AMC 美国数学竞赛 2004 AMC 10B 试题及答案解析.doc

2004 AMC 10BProblem 1 Each row of the Misty Moon Amphitheater has 33 seats. Rows 12 through 22 are reserved for a youth club. How many seats are reserved for this club? Solution There are rows of seats, giving seats.Problem 2 How many two-digit positive integers have at least one 7 as a digit? Solution Ten numbers () have as the tens digit. Nine numbers () have it as the ones digit. Number is in both sets. Thus the result is . Problem 3 At each basketball practice last week, Jenny made twice as many free throws as she made at the previous practice. At her fifth practice she made 48 free throws. How many free throws did she make at the first practice? Solution At the fourth practice she made throws, at the third one it was , then we get throws for the second practice, and finally throws at the first one.Problem 4 A standard six-sided die is rolled, and P is the product of the five numbers that are visible. What is the largest number that is certain to divide P? Solution 1The product of all six numbers is . The products of numbers that can be visible are , , ., . The answer to this problem is their greatest common divisor - which is , where is the least common multiple of . Clearly and the answer is . Solution 2Clearly, can not have a prime factor other than , and . We can not guarantee that the product will be divisible by , as the number can end on the bottom. We can guarantee that the product will be divisible by (one of and will always be visible), but not by . Finally, there are three even numbers, hence two of them are always visible and thus the product is divisible by . This is the most we can guarantee, as when the is on the bottom side, the two visible even numbers are and , and their product is not divisible by . Hence . Solution Problem 5 In the expression , the values of , , , and are , , , and , although not necessarily in that order. What is the maximum possible value of the result? Solution If or , the expression evaluates to . If , the expression evaluates to . Case remains. In that case, we want to maximize where . Trying out the six possibilities we get that the best one is , where . Problem 6 Which of the following numbers is a perfect square? Solution Using the fact that , we can write: Clearly is a square, and as , , and are primes, none of the other four are squares. Problem 7 On a trip from the United States to Canada, Isabella took U.S. dollars. At the border she exchanged them all, receiving Canadian dollars for every U.S. dollars. After spending Canadian dollars, she had Canadian dollars left. What is the sum of the digits of ? Solution Solution 1 Isabella had Canadian dollars. Setting up an equation we get , which solves to , and the sum of digits of is Solution 2 Each time Isabelle exchanges U.S. dollars, she gets Canadian dollars and Canadian dollars extra. Isabelle received a total of Canadian dollars extra, therefore she exchanged U.S. dollars times. Thus . Problem 8 Minneapolis-St. Paul International Airport is 8 miles southwest of downtown St. Paul and 10 miles southeast of downtown Minneapolis. Which of the following is closest to the number of miles between downtown St. Paul and downtown Minneapolis? Solution The directions southwest and southeast are orthogonal. Thus the described situation is a right triangle with legs 8 miles and 10 miles long. The hypotenuse length is , and thus the answer is . Without a calculator one can note that . Problem 9 A square has sides of length 10, and a circle centered at one of its vertices has radius 10. What is the area of the union of the regions enclosed by the square and the circle? Solution The area of the circle is , the area of the square is . Exactly of the circle lies inside the square. Thus the total area is . Problem 10 A grocer makes a display of cans in which the top row has one can and each lower row has two more cans than the row above it. If the display contains cans, how many rows does it contain? Solution The sum of the first odd numbers is . As in our case , we have .Problem 11 Two eight-sided dice each have faces numbered 1 through 8. When the dice are rolled, each face has an equal probability of appearing on the top. What is the probability that the product of the two top numbers is greater than their sum? Solution Solution 1We have , hence if at least one of the numbers is , the sum is larger. There such possibilities. We have . For we already have , hence all other cases are good. Out of the possible cases, we found that in the sum is greater than or equal to the product, hence in it is smaller. Therefore the answer is . Solution 2Let the two rolls be , and . From the restriction: Since and are non-negative integers between and , either , , or if and only if or . There are ordered pairs with , ordered pairs with , and ordered pair with and . So, there are ordered pairs such that . if and only if and or equivalently and . This gives ordered pair . So, there are a total of ordered pairs with . Since there are a total of ordered pairs , there are ordered pairs with . Thus, the desired probability is . Problem 12 An annulus is the region between two concentric circles. The concentric circles in the gure have radii and , with . Let be a radius of the larger circle, let be tangent to the smaller circle at , and let be the radius of the larger circle that contains . Let , , and . What is the area of the annulus? Solution The area of the large circle is , the area of the small one is , hence the shaded area is . From the Pythagorean Theorem for the right triangle we have , hence and thus the shaded area is . Problem 13 In the United States, coins have the following thicknesses: penny, mm; nickel, mm; dime, mm; quarter, mm. If a stack of these coins is exactly mm high, how many coins are in the stack? Solution All numbers in this solution will be in hundreds of a millimeter. The thinnest coin is the dime, with thickness . A stack of dimes has height . The other three coin types have thicknesses , , and . By replacing some of the dimes in our stack by other, thicker coins, we can clearly create exactly all heights in the set . If we take an odd , then all the possible heights will be odd, and thus none of them will be . Hence is even. If the stack will be too low and if it will be too high. Thus we are left with cases and . If the possible stack heights are , with the remaining ones exceeding . Therefore there are coins in the stack. Using the above observation we can easily construct such a stack. A stack of dimes would have height , thus we need to add . This can be done for example by replacing five dimes by nickels (for ), and one dime by a penny (for ). Problem 14 A bag initially contains red marbles and blue marbles only, with more blue than red. Red marbles are added to the bag until only of the marbles in the bag are blue. Then yellow marbles are added to the bag until only of the marbles in the bag are blue. Finally, the number of blue marbles in the bag is doubled. What fraction of the marbles now in the bag are blue? Solution We can ignore most of the problem statement. The only important information is that immediately before the last step blue marbles formed of the marbles in the bag. This means that there were blue and other marbles, for some . When we double the number of blue marbles, there will be blue and other marbles, hence blue marbles now form of all marbles in the bag.Problem 15 Patty has coins consisting of nickels and dimes. If her nickels were dimes and her dimes were nickels, she would have cents more. How much are her coins worth? Solution Solution 1 She has nickels and dimes. Their total cost is cents. If the dimes were nickels and vice versa, she would have cents. This value should be cents more than the previous one. We get , which solves to . Her coins are worth . Solution 2 Changing a nickel into a dime increases the sum by cents, and changing a dime into a nickel decreases it by the same amount. As the sum increased by cents, there are more nickels than dimes. As the total count is , this means that there are nickels and dimes. Problem 16 Three circles of radius are externally tangent to each other and internally tangent to a larger circle. What is the radius of the large circle? Solution The situation in shown in the picture below. The radius we seek is . Clearly . The point is clearly the center of the equilateral triangle , thus is of the altitude of this triangle. We get that . Therefore the radius we seek is . WARNING. Note that the answer does not correspond to any of the five options. Most probably there is a typo in option D. Problem 17 The two digits in Jacks age are the same as the digits in Bills age, but in reverse order. In five years Jack will be twice as old as Bill will be then. What is the difference in their current ages? Solution Solution 1 If Jacks current age is , then Bills current age is . In five years, Jacks age will be and Bills age will be . We are given that . Thus . For we get . For and the value is not an integer, and for it is more than . Thus the only solution is , and the difference in ages is . Solution 2 Age difference does not change in time. Thus in five years Bills age will be equal to their age difference. The age difference is , hence it is a multiple of . Thus Bills current age modulo must be . Thus Bills age is in the set . As Jack is older, we only need to consider the cases where the tens digit of Bills age is smaller than the ones digit. This leaves us with the options . Checking each of them, we see that only works, and gives the solution . Problem 18 In the right triangle , we have , , and . Points , , and are located on , , and , respectively, so that , , and . What is the ratio of the area of to that of ? Solution First of all, note that , and therefore . Draw the height from onto as in the picture below: Now consider the area of . Clearly the triangles and are similar, as they have all angles equal. Their ratio is , hence . Now the area of can be computed as = . Similarly we can find that as well. Hence , and the answer is . Problem 19 In the sequence , , , , each term after the third is found by subtracting the previous term from the sum of the two terms that precede that term. For example, the fourth term is . What is the term in this sequence? Solution Solution 1 We already know that , , , and . Lets compute the next few terms to get the idea how the sequence behaves. We get , , , and so on. We can now discover the following pattern: and . This is easily proved by induction. It follows that . Solution 2 Note that the recurrence can be rewritten as . Hence we get that and also From the values given in the problem statement we see that . From we get that . From we get that . Following this pattern, we get . Problem 20 In points and lie on and , respectively. If and intersect at so that and , what is ? Solution Solution (Triangle Areas) We use the square bracket notation to denote area. Without loss of generality, we can assume . Then , and . We have , so we need to find the area of quadrilateral . Draw the line segment to form the two triangles and . Let , and . By considering triangles and , we obtain , and by considering triangles and , we obtain . Solving, we get , , so the area of quadrilateral is . Therefore Solution (Mass points) The presence of only ratios in the problem essentially cries out for mass points. As per the problem, we assign a mass of to point , and a mass of to . Then, to balance and on , has a mass of . Now, were we to assign a mass of to and a mass of to , wed have . Scaling this down by (to get , which puts and in terms of the masses of and ), we assign a mass of to and a mass of to . Now, to balance and on , we must give a mass of . Finally, the ratio of to is given by the ratio of the mass of to the mass of , which is . Solution (Coordinates) Affine transformations preserve ratios of distances, and for any pair of triangles there is an affine transformation that maps the first one onto the second one. This is why the answer is the same for any , and we just need to compute it for any single triangle. We can choose the points , , and . This way we will have , and . The situation is shown in the picture below: The point is the intersection of the lines and . The points on the first line have the form , the points on the second line have the form . Solving for we get , hence . The ratio can now be computed simply by observing the coordinates of , , and : Problem 21 Let ; ; and ; ; be two arithmetic progressions. The set is the union of the first terms of each sequence. How many distinct numbers are in ? Solution The two sets of terms are and . Now . We can compute . We will now find . Consider the numbers in . We want to find out how many of them lie in . In other words, we need to find out the number of valid values of for which . The fact can be rewritten as , and . The first condition gives , the second one gives . Thus the good values of are , and their count is . Therefore , and thus . Problem 22 A triangle with sides of 5, 12, and 13 has both an inscribed and a circumscribed circle. What is the distance between the centers of those circles? Solution This is obviously a right triangle. Pick a coordinate system so that the right angle is at and the other two vertices are at and . As this is a right triangle, the center of the circumcircle is in the middle of the hypotenuse, at . The radius of the inscribed circle can be computed using the well-known identity , where is the area of the triangle and its perimeter. In our case, and , thus . As the inscribed circle touches both legs, its center must be at . The distance of these two points is then . Problem 23 Each face of a cube is painted either red or blue, each with probability 1/2. The color of each face is determined independently. What is the probability that the painted cube can be placed on a horizontal surface so that the four vertical faces are all the same color? Solution Label the six sides of the cube by numbers to as on a classic dice. Then the four vertical faces can be: , , or . Let be the set of colorings where are all of the same color, similarly let and be the sets of good colorings for the other two sets of faces. There are possible colorings, and there are good colorings. Thus the result is . We need to compute . Using the Principle of