第52届()IMO试题及解答(.7.187.19)_第39届imo试题解答

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第52届（2011）IMO试题及解答(2011.7.18-7.19)

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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

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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《中学数学信息网》系列资料 www.daodoc.com 版权所有@《中学数学信息网》

欢迎光临《中学数学信息网》 zxsx127@163.com

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 poible divisibilities are, and.Then the cases make.We have

and,.Now, aume without lo of generality

are easily got rid of.And let us try to

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.Let.Then rearranging gives

and hence,and so,..Then by the conditions, we have

.So,.This gives Indeed, if

.Now, consider the condition

.Let.Rearranging gives

.If , then and if,2.To each oriented straight line than point meeting

in exactly one point, aociate the pair of numbers

of points from , other that to each , lying on its right side, respectively left side(clearly(called anchor)corresponds some distinguished line

with).Our goal is to exhibit such a(fixed)pair

.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 paes through two points), the pair aociated 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 pa through its anchor point(because of the count of points given by

Thus, there is nothing magic about having a balanced pair containing a point(other than its centre)we can use a

;for a convex formation we can use a

pair.However, a pair known to exist for all points of discrete continuity arguments.pair;for a square

is precisely such a balanced pair, by In fact the first paragraph even provides a way to measure the correctne 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 paing through, with aociated 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 pa precisely through those points and

.gives only those as pivots, for which there exists a line having aociated the starting pair 3.Let be the aertion that

., 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.But for such, we get , we have

which is impoible, so

Now substitute.Since

into(1)to get

for all, we see that if, then, whence.Thus,for all

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whenever

Substituting gives.Thus, and

into(1)therefore yields , so whenever

.., so as well.But , so, and , so whenever

4.The problem is equivalent to counting the number of ways , and

of creating a sum is a permutation of, where

;such that it satisfies the property, for every

Notice that the signed sum of different powers of Then we can extract the term , and divide the remaining terms by

is always different than zero..(this is for, and we will have a solution for the problem with)Also the term cannot be at the begining, but otherwise,(2n-1)

.Then

can be anywhere.for every

(2n-1)！

Then we get a recuerrence

We can easily see that 5.We have implying have hence

setting gives

.This gives, setting we get , If

.This implies we may have

.Further

but from the previous inequality which means case, but this ,with ,giving and the only poibility is

.But we have

.Hence we must have, and we are done!

clearly impoible.We consider , gives, again not poible.The last case

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6.Denote Let the intersections of and

;

be the reflections of quadrilateral, respectively;wrt

.the intersections of 3 lines.be the point of contact of and be the Miquel point of the completed

We have the distances from

are equal so

is the bisector of angle

We get On the other side, let

be the projections of

onto

then

paes through the midpoint of, which follows that

concyclic.We get

.are From So

and we obtain.Similarly with

Therefore Construct a tangent of

.We will show that

is also a tangent of

iff

.But

Hence

is true.We are done.http://www.daodoc.com/Forum/viewforum.php?f=729 IMO2012将在阿根廷举行。

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