2015滑铁卢竞赛试题答案

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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.。

noip2015初赛普及组答案分析

单项选择题1.A。

计算机内部的用来传送、存贮、加工处理的数据或指令都是以二进制形式进行的。

2.A。

写这题我用的是排除法,B选项显然不对,内存在断电后数据会丢失,C选项也是,屏幕的分辨率是可以手动调整的,D选项,当年我们都用宽带连接Internet的。

3.A。

二进制小数转化为十六进制小数时,每四位二进制数转化为以为十六进制数,故0.10002可以转化为0.816。

4.D。

我的做法是将每个数都化为二进制形式,因为十六进制数和八进制数转化为二进制数很容易,最后求得答案是D。

5.D。

在链表中,每个结点包括两个部分：一个是存储数据元素的数据域,另一个是存储下一个结点地址的指针域,结点与结点之间是用指针连接的,故地址不必连续。

6.B。

模拟一下进栈出栈的过程就行了,共有6次操作:进栈,进栈,出栈,进栈,进栈,出栈,每次操作后栈内元素分别为”a”,”ab”,”a”,”a b c”,”a b c d”,”a b c”,故最后栈顶元素是c。

7.B。

前序遍历的顺序是”根->左->右”,后序遍历的顺序是”左->右->根”,对照四个答案,只有B能满足题目要求。

8.B。

我们知道树高为n的满二叉树的结点个数为2n−1,当树高为5时结点个数为31,当树高为6时结点个数为63,故答案是B。

9.B。

画一张图的事情,就不说了。

10.D。

由递推公式可得T(n)=1+(1+2+…+n)=n2+n2+1,故算法时间的复杂度为O(n2)。

11.D。

用vector存边,由一个顶点的边引到另一个顶点,再不断引出别的顶点,过程中每个顶点和每条边都只用到一遍,故复杂度为O(n+e)。

12.A。

哈夫曼算法用来求哈夫曼树,此树的特点就是引出的路程最短,求的过程运用到贪心思想,具体的请参考一下别的文章。

13.D。

llink和rlink分别指向前驱和后继,不妨设p的前驱为o,在未插入前p->llink就是o,o->rlink就是p,插入时,先将o->rlink赋为q,再将q->rlink赋为p,然后将q->llink赋为o,最后将p->llink赋为q。

2011滑铁卢竞赛试题答案

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

2010PascalSolution滑铁卢竞赛题答案

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

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。

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

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。

欧几里得滑铁卢数学竞赛_2006EuclidSolution

y = x2−2(21)x+212−212+440 = (x−21)2−212+440 = (x−21)2−441+440 = (x−21)2−1
From this form, we immediately see that the coordinates of the vertex are (21, −1).
(c) Solution 1 To determine B, the point of intersection of the lines y = x and x + 2y = 12, we set y = x in the second equation to obtain x + 2x = 12 or 3x = 12 or x = 4.
3. (a) Answer: (21, −1)
Solution 1
The x-intercepts of the given parabola are x = 20 and x = 22.
The x-coordinate of the vertex of the parabola is the average of the x-intercepts, or
10A + B
resulting number is
.
10
So we want to determine A and B so that
10A + B
A+B
=
10
2
10A + B = 5(A + B)
5A = 4B
Since A and B are digits such that 5A = 4B, then A = 4 and B = 5 is the only possibility. Therefore, n = 45. (We can quickly check that the average of the digits of n is 4.5, the number obtained by putting a decimal point between the digits of n.)

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

加拿大数学竞赛历年试题(滑铁卢大学)

Each question is worth 10 marks
Calculating devices are allowed, provided that they do not have any of the following features: (i) internet access, (ii) the ability to communicate with other devices, (iii) information previously stored by students (such as formulas, programs, notes, etc.), (iv) a computer algebra system, (v) dynamic geometry software.
Parts of each question can be of two types: 1. SHORT ANSWER parts indicated by
• worth 3 marks each • full marks given for a correct answer which is placed in the box • part marks awarded only if relevant work is shown in the space provided
WRITE ALL ANSWERS IN THE ANSWER BOOKLET PROVIDED. • Extra paper for your ﬁnished solutions must be supplied by your supervising teacher and inserted into your answer booklet. Write your name, school name, and question number on any inserted pages. • Express answers as simpli√ﬁed exact numbers except where otherwise indicated. For example, π + 1 and 1 − 2 are simpliﬁed exact numbers.

2015GaloisContest滑铁卢竞赛题

1.(a)In the diagram,line 1has equation y =2x +6and crosses the x -axis at P .What is thex -intercept of line 1?(b)Line 2has slope −3and intersects line 1atQ (3,12),as shown.Determine the equationof line 2.(c)Line 2crosses the x -axis at R ,as shown.Determine the area of P QR .y 2.On Wednesday,students at six diﬀerent schools were asked whether or not they receiveda ride to school that day.(a)At School A,there were 330students who received a ride and 420who did not.What percentage of the students at School A received a ride?(b)School B has 240students,of whom 30%received a ride.How many more of the240students in School B needed to receive a ride so that 50%of the students inSchool B got a ride?(c)School C has 200students,of whom 45%received a ride.School D has300students.When School C and School D are combined,the resulting grouphas 57.6%of students who received a ride.If x %of the students at School Dreceived a ride,determine x .(d)School E has 200students,of whom n %received a ride.School F has250students,of whom 2n %received a ride.When School E and School F arecombined,between 55%and 60%of the resulting group received a ride.If n is apositive integer,determine all possible values of n .3.is an even integer,state whether the integer n is even or odd.(b)If c and d are integers,explain why cd (c +d )is always an even integer.(c)Determine the number of ordered pairs (e,f )of positive integers where •e <f ,•e +f is odd,and •ef =300.(d)Determine the number of ordered pairs (m,n )of positive integers such that(m +1)(2n +m )=9000.4.In the diagram,square BCDE has side length 2.Equilateral XY Z has side length 1.Vertex Z coincides with D and vertex X is on ED.(a)What is the measure of ∠Y XE ?B C D E X Y Z (b)A move consists of rotating the square clockwise around a vertex of the triangleuntil a side of the square ﬁrst meets a side of the triangle.The ﬁrst move is a rotation about X and the second move is a rotation about Y ,as shown in the diagrams.(Note that the vertex of the triangle about which the square rotates remains in contact with the square during the rotation.)BC D E X Y Z B C D E EX YZ Beginning of First Move During First Move End of First Move B C D E X YZ End of Second Move In subsequent moves,the square rotates about vertex Z ,then X ,then Y ,and so on.Determine,with justiﬁcation,the total number of moves made from the beginning of the ﬁrst move to when vertex D next coincides with a vertex of the triangle.(c)Determine the length of the path travelled by point E from the beginning of theﬁrst move to when square BCDE ﬁrst returns to its original position (that is,when D next coincides with Z and XZ lies along ED ).Galois Contest(English) 2015。

2012滑铁卢竞赛试题答案

4 12
=1 . 3
(b) Since the two lines are parallel, then their slopes are equal. We re-write the given equations in the form “y = mx + b”. 2 The ﬁrst equation becomes 6y = −4x + 5 or y = − 4 x+ 5 or y = − 3 x+ 5 . 6 6 6 Since the ﬁrst line is not vertical, then the second line is not vertical, and so k = 0. 3 6 x+ k . The second equation becomes ky = −6x + 3 or y = − k 2 6 k 3 3 Therefore, − 3 = − k and so 6 = 2 or k = 6 × 2 = 9. (c) Adding the two equations, we obtain x + x2 = 2 or x2 + x − 2 = 0. Factoring, we obtain (x + 2)(x − 1) = 0, and so x = −2 or x = 1. From the ﬁrst equation, y = −x. If x = −2, then y = 2 and if x = 1, then y = −1. Therefore, the solutions are (x, y ) = (−2, 2) and (x, y ) = (1, −1). (We can check that each of these solutions satisﬁes both equations.) of the mass (or 50 g) is salt and the 3. (a) Since the 200 g solution is 25% salt by mass, then 1 4 rest (150 g) is water. When water is added, the mass of salt does not change. Therefore, the 50 g of salt initially 1 in the solution becomes 10% (or 10 ) of the ﬁnal solution by mass. Therefore, the total mass of the ﬁnal solution is 10 × 50 = 500 g. Thus, the mass of water added is 500 − 200 = 300 g.

初二语文滑铁卢之战练习题及答案

滑铁卢之战一、按所给的拼音写出词语妨ài______ 阴yù_________ pìn________马shū_______然疲惫不kān_______焦zhuó____ róu________皱zhào_______唤二、给加粗词语注音岿然不动濒于殆尽箍桶步履艰难喘咻咻厄运目眩忐忑不安两鬓殉难三、阅读选文回答问题决定世界历史的一瞬间然而格鲁希考虑的这一秒钟却决定了他自己的命运、拿破仑的命运和世界的伞运。

（a）在瓦尔海姆的一家农舍里逝去的这一秒钟决定了整个十九世纪。

而这一秒钟全取决于这个迂腐庸人的一张嘴巴。

这一秒钟全掌握在这双神经质地揉皱了皇帝命令的手中。

──这是多么不幸！倘若格鲁希在这刹那之间有勇气、有魄力、不拘泥于皇帝的命令，而是相信自己、相信显而易见的信号，那么法国也就得救了。

可惜（b）这个毫无主见的家伙只会始终听命于写在纸上的条文，而从不会听从命运的召唤。

格鲁希使劲地摇了摇手。

他说，把这样一支小部队再分散兵力是不负责的，他的任务是追击普军，而不是其他。

就这样，他拒绝了这一违背皇帝命令的行动。

军官们闷闷不乐地沉默了。

在他周围鸦雀无声。

而决定性的一秒钟就这样在静默之中消逝了，它一去不复返，以后，无论用怎样的言辞和行动都无法弥补这一秒钟。

──威灵顿胜利了。

格鲁希的部队继续往前走。

热拉尔和旺达姆愤怒地紧握着拳头。

不久，格鲁希自己也不安起来，随着一小时一小时的过去，他越来越没有把握，因为令人奇怪的是，普军始终没有出现。

显然，他们离开了退往布鲁塞尔去的方向。

接着，情报人员报告了种种可疑的迹象，说明普军在撤退过程中已分几路转移到了正在激战的战场。

如果这时候格鲁希赶紧率领队伍去增援皇帝，还是来得及的。

但他只是怀着愈来愈不安的心情，依然等待着消息，等待着皇帝要他返回的命令。

可是没有消息来。

只有低沉的隆隆炮声震颤着大地，炮声却愈来愈远。

（c）孤注一掷的滑铁卢搏斗正在进行，炮弹便是投下来的铁色子。

学科竞赛-noip2015普及组初赛试题+答案

noip2015普及组初赛试题+答案NOIP2015第二十一届全国青少年信息学奥林匹克联赛初赛普及组C++语言试题竞赛时间：2015年10月11日14:30～16:30选手注意：1、试题纸共有5 页，答题纸共有2 页，满分100 分。

请在答题纸上作答，写在试题纸上的一律无效。

2、不得使用任何电子设备（如计算器、手机、电子词典等）或查阅任何书籍资料。

一、单项选择题（共20题，每题1.5分，共计30分；每题有且仅有一个正确选项）1. 1MB 等于（）。

A. 1000 字节B. 1024 字节C.1000 X 1000 字节D. 1024 X 1024 字节2. 在P C 机中，PENTIUM (奔腾）、酷睿、赛扬等是指（）。

A.生产厂家名称B.硬盘的型号C. CPU 的型号D.显示器的型号3. 操作系统的作用是（）。

A.把源程序译成目标程序B.便于进行数据管理C. 控制和管理系统资源D.实现硬件之间的连接4. 在计算机内部用来传送、存贮、加工处理的数据或指令都是以（）形式进行的。

A. 二进制码B.八进制码C.十进制码D.智能拼音码5. 下列说法正确的是（）。

A. CPU 的主要任务是执行数据运算和程序控制B. 存储器具有记忆能力，其中信息任何时候都不会丢失C. 两个显示器屏幕尺寸相同，则它们的分辨率必定相同D. 个人用户只能使用W ifi 的方式连接到I nternet6. 二进制数00100100 和00010100 的和是（）。

A. 00101000B. 01000001C. 01000100D. 001110007. 与二进制小数0.1 相等的十六进制数是（）A. 0.8B. 0.4C. 0.2D. 0.18. 所谓的“中断”是指（）。

A. 操作系统随意停止一个程序的运行B. 当出现需要时，CPU 暂时停止当前程序的执行转而执行处理新情况的过程C. 因停机而停止一个程序的运行D. 电脑死机9. 计算机病毐是（）。

Pascal滑铁卢数学竞赛(Grade 9)-数学Mathematics-2008-试题 exam

CharteredAccountantsminutes c 2008are permittedopen the Contest until you are told toappropriate7.Scoring:8.Diagrams9.WhenThe nameshttp://www.cemc.uwaterloo.ca.Scoring:There is no penalty for an incorrect answer.Each unanswered question is worth 2,to a maximum of 10unanswered questions.Part A:Each correct answer is worth 5.1.The value of 2+3+42×3×4is(A)1(B)56(C)712(D)3(E)382.If 3x −9=12,then the value of 6x is (A)42(B)24(C)6(D)32(E)523.√52−42is equal to(A)1(B)3(C)5(D)4(E)24.In the diagram,JLMR and JKQR are rectangles.Also,JR =2,RQ =3and JL =8.What is the area of rectangle KLMQ ?(A)6(B)16(C)10(D)15(E)24J KLMQR235.If x =12and y =−6,then the value of 3x +yx −yis(A)3(B)7(C)53(D)5(E)736.In the diagram,P QR is a straight line.The value of x is (A)72(B)44(C)58(D)64(E)52R7.A bag contains 5red,6green,7yellow,and 8blue jelly beans.A jelly bean is selected at random.What is the probability that it is blue?(A)526(B)313(C)726(D)413(E)6138.Olave sold 108apples at a constant rate over 6hours.If she continues to sell apples at the same rate,how many apples will she sell in the next 1hour and 30minutes?(A)27(B)33(C)45(D)36(E)219.In the diagram,the rectangular wire grid contains 15identical squares.The length of the rectangular grid is 10.What is thelength of wire needed toconstruct thegrid?(A)60(B)70(C)120(D)66(E)761010.On the number line,S is three-quarters of the way fromP to Q .Also,T is one-third of the way from P to Q .What is the distance along the number line from T to S ?(A)20(B)15(C)6(D)25(E)3114Part B:Each correct answer is worth 6.11.At Mathville Junior High School,30boys and 20girls wrote the Pascal Contest.Certiﬁcates were awarded to 30%of the boys and 40%of the girls.What percentage of all of the participating students received certiﬁcates?(A)34(B)35(C)36(D)17(E)7012.In the diagram,the perimeter of the rectangle is 56.What is its area?(A)247(B)187(C)169(D)135(E)775x 2x + 413.23×22×33×32is equal to(A)65(B)66(C)610(D)3610(E)363614.Two 3-digit numbers,abc and def ,have the following property:a b c+d e f1000None of a ,b ,c ,d ,e ,or f is 0.What is a +b +c +d +e +f ?(A)10(B)19(C)21(D)28(E)3015.In the diagram,what is the perimeter of P QR ?(A)63(B)60(C)55(D)85(E)70PQRS8252016.A circle has an area of M cm 2and a circumference of N cm.If MN=20,what is theradius of the circle,in cm?(A)10(B)20(C)40(D)110(E)12017.The surface area of a large cube is 5400cm 2.This cube is cut into a number ofidentical smaller cubes.Each smaller cube has a volume of 216cm 3.How many smaller cubes are there?(A)25(B)125(C)164(D)180(E)21618.Alex has $2.65.He has only dimes (worth $0.10each)and quarters (worth $0.25each).He has more dimes than quarters.What is the smallest number of coins that Alex could have?(A)25(B)16(C)13(D)19(E)2219.An integer is deﬁned to be upright if the sum of its ﬁrst two digits equals its third digit.For example,145is an upright integer since 1+4=5.How many positive 3-digit integers are upright?(A)28(B)39(C)36(D)45(E)5020.Four of the six numbers 1867,1993,2019,2025,2109,and 2121have a mean (average)of 2008.What is the mean (average)of the other two numbers?(A)1994(B)2006(C)2022(D)2051(E)2064Part C:Each correct answer is worth 8.21.If 3≤p ≤10and 12≤q ≤21,then the diﬀerence between the largest and smallestpossible values of pqis(A)2942(B)295(C)1970(D)1912(E)198422.Ginger walks at 4km/h and runs at 6km/h.She saves 334minutes by running instead of walking from her home to her school.What is the distance,in kilometres,from her home to her school?(A)712(B)334(C)178(D)114(E)3423.Four pieces of lumber are placed in parallel positions,as shown,perpendicular to line M :•Piece W is 5m long•Piece X is 3m long and its left end is 3m from line M•Piece Y is 5m long and is 2m from line M•Piece Z is 4m long and is 1.5m from from line M A single cut,perpendicular to the pieces of lumber,is made along the dotted line L .The total length of lumber on each side of L is the same.What is the length,in metres,of the part of piece W to the left of the cut?(A)4.25(B)3.5(C)3.25(D)3.75(E)4.0LW XY ZM24.Five circles are drawn on a piece of paper and connectedas shown.Each circle must be coloured red,blue or green.Two circles connected by a straight line may not be coloured the same.How many diﬀerent ways are there to colour the circles?(A)24(B)60(C)72(D)36(E)4825.In the diagram, P QR is right-angled at P and hasP Q =2and P R =2√3.Altitude P L intersects median RM at F .What is the length of P F ?(A)√32(B)3√37(C)4√37(D)5√39(E)3√35RPQML FPascal Contest(English)2008•••。

2017滑铁卢竞赛试题答案

9559 = (100 + 21)(100 − 21) = 121 · 79 Therefore, (a, b) = (121, 79) satisﬁes the conditions. (In addition, the pairs (a, b) = (79, 121), (869, 11), (11, 869) satisfy the conditions. The last two of these pairs cannot be obtained in the same way.) 3. (a) The area of quadrilateral ABCD is the sum of the areas of ABC and ACD. 1 1 Since ABC is right-angled at B , its area equals 2 (AB )(BC ) = 2 (3)(4) = 6. Since ABC is right-angled at B , then by the Pythagorean Theorem, √ √ √ AC = AB 2 + BC 2 = 32 + 42 = 25 = 5 because AC > 0. (We could have also observed that ABC must be a “3-4-5” triangle.) Since ACD is right-angled at A, then by the Pythagorean Theorem, √ √ √ AD = CD2 − AC 2 = 132 − 52 = 144 = 12 because AD > 0. (We could have also observed that ACD must be a “5-12-13” triangle.) 1 (AC )(AD) = 2 (5)(12) = 30. Thus, the area of ACD equals 1 2 Finally, the area of quadrilateral ABCD is thus 6 + 30 = 36. (b) Let the width of each of the identical rectangles be a. In other words, QP = RS = T W = W X = U V = V Y = a. Let the height of each of the identical rectangles be b. In other words, QR = P S = T U = W V = XY = b. The perimeter of the whole shape equals QP + P S + SX + XY + V Y + U V + T U + T R + QR Substituting for known lengths, we obtain a + b + SX + b + a + a + b + T R + b or 3a + 4b + (SX + T R). But SX + T R = (T R + RS + SX ) − RS = (T W + W X ) − RS = a + a − a = a. Therefore, the perimeter of the whole shape equals 4a + 4b. The perimeter of one rectangle is 2a + 2b, which we are told equals 21 cm. Finally, the perimeter of the whole shape is thus 2(2a + 2b) which equals 42 cm. (c) Solution 1 Suppose that the rectangular prism has dimensions a cm by b cm by c cm. Suppose further that one of the faces that is a cm by b cm is the face with area 27 cm2 and that one of the faces that is a cm by c cm is the face with area 32 cm2 . (Since every pair of non-congruent faces shares exactly one side length, there is no loss of generality in picking these particular variables for these faces.) Therefore, ab = 27 and ac = 32. Further, we are told that the volume of the prism is 144 cm3 , and so abc = 144.

2016滑铁卢竞赛试题答案

13, 2016
(outside of North America and South America)
Solutions
©2016 University of Waterloo

1. (a) The average is 5 + 15 + 25 + 35 + 45 + 55 (5 + 55) + (15 + 45) + (25 + 35) 60 + 60 + 60 = = = 30 6 6 6 (b) Since x2 = 2016, then (x + 2)(x − 2) = x2 − 4 = 2016 − 4 = 2012. (c) Since points P , Q and R lie on a straight line, then the slope of P Q equals the slope of P R. 30 − 5 25 2a − 5 and the slope of P R equals = = 5. The slope of P Q equals a−7 12 − 7 5 2a − 5 = 5 and so 2a − 5 = 5(a − 7). Therefore, a−7 This gives 2a − 5 = 5a − 35 or 3a = 30, and so a = 10. 2. (a) If n 25 = , then n2 = 25(9) = 225. Therefore, n = 15 or n = −15. 9 n (We can check by substitution that each of these values satisﬁes the original equation.)

2008PascalSolution滑铁卢竞赛题答案

1.Calculating,2+3+42×3×4=924=38.Answer:(E)2.Since3x−9=12,then3x=12+9=21.Since3x=21,then6x=2(3x)=2(21)=42.(Note that we did not need to determine the value of x.)Answer:(A)3.Calculating,√52−42=√25−16=√9=3.Answer:(B)4.Solution1Since JLMR is a rectangle and JR=2,then LM=2.Similarly,since JL=8,then RM=8.Since RM=8and RQ=3,then QM=8−3=5.Since KLMQ is a rectangle with QM=5and LM=2,its area is5(2)=10.Solution2Since JL=8and JR=2,then the area of rectangle JLMR is2(8)=16.Since RQ=3and JR=2,then the area of rectangle JKQR is2(3)=6.The area of rectangle KLMQ is the diﬀerence between these areas,or16−6=10.Answer:(C) 5.Since x=12and y=−6,then3x+y x−y =3(12)+(−6)12−(−6)=3018=53Answer:(C)6.Solution1Since∠P QS is an exterior angle of QRS,then∠P QS=∠QRS+∠QSR,so136◦=x◦+64◦or x=136−64=72.Solution2Since∠P QS=136◦,then∠RQS=180◦−∠P QS=180◦−136◦=44◦.Since the sum of the angles in QRS is180◦,then44◦+64◦+x◦=180◦or x=180−44−64=72.Answer:(A) 7.In total,there are5+6+7+8=26jelly beans in the bag.Since there are8blue jelly beans,the probability of selecting a blue jelly bean is826=413.Answer:(D)8.Since Olave sold108apples in6hours,then she sold108÷6=18apples in one hour.A time period of1hour and30minutes is equivalent to1.5hours.Therefore,Olave will sell1.5×18=27apples in1hour and30minutes.Answer:(A)9.Since the length of the rectangular grid is 10and the grid is 5squares wide,then the side length of each square in the grid is 10÷5=2.There are 4horizontal wires,each of length 10,which thus have a total length of 4×10=40.Since the side length of each square is 2and the rectangular grid is 3squares high,then the length of each vertical wire is 3×2=6.Since there are 6vertical wires,the total length of the vertical wires is 6×6=36.Therefore,the total length of wire is 40+36=76.Answer:(E)10.Solution 1Since Q is at 46and P is at −14,then the distance along the number line from P to Q is 46−(−14)=60.Since S is three-quarters of the way from P to Q ,then S is at −14+34(60)=−14+45=31.Since T is one-third of the way from P to Q ,then T is at −14+13(60)=−14+20=6.Thus,the distance along the number line from T to S is 31−6=25.Solution 2Since Q is at 46and P is at −14,then the distance along the number line from P to Q is 46−(−14)=60.Since S is three-quarters of the way from P to Q and T is one-third of the way from P to Q ,then the distance from T to S is 60 34−13 =60 912−412 =60 512 =25.Answer:(D)11.In total,30+20=50students wrote the Pascal Contest at Mathville Junior High.Since 30%(or 310)of the boys won certiﬁcates and 40%(or 410)of the girls won certiﬁcates,then the total number of certiﬁcates awarded was 310(30)+410(20)=9+8=17.Therefore,17of 50participating students won certiﬁcates.In other words,1750×100%=34%of the participating students won certiﬁcates.Answer:(A)12.Since the perimeter of the rectangle is 56,then2(x +4)+2(x −2)=562x +8+2x −4=564x +4=564x=52x =13Therefore,the rectangle is x +4=17by x −2=11,so has area 17(11)=187.Answer:(B)ing exponent rules,23×22×33×32=23+2×33+2=25×35=(2×3)5=65.Answer:(A)14.Solution 1The wording of the problem tells us that a +b +c +d +e +f must be the same no matter what numbers abc and def are chosen that satisfy the conditions.An example that works is 889+111=1000.In this case,a +b +c +d +e +f =8+8+9+1+1+1=28,so this must always be the value.Solution 2Consider performing this “long addition”by hand.Consider ﬁrst the units column.Since c +f ends in a 0,then c +f =0or c +f =10.The value of c +f cannot be 20or more,as c and f are digits.Since none of the digits is 0,we cannot have c +f =0+0so c +f =10.(This means that we “carry”a 1to the tens column.)Since the result in the tens column is 0and there is a 1carried into this column,then b +e ends in a 9,so we must have b +e =9.(Since b and e are digits,b +e cannot be 19or more.)In the tens column,we thus have b +e =9plus the carry of 1,so the resulting digit in the tens column is 0,with a 1carried to the hundreds column.Using a similar analysis in the hundreds column to that in the tens column,we must have a +d =9.Therefore,a +b +c +d +e +f =(a +d )+(b +e )+(c +f )=9+9+10=28.Answer:(D)15.Each of P SQ and RSQ is right-angled at S ,so we can use the Pythagorean Theorem inboth triangles.In RSQ ,we have QS 2=QR 2−SR 2=252−202=625−400=225,so QS =√225=15since QS >0.In P SQ ,we have P Q 2=P S 2+QS 2=82+225=64+225=289,so P Q =√289=17since P Q >0.Therefore,the perimeter of P QR is P Q +QR +RP =17+25+(20+8)=70.Answer:(E)16.Suppose the radius of the circle is r cm.Then the area M is πr 2cm 2and the circumference N is 2πr cm.Thus,πr 22πr =20or r 2=20or r =40.Answer:(C)17.Solution 1The large cube has a total surface area of 5400cm 2and its surface is made up of 6identical square faces.Thus,the area of each face,in square centimetres,is 5400÷6=900.Because each face is square,the side length of each face is √900=30cm.Therefore,each edge of the cube has length 30cm and so the large cube has a volume of 303=27000cm 3.Because the large cube is cut into small cubes each having volume 216cm 3,then the number of small cubes equals 27000÷216=125.Solution 2Since the large cube has 6square faces of equal area and the total surface area of the cube is 5400cm 2,then the surface area of each face is 5400÷6=900cm 2.Since each face is square,then the side length of each square face of the cube is √900=30cm,and so the edge length of the cube is 30cm.Since each smaller cube has a volume of 216cm 3,then the side length of each smaller cube is 3√216=6cm.Since the side length of the large cube is 30cm and the side length of each smaller cube is 6cm,then 30÷6=5smaller cubes ﬁt along each edge of the large cube.Thus,the large cube is made up of 53=125smaller cubes.Answer:(B)18.Solution1Alex has265cents in total.Since265is not divisible by10,Alex cannot have only dimes,so must have at least1quarter.If Alex has1quarter,then he has265−25=240cents in dimes,so24dimes.Alex cannot have2quarters,since265−2(25)=215is not divisible by10.If Alex has3quarters,then he has265−3(25)=190cents in dimes,so19dimes.Continuing this argument,we can see that Alex cannot have an even number of quarters,since the total value in cents of these quarters would end in a0,making the total value of the dimes end in a5,which is not possible.If Alex has5quarters,then he has265−5(25)=140cents in dimes,so14dimes.If Alex has7quarters,then he has265−7(25)=90cents in dimes,so9dimes.If Alex has9quarters,then he has265−9(25)=40cents in dimes,so4dimes.If Alex has more than9quarters,then he will have even fewer than4dimes,so we do not need to investigate any more possibilities since we are told that Alex has more dimes than quarters.So the possibilities for the total number of coins that Alex has are1+24=25,3+19=22, 5+14=19,and7+9=16.Therefore,the smallest number of coins that Alex could have is16.(Notice that each time we increase the number of quarters above,we are in eﬀect exchanging 2quarters(worth50cents)for5dimes(also worth50cents).)Solution2Suppose that Alex has d dimes and q quarters,where d and q are non-negative integers.Since Alex has$2.65,then10d+25q=265or2d+5q=53.Since the right side is odd,then the left side must be odd,so5q must be odd,so q must be odd.If q≥11,then5q≥55,which is too large.Therefore,q<11,leaving q=1,3,5,7,9which give d=24,19,14,9,4.The solution with d>q and d+q smallest is q=7and d=9,giving16coins in total.Answer:(B) 19.From the deﬁnition,theﬁrst and second digits of an upright integer automatically determinethe third digit,since it is the sum of theﬁrst two digits.Considerﬁrst those upright integers beginning with1.These are101,112,123,134,145,156,167,178,and189,since1+0=1,1+1=2,and so on.(The second digit cannot be9,otherwise the last“digit”would be1+9=10,which is impossible.)There are9such numbers.Beginning with2,the upright integers are202,213,224,235,246,257,268,and279.There are8of them.We can continue the pattern and determine the numbers of the upright integers beginning with 3,4,5,6,7,8,and9to be7,6,5,4,3,2,and1.Therefore,there are9+8+7+6+5+4+3+2+1=45positive3-digit upright integers.Answer:(D) 20.The sum of the six given integers is1867+1993+2019+2025+2109+2121=12134.The four of these integers that have a mean of2008must have a sum of4(2008)=8032.(We do not know which integers they are,but we do not actually need to know.)Thus,the sum of the remaining two integers must be12134−8032=4102.=2051.Therefore,the mean of the remaining two integers is41022(We can verify that1867,2019,2025and2121do actually have a mean of2008,and that1993 and2109have a mean of2051.)Answer:(D)21.The maximum possible value of pqis when p is as large as possible(that is,10)and q is as smallas possible(that is,12).Thus,the maximum possible value of pqis1012=56.The minimum possible value of pqis when p is as small as possible(that is,3)and q is as largeas possible(that is,21).Thus,the maximum possible value of pqis321=17.The diﬀerence between these two values is 56−17=3542−642=2942.Answer:(A)22.Suppose that the distance from Ginger’s home to her school is d km.Since there are60minutes in an hour,then33minutes(or15minutes)is15×1=1of anhour.Since Ginger walks at4km/h,then it takes her d4hours to walk to school.Since Ginger runs at6km/h,then it takes her d6hours to run to school.Since she saves116of an hour by running,then the diﬀerence between these times is116of anhour,sod 4−d6=1163d 12−2d12=116d12=116d=1216=34Therefore,the distance from Ginger’s home to her school is34km.Answer:(E)23.Suppose that the distance from line M to line L is d m.Therefore,the total length of piece W to the left of the cut is d m.Since piece X is3m from line M,then the length of piece X to the left of L is(d−3)m, because3of the d m to the left of L are empty.Similarly,the lengths of pieces Y and Z to the left of line L are(d−2)m and(d−1.5)m.Therefore,the total length of lumber to the left of line L isd+(d−3)+(d−2)+(d−1.5)=4d−6.5mSince the total length of lumber on each side of the cut is equal,then this total length is1 2(5+3+5+4)=8.5m.(We could insteadﬁnd the lengths of lumber to the right of line L to be5−d,6−d,7−d,and 5.5−d and equate the sum of these lengths to the sum of the lengths on the left side.) Therefore,4d−6.5=8.5or4d=15or d=3.75,so the length of the part of piece W to the left of L is3.75m.Answer:(D)24.We label the ﬁve circles as shown in the diagram.P QRS TWe note that there are 3possible colours and that no two adjacent circles can be coloured the same.Consider circle R .There are three possible colours for this circle.For each of these colours,there are 2possible colours for T (either of the two colours that R is not),since it cannot be the same colour as R .Circles Q and S are then either the same colour as each other,or are diﬀerent colours.Case 1:Q and S are the same colourIn this case,there are 2possible colours for Q (either of the colours that R is not)and 1pos-sibility for S (the same colour as Q ).For each of these possible colours for Q /S ,there are two possible colours for P (either of the colours that Q and S are not).P Q R S T3 choices2 choices2 choices1 choice2 choices In this case,there are thus 3×2×2×1×2=24possible ways of colouring the circles.Case 2:Q and S are diﬀerent coloursIn this case,there are 2possible colours for Q (either of the colours that R is not)and 1possibility for S (since it must be diﬀerent from R and diﬀerent from Q ).For each of these possible colourings of Q and S ,there is 1possible colour for P (since Q and S are diﬀerent colours,P is diﬀerent from these,and there are only 3colours in total).P Q R S T3 choices2 choices1 choice1 choice In this case,there are thus 3×2×2×1×1=12possible ways of colouring the circles.In total,there are thus 24+12=36possible ways to colour the circles.Answer:(D)25.Since P Q =2and M is the midpoint of P Q ,then P M =MQ =12(2)=1.Since P QR is right-angled at P ,then by the Pythagorean Theorem,RQ = P Q 2+P R 2= 22+(2√3)2=√4+12=√16=4(Note that we could say that P QR is a 30◦-60◦-90◦triangle,but we do not actually need this fact.)Since P L is an altitude,then ∠P LR =90◦,so RLP is similar to RP Q (these triangles have right angles at L and P respectively,and a common angle at R ).Therefore,P L QP =RP RQ or P L =(QP )(RP )RQ =2(2√3)4=√3.Similarly,RL RP =RP RQ so RL =(RP )(RP )RQ =(2√3)(2√3)4=3.Therefore,LQ =RQ −RL =4−3=1and P F =P L −F L =√3−F L .So we need to determine the length of F L .Drop a perpendicular from M to X on RQ .RP QM LFX Then MXQ is similar to P LQ ,since these triangles are each right-angled and they share a common angle at Q .Since MQ =12P Q ,then the corresponding sides of MXQ are half as long as those of P LQ .Therefore,QX =12QL =12(1)=12and MX =12P L =12(√3)=√32.Since QX =12,then RX =RQ −QX =4−12=72.Now RLF is similar to RXM(they are each right-angled and share a common angle at R).Therefore,F LMX=RLRXso F L=(MX)(RL)RX=√32(3)72=3√3.Thus,P F=√3−3√37=4√37.Answer:(C)。

Hypatia滑铁卢数学竞赛(Grade 11)-数学Mathematics-2005-试题 exam

2005Hypatia Contest(Grade11)Wednesday,April20,20051.For numbers a and b,the notation a b means a2−4b.For example,5 3=52−4(3)=13.(a)Evaluate2 3.(b)Find all values of k such that k 2=2 k.(c)The numbers x and y are such that3 x=y and2 y=8x.Determine the values of x and y.2.Gwen and Chris are playing a game.They begin with a pile of toothpicks,and use the followingrules:•The two players alternate turns•On any turn,the player can remove1,2,3,4,or5toothpicks from the pile•The same number of toothpicks cannot be removed on two diﬀerent turns•The last person who is able to play wins,regardless of whether there are any toothpicks remaining in the pileFor example,if the game begins with8toothpicks,the following moves could occur: Gwen removes1toothpick,leaving7in the pileChris removes4toothpicks,leaving3in the pileGwen removes2toothpicks,leaving1in the pileGwen is now the winner,since Chris cannot remove1toothpick.(Gwen already removed 1toothpick on one of her turns,and the third rule says that1toothpick cannot be removed on another turn.)(a)Suppose the game begins with11toothpicks.Gwen begins by removing3toothpicks.Chris follows and removes1.Then Gwen removes4toothpicks.Explain how Chris can win the game.(b)Suppose the game begins with10toothpicks.Gwen begins by removing5toothpicks.Explain why Gwen can always win,regardless of what Chris removes on his turn.(c)Suppose the game begins with9toothpicks.Gwen begins by removing2toothpicks.Explain how Gwen can always win,regardless of how Chris plays.2005Hypatia Contest Page 23.In the diagram, ABC is equilateral with side length4.Points P ,Q and R are chosen on sides AB ,BC and CA ,respectively,such that AP =BQ =CR =1.BQR CA P(a)Determine the exact area of ABC .Explain how you got your answer.(b)Determine the exact areas of P BQ and P QR .Explain how you got your answers.4.An arrangement of a set is an ordering of all of the numbers in the set,in which each number appears exactly once.For example,312and 231are two of the possible arrangements of {1,2,3}.(a)Determine the number of triples (a,b,c )where a ,b and c are three diﬀerent numberschosen from {1,2,3,4,5}with a <b and b >c .Explain how you got your answer.(b)How many arrangements of {1,2,3,4,5,6}contain the digits 254consecutively in thatorder?Explain how you got your answer.(c)A local peak in an arrangement occurs where there is a sequence of 3numbers in thearrangement for which the middle number is greater than both of its neighbours.For example,the arrangement 35241of {1,2,3,4,5}contains 2local peaks.Determine,with justiﬁcation,the average number of local peaks in all 40320possible arrangements of {1,2,3,4,5,6,7,8}.。

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(b) Points N (5, 3) and P (5, c) lie on the same vertical line. We can consider N P as the base of M N P . Suppose that the length of this base is b. The corresponding height of M N P is the distance from M (1, 4) to the line through N and P . Since M lies on the vertical line x = 1 and N and P lie on the vertical line x = 5, then the height is h = 4.
Therefore x = 0 or x2 − (a + 1)x + 4 = 0. Note that x = 0 is not a solution to x2 − (a + 1)x + 4 = 0, since when x = 0 is substitutedA2AD2 = = = =
B
E
60º 120º
C
2
D
Since AC = CD, then ACD is isosceles with ∠CDA = ∠CAD. 1 Each of these angles equals 2 (180◦ − ∠ACD) = 1 (180◦ − 120◦ ) = 30◦ . 2 √ But DAE is then a 30◦ -60◦ -90◦ triangle, so AD = 2AE = 2 3.
y
P1(5, c1) M (1, 4)
N (5, 3)
x
P2(5, c2)
bh = 14. Since the area of M N P is 14, then 1 2 1 Since h = 4, then 2 b(4) = 14 or 2b = 14 and so b = 7. Therefore, P (5, c) is a distance of 7 units away from N (5, 3). Since N P is a vertical line segment, then c = 3 + 7 or c = 3 − 7, and so c = 10 or c = −4. The sum of these two values is 10 + (−4) = 6. (We could also have noted that, since the two values of c will be symmetric about y = 3, then the average of their values is 3 and so the sum of their values is 2 · 3 = 6.) 4. (a) To ﬁnd the y -intercept, we set x = 0 and obtain y = (−1)(−2)(−3) − (−2)(−3)(−4) = (−6) − (−24) = 18 . To ﬁnd the x-intercepts, we ﬁrst simplify using common factors: y = (x−1)(x−2)(x−3)−(x−2)(x−3)(x−4) = (x−2)(x−3) ((x − 1) − (x − 4)) = 3(x−2)(x−3) To ﬁnd the x-intercepts, we set y = 0 and obtain 3(x − 2)(x − 3) = 0 which gives x = 2 or x = 3. Therefore, the y -intercept is 18 and the x-intercepts are 2 and 3. (b) To ﬁnd the points of intersection of the graphs with equations y = x3 − x2 + 3x − 4 and y = ax2 − x − 4, we equate values of y and solve for x. We want to ﬁnd all values of a for which there are exactly two values of x which are solutions to x3 − x2 + 3x − 4 = ax2 − x − 4. Solving, we obtain x3 − x2 + 3 x − 4 x3 − x2 − ax2 + 4x x3 − (a + 1)x2 + 4x x(x2 − (a + 1)x + 4) = = = = ax2 − x − 4 0 0 0

3. (a) Solution 1 Join AD. Since BC = CD and BD = 4, then BC = CD = 2. Also, AB = BC = 2. Since ABC is equilateral, then ∠ABC = ∠ACB = 60◦ . Since ∠ACB = 60◦ , then ∠ACD = 180◦ − ∠ACB = 180◦ − 60◦ = 120◦ .

The CENTRE for EDUCATION in MATHEMATICS and COMPUTING cemc.uwaterloo.ca
2015 Euclid Contest
Wednesday, April 15, 2015
(in North America and South America)
A
2
B
60º 120º 2
C
2
D
Since AC = CD, then ACD is isosceles with ∠CDA = ∠CAD. 1 (180◦ − ∠ACD) = 1 (180◦ − 120◦ ) = 30◦ . Each of these angles equals 2 2 Since ∠ABD = 60◦ and ∠ADB = 30◦ , then ∠BAD = 90◦ and DBA is a 30◦ -60◦ -90◦ triangle. √ √ Therefore, AD = 3AB = 2 3. Solution 2 Join AD. Since BC = CD and BD = 4, then BC = CD = 2. Also, AC = CD = 2. Since ∠ACB = 60◦ , then ∠ACD = 180◦ − ∠ACB = 180◦ − 60◦ = 120◦ . By the cosine law in ACD, AC 2 + CD2 − 2(AC )(CD) cos(∠ACD) 22 + 22 − 2(2)(2) cos 120◦ 1 ) 4 + 4 − 8(− 2 12 √ √ Since AD2 = 12 and AD > 0, then AD = 12 = 2 3. Solution 3 Join AD and drop a perpendicular from A to E on BC . Since BC = CD and BD = 4, then BC = CD = 2. Also, AB = BC = 2. Since ABC is equilateral, then ∠ABC = ∠ACB = 60◦ . Since ∠ABC √ = 60◦ and ∠AEB = 90◦ , then ABE is a 30◦ -60◦ -90◦ triangle. √ Thus, AE = 23 AB = 3. Since ∠ACB = 60◦ , then ∠ACD = 180◦ − ∠ACB = 180◦ − 60◦ = 120◦ .
(b) Let the radius of the larger circle be r. Since the radius of the smaller circle is 1, then its area is π · 12 = π . Since the area between the circles is equal to the area of the smaller circle, then the area of the larger circle is π + π = 2π . √ Thus, πr2 = 2π or r2 = 2. Since r > 0, then r = 2. (c) Since 30 students had an average mark of 80, then the sum of the marks of these 30 students was 30 · 80 = 2400. After 2 students dropped the class, there were 28 students left. Their average mark was 82. Thus, the sum of the marks of the remaining 28 students was 28 · 82 = 2296. Therefore, the sum of the marks of the 2 students who dropped the class was 2400 − 2296 104 = 52. or 104, and so their average mark was 2
Thursday, April 16, 2015
(outside of North America and South America)
Solutions
©2015 University of Waterloo

100 − 81 19 102 − 92 = = = 1. 1. (a) Evaluating, 10 + 9 19 19 2 2 Alternatively, we could factor 10 − 9 as a diﬀerence of squares to obtain (10 + 9)(10 − 9) 102 − 92 = = 10 − 9 = 1 10 + 9 10 + 9 noting that 10 + 9, which we divided from the numerator and denominator, is not equal to 0. x+1 (b) Since = 4, then x + 1 = 4(x + 4) and so x + 1 = 4x + 16 or 3x = −15. x+4 Therefore, 3x + 8 = −15 + 8 = −7. Alternatively, we could note that since 3x = −15, then x = −5. Thus, 3x + 8 = 3(−5) + 8 = −15 + 8 = −7. (c) Since f (x) = 2x − 1, then f (3) = 2(3) − 1 = 5. Therefore, (f (3))2 + 2(f (3)) + 1 = 52 + 2(5) + 1 = 25 + 10 + 1 = 36. Alternatively, we could note that since f (x) = 2x − 1, then (f (x))2 + 2(f (x)) + 1 = (f (x) + 1)2 = (2x − 1 + 1)2 = 4x2 and so (f (3))2 + 2(f (3)) + 1 = 4(32 ) = 36. 2. (a) Since √ √ √ √ a + a = 20, then 2 a = 20 or a = 10, and so a = 102 = 100.