INTERNATIONAL MATHEMATICS AND SCIENCE OLYMPIAD FOR PRIMARY SCHOOLS (IMSO) 2009 Mathematics Contest in Taiwan Name:111
1
School: 111111
Grade: 1111
Number: 11111
Short Answer: there are 12 questions, fill in the correct answers in the answer sheet. Each correct answer is worth 10 points. Time limit: 90 minutes. 1. Find the largest possible divisor of the three numbers, 480608, 508811 and 723217, so that the reminder is the same in each case. 2. In a small group of people it was found that all of the following relationships were present: father, mother, son, daughter, brother, sister, cousin, nephew, niece, uncle and aunt. What is the smallest group of people for which this is possible? 3. Seven cubes are glued together face to face as shown in the diagram. The volume of the solid formed in this way is 189 cm3. Find the surface area of the solid.
4. Jack said to Jim: “If I give you 6 pigs for one horse, then you will own twice as many animals as I own.” Dan said to Jack: “If I give you 14 sheep for one horse, then you will own three times as many animals as I own.” Jim said to Dan: “If I give you 4 cows for one horse, then you’ll own six times as many animals as I own.” How many animals in total do Jack, Jim and Dan own? 5. By adding brackets in various ways to the expression 1÷3÷5÷7÷11÷13, what is the maximum number of different values which the expression can have? 6. Replace the asterisks with digits so that the multiplication below is correct: × * * * * * * * * *
* * * *
* * * 3 3 * * * * * * 2 0
What is the product?
1
* * 3 * *
* * * * 3 7 *
0 9 *
7. Tom has a contract to dig out some foundations and it must be done in 30 days. His own machine, which he wishes to use as much as possible, would take 48 days to do all the work. He can hire a bigger machine which would do the complete job in 21 days, but it costs $300 a day. There is only enough room for one machine at a time. What is the least number of days for which he will have to hire the larger machine? 8. Four different right-angled triangles all have sides which are of integral length and their perimeters are the same length. Find the smallest perimeter for which this is possible. 9. The diagram is of an irregular pentagon with all 5 of its diagonals drawn in. How many distinct triangles (not necessarily different) can be found, using only the lines (or parts of lines) shown in the diagram?
10. I have a rectangular picture whose edges are each an exact number of centimeters in length. At a quick glance it could be mistaken for a square, but it is not a square. It is placed inside a black border which is 3 cm wide all the way around the picture. The area of the border is exactly equal to the area of the picture. What is the area, in cm2, of the picture alone?
11. A combination lock on a safe needs a 6-letter sequence to open the safe. This is made from the letters A, B, C, D, E, F with none of them being used twice. Here are three guesses at the combination CBADF E AE D C B F E D FAC B In the FIRST guess only ONE letter is in its correct place. In the SECOND guess only TWO letters are in their correct places and those two correct places are not next to each other. In the THIRD guess only THREE letters are in their correct places. Each of the 6 letters is in its correct place once. What is the correct combination? 12. Given that ABCD is a square and the lengths EA, EB, EC are in the ratio EA:EB:EC=1:2:3, determine the size of the angle AEB, in degree.
A
B a
2a
E 3a
D 2
C
INTERNATIONAL MATHEMATICS AND SCIENCE OLYMPIAD FOR PRIMARY SCHOOLS (IMSO) 2009 Mathematics Contest (Second Round) in Taiwan Short Answer Problems School: Grade: Number: Name: Short Answer: there are 20 questions, fill in the correct answers in the space provided after each question. Each correct answer is worth 2 points. Time limit: 60 minutes. 1. Henry was given a certain number to multiply by 507, but he placed the first figure of his product by 5 below the second figure from the right instead of below the third. The result of Henry’s mistake was that his answer was wrong by 382050. What was the multiplicand? ANS: 2. Mice have 4 legs, ants have 6 legs and spiders have 8 legs. Eddy has twice as many mice as spiders, and three times as many ants as spiders. The number of legs adds up to 68. How many spiders does he have? ANS: 3. There are 100 nuts in five bags. In the first and second bags, there are altogether 42 nuts; in the second and third bags, there are 43 nuts; in the third and fourth bags, there are 34 nuts; in the fourth and fifth bags, there are 30 nuts. How many nuts are there in the first bag? ANS: 4. A student had to multiply 169 by a two-digit number whose second digit is twice as big as the first digit. Accidentally he changed the places of the two digits and obtained a product that differed from the correct one by 4563. Find the two-digit number. ANS: 5. How many days is it from Wednesday the 1st August to the first Saturday in September? (Inclusive of both dates) ANS: 6. If 6 cats can catch 6 rats in 6 minutes, how many cats are needed to catch 12 rats in 12 minutes? ANS: 7. A collection of sheep and turkeys have a total of 99 heads and legs between them. There are twice as many turkeys as there are sheep. How many turkeys are there? ANS:
8. Find the smallest possible number that leaves a remainder of 1 when divided by 2, 3, 4, 5 or 6, and which can be divided by 7 exactly. ANS: 9. Each of the letters A, B, C, D, E is used to represent a single digit in these two statements.(Same letter = same digit.) A + B = C, C + D = EA (Note EA is a 2-digit number) What is the value of B + D ? ANS: 10. Ann, Ben and Carol each have some money. If Ann gave Ben $30, then Ben would have twice as much as Ann. If Ben gave Carol $30, then Carol would have twice as much as Ben. If Carol gave Ann $30, then they would both have the same amount. How much money did Ann have? ANS:
$
11. A square lawn has a path 1 m wide which goes around the outside of all the four edges. The area of the path is 40 m2. What is the area of the lawn?
m ANS: 12. Four consecutive odd numbers add up to a total of 80. What is the smallest of those four numbers.
2
ANS: 13. A cube with an edge length of 10 cm is resting on a horizontal table. An insect starts crawling from the table at an angle of 30 degrees to the horizontal. How far will it have crawled on the cube by the time it gets to the top? ANS:
cm
14. On this diagram you may start at any square and move up or down or across (but NOT diagonally) into the next square. No square may be used twice. The digits in each square are written down in the order they are used to form a number. What is the largest number that can be made? 5
9
1
8
4
7
3
6
2 ANS:
15. A new monument is to be made in the shape of a cuboid. Only three of the faces are to be decorated. To allow for this: one face has to have an area of 48 m2; another is to have an area of 72 m2; and another of 96 m2. What will be the volume of the monument? ANS:
m3
16. Arrange the numbers 1 to 9, using each number only once and placing only one number in each cell so that the totals in both directions (vertically and horizontally) are the same. How many different sums are there?
ANS: 17. How many distinct squares (not necessarily different in size) can be traced out following only the lines of the grid drawn below?
ANS: 18. A 4-wheeler car has travelled 24,000 km and, in that distance, has worn out 6 tyres. Each tyre travelled the same distance. How far did each separate tyre travel? ANS:
km
19. A, B and C are three villages near to each other, shown in the diagram below, where the straight lines represent the only roads joining the villages. The figures give the distances in km between villages. B 8 7 C 12 A
A new fire station is to be built to serve all three villages. It is to be on a roadside at such a position that the greatest distance that the fire-engine has to travel along the roads in an emergency at one of the villages is as small as it can be. What is this smallest distance? ANS:
km
20. The diagonal of this 5×3 rectangle passes through 7 squares.
The diagonal of this 6×4 rectangle passes through 8 squares.
What is the number of squares passed through by the diagonal of a 360×2009 rectangle? ANS:
INTERNATIONAL MATHEMATICS AND SCIENCE OLYMPIAD FOR PRIMARY SCHOOLS (IMSO) 2009 Mathematics Contest (Second Round) in Taiwan, Essay Problems Name:
School:
Grade:
Number:
Answer the following 10 questions, and show your detailed solution in the space provided after each question. Each question is worth 4 points. Time limit: 60 minutes. 1. There are two clocks. One of them gains 6 seconds in every hour, while the other loses 9 seconds in every hour. If they are both set to show the same time, and then set going, how long will it be before the time displayed on them is exactly 1 hour apart?
2. Replace the asterisks in 86**** with the digits 1, 2, 3 and 4. Using each of them once so that the six-digit number obtained is the largest possible number divisible by 132.
3. There are two isosceles triangles. They are equal in area. In both triangles all edges measure an exact number of cm, and the two edges of equal length are 13 cm. In one of them the third edge measures 10 cm. What is the length of the third edge of the other?
4. In a quadrilateral ABCD, BC is parallel to AD. E is the foot of the perpendicular from B to AD. Find BE if AB=17, BC=16, CD=25 and AD=44.
B
16
25
17 A
C
E
44
5. Three different numbers from 1 to 10 were written on three cards. The cards were shuffled and dealt to three players. Each player got one card and wrote down the number of his card. Then the cards were collected and dealt again. After several deals the three players reported the totals of their written numbers, which were 13, 15, and 23. What numbers were written down on the cards at the beginning?
6. Five students A, B, C, D, and E competed in solving a math problem. The complete solution to the problem was awarded 10 points and a partial solution – an integer between 2 and 9. Each student scored some number of points so that : A, B, and C were awarded 15 points together; and B, C and D were awarded 12 points together. All students got different scores. The student A had the highest score and student E who scored 6 points, was placed third. What was the score of student D?
D
7. A grandmother has two grandsons. Her age is a two-digit number. The first digit is equal to the age of the first grandson, and the second digit is equal to the age of the second grandson. If the sum of their ages is 69, how old is the grandmother?
8. A 'Lucky number' has been defined as a number which can be divided exactly by the sum of its digits. For example: 1729 is a Lucky number since 1 + 7 + 2 + 9 = 19 and 1729 can be divided exactly by 19. Find the smallest Lucky number which is divisible by 13
9. In the middle of a large field there is a wooden hut on a rectangular base measuring 10 m by 6 m. Outside the hut, and tethered by a chain to one corner is a goat. Over what area can the goat graze if the tether is 15 m long? (Usingπ＝3.14)
10. A chess-board is made up of 64 black and white squares in the normal way, each having an edge length of 10 cm. On this board the largest possible circle is drawn so that it’s circumference does not pass through a black square. What is the radius of the circle?
INTERNATIONAL MATHEMATICS AND SCIENCE OLYMPIAD FOR PRIMARY SCHOOLS (IMSO) 2009 Mathematics Contest in Taiwan, Exploration Problems Name:
School:
Grade:
Number:
Answer the following 5 questions. Write down your answer in the space provided after each question. Each question is worth 8 points. Time limit: 60 minutes. 1. At a recent athletics meeting, five old acquaintances: Fred, Greta, Hans, Iolo and Jan met together for the first time since leaving college, so they had a lot of news to catch up on. It seemed they all lived in different towns: Acton, Buswick, Coalford, Derby and Eccles; and that they all had different jobs which were, in no particular order: an engineer, a lawyer, a teacher, a doctor and a shopkeeper. Also, each one was the winner in just one event at the meeting. These events were: 100 m, 400 m, 1500 m, High Jump and Javelin. The following facts were also known: (1) Hans the shopkeeper from Derby won the High Jump. (2) The lawyer was from Eccles and said he was not a runner. (3) Greta was a P.E. teacher from Buswick and won the 1500 m. (4) The doctor, who came from Acton, did not win the 100 m. (5) The person from Derby was not an engineer. (6) Iolo was an engineer from Coalford and did not win the 400 m. (7) Jan was not a lawyer, but did win the 400 m. (8) Fred did not come from Acton and was not a runner. (a) Which event did the person from Coalford win? (1 point) (b) Which town did Jan come from? (1 point) (c) What was the name of the lawyer? (2 point) (d) Which event did the engineer win? (2 point) (e) Which event did Fred win? (2 point)
ANS: (a)
(b)
(c)
(d)
(e)
2. In 2008, a school had 450 new pupils, making a total of 1600 pupils. In 2009, there were 504 new pupils, 8% of the number of the previous year’s pupils left the school. (a) Find the percentage increase in the number of new pupils from 2008 to 2009. (2 point) (b) How many pupils left the school? (3 point) (c) What was the total number of pupils in the school in 2009? (3 point)
ANS: (a)
(b)
(c)
3. The diagram represents a small sheet of 12 postage stamps, as they are usually sold, all perforated at the edges and all of the same value. (The letters are only there to identify the separate stamps). You need 4 of the stamps in order to post a letter but would like all 4 to be properly joined together at their edges (not at their corners). For example: ABCD, EFGH, JKLM, FGHL would all do, but NOT EFLM. In how many different ways can you get such a group of 4? Write down this number. A B C D E
F G H
J
K L M
ANS:
4. A domino consists of two unit squares joined edge to edge, each with a number on it. Fifteen dominoes, numbered 11, 12, 13, 14, 15, 22, 23, 24, 25, 33, 34, 35, 44, 45, and 55, are assembled into the 6 by 5 rectangle shown in the diagram below. However, the boundary of the individual dominoes has been erased. Reconstruct the dominoes by drawing in the boundary lines. 5
3
5
2
2
3
2
3
4
4
4
5
3
1
3
4
1
3
2
1
5
2
4
5
4
2
1
1
1
5
5. We can make shapes by joining diamonds together edge to edge. There are exactly two different shapes that can be made in this way from two diamonds. We call them bimonds. Bimond Diamon
120˚ 60˚ Sides equal in length, one angle 60˚, adjacent angle 120˚; area 1 unit2.
The two different bimonds, each with area 2 units2.
Each of the following shapes is a trimond made from three diamonds joined edge to edge; each has area 3 units2.
Note: The three trimonds above are all the same — rotations or reflections of one will produce the others.
(1) From a set of one diamond, two different bimonds and your collection of different trimonds, fit some of these together to make a parallelogram with area 12 units2. Draw this parallelogram, showing the pieces you have used. (2 point) (2) From the same set of pieces, make another parallelogram with area 12 units2, but with a different perimeter. Draw this parallelogram. (2 point) (3) How many different shapes can be made from four diamonds joined edge to edge each having an area of 4 units2. (4 point)
ANS: (1)
ANS: (2)
ANS: (3)
INTERNATIONAL MATHEMATICS AND SCIENCE OLYMPIAD FOR PRIMARY SCHOOLS (IMSO) 2009 Mathematics Contest in Taiwan Name:111
1
School: 111111
Grade: 1111
Number: 11111
Short Answer: there are 12 questions, fill in the correct answers in the answer sheet. Each correct answer is worth 10 points. Time limit: 90 minutes. 1. Find the largest possible divisor of the three numbers, 480608, 508811 and 723217, so that the reminder is the same in each case. 【Solution】 Since the reminder is the same in each case, the largest possible divisor must be a common factor of 508811－480608=28203=3×7×17×79, 723217－508811=214406=2×23×59×79, 723217－480608=242609=37×79×83. Hence the answer is 79. ANS:79 2. In a small group of people it was found that all of the following relationships were present: father, mother, son, daughter, brother, sister, cousin, nephew, niece, uncle and aunt. What is the smallest group of people for which this is possible? 【Solution】 Since father and son must be 2 different men and mother and daughter must be 2 different women, there are at least 4 people. We can find the following situation can satisfy the conditions: A brother and a sister. The brother has a son. The sister has a daughter. ANS: 4 people 3. Seven cubes are glued together face to face as shown in the diagram. The volume of the solid formed in this way is 189 cm3. Find the surface area of the solid. 【Solution】 Since the volume of the solid formed in this way is 189 cm3, the volume of a cube is 189÷7=27 cm3 and hence the area of a face is 9 cm2. The number of the faces of the solid is 6×6－6=30. So the surface area of the solid is 9×30=270 cm2. ANS: 270 cm2 4. Jack said to Jim: “If I give you 6 pigs for one horse, then you will own twice as many animals as I own.” Dan said to Jack: “If I give you 14 sheep for one horse, then you will own three times as many animals as I own.” Jim said to Dan: “If I give you 4 cows for one horse, then you’ll own six times as many animals as I own.” How many animals in total do Jack, Jim and Dan own? 1
【Solution】 Assume Jack has a animals, Jim has b animals, and Dan has c animals. Thus we have the following equations: ⎧ 2( a − 6 + 1) = b + 6 − 1 ⎧2a − 10 = b + 5 ⎧2a − 15 = b ⎪ ⎪ ⎪ ⎨3( c − 14 + 1) = a + 14 − 1 ⇔ ⎨ 3c − 39 = a + 13 ⇔ ⎨ 3c − 52 = a ⎪ 6(b − 4 + 1) = c + 4 − 1 ⎪ 6b − 18 = c + 3 ⎪ 6b − 21 = c ⎩ ⎩ ⎩ Hence a=11, b=7, and c=21. So there are 11+7+21=39 animals in total. ANS: 39 5. By adding brackets in various ways to the expression 1÷3÷5÷7÷11÷13, what is the maximum number of different values which the expression can have? 【Solution】 No matter how the brackets are added, 1 is always part of the numerator and 3 is always part of the denominator. Each other number may be in either the numerator or the denominator, we will have 24=16 different values. ANS:24=16 6. Replace the asterisks with digits so that the multiplication below is correct:
× * * * * * * * * *
* * * *
* * * 3 3 * * * * * * 2 0
* * 3 * *
* * * * 3 7 *
0 9 *
What is the product? 【Solution】 Since 33337 is not divisible by 2 and 3, we can set the multiplication as following: 3 3 3 3 7 × A B C 1 3 3 3 3 7 * * * * * D * * * * * * * * * * * * * * * * 2 0 0 9 7 Thus D must be 6 and hence C must be 8. We can set the multiplication again as following: 3 3 3 3 7 × A B 8 1 3 3 3 3 7 2 6 6 6 9 6 * * * * * E * * * * * * * * * * 2 0 0 9 7 2
Thus E must be 8 and hence B must be 4. We can set the multiplication again as following: 3 3 3 3 7 × A 4 8 1 3 3 3 3 7 2 6 6 6 9 6 1 3 3 3 4 8 * * * * * F * * * * 2 0 0 9 7 Thus F must be 5 and hence A must be 5. We can get the multiplication as following: 3 3 3 3 7 × 5 4 8 1 3 3 3 3 7 2 6 6 6 9 6 1 3 3 3 4 8 1 6 6 6 8 5 1 8 2 7 2 0 0 9 7 Hence the product is 182720097. ANS: 182720097 7. Tom has a contract to dig out some foundations and it must be done in 30 days. His own machine, which he wishes to use as much as possible, would take 48 days to do all the work. He can hire a bigger machine which would do the complete job in 21 days, but it costs $300 a day. There is only enough room for one machine at a time. What is the least number of days for which he will have to hire the larger machine? 【Solution】 1 of the work per day Set the amount of work to 1. Thus Tom’s machine does 48 1 of the work per day. Hence Tom must hire the and the larger machine does 21 1 1 1 ANS:14 days larger machine for at least (1 − × 30) ÷ ( − ) = 14 days. 48 21 48 8. Four different right-angled triangles all have sides which are of integral length and their perimeters are the same length. Find the smallest perimeter for which this is possible. The right-angled triangles with smaller integral sides have sides of length: (3, 4, 5), (5, 12, 13), (7, 24, 25), (8, 15, 17), (9, 40, 41), (11, 60, 61), (13, 84, 85) (15, 112, 113), (20, 21, 29),…. Their perimeters are 12, 30, 56, 40, 90, 132, 182, 240, 70,….We can find right-angled triangles (60, 80, 100), (40, 96, 104), (48, 90, 102), (15, 112, 113) with same perimeter. Hence the smallest perimeter is 240. ANS:240 3
9. The diagram is of an irregular pentagon with all 5 of its diagonals drawn in. How many distinct triangles (not necessarily different) can be found, using only the lines (or parts of lines) shown in the diagram? 【Solution】 There are 5 triangles with two edges of the pentagon, 5×4=20 triangles with exactly one edge of the pentagon, 5 triangles with exactly one full diagonal and without edges of the pentagon, and 5 triangles without full diagonals and without edges of the pentagon. There are in total 5+20+5+5=35 distinct diagonals. ANS:35 10. I have a rectangular picture whose edges are each an exact number of centimeters in length. At a quick glance it could be mistaken for a square, but it is not a square. It is placed inside a black border which is 3 cm wide all the way around the picture. The area of the border is exactly equal to the area of the picture. What is the area, in cm2, of the picture alone? 【Solution】 Assume the length of the picture is a and the width of the picture is b. We can suppose that b ≥ a . Thus we can get the following equation: ab=4×3×3+2×3a+2×3b=36+6a+6b ab－6a－6b+36=36+36 a(b－6)－6(b－6)=72 (a－6)(b－6)=72 Since a and b are positive integers and at a quick glance it could be mistaken for a square, there is only one possible factorization of 72 which is 8×9 and hence the solution is a=14 and b=15, the area of the picture alone is 210 cm2. ANS:210 cm2 11. A combination lock on a safe needs a 6-letter sequence to open the safe. This is made from the letters A, B, C, D, E, F with none of them being used twice. Here are three guesses at the combination CBADF E AE D C B F E D FAC B In the FIRST guess only ONE letter is in its correct place. In the SECOND guess only TWO letters are in their correct places and those two correct places are not next to each other. In the THIRD guess only THREE letters are in their correct places. Each of the 6 letters is in its correct place once. What is the correct combination? 【Solution】 (1) (2) (3) (4) (5) (6) FIRST C B A D F E SECOND A E D C B F THIRD E D F A C B 4
There are 5 mistakes in the FIRST guess, 4 mistakes in the SECOND guess, and 3 mistakes in the THIRD guess. (a) If the letter C is on the correct place in the FIRST guess, then the positions (4) and (5) can’t be C. Thus we have: (1) (2) (3) (4) (5) (6) FIRST C B A D F E SECOND A E D C B F THIRD E D F A C B Hence E is on (2) and A is on (4). Thus we have: (1) (2) (3) (4) (5) (6) FIRST C B A D F E SECOND A E D C B F THIRD E D F A C B Now there are 3 mistakes in the THIRD guess, so we can finish the table as follows: (1) (2) (3) (4) (5) (6) FIRST C B A D F E SECOND A E D C B F THIRD E D F A C B Thus B is on position (5) and (6), which is impossible! (b) If the letter B is on the correct place in the FIRST guess, then the positions (5) and (6) can’t be B. Thus we have: (1) (2) (3) (4) (5) (6) FIRST C B A D F E SECOND A E D C B F THIRD E D F A C B Hence D is on (3), C is on (5)and E is on (1). Thus we have: (1) (2) (3) (4) (5) (6) FIRST C B A D F E SECOND A E D C B F THIRD E D F A C B Thus we got one possible answer, EBDACF. (c) If the letter A is on the correct place in the FIRST guess, then the positions (1) and (4) can’t be A. Thus we have: (1) (2) (3) (4) (5) (6) FIRST C B A D F E SECOND A E D C B F THIRD E D F A C B Hence D is on (2) and F is on (6). Thus we have: (1) (2) (3) (4) (5) (6) FIRST C B A D F E SECOND A E D C B F THIRD E D F A C B Now there are 3 mistakes in the THIRD guess, so we can finish the table as follows: 5
(1) (2) (3) (4) (5) (6) FIRST C B A D F E SECOND A E D C B F THIRD E D F A C B Thus C is on position (4) and (5), which is impossible! (d) If the letter D is on the correct place in the FIRST guess, then the positions (3) and (2) can’t be D. Thus we have: (1) (2) (3) (4) (5) (6) FIRST C B A D F E SECOND A E D C B F THIRD E D F A C B Hence A is on (1) , F is on (3)and C is on (5). Thus we have: (1) (2) (3) (4) (5) (6) FIRST C B A D F E SECOND A E D C B F THIRD E D F A C B But the correct places of A and E are next to each other in the SECOND guess, hence it is not an answer. (e) If the letter F is on the correct place in the FIRST guess, then the positions (6) and (3) can’t be F. Thus we have: (1) (2) (3) (4) (5) (6) FIRST C B A D F E SECOND A E D C B F THIRD E D F A C B Hence B is on (6) and C is on (4). Thus we have: (1) (2) (3) (4) (5) (6) FIRST C B A D F E SECOND A E D C B F THIRD E D F A C B There are 3 mistakes in the THIRD guess, so we can finish the table as: (1) (2) (3) (4) (5) (6) FIRST C B A D F E SECOND A E D C B F THIRD E D F A C B Thus D is on position (3) and (2), which is impossible! (f) If the letter E is on the correct place in the FIRST guess, then the positions (2) and (1) can’t be E. Thus we have: (1) (2) (3) (4) (5) (6) FIRST C B A D F E SECOND A E D C B F THIRD E D F A C B Hence B is on (5) and F is on (3). Thus we have: (1) (2) (3) (4) (5) (6) FIRST C B A D F E SECOND A E D C B F THIRD E D F A C B 6
Now there are 3 mistakes in the THIRD guess, so we can finish the table as follows: (1) (2) (3) (4) (5) (6) FIRST C B A D F E SECOND A E D C B F THIRD E D F A C B Thus A is on position (1) and (4), which is impossible! ANS: EBDACF 12. Given that ABCD is a square and the lengths EA, EB, A B EC are in the ratio EA:EB:EC=1:2:3, determine the size a 2a E of the angle AEB, in degree. 2a 【Solution】 3a Rotate △AEB to △CE’B with center B, as the right E’ figure. Connect EE’. a Thus △BEE’ is an isosceles right triangle and hence D C ∠BE’E=45° and the length of EE’ is 2 2a . Hence △CEE’ is a right triangle by Pythagorean theorem and we get ∠CE’E=90°. We have ∠AEB=∠BE’C=∠CE’E+∠BE’E =90°+45°=135°. ANS: 135°
7
INTERNATIONAL MATHEMATICS AND SCIENCE OLYMPIAD FOR PRIMARY SCHOOLS (IMSO) 2009 Mathematics Contest (Second Round) in Taiwan Short Answer Problems Name:
School:
Grade:
ID number:
Short Answer: there are 20 questions, fill in the correct answers on the answer sheet. Each correct answer is worth 2 points. Time limit: 60 minutes. 1. Henry was given a certain number to multiply by 507, but he placed the first figure of his product by 5 below the second figure from the right instead of below the third. The result of Henry’s mistake was that his answer was wrong by 382050. What was the multiplicand? 【Solution】 Assume the number is a. Then we can get the following equation from the situation: 507a－57a=382050. Hence a=849. ANS:849 2. Mice have 4 legs, ants have 6 legs and spiders have 8 legs. Eddy has twice as many mice as spiders, and three times as many ants as spiders. The number of legs adds up to 68. How many spiders does he have? 【Solution】 Since Eddy has twice as many mice as spiders and three times as many ants as spiders, we can assume he has x spiders and hence has 2x mice and 3x ants. Thus we have 8×x+6×3x+4×2x=68, i.e. 34x=68. So x=2. ANS:2 3. There are 100 nuts in five bags. In the first and second bags, there are altogether 42 nuts; in the second and third bags, there are 43 nuts; in the third and fourth bags, there are 34 nuts; in the fourth and fifth bags, there are 30 nuts. How many nuts are there in the first bag? 【Solution】 Assume there are a nuts in the first bag, b nuts in the second bag, c nuts in the third bag, d nuts in the fourth bag and e nuts in the fifth bag. Thus we have the following equations:
⎧a + b + c + d + e = 100 ⎪ a + b = 42 a + b + c + d + e = 100 ⎧b + c + d = 49 ⎧ ⎪⎪ ⇔⎨ b + c = 43 ⇔ ⎨ ⎨ + + + + = 2 2 2 149 a b c d e a + e = 51 ⎩ ⎩ ⎪ c + d = 34 ⎪ d + e = 30 ⎪⎩ Hence d=49－43=6, e=30－d=24, a=51－e=27, b=42－a=15, c=43－b=28. ANS:27
4. A student had to multiply 169 by a two-digit number whose second digit is twice as big as the first digit. Accidentally he changed the places of the two digits and obtained a product that differed from the correct one by 4563. Find the two-digit number. 【Solution】
Assume the number is ab , where 2a=b. Thus we can get the following equation from the situation: (10b+a)×169－(10a+b)×169=4563, i.e. b－a=3. So a=3 and b=6 and hence the number is 36. ANS: 36 5. How many days is it from Wednesday the 1 August to the first Saturday in September? (Inclusive of both dates) st
【Solution】
Since there are 31 days in August and the 1st August is Wednesday, the 31st August is Friday and hence the 1st September is Saturday. So the answer is 31+1=32 days. ANS:32 6. If 6 cats can catch 6 rats in 6 minutes, how many cats are needed to catch 12 rats in 12 minutes? 【Solution】
Since 6 cats can catch 6 rats in 6 minutes, 1 cat catches 6 rats in 36 minutes and hence 1 cat catches 1 rat in 6 minutes. So 1 cat can catch 2 rats in 12 minutes, i.e. 6 cats can catch 12 rats in 12 minutes. ANS:6 cats 7. A collection of sheep and turkeys have a total of 99 heads and legs between them. There are twice as many turkeys as there are sheep. How many turkeys are there? 【Solution】
We know that a turkey has a total of 3 heads and legs and a sheep has a total of 5 heads and legs. Assume there are a turkeys and b sheep, then we have the following equations: ⎧3a + 5b = 99 ⎨ a = 2b ⎩
Hence a=18 and b=9. ANS: 18 turkeys 8. Find the smallest possible number that leaves a remainder of 1 when divided by 2, 3, 4, 5 or 6, and which can be divided by 7 exactly. 【Solution】
Since L.C.M [2, 3, 4, 5, 6]=60, the numbers that leave a remainder of 1 when divided by 2, 3, 4, 5 or 6 are all of the form 60k+1 for some k. They are 1, 61, 121, 181, 241, 301, 361,…The smallest number which can be divided by 7 exactly is 301. ANS:301 9. Each of the letters A, B, C, D, E is used to represent a single digit in these two statements.(Same letter = same digit.) A + B = C, C + D = EA (Note EA is a 2-digit number) What is the value of B + D ? 【Solution】
Because 10E+A=C+D=A+B+D, B+D=10E. Since B and D are both single digits, B+D<8+9=17. So E=1 and hence the only possible value of B + D is 10. ANS:10 10. Ann, Ben and Carol each have some money. If Ann gave Ben $30, then Ben would have twice as much as Ann. If Ben gave Carol $30, then Carol would have twice as much as Ben. If Carol gave Ann $30, then they would both have the same amount. How much money did Ann have? 【Solution】
Assume Ann has $a, Ben has $b, and Carol has $c . Thus we have the following equations: ⎧2(a − 30) = b + 30 ⎧2a − 60 = b + 30 ⎧2a − 90 = b ⎪ ⎪ ⎪ ⎨ 2(b − 30) = c + 30 ⇔ ⎨ 2b − 60 = c + 30 ⇔ ⎨ 2b − 90 = c ⎪ ⎪ ⎪ ⎩ c − 30 = a + 30 ⎩ c − 30 = a + 30 ⎩ c − 60 = a Hence a=110, b=130, and c=170. Ann has $110. ANS:$110 11. A square lawn has a path 1 m wide which goes around the outside of all the four edges. The area of the path is 40 m2. What is the area of the lawn?
【Solution】
1
Assume the length of an edge of the lawn is a. From the right figure, we know that (1×1+1×a)×4=40, i.e. a=9. Hence the area of the lawn is 9×9=81 m2.
a a
ANS: 81 m2 12. Four consecutive odd numbers add up to a total of 80. What is the smallest of those four numbers. 【Solution】
Assume the four consecutive odd numbers are a－3, a－1, a＋1, and a＋3. Thus we have 4a=80, i.e. a=20. So the four consecutive odd numbers are 17, 19, 21 and 23, the smallest number is 17. ANS: 17 13. A cube with an edge length of 10 cm is resting on a horizontal table. An insect starts crawling from the table at an angle of 30 degrees to the horizontal. How far will it have crawled on the cube by the time it gets to the top? 【Solution】
Since the angle is 30˚and the height the insect increases is 10 cm, the insect has crawled 10×2=20 cm. ANS: 20 cm 14. On this diagram you may start at any square and move up or down or across (but NOT diagonally) into the next square. No square may be used twice. The digits in each square are written down in the order they are used to form a number. What is the largest number that can be made?
5 9 1 8 4 7 3 6 2
【Solution】
If you star at the red squares, then the number of the digits of your 5 9 1 number is at most 8 since the number of red squares is less than the 8 4 7 number of white squares. If you star at the white squares, then the number of the digits of your 3 6 2 number is at most 9. So we start at 5 and get the number 594836271 ANS:594836271 15. A new monument is to be made in the shape of a cuboid. Only three of the faces are to be decorated. To allow for this: one face has to have an area of 48 m2; another is to have an area of 72 m2; and another of 96 m2. What will be the volume of the monument? 【Solution】
Let the sides of the cuboid be a, b and c. Since ab=48= 24 × 3 , bc=72= 23 × 32 and
ac=96= 25 × 3 , a2b2c2=48×72×96= 212 × 34 , hence the volume abc is
26 × 32 = 576 . ANS:576 m3 16. Arrange the numbers 1 to 9, using each number only once and placing only one number in each cell so that the totals in both directions (vertically and horizontally) are the same. How many different sums are there? e f a b
c d g h
【Solution】
Since the totals in both directions are the same, a+b+c+d=e+f+g+h, i.e. the sum of the eight numbers, a+b+c+d+e+f+g+h, must be an even number. Because 1+2+3+4+5+6+7+8+9=45 is an odd number, the number in the center square must be an odd number and hence there are 5 possibilities: If 1 is placed in the center square, then the sum is 23 and we can divide the (i) other eight number into two sets, {2, 3, 8, 9} and {4, 5, 6, 7}. (ii) If 3 is placed in the center square, then the sum is 24 and we can divide the other eight number into two sets, {1, 4, 7, 9} and {2, 5, 6, 8}. (iii) If 5 is placed in the center square, then the sum is 25 and we can divide the other eight number into two sets, {1, 2, 8, 9} and {3, 4, 6, 7}. (iv) If 7 is placed in the center square, then the sum is 26 and we can divide the other eight number into two sets, {1, 3, 6, 9} and {2, 4, 5, 8}. (v) If 9 is placed in the center square, then the sum is 27 and we can divide the other eight number into two sets, {1, 2, 7, 8} and {3, 4, 5, 6}. ANS:5 17. How many distinct squares (not necessarily different in size) can be traced out following only the lines of the grid drawn on the right? 【Solution】
There are 9×7=63 unit squares, 8×6=48 2×2 squares, 7×5=35 3×3 squares, 6×4=24 4×4 squares, 5×3=15 5×5 squares, 4×2=8 6×6 squares, and 3×1=3 7×7 squares. So there are 63+48+35+24+15+8+3=196 distinct squares. ANS:196 18. A 4-wheeler car has travelled 24,000 km and, in that distance, has worn out 6 tyres. Each tyre travelled the same distance. How far did each separate tyre travel? 【Solution】
If the car didn’t change the tyres, then the total distance of the four tyres travelled is 24,000×4=96,000 km. But it has used 6 tyres, so the answer is 96,000÷6=16,000 km. ANS:16,000 km 19. A, B and C are three villages near to each other, shown in the diagram below, where the straight lines represent the only roads joining the villages. The figures give the distances in km between villages. B
Fire Station 0.5
8
7 C 12 A
A new fire station is to be built to serve all three villages. It is to be on a roadside at such a position that the greatest distance that the fire-engine has to travel along the roads in an emergency at one of the villages is as small as it can be. What is this smallest distance? 【Solution】
We are told that the greatest distance the fire-engine has to travel to any of the villages is to be as small as it can be. At the position show in the diagram above, 0.5 km from B, the shortest distance to A, B and C are 7.5 km, 0.5 km and 7.5 km respectively. To show that no other position is satisfactory z At any other position on the road joining B to C, the shortest distance to either A or C will be more than 7.5 km. z At any other position on the road joining A to B, the shortest distance to C will be more than 8 km. z At any other position on the road joining A to C, the shortest distance to either B or C will be more than 7.5 km. Hence the fire station should be on the road from B to C, 0.5 km from B. ANS:7.5 km 20. The diagonal of this 5×3 rectangle passes through 7 squares.
The diagonal of this 6×4 rectangle passes through 8 squares.
What is the number of squares passed through by the diagonal of a 360×2009 rectangle? 【Solution】
Observe that in passing from one shaded square to the next, the diagonal passes from one vertical line to the next or from one horizontal line to the next, but not both unless it passes through a ‘corner.’ Since the greatest common factor of 360 and 2009 is 1, the diagonal doesn’t pass corners. Thus the diagonal passes through the top left square and then 360－1=359 squares (one for each of the remaining rows) and 2009－1=2008 squares (one for each of the remaining columns). This gives a total of 1+359+2008=2368 squares. ANS:2368
INTERNATIONAL MATHEMATICS AND SCIENCE OLYMPIAD FOR PRIMARY SCHOOLS (IMSO) 2009 Mathematics Contest (Second Round) in Taiwan, Essay Problems Name:
School:
Grade:
ID number:
Answer the following 10 questions, and show your detailed solution in the space provided after each question. Each question is worth 4 points. Time limit: 60 minutes. 1. There are two clocks. One of them gains 6 seconds in every hour, while the other loses 9 seconds in every hour. If they are both set to show the same time, and then set going, how long will it be before the time displayed on them is exactly 1 hour apart? 【Solution】 In 1 hour, the difference in time between them is 6+9=15 seconds. Since 1 hour is equal to 3600 seconds, they need 3600÷15=240 hours. ANS:240 hours 2. Replace the asterisks in 86**** with the digits 1, 2, 3 and 4. Using each of them once so that the six-digit number obtained is the largest possible number divisible by 132. 【Solution】 Assume the six-digit number is 86abcd . Since 132=3×4×11, (8+a+c)－(6+b+d)=2+(a+c)－(b+d) must be a multiple of 11. Since the four digits are 1, 2, 3, and 4, 2+(a+c)－(b+d)=0, i.e. a and c are 1 and 3 and b and d are 2 and 4. Since the six-digit number is divisible by 4, the last two digits can only be 12 or 32, hence the largest number is 863412. ANS:863412 3. There are two isosceles triangles. They are equal in area. In both triangles all edges measure an exact number of cm, and the two edges of equal length are 13 cm. In one of them the third edge measures 10 cm. What is the length of the third edge of the other? 【Solution】 13 h a
a
Let the above triangle be the other triangle. From the conditions, we know that the 1 1 1 area is × 10 × 132 − ( × 10) 2 = × 10 × 12 =60 cm2. And since 132 = a 2 + h 2 , we 2 2 2 know 13>h and 13>a. Thus (h, a)=(5, 12) or (12, 5) because 60=ha. If (h, a)=(12, 5),
then the triangle is the known triangle. So (h, a)=(5, 12) and the answer is 12×2=24 cm. ANS:24 cm. 4. In a quadrilateral ABCD, BC is parallel to AD. E is the foot of the perpendicular from B to AD. Find BE if AB=17, BC=16, CD=25 and AD=44. B
16
C 25
17 A
E
44
F
D
【Solution】 Let point F lie on AD such that BF//CD, thus quadrilateral FBCD is a parallelogram and hence BF=25 and AF=44－16=28. Since △ABE and △FBE are right triangles,
BE = AB 2 − AE 2 = BF 2 − EF 2 , i.e. 17 2 − AE 2 = 252 − (28 − AE ) 2 . Solve the equation to get AE=8. So BE= 17 2 − 82 =15. ANS:15 5. Three different numbers from 1 to 10 were written on three cards. The cards were shuffled and dealt to three players. Each player got one card and wrote down the number of his card. Then the cards were collected and dealt again. After several deals the three players reported the totals of their written numbers, which were 13, 15, and 23. What numbers were written down on the cards at the beginning? 【Solution】 Assume the numbers written on the cards are x, y and z, where x17= x+y+z, C got one x and two z’s or one y and two z’s. When C got one x and two z’s, then there are two cases: If B got one x and two y’s, i.e. B got in total x+2y, then A got one x, one y, one (i) z and hence A got in total x+y+z=17, this is impossible. (ii) If B got two x’s and one y, i.e. B got in total 2x+y, then A got two y’s, one z and hence A got in total 13=2y+z＞2x+y=15, this is impossible. When C got one y and two z’s, then there are two cases: If B got two x’s and one y, i.e. B got in total 2x+y, then A got one x, one y, one (i) z and hence A got in total x+y+z=17, this is impossible. If B got one x and two y’s, i.e. B got in total x+2y, then A got two x’s, one z and hence A got in total 2x+y. Thus we have x=3, y=5 and z=9, i.e. the numbers written on cards are 3, 5 and 9. ANS: 3, 5 and 9 6. Five students A, B, C, D, and E competed in solving a math problem. The complete solution to the problem was awarded 10 points and a partial solution – an integer between 2 and 9. Each student scored some number of points so that : A, B, and C were awarded 15 points together; and B, C and D were awarded 12
points together. All students got different scores. The student A had the highest score and student E who scored 6 points, was placed third. What was the score of student D? 【Solution】 Let a, b, c, d and e be the scores of A, B, C, D and E, respectively. Thus we know 2 ≤ a, b, c, d , e ≤ 10 , a+b+c=15 and b+c+d=12. If a>b>e=6, then a and b are two of {7, 8, 9, 10} and hence a+b ≥ 15 , which contradicts with a+b+c=15. So e>b. If a>c>e=6, then a and c are two of {7, 8, 9, 10} and hence a+c ≥ 15 , which contradicts with a+b+c=15. So e>c. Hence we know a>d>e=6. Because a+b+c=15 and b+c+d=12, a－d=3 and the only possible solution is a=10, d=7. ANS:7 7. A grandmother has two grandsons. Her age is a two-digit number. The first digit is equal to the age of the first grandson, and the second digit is equal to the age of the second grandson. If the sum of their ages is 69, how old is the grandmother? 【Solution】 Assume the age of the grandmother is ab , where a and b are digits, then we get 10a+b+a+b=69, i.e. 11a+2b=69. Thus we know that 0 < a ≤ 6 and a must be an odd number, hence a=1, 3 or 5. If a=1, then b=29. If a=3, then b=18. If a=5, then b=7. Only (a, b)=(5, 7) satisfies the conditions. ANS:57 8. A 'Lucky number' has been defined as a number which can be divided exactly by the sum of its digits. For example: 1729 is a Lucky number since 1 + 7 + 2 + 9 = 19 and 1729 can be divided exactly by 19. Find the smallest Lucky number which is divisible by 13 【Solution】 The multiples of 13 are 13, 26, 39, 52, 65, 78, 91, 104, 117, …. So the smallest Lucky number which is divisible by 13 is 117. ANS:117 9. In the middle of a large field there is a wooden hut on a rectangular base measuring 10 m by 6 m. Outside the hut, and tethered by a chain to one corner is a goat. Over what area can the goat graze if the tether is 15 m long? (Usingπ＝ 3.14) 【Solution】 In the figure on the right, we need to find the area of the red zone. The area is 15 3 1 1 × 3.14 × 152 + × 3.14 × (15 − 6) 2 + × 3.14 × (15 − 10) 2 6 10 4 4 4 =613.085. ANS:613.085
10. A chess-board is made up of 64 black and white squares in the normal way, each having an edge length of 10 cm. On this board the largest possible circle is drawn so that it’s circumference does not pass through a black square. What is the radius of the circle? 【Solution】 Let O be the center of a circle and r be the radius. We observe that if the circumference of the circle does not pass through a black square, then the circumference can only intersect the black squares at r the grid points or touch an edge of a black square. If O is on a grid point of the chessboard or on (i) an edge of a square, by symmetry, the circumference must pass through a black square (ii) If O is inside a square, then O should be at the center of the square. Now we’ll consider the distances between O and the grid points. By symmetry, we just consider the upper right vertex of a black square. (1) If O is inside a black square, then there are at least two kinds of distances between the grid points and O, one is of the form (5 + 10k )2 + (5 + 10k )2 which is the distance between O and the grid points on the extension of the diagonal (upper-right to bottom-left) of the black square and the other one is
(2) (3)
of the form 52 + (5 + 20k ) 2 which is the distance between O and the grid points on the vertical line which is next to O. They will be the same as k=0 and hence r= 5 2 cm and we can plot the green circle in the figure. If O is inside a white square and the circumference of the circle does not pass through the grid points, the r=5 and we can plot the blue circle in the figure. If O is inside a white square and the circumference of the circle passes some grid points, then there are at least two kinds of distances between the grid points and O, one is of the form (5 + 10k ) 2 + (5 + 10(k + 1))2 which is the distance between O and the grid points on the extend of the diagonal (upper-right to bottom-left) of the black square which is on the right and the other one is of the form 52 + (15 + 20k )2 which is the distance between O and the grid points on the vertical line which is next to O. They will be the same as k=0 and hence r= 5 10 cm and we can plot the red circle in the figure. So the largest possible radius is 5 10 cm ANS: 5 10 cm
INTERNATIONAL MATHEMATICS AND SCIENCE OLYMPIAD FOR PRIMARY SCHOOLS (IMSO) 2009 Mathematics Contest in Taiwan, Exploration Problems Name:
School:
Grade:
ID number:
Answer the following 5 questions. Write down your answer in the space provided after each question. Each question is worth 8 points. Time limit: 60 minutes. 1. At a recent athletics meeting, five old acquaintances: Fred, Greta, Hans, Iolo and Jan met together for the first time since leaving college, so they had a lot of news to catch up on. It seemed they all lived in different towns: Acton, Buswick, Coalford, Derby and Eccles; and that they all had different jobs which were, in no particular order: an engineer, a lawyer, a teacher, a doctor and a shopkeeper. Also, each one was the winner in just one event at the meeting. These events were: 100 m, 400 m, 1500 m, High Jump and Javelin. The following facts were also known: (1) Hans the shopkeeper from Derby won the High Jump. (2) The lawyer was from Eccles and said he was not a runner. (3) Greta was a P.E. teacher from Buswick and won the 1500 m. (4) The doctor, who came from Acton, did not win the 100 m. (5) The person from Derby was not an engineer. (6) Iolo was an engineer from Coalford and did not win the 400 m. (7) Jan was not a lawyer, but did win the 400 m. (8) Fred did not come from Acton and was not a runner. (a) Which event did the person from Coalford win? (1 point) (b) Which town did Jan come from? (1 point) (c) What was the name of the lawyer? (2 point) (d) Which event did the engineer win? (2 point) (e) Which event did Fred win? (2 point) 【Solution】 From (1), we know: Name Town Job Event Hans Derby Shopkeeper High Jump From (2) and (1), we know: Name Town Job Event Eccles Lawyer Javelin From (3), we know: Name Town Job Event Greta Buswick Teacher 1500 m From (4) and (3), we know: Name Town Job Event Acton Doctor 400 m
(5) can be known from (1). From (6) and (3), (4), (7), we know: Name Town Job Event Iolo Coalford Engineer 100 m From (7), we know: Name Town Job Event Jan Acton Doctor 400 m From (8), we know: Name Town Job Event Fred Eccles Lawyer Javelin (a) The person from Coalford won the 100 m race. (b) Jan came from Acton. (c) The name of the lawyer is Fred. (d) The engineer won the 100 m race. (e) Fred won the Javelin contest ANS:(a)100 m (b)Acton (c)Fred (d)100 m (e)Javelin 2. In 2008, a school had 450 new pupils, making a total of 1600 pupils. In 2009, there were 504 new pupils, 8% of the number of the previous year’s pupils left the school. (1) Find the percentage increase in the number of new pupils from 2008 to 2009. (2 point) (2) How many pupils left the school? (3 point) (3) What was the total number of pupils in the school in 2009? (3 point) 【Solution】 (a) In 2008, the school had 450 new pupils. In 2009, the school had 504 new pupils. So the percentage increase in the number of new pupils from 2008 to 2009 is (504－450)÷450×100%=12%. (b) Since we know 8% of the number of 2008’s pupils left the school, there are 1600×8%=128 pupils left the school. (c) Since there were 504 new pupils and 128 pupils left the school, the total number of pupils in 2009 is 1600－128+504=1976. ANS:(a) 12% (b) 128 pupils (c) 1976 3. The diagram represents a small sheet of 12 postage stamps, as they are usually sold, all perforated at the edges and all of the same value. (The letters are only there to identify the separate stamps). You need 4 of the stamps in order to post a letter but would like all 4 to be properly joined together at their edges (not at their corners). For example: ABCD, EFGH, JKLM, FGHL would all do, but NOT EFLM. In how many different ways can you get such a group of 4? Write down this number. A B C D E
F G H
J
K L M
【Solution】 It is necessary to find the number of tetrominoes in the figure. There are five kinds of tetromino: ABCD, EFGH, JKLM (i) I There are in total 3 I-tetrominoes. ABCG, EABC, BCDH, FBCD, EFGC, EFGL, AEFG, JEFG, FGHD, FGHM, BFGH, KFGH, JKLG, EJKL, KLMH, FKLM, AEJK, BAEJ, CBFK, ABFK, BFKL, BFKJ, BCGL, DCGL, CGLM, CGLK, DHML, CDHM There are in total 28 L-tetrominoes. ABFG, BCGH, EFKL, FGLM, EFBC, FGCD, JKFG, KLGH, AEFK, BFGL, CGHM, BFEJ, CGFK, DHGL There are in total 14 N-tetrominoes.
(ii) L
(iii) N
(iv) O
ABFE, BCGF, CDHG, EFKJ, FGLK, GHLM There are in total 6 O-tetrominoes.
(v) T
ABCF, BCDG, EFGK, FGHL, EFGB, FGHC, JKLF, KLMG, AEJF, BFKG, CGLH, BFKE, CGLF, DHMG There are in total 14 T-tetrominoes.
Hence there are 3+28+14+6+14=65 tetrominoes. ANS:65 4. A domino consists of two unit squares joined edge to edge, each with a number on it. Fifteen dominoes, numbered 11, 12, 13, 14, 15, 22, 23, 24, 25, 33, 34, 35, 44, 45, and 55, are assembled into the 6 by 5 rectangle shown in the diagram below. However, the boundary of the individual dominoes has been erased. Reconstruct the dominoes by drawing in the boundary lines. 【Solution】 5 2
3 3
5 4
2 4
2 4
3 5
3
1
3
4
1
3
2
1
5
2
4
5
4
2
1
1
1
5
5
3
5
2
2
3
2
3
4
4
4
5
3
1
3
4
1
3
2
1
5
2
4
5
4
2
1
1
1
5
5. We can make shapes by joining diamonds together edge to edge. There are exactly two different shapes that can be made in this way from two diamonds. We call them bimonds. Bimond
Diamon
120˚ 60˚ Sides equal in length, one angle 60˚, adjacent angle 120˚; area 1 unit2.
The two different bimonds, each with area 2 units2.
Each of the following shapes is a trimond made from three diamonds joined edge to edge; each has area 3 units2.
Note: The three trimonds above are all the same — rotations or reflections of one will produce the others. (1) From a set of one diamond, two different bimonds and your collection of different trimonds, fit some of these together to make a parallelogram with area 12 units2. Draw this parallelogram, showing the pieces you have used. (2 point) (2) From the same set of pieces, make another parallelogram with area 12 units2, but with a different perimeter. Draw this parallelogram. (2 point) (3) How many different shapes can be made from four diamonds joined edge to edge each having an area of 4 units2. (4 point) 【Solution】 (1)
(2)
In (1) and (2), other arrangements exit. (3) There are 9 different trimonds altogether.
To find all different shapes which can be made from four diamonds joined edge to edge, each having an area of 4 units2, we can add one diamond to each trimond. (i)
(ii)
(iii)
(iv)
(v)
(vi)
(vii)
(viii)
There are in total 8+7+7+8+6+4+4+1=45 ways. ANS:45
Microsoft Word - 2009IMSO math-Shortanswer Solutions.doc - 2009-2...
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http://www.chiuchang.org.tw/download/imso/2009-2-2.pdf
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Microsoft Word - 2009IMSO math-Shortanswer Solutions.doc - 2009-2...
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http://www.chiuchang.org.tw/download/imso/2009-2-2.pdf
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