Pascal滑铁卢数学竞赛(Grade 9)-数学Mathematics-2013-试题 exam

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Hypatia滑铁卢数学竞赛(Grade 11)-数学Mathematics-2008-试题 exam

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

滑铁卢竞赛数学题

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

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

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

如此重复，得到第五、第六个正方形，求第五个正方形的面积。

3. 一个圆被分成n个相等的扇形，其中一个是空心的，其他n-1个是实心的。

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

经过若干轮后，只剩下一个人。

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

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

滑铁卢大学欧几里得数学竞赛

该考试是学生申请滑铁卢大学数学学院本科专业的重要参考。

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

考试范围：大部分的题目基于高三或者12年级数学课学习的内容。

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

每题 10 分，共计 100 分。

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

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

Pascal滑铁卢数学竞赛(Grade 9)-数学Mathematics-2004-试题 exam

(A) 21 (D) 13
(B) 40 (E) 34
(C) 19
20. At the beginning of the game “Clock 7”, the arrow points to one of the seven numbers. On each turn, the arrow is rotated clockwise by the number of spaces indicated by the arrow at the beginning of the turn. For example, if “Clock 7” starts with the arrow pointing at 4, then on the first turn, the arrow is rotated clockwise 4 spaces so that it now points at 1. The arrow will then move 1 space on the next turn, and so on. If the arrow points at 6 after the 21st turn, at which number did the arrow point after the first turn?
1. To win a skateboard, the skill testing question is 5 × (10 – 6) ÷ 2 . The correct answer is
(A)ቤተ መጻሕፍቲ ባይዱ10
(B) 35
(C) 32
(D) 22
(E) 40
2. The average of 2, x and 12 is 8. What is the value of x?

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)。

国际中学生数学竞赛含金量排行榜

国际中学生数学竞赛含金量排行榜
国际中学生数学竞赛的含金量因赛事的权威性、参赛选手水平、奖项设置等因素而有所不同。

以下是部分国际中学生数学竞赛及简要介绍：
1. 国际数学奥林匹克（IMO）：这是最具权威性的国际中学生数学竞赛之一，每年有来自世界各地的参赛选手。

IMO的奖项分为金牌、银牌和铜牌，其
中金牌是最高荣誉。

获奖者将获得世界范围内的认可和奖励，对于未来的学术和职业发展具有重要意义。

2. 亚洲太平洋数学奥林匹克（APMO）：这是亚太地区最高水平的数学竞赛，每年有来自多个国家和地区的代表队参赛。

APMO的奖项分为金、银、铜
牌和优秀奖，其中金牌是最高荣誉。

获奖者将获得国际范围内的认可和奖励，对于未来的学术和职业发展具有重要意义。

3. 英国数学奥林匹克（BMO）：这是英国最高水平的数学竞赛，每年有来
自全国各地的参赛选手。

BMO的奖项分为金、银、铜牌和优秀奖，其中金
牌是最高荣誉。

获奖者将获得英国范围内的认可和奖励，对于未来的学术和职业发展具有重要意义。

4. 罗马尼亚数学奥林匹克（RMO）：这是罗马尼亚最高水平的数学竞赛，
每年有来自全国各地的参赛选手。

RMO的奖项分为金、银、铜牌和优秀奖，其中金牌是最高荣誉。

获奖者将获得国际范围内的认可和奖励，对于未来的学术和职业发展具有重要意义。

总的来说，这些国际中学生数学竞赛都具有较高的含金量，获奖者将获得国际范围内的认可和奖励，对于未来的学术和职业发展具有重要意义。

Fryer滑铁卢数学竞赛(Grade 9)-数学Mathematics-2009-试题 exam

2009Fryer Contest(Grade9)Wednesday,April8,20091.Emily sets up a lemonade stand.She has set-up costs of$12.00and each cup of lemonade costsher$0.15to make.She sells each cup of lemonade for$0.75.(a)What is the total cost,including the set-up,for her to make100cups of lemonade?(b)What is her proﬁt(money earned minus total cost)if she sells100cups of lemonade?(c)What is the number of cups that she must sell to break even(that is,to have a proﬁtof$0)?(d)Why is it not possible for her to make a proﬁt of exactly$17.00?2.If a>0and b>0,a new operation∇is deﬁned as follows:a∇b=a+b 1+ab.For example,3∇6=3+61+3×6=919.(a)Calculate2∇5.(b)Calculate(1∇2)∇3.(c)If2∇x=57,what is the value of x?(d)For some values of x and y,the value of x∇y is equal to x+y17.Determine all possibleordered pairs of positive integers x and y for which this is true.3.In the diagram,K,O and M are the centres of the three semi-circles.Also,OC=32and CB=36.(a)What is the length of AC?(b)What is the area of the semi-circle with centre K?(c)What is the area of the shadedregion?(d)Line l is drawn to touch the smaller semi-circlesat points S and E so that KS and ME are both perpendicular to l.Determine the area of quadrilateral KSEM.2009Fryer Contest Page24.The addition shown below,representing2+22+222+2222+···,has101rows and the lastterm consists of1012’s:2222222222...22 (2222)+222 (2222)···C B A(a)Determine the value of the ones digit A.(b)Determine the value of the tens digit B and the value of the hundreds digit C.(c)Determine the middle digit of the sum.。

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。

Cayley滑铁卢数学竞赛(Grade 10)-数学Mathematics-1998-试题 exam

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

2009PascalSolution滑铁卢竞赛题答案

1.Calculating,2×9−√36+1=18−6+1=13.Answer:(D)2.On Saturday,Deepit worked6hours.On Sunday,he worked4hours.Therefore,he worked6+4=10hours in total on Saturday and Sunday.Answer:(E) 3.Since1piece of gum costs1cent,then1000pieces of gum cost1000cents.Since there are100cents in a dollar,the total cost is$10.00.Answer:(D) 4.Since each of the18classes has28students,then there are18×28=504students who attendthe school.On Monday,there were496students present,so504−496=8students were absent.Answer:(A) 5.The sum of the angles around any point is360◦.Therefore,5x◦+4x◦+x◦+2x◦=360◦or12x=360or x=30.Answer:(D) 6.When−1is raised to an even exponent,the result is1.When−1is raised to an odd exponent,the result is−1.Thus,(−1)5−(−1)4=−1−1=−2.Answer:(A) 7.Since P Q is horizontal and the y-coordinate of P is1,then the y-coordinate of Q is1.Since QR is vertical and the x-coordinate of R is5,then the x-coordinate of Q is5.Therefore,the coordinates of Q are(5,1).Answer:(C)8.When y=3,we have y3+yy2−y=33+332−3=27+39−3=306=5.Answer:(D)9.Since there are4♣’s in each of theﬁrst two columns,then at least1♣must be moved out ofeach of these columns to make sure that each column contains exactly three♣’s.Therefore,we need to move at least2♣’s in total.If we move the♣from the top left corner to the bottom right corner♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣and the♣from the second row,second column to the third row,ﬁfth column♣♣♣♣♣♣♣♣♣♣♣♣♣♣♣then we have exactly three♣’s in each row and each column.Therefore,since we must move at least2♣’s and we can achieve the conﬁguration that we want by moving2♣’s,then2is the smallest number.(There are also other combinations of moves that will give the required result.)Answer:(B) 10.Solution1Since z=4and x+y=7,then x+y+z=(x+y)+z=7+4=11.Solution2Since z=4and x+z=8,then x+4=8or x=4.Since x=4and x+y=7,then4+y=7or y=3.Therefore,x+y+z=4+3+4=11.Answer:(C) 11.We write out theﬁve numbers to5decimal places each,without doing any rounding:5.076=5.07666...5.076=5.07676...5.07=5.070005.076=5.076005.076=5.07607...We can use these representations to order the numbers as5.07000,5.07600,5.07607...,5.07666...,5.07676...so the number in the middle is5.076.Answer:(E) 12.Solution1Since there are24hours in a day,Francis spends13×24=8hours sleeping.Also,he spends14×24=6hours studying,and18×24=3hours eating.The number of hours that he has left is24−8−6−3=7hours. Solution2Francis spends13+14+18=8+6+324=1724of a day either sleeping,studying or eating.This leaves him1−1724=724of his day.Since there are24hours in a full day,then he has7hours left.Answer:(D) 13.Solution1Since the sum of the angles in a triangle is180◦,then∠QP S=180◦−∠P QS−∠P SQ=180◦−48◦−38◦=94◦Therefore,∠RP S=∠QP S−∠QP R=94◦−67◦=27◦.Solution2Since the sum of the angles in a triangle is180◦,then∠QRP=180◦−∠P QR−∠QP R=180◦−48◦−67◦=65◦Therefore,∠P RS=180◦−∠P RQ=180◦−65◦=115◦.Using P RS,∠RP S=180◦−∠P RS−∠P SR=180◦−115◦−38◦=27◦Answer:(A)14.The perimeter of the shaded region equals the sum of the lengths of OP and OQ plus the lengthof arc P Q.Each of OP and OQ has length5.Arc P Q forms34of the circle with centre O and radius5,because the missing portion corre-sponds to a central angle of90◦,and so is14of the total circle.Thus,the length of arc P Q is34of the circumference of this circle,or34(2π(5))=152π.Therefore,the perimeter is5+5+152π≈33.56which,of the given answers,is closest to34.Answer:(A)15.After some trial and error,we obtain the two lists{4,5,7,8,9}and{3,6,7,8,9}.Why are these the only two?If the largest number of theﬁve integers was8,then the largest that the sum could be would be8+7+6+5+4=30,which is too small.This tells us that we must include one9in the list.(We cannot include any number larger than9,since each number must be a single-digit number.)Therefore,the sum of the remaining four numbers is33−9=24.If the largest of the four remaining numbers is7,then their largest possible sum would be 7+6+5+4=22,which is too small.Therefore,we also need to include an8in the list.Thus,the sum of the remaining three numbers is24−8=16.If the largest of the three remaining numbers is6,then their largest possible sum would be 6+5+4=15,which is too small.Therefore,we also need to include an7in the list.Thus,the sum of the remaining two numbers is16−7=9.This tells us that we need two diﬀerent positive integers,each less than7,that add to9.These must be3and6or4and5.This gives us the two lists above,and shows that they are the only two such lists.Answer:(B) 16.The area of the entire grid is4×9=36.The area of P QR is12(QR)(P Q)=12(3)(4)=6.The area of ST U is12(ST)(UT)=12(4)(3)=6.The area of the rectangle with base RS is2×4=8.Therefore,the total shaded area is6+6+8=20and so the unshaded area is36−20=16.The ratio of the shaded area to the unshaded area is20:16=5:4.Answer:(E) 17.We can suppose that each test is worth100marks.Since the average of herﬁve test marks is73%,then the total number of marks that she received is5×73=365.Once her teacher removes a mark,her new average is76%so the sum of the remaining four marks is4×76=304.Since365−304=61,then the mark removed was61%.Answer:(B) 18.Solution1From December31,1988to December31,2008,a total of20years have elapsed.A time period of20years is the same asﬁve4year periods.Thus,the population of Arloe has doubled 5times over this period to its total of 3456.Doubling 5times is equivalent to multiplying by 25=32.Therefore,the population of Arloe on December 31,1988was 345632=108.Solution 2The population doubles every 4years going forward,so is halved every 4years going backwards in time.The population on December 31,2008was 3456.The population on December 31,2004was 3456÷2=1728.The population on December 31,2000was 1728÷2=864.The population on December 31,1996was 864÷2=432.The population on December 31,1992was 432÷2=216.The population on December 31,1988was 216÷2=108.Answer:(D)19.Since Pat drives 60km at 80km/h,this takes him 60km 80km/h =34h.Since Pat has 2hours in total to complete the trip,then he has 2−34=54hours left to complete the remaining 150−60=90km.Therefore,he must travel at 90km 54h =360km/h =72km/h.Answer:(C)20.Since the three numbers in each straight line must have a product of 3240and must include 45,then the other two numbers in each line must have a product of 3240=72.The possible pairs of positive integers are 1and 72,2and 36,3and 24,4and 18,6and 12,and 8and 9.The sums of the numbers in these pairs are 73,38,27,22,18,and 17.To maximize the sum of the eight numbers,we want to choose the pairs with the largest possible sums,so we choose the ﬁrst four pairs.Thus,the largest possible sum of the eight numbers is 73+38+27+22=160.Answer:(E)21.Since each of Alice and Bob rolls one 6-sided die,then there are 6×6=36possible combinationsof rolls.Each of these 36possibilities is equally likely.Alice wins when the two values rolled diﬀer by 1.The possible combinations that diﬀer by 1are (1,2),(2,3),(3,4),(4,5),(5,6),(2,1),(3,2),(4,3),(5,4),and (6,5).Therefore,there are 10combinations when Alice wins.Thus her probability of winning is 1036=518.Answer:(C)22.Diameters P Q and RS cross at the centre of the circle,which we call O .The area of the shaded region is the sum of the areas of P OS and ROQ plus the sum of the areas of sectors P OR and SOQ .Each of P OS and ROQ is right-angled and has its two perpendicular sides of length 4(the radius of the circle).Therefore,the area of each of these triangles is 12(4)(4)=8.Each of sector P OR and sector SOQ has area 14of the total area of the circle,as each hascentral angle 90◦(that is,∠P OR =∠SOQ =90◦)and 90◦is one-quarter of the total central angle.Therefore,each sector has area 14(π(42))=14(16π)=4π.Thus,the total shaded area is 2(8)+2(4π)=16+8π.Answer:(E)23.The maximum possible mass of a given coin is 7×(1+0.0214)=7×1.0214=7.1498g.The minimum possible mass of a given coin is 7×(1−0.0214)=7×0.9786=6.8502g.What are the possible numbers of coins that could make up 1000g?To ﬁnd the largest number of coins,we want the coins to be as light as possible.If all of the coins were as light as possible,we would have 10006.8502≈145.98coins.Now,we cannot have a non-integer number of coins.This means that we must have at most 145coins.(If we had 146coins,the total mass would have to be at least 146×6.8502=1000.1292g,which is too heavy.)Practically,we can get 145coins to have a total mass of 1000g by taking 145coins at the minimum possible mass and making each slightly heavier.To ﬁnd the smallest number of coins,we want the coins to be as heavy as possible.If all of the coins were as heavy as possible,we would have 10007.1498≈139.86coins.Again,we cannot have a non-integer number of coins.This means that we must have at least 140coins.(If we had 139coins,the total mass would be at most 139×7.1498=993.8222g,which is too light.)Therefore,the diﬀerence between the largest possible number and smallest possible number of coins is 145−140=5.Answer:(B)24.Divide the large cube of side length 40into 8smaller cubes of side length 20,by making threecuts of the large cube through its centre using planes parallel to the pairs of faces.Each of these small cubes has the centre of the large cube as its vertex.Each of these small cubes also just encloses one of the large spheres,in the sense that the sphere just touches each of the faces of the small cube.We call the sphere that ﬁts in the central space the inner sphere.To make this sphere as large possible,its centre will be at the centre of the large cube.(If this was not the case,the centre be outside one of the small cubes,and so would be farther away from one of the large spheres than from another.)To ﬁnd the radius of the inner sphere,we must ﬁnd the shortest distance from the centre of the large cube (that is,the centre of the inner sphere)to one of the large spheres.(Think of starting the inner sphere as a point at the centre of the cube and inﬂating it until it just touches the large spheres.)Consider one of these small cubes and the sphere inside it.Join the centre of the small cube to one of its vertices.Since the small cube has side length 20,then this new segment has length √102+102+102or √300,since to get from the centre to a vertex,we must go over 10,down 10and across 10.(See below for an explanation of why this distance is thus √300.)The inner sphere will touch the large sphere along this segment.Thus,the radius of the inner sphere will be this distance (√300)minus the radius of the large sphere (10),and so is √300−10≈7.32.Of the given answers,this is closest to 7.3.(We need to justify why the distance from the centre of the small cube to its vertex is √102+102+102.Divide the small cube into 8tiny cubes of side length 10each.The distance from the centre of the small cube to its vertex is equal to the length of a diagonal of one of the tiny cubes.Consider a rectangular prism with edge lengths a ,b and c .What is the length,d ,of the diag-onal inside the prism?By the Pythagorean Theorem,a face with side lengths a and b has a diagonal of length √a 2+b 2.Consider the triangle formed by this diagonal,the diagonal of the prism and one of the vertical edges of the prism,of length c .cThis triangle is right-angled since the vertical edge is perpendicular to the top face.By the Pythagorean Theorem again,d 2=(√a 2+b 2)2+c 2,so d 2=a 2+b 2+c 2or d =√a 2+b 2+c 2.)Answer:(B)25.The three machines operate in a way such that if the two numbers in the output have a commonfactor larger than 1,then the two numbers in the input would have to have a common factor larger than 1.To see this,let us look at each machine separately.We use the fact that if two numbers are each multiples of d ,then their sum and diﬀerence are also multiples of d .Suppose that (m,n )is input into Machine A.The output is (n,m ).If n and m have a common factor larger than 1,then m and n do as well.Suppose that (m,n )is input into Machine B.The output is (m +3n,n ).If m +3n and n have a common factor d ,then (m +3n )−n −n −n =m has a factor of d as each part of the subtraction is a multiple of d .Therefore,m and n have a common factor of d .Suppose that (m,n )is input into Machine C.The output is (m −2n,n ).If m −2n and n have a common factor d ,then (m −2n )+n +n =m has a factor of d as each part of the addition is a multiple of d .Therefore,m and n have a common factor of d .In each case,any common factor that exists in the output is present in the input.Let us look at the numbers in the ﬁve candidates.After some work,we can ﬁnd the prime factorizations of the six integers:2009=7(287)=7(7)(41)1016=8(127)=2(2)(2)(127)1004=4(251)=2(2)(251)1002=2(501)=2(3)(167)1008=8(126)=8(3)(42)=16(3)(3)(7)=2(2)(2)(2)(3)(3)(7)1032=8(129)=8(3)(43)=2(2)(2)(3)(43)Therefore,the only one of 1002,1004,1008,1016,1032that has a common factor larger than 1with 2009is 1008,which has a common factor of 7with 2009.How does this help?Since2009and1008have a common factor of7,then whatever pair was input to produce(2009,1008)must have also had a common factor of7.Also,the pair that was input to create this pair also had a common factor of7.This can be traced back through every step to say that the initial pair that produces the eventual output of(2009,1008)must have a common factor of7.Thus,(2009,1008)cannot have come from(0,1).Notes:•This does not tell us that the other four pairs necessarily work.It does tell us,though, that(2009,1008)cannot work.•We can trace the other four outputs back to(0,1)with some eﬀort.(This process is easier to do than it is to describe!)To do this,we notice that if the output of Machine A was(a,b),then its input was(b,a), since Machine A switches the two entries.Also,if the output of Machine B was(a,b),then its input was(a−3b,b),since MachineB adds three times the second number to theﬁrst.Lastly,if the output of Machine C was(a,b),then its input was(a+2b,b),since MachineC subtracts two times the second number from theﬁrst.Consider(2009,1016)for example.We try toﬁnd a way from(2009,1016)back to(0,1).We only need toﬁnd one way that works,rather than looking for a speciﬁc way.We note before doing this that starting with an input of(m,n)and then applying MachineB then MachineC gives an output of((m+3n)−2n,n)=(m+n,n).Thus,if applyingMachine B then Machine C(we call this combination“Machine BC”)gives an output of (a,b),then its input must have been(a−b,b).We can use this combined machine to try to work backwards and arrive at(0,1).This will simplify the process and help us avoid negative numbers.We do this by making a chart and by attempting to make the larger number smaller wher-ever possible:Output Machine Input(2009,1016)BC(993,1016)(993,1016)A(1016,993)(1016,993)BC(23,993)(23,993)A(993,23)(993,23)BC,43times(4,23)(4,23)A(23,4)(23,4)BC,5times(3,4)(3,4)A(4,3)(4,3)BC(1,3)(1,3)A(3,1)(3,1)B(0,1)Therefore,by going up through this table,we can see a way to get from an initial input of(0,1)to aﬁnal output of(2009,1016).In a similar way,we can show that we can obtainﬁnal outputs of each of(2009,1004), (2009,1002),and(2009,1032).Answer:(D)。

Fryer滑铁卢数学竞赛(Grade 9)-数学Mathematics-2004-试题 exam

(c) If the game is in the position shown and Bob goes next, show that however Bob plays, Avril can win this game.
4. Triangular numbers can be calculated by counting the dots in the following triangular shapes:
x →
R
→
A
→
R
→
A
→
14 27 ?
R
(c) Using the five steps from (b), what input should you begin with to get a final result of Justify your answer.
2. The Fryer Foundation is giving out four types of prizes, valued at $5, $25, $125 and $625. (a) The Foundation gives out at least one of each type of prize. If five prizes are given out with a total value of $905, how many of each type of prize is given out? Explain how you got your answer. (b) If the Foundation gives out at least one of each type of prize and five prizes in total, determine the other three possible total values it can give out. Explain how you got your answer. (c) There are two ways in which the Foundation could give away prizes totalling $880 while making sure to give away at least one and at most six of each prize. Determine the two ways of doing this, and explain how you got your answer. € 3. In “The Sun Game”, two players take turns placing discs numbered 1 to 9 in the circles on the board. Each number can only be used once. The object of the game is to be the first to place a disc so that the sum of the 3 numbers along a line through the centre circle is 15.

滑铁卢数学竞赛

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

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

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

【比赛形式】注：上述数学竞赛都可以使用计算器。

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

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

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

滑铁卢数学竞赛高中试题

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

求$ a $的值。

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

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

二、填空题1. 计算$ \int_{0}^{1} x^2 dx $。

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

3. 一个等差数列的前三项分别为2，5，8，求该数列的第10项。

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

2. 一个矩形的长是宽的两倍，若矩形的周长为24，求矩形的面积。

3. 已知一个等比数列的前三项分别为3，9，27，求该数列的第5项。

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

如果每天生产150个零件，工厂可以在20天内完成订单。

求工厂每天实际生产的零件数量。

2. 一个圆环的外圆半径是内圆半径的两倍，且圆环的面积为π。

求外圆的半径。

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

六、开放性问题1. 考虑一个无限大的棋盘，每个格子可以放置一个硬币。

欧几里得滑铁卢数学竞赛_2001EuclidSolution

Solution
If f (k) = k , then k2 – 3k – 5 = k
k2 – 4k – 5 = 0
(k – 5)(k + 1) = 0
so
k = 5 or k = –1.
(c) Determine all (x, y) such that x2 + y2 = 25 and x – y = 1.
25
Therefore BC = 59 m to the nearest metre.
Solution 2
Since QR AC, ∠QRP = ∠ BAC (alternating angles).
B
This means ∆ ABC ~ ∆ RPQ (two equal angles).
By Pythagoras, PR2 = QP2 + QR2
2. (a) The vertex of the parabola y = (x – b)2 + b + h has coordinates (2, 5). What is the value of h?
Solution Since the x-coordinate of the vertex is 2, then b = 2. Since the y-coordinate of the vertex is 5, then b + h = 5. Since b = 2, then h = 3.
Solution
Let n be the smallest integer in one of these sequences.
So we want to solve the equation n2 + (n + 1)2 + (n + 2)2 = (n + 3)2 + (n + 4)2 (translating the

滑铁卢大学数学竞赛

加拿大滑铁卢大学举办费马国际数学竞赛及欧几里得数学竞赛。

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

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

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

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

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

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

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

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

考试的及格分数每年大概在40分左右。

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

滑铁卢大学数学竞赛。

Fryer滑铁卢数学竞赛(Grade 9)-数学Mathematics-2012-试题 exam

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

Fryer滑铁卢数学竞赛(Grade 9)-数学Mathematics-2012-试题 exam

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)。

滑铁卢euclid考试常用公式

滑铁卢euclid考试常用公式滑铁卢Euclid考试常用公式。

一、代数部分。

1. 二次函数公式。

二次函数在Euclid考试里那可是常客呢！对于二次函数y = ax^2+bx + c（a≠0），它的顶点坐标公式可太重要啦。

顶点的横坐标x =-(b)/(2a)，纵坐标y=frac{4ac - b^2}{4a}。

就像这个函数是一个小怪兽，这个公式就是找到它最特别的点（顶点）的魔法咒语。

还有求根公式x=frac{-b±√(b^2)-4ac}{2a}，当我们想知道这个二次函数和x轴交点的时候，就靠它啦。

比如说y = x^2-2x - 3，a = 1，b=-2，c=-3，用求根公式就能算出x = 3或者x=-1，是不是很神奇呢？2. 等差数列公式。

等差数列也经常出现在考试中哦。

如果一个数列{a_n}是等差数列，首项是a_1，公差是d，那么通项公式a_n=a_1+(n - 1)d。

这个公式就像是数列的身份证，知道了首项和公差，就能把数列里的任何一项都找出来。

比如说首项a_1=2，公差d = 3，那么第5项a_5=2+(5 - 1)×3=14。

还有求和公式S_n=frac{n(a_1+a_n)}{2}=na_1+(n(n -1)d)/(2)。

这个求和公式就像是一个小钱包，把数列里的数都加起来，然后把总和放在这个小钱包里。

3. 等比数列公式。

等比数列也不甘示弱呢。

等比数列{b_n}，首项b_1，公比q（q≠0），通项公式b_n=b_1q^n - 1。

比如说首项b_1=3，公比q = 2，那么第4项b_4=3×2^4 - 1=24。

求和公式呢，当q≠1时，S_n=frac{b_1(1 - q^n)}{1 - q}；当q = 1时，S_n=nb_1。

这就像是等比数列自己的小账本，把数列的和算得清清楚楚。

二、几何部分。

1. 三角形面积公式。

三角形可是几何里的小明星呢。

最常见的面积公式S=(1)/(2)ah，这里a是三角形的底，h是这条底对应的高。

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1. The value of (4 + 44 + 444) ÷ 4 is
(A) 111
(B) 123
(C) 459
(D) 489
(E) 456
2. Jing purchased eight identical items. If the total cost was $26, then the cost per item, in dollars, was
7. Scoring: Each correct answer is worth 5 in Part A, 6 in Part B, and 8 in Part C. There is no penalty for an incorrect answer. Each unanswered question is worth 2, to a maximum of 10 unanswered questions.
Friday, February 22, 2013
(outside of North America and South America)
Time: 60 minutes Calculators are permitted Instructions
©2012 University of Waterloo
Mass of Jeff s Pet Atlantic Cod
20
Mass in kg
10
00
2
4
6
8
Age in Years
(A) 3
(B) 7
(C) 4
(D) 6
(E) 5
5. What is the value of 13 + 23 + 33 + 43?
(A) 101
(B) 103
(C) 102
(A) 28
(B) 38
(C) 26
(D) 152
(E) 45
R
Q 3x
x P
S
T
15. If 4n = 642, then n equals
(A) 3
(B) 5
(C) 6
(D) 8
(E) 12
16. An integer x is chosen so that 3x + 1 is an even integer. Which of the following must be an odd integer?
(A) $4.80
(B) $1.50
(C) $4.50
(D) $6.00
(E) $7.50
10. The time on Байду номын сангаас cell phone is 3:52. How many minutes will pass before the phone next shows a time using each of the digits 2, 3 and 5 exactly once?
(A) x + 3
(B) x − 3
(C) 2x
(D) 7x + 4 (E) 5x + 3
17. The graph shows styles of music on a playlist. Country music songs are added to the playlist so that now 40% of the songs are Country. If the ratio of Hip Hop songs to Pop songs remains the same, what percentage of the total number of songs are now Hip Hop?
6. This is a multiple-choice test. Each question is followed by ﬁve possible answers marked A, B, C, D, and E. Only one of these is correct. After making your choice, ﬁll in the appropriate circle on the response form.
(A) −6
(B) 13
(C) 54
(D)
62 3
(E)
−
71 3
13. Which number from the set {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11} must be removed so that the mean (average) of the numbers remaining in the set is 6.1?
1. Do not open the Contest booklet until you are told to do so.
2. You may use rulers, compasses and paper for rough work.
3. Be sure that you understand the coding system for your response form. If you are not sure, ask your teacher to clarify it. All coding must be done with a pencil, preferably HB. Fill in circles completely.
(A)
(B)
(C)
(D)
(E)
4. The graph shows the mass, in kilograms, of Jeﬀ’s pet Atlantic cod, given its age in years. What is the age of the cod when its mass is 15 kg?
The CENTRE for EDUCATION in MATHEMATICS and COMPUTING
cemc.uwaterloo.ca
Pascal Contest
(Grade 9)
Thursday, February 21, 2013
(in North America and South America)
(D) 105
(E) 104
6.
Erin walks
3 5
of the way home in 30 minutes.
If she continues to walk at the same
rate, how many minutes will it take her to walk the rest of the way home?
4. On your response form, print your school name and city/town in the box in the upper right corner.
5. Be certain that you code your name, age, sex, grade, and the Contest you are writing in the response form. Only those who do so can be counted as eligible students.
(A) 4
(B) 5
(C) 6
(D) 7
(E) 8
14. In the diagram, P QR is a straight line segment and QS = QT . Also, ∠P QS = x◦ and ∠T QR = 3x◦. If ∠QT S = 76◦, the value of x is
SR = 3. Point U is on QR with QU = 2. Point T is
6
on P S with ∠T UR = 90◦. What is the length of T R? P
T
S
(A) 3
(B) 4
(C) 5
(D) 6
(E) 7
3
Q 2U
R
9. Owen spends $1.20 per litre on gasoline. He uses an average of 1 L of gasoline to drive 12.5 km. How much will Owen spend on gasoline to drive 50 km?
How many times does the symbol ♥ occur within the ﬁrst 53 symbols of the pattern?
(A) 25
(B) 26
(C) 27
(D) 28
(E) 29
12. If x = 11, y = −8, and 2x − 3z = 5y, what is the value of z?
(A) 27
(B) 59
(C) 77
(D) 91
(E) 171
Part B: Each correct answer is worth 6.
11. The same sequence of four symbols repeats to form the following pattern:
♥ ♣ ♠ ♥ ♥ ♣ ♠ ♥ ♥ ♣ ♠ ♥ ...
radius 1. What is the area of the shaded region?
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(A) 16 − π2 (B) 16 − 4π (C) 4 − 4π
(D) 4 − 4π2 (E) 4 − π
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