Problem. a, b, c, d<\/em> pozitif tamsay\u0131lar\u0131n\u0131n toplam\u0131 63 ise ab + bc + cd\u00a0<\/em>de\u011ferinin alabilece\u011fi en b\u00fcy\u00fck de\u011fer ka\u00e7t\u0131r?<\/p>\n Bu problem, belki sezgisel yolla deneme yap\u0131larak \u201c\u00e7\u00f6z\u00fclebilir\u201d, ama biliyoruz ki b\u00f6ylesi \u00e7\u00f6z\u00fcmlerin hi\u00e7bir matematiksel de\u011feri yok. A\u015fa\u011f\u0131da birbirinden g\u00fczel iki \u00e7\u00f6z\u00fcm verece\u011fiz, ama yarat\u0131c\u0131l\u0131k ve zarafet \u00f6l\u00e7\u00fctlerine g\u00f6re benim se\u00e7imim birinciden yana.<\/p>\n Ayr\u0131ca, \u00e7\u00f6z\u00fcmlerde kullan\u0131lan bilgilerin ortaokul d\u00fczeyinde oldu\u011funu belirtmek isterim.<\/p>\n Birinci \u00e7\u00f6z\u00fcm:<\/strong> A\u015fa\u011f\u0131daki \u015fekilde kenar uzunluklar\u0131 a, b, c, d<\/em> ve alanlar\u0131\u00a0 b x c= bc, a x b= ab, c x d= cd, a x d= ad olan dikd\u00f6rtgenler \u00e7izilmi\u015ftir.<\/em><\/p>\n ab + bc + cd<\/em> toplam\u0131n\u0131n en b\u00fcy\u00fck de\u011ferini istedi\u011fimiz i\u00e7in do\u011fal olarak, yukar\u0131daki \u015fekilde griye boyanm\u0131\u015f b\u00f6lgenin alan\u0131n\u0131 b\u00fcy\u00fctmemiz gerekiyor. O zaman ad<\/em>‘yi k\u00fc\u00e7\u00fcltmeliyiz. a<\/em> ve d <\/em>pozitif tamsay\u0131 oldu\u011fundan alabilecekleri en k\u00fc\u00e7\u00fck de\u011fer 1’dir. Dolay\u0131s\u0131yla ad<\/em>= 1 olur. A\u015fa\u011f\u0131daki \u015fekil a= d=<\/em> 1 durumunu g\u00f6stermektedir.<\/p>\n Yukar\u0131daki en b\u00fcy\u00fck dikd\u00f6rtgenin uzun kenar\u0131n\u0131 u <\/em>ile, k\u0131sa kenar\u0131n\u0131 da v<\/em> ile g\u00f6zterelim ve u= a + c <\/em>ve v= b + d <\/em>e\u015fitliklerini yazal\u0131m. \u00d6te yandan a= d= 1<\/em> ve u= a + 1<\/em>, v= b + 1<\/em> ve u + v =<\/em> 63 oldu\u011funu biliyoruz.<\/p>\n \u015eimdi, en b\u00fcy\u00fck de\u011ferini bulmak istedi\u011fimiz griye boyal\u0131 b\u00f6lgenin alan\u0131n\u0131 S<\/em> ile g\u00f6sterelim ve bu alan\u0131 en b\u00fcy\u00fck dikd\u00f6rtgenin alan\u0131ndan kenar uzunluklar\u0131 1 birim olan karenin alan\u0131n\u0131 \u00e7\u0131kartarak ifade edelim:<\/p>\n S = uv –<\/em> 1.<\/p>\n Bu ifadede uv<\/em>\u2019nin en b\u00fcy\u00fck olmas\u0131n\u0131 istiyoruz ve ayr\u0131ca u + v<\/em>\u2019nin de 63 oldu\u011funu biliyoruz. Yapaca\u011f\u0131m\u0131z birka\u00e7 denemeyle toplamlar\u0131 sabit iki pozitif tamsay\u0131n\u0131n \u00e7arp\u0131mlar\u0131n\u0131n en b\u00fcy\u00fck olmas\u0131 i\u00e7in bu say\u0131lar\u0131n birbirlerine en yak\u0131n, hatta e\u015fit olmas\u0131 gerekti\u011fini g\u00f6rebiliriz. (Bu sonu\u00e7 parabol veya t\u00fcrev bilgisiyle genel olarak kan\u0131tlanabilir.)<\/p>\n uv<\/em>\u2019nin en b\u00fcy\u00fck de\u011ferini bulabilmemiz i\u00e7in u = v =<\/em> 63\/2 olmal\u0131, ama u<\/em> ve v<\/em> pozitif tam say\u0131 oldu\u011fundan u<\/em> ve v<\/em> say\u0131lar\u0131ndan birini 31 di\u011ferini de 32 almal\u0131y\u0131z. Bu durumda en b\u00fcy\u00fck olmas\u0131n\u0131 istedi\u011fimiz S<\/em> de\u011feri 31 x 32 – 1 = 991 olur ki, bu say\u0131 ab + bc + cd\u00a0<\/em>toplam\u0131n\u0131n alabilece\u011fi en b\u00fcy\u00fck de\u011ferdir.<\/p>\n \u0130kinci \u00e7\u00f6z\u00fcm: <\/strong>Bir reel say\u0131n\u0131n karesinin ya pozitif bir say\u0131ya ya da s\u0131f\u0131ra e\u015fit oldu\u011funu biliyoruz. Bu y\u00fczden olur. Yukar\u0131daki e\u015fitsizli\u011fin her iki yan\u0131na 2xy<\/em> terimini eklersek<\/p>\n elde edilir. \u015eimdi, a + c = x<\/em> ve b + d = y<\/em> alal\u0131m, ve yukar\u0131daki e\u015fitsizlikte x<\/em> ve y<\/em> yerine bu de\u011ferleri yazal\u0131m.<\/p>\n Bu e\u015fitsizli\u011fin sol taraf\u0131n\u0131 da\u011f\u0131lma \u00f6zelli\u011fini kullanarak d\u00fczenleyelim ve a + c + b + d<\/em> yenine 63 yazal\u0131m:<\/p>\n a, b, c, d<\/em> pozitif tamsay\u0131 oldu\u011fundan 992’nin k\u00fcsurat\u0131n\u0131 atabiliriz:<\/p>\n Bu e\u015fitsizlikte ab + bc + cd<\/em> toplam\u0131n\u0131n en b\u00fcy\u00fck de\u011ferini bulmak istedi\u011fimiz i\u00e7in ad<\/em>\u2019ye en k\u00fc\u00e7\u00fck de\u011ferini vermeliyiz. Ki bu de\u011fer de a = d = <\/em>1 iken 1\u2019dir. Dolay\u0131s\u0131yla<\/p>\n e\u015fitsizli\u011fi elde edilir. O halde ab + bc + cd<\/em> toplam\u0131n\u0131n en b\u00fcy\u00fck de\u011feri 991\u2019dir.<\/p>\n KAYNAK<\/strong><\/p>\n 1) http:\/\/alekdimitrov.com\/downloads\/aimo_test.pdf<\/p>\n","protected":false},"excerpt":{"rendered":" Matematik olimpiyatlar\u0131nda sorulan bir\u00e7ok problemin \u00e7\u00f6z\u00fcm\u00fc s\u0131ra d\u0131\u015f\u0131, ak\u0131l dolu ve \u00e7ok ilgin\u00e7 ad\u0131mlardan olu\u015fur. \u00c7\u00f6z\u00fcm\u00fcn sonunda adeta usta bir satran\u00e7 oyuncusunun birka\u00e7 hamle sonra ortaya \u00e7\u0131kacak olan o p\u0131r\u0131lt\u0131l\u0131 hamlesi gibi bir s\u00fcrprizle kar\u015f\u0131la\u015f\u0131rs\u0131n\u0131z. A\u015fa\u011f\u0131da ele alaca\u011f\u0131m\u0131z olimpiyat sorusunun \u00e7\u00f6z\u00fcm\u00fcndeki y\u00f6ntemin de bu \u00f6zelliklere sahip oldu\u011funu ve merakl\u0131 okuru heyecanland\u0131raca\u011f\u0131n\u0131 d\u00fc\u015f\u00fcn\u00fcyorum. Bu problem 2013\u2019te […]<\/p>\n","protected":false},"author":375,"featured_media":41467,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"_acf_changed":false,"footnotes":""},"categories":[514,5947,510],"tags":[3562,1008],"class_list":["post-41466","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-matematik-sohbetleri","category-olaganustu-nisan-sayisi","category-surekli-bolumler","tag-ali-torun","tag-matematik-sohbetleri"],"acf":[],"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/bilimvegelecek.com.tr\/index.php\/wp-json\/wp\/v2\/posts\/41466","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/bilimvegelecek.com.tr\/index.php\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/bilimvegelecek.com.tr\/index.php\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/bilimvegelecek.com.tr\/index.php\/wp-json\/wp\/v2\/users\/375"}],"replies":[{"embeddable":true,"href":"https:\/\/bilimvegelecek.com.tr\/index.php\/wp-json\/wp\/v2\/comments?post=41466"}],"version-history":[{"count":0,"href":"https:\/\/bilimvegelecek.com.tr\/index.php\/wp-json\/wp\/v2\/posts\/41466\/revisions"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/bilimvegelecek.com.tr\/index.php\/wp-json\/wp\/v2\/media\/41467"}],"wp:attachment":[{"href":"https:\/\/bilimvegelecek.com.tr\/index.php\/wp-json\/wp\/v2\/media?parent=41466"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/bilimvegelecek.com.tr\/index.php\/wp-json\/wp\/v2\/categories?post=41466"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/bilimvegelecek.com.tr\/index.php\/wp-json\/wp\/v2\/tags?post=41466"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}![\"\"](\"https:\/\/bilimvegelecek.com.tr\/wp-content\/uploads\/2020\/04\/\u015fekil-1-300x206.jpg\")
<\/p>\n![\"\"](\"https:\/\/bilimvegelecek.com.tr\/wp-content\/uploads\/2020\/04\/\u015fekil-2-300x191.jpg\")
<\/p>\n![\"\"](\"https:\/\/bilimvegelecek.com.tr\/wp-content\/uploads\/2020\/04\/1-7.jpg\")
. Parantez kareyi a\u00e7arsak,<\/p>\n![\"\"](\"https:\/\/bilimvegelecek.com.tr\/wp-content\/uploads\/2020\/04\/2-3.jpg\")
<\/p>\n![\"\"](\"https:\/\/bilimvegelecek.com.tr\/wp-content\/uploads\/2020\/04\/3-4.jpg\")
<\/p>\n![\"\"](\"https:\/\/bilimvegelecek.com.tr\/wp-content\/uploads\/2020\/04\/4-1.jpg\")
<\/p>\n![\"\"](\"https:\/\/bilimvegelecek.com.tr\/wp-content\/uploads\/2020\/04\/5-3.jpg\")
<\/p>\n![\"\"](\"https:\/\/bilimvegelecek.com.tr\/wp-content\/uploads\/2020\/04\/6-300x50.jpg\")
<\/p>\n![\"\"](\"https:\/\/bilimvegelecek.com.tr\/wp-content\/uploads\/2020\/04\/7-300x41.jpg\")
<\/p>\n![\"\"](\"https:\/\/bilimvegelecek.com.tr\/wp-content\/uploads\/2020\/04\/8.jpg\")
<\/p>\n![\"\"](\"https:\/\/bilimvegelecek.com.tr\/wp-content\/uploads\/2020\/04\/9.jpg\")
<\/p>\n