2009年西部数学奥林匹克试题及解答

首届中国大学生数学竞赛赛区赛试卷解答（非数学类，2009）考试形式： 闭卷 考试时间： 120 分钟 满分： 100 分.注意：1、所有答题都须写在此试卷纸密封线右边，写在其它纸上一律无效。

2、密封线左边请勿答题，密封线外不得有姓名及相关标记。

一、 填空题（每小题5分，共20分）（1）计算=--++⎰⎰dxdy yx x yy x D1)1ln()(_____________，其中区域D 由直线x + y = 1与两坐标轴所围三角形区域。

（2） 设f (x ) 是连续函数， 满足⎰--=2022)(3)(dx x f x x f ，则=)(x f _ __ __。

（3）曲面2222-+=y x z 平行平面 2x + 2y − z = 0 的切平面方程是_ __ _。

（4）设函数y = y (x )由方程29ln )(y y f e xe =确定，其中f 具有二阶导数，且1≠'f ，则22dxyd =_________________。

答案：1516，31032-x ，0522=--+z y x ，322)](1[)()](1[y f x y f y f '-''-'--。

二、（5 分）求极限xenx x x x ne e e )(lim 20+++→ ，其中n 是给定的正整数。

解：原式=)}ln(ex p{lim 20ne e e x e nxx x x +++→ =}ln )ln({lim ex p{20xne e e e nx x x x -+++→ ………………….….…（2 分）其中大括号内的极限是型未定式，由 L ′Hospital 法则，有nxx x x x x x nx x x x e e e ne e e e x n e e e e ++++++=-+++→→ 2020)2(lim }ln )ln({lim e n n n e )21()21(+=+++=于是 原式=e n e )21(+ …….…. . …………………………….………………（5 分)三 、（15 分） 设函数 f (x) 连续, ⎰=1)()(dt xt f x g ，且A xx f x =→)(lim， A 为常数，求)(x g '并讨论)(x g '在x = 0处的连续性。

2009届小学数学奥林匹克竞赛预赛试题及答案2009届小学数学奥林匹克竞赛预赛试题及答案时间：2012-12-06 11:18 来源：世奥赛资讯站作者：世奥赛小编阅读：175次2009年小学数学奥林匹克预赛试卷及参考答案(本卷共12个题，每题10分，总分120分)1、23×( ＋)＋13×( －)－15×( ＋)=( )解：原式=69/11+11+13×15/23-39/11-30/11-15×13/23=112、(1－)(1－)…(1－)=( )解：原式=1/2×2/3×3/4×4/5×……×2007/2008×2008/2009=1/20093、两个整数相除，商数=4，余数=7。

已知被除数比除数大58，那么除数是( )。

解：设除数为x。

则x+58=4x+7 x=174、四位数- =5904，如果是偶数，那么=( 8892 )。

解：8892-2988=59045、右图中的三角形都是等腰直角三角形。

图中阴影部分的面积=( )。

解：5×5÷2÷2-2×2÷2=4.256、下面是一个乘法算式，它的得数是(69104 )。

１２□□×５□□□０４□□７０□□□□□解：1234×56=69014７、一个泉水池，每分钟涌出的泉水量不变。

如果用8台抽水机工作，10小时能把水抽干；如果用12台抽水机工作，6小时能把水抽干。

那么，用14台抽水机把水抽干，需要工作( )小时。

解：设1台抽水机1小时抽的水为1份。

则每小时涌出的泉水量为（8×10-12×6）÷（10-6）=2（份）原有的水量为8×10-10×2=60（份）用14台抽水机把水抽干，需要工作60÷（14-2）=5（小时）。

首届中国大学生数学竞赛赛区赛试卷解答（非数学类，2009）考试形式： 闭卷 考试时间： 120 分钟 满分： 100 分.注意：1、所有答题都须写在此试卷纸密封线右边，写在其它纸上一律无效。

2、密封线左边请勿答题，密封线外不得有姓名及相关标记。

一、 填空题（每小题5分，共20分）（1）计算=--++⎰⎰dxdy yx x y y x D1)1ln()(_____________，其中区域D 由直线x + y = 1与两坐标轴所围三角形区域。

（2） 设f (x ) 是连续函数， 满足⎰--=222)(3)(dx x f x x f ，则=)(x f _ __ __。

（3）曲面2222-+=yxz平行平面 2x + 2y − z = 0 的切平面方程是_ __ _。

（4）设函数y = y (x )由方程29ln )(yy f e xe =确定，其中f 具有二阶导数，且1≠'f ，则22dxy d =_________________。

答案：1516，31032-x，0522=--+z y x，322)](1[)()](1[y f x y f y f '-''-'--。

二、（5 分）求极限xenxxxx ne ee)(lim 20+++→ ，其中n 是给定的正整数。

解：原式=)}ln(exp{lim20ne eexe nxxxx +++→=}ln )ln({lim exp{20xne eee nxxxx -+++→ ………………….….…（2 分）其中大括号内的极限是0型未定式，由 L ′Hospital 法则，有nxxxxxx x nxxxx eeene e e e xne eee ++++++=-+++→→ 2020)2(lim}ln )ln({lime n nn e )21()21(+=+++=于是 原式=en e )21(+ …….…. . …………………………….………………（5 分)三 、（15 分） 设函数 f (x) 连续, ⎰=1)()(dtxt f x g ，且Axx f x =→)(lim， A为常数，求)(x g '并讨论)(x g '在x = 0处的连续性。

2009 AMC 10B1 、Each morning of her five-day workweek, Jane bought either a 50-cent muffin or a 75-cent bagel. Her total cost for the week was a whole number of dollars, How many bagels did she buy?SolutionThe only combination of five items with total cost a whole number ofdollars is 3 muffins and bagels. The answer is .2 、Which of the following is equal to ?SolutionMultiplying the numerator and the denumerator by the same value does not change the value of the fraction. We can multiply both by ,getting .Alternately, we can directly compute that the numerator is , thedenumerator is , and hence their ratio is .3 、Paula the painter had just enough paint for identically sizedrooms. Unfortunately, on the way to work, three cans of paint fell off her truck, so she had only enough paint for rooms. How many cansof paint did she use for the rooms?SolutionLosing three cans of paint corresponds to being able to paint fivefewer rooms. So . The answer is .4 、A rectangular yard contains two flower beds in the shape of congruent isosceles right triangles. The remainder of the yard has a trapezoidal shape, as shown. The parallel sides of the trapezoid have lengths and meters. What fraction of the yard is occupied by theflower beds?SolutionEach triangle has leg length meters and areasquare meters. Thus the flower beds have a total area of 25 square meters. The entire yard has length 25 m and width 5 m, so its area is 125 square meters. The fraction of the yard occupied by theflower beds is . The answer is .5 、Twenty percent less than 60 is one-third more than what number?SolutionTwenty percent less than 60 is . One-third more than anumber n is . Therefore and the number is . The answeris .6 、Kiana has two older twin brothers. The product of their three ages is 128. What is the sum of their three ages?SolutionThe age of each person is a factor of . So the twins could beyears of age and, consequently Kiana could be 128, 32, 8 or 2 years old, respectively. Because Kiana is younger than her brothers, she must be 2 years old. So the sum of their agesis . The answer is .7 、By inserting parentheses, it is possible to give the expressionseveral values. How many different values can beobtained?SolutionThe three operations can be performed on any of orders.However, if the addition is performed either first or last, then multiplying in either order produces the same result. So at most four distinct values can be obtained. It is easy to check that the values ofthe four expressions are in fact all distinct. Sothe answer is , which is choice .8 、In a certain year the price of gasoline rose by during January,fell by during February, rose by during March, and fell byduring April. The price of gasoline at the end of April was the sameas it had been at the beginning of January. To the nearest integer, what isSolutionLet be the price at the beginning of January. The price at the end ofMarch was Because the price at the of April was , the price decreased by during April, and the percent decreasewas So to the nearest integer is . Theanswer is .9 、Segment and intersect at , as shown,, and . What is the degree measureof ?Solutionis isosceles, hence .The sum of internal angles of can now be expressed as, hence , and each of the other twoangles is .Now we know that .Finally, is isosceles, hence each of the two remaining angles(and ) is equal to .10 、A flagpole is originally meters tall. A hurricane snaps theflagpole at a point meters above the ground so that the upper part, still attached to the stump, touches the ground meter away from thebase. What is ?SolutionThe broken flagpole forms a right triangle with legs and , andhypotenuse . The Pythagorean theorem now states that, hence , and .(Note that the resulting triangle is the well-known righttriangle, scaled by .)11 、How many -digit palindromes (numbers that read the samebackward as forward) can be formed using the digits , , , , , ,?SolutionA seven-digit palindrome is a number of the form . Clearly,must be , as we have an odd number of fives. We are then left with. Each of the permutations of the set will give us one palindrome.12、Distinct points , , , and lie on a line, with. Points and lie on a second line, parallel to thefirst, with . A triangle with positive area has three of the sixpoints as its vertices. How many possible values are there for the area of the triangle?SolutionConsider the classical formula for triangle area: . Each of the triangles that we can make has exactly one side lying on one of the two parallel lines. If we pick this side to be the base, the height will always be the same - it will be the distance between the two lines.Hence each area is uniquely determined by the length of the base. And it can easily be seen, that the only possible base lengths are , ,and . Therefore there are only possible values for the area.(To be more precise in the last step, the possible base lengths are, , and .)13 、As shown below, convex pentagon has sides ,, , , and . The pentagon is originallypositioned in the plane with vertex at the origin and vertex on thepositive -axis. The pentagon is then rolled clockwise to the rightalong the -axis. Which side will touch the point on the-axis?SolutionThe perimeter of the polygon is . Hence as we rollthe polygon to the right, every units the side will be the bottomside.We have . Thus at some point in time we will get thesituation when and is the bottom side. Obviously, atthis moment .After that, the polygon rotates around until point hits the axis at.And finally, the polygon rotates around until point hits the axisat . At this point the side touches the point .14 、On Monday, Millie puts a quart of seeds, of which are millet,into a bird feeder. On each successive day she adds another quart of the same mix of seeds without removing any seeds that are left. Each day the birds eat only of the millet in the feeder, but they eat all ofthe other seeds. On which day, just after Millie has placed the seeds, will the birds find that more than half the seeds in the feeder are millet?SolutionOn Monday, day 1, the birds find quart of millet in the feeder. OnTuesday they find quarts of millet. On Wednesday, day 3,they find quarts of millet. The number of quarts of millet they find on day isThe birds always find quart of other seeds, so more than half theseeds are millet if , that is, when . Becauseand , this will first occur on daywhich is . The answer is .15 、When a bucket is two-thirds full of water, the bucket and water weigh kilograms. When the bucket is one-half full of water the total weight is kilograms. In terms of and , what is the total weight inkilograms when the bucket is full of water?SolutionSolution 1Let be the weight of the bucket and let be the weight of the waterin a full bucket. Then we are given that and .Hence , so . Thus .Finally . The answer is .Solution 2Imagine that we take three buckets of the first type, to get rid of the fraction. We will have three buckets and two buckets' worth of water. On the other hand, if we take two buckets of the second type, we will have two buckets and enoung water to fill one bucket.The difference between these is exactly one bucket full of water, hence the answer is .Solution 3We are looking for an expression of the form .We must have , as the desired result contains exactly onebucket. Also, we must have , as the desired result contains exactly one bucket of water.At this moment, it is easiest to check that only the options (A), (B), and (E) satisfy , and out of these only (E) satisfies the secondequation.Alternately, we can directly solve the system, getting and.16 、Points and lie on a circle centered at , each of andare tangent to the circle, and is equilateral. The circleintersects at . What is ?SolutionSolution 1As is equilateral, we have , hence. Then , and from symmetry we have. Finally this gives us .We know that , as lies on the circle. From we alsohave , Hence , therefore, and .Solution 2As in the previous solution, we find out that .Hence and are both equilateral.We then have , hence is the incenter of, and as is equilateral, is also its centroid. Hence, and as , we have ,therefore , and as before we conclude that .17、Five unit squares are arranged in the coordinate plane as shown, with the lower left corner at the origin. The slanted line, extendingfrom to , divides the entire region into two regions of equalarea. What is ?SolutionSolution 1For the shaded area is at most , which is too little. Hence, and therefore the point is indeed inside the shaded part,as shown in the picture.Then the area of the shaded part is one less than the area of thetriangle with vertices , , and . Its area is obviously, therefore the area of the shaded part is .The entire figure has area , hence we want the shaded part to havearea . Solving for , we get . The answer is .Solution 2The total area is 5, so the area of the shaded area is . If we add aunit square in the lower right corner, the area is . Therefore, or . Therefore .18 、Rectangle has and . Point is themidpoint of diagonal , and is on with . What is thearea of ?SolutionBy the Pythagorean theorem we have , hence .The triangles and have the same angle at and a rightangle, thus all their angles are equal, and therefore these two triangles are similar.The ratio of their sides is , hence the ratio of their areas is.And as the area of triangle is , the area of triangleis .19 、A particular -hour digital clock displays the hour and minute ofa day. Unfortunately, whenever it is supposed to display a , itmistakenly displays a . For example, when it is 1:16 PM the clockincorrectly shows 9:96 PM. What fraction of the day will the clock show the correct time?SolutionSolution 1The clock will display the incorrect time for the entire hours ofand . So the correct hour is displayed of the time. The minutes willnot display correctly whenever either the tens digit or the ones digit is a , so the minutes that will not display correctly areand and . This amounts to fifteen of the sixty possibleminutes for any given hour. Hence the fraction of the day that theclock shows the correct time is . The answeris .Solution 2The required fraction is the number of correct times divided by the total times. There are 60 minutes in an hour and 12 hours on a clock, so there are 720 total times.We count the correct times directly; let a correct time be , whereis a number from 1 to 12 and and are digits, where . Thereare 8 values of that will display the correct time: 2, 3, 4, 5, 6, 7, 8, and 9. There are five values of that will display the correct time: 0, 2, 3, 4, and 5. There are nine values of that will display the correct time: 0, 2, 3, 4, 5, 6, 7, 8, and 9. Therefore there arecorrect times.Therefore the required fraction is .20 、Triangle has a right angle at , , and . Thebisector of meets at . What is ?Solution、By the Pythagorean Theorem, . The Angle Bisector Theorem now yields that21 、What is the remainder when is divided by 8?SolutionSolution 1The sum of any four consecutive powers of 3 is divisible byand hence is divisible by 8. Thereforeis divisible by 8. So the required remainder is . Theanswer is .Solution 2We have . Hence for any we have, and then . Therefore our sum gives the same remainder modulo as. There are terms in the sum, hencethere are pairs , and thus the sum is.22 、A cubical cake with edge length inches is iced on the sides andthe top. It is cut vertically into three pieces as shown in this top view, where is the midpoint of a top edge. The piece whose top is trianglecontains cubic inches of cake and square inches of icing. What is?SolutionLet's label the points as in the picture above. Let be the area of. Then the volume of the corresponding piece is . This cake piece has icing on the top and on the vertical side thatcontains the edge . Hence the total area with icing is. Thus the answer to our problem is, and all we have to do now is to determine . Solution 1Introduce a coordinate system where , and.In this coordinate system we have , and the line has theequation .As the line is orthogonal to , it must have the equationfor some suitable constant . As this line contains thepoint , we have .Substituting into , we get , and then .We can note that in is the height from onto , hence itsarea is , and therefore the answer is.Solution 2Extend to intersect at :It is now obvious that is the midpoint of . (Imagine rotating thesquare by clockwise around its center. This rotation willmap the segment to a segment that is orthogonal to ,contains and contains the midpoint of .)From we can compute that .Observe that and have the same angles and thereforethey are similar. The ratio of their sides is .Hence we have , and .Knowing this, we can compute the area of as.Finally, we compute , andconclude that the answer is .You could also notice that the two triangles in the original figure are similar.Solution 3Use trigonometry.The length of and is and respectively. So ,and .From the right-angled triangle , the hypotenuse, So, andKnowing this, . So we proceed as follows:So the answer is .Note that we didn't use a calculator, but we used trigonometric identities23 、Rachel and Robert run on a circular track. Rachel runs counterclockwise and completes a lap every 90 seconds, and Robert runs clockwise and completes a lap every 80 seconds. Both start from the same line at the same time. At some random time between 10 minutes and 11 minutes after they begin to run, a photographer standing inside the track takes a picture that shows one-fourth of the track, centered on the starting line. What is the probability that both Rachel and Robert are in the picture?SolutionAfter 10 minutes (600 seconds), Rachel will have completed 6 laps and be 30 seconds from completing her seventh lap. Because Rachel runs one-fourth of a lap in 22.5 seconds, she will be in the picture between 18.75 seconds and 41.25 seconds of the tenth minute. After 10 minutes Robert will have completed 7 laps and will be 40 seconds past the starting line. Because Robert runs one-fourth of a lap in 20 seconds, he will be in the picture between 30 and 50 seconds of the tenth minute. Hence both Rachel and Robert will be in the picture if it is taken between 30 and 41.25 seconds of the tenth minute. So theprobability that both runners are in the picture is .The answer is .24 、The keystone arch is an ancient architectural feature. It is composed of congruent isosceles trapezoids fitted together along the non-parallel sides, as shown. The bottom sides of the two end trapezoids are horizontal. In an arch made with trapezoids, let bethe angle measure in degrees of the larger interior angle of the trapezoid. What is ?SolutionExtend all the legs of the trapezoids. They will all intersect in the middle of the bottom side of the picture, forming the situation shown below.Each of the angles at has . From , the size of thesmaller internal angle of the trapezoid is , hence thesize of the larger one is .Proof that all the extended trapezoid legs intersect in the same point: It is sufficient to prove this for any pair of neighboring trapezoids. For two neighboring trapezoids, the situation is symmetric according to their common leg, therefore the extensions of both outside legs intersect the extension of the common leg in the same point, q.e.d. Knowing this, we can now easily see that the intersection point must be on the bottom side of our picture, as it lies on the bottom leg of the rightmost trapezoid. And by symmetry the point must be in the center of this side.25 、Each face of a cube is given a single narrow stripe painted from the center of one edge to the center of the opposite edge. The choice of the edge pairing is made at random and independently for each face. What is the probability that there is a continuous stripe encircling the cube?SolutionSolution 1There are two possible stripe orientations for each of the six faces ofthe cube, so there are possible stripe combinations. There are three pairs of parallel faces so, if there is an encircling stripe, then the pair of faces that do not contribute uniquely determine the stripeorientation for the remaining faces. In addition, the stripe on each face that does not contribute may be oriented in either of two different ways. So a total of stripe combinations on the cube resultin a continuous stripe around the cube. The required probability is.Here's another way similar to this:So there are choices for the stripes as mentioned above. Now, let'sjust consider the "view point" of one of the faces. We can choose any of the 2 orientation for the stripe (it can go from up to down, or from right to left). Once that orientation is chosen, each of the other faces that contribute to that loop only have 1 choice, which is to go in the direction of the loop. That gives us a total count of 2 possibilities for any one of the faces. Since there are six faces, and this argument is valid for all of them, we conclude that there are 2(6) = 12 total ways to have the stripe. Therefore, the probability is 12/64 = 3/16. Solution 2Without loss of generality, orient the cube so that the stripe on the top face goes from front to back. There are two mutually exclusive ways for there to be an encircling stripe: either the front, bottom and back faces are painted to complete an encircling stripe with the top face's stripe or the front, right, back and left faces are painted to form anencircling stripe. The probability of the first case is , and theprobability of the second case is . The cases are disjoint, sothe probabilities sum .Solution 3There are three possible orientations of an encircling stripe. For any one of these to appear, the stripes on the four faces through which the continuous stripe is to pass must be properly aligned. The probabilityof each such stripe alignment is . Since there are three such- 21 -possibilities and they are disjoint, the total probability is .The answer is. Solution 4 Consider a vertex on the cube and the three faces that are adjacent to that vertex. If no two stripes on those three faces are aligned, then there is no stripe encircling the cube. The probability that the stripes aren't aligned is , since for each alignment of one stripe, there is one and only one way to align the other two stripes out of four total possibilities. therefore there is a probability of that two stripes are aligned.Now consider the opposing vertex and the three sides adjacent to it. Given the two connected stripes next to our first vertex, we have two more that must be connected to make a continuous stripe. There is a probability ofthat they are aligned, so there is a probability ofthat there is a continuous stripe.。

《数学奥林匹克报》Mathematical Olympiad Express2009 第 38 届美国数学奥林匹克USA Mathematical Olympiad2009 年 4 月 28 日美东时间 12:30～17:001,Given circles intersectingω1 and ω2 intersecting at points X and Y ,let21be a line through the center ofω1ω2 at points P and Q and letbe a line through the center ofω2 intersecting ω1 atpoints R and S . Prove that if P , Q , R and S lie on a circle then the center of this circle lies on lineXY .汉译:两给定的圆 ω1 ,ω2 交于点 X ,Y .设过⊙ ω1 的圆心的直线 圆心的直线2 交⊙ 1 交⊙ω2 于 P ,Q 两点;设过⊙ ω2 的ω1 于 R ,S 两点.求证:P ,Q ,R ,S 四点共圆且所共圆的圆心在直线 XY 上.2,Let n be a positive integer. Determine the size of the largest subset of{n, n + 1,, n 1, n}which does not contain three elements a , b , c (not necessarily distinct) satisfying a + b + c = 0 . 汉译:设 n 是正整数,在集合 {n, n + 1,, n 1, n} 的所有子集中,求元素个数最多那个子集有多少个元素,使得该子集中任意三个数之和均不为零(此三数不必不同) . 3, define a chessboard polygon to be a polygon whose edges are situated along lines of the form x = a or Wey = b , where a and b are integers. These lines divide the interior into unit squares, which are shadedalternately grey and white so that adjacent squares have different colors. To tile a chessboard polygon by dominoes is to exactly cover the polygon by non-overlapping 1×2 rectangles. Finally, a tasteful tiling is one which avoids the two configurations of dominoes shown on the left below. Two tilings of a 3×4 rectangle are shown; the first one is tasteful, while the second is not, due to the vertical dominoes in the upper right corner.Distasteful tilings普及数学知识,传播奥林文化,快递竞赛信息.《数学奥林匹克报》Mathematical Olympiad Expressa ) Prove that if a chessboard polygon can be tiled by dominoes, then it can be done so tastefully. b ) Prove that such a tasteful tiling is unique.汉译:定义多边形棋盘是这样一个多边形:它的边由直线 x = a 或 y = b 形成(这里 a , b 是整数)这些线 把此棋盘分割为一些小方格 (小方格的意思是指 1×1 的正方形) 把这些小方格间隔染成黑白两色, , 使得相邻的两个小方格不同色.把这样的棋盘用 1×2 的多米诺骨牌可以不重不漏的"贴瓷砖"完 全覆盖,如果在覆盖后的图中不出现如图左的两种情况则称为是"完美覆盖" (即两块 1×2 的多米 诺骨牌不能按如图方式并排放置) .在两个 3×4 的棋盘的多米诺骨牌的"贴瓷砖"覆盖图中,前图 是"完美覆盖" ,后图不是"完美覆盖" ,因为左上角出现的两个竖直的多米诺骨牌是不完美的.不完美覆盖a ) 求证:如果一个棋盘能被这样的多米诺骨牌"贴瓷砖"覆盖,那么它就能被"完美覆盖" . b ) 求证:这样的"完美覆盖"是唯一的.2009 年 4 月 29 日美东时间 12:30～17:004,For n ≥2 let a1 , a2 ,……, an be positive real numbers such that( a1 + a2 +Prove that max ( a1 , a2 ,1 1 + an ) + + a1 a22 1 1 + ≤n + . an 2, an ) ≤ 4 min ( a1 , a2 ,, an ) .汉译:正整数 n ≥2,令正实数 a1 , a2 ,……, an 满足( a1 + a2 +求证: max ( a1 , a2 ,1 1 + an ) + + a1 a2+2 1 1 ≤n + . an 2, an ) ≤ 4 min ( a1 , a2 ,, an ) .普及数学知识,传播奥林文化,快递竞赛信息.《数学奥林匹克报》Mathematical Olympiad Express 5,Trapezoid ABCD , with AB ‖ CD , is inscribed in circle Rays AG and BG meetω and point G lies inside triangle BCD .ω again at points P and Q , respectively. Let the line through G parallelto AB intersects BD and BC at points R and S , respectively. Prove that quadrilateral PQRS is cyclic if and only if BG bisects ∠ CBD . 点 射线 AG 与 BG 分别交⊙ ω 于点 P ,Q . 汉译: 梯形 ABCD 内接于圆 ω ,AB ‖ CD , G 在△ BCD 内. 过点 G 且平行于 AB 的直线分别交 BD , BC 于点 R , S .求证: P , Q , R , S 四点共圆当且 仅当 BG 平分∠ CBD . 6,Let s1 ,s2 ,s3 ,……be an infinite, nonconstant sequence of rational numbers, meaning it is not the case thats1 = s2 = s3 =……. Suppose that t1 , t2 , t3 ,……is also an infinite, nonconstant sequence of rationalnumbers with the property that rational number r such that( s s )( t t )i j i j i jis an integer for all i and j . Prove that there exists a(s s )rand(t t )i jr are integers for all i and j .汉译:设有理数 s1 , s2 , s3 ,……是无穷非常数数列.设有理数 t1 , t2 , t3 ,……也是无穷非常数数列, 且对所有 i , j 使得 si s j()( t t ) 都是整数.求证:存在有理数 r 使得 ( s s ) r 与 ( t t )i j i j i jr都是整数对所有 i , j 都成立.普及数学知识,传播奥林文化,快递竞赛信息.。

2009年第50届国际数学奥林匹克竞赛试题(中文版)与参考答案2009年第50届IMO解答2009年7月15日1、是一个正整数，是n12,,...,(2)kaaak≥{}1,2,...,n中的不同整数，并且1(1iinaa+.对于所有都成立，证明：1,2,...,1ik=1(1kaa.不能被n整除。

证明1：由于12(1naa.，令1(,)nap=，nqp=也是整数，则npq=，并且1pa，21qa.。

因此，由于2(,)1qa=23(1npqaa=.，故31qa.；同理可得41qa.，。

，因此对于任意都有2i≥1iqa.，特别的有1kqa.，由于1pa，故1(1knpqaa=.(*)。

若结论不成立，则1(1knpqaa=，与(*)相减可得1(knaa.，矛盾。

综上所述，结论成立。

此题平均得分：4.804分2、外接圆的圆心为O，分别在线段上，ABCΔ,PQ,CAAB,,KLM分别是,,BPCQPQ的中点，圆过Γ,,KLM并且与相切。

证明：OPPQOQ=。

*****QP证明：由已知*****QP∠=∠=∠，*****PQ∠=∠=∠，因此APQMKLΔΔ～。

所以*****QMLCP==，故*****Q.=.(*)。

设圆O的半径为R，则由(*)有222ROPROQ.=.，因此OPOQ=。

不难发现OP也是圆Γ与相切的充分条件。

OQ=PQ此题平均得分：3.710分3、是严格递增的正整数数列，并且它的子数列和都是等差数列。

证明：是一个等差数列。

123,,,...SSS123,,,...SSSSSS*****,,,.SSSSSS+++123,,,...SSS问题等价于：:fZZ+→是一个严格递增的函数。

()()nbffn=是一个等差数列，也是一个等差数列。

证明：(()1ncffn=()nafn=也是等差数列。

证明：由于是一个严格递增的整值函数，所以对于任意f,xy均有()()fxfy xy.≥.。

令{}{},nnbc的公差分别为，则有,de()()(1)()(1)(dffnffnfnfn=+.≥+.，将可得()nfn→()()()1()0nndffnffncb≥+.=.，因此对于任意都有kZ+∈()()*****kkdcbcbkde++≥.=.+.故只能有，也即两个等差数列公差相等，故可设de=nncbg.=是一个为常数。

目录2001年西部数学奥林匹克 (2)2002年西部数学奥林匹克 (4)2003年西部数学奥林匹克 (6)2004年西部数学奥林匹克 (7)2005年西部数学奥林匹克 (8)2006年西部数学奥林匹克 (10)2007年西部数学奥林匹克 (12)2008年西部数学奥林匹克 (14)2009年西部数学奥林匹克 (16)2010年西部数学奥林匹克 (18)2011年西部数学奥林匹克 (21)2012年西部数学奥林匹克 (23)2001年西部数学奥林匹克1.设数列{x n}满足x1=12，x n+1=x n+x n2n2.证明：x2001<1001.（李伟固供题）2.设ABCD是面积为2的长方形，P为边CD上的一点，Q为△P AB的内切圆与边AB的切点.乘积PP⋅PP的值随着长方形ABCD及点P 的变化而变化，当PP⋅PP取最小值时，（1）证明：PP≥2PB；（2）求PQ⋅PQ的值.（罗增儒供题）3.设n、m是具有不同奇偶性的正整数，且n>m.求所有的整数x，使得x2n−1x m−1是一个完全平方数.（潘曾彪供题）4.设x、y、z为正实数，且x+y+z≥xyz.求x2+y2+z2xyz的最小值.（冯志刚供题）5.求所有的实数x，使得[x3]=4x+3.这里[y]表示不超过实数y的最大整数.（杨文鹏供题）6.P为⊙O外一点，过P作⊙O的两条切线，切点分别为A、B.设Q为PO与AB的交点，过Q作⊙O的任意一条弦CD.证明：△PAB与△PCD有相同的内心. （刘康宁供题）7.求所有的实数x∈�0,π2�，使得(2−sss2x)sss�x+π4�=1，并证明你的结论.（李胜宏供题）8.我们称P1,P2,⋯,P n为集合A的一个n分划，如果（1）P1∪P2∪⋯∪P n=P；（2）P i∩P j≠Φ,1≤s<j≤s.求最小正整数m，使得对P={1,2,⋯,m}的任意一个14分划P1,P2,⋯,P14，一定存在某个集合P i(1≤s≤14)，在P i中有两个元素a、b满足b<a≤43b. （冷岗松供题）2002年西部数学奥林匹克1.求所有的正整数n，使得s4−4s3+22s2−36s+18是一个完全平方数.2.设O为锐角△ABC的外心，P为△AOB内部一点，P在△ABC的三边BC、CA、AB上的射影分别为D、E、F.求证：以FE、FD为邻边的平行四边形位于△ABC内.3.考虑复平面上的正方形，它的4个顶点所对应的复数恰好是某个整系数一元四次方程x4+px3+qx2+rx+s=0的4个根.求这种正方形面积的最小值.4.设n为正整数，集合P1,P2,⋯,P n+1是集合{1,2,⋯,s}的n+1个非空子集.证明：存在{1,2,⋯,s+1}的两个不交的非空子集{s1,s2,⋯,s k}和{j1,j2,⋯,j m}，使得P i1∪P i2∪⋯∪P i k=P j1∪P j2∪⋯∪P j m.5.在给定的梯形ABCD中，AD∥BC，E是边AB上的动点，O1、O2分别是△AED、△BEC的外心.求证：O1O2的长为一定值.6.设s(s≥2)是给定的正整数，求所有整数组(a1,a2,⋯,a n)满足条件：（1）a1+a2+⋯+a n≥s2；（2）a12+a22++a n2≤s3+1.7.设α、β为方程x2−x−1=0的两个根，令a n=αn−βnα−β,s=1,2,⋯.（1）证明：对任意正整数n，有a n+2=a n+1+a n；（2）求所有正整数a、b，a<b，满足对任意正整数n，有b整除a n−2sa n.8.设S=(a1,a2,⋯,a n)是一个由0，1组成的满足下述条件的最长的数列：数列S中任意两个连续5项不同，即对任意1≤s<j≤s−4，a i,a i+1,a i+2,a i+3,a i+4与a j,a j+1,a j+2,a j+3,a j+4不相同.证明：数列S 最前面的4项与最后面的4项相同.1. 将1,2,3,4,5,6,7,8分别放在正方体的八个顶点上，使得每一个面上的任意三个数之和均不小于10.求每一个面上四个数之和的最小值.2. 设2n 个实数a 1,a 2,⋯,a 2n 满足条件∑(a i+1−a i )2=12n−1i=1.求(a n+1+a n+2+⋯+a 2n )−(a 1+a 2+⋯+a n )的最大值.3. 设n 为给定的正整数.求最小的正整数u n ，满足：对每一个正整数d ，任意u n 个连续的正奇数中能被d 整除的数的个数不少于奇数1,3,5,⋯,2s −1中能被d 整除的数的个数.4. 证明：若凸四边形ABCD 内任意一点P 到边AB 、BC 、CD 、DA 的距离之和为定值，则ABCD 是平行四边形.5. 已知数列{a n }满足：a 0=0，a n+1=ka n +�(k 2−1)a n 2+1,s =0,1,2,⋯,其中k 为给定的正整数.证明：数列{a n }的每一项都是整数，且2k |a 2n ,s =0,1,2,⋯. 6. 凸四边形ABCD 有内切圆，该内切圆切边AB 、BC 、CD 、DA 的切点分别为A 1、B 1、C 1、D 1，连结A 1B 1、B 1C 1、C 1D 1、D 1A 1，点E 、F 、G 、H 分别为A 1B 1、B 1C 1、C 1D 1、D 1A 1的中点.证明：四边形EFGH 为矩形的充分必要条件是A 、B 、C 、D 四点共圆.7. 设非负实数x 1、x 2、x 3、x 4、x 5满足∑11+x i =15i=1.求证：∑x i4+x i 25i=1≤1. 8. 1650个学生排成22行、75列.已知其中任意两列处于同一行的两个人中，性别相同的学生都不超过11对.证明：男生的人数不超过928.1.求所有的整数n，使得s4+6s3+11s2+3s+31是完全平方数.2.四边形ABCD为一凸四边形，I1、I2分别为△ABC、△DBC的内心，过点I1、I2的直线分别交AB、DC于点E、F，分别延长AB、DC，它们相交于点P，且PE=PF.求证：A、B、C、D四点共圆.3.求所有的实数k，使得不等式a3+b3+c3+d3+1≥k(a+b+c+d)对任意a、b、c、d∈[−1,+∞)都成立.4.设s∈N+，用d(s)表示n的所有正约数的个数，ϕ(s)表示1,2,⋯,s 中与n互质的数的个数.求所有的非负整数c，使得存在正整数n，满足d(s)+ϕ(s)=s+c，且对这样的每一个c，求出所有满足上式的正整数n.5.设数列{a n}满足a1=a2=1，且a n+2=1a n+1+a n,s=1,2,⋯.求a2004.6.将m×s棋盘（由m行n列方格构成，m≥3,s≥3）的所有小方格都染上红蓝两色之一.如果2个相邻（有公共变）的小方格异色，则称这2个小方格为1个“标准对”.设期盼中“标准对”的个数为S.试问：S是奇数还是偶数有哪些方格的颜色确定？什么情况下S为奇数？什么情况下S为偶数？说明理由.7.已知锐角△ABC的三边长不全相等，周长为l，P是其内部一动点，点P在边BC、CA、AB上的射影分别为D、E、F.求证：2（PB+PD+ BB）=l的充分必要条件是：点P在△ABC的内心与外心的连线上.8.求证：对任意正实数a、b、c，都有1<a√a2+b2+b√b2+c2+c√c2+a2≤3√22.1. 已知α2005+β2005可表示成以α+β、αβ为变元的二元多项式.求这个多项式的系数之和.2. 如图1，过圆外一点P 作圆的两条切线P A 、PB ，A 、B 为切点，再过点P 作圆的一条割线分别与圆交于C 、D 两点，过切点B 作P A 的平行线分别交直线AC 、AD 于E 、F .求证：PB =PB .图13. 设S ={1,2,⋯,2005}.若S 中任意n 个两两互质的数组成的集合中都至少有一个质数，试求n 的最小值.4. 已知实数x 1,x 2,⋯,x n (s >2)满足|∑x i n i=1|>1，|x i |≤1(s =1,2,⋯,s ).求证：存在正整数k ，使得�∑x i k i=1−∑x i n i=k+1�≤1 5. 如图2，⊙O 1、⊙O 2交于A 、B 两点.过点O 1的直线DC 交⊙O 1于点D 且切⊙O 2于点C ，CA 且⊙O 1于点A ，⊙O 1的弦AE 与直线DC 垂直.过点A 作AF 垂直于DE ，F 为垂足.求证：BD 平分线段AF .图2P6.在等腰Rt△ABC中，BP=BP=1，P是△ABC边界上任意一点.求PP⋅PP⋅PB的最大值.7.设正实数a、b、c满足a+b+c=1.证明：10(a3+b3+c3)−9(a5+b5+c5)≥1.8.设n个新生汇总，任意3个人中有2个人互相认识，任意4个人中有2个人互不任何.试求n的最大值.2006年西部数学奥林匹克1. 设s (s ≥2)是给定的正整数，a 1,a 2,⋯,a n ∈(0,1).求∑�a i (1−a i+1)6n i=1的最大值，这里a n+1=a 1. 2. 求满足下述条件的最小正实数k ：对任意不小于k 的4个互不相同的实数a 、b 、c 、d ，都存在a 、b 、c 、d 的一个排列p 、q 、r 、s ，使得方程(x 2+px +q )(x 2+rx +s )=0有4个互不相同的实数根. 3. 如图1，在△ABC 中，∠PPB =60°，过点P 作△PBC 的外接圆⊙O 的切线，与CA 的延长线交于点A .点D 、E 分别在线段PA 和⊙O 上，使得∠DPB =90°，PD =PE .连结BE 与PC 相交于点F .已知AF 、BP 、CD 三线共点.（1） 求证：BF 是∠PPB 的角平分线；（2） 求tas ∠PBP 的值.图14. 设正整数a 不是完全平方数.求证：对每一个正整数n ，S n =�√a�+�√a�2+⋯+�√a�n的值都是无理数.这里{x }=x −[x ]，其中，[x ]表示不超过x 的最大整数.5. 设S =�s�s −1,s ,s +1都可以表示为两个正整数的平方和�.证明：若s ∈S ，则s 2∈S .C6. 如图2，AB 是⊙O 的直径，C 为AB 延长线上的一点，过点C 作⊙O 的割线，与⊙O 交于点D 、E ，OF 是△BOD 的外接圆⊙O 1的直径，连结CF 并延长交⊙O 1于点G .求证：O 、A 、E 、G 四点共圆.图27. 设k 是一个不小于3的正整数，θ是一个实数.证明：如果cms (k −1)θ和cms kθ都是有理数，那么，存在正整数s (s >k )，使得cms (s −1)θ和cms sθ都是有理数. 8. 给定正整数s (s ≥2)，求|X |的最小值，使得对集合X 的任意n 个二元子集P 1,P 2,⋯,P n ，都存在集合X 的一个子集Y ，满足：(1)|Y |=s ;(2) 对s =1,2,⋯,s ，都有|Y ∩P i |≤1.这里，|P |表示有限集合A 的元素个数.A2007年西部数学奥林匹克1. 已知T ={1,2,⋯,8}.对于P ⊆T ,P ≠Φ，定义S (P )为A 中所有元素之和.问：T 有多少个非空子集A ，使得S (P )是3的倍数，但不是5的倍数？2. 如图1，⊙O 1、⊙O 2交于点C 、D ，过D 的一条直线分别与⊙O 1、⊙O 2交于点A 、B ，点P 在⊙O 1的AD 弧上，PD 与线段AC 的延长线交于点M ，点Q 在⊙O 2的BD 弧上，QD 与线段BC 的延长线交于点N ，O 是△ABC 的外心.求证：OD ⊥MN 的充要条件为P 、Q 、M 、N 四点共圆.图13. 设实数a 、b 、c 满足a +b +c =3.求证：15a −4a+11+15b −4b+11+15c −4c+11≤14. 4. 设O 是△ABC 内部一点.证明：存在正整数p 、q 、r ，使得|pOP +qOP +rOB |<12007.5. 是否存在三边长都为整数的三角形，满足以下条件：最短边长为2007，且最大的角等于最小角的两倍？O6.求所有的正整数n，使得存在非零整数x1,x2,⋯,x n,y,满足�x1+x2+⋯+x n=0,x12+x22+⋯+x n2=sy2.7.设P是锐角△ABC内一点，AP、BP、CP分别与边BC、CA、AB 交于点D、E、F，已知△DBB∼△PPB.求证：P是△ABC的重心. 8.将n枚白子与n枚黑子任意地放在一个圆周上.从某枚白子起，按顺时针方向依次将白子标以1,2,⋯,s.在从某枚黑子起，按逆时针方向依次将黑子标以1,2,⋯,s.证明：存在连续n枚棋子（不计黑白），它们的标号组成的集合为{1,2,⋯,s}.2008年西部数学奥林匹克1.实数数列{a n}满足a0≠0,1,a1=1−a0，a n+1=1−a n(1−a n)(s=1,2,⋯).证明：对任意的正整数n，都有a0a1⋯a n�1a0+1a1+⋯+1a n�=1.2.如图1，在△ABC中，AB=AC，其内切圆⊙I分别切边BC、CA、AB于点D、E、F，P为弧EF（不含点D的弧）上一点.设线段BP交⊙I于另一点Q，直线EP、EQ分别交BC于点M、N.证明：(1)P、F、B、M四点共圆；(2)EE EE=BB BB.图13.设整数m(m≥2)，a1,a2,⋯,a m都是正整数.证明：存在无穷多个正整数n，使得数a1×1n+a2×2n+⋯+a m×m n都是合数.4.设整数m(m≥2)，a为正实数，b为非零实数，数列{x n}定义如下：x1=b,x n+1=ax n m+b(s=1,2,⋯).证明：(1)当b<0且m为偶数时，数列{x n}有界的充要条件是ab m−1≥−2；(2)当b<0且m为奇数，或b>0时，数列{x n}有界的充要条件是ab m−1≤(m−1)m−1m m.5.在一直线上相邻的距离都等于1的四个点上各有一只青蛙，允许任意一只青蛙以其余三只青蛙中的某一只为中心跳到其对称点上.证明：无论跳动多少次后，四只青蛙所在的点中相邻两点之间的距离不能都等于2008.6.设x、y、z∈(0,1)，满足�1−x yz+�1−y zx+�1−z xy=2.求xyz的最大值.7.设n为给定的正整数.求最大的正整数k，使得存在三个由非负整数组成的k元集P={x1,x2,⋯,x k}，P={y1,y2,⋯,y k}，B= {z1,z2,⋯,z k}满足对任意的j(1≤j≤k)，都有x j+y j+z j=s.8.设P为正n边形P1P2⋯P n内的任意一点，直线P i P(s=1,2,⋯s)交正n边形P1P2⋯P n的边界于另一点P i.证明：∑PP i n i=1≥∑PP i n i=1.2009年西部数学奥林匹克1.设M是一个由实数集R去掉有限个元素后得到的集合.证明：对任意正整数n，都存在n次多项式f(x)，使得f(x)的所有系数及n个实根都属于M.2.给定整数s≥3.求最小的正整数k，使得存在一个k元集合A和n 个两两不同的实数x1,x2,⋯,x n，满足x1+x2,x2+x3,⋯,x n−1+x n,x n+x1均属于A.3.设H为锐角△ABC的垂心，D为边BC的中点.过点H的直线分别交边AB、AC于点F、E，使得AE=AF，射线DH与△ABC的外接圆交于点P.求证：P、A、E、F四点共圆.4.求证：对任意给定的正整数k，总存在无穷多个正整数n，使得2n+3n−1,2n+3n−2,⋯,2n+3n−k均为合数.5.设数列{x n}满足x1∈{5,7}及当k≥1时，有x k+1∈{5x k,7x k}.试确定x2009的末两位数字的所有可能值.6.如图1，设D是锐角△ABC的边BC上一点，以线段BD为直径的圆分别交直线AB、AD于点X、P（异于点B、D），以线段CD为直径的元分别交直线AC、AD于点Y、Q（异于点C、D）.过点A作直线PX、QY的垂线，垂足分别为M、N.求证△PMN∼△PPB的充分必要条件是直线AD过△ABC的外心.图17. 有s (s >12)个人参加某次数学邀请赛，试卷由十五道填空题组成，每答对一题得1分，不答或答错得0分.分析每一种可能的得分情况发现：只要其中任意12个人得分之和不少于36分，则这n 个人中至少有3个人答对了至少三道同样的题.求n 的最小可能值.8. 实数a 1,a 2,⋯,a n (s ≥3)满足a 1+a 2+⋯+a n =0，且2a k ≤a k−1+a k+1(k =2,3,⋯,s −1).求最小的λ(s )，使得对所有的k ∈{1,2,⋯s }，都有|a k |≤λ(s )⋅max {|a 1|,|a n |}.B2010年西部数学奥林匹克1. 设m 、k 为给定的非负整数，p =22m +1为质数.求证： (1) 22m+1p k ≡1(mmd p k+1);(2) 满足同余方程2n ≡1(mmdp k+1) 的最小正整数n 为2m+1p k . （靳 平 供题）2. 如图1，已知AB 是⊙O 的直径，C 、D 是圆周上异于点A 、B 且在AB 同侧的两点，分别过点C 、D 作圆的切线，它们交于点E ，线段AD 与BC 的交点为F ，直线EF 与AB 交于点M .求证：E 、C 、M 、D 四点共圆.图1（刘诗雄 供题）3. 求所有的正整数n ，使得集合{1,2,⋯,s }有n 个两两不同的三元子集P 1,P 2,⋯,P n ，满足对任意的k (1≤s <j ≤s )，都有�P i ∩P j �≠1.（冯志刚 供题）4. 设非负实数a 1,a 2,⋯,a n 与b 1,b 2,⋯,b n 满足以下条件： （1） ∑a i +b i n i=1=1; （2） ∑s (a i −b i )n i=1=0; （3） ∑s 2(a i +b i )n i=1=10.求证：对任意的k(1≤k≤s)，都有max{a k,b k}≤1010+k2. （李胜宏供题）5.设k为大于1的整数，数列{a n}定义如下：a0=0,a1=1,a n+1=ka n+a n−1(s=1,2,⋯).求所以满足如下条件的k：存在非负整数l、m(l≠m)，及正整数p、q，使得a l+ka p=a m+ka q. （熊斌供题）6.如图2，在△ABC中，∠PBP=90°，以B为圆心、BC为半径作圆，点D在边AC上，直线DE切⊙B于点E，过点C垂直于AB的直线于直线BE交于点F，AF与DE交于点G，作AH∥BG于DE交于点H.求证GE=GH.图2（边红平供题）7.有s(s≥3)名选手参加乒乓球比赛，每两名选手之间恰比赛一场且没有平局.若选手A的手下败将不都是B的手下败将，则称A不亚于B.试求所有可能的n，使得存在一种比赛结果，其中每一名选手都不亚于其他任何一名选手.（李秋生供题）8.求所有的整数k，使得存在正整数a和b，满足b+1a+a+1b=k.（陈永高供题）2011年西部数学奥林匹克1. 已知0<x 、y <1.求xy (1−x−y )(x+y )(1−x )(1−y )的最大值.2. 设集合满足：M ⊆{1,2,⋯,2011}在M 的任意三个元素中都可以找到两个元素a 、b ，使得a |b 或b |a .求|M |的最大值（|M |表示集合M 的元素个数）.3. 给定整数s ≥2.(1) 证明：可以将集合{1,2,⋯,s }的左右子集适当地排列为P 1,P 2,⋯,P 2n ，使得P i 与P i+1(s =1,2,,2n ,且P 2n +1=P 1)的元素个数恰相差1.(2) 对于满足(1)中条件的子集P 1,P 2,⋯,P 2n ，求∑(−1)i S (P i )2n i=1的所以可能值，其中，S (P i )=∑x x∈A i ,S (∅)=0. 4. 如图1，AB 、CD 是⊙O 中长度不相等的两条弦，AB 与CD 交于点E ，⊙I 内切⊙O 于点F ，且分别与弦AB 、CD 切于点G 、H .过点O 的直线l 分别于AB 、CD 交于点P 、Q ，使得EP =EQ ，直线EF 于直线l 交于点M .证明：过点M 且与AB 平行的直线是⊙O 的切线.图15. 是否存在奇数s (s ≥3)及n 个互不相同的质数p 1,p 2,⋯,p n ，使得p i +p i+1(s =1,2,⋯,s ,p n+1=p 1)都是完全平方数？请证明你的结论.6.设a、b、c>0.证明：(a−b)2(c+a)(c+b)+(b−c)2(a+b)(a+c)+(c−a)2(b+c)(b+a)≥(a−b)2a+b+c.7.在△ABC中，PP>PB内切圆⊙I与边BC、CA、AB分别切于点D、E、F，M是边BC的中点，PH⊥PB于点H，∠PPB的平分线AI分别与直线DE、DF交于点K、L.证明：M、L、H、K四点共圆. 8.求所有的整数对(a,b)，使得对任意的正整数n都有s|(a n+b n+1).2012年西部数学奥林匹克1.求最小的正整数m，使得对任意大于3的质数p，都有：105|9p2−29p+m.2.证明：在正2s−1边形(s≥3)的顶点中，任意取出s个点，其中必有3个点，以它们为顶点的三角形为等腰三角形。

第五届北方数学奥林匹克邀请赛试卷第 一 天(2009年7月30日 8:40 —11:40)一、(25分)设数列{}n x 满足111,(2)−==+≥n n x x x n .求数列}{n x 的通项公式.二、(25分)如图，在锐角△ABC 中，AC AB >，1cos cos =+C B .F E 、分别是AB 、AC 延长线上的点，且 90=∠=∠ACE ABF .（1）求证：EF CF BE =+；（2）设EBC ∠的平分线与EF 交于点P ，求证：CP 平分BCF ∠.三、(25分)已知有26个互不相等的正整数，其中任意六个数中都至少有两个数，一个数整除另一个数.证明：一定存在六个数，其中一个数能被另外五个数整除.四、(25分)船长和三位水手共得到2 009枚面值相同的金币，四人商定按照如下规则对金币进行分配：水手1、水手2、水手3每人写下一个正整数分别为1b 、2b 、3b ，满足123≥≥b b b ，且123++=b b b 2 009；船长在不知道水手写的数的情况下，将2 009枚金币分成3堆,各堆数量分别1a 、2a 、3a ，且123≥≥a a a .对于水手k (1,2,3=k ),当<k k b a 时，可以从第k 堆拿走k b 枚金币，否则不能拿.最后所有余下的金币归船长所有.若无论三位水手怎样写数，船长总可以确保自己拿到n 枚金币.试确定n 的最大值，并证明你的结论.F第 二 天(2009年7月31日 8:40 —11:40)五、（25分）如图，C 为扇形AOB 的弧AB 上一点，在射线OC 上任取一点P ,连结AP ，过点B 作直线BQ ∥AP 交OC 于点Q .证明：五边形OAQPB 的面积与点C 、P 的选取无关。

六、（25分）若0>、、x y z 且3222=++z y x ，求证：++−−z y x x )1(20082009x z y y +−−)1(20082009)(21)1(20082009z y x y x z z ++≥+−−+.七、（25分）记[]m 为不超过实数m 的最大整数.设x 、y 均为正实数，且对所有的正整数n ，都有[]1 =−x ny n 成立.证明：1,=xy 且y 是大于1的无理数.八、（25分）求能被209整除且各位数字之和等于209的最小正整数.。