滑铁卢大学欧几里得数学竞赛

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欧几里得数学竞赛奖项设置

欧几里得数学竞赛奖项设置全文共四篇示例，供读者参考第一篇示例：欧几里得数学竞赛是一个广受欢迎的数学竞赛，旨在挑战学生的数学思维和解题能力。

为了激励学生参与竞赛并取得优异成绩，欧几里得数学竞赛设立了多个奖项，以表彰在竞赛中表现出色的学生。

欧几里得数学竞赛设立了优胜奖，对取得竞赛前几名的学生进行奖励。

这些优胜奖通常包括奖杯、奖状以及现金奖励，同时也为学生的学业成绩和未来的发展增添了荣誉。

优胜奖的设置旨在激励学生在竞赛中发挥自己最好的水平，争取取得好成绩。

欧几里得数学竞赛还设立了进步奖，以奖励在竞赛中有所进步并尽力发挥自己潜力的学生。

这些进步奖可以帮助学生建立自信，鼓励他们在数学学习上不断努力，不断提高自己的水平。

进步奖的设置体现了竞赛的公平性和包容性，让更多的学生感受到参与竞赛的乐趣和意义。

欧几里得数学竞赛还设立了团体奖项，以表彰在竞赛中表现出色的团队。

团体奖通常根据团队在竞赛中的总成绩来评定，对团队成绩最优异的队伍进行奖励。

团体奖的设置有助于培养学生的团队合作精神和集体荣誉感，激发学生共同努力、共同进步的动力。

在欧几里得数学竞赛中还设立了一些特别奖项，如最佳答题奖、最佳团队合作奖等，以表彰在某些特定方面表现突出的学生。

这些特别奖项旨在鼓励学生在竞赛中展现出色的特长和才能，帮助他们更全面地发展自己的能力和潜力。

欧几里得数学竞赛奖项的设置充分考虑了学生的学习需求和成长动力，通过不同类型的奖项激励学生参与竞赛、勇于挑战自我，并在竞赛中发挥出色的表现。

这些奖项的设立有助于培养学生的自信心、竞争意识和团队精神，为他们的未来学习和发展奠定了良好的基础。

希望欧几里得数学竞赛能够继续发扬光大，激励更多学生热爱数学、热爱竞赛，为数学事业的发展贡献自己的力量。

第二篇示例：欧几里得数学竞赛是一项著名的数学竞赛，旨在鼓励和推广对数学的兴趣和研究。

该竞赛以古希腊数学家欧几里得的名字命名，他被认为是几何学之父，开创了几何学的先河。

欧几里得数学竞赛_

欧几里得数学竞赛_摘要：I.欧几里得数学竞赛概述- 竞赛起源与发展- 竞赛难度与影响力II.欧几里得数学竞赛适合人群- 参赛对象与报名方式- 竞赛对申请大学的帮助III.欧几里得数学竞赛考试内容与形式- 竞赛知识点覆盖范围- 考试时间与题型- 评分标准与奖项设置IV.欧几里得数学竞赛备考策略- 备考时间安排- 推荐教材与学习资源- 真题练习与模拟考试V.欧几里得数学竞赛在中国的发展- 我国学生参赛情况- 相关培训机构与课程- 对我国数学教育的启示与影响正文：欧几里得数学竞赛（Euclid Mathematics Contest）是由加拿大滑铁卢大学（University of Waterloo）数学与计算机学院主办的面向全球高中生的数学竞赛，被誉为数学界的托福。

竞赛始于1963年，每年有来自10多个国家和地区、1850多所学校的2万多名学生参加。

该竞赛在数学界中已经得到广泛认可，对学生的申请大学具有很大的帮助。

欧几里得数学竞赛适合人群广泛，参赛对象为全球各地的高中生，报名方式一般由学校统一组织。

竞赛难度较高，知识点覆盖范围广泛，对学生的逻辑思维能力和数学素养有很高的要求。

在我国，许多学生通过参加欧几里得数学竞赛，提高了自身的数学能力，为申请国内外知名大学提供了有力的砝码。

欧几里得数学竞赛的考试内容主要包括代数、几何、组合、数论等多个方面，考试形式为笔试，分为简答题和解答题。

评分标准根据解题过程的准确性、完整性和创新性来评判，奖项分为金、银、铜三个等级。

对于如何备考欧几里得数学竞赛，建议学生合理安排时间，提前准备。

推荐使用一些经典的数学竞赛教材和在线学习资源，如《数学竞赛题型解析》、《欧几里得数学竞赛真题详解》等。

在备考过程中，要注重真题练习和模拟考试，以检验自己的学习效果，逐步提高自己的解题能力。

近年来，随着我国学生对国际数学竞赛的热情逐渐高涨，欧几里得数学竞赛在我国也得到了广泛关注。

越来越多的学生通过参加欧几里得数学竞赛，提升了自己的数学素养，为我国数学教育的发展带来了新的启示和影响。

加拿大欧几里得数学竞赛真题解析2020EuclidSolution

1.(a)Solution1If x=−2,then 3x+6x+2=3(x+2)x+2=3.In other words,for every x=−2,the expression is equal to3.Therefore,when x=11,we get 3x+6x+2=3.Solution2When x=11,we obtain 3x+6x+2=3(11)+611+2=3913=3.(b)Solution1The point at which a line crosses the y-axis has x-coordinate0.Because A has x-coordinate−1and B has x-coordinate1,then the midpoint of AB is on the y-axis and is on the line through A and B,so is the point at which this line crossesthe x-axis.The midpoint of A(−1,5)and B(1,7)is 12(−1+1),12(5+7)or(0,6).Therefore,the line that passes through A(−1,5)and B(1,7)has y-intercept6.Solution2The line through A(−1,5)and B(1,7)has slope7−51−(−1)=22=1.Since the line passes through B(1,7),its equation can be written as y−7=1(x−1)or y=x+6.The line with equation y=x+6has y-intercept6.(c)First,weﬁnd the coordinates of the point at which the lines with equations y=3x+7and y=x+9intersect.Equating values of y,we obtain3x+7=x+9and so2x=2or x=1.When x=1,we get y=x+9=10.Thus,these two lines intersect at(1,10).Since all three lines pass through the same point,the line with equation y=mx+17 passes through(1,10).Therefore,10=m·1+17which gives m=10−17=−7.2.(a)Suppose that m has hundreds digit a,tens digit b,and ones(units)digit c.From the given information,a,b and c are distinct,each of a,b and c is less than10, a=bc,and c is odd(since m is odd).The integer m=623satisﬁes all of these conditions.Since we are told there is only one such number,then623must be the only answer.Why is this the only possible value of m?We note that we cannot have b=1or c=1,otherwise a=c or a=b.Thus,b≥2and c≥2.Since c≥2and c is odd,then c can equal3,5,7,or9.Since b≥2and a=bc,then if c equals5,7or9,a would be larger than10,which is not possible.Thus,c=3.Since b≥2and b=c,then b=2or b≥4.If b≥4and c=3,then a>10,which is not possible.Therefore,we must have c=3and b=2,which gives a=6.(b)Since Eleanor has 100marbles which are black and gold in the ratio 1:4,then 15of her marbles are black,which means that she has 15·100=20black marbles.When more gold marbles are added,the ratio of black to gold is 1:6,which means thatshe has 6·20=120gold marbles.Eleanor now has 20+120=140marbles,which means that she added 140−100=40gold marbles.(c)First,we see that n 2+n +15n =n 2n +n n +15n =n +1+15n .This means that n 2+n +15n is an integer exactly when n +1+15nis an integer.Since n +1is an integer,then n 2+n +15n is an integer exactly when 15n is an integer.The expression 15nis an integer exactly when n is a divisor of 15.Since n is a positive integer,then the possible values of n are 1,3,5,and 15.3.(a)First,we note that a triangle with one right angle and one angle with measure 45◦isisosceles.This is because the measure of the third angle equals 180◦−90◦−45◦=45◦which means that the triangle has two equal angles.In particular, CDE is isosceles with CD =DE and EF G is isosceles with EF =F G .Since DE =EF =1m,then CD =F G =1m.Join C to G .ACB D FG HE 45°45°45°1 m 1 m 1 m 1 m Consider quadrilateral CDF G .Since the angles at D and F are right angles and since CD =GF ,it must be the case that CDF G is a rectangle.This means that CG =DF =2m and that the angles at C and G are right angles.Since ∠CGF =90◦and ∠DCG =90◦,then ∠BGC =180◦−90◦−45◦=45◦and ∠BCG =90◦.This means that BCG is also isosceles with BC =CG =2m.Finally,BD =BC +CD =2m +1m =3m.(b)We apply the process two more times:x y Before Step 1243After Step 1273After Step 2813After Step 3814x y Before Step 1814After Step 1854After Step 23404After Step 33405Therefore,the ﬁnal value of x is 340.(c)The parabola with equation y =kx 2+6x +k has two distinct x -intercepts exactly whenthe discriminant of the quadratic equation kx 2+6x +k =0is positive.Here,the disciminant equals ∆=62−4·k ·k =36−4k 2.The inequality 36−4k 2>0is equivalent to k 2<9.Since k is an integer and k =0,then k can equal −2,−1,1,2.(If k ≥3or k ≤−3,we get k 2≥9so no values of k in these ranges give the desired result.)4.(a)Since a b <47and 47<1,then a b<1.Since a and b are positive integers,then a <b .Since the diﬀerence between a and b is 15and a <b ,then b =a +15.Therefore,we have 59<a a +15<47.We multiply both sides of the left inequality by 9(a +15)(which is positive)to obtain 5(a +15)<9a from which we get 5a +75<9a and so 4a >75.From this,we see that a >754=18.75.Since a is an integer,then a ≥19.We multiply both sides of the right inequality by 7(a +15)(which is positive)to obtain 7a <4(a +15)from which we get 7a <4a +60and so 3a <60.From this,we see that a <20.Since a is an integer,then a ≤19.Since a ≥19and a ≤19,then a =19,which means that a b =1934.(b)The ﬁrst 6terms of a geometric sequence with ﬁrst term 10and common ratio 12are 10,5,52,54,58,516.Here,the ratio of its 6th term to its 4th term is 5/165/4which equals 14.(We could have determined this without writing out the sequence,since moving from the 4th term to the 6th involves multiplying by 12twice.)The ﬁrst 6terms of an arithmetic sequence with ﬁrst term 10and common diﬀerence d are 10,10+d,10+2d,10+3d,10+4d,10+5d .Here,the ratio of the 6th term to the 4th term is 10+5d 10+3d .Since these ratios are equal,then 10+5d 10+3d =14,which gives 4(10+5d )=10+3d and so 40+20d =10+3d or 17d =−30and so d =−3017.5.(a)Let a=f(20).Then f(f(20))=f(a).To calculate f(f(20)),we determine the value of a and then the value of f(a).By deﬁnition,a=f(20)is the number of prime numbers p that satisfy20≤p≤30.The prime numbers between20and30,inclusive,are23and29,so a=f(20)=2.Thus,f(f(20))=f(a)=f(2).By deﬁnition,f(2)is the number of prime numbers p that satisfy2≤p≤12.The prime numbers between2and12,inclusive,are2,3,5,7,11,of which there are5.Therefore,f(f(20))=5.(b)Since(x−1)(y−2)=0,then x=1or y=2.Suppose that x=1.In this case,the remaining equations become:(1−3)(z+2)=01+yz=9or−2(z+2)=0yz=8From theﬁrst of these equations,z=−2.From the second of these equations,y(−2)=8and so y=−4.Therefore,if x=1,the only solution is(x,y,z)=(1,−4,−2).Suppose that y=2.In this case,the remaining equations become:(x−3)(z+2)=0x+2z=9From theﬁrst equation x=3or z=−2.If x=3,then3+2z=9and so z=3.If z=−2,then x+2(−2)=9and so x=13.Therefore,if y=2,the solutions are(x,y,z)=(3,2,3)and(x,y,z)=(13,2,−2).In summary,the solutions to the system of equations are(x,y,z)=(1,−4,−2),(3,2,3),(13,2,−2)We can check by substitution that each of these triples does indeed satisfy each of the equations.6.(a)Draw a perpendicular from S to V on BC .Since ASV B is a quadrilateral with three right angles,then it has four right angles and so is a rectangle.Therefore,BV =AS =r ,since AS is a radius of the top semi-circle,and SV =AB =4.Join S and T to P .Since the two semi-circles are tangent at P ,then SP T is a straight line,which means that ST =SP +P T =r +r =2r .64Consider right-angled SV T .We have SV =4and ST =2r .Also,V T =BC −BV −T C =6−r −r =6−2r .By the Pythagorean Theorem,SV 2+V T 2=ST 242+(6−2r )2=(2r )216+36−24r +4r 2=4r 252=24rThus,r =5224=136.(b)Since ABE is right-angled at A and is isosceles with AB =AE =7√2,then ABE is a 45◦-45◦-90◦triangle,which means that ∠ABE =45◦and BE =√2AB =√2·7√2=14.Since BCD is right-angled at C with DB DC =8x 4x =2,then BCD is a 30◦-60◦-90◦triangle,which means that ∠DBC =30◦.Since ∠ABC =135◦,then ∠EBD =∠ABC −∠ABE −∠DBC =135◦−45◦−30◦=60◦.Now consider EBD .We have EB =14,BD =8x ,DE =8x −6,and ∠EBD =60◦.Using the cosine law,we obtain the following equivalent equations:DE 2=EB 2+BD 2−2·EB ·BD ·cos(∠EBD )(8x −6)2=142+(8x )2−2(14)(8x )cos(60◦)64x 2−96x +36=196+64x 2−2(14)(8x )·12−96x =160−14(8x )112x −96x =16016x =160x =10Therefore,the only possible value of x is x =10.7.(a)Solution 1Since the function g is linear and has positive slope,then it is one-to-one and so invertible.This means that g −1(g (a ))=a for every real number a and g (g −1(b ))=b for every real number b .Therefore,g (f (g −1(g (a ))))=g (f (a ))for every real number a .This means thatg (f (a ))=g (f (g −1(g (a ))))=2(g (a ))2+16g (a )+26=2(2a −4)2+16(2a −4)+26=2(4a 2−16a +16)+32a −64+26=8a 2−6Furthermore,if b =f (a ),then g −1(g (f (a )))=g −1(g (b ))=b =f (a ).Therefore,f (a )=g −1(g (f (a )))=g −1(8a 2−6)Since g (x )=2x −4,then y =2g −1(y )−4and so g −1(y )=12y +2.Therefore,f (a )=12(8a 2−6)+2=4a 2−1and so f (π)=4π2−1.Solution 2Since the function g is linear and has positive slope,then it is one-to-one and so invertible.To ﬁnd a formula for g −1(y ),we start with the equation g (x )=2x −4,convert to y =2g −1(y )−4and then solve for g −1(y )to obtain 2g −1(y )=y +4and so g −1(y )=y +42.We are given that g (f (g −1(x )))=2x 2+16x +26.We can apply the function g −1to both sides to obtain successively:f (g −1(x ))=g −1(2x 2+16x +26)f (g −1(x ))=(2x 2+16x +26)+42(knowing a formula for g −1)f (g −1(x ))=x 2+8x +15f x +42 =x 2+8x +15(knowing a formula for g −1)f x +42 =x 2+8x +16−1f x +42=(x +4)2−1We want to determine the value of f (π).Thus,we can replace x +42with π,which is equivalent to replacing x +4with 2π.Thus,f (π)=(2π)2−1=4π2−1.(b)Solution 1Using logarithm laws,the given equations are equivalent tolog 2(sin x )+log 2(cos y )=−32log 2(sin x )−log 2(cos y )=12Adding these two equations,we obtain 2log 2(sin x )=−1which gives log 2(sin x )=−12and so sin x =2−1/2=121/2=1√2.Since 0◦≤x <180◦,then x =45◦or x =135◦.Since log 2(sin x )+log 2(cos y )=−32and log 2(sin x )=−12,then log 2(cos y )=−1,which gives cos y =2−1=12.Since 0◦≤y <180◦,then y =60◦.Therefore,(x,y )=(45◦,60◦)or (x,y )=(135◦,60◦).Solution 2First,we note that 21/2=√2and 2−3/2=123/2=12121/2=12√2.From the given equations,we obtainsin x cos y =2−3/2=12√2sin x cos y=21/2=√2Multiplying these two equations together,we obtain (sin x )2=12which gives sin x =±1√2.Since 0◦≤x <180◦,it must be the case that sin x ≥0and so sin x =1√2.Since 0◦≤x <180◦,we obtain x =45◦or x =135◦.Since sin x cos y =12√2and sin x =1√2,we obtain cos y =12.Since 0◦≤y <180◦,then y =60◦.Therefore,(x,y )=(45◦,60◦)or (x,y )=(135◦,60◦).8.(a)Solution1Let x be the probability that Bianca wins the tournament.Because Alain,Bianca and Chen are equally matched and because their roles in the tour-nament are identical,then the probability that each of them wins will be the same.Thus,the probability that Alain wins the tournament is x and the probability that Chen wins the tournament is x.Let y be the probability that Dave wins the tournament.Since exactly one of Alain,Bianca,Chen,and Dave wins the tournament,then3x+y=1and so x=1−y3.We can calculate y in terms of p.In order for Dave to win the tournament,he needs to win two matches.No matter who Dave plays,his probability of winning each match is p.Thus,the probability that he wins his two consecutive matches is p2and so the probability that he wins the tournament is y=p2.Thus,the probability that Bianca wins the tournament is 1−p23.(We could rewrite this as −p2+0p+13to match the desired form.)Solution2Let x be the probability that Bianca wins the tournament. There are three possible pairings for theﬁrst two matches: (i)Bianca versus Alain,and Chen versus Dave(ii)Bianca versus Chen,and Alain versus Dave(iii)Bianca versus Dave,and Alain versus ChenEach of these three pairings occurs with probability13.In(i),Bianca wins either if Bianca beats Alain,Chen beats Dave,and Bianca beats Chen, or if Bianca beats Alain,Dave beats Chen,and Bianca beats Dave.Since Bianca beats Alain with probability12,Chen beats Dave with probability1−p,andBianca beats Chen with probability12,then theﬁrst possibility has probability12·(1−p)·12.Since Bianca beats Alain with probability12,Dave beats Chen with probability p,andBianca beats Dave with probability1−p,then the second possibility has probability1 2·p·(1−p).Therefore,the probability of Bianca winning,given that possibility(i)occurs,is12·(1−p)·12+12·p·(1−p).In(ii),Bianca wins either if Bianca beats Chen,Alain beats Dave,and Bianca beats Alain, or if Bianca beats Alain,Dave beats Alain,and Bianca beats Dave.The combined probability of these is12·(1−p)·12+12·p·(1−p).In(iii),Bianca wins either if Bianca beats Dave,Alain beats Chen,and Bianca beats Alain,or if Bianca beats Dave,Chen beats Alain,and Bianca beats Chen.The combined probability of these is(1−p)·12·12+(1−p)·12·12.Therefore,x=13 14(1−p)+12p(1−p)+14(1−p)+12p(1−p)+14(1−p)+14(1−p)=13(p(1−p)+(1−p))=13(p−p2+1−p)Thus,the probability that Bianca wins the tournament is 1−p23.(b)Throughout this solution,we will mostly not include units,but will assume that all lengthsare in kilometres,all times are in seconds,and all speeds are in kilometres per second.We place the points in the coordinate plane with B at(0,0),A on the negative x-axis, and C on the positive x-axis.We put A at(−1,0)and C at(2,0).Suppose that P has coordinates(x,y)and that the distance from P to B is d km.Since the sound arrives at A12s after arriving at B and sound travels at13km/s,then Ais(12s)·(13km/s)=16km farther from P than B is.Thus,the distance from P to A is(d+16)km.Since the sound arrives at C an additional1second later,then C is an additional13kmfarther,and so is(d+16)km+(13km)=(d+12)km from P.Since the distance from P to B is d km,then(x−0)2+(y−0)2=d2.Since the distance from P to A is(d+16)km,then(x+1)2+(y−0)2=(d+16)2.Since the distance from P to C is(d+12)km,then(x−2)2+(y−0)2=(d+12)2.When these equations are expanded and simpliﬁed,we obtainx2+y2=d2x2+2x+1+y2=d2+13d+136x2−4x+4+y2=d2+d+14 Subtracting theﬁrst equation from the second,we obtain2x+1=13d+136Subtracting theﬁrst equation from the third,we obtain−4x+4=d+14 Therefore,2(2x+1)+(−4x+4)=2(13d+136)+(d+14)6=23d+118+d+14216=24d+2+36d+9(multiplying by36)205=60dd=4112Therefore,the distance from B to P is4112km.9.(a)After each round,each L shape is divided into4smaller L shapes.This means that the number of L shapes increases by a factor of4after each round.After1round,there are4L shapes.After2rounds,there are42=16L’s of the smallest size.After3rounds,there are43=64L’s of the smallest size.(b)There are four orientations of L shapes of a given size:,,,.When an L of each orientation is subdivided,the followingﬁgures are obtained:From theseﬁgures,we can see that after each subsequent round,•Each produces2,0,1,and1of the smallest size.•Each produces0,2,1,and1.•Each produces1,1,2,and0.•Each produces1,1,0,and2.After1round,there are2,0,1,and1.After2rounds,the number of L’s of each orientation are as follows:•:2·2+0·0+1·1+1·1=6•:2·0+0·2+1·1+1·1=2•:2·1+0·1+1·2+1·0=4•:2·1+0·1+1·0+1·2=4After3rounds,the number of L’s of each orientation are as follows:•:6·2+2·0+4·1+4·1=20•:6·0+2·2+4·1+4·1=12•:6·1+2·1+4·2+4·0=16•:6·1+2·1+4·0+4·2=16Where do these numbers come from?For example,to determine the number of after2rounds,we look at the number of L’s of each orientation after round1(2,0,1,1)and ask how many each of these produces at the next level.Since the four types each produce2,0,1,and1,then the total number of after2rounds equals2·2+0·0+1·1+1·1which equals6.As a second example,to determine the number of after3rounds,we note that after2 rounds the number of L’s of the four diﬀerent orientations are6,2,4,4and that each L of each of the four types produces0,2,1,1.This means that the total number ofafter3rounds is6·0+2·2+4·1+4·1=12.Putting all of this together,the number of L’s of the smallest size in the same orientation as the original L is20.(c)In(b),we determined the number of L’s of the smallest size in each orientation after1,2and3rounds.We continue to determine the number of L’s of the smallest size after4rounds.After4rounds,the number of L’s of each orientation are as follows:•:20·2+12·0+16·1+16·1=72•:20·0+12·2+16·1+16·1=56•:20·1+12·1+16·2+16·0=64•:20·1+12·1+16·0+16·2=64This gives us the following tables of the numbers of L’s of the smallest size in each orien-tation after1,2,3,and4rounds:After Round1201126244320121616472566464We re-write these numbers in the third row as16+4,16−4,16,16and the numbers in the fourth row as64+8,64−8,64,64.Based on this,we might guess that the numbers of L’s of the smallest size in each orien-tation after n rounds are4n−1+2n−1,4n−1−2n−1,4n−1,4n−1.If this guess is correct,then,after2020rounds,the number of L’s of the smallest size in the same orientation as the original L is42019+22019.We prove that these guesses are right by using an inductive process.First,we note that the table above shows that our guess is correct when n=1,2,3,4.Next,if we can show that our guess being correct after a given number of rounds implies that it is correct after the next round,then it will be correct after every round.This is because being correct after4rounds will mean that it is correct after5rounds,being correct after5rounds will mean that it is correct after6rounds,and so on to be correct after any number of rounds.Suppose,then,that after k rounds the numbers of L’s of the smallest size in each orienta-tion are4k−1+2k−1,4k−1−2k−1,4k−1,4k−1.After k+1rounds(that is,after the next round),the number of L’s of each orientation is:•:(4k−1+2k−1)·2+(4k−1−2k−1)·0+4k−1·1+4k−1·1=4·4k−1+2·2k−1=4k+2k•:(4k−1+2k−1)·0+(4k−1−2k−1)·2+4k−1·1+4k−1·1=4·4k−1−2·2k−1=4k−2k•:(4k−1+2k−1)·1+(4k−1−2k−1)·1+4k−1·2+4k−1·0=4·4k−1=4k•:(4k−1+2k−1)·1+(4k−1−2k−1)·1+4k−1·0+4k−1·2=4·4k−1=4kSince k=(k+1)−1,these expressions match our guess.This means that our guess is correct after every number of rounds.Therefore,after2020rounds,the number of L’s of the smallest size in the same orientation as the original L is42019+22019.10.(a)Here,the pairwise sums of the numbers a1≤a2≤a3≤a4are s1≤s2≤s3≤s4≤s5≤s6.The six pairwise sums of the numbers in the list can be expressed asa1+a2,a1+a3,a1+a4,a2+a3,a2+a4,a3+a4Since a1≤a2≤a3≤a4,then the smallest sum must be the sum of the two smallest numbers.Thus,s1=a1+a2.Similarly,the largest sum must be the sum of the two largest numbers,and so s6=a3+a4.Since a1≤a2≤a3≤a4,then the second smallest sum is a1+a3.This is because a1+a3 is no greater than each of the four sums a1+a4,a2+a3,a2+a4,and a3+a4: Since a3≤a4,then a1+a3≤a1+a4.Since a1≤a2,then a1+a3≤a2+a3.Since a1≤a2and a3≤a4,then a1+a3≤a2+a4.Since a1≤a4,then a1+a3≤a3+a4.Thus,s2=a1+a3.Using a similar argument,s5=a2+a4.So far,we have s1=a1+a2and s2=a1+a3and s5=a2+a4and s6=a3+a4.This means that s3and s4equal a1+a4and a2+a3in some order.It turns out that either order is possible.Case1:s3=a1+a4and s4=a2+a3Here,a1+a2=8and a1+a3=104and a2+a3=110.Adding these three equations gives(a1+a2)+(a1+a3)+(a2+a3)=8+104+110and so2a1+2a2+2a3=222or a1+a2+a3=111.Since a2+a3=110,then a1=(a1+a2+a3)−(a2+a3)=111−110=1.Since a1=1and a1+a2=8,then a2=7.Since a1=1and a1+a3=104,then a3=103.Since a3=103and a3+a4=208,then a4=105.Thus,(a1,a2,a3,a4)=(1,7,103,105).Case2:s3=a2+a3and s4=a1+a4Here,a1+a2=8and a1+a3=104and a2+a3=106.Using the same process,a1+a2+a3=109.From this,we obtain(a1,a2,a3,a4)=(3,5,101,107).Therefore,Kerry’s two possible lists are1,7,103,105and3,5,101,107.(b)Suppose that the values of s1,s2,s3,s4,s5,s6,s7,s8,s9,s10areﬁxed,but unknown.In terms of the numbers a1≤a2≤a3≤a4≤a5,the ten pairwise sums are a1+a2,a1+a3,a1+a4,a1+a5,a2+a3,a2+a4,a2+a5,a3+a4,a3+a5,a4+a5 These will equal s1,s2,s3,s4,s5,s6,s7,s8,s9,s10in some order.Using a similar analysis to that in(a),the smallest sum is a1+a2and the largest sum is a4+a5.Thus,s1=a1+a2and s10=s4+s5.Also,the second smallest sum will be s2=a1+a3and the second largest sum will be s9=a3+a5.We letS=s1+s2+s3+s4+s5+s6+s7+s8+s9+s10Note that S has aﬁxed,but unknown,value.Even though we do not know the order in which these pairwise sums are assigned to s1 through s10,the value of S will equal the sum of these ten pairwise expressions.In other words,S=4a1+4a2+4a3+4a4+4a5,since each of the numbers in the list occurs in four sums.Thus,a1+a2+a3+a4+a5=14S and so(a1+a2)+a3+(a4+a5)=14S.This means that s1+a3+s10=14S and so a3=14S−s1−s10.Since the values of s1,s10and S areﬁxed,then we are able to determine the value of a3 from the list of sums s1through s10.Using the value of a3,the facts that s2=a1+a3and s9=a3+a5,and that s2and s9are known,we can determine a1and a5.Finally,using s1=a1+a2and s10=a4+a5and the values of a1and a5,we can determine a2and a4.Therefore,given the ten sums s1through s10,we can determine the values of a3,a1,a5, a2,a4and so there is only one possibility for the list a1,a2,a3,a4,a5.(Can you write out expressions for each of a1through a5in terms of s1through s10only?)(c)Suppose that the lists a1,a2,a3,a4and b1,b2,b3,b4produce the same list of sums s1,s2,s3,s4,s5,s6.(Examples of such lists can be found in(a).)Let x be a positive integer.Consider the following list with8entries:a1,a2,a3,a4,b1+x,b2+x,b3+x,b4+xFrom this list,there are three categories of pairwise sums:(i)a i+a j,1≤i<j≤4:these give the sums s1through s6(ii)(b i+x)+(b j+x),1≤i<j≤4:each of these is2x greater than the six sums s1 through s6because the pairwise sums b i+b j give the six sums s1through s6 (iii)a i+(b j+x),1≤i≤4and1≤j≤4Consider also the list with8entries:a1+x,a2+x,a3+x,a4+x,b1,b2,b3,b4From this list,there are again three categories of pairwise sums:(i)b i+b j,1≤i<j≤4:these give the sums s1through s6(ii)(a i+x)+(a j+x),1≤i<j≤4:each of these is2x greater than the six sums s1 through s6because the pairwise sums a i+a j give the six sums s1through s6 (iii)(a i+x)+b j,1≤i≤4and1≤j≤4Thus,the28pairwise sums in each case are the same.In each case,there are6sums in (i),6sums in(ii),and16sums in(iii).If we choose the initial lists to have the same pairwise sums and choose the value of x to be large enough so that a i+x is not equal to any b j and b i+x is not equal to any a j,we obtain two diﬀerent lists of8numbers that each produce the same list of28sums.For example,if we choose a1,a2,a3,a4to be1,7,103,105and b1,b2,b3,b4to be3,5,101,107 and x=10000,we get the lists1,7,103,105,10003,10005,10101,10107and3,5,101,107,10001,10007,10103,10105Using a similar analysis to that above,if the lists a1,a2,a3,a4,a5,a6,a7,a8and b1,b2, b3,b4,b5,b6,b7,b8have the same set of pairwise sums,then the listsa1,a2,a3,a4,a5,a6,a7,a8,b1+y,b2+y,b3+y,b4+y,b5+y,b6+y,b7+y,b8+y anda1+y,a2+y,a3+y,a4+y,a5+y,a6+y,a7+y,a8+y,b1,b2,b3,b4,b5,b6,b7,b8will also have the same pairwise sums.Therefore,setting y=1000000,we see that the lists1,7,103,105,10003,10005,10101,10107,1000003,1000005,1000101,1000107,1010001,1010007,1010103,1010105and3,5,101,107,10001,10007,10103,10105,1000001,1000007,1000103,1000105,1010003,1010005,1010101,1010107have the same list of sums s1,s2,...,s120,as required.。

2011EuclidSolution

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 sh.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 −B 2 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.。

欧几里得滑铁卢数学竞赛_2010EuclidSolution

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Solutions
©2010 Centre for Education in Mathematics and Computing
2010 Euclid Contest Solutions
Page 2
1. (a) Solution 1 Since 3x = 27, then 3x+2 = 3x32 = 27 · 9 = 243.

欧几里得竞赛试题

欧几里得竞赛试题欧几里得数学竞赛（Euclid Mathematics Contest）是由加拿大滑铁卢大学主办的一项面向高中生的数学竞赛，其目的是提高学生的数学问题解决能力。

这项竞赛通常包括一系列选择题和解答题，题目涉及代数、几何、组合数学和数论等不同领域。

# 题目1：代数问题给定一个二次方程 \( ax^2 + bx + c = 0 \)，其中 \( a \)、\( b \) 和 \( c \) 是已知的实数，且 \( a \neq 0 \)。

求方程的根，并证明根的和与积与 \( a \)、\( b \)、\( c \) 的关系。

# 题目2：几何问题考虑一个直角三角形，其中直角边长分别为 \( 3 \) 和 \( 4 \)，斜边长为 \( 5 \)。

求以斜边为轴旋转一周形成的圆锥体的体积。

# 题目3：组合问题一个班级有 50 名学生，需要从中选出一个由 5 人组成的委员会。

如果委员会中必须包括至少一名女生，那么有多少种不同的委员会组成方式？# 题目4：数论问题证明对于任意的正整数 \( n \)，\( n^5 - n \) 总是能被 30 整除。

# 题目5：逻辑推理在一个数学竞赛中，有三名选手 A、B 和 C。

已知 A 没有获得第一名，B 没有获得第二名，C 获得了第三名。

根据这些信息，推断出 A、B和 C 的名次。

# 题目6：概率问题一个袋子里有 5 个红球和 3 个蓝球。

随机取出一个球，观察颜色后放回，重复这个过程两次。

求两次都取出红球的概率。

# 结束语这些题目覆盖了欧几里得竞赛中常见的数学问题类型。

解决这些问题需要学生具备扎实的数学基础和良好的逻辑思维能力。

希望这些题目能够激发学生对数学的兴趣，并帮助他们在竞赛中取得优异的成绩。

Euclid欧几里得数学竞赛(Grade12)-数学Mathematics-2001-试题 exam

4.
appropriate box in the answer booklet. Marks may be given for work shown. Students are strongly encouraged to show their work. It is expected that all calculations and answers will be expressed as exact numbers such as 4 T , 2 7 , etc., except where otherwise indicated. What are the values of x such that 2 x – 3 2 ! 9 ? If f x ! x 2 – 3 x – 5 , what are the values of k such that f k ! k ? Determine all x, y such that x 2 y 2 ! 25 and x – y ! 1. The vertex of the parabola y ! x – b 2 b h has coordinates 2, 5 . What is the value of h? In the isosceles triangle ABC , AB ! AC and BAC ! 40r . Point P is on AC such that BP is the bisector of ABC . Similarly, Q is on AB such that CQ bisects ACB . What is the size of APB , in degrees?
Canadian Institute of Actuaries

欧几里得数学竞赛试题

欧几里得数学竞赛试题数学竞赛一直是评价学生数学能力的重要指标之一，其中欧几里得数学竞赛更是备受青睐。

欧几里得数学竞赛试题通常涉及数论、几何、代数等多个数学领域，要求学生在有限的时间内解决一系列复杂而富有挑战性的问题。

下面我们来看几个典型的欧几里得数学竞赛试题。

1. 问题一已知正整数a、b、c满足条件：a>b>c>1，且ab=bc+1。

求证：abc不可能是完全平方数。

解：假设abc是完全平方数，即存在正整数d，使得abc=d^2。

展开等式ab=bc+1，得到ab-bc=1。

可进一步得到b(a-c)=1，由于a、b、c均为正整数，所以a与c的差值只能是1。

设a=c+1，则上述等式变为b=1/(c+1)。

显然，b不可能是整数，因此矛盾。

由此可证明abc不可能是完全平方数。

2. 问题二已知正整数a、b、c满足条件：a+b+c=1000，且abc为最大的正整数。

求abc的值。

解：首先，根据给定条件可得 a = 1000 - b - c。

设函数 f(b, c) = b * c * (1000 - b - c)，要求abc的最大值，即需要求f(b, c) 的最大值。

对 f(b, c) 进行求导，得到 f'(b, c) = c - 2b + 1000 - c = -2b + 1000，令其等于0，可以得到 b = 500。

将 b = 500 代入 a = 1000 - b - c，可以得到 a + c = 500。

因此，满足条件 a + c = 500 的值对应的 b 为最大值的情况。

由于 a、b、c 均为正整数，不妨设 a = 499, c = 1，则 b = 500。

因此，abc的最大值为 499 * 500 * 1 = 249500。

3. 问题三已知正整数 n 的平方是一个五位数，且满足 n^2 % 10000 = n。

求 n的值。

解：设 n = 100a + 10b + c（其中 a、b、c 均为整数）。

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.)

欧几里得滑铁卢数学竞赛_2008EuclidSolution

(We can check this using the given example. We might also have been able to see initially
that the average of the numbers inside the square frame is always the middle number.)
Tuesday, April 15, 2008
Solutions
c 2008 Centre for Education in Mathematics and Computing
2008 Euclid Contest Solutions
1. (a) Solution 1 By the Pythagorean Theorem in ADB,
Since A(5, 0) lies on the line, then 0 = 5 + b so b = −5, so the line has equation y = x − 5.
To ﬁnd the intersection of the line with parabola, we use the two equations and equate:
Therefore, the sum of the numbers in the frame is
x + x − 1 + x + 1 + x − 8 + x − 7 + x − 6 + x + 6 + x + 7 + x + 8 = 9x
Thus, the sum of numbers in the frame is 9 times the middle number.

欧几里得滑铁卢数学竞赛_2009EuclidSolution

Solution 2
We label the point (k, k) as K.
Since K lies on the line segment AB, then the slope of AK equals the slope of AB.
4 − (−4) 8
Line segment AB joins (0, 4) to (8, −4), so has slope
Since this equation is true for all values of x, then the coeﬃcients on the left side and right side must be equal, so a + ab = 2 and 3ab = 12. From the second equation, ab = 4 so the ﬁrst equation becomes a + 4 = 2 or a = −2. Since ab = 4, then −2b = 4 and so b = −2. Thus, a = b = −2.
Canadian Mathematics Competition
An activity of the Centre for Education in Mathematics and Computing, University of Waterloo, Waterloo, Ontario
2009 Euclid Contest
Solution 2
Since x3 − 6x2 + 5x = 0, then x(x2 − 6x + 5) = 0 or x(x − 5)(x − 1) = 0.
The three roots of this equation are x = 0, x = 1 and x = 5.

欧几里得数学竞赛奖项设置

欧几里得数学竞赛奖项设置
欧几里得数学竞赛（Euclid Mathematics Contest）是由加拿大滑铁卢大学的数学院（Centre for Education in Mathematics and Computing, CEMC）主办的一项国际性高中数学竞赛。

该竞赛为全球高中生提供了一个展示数学才能的平台，并设置了以下奖项：
个人奖项：
Certificate of Distinction：颁发给在全球参赛者中排名前25%的学生。

Contest Medal：由CEMC决定，通常授予每个学校表现最优秀的学生。

Honour Rolls：根据成绩分设不同的荣誉榜，如全国荣誉榜、省级荣誉榜等。

团队奖项：
虽然主要以个人形式参加，但竞赛可能也会基于学校或地区团队整体成绩进行评价，并设立相应的团队奖项。

区域奖项：
根据成绩，可能会评出不同等级的奖项，比如针对加拿大区域的Zone、Provincial和National级别奖项。

其他表彰：
高分选手可能还会获得额外的证书或其他形式的表彰。

需要注意的是，具体的奖项设置以及获奖标准可能会随着年份的不同有所调整，请参考当年竞赛官方发布的最新公告和规则。

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

欧几里得数学竞赛_

欧几里得数学竞赛_（实用版）目录1.欧几里得数学竞赛简介2.竞赛的含金量和认可度3.竞赛的难度和价值4.如何准备和报名欧几里得数学竞赛5.竞赛对留学申请的影响正文欧几里得数学竞赛是全球三大数学竞赛之一，由加拿大滑铁卢大学与计算教育中心（CEMC）主办，面向全球高中生举办。

该竞赛自 1963 年创办以来，每年都有来自 10 多个国家和地区、1850 多所学校的 2 万多名学生报名参加。

竞赛成绩在数学界中已经得到广泛认可，被誉为数学届托福。

欧几里得数学竞赛的含金量和认可度非常高，被誉为全球高含金量高认可度的数学竞赛之一。

在北美地区，该竞赛的知名度与 AMC 和 BMO 齐名。

每年成绩排名前 25% 的学生会获得滑铁卢大学颁发的证书，将为留学申请履历添上丰富的一笔。

虽然欧几里得数学竞赛的含金量和认可度非常高，但是参赛者需要面对一定的难度。

竞赛的知识点考察非常平稳，主要集中在几何和代数模块。

在几何题中，参赛者可能需要运用、探索较为抽象和复杂的几何关系，从切入解决问题。

对于代数题，参赛者要求理解和的问题。

因此，参赛者需要具备较强的数学基础和解题能力。

对于如何准备和报名欧几里得数学竞赛，首先可以找负责数学部门的老师进行咨询。

此外，可以参加专门的欧几里得数学竞赛培训班，了解更多关于竞赛的详细信息和解题技巧。

在准备过程中，可以多做历年真题，提高自己的应试能力。

参加欧几里得数学竞赛对于留学申请有着非常积极的影响。

竞赛成绩可以为留学申请履历增色，提高被名校录取的几率。

此外，在竞赛中获得优异成绩的学生，更容易获得国外大学奖学金和助学金。

欧几里得奖项设置

欧几里得奖项的设立旨在表彰在数学领域有杰出贡献的个人或团队。

该奖项将分为以下几个部分：1. 基础数学奖：奖励在基础数学领域有杰出研究成果的个人或团队。

该奖项将由一个由数学家和相关领域专家组成的评审委员会评选，以确定获奖者。

2. 应用数学奖：奖励在应用数学领域有杰出研究成果的个人或团队。

该奖项将由相关领域的实践者或企业家组成的评审委员会评选，以确定获奖者。

3. 青年数学家奖：奖励在数学领域有杰出表现和潜力的青年个人。

该奖项将由一个由数学家和相关领域专家组成的评审委员会评选，以确定获奖者。

4. 团队奖：奖励在数学领域有杰出研究成果的团队。

该奖项将由一个由相关领域的实践者或企业家组成的评审委员会评选，以确定获奖者。

5. 特殊贡献奖：奖励对数学发展做出特殊贡献的个人或团队。

该奖项将由一个由数学家和相关领域专家组成的评审委员会评选，以确定获奖者。

欧几里得奖项的评选标准包括研究工作的原创性、对数学发展的贡献、研究成果的影响力以及对年轻一代学者的激励作用等方面。

评选过程将遵循公平、公正和公开的原则，确保所有参与者都有平等的机会展示自己的研究成果和潜力。

欧几里得奖项的颁奖仪式将在每年的国际数学家年会期间举行，届时将邀请获奖者、评委和相关领域的专家学者共同参加。

颁奖仪式将包括获奖者的演讲、颁奖典礼和晚宴等活动，旨在为获奖者和相关人士提供一个交流和学习的平台。

欧几里得奖项的设立旨在激励更多的人投身数学研究，推动数学的发展和应用，为人类社会的发展和进步做出更大的贡献。

我们相信，通过设立欧几里得奖项，我们可以吸引更多优秀的人才加入数学领域，推动数学的发展和应用，为人类社会的发展和进步做出更大的贡献。

同时，我们也期待更多的学者和相关人士能够关注和支持欧几里得奖项的设立和评选工作，共同为数学的发展和应用贡献力量。

欧几里得数学竞赛考点

欧几里得数学竞赛考点欧几里得（Euclid）是著名的古希腊数学家，他给我们留下了许多有关几何的宝贵知识。

这些知识被用于数学竞赛考点。

在数学竞赛的时候，许多学生都期待着某种方式可以贴近欧几里得的成果，甚至希望有一个可以学习欧几里得的概念的环境。

本文旨在介绍数学竞赛的欧几里得考点，以及学生在竞赛时可以掌握的技巧。

首先，我们需要知道，欧几里得考点通常都是几何考点，考生们可能会遇到多边形面积的计算、平面几何里的角的大小测量等问题。

其次，考生们需要熟悉欧几里得的几何原理，包括其最著名的《几何原本》，以及其他著作中提出的概念。

另外，考生们还可以练习这些概念的应用，比如几何构造、向量计算和平面几何等等。

考生们能够有效地练习欧几里得原理，并在考试中灵活运用，才能够取得好成绩。

第三，考生们需要学习解决欧几里得考点所需的技巧。

这些技巧包括图形构造、背影几何、几何证明和齐次几何等。

在使用这些技巧时，考生可以将欧几里得的原理运用到自己的知识中，进一步提高自己的解题能力。

在通过以上技巧练习之后，考生们就可以开始尝试解决欧几里得考点上的实际问题了。

对于每一个问题，考生需要先读懂题目，然后正确分解问题，掌握解题的思路，并且有效地将欧几里得的原理运用于解决问题。

最后，欧几里得考点也可以利用计算机来解决。

当考生完成解决欧几里得考点的思考过程之后，他们可以选择使用计算机对这些问题进行运算求解，以获得满意的结果。

综上所述，欧几里得考点是一个复杂但有趣的环境，可以给学生带来充实的学习体验。

考生们需要熟悉欧几里得的原理，并熟练掌握多种技巧，才能在数学竞赛中取得好的成绩。

此外，考生们也可以利用计算机辅助解决欧几里得考点，提高解题效率。

最后，希望考生们可以在竞赛中发挥出色，取得更好的成绩。

加拿大滑铁卢大学欧几里德数学竞赛考试在哪考？

加拿大滑铁卢大学欧几里德数学竞赛考试在哪考？滑铁卢大学欧几里德数学竞赛考试地点。

一年一度欧几里德考试（Euclid Contest）即“数学托福考试”，是由世界上最大数学院，滑铁卢大学数学学院主办（该学院堪称为加拿大MIT即麻省理工学院），至今已经成功举办过35届，每年大约有15,000人参加，目的是为测试学生解决数学问题能力。

同时也是该学院在录取和发放奖学金时重要参考标准之一，同时也是其他数学院录取时参考标准。

考试形式考试时间为2.5小时，总共包括10道题，每题10分，总共100分。

考题分为二部分，一部分为简答题，另一部分为大答题，没有选择题。

评分标准不仅是以最终结果正确与否给分，也会根据答题步骤及方式来给分。

如果答题步骤或方式过为散乱，即使最终结果是正确的也不会给予满分。

考试范围考题内容以高中数学课程为主包括高中最后一年课程内容，包括以下主题：• Euclidea n and analytic geometry 解析几何• Trigonometry, including functions, graphs, identities, sine and cosine laws 三角函数• Exponential and logarithmic functions 指数和对数函数• Functional notation 函数记号• Systems of equations 方程组• Polynomials, including relationships involving the roots of quadratic and cubic equations, the remainder theorem 多项式，包括二次求根，三次方程式，余部定理• Sequences and series 数列和级数• Simple counting problems 简单计算题• Properties of numbers 数字的性质考试优势成绩优秀者不仅可以提高被滑铁卢大学数学系和软件工程系的录取机率，而且还可以获得相应的入学奖学金（加元1000—8000不等）。

欧几里得竞赛

欧几里得竞赛
欧几里得竞赛是一种传统的数学竞赛，传统上由学校或个人进行组织，主要针对数学初学者或有资历的数学专业学生。

竞赛名称来源于古希腊数学家欧几里得，他是最具有影响力的古典数学家，他的研究又是数学的根源。

竞赛的支持者认为，欧几里得竞赛有助于促进学生的数学能力，并帮助学生了解古代数学历史，比如欧几里得的研究，推理和定理。

它还培养学生迅速解决疑难问题的能力，对日后从事数学行业有大量的帮助。

欧几里得竞赛一般分为三级，分别是初级、中级和高级，具体的内容根据竞赛的类型和时间而定。

多学校和机构会给竞赛提供一些礼物，比如书籍、学习材料或出国旅行的机会。

欧几里得竞赛的目标是让参赛者了解古典数学，培养他们的分析性思维力，提高他们的解决问题的能力，培养他们的学习能力和独立思考能力。

参赛者可以根据自己的能力练习和完成作业题，比如数学推理、定理、概率论等，并给出有解释的解答。

总体而言，欧几里得竞赛是一个值得参与的有趣而有价值的活动。

它既有助于提高参赛者的数学能力，也帮助学生了解古代数学历史，促进他们的学习能力。

此外，竞赛还可以提高参赛者的思维和分析能力，激发他们的热情，锻炼他们的独立思考能力。

希望更多的学校和机构可以组织欧几里得竞赛，数学才华的学生可以拥有更多的机会，去展示他们的数学天赋，实现他们的梦想。

欧几里得数学竞赛中国

欧几里得数学竞赛中国篇一：欧几里得数学竞赛中国是由中国举办的数学竞赛,旨在激发小学生对数学的兴趣,提高其数学能力和解决问题的能力。

该竞赛每年举办一次,参赛者包括来自全国各地的小学生。

竞赛的内容主要包括数学基础知识和欧几里得几何学。

欧几里得几何学是数学中的一个分支,它研究平面上的几何图形及其之间的关系。

竞赛中的题目涉及欧几里得几何学的基本概念、运算、证明等方面。

竞赛的难度较高,需要参赛者具备扎实的数学基础和严密的思维逻辑能力。

竞赛中的题目涵盖了小学至高中的数学知识,甚至有一些题目涉及到大学数学的知识。

除了竞赛本身,欧几里得数学竞赛中国还组织了一系列的培训和活动,旨在帮助参赛者更好地理解数学知识和提高解题能力。

这些活动包括数学讲座、竞赛模拟考试、数学竞赛讲解等,吸引了大量的学生参与。

欧几里得数学竞赛中国是一个面向全国各地小学生的数学竞赛,不仅有助于提高学生的数学素养和解题能力,还能增强他们对数学的兴趣和信心。

通过竞赛中国这个平台,学生可以与其他地区的参赛者进行交流和学习,互相借鉴和启发。

篇二：欧几里得数学竞赛中国是由中国数学会主办的一项数学竞赛,旨在激发青少年学生对数学的兴趣和爱好,提高其数学思维和解决问题的能力。

该竞赛创办于2000年,每年举办一次,目前已经成为中国数学界最具影响力的竞赛之一。

欧几里得数学竞赛中国的内容主要包括数学分析和代数、几何、三角函数、微积分等多个学科,竞赛题型多样,包括选择题、填空题、计算题、应用题等。

参赛者需要在规定时间内完成各种题目,并提交最终答案。

竞赛的难度和深度都相当高,旨在考察参赛者的数学思维和解决问题的能力。

除了竞赛本身,欧几里得数学竞赛中国还邀请了多位国内和国际著名的数学家担任竞赛的评审,对参赛者提交的论文进行严格评审和评分,选出优秀的参赛者和论文。

竞赛的结果不仅有利于参赛者个人的成长,也可以为相关学科的发展做出贡献。

在欧几里得数学竞赛中国的过程中,不仅考验了参赛者的数学能力,也促进了数学教育的普及和推广。

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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。

学校可以用学校注册号以及
密码登录系统查询成绩并且下载电子版获奖证书；
Ø 欧几里德竞赛没有纸质成绩单，只发放电子版或者纸质的前25%的证书；每个考点的第一
名的学生会有竞赛的奖牌。

Ø 欧几里德竞赛成绩一般在竞赛结束之后3个星期左右公布。

为何参加：
Ø 喜欢数学、对数学解题感兴趣；
Ø 如果学生申请滑铁卢大学数学院的专业，鼓励学生参加欧几里德数学竞赛；
Ø 竞赛的成绩对学生的录取没有直接的影响；数学院的录取不要求学生参加欧几里德数学竞赛；
Ø 如果学生在欧几里德竞赛中成绩优异，会对他们的申请有很大的帮助，并且有获得入学奖学金的机会；
Ø 数学院入学奖学金要求学生必须参加欧几里德竞赛；
Ø 如果学生在欧几里德竞赛中成绩不够好，不会影响录取
Ø 数学院入学奖学金的发放都需要根据学生申请当年的成绩，即高三或者12年级当年的竞赛成绩；11年级或者之前的竞赛成绩仅供参考。

关于报名;
Ø 学生首先询问学校是否可以申请注册考点、安排学生参加竞赛；
Ø 学校注册程序请参见我们网站信息或者联系：cemc@uwaterloo.ca
Ø 如果学校不能注册考点，可以报名滑铁卢大学在北京或者上海的考点参加欧几里德数学竞赛；
滑铁卢大学数学学院提供的本科专业主要有：• Actuarial Science 精算学• Applied Mathematics 应用数学• Bioinformatics 生物信息学• BBA/BMath Double Degree 商业管理和数学双学位• Business Administration 商业管理• Chartered Accountancy 会计• Combinatori cs and Optimization 组合和优化科学• Computational Mathematics 计算数学• Computer Scienc 计算机科学• Mathematics/Business Administration 数学 / 商业管理• Mathematics/Financial Analysis and Risk Management 数学 / 金融分析险管理• Mathematical Sciences 数学科学• Mathematical Physics 数学物理学• Operations Research 运营研究• Pure Mathematics 纯数学• Software Engineering 软件工程• Statistics 统计学
滑铁卢大学数学学院入学要求：高中毕业以上高中三年的平均成绩（数理化） 90 分以上单科数学平均 90 分以上。

由该数学学院每年 4 月举行的“欧几里德”数学考试为评估国际学生数学水平及入学的重要依据同时该考试成绩也可作为申请加拿大和美国名校的重要参考。