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加拿大国家中小学数学竞赛( kangaroo math 袋鼠竞赛)2017年五六年级(含答案)

加拿大国家中小学数学竞赛( kangaroo math 袋鼠竞赛)2017年五六年级(含答案)

I N T ER N A T I ON A L CO N T E S T-GA M EM A TH KA N GA RO OC A N A DA, 2017INSTRUCTIONSGRADE 5-61.You have 75 minutes to solve 30 multiple choice problems. For each problem, circle onlyone of the proposed five choices. If you circle more than one choice, your response will be marked as wrong.2.Record your answers in the response form. Remember that this is the only sheet that ismarked, so make sure you have all your answers transferred here by the end of the contest.3.The problems are arranged in three groups. A correct answer of the first 10 problems isworth 3 points. A correct answer of problems 11-20 is worth 4 points. A correct answer of problems 21-30 is worth 5 points. For each incorrect answer, one point is deducted from your score. Each unanswered question is worth 0 points. To avoid negative scores, you start from 30 points. The maximum score possible is 150.4.Calculators and graph paper are not permitted. You are allowed to use rough paper for draftwork.5.The figures are not drawn to scale. They should be used only for illustration.6.Remember, you have about 2-3 minutes for each problem; hence, if a problem appears tobe too difficult, save it for later and move on to the other problems.7.At the end of the allotted time, please submit the response form to the contest supervisor.Please do not forget to pick up your Certificate of Participation!Good luck! Canadian Math Kangaroo Contest team2017 CMKC locations: Algoma University; Bishop's University; Brandon University; Brock University; Carlton University; Concordia University; Concordia University of Edmonton; Coquitlam City Library; Dalhousie University; Evergreen Park School; F.H. Sherman Recreation & Learning Centre; GAD Elementary School; Grande Prairie Regional College; Humber College; Lakehead University (Orillia and Thunder Bay); Laurentian University; MacEwan University; Memorial University of Newfoundland; Mount Allison University; Mount Royal University; Nipissing University; St. Mary’s University (Calgary); St. Peter’s College; The Renert School at Royal Vista; Trent University; University of Alberta-Augustana Campus; University of British Columbia (Okanagan); University of Guelph; University of Lethbridge; University of New Brunswick; University of Prince Edward Island; University of Quebec at Chicoutimi; University of Quebec at Rimouski; University of Regina; University of Toronto Mississauga; University of Toronto Scarborough; University of Toronto St. George; University of Windsor; The University of Western Ontario; University of Winnipeg; Vancouver Island University; Walter Murray Collegiate, Wilfrid Laurier University; YES Education Centre; York University; Yukon College.2017 CMKC supporters: Laurentian University; Canadian Mathematical Society; IEEE; PIMS.Canadian Math Kangaroo ContestPart A: Each correct answer is worth 3 points1.A fly has 6 legs, a spider has 8 legs. Together, 3 flies and 2 spiders have as many legs as 9 chickens andseveral cats. How many cats are there?(A) 2 cats (B) 3 cats (C) 4 cats (D) 5 cats (E) 6 cats2.Alice has 4 pieces of this shape: . Which picture can she not make from these four pieces?(A) (B) (C)(D) (E)3.Kalle knows that 1111 × 1111 = 1234321. What is the answer of 1111 × 2222?(A) 3456543 (B) 2346642 (C)2457642 (D) 2468642 (E) 43212344.There are 10 islands and 12 bridges, as depicted in the figure. All bridgesare open for traffic right now. What is the smallest number of bridges thatmust be closed in order to stop the traffic between A and B?(A) 1 (B) 2 (C) 3 (D) 4 (E) 55.Martin wants to colour the squares of the rectangle so that 1/3 of allsquares are blue and half of all squares are yellow. The rest of the squaresare to be coloured red.How many squares will he colour red?(A) 1 (B) 2 (C) 3 (D) 4 (E) 56.When the car wheels make one full rotation the car moves forward by about 1.8 meters. Approximatelyhow many kilometres will the car move forward after 10,000 full rotations of the wheels?(A) 1.8 (B) 18 (C) 180 (D) 1 800 (E) 18 0007.There are 32 students in Mrs. Vicky’s class. Part of the students took one pencil each from the box withpencils on the teacher’s desk. Then a third of the remaining students took 3 pencils each, and there were no more pencils left in the box. How many pencils were there in the box at first?(A) 16 (B) 24 (C) 32 (D) 43 (E) 648.Three rhinoceroses Jane, Kate and Lynn go for a walk: Jane first, Kate in the middle, and Lynn – last. Janeweighs 500 kg more than Kate. Kate weighs 1000 kg less than Lynn. Which of the following pictures may show Jane, Kate and Lynn in the order they walked?(A) (B)(C) (D)(E)9.Peter and Nick are both working on "Kangaroo" contest problems. For every two problems that Petersolves, Nick manages to solve three problems. In total, the boys solved 30 problems. How many problems did Nick solve more than Peter?(A) 5 (B) 6 (C) 7 (D) 8 (E) 910.Bob folded a piece of paper, used a hole puncher and punched exactly one hole in the folded paper.Then, he unfolded the piece of paper, which looked as shown below.Which of the following pictures shows the lines along which Bob folded the piece of paper?(A) (B) (C) (D) (E)Part B: Each correct answer is worth 4 points11.A special die has a number on each of its six faces. The sums of the numbers on opposite faces are all equal. Five of the numbers are 5, 6, 9, 11 and 14. What number is on the sixth face? (A) 4 (B) 7 (C) 8 (D) 13 (E) 15 12.Tom wrote all the numbers from 1 to 20 in a row and obtained the 31‐digit number1234567891011121314151617181920.Then he deleted 24 of the 31 digits, so that the remaining number was as large as possible. Which number was it? (A) 9671819 (B) 9567892 (C) 9781920 (D) 9912345 (E) 981819213.Peter went hiking in the mountains for 5 days. He started on Monday and his last trip was on Friday. Each day he walked 2km more than the day before. The total distance he walked during the five days was 70km. What distance did Peter walk on Thursday? (A) 12 km (B) 13 km (C) 14 km (D) 15 km (E) 16 km14.In a chocolate store, one chocolate costs $3. One day the store had a deal: “Buy two and get a third one free” and Adam decided to take 49 chocolates. How much did he pay for the chocolates? (A) $75 (B) $98 (C) $99 (D) $102 (E) $14715.Eight kangaroos stood in a line as shown in the diagram.At some point, two kangaroos standing side by side and facing each other exchanged places by jumping past each other. This was repeated until no further jumps were possible. How many exchanges were made? (A) 2 (B) 10 (C) 12 (D) 13 (E) 1616.The Modern Furniture store is selling sofas, loveseats, and chairs made from identical modular pieces as shown in the picture. Including the armrests, the width of the sofa is 220 cm and the width of the loveseat is 160 cm.What is the width of the chair? (A) 60 cm (B) 80 cm (C) 90 cm(D) 100 cm(E) 120 cmsofa loveseatchair220 cm160cm17.There are five padlocks and 5 keys – one for each of them (see the figure). The number code on each key has been modified into a letter code on the corresponding padlock. Equal digits have been replaced by the same letter, and different digits – by different letters. What is the number code on the fifth key?(A) 382(B) 282 (C) 284 (D) 823 (E) 82418.Boris has an amount of money and three magic wands that he can use only once. Wand A adds $1. Wand S subtracts $1. Wand D doubles the amount. In which order must he use these wands to obtain the largest amount of money? (A) DAS (B) ASD (C) DSA (D) ADS (E) SAD19.A vase weighs 600 g when one third of it is filled with water. The same vase weighs 800 g when two thirds of it are filled with water. What is the weight of the vase when it is empty? (A) 100 g (B) 200 g (C) 300 g (D) 400 g (E) 500 g20.Rafael has three squares. The first one has side length 2 cm. The second one has side length 4 cm and a vertex is placed in the centre of the first square. The last one has side length 6 cm and a vertex is placed in the centre of the second square, as shown in the picture. What is the area of the figure? (A) 32 cm 2 (B) 51 cm 2 (C) 27 cm 2 (D) 16 cm 2 (E) 6 cm 2Part C: Each correct answer is worth 5 points21.The natural numbers are arranged in the form of a triangle: 1 is in the first row, 2 and 3 are in the second row, 4, 5 and 6 are in the third row, and so on. What is the sum of the numbers written in the 10‐th row?(A) 490(B) 495 (C) 500(D) 505 (E) 5101 2 3 456.. .22.There are eight balls numbered with the numbers 40, 80, 100, 101, 190, 200, 260 and 292 in a bag.Martina takes four balls out of the bag and calculates the sum of the numbers on these balls. It appears that this sum is half of the sum of the numbers on the balls that remain in the bag. What is the greatest number written on the balls taken out?(A) 101 (B) 200 (C) 260 (D) 190 (E) 29223.The structure on the figure is made of unit cubes glued together. Morten wants toput it into a rectangular box. What are the dimensions (length, width and height)of the smallest box he can use?(A) 3 × 3 × 4 (B) 3 × 5 × 5 (C) 3 × 4 × 5 (D) 4 × 4 × 4 (E) 4 × 4 × 524.Four players scored goals in a handball game. All of them scored a different number of goals. One of theplayers, Mike, scored the least number of goals. The other three players scored 20 goals in total. What is the largest number of goals Mike could have scored?(A) 2 (B) 3 (C) 4 (D) 5 (E) 625.Ala likes even numbers, Beata likes numbers divisible by 3, Celina likes numbers divisible by 5. Each ofthese three girls went separately to a basket containing 8 balls with numbers written on them, and took all the balls with numbers she liked. It turned out that Ala collected balls with numbers 32 and 52, Beata ‐ 24,33 and 45, Celina ‐ 20, 25 and 35. In what order did the girls approach the basket?(A) Ala, Celina, Beata (B) Celina, Beata, Ala (C) Beata, Ala, Celina(D) Beata, Celina, Ala (E) Celina, Ala, Beata26.The picture of a kangaroo in the first (leftmost) triangle was reflected across the dotted lines, as in mirrors.The first two reflections are shown.What does the reflection look like in the shaded triangle?(A) (B) (C) (D) (E)27.The numbers 1, 2, 3, 4, and 5 must be written in the five cells in the figure, respecting the following rules:-If a number is just below another number, it must be greater.-If a number is just to the right of another number, it must be greater.In how many ways can this be done?(A) 3 (B) 4 (C) 5 (D) 6 (E) 828.John wrote a natural number in each of the four boxes in the bottom row of the diagram. Then he wrote ineach of the other boxes the sum of the two numbers in the boxes immediately underneath. What is the largest number of odd numbers that could appear in the completed diagram?(A) 4 (B) 5 (C) 6 (D) 7 (E) 829.Julia has four pencils of different colours and wants to use some or all of them to paint the map of anisland divided into four countries, as in the picture. Any two countries with a common border must be coloured differently on the map. How many different colourings of this map are possible? (Twocolourings are considered different if at least one of the countries is coloured differently).(A) 12 (B) 18 (C) 24 (D) 36 (E) 4830.A bar consists of two grey cubes and one white cube glued together as shown in the figure.Which cube can be built from nine such bars?(A) (B) (C) (D) (E)International Contest-GameMath Kangaroo Canada, 2017Answer KeyGrade 5-61 A B C D E 11 A B C D E21 A B C D E2 A B C D E12 A B C D E 22 A B C D E3 A B C D E 13 A B C D E23 A B C D E4 A B C D E 14 A B C D E 24 A B C D E5 A B C D E 15 A B C D E 25 A B C D E6 A B C D E 16 A B C D E 26 A B C D E7 A B C D E 17 A B C D E 27 A B C D E8 A B C D E 18 A B C D E 28 A B C D E9 A B C D E 19 A B C D E 29 A B C D E10 A B C D E 20 A B C D E 30 A B C D E。

加拿大欧几里得数学竞赛真题解析2020EuclidSolution

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.。

温哥华高中课程内容

温哥华高中课程内容

温哥华高中课程内容温哥华作为加拿大著名的城市,拥有世界一流的教育资源。

在这里,高中教育注重培养学生的全面发展,课程内容丰富多样,旨在帮助学生掌握知识,培养技能,为未来大学和职业生涯打下坚实基础。

本文将为您详细介绍温哥华高中课程内容。

一、学术课程1.必修课程(1)英语:培养学生的阅读、写作、听说和语法能力。

(2)数学:包括代数、几何、三角学等基本数学知识。

(3)科学:涉及物理、化学、生物等领域的知识。

(4)社会科学:包括加拿大历史、地理、政治、经济学等。

2.选修课程(1)高级数学:如微积分、线性代数等。

(2)高级科学:如物理学、化学、生物学等。

(3)外语:如法语、西班牙语、中文等。

(4)艺术:如音乐、舞蹈、戏剧、视觉艺术等。

二、实践课程1.技术教育:培养学生掌握实用技能,如计算机编程、汽车维修、木工等。

2.社区服务:鼓励学生参与社区服务活动,提高社会责任感。

3.体育运动:提供丰富的体育项目,如篮球、足球、游泳、田径等。

三、特色课程1.国际文凭课程(IB):为优秀学生提供挑战性的学术课程,帮助学生适应全球化的大学和职场环境。

2.大学先修课程(AP):让学生在高中阶段提前学习大学课程,提高学术竞争力。

3.职业教育课程:为学生提供职业技能培训,如烹饪、美容、护理等。

四、课外活动1.学术竞赛:如数学竞赛、科学竞赛、辩论赛等。

2.社团活动:如音乐社团、舞蹈社团、志愿者社团等。

3.学术讲座和研讨会:邀请知名人士和专家分享知识,拓宽学生视野。

总结:温哥华高中课程内容丰富多样,旨在培养学生的学术素养、实践能力和综合素质。

caribou数学竞赛介绍

caribou数学竞赛介绍

caribou数学竞赛介绍
Caribou数学竞赛是一项面向学生的国际性数学竞赛,旨在激发学生对数学的兴趣,并提高他们的数学技能和解决问题的能力。

该竞赛由加拿大Caribou Mathematics Competition组织,每年举办一次。

Caribou数学竞赛分为多个级别,适合不同年级的学生参与,从幼儿园到高中都有相应的竞赛级别。

竞赛题目设计丰富多样,包括选择题、填空题、解答题等形式,涵盖了各个数学领域的知识和技能。

参加Caribou数学竞赛有许多好处。

首先,它可以培养学生的数学思维和解决问题的能力。

通过面对不同类型的数学问题,学生需要思考和分析,找到解题的方法和策略。

其次,竞赛可以激发学生对数学的兴趣和热爱,让他们深入了解数学的魅力和应用。

此外,竞赛还提供了一个与全球各地学生交流和比较的平台,激发学生的竞争意识和合作精神。

参加Caribou数学竞赛需要学校或个人报名,并按照竞赛规定的时间和方式进行答题。

竞赛结果将根据学生的得分和排名进行评定,并颁发相应的证书和奖励。

总而言之,Caribou数学竞赛是一项激发学生数学兴趣和提高数学能力的国际性竞赛,为学生提供了锻炼和展示自己数学才能的机会。

通过参与竞赛,学生可以发展数学思维,提高解决问题的能力,并与来自世界各地的学生交流和比较。

奥数简介

奥数简介

奥数简介“奥数”是奥林匹克数学竞赛的简称。

1934年—1935年,前苏联开始在列宁格勒和莫斯科举办中学数学竞赛,并冠以数学奥林匹克竞赛的名称,1959年在布加勒斯特举办第一届国际数学奥林匹克竞赛。

国际数学奥林匹克作为一项国际性赛事,由国际数学教育专家命题,出题范围超出了所有国家的义务教育水平,难度大大超过大学入学考试。

有关专家认为,只有5%的智力超常儿童适合学奥林匹克数学,而能一路过关斩将冲到国际数学奥林匹克顶峰的人更是凤毛麟角。

1934年和1935年苏联开始在列宁格勒和莫斯科举办中学数学竞赛,并冠以数学奥林匹克的名称。

1959年罗马尼亚数学物理学会邀请东欧国家中学生参加,在布加勒斯特举办了第一届国际数学奥林匹克竞赛,从此每年举办一次,至今已举办了43届。

近年来中国代表在数学奥林匹克上的成绩就像中国健儿在奥运会的成绩一样,突飞猛进,从40届到第43届,中国代表队连续四年总分第一。

奥数分类为:浓度问题、分数比大小问题、行程问题、分数巧算、逻辑推理、工程问题、牛顿问题、数字的巧算问题。

奥数与一般数学有一定的区别:奥数相对比较深.小学数学奥林匹克活动的蓬勃发展,极大地激发了广大少年儿童学习数学的兴趣,成为引导少年积极向上,主动探索,健康成长的一项有益活动.国际奥林匹克数学竞赛奖项名称: 国际奥林匹克数学竞赛其他名称: International Mathematics Olympiad创办时间: 1959年主办单位: 由参赛国轮流主办奖项介绍:国际奥林匹克数学竞赛是国际中学生数学大赛,在世界上影响非常之大。

国际奥林匹克竞赛的目的是:发现鼓励世界上具有数学天份的青少年,为各国进行科学教育交流创造条件,增进各国师生间的友好关系。

这一竞赛1959年由东欧国家发起,得到联合国教科文组织的资助。

第一届竞赛由罗马尼亚主办,1959年7月22日至30日在布加勒斯特举行,保加利亚、捷克斯洛伐克、匈牙利、波兰、罗马尼亚和苏联共7个国家参加竞赛。

cmo数学竞赛

cmo数学竞赛

cmo数学竞赛作为全球最具有影响力的数学竞赛之一,CMO(Canadian Mathematical Olympiad)也就是加拿大数学奥林匹克竞赛,一直以来都备受备受广大数学爱好者的热爱和追捧。

CMO数学竞赛自1980年开始举办,是加拿大最优秀的年度学科竞赛之一。

而作为具有世界影响力的竞赛,CMO每年都吸引着来自全球各地的顶尖数学选手参加。

一、竞赛概况竞赛背景CMO数学竞赛是由加拿大教育局主办并负责管理的,这项竞赛是有关方面为了鼓励学生热爱数学、提高学生的数学能力而举办的。

CMO竞赛难度极大,不仅需要参赛者具有超强的计算、分析和推理能力,更要求其具备优秀的解决问题的能力和出色的数学才能。

竞赛组织每年一次的CMO竞赛都会在加拿大各个省级地区进行,在每个地区中的具体机构组织相应的比赛活动,官方网站www.cms.math.ca 会发布该年度竞赛活动的时间、赛制和场地信息。

而整个竞赛过程,由加拿大数学协会负责组织和管理,同时全球各地的加拿大驻外使领馆也会在本地为参赛选手提供相应的帮助和协助。

竞赛难度在CMO数学竞赛中,因为竞赛的内容极为复杂,加上竞赛设置了多个级别,所以它的难度从高到低分别为:A级、B级、C级、D级。

而且不同等级的题目会出现不同的难度程度。

其中A级竞赛是难度最大的,除了试题难度更高之外,时间限制也更严格,竞赛难度被认为是全世界竞赛中最高的之一。

竞赛奖项在CMO数学竞赛中,除了最终的成绩排名之外,还设置了一些重要的荣誉奖项。

比如金、银、铜奖、优异奖、最佳小学奖和五获大赛奖等。

而且一些得奖的选手还可以被选拔参加国际数学奥林匹克竞赛。

二、竞赛流程参赛条件CMO竞赛面向各个年龄段的选手,可以是在校中小学生、初中生或者高中生,但是要求其具备较为扎实的数学基础和优秀的数学综合能力,一般来说,前三年的参赛者可以由学校老师推荐参加,而在高中阶段之后,他们可以直接通过自报名来参加竞赛。

初赛CMO数学竞赛的初赛主要是一个笔试环节,每组的试卷难度、长度、时间(参赛者有120分钟来完成试题)、总分和比例都不完全相同。

canadamo数学竞赛知识点

canadamo数学竞赛知识点

在撰写这篇文章之前,我首先要对“canadamo数学竞赛知识点”进行全面评估,以确保文章的深度和广度兼具。

在这篇文章中,我将从简到繁地分析并探讨canadamo数学竞赛的知识点,帮助你更深入地理解这个主题。

canadamo是加拿大数学奥林匹克(Canadian Mathematical Olympiad)的缩写,是加拿大国内最具权威性和影响力的数学竞赛之一。

参加canadamo数学竞赛,不仅能锻炼学生的数学能力,更能培养学生的逻辑思维和解决问题的能力。

我将从基础知识点开始,逐步深入,全面探讨canadamo数学竞赛的重要知识点。

1. 数论- 数论是canadamo数学竞赛中的重要知识点之一。

它涉及整数的性质、因数分解、同余方程等内容。

在canadamo数学竞赛中,数论题目常常涉及数字性质的推导和证明,考查选手的数学逻辑推理能力。

2. 几何- 几何是canadamo数学竞赛的另一个重要知识点。

它包括平面几何和立体几何两部分,涉及角度、边长、面积、体积等概念。

在canadamo数学竞赛中,几何题目常常涉及图形的性质和相似性的判断,考查选手的几何分析能力和空间想象能力。

3. 代数- 代数是canadamo数学竞赛的核心知识点之一。

它涉及方程、不等式、多项式、数列等内容。

在canadamo数学竞赛中,代数题目常常涉及函数的性质和变量的关系,考查选手的代数运算能力和推理能力。

4. 组合数学- 组合数学是canadamo数学竞赛的另一个重要知识点。

它包括排列、组合、概率等内容。

在canadamo数学竞赛中,组合数学题目常常涉及排列组合的计算和概率问题的推导,考查选手的组合分析能力和概率计算能力。

总结回顾:通过对canadamo数学竞赛知识点的全面评估,我们可以看到,数论、几何、代数和组合数学是其重要的知识点。

参加canadamo数学竞赛不仅需要掌握这些知识点,还需要灵活运用,并具备深入思考和解决问题的能力。

历年中国参加国际数学奥林匹克竞赛选手详细去向第26届IMO

历年中国参加国际数学奥林匹克竞赛选手详细去向第26届IMO

历年中国参加国际数学奥林匹克竞赛选手详细去向第26届IMO(1985年,芬兰赫尔辛基)吴思皓(男)上海向明中学确规定铜牌上海交通大学王锋(男)北京大学(根据yongcheng先生提供的信息修订)目前作企业软件第27届IMO(1986年,波兰华沙)李平立(男)天津南开中学金牌北京大学方为民(男)河南实验中学金牌北京大学张浩(男)上海大同中学金牌复旦大学荆秦(女)陕西西安八十五中银牌北京大学,现在美国哈佛大学任教林强(男)湖北黄冈中学铜牌中国科技大学第28届IMO(1987年,古巴哈瓦那)刘雄(男)湖南湘阴中学金牌南开大学滕峻(女)北京大学附中金牌北京大学林强(男)湖北黄冈中学银牌中国科技大学潘于刚(男)上海向明中学银牌北京大学何建勋(男)广东华南师范大学附中铜牌中国科技大学高峡(男)北京大学附中铜牌北京大学,现在北大任教第29届IMO(1988年,澳大利亚堪培拉)团体总分第二陈晞(男)上海复旦大学附中金牌复旦大学,美国密苏里大学,美国哈佛大学,现在加拿大Alberta大学数学系任教授韦国恒(男)湖北武汉武钢三中银牌北京大学查宇涵(男)南京十中银牌北京大学,在中科院数学所任副研究员邹钢(男)江苏镇江中学银牌北京大学王健梅(女)天津南开中学银牌北京大学何宏宇(男)以满分成绩获第29届国际数学奥林匹金牌,1993年破格列入美国数学家协会会员,1994年获博士学位,现任亚特兰大乔治大学教授、博士生导师,从事现代数学研究前沿的《李群》《微分几何》等方向的研究,在《李群》的研究上已有重大突破。

第30届IMO(1989年,原德意志联邦共和国布伦瑞克)团体总分第一罗华章(男)重庆水川中学金牌北京大学俞扬(男)吉林东北师范大学附中金牌吉林大学霍晓明(男)江西景德镇景光中学金牌中国科技大学唐若曦(男)四川成都九中银牌中国科技大学颜华菲(女)北京中国人民大学附中银牌北京大学本科,1997年获美国麻省理工博士,现任Texax A&M Uneversity 数学系教授,美国数学会常务理事会成员,Mathematical Reviews评论员。

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在加拿大教“奥数”
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适应北美高校的录取标准,依然是华裔学生考虑的主要问题
十几年前,在加拿大华人社区刷电线杆小广告的江亚东怎么也没想到,最初只有6个学生的奥数数学班能发展成为注册学生超过1000人的课外培训学校。

这所课外培训学校坐落在多伦多城区,学生大部分来自亚裔家庭,其中以第二代华裔居多。

在过去十余年里,数十名在此学习过的亚裔青年被哈佛、普林斯顿、麻省理工等美国常青藤大学和其他名校录取,甚至有美国的华裔家长让孩子从美国坐飞机来这里上课。

这所学校的发展过程本身,投射出亚裔、尤其是华裔群体与西方教育观念的巨大差异。

最初都是亚裔学生
出国前,江亚东是中国一所高校教授计算机和数学建模课程。

2002年,45岁的他来到加拿大工作后,不久便失业了,拥有博士学历的他开始为生计作打算。

江亚东告诉《瞭望东方周刊》,当时身边的华人朋友们都认为孩子所在的中学数学教学过于简单。

“加拿大公立学校的老师只管教学,没有学生升学业绩压力,所以对于学生是否能上重点大学并不十分上心。

”江亚东说,由此,他萌生了开个数学课外培训班的想法。

而就在这段时间,加拿大的第二代华裔数量正在不断增长。

根据《中国国际移民报告(2015)》数据显示,过去15年里,中国一直是加拿大永久居民最大来源国之一。

2001年到2011年10年间,加拿大的中国移民人数增长率超过60%,人口增加超过20万。

2011年的数据显示,在加拿大,近50%的中国移民居住在安大略省,其中超过八成、约为22万人居住在大多伦多地区。

所以,在这座城市开设针对华裔孩子的课外培训,市场足够大。

另一方面,江亚东发现,整个北美地区的中学生竞赛制度十分健全,很多学科的竞赛都是面对全体学生开设的。

北美名校对于竞赛成绩也同样认可。

比如,麻省理工学院的官网就为准备申请该校的高中生列出了六项国际竞赛(国际生物奥林匹克竞赛、国际化学奥林匹克竞赛、国际语言学奥林匹克竞赛、国际信息学奥林匹克竞赛、国际数学奥林匹克竞赛、国际物理奥林匹克竞赛)作为申请内容,建议学生把这些竞赛作为适应未来校园生活的重要准备。

不过,当地并没有针对竞赛教学的培训班,所以江亚东选择了以竞赛培训作为切入点。

在决定开设奥数培训班后,江亚东在多伦多各大书店转遍了也没有找到一本奥数课外培训书。

于是,2004年的春天,他回国购买了大量中学奥数竞赛教材,带回加拿大钻研。

半年之后,他在多伦多注册了培训机构。

2005年
1月,他终于招来了6个学生,全是华裔,授课地点就是家里的客厅。

虽然有过大学执教经验,但如何使学生对奥数产生兴趣成为他面临的最大难题。

“如果把国内的教材直接翻译成英语,不仅工作量大,而且很难被理解。

对于在加拿大长大的孩子,国内的奥数题目拐弯太多了。

所以必须结合加拿大中学课本难度重新编写英语讲义,抛弃太偏、太复杂的题。

”他说。

与一位“虎妈”的吵架
2006年2月,江亚东首次带领14名学生来到美国参加“哈佛—麻省理工大学数学竞赛”。

这个竞赛是全美国影响力最大、名校理工科专业认可程度最高的高中数学竞赛之一。

最终,两名学生分别获得几何组第二名、代数组第五名的好成绩。

江亚东告诉本刊记者,前者次年就被斯坦福大学录取了。

这次比赛后,江亚东的奥数班在华人社区开始小
有名气,学生也多了起来。

江亚东开始把培训班的课程从数学拓展到物理和化学。

此时,如何招聘到合适的培训老师成了难题。

江亚东说,多数加拿大本地老师的教育理念都是“学生快乐最重要”。

但是,这招致了很多华裔家长的不满,他们希望孩子能学到“干货”,而不是空有活泼的教学形式。

而他招聘的华人老师,有不少却因为英语口音太重而被学生投诉。

经过不断的面试、试听和培训,一些英语好的华人老师和课堂上“不扯闲篇”的本地老师被留了下来。

如今28岁的崔唯已经在奥林匹克学校兼职教竞赛物理六年了。

北大物理系本科毕业后,崔唯来到多伦多大学攻读量子物理学博士,2010年读博期间,他来到奥林匹克学校当兼职讲师。

崔唯告诉《瞭望东方周刊》,亚裔孩子的家长们都很重视竞赛成绩。

如果某次竞赛孩子表现不太好,家长就会与老师交流,细致分析孩子的问题。

不仅在加拿大,在美国亚裔美籍家长对于子女的教育也非常重视。

纽约市立大学皇后学院社会学系教授艾米·辛(Amy Hsin)和密歇根大学教授谢宇发表在《美国科学院院报》上的研究成果指出,刻苦努力是亚裔美籍学生学业优异的主要原因,亚裔美籍学生努力的背后是亚裔美籍学生的父母更加重视学校教育,对孩子的学业期望较高;移民身份使得亚裔美籍学生认识到,只有努力学习才能有效地改善生活地位。

崔唯认为,让学生提高竞赛能力最有效的办法还是“题海战术”。

“虽然‘题海战术’在国内广受诟病,但无论是在奥林匹克学校,还是在多伦多大学当助教的经验都证明,让学生做练习都是最有效的方法。


很多华裔家长也坚持,“必须要有课后作业”。

学校运营初期,除了编写教材讲义外,江亚东并没有时间来编写配套练习册,很少给学生留家庭作业,一位华人“虎妈”还为此与他吵了起来。

这位“虎妈”认为,奥数课程对孩子来说并不简单,没有家庭作业,就无法消化与复习。

此后,江亚东开始投入更多的时间、精力和资金来编辑各个科目的配套习题册。

“潜规则”依然在
江亚东遭遇华人“虎妈”一点也不意外。

几年前,美籍华人蔡美儿出版的《虎妈战歌》曾引起舆论对东西方教育观点的广泛争论。

在北美本地,华裔对于当地教育的影响也开始显现。

2015年末,美国新泽西州格罗夫尔中学的董事会曾就是否要降低四、五年级的数学教学标准进行过一场讨论。

讨论中,一边是白人家长对于减负的支持,一边是亚裔家长们的反对。

学校董事会最终采取了白人家长的意见。

但是,亚裔对当地人教育态度的改变正在发生。

江亚东说,他的学校如今有几十名学生不是亚裔,很多白人学生就是家长向亚裔家长“取经”时被推荐来的。

崔唯也说,一些白人家庭对这种模式的课外培训还是很感兴趣的。

但是,适应北美高校的录取标准,依然是华裔学生要考虑的主要问题。

一些家长向江亚东反映,虽然孩子在学校理科成绩很好,但语言成绩和演讲技能都很一般。

即便是出生在加拿大的第二代移民,由于家长的母语不是英语,孩子的语言和文化知识还是少于本地孩子。

于是,江亚东决定请有语言经验的本地老师开设演讲课和辩论课。

据江亚东介绍,辩论课开课之初效果并不好,但是,通过招聘本地辩论教练授课、增加学生的阅读量、总结辩论技巧等方法,一些学生开始在辩论竞赛中取
得优异成绩。

2014年,一名学生在约克大学举办的中学生辩论赛中获得冠军,次年她凭借中学阶段的优异成绩和辩论特长被哈佛大学录取。

不过,一些证据表明,亚裔至今仍然在申请北美顶尖大学时处于劣势。

2015年10月,英国《经济学人》杂志发表的一篇题为《美国亚裔,不再沉默》的文章指出,美国亚裔要在SAT考试中,比白人学生高出140分才能进入同样的私立大学,而黑人学生的考分可以比华裔学生低310分。

其中的“潜规则”就是,顶尖学校录取黑人学生和西班牙裔学生的分数比较低,这是考虑到这些族裔在过去的遭遇;贵族、体育明星、政界家庭和富豪(这里很少有亚洲人)给学校捐了大楼,他们的孩子就可能占去录取名额,因此,光有成绩的华裔进名校的机会就更小了。

文章指出,尽管美国华裔鼓励孩子多元化发展——参加辩论赛、公益活动、体育运动,学习音乐,参与任何可以增加录取率的活动,但很多在这些领域表现出色的学生,还是被常春藤大学拒之门外。

本文从网络收集而来,上传到平台为了帮到更多的人,如果您需要使用本文档,请点击下载按钮下载本文档(有偿下载),另外祝您生活愉快,工作顺利,万事如意!。

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