加拿大的数学竞赛

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加拿大国际袋鼠数学竞赛试题及答案-2018年

I N T E R N A T I O N A L C O N T E S T-G A M EM A T H K A N G A R O OC A N AD A,2018I N S T R U C T I O N SG R A D E1-21.You have 45 minutes to solve 18 multiple choice problems. For eachproblem, circle only one 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 onlysheet that is marked, so make sure you have all your answers transferred to the response form before giving it back to the contest supervisor.3.The problems are arranged in three groups. A correct answer of the first 6problems is worth 3 points. A correct answer of the problems 7-12 is worth4 points. A correct answer of the problems 13-18 is worth5 points. Foreach incorrect answer, one point is deducted from your score. Each unanswered question is worth 0 points. To avoid negative scores, you start from 18 points. The maximum score possible is 90.4.The use of external material or aid of any kind is not permitted.5.The figures are not drawn to scale. They should be used only for illustrationpurposes.6.Remember, you have about 2 to 3 minutes for each problem; hence, if aproblem appears to be too difficult, save it for later and move on to another problem.7.At the end of the allotted time, please give the response form to thecontest supervisor.8.Do not forget to pick up your Certificate of Participation on your way out!Good luck!Canadian Math Kangaroo Contest teamCanadian Math Kangaroo ContestPart A: Each correct answer is worth 3 points1.Which shape cannot be formed using and ?(A) (B) (C) (D) (E)2.At least how many 4-ray stars like this are glued together tomake this shape ?(A) 5 (B) 6 (C) 7 (D) 8 (E) 93.This pizza was divided into equal slices.How many slices are missing?(A) 1 (B) 2 (C) 3 (D) 4 (E) 54.How many kangaroos must be moved from one park to the other in order toget the same number of kangaroos in each park?(A) 4 (B) 5 (C) 6 (D) 8 (E) 95.Which of these ladybugs has to fly away so that the rest of them have 20dots in total?(A) (B) (C) (D) (E)6.Emilie builds towers in the following patternWhich one will be the tower number 6?(A) (B) (C) (D) (E)Part B: Each correct answer is worth 4 points7.If ◊+ ◊ = 4 and ∆ + ∆ + ∆ = 9, what is the value of ◊ + ∆ = ?(A) 2 (B) 3 (C) 4 (D) 5 (E) 68.Lisa has 4 pieces , but she only needs 3 forcompleting her puzzle frame . Which piece will be left over?(A)(B)(C)(D) (E)or9.How many right hands are in this picture?(A) 3 (B) 4 (C) 5 (D) 6 (E) 710.The dog went to its food following a path. In total it made 3 right turns and2 left turns. Which path did the dog follow?(A) (B) (C)(D) (E)11.What number is in the box marked "?" ?(A) 6 (B) 13 (C) 24 (D) 29 (E) Some other number12.Charles cut a rope in three equal pieces and then made some equal knotswith them. Which figure correctly shows the three pieces with the knots?(A) (B)(C) (D)(E)Part C: Each correct answer is worth 5 points13.How many circles and how many squares are covered by the blot in thepicture?(A) 1 circle and 3 squares(B) 2 circles and 1 square(C) 3 circles and 1 square(D) 1 circles and 2 squares(E) 2 circles and 2 squares14.Diana shoots three arrows at a target.On her first try, she gets 6 points and the arrows land like this: 6 pointsOn her second try, she gets 8 points and the arrows land like this: 8 pointsOn her third try, the arrows land like this:? points How many points will she get the third time?(A) 8 (B) 10 (C) 12 (D) 14 (E) 1615.How many different numbers greater than 10 and smaller than 25 with distinct digits can we make by using any two of the digits 2, 0, 1, and 8?(A) 4 (B) 5 (C) 6 (D) 7 (E) 816.Mark had some sticks of length 5 cm and width 1 cm.With the sticks he constructed the fence below.What is the length of the fence?(A) 20 cm(B) 21 cm(C) 22 cm (D) 23 cm (E) 25 cmlength17.The road from Anna's house to Mary's house is 16 km long.The road from Mary's house to John's house is 20 km long.The road from the crossroad to Mary's house is 9 km long.How long is the road from Anna’s house to John's house?(A) 7 km (B) 9 km (C) 11 km (D) 16 km (E) 18 km18.There are four ladybugs on a 4×4 board. Two are asleep and do not move.The other two ladybugs move one square every minute (up, down, left, or right). Here are pictures of the board for the first four minutes:Minute 1 Minute 2 Minute 3 Minute 4Which of these is a picture of the fifth minute (Minute 5)?(A) (B) (C) (D) (E)International Contest-GameMath Kangaroo Canada, 2018Answer KeyGrade 1-21 A B C D E 7 A B C D E 13 A B C D E2 A B C D E 8 A B C D E 14 A B C D E3 A B C D E 9 A B C D E 15 A B C D E4 A B C D E 10 A B C D E 16 A B C D E5 A B C D E 11 A B C D E 17 A B C D E6 A B C D E 12 A B C D E 18 A B C D EI N T E R N A T I O N A L C O N T E S T-G A M EM A T H K A N G A R O OC A N AD A,2018I N S T R U C T I O N SG R A D E3-41.You have 60 minutes to solve 24 multiple choice problems. For each problem,circle only one 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 sheetthat is marked, so make sure you have all your answers transferred to the response form before giving it back to the contest supervisor.3.The problems are arranged in three groups. A correct answer of the first 8problems is worth 3 points. A correct answer of the problems 9-16 is worth 4 points. A correct answer of the problems 17-24 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 24 points. The maximum score possible is 120.4.The use of external material or aid of any kind is not permitted.5.The figures are not drawn to scale. They should be used only for illustrationpurposes.6.Remember, you have about 2 to 3 minutes for each problem; hence, if a problemappears to be too difficult, save it for later and move on to another problem.7.At the end of the allotted time, please give the response form to the contestsupervisor.8.Do not forget to pick up your Certificate of Participation on your way out!Good luck!Canadian Math Kangaroo Contest teamGrade 3-42018 Canadian Math Kangaroo ContestPart A: Each correct answer is worth 3 points1.Lea has 10 rubber stamps. Each stamp has one of the digits:0, 1, 2, 3, 4, 5, 6, 7, 8, 9.She prints the date of St. Patrick’s Day 2018:How many different stamps does she use?(A) 5(B) 6 (C) 7 (D) 9 (E) 102.The picture shows three flying arrows and nine fixedballoons. When an arrow hits a balloon, it bursts,and the arrow flies further in the same direction.How many balloons will be hit by the arrows?(A) 2 (B) 3 (C) 4(D) 5 (E) 63.Susan is six years old. Her sister is one year younger, and her brother is one yearolder. What is the sum of the ages of the three siblings?(A) 10 (B) 15 (C) 18 (D) 21 (E) 304.Here is a picture of Sophie the ladybug. She turns around. Which picture ofthe ladybugs below is not Sophie?(A)(B)(C)(D)(E)5.Lucy folds a sheet of paper in half. Then she cuts a piece out of it. What willshe see when she unfolds the paper?(A)(B)(C) (D)(E)1 70320186. A table is set for 8 people.How many settings have the fork to the left of the plate and the knife to the right of the plate?(A) 5(B) 4 (C) 6 (D) 2 (E) 3 7.Emily added two 2-digit numbers correctly on paper. Then she painted out two cells,as shown below.What is the sum of two digits in the painted cells?(A) 5(B) 7 (C) 8 (D) 9 (E) 13 8.First, Diana scores 12 points in total with three arrows. On her second turn shescores 15 points.How many points does she score on her third turn?(A) 18 (B) 19 (C) 20 (D) 21 (E) 22 Part B: Each correct answer is worth 4 points9.How many different numbers greater than 12 and smaller than 58 with distinct digitscan we make by using any two of the digits 0, 1, 2, 5, and 8?(A) 3(B) 5 (C) 7(D) 8 (E) 912 points15 points ? points10.Roberto makes designs using tiles like this .How many of the following five designs can he make?(A) 1 (B) 2 (C) 3 (D) 4 (E) 511.Each of these five figures ,, , , , appears exactly once in everycolumn and every row of the given table.Which figure must we put in the cell with the question mark?(A) (B) (C) (D) (E)12.Toby glues 10 cubes together to make the structure shown.He paints the whole structure, even the bottom.How many cubes are painted on exactly four of their faces?(A) 6 (B) 7 (C) 8 (D) 9 (E) 1013.The opposite faces of a cube are identical, being dark, bright or patterned.Which picture below is the unfolded net of this cube?(A)14.Tom cuts two types of pieces out of grid paper.What is the smallest number of pieces identical to the ones shown that Tom needs to build the boat in the picture?(A) 5 (B) 6 (C) 7 (D) 8 (E) 915.The rooms in Kanga's house are numbered. Baby Roo entersthe main door, passes through some rooms and leaves thehouse. The numbers of the rooms that he visits are alwaysincreasing. Through which door does he leave the house?(A) A (B) B (C) C (D) D (E) E16.Peta rabbit had 20 carrots. She ate two carrots every day. She ate the twelfth carroton Wednesday. On which day did she start eating the carrots?(A) Monday (B) Tuesday (C) Wednesday (D) Thursday (E) FridayPart C: Each correct answer is worth 5 points17.The belt shown in the drawing can be fastened in five ways.How much longer is the belt fastened in one hole than the belt fastened in all five holes?(A) 4 cm (B) 8 cm (C) 10 cm (D) 16 cm (E) 20 cm18.In an ancient writing the symbols represent thenumbers 1, 2, 3, 4, and 5. Nobody knows which symbol represents which number.We know thatWhich symbol represents the number 3?(A)(B) (C) (D) (E)19. A stained-glass tile is flipped along the black line. The figure shows the tile after thefirst flip.What will the stained-glass tile look like after the third flip (at the far right)?(A)(B)(C)(D)(E)20.The large rectangle is made up of squares of varied sizes. The three smallest squareseach have an area of 1, as shown.What is the area of the largest square?(A) 81 (B) 100 (C) 110 (D) 121 (E) 14421.Five ducklings walk behind the mother duck in a row from the oldest to the youngestlike this: Dina and Becca walk right one after the other, Mingo walks behind Lisa but in front of Becca, Becca walks directly in front of Pip. What is the name of theyoungest duckling?(A) Dina (B) Pip (C) Becca (D) Lisa (E) Mingo22.Four balls each weigh 10, 20, 30 and 40 grams. Which ball weighs 30 grams?(A) A (B) B (C) C (D) D (E) it could be A or B23.Lois wants to write the numbers from 1 to 7 in the grid shown.Two consecutive numbers cannot be written in two neighbouringcells. Neighbouring cells meet at the edge or at a corner. Whatnumbers can she write in the cell marked with a question mark?(A) all seven numbers (B) only odd numbers(C) only even numbers (D) only number 4(E) only the numbers 1 or 7 24.The distance from Anna's to Mary's house is 16 kilometers along the shown road.The distance from Mary's to Nick's house is 20 kilometers.The distance from Nick's to John's house is 19 kilometers.How far is Anna's house from John's?(A) 15 (B) 16(C) 18(D) 19 (E) 20 ?International Contest-GameMath Kangaroo Canada, 2018Answer KeyGrade 3-41 A B C D E 9 A B C D E17 A B C D E2 A B C D E10 A B C D E 18 A B C D E3 A B C D E 11 A B C D E 19 A B C D E4 A B C D E 12 A B C D E 20 A B C D E5 A B C D E 13 A B C D E21 A B C D E6 A B C D E 14 A B C D E 22 A B C D E7 A B C D E 15 A B C D E 23 A B C D E8 A B C D E 16 A B C D E24 A B C D EI N T E R N A T I O N A L C O N T E S T-G A M EM A T H K A N G A R O OC A N AD A,2018I N S T R U C T I O N SG R A D E5-121.You have 75 minutes to solve 30 multiple choice problems. For eachproblem, circle only one 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 theonly sheet that is marked, so make sure you have all your answers transferred to that form before giving it back to the contest supervisor. 3.The problems are arranged in three groups. A correct answer of the first10 problems is worth 3 points. A correct answer of problems 11 -20 isworth 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.The use of external material or aid of any kind is not permitted.5.The figures are not drawn to scale. They should be used only forillustration purposes.6.Remember, you have about 2 to 3 minutes for each problem; hence, if aproblem appears to be too difficult, save it for later and move on to another problem.7.At the end of the allotted time, please give the response form to thecontest supervisor.8.Do not forget to pick up your Certificate of Participation on your way out!Good luck!Canadian Math Kangaroo Contest teamCanadian Math Kangaroo ContestPart A: Each correct answer is worth 3 points1.The drawing shows 3flying arrows and 9fixed balloons. Whenan arrow hits a balloon, it bursts, and the arrow flies further inthe same direction. How many balloons will not be hit byarrows?(A) 3 (B) 2(C) 6(D) 5(E) 42.The image shows a structure made of three objects.What does Peter see if he looks at the structure from above?(A)(B)(C) (D) (E)3.Diana played darts throwing arrows toward a target with three sections. First she got 14 points with twoarrows on the target. The second time she got 16 points. How many points did she get the third time?(A) 17(B) 18(C) 19 (D) 20 (E) 22 4. A garden is divided into identical squares. A fast snail and a slow snail move along the perimeter of thegarden starting simultaneously from the corner S but in different directions. The slow snail moves at the speed of 1 metre per hour (1 m/h) and the fast one at 2 metres per hour (2 m/h).At what point will the two snails meet?(A) A (B) B (C) C (D) D(E) E 14 points16 points ? A B CDE S 1 m/h2 m/h5.In which of the four squares is the fraction of the black area the largest?(A) A (B) B (C) C (D) D (E) they are all the same6. A star is made out of four equilateral triangles and a square. The perimeter of thesquare is 36 cm. What is the perimeter of the star?(A) 144 cm (B) 120 cm (C) 104 cm (D) 90 cm (E) 72 cm7.From the list 3, 5, 2, 6, 1, 4, 7 Masha chose 3 different numbers whose sum is 8. From the same list Dashachose 3 different numbers whose sum is 7. How many common numbers have been chosen by both girls?(A) none (B) 1 (C) 2 (D) 3 (E) impossible to determine8.We move a bead along a piece of wire. What shall we see when the beadcomes to the end of the wire?(A) (B) (C)(D) (E)9.There are 3squares in the figure. The side length of the smallest square is 6 cm.What is the side length of the biggest square?(A) 8(B) 10(C) 12(D) 14(E) 1610.In the following figure, the circles are light bulbs connected to some other lightbulbs. Initially, all light bulbs are off. When you touch a light bulb, this light bulband all its neighbours (e.g., the light bulbs connected to it) are lit.At least how many light bulbs do you have to touch to turn on all the light bulbs?(A) 2 (B) 3 (C) 4 (D) 5 (E) 6Part B: Each correct answer is worth 4 points11.Each square contains one of the numbers 1, 2, 3, 4, or 5, so that both of thecalculations following the arrows are correct. A number may be used morethan once. What number goes into the box with the question mark?(A) 1 (B) 2 (C) 3 (D) 4 (E) 5 12. Nine cars arrive at a crossroads and drive off as indicated by the arrows. Which figure shows these cars after leaving the crossroads?(A)(B) (C) (D) (E) 13. The faces of a cube are painted black, white or grey so that opposite faces are of different colour. Which of the following is not a possible net of this cube?(A)(B) (C) (D) (E)14.In a box there are many one-euro, two-euro and five-euro coins. A dispenser draws coins out of the box – one at a time, and stops when three identical coins are taken out. What is the largest possible amount that can be withdrawn? (A) 24 (B) 23 (C) 22 (D) 21 (E) 1515.Two girls, Eva and Olga and three boys, Adam, Isaac and Urban play with a ball. When a girl has the ball, she throws it to the other girl or to a boy. When a boy has the ball, he throws it to another boy but never to the boy from whom he just received it. Eva starts by throwing the ball to Adam. Who will do the fifth throw?(A) Adam (B) Eva (C) Isaac (D) Olga (E) Urban16.Emily wants to enter a number into each cell of the triangular table. The sum of thenumbers in any two cells with a common edge must be the same. She has alreadyentered two numbers. What is the sum of all the numbers in the table?(A) 18 (B) 20 (C) 21 (D) 22 (E) impossible to determine17.John coded a correct addition calculation naming the digits AA , BB , CC and DD .Which digit is represented by BB ?(A) 0 (B) 2 (C) 4 (D) 5(E) 6 + A B C C B A D D DD18.On Monday Alexandra shares a picture with 5 friends. For several days, everybody who receives thepicture, sends it once on the next day to two friends. On which day does the number of people who have seen the picture (including Alexandra) become greater than 75, if it is known that no one receives the picture more than once?(A) Wednesday (B) Thursday (C) Friday (D) Saturday (E) Sunday 19.The sum of the ages of Kate and her mother is 36, and the sum of the ages of her mother and her grandmother is 81. How old was the grandmother when Kate was born? (A) 28 (B) 38 (C) 45 (D) 53 (E) 56 20.Annie replaced the letters with numbers in the word KANGAROO (identical letters with the same digits, different letters with different digits) so that she got the largest possible 8-digit number, which is not a multiple of 4. What is the sum of the last three digits replacing the word ROO? (A) 13 (B) 14 (C) 12 (D) 15 (E) 11Part C: Each correct answer is worth 5 points21.Captain Hook has plundered a safe that contains 2520 gold coins. During the night, each of his pirates secretly took out some coins just for themselves. The first one took out �12�of the coins, the second one�13�of the remaining coins, the third one �14�of the remaining coins and so on. When Captain Hook opened the safe in the morning, he found only 252 coins inside. How many pirates are commanded by Captain Hook?(A) 8 (B) 9 (C) 10 (D) 11 (E) 12 22.In the figure on the right, the five balls A, B, C, D and E weigh 30, 50, 50, 50 and 80 grams, but not necessarily in this order. Which ball weighs 30 grams? (A) A (B) B (C) C (D) D (E) E23.If A, B, C are distinct digits, which of the following numbers cannot be the largest possible 6-digit number written using three digits A, two digits B, and one digit C? (A) AAABBC (B) CAAABB (C) BBAAAC (D) AAABCB (E) AAACBB 24.In the World of Numbers, there are many number-machines, which work in the following way: the machine adds the two beginning digits of the number and replaces them by their sum. For example, beginning with the number 87312 and using six such machines we obtain:How many such machines should be used in order to get the number times509...9 from the numbertimes1009...9? (A) 50(B) 60(C) 100(D) 80(E) Not possible to obtain this number8731215312 6312 91210212 3Page 525.Nick wants to arrange the numbers 2, 3, 4, ..., 10 into several groups such that the sum of the numbers in each group is the same. What is the largest number of groups he can get?(A) 2 (B) 3 (C) 4 (D) 6 (E) other answer 26.Peter cut an 8-cm wide wooden plank with a saw into 9 parts across the width of the plank.One piece was a square, the other were rectangles. Then he arranged all the pieces together as shown in the picture. What was the length of the plank?(A) 150 cm (B) 168 cm (C) 196 cm (D) 200 cm (E) 232 cm 27.Write 0 or 1 in each cell of the 5×5 table so that each 2×2 square of the 5×5 table contains exactly 3 equal numbers. What is the largest possible sum of all the numbers in the table?(A) 22 (B) 21 (C) 19 (D) 17 (E) 1528.14 people are seated at a round table.Each person is either a liar or tells the truth. Everybody says: "Both my neighbours are liars". What is themaximum number of liars at the table?(A) 7 (B) 8 (C) 9(D) 10(E) 1429.There are eight domino tiles on the table (pic 1). One half of one tile is covered. The 8 tiles can be arranged into a 4×4 square (pic 2), so that the number of dots in each row and column is the same.How many dots are on the covered part? (A) 1 (B) 2 (C) 3 (D) 4(E) 530.Four ladybugs sit on different cells of a 4×4 grid.One of them is sleeping and does not move. Each time you whistle, the other three ladybugs move toa free neighbouring cell. They can move up, down,right or left but they are not allowed to go back tothe cell they just came from. Which of the following images might show the result after the fourth whistle?(A)(B)(C)(D)(E)pic 1pic 2initial position after firstwhistleafter second whistle after third whistle Both my neighboursare liars.International Contest-GameMath Kangaroo Canada, 2018Answer KeyGrade 5-61 A B C D E 11 A B C D E21 A B C D E2 A B C D E 12 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 E15 A B C D E 25 A B C D E6 A B C D E16 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 E30 A B C D EI N T E R N A T I O N A L C O N T E S T-G A M EM A T H K A N G A R O OC A N AD A,2018I N S T R U C T I O N SG R A D E5-121.You have 75 minutes to solve 30 multiple choice problems. For eachproblem, circle only one 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 theonly sheet that is marked, so make sure you have all your answers transferred to that form before giving it back to the contest supervisor. 3.The problems are arranged in three groups. A correct answer of the first10 problems is worth 3 points. A correct answer of problems 11 -20 isworth 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.The use of external material or aid of any kind is not permitted.5.The figures are not drawn to scale. They should be used only forillustration purposes.6.Remember, you have about 2 to 3 minutes for each problem; hence, if aproblem appears to be too difficult, save it for later and move on to another problem.7.At the end of the allotted time, please give the response form to thecontest supervisor.8.Do not forget to pick up your Certificate of Participation on your way out!Good luck!Canadian Math Kangaroo Contest teamPage 1Canadian Math Kangaroo ContestPart A: Each correct answer is worth 3 points1.When the letters of the word MAMA are written vertically above one another, the word has a vertical line of symmetry. Which of these words also has a vertical line of symmetry when written in the same way?(A) ROOT (B) BOOM (C) BOOT (D) LOOT (E) TOOT2.A triangle has sides of length 6, 10 and 11. An equilateral triangle has the same perimeter. What is the length of each side of the equilateral triangle?(A) 6 (B) 9 (C) 10 (D) 11 (E) 273.Which number should replace ∗in the equation 2 ∙ 18 ∙ 14 = 6 ∙ ∗ ∙ 7to make it correct?(A) 8 (B) 9 (C) 10 (D) 12 (E) 154.The panels of Fergus' fence are full of holes. One morning, one of the panels fell flat on the floor.Which of the following could Fergus see as he approaches his fence?(A) (B) (C) (D) (E)5.How many possible routes are there to go from A to B in the direction indicated by the arrows?(A) 2 (B) 3 (C) 4 (D) 5 (E) 66.Martha multiplied two 2-digit numbers correctly on a piece of paper.Then she scribbled out three digits as shown.What is the sum of the three digits she scribbled out? (A) 5 (B) 6 (C) 9 (D) 12 (E) 14 7.A large rectangle is made up of nine identical rectangles whose longest sides are 10 cm long. What is the perimeter of the large rectangle?(A) 40 cm(B) 48 cm(C) 76 cm(D) 81 cm(E) 90 cm8. A hotel on an island in the Caribbean advertises using the slogan "350 days of sun every year!''. According tothe advert, what is the smallest number of days Willi Burn has to stay at the hotel in 2018 to be certain of having two consecutive days of sun?(A) 17 (B) 21 (C) 31 (D) 32 (E) 359.The diagram shows a rectangle of dimensions 7 × 11 containing two circles eachtouching three of the sides of the rectangle. What is the distance between the centres of the two circles?(A) 1 (B) 2(C) 3(D) 4 (E) 510.Only one of the digits in the year 2018 is a prime number. How many years will pass till the next year whenall of the digits in the year number are prime numbers?(A) 201 (B) 202 (C) 203 (D) 204 (E) 205Part B: Each correct answer is worth 4 points11.Square AAAAAAAA has sides of length 3 cm. The points MM and NN lie on AAAA and AAAA so that AAMMand AANN split the square into three pieces of the same area. What is the length of AAMM?(A) 0.5 cm (B) 1 cm (C) 1.5 cm (D) 2 cm (E) 2.5 cm12.A rectangle is divided into 40 identical squares. The rectangle contains more than one row of squares. Avacoloured the middle row. What is the largest possible number of squares that remain uncoloured?(A) 20 (B) 30 (C) 32 (D) 35 (E) 3913.A lion is hidden in one of three rooms. A note on the door of room 1 reads "The lion is here". A note on thedoor of room 2 reads "The lion is not here". A note on the door of room 3 reads "2+3=2×3". Only one of these statements is true. In which room is the lion hidden?(A) In room 1 (B) In room 2 (C) In room 3 (D) It may be in any room(E) It may be in either room 1 or room 214.Valeriu draws a zig-zag line inside a rectangle, creating angles of 10°,14°,33°, and 26°as shown.What is the size of angle θθ?(A) 11°(B) 12°(C) 16°(D) 17°(E)33°。

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.。

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。

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。

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

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。

加拿大初中数学竞赛试卷

一、选择题（每题5分，共20分）
1. 下列各数中，哪个是质数？
A. 4
B. 9
C. 15
D. 17
2. 下列哪个图形的面积是24平方厘米？
A. 正方形
B. 长方形
C. 三角形
D. 梯形
3. 一个长方形的长是8厘米，宽是5厘米，它的周长是多少厘米？
A. 20
B. 25
C. 30
D. 40
4. 下列哪个数是3的倍数？
A. 14
B. 21
C. 27
D. 33
5. 下列哪个图形的对称轴最多？
A. 正方形
B. 矩形
C. 等腰三角形
D. 菱形
二、填空题（每题5分，共20分）
6. 3的4次方等于多少？
7. 一个圆的半径是5厘米，它的面积是多少平方厘米？
8. 一个正方形的边长是6厘米，它的周长是多少厘米？
9. 下列各数中，哪个数是偶数？
10. 下列哪个数是奇数？
三、解答题（每题10分，共30分）
11. （10分）一个长方形的长是12厘米，宽是8厘米，求它的面积和周长。

12. （10分）一个等边三角形的边长是10厘米，求它的面积。

13. （10分）一个梯形的上底是6厘米，下底是12厘米，高是5厘米，求它的面积。

四、附加题（每题15分，共30分）
14. （15分）一个长方体的长、宽、高分别是4厘米、3厘米、2厘米，求它的体积。

15. （15分）一个正方体的对角线长是9厘米，求它的边长。

加拿大数学竞赛历年试题(滑铁卢大学)

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

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加拿大的數學競賽
(1)加拿大公開數學挑戰賽
加拿大公開數學挑戰賽Canadian Open Mathematics Challenge (COMC):一般在每年的11月份舉行。

學生在這個競賽裏獲得高分可以得到邀請去參加一些更高級別的競賽，比如加拿大數學奧林匹克(CMO),亞太數學奧林匹克(APMO),美國數學奧林匹克(USAMO)和國際數學奧林匹克(IMO)。

11月COMC活動可以提高學生的學習興趣，激發他們的”鬥志”。

一旦他們能主動地學習，那就什麼困難都不在話下。

其次，競賽對學生沒有壓力;因為考不好沒關係，而考好了就有關係：它對你進入好的大學，好的高中都有幫助。

第三，有的競賽(如COMC, IMC等)還直接發獎金給優勝者。

也會有單位邀請你去參加夏令營。

此項競賽有兩個目的：(1) 為秋季學期提供一項數學課外活動，對那些想豐富自己數學知識的學生大有幫助。

(2) 為加拿大數學奧林匹克(CMO)選拔人才。

此外，有的問題是對十年級數學課程的檢驗。

誰能參加這個競賽?(1)19歲以下;(2)加拿大公民或永久居民，並在公/私立中學註冊;(3)未在大學註冊;(4)未參加過PUTNAM數學競賽。

評獎：(1)大約50名優勝者會被邀請參加CMO;(2)每個省或地區的第一名會獲得一塊匾，他/她所在的學校也會獲得一塊匾;(3)每個省或地區的前九名獲得金牌，(4)前25%獲得證書。

考試時間為2.5小時，滿分40分。

A部分8道題，每題5分;解答過程部分正確也可得一些分。

B部分4道題，每題10分;即使答案正確但表述不清，也會被扣分。

考試中不得使用計算器。

考試內容大致如下：(1)EUCLID幾何，(2)解析幾何，(3)三角學，(4)函數，(5)方程組，(6)多項式，(7)數列與求和，(8)計數，(9)初等數論。

如何準備競賽?你可以去WWW.CEMC.UWATERLOO.CA找以前的考題做.有的較難，比如遊戲題, 它要求你會把遊戲問題數學化，知道如何找序列的規律，如何把特例推廣到一般情況。

如果沒有接觸過這種問題，是很難在兩個小時內給出必勝的策略來。

另外還要明白已考過的題近期內是不會重複的;再次，問題是永遠做不完的，關鍵在於掌握解題方法與技巧;而任何解題方法與技巧又代替不了INGENUITY(獨創性)和INSIGHT(洞察力)。

事實上，競賽的目的也正是為了培養這兩種能力。

相關網站-》http://cemc.math.uwaterloo.ca/
(2)加拿大全國性的數學競賽
加拿大全國性的數學競賽(CMC)是由waterloo大學舉辦的.從Grade 7 到Grade 12, 每個年級每年一次; 低年級學生可以參加高年級的競賽, 但高年級學生不能參加低年級的競賽. 各年級的競賽時間及名稱都不一樣: Grade 7-8的競賽叫GAUSS CONTEST, 每年5月中旬舉行; Grade 9的叫PASCAL CONTEST, Grade 10的叫CAYLEY CONTEST, GRADE 11的叫FERMAT CONTEST, 每年2月中旬舉行. 以上競賽為25道多項選擇題, 限時一個小時.
從2003年開始,CMC又為9至11年級學生增設了一種要寫過程的數學競賽: FRYER CONTEST (GRADE 9), GALOIS CONTEST (GRADE 10)和HYPATIA CONTEST (GRADE 11), 每年4月中旬舉行; 4道題，兩個半小時. 既要寫過程, 你就不能像多項選擇題那樣去“猜”答案了; 而且對你的寫作和表達能力也是一個很好的檢驗. 其中, 遊戲題是最讓人頭疼的，它要求你會把遊戲數學化，知道如何找序列的規律, 如何把特例推廣到一般情況. 如果沒有接觸過這種問題，是很難在兩個小時內給出必勝的策略來.
PASCAL競賽是為九年級及以下的學生舉辦的, CAYLEY競賽的參賽隊像是十年級及以下的學生, 而FERMAT競賽則是針對十一年級及以下的學生. 這些競
賽的目的是為學生們提高數學解題能力. 競賽不是考試, 沒有PASS或FAIL之說. 所有競賽都是這是這樣的: 考不好沒關係, 考好了就大有關係榮譽自不必說,更
重要的是你有了自信.
競賽內容為25道多項選擇題, 分為A(10道), B(10道), C(5道)三部分. 問題由易到難, 前兩部分不超出各省的教學大綱, PART C 考靈活性和技巧性. 競賽不
注重內容, 重在邏輯思維和數學解題方法. 每項競賽滿分150分, 限時一個小時. 每人每年只能參加一項(都在同一天舉行).
獲獎: 全國前五名獲得金牌; 前25%獲得證書; 各校的第一名獲得獎牌. 對於CAYLEY競賽和FERMAT 競賽的成績優異者會被邀請到WATERLOO大學參加六月的競賽培訓班. 對於FERMAT 競賽的各省第一名,可獲得$200的獎金.
(3)Euclid Contest等12年級數學競賽
對12年級學生，有多種競賽可以參加：EUCLID CONTEST (10道題) 和DESCARTES CONTEST(3部分，16道題), 每年4月中旬舉行, 兩個半小時，大題下還有小題. 這些競賽都要寫過程.
Euclid Contest:這個比賽是在4月舉行的，由加拿大滑鐵盧大學舉辦。

這個競賽有助於獲得該校的入學獎學金。

相關網站-》http://cemc.math.uwaterloo.ca/
(4)加拿大數學奧林匹克競賽
加拿大數學奧林匹克Canadian Mathematics Olympiad (CMO):在3月舉行，只有在COMC競賽中成績突出並收到邀請函的才可參加。

相關網站-》http://cms.math.ca/Competitions/CMO/
(5)魁省數學競賽
Concours de l'Association Mathématique du Québec (AMQ):這個比賽是2月舉行的，專門針對魁省CEGEP學生。

成功的候選人被邀請到Laval大學舉辦的數學夏令營。

相關網站：http://newton.mat.ulaval.ca/amq/ConcoursCollegiaux.html。