Fermat滑铁卢数学竞赛(Grade 11)-数学Mathematics-2002-试题 exam

2008Hypatia Contest(Grade11)Wednesday,April16,20081.For numbers a and b,the notation a∇b means2a+b2+ab.For example,1∇2=2(1)+22+(1)(2)=8.(a)Determine the value of3∇2.(b)If x∇(−1)=8,determine the value of x.(c)If4∇y=20,determine the two possible values of y.(d)If(w−2)∇w=14,determine all possible values of w.2.(a)Determine the equation of the line through the points A(7,8)and B(9,0).(b)Determine the coordinates of P,the point of intersection of the line y=2x−10and theline through A and B.(c)Is P closer to A or to B?Explain how you obtained your answer.3.In the diagram,ABCD is a trapezoid with AD parallel to BC andBC perpendicular to AB.Also,AD=6,AB=20,and BC=30.(a)Determine the area of trapezoid ABCD.(b)There is a point K on AB such that the area of KBCequals the area of quadrilateral KADC.Determine the length of BK.(c)There is a point M on DC such that the area of MBCequals the area of quadrilateral MBAD.Determine the length of MC.C4.The peizi-sum of a sequence a1,a2,a3,...,a n is formed by adding the products of all of thepairs of distinct terms in the sequence.For example,the peizi-sum of the sequence a1,a2,a3,a4 is a1a2+a1a3+a1a4+a2a3+a2a4+a3a4.(a)The peizi-sum of the sequence2,3,x,2x is−7.Determine the possible values of x.(b)A sequence has100terms.Of these terms,m are equal to1and n are equal to−1.Therest of the terms are equal to2.Determine,in terms of m and n,the number of pairs of distinct terms that have a product of1.(c)A sequence has100terms,with each term equal to either2or−1.Determine,withjustification,the minimum possible peizi-sum of the sequence.。

2007Galois Contest (Grade 10)Wednesday,April 18,20071.Jim shops at a strange fruit store.Instead of putting prices on each item,the mathematical store owner will answer questions about combinations of items.(a)In Aisle 1,Jim receives the following answers to his questions:Jim’s Question AnswerWhat is the sum of the prices of an Apple and a Cherry?62centsWhat is the sum of the prices of a Banana and a Cherry?66centsWhat is difference between the prices of an Apple and a Banana?Which has a higher price?Explain how you obtained your answer.(b)In Aisle 2,Jim receives the following answers to his questions:Jim’s Question AnswerWhat is the sum of the prices of a Mango and a Nectarine?60centsWhat is the sum of the prices of a Pear and a Nectarine?60centsWhat is the sum of the prices of a Mango and a Pear?68centsWhat is the price of a Pear?Explain how you obtained your answer.(c)In Aisle 3,Jim receives the following answers to his questions:Jim’s Question AnswerWhat is the sum of the prices of a Tangerine and a Lemon?60centsHowmuch more does a Tangerine cost than a Grapefruit?6centsWhat is the sum of the prices of Grapefruit,a Tangerine and a Lemon?94centsWhat is the price of a Lemon?Explain how you obtained youranswer.2.(a)In the diagram,what is the perimeter of the sector of the circle with radius 12?Explain how you obtained your answer.(b)Two sectors of a circle of radius 12are placed side by side,as shown.Determine the area of figure ABCD .Explain how you obtainedyour answer.A (c)In the diagram,AOB is a sector of a circle with ∠AOB =60◦.OY is drawn perpendicular to AB and intersects AB at X .What is the length of XY ?Explain how you obtained your answer.A O BX Y1212(d)See over...2007Galois Contest Page2(d)Two sectors of a circle of radius12overlap as shown.Determine the area of the shaded region.Explain how youobtained your answer.R3.(a)Each face of a5by5by5wooden cube is divided into1by1squares.Each square is painted black or white,asshown.Next,the cube is cut into1by1by1cubes.Howmany of these cubes have at least two painted faces?Explain how you obtained youranswer.(b)A(2k+1)by(2k+1)by(2k+1)cube,where k is a in thesame manner as the5by5by5cube with white squares in the corners.Again,the cube is cut into1by1by1cubes.i.In terms of k,how many of these cubes have exactly two white faces?Explain howyou obtained your answer.ii.Prove that there is no value of k for which the number of cubes having at least two white faces is2006.4.Jill has a container of small cylindrical rods in six different colours.Each colour of rod has adifferent length as summarized in the chart.Colour LengthGreen3cmPink4cmYellow5cmBlack7cmViolet8cmRed9cmThese rods can be attached together to form a pole.There are2ways to choose a set of yellow and green rods that will form a pole29cm in length: 8green rods and1yellow rod OR3green rods and4yellow rods.(a)How many different sets of yellow and green rods can be chosen that will form a pole62cm long?Explain how you obtained your answer.(b)Among the green,yellow,black and red rods,find,with justification,two colours for whichit is impossible to make a pole62cm in length using only rods of those two colours.(c)If at least81rodsof each of the colours green,pink,violet,and red must be used,howmany different sets of rods of these four colours can be chosen that will form a pole2007cm in length?Explain how you got your answer.。

滑铁卢竞赛数学题概述
滑铁卢竞赛数学题通常比较难，涉及的知识点广泛，包括代数、几何、数论、组合数学等多个领域。

以下是一些滑铁卢竞赛数学题的示例：
1. 有100个球，其中有一个与其他99个重量不同，但外观相同。

用一个天平，最少需要称多少次才能确定这个重量不同的球？
2. 一个正方形的面积为1，将其四边中点连接起来，形成另一个正方形。

如此重复，得到第五、第六个正方形，求第五个正方形的面积。

3. 一个圆被分成n个相等的扇形，其中一个是空心的，其他n-1个是实心的。

求空心扇形的圆心角是多少度？
4. 有100个人站成一排，从第1个人开始报数，每次报到奇数的人离开队伍。

经过若干轮后，只剩下一个人。

求这个人最初站在第几位？
5. 有5个不同质因数的最小正整数是多少？
以上仅是滑铁卢竞赛数学题的一些示例，实际上还有更多难题和技巧题。

如果想要深入了解滑铁卢竞赛数学题的解题技巧和策略，建议参考相关的竞赛书籍和资料，或者参加专业的数学竞赛培训课程。

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。

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.。

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。

滑铁卢数学竞赛1、21.|x|>3表示的区间是（）[单选题] *A.（-∞，3）B.(-3,3)C. [-3,3]D. （-∞，-3）∪（3，+ ∞）(正确答案)2、15.下列数中，是无理数的为（）[单选题] *A.-3.14B.6/11C.√3(正确答案)D.03、9.一棵树在离地5米处断裂，树顶落在离树根12米处，问树断之前有多高（）[单选题] *A. 17(正确答案)B. 17.5C. 18D. 204、若tan（π-α）>0且cosα>0，则角α的终边在（）[单选题] *A.第一象限B.第二象限C.第三象限D.第四象限(正确答案)5、下列说法正确的是[单选题] *A.带“+”号和带“-”号的数互为相反数B.数轴上原点两侧的两个点表示的数是相反数C.和一个点距离相等的两个点所表示的数一定互为相反数D.一个数前面添上“-”号即为原数的相反数(正确答案)6、16、在中,则( ). [单选题] *A. AB<2AC (正确答案)B. AB=2ACC. AB>2ACD. AB与2AC关系不确定7、260°是第（）象限角？[单选题] *第一象限第二象限第三象限(正确答案)第四象限8、4.已知两圆的半径分别为3㎝和4㎝，两个圆的圆心距为10㎝,则两圆的位置关系是（）[单选题] *Ａ.内切Ｂ.相交Ｃ.外切Ｄ.外离(正确答案)9、下列各角中，是界限角的是（）[单选题] *A. 1200°B. -1140°C. -1350°(正确答案)D. 1850°10、下列各对象可以组成集合的是（）[单选题] *A、与1非常接近的全体实数B、与2非常接近的全体实数(正确答案)C、高一年级视力比较好的同学D、与无理数相差很小的全体实数11、若2?＝a2＝4 ?，则a?等于( ) [单选题] *A. 43B. 82C. 83(正确答案)D. 4?12、47、若△ABC≌△DEF，AB＝2，AC＝4，且△DEF的周长为奇数，则EF的值为（）[单选题] *A．3B．4C．1或3D．3或5(正确答案)13、8. 下列事件中，不可能发生的事件是(? ? )．[单选题] *A．明天气温为30℃B．学校新调进一位女教师C．大伟身长丈八(正确答案)D．打开电视机，就看到广告14、3．中国是最早采用正负数表示相反意义的量，并进行负数运算的国家．若零上10℃记作+10℃，则零下10℃可记作（）[单选题] *A．10℃B．0℃C.-10 ℃(正确答案)D．-20℃15、19．对于实数a、b、c,“a>b”是“ac2(c平方)>bc2(c平方) ; ”的（）[单选题] * A．充分不必要条件B．必要不充分条件(正确答案)C．充要条件D．既不充分也不必要条件16、22．如图棋盘上有黑、白两色棋子若干，找出所有使三颗颜色相同的棋在同一直线上的直线，满足这种条件的直线共有（）[单选题] *A．5条(正确答案)B．4条C．3条D．2条17、f（x）=-2x+5在x=1处的函数值为（）[单选题] *A、-3B、-4C、5D、3(正确答案)18、10. 如图所示，小明周末到外婆家，走到十字路口处，记不清哪条路通往外婆家，那么他一次选对路的概率是(? ? ?)．[单选题] *A．1/2B．1/3(正确答案)C．1/4D．119、15．如图所示，下列数轴的画法正确的是（）[单选题] *A．B．C．(正确答案)D．20、-950°是（）[单选题] *A. 第一象限角B. 第二象限角(正确答案)C. 第三象限角D. 第四象限角21、1．如图，∠AOB＝120°，∠AOC＝∠BOC，OM平分∠BOC，则∠AOM的度数为（）[单选题] *A．45°B．65°C．75°(正确答案)D．80°22、27．下列计算正确的是（）[单选题] *A．（﹣a3）2＝a6(正确答案)B．3a+2b＝5abC．a6÷a3＝a2D．（a+b）2＝a2+b223、16．“x2（x平方）－4x－5=0”是“x=5”的( ) [单选题] *A．充分不必要条件B．必要不充分条件(正确答案)C．充要条件D．既不充分也不必要条件24、19．下列两个数互为相反数的是（）[单选题] *A．（﹣）和﹣（﹣）B．﹣5和(正确答案)C．π和﹣14D．+20和﹣（﹣20）25、13．在数轴上，下列四个数中离原点最近的数是（）[单选题] *A．﹣4(正确答案)B．3C．﹣2D．626、14．命题“?x∈R，?n∈N*，使得n≥x2（x平方）”的否定形式是（）[单选题] * A．?x∈R，?n∈N*，使得n<x2B．?x∈R，?x∈N*，使得n<x2C．?x∈R，?n∈N*，使得n<x2D．?x∈R，?n∈N*，使得n<x2(正确答案)27、8．(2020·课标Ⅱ)已知集合U={－2，－1,0,1,2,3}，A={－1,0,1}，B={1,2}，则?U(A∪B)=( ) [单选题] *A．{－2,3}(正确答案)B．{－2,2,3}C．{－2，－1,0,3}D．{－2，－1,0,2,3}28、7人小组选出2名同学作正副组长，共有选法（）种。

专题03 高考英语时文阅读热点主题-校园生活一、阅读理解（2022秋·江苏无锡·高三统考期末）AOn Thursday 24th March, the University of Waterloo will be hosting the 2022 Canadian mathematics competition!These are common multiple-choice exam-style contests. Students who enter will sit the exam during school time. The contests last for 60 minutes and attention: calculating devices are allowed.The Pascal contest — Intended for Grade 9 students, however younger students are allowed to enter it. Students below Grade 7 are generally not recommended.How much to pay?$40 per student, which takes into consideration the entrance fee per student, the extra registration fees as a non-Canadian school, as well as postage fees. The school finance department will automatically charge this fee to the parent’s account.How to register?To register, please follow the link below: https:// forms Pages Response Page. asps. The deadline of registration is April 29th. We do not accept late registration.If you have registered, you can try practicing some past papers by visiting the following websites:1．What’s the purpose of the text?A．To advertise the Canadian university of Waterloo.B．To introduce the Canadian mathematics competition.C．To call on students to enter the 2022 Canadian mathematics competition.D．To arrange for students to learn mathematics in Waterloo university.2．What is special about the mathematics contest?A．It is a multiple-choice exam1-styie contest.B．Calculating devices are allowed in its exam.C．It is designed specially for senior students.D．The cost is low for both Canadians and foreigners.3．Which of the following is the biggest advantage of the competition?A．It is useful in university application.B．It is friendly to students’ summer vacation.C．it helps students to challenge themselves in mathematics.D．It helps students to collect the past papers of the competition.（2022秋·广东深圳·高三统考期末）BWorking out which colleges to apply for can be one of the most difficult decisions any high schooler can face, but Alena got it all out of the way early as she’s only 13 and already has her medical school offer.The application process for the medical school was “super stressful”, but she was adamant to follow her dreams regardless of what anyone said, even if it was “no”. “If I could say anything to another little girl that looks like me: never stop believing and don’t give up on your dreams just because somebody tells you ‘no’. You can do anything that you put your mind and your heart to,” she said.Alena is the youngest African American person to get into a US medical school, and the second-youngest person overall.Alena didn’t just wake up one day and decide to go to the medical school while in middle school — she’s a child genius. By the time she was 11 she’d already taken several high school courses. When Alena turned 12, she graduated from high school and is currently enrolled (注册) in online classes at Arizona State University.If all goes to plan for Alena, she’ll start studying to be a doctor at age 15, in 2024, and graduate when she’s 18 when other kids her age are preparing for their freshmen year at college. Although she’s so young, age is “not a big deal” to her and she is just a normal kid.She explained, “I’m in honors choir, I’m in cheerleading and I play soccer. I hang out at the mall with my friends like a 13-year-old does, and I also go to the movies and go swimming with them. I’m very structured and very disciplined.”4．Why does the author mention the high schoolers’ college application?A．To show how fortunate Alena is.B．To explain high schoolers’ difficult life.C．To present preferences of high schoolers.D．To stress how unusual Alena is to get the offer.5．What does the underlined word “adamant” in paragraph 2 mean?A．Determined.B．Qualified.C．Innocent.D．Patient.6．What can be inferred about Alena?A．She is now studying in a high school.B．She is taking online university classes.C．She will become a doctor at the age of 15.D．She is the youngest to get into a US medical school.7．Which of the following can best describe Alena’s life?A．Stressful but hopeful.B．Cheerful but meaningless.C．Rich and colorful.D．Simple and free.（2022秋·安徽六安·高三统考期末）COnline classes began to be popularized just a few decades ago. They are advertised as a way for adults to finish their education and students to learn the material at their own pace—it is far more suitable for people with busy schedules.But after being enrolled in an online course last fall semester, I came to realize online classes were merely a means to fulfil course requirements.First of all, students lack the desire to learn, and they simply complete their assignments to receive credit for a passing grade rather than genuinely engage with the course material.As online courses tend to have more than 100 students, most of the assignments are short and simple. They are not designed for students to interact with the material in depth but designed to be graded easily to accommodate such a large number of students.Perhaps the biggest disadvantage of taking an online class is the absence of face-to-face interaction between the teacher and their students. Live sessions are infrequent and are often scheduled during the middle of the day when students have to attend other classes or work. The office hours of the professor may also be during inconvenient times for many students as well. Most interaction with the professor has to be through email which is often impersonal. It is nearly impossible for students to build a relationship with their professor.There is also little interaction among students. It can be harder for students to create study groups and form relationships with their peers.Online classes also require either a computer or laptop and a reliable Internet connection. Not all students have access to these types of resources, whether it is for financial or other reasons, and some students can be put at a disadvantage. Offering online classes certainly helps students who would otherwise not be able to attend classroom sessions. However, they fail to provide a true education with an emphasis on convenience rather than critical thinking. We need restructure online classes in which students can have a learning experience that will actually provide quality education.8．What can we infer about students enrolled in online classes?A．They are unmotivated to learn,B．They can access course materials easily.C．They rarely fulfil the course requirements.D．They can learn at their own pace.9．What is the author’s opinion about online course assignments? ‘A．They are meant to facilitate interaction.B．They are made convenient to mark.C．They are given to accommodate students` needs.D．They are based on easily accessible material.10．From the passage, which is one disadvantage of online classes?A．They make professors’ offices much less accessible.B．They are frequently scheduled at irregular times.C．They provide little chance for students to build relationships with each other.D．They tend to increase professors’ burden of replying students’ emails.11．What problem may arise if classes go online?A．Teachers will worry about poor Internet connections.B．Most students may get critical thinking skills from it.C．Schools with limited resources will be at a disadvantage.D．Some students may have difficulty attending them.（2022秋·江苏南通·高三统考期末）DIn the coming era of budget cuts to education, distance learning could become a common thing.The appeal to those in charge of education budgets to trade teachers for technology is so strong that they tend to ignore the disadvantages of distance learning. School facilities are expensive to build and maintain, and teachers are expensive to employ. It’s true that online classes do not require buildings and each class can host hundreds of people, which can result in greater savings, but moving away from a traditional classroom in which a living, breathing human being teaches and interacts with students daily would be a disaster.Physically attending school has hidden benefits: getting up every morning interacting with peers, and building relationships with teachers are important skills to cultivate in young people. Moreover, schools are more than simple places of traditional learning. They are also places that provide meals, places where students receivemental help and other support.Those policy-makers are often fascinated by the latest technology in education and its potential to transform education overnight. But online education does not allow a teacher to keep a struggling student after class and offer help. Educational videos may deliver academic content, but they are unable to make eye contact or assess a student’s level of engagement. Distance education will never match the personal teaching in a traditional classroom. In their first 18 years of life, American children spend only 9% of their time in school. Yet teachers are expected to prepare them to be responsible citizens, cultivate their social skills, encourage successful time management, and enhance their capacity to compete in a competitive job market. Given these expectations, schools should not become permanently “remote”.The power of the classroom is rooted in the qualities of the people gathered in the same place, at the same time, including their nature, empathy, devotion and so on. Technology, no matter how advanced, should simply be a tool of a good teacher.12．What is one possible benefit of students attending school physically?A．Forming the habit of getting up early.B．Eating nutritionally well-balanced meals.C．Growing into living and breathing human beings.D．Developing relationships with peers and teachers.13．What does the author think of the latest technology in education?A．It may reduce face-to-face interaction.B．It may make many teachers jobless.C．It may add to student’s financial burden.D．It may revolutionize classroom teaching.14．What does the author say teachers are expected to do?A．To enhance student’s leadership capacity.B．To teach students skills in applying for jobs.C．To enable students to adapt to the changes in life.D．To prepare students to be competitive in the future.15．Why couldn’t technology replace a good teacher?A．It lacks humanity.B．It can’t meet personal needs.C．It is still not advanced.D．It can’t track students’ growth.二、七选五（2023·山东枣庄·统考一模）It’s the start of a new academic year. There’s every reason to go back to school—or at least acquire a new skill.Start with art. If you’ve always fancied yourself as a bit of a Picasso, consider joining an art class. As learning anything new in a group, you’ll meet like-minded people and boost your self-esteem. ___16___ Research suggests that it has a positive effect on body image, especially for women.Speak a foreign language. Just back from holiday and feeling ashamed of your halting attempts to speak Spanish to waiters? ___17___ Learning a foreign language can keep your brain in trim. There was still a benefit even if you took up a foreign tongue later in life.Take photos to feel good. Enrolling on a course to take better pictures could improve your mental health. Taking a photo a day and posting it online is beneficial. ___18___Know how to knit. It’s sociable, calming and might help you to avoid mild cognitive impairment. ___19___ It also help distract people from chronic pain, according to Knit for Peace, an initiative from the Charities Advisory Trust.___20___ Research from the University of California Riverside found that when older people learned several skills at a time, their cognitive abilities increased to the level of people 30 years younger.A．It’s never too late to learn a new skill.B．It’s even been shown to lower blood pressure.C．And think about learning several new things at once.D．In that case, you cater to others and build up your strength.E．Trying your hand at life drawing will help your motor skills.F．Brushing up your language skills could be just what you need.G．Actually, it helps people to connect with others, and see the world differently.三、完形填（2022秋·广东·高三校联考阶段练习）When I saw a name I’ve not seen for three years next to the words “Teacher of the Year” on Wednesday night, I wasn’t at all ____21____ . The woman who helped me achieve my dreams of working as a ____22____ had finally got the ____23____ she deserved.After graduating from Miss Chase’s English and media classes in 2018, one year later her ____24____ helped me get on an apprenticeship (学徒工作) course for the BBC website I’m ____25____ for now as my job. If I wasn’t taught by the teacher of the year, who knows where I’d be today, as I was such a timid teen.Meeting Miss Chase for the first time as a very ____26____ 11-year-old at Newport’s Bassaleg School, it was clear she was the type of teacher that always went above and beyond. She ____27____ against exam boards when she thought I deserved a ____28____ score and she told me “You can do it” when the ____29____ of exams got too much. Also, giving me probably too many extensions for my media coursework ____30____ showed her generosity and patience. It’s rare having a teacher that ____31____ such a lasting impact on you after school.Even when pupils have left school, Miss Chase doesn’t ____32____ them. My brother Brandon, who is now in Year 12 at Bassaleg School, ____33____ comes home and says:“Miss Chase was asking about you today. She said she saw you on the BBC again.”For Miss Chase, developing ____34____ good relationships with her pupils, as well as “making them feel safe and making them feel like they can be ____35____ ” is what it takes to deserve this award. 21．A．excited B．frightened C．interested D．surprised22．A．scientist B．journalist C．teacher D．director23．A．money B．promotion C．recognition D．return24．A．care B．reference C．knowledge D．choice25．A．showing B．keeping C．writing D．recording26．A．adorable B．unreasonable C．fantastic D．shy27．A．played B．leaned C．appealed D．guarded28．A．higher B．lower C．fuller D．fewer29．A．praise B．pressure C．control D．honour30．A．deadline B．milestone C．keyboard D．headache31．A．engages B．falls C．influences D．leaves32．A．help B．forget C．contact D．find33．A．initially B．warmly C．probably D．frequently34．A．really B．hardly C．quickly D．simply35．A．artful B．energetic C．successful D．wise四、语法填空（2022秋·山东泰安·高三泰安一中统考期中）阅读下面短文，在空白处填入1个适当单词或括号内单词的正确形式。

Chartered Accountants SybaseInc. (Waterloo) IBMCanada Ltd.Canadian Institute of ActuariesDo not open the contest booklet until you are told to do so.You may use rulers, compasses and paper for rough work.Calculators are permitted, providing they are non-programmable and without graphic displays.Part A: Each question is worth 5 credits.1.The value of 03012..()+ is(A ) 0.7(B ) 1(C ) 0.1(D ) 0.19(E ) 0.1092.The pie chart shows a percentage breakdown of 1000 votesin a student election. How many votes did Sue receive?(A ) 550(B ) 350(C ) 330(D ) 450(E ) 9353.The expression a a a 9153× is equal to(A ) a 45(B ) a 8(C ) a 18(D ) a 14(E ) a 214.The product of two positive integers p and q is 100. What is the largest possible value of p q +?(A ) 52(B ) 101(C ) 20(D ) 29(E ) 255.In the diagram, ABCD is a rectangle with DC =12. If the area of triangle BDC is 30, what is the perimeter ofrectangle ABCD ?(A ) 34(B ) 44(C ) 30(D ) 29(E ) 606.If x =2 is a solution of the equation qx –311=, the value of q is (A ) 4(B ) 7(C ) 14(D ) –7(E ) –47.In the diagram, AB is parallel to CD . What is the value ofy ?(A ) 75(B ) 40(C ) 35(D ) 55(E ) 508.The vertices of a triangle have coordinates 11,(), 71,() and 53,(). What is the area of this triangle?(A ) 12(B ) 8(C ) 6(D ) 7(E ) 99.The number in an unshaded square is obtained by adding thenumbers connected to it from the row above. (The ‘11’ is one such number.) The value of x must be (A ) 4(B ) 6(C ) 9(D ) 15(E) 10Scoring:There is no penalty for an incorrect answer.Each unanswered question is worth 2 credits, to a maximum of 20 credits.A BCD DAC B10.The sum of the digits of a five-digit positive integer is 2. (A five-digit integer cannot start with zero.)The number of such integers is(A ) 1(B ) 2(C ) 3(D ) 4(E ) 5Part B: Each question is worth 6 credits.11.If x y z ++=25, x y +=19 and y z +=18, then y equals(A ) 13(B ) 17(C ) 12(D ) 6(E ) –612. A regular pentagon with centre C is shown. The value of xis(A ) 144(B ) 150(C ) 120(D ) 108(E ) 7213.If the surface area of a cube is 54, what is its volume?(A ) 36(B ) 9(C ) 8138(D ) 27(E ) 162614.The number of solutions x y ,() of the equation 3100x y +=, where x and y are positive integers, is(A ) 33(B ) 35(C ) 100(D ) 101(E ) 9715.If y –55= and 28x =, then x y + equals(A ) 13(B ) 28(C ) 3316.Rectangle ABCDhas length 9 and width 5. Diagonal is divided into 5 equal parts at W , X , Y , and Z area of the shaded region.(A ) 36(B ) 365(C ) 18(D ) 41065(E ) 2106517.If N p q =()()()+75243 is a perfect cube, where p and q are positive integers, the smallest possible valueof p q + is(A ) 5(B ) 2(C ) 8(D ) 6(E ) 1218.Q is the point of intersection of the diagonals of one face ofa cube whose edges have length 2 units. The length of QRis(A ) 2(B ) 8(C ) 5(D ) 12(E ) 619.Mr. Anderson has more than 25 students in his class. He has more than 2 but fewer than 10 boys andmore than 14 but fewer than 23 girls in his class. How many different class sizes would satisfy these conditions?(A ) 5(B ) 6(C ) 7(D ) 3(E ) 420.Each side of square ABCD is 8. A circle is drawn through A and D so that it is tangent to BC . What is the radius of thiscircle?(A ) 4(B ) 5(C ) 6(D ) 42(E ) 5.25Part C: Each question is worth 8 credits.21.When Betty substitutes x =1 into the expression ax x c 32–+ its value is –5. When she substitutesx =4 the expression has value 52. One value of x that makes the expression equal to zero is(A ) 2(B ) 52(C ) 3(D ) 72(E ) 422. A wheel of radius 8 rolls along the diameter of a semicircleof radius 25 until it bumps into this semicircle. What is thelength of the portion of the diameter that cannot be touchedby the wheel?(A ) 8(B ) 12(C ) 15(D ) 17(E ) 2023.There are four unequal, positive integers a , b , c , and N such that N a b c =++535. It is also true thatN a b c =++454 and N is between 131 and 150. What is the value of a b c ++?(A ) 13(B ) 17(C ) 22(D ) 33(E ) 3624.Three rugs have a combined area of 2002m . By overlapping the rugs to cover a floor area of 1402m ,the area which is covered by exactly two layers of rug is 242m . What area of floor is covered by three layers of rug?(A ) 122m (B ) 182m (C ) 242m (D) 362m (E ) 422m 25.One way to pack a 100 by 100 square with 10000 circles, each of diameter 1, is to put them in 100rows with 100 circles in each row. If the circles are repacked so that the centres of any three tangent circles form an equilateral triangle, what is the maximum number of additional circles that can be packed?(A ) 647(B ) 1442(C ) 1343(D) 1443(E ) 1344。

滑铁卢数学竞赛年1月10日更新这篇主要来介绍一下滑铁卢系列数学竞赛，下面是该项赛事的特点以及备考建议：【特点】（1）全年龄段，从7年级到12年级都有，建议参加12年级Euclid 竞赛；（2）全球统考，高含金量，对于申请有帮助；（2）有些比赛项目并不是选择题，是填空、简答题，对于英语表达有一定要求；（3）难度相对比较小，获个奖比较容易；【建议】做真题！做真题！做真题！真题网址：下面是Waterloo数学竞赛的详细介绍以及21年Euclid真题卷：滑铁卢大学始建于1957年，在加拿大最权威的教育杂志Maclean's （麦克林）的排名榜上，连续五年综合排名第一第二。

滑铁卢大学设有加拿大唯一一所数学学院，这也是北美乃至全世界最大的数学学院，因滑铁卢大学在数学领域的优良声誉及传统，以及欧几里德数学竞赛考察标准的严格性和专业性，该竞赛成绩在加拿大和美国大学中已经得到广泛认可，被誉为类似加拿大“数学托福”的考试。

官方网址：【比赛特点】（1）可选择的比赛种类非常多，有适合各种不同年龄段学生的比赛；（2）比赛难度相对较小，比较适合想参加数学类竞赛，但本身数学程度并不是特别出挑的学生；（3）在加拿大的高校中有比较大的影响力，尤其是Euclid的比赛成绩；（4）部分赛题以测试学生的分析能力、逻辑思维能力为主，在2021年MAT考试中出现了21年Euclid的类似试题。

【比赛形式】注：上述数学竞赛都可以使用计算器。

【2021-2022赛程】大家可以看到Waterloo数学竞赛各项赛事跨度比较大，主要集中在四、五月份，特别是Euclid数学竞赛，是一项含金量比较高的全球数学竞赛。

【报名方式】当地可提供报名的机构或以自己学校的名义注册报名注：Waterloo报名时需要填写一个学校代码，所以一般需要通过学校报名。

【奖项设置】（1）国际生全球排名前25%的学生将获得杰出荣誉证书，2021年大概68分；（2）在参赛学校中成绩最高的学生会得到一个校级冠军的奖牌；（3）每位参赛者都可以获得一个参赛证书；从上述奖项的设置可以看出，我们参加Waterloo数学竞赛获一个奖相对而言是比较容易的。

滑铁卢数学竞赛高中试题一、选择题1. 已知函数\( f(x) = ax^2 + bx + c \)，其中\( a, b, c \)为实数，且\( f(1) = 2 \)，\( f(-1) = 0 \)，\( f(2) = 6 \)。

求\( a \)的值。

2. 一个圆的半径为5，圆心位于原点，求圆上点\( P(3,4) \)到圆心的距离。

3. 若\( \sin(\alpha + \beta) = \frac{1}{2} \)，\( \cos(\alpha + \beta) = \frac{\sqrt{3}}{2} \)，且\( \alpha \)在第二象限，\( \beta \)在第一象限，求\( \sin(\alpha) \)的值。

二、填空题1. 计算\( \int_{0}^{1} x^2 dx \)。

2. 若\( \log_{2}8 = n \)，则\( n \)的值为______。

3. 一个等差数列的前三项分别为2，5，8，求该数列的第10项。

三、解答题1. 证明：对于任意正整数\( n \)，\( 1^3 + 2^3 + ... + n^3 =\frac{n^2(n+1)^2}{4} \)。

2. 一个矩形的长是宽的两倍，若矩形的周长为24，求矩形的面积。

3. 已知一个等比数列的前三项分别为3，9，27，求该数列的第5项。

四、应用题1. 一个工厂每天生产相同数量的零件，如果每天生产100个零件，工厂可以在30天内完成订单。

如果每天生产150个零件，工厂可以在20天内完成订单。

求工厂每天实际生产的零件数量。

2. 一个圆环的外圆半径是内圆半径的两倍，且圆环的面积为π。

求外圆的半径。

五、证明题1. 证明：对于任意实数\( x \)，\( \cos(x) + \cos(2x) + \cos(3x) \)可以表示为一个单一的余弦函数。

六、开放性问题1. 考虑一个无限大的棋盘，每个格子可以放置一个硬币。

滑铁卢数学竞赛滑铁卢数学竞赛是加拿大一项著名的数学竞赛活动，每年都吸引了许多有志于挑战自己数学能力的学生参加。

它的历史可以追溯到1967年，从那以后，滑铁卢数学竞赛已经成为了全球最重要的数学竞赛之一。

滑铁卢数学竞赛分为不同的级别，包括高中水平的离散数学竞赛和全国高中生数学竞赛等。

它不仅仅考察了学生的计算能力，更注重培养学生的数学思维能力和解决问题的能力。

滑铁卢数学竞赛的题目涵盖了数学的各个领域，如代数、几何、数论和组合数学等。

这些题目往往以形式化和抽象的方式出现，需要参赛学生进行深入的分析和推理。

竞赛要求学生在有限的时间内回答一系列问题，并用严谨的数学推导来解决问题。

滑铁卢数学竞赛的题目难度非常高，需要具备扎实的数学基础和高超的解题能力。

参赛学生需要理解问题的本质，并能够找到解题的关键步骤。

在竞赛中，学生往往需要面对复杂的数学问题，需要运用各种数学知识和技巧来解答。

参加滑铁卢数学竞赛对于学生的数学能力和素质有着很高的要求。

它不仅考察了学生的记忆和计算能力，更注重培养学生的数学思维和解决问题的能力。

通过参加竞赛，学生可以提高自己的数学水平，拓宽数学思维的广度和深度。

滑铁卢数学竞赛为学生提供了一个展示自己数学才能的舞台。

通过竞赛，学生可以与其他优秀的数学爱好者交流和切磋，共同进步。

竞赛的结果不仅是学生们的荣誉和成绩，更是他们自信心的提升和未来学习、发展的动力。

作为一项重要的学科竞赛，滑铁卢数学竞赛不仅在加拿大，而且在国际上都享有较高的声誉。

许多优秀的数学家和科学家都曾经参加过滑铁卢数学竞赛，这为他们日后的学术研究和职业发展奠定了坚实的基础。

总之，滑铁卢数学竞赛是一个激励学生充分发挥数学潜能、提高数学能力和培养创新思维的重要平台。

通过参加竞赛，学生可以不断挑战自我，锻炼解决问题的能力，为未来的学习和职业发展奠定良好的基础。

滑铁卢数学竞赛的成功举办，为数学教育和科学研究做出了重要贡献。

2014-2012加拿⼤滑铁卢⼤学11年级数学竞赛试题1.For real numbers a and b with a≥0and b≥0,the operation is de?ned bya b=√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 justi?cation,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 been?lled 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 exactly?ve 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 the?rst 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 the? rst 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,?nd 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.。

十一届数学竞赛试题及答案在十一届数学竞赛中，我们设计了一系列试题，旨在测试参赛者的数学知识和解题技巧。

以下是部分试题及答案。

试题一：已知函数\( f(x) = 3x^2 - 2x + 1 \)，求该函数在区间[1, 3]上的最大值和最小值。

答案一：首先，我们可以通过求导数来找到函数的极值点。

函数\( f(x) \)的导数为\( f'(x) = 6x - 2 \)。

令\( f'(x) = 0 \)，解得\( x =\frac{1}{3} \)。

然而，\( x = \frac{1}{3} \)不在区间[1, 3]内，因此我们需要检查区间端点的函数值。

计算\( f(1) = 3(1)^2 - 2(1) + 1 = 2 \)和\( f(3) = 3(3)^2 - 2(3) + 1 = 22 \)。

由于\( f(x) \)是一个二次函数，且二次项系数为正，函数开口向上，因此\( f(1) \)是最小值，\( f(3) \)是最大值。

所以，最大值为22，最小值为2。

试题二：解不等式\( |x - 3| + |x + 1| \geq 4 \)。

答案二：我们可以将不等式分为三个区间来解决：1. 当\( x < -1 \)时，不等式变为\( -(x - 3) - (x + 1) \geq 4 \)，简化后得到\( -2x + 2 \geq 4 \)，即\( x \leq -1 \)。

2. 当\( -1 \leq x < 3 \)时，不等式变为\( -(x - 3) + (x + 1)\geq 4 \)，简化后得到\( 4 \geq 4 \)，这是一个恒成立的不等式。

3. 当\( x \geq 3 \)时，不等式变为\( (x - 3) + (x + 1) \geq 4 \)，简化后得到\( 2x - 2 \geq 4 \)，即\( x \geq 3 \)。

综合以上三个区间，解集为\( x \in (-\infty, -1] \cup [3,+\infty) \)。

加拿大滑铁卢大学举办费马国际数学竞赛及欧几里得数学竞赛。

滑铁卢大学是加拿大综合排名第三的大学，其创新精神在加拿大排名第一，该大学面向青少年举办的数学竞赛，在全球具有影响力。

费马国际数学竞赛面向高一学生的费马数学竞赛，考试时间60为分钟，满分为150分，其间学生必须面对全英语试卷解答问题。

欧几里得数学竞赛据360教育集团介绍，面向高二学生的欧几里得数学竞赛，考试时间150为分钟，满分为100分，其间学生必须面对全英语试卷解答问题。

最初的数学考试是由安大略省西南部的几个高中老师联合创办的，从六十年代初年每年300人参加考试到今天，累计已经有21万名学生参加了这个考试。

根据滑铁卢大学的校方统计资料：21万名学生中有40%是来自安大略省的学生，20%是来自英属哥伦比亚省的学生，35%是来自加拿大其他省份的，还有5%是来自国际学生，包括美国、英国、中国等世界各国的学生。

2003年因为安大略省取消13年级，部分涉及微积分的试题不再使用，于是将迪卡尔（法国著名数学家）数学竞赛（DescartesContest）更名为（欧几里德数学竞赛）。

现在，欧几里德数学竞赛的分数已经成为Waterloo数学学院各专业以及“软件工程”专业入学录取的重要指标，更成为学生申请该学院奖学金的重要考核标准。

欧几里德数学竞赛（EuclidContest）主要是为高二年级（加拿大11年级）的高中学生提供的考试，考试内容主要包括：代数（函数、三角、排列、组合）、平面组合、解析几何等，他不仅仅看的是结果，更看重的是学生的解题思路和技巧。

考试的及格分数每年大概在40分左右。

因滑铁卢大学在数学领域的优良声誉及传统，以及欧几里德数学竞赛考察标准的严格性和专业性，该竞赛成绩在加拿大大学中已经得到广泛认可，被誉为类似加拿大“数学托福”的考试。

滑铁卢大学数学竞赛。

1.(a)In Carrotford,candidate A ran for mayor and received 1008votes out of a totalof 5600votes.What percentage of all votes did candidate A receive?(b)In Beetland,exactly three candidates,B,C and D,ran for mayor.Candidate Bwon the election by receiving 35of all votes,while candidates C and D tied withthe same number of votes.What percentage of all votes did candidate C receive?(c)In Cabbagetown,exactly two candidates,E and F,ran for mayor and 6000votes were cast.At 10:00p.m.,only 90%of these votes had been counted.Candidate E received 53%of those votes.How many more votes had been countedfor candidate E than for candidate F at 10:00p.m.?(d)In Peaville,exactly three candidates,G,H and J,ran for mayor.When all of thevotes were counted,G had received 2000votes,H had received 40%of the votes,and J had received 35%of the votes.How many votes did candidate H receive?2.The prime factorization of 144is 2×2×2×2×3×3or 24×32.Therefore,144is a perfect square because it can be written in the form (22×3)×(22×3).The prime factorization of 45is 32×5.Therefore,45is not a perfect square,but 45×5is a perfect square,because 45×5=32×52=(3×5)×(3×5).(a)Determine the prime factorization of 112.(b)The product 112×u is a perfect square.If u is a positive integer,what is thesmallest possible value of u ?(c)The product 5632×v is a perfect square.If v is a positive integer,what is thesmallest possible value of v ?(d)A perfect cube is an integer that can be written in the form n 3,where n is aninteger.For example,8is a perfect cube since 8=23.The product 112×w is aperfect cube.If w is a positive integer,what is the smallest possible value of w ?3.The positive integers are arranged in rows and columns,as shown,and described below.A B C D E F GRow1123456Row2121110987Row3131415161718Row4242322212019...The odd numbered rows list six positive integers in order from left to right beginning in column B.The even numbered rows list six positive integers in order from right toleft beginning in columnF.(a)Determine the largest integer in row30.(b)Determine the sum of the six integers in row2012.(c)Determine the row and column in which the integer5000appears.(d)For how many rows is the sum of the six integers in the row greater than10000and less than20000?4.The volume of a cylinder with radius r and height h equalsπr2h.The volume of a sphere with radius r equals43πr3.(a)The diagram shows a sphere thatfits exactlyinside a cylinder.That is,the top and bottomfaces of the cylinder touch the sphere,and thecylinder and the sphere have the same radius,r.State an equation relating the height of thecylinder,h,to the radius of the sphere,r.(b)Forthe cylinder and sphere given in part(a),determine the volume of the cylinder if the volume of the sphere is288π.(c)A solid cube with edges of length1km isfixed in outer space.Darla,the babyspace ant,travels on this cube and in the space around(but not inside)this cube.If Darla is allowed to travel no farther than1km from the nearest point on the cube,then determine the total volume of space that Darla can occupy.Fryer Contest(English) 2012。

滑铁卢数学竞赛滑铁卢数学竞赛是一项受广泛关注的数学竞赛活动。

作为加拿大著名的数学竞赛之一，滑铁卢数学竞赛每年吸引了来自全国各地高中生的参与。

本文将介绍滑铁卢数学竞赛的背景、组织方式、题型以及参赛经验等方面内容。

滑铁卢数学竞赛由滑铁卢大学主办，旨在鼓励和推广数学学科的学习和研究。

该竞赛已经举办了多年，吸引了大量对数学感兴趣的学生参与。

这项竞赛不仅考察学生的数学知识和解题能力，还培养了学生的逻辑思维、问题解决和团队协作能力。

滑铁卢数学竞赛一般分为初赛和决赛两个阶段。

初赛通常在本校进行，学生需接受一场笔试考试。

考试题目涵盖了数学的各个领域，包括代数、几何、概率与统计等。

考试时间为3个小时，学生需要在规定时间内回答一系列选择题和解答题。

初赛结束后，滑铁卢数学竞赛组织方将选取成绩优秀的学生进入决赛。

决赛通常在滑铁卢大学校园内举行，参赛学生将进行更加复杂和综合的数学题目的解答。

决赛题目的难度较高，往往需要学生灵活运用各种数学方法和技巧来解答。

参加滑铁卢数学竞赛对学生而言是一次宝贵的学习和锻炼机会。

不仅可以巩固和应用所学的数学知识，还能提高逻辑思维和解决问题的能力。

同时，参与竞赛还有利于学生将所学的数学知识与实际问题相结合，培养创新和发散思维能力。

在准备滑铁卢数学竞赛时，学生需要进行系统的复习和训练。

可以通过参加数学讲座和培训班来提高数学素养和解题能力。

此外，参考往年的竞赛试题，分析解题思路和方法也是非常重要的。

通过多做题、多联系，不断提升自己的数学水平和解题技巧。

总之，滑铁卢数学竞赛是一项受广泛关注的数学竞赛活动。

通过参与竞赛，学生可以提高数学素养、锻炼解题能力，并为将来的学习和科研打下坚实的数学基础。

希望更多的学生参与到这项活动中来，探索数学的魅力，享受数学竞赛的乐趣。