第52届(2011)IMO试题及解答(2011.7.18-7.19).doc

欢迎光临中学数学信息网 第52届（2011）IMO试题及解答(2011.7.18-7.19) /html/2011/07/2371.html在荷兰阿姆斯特丹举行的第52届IMO（2011）中，中国队获六块金牌，总分第一！举小旗的是陈麟，这是2011年第52届国际奥林匹克数学竞赛（IMO2011）的开幕式。陈 麟，38分， 总分第3名，金牌！周天佑，34分，总分第10名，金牌！姚博文，30分，总分第14名，金牌！龙子超，30分，总分第14名，金牌！靳兆融，29分，总分第25名，金牌！吴梦希，28分，总分第39名，金牌！18日的比赛陈麟中国人民大学附属中学 高二吴梦希江苏南菁高级中学 高二龙子超湖南师大附中 高二靳兆融中国人民大学附属中学 高三周天佑上海中学 高三姚博文河南省实验中学 高三IMO2011，第14-24名都是30分，第25-38名都是29分，第39-54名都是28分。1分是很精贵的。金牌分数线是28分，大致得做对4道题才行，这也是IMO常见的情况。2011年IMO共有54块金牌，占参赛总人数的约10%！这也是历年的惯例。银牌分数线是22分，得做对3道题，这也是IMO常见的情况。2011年IMO共有90块银牌，占参赛总人数的约20%！这也是历年的惯例。 参加本次数学奥赛的6名中国选手均获得金牌，这使得中国队的团体总分名列第一。美国队的6名选手也均获得金牌，但因总分略低于中国队而获得团体第二名，新加坡队则以4金、1银、1铜的成绩名列团体第三名。 International Mathematical Olympiad是世界上规模最大的奥林匹克数学竞赛。在两天的竞赛中，参赛学生每天需要解答3道题，每道题7分，满分为42分。 赛事分两日进行，每日参赛者有4.5小时来解决三道问题（由上午9时到下午1时30分）。通常每天的第1题（即第1、4题）最浅，第2题（即第2、5题）中等，第3题（即第3、6题）最深。所有问题是由中学数学课程中的不同范畴中选出，通常是组合数学（Combinatorics）、数论（Number Theory）、几何（Geometry）和代数（Algebra）、不等式。解决这些问题，参赛者通常不需要更深入的数学知识（虽然大部分参赛者都有，而且实际上需要很多课程以外的数学知识和技巧），但通常要有异想天开的思维和良好的数学能力，才能找出解答。 共有来自101个国家和地区的564位选手参加了本次大赛。18岁的德国天才少女Lisa Sauermann，凭藉42分的满分，摘得2011年国际奥林匹克数学竞赛金牌，成为历史上获金牌数第一的选手。（2007IMO银牌，2008IMO金牌，2009IMO金牌，2010IMO金牌，2011IMO金牌）美国队自冯祖念做领队以来，几年来首次获得6块金牌。1. All the possible divisibilities are , , and . Now, assume without loss of generality . Then the cases and are easily got rid of. And let us try to make . We have . Now, we still have . Let . Then rearranging gives and so, and hence, . Indeed, if . Then by the conditions , we have . So, . Now, consider the condition . This gives . Let . Rearranging gives . If , then and if , 2.To each oriented straight line meeting in exactly one point , associate the pair of numbers of points from , other than , lying on its right side, respectively left side (clearly ). Our goal is to exhibit such a (fixed) pair that to each point (called anchor) corresponds some distinguished line with . Then starting the windmill on one of these lines will fulfill the requirements, since except at the moment of changing pivots (when the line of the windmill passes through two points), the pair associated to the windmill line stays (this is easily seen, by the very movement of the windmill). As said above, when the windmill line becomes parallel (and similarly oriented) with one of the distinguished lines, it will have to pass through its anchor point (because of the count of points given by ).Thus, there is nothing magic about having a balanced pair ; for a convex formation we can use a pair; for a square containing a point (other than its centre) we can use a pair. However, a pair known to exist for all points of is precisely such a balanced pair, by discrete continuity arguments.In fact the first paragraph even provides a way to measure the correctness of a different approach. Clearly, the lines of every windmill will (except at changing pivots) share a same pair , so for each of the points of there must exist some line passing through, with associated pair that of the starting line of the windmill, otherwise such a faulty point will be skipped, and the proof fail.A final word. A windmill starting on a line through a point , corresponding to the pair , will pass precisely through those points and only those as pivots, for which there exists a line having associated the starting pair .3. Let be the assertion that . gives , whence Suppose for the sake of contradiction that for some . Then by taking to be very small (i.e., negative and having large absolute value) and setting in (1), we get that . Thus, for sufficiently small , we have . But for such , we get which is impossible, so for all . Now substitute into (1) to get , whence . Since for all , we see that if , then . Thus, whenever . Substituting and into (1) therefore yields , so as well. But gives , so , so . Thus, whenever , and , so whenever .4. The problem is equivalent to counting the number of ways of creating a sum , where , and is a permutation of ; such that it satisfies the property , for every Notice that the signed sum of different powers of is always different than zero.Then we can extract the term , and divide the remaining terms by , and we will have a solution for the problem with . (this is for )Also the term cannot be at the begining, but otherwise, can be anywhere.Then we get a recuerrence (2n-1) for every We can easily see that . Then (2n-1)!5. We have setting gives , setting we get implying hence .This gives , If we have .Further .This implies we may have but from the previous inequality which means clearly impossible.We consider case , but this ,with , gives ,giving , again not possible.The last case and the only possibility is .But we have implying . Hence we must have , and we are done!6. Denote the intersections of and , respectively; the intersections of 3 lines .Let be the point of contact of and ; be the reflections of wrt be the Miquel point of the completed quadrilateral .We have the distances from to are equal so is the bisector of angle We get On the other side, let be the projections of onto then passes through the midpoint of . , which follows that are concyclic. We get From and we obtain . Similarly with So There