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

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Hypatia滑铁卢数学竞赛(Grade 11)-数学Mathematics-2008-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.。

Galois滑铁卢数学竞赛(Grade 10)-数学Mathematics-2007-试题 exam

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

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。

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。

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.。

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。

滑铁卢数学竞赛

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。

高考英语时文阅读热点主题题型组合练：专题03 校园生活 (全国通用)

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。

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Canadian Instituteof Actuaries Chartered AccountantsSybasei Anywhere SolutionsScoring:There is no penalty for an incorrect answer.Each unanswered question is worth 2, to a maximum of 10 unanswered questions.Part A: Each correct answer is worth 5.1.If x =3, the numerical value of 522–x is(A ) –1(B ) 27(C ) –13(D )–31(E ) 32.332232++ is equal to(A ) 3(B ) 6(C ) 2(D )32(E ) 53.If it is now 9:04 a.m., in 56 hours the time will be(A ) 9:04 a.m.(B ) 5:04 p.m.(C ) 5:04 a.m.(D ) 1:04 p.m.(E ) 1:04 a.m.4.Which one of the following statements is not true?(A ) 25 is a perfect square.(B ) 31 is a prime number.(C ) 3 is the smallest prime number.(D ) 8 is a perfect cube.(E ) 15 is the product of two prime numbers.5. A rectangular picture of Pierre de Fermat, measuring 20 cmby 40 cm, is positioned as shown on a rectangular postermeasuring 50 cm by 100 cm. What percentage of the areaof the poster is covered by the picture?(A ) 24%(B ) 16%(C ) 20%(D ) 25%(E ) 40%6.Gisa is taller than Henry but shorter than Justina. Ivan is taller than Katie but shorter than Gisa. Thetallest of these five people is(A ) Gisa (B ) Henry (C ) Ivan (D ) Justina (E ) Katie7. A rectangle is divided into four smaller rectangles. Theareas of three of these rectangles are 6, 15 and 25, as shown.The area of the shaded rectangle is(A ) 7(B ) 15(C ) 12(D ) 16(E) 108.In the diagram, ABCD and DEFG are squares with equal side lengths, and ∠=°DCE 70. The value of y is (A ) 120(B ) 160(C ) 130(D ) 110(E ) 1409.The numbers 1 through 20 are written on twenty golf balls, with one number on each ball. The golfballs are placed in a box, and one ball is drawn at random. If each ball is equally likely to be drawn,what is the probability that the number on the golf ball drawn is a multiple of 3?(A )320(B )620(C )1020(D )520(E )12010.ABCD is a square with AB x =+16 and BC x =3, as shown.The perimeter of ABCD is(A ) 16(B ) 32(C ) 96(D ) 48(E ) 24Part B: Each correct answer is worth 6.11. A line passing through the points 02,−() and 10,() also passes through the point 7,b (). The numericalvalue of b is(A ) 12(B )92(C ) 10(D ) 5(E ) 1412.How many three-digit positive integers are perfect squares?(A ) 23(B ) 22(C ) 21(D ) 20(E ) 1913. A “double-single” number is a three-digit number made up of two identical digits followed by adifferent digit. For example, 553 is a double-single number. How many double-single numbers are there between 100 and 1000?(A ) 81(B ) 18(C ) 72(D ) 64(E ) 9014.The natural numbers from 1 to 2100 are entered sequentially in 7 columns, with the first 3 rows asshown. The number 2002 occurs in column m and row n . The value of m n + isColumn 1Column 2Column 3Column 4Column 5Column 6Column 7Row 1 1 2 3 4 5 6 7Row 2 8 91011121314Row 315161718192021M M M M M M M M(A ) 290(B ) 291(C ) 292(D ) 293(E ) 294x + 163xA BD C15.In a sequence of positive numbers, each term after the first two terms is the sum of all of the previousterms . If the first term is a ,the second term is 2, and the sixth term is 56, then the value of a is(A ) 1(B ) 2(C ) 3(D ) 4(E ) 516.If ac ad bc bd +++=68 and c d +=4, what is the value of a b c d +++?(A ) 17(B ) 85(C ) 4(D ) 21(E ) 6417.The average age of a group of 140 people is 24. If the average age of the males in the group is 21 andthe average age of the females is 28, how many females are in the group?(A ) 90(B ) 80(C ) 70(D ) 60(E ) 5018. A rectangular piece of paper AECD has dimensions 8 cm by 11 cm. Corner E is folded onto point F , which lies on DC ,as shown. The perimeter of trapezoid ABCD is closest to (A ) 33.3 cm (B ) 30.3 cm (C ) 30.0 cm(D ) 41.3 cm (E ) 35.6 cm 19.If 238610a b =(), where a and b are integers, then b a − equals(A ) 0(B ) 23(C )−13(D )−7(E )−320.In the diagram, YQZC is a rectangle with YC =8 and CZ = 15. Equilateral triangles ABC and PQR , each withside length 9, are positioned as shown with R and B on sidesYQ and CZ , respectively. The length of AP is (A ) 10(B )117(C ) 9(D ) 8(E )72Part C: Each correct answer is worth 8.21.If 31537521219⋅⋅⋅⋅+−=L n n , then the value of n is(A ) 38(B ) 1(C ) 40(D ) 4(E ) 3922.The function f x () has the property that f x y f x f y xy +()=()+()+2, for all positive integers x and y .If f 14()=, then the numerical value of f 8() is(A ) 72(B ) 84(C ) 88(D ) 64(E ) 80continued ...Figure 1Figure 223.The integers from 1 to 9 are listed on a blackboard. If an additional m eights and k nines are added tothe list, the average of all of the numbers in the list is 7.3. The value of k m + is(A ) 24(B ) 21(C ) 11(D ) 31(E ) 8924. A student has two open-topped cylindrical containers. (Thewalls of the two containers are thin enough so that theirwidth can be ignored.) The larger container has a height of20 cm, a radius of 6 cm and contains water to a depth of 17cm. The smaller container has a height of 18 cm, a radius of5 cm and is empty. The student slowly lowers the smallercontainer into the larger container, as shown in the cross-section of the cylinders in Figure 1. As the smaller container is lowered, the water first overflows out of the larger container (Figure 2) and then eventually pours into thesmaller container. When the smaller container is resting onthe bottom of the larger container, the depth of the water in the smaller container will be closest to(A ) 2.82 cm (B ) 2.84 cm (C ) 2.86 cm(D ) 2.88 cm (E ) 2.90 cm25.The lengths of all six edges of a tetrahedron are integers. The lengths of five of the edges are 14, 20,40, 52, and 70. The number of possible lengths for the sixth edge is(A ) 9(B ) 3(C ) 4(D ) 5(E ) 6。