﮔﺮ ﺑﻪ ﻳﻚ ﺳــﺘﺎﺭﻩﻯ ﺩﺭﻳﺎﻳﻰ ﻛﻪ ﻫﻤﺮﺍﻩ ﺑــﺎ ﺍﻣﻮﺍﺝ ﺁﺏ ﺑﻪ ﻣﻨﺎﻃﻖ ﻛﻢﻋﻤﻖ ﺩﺭﻳﺎ ﺁﻣﺪﻩ ﺍﺳﺖ ﺑﺮﺧﻮﺭﺩ ﻛﻨﻴﺪ ،ﻣﺘﻮﺟﻪ ﺧﻮﺍﻫﻴﺪ ﺷﺪ ﻛﻪ ﺍﻏﻠﺐ ﺁﻥﻫﺎ ﭘﻨﺞ ﮔﻮﺷﻪ ﻫﺴﺘﻨﺪ .ﻫﻤﻪ ﺟﺎﻯ ﻃﺒﻴﻌﺖ ﭘﺮ ﺍﺯ ﺗﺮﻛﻴﺐﻫﺎﻳﻰ ﺍﺳﺖ ﻛﻪ ﻣﺎ ﺁﻥﻫﺎ ﺭﺍ ﺍﺷــﻜﺎﻝ ﺳﺎﺩﻩ ﻫﻨﺪﺳــﻰ ﻣﻰﻧﺎﻣﻴﻢ .ﺑﺎ ﻭﺟﻮﺩ ﺗﻔﺎﻭﺕﻫﺎﻯ ﻇﺎﻫﺮﻯ ﺩﺭ ﻃﺒﻴﻌﺖ ﻭ ﺩﺭ ﺗﻤﺎﻣﻰ ﻋﺎﻟﻢ ﻭ ﺩﺭ ﺟﺰﺋﻴﺎﺕ ﻳﮕﺎﻧﮕﻰ ﻭ ﻭﺣﺪﺕ ﻣﻮﺝ ﻣﻰﺯﻧﺪ. ﺑﻪ ﻳﻚ ﺩﺍﻧﻪﻯ ﺑﺮﻑ ﺑﻴﻨﺪﻳﺸﻴﺪ ﺍﻳﻦ ﮔﻞﻫﺎﻯ ﻳﺨﻰ ﺷﺶ ﺿﻠﻌﻰ ﻛﻪ ﺩﺭ ﺍﺭﺗﻔﺎﻉ ﺑﺎﻻ ﺑﺮ ﺍﺛﺮ ﻧﻴﺮﻭﻯ ﺑﺎﺩ ﻭ ﺳﺮﻣﺎ ﻓﺸﺮﺩﻩ ﻣﻰﺷﻮﻧﺪ ﻭ ﻳﺦ ﻣﻰﺯﻧﻨﺪ ﻭ ﺑﻪ ﺻﻮﺭﺕ ﺷــﺶﺿﻠﻌﻰ ﺑﺎﻗﻰ ﻣﻰﻣﺎﻧﻨﺪ .ﻣﻄﺎﻟﻌــﻪ ﺩﺭ ﻗﺎﻧﻮﻥ ﻃﺒﻴﻌﺖ ﺍﺳﺖ ﻛﻪ ﺭﻳﺎﺿﻴﺎﺕ ﺭﺍ ﺟﺬﺍﺏ ﻣﻰﻛﻨﻨﺪ. ﺍﻧﺴﺎﻥﻫﺎﻯ ﻫﻮﻟﻨﺎﻙ ﻭ ﺣﺘﻰ ﻫﺮﺍﺱﺁﻭﺭﻯ ﺑﻪ ﻃﺒﻴﻌﺖ ﻭ ﻧﻴﺮﻭﻫﺎﻯ ﺁﻥ ﻧﺰﺩﻳﻚ ﺑﻮﺩﻧﺪ ﺁﻥﻫﺎ ﺍﻋﺠﺎﺯ ﻣﻮﺟﻮﺩ ﺩﺭ ﻃﺒﻴﻌﺖ ﺭﺍ ﺑﻪ ﺷــﺪﺕ ﻣﻰﺩﻳﺪﻧﺪ ﻭ ﻟﻤﺲ ﻣﻰﻛﺮﺩﻧﺪ .ﺑﻪ ﻫﻤﻴﻦ ﺟﻬﺖ ﺑﻮﺩ ﻛﻪ ﺍﻧﺴــﺎﻥ ﺍﻭﻟﻴﻪ ﺍﺯ ﻧﻤﺎﻳﺸﮕﺎﻩ ﺁﻓﺮﻳﻨﺶ ﻫﻨﺮﻯ ﺟﻬﺎﻥ ،ﻫﻨﺪﺳﻪ ﺁﻣﻮﺧﺖ.
ﺭﻳﺎﺿﻴﺎﺕ ﺍﻣﻜﺎﻥ ﺷﺎﻧﺲ ﺑﺮﻧﺪﻩ ﺷﺪﻥ ﻛﻢ ﺍﺳﺖ ﻳﺎ ﺯﻳﺎﺩ؟ ﺁﻳﺎ ﺷﻤﺎ ﻣﻰﺗﻮﺍﻧﻴﺪ ﺍﺯ ﻗﻮﺍﻧﻴﻦ ﺷــﺎﻧﺲ ﭘﻴﺶﺑﻴﻨﻰ ﺍﺳــﺘﻔﺎﺩﻩ ﻛﻨﻴﺪ؟ ﺁﻳﺎ ﻣﻰﺩﺍﻧﻴﺪ ﻛــﻪ ﻭﻗﺘﻰ ﺩﺭ ﻳﻚ
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ﺭﺍﻫﻨﻤﺎﻳﻲ
ﺩﻭﺭﺓ ﺷﺎﻧﺰﺩﻫﻢ ،ﺷﻤﺎﺭﺓ ،4ﺗﺎﺑﺴﺘﺎﻥ 1390
ﺩﺍﻧﺶﺍﻓﺰﺍﻳﻲ
﹞︧﹚﹤﹨︀ى وا﹇︺﹩ ﺍﻋﻈﻢ ﭘﻮﺭﭘﺮﻭﻳﻦ
ﻛﻠﻴﺪﻭﺍژﻩﻫﺎ :ﻣﺴﺌﻠﻪﻫﺎﻱ ﻭﺍﻗﻌﻲ ،ﺳﺮﻋﺖ ،ﻣﺴﺎﺣﺖ. ﺩﺭ ﺍﻳﻦ ﺳـﺘﻮﻥ ﻗﺼـﺪ ﺩﺍﺭﻳﻢ ﻣﺴـﺌﻠﻪﻫﺎﻱ ﻭﺍﻗﻌﻲ ﻃـﺮﺡ ﻛﻨﻴﻢ؛ ﻣﺴـﺌﻠﻪﻫﺎﻳﻲ ﻛﻪ ﺍﻳﻦ ﻃـﺮﻑ ﻭ ﺁﻥ ﻃﺮﻑ ﺷـﻨﻴﺪﻩﺍﻳﻢ ﻭ ﺩﻳﺪﻩﺍﻳﻢ ﻛﻪ ﭼﮕﻮﻧﻪ ﺩﺍﻧﺶﺁﻣﻮﺯﺍﻥ ﺭﺍﻫﻨﻤﺎﻳﻲ ﺗﻮﺍﻧﺴﺘﻪﺍﻧﺪ ﺑﺎ ﺍﻃﻼﻋﺎﺕ ﻭ ﻣﻬﺎﺭﺕﻫﺎﻱ ﺭﻳﺎﺿﻲ ﺧﻮﺩ ،ﺁﻥﻫﺎ ﺭﺍ ﺣﻞ ﻛﻨﻨﺪ .ﺷﻤﺎ ﻫﻢ ﻣﺴﺌﻠﻪﻫﺎﻱ ﺭﻳﺎﺿﻲ ﻭﺍﻗﻌﻲ
ﺗﺎﺑﻠﻮﻳﻲ ﺩﺭ ﻛﻨﺎﺭ ﺟﺎﺩﻩ ﻧﺼﺐ ﺷــﺪﻩ ﺍﺳﺖ ﻛﻪ ﻓﺎﺻﻠﻪﻱ ﻣﺎ ﺗﺎ ﻛﺎﺷﺎﻥ ﺭﺍ ﻧﺸــﺎﻥ ﻣﻲﺩﻫﺪ .ﻣﻦ ﻫﻢ ﺳﺎﻋﺖ ﺩﺍﺭﻡ .ﺳــﺮﻋﺖ ﺭﺍ ﻣﺤﺎﺳﺒﻪ ﻣﻲﻛﻨﻢ«. ﺑﻌﺪ ﺍﺯ ﭼﻨﺪ ﺩﻗﻴﻘﻪ ﺻﺎﻟﺤﻪ ﮔﻔﺖ ﻛﻪ 5ﻛﻴﻠﻮﻣﺘﺮ ﺭﺍ ﺩﺭ 2ﺩﻗﻴﻘﻪ ﻭ 40 ﺛﺎﻧﻴﻪ ﻃﻲ ﻛﺮﺩﻩﺍﻳﻢ ﻭ ﺗﻮﺍﻧﺴﺖ ﺳﺮﻋﺖ ﻣﺎﺷﻴﻦ ﺭﺍ ﺣﺴﺎﺏ ﻛﻨﺪ.
ﺭﺍ ﻛـﻪ ﺑﺎ ﺁﻥﻫﺎ ﻣﻮﺍﺟﻪ ﺷـﺪﻩﺍﻳﺪ ﻳﺎ ﺩﺭ ﺍﻃﺮﺍﻑ ﺧـﻮﺩ ﺩﻳﺪﻩﺍﻳﺪ ،ﺑﺮﺍﻱ ﻣﺠﻠﻪ ﺑﻔﺮﺳـﺘﻴﺪ ﺗﺎ ﺑﺎ ﻧﺎﻡ ﺧﻮﺩﺗﺎﻥ ﺩﺭ ﺍﻳﻦ ﺳـﺘﻮﻥ ﻃﺮﺡ ﺷـﻮﻧﺪ .ﺩﺭ ﺿﻤﻦ ،ﻣﻲﺗﻮﺍﻧﻴﺪ ﻣﺴـﺌﻠﻪﻫﺎﻱ ﻃﺮﺡﺷـﺪﻩ ﺩﺭ ﺍﻳﻦ ﺷﻤﺎﺭﻩ ﺭﺍ ﻫﻢ ﺣﻞ ﻛﻨﻴﺪ ﻭ ﺭﺍﻩﺣﻞﻫﺎﻳﺘﺎﻥ ﺭﺍ ﺑﺮﺍﻱ ﻣﺠﻠﻪ ﺑﻔﺮﺳﺘﻴﺪ.
ﻣﺴﺌﻠﻪﻱ 1
ﻧﻮﻳﺴﻨﺪﻩ :ﺷﻴﺮﻳﻦ ﺣﺠﺎﺯﻱ
ﻣﺎﺩﺭﻡ ﻳﻚ ﺗﺎﺑﻠﻮ ﻓﺮﺵ ﺯﻳﺒﺎ ﺧﺮﻳﺪﻩ ﺑﻮﺩ ﻭ ﻣﻲﺧﻮﺍﺳﺖ ﺁﻥ ﺭﺍ ﻭﺳﻂ ﺩﻳــﻮﺍﺭ ﺍﻃﺎﻕ ﭘﺬﻳﺮﺍﻳﻲ ﻧﺼﺐ ﻛﻨﺪ .ﺍﻭ ﻣﻲﺧﻮﺍﺳــﺖ ﺗﺎﺑﻠﻮ ﺩﻗﻴﻘﺎً ﻭﺳــﻂ ﺩﻳﻮﺍﺭ ﻗﺮﺍﺭ ﺑﮕﻴﺮﺩ ﻭ ﺍﺯ ﻣﻦ ﺧﻮﺍﺳﺖ ﺗﺎ ﻓﺎﺻﻠﻪﻱ ﺗﺎﺑﻠﻮ ﺗﺎ ﻫﺮ ﻃﺮﻑ ﺩﻳﻮﺍﺭ ﺭﺍ ﭘﻴــﺪﺍ ﻛﻨﻢ .ﻣﻦ ﺍﻳﻦ ﻣﺴــﺌﻠﻪ ﺭﺍ ﺑﺎ ﺍﻧﺪﺍﺯﻩﮔﻴﺮﻱ ﻃــﻮﻝ ﺗﺎﺑﻠﻮ ﻭ ﻃﻮﻝ ﺩﻳﻮﺍﺭ ،ﺣﻞ ﻛﺮﺩﻡ.
ﻣﺴﺌﻠﻪﻱ 3
ﻧﻮﻳﺴﻨﺪﻩ :ﺁﻗﺎﻱ ﺣﺒﻴﺒﻲ
ﻣﻦ ﺩﺭ ﺭﻭﺳــﺘﺎ ﺑﻪ ﺩﻧﻴﺎ ﺁﻣﺪﻩﺍﻡ .ﺍﻣﺎ ﺣﺎﻻ ﺳﺎﻝﻫﺎﺳــﺖ ﻛﻪ ﺩﺭ ﺷﻬﺮ ﺯﻧﺪﮔﻲ ﻣﻲﻛﻨﻢ .ﻫﻤﻴﺸﻪ ﺩﻟﻢ ﻣﻲﺧﻮﺍﺳﺖ ﺑﻪ ﻳﺎﺩ ﻛﻮﺩﻛﻲﻫﺎﻳﻢ ﺩﺭ ﺧﺎﻧﻪ ﻣﺮﻍ ﻭ ﺧﺮﻭﺱ ﺩﺍﺷــﺘﻪ ﺑﺎﺷــﻢ ،ﺍﻣﺎ ﺯﻧﺪﮔﻲ ﺩﺭ ﺁﭘﺎﺭﺗﻤﺎﻥ ﺍﻳﻦ ﺍﻣﻜﺎﻥ ﺭﺍ ﺑﻪ ﻣﻦ ﻧﻤﻲﺩﺍﺩ .ﭼﻨﺪﻱ ﭘﻴﺶ ﺗﻮﺍﻧﺴﺘﻢ ﻳﻚ ﺧﺎﻧﻪﻱ ﻛﻮﭼﻚ ﺣﻴﺎﻁﺩﺍﺭ ﺑﺨﺮﻡ ﻭ ﺑﺎ ﺧﺎﻧﻮﺍﺩﻩﺍﻡ ﺩﺭ ﺍﻳﻦ ﺧﺎﻧﻪ ﺯﻧﺪﮔﻲ ﻛﻨﻢ .ﺭﻭﺯﻱ ﺗﺼﻤﻴﻢ ﮔﺮﻓﺘﻢ ﻛﻪ ﺩﺭ ﻛﻨﺎﺭ ﺣﻴﺎﻁ ﺍﻳﻦ ﺧﺎﻧﻪ ﻳﻚ ﺣﺼﺎﺭ ﺑﺮﺍﻱ ﻧﮕﻪﺩﺍﺭﻱ ﻣﺮﻍ ﻭ ﺧﺮﻭﺱ ﺩﺭﺳــﺖ ﻛﻨﻢ 10 .ﻣﺘﺮ ﺣﺼﺎﺭ ﺧﺮﻳﺪﻡ ﺗﺎ ﺩﺭ ﻛﻨــﺎﺭ ﻳﻜﻲ ﺍﺯ ﺩﻳﻮﺍﺭﻫﺎﻱ ﺣﻴﺎﻁ ﺳــﻪ ﺩﻳﻮﺍﺭ ﺑﺴــﺎﺯﻡ ﻭ ﻳﻚ ﭼﻬﺎﺭ ﺩﻳﻮﺍﺭﻱ ﻣﺴﺘﻄﻴﻞ ﺷﻜﻞ ﺑﺮﺍﻱ ﻣﺮﻍ ﻭ ﺧﺮﻭﺱﻫﺎ ﺁﻣﺎﺩﻩ ﻛﻨﻢ.
41 4
11 2
ﻣﺴﺌﻠﻪﻱ 2
ﻧﻮﻳﺴﻨﺪﻩ :ﺁﻗﺎﻱ ﺻﺎﺩﻗﻲ
ﻫﻨﮕﺎﻡ ﺳــﻔﺮ ﺑﻪ ﻛﺎﺷﺎﻥ ﺳﺮﻋﺖ ﺳﻨﺞ ﻣﺎﺷــﻴﻨﻢ ﺧﺮﺍﺏ ﺷﺪﻩ ﺑﻮﺩ. ﺩﺧﺘﺮﻡ ﺻﺎﻟﺤﻪ ﻛﻪ ﺩﺍﻧﺶﺁﻣﻮﺯ ﺍﻭﻝ ﺭﺍﻫﻨﻤﺎﻳﻲ ﺍﺳــﺖ ﺩﺭ ﺻﻨﺪﻟﻲ ﻋﻘﺐ ﻣﺎﺷﻴﻦ ﻧﺸﺴﺘﻪ ﺑﻮﺩ .ﺍﺯ ﺍﻭ ﭘﺮﺳﻴﺪﻡ ،ﺁﻳﺎ ﻣﻲﺗﻮﺍﻧﻲ ﺳﺮﻋﺖ ﺣﺮﻛﺘﻤﺎﻥ ﺭﺍ ﻣﺤﺎﺳﺒﻪ ﻛﻨﻲ؟ ﺍﻭ ﻧﮕﺎﻫﻲ ﺑﻪ ﻛﻨﺎﺭ ﺟﺎﺩﻩ ﻛﺮﺩ ﻭ ﮔﻔﺖ» :ﻫﺮ 5ﻛﻴﻠﻮﻣﺘﺮ
ﻣﺴــﺌﻠﻪ ﺍﻳﻦ ﺑﻮﺩ ﻛﻪ ﻧﻤﻲﺩﺍﻧﺴــﺘﻢ ﻃﻮﻝ ﻭ ﻋــﺮﺽ ﻻﻧﻪﻱ ﻣﺮﻍ ﻭ ﺧﺮﻭﺱﻫﺎ ﺭﺍ ﭼﻪﻗﺪﺭ ﺑﺎﻳﺪ ﺩﺭ ﻧﻈﺮ ﺑﮕﻴﺮﻡ ﺗﺎ ﺑﺘﻮﺍﻧﻢ ﺑﺎ 10ﻣﺘﺮ ﺣﺼﺎﺭﻱ ﻛﻪ ﺩﺍﺷﺘﻢ ﺑﻴﺶﺗﺮﻳﻦ ﻣﺴﺎﺣﺖ ﻣﻤﻜﻦ ﺭﺍ ﺑﻪ ﺩﺳﺖ ﺁﻭﺭﻡ .ﭘﺴﺮﻡ ﺣﺎﻣﺪ ﻛﻪ ﺩﺍﻧﺶﺁﻣﻮﺯ ﺩﻭﻡ ﺭﺍﻫﻨﻤﺎﻳﻲ ﺍﺳﺖ ،ﺍﻳﻦ ﻣﺴﺌﻠﻪ ﺭﺍ ﺣﻞ ﻛﺮﺩ. ﺩﻭﺭﺓ ﺷﺎﻧﺰﺩﻫﻢ ،ﺷﻤﺎﺭﺓ ،4ﺗﺎﺑﺴﺘﺎﻥ 1390
ﺭﺍﻫﻨﻤﺎﻳﻲ
25
ﻣﺴﺌﻠﻪﻱ 4 ﺑــﺮﺍﻱ ﺧﻮﺍﺳــﺖ ﺑ ﺮ ﻱ ﺖ ﻣﺎﻫــﺎﻥ ﻣﻲ ﻼﻩ ﺍﻳﻤﻨﻲ ﺩﻭﭼﺮﺧﻪﺍﺵ ،ﻗﻔﻞ ،ﺯﻧﮓ ﻭ ﻛﻼﻩ ﻗﻴﻤﺖﻫﺎﻳﺸﺎﻥ ﺑﺨﺮﺩ .ﺑﺴــﺘﻪﻫﺎﻱ ﺯﻳﺮ ﺑﺎ ﻤﺖ ﺷــﺪ .ﻣﺎﻫﺎﻥ ﺩﺭ ﻓﺮﻭﺷــﮕﺎﻩ ﺩﻳﺪﻩ ﻣﻲ ﺪ. ﻚ ﻗﻔﻞ، ﻣﻲﺧﻮﺍﺳــﺖ ﺑﺪﺍﻧﺪ ﺑﺮﺍﻱ ﻳﻚ ﺍﻳﻤﻨﻲ ﺭﻭﻱ ﻳﻚ ﺯﻧﮓ ﻭ ﻳــﻚ ﻛﻼﻩ ﻲ ﻫﻢ ﭼﻪﻗﺪﺭ ﺑﺎﻳــﺪ ﭘﺮﺩﺍﺧﺖ ﻛﻨﺪ .ﺍﻭ ﺭﺍﻫﻨﻤﺎﻳﻲ ﺩﺭﺱ ﻲ ﻛﻪ ﺩﺭ ﻛﻼﺱ ﺳــﻮﻡ ﺌﻠﻪ ﺭﺍ ﺣﻞ ﻣﻲﺧﻮﺍﻧﺪ ﺗﻮﺍﻧﺴــﺖ ﺍﻳﻦ ﻣﺴﺌﻠﻪ ﻛﻨﺪ. 4400ﺗﻮﻣﺎﻥ 5300ﺗﻮﻣﺎﻥ 3300ﺗﻮﻣﺎﻥ
ﻣﺴﺌﻠﻪﻱ 5 ﭘــﺲ ﺍﺯ ﻳــﻚ ﻣﻬﻤﺎﻧﻲ ﺍﺯ ﻳﻚ ﺟﻌﺒﻪ ﺷــﻜﻼﺕ 72ﺗﺎﻳﻲ ﻓﻘﻂ 25 ﺷﻜﻼﺕ ﺯﻳﺮ ﺑﺎﻗﻲﻣﺎﻧﺪﻩ ﺑﻮﺩ. ﻣــﺎﺩﺭ ﺗﻴﻨﺎ ﺭﻭﻱ ﺟﻌﺒﻪ ﺷــﻜﻼﺕ ﺭﺍ ﺧﻮﺍﻧﺪ؛ ﺭﻭﻱ ﺁﻥ ﻧﻮﺷــﺘﻪ ﺑﻮﺩ »ﺗﻌﺪﺍﺩ ﺷﻜﻼﺕ ﻗﻬﻮﻩﺍﻱ ﺩﺭ ﺍﻳﻦ ﺑﺴﺘﻪ ﺩﻭ ﺑﺮﺍﺑﺮ ﺷﻜﻼﺕﻫﺎﻱ ﺷﻴﺸﻪﺍﻱ ﺍﺳــﺖ «.ﺍﻭ ﭘﺮﺳﻴﺪ» :ﺗﻴﻨﺎ ﻓﻜﺮ ﻣﻲﻛﻨــﻲ ﻣﻬﻤﺎﻥﻫﺎ ﺑﻴﺶﺗﺮ ﺷــﻜﻼﺕ ﺷﻴﺮﻱ ﺩﻭﺳــﺖ ﺩﺍﺷــﺘﻪﺍﻧﺪ ﻳــﺎ ﻗﻬــﻮﻩﺍﻱ؟« ﺗﻴﻨﺎ ﺑﺎ ﺗﻮﺟﻪ ﺑــﻪ ﻫﻤــﻪﻱ ﺍﻃﻼﻋــﺎﺕ، ﺳــﺆﺍﻝ ﻣــﺎﺩﺭ ﺭﺍ ﭘﺎﺳــﺦ ﺩﺍﺩ .ﺷــﻤﺎ ﻫﻢ ﻣﻲﺗﻮﺍﻧﻴﺪ ﺳﺆﺍﻝ ﻣﺎﺩﺭ ﺗﻴﻨﺎ ﺭﺍ ﭘﺎﺳﺦ ﺩﻫﻴﺪ؟
26 26
1390 1390 ﺗﺎﺑﺴﺘﺎﻥ ﺗﺎﺑﺴﺘﺎﻥ ﺷﻤﺎﺭﺓ ،4 ﺷﻤﺎﺭﺓ ،4 ﺷﺎﻧﺰﺩﻫﻢ، ﺷﺎﻧﺰﺩﻫﻢ، ﺩﻭﺭﺓ ﺩﻭﺭﺓ
ﺭﺍﻫﻨﻤﺎﻳﻲ ﺭﺍﻫﻨﻤﺎﻳﻲ
ﻣﺴﺌﻠﻪﻱ 6 ﭘﺪﺭ ﺳــﻴﻨﺎ ﺍﺯ ﻣﺸــﺘﺮﻳﺎﻥ ﺩﺍﺋﻤﻲ ﻳﻚ ﻛﻔﺶﻓﺮﻭﺷــﻲ ﺍﺳــﺖ .ﺑﻪ ﻫﻤﻴﻦ ﺟﻬﺖ ،ﺍﻳﻦ ﻛﻔﺶﻓﺮﻭﺷــﻲ ﺩﺭ ﺭﻭﺯ ﺗﻮﻟﺪ ﭘﺪﺭ ﺳــﻴﻨﺎ ﻳﻚ ﻛﺎﺭﺕ ﺗﺨﻔﻴﻒ 40ﺩﺭﺻﺪﻱ ﺑﺮﺍﻱ ﺍﻭ ﻓﺮﺳﺘﺎﺩ .ﭘﺪﺭ ﻫﻤﺮﺍﻩ ﺳﻴﻨﺎ ﺑﺮﺍﻱ ﺧﺮﻳﺪ ﺑﻪ ﻛﻔﺶﻓﺮﻭﺷﻲ ﺭﻓﺘﻨﺪ ﻭ ﻣﺘﻮﺟﻪ ﺷﺪﻧﺪ ﻛﻪ ﻫﻢﺯﻣﺎﻥ ﺩﺭ ﺍﻳﻦ ﻛﻔﺶﻓﺮﻭﺷﻲ ﻫﻤﻪﻱ ﻛﻔﺶﻫﺎ ﺑﺎ ٪15ﺗﺨﻔﻴﻒ ﺑﺮﺍﻱ ﻫﻤﻪ ﻓﺮﻭﺧﺘﻪ ﻣﻲﺷــﻮﺩ .ﭘﺪﺭ ﺍﺯ ﺳﻴﻨﺎ ﭘﺮﺳﻴﺪ ﻓﻜﺮ ﻣﻲﻛﻨﻲ ﺑﻬﺘﺮ ﺍﺳﺖ ﺍﻭﻝ ٪ 15ﺧﺮﻳﺪﻣﺎﻥ ﺭﺍ ﺑﮕﻴﺮﻳﻢ ﻭ ﺑﻌــﺪ ﻛﺎﺭﺕ ﺗﺨﻔﻴــﻒ 40ﺩﺭﺻﺪﻱ ﺭﺍ ﻧﺸــﺎﻥ ﺑﺪﻫﻴﻢ ﻭ ﺍﺯ ﻗﻴﻤﺘﻲ ﻛﻪ ﺑﺎﻳﺪ ﭘﺮﺩﺍﺧﺖ ﻛﻨﻴﻢ 40 ،ﺩﺭﺻﺪ ﻛﻢ ﻛﻨﻴﻢ ﻳﺎ ﺍﻳﻦﻛﻪ ﺍﻭﻝ ٪40ﺗﺨﻔﻴﻒ ﺑﮕﻴﺮﻳﻢ ﻭ ﺑﻌﺪ 15ﺩﺭﺻﺪ؟ ﺳــﻴﻨﺎ ﻛﻤﻲ ﻓﻜﺮ ﻛﺮﺩ ﻭ ﭘﺎﺳﺦ ﺳﺆﺍﻝ ﭘﺪﺭ ﺭﺍ ﺩﺍﺩ .ﺁﻳﺎ ﺷﻤﺎ ﻫﻢ ﻣﻲﺗﻮﺍﻧﻴﺪ ﺳﺆﺍﻝ ﭘﺪﺭ ﺳﻴﻨﺎ ﺭﺍ ﭘﺎﺳﺦ ﺩﻫﻴﺪ؟
ﻫﻤﺮﺍﻩ ﺑﺎ ﻛﺘﺎﺏ
﹡﹍︀﹨﹢︧﹆﹞ ﹤︋ ﹢﹡ ﹩م ︻﹚﹫﹤ ﻣﺠﻴﺪ ﻣﻨﺸﻮﺭﻱ
ﻛﻠﻴﺪﻭﺍژﻩﻫﺎ :ﻣﻘﺴﻮﻡﻋﻠﻴﻪ ،ﻣﻘﺴﻮﻡﻋﻠﻴﻪﻫﺎﻱ ﺩﻳﮕﺮ ،ﻋﺪﺩ ،1ﺍﻋﺪﺍﺩ ﺍﻭﻝ ،ﻧﻤﻮﺩﺍﺭ ﺩﺭﺧﺘﻲ.
ﻛﻞ ﻣﺎﺟﺮﺍﻳﻲ ﻛﻪ ﻣﻲﺧﻮﺍﻫﻢ ﺑﺮﺍﻳﺘﺎﻥ ﺗﻌﺮﻳﻒ ﻛﻨﻢ ﻣﺮﺑﻮﻁ ﻣﻲﺷــﻮﺩ ﺑﻪ ﭘﺮﺳﺸﻲ ﻛﻪ ﺩﺭ ﺯﻧﮓ ﺣﺴﺎﺏ ،ﻣﻌﻠﻢ ﺭﻭﻱ ﺗﺎﺑﻠﻮ ﻧﻮﺷﺖ: »ﺍﮔﺮ 14ﻭ 15ﺩﻭ ﻣﻘﺴﻮﻡ ﻋﻠﻴﻪ ﻳﻚ ﻋﺪﺩ ﺑﺎﺷﻨﺪ ،ﻣﻘﺴﻮﻡ ﻋﻠﻴﻪﻫﺎﻱ ﺩﻳﮕﺮ ﺁﻥ ﻋﺪﺩ ﺭﺍ ﺑﻨﻮﻳﺴﻴﺪ«. ً ﺑﻌﺪ ﺍﺯ ﻛﻤﻲ ﻫﻤﻬﻤﻪ ﺗﻘﺮﻳﺒﺎ ﺗﻤﺎﻡ ﺩﺍﻧﺶﺁﻣﻮﺯﺍﻥ ﻣﺸــﻐﻮﻝ ﻧﻮﺷــﺘﻦ ﺷــﺪﻧﺪ .ﻣﻌﻠﻢ ﺣﺴــﺎﺏ ،ﺑﭽﻪﻫــﺎﻱ ﻛﻼﺱ ﺭﺍ ﺑﻪ ﮔﺮﻭﻩﻫﺎﻱ ﺳــﻪ ﻧﻔﺮﻱ ﺩﺳﺘﻪﺑﻨﺪﻱ ﻛﺮﺩﻩ ﺑﻮﺩ ﻭ ﻣﻦ ﻭ ﺳﻌﻴﺪ ﻭ ﻣﺤﻤﺪ ﻫﻢﮔﺮﻭﻩ ﺑﻮﺩﻳﻢ. ﻣﺤﻤــﺪ ﺳــﺮﺵ ﺭﺍ ﺍﺯ ﺭﻭﻱ ﻛﺎﻏﺬ ﺑﻠﻨــﺪ ﻛﺮﺩ ﻭ ﮔﻔــﺖ» :ﺍﻭﻝ ﺑﺎﻳﺪ 14ﻭ 15ﺭﺍ ﺩﺭ ﻫــﻢ ﺿــﺮﺏ ﻛﻨﻴﻢ ﻛﻪ ﻣﻲﺷــﻮﺩ 210ﻋﺪﺩ ﻣﻮﺭﺩ ﻧﻈﺮ ﺑﺎﺷﻪ؟ 420ﻫﻢ ﻣﻲﺗﻮﻧﻪ ﻋﺪﺩ ﻣﻮﺭﺩﻧﻈﺮ ﺑﺎﺷﻪ ،ﭼﻮﻥ 14ﻭ 15ﻣﻘﺴﻮﻡ ﻋﻠﻴﻪﻫﺎﻱ 420ﻫﻢ ﻫﺴــﺘﻨﺪ .ﻳﺎ 630ﻳﺎ 840ﻳﺎ « ....ﻗﺮﺍﺭ ﺷﺪ ﺩﻭﺑﺎﺭﻩ ﺍﺯ ﺍﻭﻝ ﺷــﺮﻭﻉ ﻛﻨﻴﻢ .ﻣﻌﻠــﻢ ﺑﻪ ﻣﺎ ﻳﺎﺩ ﺩﺍﺩﻩ ﺑﻮﺩ ﻛﻪ ﻫﺮﺟﺎ ﺑﻪ ﻣﺸــﻜﻞ ﺑﺮﺧﻮﺭﺩﻳﻢ ﻳﻚ ﻗﺪﻡ ﺑﻪ ﻋﻘﺐ ﺑﺮﮔﺮﺩﻳﻢ ﻭ ﺍﺯ ﻧﻮ ﺷــﺮﻭﻉ ﻛﻨﻴﻢ )ﺑﺎﺯﮔﺸﺖ ﺑﻪ ﻋﻘﺐ ﻭ ﺷﺮﻭﻉ ﺩﻭﺑﺎﺭﻩ!( ﻣﻦ ﮔﻔﺘﻢ» :ﻣﮕﻪ 1ﻣﻘﺴــﻮﻡ ﻋﻠﻴﻪ ﻫﻤﻪﻱ ﺍﻋﺪﺍﺩ ﻧﻴﺴﺖ؛ ﺧﻮﺏ 1
ﺭﺍ ﺑﻨﻮﻳﺴﻴﻢ«. ﺳــﻌﻴﺪ ﮔﻔﺖ ،15=3×5 ، 14=2×7» :ﭘﺲ ﻣﻌﻠﻮﻡ ﻣﻴﺸﻪ ،3 ،2 5ﻭ 7ﻫﻢ ﻣﻘﺴﻮﻡﻋﻠﻴﻪﻫﺎﻱ ﺁﻥ ﻋﺪﺩ ﻫﺴﺘﻨﺪ«. ﻣﺤﻤﺪ ﺑﺎ ﻧﺸﺎﻥ ﺩﺍﺩﻥ ﻧﻤﻮﺩﺍﺭ ﺩﺭﺧﺘﻲ ﺍﻋﺪﺍﺩ 14ﻭ 15ﮔﻔﺖ» :ﻣﻦ ﻫﻢ ﺑﻪ ﺍﻳﻦ ﻧﺘﻴﺠﻪ ﺭﺳﻴﺪﻡ ﻛﻪ 5 ،3 ،2ﻭ 7ﻣﻘﺴﻮﻡ ﻋﻠﻴﻪﻫﺎﻱ ﺁﻥ ﻋﺪﺩ ﻫﺴﺘﻨﺪ«. 15 14 3
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ﺗﺎ ﺍﻳﻦﺟﺎ ﻫﻤﻪﭼﻴﺰ ﺩﺍﺷﺖ ﺧﻮﺏ ﭘﻴﺶ ﻣﻲﺭﻓﺖ .ﻣﺎ ﺗﻮﺍﻧﺴﺘﻪ ﺑﻮﺩﻳﻢ ﭼﻬــﺎﺭ ﺗﺎ ﻋﺪﺩ ﺍﻭﻝ ﭘﻴﺪﺍ ﻛﻨﻴﻢ ﻛﻪ ﺩﺭ ﺳــﺎﺧﺘﻦ ﻋــﺪﺩ ﻣﻮﺭﺩﻧﻈﺮ ﺑﻪ ﻛﺎﺭ ﺭﻓﺘﻪ ﺑﻮﺩﻧﺪ. ﻣــﻦ ﻳﺎﺩﻡ ﺁﻣﺪ ﻛﻪ ﻣﻌﻠﻢ ﻳﻚ ﺭﻭﺯ ﺩﺭ ﻣﻮﺭﺩ ﻛﺎﺭﺧﺎﻧﻪﻱ ﻋﺪﺩﺳــﺎﺯﻱ ﺑﺮﺍﻱ ﻣﺎ ﺻﺤﺒﺖ ﻛﺮﺩﻩ ﺑﻮﺩ. ﺑﺮﺍﻱ ﻣﺜﺎﻝ ،ﺍﮔﺮ ﻣﺎﺩﻩﻱ ﺍﻭﻟﻴﻪﻱ ﻛﺎﺭﺧﺎﻧﻪﻱ ﻋﺪﺩﺳﺎﺯﻱ ،ﻋﺪﺩ ﺍﻭﻝ 2 ﺩﻭﺭﺓ ﺷﺎﻧﺰﺩﻫﻢ ،ﺷﻤﺎﺭﺓ ،4ﺗﺎﺑﺴﺘﺎﻥ 1390
ﺭﺍﻫﻨﻤﺎﻳﻲ
27
ﺑﺎﺷﺪ ﻣﺤﺼﻮﻻﺕ ﺍﻳﻦ ﻛﺎﺭﺧﺎﻧﻪ ﺑﻪ ﺷﻜﻞ ﺯﻳﺮ ﺗﻮﻟﻴﺪ ﻣﻲﺷﻮﻧﺪ: 2 2×2 2×2×2 2×2×2×2 ...
2 ، 4 ، 8 ،16 ، ...
ﻭ ﺍﮔــﺮ ﻣــﺎﺩﻩﻱ ﺍﻭﻟﻴﻪﻱ ﻛﺎﺭﺧﺎﻧﻪﻱ ﻋﺪﺩﺳــﺎﺯﻱ ،ﺍﻋﺪﺍﺩ ﺍﻭﻝ 2ﻭ 3 ﺑﺎﺷﻨﺪ ،ﻣﺤﺼﻮﻻﺕ ﻛﺎﺭﺧﺎﻧﻪ ﺑﻪ ﺷﻜﻞ ﺯﻳﺮ ﺗﻮﻟﻴﺪ ﻣﻲﺷﻮﻧﺪ:
2×3 2×3×3 2×3×3×3 2×2×3 2×2×3×3 2×2×3×3×3 2×3×2×3 2×2×2×3×3 2×2×2×3×3×3 ...
...
...
... ... ...
6 ،12 ،24 ، . . . ، 18 ،36 ، 72، . . .
ﺑﻌﺪ ﺍﺯ ﺍﻳﻦ ﻳﺎﺩﺁﻭﺭﻱ ﺑﺮﻣﻲﮔﺮﺩﻳﻢ ﺑﻪ ﺣﻞ ﻣﺴﺌﻠﻪﻱ ﺧﻮﺩﻣﺎﻥ. ﻣﺎ ﺑﺎ ﻳﻚ ﻛﺎﺭﺧﺎﻧﻪﺍﻱ ﺳﺮ ﻭ ﻛﺎﺭ ﺩﺍﺭﻳﻢ ﻛﻪ ﻣﻮﺍﺩ ﺍﻭﻟﻴﻪﻱ ﺁﻥ ﺍﻋﺪﺍﺩ ﺍﻭﻝ 5 ،3 ،2ﻭ 7ﻫﺴﺘﻨﺪ ﻭ ﻣﺤﺼﻮﻻﺕ ﺁﻥ ﺑﻪ ﺷﻜﻞ ﺯﻳﺮ ﺗﻮﻟﻴﺪ ﻣﻲﺷﻮﻧﺪ:
...
...
7
28
5
ﺭﺍﻫﻨﻤﺎﻳﻲ
ﺩﻭﺭﺓ ﺷﺎﻧﺰﺩﻫﻢ ،ﺷﻤﺎﺭﺓ ،4ﺗﺎﺑﺴﺘﺎﻥ 1390
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ﺩﺭ ﺍﺩﺍﻣــﻪﻱ ﺻﺤﺒﺖﻫــﺎﻱ ﻣــﻦ ﻣﺤﻤﺪ ﮔﻔﺖ ﻛــﻪ ﺩﺭ ﺗﻮﻟﻴﺪ ﻋﺪﺩ ﻣﻮﺭﺩﻧﻈــﺮ ﺍﺯ ﻣﻘﺴــﻮﻡ ﻋﻠﻴﻪﻫــﺎﻱ ﺍﻭﻝ 5 ،3 ،2ﻭ 7ﺣﺪﺍﻗــﻞ ﻳﻚ ﺑﺎﺭ ﺍﺳﺘﻔﺎﺩﻩ ﺷﺪﻩ ﺍﺳﺖ ﻭ ﺑﺎ ﻧﺸﺎﻥ ﺩﺍﺩﻥ ﺷﻜﻞ ﺯﻳﺮ ﺑﺮﺍﻱ ﻣﺎ ﺗﻮﺿﻴﺢ ﺩﺍﺩ ﻛﻪ ﭼﮕﻮﻧﻪ ﺗﻮﺍﻧﺴﺘﻪ ﺍﺳﺖ ﻣﻘﺴﻮﻡ ﻋﻠﻴﻪﻫﺎﻱ ﺩﻳﮕ ِﺮ ﻋﺪﺩ ﺭﺍ ﺑﻪ ﺩﺳﺖ ﺁﻭﺭﺩ:
3
27 3
...
.... .... .... ....
2×2×2×3×5×7 2×3×3×3×5×7 2×3×5×5×5×7 2×3×5×7×7×7
2×2×3×5×7 2×3×3×5×7 2×3×5×5×7 2×3×5×7×7
2×3×5×7
ﺳــﻌﻴﺪ ﻫﻢ ﻛﻪ ﺑﻴﻜﺎﺭ ﻧﺸﺴﺘﻪ ﺑﻮﺩ ،ﻧﺸﺎﻥ ﺩﺍﺩ ﻛﻪ ﺑﻪ ﻛﻤﻚ ﺟﺪﻭﻝ ﻧﻈﺎﻡﺩﺍﺭ ،ﻫﻤﻴﻦ ﻣﻘﺴﻮﻡ ﻋﻠﻴﻪﻫﺎ ﺭﺍ ﺑﻪ ﺩﺳﺖ ﺁﻭﺭﺩﻩ ﺍﺳﺖ: 2 2×3 2×3×5 2×3×5×7 3 2×5 2×3×7 5 2×7 2×5×7 7 3×5 3×5×7 3×7 5×7 ﻭ ﺍﻟﺒﺘﻪ ﻋﺪﺩ 1ﻛﻪ ﺩﺭ ﺷﺮﻭﻉ ﺣﻞ ﻣﺴﺌﻠﻪ ﺁﻥﺭﺍ ﻧﻮﺷﺘﻪ ﺑﻮﺩﻳﻢ. ﺍﻳﻦﺟﺎ ﺑﻮﺩ ﻛﻪ ﻣﻦ ﺑﺎ ﺑﻠﻨﺪ ﻛﺮﺩﻥ ﺩﺳــﺖ ﺍﺯ ﻣﻌﻠﻢ ﺧﻮﺍﺳﺘﻢ ﺩﺭﺳﺘﻲ ﺟﻮﺍﺏ ﻣﺎ ﺭﺍ ﺑﺮﺭﺳﻲ ﻛﻨﺪ. ﺁﻗﺎﻱ ﻣﻌﻠﻢ ﺑﻌﺪﺍﺯ ﺍﻳﻦﻛﻪ ﻣﺎ ﺭﺍ ﺗﺸــﻮﻳﻖ ﻛﺮﺩ ،ﮔﻔﺖ :ﺣﺎﻻ ﺑﻪ ﻋﻨﻮﺍﻥ ﺟﺎﻳﺰﻩ ﺑﻪ ﺍﻳﻦ ﺳﺆﺍﻝ ﭘﺎﺳﺦ ﺩﻫﻴﺪ: »ﺍﮔﺮ 6ﻭ 9ﺩﻭ ﻣﻘﺴــﻮﻡ ﻋﻠﻴﻪ ﻳﻚ ﻋﺪﺩ ﺑﺎﺷﻨﺪ ،ﻣﻘﺴﻮﻡ ﻋﻠﻴﻪﻫﺎﻱ ﺩﻳﮕﺮ ﺁﻥ ﺭﺍ ﺑﻨﻮﻳﺴﻴﺪ«. ﻣﺎ ﺩﻭﺑﺎﺭﻩ ﻣﺸﻐﻮﻝ ﻧﻮﺷﺘﻦ ﺟﻮﺍﺏ ﺷﺪﻳﻢ. ﺑﻌــﺪ ﺍﺯ ﻣﺪﺕ ﻛﻮﺗﺎﻫﻲ ﻣﺤﻤﺪ ﺟﻮﺍﺑﺶ ﺭﺍ ﺑﺎ ﻣﺎ ﻧﺸــﺎﻥ ﺩﺍﺩ .ﺍﻭ ﺍﻭﻝ 6ﻭ 9ﺭﺍ ﺩﺭ ﻫــﻢ ﺿﺮﺏ ﻛﺮﺩ (6×9=54) .ﻭ ﺳــﭙﺲ ﺑﻪ ﻛﻤﻚ ﻧﻤﻮﺩﺍﺭ ﺩﺭﺧﺘﻲ ،ﻣﻘﺴــﻮﻡ ﻋﻠﻴﻪﻫﺎﻱ ﺍﻭﻝ ﻋﺪﺩ 54ﺭﺍ ﺑﻪ ﺩﺳــﺖ ﺁﻭﺭﺩﻩ ﺑﻮﺩ ﻛﻪ ﻋﺒﺎﺭﺕ ﺑﻮﺩﻧﺪ ﺍﺯ 2ﻭ .3 54
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ﺑﻌﺪ ﺑﻪ ﻛﻤﻚ ﻛﺎﺭﺧﺎﻧﻪﻱ ﻋﺪﺩﺳــﺎﺯﻱ ﻭ ﺑﺎ ﺗﻮﺟﻪ ﺑﻪ ﺍﻳﻦﻛﻪ ﺍﺯ ﻋﺪﺩ 2ﺑﻴﺶ ﺍﺯ ﻳﻚ ﺑﺎﺭ ﻭ ﺍﺯ ﻋﺪﺩ 3ﺑﻴﺶ ﺍﺯ ﺳﻪ ﺑﺎﺭ ﺍﺳﺘﻔﺎﺩﻩ ﻧﺸﻮﺩ ،ﻣﻘﺴﻮﻡ ﻋﻠﻴﻪﻫﺎﻱ ﺩﻳﮕﺮ ﻋﺪﺩ ﻣﻮﺭﺩﻧﻈﺮ ﺭﺍ ﺑﻪ ﺷﻜﻞ ﺯﻳﺮ ﺗﻮﻟﻴﺪ ﻛﺮﺩﻩ ﺑﻮﺩ: 1 2 3 2×3 3×3 2×3×3 3×3×3 2×3×3×3 ﻭ ﺍﻟﺒﺘﻪ ﻋﺪﺩ ﻳﻚ ﺭﺍ ﻫﻢ ﻓﺮﺍﻣﻮﺵ ﻧﻜﺮﺩﻩ ﺑﻮﺩ! ﻣﻦ ﻫﻨﮕﺎﻡ ﺩﻳﺪﻥ ﺟﻮﺍﺏ ﻣﺤﻤﺪ ﺑﻪ ﻳﺎﺩ ﺣﺮﻑ ﺳــﻌﻴﺪ ﺩﺭ ﻣﻮﺭﺩ 14 ﻭ 15ﺩﺭ ﻣﺴــﺌﻠﻪﻱ ﻗﺒﻞ ﺍﻓﺘﺎﺩﻡ ﻭ ﺑﻪ ﻣﺤﻤــﺪ ﮔﻔﺘﻢ» :ﻣﺤﻤﺪ ﺟﺎﻥ ،ﺍﺯ ﻛﺠﺎ ﻣﻌﻠﻮﻡ ﻛﻪ 54ﻋﺪﺩ ﻣﻮﺭﺩﻧﻈﺮ ﺑﺎﺷــﺪ؟ ﺷــﺎﻳﺪ ﻋــﺪﺩ ﻣﻮﺭﺩﻧﻈﺮ 18
ﺑﺎﺷــﺪ ،ﭼﻮﻥ 6ﻭ 9ﻣﻘﺴﻮﻡ ﻋﻠﻴﻪﻫﺎﻱ 18ﻫﺴﺘﻨﺪ ﻳﺎ ﺷﺎﻳﺪ 36ﻳﺎ 54 ﻳﺎ 72ﻳﺎ « .... ﺳــﻌﻴﺪ ﺑﺎ ﺗﻜﺎﻥ ﺩﺍﺩﻥ ﺳــﺮﺵ ﺩﺭ ﺗﺄﻳﻴﺪ ﺣﺮﻑ ﻣﻦ ،ﻧﻮﺷﺘﻪﻫﺎﻳﺶ ﺭﺍ ﺑﻪ ﻣﺎ ﻧﺸــﺎﻥ ﺩﺍﺩ ﻭ ﮔﻔﺖ ﻛﻪ ﭼﺮﺍ 54ﻣﻤﻜﻦ ﺍﺳــﺖ ﻋﺪﺩ ﻣﻮﺭﺩﻧﻈﺮ ﻧﺒﺎﺷﺪ. 6 9 3
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3
2
ﺍﻭ ﺍﺑﺘﺪﺍ ﻧﻤﻮﺩﺍﺭ ﺩﺭﺧﺘﻲ 6ﻭ 9ﺭﺍ ﺭﺳﻢ ﻛﺮﺩﻩ ﻭ ﻣﻌﻠﻮﻡ ﺷﺪﻩ ﺑﻮﺩ ﻛﻪ 3ﻣﻘﺴﻮﻡ ﻋﻠﻴﻪ ﻣﺸﺘﺮﻙ 6ﻭ 9ﺍﺳﺖ ﻭ ﺑﻪ ﻫﻤﻴﻦ ﺩﻟﻴﻞ ﺣﺎﺻﻞ ﺿﺮﺏ 6ﻭ 9ﻳﻌﻨﻲ 54ﻣﻤﻜﻦ ﺍﺳﺖ ﻋﺪﺩ ﻣﻮﺭﺩﻧﻈﺮ ﻧﺒﺎﺷﺪ. ﺑﻌﺪ ﺍﺯ ﺍﻳﻦ ﺗﻮﺿﻴﺢ ﻛﻮﺗﺎﻩ ﺳــﻌﻴﺪ ﻣﻦ ﺍﻳﻦﻃﻮﺭ ﺍﺩﺍﻣﻪ ﺩﺍﺩﻡ ﻛﻪ ﺍﻭﻝ ﻧﻤﻮﺩﺍﺭ ﺩﺭﺧﺘﻲ 6ﻭ 9ﺭﺍ ﺭﺳــﻢ ﻛﺮﺩﻩﺍﻡ ﻭ ﺑﻌﺪ ﺍﺯ ﺭﻭﻱ ﻧﻤﻮﺩﺍﺭ ﻣﺘﻮﺟﻪ ﺷــﺪﻡ ﻛﻪ 3ﺩﺭ ﺍﻳﻦ ﺩﻭ ﺩﺭﺧﺖ ﻣﻴﻮﻩﻱ ﻣﺸﺘﺮﻙ ﺍﺳﺖ ،ﭘﺲ ﺩﺭ ﺗﻮﻟﻴﺪ ﻋــﺪﺩ ﻣﻮﺭﺩﻧﻈــﺮ ﺍﺯ ﺑﻴﺶ ﺍﺯ ﻳــﻚ 2ﻭ ﺩﻭ ﺗﺎ 3ﺍﺳــﺘﻔﺎﺩﻩ ﻧﻤﻲﻛﻨﻢ ﻭ ﺑﻪ ﻛﻤﻚ ﻛﺎﺭﺧﺎﻧﻪﻱ ﻋﺪﺩﺳــﺎﺯﻱ ﻭ ﺟــﺪﻭﻝ ﻧﻈﺎﻡﺩﺍﺭ ﺑﻪ ﺟﻮﺍﺏ ،ﻳﻌﻨﻲ ﻣﻘﺴﻮﻡ ﻋﻠﻴﻪﻫﺎﻱ ﺩﻳﮕﺮ ﻋﺪﺩ ﻣﻮﺭﺩﻧﻈﺮ ،ﺭﺳﻴﺪﻩﺍﻡ. 9 3
6 3
ﻋﻠﻴﻪ ﻣﺸــﺘﺮﻙ 6ﻭ 9ﺑﺎ ﺭﺳﻢ ﻧﻤﻮﺩﺍﺭ ﻣﻘﺴﻮﻡ ﻋﻠﻴﻪ ﺑﻪ ﺟﻮﺍﺏ ﺭﺳﻴﺪﻩ ﺑﻮﺩ! ﻣﺎ ﻫﻤﻴﺸــﻪ ﺩﺭ ﺭﺳﻢ ﻧﻤﻮﺩﺍﺭ ﻣﻘﺴﻮﻡ ﻋﻠﻴﻪ ﻳﻚ ﻋﺪﺩ ﺍﺯ ﺧﻮﺩ ﻋﺪﺩ 3
6
12
1
2
4
÷2 ÷3
ﺷــﺮﻭﻉ ﻣﻲﻛﻨﻴﻢ ﺗﺎ ﺑﻪ ﻳﻚ ﺑﺮﺳﻴﻢ .ﺑﺮﺍﻱ ﻣﺜﺎﻝ ﺩﺭ ﻣﻮﺭﺩ ﻋﺪﺩ 12ﺑﻌﺪ ﺍﺯ ﺍﻳﻦﻛﻪ ﻓﻬﻤﻴﺪﻳﻢ ﻣﻘﺴﻮﻡ ﻋﻠﻴﻪﻫﺎﻱ ﺍﻭﻝ ،12ﺍﻋﺪﺍﺩ 2ﻭ 3ﻫﺴﺘﻨﺪ، ﻧﻤﻮﺩﺍﺭ ﺭﺍ ﺍﺯ ﮔﻮﺷﻪﻱ ﺑﺎﻻ ﺳﻤﺖ ﭼﭗ ﻭ ﺍﺯ ﺧﻮﺩ ﻋﺪﺩ ﻳﻌﻨﻲ 12ﺷﺮﻭﻉ ﻭ ﺑﻪ ﮔﻮﺷﻪﻱ ﭘﺎﻳﻴﻦ ﺳﻤﺖ ﺭﺍﺳﺖ ﻣﻲﺭﻭﻳﻢ ﺗﺎ ﺑﻪ ﻋﺪﺩ 1ﺑﺮﺳﻴﻢ. ﻭ ﻣﻲﺩﺍﻧﻴﻢ ﺍﻋﺪﺍﺩ ﻧﻮﺷﺘﻪ ﺷﺪﻩ ﺩﺭ ﻧﻤﻮﺩﺍﺭ ،ﻳﻌﻨﻲ 6 ،3 ،4 ،2 ،1ﻭ 12ﻣﻘﺴﻮﻡ ﻋﻠﻴﻪﻫﺎﻱ ﻋﺪﺩ 12ﻫﺴﺘﻨﺪ. ﺍﻣﺎ ﺩﺭ ﺣﻞ ﺍﻳﻦ ﻣﺴﺌﻠﻪ ،ﺳﻌﻴﺪ ﺍﺯ ﮔﻮﺷﻪﻱ ﭘﺎﻳﻴﻦ ﺳﻤﺖ ﭼﭗ ﻭ ﺍﺯ ﻋﺪﺩ 1ﺷﺮﻭﻉ ﻛﺮﺩﻩ ﻭ ﺑﻪ ﺳﻤﺖ ﺑﺎﻻ ﻭ ﺑﻪ ﺳﻤﺖ ﭼﭗ ﺭﻓﺘﻪ ﺑﻮﺩ. ﺑﺎ ﻧﮕﺎﻩ ﻛﺮﺩﻥ ﺑﻪ ﻧﻤﻮﺩﺍﺭ ﻣﻌﻠﻮﻡ ﻣﻲﺷﺪ ﻛﻪ ﻫﺮﻳﻚ ﺍﺯ ﺍﻋﺪﺍﺩ ﺑﻌﺪ ﺍﺯ 18ﭼﻪ ﺍﺯ ﺳــﻤﺖ ﭼﭗ ﻭ ﭼﻪ ﺭﻭﺑﻪ ﺑﺎﻻ ﻣﻲﺗﻮﺍﻧﺪ ﻋﺪﺩ ﻣﻮﺭﺩﻧﻈﺮ ﺑﺎﺷﺪ، ﻭﻟﻲ 18ﻛﻮﭼﻚﺗﺮﻳﻦ ﻭ ﻟﺬﺍ ﻣﻄﻤﺌﻦﺗﺮﻳﻦ ﺟﻮﺍﺏ ﺑﺮﺍﻱ ﺳﺆﺍﻝ ﺍﺳﺖ.
3
2 3
2
2×3 2×3×3 3×3 1 ﺳﻌﻴﺪ ﻭ ﻣﺤﻤﺪ ،ﻫﺮ ﺩﻭ ﺣﺮﻑ ﻣﺮﺍ ﺗﺄﻳﻴﺪ ﻛﺮﺩﻧﺪ. ﺳــﻌﻴﺪ ﺑﺎ ﻟﺒﺨﻨﺪ ﻣﻌﻨﺎﺩﺍﺭﻱ ﻛﻪ ﻧﺸــﺎﻥ ﻣــﻲﺩﺍﺩ ﺭﺍﻩ ﺟﺪﻳﺪﻱ ﺑﻪ ﺫﻫﻨﺶ ﺭﺳــﻴﺪﻩ ﺍﺳــﺖ ،ﺑﻌﺪ ﺍﺯ ﻣﺪﺕ ﻛﻮﺗﺎﻫﻲ ﻧﻮﺷﺘﻪﻱ ﺧﻮﺩ ﺭﺍ ﺑﻪ ﻣﺎ ﻧﺸﺎﻥ ﺩﺍﺩ. ﺍﻭ ﭘﺲ ﺍﺯ ﺭﺳﻢ ﻧﻤﻮﺩﺍﺭ ﺩﺭﺧﺘﻲ ﻭ ﻭﺟﻮﺩ ﻋﺪﺩ 3ﺑﻪ ﻋﻨﻮﺍﻥ ﻣﻘﺴﻮﻡ
27
54
108
....
....
9
18
36
72
....
3
6
12
24
....
1
2
4
8
....
ﻭ ﺍﺯ ﺭﻭﻱ ﻧﻤﻮﺩﺍﺭ ،ﻣﻘﺴﻮﻡﻋﻠﻴﻪﻫﺎﻱ 18ﻳﺎ ﻫﻤﺎﻥ ﻋﺪﺩ ﻣﻮﺭﺩﻧﻈﺮ ﺭﺍ ﻣﻲﺗﻮﺍﻥ ﻧﻮﺷﺖ ،ﻳﻌﻨﻲ: 18ﻭ 9ﻭ 6ﻭ 3ﻭ 2ﻭ 1 ﺍﻳﻦﺑﺎﺭ ﻣﻦ ﺩﻭﺑﺎﺭﻩ ﺩﺳﺘﻢ ﺭﺍ ﺑﻠﻨﺪ ﻛﺮﺩﻡ ﻭ ﺑﻪ ﺁﻗﺎﻱ ﻣﻌﻠﻢ ﮔﻔﺘﻢ ﻛﻪ ﻣﺎ ﺑﻪ ﺟﻮﺍﺏ ﺭﺳﻴﺪﻩﺍﻳﻢ. ﻟﺒﺨﻨﺪ ﺭﺿﺎﻳﺖ ﻣﻌﻠﻢ ﻧﺸــﺎﻥ ﻣﻲﺩﺍﺩ ﻛﻪ ﻫﺮﺩﻭ ﺭﺍﻩ ﻣﺎ ﺩﺭﺳﺖ ﺍﺳﺖ ﻭ ﺑﺮﻕ ﺷﺎﺩﻱ ﺩﺭ ﭼﺸﻤﺎﻥ ﻣﺎ ﺩﻳﺪﻩ ﻣﻲﺷﺪ. ﺩﻭﺭﺓ ﺷﺎﻧﺰﺩﻫﻢ ،ﺷﻤﺎﺭﺓ ،4ﺗﺎﺑﺴﺘﺎﻥ 1390
ﺭﺍﻫﻨﻤﺎﻳﻲ
29
ﺩﺍﻧﺶﺍﻓﺰﺍﻳﻲ
واژه﹨︀ی ر︀︲﹩ »︨︀ده ﹋︣دن ︻︊︀رت«︀︧︑» ،وی =« ﺳﭙﻴﺪﻩ ﭼﻤﻦﺁﺭﺍ
ﻛﻠﻴﺪﻭﺍژﻩﻫﺎ :ﻭﺍژﻩﻫﺎﻱ ﺭﻳﺎﺿﻲ ،ﺳﺎﺩﻩ ﻛﺮﺩﻥ ﻋﺒﺎﺭﺕ ،ﺗﺴﺎﻭﻱ = . ﺩﺭ ﺍﻳــﻦ ﺑﺨﺶ ﺍﺯ ﻭﺍژﻩﻫﺎﻱ ﺭﻳﺎﺿﻲ ﺑﻪ ﺩﻭ ﻣﻮﺿﻮﻉ ﻣﻲﭘﺮﺩﺍﺯﻳﻢ :ﻳﻜﻲ ﺍﺻﻄﻼﺡ »ﺳــﺎﺩﻩ ﻛﺮﺩﻥ ﻋﺒــﺎﺭﺕ« ﻭ ﺩﻳﮕﺮﻱ ﻣﻌﻨﺎﻫﺎﻱ ﻣﺨﺘﻠﻒ ﻣﻌﻨﺎﻱ ِ ﻋﻼﻣﺖ.« = » : ﻫﻤﻪﻱ ﻣﺎ ﺍﻭﻟﻴﻦ ﺑﺎﺭ ﺍﺻﻄﻼﺡ »ﺳــﺎﺩﻩ ﻛﺮﺩﻥ« ﺭﺍ ﺩﺭ ﺩﺑﺴــﺘﺎﻥ ﻭ ﺩﺭ ﻛﺎﺭ ﺑﺎ ﻛﺴﺮﻫﺎﻱ ﻣﺴﺎﻭﻱ ﺷﻨﻴﺪﻩﺍﻳﻢ: »ﻛﺴﺮﻫﺎﻱ ﺯﻳﺮ ﺭﺍ ﺗﺎ ﺣﺪ ﺍﻣﻜﺎﻥ ﺳﺎﺩﻩ ﻛﻨﻴﺪ: 4 8
,
3 15
,
21 56
.... ,
ﺗﻮﺟــﻪ ﻛﻨﻴﻢ ﻛﻪ ﺑﺎ ﺗﻮﺟﻪ ﺑﻪ ﻣﺤﻞ » =« ،ﺧﻂ ﻛﺴــﺮﻱ ﻣﻘﺎﺑﻞ ﻋﻼﻣﺖ = ،ﺧﻂ ﻛﺴــﺮﻱ ﺍﺻﻠﻲ ﺍﺳــﺖ؛ ﻭﻟــﻲ ﺻﻮﺭﺕ ﻭ ﻣﺨﺮﺝ ﺁﻥ ،ﺧﻮﺩﺷــﺎﻥ ﻋﺒﺎﺭﺕﻫﺎﻱ ﺭﻳﺎﺿﻲ ﻫﺴــﺘﻨﺪ ﻭ ﻳﻚ ﻋﺪﺩ ﻃﺒﻴﻌﻲ ﻧﻴﺴــﺘﻨﺪ ﻛﻪ »ﻛﺴﺮ ﺭﺍ ﺳــﺎﺩﻩ ﻛﻨﻴﻢ« .ﺑﻨﺎﺑﺮﺍﻳﻦ ﺑﺎﻳﺪ ﻧﺨﺴــﺖ ﻋﺒﺎﺭﺕِ ) (1ﺭﺍ ﻣﺤﺎﺳﺒﻪ ﻛﻨﻴﻢ ﻭ ﺣﺎﺻــﻞ ﺁﻥ ﺭﺍ ﻛــﻪ ﺍﺣﺘﻤﺎﻻً ﻳﻚ ﻋﺪﺩ ﻛﺴــﺮﻱ ﺍﺳــﺖ ،ﺑﻴﺎﺑﻴﻢ ﻭ ﺁﻥ ﻛﺴــﺮ ﺭﺍ ﺗﺎ ﺣﺪ ﺍﻣﻜﺎﻥ ﺳـﺎﺩﻩ ﻛﻨﻴﻢ .ﭘﺲ ﺩﺭ ﺍﻳﻦﺟﺎ ،ﺳـﺎﺩﻩ ﻛﺮﺩﻥ، ﺷــﺎﻣﻞ ﻣﺤﺎﺳﺒﺎﺕ ﻭ ﺳﺎﺩﻩ ﻛﺮﺩﻥ ﻛﺴﺮ ﺣﺎﺻﻞ ﻣﻲﺷﻮﺩ ﻛﻪ ﻗﺪﺭﻱ ﺑﺎ ﻣﻌﻨﺎﻱ ﻗﺒﻠﻲ ﺗﻔﺎﻭﺕ ﺩﺍﺭﺩ .ﺑﮕﺬﺍﺭﻳﺪ ﻋﻤﻠﻴﺎﺕ ﺯﻳﺮ ﺭﺍ ﺍﻧﺠﺎﻡ ﺩﻫﻴﻢ:
ﻭ ﻫﻤﻪ ﺑﻪ ﺧﻮﺑﻲ ﻣﻲﺩﺍﻧﻴﺪ ﻛﻪ »ﺳــﺎﺩﻩ ﻛــﺮﺩﻥ« ﺩﺭ ﺍﻳﻦ ﺟﺎ ،ﻳﻌﻨﻲ ﻳﺎﻓﺘﻦ ﻛﺴـﺮﻱ ﻣﺴـﺎﻭﻱ ﺑﺎ ﻛﺴـﺮ ﻣﻮﺭﺩﻧﻈﺮ ﻛﻪ ﺑﺰﺭگﺗﺮﻳﻦ ﻣﻘﺴﻮﻡ ﻋﻠﻴﻪ ﻣﺸﺘﺮﻙ ﺻﻮﺭﺕ ﻭ ﻣﺨﺮﺝ ﺁﻥ ،ﻋﺪﺩ 1ﺑﺎﺷﺪ .ﻳﻌﻨﻲ ﺻﻮﺭﺕ ﻣﺨﺮﺝ ﺁﻥ ﻛﺴــﺮ ﻫﻢﺯﻣﺎﻥ ﺑﻪ ﻫﻴﭻ ﻋﺪﺩﻱ ﺟﺰ 1ﺗﻘﺴﻴﻢﭘﺬﻳﺮ ﻧﺒﺎﺷﻨﺪ .ﺣﺘﻲ ﺩﺭ ﺗﺎ ﺍﻳﻦﺟﺎ ﺑﺎ ﺍﻧﺠﺎﻡ ﻣﺤﺎﺳــﺒﺎﺕ ﻭ ﭼﻬﺎﺭ ﻋﻤﻞ ﺍﺻﻠﻲ ،ﺣﺎﺻﻞ ﻋﺒﺎﺭﺕ ﻛﺘﺎﺏ ﺭﻳﺎﺿﻲ ﭘﺎﻳﻪﻱ ﺍﻭﻝ ﺭﺍﻫﻨﻤﺎﻳﻲ ،ﺗﻤﺮﻳﻦﻫﺎﻳﻲ ﺑﻪ ﺍﻳﻦ ﺻﻮﺭﺕ ﺩﺍﺭﻳﻢ 28 ﺍﺳﺖ .ﺣﺎﻝ ﺍﻳﻦ ﻛﺴﺮ ﺭﺍ )ﺑﻪ ﻣﻌﻨﺎﻱ ﺍﻭﻝ( ﺭﺍ ﭘﻴﺪﺍ ﻛﺮﺩﻳﻢ ﻛﻪ ﻛﺴ ِﺮ ﻛﻪ ﺑﻪ ﻃﻮﺭ ﻣﺴﺘﻘﻴﻢ ﺑﻪ ﺍﻳﻦ ﻣﻮﺿﻮﻉ ﺍﺷﺎﺭﻩ ﻣﻲﻛﻨﺪ: 30 ﻛﺴــﺮﻫﺎﻱ ﺯﻳــﺮ ﺭﺍ ﺑﺎ ﺍﺳــﺘﻔﺎﺩﻩ ﺍﺯ ﺏ.ﻡِ .ﻡ ﺻﻮﺭﺕ ﻭ ﻣﺨﺮﺝ ،ﺳــﺎﺩﻩ 28 14 ﺳﺎﺩﻩ ﻣﻲﻛﻨﻴﻢ: = 30 15 ﻛﻨﻴﺪ: 6 3 4 7 1 2 1− + − + 2 3 = 6 6 6 = 6 = 7 ÷ 5 = 7 × 4 = 28 1 5 3 6 4 6 5 30 1 2− 4 4 4
51
121
14
.... , , ﺍﺳﺖ. ﭘﺲ ﭘﺎﺳﺦ ﺳﺆﺍﻝ ﻓﻮﻕ، 99 68 15 ﺍ ّﻣــﺎ ﻫﻨﻮﺯ ﻣﺎﺟﺮﺍ ﺗﻤﺎﻡ ﻧﺸــﺪﻩ ﺍﺳــﺖ .ﺯﻣﺎﻧﻲ ﻛــﻪ ﺩﺭ ﭘﺎﻳﻪﻱ ﺩﻭﻡ ﺍ ّﻣﺎ ﺳـﺎﺩﻩ ﻛﺮﺩﻥ ﻛﺴـﺮ ﻭ ﻳﺎﻓﺘﻦ ﻛﺴﺮﻫﺎﻱ ﻣﺴــﺎﻭﻱ ،ﺗﻨﻬﺎ ﺟﺎﻳﻲ ﺭﺍﻫﻨﻤﺎﻳﻲ ﺑﺎ ﻋﺒﺎﺭﺕﻫﺎﻱ ﺟﺒﺮﻱ ﺁﺷــﻨﺎ ﻣﻲﺷﻮﻳﻢ ﺍﺯ ﻣﺎ ﻣﻲﺧﻮﺍﻫﻨﺪ ﻛﻪ ﻧﻴﺴﺖ ﻛﻪ ﺑﻪ ﻣﺎ ﮔﻔﺘﻪ ﻣﻲﺷﻮﺩ» :ﺳﺎﺩﻩ ﻛﻨﻴﺪ« ! ﻋﺒﺎﺭﺕﻫﺎﻳﻲ ﻣﺎﻧﻨﺪ ﻋﺒﺎﺭﺕ ﺯﻳﺮ ﺭﺍ ﺳﺎﺩﻩ ﻛﻨﻴﻢ: ﺑﻪ ﻣﺜﺎﻝ ﺯﻳﺮ ﺗﻮﺟﻪ ﻛﻨﻴﺪ: = ) x + 2y − 1 + 3 x − yﺍﻟﻒ »ﻋﺒﺎﺭﺕ ﺯﻳﺮ ﺭﺍ ﺗﺎ ﺣﺪ ﺍﻣﻜﺎﻥ ﺳﺎﺩﻩ ﻛﻨﻴﺪ: = ) 3aa − 4b + 2a + b − 7ﺏ 1 2 1− + ﺍ ّﻣﺎ ﺩﻳﮕﺮ ﻧﻪ ﻛﺴﺮﻱ ﺍﺳﺖ ،ﻧﻪ ﺧﻂ ﻛﺴﺮﻱ! ﺩﺭ ﭼﻨﻴﻦ ﻋﺒﺎﺭﺕﻫﺎﻳﻲ، = 2 3 )(1 3 ﺗﻔﺮﻳﻖ ﺟﻤﻠﻪﻫﺎﻱ ﻣﺸﺎﺑﻪ ﺑﺎ ﻳﻜﺪﻳﮕﺮ ﻣﻨﻈﻮﺭ ﺍﺯ ﺳـﺎﺩﻩ ﻛﺮﺩﻥ ،ﺟﻤﻊ ﻳﺎ ِ 2− 4 ﺍﺳﺖ .ﺣﺎﻝ ﺑﻪ ﻣﺜﺎﻝ ﺯﻳﺮ ﺗﻮﺟﻪ ﻛﻨﻴﺪ: ﺑﻪ ﻇﺎﻫﺮ ﺧﻂﻫﺎﻱ ﻛﺴــﺮﻱ ﺯﻳﺎﺩﻱ ﺩﺭ ﺍﻳﻦ ﻋﺒﺎﺭﺕ ﻫﺴﺖ ،ﻭﻟﻲ ﺑﺎﻳﺪ ﻋﺒﺎﺭﺕﻫﺎﻱ ﺯﻳﺮ ﺭﺍ ﺳﺎﺩﻩ ﻛﻨﻴﺪ: 30
ﺭﺍﻫﻨﻤﺎﻳﻲ
ﺩﻭﺭﺓ ﺷﺎﻧﺰﺩﻫﻢ ،ﺷﻤﺎﺭﺓ ،4ﺗﺎﺑﺴﺘﺎﻥ 1390
1 3 2 1 x + 3y − 5 + − y + x 2 2 3 2
ﺍ ّﻣﺎ ﺑﺎﺯ ﺑﻪ ﺳــﺮﺍﻍ ﻋﺒﺎﺭﺕﻫﺎﻱ ﺟﺒﺮﻱ ﻣﻲﺭﻭﻳﻢ ﻭ ﺍﺯ ﻫﻤﺎﻥ ﻣﺜﺎﻝﻫﺎﻱ ﺑﺨﺶ ﻗﺒﻞ ﺍﺳﺘﻔﺎﺩﻩ ﻣﻲﻛﻨﻴﻢ:
ﺩﺭ ﺍﻳﻦ ﻋﺒﺎﺭﺕ ،ﻛﺴــﺮﻫﺎﻳﻲ ﻭﺟــﻮﺩ ﺩﺍﺭﺩ )ﻣﺜﻞ ( 2 ، 3 ، 1ﺍ ّﻣﺎ 2
2
3
ﺍﻳﻦﺟﺎ ﻣﻨﻈﻮﺭ ﺍﺯ ﺳــﺎﺩﻩ ﻛﺮﺩﻥ ،ﺳــﺎﺩﻩ ﻛﺮﺩﻥ ﻛﺴــﺮ ﻧﻴﺴﺖ ،ﺯﻳﺮﺍ ﻳﻚ ﻋﺒﺎﺭﺕ ﺟﺒﺮﻱ ﺍﺳﺖ ) (1ﻭ ﻟﺬﺍ ﺑﺎﻳﺪ ﺟﻤﻼﺕ ﻣﺸﺎﺑﻪ ﺭﺍ ﺗﺸﺨﻴﺺ ﺩﻫﻴﻢ ﻭ ﺁﻥﻫﺎ ﺭﺍ ﺑﺎ ﻫﻢ ﺟﻤﻊ ﻳﺎ ﺗﻔﺮﻳﻖ ﻛﻨﻴﻢ. 1 3 2 2 12 9 9 x + 3y − 5 + − y + x = x + y − 2 4 15 3 15 4 2
ﺧﻮﺏ ،ﺑﻪ ﻧﻈﺮ ﻣﻲﺭﺳــﺪ ﻛﻪ ﻛﺎﺭ ﺗﻤﺎﻡ ﺷﺪﻩ ﺍﺳــﺖ ،ﺍ ّﻣﺎ ﺍﮔﺮ ﺧﻮﺏ ﺩﻗﺖ ﻛﻨﻴﺪ ،ﺿﺮﻳﺐ xﻳﻌﻨﻲ 12ﻛﺴﺮﻱ ﺍﺳﺖ ﻛﻪ ﺳﺎﺩﻩ ﻣﻲﺷﻮﺩ :ﭘﺲ 15 ﺣﺎﺻﻞ ﻋﺒﺎﺭﺕ ﺭﺍ ﺑﻪ ﺻﻮﺭﺕ ﺯﻳﺮ ﻧﻴﺰ ﻣﻲﺗﻮﺍﻥ ﻧﻮﺷﺖ: 4 9 9 x+ y− 5 4 2
ﻟــﺬﺍ ﺩﺭ ﺍﻳﻦ ﻣﺜــﺎﻝ ،ﻋﻼﻭﻩ ﺑﺮ ﺳــﺎﺩﻩ ﻛــﺮﺩﻥ ﻋﺒﺎﺭﺕﻫﺎﻱ ﺟﺒﺮﻱ )ﺑﻪ ﻣﻌﻨﺎﻱ ﺟﻤﻊ ﻳﺎ ﺗﻔﺮﻳﻖ ﺟﻤﻼﺕ ﻣﺸــﺎﺑﻪ( ،ﺳــﺎﺩﻩ ﻛﺮﺩﻥ ﻛﺴﺮ ﻧﻴﺰ ﺩﺍﺷﺘﻴﻢ! ﺣﺎﻝ ﺷــﻤﺎ ﺑﮕﻮﻳﻴﺪ ﻛﻪ ﺩﺭ ﺳــﺆﺍﻝ ﺯﻳﺮ ،ﻣﻨﻈﻮﺭ ﺍﺯ »ﺳــﺎﺩﻩ ﻛﻨﻴﺪ« ﭼﻴﺴﺖ؟ ﻋﺒﺎﺭﺕ ﺯﻳﺮ ﺭﺍ ﺗﺎ ﺣﺪ ﺍﻣﻜﺎﻥ ﺳﺎﺩﻩ ﻛﻨﻴﺪ: 21× ( −14) × 55 × 4 ( −8) × ( −49) × 11
ﭘﺎﺳــﺦ :ﺑﻪ ﻣﻌﻨﺎﻱ ﺳﺎﺩﻩ ﻛﺮﺩﻥ ﻛﺴﺮ ﺍﺳﺖ ،ﺯﻳﺮﺍ ﻋﺒﺎﺭﺕ ﺩﺍﺩﻩ ﺷﺪﻩ ﺩﺭ ﻭﺍﻗﻊ ﻳﻚ ﻛﺴــﺮ ﺍﺳــﺖ ﻛﻪ ﺍﻋﺪﺍﺩ ﺻﻮﺭﺕ ﻭ ﻣﺨﺮﺝ ﺑﻪ ﺣﺎﺻﻞ ﺿﺮﺏ ﺍﻋﺪﺍﺩﻱ ﺩﻳﮕﺮ ﺗﺠﺰﻳﻪ ﺷــﺪﻩﺍﻧﺪ .ﺍﮔﺮ ﺑﺨﻮﺍﻫﻴﻢ ﺩﻗﻴﻖﺗﺮ ﺑﮕﻮﻳﻴﻢ ،ﺩﺭ ﻭﺍﻗﻊ ﻣﻌﻨﺎﻱ ﻣﺤﺎﺳﺒﻪ ﻭ ﺳﭙﺲ ﺳﺎﺩﻩ ﻛﺮﺩﻥ ﻛﺴﺮ ﺣﺎﺻﻞ ﺍﺳﺖ.
ﻋﻼﻣﺖ » =«
ﺣﺘﻤﺎً ﻫﻤﻪﻱ ﺷــﻤﺎ ﻋﻼﻣﺖ » =« ﺭﺍ ﻣﻲﺷﻨﺎﺳــﻴﺪ .ﺍﻳﻦ ﻋﻼﻣﺖ ﺭﺍ ﺍﻭﻟﻴﻦ ﺑﺎﺭ ﺩﺭ ﺭﻳﺎﺿﻲ ﻛﻼﺱ ﺍﻭﻝ ﺩﺑﺴﺘﺎﻥ ﺩﻳﺪﻳﻢ: ? =1+1 ﺩﺭ ﻋﺒﺎﺭﺕِ = ،1+1ﺑﺎﻳﺪ ﺣﺎﺻﻞ ﺟﻤﻊ ﺳﻤﺖ ﭼﭗ ﺗﺴﺎﻭﻱ ﺭﺍ ﺑﻴﺎﺑﻴﻢ ﻭ ﺩﺭ ﺳــﻤﺖ ﺭﺍﺳﺖ ﺗﺴﺎﻭﻱ ﺑﻨﻮﻳﺴﻴﻢ .ﺍﻳﻦ ﻣﻌﻨﺎ ﺑﺮﺍﻱ ﺗﻤﺎﻡ ﻋﺒﺎﺭﺕﻫﺎﻱ ﻋﺪﺩﻱ ﻛﻪ ﻋﻤﻠﻴﺎﺕ ﻣﺸﺎﺑﻪ ﺩﺭ ﺁﻥ ﺭﺍ ﺑﺪﺍﻧﻴﻢ ،ﻳﻜﺴﺎﻥ ﺍﺳﺖ: 1 3 2 = ) −(2 − 3 + 7× 5 − (4 − ( −8)) ) ÷ (1 − 8
ﺧــﺪﺍﻱ ﻣﻦ! ﻋﺒﺎﺭﺕ ﺧﻴﻠﻲ ﻃﻮﻻﻧﻲ ﺷــﺪ! ﺣﺘﻲ ﺣﺴــﺎﺏ ﻧﻜﺮﺩﻳﻢ ﺟﻮﺍﺑﺶ ﭼﻪ ﻋﺪﺩﻱ ﻣﻲﺷﻮﺩ؟ ﺍ ّﻣﺎ ﺑﻪ ﻫﺮ ﺣﺎﻝ ﻣﻌﻨﺎﻱ = ﺩﺭ ﻋﺒﺎﺭﺕ ﺍﺧﻴﺮ، ﻫﻤﺎﻥ ﻳﺎﻓﺘﻦ ﺣﺎﺻﻞ ﻋﺒﺎﺭﺕِ ﺳــﻤﺖ ﭼﭗ ﺍﺳﺖ .ﺩﺭ ﺍﻳﻦ ﻣﻮﺍﺭﺩ ،ﺁﻥﭼﻪ ﺳــﻤﺖ ﺭﺍﺳﺖ = ﻣﻲﻧﻮﻳﺴﻴﻢ ،ﻳﻚ ﻋﺪﺩ ﺍﺳــﺖ )ﮔﻮﻳﺎ ،ﮔﻨﮓ ﻳﺎ ﺻﺤﻴﺢ )ﻣﻨﻔﻲ ﻳﺎ ﻣﺜﺒﺖ( ،ﻓﺮﻗﻲ ﻧﻤﻲﻛﻨﺪ ،ﺑﻪ ﻫﺮ ﺣﺎﻝ ﻳﻚ ﻋﺪﺩ ﺍﺳﺖ(.
? = x + 2y − 1 + 3 x − y
ﺣﺎﺻــﻞ ﻋﺪﺩﻱ ﺑﺮﺍﻱ ﺩﺭ ﺍﻳﻦﺟــﺎ ،ﻣﻌﻨﺎﻱ = ﺍﻳﻦ ﻧﻴﺴــﺖ ﻛﻪ ﻳﻚ ِ ﻋﺒﺎﺭﺕِ ﺳﻤﺖ ﭼﭗ ﺑﻪ ﺩﺳــﺖ ﺁﻭﺭﻳﻢ ،ﺑﻠﻜﻪ ﺍﻳﻦﺟﺎ ﺑﺎﻳﺪ ﻋﺒﺎﺭﺕ ﺟﺒﺮﻱ ﺭﺍ ﺑﺎ ﺍﺳﺘﻔﺎﺩﻩ ﺍﺯ ﺟﻤﻊ ﻳﺎ ﺗﻔﺮﻳﻖ ﺟﻤﻼﺕ ﻣﺸﺎﺑﻪ ،ﺳﺎﺩﻩﺗﺮ ﻛﻨﻴﻢ .ﭘﺲ = ﺩﺭ ﺍﻳﻦ ﻋﺒﺎﺭﺕ ﺑﻪ ﻣﻌﻨﺎﻱ ﺩﻳﮕﺮﻱ ﺁﻣﺪﻩ ﺍﺳــﺖ ﻭ ﺩﺭ ﻭﺍﻗﻊ ﺍﺯ ﺁﻥﺟﺎ ﻛﻪ ﻋﺒﺎﺭﺕ xﻭ yﻭ ﻋﺪﺩﻱ ﺩﺭ ﺍﻳﻦ ﻋﺒﺎﺭﺕ ﻣﺸــﺎﺑﻪ ﻧﻴﺴــﺘﻨﺪ ،ﻫﺮﮔﺰ ﺳﻤﺖ ﭼﭗ ﺗﺴﺎﻭﻱ ﺑﺎ ﻳﻚ ﻋﺪﺩ ﺑﺮﺍﺑﺮ ﻧﻴﺴﺖ. x + 2y − 1 + 3 x − y = 4x + y − 1 ﺳﻤﺖ ﺭﺍﺳﺖ ﺗﺴﺎﻭﻱ ﻛﻪ ﻳﻚ ﻋﺒﺎﺭﺕ ﺟﺒﺮﻱ ﺟﺪﻳﺪ ﺍﺳﺖ.
ﺩﺭ ﻋﺒﺎﺭﺕ ﺯﻳﺮ ،ﻣﻌﻨﺎﻱ = ﭼﻴﺴﺖ؟
? = a + 2b
ﺩﺭ ﺍﻳﻦﺟﺎ ﻧﻴﺰ ﺑﻪ ﻣﻌﻨﺎﻱ ﻳﺎﻓﺘﻦ ﻳﻚ ﻋﺒﺎﺭﺕ ﺟﺒﺮﻱ ﺳــﺎﺩﻩﺗﺮ ﺍﺳﺖ )ﻛﻪ ﺟﻤﻠﻪﻫﺎﻱ ﻣﺸﺎﺑﻪ ﻧﺪﺍﺷﺘﻪ ﺑﺎﺷﺪ( .ﻭﻟﻲ ﺍﺯ ﺁﻥﺟﺎ ﻛﻪ aﻭ 2bﻣﺸﺎﺑﻪ ﻧﻴﺴــﺘﻨﺪ ،ﭘﺲ ﺩﺭ ﻭﺍﻗﻊ ﻫﻴﭻ ﻋﺒﺎﺭﺕ ﺟﺪﻳﺪﻱ ﺟﻠﻮ ﻣﺴﺎﻭﻱ ﻧﻤﻲﺗﻮﺍﻧﻴﻢ ﺑﻨﻮﻳﺴﻴﻢ! ﺍﻳﻦ ،ﺍﺯ ﺁﻥ =ﻫﺎﻳﻲ ﺍﺳﺖ ﻛﻪ ﺳﻤﺖ ﺭﺍﺳﺘﺶ ﻋﺒﺎﺭﺕ ﺳﺎﺩﻩﺗﺮﻱ ﻧﻤﻲﺗﻮﺍﻥ ﻧﻮﺷﺖ. ﺩﺭ ﺗﺴﺎﻭﻱ ﺯﻳﺮ = ،ﺑﻪ ﭼﻪ ﻣﻌﻨﺎﺳﺖ؟ 2
? = ) 2x ( x − 4 y
ﺩﺭ ﺍﻳﻦﺟــﺎ ﺑﺎﻳﺪ 2xﺭﺍ ﺩﺭ ﻋﺒــﺎﺭﺕ ﺩﻭﺟﻤﻠﻪﺍﻱ ، x 2 − 4 yﭘﺨﺶ ِ ﺧﺎﺻﻴﺖ ﭘﺨﺸﻲ ،ﺿﺮﺏ ﻛﻨﻴﻢ( ﻭ ﻳﻚ ﺩﻭ ﻛﻨﻴﻢ )ﻳﻌﻨﻲ ﺑﺎ ﺍﺳــﺘﻔﺎﺩﻩ ﺍﺯ ﺟﻤﻠﻪﺍﻱ ﺑﻪ ﺩﺳﺖ ﺁﻭﺭﻳﻢ: 3
2
2x ( x − 4 y) = 2x − 8 xy
ﻭ ﻋﺒــﺎﺭﺕِ ، 2x 3 − 8 xyﺩﻭ ﺟﻤﻠﻪﻱ ﻏﻴﺮ ﻣﺸــﺎﺑﻪ ﺩﺍﺭﺩ ﻛﻪ ﺑﺎ ﻫﻢ ﺟﻤﻊ )ﻳﺎ ﺗﻔﺮﻳﻖ( ﻧﻤﻲﺷﻮﻧﺪ .ﭼﻨﻴﻦ ﺣﺎﻟﺘﻲ ﺑﻪ ﺍﻳﻦ ﺩﻟﻴﻞ ﭘﻴﺶ ﻣﻲﺁﻳﺪ ﻛﻪ ) x 2 − 4 yﻋﺒﺎﺭﺕ ﺩﺍﺧﻞ ﭘﺮﺍﻧﺘﺰ( ﻧﻴﺰ ﺍﺯ ﺩﻭ ﺟﻤﻠﻪﻱ x2ﻭ -4yﻛﻪ ﻣﺸﺎﺑﻪ ﻧﺒﻮﺩﻧﺪ ،ﺗﺸﻜﻴﻞ ﺷﺪﻩ ﺑﻮﺩ. ﭘﺲ ﺩﺭ ﻣﺜﺎﻝ ﺍﺧﻴﺮ = ،ﺑﻪ ﻣﻌﻨﺎﻱ ﻳﺎﻓﺘﻦ ﺣﺎﺻﻞ ﺿﺮﺏِ ﺗﻚﺟﻤﻠﻪﺍﻱ 2xﺩﺭ ﺟﻤﻠﻪﻱ x 2 − 4 yﺍﺳــﺖ .ﺩﺭ ﻣﺜــﺎﻝ ﺯﻳﺮ ،ﺑﺎﺯ ﻫﻢ = ﺑﻪ ﻣﻌﻨﺎﻱ ﻳﺎﻓﺘﻦ ﺣﺎﺻﻞ ﺿﺮﺏ ﺍﺳــﺖ ،ﺍﻣﺎ ﺣﺎﺻﻞ ﺿﺮﺏِ ﺩﻭ ﺟﻤﻠﻪﻱ x-yﺩﺭ ﺩﻭ ِ ﺟﻤﻠﻪﻱ x+y؛ ﻭ ﺍﻟﺒﺘﻪ ﺩﺭ ﺍﻳﻦ ﻣﺜﺎﻝ ﻭ ﻣﺜﺎﻝ ﻗﺒﻠﻲ ،ﻫﻴﭻ ﻳﻚ ﺍﺯ ﺣﺎﺻﻞ ﺿﺮﺏﻫﺎ ﻋﺪﺩ ﻧﻴﺴــﺘﻨﺪ ،ﺑﻠﻜــﻪ ﺑﺎﺯ ﻫﻢ ﻳﻚ ﻋﺒﺎﺭﺕ ﺟﺒﺮﻱ ﺑﻪ ﺩﺳــﺖ ﻣﻲﺁﻳﺪ:
ﺩﻭﺭﺓ ﺷﺎﻧﺰﺩﻫﻢ ،ﺷﻤﺎﺭﺓ ،4ﺗﺎﺑﺴﺘﺎﻥ 1390
ﺭﺍﻫﻨﻤﺎﻳﻲ
31
2
2
( x − y)( x + y) = x − xy + xy − y
ﺧﻮﺏ ،ﺑﻪ ﻧﻈﺮ ﻣﻲﺭﺳــﺪ ﻛﺎﺭﻣﺎﻥ ﺗﻤﺎﻡ ﺷــﺪﻩ ﺍﺳﺖ؛ ﺍ ّﻣﺎ ﻳﻚ ﺩﻗﻴﻘﻪ ﺻﺒﺮ ﻛﻨﻴﺪ! ﺩﺭ ﻋﺒﺎﺭﺕ ﺳﻤﺖ ﺭﺍﺳﺖ ،ﺟﻤﻼﺕ xyﻭ -xyﺑﺎ ﻫﻢ ﻣﺸﺎﺑﻪ ﻫﺴﺘﻨﺪ ﻭ ﺣﺎﺻﻞ ﺟﻤﻊ ﺁﻥﻫﺎ ،ﺻﻔﺮ ﺍﺳﺖ xy − xy = 0 :ﭘﺲ ﻣﻲﺗﻮﺍﻥ ﻳﻚ = ﺩﻳﮕﺮ ﮔﺬﺍﺷﺖ ﻭ ﻋﺒﺎﺭﺕ ﺭﺍ ﺳﺎﺩﻩﺗﺮ ﻛﺮﺩ: 2
2
2
2
(xx − y )( x + y ) = x − xy + xy − y = x − y
ﻭ ﻋﺒﺎﺭﺕ ﺳــﺎﺩﻩﺗﺮﻱ ﻧﻤﻲﺗﻮﺍﻥ ﺟﻠﻮ ﺁﻥ ﻧﻮﺷﺖ )ﻣﺜﻞ ﻳﻜﻲ ﺍﺯ ﻣﺜﺎﻝﻫﺎﻱ ﻗﺒــﻞ( .ﺍ ّﻣﺎ ﺍﮔﺮ ﺩﺍﻧﺶﺁﻣﻮﺯ ﺳــﻮﻡ ﺭﺍﻫﻨﻤﺎﻳﻲ ﺑﺎﺷــﻴﺪ ﻭ ﻓﺎﻛﺘﻮﺭﮔﻴﺮﻱ ﻳﺎ ﺗﺒﺪﻳﻞ ﭼﻨﺪ ﺟﻤﻠﻪﺍﻱ ﺑﻪ ﺣﺎﺻﻞ ﺿﺮﺏ ﺭﺍ ﻳﺎﺩ ﮔﺮﻓﺘﻪ ﺑﺎﺷــﻴﺪ ،ﻣﻲﺑﻴﻨﻴﺪ ﺍﺯ 2xyﻣﻲﺗﻮﺍﻥ ﺩﺭ ﺍﻳﻦ ﺳﻪ ﺟﻤﻠﻪ ﻓﺎﻛﺘﻮﺭ ﮔﺮﻓﺖ ﻭ ﻋﺒﺎﺭﺕ ﺳﻤﺖ ﭼﭗ ﻣﺜﺎﻝ ﺭﺍ ﺑﻪ ﺣﺎﺻﻞ ﺿﺮﺏ ﺗﺒﺪﻳﻞ ﻛﺮﺩ ،ﻳﻌﻨﻲ ﺑﺮﻋﻜﺲ ﺁﻥ ﻛﺎﺭﻱ ﻛﻪ ﺩﺭ ِ ) 2x ( x 2 − 4 yﻛﺮﺩﻳﻢ! ﭘﺲ ﺩﺍﺭﻳﻢ: 2
ﺍﻳﻦ ﻣﺴﺎﻭﻱ ﺑﻪ ﻣﻌﻨﺎﻱ ﺳﺎﺩﻩ ﻛﺮﺩﻥ ﻋﺒﺎﺭﺕ ﺟﺒﺮﻱ ﺍﺳﺖ
ﺍﻳﻦ ﻣﺴﺎﻭﻱ ﺑﻪ ﻣﻌﻨﺎﻱ ﺿﺮﺏ ﻛﺮﺩﻥ ﺩﻭ ﻋﺒﺎﺭﺕ ﺳﻤﺖ ﭼﭗ ﺩﺭ ﻳﻜﺪﻳﮕﺮ ﺍﺳﺖ )ﺍﻟﺒﺘﻪ ﺑﺎ ﺍﺳﺘﻔﺎﺩﻩ ﺍﺯ ﺧﺎﺻﻴﺖ ﭘﺨﺸﻲ(
ﺧﻮﺏ ،ﺑﻪ ﻧﻈﺮ ﻣﻲﺭﺳﺪ » =« ﺩﺭ ﻋﺒﺎﺭﺕﻫﺎﻱ ﻣﺨﺘﻠﻒ ،ﻧﺸﺎﻥﺩﻫﻨﺪﻩﻱ »ﺩﺭﺧﻮﺍﺳــﺖﻫﺎﻱ« ﻣﺨﺘﻠﻒ ﺍﺳﺖ! ﻣﺎ ﺑﺎﻳﺪ ﺍﺯ ﺷــﻜﻞ ﻭ ﺍﺟﺰﺍﻱ ﻋﺒﺎﺭﺕ ﺩﺍﺩﻩ ﺷﺪﻩ ،ﻣﺘﻮﺟﻪ ﺷﻮﻳﻢ ﻛﻪ ﭼﻪ ﻧﻮﻉ ﻋﻤﻠﻴﺎﺗﻲ ﺭﺍ ﺍﻧﺠﺎﻡ ﺩﻫﻴﻢ ﻭ ﺳﻤﺖ ﺭﺍﺳــﺖ ﺗﺴﺎﻭﻱ ،ﭼﻪ ﺑﻨﻮﻳﺴﻴﻢ .ﺍﮔﺮ ﻓﻜﺮ ﻣﻲﻛﻨﻴﺪ ﻣﻌﻨﺎﻱ ﺩﻳﮕﺮﻱ ﺑﺮﺍﻱ = ﺩﺭ ﻋﺒﺎﺭﺕﻫﺎﻳﻤﺎﻥ ﻧﺪﺍﺭﻳﻢ ،ﺑﻪ ﻣﺜﺎﻝ ﺯﻳﺮ ﺗﻮﺟﻪ ﻛﻨﻴﺪ ﻭ ﺗﻮﺿﻴﺢ ﺩﻫﻴﺪ ﭼﻪ ﺍﺗﻔﺎﻗﻲ ﺍﻓﺘﺎﺩﻩ ﻭ ﻫﺮ = ﺑﻪ ﻣﻌﻨﺎﻱ ﭼﻴﺴﺖ؟ )(3
)( 2
)(1
3 x − 1 = 3 ( −2) − 1 = − 6 − 1 = − 7
ﺑﻠﻪ ،ﺩﺭﺳﺖ ﺍﺳﺖ = :ﺷﻤﺎﺭﻩ ) ،(1ﻧﺸﺎﻥ ﻣﻲﺩﻫﺪ ﻋﺒﺎﺭﺕ ،3x-1ﺑﻪ ﺍﺯﺍﻱ )ﺍﻳﻦ ﻣﺴﺎﻭﻱ x=-2ﻳﻌﻨﻲ xﻫﻤﺎﻥ ﻋﺪﺩ -2ﺍﺳﺖ ﻭ ﺑﺎﺯ ﻣﻌﻨﻲﺍﺵ ﻓﺮﻕ ﻣﻲﻛﻨﺪ! ﻭﺍﻱ ﺧﺪﺍﻱ ﻣﻦ ! ! ! ( ﺑﺎ ﻋﺒﺎﺭﺕ ﻋﺪﺩﻱ 3 ( −2) − 1 ﺑﺮﺍﺑﺮ ﺍﺳﺖ. = ﺷﻤﺎﺭﻩﻱ ) (2ﻳﻌﻨﻲ ﺣﺎﺻﻞ ﻋﺒﺎﺭﺕ ﻋﺪﺩﻱ 3( −2) − 1ﺑﺎ ﺣﺎﺻﻞ ﻋﺒﺎﺭﺕ ﻋﺪﺩﻱ -6-1ﺑﺮﺍﺑﺮ ﺍﺳــﺖ ﻭ ﺑﺎﻻﺧﺮﻩ = ﺷﻤﺎﺭﻩﻱ (3ﺑﻪ ﻣﻌﻨﺎﻱ ﺍﻳﻦ ﺍﺳــﺖ ﻛﻪ ﺣﺎﺻﻞ ﻋﺒﺎﺭﺕ ،-6-1ﻋﺪﺩ -7ﺍﺳﺖ .ﭘﺲ = ﺷﻤﺎﺭﻩﻱ ) (1ﺩﺭ ﺍﻳﻦﺟﺎ ﺑﻪ ﻣﻌﻨﺎﻱ ﺟﺎﮔﺬﺍﺭﻱ ﻳﻚ ﻋﺪﺩ ﺩﺭ ﻋﺒﺎﺭﺕ ﺳﻤﺖ ﭼﭗ ﺑﻪ ﺟﺎﻱ ﻣﺠﻬﻮﻝ )ﻳﺎ ﻣﺘﻐﻴﺮ( ﺁﻥ ﻋﺒﺎﺭﺕ ﻭ ﻳﺎﻓﺘﻦ ﻋﺒﺎﺭﺕ ﻋﺪﺩﻱ ﺑﻮﺩ ﻭ ﺍﻳﻦ ﻣﻌﻨﺎ ﺑﺎ ﺳﺎﻳﺮ ﻣﻌﺎﻧﻲ ﻛﻪ ﺑﺮﺭﺳﻲ ﻛﺮﺩﻳﻢ ،ﻣﺘﻔﺎﻭﺕ ﺍﺳﺖ. ﺣﺎﻝ ﺷــﻤﺎ ﺑﮕﻮﻳﻴﺪ ﻛﻪ ﻣﻌﻨــﺎﻱ = ﺩﺭ ﻋﺒﺎﺭﺕ ﺯﻳﺮ ﭼﻴﺴــﺖ؟ )ﺩﺭ ﻭﺍﻗﻊ» ،ﺩﺭﺧﻮﺍﺳــﺖ« ﭼﻴﺴــﺖ ﻭ ﭼﻪ ﭼﻴﺰﻱ ﺑﺎﻳﺪ ﺩﺭ ﺳــﻤﺖ ﺭﺍﺳﺖ = ﺑﻨﻮﻳﺴﻴﻢ؟( 2
ﻓﻘﻂ ﺣﻮﺍﺳــﺘﺎﻥ ﺑﺎﺷــﺪ ﻛﻪ ﺩﻭﺑﺎﺭﻩ ﻳﻚ ﻣﺴــﺎﻭﻱ ﺩﺭ ﺳﻤﺖ ﺭﺍﺳﺖ ﻧﮕﺬﺍﺭﻳﺪ ﻭ 2xyﺭﺍ ﺩﻭﺑﺎﺭﻩ ﺩﺭ 2x − 3 + yﭘﺨﺶ ﻧﻜﻨﻴﺪ ،ﭼﻮﻥ ﺩﻭﺑﺎﺭﻩ ﻫﻤــﺎﻥ ﻋﺒﺎﺭﺕِ 4x 2 y − 6 xy + 2xy2ﺑﻪ ﺩﺳــﺖ ﻣﻲﺁﻳــﺪ! )ﺧﻴﻠﻲ ﺍﺯ ﺑﭽﻪﻫﺎ ﺑﻪ ﻋﺎﺩﺕ ،ﺍﻳﻦ ﻛﺎﺭ ﺭﺍ ﻣﻲﻛﻨﻨﺪ(. ﻧﻜﻨﺪ ﺍﺯ ﺍﻳﻦ ﻫﻤﻪ ﺗﻨﻮﻉ ﺩﺭ ﻧﻮﻉ ﺩﺭﺧﻮﺍﺳــﺖﻫﺎ ﺑﺮﺍﻱ ﻧﻮﺷﺘﻦ ﭼﻴﺰﻱ ﺩﺭ ﺳــﻤﺖ ﺭﺍﺳﺖ = ﮔﻴﺞ ﺷﺪﻩ ﺑﺎﺷﻴﺪ؟ ! ﺑﺮﺍﻱ ﺍﻳﻦ ﻛﻪ ﻳﻚ ﺟﻤﻊﺑﻨﺪﻱ ﻛﻨﻴﻢ ،ﻳﻚ ﺩﻭﺭ ﺩﻳﮕﺮ ﻣﻄﻠﺐ ﺭﺍ ﺍﺯ ﺁﻥﺟﺎ ﻛﻪ ﺩﺭﺑﺎﺭﻩﻱ ﻋﻼﻣﺖ = ﻧﻮﺷﺘﻪ ﺷﺪﻩ ﺍﺳﺖ ﺑﺨﻮﺍﻧﻴﺪ ﻭ ﺳﭙﺲ ﺍﺯ ﺍﻳﻦﺟﺎ ﺑﻪ ﺑﻌﺪ ِ ﻣﻄﻠﺐ ﺭﺍ ﺍﺩﺍﻣﻪ ﺩﻫﻴﺪ. ﺑﺎ ﻣﺮﻭﺭ ﻛﻠﻲ ﻣﺜﺎﻝﻫﺎ ،ﻣﺘﻮﺟﻪ ﻣﻲﺷﻮﻳﻢ ﻛﻪ ﺩﺭ ﻛﻞ ﻋﻼﻣﺖ ﺗﺴﺎﻭﻱ ) =( ﺩﺭ ﻣﻮﺍﺭﺩ ﺯﻳﺮ ﻣﻮﺭﺩ ﺍﺳﺘﻔﺎﺩﻩ ﻗﺮﺍﺭ ﻣﻲﮔﻴﺮﺩ: (1ﺑﻴــﻦ ﺣﺎﺻﻞ ﻋﺪﺩﻱ ﻳــﻚ ﻋﺒﺎﺭﺕ ﻋﺪﺩﻱ ﻭ ﺧــﻮ ِﺩ ﺁﻥ ﻋﺒﺎﺭﺕ ﻋﺪﺩﻱ ،ﻣﺜﻞ 1 3 1 3 + − =1 2 2 4 4
(2ﺑﻴﻦ ﺩﻭ ﻋﺒﺎﺭﺕ ﻋﺪﺩﻱ ﻛﻪ ﺣﺎﺻﻞ ﻳﻜﺴﺎﻥ ﺩﺍﺭﻧﺪ ،ﻣﺜﻞ
1+22 +4×5 = 1+4+2 0
(3ﺑﻴﻦ ﺩﻭ ﻋﺒﺎﺭﺕ ﺟﺒﺮﻱ ﻛﻪ ﻳﻜﻲ ﺑﺎ ﻋﻤﻠﻴﺎﺕ ُﻣﺠﺎﺯ ﺍﺯ ﺩﻳﮕﺮﻱ ﺑﻪ ﻣﺜﻞ ﺍﻧﻮﺍﻉ ﻧﻤﻮﻧﻪﻫﺎﻱ ﺯﻳﺮ: ﺩﺳﺖ ﺁﻣﺪﻩ ﺍﺳﺖِ ، )ﺟﻤﻊ ﺟﻤﻼﺕ ﻣﺸﺎﺑﻪ( 4a − 6 b + a = 5a − 6 b 2( x − 5) + 3 (2 − x ) = 2x − 10 + 6 − 3 x = − x − 4
2
? = 4 x y − 6 xy + 2xy
ﺍﮔﺮ ﻫﻨﻮﺯ ﺣﺪﺱ ﻧﺰﺩﻩﺍﻳﺪ ﻛﻪ ﺑﺎﻳﺪ ﭼﻪ ﻛﺎﺭ ﻛﻨﻴﺪ ،ﺑﻪ ﺍﻳﻦ ﺗﻮﺟﻪ ﻛﻨﻴﺪ ﻛﻪ ﻋﺒﺎﺭﺕ ﺳــﻤﺖ ﭼﭗ ،ﻳﻚ ﻋﺒﺎﺭﺕ ﺟﺒﺮﻱ ﺍﺳــﺖ ،ﻭﻟﻲ ﺩﺭﺧﻮﺍﺳﺖ ﺿــﺮﺏ ﺩﺭ ﺁﻥ ﻭﺟﻮﺩ ﻧﺪﺍﺭﺩ ،ﺯﻳﺮﺍ ﻓﻘﻂ ﺳــﻪ ﺟﻤﻠﻪﻱ 4 x 2 yﻭ − 6 xy ﻭ + 2xy2ﺭﺍ ﺩﺍﺭﺩ .ﭘﺲ ﺩﺭ ﻭﺍﻗﻊ ﻳﻚ ﺳــﻪ ﺟﻤﻠﻪﺍﻱ ﺍﺳــﺖ ﻭ ﺍﻳﻦ ﺳﻪ ﺟﻤﻠﻪ ﺑﺎ ﻫﻢ ﻣﺸﺎﺑﻪ ﻧﻴﺴﺘﻨﺪ .ﭘﺲ ﭼﻪ ﻛﺎﺭ ﻛﻨﻴﻢ؟ ﻗﻄﻌﺎً ﺑﻪ ﻧﻈﺮ ﻣﻲﺭﺳﺪ ﻛﻪ ﺍﻳﻦ = ﺍﺯ ﺁﻥ ﻣﺴــﺎﻭﻱﻫﺎﻳﻲ ﺑﺎﺷﺪ ﻛﻪ ﺳﻤﺖ ﺭﺍﺳﺘﺶ ﺧﺎﻟﻲ ﻣﻲﻣﺎﻧﺪ
2
) 4 x y − 6 xy + 2xy = 2xy (2x − 3 + y
ﺳﻤﺖ ﺭﺍﺳﺖ ،ﺑﺎ ﺟﻤﻊ ﻣﺸﺎﺑﻪ ﺑﻪ ﺩﺳﺖ ﻣﻲ ﺁﻳﺪ
ﺳﻤﺖ ﺭﺍﺳﺖ ،ﺑﺎ ﭘﺨﺸﻲ ﻋﺪﺩ ﺩﺭ ﻋﺒﺎﺭﺕﻫﺎﻱ ﺩﺍﺧﻞ ﭘﺮﺍﻧﺘﺰ ﺑﻪ ﺩﺳﺖ ﻣﻲﺁﻳﺪ.
)ﺑﺎ ﻓﺎﻛﺘﻮﺭﻱ ﺳﻤﺖ ﺭﺍﺳﺖ ﺑﻪ ﺩﺳﺖ ﻣﻲﺁﻳﺪ( 2
)6 x − 3 x = 3 x((2x − 1
(4ﺩﺭ ﻋﺒﺎﺭﺕﻫــﺎﻱ ﺟﺒﺮﻱ ،ﻭﻗﺘﻲ ﻣﺘﻐﻴﺮﻫــﺎﻱ ﺁﻥ ﻋﺒﺎﺭﺕ ،ﺍﻋﺪﺍﺩ ﻣﻮﺭﺩﻧﻈﺮ ﺭﺍ ﻣﻲﮔﺰﺍﺭﻳﻢ ﻭ ﺁﻥ ﺭﺍ ﺑﻪ ﻳﻚ ﻋﺒﺎﺭﺕ ﻋﺪﺩﻱ ﺗﺒﺪﻳﻞ ﻣﻲﻛﻨﻴﻢ، ﻣﺜﻞ ِ 2
2
)x − 8 + y = 3 − 8 + ( −5 32
ﺭﺍﻫﻨﻤﺎﻳﻲ
ﺩﻭﺭﺓ ﺷﺎﻧﺰﺩﻫﻢ ،ﺷﻤﺎﺭﺓ ،4ﺗﺎﺑﺴﺘﺎﻥ 1390
ﻛﻪ ﺩﺭ ﺁﻥ ﺑﻪ ﺟﺎﻱ ،xﻋﺪﺩ 3ﻭ ﺑﻪ ﺟﺎﻱ ،yﻋﺪﺩ -5ﺭﺍ ﮔﺬﺍﺷﺘﻪﺍﻳﻢ، ﻳﻌﻨﻲ x=3ﻭ ،y=-5ﻛﻪ ﺍﻳﻦ ﺩﻭ ﺗﺴﺎﻭﻱ ﺍﺧﻴﺮ ﺑﻪ ﻣﻌﻨﺎﻱ ﺑﺮﺍﺑﺮﻱ ﻣﻘﺎﺩﻳﺮ xﻭ yﺑﺎ ﺍﻋﺪﺍﺩ ﻣﻌﻴﻨﻲ ﻫﺴﺘﻨﺪ ،ﻳﻌﻨﻲ: (5ﺑﺮﺍﺑﺮﻱ ﺩﻭ ﺷﻲء ﺍﺯ ﻳﻚ ﺟﻨﺲ ﺑﺎ ﻫﻢ ،ﻣﺜﻞ ﻫﻤﺎﻥ x=3ﻭ y=-5 ﺩﺭ ﻣﺜﺎﻝ ﻗﺒﻞ. ﺍﻟﺒﺘﻪ ﺗﻮﺟﻪ ﻛﻨﻴﺪ ﻛﻪ ﺑﺎﻳﺪ ﺟﻨﺲ ﺩﻭ ﺷــﻲء ﺩﻭ ﻃﺮﻑ ﺗﺴﺎﻭﻱ ﻣﺜﻞ ﻫﻢ ﺑﺎﺷﺪ .ﺩﺭ ﻣﺠﻤﻮﻋﻪﻫﺎ ﻳﺎ ﺩﺭ ﺑﺤﺚ ﺑﺮﺩﺍﺭﻫﺎ ﻧﻴﺰ ﺗﺴﺎﻭﻱﻫﺎﻳﻴﻲ ﺍﺯ ﺍﻳﻦ ﺩﺳﺖ ﺩﺍﺭﻳﻢ: 0 ﺯﻭﺝ ﻛﻢﺗﺮ ﺍﺯ 1 ﻣﺜــﺎﻝ (1ﺍﮔﺮ =Aﻣﺠﻤﻮﻋﻪﻱ ﻋﺪﺩﻫﺎﻱ ﻃﺒﻴﻌﻲ ِ ﻭ =Bﻣﺠﻤﻮﻋﻪﻱ ﻣﻀﺎﺭﺏ ﻃﺒﻴﻌﻲ 2ﻛﻪ ﻳﻚ ﺭﻗﻤﻲ ﺑﺎﺷﻨﺪ، ﺁﻥﮔﺎﻩ A=B ﻣﺜﺎﻝ (2ﺩﺭ ﺷﻜﻞ ﺯﻳﺮ ،ﺑﺮﺩﺍﺭﻫﺎﻱ aﻭ bﺑﺮﺍﺑﺮﻧﺪ ،ﻳﻌﻨﻲ
a=b
ﺩﻓﺘﺮ ﺍﻧﺘﺸﺎﺭﺍﺕ ﻛﻤﻚ ﺁﻣﻮﺯﺷﻰ
ﺑﺎ ﻣﺠﻠﻪﻫﺎﻯ ﺭﺷﺪﺁﺷﻨﺎ ﺷﻮﻳﺪ ﻣﺠﻠﻪﻫﺎﻯ ﺭﺷـﺪ ﺗﻮﺳـﻂ ﺩﻓﺘﺮ ﺍﻧﺘﺸـﺎﺭﺍﺕ ﻛﻤﻚﺁﻣﻮﺯﺷﻰ ﺳـﺎﺯﻣﺎﻥ ﭘﮋﻭﻫـﺶ ﻭ ﺑﺮﻧﺎﻣﻪﺭﻳﺰﻯ ﺁﻣﻮﺯﺷـﻰ ﻭﺍﺑﺴـﺘﻪ ﺑﻪ ﻭﺯﺍﺭﺕ ﺁﻣـﻮﺯﺵ ﻭ ﭘـﺮﻭﺭﺵ ﺗﻬﻴـﻪ ﻭ ﻣﻨﺘﺸـﺮ ﻣﻰﺷـﻮﻧﺪ:
ﻣﺠﻠﻪ ﻫﺎﯼ ﺩﺍﻧﺶ ﺁﻣﻮﺯﯼ ) ﺑﻪ ﺻﻮﺭﺕ ﻣﺎﻫﻨﺎﻣﻪ ﻭ 8ﺷﻤﺎﺭﻩ ﺩﺭ ﻫﺮ ﺳﺎﻝ ﺗﺤﺼﻴﻠﻰ ﻣﻨﺘﺸﺮ ﻣﻰﺷﻮﻧﺪ ( :
)ﺑﺮﺍﻯ ﺩﺍﻧﺶﺁﻣﻮﺯﺍﻥ ﺁﻣﺎﺩﮔﻰ ﻭ ﭘﺎﻳﻪﻯ ﺍﻭﻝ ﺩﻭﺭﻩﻯ ﺩﺑﺴﺘﺎﻥ(
a=b
b
)ﺑﺮﺍﻯ ﺩﺍﻧﺶﺁﻣﻮﺯﺍﻥ ﭘﺎﻳﻪﻫﺎﻯ ﺩﻭﻡ ﻭ ﺳﻮﻡ ﺩﻭﺭﻩﻯ ﺩﺑﺴﺘﺎﻥ(
)ﺑﺮﺍﻯ ﺩﺍﻧﺶﺁﻣﻮﺯﺍﻥ ﭘﺎﻳﻪﻫﺎﻯ ﭼﻬﺎﺭﻡ ﻭ ﭘﻨﺠﻢ ﺩﻭﺭﻩﻯ ﺩﺑﺴﺘﺎﻥ(
a
)ﺑﺮﺍﻯ ﺩﺍﻧﺶﺁﻣﻮﺯﺍﻥ ﺩﻭﺭﻩﻯ ﺭﺍﻫﻨﻤﺎﻳﻰ ﺗﺤﺼﻴﻠﻰ(
ﻣﺜﻞ (6ﺑﻴﻦ ﻧﺎﻡ ﺷﻲء ﻭ ﺧﻮﺩ ﺷﻲء ﻗﺮﺍﺭ ﻣﻲﮔﻴﺮﺩِ ، ﻳﺎ
)ﺑﺮﺍﻯ ﺩﺍﻧﺶﺁﻣﻮﺯﺍﻥ ﺩﻭﺭﻩﻯ ﻣﺘﻮﺳﻄﻪﻭﭘﻴﺶﺩﺍﻧﺸﮕﺎﻫﻰ(
} = { 1 , 2 , 3 , 4 , ....
⎤ ⎡1
⎥ ⎢ =¡ ⎦⎣0
(7ﺩﺭ ﻫﻨﺪﺳــﻪ ،ﺑﻴﻦ ﻧﺎﻡ ﻳﻚ ﺷــﻲء ﻭ ﺍﻧﺪﺍﺯﻩﻱ ﺁﻥ ﻗﺮﺍﺭ ﻣﻲﮔﻴﺮﺩ، ﻣﺜﻞ ِ = 45 AB = 3cmﻳﺎ AB (8ﺁﻳﺎ ﺑﺎﺯ ﻫﻢ ﻣﻮﺍﺭﺩ ﺩﻳﮕﺮﻱ ﻫﺴــﺖ ﻛﻪ ﻣﻦ ﻓﺮﺍﻣﻮﺵ ﻛﺮﺩﻩﺍﻡ ﻧﺎﻡ ﺑﺒــﺮﻡ؟ ﺑﻪ ﻫــﺮ ﺣﺎﻝ ﺑﻪ ﺍﻳﻦ ﺟﻤﻊﺑﻨﺪﻱ ﺭﺳــﻴﺪﻳﻢ ﻛﻪ ﻭﻗﺘﻲ ﻋﻼﻣﺖ = ﺭﺍ ﺩﻳﺪﻳﻢ ﻭ ﺳــﻤﺖ ﺭﺍﺳــﺖ ﺁﻥ ﺧﺎﻟﻲ ﺑﻮﺩ ،ﺣﺘﻤﺎً ﻧﺒﺎﻳﺪ ﻳﻚ ﻋﻤﻠﻴﺎﺕ ﻳﺎ ﻣﺤﺎﺳــﺒﺎﺗﻲ ﺍﻧﺠﺎﻡ ﺩﻫﻴﻢ ﺗﺎ ﺩﺭ ﺳﻤﺖ ﺭﺍﺳﺖ ﺗﺴﺎﻭﻱ ﭼﻴﺰﻱ ﺑﻨﻮﻳﺴﻴﻢ. ﭼﮕﻮﻧﮕﻲ ﻭ ﻧﻮﻉ ﻧﻮﺷــﺘﻦ ﭼﻴﺰﻱ ﺩﺭ ﺳﻤﺖ ﺭﺍﺳﺖ ﺗﺴﺎﻭﻱ ﺑﺴﺘﮕﻲ ﺩﺍﺭﺩ ﺑﻪ ﻣﻌﻨﺎﻳﻲ ﻛﻪ ﺁﻥ ﻋﺒﺎﺭﺕﻫﺎ ﻭ ﺗﺴﺎﻭﻱﻫﺎ ﻣﻲﺩﻫﻨﺪ. ﺣﺎﻝ ﺁﻳﺎ ﻣﻲﺗﻮﺍﻧﻴﺪ ﺑﮕﻮﻳﻴﺪ ﺗﺴﺎﻭﻱ ﺯﻳﺮ ﭼﻴﺴﺖ؟ 4x-6=12 ﺑﻠﻪ ،ﺩﺭﺳــﺖ ﺍﺳﺖ؛ ﺍﻳﻦ ﺗﺴــﺎﻭﻱ ﻳﻚ ﻣﻌﺎﺩﻟﻪ ﺍﺳــﺖ ﻛﻪ ﻣﺎ ﺑﺮﺍﻱ ﺗﻜﻤﻴــﻞ ﺗﺴــﺎﻭﻱ ﻛﺎﺭﻱ ﺍﻧﺠﺎﻡ ﻧﻤﻲﺩﻫﻴﻢ ،ﺑﻠﻜﻪ ﺗﺴــﺎﻭﻱ ﺭﺍ ﺷــﺨﺺ ﺩﻳﮕﺮﻱ ﭘﻴﺪﺍ ﻛﺮﺩﻩ ﺍﺳــﺖ ﻭ ﻣﺎ ﺑﻪ ﻛﻤﻚ ﺁﻥ ،ﻣﻘﺪﺍﺭ ﻣﺠﻬﻮﻝ )ﻳﻌﻨﻲ (x ﺭﺍ ﭘﻴﺪﺍ ﻣﻲﻛﻨﻴﻢ .ﭘﺲ: (9ﺑﻴﻦ ﺩﻭ ﻋﺒﺎﺭﺕ ﺟﺒﺮﻱ ﻛﻪ ﺩﺳﺖ ﻛﻢ ﻳﻜﻲ ﺍﺯ ﺁﻥﻫﺎ ﺣﺪﺍﻗﻞ ﻳﻚ ﻣﺠﻬﻮﻝ ﺩﺍﺭﺩ ﻭ ﺑﻪ ﺁﻥ ﺗﺴﺎﻭﻱ ،ﻣﻌﺎﺩﻟﻪ ﮔﻮﻳﻴﻢ.
ﻣﺠﻠﻪ ﻫﺎﯼ ﺑﺰﺭﮔﺴﺎﻝ ﻋﻤﻮﻣﯽ )ﺑﻪ ﺻﻮﺭﺕ ﻣﺎﻫﻨﺎﻣﻪ ﻭ 8ﺷﻤﺎﺭﻩ ﺩﺭ ﻫﺮ ﺳﺎﻝ ﺗﺤﺼﻴﻠﻰ ﻣﻨﺘﺸﺮ ﻣﻰﺷﻮﻧﺪ(: ﺭﺷﺪ ﺁﻣـﻮﺯﺵ ﺍﺑﺘــﺪﺍﻳﯽ ﺁﻣﻮﺯﺷﯽ
ﺭﺷﺪ ﺁﻣﻮﺯﺵ ﺭﺍﻫﻨﻤـﺎﻳﯽ ﺗﺤﺼﻴﻠﯽ
ﺭﺷﺪ ﻣﺪﺭﺳﻪ ﻓﺮﺩﺍ
ﺭﺷﺪ ﻣﺪﻳﺮﻳﺖ ﻣﺪﺭﺳﻪ
ﺭﺷﺪ ﺗﻜﻨﻮﻟﻮﮊﯼ
ﺭﺷﺪ ﻣﻌﻠﻢ
ﻣﺠﻠﻪ ﻫﺎﯼ ﺑﺰﺭﮔﺴﺎﻝ ﺍﺧﺘﺼﺎﺻﯽ
)ﺑﻪ ﺻﻮﺭﺕ ﻓﺼﻠﻨﺎﻣﻪ ﻭ 4ﺷﻤﺎﺭﻩ ﺩﺭ ﻫﺮ ﺳﺎﻝ ﺗﺤﺼﻴﻠﻰ ﻣﻨﺘﺸﺮ ﻣﻰﺷﻮﻧﺪ(:
ﺭﺷﺪ ﺑﺮﻫﺎﻥ ﺭﺍﻫﻨﻤﺎﻳﻰ )ﻣﺠﻠﻪ ﺭﻳﺎﺿﻰ ﺑﺮﺍﻯ ﺩﺍﻧﺶﺁﻣﻮﺯﺍﻥ ﺩﻭﺭﻩﻯ ﺭﺍﻫﻨﻤﺎﻳﻰ ﺗﺤﺼﻴﻠﻰ( ﺭﺷﺪ ﺑﺮﻫﺎﻥ ﻣﺘﻮﺳﻄﻪ )ﻣﺠﻠﻪ ﺭﻳﺎﺿﻰ ﺑﺮﺍﻯ ﺩﺍﻧﺶﺁﻣﻮﺯﺍﻥ ﺩﻭﺭﻩﻯ ﻣﺘﻮﺳﻄﻪ(
ﺭﺷﺪ ﺁﻣﻮﺯﺵ
ﻗﺮﺁﻥ ﺭﺷﺪ ﺁﻣﻮﺯﺵ ﻣﻌﺎﺭﻑ ﺍﺳﻼﻣﻰ ﺭﺷﺪ ﺁﻣﻮﺯﺵ ﺯﺑﺎﻥ ﻭ ﺍﺩﺏ ﻓﺎﺭﺳﻰ ﺭﺷﺪ ﺁﻣﻮﺯﺵ ﻫﻨﺮ ﺭﺷـﺪ ﻣﺸـﺎﻭﺭ ﻣﺪﺭﺳﻪ ﺭﺷﺪ ﺁﻣﻮﺯﺵ ﺗﺮﺑﻴﺖﺑﺪﻧﻰ ﺭﺷﺪ ﺁﻣﻮﺯﺵ ﻋﻠﻮﻡ ﺍﺟﺘﻤﺎﻋﻰ ﺭﺷـﺪ ﺁﻣﻮﺯﺵ ﺗﺎﺭﻳﺦ ﺭﻳﺎﺿﻰ
ﺭﺷـﺪ ﺁﻣﻮﺯﺵ ﺟﻐﺮﺍﻓﻴﺎ ﺭﺷـﺪ ﺁﻣﻮﺯﺵ ﺯﺑﺎﻥ ﺭﺷﺪ ﺁﻣﻮﺯﺵ
ﺭﺷـﺪ ﺁﻣﻮﺯﺵ ﻓﻴﺰﻳﻚ
ﺭﺷﺪ ﺁﻣﻮﺯﺵ ﺷﻴﻤﻰ
ﺭﺷﺪ ﺁﻣﻮﺯﺵ ﺯﻳﺴﺖﺷﻨﺎﺳﻰ
ﺭﺷﺪ ﺁﻣﻮﺯﺵ ﺯﻣﻴﻦﺷﻨﺎﺳﻰ ﺭﺷﺪ ﺁﻣﻮﺯﺵ ﻓﻨﻰﻭﺣﺮﻓﻪﺍﻯ ﺭﺷﺪ ﺁﻣﻮﺯﺵ ﭘﻴﺶ ﺩﺑﺴﺘﺎﻧﻰ
ﻣﺠﻠﻪﻫﺎﻯ ﺭﺷــﺪ ﻋﻤﻮﻣــﻰ ﻭ ﺍﺧﺘﺼﺎﺻﻰ ﺑــﺮﺍﻯ ﺁﻣﻮﺯﮔﺎﺭﺍﻥ ،ﻣﻌﻠﻤــﺎﻥ ،ﻣﺪﻳﺮﺍﻥ ﻭ ﻛﺎﺭﻛﻨــﺎﻥ ﺍﺟﺮﺍﻳــﻰ ﻣــﺪﺍﺭﺱ ،ﺩﺍﻧﺶﺟﻮﻳﺎﻥ ﻣﺮﺍﻛــﺰ ﺗﺮﺑﻴﺖﻣﻌﻠﻢ ﻭ ﺭﺷــﺘﻪﻫﺎﻯ ﺩﺑﻴﺮﻯ ﺩﺍﻧﺸــﮕﺎﻩﻫﺎ ﻭ ﻛﺎﺭﺷﻨﺎﺳــﺎﻥ ﺗﻌﻠﻴﻢ ﻭ ﺗﺮﺑﻴﺖ ﺗﻬﻴﻪ ﻭ ﻣﻨﺘﺸــﺮ ﻣﻰﺷﻮﻧﺪ.
ﻧﺸـﺎﻧﻰ :ﺗﻬــﺮﺍﻥ ،ﺧﻴﺎﺑــﺎﻥ ﺍﻳﺮﺍﻧﺸــﻬﺮ ﺷﻤﺎﻟﻰ،ﺳــﺎﺧﺘﻤﺎﻥ ﺷــﻤﺎﺭﻩﻯ4 ﺁﻣﻮﺯﺵﻭﭘﺮﻭﺭﺵ ،ﭘﻼﻙ ،266ﺩﻓﺘﺮ ﺍﻧﺘﺸﺎﺭﺍﺕ ﻛﻤﻚﺁﻣﻮﺯﺷﻰ. ﺗﻠﻔﻦ ﻭ ﻧﻤﺎﺑﺮ 88301478 :ـ 021
ﺩﻭﺭﺓ ﺷﺎﻧﺰﺩﻫﻢ ،ﺷﻤﺎﺭﺓ ،4ﺗﺎﺑﺴﺘﺎﻥ 1390
ﺭﺍﻫﻨﻤﺎﻳﻲ
33
ﺷﺮﺍﻳﻂ: .1ﭘﺮﺩﺍﺧﺖ ﻣﺒﻠﻎ 70/000ﺭﻳﺎﻝ ﺑﻪ ﺍﺯﺍﻯ ﻳﻚ ﺩﻭﺭﻩ ﻳﻚ ﺳﺎﻟﻪ ﻣﺠﻠﻪﻯ ﺩﺭﺧﻮﺍﺳﺘﻰ، ﺑﻪ ﺻﻮﺭﺕ ﻋﻠﻰﺍﻟﺤﺴﺎﺏ ﺑﻪ ﺣﺴﺎﺏ ﺷﻤﺎﺭﻩﻯ 39662000ﺑﺎﻧﻚ ﺗﺠﺎﺭﺕ ﺷﻌﺒﻪﻯ ﺳﻪ ﺭﺍﻩ ﺁﺯﻣﺎﻳﺶ )ﺳﺮﺧﻪﺣﺼﺎﺭ( ﻛﺪ 395ﺩﺭ ﻭﺟﻪ ﺷﺮﻛﺖ ﺍﻓﺴﺖ. .2ﺍﺭﺳﺎﻝ ﺍﺻﻞ ﻓﻴﺶ ﺑﺎﻧﻜﻰ ﺑﻪ ﻫﻤﺮﺍﻩ ﺑﺮگ ﺗﻜﻤﻴﻞ ﺷﺪﻩ ﻯ ﺍﺷﺘﺮﺍﻙ ﺑﺎﭘﺴﺖﺳﻔﺎﺭﺷﻰ) .ﻛﭙﻰﻓﻴﺶﺭﺍﻧﺰﺩﺧﻮﺩﻧﮕﻪ ﺩﺍﺭﻳﺪ(. ﻧﺎﻡ ﻣﺠﻠﻪ ﻫﺎﻯﺩﺭﺧﻮﺍﺳﺘﻰ: ...................................................................................... ...................................................................................... ......................................................................................
ﻧﺎﻡ ﻭ
ﻧﺎﻡﺧﺎﻧﻮﺍﺩﮔﻰ. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .:
ﺗﺎﺭﻳﺦ
ﺗﻮﻟﺪ. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .:
ﻣﻴﺰﺍﻥ ﺗﺤﺼﻴﻼﺕ:
.................................................................
ﺗﻠﻔﻦ. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .:
ﻧﺸﺎﻧﻰ ﻛﺎﻣﻞ ﭘﺴﺘﻰ:
ﺍﺳﺘﺎﻥ . . . . . . . . . . . . . . . . . . . . . . . . . . . . :ﺷﻬﺮﺳﺘﺎﻥ: ﺧﻴﺎﺑﺎﻥ. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . : ﻛﺪﺍﺷﺘﺮﺍﻙ. . . . . . . . . . . . . . . . . . . . . . . . : ﭘﻼﻙ . . . . . . . . . . . . . . . . . . . :ﺷﻤﺎﺭﻩﻯ ﭘﺴﺘﻰ. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . : .............................
ﺩﺭ ﺻﻮﺭﺗﻰ ﻛﻪ ﻗﺒ ًﻼ ﻣﺸﺘﺮﻙ ﻣﺠﻠﻪ ﺑﻮﺩﻩﺍﻳﺪ ،ﺷﻤﺎﺭﻩﻯ ﺍﺷﺘﺮﺍﻙ ﺧﻮﺩ ﺭﺍ ﺑﻨﻮﻳﺴﻴﺪ:
ﺍﻣﻀﺎ:
ﺻﻨـﺪﻭﻕ ﭘﺴﺘﻰ ﻣﺮﻛﺰﺑﺮﺭﺳﻰﺁﺛﺎﺭ: ﺻﻨـﺪﻭﻕ ﭘﺴﺘﻰ ﺍﻣﻮﺭﻣﺸﺘﺮﻛﻴﻦ:
15875/6567 16595/111
ﻧﺸﺎﻧﻰ ﺍﻳﻨﺘﺮﻧﺘﻰ:
www.roshdmag.ir
ﺍﻣﻮﺭ ﻣﺸﺘﺮﻛﻴﻦ:
77335110ـ 77336656ـ021
ﭘﻴﺎﻡﮔﻴﺮ ﻣﺠﻠﻪ ﻫﺎﻯ ﺭﺷﺪ:
88301482ـ021
ﻳﺎﺩﺁﻭﺭﻯ:
ﻫﺰﻳﻨﻪﻯ ﺑﺮﮔﺸﺖ ﻣﺠﻠﻪ ﺩﺭ ﺻﻮﺭﺕ ﺧﻮﺍﻧﺎ ﻭ ﻛﺎﻣﻞ ﻧﺒﻮﺩﻥ ﻧﺸﺎﻧﻰ ﻭ ﻋﺪﻡ ﺣﻀﻮﺭ ﮔﻴﺮﻧﺪﻩ ،ﺑﺮﻋﻬﺪﻩﻯ ﻣﺸﺘﺮﻙ ﺍﺳﺖ. ﻣﺒﻨﺎﻯ ﺷﺮﻭﻉ ﺍﺷﺘﺮﺍﻙ ﻣﺠﻠﻪ ﺍﺯ ﺯﻣﺎﻥ ﺩﺭﻳﺎﻓﺖ ﺑﺮگ ﺍﺷﺘﺮﺍﻙ ﺧﻮﺍﻫﺪ ﺑﻮﺩ.
34
ﺭﺍﻫﻨﻤﺎﻳﻲ
ﺩﻭﺭﺓ ﺷﺎﻧﺰﺩﻫﻢ ،ﺷﻤﺎﺭﺓ ،4ﺗﺎﺑﺴﺘﺎﻥ 1390
ﺩﺍﻧﺶﺍﻓﺰﺍﻳﻲ
ﺑﺮگ ﺍﺷﺘﺮﺍﻙ ﻣﺠﻠﻪﻫﺎﻯ ﺭﺷﺪ
ﺷﺎﺩﻱ ﺑﻬﺎﺭﻱ
ﻛﻠﻴﺪﻭﺍژﻩﻫـﺎ :ﺗﻘﺴــﻴﻢ ،ﺑﺎﻗﻲﻣﺎﻧﺪﻩ ،ﺗﻘﺴــﻴﻢ ﺷﻜﻼﺕ ،ﺳﻬﻢ ﻓﺮﺯﻧﺪﺍﻥ.
دو ﹞︧﹚﹤ی ︗︀﹛︉ ﻣﺴﺌﻠﻪﻱ ﺍﻭﻝ :ﻣﺎﺩﺭﻱ 25ﺷــﻜﻼﺕ ﺩﺍﺷﺖ .ﺍﻭ ﻣﻰﺧﻮﺍﺳﺖ ﺍﻳﻦ 25ﺷــﻜﻼﺕ ﺭﺍ ﺑﻴﻦ 5ﻓﺮﺯﻧﺪﺵ ﺗﻘﺴــﻴﻢ ﻛﻨﺪ .ﺍﻭ ﺑﻪ ﺗﺮﺗﻴﺐ ﺯﻳﺮ ﻋﻤﻞ ﻛﺮﺩ: 1 ﻳﻚ ﺷــﻜﻼﺕ ﺑﻪ ﻫﻤﺮﺍﻩ ﺷﻜﻼﺕﻫﺎﻯ ﺑﺎﻗﻰﻣﺎﻧﺪﻩ ﺑﺮﺍﻯ ﻓﺮﺯﻧﺪ 6 ﺍﻭﻝ. 1 ﺩﻭ ﺷــﻜﻼﺕ ﺑﻪ ﻫﻤﺮﺍﻩ ﺷــﻜﻼﺕﻫﺎﻯ ﺑﺎﻗﻰﻣﺎﻧﺪﻩ ﺑﺮﺍﻯ ﻓﺮﺯﻧﺪ 6 ﺩﻭﻡ. 1 ﺳﻪ ﺷﻜﻼﺕ ﺑﻪ ﻫﻤﺮﺍﻩ ﺷــﻜﻼﺕﻫﺎﻯ ﺑﺎﻗﻰﻣﺎﻧﺪﻩ ﺑﺮﺍﻯ ﻓﺮﺯﻧﺪ 6 ﺳﻮﻡ. 1 ﭼﻬﺎﺭ ﺷﻜﻼﺕ ﺑﻪ ﻫﻤﺮﺍﻩ ﺷﻜﻼﺕﻫﺎﻯ ﺑﺎﻗﻰﻣﺎﻧﺪﻩ ﺑﺮﺍﻯ ﻓﺮﺯﻧﺪ 6 ﭼﻬﺎﺭﻡ. 1 ﭘﻨﺞ ﺷــﻜﻼﺕ ﺑﻪ ﻫﻤﺮﺍﻩ ﺷﻜﻼﺕﻫﺎﻯ ﺑﺎﻗﻰﻣﺎﻧﺪﻩ ﺑﺮﺍﻯ ﻓﺮﺯﻧﺪ 6 ﭘﻨﺠﻢ. ﺍﺑﺘﺪﺍ ﺑﺪﻭﻥ ﻣﺤﺎﺳﺒﻪ ﺣﺪﺱ ﺑﺰﻧﻴﺪ ﭼﻪ ﻛﺴﻰ ﺷﻜﻼﺕﻫﺎﻯ ﺑﻴﺶﺗﺮﻯ ﮔﺮﻓﺘﻪ ﺍﺳــﺖ .ﺳــﭙﺲ ﺑﺎ ﻣﺤﺎﺳﺒﻪ ،ﺩﺭﺳﺘﻰ ﻳﺎ ﻧﺎﺩﺭﺳــﺘﻰ ﺣﺪﺳﺘﺎﻥ ﺭﺍ ﺑﺮﺭﺳﻰ ﻛﻨﻴﺪ. ﻣﺴـﺌﻠﻪﻱ ﺩﻭﻡ :ﭘﺪﺭﻱ ﺗﻌﺪﺍﺩﻯ ﺷﻜﻼﺕ ﺩﺍﺷﺖ .ﺍﻭ ﺷﻜﻼﺕﻫﺎ ﺭﺍ ﺑﻪ ﺗﺮﺗﻴﺐ ﺯﻳﺮ ﺑﻴﻦ ﻓﺮﺯﻧﺪﺍﺵ ﺗﻘﺴﻴﻢ ﻛﺮﺩ. ﻳﻚ ﺷــﻜﻼﺕ ﺑﻪ ﻫﻤﺮﺍﻩ 1ﺷﻜﻼﺕﻫﺎﻯ ﺑﺎﻗﻰﻣﺎﻧﺪﻩ ﺑﺮﺍﻯ ﻓﺮﺯﻧﺪ 5 ﺍﻭﻝ.
ﺩﻭ ﺷــﻜﻼﺕ ﺑﻪ ﻫﻤﺮﺍﻩ ﺩﻭﻡ. ﺳﻪ ﺷــﻜﻼﺕ ﺑﻪ ﻫﻤﺮﺍﻩ ﺳﻮﻡ.
1 5
ﺷــﻜﻼﺕﻫﺎﻯ ﺑﺎﻗﻰﻣﺎﻧﺪﻩ ﺑﺮﺍﻯ ﻓﺮﺯﻧﺪ 1 5
ﺷﻜﻼﺕﻫﺎﻯ ﺑﺎﻗﻰﻣﺎﻧﺪﻩ ﺑﺮﺍﻯ ﻓﺮﺯﻧﺪ
ﻭ ﺑﻪ ﻫﻤﻴﻦ ﺗﺮﺗﻴﺐ ﺑﺮﺍﻯ ﻓﺮﺯﻧﺪﺍﻥ ﺑﻌﺪﻯ! ﺩﺭ ﭘﺎﻳﺎﻥ ،ﺗﻌﺪﺍﺩ ﺷــﻜﻼﺕﻫﺎﻳﻰ ﻛﻪ ﻫﺮ ﻳــﻚ ﺍﺯ ﻓﺮﺯﻧﺪﺍﻧﺶ ﮔﺮﻓﺘﻪ ﺑﻮﺩﻧﺪ ﺑﺎ ﺑﻘﻴﻪ ﻣﺴﺎﻭﻯ ﺑﻮﺩ!!! ﻓﻜﺮ ﻣﻰﻛﻨﻴﺪ ﺍﻭ ﭼﻨﺪ ﺷﻜﻼﺕ ﺭﺍ ﺑﻴﻦ ﭼﻨﺪ ﻓﺮﺯﻧﺪﺵ ﺗﻘﺴﻴﻢ ﻛﺮﺩﻩ ﺍﺳﺖ؟ ﭘﺎﺳﺦ ﻣﺴﺌﻠﻪﻱ ﺍﻭﻝ: ﺑﺎﻗﻰﻣﺎﻧﺪﻩ
⇒ 25 − 5 = 20
24 =5 6
1+
ﺑﺎﻗﻰﻣﺎﻧﺪﻩ
20 − 5 = 15
⇒
18 =5 6
2+
ﻓﺮﺯﻧﺪ ﺩﻭﻡ
ﺑﺎﻗﻰﻣﺎﻧﺪﻩ
⇒ 15 − 5 = 10
12 =5 6
3+
ﻓﺮﺯﻧﺪ ﺳﻮﻡ
ﺑﺎﻗﻰﻣﺎﻧﺪﻩ
10 − 5 = 5
4+
ﻓﺮﺯﻧﺪ ﭼﻬﺎﺭﻡ
ﺑﺎﻗﻰﻣﺎﻧﺪﻩ
=5
0 6
ﻧﺘﻴﺠﻪ ﺳﻬﻢ
ﻗﺎﺑﻞ ﻣﺤﺎﺳﺒﻪ ﻧﻴﺴﺖ ×
ﻓﺮﺯﻧﺪ ﺍﻭﻝ
5+
ﭘﺎﺳﺦ ﻣﺴﺌﻠﻪﻱ ﺩﻭﻡ: 1 ﺑﻪ ﻓﺮﺯﻧﺪ ﺍﻭﻟﺶ ﻳﻚ ﺷﻜﻼﺕ ﺑﻪ ﻫﻤﺮﺍﻩ ﺷﻜﻼﺕﻫﺎﻯ ﺑﺎﻗﻰﻣﺎﻧﺪﻩ 5 ﺩﺍﺩﻩ ﺍﺳــﺖ ،ﭘﺲ ﭘﺎﺳــﺦ ﺍﺯ ﻳﻜﻰ ﺍﺯ ﻣﻀﺮﺏﻫﺎﻯ 5ﻳﻚ ﻭﺍﺣﺪ ﺑﻴﺶﺗﺮ ﺍﺳﺖ. ﺩﺭ ﺿﻤﻦ ،ﺗﻌﺪﺍﺩ ﻛﻞ ﺷــﻜﻼﺕﻫﺎ ﺑﺎﻳﺪ ﻣﻀﺮﺑﻰ ﺍﺯ ﻓﺮﺯﻧﺪ ﺍﻭﻝ ﺑﺎﺷﺪ، ﺯﻳﺮﺍ ﺳﻬﻢ ﻫﻤﻪﻯ ﻓﺮﺯﻧﺪﺍﻥ ﻣﺴﺎﻭﻯ ﺑﻮﺩﻩ ﺍﺳﺖ. ﺑﻪ ﺟﺪﻭﻝ ﺯﻳﺮ ﻭ ﺣﺪﺱﻫﺎﻳﻰ ﻛﻪ ﺯﺩﻩﺍﻳﻢ ﻧﮕﺎﻩ ﻛﻨﻴﺪ:
⇒
6 =5 6
ﻓﺮﺯﻧﺪ ﺩﻭﻡ ﻧﺘﻴﺠﻪ
2 5
2+ ×
ﺳﻬﻢ ﻓﺮﺯﻧﺪﺍﻥ 10 2+ ﺍﻭﻝ ﻭ ﺩﻭﻡ = 4 5 ﻣﺴﺎﻭﻱ ﺍﺳﺖ
ﺳﻬﻢ ﻓﺮﺯﻧﺪ ﺍﻭﻝ
6ﻣﻀﺮﺏ 2 ﺍﺳﺖ
5 =2 5
11ﻣﻀﺮﺏ 3ﻧﻴﺴﺖ
10 =3 5
16ﻣﻀﺮﺏ 4ﻫﺴﺖ
ﺗﻌﺪﺍﺩ ﺷﻜﻼﺕﻫﺎ
1+
6
1+
11
15 1+ =4 5
16
ﺑﺎ ﺗﻮﺟﻪ ﺑﻪ ﺟﺪﻭﻝ ﺑﺎﻻ ﺑﻪ ﻧﻈﺮ ﻣﻰﺭﺳﺪ ﺣﺪﺱ 16ﺣﺪﺱ ﺩﺭﺳﺘﻰ ﺑﺎﺷﺪ. 5 0 4+ ﺳﻬﻢ ﻓﺮﺯﻧﺪ ﺳﻮﻡ ﻧﻴﺰ 3 + = 4ﻭ ﺳﻬﻢ ﻓﺮﺯﻧﺪ ﭼﻬﺎﺭﻡ = 4 5 5 ﺷﻜﻼﺕ ﺧﻮﺍﻫﺪ ﺑﻮﺩ.
ﻓﺮﺯﻧﺪ ﭘﻨﺠﻢ
ﺟﺎﻟﺐ ﻧﻴﺴﺖ .ﻫﻤﻪﻯ ﺁﻥﻫﺎ ﺑﻪ ﺗﻌﺪﺍﺩ ﻣﺴﺎﻭﻯ ﺷﻜﻼﺕ ﮔﺮﻓﺘﻪﺍﻧﺪ!!!
ﺩﻭﺭﺓ ﺷﺎﻧﺰﺩﻫﻢ ،ﺷﻤﺎﺭﺓ ،4ﺗﺎﺑﺴﺘﺎﻥ 1390
ﺭﺍﻫﻨﻤﺎﻳﻲ
35
ﺩﺍﻧﺶﺍﻓﺰﺍﻳﻲ
﹏ ﹞︧﹚﹤ ﹇︡م ︋﹤ ﹇︡م ﺳﺎﻳﻪ ﻣﻬﺮﺑﺎﻥ
ﻛﻠﻴﺪﻭﺍژﻩﻫﺎ :ﻣﺴﺎﺣﺖ ،ﺷﻜﻞﻫﺎﻱ ﻣﻌﺎﺩﻝ ،ﻧﺴﺒﺖ ﻣﺴﺎﺣﺖﻫﺎ.
ﭼﻜﻴﺪﻩ :ﻳﻚ ﻣﺴــﺌﻠﻪ ﺩﺭ ﻣﻮﺿﻮﻉ ﻣﺴــﺎﺣﺖ ﻣﺜﻠﺚ ﻗﺪﻡ ﺑﻪ ﻗﺪﻡ، ﺍﺳﺘﻔﺎﺩﻩ ﺍﺯ ﺭﺍﻫﺒﺮﺩ ﺣﻞ ﻣﺴﺌﻠﻪ ﻭ ﺑﻪ ﻛﺎﺭﮔﻴﺮﻱ ﻋﺒﺎﺭﺕﻫﺎﻱ ﺟﺒﺮﻱ ﺣﻞ ﺷﺪﻩ ﺍﺳﺖ.
ﺑﺮﺍﻱ ﺷــﺮﻭﻉ ﻣﺤﻞ ﺑﺮﺧــﻮﺭﺩ BQﻭ CPﺭﺍ Oﺑﻨﺎﻣﻴﺪ ﻭ AOﺭﺍ ﺭﺳــﻢ ﻛﻨﻴﺪ .ﻣﺴﺎﺣﺖ AOQﺭﺍ ﺑﺮﺍﺑﺮ aﻭ ﻣﺴﺎﺣﺖ AOPﺭﺍ ﺑﺮﺍﺑﺮ b A ﻓﺮﺽ ﻛﻨﻴﺪ.
ﺑﻪ ﺷﻜﻞ ﺯﻳﺮ ﻧﮕﺎﻩ ﻛﻨﻴﺪ.
a
Q
A
8
?
Q
10
P
10
5
O
B
C
5
8
b
P
B
-1ﺷﻜﻞ ﺯﻳﺮ ﻗﺴــﻤﺘﻲ ﺍﺯ ﺷﻜﻞ ﺑﺎﻻﺳﺖ .ﺑﺎ ﺗﻮﺟﻪ ﺑﻪ ﺁﻥ ،ﺟﺎﻫﺎﻱ ﺧﺎﻟﻲ ﺭﺍ ﺩﺭ ﻋﺒﺎﺭﺕ ﺯﻳﺮ ﭘﺮ ﻛﻨﻴﺪ: H
C
Q ﺩﺭ ﻣﺜﻠﺚ ABCﺩﻭ ﭘﺎﺭﻩ ﺧﻂ BQﻭ CPﺭﺳﻢ ﺷﺪﻩﺍﻧﺪ ﻭ ﻣﺜﻠﺚ ﺑﻪ ﭼﻬﺎﺭ ﺗﻜﻪ ﺗﻘﺴﻴﻢ ﺷﺪﻩ ﺍﺳﺖ .ﻣﺴﺎﺣﺖ ﺳﻪ ﺗﺎ ﺍﺯ ﺗﻜﻪﻫﺎ ﺩﺭﻭﻥ ﺁﻥﻫﺎ ﻧﻮﺷﺘﻪ ﺷﺪﻩ ﺍﺳﺖ .ﻣﺴﺎﺣﺖ ﺗﻜﻪ ﭼﻬﺎﺭﻡ ﺭﺍ ﺑﻪ ﺩﺳﺖ ﺁﻭﺭﻳﺪ. ﻣﺴــﺌﻠﻪ ﺑﻪ ﻧﻈﺮ ﻣﺸﻜﻞ ﺍﺳــﺖ! ﺍﻳﻦﻃﻮﺭ ﻧﻴﺴﺖ؟ ﻗﺪﻡ ﺑﻪ ﻗﺪﻡ ﺑﺎ ﻣﺎ ﭘﻴﺶ ﺑﻴﺎﻳﻴﺪ ﺗﺎ ﺑﺘﻮﺍﻧﻴﺪ ﺍﻳﻦ ﻣﺴﺌﻠﻪ ﺭﺍ ﺣﻞ ﻛﻨﻴﺪ. 36
ﺭﺍﻫﻨﻤﺎﻳﻲ
ﺩﻭﺭﺓ ﺷﺎﻧﺰﺩﻫﻢ ،ﺷﻤﺎﺭﺓ ،4ﺗﺎﺑﺴﺘﺎﻥ 1390
O
8 10 C
B
8 10
=
QO .......
=
QO × ...........
=
BO × ...........
ﻣﺴﺎﺣﺖ ﻣﺜﻠﺚ
OCQ
ﻣﺴﺎﺣﺖ ﻣﺜﻠﺚ
OCB
-5ﺷــﻜﻞ ﺭﻭﺑﺮﻭ ﻗﺴﻤﺖ ﺩﻳﮕﺮﻱ ﺍﺯ ﺷــﻜﻞ ﺍﺳﺖ .ﺑﺎ ﺗﻮﺟﻪ ﺑﻪ ﺁﻥ، ﺟﺎﻫﺎﻱ ﺧﺎﻟﻲ ﺭﺍ ﺩﺭ ﻋﺒﺎﺭﺕ ﺯﻳﺮ ﭘﺮ ﻛﻨﻴﺪA :
-2ﺷــﻜﻞ ﺭﻭﺑﺮﻭ ﻗﺴﻤﺖ ﺩﻳﮕﺮﻱ ﺍﺯ ﺷــﻜﻞ ﺍﺳﺖ .ﺑﺎ ﺗﻮﺟﻪ ﺑﻪ ﺁﻥ، ﺟﺎﻫﺎﻱ ﺧﺎﻟﻲ ﺭﺍ ﺩﺭ ﻋﺒﺎﺭﺕ ﺯﻳﺮ ﭘﺮ ﻛﻨﻴﺪ:
P
A
O
Q H
H
C O
QO×............. QO a = = BO×............ ....... 5+ b
ﻣﺴﺎﺣﺖ ﻣﺜﻠﺚ
-6ﺗﺴــﺎﻭﻱ ﺯﻳﺮ ﺍﺯ ﺩﻭ ﺗﺴﺎﻭﻱ ﺣﺎﺻﻞ ﺍﺯ ﺳﺆﺍﻻﺕ 4ﻭ 5ﺑﻪ ﺩﺳﺖ ﺁﻣﺪﻩ ﺍﺳﺖ: 10 a + 8
B
=
OC × ........... OC a +8 = = = OP ×........... ...... ......
ﻣﺴﺎﺣﺖ ﻣﺜﻠﺚ
AOC AOP
b
ﻣﺴﺎﺣﺖ ﻣﺜﻠﺚ ﻣﺴﺎﺣﺖ ﻣﺜﻠﺚ
OAQ OAB
-3ﺗﺴــﺎﻭﻱ ﺯﻳﺮ ﺍﺯ ﺩﻭ ﺗﺴﺎﻭﻱ ﺣﺎﺻﻞ ﺍﺯ ﺳﺆﺍﻻﺕ 1ﻭ 2ﺑﻪ ﺩﺳﺖ ﺁﻣﺪﻩ ﺍﺳﺖ: 8 a = 10 5 + b
ﺑﺎ ﺗﻮﺟﻪ ﺑﻪ ﺭﺍﺑﻄﻪﻱ ﺑﺎﻻ ،ﺗﺴﺎﻭﻱ ﺭﻭﺑﺮﻭ ﺭﺍ ﻛﺎﻣﻞ ﻛﻨﻴﺪ: 10a=40+.......
=
5
ﺑﺎ ﺗﻮﺟﻪ ﺑﻪ ﺍﻳﻦ ﺭﺍﺑﻄﻪ ،ﺗﺴﺎﻭﻱ ﺭﻭﺑﺮﻭ ﺭﺍ ﻛﺎﻣﻞ ﻛﻨﻴﺪ: 10b=5a+... -7ﺩﻭ ﺗﺴــﺎﻭﻱ ﺑﻪ ﺩﺳــﺖ ﺁﻣﺪﻩ ﺩﺭ ﺳــﺆﺍﻻﺕ 3ﻭ 6ﺭﺍ ﻛﻨﺎﺭ ﻫﻢ ﺑﮕﺬﺍﺭﻳﺪ ﻭ ﺳــﻌﻲ ﻛﻨﻴﺪ ﻣﻘﺎﺩﻳﺮﻱ ﺑــﺮﺍﻱ aﻭ bﭘﻴﺪﺍ ﻛﻨﻴﺪ ﻛﻪ ﻫﺮ ﺩﻭ ﺗﺴﺎﻭﻱ ﺑﺮﻗﺮﺍﺭ ﺑﺎﺷﻨﺪ. -8ﺍﮔﺮ ﺑﻪ a=12ﻭ b=10ﺭﺳــﻴﺪﻩﺍﻳﺪ ،ﻛﺎﺭﺗﺎﻥ ﺩﺭﺳﺖ ﺍﺳﺖ! ﺁﻳﺎ ﻣﻲﺗﻮﺍﻧﻴﺪ ﺭﻭﺵ ﺩﻳﮕﺮﻱ ﺑﺮﺍﻱ ﺣﻞ ﻣﺴــﺌﻠﻪﻱ ﺍﻭﻟﻴﻪ ﭘﻴﺪﺍ ﻛﻨﻴﺪ؟ ﺳﻌﻲ ﻛﻨﻴﺪ!
-4ﺷﻜﻞ ﺯﻳﺮ ﻗﺴﻤﺖ ﺩﻳﮕﺮﻱ ﺍﺯ ﺷﻜﻞ ﻗﺒﻠﻲ ﺍﺳﺖ .ﺑﺎ ﺗﻮﺟﻪ ﺑﻪ ﺁﻥ، ﺟﺎﻫﺎﻱ ﺧﺎﻟﻲ ﺭﺍ ﺩﺭ ﻋﺒﺎﺭﺕ ﺯﻳﺮ ﭘﺮ ﻛﻨﻴﺪ:
P
O
C
H
B
OC × ............. OC = = O OP ×............. ......
=
ﻣﺴﺎﺣﺖ ﻣﺜﻠﺚ ﻣﺴﺎﺣﺖ ﻣﺜﻠﺚ
BOC BOP
ﺩﻭﺭﺓ ﺷﺎﻧﺰﺩﻫﻢ ،ﺷﻤﺎﺭﺓ ،4ﺗﺎﺑﺴﺘﺎﻥ 1390
ﺭﺍﻫﻨﻤﺎﻳﻲ
37
ﻣﻌﻤﺎ ﻭ ﺳﺮﮔﺮﻣﻲ
﹞︺﹝︀﹨︀︣︋ ︣﹊︋ ﹩ای ︑︀︋︧︐︀ن ﻋﻠﻴﺮﺿﺎ ﻳﻮﺳﻔﻲ
ﻛﻠﻴﺪﻭﺍژﻩﻫﺎ :ﻣﻌﻤﺎ ﻭ ﺳﺮﮔﺮﻣﻲ ،ﻣﺜﻠﺚ ،ﺧﻂ ﻣﺴﺘﻘﻴﻢ ،ﻗﻄﺎﺭ ،ﺳﺎﻋﺖ. .1ﻣﺜﻠﺚ ﺟﺎﺩﻭﻳﻲ؛ ﺳــﻪ ﻋﺪﺩ 2 ،1ﻭ 3ﺩﺭ ﺭﺃﺱﻫﺎﻱ ﻳﻚ ﻣﺜﻠﺚ ﻗﺮﺍﺭ ﺩﺍﺩﻩ ﺷــﺪﻩ ﺍﺳﺖ .ﺣﺎﻝ ﺍﻋﺪﺍﺩ 8 ،7 ،6 ،5 ،4ﻭ 9ﺭﺍ ﺩﺭ ﺍﺿﻼﻉ ﺁﻥ ﻃﻮﺭﻱ ﻗﺮﺍﺭ ﺩﻫﻴﺪ ﻛﻪ ﺟﻤﻊ ﻫﺮ ﺿﻠﻊ )ﺑﺎ ﺭﺃﺱﻫﺎﻱ ﺁﻥ( 17ﺷﻮﺩ. ﻣﺴــﺌﻠﻪﻱ ﻣﺸــﻜﻞﺗﺮ :ﺣﺎﻝ ﺍﻋﺪﺍﺩ 1ﺗﺎ 9ﺭﺍ ﺑــﺪﻭﻥ ﺍﻳﻦﻛﻪ ﺑﺪﺍﻧﻴﻢ ﻛﺪﺍﻡ ﻳــﻚ ﺩﺭ ﺭﺃﺱﻫﺎﻱ ﻣﺜﻠﺚ ﻗﺮﺍﺭ ﺩﺍﺭﻧﺪ ،ﻃــﻮﺭﻱ ﺩﺭ ﺍﻃﺮﺍﻑ ﻣﺜﻠﺚ )ﺭﺃﺱﻫﺎ ﻭ ﺍﺿﻼﻉ( ﻗﺮﺍﺭ ﺩﻫﻴﺪ ﻛﻪ ﺟﻤﻊ ﺁﻥ 20ﺷﻮﺩ. .2ﺗـﻮپ ﺑﺎﺯﻱ ﺩﺧﺘﺮ ﺑﭽﻪﻫﺎ؛ 12ﺩﺧﺘﺮ ﺑﭽﻪ ﺩﺭ ﺩﺍﺧﻞ ﺩﺍﻳﺮﻩﺍﻱ ﺗﻮپ ﺭﺍ ﺑﺮﺍﻱ ﻳﻜﺪﻳﮕﺮ ﭘﺮﺗﺎﺏ ﻣﻲﻛﻨﻨﺪ؛ ﻫﺮﻳﻚ ﺑﺮﺍﻱ ﻧﻔﺮ ﺳــﻤﺖ ﭼﭙﻲ ﺧــﻮﺩ .ﺯﻣﺎﻧﻲ ﻛﻪ ﺗﻮپ ﻳــﻚ ﺩﻭﺭ ﻛﺎﻣﻞ ﺭﺍ ﺑﻪ ﺩﻭﺭ ﺩﺍﻳــﺮﻩ ﭼﺮﺧﻴﺪ ،ﺩﺭ ﺟﻬﺖ ﻋﻜﺲ ﺗــﻮپ ﺑﻪ ﻧﻘﻄﻪﻱ ﺍﻭﻝ ﺑﻪ ﻫﻤﺎﻥ ﺗﺮﺗﻴﺐ ﺑﺮ ﻣﻲﮔﺮﺩﺩ .ﺑﻌﺪ ﺍﺯ ﻣــﺪﺕ ﻛﻮﺗﺎﻫﻲ ﻳﻜﻲ ﺍﺯ ﺩﺧﺘﺮ ﺑﭽﻪﻫﺎ ﭘﻴﺸــﻨﻬﺎﺩ ﺩﺍﺩ» :ﺣﺎﻻ ﺗﻮپ ﺭﺍ ﻳــﻚ ﺩﺭ ﻣﻴــﺎﻥ ﭘﺮﺗﺎﺏ ﻛﻨﻴﻢ« .ﺍﻣﺎ ﺩﻳﮕﺮﻱ ﮔﻔــﺖ» :ﺍﮔﺮ ﻣﺎ ﺍﻳﻦ ﻛﺎﺭ ﺭﺍ ﺍﻧﺠــﺎﻡ ﺩﻫﻴﻢ ﺗﺎ ﺯﻣﺎﻧﻲ ﻛﻪ 12ﻧﻔﺮ ﻫﺴــﺘﻴﻢ ﻧﺼﻒ ﺩﺧﺘﺮﻫﺎ ﻧﻤﻲﺗﻮﺍﻧﻨﺪ ﺑﺎﺯﻱ ﻛﻨﻨﺪ«. ﺍﻭﻟﻲ» :ﺑﻨﺎﺑﺮﺍﻳﻦ 2ﺩﺭ ﻣﻴﺎﻥ ﺍﻳﻦ ﻛﺎﺭ ﺭﺍ ﺍﻧﺠﺎﻡ ﺩﻫﻴﻢ«. ﺩﻭﻣــﻲ» :ﺍﻳﻦﻛﻪ ﺑﺪﺗﺮﻩ ،ﺑــﺎ ﺍﻳﻦ ﻛﺎﺭ ﺗﻨﻬﺎ 4ﻧﻔــﺮ ﻣﻲﺗﻮﺍﻧﻨﺪ ﺑﺎﺯﻱ ﻛﻨﻨــﺪ؛ ﻣﺎ ﺑﺎﻳﺪ 4ﺩﺭﻣﻴﺎﻥ ﺗــﻮپ ﺭﺍ ﭘﺮﺗﺎﺏ ﻛﻨﻴﻢ ﺗﺎ ﻧﻔــﺮ ﭘﻨﺠﻢ ﺁﻥ ﺭﺍ ﺑﮕﻴﺮﺩ؛ ﺗﺮﻛﻴﺐ ﺩﻳﮕﺮﻱ ﻭﺟﻮﺩ ﻧﺪﺍﺭﺩ«. ﺍﻭﻟﻲ» :ﻭ ﺍﮔﺮ ﻣﺎ 6ﻧﻔﺮ ﺭﺍ ﺟﺎ ﺑﮕﺬﺍﺭﻳﻢ؟« ﺩﻭﻣﻲ» :ﺍﻳﻦ ﻛﺎﺭ ﻣﺎﻧﻨﺪ ﺍﻳﻦ ﺍﺳﺖ ﻛﻪ ﻣﺎ 4ﻧﻔﺮ ﺭﺍ ﺟﺎ ﺑﮕﺬﺍﺭﻳﻢ ،ﺗﻨﻬﺎ ﺗﻮپ ﺩﺭ ﺟﻬﺖ ﻣﺨﺎﻟﻒ ﭘﺮﺗﺎﺏ ﻣﻲﺷﻮﺩ«. ﺍﻭﻟــﻲ» :ﻭ ﺍﮔــﺮ ﻣﺎ ﻫﺮ 10ﻧﻔــﺮ ﺭﺍ ﺟﺎ ﺑﮕﺬﺍﺭﻳﻢ ،ﺑــﻪ ﻃﻮﺭﻱﻛﻪ ﻧﻔﺮ ﻳﺎﺯﺩﻫﻢ ﺗﻮپ ﺭﺍ ﺑﮕﻴﺮﺩ ﭼﻪﻃﻮﺭ؟« ﺩﻭﻣﻲ» :ﺧﻮﺏ ﻫﻤﻴﻦ ﺣﺎﻻ ﻫﻢ ﺩﺍﺭﻳﻢ ﻫﻤﻴﻦ ﻛﺎﺭ ﺭﺍ ﻣﻲﻛﻨﻴﻢ«. ﺁﻥﻫﺎ ﺷــﺮﻭﻉ ﻛﺮﺩﻧﺪ ﺑﻪ ﻛﺸﻴﺪﻥ ﺷــﻜﻞﻫﺎﻱ ﻣﺨﺘﻠﻒ ﭘﺮﺗﺎﺏ ﺗﻮپ ﺑــﻪ ﻳﻜﺪﻳﮕﺮ ﻭ ﭘﺲ ﺍﺯ ﻣﺪﺕ ﻛﻮﺗﺎﻫﻲ ﻣﺘﻮﺟﻪ ﺷــﺪﻧﺪ ﻛﻪ ﺣﻖ ﺑﺎ ﺩﻭﻣﻲ ﻳﻌﻨﻲ ﺳﺎﺭﺍ ﺑﻮﺩ .ﻳﻌﻨﻲ ﺑﺮﺍﻱ ﺍﻳﻦﻛﻪ ﻫﻤﻪ ﺩﺭ ﺑﺎﺯﻱ ﺷﺮﻛﺖ ﻛﻨﻨﺪ ،ﻋﻼﻭﻩ ﺑــﺮ ﺣﺎﻟﺖ ﺍﻭﻝ ﻳﻌﻨﻲ ﺑﻪ ﺗﺮﺗﻴﺐ ﺑﻪ ﻫﻤﻪ ﺗﻮپ ﺭﺍ ﭘﺮﺗﺎﺏ ﻛﺮﺩﻥ ،ﺗﻨﻬﺎ ﺩﺭ 38
ﺭﺍﻫﻨﻤﺎﻳﻲ
ﺩﻭﺭﺓ ﺷﺎﻧﺰﺩﻫﻢ ،ﺷﻤﺎﺭﺓ ،4ﺗﺎﺑﺴﺘﺎﻥ 1390
ﺣﺎﻟﺖ 4ﺩﺭ ﻣﻴﺎﻥ ﻳﺎ ﺣﺎﻟﺖ ﻣﻜﻤﻞ ﺁﻥ ﻳﻌﻨﻲ 6ﺩﺭ ﻣﻴﺎﻥ ﺍﺳﺖ ﻛﻪ ﺗﻮپ ﺑﻪ ﻫﻤﻪ ﻣﻲﺭﺳﺪ ﻭ ﻫﻤﻪ ﻣﻲﺗﻮﺍﻧﻨﺪ ﺩﺭ ﺑﺎﺯﻱ ﺷﺮﻛﺖ ﻛﻨﻨﺪ )ﺷﻜﻞ ﺍﻟﻒ(. ﺍﻛﻨﻮﻥ ﺍﮔﺮ ﺗﻌﺪﺍﺩ ﻧﻔﺮﺍﺕ 13ﻧﻔﺮ ﺑﺎﺷــﺪ ،ﺗﻮپ ﺭﺍ ﻣﻲﺗﻮﺍﻥ ﻳﻚ ﺩﺭ ﻣﻴﺎﻥ )ﺷﻜﻞ ﺏ( 2 ،ﺩﺭ ﻣﻴﺎﻥ )ﺷﻜﻞ ﺝ( 3 ،ﺩﺭﻣﻴﺎﻥ )ﺷﻜﻞ چ( ﻳﺎ 4ﺩﺭﻣﻴﺎﻥ )ﺷــﻜﻞ ﺡ( ﺑﻪ ﻳﻜﺪﻳﮕﺮ ﭘﺮﺗﺎﺏ ﻛﺮﺩ ﺑﺪﻭﻥ ﺍﻳﻦﻛﻪ ﻛﺴﻲ ﺟﺎ ﺑﻤﺎﻧﺪ .ﺣﺎﻻ ﺍﮔﺮ ﺗﻮپ 5ﻳﺎ 6ﺩﺭ ﻣﻴﺎﻥ ﭘﺮﺗﺎﺏ ﺷﻮﺩ ﭼﻪﻃﻮﺭ؟ ﺷﻜﻞ ﺁﻥ ﺭﺍ ﺑﻜﺸﻴﺪ. 5
13 9
4
7
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1 10 6
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8
9 13
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10
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11
9
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13
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11
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5
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3
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6 13
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12 11
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.3ﭼﻬﺎﺭ ﺧﻂ ﻣﺴـﺘﻘﻴﻢ )ﺭﺍﺳـﺖ(؛ ﻣﺮﺑﻌﻲ ﺑﺎ 9ﻧﻘﻄﻪ ﺑﻪ ﻣﺎﻧﻨﺪ ﺷﻜﻞ ﺯﻳﺮ ﺑﺴﺎﺯﻳﺪ .ﺑﺪﻭﻥ ﺍﻳﻦﻛﻪ ﻣﺪﺍﺩ ﺭﺍ ﺍﺯ ﺭﻭﻱ ﻛﺎﻏﺬ ﺑﺮﺩﺍﺭﻳﺪ ،ﺗﻨﻬﺎ ﺑﺎ ﭼﻬﺎﺭ ﺧﻂ ﺭﺍﺳﺖ ﺗﻤﺎﻡ ﻧﻘﺎﻁ ﺭﺍ ﺑﻪ ﻳﻜﺪﻳﮕﺮ ﻣﺘﺼﻞ ﻛﻨﻴﺪ.
.4ﺟـﺪﺍ ﻛﺮﺩﻥ ﺑﺰﻫﺎ ﺍﺯ ﻛﻠﻢﻫﺎ؛ ﺍﻛﻨــﻮﻥ ﺑﻪ ﺟﺎﻱ ﻣﺘﺼﻞ ﻛﺮﺩﻥ ﻧﻘﺎﻁ ،ﺗﻨﻬﺎ ﺑﺎ 3ﺧﻂ ﺭﺍﺳــﺖ ﺗﻤﺎﻡ ﺑﺰﻫﺎ ﺭﺍ ﺍﺯ ﻛﻠﻢﻫﺎ ﺩﺭ ﺷﻜﻞ ﺯﻳﺮ ﺟﺪﺍ ﻛﻨﻴﺪ.
.7ﺻﻔﺤﻪﻱ ﺳـﺎﻋﺖ ﺟﻴﺒﻲ؛ ﺁﻳﺎ ﻣﻲﺗﻮﺍﻧﻴﺪ ﻛﺎﺭﻱ ﻛﻨﻴﺪ ﺗﺎ ﺑﺎ 2 ﺧﻂ ﺭﺍﺳــﺖ ،ﺻﻔﺤﻪﻱ ﺳﺎﻋﺖ ﺟﻴﺒﻲ ﺭﺍ ﻃﻮﺭﻱ ﺗﻘﺴﻴﻢ ﻛﻨﻴﺪ ﺗﺎ ﺟﻤﻊ ﺍﻋﺪﺍﺩ ﻫﺮ ﻗﺴﻤﺖ ﺑﺎ ﻳﻜﺪﻳﮕﺮ ﺑﺮﺍﺑﺮ ﺑﺎﺷﻨﺪ؟ ﺁﻳﺎ ﻣﻲﺗﻮﺍﻧﻴﺪ ﺁﻥ ﺭﺍ ﺑﻪ 6ﻗﺴــﻤﺖ ﺗﻘﺴــﻴﻢ ﻛﻨﻴﺪ ﺑﻪ ﻃﻮﺭﻱﻛﻪ ﻫﺮ ﻗﺴــﻤﺖ ﺷﺎﻣﻞ ﺩﻭ ﻋﺪﺩ ﺑﺎﺷــﺪ ﻛﻪ ﺟﻤﻊ ﻫﺮ ﺩﻭ ﻋﺪﺩ ﺩﺭ ﻫﺮ ﻗﺴﻤﺖ ﺑﺎ ﻳﻜﺪﻳﮕﺮ ﺑﺮﺍﺑﺮ ﺑﺎﺷﻨﺪ.
11 12 1 2
10
3
9 4
.5ﺩﻭ ﻗﻄـﺎﺭ؛ ﻗﻄﺎﺭﻱ ﺑــﺪﻭﻥ ﺗﻮﻗﻒ ﺑﺎ ﺳــﺮﻋﺖ 60ﻛﻴﻠﻮﻣﺘﺮ ﺩﺭ ﺳــﺎﻋﺖ ﺍﺯ ﺗﻬﺮﺍﻥ ﺑﻪ ﻣﻘﺼﺪ ﻣﺸﻬﺪ ﺣﺮﻛﺖ ﻣﻲﻛﻨﺪ ﻭ ﻗﻄﺎﺭ ﺩﻳﮕﺮﻱ ﺩﺭ ﻫﻤﺎﻥ ﺯﻣﺎﻥ ﺑﺎ ﺳــﺮﻋﺖ 40ﻛﻴﻠﻮﻣﺘﺮ ﺩﺭ ﺳﺎﻋﺖ ﺑﺪﻭﻥ ﺗﻮﻗﻒ ﺍﺯ ﻣﺸﻬﺪ ﺑﻪ ﻣﻘﺼﺪ ﺗﻬﺮﺍﻥ ﺣﺮﻛﺖ ﻣﻲﻛﻨﺪ .ﺣﺎﻝ ﻫﺮﻳﻚ ﺍﺯ ﺩﻭ ﻗﻄﺎﺭ ﻳﻚ ﺳــﺎﻋﺖ ﻗﺒﻞ ﺍﺯ ﻋﺒﻮﺭ ﺍﺯ ﻳﻜﺪﻳﮕﺮ ،ﭼﻪ ﻣﺴﺎﻓﺘﻲ ﭘﻴﻤﻮﺩﻩﺍﻧﺪ. .6ﺟﺰﺭ ﻭ ﻣﺪ ﺩﺭﻳﺎ؛ ﻛﺸــﺘﻲ ﺩﺭ ﻛﻨﺎﺭ ﺳــﺎﺣﻞ ﻟﻨﮕــﺮ ﺍﻧﺪﺍﺧﺘﻪ ﻭ ﻧﺮﺩﺑﺎﻧﻲ ﻛﻪ ﺍﺯ ﻃﻨﺎﺏ ﺳﺎﺧﺘﻪ ﺷﺪﻩ ﺍﺳﺖ ﺍﺯ ﻛﺸﺘﻲ ﺗﺎ ﺳﻄﺢ ﺩﺭﻳﺎ ﭘﺎﻳﻴﻦ ﺍﻧﺪﺍﺧﺘﻪ ﺷﺪﻩ ﺍﺳﺖ .ﺍﻳﻦ ﻧﺮﺩﺑﺎﻥ ﺩﺍﺭﺍﻱ 10ﭘﻠﻪ ﺍﺳﺖ ﻛﻪ ﺍﺯ ﻳﻜﺪﻳﮕﺮ ﺑﻪ ﺍﻧﺪﺍﺯﻩﻱ 12ﺍﻳﻨﭻ ) 30ﺳﺎﻧﺘﻲﻣﺘﺮ( ﻓﺎﺻﻠﻪ ﺩﺍﺭﻧﺪ ،ﺁﺧﺮﻳﻦ ﭘﻠﻪ ﺑﻪ ﺳﻄﺢ ﺁﺏ ﺩﺭﻳﺎ ﺭﺳــﻴﺪﻩ ﺍﺳﺖ .ﺩﺭﻳﺎ ﺩﺭ ﺣﺎﻝ ﺣﺎﺿﺮ ﺁﺭﺍﻡ ﺍﺳﺖ .ﺑﻪ ﺩﻟﻴﻞ ﺟﺰﺭ ﻭ ﻣﺪ ﺩﺭﻳﺎ ،ﺁﺏ ﺑﺎ ﺳــﺮﻋﺖ 4ﺍﻳﻨﭻ ) 10ﺳــﺎﻧﺘﻴﻤﺘﺮ( ﺩﺭ ﺳﺎﻋﺖ ﺑﺎﻻ ﻣﻲ ﺁﻳﺪ .ﺣﺎﻝ ﺑﮕﻮﻳﻴﺪ ﭼﻪ ﻣﺪﺕ ﻃﻮﻝ ﻣﻲﻛﺸــﺪ ﺗﺎ ﺁﺏ ﺑﻪ ﺳــﻮﻣﻴﻦ ﭘﻠﻪ ﺍﺯ ﺑﺎﻻﻱ ﻧﺮﺩﺑﺎﻥ ﺑﺮﺳﺪ.
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7
8
.8ﺻﻔﺤﻪﻱ ﺳـﺎﻋﺖ ﺷﻜﺴﺘﻪ ﺷﺪﻩ؛ ﺩﺭ ﻳﻚ ﻣﻮﺯﻩﻱ ﻗﺪﻳﻤﻲ، ﺳﺎﻋﺖ ﻗﺪﻳﻤﻲ ﺭﺍ ﺑﺎ ﺍﻋﺪﺍﺩ ﺭﻭﻣﻲ ﻗﺪﻳﻤﻲ ﺩﻳﺪﻡ ﻛﻪ ﺑﻪ ﺟﺎﻱ ﻋﺪﺩ ﺭﻭﻣﻲ ،(4) IVﻋﺪﺩ ﻗﺪﻳﻤﻲ IIIIﻗﺮﺍﺭ ﺩﺍﺷــﺖ .ﺗﺮﻛﻲ ﺩﺭ ﺻﻔﺤﻪﻱ ﺳــﺎﻋﺖ ﺁﻥ ﺭﺍ ﺑﻪ 4ﻗﺴــﻤﺖ ﺗﻘﺴﻴﻢ ﻛﺮﺩﻩ ﺑﻮﺩ .ﻫﻤﺎﻥﻃﻮﺭ ﻛﻪ ﺩﺭ ﺗﺼﻮﻳﺮ ﻧﺸﺎﻥ ﺩﺍﺩﻩ ﺷﺪﻩ ﺍﺳﺖ .ﺟﻤﻊ ﺍﻋﺪﺍﺩ ﻫﺮ ﻗﺴﻤﺖ ﻛﻪ ﺍﺯ 17ﺗﺎ 21ﺍﺳﺖ ﺑﺎ ﻫﻢ ﺑﺮﺍﺑﺮ ﻧﻴﺴــﺘﻨﺪ .ﺁﻳﺎ ﻣﻲﺗﻮﺍﻧﻴﺪ ﺗﻨﻬﺎ ﺑــﺎ ﺗﻐﻴﻴﺮ ﻳﻚ ﺗﺮﻙ ﻛﺎﺭﻱ ﻛﻨﻴﺪ ﻛﻪ ﺟﻤﻊ ﻫﺮ 4ﻗﺴﻤﺖ 20ﺷﻮﺩ؟ )ﺭﺍﻫﻨﻤﺎﻳﻲ :ﻫﺮ ﺗﺮﻛﻲ ﻛﻪ ﺗﻐﻴﻴﺮ ﻛﻨﺪ ﻧﺒﺎﻳﺪ ﺍﺯ ﻭﺳــﻂ ﺳﺎﻋﺖ ﻋﺒﻮﺭ ﻛﻨﺪ(.
.9ﺳـﺎﻋﺖ ﺷـﮕﻔﺖﺍﻧﮕﻴﺰ؛ ﺳﺎﻋﺖﺳﺎﺯﻱ ﺷــﺎﮔﺮﺩ ﺧﻮﺩ ﺭ ﺍﺑﺮﺍﻱ ﺗﻌﻮﻳﺾ ﻋﻘﺮﺑﻪﻫﺎﻱ ﺷﻜﺴــﺘﻪﻱ ﻳﻚ ﺳﺎﻋﺖ ﺩﻳﻮﺍﺭﻱ ﺩﺭ ﺧﺎﻧﻪﺍﻱ ﺑﺰﺭگ ﻭ ﻗﺪﻳﻤﻲ ﻣﻲﻓﺮﺳــﺘﺪ .ﺷــﺎﮔﺮﺩ ﺍﻭ ﺩﺭ ﺗﺎﺭﻳﻜﻲ ﺷﺐ ﺑﺎ ﻋﺠﻠﻪ ﻋﻘﺮﺑﻪﻫﺎﻱ ﺁﻥ ﺳــﺎﻋﺖ ﺭﺍ ﺗﻌﻮﻳﺾ ﻭ ﺁﻥ ﺭﺍ ﺑﺎ ﺳــﺎﻋﺖ ﺟﻴﺒﻲ ﺧﻮﺩ ﺗﻨﻈﻴﻢ ﻣﻲﻛﻨﺪ ﻭ ﻋﻘﺮﺑــﻪﻱ ﺑﺰﺭگ ﺭﺍ ﺭﻭﻱ 12ﻭ ﻋﻘﺮﺑﻪﻱ ﻛﻮﭼﻚ ﺭﺍ ﺭﻭﻱ ﻋﺪﺩ 6ﻗﺮﺍﺭ ﻣﻲﺩﻫﺪ ﺗﺎ ﺯﻣﺎﻥ 6ﺷــﺐ ﺭﺍ ﻧﺸــﺎﻥ ﺩﻫﺪ ﻭ ﺑﺎ ﻋﺠﻠﻪ ﺑﺮﻣﻲﮔﺮﺩﺩ .ﻫﻨﻮﺯ ﺍﺯ ﺭﺳــﻴﺪﻥ ﺍﻭ ﭼﻨﺪ ﺩﻗﻴﻘﻪﺍﻱ ﻧﻤﻲﮔﺬﺭﺩ ﻛﻪ ﺗﻠﻔﻦ ﺑﻪ ﺻﺪﺍ ﺩﺭﻣﻲﺁﻳﺪ ﻭ ﺩﻭﺭﺓ ﺷﺎﻧﺰﺩﻫﻢ ،ﺷﻤﺎﺭﺓ ،4ﺗﺎﺑﺴﺘﺎﻥ 1390
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ﺻﺎﺣــﺐ ﺁﻥ ﺧﺎﻧﻪﻱ ﺑﺰﺭگ ﺑﺎ ﻋﺼﺒﺎﻧﻴﺖ ﻣﻲﮔﻮﻳﺪ ﻛﻪ ﺳــﺎﻋﺖ ﺩﻳﻮﺍﺭﻱ ﺍﻭ ﻛﺎﻣﻞ ﺩﺭﺳــﺖ ﻧﺸﺪﻩ ﺍﺳــﺖ ﻭ ﺯﻣﺎﻥ ﺭﺍ ﺍﺷﺘﺒﺎﻩ ﻧﺸﺎﻥ ﻣﻲﺩﻫﺪ .ﻭﻗﺘﻲ ﺷﺎﮔﺮﺩ ﺑﻪ ﺧﺎﻧﻪﻱ ﺁﻥ ﻣﺮﺩ ﻣﻲﺭﻭﺩ ﻭ ﻣﻲﺑﻴﻨﺪ ﻛﻪ ﺳﺎﻋﺖ ﭼﻨﺪ ﺛﺎﻧﻴﻪﺍﻱ ﺍﺯ 8ﮔﺬﺷــﺘﻪ ﺍﺳﺖ ﻭ ﺳﺎﻋﺖ ﺟﻴﺒﻲ ﺧﻮﺩ ﺭﺍ ﺑﻪ ﺁﻥ ﻣﺮﺩ ﻧﺸﺎﻥ ﻣﻲﺩﻫﺪ ﻭ ﻣﻲﮔﻮﻳﺪ» :ﺑﺒﻴﻨﻴﺪ ﺳﺎﻋﺖ ﺷﻤﺎ ﺧﻮﺍﺏ ﻧﺮﻓﺘﻪ ﺍﺳﺖ ﻭ ﺯﻣﺎﻥ ﺭﺍ ﺩﺭﺳﺖ ﻧﺸﺎﻥ ﻣﻲﺩﻫﺪ«. ﻣﺮﺩ ﺣﺮﻑ ﺍﻭ ﺭﺍ ﺗﺄﻳﻴﺪ ﻣﻲﻛﻨﺪ ﻭ ﺷــﺎﮔﺮﺩ ﺑــﻪ ﻛﺎﺭﮔﺎﻩ ﺑﺮﻣﻲﮔﺮﺩﺩ. ﺻﺒــﺢ ﻓــﺮﺩﺍﺭﻱ ﺁﻥ ﺭﻭﺯ ﺩﻭﺑﺎﺭﻩ ﺁﻥ ﻣﺮﺩ ﺯﻧﮓ ﻣﻲﺯﻧــﺪ ﻭ ﺑﺎ ﻧﺎﺭﺍﺣﺘﻲ ﻭ ﺷــﮕﻔﺘﻲ ﻣﻲﮔﻮﻳﺪ ﻛﻪ ﺳــﺎﻋﺖ ﺍﻭ ﺩﻳﻮﺍﻧﻪ ﺷﺪﻩ ﻭ ﻫﺮ ﺯﻣﺎﻧﻲ ﺭﺍ ﻛﻪ ﺩﻟﺶ ﺑﺨﻮﺍﻫﺪ ﻧﺸﺎﻥ ﻣﻲﺩﻫﺪ! ﺷــﺎﮔﺮ ﺑﻪ ﻣﻨﺰﻝ ﺁﻥ ﻣﺮﺩ ﻣﻲﺭﻭﺩ ﻭ ﺳــﺎﻋﺖ ﺍﻭ ﺭﺍ ﺑﺮﺭﺳﻲ ﻣﻲﻛﻨﺪ ﻭ ﭘﺲ ﺍﺯ ﭼﻚ ﻛﺮﺩﻥ ﺁﻥ ﺑﺎ ﺳــﺎﻋﺖ ﺧﻮﺩ ﺑﺎ ﻛﻤــﺎﻝ ﺗﻌﺠﺐ ﻣﻲﺑﻴﻨﺪ ﻛﻪ ﺳــﺎﻋﺖ ﺯﻣﺎﻥ ﺩﺭﺳﺖ ﺭﺍ ﻧﺸﺎﻥ ﻣﻲﺩﻫﺪ .ﺑﻠﻪ ﺳﺎﻋﺖ ﻛﻤﻲ ﺍﺯ 7ﮔﺬﺷﺘﻪ ﺭﺍ ﻧﺸﺎﻥ ﻣﻲﺩﻫﺪ. ﺷﺎﮔﺮﺩ ﺳﺎﻋﺖﺳﺎﺯ ﺑﺎ ﻋﺼﺒﺎﻧﻴﺖ ﺑﻪ ﺁﻥ ﻣﺮﺩ ﻣﻲﮔﻮﻳﺪ» :ﺳﺎﻋﺖ ﺷﻤﺎ ﻛﻪ ﺩﺭﺳﺖ ﻛﺎﺭ ﻣﻲﻛﻨﺪ ،ﻣﻦ ﺭﺍ ﺳﺮ ﻛﺎﺭ ﻣﻲﮔﺬﺍﺭﻱ؟ !«
C E B
A
F
D
ﺣﺎﻝ 3ﺩﮔﻤــﻪ ﺭﺍ ﺑﺮﺩﺍﺭﻳﺪ ﻭ 6ﺩﮔﻤﻪﻱ ﺑﺎﻗﻲﻣﺎﻧﺪﻩ ﺭﺍ ﺩﺭ 3ﺭﺩﻳﻒ ﻃــﻮﺭﻱ ﻗــﺮﺍﺭ ﺩﻫﻴﺪ ﻛﻪ ﻫﺮ ﺭﺩﻳــﻒ ﺩﺍﺭﺍﻱ 3ﺩﮔﻤﻪ ﺑﺎﺷــﺪ )ﺍﻳﻦﺑﺎﺭ ﺍﺯ ﺭﺩﻳﻒﻫﺎﻱ ﻓﺮﻋﻲ 2ﺩﮔﻤﻪﺍﻱ ﺻﺮﻑﻧﻈﺮ ﻛﻨﻴﺪ(. .11ﭼﻴﺪﻥ 16ﻣﻬﺮﻩ ﺩﺭ ﻳﻚ ﺩﻭﺯ 10ﺭﺩﻳﻔﻲ ﻛﻪ ﺩﺭ ﻫﺮ ﻛﺪﺍﻡ 4ﻣﻬﺮﻩ ﻗﺮﺍﺭ ﺩﺍﺷـﺘﻪ ﺑﺎﺷـﺪ ،ﻛﺎﺭ ﺁﺳﺎﻧﻲ ﺍﺳــﺖ ،ﻭﻟﻲ ﺳﺨﺖﺗﺮ ﺍﻳﻦ ﺍﺳﺖ ﻛﻪ 9ﻣﻬﺮﻩ ﺭﺍ ﺩﺭ ﻳﻚ ﺩﻭﺯ 6ﺭﺩﻳﻔﻲ ﻛﻪ ﻫﺮ ﻛﺪﺍﻡ ﺩﺍﺭﺍﻱ 3ﻣﻬﺮﻩ ﺍﺳﺖ ﻗﺮﺍﺭ ﺩﻫﻴﺪ. .12ﺍﻟﮕﻮ ﻭ ﻃﺮﺡ ﭼﻴﻨﺶ ﺳﻜﻪﻫﺎ؛ ﺑﺮگ ﻛﺎﻏﺬﻱ ﺑﺮﺩﺍﺭﻳﺪ ﻭ ﺷﻜﻞ ﺯﻳﺮ ﺭﺍ ﺩﺭ ﺁﻥ ﻛﭙﻲ ﻛﻨﻴﺪ ﻭ ﺁﻥﺭﺍ ﺗﺎ 3ﺑﺮﺍﺑﺮ ﺑﺰﺭگ ﻛﻨﻴﺪ ﻭ 17ﺳﻜﻪ ﺑﻪ ﺷﻜﻞ ﺯﻳﺮ ﺁﻣﺎﺩﻩ ﻛﻨﻴﺪ: 5ﻋﺪﺩ ﺳﻜﻪﻱ 20ﺭﻳﺎﻟﻲ 3ﻋﺪﺩ ﺳﻜﻪﻱ 15ﺭﻳﺎﻟﻲ 3ﻋﺪﺩ ﺳﻜﻪﻱ 10ﺭﻳﺎﻟﻲ 6ﻋﺪﺩ ﺳﻜﻪﻱ 5ﺭﻳﺎﻟﻲ ﺩﺭﻫﺮ ﺧﺎﻧﻪ ﺳــﻜﻪﻫﺎ ﺭﺍ ﻃﻮﺭﻱ ﻗﺮﺍﺭ ﺩﻫﻴﺪ )ﻫﺮ ﺧﺎﻧﻪ ﻳﻚ ﺳﻜﻪ( ﻛﻪ ﺟﻤﻊ ﺁﻥﻫﺎ ﺩﺭ ﻫﺮ ﺧﻂ ﺭﺍﺳﺖ 55ﺭﻳﺎﻝ ﺷﻮﺩ.
ﺣﺎﻝ ،ﺷﻤﺎ ﺑﮕﻮﻳﻴﺪ ﻛﻪ ﭼﻪ ﺍﺗﻔﺎﻗﻲ ﺍﻓﺘﺎﺩﻩ ﺍﺳﺖ. .10ﺳـﻪ ﺗﺎ ﺩﺭ ﻳﻚ ﺭﺩﻳﻒ؛ ﺩﺭ ﻳﻚ ﺟﺪﻭﻝ 9ﺩﮔﻤﻪ ﺭﺍ ﺑﻪ ﺷﻜﻞ ﻣﺮﺑﻊ ﺳــﻪ ﺩﺭ ﺳــﻪ ﻗﺮﺍﺭ ﺩﻫﻴــﺪ .ﺯﻣﺎﻧﻲ ﻛﻪ 2ﺩﮔﻤﻪ ﻳــﺎ ﺑﻴﺶﺗﺮ ﺍﺯ ﺁﻥ ﺩﺭ ﻳﻚ ﺧﻂ ﺭﺍﺳــﺖ ﻗﺮﺍﺭ ﮔﺮﻓﺖ ﻣﻲﮔﻮﻳﻴــﻢ ﺁﻥﻫﺎ ﺩﺭ ﻳﻚ ﺭﺩﻳﻒ ﻗﺮﺍﺭ ﮔﺮﻓﺘﻨﺪ .ﺑﻨﺎﺑﺮﺍﻳﻦ ،ﺭﺩﻳﻒﻫﺎﻱ ABﻭ CDﻫﺮ ﻛﺪﺍﻡ ﺩﺍﺭﺍﻱ ﺩﻭ ﺩﮔﻤﻪ ﻭ ﺭﺩﻳﻒ EFﺩﺍﺭﺍﻱ 2ﺩﮔﻤﻪ ﺍﺳﺖ. ﭼﻪ ﺗﻌﺪﺍﺩ ﺭﺩﻳﻒﻫﺎﻱ 2ﻭ 3ﺩﮔﻤﻪﺍﻱ ﺩﺭ ﺷﻜﻞ ﻭﺟﻮﺩ ﺩﺍﺭﺩ؟ 40
ﺭﺍﻫﻨﻤﺎﻳﻲ
ﺩﻭﺭﺓ ﺷﺎﻧﺰﺩﻫﻢ ،ﺷﻤﺎﺭﺓ ،4ﺗﺎﺑﺴﺘﺎﻥ 1390
.13ﺍﺯ ﻳـﻚ ﺗـﺎ 19؛ ﺍﻋﺪﺍﺩ 1ﺗﺎ 19ﺭﺍ ﺩﺭ ﺧﺎﻧﻪﻫﺎﻱ ﺷــﻜﻞ ﺯﻳﺮ ﻃﻮﺭﻱ ﻗﺮﺍﺭ ﺩﻫﻴﺪ ﺗﺎ ﺟﻤﻊ ﺳﻪ ﻋﺪﺩ ﻭﺍﻗﻊ ﺩﺭ ﻳﻚ ﺩﺍﻳﺮﻩ 30ﺷﻮﺩ.
.14ﺑﺎ ﺳـﺮﻋﺖ ﻭ ﺩﺭﻋﻴﻦ ﺣﺎﻝ ﻫﻮﺷﻤﻨﺪﺍﻧﻪ؛ ﻋﻨﻮﺍﻥ ﻣﺴﺌﻠﻪ ﺑﻪ ﺷﻤﺎ ﻣﻲﮔﻮﻳﺪ ﻛﻪ ﭼﻪﻃﻮﺭ ﻣﺴﺌﻠﻪ ﺭﺍ ﺣﻞ ﻛﻨﻴﺪ. ﺍﻟﻒ( ﺍﺗﻮﺑﻮﺳﻲ ﺍﺯ ﺗﻬﺮﺍﻥ ﺑﻪ ﺳﻮﻱ ﻗﻢ ﺣﺮﻛﺖ ﻣﻲﻛﻨﺪ ﻭ ﻳﻚ ﺳﺎﻋﺖ ﺑﻌﺪ ﺩﻭﭼﺮﺧﻪﺳﻮﺍﺭﻱ ﺍﺯ ﻗﻢ ﻭ ﺍﻟﺒﺘﻪ ﺑﺎ ﺳﺮﻋﺖ ﻛﻢﺗﺮ ﺍﺯ ﺍﺗﻮﺑﻮﺱ ﺑﻪ ﺳﻮﻱ ﺗﻬﺮﺍﻥ ﺣﺮﻛﺖ ﻣﻲﻛﻨﺪ .ﺯﻣﺎﻧﻲ ﻛﻪ ﻫﺮ ﺩﻭ ﺑﻪ ﻫﻢ ﻣﻲﺭﺳــﻨﺪ ﻛﺪﺍﻡ ﻳﻚ ﺍﺯ ﺗﻬﺮﺍﻥ ﺩﻭﺭﺗﺮ ﻫﺴﺘﻨﺪ. 10 ﺏ( ﻛﺪﺍﻡﻳﻚ ﺑﺎ ﺍﺭﺯﺵﺗﺮ ﺍﺳــﺖ .ﻳﻚ ﻛﻴﻠﻮ ﻃﻼﻱ ﻫﺰﺍﺭ ﺗﻮﻣﺎﻧﻲ ﻳﺎ ﻧﻴﻢﻛﻴﻠﻮ ﻃﻼﻱ 20ﻫﺰﺍﺭ ﺗﻮﻣﺎﻧﻲ؟ ﺝ( ﺳﺎﻋﺖ 6ﺯﻧﮓ ﺳﺎﻋﺖ ﺩﻳﻮﺍﺭﻱ 6ﺑﺎﺭ ﺑﻪ ﺻﺪﺍ ﺩﺭﻣﻲﺁﻳﺪ .ﺑﺎ ﻧﮕﺎﻩ ﺑﻪ ﺳــﺎﻋﺖ ﻣﭽﻲ ﺧﻮﺩ ﻣﺘﻮﺟﻪ ﺷــﺪﻡ ﻛﻪ ﺑﻴﻦ ﻫﺮ ﺑﺎﺭ ﺑﻪ ﺻﺪﺍ ﺩﺭﺁﻣﺪﻥ ﺯﻧﮓ ﺳﺎﻋﺖ 30ﺛﺎﻧﻴﻪ ﻃﻮﻝ ﻣﻲﻛﺸﺪ ﺗﺎ ﺳﺎﻋﺖ ﺩﻳﻮﺍﺭﻱ ﺩﺭ ﻧﻴﻤﻪ ﺷﺐ 12ﺑﺎﺭ ﺑﻪ ﺻﺪﺍ ﺩﺭﺁﻳﺪ؟ ﺩ( ﺳﻪ ﭘﺮﺳــﺘﻮ ﺍﺯ ﻳﻚ ﻧﻘﻄﻪ ﺑﻪ ﺑﻴﺮﻭﻥ ﭘﺮﻭﺍﺯ ﻣﻲﻛﻨﻨﺪ .ﭼﻪ ﺯﻣﺎﻧﻲ ﺁﻥﻫﺎ ﺩﺭ ﻳﻚ ﺳﻄﺢ ﺩﺭ ﻓﻀﺎ ﻗﺮﺍﺭ ﻣﻲﮔﻴﺮﻧﺪ؟ .15ﺧﺮﭼﻨـﮓ ﭘﺮ ﺍﺯ ﺍﺷـﻜﺎﻝ ﻣﺨﺘﻠﻒ؛ ﺧﺮﭼﻨــﮓ ﺯﻳﺮ ﺍﺯ 17 ﻗﻄﻌﻪﻱ ﺷﻤﺎﺭﻩﮔﺬﺍﺭﻱ ﺷﺪﻩ ﺗﺸﻜﻴﻞ ﺷﺪﻩ ﺍﺳﺖ .ﺁﻥ ﺭﺍ ﺭﻭﻱ ﻳﻚ ﻛﺎﻏﺬ ﻛﭙﻲ ﻛﻨﻴﺪ ﻭ ﺑﺎ ﻗﻴﭽﻲ ﺁﻥﻫﺎ ﺭﺍ ﺍﺯ ﻫﻢ ﺟﺪﺍ ﻛﻨﻴﺪ .ﺣﺎﻝ ﺑﺎ ﺍﺳــﺘﻔﺎﺩﻩ ﺍﺯ ﺗﻤﺎﻡ ﻧﻘﺎﻁ ﻗﻄﻌﺎﺕ ﺁﻥ ،ﻳﻚ ﺩﺍﻳﺮﻩ ﺑﺴــﺎﺯﻳﺪ ﻭ ﺑﺎ ﻛﻨﺎﺭﻩﻫﺎﻱ ﺁﻥﻫﺎ ﻳﻚ ﻣﺮﺑﻊ.
3 1
2 2
3 4
1
3
3
5 5
6 6
7 8
7
.17ﻣﮕﺲ ﺑﻲﻗﺮﺍﺭ؛ ﺩﻭ ﺩﻭﭼﺮﺧﻪﺳــﻮﺍﺭ ﻳﻜﻲ ﺍﺯ ﺗﻬﺮﺍﻥ ﺑﻪ ﺳﻤﺖ ﺳﻤﻨﺎﻥ ﻭ ﺩﻳﮕﺮﻱ ﺍﺯ ﺳﻤﻨﺎﻥ ﺑﻪ ﻃﺮﻑ ﺗﻬﺮﺍﻥ ﺑﻪ ﻃﻮﺭ ﻫﻤﺰﻣﺎﻥ ﺷﺮﻭﻉ ﺑﻪ ﺣﺮﻛﺖ ﻛﺮﺩﻧﺪ .ﻭﻗﺘﻲ ﺩﻭﭼﺮﺧﻪﺳﻮﺍﺭﺍﻥ 180ﻛﻴﻠﻮﻣﺘﺮ ﺍﺯ ﻳﻜﺪﻳﮕﺮ ﻓﺎﺻﻠﻪ ﺩﺍﺷﺘﻨﺪ ،ﻣﺎﺟﺮﺍﺟﻮﻳﻲ ﻣﮕﺲ ﺷﺮﻭﻉ ﺷﺪ .ﺍﺯ ﺷﺎﻧﻪﻱ ﺩﻭﭼﺮﺧﻪﺳﻮﺍﺭ ﺍﻭﻟﻲ ﺷــﺮﻭﻉ ﺑﻪ ﭘﺮﻭﺍﺯ ﻛﺮﺩ ﺗﺎ ﺑﻪ ﺩﻭﭼﺮﺧﻪﺳﻮﺍﺭ ﺩﻭﻣﻲ ﺑﺮﺳﺪ .ﻭﻗﺘﻲ ﻣﮕﺲ ﺑﻪ ﺩﻭﻣﻲ ﺭﺳﻴﺪ ،ﺳﺮﻳﻊ ﺑﺪﻭﻥ ﺗﻮﻗﻒ ﺑﺮﮔﺸﺖ .ﻣﮕﺲ ﻫﻤﻴﻦﻃﻮﺭ ﺑﻪ ﺣﺮﻛﺖ ﺭﻓﺖ ﻭ ﺁﻣﺪ ﺧﻮﺩ ﺍﺩﺍﻣﻪ ﺩﺍﺩ ﺗﺎ ﺩﻭ ﺩﻭﭼﺮﺧﻪﺳﻮﺍﺭ ﺑﻪ ﻳﻜﺪﻳﮕﺮ ﺭﺳﻴﺪﻧﺪ .ﺩﺭ ﺍﻳﻦ ﻫﻨﮕﺎﻡ ﻣﮕﺲ ﺭﻭﻱ ﺑﻴﻨﻲ ﻳﻜﻲ ﺍﺯ ﺩﻭﭼﺮﺧﻪﺳﻮﺍﺭﺍﻥ ﻧﺸﺴﺖ.
ﺳــﺮﻋﺖ ﻣﮕﺲ 30ﻛﻴﻠﻮﻣﺘﺮ ﺩﺭ ﺳﺎﻋﺖ ﻭ ﺳﺮﻋﺖ ﺩﻭﭼﺮﺧﻪﺳﻮﺍﺭﺍﻥ ﻧﻴﺰ 15ﻛﻴﻠﻮﻣﺘﺮ ﺩﺭ ﺳــﺎﻋﺖ ﺍﺳﺖ .ﺣﺎﻝ ﺑﮕﻮﻳﻴﺪ ﻣﮕﺲ ﭼﻪ ﻣﺴﺎﻓﺘﻲ ﺭﺍ ﭘﻴﻤﻮﺩﻩ ﺍﺳﺖ؟ .18ﺳﺎﻝ ﻭﺍﺭﻭﻧﻪ؛ ﺁﺧﺮﻳﻦ ﺳﺎﻟﻲ ﻛﻪ ﺗﺎﻛﻨﻮﻥ ﻭﺍﺭﻭﻧﻪﻱ ﺁﻥ ﺭﺍ ﻣﺎﻧﻨﺪ ﺧﻮﺩﺵ ﺍﺳﺖ ،ﻛﺪﺍﻡ ﺍﺳﺖ؟ )ﺑﻪ ﻫﺠﺮﻱ ﻗﻤﺮﻱ ،ﺷﻤﺴﻲ ﻭ ﻣﻴﻼﺩﻱ( .19ﺩﻭ ﻟﻄﻴﻔﻪ؛ ﺍﻟﻒ( ﻣﺮﺩﻱ ﺑﻪ ﭘﺴﺮﺵ ﺗﻠﻔﻦ ﻣﻲﻛﻨﺪ ﻭ ﺍﺯ ﺍﻭ ﻣﻲﺧﻮﺍﻫﺪ ﺗﺎ ﻣﻘﺪﺍﺭﻱ ﻭﺳﺎﻳﻞ ﻻﺯﻡ ﺑﺮﺍﻱ ﺍﻭ ﺑﺨﺮﺩ ﺗﺎ ﺑﺮﺍﻱ ﻣﺴﺎﻓﺮﺕ ﺁﻣﺎﺩﻩ ﺷﻮﺩ .ﭘﺪﺭ ﺑﻪ ﭘﺴﺮﺵ ﻣﻲﮔﻮﻳــﺪ ﻛﻪ ﭘﻮﻝ ﻛﺎﻓﻲ ﺑــﺮﺍﻱ ﺍﻭ ﺭﻭﻱ ﻣﻴﺰ ﺩﺭ ﺩﺍﺧﻞ ﭘﺎﻛﺖ ﻗﺮﺍﺭ ﺩﺍﺩﻩ ﺍﺳــﺖ .ﻭﻗﺘﻲ ﭘﺴﺮ ﺑﻪ ﺍﺗﺎﻕ ﭘﺪﺭ ﻣﻲﺭﻭﺩ ﭘﺎﻛﺘﻲ ﺭﺍ ﭘﻴﺪﺍ ﻣﻲﻛﻨﺪ ﻛﻪ ﺭﻭﻱ ﺁﻥ ﻋﺪﺩ 87ﻧﻮﺷﺘﻪ ﺷﺪﻩ ﺍﺳﺖ .ﺩﺭ ﻓﺮﻭﺷﮕﺎﻩ ﭘﺴﺮ ﺑﻪ ﺍﻧﺪﺍﺯﻩﻱ 80ﻫﺰﺍﺭ ﺗﻮﻣﺎﻥ ﺧﺮﻳﺪ ﻣﻲﻛﻨﺪ ،ﻭﻟﻲ ﻣﻮﻗﻊ ﭘﺮﺩﺍﺧﺖ ﻧﻪ ﺗﻨﻬﺎ 7ﻫﺰﺍﺭ ﺗﻮﻣﺎﻥ ﺍﺿﺎﻓﻪ ﻧﻤــﻲﺁﻭﺭﺩ ،ﺑﻠﻜﻪ ﺑﺪﻫﻜﺎﺭ ﻧﻴﺰ ﻣﻲﺷــﻮﺩ! ﺣﺎﻝ ﺑﮕﻮﻳﻴــﺪ ﭼﻪﻗﺪﺭ ﺑﺪﻫﻜﺎﺭ ﻣﻲﺷﻮﺩ ﻭ ﭼﺮﺍ؟ ﺏ( ﺍﻋﺪﺍﺩ 1ﺗﺎ 9ﺭﺍ ﺭﻭﻱ ﺗﻜﻪ ﻛﺎﻏﺬ ﻳﺎﺩﺩﺍﺷﺖ ﻛﺮﺩﻩ ﻭ ﺁﻥﻫﺎ ﺭﺍ ﺩﺭ ﺩﻭ ﺭﺩﻳﻒ ﻣﺎﻧﻨﺪ ﺷــﻜﻞ ﺯﻳﺮ ﻗﺮﺍﺭ ﺩﻫﻴﺪ .ﺣﺎﻝ ﺩﻭ ﺗﻜﻪ ﻛﺎﻏﺬ ﺭﺍ ﻃﻮﺭﻱ ﺟﺎﺑﻪﺟﺎ ﻛﻨﻴﺪ ﺗﺎ ﺟﻤﻊ ﺍﻋﺪﺍﺩ ﻫﺮ ﺩﻭ ﺳﺘﻮﻥ ﺑﺎ ﻫﻢ ﺑﺮﺍﺑﺮ ﺷﻮﺩ.
ﺩﻭﺭﺓ ﺷﺎﻧﺰﺩﻫﻢ ،ﺷﻤﺎﺭﺓ ،4ﺗﺎﺑﺴﺘﺎﻥ 1390
ﺭﺍﻫﻨﻤﺎﻳﻲ
41
ﭘﺎﺳﺦﻫﺎ ﺩﺭ ﺷﻤﺎﺭﻩﻱ ﺁﻳﻨﺪﻩﻱ ﻣﺠﻠﻪ
4
.16ﻗﻴﻤـﺖ ﻳﻚ ﻛﺘـﺎﺏ؛ ﻗﻴﻤﺖ ﻳﻚ ﻛﺘﺎﺏ ﺑﺮﺍﺑﺮ ﺍﺳــﺖ ﺑﺎ ﻫﺰﺍﺭ ﺗﻮﻣﺎﻥ ﺑﻪﻋﻼﻭﻩﻱ ﻧﺼﻒ ﻗﻴﻤﺖ ﺁﻥ .ﺣﺎﻝ ﻗﻴﻤﺖ ﻛﺘﺎﺏ ﭼﻪﻗﺪﺭ ﺍﺳﺖ؟
︎︀︨ ر︀︲﹫︀ت ر﹇︀︋︐﹩ ﭘﺎﺳﺦ : 1ﮔﺰﻳﻨﻪﻱ 1ﺻﺤﻴﺢ ﺍﺳﺖ. ﺍﮔﺮ ﺗﻌﺪﺍﺩ ﺳﻜﻪﻫﺎﻱ 5ﺭﻳﺎﻟﻲ ﺭﺍ xﻭ ﺗﻌﺪﺍﺩ ﺳﻜﻪﻫﺎﻱ 20ﺭﻳﺎﻟﻲ ﺭﺍ yﺩﺭ ﻧﻈﺮ ﺑﮕﻴﺮﻳﻢ ،ﺑﻨﺎﺑﺮ ﺻﻮﺭﺕ ﻣﺴﺌﻠﻪ ﺩﺍﺭﻳﻢ: − 20 y = −700
×( −2 0 ) ⎧−20 x
⎨ ⎩5 x + 20 y = 625
− 15 x = −75 ⇒ x = 5
⎧x + y = 35 ⎨ ⎩5 x + 20 y = 625
⇒
ﭘﺎﺳﺦ : 4ﮔﺰﻳﻨﻪﻱ 2ﺻﺤﻴﺢ ﺍﺳﺖ. ﺍﮔﺮ ﻣﻘﺪﺍﺭ ﭘﻮﻝ ﻧﻴﻤﺎ ﺭﺍ xﻭ ﻣﻘﺪﺍﺭ ﭘﻮﻝ ﺳﻴﻨﺎ ﺭﺍ yﺩﺭ ﻧﻈﺮ ﺑﮕﻴﺮﻳﻢ، ﺑﻨﺎﺑﺮ ﺻﻮﺭﺕ ﻣﺴﺌﻠﻪ ﺩﺍﺭﻳﻢ: ⎧ x + y =105 00 ⎧ x + y =1 0 500 ⎧⎪ x + y =1 0 5 00 ⎪ ⎪ ⎨⇒ ⇒ ⎨ 1 ⎨ 1 2 5 ⎩⎪2x −5 y = 0 ) ⎪ x − x =2( y − y ⎪ x =2× y 6 6 ⎩ 3 ⎩3
⇒ y = 30
⎧−2x −2 y =−2100 ⎨ ⎩2x −5 y = 0 −7 y=−21000⇒ y=3 000
⇒
ﭘﺎﺳﺦ : 2ﮔﺰﻳﻨﻪﻱ 2ﺻﺤﻴﺢ ﺍﺳﺖ. ﺍﮔﺮ ﻛﺴﺮ ﻣﻮﺭﺩﻧﻈﺮ ﺭﺍ ﺑﻪ ﺻﻮﺭﺕ xﺩﺭ ﻧﻈﺮ ﺑﮕﻴﺮﻳﻢ ،ﺑﻨﺎﺑﺮ ﺻﻮﺭﺕ y ﻣﺴﺌﻠﻪ ﺩﺍﺭﻳﻢ: ⎪⎧3 x −2 y= 0 ⎧3 x −2 y= 0 ⎨ ⇒ ⎨ 1 ⎪⎩− x + y =7 ⎩−3 x + 3 y =21
y=21 ⇒ x =14 ⇒ x + y=35
⎧⎪ x + y =1 0 5 00 ⎨⇒ ⎪⎩2x −5 y = 0
⇒ x = 7500
⎧x 2 = ⎪ ⇒ ⎨y 3 ⎪ y − x =7 ×3 ⎩
ﭘﺎﺳﺦ : 3ﮔﺰﻳﻨﻪﻱ 2ﺻﺤﻴﺢ ﺍﺳﺖ. ﺍﮔــﺮ ﻋﺪﺩ ﻃﺒﻴﻌﻲ ﻣﻮﺭﺩﻧﻴــﺎﺯ ﺭﺍ xﺩﺭ ﻧﻈﺮ ﺑﮕﻴﺮﻳــﻢ ،ﺍﻋﺪﺍﺩ ﺑﻌﺪﻱ ﻋﺒﺎﺭﺗﻨﺪ ﺍﺯ: x , x + 1 , x + 2 , x + 3 , x + 4 , ...
ﺑﻨﺎﺑﺮﺍﻳــﻦ ﺍﮔﺮ ﺳــﻪ ﻋﺪﺩ ﻓــﺮﺩ ﺭﺍ x+2 ،xﻭ x+3ﺩﺭ ﻧﻈﺮ ﺑﮕﻴﺮﻳﻢ، ﺩﺍﺭﻳﻢ: x + ( x + 2) + ( x + 4) = 87 ⇒ 3 x + 6 = 87 ⎪⎧ x +2=29
⎨ ⇒ ⇒ 3 x = 81 ⇒ x = 27
⎪⎩ x +4=31
⎧ x =27 ⎪ ⇒ ⎨ x +2=29 ⇒ 7+9+1=17 ⎪ x +4=31 ⎩
42
ﺭﺍﻫﻨﻤﺎﻳﻲ
ﺩﻭﺭﺓ ﺷﺎﻧﺰﺩﻫﻢ ،ﺷﻤﺎﺭﺓ ،4ﺗﺎﺑﺴﺘﺎﻥ 1390
ﭘﺎﺳﺦ : 5ﮔﺰﻳﻨﻪﻱ 4ﺻﺤﻴﺢ ﺍﺳﺖ. x ﺍﮔﺮ ﻛﺴﺮ ﻣﻮﺭﺩﻧﻈﺮ ﺭﺍ yﺩﺭ ﻧﻈﺮ ﺑﮕﻴﺮﻳﻢ ،ﺑﻨﺎﺑﺮ ﺻﻮﺭﺕ ﻣﺴﺌﻠﻪ 4
x +3
ﺩﺍﺭﻳﻢ y+3 = 5 :ﻭ
x −3 1 = y −3 2
ﺑﻨﺎﺑﺮﺍﻳﻦ ﺑﺎﻳﺪ ﺩﺳــﺘﮕﺎﻩ ﺩﻭ ﻣﻌﺎﺩﻟــﻪ ﻭ ﺩﻭ ﻣﺠﻬﻮﻟﻲ ﺣﻞ ﻛﻨﻴﻢ.
⎧ x +3 4 = ⎪ ⎪ y+3 5 ⎨ ⎪ x −3 = 1 ⎪⎩ y−3 2
ﺭﺍ
⎧ x +3 4 = ⎪ ⎪⎧5 x −4 y =−3 ) ⎪⎧5( x +3 )=4( y +3 ⎪ y+3 5 ⎨ ⇒ ⎨ ⇒ ⎨ x − 3 1 ⎩⎪2x − y =3 ⎩⎪2( x −3 )= y −3 ⎪ = ⎪⎩ y −3 2 ⎧5 x −4 y =−3 ⎧5 x −4 y =−3 ⎨⎪ ⇒ ×3−4 ⎨⇒ ⎩⎪2x − y =3 ⎩−8 x +4 y =−12
−3 x =−15 ⇒ x =5
x 5 = ⇒ x + y = 12 y 7
⇒⇒ y = 7
ﭘﺎﺳﺦ : 6ﮔﺰﻳﻨﻪﻱ 3ﺻﺤﻴﺢ ﺍﺳﺖ؟ ﺍﮔﺮ ﺍﻧﺪﺍﺯﻩﻱ ﺿﻠﻊ ﻣﺮﺑﻊ s1ﺭﺍ aﻭ ﺍﻧﺪﺍﺯﻩﻱ ﻃﻮﻝ ﻣﺴــﺘﻄﻴﻞ ﺍﻳﺠﺎﺩ ﺷــﺪﻩ ﺗﻮﺳﻂ ﻣﺴــﺘﻄﻴﻞ s1ﻭ ﻣﺮﺑﻊ s2ﺭﺍ bﺑﻨﺎﻣﻴﻢ ،ﺍﻧﺪﺍﺯﻩﻱ ﻣﺴﺎﺣﺖ 1 ﻫﺎﺷــﻮﺭﺯﺩﻩ ﺑﺮﺍﺑﺮ ﺑﺎ ) s A = × a ( b − aﻭ ﻣﺤﻴﻂ ﻣﺴﺘﻄﻴﻞ s3ﺑﺮﺍﺑﺮ 2 ﺑﺎ 2bﺧﻮﺍﻫﺪ ﺷﺪ .ﺑﻨﺎﺑﺮﺍﻳﻦ: ⎧⎪s1+s2 +s3 =37 ⎨ ⇒ 2 ⎩⎪ab−a =12
ﺍﺯ ﻃﺮﻓﻲ ،ﭼﻮﻥ ﺩﺍﺭﻳﻢ، s1 = a 2 : ﺑﻨﺎﺑﺮﺍﻳﻦ:
) s 3 = a ( b − a ) ، s2 = ( b − a
⎧ 2 2 ⎪a +( b −a ) +a ( b −a )=37 ⎨ ⎪⎩ab −a 2 =6
⇒
x = 2( y − ( x − y )) = 2y − 2x + 2y ⇒ 3 x = 4y ⇒ 3 x − 4y = 0
ﺑﻨﺎﺑﺮﺍﻳﻦ ﺑﺎﻳﺪ ﺩﺳــﺘﮕﺎﻩ ﺩﻭ ﻣﻌﺎﺩﻟﻪ ﻭ ﺩﻭ ﻣﺠﻬﻮﻟﻲ ⎪⎧⎨ x + y=49ﺭﺍ ⎩⎪3 x −4 y = 0 ﺣﻞ ﻛﻨﻴﻢ .ﭘﺲ: ⎧⎪4 x +4 y =196 ⎨ ⇒ ⎨ ⎪⎩3 x −4 y = 0 ⎪⎩3 x −4 y = 0
⎪ x + y =49 ⎧ ×4
7 x = 196 ⇒ x = 28
⎧s1+s2 +s3 =37 ⎪ ⎨1 ⎪ ×a ( b−a )=6 ⎩2 2
ﭘﺎﺳﺦ : 8ﮔﺰﻳﻨﻪﻱ 3ﺻﺤﻴﺢ ﺍﺳﺖ. ﺍﮔﺮ ﺳﻦ ﻛﺸﺘﻲ ﺭﺍ xﻭ ﻋﻤﺮ ﺩﻳﮓ ﺑﺨﺎﺭ ﺭﺍ yﺩﺭ ﻧﻈﺮ ﺑﮕﻴﺮﻳﻢ ،ﺑﻨﺎﺑﺮ ﺻﻮﺭﺕ ﻣﺴﺌﻠﻪ ﺩﺍﺭﻳﻢ.x+y=49 : ﺩﺭ ﺿﻤﻦ x-yﺳــﺎﻝ ﻗﺒﻞ ﻋﻤﺮ ﻛﺸــﺘﻲ ﺑﺮﺍﺑﺮ ﺑﺎ ﻋﻤﺮ ﻓﻌﻠﻲ ﺩﻳﮓ ﺑﺨﺎﺭ ﺑﻮﺩﻩ ﺍﺳﺖ ،ﭘﺲ ﺩﺍﺭﻳﻢ:
ﭘﺎﺳﺦ : 9ﮔﺰﻳﻨﻪﻱ 3ﺻﺤﻴﺢ ﺍﺳﺖ. ﺍﮔﺮ ﻗﻴﻤﺖ ﺩﻓﺘﺮ ﺭﺍ xﻭ ﻗﻴﻤﺖ ﺧﻮﺩﻛﺎﺭ ﺭﺍ yﺩﺭ ﻧﻈﺮ ﺑﮕﻴﺮﻳﻢ ،ﺑﻨﺎﺑﺮ ﺻﻮﺭﺕ ﻣﺴﺌﻠﻪ ﺩﺍﺭﻳﻢ 2x = 5 y :ﻭ . 3 x + 5 y + 10 = 2x + 8 y ﺑﻨﺎﺑﺮﺍﻳــﻦ ﺑﺎﻳﺪ ﺩﺳــﺘﮕﺎﻩ ﺩﻭ ﻣﻌﺎﺩﻟﻪ ﻭ ﺩﻭ ﻣﺠﻬﻮﻟﻲ ⎪⎧ x −3 y =−10 ⎨ ⎩⎪2x −5 y = 0
ﺭﺍ ﺣﻞ ﻛﻨﻴﻢ .ﭘﺲ
⎧ 2 2 2 2 ⎪a + b −2ab +a +ab −a =37
⎨ ⇒
⎪⎩ab −a 2 =12
ﺍﺯ ﺩﺳﺘﮕﺎﻩ ﺑﺎﻻ ﻧﺘﻴﺠﻪ ﻣﻲﮔﻴﺮﻳﻢ ﻛﻪ ﺟﺎﻳﮕﺰﻳﻨﻲ ﺍﻳﻦ ﻣﻘﺎﺩﻳﺮ ﺩﺭ ﺩﺍﺭﻳﻢ:
⎪⎧−2x +6 y =20 ⎨ ⇒ ⎩⎪2x −5 y = 0
⎧ 2 . ⎪⎨ab−a =12ﻛﻪ ﺍﺯ
⎪⎩a 2 −ab=−12 2
2
⇒ x = 50 2
2
a + b − 2ab + a + ab − a = 37 2
2
⎪ )×( −2 ⎧ x −3 y =−10
⎨ ⎩⎪2x −5 y = 0
y = 20 ⇒ 2x + 8 y = 260
2
2(a − ab ) + b + (ab − a ) = 37 2
2
⇒ 2 × ( −12) + b + 12 = 37 ⇒ b = 49 ⇒ b = 7
ﺑﻨﺎﺑﺮﺍﻳﻦ ،ﻣﺤﻴﻂ ﻣﺴﺘﻄﻴﻞ s3ﺑﺮﺍﺑﺮ ﺑﺎ 2b=14ﺍﺳﺖ. ﭘﺎﺳﺦ : 7ﮔﺰﻳﻨﻪﻱ 4ﺻﺤﻴﺢ ﺍﺳﺖ. ﺍﮔﺮ ﺳــﻦ ﺳــﺎﻧﺎﺯ ﺭﺍ xﻭ ﺳــﻦ ﮔﻠﻨﺎﺭ ﺭﺍ yﺩﺭ ﻧﻈــﺮ ﺑﮕﻴﺮﻳﻢ ،ﺑﻨﺎﺑﺮ ﺻﻮﺭﺕ ﻣﺴــﺌﻠﻪ ﺩﺍﺭﻳﻢ . x + y = 30 ، y = 2x :ﺑﻨﺎﺑﺮﺍﻳﻦ ﺑﺎﻳﺪ ﺩﺳﺘﮕﺎﻩ ﺩﻭ ﻣﻌﺎﺩﻟﻪ ﻭ ﺩﻭ ﻣﺠﻬﻮﻟﻲ ⎧⎪⎨2x − y= 0ﺭﺍ ﺣﻞ ﻛﻨﻴﻢ .ﭘﺲ: ⎩⎪ x + y =30
⎧2x − y = 0 ⎨ ⎩ x + y =30 ⇒ y = 20 3 x =30 ⇒ x =10 ﺩﻭﺭﺓ ﺷﺎﻧﺰﺩﻫﻢ ،ﺷﻤﺎﺭﺓ ،4ﺗﺎﺑﺴﺘﺎﻥ 1390
ﺭﺍﻫﻨﻤﺎﻳﻲ
43
ﭘﺎﺳﺦ : 10ﮔﺰﻳﻨﻪﻱ 4ﺻﺤﻴﺢ ﺍﺳﺖ. ﻃﻮﻝ ﺷﻤﻊ ﺭﺍ Lﻭ ﻣﺪﺕ ﺯﻣﺎﻥ ﻣﻮﺭﺩ ﻧﻈﺮ ﺭﺍ xﺩﺭ ﻧﻈﺮ ﻣﻲﮔﻴﺮﻳﻢ. ﺑﻨﺎﺑﺮ ﺻﻮﺭﺕ ﻣﺴﺌﻠﻪ ،ﭼﻮﻥ ﺩﺭ ﻫﺮ ﺳﺎﻋﺖ 1ﺷﻤﻊ ﺍﻭﻝ ﻣﻲﺳﻮﺯﺩ ،ﭘﺲ 4
ﺩﺭ xﺳﺎﻋﺖ xﺁﻥ ﻣﻲﺳﻮﺯﺩ ﻭ ﭼﻮﻥ ﻃﻮﻝ ﺁﻥ ﺭﺍ Lﻓﺮﺽ ﻛﺮﺩﻩﺍﻳﻢ، 4
ﺑﻨﺎﺑﺮﺍﻳﻦ ﻃﻮﻝ ﺷﻤﻊ ﺍﻭﻝ ﺑﻌﺪ ﺍﺯ xﺳﺎﻋﺖ ﺑﺮﺍﺑﺮ ﺑﺎ ﺩﻭﻡ ﺑﺮﺍﺑﺮ ﺑﺎ
x L −3 L
x ×L − L 4
ﻭ ﺷﻤﻊ
ﻣﻲﺷﻮﻧﺪ .ﺑﻨﺎﺑﺮﺍﻳﻦ ﺧﻮﺍﻫﻴﻢ ﺩﺍﺷﺖ:
1 1 1 1 ) xL = 2( L − xL ) ⇒ L (1 − x ) = 2L(1 − x 4 3 3 4 1 2 ⇒ 1− x = 2 − x ⇒ x = 2 / 4 4 3 L−
ﭘﺎﺳﺦ : 11ﮔﺰﻳﻨﻪﻱ 4ﺻﺤﻴﺢ ﺍﺳﺖ. ﻣﻘﺪﺍﺭ ﭘﻮﻝ ﺍﻭﻟﻴﻪ ﺩﺭ ﺻﻨﺪﻭﻕ ﺭﺍ xﺩﺭ ﻧﻈﺮ ﻣﻲﮔﻴﺮﻳﻢ .ﺑﻨﺎﺑﺮ ﺻﻮﺭﺕ ﻣﺴﺌﻠﻪ ﺩﺍﺭﻳﻢ: ﻣﻘﺪﺍﺭ ﭘﻮﻝ ﺻﻨﺪﻭﻕ ﭘﺲ ﺍﺯ ﻣﺮﺍﺟﻌﻪﻱ ﻧﻔﺮ ﺍﻭﻝ x + x − 40 = 2x − 40
ﻣﻘﺪﺍﺭ ﭘﻮﻝ ﺻﻨﺪﻭﻕ ﭘﺲ ﺍﺯ ﻣﺮﺍﺟﻌﻪﻱ ﻧﻔﺮ ﺩﻭﻡ (2x − 4 0 ) + (2x − 40 ) − 4 0 = 4 x − 120
ﻣﻘﺪﺍﺭ ﭘﻮﻝ ﺻﻨﺪﻭﻕ ﭘﺲ ﺍﺯ ﻣﺮﺍﺟﻌﻪﻱ ﻧﻔﺮ ﺳﻮﻡ (2x − 120 ) + (2x − 12 0 ) − 40 = 8 x − 280
ﺑﻨﺎﺑﺮﺍﻳﻦ:
8 x − 288 0 = 0 ⇒ 8 x = 28 0 ⇒ x = 35
ﭘﺎﺳﺦ : 12ﮔﺰﻳﻨﻪﻱ 4ﺻﺤﻴﺢ ﺍﺳﺖ. ﻃﻮﻝ ﻣﺴــﺘﻄﻴﻞ ﺭﺍ xﻭ ﻋﺮﺽ ﺁﻥ ﺭﺍ yﺩﺭ ﻧﻈــﺮ ﻣﻲﮔﻴﺮﻳﻢ .ﺑﻨﺎﺑﺮ ﺻﻮﺭﺕ ﻣﺴﺌﻠﻪ ﺩﺍﺭﻳﻢ: x = 2y − 4ﻭ x − 6 = y + 2 ⎧ x −2 y=−4 ⎪ ⎨ ﺭﺍ ﺑﻨﺎﺑﺮﺍﻳﻦ ﺑﺎﻳﺪ ﺩﺳــﺘﮕﺎﻩ ﺩﻭ ﻣﻌﺎﺩﻟﻪ ﻭ ﺩﻭ ﻣﺠﻬﻮﻟﻲ ⎩⎪ x − y=8 ﺣﻞ ﻛﻨﻴﻢ. ﭘﺲ ﺩﺍﺭﻳﻢ: ⎧⎪ x −2 y =−4
⎨ ⇒
⎩⎪− x + y =−8
⎧⎪ x −2 y=−4 ⎨ ⎩ )×( −1 ⎪ x − y =8
− y = −12 ⇒ y = 12 ⇒ x = 20
ﺑﻨﺎﺑﺮﺍﻳﻦ ،ﻣﺴﺎﺣﺖ ﻣﺴﺘﻄﻴﻞ ﻣﺰﺑﻮﺭ ﺑﻪ ﻃﻮﻝ x=20ﻭ ﻋﺮﺽ y=40
44
ﺭﺍﻫﻨﻤﺎﻳﻲ
ﺩﻭﺭﺓ ﺷﺎﻧﺰﺩﻫﻢ ،ﺷﻤﺎﺭﺓ ،4ﺗﺎﺑﺴﺘﺎﻥ 1390
ﺑﺮﺍﺑﺮ ﺑﺎ 20×12=240ﻭ ﻣﺴﺎﺣﺖ ﻣﺮﺑﻊ ﻣﻮﺭﺩﻧﻈﺮ ﺑﺮﺍﺑﺮ ﺑﺎ 14×14=196 ﺍﺳﺖ .ﭘﺲ ﺧﻮﺍﻫﻴﻢ ﺩﺍﺷﺖ: 240-196=44 ﭘﺎﺳﺦ : 13ﮔﺰﻳﻨﻪﻱ 1ﺻﺤﻴﺢ ﺍﺳﺖ. ﺗﻌــﺪﺍﺩ ﻣﻬﺮﻩﻫــﺎﻱ ﺩﺍﺧﻞ ﻛﻴﺴــﻪ ﺭﺍ xﺩﺭ ﻧﻈــﺮ ﻣﻲﮔﻴﺮﻳﻢ .ﺑﻨﺎﺑﺮ ﺻﻮﺭﺕ ﻣﺴﺌﻠﻪ ﺩﺍﺭﻳﻢ: ﺗﻌﺪﺍﺩ ﻣﻬﺮﻩﻫﺎﻳﻲ ﻛﻪ ﭘﺴﺮ ﺑﭽﻪﻱ ﺍﻭﻝ ﺑﺮﻣﻲﺩﺍﺭﺩ: 1 x +1 2
ﺗﻌﺪﺍﺩ ﻣﻬﺮﻩﻫﺎﻳﻲ ﻛﻪ ﭘﺴﺮﺑﭽﻪﻱ ﺩﻭﻡ ﺑﺮﻣﻲﺩﺍﺭﺩ: 1 1 1 1 ( x − ( x + 1)) = x − 2 6 3 3
ﺑﻨﺎﺑﺮﺍﻳــﻦ ﺑﺎﻳﺪ ﻣﻌﺎﺩﻟــﻪﻱ 1 x + 1 + 1 x − 1 + 4 = xﺭﺍ ﺣﻞ 2 6 3 ﻛﻨﻴﻢ .ﭘﺲ 1 1 1 2 14 x +1+ x − + 4 = x ⇒ x + = x 2 6 3 3 3 1 14 =⇒ x ⇒ x = 14 3 3 ﺑﻨﺎﺑﺮﺍﻳﻦ ﺍﮔﺮ ﺩﺭ ﻋﺒﺎﺭﺕ 1 x − 1ﺭﺍ ﻗﺮﺍﺭ ﺩﻫﻴﻢ ،ﺗﻌﺪﺍﺩ ﻣﻬﺮﻩﻫﺎﻳﻲ 6 3
ﻛﻪ ﺑﻪ ﭘﺴــﺮﺑﭽﻪﻱ ﺩﻭﻡ ﺭﺳــﻴﺪﻩ ﺍﺳــﺖ ﺑﻪ ﺩﺳــﺖ ﻣﻲﺁﻳﺪ ﻭ ﺧﻮﺍﻫﻴﻢ ﺩﺍﺷﺖ: 1 1 14 1 7 1 6 × 14 − = − = − = = 2 6 3 6 3 3 3 3
ﭘﺎﺳﺦ : 14ﮔﺰﻳﻨﻪﻱ 1ﺻﺤﻴﺢ ﺍﺳﺖ. ﺗﻌــﺪﺍﺩ ﺻﻨﺪﻟﻲﻫﺎ ﺩﺭ ﻫــﺮ ﺭﺩﻳﻒ ﺭﺍ xﻭ ﺗﻌــﺪﺍﺩ ﺻﻨﺪﻟﻲﻫﺎ ﺩﺭ ﻫﺮ ﺳﺘﻮﻥ ﺭﺍ yﺩﺭ ﻧﻈﺮ ﻣﻲﮔﻴﺮﻳﻢ .ﺑﻨﺎﺑﺮ ﺻﻮﺭﺕ ﻣﺴﺌﻠﻪ ﺩﺍﺭﻳﻢ: ( x − 3 )( y + 1) = xyﻭ (xx − 5)( y + 2) = xy ﺑﻨﺎﺑﺮﺍﻳﻦ ﺑﺎﻳﺪ ﺩﺳﺘﮕﺎﻩ ﺩﻭ ﻣﻌﺎﺩﻟﻪ ﻭ ﺩﻭ ﻣﺠﻬﻮﻟﻲ ⎪⎧( x −3 )( y +1) = xy ⎨ ⎩⎪( x −5)( y +2) = xy ﺭﺍ ﺣﻞ ﻛﻨﻴﻢ .ﭘﺲ: ⎧⎪ x −3 y =3 ⎨⇒ ⎪⎩2x −5 y =10
⎧⎪( x −3 )( y +1) = xy ⎨ ⎪⎩( x −5)( y +2) = xy
⎧⎪−2x +6 y=−6 ⎨ ⇒ ⎨ ⎪⎩2x −5 y=10 ⎩⎪2x −5 y=10
⎧ )×( −2 ⎪ x −3 y =3
⇒ x = 15
y=4
ﺑﻨﺎﺑﺮﺍﻳﻦ ﺧﻮﺍﻫﻴﻢ ﺩﺍﺷﺖ: n = xy = 4 × 15 = 6 0
⇒
︨﹣ال﹨︀ی ︎﹠︕﹎︤﹠﹤ای ︎︀﹤ی اول را﹨﹠﹝︀﹩ ﺗﺮﺟﻤﻪ
ﻛﻠﻴﺪﻭﺍژﻩﻫﺎ :ﺳــﺆﺍﻝﻫﺎﻱ ﻣﺴــﺎﺑﻘﻪﺍﻱ، ﺭﻳﺎﺿــﻲ ﺍﺳــﺘﺮﺍﻟﻴﺎ ،ﭘﻨﺞ ﮔﺰﻳﻨــﻪﺍﻱ ،ﺩﻭﺭﻩﻱ ﺭﺍﻫﻨﻤﺎﻳﻲ.
ﻱ ﺳﭙﻴﺪﻩ ﭼﻤﻦﺁﺭﺍ
.1ﻋﺪﺩ ﻳﻚ ﻫﺰﺍﺭ ﻭ ﺑﻴﺴﺖ ﻭ ﻫﻔﺖ ﻛﺪﺍﻡ ﺍﺳﺖ؟ ﺍﻟﻒ( 100027ﺏ( 10027 ﺙ( 27 ﺕ( 127 پ( 1027 .2ﺳــﺎﺭﺍ ﺩﺭﻭﻥ ﻓﺮﻭﺷــﮕﺎﻩ ﺣﻴﻮﺍﻧــﺎﺕ ﺧﺎﻧﮕﻲ ﺍﻳﺴــﺘﺎﺩﻩ ﻭ ﺍﺯ ﭘﻨﺠــﺮﻩ ﺑﻪ ﺑﻴﺮﻭﻥ ﻧﮕﺎﻩ ﻣﻲﻛﻨﺪ .ﺳﺎﺭﺍ ﻧﻮﺷﺘﻪﻱ ﺭﻭﻱ ﭘﻨﺠﺮﻩ ﺭﺍ ﭼﮕﻮﻧﻪ ﻣﻲﺑﻴﻨﺪ؟ ﺍﻟﻒ( POHS T E P ﺏ( POH S TEP
پ( T E P POH S ﺕ( POH S T E P ﺙ( P OH S T EP
PET SHOP
ﻢ ،ﺷﻤﺎﺭﺓ ،4ﺗﺎﺑﺴﺘﺎﻥ 1390 ﺷﺎﻧﺰﺩﻫﻢ، ﺰﺩﻫﻢ ﺷﺎﻧﻧﺰﺰﺩﺩﻫ ﺩﻭﺭﺓ ﺷ ﺩﻭﺭﺭﺓ ﺩﻭ
ﺭﺍﻫﻨﻤﺎﻳﻲ
45
ﺳﺆﺍﻝﻫﺎﻱ ﻣﺴﺎﺑﻘﻪﺍﻱ
)(٢٠١٠ )٢٠١٠
.3ﻓﺮﻫﺎﺩ 14ﺳﺎﻟﻪ ﺍﺳﺖ .ﻓﺮﻳﺒﺎ 10ﺳﺎﻟﻪ ﺳﻦ ﭘﺪﺭ ﻓﺮﻫﺎﺩ ﻭ ﻓﺮﻳﺒﺎ ﺩﻭ ﺑﺮﺍﺑﺮ ﻣﺠﻤﻮﻉ ﺍﺳﺖ .ﱢ ﺳﻦ ﺁﻥ ﺩﻭ ﺍﺳﺖ .ﭘﺪﺭ ﭼﻨﺪ ﺳﺎﻟﻪ ﺍﺳﺖ؟ ﱢ ﺏ( 48 ﺍﻟﻒ( 46 ﺙ( 54 ﺕ( 52 پ( 50 .4ﻧﻘﻄــﻪﻱ ﻭﺳــﻂ ﻫــﺮ ﺿﻠــﻊ ﻣﺮﺑﻊ ﺭﺍ ﻣﻄﺎﺑﻖ ﺷﻜﻞ ﺑﻪ ﻫﻢ ﻭﺻﻞ ﻛﺮﺩﻩﺍﻳﻢ ،ﻗﺴﻤﺘﻲ ﺍﺯ ﺷــﻜﻞ ﺭﻧﮓ ﺷﺪﻩ ﺍﺳﺖ .ﻗﺴﻤﺖ ﺭﻧﮓﺷﺪﻩ ﭼﻪ ﻛﺴﺮﻱ ﺍﺯ ﻣﺮﺑﻊ ﺑﺰﺭگ ﺍﺻﻠﻲ ﺍﺳﺖ؟
ﺍﻟﻒ( 7 پ( 10 ﺙ( 18
ﭘﺮﺳﺶﻫﺎﻱ 11ﺗﺎ 20 ﻫﺮ ﻛﺪﺍﻡ 4ﺍﻣﺘﻴﺎﺯ ﺩﺍﺭﺩ. .11ﺑﺮﻧﺎﻣﻪﻱ ﺭﻭﺯﺍﻧﻪﻱ ﺩﺑﺴــﺘﺎﻥ ﺑﻪ ﺷﻜﻞ ﺯﻳﺮ ﺍﺳــﺖ .ﺩﺭ ﻫــﺮ ﺭﻭﺯ ﭼﻨــﺪ ﺩﻗﻴﻘﻪ ﺑﺮﺍﻱ ﻛﻼﺱ ﺩﺭ ﻧﻈﺮ ﮔﺮﻓﺘﻪ ﺷﺪﻩ ﺍﺳﺖ؟
ﺏ( 8 ﺕ( 12
ﺍﻟﻒ( ﺕ(
1 4 1
ﺏ( ﺙ(
3
1 6
پ(
3
1 5
.5ﺩﺭ ﺣﻴﺎﻁ ﻣﺪﺭﺳﻪ ﺩﺭ ﻳﻚ ﺻﻒ ،ﺳﺎﺭﺍ ﭘﺸــﺖ ﺳﺮ ﻣﺮﻳﻢ ﺍﻳﺴﺘﺎﺩﻩ ﻭ ﺳﻤﻴﺮﺍ ﺑﻴﻦ ﺳﺎﺭﺍ ﻭ ﻣﺮﻳﻢ ﺍﺳــﺖ .ﺳــﺎﺭﺍ ﺟﻠﻮﻱ ﭘﺮﻭﺍﻧﻪ ﺍﺳﺖ ﻛﻪ ﭘﺮﻭﺍﻧﻪ ﺧﻮﺩﺵ ﺟﻠﻮﻱ ﭘﺮﺳﺘﻮ ﺍﺳﺖ .ﭼﻬﺎﺭﻣﻴﻦ ﻧﻔﺮ ﺩﺭ ﺍﻳﻦ ﺻﻒ ﻛﻴﺴﺖ؟ ﺏ( ﻣﺮﻳﻢ ﺍﻟﻒ( ﺳﺎﺭﺍ ﺕ( ﭘﺮﻭﺍﻧﻪ پ( ﺳﻤﻴﺮﺍ ﺙ( ﭘﺮﺳﺘﻮ .6ﻣﺠﻤﻮﻉ ﭘﻨﺞ ﻋﺪﺩ 2010ﺷﺪﻩ ﺍﺳﺖ. ﻳﻜﻲ ﺍﺯ ﺍﻳــﻦ ﺍﻋﺪﺍﺩ ﺍﺯ 235ﺑــﻪ 532ﺗﻐﻴﻴﺮ ﻣﻲﻛﻨﺪ .ﻣﺠﻤﻮﻉ ﺟﺪﻳﺪ ﭼﻪ ﻋﺪﺩﻱ ﺍﺳﺖ؟ ﺏ( 2542 ﺍﻟﻒ( 1723 ﺕ( 1896 پ( 2360 ﺙ( 2307 .7ﻫﺸــﺖ ﻣﻜﻌــﺐ ﺭﺍ ﺑﻪ ﺷــﻜﻞ ﺯﻳﺮ ﺑﻪ ﻫــﻢ ﭼﺴــﺒﺎﻧﺪﻩﺍﻳﻢ ،ﭼﻨﺪ ﻭﺟــﻪ ﺍﺯ ﻣﻜﻌﺐﻫﺎ ﭼﺴﺐﻛﺎﺭﻱ ﺷﺪﻩﺍﻧﺪ؟ 46
ﺭﺍﻫﻨﻤﺎﻳﻲ
ﺑﺮﻧﺎﻣﻪﻱ ﺻﺒﺤﮕﺎﻩ
9-9:10
ﻛﻼﺱ
9:10-11:00
ﺯﻧﮓ ﺗﻔﺮﻳﺢ
11:00-11:30
ﻛﻼﺱ
11:30-13:00
ﻭﻗﺖ ﻧﻬﺎﺭ
13:00-13:50
ﻛﻼﺱ
13:50-15:00
ﭘﺎﻳﺎﻥ ﻛﺎﺭ ﺩﺑﺴﺘﺎﻥ
15:00
.8ﺑﺎ ﺍﺳــﺘﻔﺎﺩﻩ ﺍﺯ ﺳــﻪ ﻛﺎﺭﺕ ﺯﻳــﺮ ،ﺍﻋﺪﺍﺩ ﺳﻪﺭﻗﻤﻲ ﻣﻲﺳﺎﺯﻳﻢ ﺑﻪ ﻃﻮﺭﻱ ﻛﻪ ﺍﺯ ﻫﺮ ﻛﺎﺭﺕ ﺩﺭ ﻫﺮ ﻋﺪﺩ ﻓﻘﻂ ﻳﻚ ﺑﺎﺭ ﺍﺳﺘﻔﺎﺩﻩ ﻛﻨﻴﻢ .ﺍﺧﺘﻼﻑ ﺑﻴــﻦ ﺑﺰﺭگﺗﺮﻳﻦ ﻭ ﻛﻮﭼﻚﺗﺮﻳﻦ ﻋﺪﺩﻱ ﻛﻪ ﺑﻪ ﺍﻳﻦ ﺗﺮﺗﻴﺐ ﻣﻲﺳﺎﺯﻳﻢ ﭼﻴﺴﺖ؟
2 7 5 2
ﻓﻌﺎﻟﻴﺖ
ﺯﻣﺎﻥ
ﺍﻟﻒ( 477 ﺕ( 1009 ﺙ( 555
ﺍﻟﻒ( 300 پ( 500 ﺙ( 240
ﺏ( 495 ﺕ( 468
.9ﺩﺭ ﻳــﻚ ﻗﺮﻋﻪﻛﺸــﻲ ،ﭘــﺪﺭﻡ 1000 ﺗﻮﻣﺎﻥ ﺑﺮﻧﺪﻩ ﺷﺪ .ﺍﻭ ﺧﻤﺲ 1 1 ) ( ﺁﻥ ﺭﺍ ﺩﺭ ﺑﺎﻧﻚ ﮔﺬﺍﺷﺖ ﻭ ﺭﺑﻊ ) ( 4 5 ﻭ ﺑﻘﻴﻪﻱ ﺁﻥ ﺭﺍ ﺑﻪ ﻣﻦ ﺩﺍﺩ ﻭ ﻫﺮﭼﻪ ﻣﺎﻧﺪﻩ ﺑﻮﺩ ﺑﻪ ﻣﺎﺩﺭﻡ ﺩﺍﺩ .ﭼﻪ ﻣﺒﻠﻐﻲ ﺑﻪ ﻣﺎﺩﺭﻡ ﺭﺳﻴﺪ؟ ﺍﻟﻒ( 400ﺗﻮﻣﺎﻥ ﺏ( 888ﺗﻮﻣﺎﻥ پ( 450ﺗﻮﻣﺎﻥ ﺕ( 550ﺗﻮﻣﺎﻥ ﺙ( 600ﺗﻮﻣﺎﻥ
ﺏ( 250 ﺕ( 270
.12ﻣﻴﺎﻧﮕﻴﻦ ﺩﻭ ﻋﺪﺩ 11ﺍﺳﺖ .ﺍﮔﺮ ﻳﻜﻲ ﺍﺯ ﻋﺪﺩﻫﺎ 6ﺗﺎ ﺍﺯ ﺩﻳﮕﺮﻱ ﺑﻴﺶﺗﺮ ﺑﺎﺷــﺪ ،ﻋﺪﺩ ﺑﺰﺭگﺗﺮ ﻛﺪﺍﻡ ﺍﺳﺖ؟ ﺏ( 8 ﺍﻟﻒ( 6 پ( 11 ﺙ( 17 ﺕ( 14 .13ﭼﻪ ﻛﺴــﺮﻱ ﺍﺯ ﻣﺴــﺘﻄﻴﻞ ﺯﻳﺮ ﺭﻧﮓ ﺷﺪﻩ ﺍﺳﺖ؟ 3
.10ﺍﮔــﺮ ﺩﺭ ﺷــﻜﻞ ﺯﻳــﺮ ﺍﺯ ﻧﻘﻄﻪﻱ A ﺣﺮﻛﺖ ﻛﻨﻴــﻢ ﻭ ﺩﻭﺑﺎﺭﻩ ﺑﻪ Aﺑﺎﺯﮔﺮﺩﻳﻢ ،ﭼﻪ ﻣﺴﺎﻓﺘﻲ ﭘﻴﻤﻮﺩﻩﺍﻳﻢ؟
2
4 12m
13m
ﺍﻟﻒ( 52ﻣﺘﺮ پ( 52ﻣﺘﺮ ﺙ( 50ﻣﺘﺮ
ﺩﻭﺭﺓ ﺷﺎﻧﺰﺩﻫﻢ ،ﺷﻤﺎﺭﺓ ،4ﺗﺎﺑﺴﺘﺎﻥ 1390
A
ﺏ( 48ﻣﺘﺮ ﺕ( 50ﻣﺘﺮ
ﺍﻟﻒ( پ(
1 2
1 3
ﺏ( ﺕ(
5 12 2 7
ﺙ(
3 8
.14ﺩﺭ ﮔﺮﻭﻫــﻲ ﺍﺯ 55ﺩﺍﻧﺶﺁﻣــﻮﺯ39 ، ﻧﻔﺮ ﺩﺭ ﻛﻼﺱ ﺭﻳﺎﺿﻲ ﺛﺒﺖﻧﺎﻡ ﻛﺮﺩﻩﺍﻧﺪ ﻭ 35
ﺏ
16
ﺏ
15
پ
14
ﺏ
13
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12
ﺕ
11
ﺕ
10
ﺙ
9
ﺏ
8
ﺙ
7
ﺙ
6
ﺕ
5
ﺍﻟﻒ
4
ﺏ
3
ﺙ
2
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1 ﺳﺆﺍﻝ
ﮔﺰﻳﻨﻪﻱ ﺻﺤﻴﺢ ﭘﺎﺳﺦ ﭘﺮﺳﺶﻫﺎ:
.20ﺩﺭ ﺷﻜﻞ ﺯﻳﺮ ،ﻣﺴﺎﺣﺖ ﺳﻪ ﻣﺴﺘﻄﻴﻞ ﺑﺮ ﺣﺴﺐ ﺳــﺎﻧﺘﻲﻣﺘﺮ ﻣﺮﺑﻊ ﺩﺍﺩﻩ ﺷﺪﻩ ﺍﺳﺖ. ﻣﺴﺎﺣﺖ ﻗﺴﻤﺖ ﺭﻧﮕﻲ ﭼﻨﺪ ﺳﺎﻧﺘﻲﻣﺘﺮ ﻣﺮﺑﻊ ﺍﺳﺖ؟
17
.17ﻫــﺮ ﻳﻚ ﺍﺯ ﺍﻋــﺪﺍﺩ 5 ،4 ،3 ،2 ،1ﺭﺍ ﺩﺭ ﻳﻜﻲ ﺍﺯ ﺩﺍﻳﺮﻩﻫﺎﻱ ﺷﻜﻞ ﺯﻳﺮ ﻗﺮﺍﺭ ﻣﻲﺩﻫﻴﻢ ﺗــﺎ ﺍﻋﺪﺍﺩﻱ ﻛــﻪ ﺑﺎ ﻳﻚ ﺧﻂ ﺑــﻪ ﻫﻢ ﻣﺘﺼﻞ ﻣﻲﺷﻮﻧﺪ ،ﭘﺸﺖ ﺳــﺮ ﻫﻢ ﺑﺎﺷﻨﺪ .ﻣﺠﻤﻮﻉ X ﻭ Yﭼﻨﺪ ﻣﻲﺗﻮﺍﻧﺪ ﺑﺎﺷﺪ؟
.19ﻛﻼﻩﻓﺮﻭﺷــﻲ ﺩﺭ ﻳــﻚ ﺭﻭﺯ ﺣﺮﺍﺟﻲ، ﻛﻼﻩﻫــﺎﻱ 1200ﺗﻮﻣﺎﻧﻲ ﺭﺍ ﺻﺒﺢ ﻓﺮﻭﺧﺖ ﻭ ﻛ ً ﻼ 72000ﺗﻮﻣــﺎﻥ ﻓﺮﻭﺵ ﻛﺮﺩ .ﺑﻌﺪ ﺍﺯﻇﻬﺮ، ﻗﻴﻤﺖ ﺑﺎﻗﻲﻣﺎﻧــﺪﻩﻱ ﻛﻼﻩﻫﺎ ﺭﺍ 900ﻛﺎﻫﺶ ﺩﺍﺩ ﻭ ﺑــﻪ ﺗﻌﺪﺍﺩ ﺩﻭ ﺑﺮﺍﺑــﺮ ﻓﺮﻭﺵ ﺻﺒﺢ ،ﻛﻼﻩ ﻓﺮﻭﺧﺘﻪ ﺷﺪ .ﻛﻞ ﺩﺭﺁﻣﺪ ﺍﻳﻦ ﻛﻼﻩﻓﺮﻭﺷﻲ ﺩﺭ ﺁﻥ ﺭﻭﺯ ﭼﻨﺪﺗﻮﻣﺎﻥ ﺑﻮﺩﻩ ﺍﺳﺖ؟ ﺍﻟﻒ( 180000 ﺏ( 90000 پ( 126000 ﺕ( 144000 ﺙ( 288000
ﺕ
ﺍﻟﻒ( 100ﻭ 60 پ( 90ﻭ 70 ﺙ( 75ﻭ 50
ﺏ( 60ﻭ 90 ﺕ( 86ﻭ 36
ﺍﻟﻒ( 216 ﺕ( 207
ﺏ( 54 ﺙ( 200
پ( 181
18
.16ﺑﺮﺍﻱ ﺳــﺎﺧﺖ ﻳﻚ ﭘﺘﻮﻱ ﻣﺴﺘﻄﻴﻞ ﺷــﻜﻞ ﺑــﻪ ﺍﺑﻌــﺎﺩ 120ﺳــﺎﻧﺘﻲﻣﺘﺮ ﺩﺭ 90 ﺳــﺎﻧﺘﻲﻣﺘﺮ ،ﺣﺎﺷــﻴﻪﻫﺎﻳﻲ ﻣﺎﻧﻨﺪ ﺷــﻜﻞ ﺑﻪ ﻳــﻚ ﺗﻜﻪ ﭘﺘــﻮﻱ ﻛﻮﭼﻚﺗﺮ ﺍﺿﺎﻓــﻪ ﻛﺮﺩﻳﻢ. ﺍﮔﺮ ﺿﺨﺎﻣﺖ ﺣﺎﺷــﻴﻪﻫﺎ ﺩﻭﺭ ﺗﺎ ﺩﻭﺭ ﭘﺘﻮ ﻳﻚ ﺍﻧﺪﺍﺯﻩ ﺑﺎﺷــﺪ ،ﺍﺑﻌﺎﺩ ﭘﺘﻮﻱ ﻣﺮﻛﺰﻱ )ﺑﺮ ﺣﺴﺐ ﺳﺎﻧﺘﻲﻣﺘﺮ( ﭼﻪ ﺍﻧﺪﺍﺯﻩﺍﻱ ﻣﻲﺗﻮﺍﻧﺪ ﺑﺎﺷﺪ؟
.18ﺷﻜﻞ ﺯﻳﺮ ،ﻧﻤﻮﺩﺍﺭ ﻳﻚ ﺑﺎﻏﭽﻪ ﺍﺳﺖ. ﻗﺴــﻤﺘﻲ ﺍﺯ ﺑﺎﻏﭽﻪ ،ﭼﻤﻦ ﻛﺎﺷــﺘﻪ ﺷــﺪﻩ ﻭ ﻗﺴــﻤﺘﻲ ﺍﺯ ﺁﻥ ﺑﺎ ﺳــﻨﮓﻫﺎﻱ ﻣﺮﺑﻊ ﺷــﻜﻞ، ﺳﻨﮓﻓﺮﺵ ﺷﺪﻩ ﺍﺳﺖ. ِ ﻗﺴﻤﺖ ﭼﻤﻦ ﻛﺎﺷﺘﻪ ﺷﺪﻩ، ﻣﺴــﺎﺣﺖ ﻛﻞ 108ﻣﺘﺮ ﻣﺮﺑﻊ ﺍﺳﺖ. ﻣﺴــﺎﺣﺖ ﻗﺴﻤﺖ ﺳــﻨﮓﻓﺮﺵ ﭼﻨﺪ ﻣﺘﺮ ﻣﺮﺑﻊ ﺍﺳﺖ؟
ﺕ
2
ﺙ( 3
پ( 6
ﺏ( 48 ﺙ( 70
19
ﺕ(
1
4
6
3
ﺍﻟﻒ( 3 ﺕ( 7
ﺏ( 4 ﺙ( 8
ﺍﻟﻒ( 36 ﺕ( 60
پ( 56
ﺍﻟﻒ
ﺍﻟﻒ( 4
ﺏ(
1
4
پ(
1
4
X
٢٠
20
.15ﻣﺴﺎﺣﺖ ﻣﺜﻠﺚ ﺑﺮ ﺣﺴﺐ ﺷﺶﺿﻠﻌﻲ ﻭﺍﺣﺪ ﻧﺸﺎﻥ ﺩﺍﺩﻩ ﺷﺪﻩ ﺩﺭ ﺷﻜﻞ ﭼﻴﺴﺖ؟
٢٥
پ
ﻧﻔــﺮ ﺩﺭ ﻛﻼﺱ ﻋﻠﻮﻡ .ﭼﻨﺪ ﺩﺍﻧﺶﺁﻣﻮﺯ ﺩﺭ ﻫﺮ ﺩﻭ ﻛﻼﺱ ﺭﻳﺎﺿﻲ ﻭ ﻋﻠﻮﻡ ﺛﺒﺖﻧﺎﻡ ﻛﺮﺩﻩﺍﻧﺪ؟ پ( 19 ﺏ( 16 ﺍﻟﻒ( 20 ﺙ( 55 ﺕ( 4
Y
٧٠
ﺩﻭﺭﺓ ﺷﺎﻧﺰﺩﻫﻢ ،ﺷﻤﺎﺭﺓ ،4ﺗﺎﺑﺴﺘﺎﻥ 1390
ﺭﺍﻫﻨﻤﺎﻳﻲ
47
ﻣﻌﺮﻓﻲ ﻛﺘﺎﺏ
ر︀︲﹫︀ت ز︊︀ و دو︨️دا︫︐﹠﹩ ﺟﻌﻔﺮ ﺭﺑﺎﻧﻲ
ﻧﻮﻳﺴﻨﺪﻩ :ﻣﺎﺭﺗﻴﻦ ﻟﻮ ﺗﺮﺟﻤﻪ :ﻛﺎﻇﻢ ﻓﺎﺋﻘﻲ
ﻧﺎﺷﺮ :ﻟﻮﺡ ﺩﺍﻧﺶ1379 ، ]ﺗﻬﺮﺍﻥ66403512 :ـ[021
ﻣﺎ ﺗﺎﻛﻨﻮﻥ ﭼﻨﺪ ﻛﺘﺎﺏ ﻣﺨﺘﻠﻒ ﺭﺍ ﻛﻪ ﺭﻳﺎﺿﻲ ﺭﺍ ﺍﺯ ﻃﺮﻳﻖ ﺑﺎﺯﻱ، ﻣﻌﻤﺎ ،ﻟﻄﻴﻔﻪ ﻭ ﺳــﺮﮔﺮﻣﻲ ﺑﺮﺍﻱ ﺷﻤﺎ ﺁﺳﺎﻥ ﻣﻲﻛﻨﺪ ﻭ ﻣﻬﻢﺗﺮ ّ ﺍﺯ ﺍﻳﻦ ،ﺑﺎﻋﺚ ﻣﻲﺷــﻮﺩ ﺷــﻤﺎ ﺑﻪ ﺍﻳﻦ ﺩﺭﺱ ﻋﻼﻗﻤﻨﺪ ﺷﻮﻳﺪ ﻣﻌﺮﻓﻲ ﻛﺮﺩﻩﺍﻳﻢ .ﻛﺘﺎﺏ »ﺭﻳﺎﺿﻴﺎﺕ ﺯﻳﺒﺎ ﻭ ﺩﻭﺳﺖﺩﺍﺷﺘﻨﻲ« ﻧﻴﺰ ﻳﻜﻲ ﺍﺯ ﺍﻳﻦ ﻛﺘﺎﺏﻫﺎﺳﺖ .ﺍﻳﻦ ﻛﺘﺎﺏ ﺷﺎﻣﻞ 145ﻣﺴﺌﻠﻪ ﺩﺭ ﺍﻧﻮﺍﻉ ﻣﺨﺘﻠﻒ ﺍﺳــﺖ ﻛﻪ ﺑﺪﻭﻥ ﻧﻈﻢ ﻭ ﺗﺮﺗﻴﺐ ﻣﺸﺨﺼﻲ، ﺑﻪ ﺩﻧﺒﺎﻝ ﻳﻜﺪﻳﮕﺮ ﻗﺮﺍﺭ ﮔﺮﻓﺘﻪﺍﻧﺪ .ﺩﺭ ﭘﺎﻳﺎﻥ ﻛﺘﺎﺏ ﻧﻴﺰ ﭘﺎﺳﺦ ﻫﻤﻪ ﻣﺴﺎﺋﻞ ﺁﻣﺪﻩ ﺍﺳﺖ .ﺩﺭ ﺍﻳﻨﺠﺎ ﺩﻭ ﻧﻤﻮﻧﻪ ﺍﺯ ﺍﻳﻦ ﻣﺴﺎﺋﻞ ﺭﺍ ﺑﺮﺍﻱ ﺷــﻤﺎ ﻣﻲﺁﻭﺭﻳﻢ .ﻳﻜﻲ ﺑــﺪﻭﻥ ﺟﻮﺍﺏ ﻭ ﺩﻳﮕﺮﻱ ﺑﺎ ﺟﻮﺍﺏ. • 6ﮔﺮﺑــﻪ 6ﻣــﻮﺵ ﺭﺍ ﺩﺭ 6ﺩﻗﻴﻘﻪ ﻣﻲﺧﻮﺭﻧﺪ .ﭼﻨﺪ ﮔﺮﺑﻪ 60ﻣﻮﺵ ﺭﺍ ﺩﺭ 60ﺩﻗﻴﻘﻪ ﻣﻲﺧﻮﺭﻧﺪ؟ • ﺍﺯ ﻳﻚ ﻧﻔﺮ ﭘﺮﺳﻴﺪﻧﺪ :ﭼﻬﻞ ﺳﺎﻝ ﭘﻴﺶ ﭼﻨﺪ ﺑﻬﺎﺭ ﺍﺯ ﻋﻤﺮ ﺗﻮ ﻣﻲﮔﺬﺷﺖ؟ ﻭ ﺍﻭ ﭼﻨﻴﻦ ﭘﺎﺳﺦ ﺩﺍﺩ :ﺍﮔﺮ ﺑﻪ ﺳــﻦ ﻣﻦ ﺩﺭ ﺁﻥ ﺯﻣﺎﻥ ﺣﺎﺻﻠﻀﺮﺏ 5ﺩﺭ 7ﻭ ﻫﻢﭼﻨﻴــﻦ 7ﺩﺭ 3ﺭﺍ ﻣﻲﺍﻓﺰﻭﺩﻳﺪ ﻭ ﺣﺎﺻﻠﻀﺮﺏ 6ﺩﺭ 9ﺑــﻪ ﺍﺿﺎﻓــﻪ 4ﺭﺍ ﺍﺯ ﺁﻥ ﻛﻢ ﻣﻲﻛﺮﺩﻳﺪ، ﺳــﻦ ﺁﻥ ﺯﻣﺎﻥ ﻣﻦ ﺣﺎﺻﻞ ﻣﺴــﺎﻭﻱ ﺑﺎ ﺩﻭﺑﺮﺍﺑﺮ ّ ﻣﻨﻬــﺎﻱ 20ﺑﻮﺩ .ﺍﮔﺮ ﺍﻳــﻦ ﮔﻔﺘﻪ ﻭﻱ ﺻﺤﻴﺢ ﺑﺎﺷﺪ ،ﺍﻭ ﺩﺭ ﺣﺎﻝ ﺣﺎﺿﺮ ﭼﻨﺪ ﺳﺎﻝ ﺩﺍﺭﺩ؟ ﺟﻮﺍﺏ ﺭﺍ ﺍﺯ ﺍﻳﻦ ﻣﻌﺎﺩﻟﻪ ﭘﻴﺪﺍ ﻛﻨﻴﺪ: x+(5×7)+(7×3)-(6×9+4)=2x-20 48
ﺭﺍﻫﻨﻤﺎﻳﻲ
ﺩﻭﺭﺓ ﺷﺎﻧﺰﺩﻫﻢ ،ﺷﻤﺎﺭﺓ ،4ﺗﺎﺑﺴﺘﺎﻥ 1390