2014-2012加拿大滑铁卢大学11年级数学竞赛试题

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国际数学竞赛试题及答案

国际数学竞赛试题及答案一、选择题（每题3分，共15分）1. 下列哪个数是最小的正整数？A. 0B. 1C. 2D. 3答案：B2. 一个圆的半径为5，它的面积是多少？A. 25πB. 50πC. 100πD. 25答案：B3. 如果一个数的平方等于它本身，这个数可能是：A. 1B. -1C. 2D. 3答案：A B4. 一个数列的前三项是1, 1, 2，如果这个数列是等差数列，那么第四项是：A. 3B. 4C. 5D. 6答案：A5. 以下哪个表达式是正确的？A. sin²θ + cos²θ = 1B. sinθ + cosθ = 1C. sinθ * cosθ = 1D. tanθ = sinθ / cosθ答案：A D二、填空题（每题4分，共20分）6. 如果一个三角形的两边长分别是3和4，且这两边夹角为60度，那么第三边的长度是________。

答案：√137. 一个函数f(x) = 2x³ - 3x² + x - 5，求导后得到的导函数是________。

答案：6x² - 6x + 18. 一个数的立方根等于它本身，这个数可以是________。

答案：1, -1, 09. 一个正六边形的内角和是________。

答案：720°10. 如果一个等差数列的首项是2，公差是3，那么第10项是________。

答案：37三、解答题（每题10分，共30分）11. 证明：对于任意实数x，等式e^x ≥ x + 1成立。

证明：考虑函数f(x) = e^x - (x + 1)。

对f(x)求导得到f'(x) = e^x - 1。

当x > 0时，f'(x) > 0，说明f(x)在此区间单调递增；当x < 0时，f'(x) < 0，说明f(x)在此区间单调递减。

因此，f(x)的最小值出现在x = 0处，此时f(0) = e^0 - (0 + 1) = 1 - 1 = 0。

2013滑铁卢竞赛试题答案

The CENTRE for EDUCATION in MATHEMATICS and COMPUTING cemc.uwaterloo.ca
2013 Euclid Contest
Wednesday, April 17, 2013
(in North America and South America)
since AC > 0. Finally, the perimeter of
3. (a) The parity of an integer is whether it is even or odd. Since the Fibonacci sequence begins 1, 1, 2, 3, 5, 8, 13, 21, . . ., then the parities of the ﬁrst eight terms are Odd, Odd, Even, Odd, Odd, Even, Odd, Odd. In the sequence, if x and y are consecutive terms, then the next term is x + y . In general, suppose that x and y are integers. If x is even and y is even, then x + y is even. If x is even and y is odd, then x + y is odd. If x is odd and y is even, then x + y is odd. If x is odd and y is odd, then x + y is even. Therefore, the parities of two consecutive terms x and y in the Fibonacci sequence determine the parity of the following term x + y . Also, once there are two consecutive terms whose parities match the parities of two earlier consecutive terms in the sequence, then the parities will repeat in a cycle. In particular, the parities of the fourth and ﬁfth terms (Odd, Odd) are the same as the parities of the ﬁrst and second terms (Odd, Odd). Therefore, the parities in the sequence repeat the cycle Odd, Odd, Even. This cycle has length 3. Therefore, the 99th term in the Fibonacci sequence ends one of these cycles, since 99 is a multiple of 3. In particular, the 99th term ends the 33rd cycle. Each cycle contains two odd terms. Therefore, the ﬁrst 99 terms in the sequence include 2 × 33 = 66 odd terms. Finally, the 100th term in the sequence begins a new cycle, so is odd. Therefore, the ﬁrst 100 terms include 66 + 1 = 67 odd terms. (b) Suppose that the ﬁrst term in the given sequence is a and the common diﬀerence is d. Then the ﬁrst four terms are a, a + d, a + 2d, a + 3d. From the given information, a + (a + 2d) = 6 and (a + d) + (a + 3d) = 20. The ﬁrst equation simpliﬁes to 2a + 2d = 6 or a + d = 3. The second equation simpliﬁes to 2a + 4d = 20 or a + 2d = 10. Therefore, (a + 2d) − (a + d) = 10 − 3 or d = 7. Since a + d = 3 and d = 7, then a = −4. Therefore, the tenth term in the sequence is a + 9d = −4 + 9(7) = 59. 4. (a) There are ﬁve odd digits: 1, 3, 5, 7, 9. We consider the positive integers less than 1000 in three sets: those with one digit, those with two digits, and those with three digits. There are 5 positive one-digit integers with one odd digit (namely 1, 3, 5, 7, 9). Consider the two-digit positive integers with only odd digits. Such an integer has the form XY where X and Y are digits. There are ﬁve possibilities for each of X and Y (since each must be odd). Therefore, there are 5 × 5 = 25 two-digit positive integers with only odd digits. Consider the three-digit positive integers with only odd digits. Such an integer has the form XY Z where X , Y and Z are digits.

滑铁卢大学欧几里得数学竞赛

该考试是学生申请滑铁卢大学数学学院本科专业的重要参考。

众所周知滑铁卢大学数学学院是全球最大的数学、统计学、计算机科学等学科教学中心比尔•盖茨曾于 2005 年、 2008 年两度造访该大学是比尔•盖茨大学巡回讲座的北美5 所大学之一也是唯一的一所加拿大大学。

考试范围：大部分的题目基于高三或者12年级数学课学习的内容。

我们的竞赛题目主要包括以下的数学内容：Ø 欧几里德几何和解析几何Ø 三角函数，包括函数、图像、性质、正弦余弦定理Ø 指数和对数函数Ø 函数符号Ø 方程组Ø 多项式，包括二次三次方程根的关系、余数定理Ø 数列、数列求和Ø 简单的计算问题Ø 数字的性质考试时间为 2.5 个小时， 10 道题。

每题 10 分，共计 100 分。

考试题有两种，一种只需要给出答案，另一种则需要写出整个解题过程，这种题的最终得分不仅取决于结果正确与否，还与解题思路有关。

Ø 笔试Ø 10道题：大部分要求写出完整的解题步骤；Ø 根据解题的方法和步骤获得相应的分数；Ø 步骤不完整的解题无法得到全部的分数；Ø 竞赛时长为2.5小时；Ø 共100分；Ø 可以使用无编程无绘图功能的计算器；Ø 不可以使用任何可接入互联网的设备，如手机、平板电脑等均不能携带如何准备：Ø CEMC官网可以免费下载历年的竞赛原题以及标准答案；Ø CEMC官网提供各种免费的数学资源；Ø www.cemc.uwaterloo.ca；如何参加：Ø 学校可以申请注册为考点，安排组织欧几里德数学竞赛；Ø 学生需要通过自己所在的学校报名参加欧几里德数学竞赛；Ø 如果学生所在学校未注册考点，学生可以报名在我们北京或者上海的考点参加欧几里德数学竞赛；Ø 竞赛结束之后，学校需要将全部的试卷寄回滑铁卢大学；Ø 改卷结束之后，滑铁卢大学会在CEMC官网录入学生的成绩。

Fryer滑铁卢数学竞赛(Grade 9)-数学Mathematics-2010-试题 exam

2010Fryer Contest(Grade9)Friday,April9,20101.Consider the following sequence ofﬁgures showing arrangements of square tiles:Figure 1Figure 2Figure 3Figure 4Moreﬁgures can be drawn,each having one row of tiles more than the previousﬁgure.This new bottom row is constructed using two tiles more than the number of tiles in the bottom row of the previousﬁgure.(a)Figure4is cut into two pieces as shown.Draw arearrangement of these two pieces showing how they canbe formed into a square having42=16tiles.(b)Determine the number of tiles in Figure5.(c)Determine the number of tiles in the bottom row of Figure10.(d)Determine the diﬀerence between the total number of tiles in Figure11and the totalnumber of tiles in Figure9.2.(a)Determine the average of the integers71,72,73,74,75.(b)Suppose that n,n+1,n+2,n+3,n+4areﬁve consecutive integers.(i)Determine a simpliﬁed expression for the sum of theseﬁve consecutive integers.(ii)If the average of theseﬁve consecutive integers is an odd integer,explain why n must be an odd integer.(c)Six consecutive integers can be represented by n,n+1,n+2,n+3,n+4,n+5,wheren is an integer.Explain why the average of six consecutive integers is never an integer.2010Fryer Contest Page23.Train1is travelling from Amville to Battonat a constant speed.Train2is travelling from Batton to Amville at a constantspeed.Train 1Train 2(a)Train1travels at60km/h and travels23of the distance to Batton in9hours.Determine the distance from Amville to Batton.(b)Train2travels23of the distance to Amville in6hours.How fast is the train going?(c)Train2started its trip312hours after Train1started its trip.Both trains arrived atCuford at9:00p.m.What time did Train1leave Amville?4.A palindrome is a positive integer that is the same when read forwards or backwards.Forexample,three palindromes are7,121and7739377.(a)Determine the number of palindromes less than1000.(b)Determine the number of palindromes with7digits.(c)If the palindromes in part(b)are written in increasing order,determine the2125thpalindrome in the list.(d)Determine the number of six-digit palindromes that are divisible by91.。

2011滑铁卢竞赛试题答案

1.(a)Since (x +1)+(x +2)+(x +3)=8+9+10,then 3x +6=27or 3x =21and so x =7.(b)Since 25+√x =6,then squaring both sides gives 25+√x =36or √x =11.Since √x =11,then squaring both sides again,we obtain x =112=121.Checking, 25+√121=√25+11=√36=6,as required.(c)Since (a,2)is the point of intersection of the lines with equations y =2x −4and y =x +k ,then the coordinates of this point must satisfy both equations.Using the ﬁrst equation,2=2a −4or 2a =6or a =3.Since the coordinates of the point (3,2)satisfy the equation y =x +k ,then 2=3+k or k =−1.2.(a)Since the side length of the original square is 3and an equilateral triangle of side length 1is removed from the middle of each side,then each of the two remaining pieces of each side of the square has length 1.Also,each of the two sides of each of the equilateral triangles that are shown has length 1.1111Therefore,each of the 16line segments in the ﬁgure has length 1,and so the perimeter of the ﬁgure is 16.(b)Since DC =DB ,then CDB is isosceles and ∠DBC =∠DCB =15◦.Thus,∠CDB =180◦−∠DBC −∠DCB =150◦.Since the angles around a point add to 360◦,then∠ADC =360◦−∠ADB −∠CDB =360◦−130◦−150◦=80◦.(c)By the Pythagorean Theorem in EAD ,we have EA 2+AD 2=ED 2or 122+AD 2=132,and so AD =√169−144=5,since AD >0.By the Pythagorean Theorem in ACD ,we have AC 2+CD 2=AD 2or AC 2+42=52,and so AC =√25−16=3,since AC >0.(We could also have determined the lengths of AD and AC by recognizing 3-4-5and 5-12-13right-angled triangles.)By the Pythagorean Theorem in ABC ,we have AB 2+BC 2=AC 2or AB 2+22=32,and so AB =√9−4=√5,since AB >0.3.(a)Solution 1Since we want to make 15−y x as large as possible,then we want to subtract as little as possible from 15.In other words,we want to make y x as small as possible.To make a fraction with positive numerator and denominator as small as possible,wemake the numerator as small as possible and the denominator as large as possible.Since 2≤x ≤5and 10≤y ≤20,then we make x =5and y =10.Therefore,the maximum value of 15−y x is 15−105=13.Solution2Since y is positive and2≤x≤5,then15−yx≤15−y5for any x with2≤x≤5andpositive y.Since10≤y≤20,then15−y5≤15−105for any y with10≤y≤20.Therefore,for any x and y in these ranges,15−yx≤15−105=13,and so the maximumpossible value is13(which occurs when x=5and y=10).(b)Solution1First,we add the two given equations to obtain(f(x)+g(x))+(f(x)−g(x))=(3x+5)+(5x+7)or2f(x)=8x+12which gives f(x)=4x+6.Since f(x)+g(x)=3x+5,then g(x)=3x+5−f(x)=3x+5−(4x+6)=−x−1.(We could alsoﬁnd g(x)by subtracting the two given equations or by using the second of the given equations.)Since f(x)=4x+6,then f(2)=14.Since g(x)=−x−1,then g(2)=−3.Therefore,2f(2)g(2)=2×14×(−3)=−84.Solution2Since the two given equations are true for all values of x,then we can substitute x=2to obtainf(2)+g(2)=11f(2)−g(2)=17Next,we add these two equations to obtain2f(2)=28or f(2)=14.Since f(2)+g(2)=11,then g(2)=11−f(2)=11−14=−3.(We could alsoﬁnd g(2)by subtracting the two equations above or by using the second of these equations.)Therefore,2f(2)g(2)=2×14×(−3)=−84.4.(a)We consider choosing the three numbers all at once.We list the possible sets of three numbers that can be chosen:{1,2,3}{1,2,4}{1,2,5}{1,3,4}{1,3,5}{1,4,5}{2,3,4}{2,3,5}{2,4,5}{3,4,5} We have listed each in increasing order because once the numbers are chosen,we arrange them in increasing order.There are10sets of three numbers that can be chosen.Of these10,the4sequences1,2,3and1,3,5and2,3,4and3,4,5are arithmetic sequences.Therefore,the probability that the resulting sequence is an arithmetic sequence is410or25.(b)Solution 1Join B to D .AConsider CBD .Since CB =CD ,then ∠CBD =∠CDB =12(180◦−∠BCD )=12(180◦−60◦)=60◦.Therefore, BCD is equilateral,and so BD =BC =CD =6.Consider DBA .Note that ∠DBA =90◦−∠CBD =90◦−60◦=30◦.Since BD =BA =6,then ∠BDA =∠BAD =12(180◦−∠DBA )=12(180◦−30◦)=75◦.We calculate the length of AD .Method 1By the Sine Law in DBA ,we have AD sin(∠DBA )=BA sin(∠BDA ).Therefore,AD =6sin(30◦)sin(75◦)=6×12sin(75◦)=3sin(75◦).Method 2If we drop a perpendicular from B to P on AD ,then P is the midpoint of AD since BDA is isosceles.Thus,AD =2AP .Also,BP bisects ∠DBA ,so ∠ABP =15◦.Now,AP =BA sin(∠ABP )=6sin(15◦).Therefore,AD =2AP =12sin(15◦).Method 3By the Cosine Law in DBA ,AD 2=AB 2+BD 2−2(AB )(BD )cos(∠ABD )=62+62−2(6)(6)cos(30◦)=72−72(√32)=72−36√3Therefore,AD = 36(2−√3)=6 2−√3since AD >0.Solution 2Drop perpendiculars from D to Q on BC and from D to R on BA .AThen CQ =CD cos(∠DCQ )=6cos(60◦)=6×12=3.Also,DQ =CD sin(∠DCQ )=6sin(60◦)=6×√32=3√3.Since BC =6,then BQ =BC −CQ =6−3=3.Now quadrilateral BQDR has three right angles,so it must have a fourth right angle and so must be a rectangle.Thus,RD =BQ =3and RB =DQ =3√3.Since AB =6,then AR =AB −RB =6−3√3.Since ARD is right-angled at R ,then using the Pythagorean Theorem and the fact that AD >0,we obtain AD =√RD 2+AR 2= 32+(6−3√3)2= 9+36−36√3+27= 72−36√3which we can rewrite as AD = 36(2−√3)=6 2−√3.5.(a)Let n be the original number and N be the number when the digits are reversed.Sincewe are looking for the largest value of n ,we assume that n >0.Since we want N to be 75%larger than n ,then N should be 175%of n ,or N =74n .Suppose that the tens digit of n is a and the units digit of n is b .Then n =10a +b .Also,the tens digit of N is b and the units digit of N is a ,so N =10b +a .We want 10b +a =74(10a +b )or 4(10b +a )=7(10a +b )or 40b +4a =70a +7b or 33b =66a ,and so b =2a .This tells us that that any two-digit number n =10a +b with b =2a has the required property.Since both a and b are digits then b <10and so a <5,which means that the possible values of n are 12,24,36,and 48.The largest of these numbers is 48.(b)We “complete the rectangle”by drawing a horizontal line through C which meets they -axis at P and the vertical line through B at Q .x A (0,Since C has y -coordinate 5,then P has y -coordinate 5;thus the coordinates of P are (0,5).Since B has x -coordinate 4,then Q has x -coordinate 4.Since C has y -coordinate 5,then Q has y -coordinate 5.Therefore,the coordinates of Q are (4,5),and so rectangle OP QB is 4by 5and so has area 4×5=20.Now rectangle OP QB is made up of four smaller triangles,and so the sum of the areas of these triangles must be 20.Let us examine each of these triangles:• ABC has area 8(given information)• AOB is right-angled at O ,has height AO =3and base OB =4,and so has area 12×4×3=6.• AP C is right-angled at P ,has height AP =5−3=2and base P C =k −0=k ,and so has area 1×k ×2=k .• CQB is right-angled at Q ,has height QB =5−0=5and base CQ =4−k ,andso has area 12×(4−k )×5=10−52k .Since the sum of the areas of these triangles is 20,then 8+6+k +10−52k =20or 4=32k and so k =83.6.(a)Solution 1Suppose that the distance from point A to point B is d km.Suppose also that r c is the speed at which Serge travels while not paddling (i.e.being carried by just the current),that r p is the speed at which Serge travels with no current (i.e.just from his paddling),and r p +c his speed when being moved by both his paddling and the current.It takes Serge 18minutes to travel from A to B while paddling with the current.Thus,r p +c =d 18km/min.It takes Serge 30minutes to travel from A to B with just the current.Thus,r c =d 30km/min.But r p =r p +c −r c =d 18−d 30=5d 90−3d 90=2d 90=d 45km/min.Since Serge can paddle the d km from A to B at a speed of d 45km/min,then it takes him 45minutes to paddle from A to B with no current.Solution 2Suppose that the distance from point A to point B is d km,the speed of the current of the river is r km/h,and the speed that Serge can paddle is s km/h.Since the current can carry Serge from A to B in 30minutes (or 12h),then d r =12.When Serge paddles with the current,his speed equals his paddling speed plus the speed of the current,or (s +r )km/h.Since Serge can paddle with the current from A to B in 18minutes (or 310h),then d r +s =310.The time to paddle from A to B with no current would be d s h.Since d r =12,then r d =2.Since d r +s =310,then r +s d =103.Therefore,s d =r +s d −r d =103−2=43.Thus,d s =34,and so it would take Serge 34of an hour,or 45minutes,to paddle from A to B with no current.Solution 3Suppose that the distance from point A to point B is d km,the speed of the current of the river is r km/h,and the speed that Serge can paddle is s km/h.Since the current can carry Serge from A to B in 30minutes (or 12h),then d r =12or d =1r .When Serge paddles with the current,his speed equals his paddling speed plus the speed of the current,or (s +r )km/h.Since Serge can paddle with the current from A to B in 18minutes (or 310h),then d r +s =310or d =310(r +s ).Since d =12r and d =310(r +s ),then 12r =310(r +s )or 5r =3r +3s and so s =23r .To travel from A to B with no current,the time in hours that it takes is d s =12r 2r =34,or 45minutes.(b)First,we note that a =0.(If a =0,then the “parabola”y =a (x −2)(x −6)is actuallythe horizontal line y =0which intersects the square all along OR .)Second,we note that,regardless of the value of a =0,the parabola has x -intercepts 2and 6,and so intersects the x -axis at (2,0)and (6,0),which we call K (2,0)and L (6,0).This gives KL =4.Third,we note that since the x -intercepts of the parabola are 2and 6,then the axis ofsymmetry of the parabola has equation x =12(2+6)=4.Since the axis of symmetry of the parabola is a vertical line of symmetry,then if theparabola intersects the two vertical sides of the square,it will intersect these at the same height,and if the parabola intersects the top side of the square,it will intersect it at two points that are symmetrical about the vertical line x =4.Fourth,we recall that a trapezoid with parallel sides of lengths a and b and height h hasarea 12h (a +b ).We now examine three cases.Case1:a<0Here,the parabola opens downwards.Since the parabola intersects the square at four points,it must intersect P Q at points M and N.(The parabola cannot intersect the vertical sides of the square since it gets “narrower”towards the vertex.)xx =4Since the parabola opens downwards,then MN<KL=4.Since the height of the trapezoid equals the height of the square(or8),then the area of the trapezoid is1h(KL+MN)which is less than1(8)(4+4)=32.But the area of the trapezoid must be36,so this case is not possible.Case2:a>0;M and N on P QWe have the following conﬁguration:xx =4Here,the height of the trapezoid is8,KL=4,and M and N are symmetric about x=4.Since the area of the trapezoid is36,then12h(KL+MN)=36or12(8)(4+MN)=36or4+MN=9or MN=5.Thus,M and N are each52units from x=4,and so N has coordinates(32,8).Since this point lies on the parabola with equation y=a(x−2)(x−6),then8=a(32−2)(32−6)or8=a(−12)(−92)or8=94a or a=329.Case3:a>0;M and N on QR and P Oxx =4Here,KL=4,MN=8,and M and N have the same y-coordinate.Since the area of the trapezoid is36,then12h(KL+MN)=36or12h(4+8)=36or6h=36or h=6.Thus,N has coordinates(0,6).Since this point lies on the parabola with equation y=a(x−2)(x−6),then 6=a(0−2)(0−6)or6=12a or a=12.Therefore,the possible values of a are329and12.7.(a)Solution1Consider a population of100people,each of whom is75years old and who behave ac-cording to the probabilities given in the question.Each of the original100people has a50%chance of living at least another10years,so there will be50%×100=50of these people alive at age85.Each of the original100people has a20%chance of living at least another15years,so there will be20%×100=20of these people alive at age90.Since there is a25%(or14)chance that an80year old person will live at least another10years(that is,to age90),then there should be4times as many of these people alive at age80than at age90.Since there are20people alive at age90,then there are4×20=80of the original100 people alive at age80.In summary,of the initial100people of age75,there are80alive at age80,50alive at age85,and20people alive at age90.Because50of the80people alive at age80are still alive at age85,then the probability that an80year old person will live at least5more years(that is,to age85)is50=5,or 62.5%.Solution2Suppose that the probability that a75year old person lives to80is p,the probability that an80year old person lives to85is q,and the probability that an85year old person lives to90is r.We want to the determine the value of q.For a75year old person to live at least another10years,they must live another5years (to age80)and then another5years(to age85).The probability of this is equal to pq. We are told in the question that this is equal to50%or0.5.Therefore,pq=0.5.For a75year old person to live at least another15years,they must live another5years (to age80),then another5years(to age85),and then another5years(to age90).The probability of this is equal to pqr.We are told in the question that this is equal to20% or0.2.Therefore,pqr=0.2Similarly,since the probability that an80year old person will live another10years is25%,then qr=0.25.Since pqr=0.2and pq=0.5,then r=pqrpq=0.20.5=0.4.Since qr=0.25and r=0.4,then q=qrr=0.250.4=0.625.Therefore,the probability that an80year old man will live at least another5years is0.625,or62.5%.(b)Using logarithm rules,the given equation is equivalent to22log10x=3(2·2log10x)+16or(2log10x)2=6·2log10x+16.Set u=2log10x.Then the equation becomes u2=6u+16or u2−6u−16=0.Factoring,we obtain(u−8)(u+2)=0and so u=8or u=−2.Since2a>0for any real number a,then u>0and so we can reject the possibility that u=−2.Thus,u=2log10x=8which means that log10x=3.Therefore,x=1000.8.(a)First,we determine theﬁrst entry in the50th row.Since theﬁrst column is an arithmetic sequence with common diﬀerence3,then the50th entry in theﬁrst column(theﬁrst entry in the50th row)is4+49(3)=4+147=151.Second,we determine the common diﬀerence in the50th row by determining the second entry in the50th row.Since the second column is an arithmetic sequence with common diﬀerence5,then the 50th entry in the second column(that is,the second entry in the50th row)is7+49(5) or7+245=252.Therefore,the common diﬀerence in the50th row must be252−151=101.Thus,the40th entry in the50th row(that is,the number in the50th row and the40th column)is151+39(101)=151+3939=4090.(b)We follow the same procedure as in(a).First,we determine theﬁrst entry in the R th row.Since theﬁrst column is an arithmetic sequence with common diﬀerence3,then the R th entry in theﬁrst column(that is,theﬁrst entry in the R th row)is4+(R−1)(3)or 4+3R−3=3R+1.Second,we determine the common diﬀerence in the R th row by determining the second entry in the R th row.Since the second column is an arithmetic sequence with common diﬀerence5,then the R th entry in the second column(that is,the second entry in the R th row)is7+(R−1)(5) or7+5R−5=5R+2.Therefore,the common diﬀerence in the R th row must be(5R+2)−(3R+1)=2R+1.Thus,the C th entry in the R th row(that is,the number in the R th row and the C th column)is3R+1+(C−1)(2R+1)=3R+1+2RC+C−2R−1=2RC+R+C(c)Suppose that N is an entry in the table,say in the R th row and C th column.From(b),then N=2RC+R+C and so2N+1=4RC+2R+2C+1.Now4RC+2R+2C+1=2R(2C+1)+2C+1=(2R+1)(2C+1).Since R and C are integers with R≥1and C≥1,then2R+1and2C+1are each integers that are at least3.Therefore,2N+1=(2R+1)(2C+1)must be composite,since it is the product of two integers that are each greater than1.9.(a)If n=2011,then8n−7=16081and so √8n−7≈126.81.Thus,1+√8n−72≈1+126.812≈63.9.Therefore,g(2011)=2(2011)+1+8(2011)−72=4022+ 63.9 =4022+63=4085.(b)To determine a value of n for which f(n)=100,we need to solve the equation2n−1+√8n−72=100(∗)Weﬁrst solve the equation2x−1+√8x−72=100(∗∗)because the left sides of(∗)and(∗∗)do not diﬀer by much and so the solutions are likely close together.We will try integers n in(∗)that are close to the solutions to(∗∗). Manipulating(∗∗),we obtain4x−(1+√8x−7)=2004x−201=√8x−7(4x−201)2=8x−716x2−1608x+40401=8x−716x2−1616x+40408=02x2−202x+5051=0By the quadratic formula,x=202±2022−4(2)(5051)2(2)=202±√3964=101±√992and so x≈55.47or x≈45.53.We try n=55,which is close to55.47:f(55)=2(55)−1+8(55)−72=110−1+√4332Since √433≈20.8,then1+√4332≈10.9,which gives1+√4332=10.Thus,f(55)=110−10=100.Therefore,a value of n for which f(n)=100is n=55.(c)We want to show that each positive integer m is in the range of f or the range of g ,butnot both.To do this,we ﬁrst try to better understand the “complicated”term of each of the func-tions –that is,the term involving the greatest integer function.In particular,we start witha positive integer k ≥1and try to determine the positive integers n that give 1+√8n −72 =k .By deﬁnition of the greatest integer function,the equation 1+√8n −72 =k is equiv-alent to the inequality k ≤1+√8n −72<k +1,from which we obtain the following set of equivalent inequalities 2k ≤1+√8n −7<2k +22k −1≤√8n −7<2k +14k 2−4k +1≤8n −7<4k 2+4k +14k 2−4k +8≤8n <4k 2+4k +812(k 2−k )+1≤n <12(k 2+k )+1If we deﬁne T k =1k (k +1)=1(k 2+k )to be the k th triangular number for k ≥0,thenT k −1=12(k −1)(k )=12(k 2−k ).Therefore, 1+√8n −72 =k for T k −1+1≤n <T k +1.Since n is an integer,then 1+√8n −72=k is true for T k −1+1≤n ≤T k .When k =1,this interval is T 0+1≤n ≤T 1(or 1≤n ≤1).When k =2,this interval is T 1+1≤n ≤T 2(or 2≤n ≤3).When k =3,this interval is T 2+1≤n ≤T 3(or 4≤n ≤6).As k ranges over all positive integers,these intervals include every positive integer n and do not overlap.Therefore,we can determine the range of each of the functions f and g by examining the values f (n )and g (n )when n is in these intervals.For each non-negative integer k ,deﬁne R k to be the set of integers greater than k 2and less than or equal to (k +1)2.Thus,R k ={k 2+1,k 2+2,...,k 2+2k,k 2+2k +1}.For example,R 0={1},R 1={2,3,4},R 2={5,6,7,8,9},and so on.Every positive integer occurs in exactly one of these sets.Also,for each non-negative integer k deﬁne S k ={k 2+2,k 2+4,...,k 2+2k }and deﬁne Q k ={k 2+1,k 2+3,...,k 2+2k +1}.For example,S 0={},S 1={3},S 2={6,8},Q 0={1},Q 1={2,4},Q 2={5,7,9},and so on.Note that R k =Q k ∪S k so every positive integer occurs in exactly one Q k or in exactly one S k ,and that these sets do not overlap since no two S k ’s overlap and no two Q k ’s overlap and no Q k overlaps with an S k .We determine the range of the function g ﬁrst.For T k −1+1≤n ≤T k ,we have 1+√8n −72=k and so 2T k −1+2≤2n ≤2T k 2T k −1+2+k ≤2n + 1+√8n −72 ≤2T k +k k 2−k +2+k ≤g (n )≤k 2+k +k k 2+2≤g (n )≤k 2+2kNote that when n is in this interval and increases by 1,then the 2n term causes the value of g (n )to increase by 2.Therefore,for the values of n in this interval,g (n )takes precisely the values k 2+2,k 2+4,k 2+6,...,k 2+2k .In other words,the range of g over this interval of its domain is precisely the set S k .As k ranges over all positive integers (that is,as these intervals cover the domain of g ),this tells us that the range of g is precisely the integers in the sets S 1,S 2,S 3,....(We could also include S 0in this list since it is the empty set.)We note next that f (1)=2− 1+√8−72 =1,the only element of Q 0.For k ≥1and T k +1≤n ≤T k +1,we have 1+√8n −72=k +1and so 2T k +2≤2n ≤2T k +12T k +2−(k +1)≤2n − 1+√8n −72 ≤2T k +1−(k +1)k 2+k +2−k −1≤f (n )≤(k +1)(k +2)−k −1k 2+1≤f (n )≤k 2+2k +1Note that when n is in this interval and increases by 1,then the 2n term causes the value of f (n )to increase by 2.Therefore,for the values of n in this interval,f (n )takes precisely the values k 2+1,k 2+3,k 2+5,...,k 2+2k +1.In other words,the range of f over this interval of its domain is precisely the set Q k .As k ranges over all positive integers (that is,as these intervals cover the domain of f ),this tells us that the range of f is precisely the integers in the sets Q 0,Q 1,Q 2,....Therefore,the range of f is the set of elements in the sets Q 0,Q 1,Q 2,...and the range of g is the set of elements in the sets S 0,S 1,S 2,....These ranges include every positive integer and do not overlap.10.(a)Suppose that ∠KAB =θ.Since ∠KAC =2∠KAB ,then ∠KAC =2θand ∠BAC =∠KAC +∠KAB =3θ.Since 3∠ABC =2∠BAC ,then ∠ABC =23×3θ=2θ.Since ∠AKC is exterior to AKB ,then ∠AKC =∠KAB +∠ABC =3θ.This gives the following conﬁguration:BNow CAK is similar to CBA since the triangles have a common angle at C and ∠CAK =∠CBA .Therefore,AKBA=CACBordc=baand so d=bca.Also,CKCA=CACBora−xb=baand so a−x=b2aor x=a−b2a=a2−b2a,as required.(b)From(a),bc=ad and a2−b2=ax and so we obtainLS=(a2−b2)(a2−b2+ac)=(ax)(ax+ac)=a2x(x+c) andRS=b2c2=(bc)2=(ad)2=a2d2In order to show that LS=RS,we need to show that x(x+c)=d2(since a>0).Method1:Use the Sine LawFirst,we derive a formula for sin3θwhich we will need in this solution:sin3θ=sin(2θ+θ)=sin2θcosθ+cos2θsinθ=2sinθcos2θ+(1−2sin2θ)sinθ=2sinθ(1−sin2θ)+(1−2sin2θ)sinθ=3sinθ−4sin3θSince∠AKB=180◦−∠KAB−∠KBA=180◦−3θ,then using the Sine Law in AKB givesx sinθ=dsin2θ=csin(180◦−3θ)Since sin(180◦−X)=sin X,then sin(180◦−3θ)=sin3θ,and so x=d sinθsin2θandc=d sin3θsin2θ.This givesx(x+c)=d sinθsin2θd sinθsin2θ+d sin3θsin2θ=d2sinθsin22θ(sinθ+sin3θ)=d2sinθsin22θ(sinθ+3sinθ−4sin3θ)=d2sinθsin22θ(4sinθ−4sin3θ)=4d2sin2θsin22θ(1−sin2θ)=4d2sin2θcos2θsin22θ=4d2sin2θcos2θ(2sinθcosθ)2=4d2sin2θcos2θ4sin2θcos2θ=d2as required.We could have instead used the formula sin A +sin B =2sinA +B 2 cos A −B2 toshow that sin 3θ+sin θ=2sin 2θcos θ,from which sin θ(sin 3θ+sin θ)=sin θ(2sin 2θcos θ)=2sin θcos θsin 2θ=sin 22θMethod 2:Extend ABExtend AB to E so that BE =BK =x and join KE .ENow KBE is isosceles with ∠BKE =∠KEB .Since ∠KBA is the exterior angle of KBE ,then ∠KBA =2∠KEB =2θ.Thus,∠KEB =∠BKE =θ.But this also tells us that ∠KAE =∠KEA =θ.Thus, KAE is isosceles and so KE =KA =d.ESo KAE is similar to BKE ,since each has two angles equal to θ.Thus,KA BK =AE KE or d x =c +x dand so d 2=x (x +c ),as required.Method 3:Use the Cosine Law and the Sine LawWe apply the Cosine Law in AKB to obtainAK 2=BK 2+BA 2−2(BA )(BK )cos(∠KBA )d 2=x 2+c 2−2cx cos(2θ)d 2=x 2+c 2−2cx (2cos 2θ−1)Using the Sine Law in AKB ,we get x sin θ=d sin 2θor sin 2θsin θ=d x or 2sin θcos θsin θ=d x and so cos θ=d 2x.Combining these two equations,d2=x2+c2−2cx2d24x2−1d2=x2+c2−cd2x+2cxd2+cd2x=x2+2cx+c2d2+cd2x=(x+c)2xd2+cd2=x(x+c)2d2(x+c)=x(x+c)2d2=x(x+c)as required(since x+c=0).(c)Solution1Our goal is toﬁnd a triple of positive integers that satisfy the equation in(b)and are the side lengths of a triangle.First,we note that if(A,B,C)is a triple of real numbers that satisﬁes the equation in(b)and k is another real number,then the triple(kA,kB,kC)also satisﬁes the equationfrom(b),since(k2A2−k2B2)(k2A2−k2B2+kAkC)=k4(A2−B2)(A2−B2+AC)=k4(B2C2)=(kB)2(kC)2 Therefore,we start by trying toﬁnd a triple(a,b,c)of rational numbers that satisﬁes the equation in(b)and forms a triangle,and then“scale up”this triple to form a triple (ka,kb,kc)of integers.To do this,we rewrite the equation from(b)as a quadratic equation in c and solve for c using the quadratic formula.Partially expanding the left side from(b),we obtain(a2−b2)(a2−b2)+ac(a2−b2)=b2c2which we rearrange to obtainb2c2−c(a(a2−b2))−(a2−b2)2=0By the quadratic formula,c=a(a2−b2)±a2(a2−b2)2+4b2(a2−b2)22b2=a(a2−b2)±(a2−b2)2(a2+4b2)2b2Since∠BAC>∠ABC,then a>b and so a2−b2>0,which givesc=a(a2−b2)±(a2−b2)√a2+4b22b2=(a2−b2)2b2(a±√a2+4b2)Since a2+4b2>0,then √a2+4b2>a,so the positive root isc=(a2−b2)2b2(a+a2+(2b)2)We try to ﬁnd integers a and b that give a rational value for c .We will then check to see if this triple (a,b,c )forms the side lengths of a triangle,and then eventually scale these up to get integer values.One way for the value of c to be rational (and in fact the only way)is for a 2+(2b )2to be an integer,or for a and 2b to be the legs of a Pythagorean triple.Since √32+42is an integer,then we try a =3and b =2,which givesc =(32−22)2·22(3+√32+42)=5and so (a,b,c )=(3,2,5).Unfortunately,these lengths do not form a triangle,since 3+2=5.(The Triangle Inequality tells us that three positive real numbers a ,b and c form a triangle if and only if a +b >c and a +c >b and b +c >a .)We can continue to try small Pythagorean triples.Now 152+82=172,but a =15and b =4do not give a value of c that forms a triangle with a and b .However,162+302=342,so we can try a =16and b =15which givesc =(162−152)2·152(16+√162+302)=31450(16+34)=319Now the lengths (a,b,c )=(16,15,319)do form the sides of a triangle since a +b >c and a +c >b and b +c >a .Since these values satisfy the equation from (b),then we can scale them up by a factor of k =9to obtain the triple (144,135,31)which satisﬁes the equation from (b)and are the side lengths of a triangle.(Using other Pythagorean triples,we could obtain other triples of integers that work.)Solution 2We note that the equation in (b)involves only a ,b and c and so appears to depend only on the relationship between the angles ∠CAB and ∠CBA in ABC .Using this premise,we use ABC ,remove the line segment AK and draw the altitude CF .CBA 3θ2θb aa c os 2θbc os 3θF Because we are only looking for one triple that works,we can make a number of assump-tions that may or may not be true in general for such a triangle,but which will help us ﬁnd an example.We assume that 3θand 2θare both acute angles;that is,we assume that θ<30◦.In ABC ,we have AF =b cos 3θ,BF =a cos 2θ,and CF =b sin 3θ=a sin 2θ.Note also that c =b cos 3θ+a cos 2θ.One way to ﬁnd the integers a,b,c that we require is to look for integers a and b and an angle θwith the properties that b cos 3θand a cos 2θare integers and b sin 3θ=a sin 2θ.Using trigonometric formulae,sin 2θ=2sin θcos θcos 2θ=2cos 2θ−1sin 3θ=3sin θ−4sin 3θ(from the calculation in (a),Solution 1,Method 1)cos 3θ=cos(2θ+θ)=cos 2θcos θ−sin 2θsin θ=(2cos 2θ−1)cos θ−2sin 2θcos θ=(2cos 2θ−1)cos θ−2(1−cos 2θ)cos θ=4cos 3θ−3cos θSo we can try to ﬁnd an angle θ<30◦with cos θa rational number and then integers a and b that make b sin 3θ=a sin 2θand ensure that b cos 3θand a cos 2θare integers.Since we are assuming that θ<30◦,then cos θ>√32≈0.866.The rational number with smallest denominator that is larger than √32is 78,so we try the acute angle θwith cos θ=7.In this case,sin θ=√1−cos 2θ=√158,and sosin 2θ=2sin θcos θ=2×78×√158=7√1532cos 2θ=2cos 2θ−1=2×4964−1=1732sin 3θ=3sin θ−4sin 3θ=3×√158−4×15√15512=33√15128cos 3θ=4cos 3θ−3cos θ=4×343512−3×78=7128To have b sin 3θ=a sin 2θ,we need 33√15128b =7√1532a or 33b =28a .To ensure that b cos 3θand a cos 2θare integers,we need 7128b and 1732a to be integers,andso a must be divisible by 32and b must be divisible by 128.The integers a =33and b =28satisfy the equation 33b =28a .Multiplying each by 32gives a =1056and b =896which satisfy the equation 33b =28a and now have the property that b is divisible by 128(with quotient 7)and a is divisible by 32(with quotient 33).With these values of a and b ,we obtain c =b cos 3θ+a cos 2θ=896×7128+1056×1732=610.We can then check that the triple (a,b,c )=(1056,896,610)satisﬁes the equation from(b),as required.As in our discussion in Solution 1,each element of this triple can be divided by 2to obtain the “smaller”triple (a,b,c )=(528,448,305)that satisﬁes the equation too.Using other values for cos θand integers a and b ,we could obtain other triples (a,b,c )of integers that work.。

2010PascalSolution滑铁卢竞赛题答案

1.In cents,theﬁve given choices are50,90,95,101,and115cents.The diﬀerences between each of these and$1.00(or100cents),in cents,are100−50=50100−90=10100−95=5101−100=1115−100=15 The diﬀerence between$1.01and$1.00is the smallest(1cent),so$1.01is closest to$1.00.Answer:(D) ing the correct order of operations,(20−16)×(12+8)4=4×204=804=20Answer:(C)3.We divide the750mL ofﬂour into portions of250mL.We do this by calculating750÷250=3.Therefore,750mL is three portions of250mL.Since50mL of milk is required for each250mL ofﬂour,then3×50=150mL of milk is required in total.Answer:(C) 4.There are8ﬁgures in total.Of these,3are triangles.Therefore,the probability is3.Answer:(A) 5.We simplify the left side and express it as a fraction with numerator1:1 9+118=218+118=318=16Therefore,the number that replaces the is6.Answer:(C) 6.There are16horizontal segments on the perimeter.Each has length1,so the horizontalsegments contribute16to the perimeter.There are10vertical segments on the perimeter.Each has length1,so the vertical segments contribute10to the perimeter.Therefore,the perimeter is10+16=26.(We could arrive at this total instead by starting at aﬁxed point and travelling around the outside of theﬁgure counting the number of segments.)Answer:(E) 7.Since33=3×3×3=3×9=27,then√33+33+33=√27+27+27=√81=9Answer:(B)8.The diﬀerence between the two given numbers is7.62−7.46=0.16.This length of the number line is divided into8equal segments.The length of each of these segments is thus0.16÷8=0.02.Point P is three of these segments to the right of7.46.Thus,the number represented is7.46+3(0.02)=7.46+0.06=7.52.Answer:(E)9.A 12by 12grid of squares will have 11interior vertical lines and 11interior horizontal lines.(In the given 4by 4example,there are 3interior vertical lines and 3interior horizontal lines.)Each of the 11interior vertical lines intersects each of the 11interior horizontal lines and creates an interior intersection point.Thus,each interior vertical line accounts for 11intersection points.Therefore,the number of interior intersection points is 11×11=121.Answer:(B)10.Because the central angle for the interior sector “Less than 1hour”is 90◦,then the fraction of the students who do less than 1hour of homework per day is 90◦360◦=14.In other words,25%of the students do less than 1hour of homework per day.Therefore,100%−25%=75%of the students do at least 1hour of homework per day.Answer:(E)11.Solution 1Since there is more than 1four-legged table,then there are at least 2four-legged tables.Since there are 23legs in total,then there must be fewer than 6four-legged tables,since 6four-legged tables would have 6×4=24legs.Thus,there are between 2and 5four-legged tables.If there are 2four-legged tables,then these tables account for 2×4=8legs,leaving 23−8=15legs for the three-legged tables.Since 15is divisible by 3,then this must be the solution,so there are 15÷3=5three-legged tables.(We can check that if there are 3or 4four-legged tables,then the number of remaining legs is not divisible by 3,and if there are 5four-legged tables,then there is only 1three-legged table,which is not allowed.)Solution 2Since there is more than 1table of each type,then there are at least 2three-legged tables and 2four-legged tables.These tables account for 2(3)+2(4)=14legs.There are 23−14=9more legs that need to be accounted for.These must come from a combination of three-legged and four-legged tables.The only way to make 9from 3s and 4s is to use three 3s.Therefore,there are 2+3=5three-legged tables and 2four-legged tables.Answer:(E)12.Solution 1The total area of the rectangle is 3×4=12.The total area of the shaded regions equals the total area of the rectangle (12)minus the area of the unshaded region.The unshaded region is a triangle with base of length 1and height 4;the area of this region is 12(1)(4)=2.Therefore,the total area of the shaded regions is 12−2=10.Solution 2The shaded triangle on the left has base of length 2and height of length 4,so has an area of 12(2)(4)=4.The shaded triangle on the right has base of length3(at the top)and height of length4,sohas an area of12(3)(4)=6.Therefore,the total area of the shaded regions is4+6=10.Answer:(C)13.Since the ratio of boys to girls at Cayley H.S.is3:2,then33+2=35of the students at CayleyH.S.are boys.Thus,there are35(400)=12005=240boys at Cayley H.S.Since the ratio of boys to girls at Fermat C.I.is2:3,then22+3=25of the students at FermatC.I.are boys.Thus,there are25(600)=12005=240boys at Fermat C.I.There are400+600=1000students in total at the two schools.Of these,240+240=480are boys,and so the remaining1000−480=520students are girls.Therefore,the overall ratio of boys to girls is480:520=48:52=12:13.Answer:(B) 14.When the given net is folded,the face numbered5will be opposite the face numbered1.Therefore,the remaining four faces share an edge with the face numbered1,so the product of the numbers is2×3×4×6=144.Answer:(B)15.The percentage10%is equivalent to the fraction110.Therefore,t=110s,or s=10t.Answer:(D)16.Since the base of the folded box measures5cm by4cm,then the area of the base of the boxis5(4)=20cm2.Since the volume of the box is60cm3and the area of the base is20cm2,then the height of the box is60=3cm.Therefore,each of the four identical squares has side length3cm,because the edges of these squares form the vertical edges of thebox.Therefore,the rectangular sheet measures3+5+3=11cm by3+4+3=10cm,and so has area11(10)=110cm2.Answer:(B) 17.Solution1Since SUR is a straight line,then∠RUV=180◦−∠SUV=180◦−120◦=60◦.Since P W and QX are parallel,then∠RV W=∠V T X=112◦.Since UV W is a straight line,then∠RV U=180◦−∠RV W=180◦−112◦=68◦.Since the measures of the angles in a triangle add to180◦,then∠URV=180◦−∠RUV−∠RV U=180◦−60◦−68◦=52◦Solution2Since SUR is a straight line,then∠RUV=180◦−∠SUV=180◦−120◦=60◦.Since P W and QX are parallel,then∠RST=∠RUV=60◦.Since ST X is a straight line,then∠RT S=180◦−∠V T X=180◦−112◦=68◦.Since the measures of the angles in a triangle add to180◦,then∠URV=∠SRT=180◦−∠RST−∠RT S=180◦−60◦−68◦=52◦Answer:(A) 18.Solution1When Catherine adds30litres of gasoline,the tank goes from18full to34full.Since34−18=68−18=58,then58of the capacity of the tank is30litres.Thus,18of the capacity of the tank is30÷5=6litres.Also,the full capacity of the tank is8×6=48litres.Toﬁll the remaining14of the tank,Catherine must add an additional14×48=12litres of gas.Because each litre costs$1.38,it will cost12×$1.38=$16.56toﬁll the rest of the tank. Solution2Suppose that the capacity of the gas tank is x litres.Starting with1of a tank,30litres of gas makes the tank3full,so1x+30=3x or5x=30or x=48.The remaining capacity of the tank is14x=14(48)=12litres.At$1.38per litre,it will cost Catherine12×$1.38=$16.56toﬁll the rest of the tank.Answer:(C) 19.The area of a semi-circle with radius r is1πr2so the area of a semi-circle with diameter d is1 2π(12d)2=18πd2.The semicircles with diameters UV,V W,W X,XY,and Y Z each have equal diameter andthus equal area.The area of each of these semicircles is18π(52)=258π.The large semicircle has diameter UZ=5(5)=25,so has area18π(252)=6258π.The shaded area equals the area of the large semicircle,minus the area of two small semicircles, plus the area of three small semicircles,which equals the area of the large semicircle plus the area of one small semicircle.Therefore,the shaded area equals6258π+258π=6508π=3254π.Answer:(A)20.The sum of the odd numbers from5to21is5+7+9+11+13+15+17+19+21=117Therefore,the sum of the numbers in any row is one-third of this total,or39.This means as well that the sum of the numbers in any column or diagonal is also39.Since the numbers in the middle row add to39,then the number in the centre square is 39−9−17=13.Since the numbers in the middle column add to39,then the number in the middle square in the bottom row is39−5−13=21.591317x21Since the numbers in the bottom row add to39,then the number in the bottom right square is39−21−x=18−x.Since the numbers in the bottom left to top right diagonal add to39,then the number in the top right square is39−13−x=26−x.Since the numbers in the rightmost column add to39,then(26−x)+17+(18−x)=39or 61−2x=39or2x=22,and so x=11.We can complete the magic square as follows:195159131711217Answer:(B) 21.We label the numbers in the empty boxes as y and z,so the numbers in the boxes are thus8,y,z,26,x.Since the average of z and x is26,then x+z=2(26)=52or z=52−x.We rewrite the list as8,y,52−x,26,x.Since the average of26and y is52−x,then26+y=2(52−x)or y=104−26−2x=78−2x.We rewrite the list as8,78−2x,52−x,26,x.Since the average of8and52−x is78−2x,then8+(52−x)=2(78−2x)60−x=156−4x3x=96x=32Therefore,x=32.Answer:(D) 22.Since JKLM is a rectangle,then the angles at J and K are each90◦,so each of SJP andQKP is right-angled.By the Pythagorean Theorem in SJP,we haveSP2=JS2+JP2=522+392=2704+1521=4225Since SP>0,then SP=√4225=65.Since P QRS is a rhombus,then P Q=P S=65.By the Pythagorean Theorem in QKP,we haveKP2=P Q2−KQ2=652−252=4225−625=3600Since KP>0,then KP=√3600=60.(Instead of using the Pythagorean Theorem,we could note instead that SJP is a scaled-up version of a3-4-5right-angled triangle and that QKP is a scaled-up version of a5-12-13 right-angled triangle.This would allow us to use the known ratios of side lengths to calculate the missing side length.)Since KQ and P Z are parallel and P K and W Q are parallel,then P KQW is a rectangle,andso P W=KQ=25.Similarly,JP ZS is a rectangle and so P Z=JS=52.Thus,W Z=P Z−P W=52−25=27.Also,SY RM is a rectangle.Since JM and KL are parallel(JKLM is a rectangle),JK and ML are parallel,and P Q and SR are parallel(P QRS is a rhombus),then∠MSR=∠KQP and∠SRM=∠QP K.Since SMR and QKP have two equal angles,then their third angles must be equal too.Thus,the triangles have the same proportions.Since the hypotenuses of the triangles are equal, then the triangles must in fact be exactly the same size;that is,the lengths of the corresponding sides must be equal.(We say that SMR is congruent to QKP by“angle-side-angle”.) In particular,MR=KP=60.Thus,ZY=SY−SZ=MR−JP=60−39=21.Therefore,the perimeter of rectangle W XY Z is2(21)+2(27)=96.Answer:(D) 23.First,we note that2010=10(201)=2(5)(3)(67)and so20102=223252672.Consider N consecutive four-digit positive integers.For the product of these N integers to be divisible by20102,it must be the case that two diﬀerent integers are divisible by67(which would mean that there are at least68integers in the list)or one of the integers is divisible by672.Since we want to minimize N(and indeed because none of the answer choices is at least68), we look for a list of integers in which one is divisible by672=4489.Since the integers must all be four-digit integers,then the only multiples of4489the we must consider are4489and8978.First,we consider a list of N consecutive integers including4489.Since the product of these integers must have2factors of5and no single integer within10 of4489has a factor of25,then the list must include two integers that are multiples of5.To minimize the number of integers in the list,we try to include4485and4490.Thus our candidate list is4485,4486,4487,4488,4489,4490.The product of these integers includes2factors of67(in4489),2factors of5(in4485and 4490),2factors of2(in4486and4488),and2factors of3(since each of4485and4488is divisible by3).Thus,the product of these6integers is divisible by20102.Therefore,the shortest possible list including4489has length6.Next,we consider a list of N consecutive integers including8978.Here,there is a nearby integer containing2factors of5,namely8975.So we start with the list8975,8976,8977,8978and check to see if it has the required property.The product of these integers includes2factors of67(in8978),2factors of5(in8975),and2 factors of2(in8976).However,the only integer in this list divisible by3is8976,which has only1factor of3.To include a second factor of3,we must include a second multiple of3in the list.Thus,we extend the list by one number to8979.Therefore,the product of the numbers in the list8975,8976,8977,8978,8979is a multiple of 20102.The length of this list is5.Thus,the smallest possible value of N is5.(Note that a quick way to test if an integer is divisible by3is to add its digit and see if this total is divisible by3.For example,the sum of the digits of8979is33;since33is a multiple of3,then8979is a multiple of3.)Answer:(A)24.We label the terms x1,x2,x3,...,x2009,x2010.Suppose that S is the sum of the odd-numbered terms in the sequence;that is,S=x1+x3+x5+···+x2007+x2009We know that the sum of all of the terms is5307;that is,x1+x2+x3+···+x2009+x2010=5307Next,we pair up the terms:each odd-numbered term with the following even-numbered term.That is,we pair theﬁrst term with the second,the third term with the fourth,and so on,until we pair the2009th term with the2010th term.There are1005such pairs.In each pair,the even-numbered term is one bigger than the odd-numbered term.That is, x2−x1=1,x4−x3=1,and so on.Therefore,the sum of the even-numbered terms is1005greater than the sum of the odd-numbered terms.Thus,the sum of the even-numbered terms is S+1005.Since the sum of all of the terms equals the sum of the odd-numbered terms plus the sum of the even-numbered terms,then S+(S+1005)=5307or2S=4302or S=2151.Thus,the required sum is2151.Answer:(C) 25.Before we answer the given question,we determine the number of ways of choosing3objectsfrom5objects and the number of ways of choosing2objects from5objects.Consider5objects labelled B,C,D,E,F.The possible pairs are:BC,BD,BE,BF,CD,CE,CF,DE,DF,EF.There are10such pairs.The possible triples are:DEF,CEF,CDF,CDE,BEF,BDF,BDE,BCF,BCE,BCD.There are10such triples.(Can you see why there are the same number of pairs and triples?)Label the six teams A,B,C,D,E,F.We start by considering team A.Team A plays3games,so we must choose3of the remaining5teams for A to play.As we saw above,there are10ways to do this.Without loss of generality,we pick one of these sets of3teams for A to play,say A plays B,C and D.We keep track of everything by drawing diagrams,joining the teams that play each other witha line.Thus far,we haveAB C DThere are two possible cases now–either none of B,C and D play each other,or at least one pair of B,C,D plays each other.Case1:None of the teams that play A play each otherIn the conﬁguration above,each of B,C and D play two more games.They already play A and cannot play each other,so they must each play E and F.This givesAE FB C DNo further choices are possible.There are10possible schedules in this type of conﬁguration.These10combinations come from choosing the3teams that play A.Case2:Some of the teams that play A play each otherHere,at least one pair of the teams that play A play each other.Given the teams B,C and D playing A,there are3possible pairs(BC,BD,CD).We pick one of these pairs,say BC.(This gives10×3=30conﬁgurations so far.)AB C DIt is now not possible for B or C to also play D.If it was the case that C,say,played D,then we would have the conﬁgurationAB C DE FHere,A and C have each played3games and B and D have each played2games.Teams E and F are unaccounted for thus far.They cannot both play3games in this conﬁguration as the possible opponents for E are B,D and F,and the possible opponents for F are B,D and E,with the“B”and“D”possibilities only to be used once.A similar argument shows thatB cannot play D.Thus,B or C cannot also play D.So we have the conﬁgurationAB C DHere,A has played3games,B and C have each played2games,and D has played1game.B andC must play1more game and cannot playD or A.They must play E and F in some order.There are2possible ways to assign these games(BE and CF,or BF and CE.)This gives30×2=60conﬁgurations so far.Suppose that B plays E and C plays F.AB C DE FSo far,A,B and C each play3games and E,F and D each play1game.The only way to complete the conﬁguration is to join D,E and F.AB C DE FTherefore,there are60possible schedules in this case.In total,there are10+60=70possible schedules.Answer:(E)。

第十一届全国大学生数学竞赛(非数学类)试题

第十一届全国大学生数学竞赛(非数学类)试题参考解答及评分标准一、填空题（每小题6分）1. sin 014x x →=.解：sin sin 00x x x x x →→→=- sin 1/31/30022(e 1)1sin 1limlim 444422x x x x x x →→-=+-=⎛⎫⎛⎫ ⎪ ⎪⎝⎭⎝⎭. 2. 设隐函数()y y x =由方程22()y x y x -=所确定，则232ln ||dx y y C y x x=-+⎰. 解：令y tx =，则21(1)x t t =-，1(1)y t t =-，3223(1)tdx dt t t -+=-，这样，223332ln ||2ln ||dx t y ydt t t C C y t x x-+==-+=-+⎰⎰. 3. 定积分220(1sin )1cos x e x dx e xππ+=+⎰.解：222000(1sin )sin 1cos 1cos 1cos x xx e x e xdx dx de xx x πππ+=++++⎰⎰⎰ 2222200sin cos (1cos )+sin 1cos 1cos (1cos )xxxe xe x x x dx e dx x x x πππ+=+-+++⎰⎰2222000sin 1cos 1cos 1cos xxx e xe edx dx e x x x ππππ=+-=+++⎰⎰. 4. 已知22(,)323ydx xdy du x y x xy y -=-+，则1(,)()C 3x u x y y =-+. 解：22(,)323ydx xdy du x y x xy y -=-+21()233()3xd x yx x y y y ==--+（）.所以，1(,)()C 3x u x y y =-+.5. 设,,,0a b c μ>，曲面xyz μ=与曲面2222221x y z a b c ++=相切，则μ=.解：根据题意有：22x yz a λ=,22y xz b λ=，22zxy c λ=，以及 222x a μλ=，222y b μλ=，222z c μλ=，从而得：32228a b cλμ=，32μλ=，联立解得：μ=二、（14分）计算三重积分22d d d Ω+⎰⎰⎰xyzx y z x y，其中Ω是由曲面2222()2++=x y z xy 围成的区域在第一卦限部分.解：采用“球面坐标”计算，并利用对称性，得ππ3224222sin cos sin cos 2d d sin d sin I ρϕθθϕθϕρϕρρϕ=⎰⎰ -------5分ππ342002sin cos d sin cos d d θθθϕϕϕρρ=⎰⎰ππ3354202sin cos d sin cos d θθθϕϕϕ=⎰⎰ -------10分ππ354201sin 2d sin d(sin )4θθϕϕ=⎰⎰π3201121sin d 4848372t t ==⋅=⎰. -------14分三、（14分）设()f x 在[0,)+∞上可微，(0)0f =，且存在常数0A >，使得|()||()|f x A f x '≤在[0,)+∞上成立，试证明：在(0,)+∞上有()0f x ≡.证明：设01[0,]2x A ∈，使得01|()|max |()|[0,]2f x f x x A ⎧⎫=∈⎨⎬⎩⎭， -------5分 000011|()||(0)+()||()||()|22f x f f x A f x f x A ξ'=≤=，只有0|()|0f x =. 故当 1[0,]2x A∈时，()0f x ≡. -------12分递推可得，对所有的1[,]22k kx A A-∈，1,2,k =，均有()0f x ≡. -------14分四、（14分）计算积分2sin (cos sin )0sin I d e d ππθφφφθθ-=⎰⎰解：设球面 Σ:x 2+y 2+z 2=1, 由球面参数方程sin cos x θφ=，sin sin y θφ=，cos z θ=知sin dS d d θθφ=，所以，所求积分可化为第一型曲面积分I =∬e x−ydS Σ-------4分设平面P t :√2=t,−1≤t ≤1，其中t 为平面P t 被球面截下部分中心到原点距离.用平面P t 分割球面Σ，球面在平面P t ,P t+dt 之间的部分形如圆台外表面状，记为Σt,dt .被积函数在其上为 e x−y =e √2t . -------8分由于Σt,dt 半径为r t =√1−t 2，半径的增长率为 d√1−t 2=√1−t 2 就是 Σt,dt 上下底半径之差. 记圆台外表面斜高为ℎt ，则由微元法知 dt 2+(d √1−t 2)2=ℎt 2, 得到ℎt =√1−t 2 ，所以 Σt,dt 的面积为 dS =2πr t ℎt =2πdt, -------12分I =∫e √2t 1−12πdt =√2√2t |−11=√2π(e √2−e −√2). -------14分五、（14分）设()f x 是仅有正实根的多项式函数，满足 0()()n n n f x c x f x +∞='=-∑. 试证：0n c >，(0n ≥），极限lim n ()f x 的最小根. 证明：由f (x )为仅有正实根的多项式，不妨设()f x 的全部根为 0<a 1<a 2<⋯<a k ，这样，f (x )=A (x −a 1)r 1⋯(x −a k )r k ，其中 r i 为对应根a i 的重数 (i =1,⋯,k,r k ≥1). -------2分f ′(x )=Ar 1(x −a 1)r 1−1⋯(x −a k )r k +⋯+Ar k (x −a 1)r 1⋯(x −a k )r k −1，所以，f ′(x )=f (x )(r 1x−a 1+⋯+rkx−a k)，从而， −f ′(x)f(x)=r 1a 1∙11−xa 1+⋯+r k a k∙11−x a k.-------6分若|x |<a 1, 则 −f ′(x)f(x)=r 1a 1∙∑(xa1)n∞n=0+⋯+r k a k∙∑(xak)n∞n=0=∑(r 1a 1n+1+⋯+r k a kn+1)∞n=0x n .而 −f ′(x)f(x)=∑c n x n∞n=0，由幂级数的唯一性知c n =r 1a 1n+1+⋯+r kak n+1>0， ------9分c ncn+1=r 1a 1n+1+⋯+r k a kn+1r 1a 1n+2+⋯+r k a kn+2=a 1∙r 1+⋯+(a1a k)n+1r kr 1+⋯+(a 1a k)n+2r k.limn→∞c nc =a 1∙r 1+0+⋯+0r +0+⋯+0=a 1>0， limn→∞c n+1c =1a ， -----12分limn→∞1n ∙(ln c2c1+⋯+ln c n+1c n)=ln 1a 1,√c n n=elnc nn=elnc 1n +1n (ln c 2c 1+⋯+ln cn+1c n)→eln1a 1=1a 1.从而，lim√c nn=a 1,即f (x )的最小正根. -----14分六、（14分）设函数()f x 在[0, )+∞上具有连续导数，满足22223[3()]()2[1()]-'+=+x f x f x f x e ，且(0)1≤f .证明：存在常数0>M ，使得[0,)∈+∞x 时，恒有()≤f x M .证明：由于()0'>f x ，所以()f x 是[0, )+∞上的严格增函数，故+lim ()→∞=x f x L (有限或为+∞). 下面证明 ≠+∞L . -----2分记()=y f x ，将所给等式分离变量并积分得 222232d d (1)3-+=+⎰⎰x y y e x y ，即 2222arctan d 13-+=++⎰x t y y e t C y ， ------6分其中2(0)2arctan (0)1(0)=++f C f f . ------8分若=+∞L ，则对上式取极限→+∞x ，并利用2d 2+∞-=⎰t e t ，得π3=-C .-----10分另一方面，令2()2arctan 1=++ug u u u ，则2223()>0(1)+'=+u g u u ，所以函数()g u 在(, )-∞+∞上严格单调增加. 因此，当(0)1≤f 时，1π((0))(1)2+=≤=C g f g ，但2π1π22+>>C ，矛盾，这就证明了+lim ()→∞=x f x L 为有限数.最后，取max{(0),}=M f L ，则|()|≤f x M ，[0,)∀∈+∞x . -----14分。

滑铁卢数学竞赛

滑铁卢数学竞赛滑铁卢数学竞赛是加拿大滑铁卢大学举办的一项年度数学竞赛活动。

该竞赛旨在通过一系列难度不断增加的数学问题，考察参赛者的数学思维能力、解题能力以及创造力。

每年都有来自世界各地的学生参加该比赛，其中包括来自中小学的学生以及大学生。

滑铁卢数学竞赛分为两个阶段，第一阶段为全球性选拔赛，任何人都可以参加。

参赛者需要在线完成一套由滑铁卢大学编制的数学测试，题型涵盖代数、几何、组合数学等多个数学领域。

根据第一阶段的成绩，滑铁卢大学将选拔出前几百名成绩优异的参赛者晋级到第二阶段。

第二阶段为面试阶段，只有第一阶段晋级的学生才可以参加。

参赛者需要前往滑铁卢大学进行现场的笔试和面试。

笔试部分主要考察参赛者的数学基础知识和解题能力，而面试部分则更加注重参赛者的思维过程和解题思路。

面试时，学生需要与评委进行面对面的交流，展示自己的数学思考能力。

滑铁卢数学竞赛的题目通常非常有难度，涉及到一些高级数学概念和方法。

参赛者需要具备扎实的数学基础知识，并且具备独立思考和解决问题的能力。

竞赛的目的不仅是测试学生的数学水平，更重要的是培养他们解决问题的能力和数学思维方式。

参加滑铁卢数学竞赛对于学生来说是一次宝贵的经历。

这个竞赛可以提供一个展示自己数学才能的平台，也可以锻炼参赛者的思维能力和团队合作精神。

在竞赛中，学生们可以结识来自不同国家和地区的志同道合的数学爱好者，分享彼此的数学体验和解题方法。

滑铁卢数学竞赛也为参赛者提供了一些奖励和机会。

根据参赛者在竞赛中的表现，滑铁卢大学会为他们颁发证书和奖状，并且可以获得一些奖金和奖品。

此外，优秀的参赛者还有机会获得滑铁卢大学的奖学金和入学机会，为他们的未来发展开启了一扇大门。

总之，滑铁卢数学竞赛是一个非常有挑战性和有意义的数学竞赛活动。

通过参加这个竞赛，学生们可以提升自己的数学能力，拓展自己的数学视野，同时也能够展示自己的才能和潜力。

无论是对于中小学生还是大学生，参加滑铁卢数学竞赛都是一个值得鼓励和支持的选择。

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1.For real numbers a and b with a≥0and b≥0,the operation is deﬁned bya b=√a+4b.For example,5 1=5+4(1)=√9=3.(a)What is the value of8 7?(b)If16 n=10,what is the value of n?(c)Determine the value of(9 18) 10.(d)With justiﬁcation,determine all possible values of k such that k k=k.2.Each week,the MathTunes Music Store releases a list of the Top200songs.A newsong“Recursive Case”is released in time to make it onto the Week1list.The song’s position,P,on the list in a certain week,w,is given by the equation P=3w2−36w+110.The week number w is always a positive integer.(a)What position does the song have on week1?(b)Artists want their song to reach the best position possible.The closer that theposition of a song is to position#1,the better the position.(i)What is the best position that the song“Recursive Case”reaches?(ii)On what week does this song reach its best position?(c)What is the last week that“Recursive Case”appears on the Top200list?3.A pyramid ABCDE has a square base ABCD of side length 20.Vertex E lies on theline perpendicular to the base that passes through F ,the centre of the base ABCD .It is given that EA =EB =EC =ED =18.(a)Determine the surface area of the pyramidABCDEincluding its base.(b)Determine the height EF of the pyramid.A B C D EF 2018(c)G and H are the midpoints of ED and EA ,respectively.Determine the area of thequadrilateral BCGH .4.The triple ofpositive integers (x,y,z )is called an Almost Pythagorean Triple (or APT)if x >1and y >1and x 2+y 2=z 2+1.For example,(5,5,7)is an APT.(a)Determine the values of y and z so that (4,y,z )is an APT.(b)Prove that for any triangle whose side lengths form an APT,the area of thetriangle is not an integer.(c)Determine two 5-tuples (b,c,p,q,r )of positive integers with p ≥100for which(5t +p,bt +q,ct +r )is an APT for all positive integers t .1.At the JK Mall grand opening,some lucky shoppers are able to participate in a moneygiveaway.A large box has beenﬁlled with many$5,$10,$20,and$50bills.The lucky shopper reaches into the box and is allowed to pull out one handful of bills.(a)Rad pulls out at least two bills of each type and his total sum of money is$175.What is the total number of bills that Rad pulled out?(b)Sandy pulls out exactlyﬁve bills and notices that she has at least one bill of eachtype.What are the possible sums of money that Sandy could have?(c)Lino pulls out six or fewer bills and his total sum of money is$160.There areexactly four possibilities for the number of each type of bill that Lino could have.Determine these four possibilities.2.A parabola has equation y=(x−3)2+1.(a)What are the coordinates of the vertex of the parabola?(b)A new parabola is created by translating the original parabola3units to the leftand3units up.What is the equation of the translated parabola?(c)Determine the coordinates of the point of intersection of these two parabolas.(d)The parabola with equation y=ax2+4,a<0,touches the parabola withequation y=(x−3)2+1at exactly one point.Determine the value of a.3.A sequence of m P’s and n Q’s with m>n is called non-predictive if there is some pointin the sequence where the number of Q’s counted from the left is greater than or equal to the number of P’s counted from the left.For example,if m=5and n=2the sequence PPQQPPP is non-predictive because in counting theﬁrst four letters from the left,the number of Q’s is equal to the number of P’s.Also,the sequence QPPPQPP is non-predictive because in counting theﬁrst letterfrom the left,the number of Q’s is greater than the number ofP’s.(a)If m=7and n=2,determine the number of non-predictive sequences that beginwith P.(b)Suppose that n=2.Show that for every m>2,the number of non-predictivesequences that begin with P is equal to the number of non-predictive sequences that begin with Q.(c)Determine the number of non-predictive sequences with m=10and n=3.4.(a)Twenty cubes,each with edge length1cm,are placed together in4rows of5.What isthe surface area of this rectangularprism?(b)A number of cubes,each with edge length1cm,are arranged to form a rectangularprism having height1cm and a surface area of180cm2.Determine the number of cubes in the rectangular prism.(c)A number of cubes,each with edge length1cm,are arranged to form a rectangularprism having length l cm,width w cm,and thickness1cm.A frame is formed byremoving a rectangular prism with thickness 1cm located k cm from each of the sides of the original rectangular prism,as shown. Each of l,w and k is a positive integer.If the frame has surface area532cm2,determine all possible values for l and w such that l≥w.l cmw cmk cmk cmk cmk cm1 cm1.Quadrilateral P QRS is constructed with QR =51,as shown.The diagonals of P QRS intersect at 90◦at point T ,such that P T =32and QT =24.322451P QRST (a)Calculate the length of P Q.(b)Calculate the area of P QR .(c)If QS :P R =12:11,determine the perimeter of quadrilateral P QRS .2.(a)Determine the value of (a +b )2,given that a 2+b 2=24and ab =6.(b)If (x +y )2=13and x 2+y 2=7,determine the value of xy .(c)If j +k =6and j 2+k 2=52,determine the value of jk .(d)If m 2+n 2=12and m 4+n 4=136,determine all possible values of mn .3.(a)Points M (12,14)and N (n,n 2)lie on theparabola with equation y =x 2,as shown.Determine the value of n such that∠MON =90◦.yx(b)Points A (2,4)and B (b,b 2)are the endpointsofa chord of the parabola with equationy =x 2,as shown.Determine the value of bso that ∠ABO =90◦.y x(c)Right-angled triangle P QR is inscribed inthe parabola with equation y =x 2,asshown.Points P,Q and R have coordinates(p,p 2),(q,q 2)and (r,r 2),respectively.If p ,qand r are integers,show that 2q +p +r =0.y x4.The positive divisors of 21are 1,3,7and 21.Let S (n )be the sum of the positive divisors of the positive integer n .For example,S (21)=1+3+7+21=32.(a)If p is an odd prime integer,ﬁnd the value of p such that S (2p 2)=2613.(b)The consecutive integers 14and 15have the property that S (14)=S (15).Determine all pairs of consecutive integers m and n such that m =2p and n =9q for prime integers p,q >3,and S (m )=S (n ).(c)Determine the number of pairs of distinct prime integers p and q ,each less than 30,with the property that S (p 3q )is not divisible by 24.。