2010 Short Answer IMSO (+solution)
International Mathematics and Science Olympiad 2010
SHORT ANSWER PROBLEMS 1. E is the midpoint of the side BC of a triangle ABC. F is on AC, so that AC = 3 FC.
The ratio of the area of quadrilateral ABEF to the area of triangle ABC is . . . : . . . . Answer: 5:6 Since E midpoint BC then area AEB = area AEC = 1/2 area ABC F C = 1/3AC then area CEF = 1/3 area AEC = 1/6 area ABC Area of quadrilateral ABEF is 5/6 area ABC Ratio area ABEF to CEF would be 5 to 1
2. A rectangle is divided into four rectangles, as shown in the figure below. The areas of 2 2 three of them are given in cm . The area of the original rectangle is . . . cm .
Answer: 84 21 28 7 28 15 = , = , 7A = 140, A = 20 A A 5 The area of the original rectangle is 15 + 21 + 28 + 20 = 84
3. A shape on the right is constructed by putting two half circles with diameter 2 cm on the top and on the bottom of a square with 2 cm sides. A circle is inscribed in the 2 square as shown in the figure. The area of the shaded region is . . . cm .
Answer: 4 Area A = Area B ; Area C= Area D Area of the shaded region is equal to area of the square.
4. The least positive integer a so that 490 × a is a perfect cube number is . . . .
Answer: 700 2 In order to get a Perfect Cube number, since 490 = 2 × 5 × 7 , then it has to be multiple 2 2 × × by 2
5 7 = 700
5. A number of cubes are stacked as shown seen in the figure below. The highest level consists of one cube, the second highest level consists of 3 × 3 cubes, the third highest level consists of 5 × 5 cubes, and so on.
If there are 2010 cubes to be stacked in this way, such that all levels are complete, some cubes may not be used. There can be as many as . . . levels. Answer: 11 st 2 2
11 Number of cube needed for the 1 level (from the top) is 1 , 2nd level is 3 , 3rd 2 2 level is 5 according to order (2n − 1) , with n refer to level.
Having calculated: 2 2 2 2 2 2 2 2 2 2 2 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 = 1771 and 2 2 2 2 2 2 2 2 2 2 2 2 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23 = 2300 2 giving the number of cube needed at the lowest level equal to 21 and giving the 11 stage of level.
6. Draw a square inside a circle of radius 1 so that all four vertices lie on the circle. The 22 ratio of the area of the square to the area of the circle is . . . : . . . . [Use π = .] 7 Answer: 7:11
See the figure below Diameter of the circle is 2 unit length or equal to diagonal square. Since the square transfered to rhombus, the area of the rhombus would be :
1
1 × ×
L = diagonal 1 × diagonal 2 = 2 × 2 = 2
2
2 Area of the circle with radius 1 unit length is π, hence the area ratio between square and circle is 2 : π.
7. If we move from point A to B along the directed lines shown in the figure, then the number of different routes is . . . .
Answer: 34 see figure
2
8. The areas of three faces of a rectangular box are 35, 55 and 77 cm . If the length, width 3 and height of the box are integers, then the volume of the box is . . . cm .
Answer: 385 Assume q = length, h=height, w=width q × w = 55 q × h = 77 w × h = 35 w/h = 55/77 = 5/7 q/w = 77/35 = 11/5 3 The volume of the box is length × width × height = 11 × 5 × 7 = 385 cm
9. In a certain year, the month of January has exactly 4 Mondays and 4 Thursdays. The day on January 1st of that year is . . . .
Answer: Friday 10. In an isosceles triangle, the measure of one of its angles is four times the other angle.
The measure of the largest possible angle of the triangle, in degree, is . . . . o Answer: 120 For ease those two angles are x and 4x. o
If the third angle is x, the sum of the internal angle is x + x + 4x = 6x or x = 30 and o 4x = 120 . o If the third angle is 4x, the sum of the internal angle is x + 4x + 4x = 9x or x = 20 o and 4x = 80 . o The largest possible angle of the triangle is 120
11. The average of 2010 consecutive integers is 1123.5. The smallest of the 2010 integers is . . . .
Answer: 119 Number of integers which are less than 1123.5 are 2010 : 2 = 1005 numbers. Since those numbers are consecutive, then the smallest integer is 1123 − (1005 − 1) = 119.
12. If the length of each side of a cube is decreased by 10%, then the volume of the cube is decreased by . . .%.
Answer: 27.1 % Suppose the side of original square is 10 cm, hence the original volume would be 103 = 3 1000 cm . 3 Shorten the side to 10%, hence the volume become 93 = 729 cm
The decrease of the volume is (1000 − 729)/1000 × 100% = 271/1000 × 100% = 27.1% 13. The operations ◦ and are defined by the following tables.
For example, 2 ◦ 3 = 2 and 2 3 = 3 . The value of (1 2) ◦ 3 is . . . . Answer: (1 2) ◦ 3 = 3 ◦ 3 = 3
14. The number of zeros in the end digits in the product of 1×5×10×15×20×25×30×35×40×45×50×55×60×65×70×75×80×85×90×95 is . . . .
Answer : 16 We have 1 × 5 × 10 × 15 × 20× 25 × 30 × 35 × 40 × 45 × 50 × 55× 60 × 65 × 70 × 75 × 80 × 85 × 90× 95 = 5 × 10 × 3 × 5 × 2 × 10 × 5 × 5 × 3 × 10 × 7 × 5 × 4 × 10 × 9 × 5 × 5 × 10 × 11 × 5 × 6 × 10 × 13 × 5 × 7 × 10 × 15 × 5 × 8 × 10 × 17 × 5 × 9 × 10 × 19 × 5 10 9
= 5 × 11 10 × 1 × 2 × 3 × 4 × 5 × 6 × 7 × 8 × 9 9 × ×
= 5 11 10 2 × 3 × 2 × 2 × 2 × 3 × 7 × 2 × 2 × 2 × 9 9 7 × × ×
= 5
10 2 3 × 5 × 3 × 7 × 9 9 From the product factors above, 10 has 9 zero digits.
Product of 2 × 2 × 5 × 5 results 2 zero digits. There are 2 pairs of 2 × 2 × 5 × 5, so that we have 4 zero digits. Product of 2 × 5 gives 1 zero digit. There are 3 pairs of 2 × 5, then we have 3 digit zeros. Then we totally have 9 + 2 × 2 + 3 = 16 zero digits.
15. In the addition sentence below each letter represents a different digit. The number of all possible digits represented by ’E’ is . . . . Answer: 4
E O E O O
Since 2 = or 2 = 10 + , must be even. Furthermore , since • 2O ≤ T ≤ 9 then O ≤ 4. Clearly, O 6= 0 .
- E = 6 .
For case O = 2, we have 2E = 2 or 2E = 12 . Thus we have E = 1 or
For case O = 4, we have 2E = 4 or 2E = 42 . Thus we have E = 2 or • E = 7 .
16. A piece of rod is 16 meters long. There is a device that can divide any piece of rod into two equal pieces. We can use this device 8 times. In the end, we will have 9 pieces of rod. The maximum possible difference between the length of the longest piece and the length of shortest piece is . . . meters. 15 Answer: 7 th 16 16 n After the n cut, the length of the shortest piece is m. 2 Hence, the maximum difference of the length of the longest piece and the length of the 1 15 shortest piece is 8 − = 7 m. 16 16
17. Rearrange the twelve numbers of the clock 1, 2, 3, . . . , 12 around its face so that any two adjacent numbers add up to a triangle number. The triangle numbers are 1, 3, 6, 10, 15, 21 and so on. If 12 is placed in its original position, then the number that should be placed in the opposite position of 12 must be . . . .
Answer: 5 2
18. In the figure, A, B and C are circles of area of 60 cm . One-half the area 1 1 of A is shaded, area of B is shaded, and area of C 3 4 2 is shaded. The total area of the shaded regions is . . . cm .
65
Answer: 1 2 = 32.5
I + II = × 60 = 30 2 1 I + III = × 60 = 20 3 1 II + III = × 60 = 15 4
2I + 2II + 2III = 30 + 20 + 15 = 65 65 I + II + III = = 32.5 2 19. For any positive integer n, let d(n) be the sum of digits in n.
For example, d(123) = 1 + 2 + 3 = 6 and d(7879) = 7 + 8 + 7 + 9 = 31. The value of d (d (999 888 777 666 555 444 333 222 111)) is . . . .
Answer: 9 Let M = 999 888 777 666 555 444 333 222 111 Then d(M ) = 3 × (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9) = 3 × 45 = 135 Then, d(d(M )) = 1 + 3 + 5 = 9.
20. The numbers 1447, 1005 and 1231 have something in common. Each is a four-digit number beginning with 1 that has exactly two identical digits. There are . . . such num- bers. Answer: 432 We cluster the set of such numbers into two subsets : ′
- Those numbers having two identical digits other than 1. Let 1xyz be the number. The number of such numbers in subset (1): The second digit 1 may be placed in x or y or z. (There are three possibilities). The others two digits must be distinct, so that there are 9 × 8 possibilities. Thus, there are 3 × 9 × 8 = 216 such numbers. The number of such numbers in subset (2):
Those numbers having two 1 s.
There are nine possibilities for the two identical digits, and there are three possibil- ities for its position (xy, xz, and yz). There are eight possibilities for the remaining digit Thus there are 9 × 3 × 8 = 216 such numbers. Thus, the answer is 216 + 216 = 432.
21. Let n be a positive integer greater than 1. By the length of n, we mean the number of factors in the representation of n as a product of prime numbers. For example, the ’length’ of the number 90 is 4, since 90 = 2 × 3 × 3 × 5. The number of odd numbers between 2 and 100 having ’length’ 3 is . . . .
Answer: 5 One of the prime factor is 3, because 5 × 5 × 5 > 100. This allows only five numbers: 3 × 3 × 3 = 27, 3 × 3 × 5 = 45, 3 × 3 × 7 = 63, 3 × 5 × 5 = 75 and 3 × 3 × 11 = 99.
22. A rectangle intersects a circle as shown: AB = 8cm, BC = 9cm and DE = 6cm. The length of EF is . . .cm.
Answer: 13 Draw in the perpendicular bisector of chord EF , as shown in the figure. This bisector must go through the center O of the circle. Since P Q is perpendicular to chord BC as well, it bisects BC. Since P Q//AD, we have 9 25 DQ = AP = 8 + = . 2 2 Thus EF = 2EQ = 2(DQ − DE) = 13.
Alternate Solution Drop perpendiculars BR and CS as shown in the second figure, yielding rectangles BCSR and ABRD.
Then ER = 2, and by symmetry, SF = 2. Also, RS = BC = 9. So EF = 2 + 9 + 2 = 13.
23. The numbers on each pair of opposite faces on a die add up to 7. A die is rolled without slipping around the circuit shown. At the start the top face is 3.
At the end point, the number displayed on the top face is . . . .
Answer: 6 As the diagram in the question shows, after one rotation a 6 is on the top face. After a second rotation a 4 is on the top face. The next rotation brings the face that was on the back of the die to the top; since 2 was on the front this is 5. A further rotation brings 3 to the top. With another rotation along the path, 1 is displayed, followed by 4, and finally 6.
24. Let A, B and C be three distinct prime numbers. If A × B × C is even and A × B × C > 100, then the smallest possible value of A + B + C is . . . .
Answer: 18 Since the product of the three numbers is even, then one of them is 2. Therefore, the sum of the remaining two is as small as possible, whereas the product is at least 51. Let the two prime numbers be A and B. If A = 3 then the possible value for B is 17. Thus, A + B = 20. If A = 5 then the possible value for B is 11. Thus, A + B = 16. If A = 7 then the possible value for B is 11. Thus, A + B = 18. If A = 11 then the possible value for B is 5. Thus, A + B = 16. If A = 13 then the possible value for B is 5. Thus, A + B = 18. If A > 13 then A + B > 16. Thus the smallest possible value of A + B is 16, and the smallest possible value of A + B + C is 18.
25. The product of a × b × c × d = 2010, where a, b, c and d are positive integers and a < b < c < d. There are . . . different solutions for a, b, c and d.
Answer: 7 We know that 2010 = 1 × 2 × 3 × 5 × 67, then we have list of all possible numbers.