加载中…[转载]2013年湖北省数学预赛第13题(高一)的命制背景
(2013-08-12 15:03:31)
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题目：设http://data.artofproblemsolving.com/images/latex/3/7/a/37af1f49ab17dd33465780099f8cf172478970c4.gif，
求四边形http://data.artofproblemsolving.com/images/latex/3/0/2/302be1442ed0f92a3a73687e7de9b17faadd1015.gif的面积的最大值.
背景分析：个人感觉该试题是根据如下的两个常见结论命制的.
结论1：设http://data.artofproblemsolving.com/images/latex/0/c/2/0c2a050bd035376f54ec47012ba5c5d76d3daa07.gif.
结论2：(阿波罗尼奥斯圆)若一动点到两定点的距离之比是一个不等于http://data.artofproblemsolving.com/images/latex/3/5/6/356a192b7913b04c54574d18c28d46e6395428ab.gif的常数，则该动点的轨迹是一个圆.
若熟悉上述两个几何结论，则解决该题并不困难.
解：记http://data.artofproblemsolving.com/images/latex/0/5/8/0582ab4d28e5198f79a9a0955636abd6111ef183.gif.
显然http://data.artofproblemsolving.com/images/latex/1/3/d/13db13a0215f7af6ac6c1124f19edc7e939b072c.gif.
由http://data.artofproblemsolving.com/images/latex/8/a/3/8a39e969831d9b367ff829e0363e50ca409fec47.gif.
故http://data.artofproblemsolving.com/images/latex/e/0/f/e0f5d54a0b0908dbaffe996e42697291d39d2e4c.gif时，等号成立，
于是即求http://data.artofproblemsolving.com/images/latex/2/1/5/215523d3212aa2b8b7b4e6bec3478ba3a17fce03.gif的最大值即可.
建立直角坐标系如图所示，则http://data.artofproblemsolving.com/images/latex/7/4/a/74a744ab3baf51f47b28d2e75ee854b2881ac558.gif，
则由http://data.artofproblemsolving.com/images/latex/f/c/8/fc8846dd4cb064206dc662bce2b5a00be295a9bc.gif，
化简得http://data.artofproblemsolving.com/images/latex/e/d/8/ed8ea5d2b129990a6c2495d7f4224ad72605b9de.gif为圆心，
http://data.artofproblemsolving.com/images/latex/0/8/9/089fd7a0a9d12dee7b00c48254726b12bcb212b7.gif 为半径的圆，所以http://data.artofproblemsolving.com/images/latex/8/b/8/8b84b3598c02e4c28e7ab7c01f0e3ed137b4eccc.gif
故http://data.artofproblemsolving.com/images/latex/c/3/f/c3f2a0a17ff148464f1c0ffce3bc7bb7385d6f87.gif.
注：偶尔写一次博客，主要是没时间，欢迎大家转载.
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