2022年IMO试题及解答.docx

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1、名师归纳总结 精品学习资料 - - - - - - - - - - - - - - -IMO2022试题及解答第一天, 20XX年 7 月 18 日,星期一1. 对任意由 4 个不同正整数组成的集合,记, are ,设是满意整除的数对的个数求全部由4 个不同正整数组成的集合,使得达到最大值All the possible divisibilities are . Now, assume without loss of , and generality . Then the cases and easily got rid of. And let us try to make . We have

2、 . Now, we still have . Let . Then rearranging gives and so, and hence, . Indeed, if . Then by the conditions gives , . If , we have . So, . This . Now, consider the condition . Let . Rearranging gives and if , then 2. 设 是平面上包含至少两个点的一个有限点集,其中没有三点在同一条直线上所谓一个“ 风车” 是指这样一个过程：从经过 中单独一点 的一条直线 开头,以 为旋转中心顺时

3、针旋转,直至首次遇到 中的另一点,记为点接着这条直线以 为新的旋转中心顺时针旋转,直到再次遇到 中的某一点,这样的过程无限连续下去证明：可以适当选取中的一点,以及过的一条直线 ,使得由此产生的“ 风 第 1 页,共 4 页 - - - - - - - - - 车” 将中的每一点都无限多次用作旋转中心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, res

4、pectively 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 细心整理归纳 精选学习资料 - - - - - - - - - - - - - - -名师归纳总结 精品学习资料 - - - - - - - - - - - - - - -the windmill on one of these lines will fulfill the r

5、equirements, 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

6、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.

7、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 p

8、air , 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 , w

9、ill pass precisely through those points and only those as pivots, for which there exists a line having associated the starting pair . 3. 设 是一个定义在实数集上的实值函数,满意对全部实数, ,都有证明：对全部实数,有Let be the assertion that . gives , whence Suppose for the sake of contradiction that for some . Then by taking to be very

10、第 2 页,共 4 页 small i.e., negative and having large absolute value and setting in 1, we get that which is impossible, so . Thus, for sufficiently small , we have . But for such , we get for all . Now substitute into 1 to get , whence . Since for all , we see that if , then . Thus, whenever . , so Subs

11、tituting and into 1 therefore yields 细心整理归纳 精选学习资料 - - - - - - - - - - - - - - - - - - - - - - - - 名师归纳总结 精品学习资料 - - - - - - - - - - - - - - -as well. But gives , so whenever , and , so whenever . Thus, , so . 其次天, 20XX年 7 月 19 日,星期二4. 给定整数有一个天平和个重量分别为的砝码现通过步操作逐个将全部砝码都放上天平,使得在操作过程中,右边的重量总不超过左边的重量 每一

12、步操作是从尚未放上天平的砝码中挑选一个砝码,将其放到天平的左边或右边,直至全部砝码都被放上天平 求整个操作过程的不同方法个数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

13、. 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. 设 是一个定义在整数集上取值为正

14、整数的函数,已知对任意两个整数, ,差 能被 整除证明：对全部整数, ,如,就 被 整除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 las

15、t case and the only possibility is .But we have implying . Hence we must have , and we are done. 细心整理归纳 精选学习资料 - - - - - - - - - - - - - - - 第 3 页,共 4 页 - - - - - - - - - 名师归纳总结 精品学习资料 - - - - - - - - - - - - - - -6. 设锐角三角形的外接圆为, 是圆的一条切线记切线关于直线,和的对称直线分别为,和证明：由直线,和构成的三角形的外接圆与圆相切of and , respectively;

16、 the Denote the intersections intersections of 3 lines . ; be the reflections of wrt Let be the point of contact of and 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 th

17、en passes through the that midpoint of . , which follows From and we obtain are concyclic. We get . Similarly with So Therefore Construct a tangent of . We will show that is also a tangent of iff . But Hence is true. We are done.细心整理归纳 精选学习资料 - - - - - - - - - - - - - - - 第 4 页,共 4 页 - - - - - - - - -