美国数学竞赛答案
2019年美国数学竞赛(AMC10A)的试题与解答
(
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0(19 ) 译文: 2
+ ((20)1)9 的值是多少?
解 原式 = 20 + 19 = 2, 故 (C) 正确.
2. What is the hundreds digit of (20! − 15!)?
(A) 0 (B) 1 (C) 2 (D) 4 (E) 5
译文: (20! − 15!) 的百位数是多[ 少]?
years, n years apart. year Ana was 5 times as old as Bonita.
This year Ana’s age is the square of Bonita’s age. What is n?
(A) 3 (B) 5 (C) 9 (D) 12 (E) 15
equidistant from all four vertices of the quadrilateral?
— a square
— a rectangle that is not a square
— a rhombus that is not a square
— a parallelogram that is not a rectangle or a rhombus
13 个蓝球, 11 个白球和 9 个黑球. 在不放回的情况下, 必须
至少从盒子中取出多少个球才能确保会取出 15 个同种颜色
的球?
解 依题意, 红球、绿球、黄球得取出 14 个, 蓝球、白球、
黑球全部取出, 再加任意 1 个就符合条件, 即至少需要取出
14 × 3 + 13 + 11 + 9 + 1 = 76 个球, 故 (B) 正确.
AMC 美国数学竞赛 2004 AMC 10B 试题及答案解析
2004 AMC 10BProblem 1Each 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?SolutionThere are rows of seats, giving seats.Problem 2How many two-digit positive integers have at least one 7 as a digit?SolutionTen numbers () have as the tens digit. Nine numbers () have it as the ones digit. Number is in both sets. Thus the result is .Problem 3At 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?SolutionAt 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 4A 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 .SolutionProblem 5In the expression , the values of , , , and are , , , and , although not necessarily in that order. What is the maximum possible value of the result?SolutionIf 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 6Which of the following numbers is a perfect square?SolutionUsing the fact that , we can write:▪▪▪▪▪Clearly is a square, and as , , and are primes, none of the other four are squares.Problem 7On 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 ?SolutionSolution 1Isabella had Canadian dollars. Setting up an equation we get , which solves to , and the sum of digits of isSolution 2Each 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 8Minneapolis-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?SolutionThe 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 answeris .Without a calculator one can note that . Problem 9A 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?SolutionThe 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 10A 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?SolutionThe sum of the first odd numbers is . As in our case , we have .Problem 11Two 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?SolutionSolution 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 , , orif and only if or .There are ordered pairs with , ordered pairs with , and ordered pair with and . So, there areordered 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 12An annulus is the region between two concentric circles. The concentric circles in the figure 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?SolutionThe 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 13In 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?SolutionAll 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 14A bag initially contains red marbles and blue marbles only, with moreblue 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 baguntil 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?SolutionWe 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 ofblue marbles, there will be blue and other marbles, hence blue marbles now form of all marbles in the bag.Problem 15Patty 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?SolutionSolution 1She has nickels and dimes. Their total cost iscents. If the dimes were nickels and vice versa, she would havecents. This value should be cents more than the previous one. We get , which solves to . Her coins are worth .Solution 2Changing 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 16Three circles of radius are externally tangent to each other and internally tangent to a larger circle. What is the radius of the large circle?SolutionThe 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 17The two digits in Jack's age are the same as the digits in Bill's 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?SolutionSolution 1If Jack's current age is , then Bill's current age is .In five years, Jack's age will be and Bill's 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 2Age difference does not change in time. Thus in five years Bill's age will be equal to their age difference.The age difference is , hence it is a multiple of . Thus Bill's current age modulo must be .Thus Bill's age is in the set .As Jack is older, we only need to consider the cases where the tens digit of Bill's 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 18In 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 ?SolutionFirst 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 19In 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?SolutionSolution 1We already know that , , , and . Let's 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 2Note 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 20In points and lie on and , respectively. If and intersect at so that and , what isSolutionSolution (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 trianglesand , we obtain , and by considering triangles and , we obtain . Solving, we get , , so the area of quadrilateral is .ThereforeSolution (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 , we'd 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 tothe 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 21Let ; ; and ; ; be two arithmetic progressions. The set is the union of the first terms of each sequence. How many distinct numbers are in ?SolutionThe 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 22A 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?SolutionThis 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 23Each 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?SolutionLabel 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 goodcolorings. Thus the result is . We need to compute .Using the Principle of Inclusion-Exclusion we can writeClearly , as we have two possibilities for the common color of the four vertical faces, and two possibilities for each of the horizontal faces.What is ? The faces must have the same color, and at the same time faces must have the same color. It turns out thatthe set containing just the two cubes where all six faces have the same color.Therefore , and the result is .Problem 24In we have , , and . Point is on the circumscribed circle of the triangle so that bisects . What is the value of ?SolutionProblem 25A circle of radius is internally tangent to two circles of radius at points and , where is a diameter of the smaller circle. What is the area of the region, shaded in the picture, that is outside the smaller circle and inside each of the two larger circles?SolutionThe area of the small circle is . We can add it to the shaded region, compute the area of the new region, and then subtract the area of the small circle from the result.Let and be the intersections of the two large circles. Connect them to and to get the picture below:Now obviously the triangles and are equilateral with side .Take a look at the bottom circle. The angle is , hence the sector is of the circle. The same is true for the sector of the bottom circle, and sectors and of the top circle.If we now sum the areas of these four sectors, we will almost get the area of the new shaded region - except that each of the two equilateral triangles will be counted twice.Hence the area of the new shaded region is, and the area of the original shared region is .- 21 -。
AMC 美国数学竞赛 2000 AMC 10 试题及答案解析
USA AMC 10 20001In the year , the United States will host the International Mathematical Olympiad. Let , , and be distinct positive integers such that the product . What's the largest possible value of the sum ?SolutionThe sum is the highest if two factors are the lowest.So, and .2Solution.3Each day, Jenny ate of the jellybeans that were in her jar at the beginning of the day. At the end of the second day, remained. How many jellybeans were in the jar originally?Solution4Chandra pays an online service provider a fixed monthly fee plus an hourly charge for connect time. Her December bill was , but in January her bill was because she used twice as much connect time as in December. What is the fixxed monthly fee?SolutionLet be the fixed fee, and be the amount she pays for the minutes she used in the first month.We want the fixed fee, which is5Points and are the midpoints of sides and of . As moves along a line that is parallel to side , how many of the four quantities listed below change?(a) the length of the segment(b) the perimeter of(c) the area of(d) the area of trapezoidSolution(a) Clearly does not change, and , so doesn't change either.(b) Obviously, the perimeter changes.(c) The area clearly doesn't change, as both the base and its corresponding height remain the same.(d) The bases and do not change, and neither does the height, so the trapezoid remains the same.Only quantity changes, so the correct answer is .6The Fibonacci Sequence starts with two 1s and each term afterwards is the sum of its predecessors. Which one of the ten digits is the last to appear in thet units position of a number in the Fibonacci Sequence?SolutionThe pattern of the units digits areIn order of appearance:.is the last.7In rectangle , , is on , and and trisect . What is the perimeter of ?Solution.Since is trisected, .Thus,.Adding, .8At Olympic High School, of the freshmen and of the sophomores took the AMC-10. Given that the number of freshmen and sophomore contestants was the same, which of the following must be true?There are five times as many sophomores as freshmen.There are twice as many sophomores as freshmen.There are as many freshmen as sophomores.There are twice as many freshmen as sophomores.There are five times as many freshmen as sophomores.SolutionLet be the number of freshman and be the number of sophomores.There are twice as many freshmen as sophomores.9If , where , thenSolution, so ...10The sides of a triangle with positive area have lengths , , and . The sides of a second triangle with positive area have lengths , , and . What is the smallest positive number that is not a possible value of ?SolutionFrom the triangle inequality, and . The smallest positive number not possible is , which is .11Two different prime numbers between and are chosen. When their sum is subtracted from their product, which of the following numbers could be obtained?SolutionTwo prime numbers between and are both odd.Thus, we can discard the even choices.Both and are even, so one more than is a multiple of four.is the only possible choice.satisfy this, .12Figures , , , and consist of , , , and nonoverlapping unit squares, respectively. If the pattern were continued, how many nonoverlapping unit squares would there be in figure 100?SolutionSolution 1We have a recursion:.I.E. we add increasing multiples of each time we go up a figure. So, to go from Figure 0 to 100, we add.Solution 2We can divide up figure to get the sum of the sum of the firstodd numbers and the sum of the first odd numbers. If you do not see this, here is the example for :The sum of the first odd numbers is , so for figure , there are unit squares. We plug in to get , which is choice13There are 5 yellow pegs, 4 red pegs, 3 green pegs, 2 blue pegs, and 1 orange peg to be placed on a triangular peg board. In how many ways can the pegs be placed so that no (horizontal) row or (vertical) column contains two pegs of the same color?SolutionIn each column there must be one yellow peg. In particular, in the rightmost column, there is only one peg spot, therefore a yellow peg must go there.In the second column from the right, there are two spaces for pegs. One of them is in the same row as the corner peg, so there is only one remaining choice left for the yellow peg in this column.By similar logic, we can fill in the yellow pegs as shown:After this we can proceed to fill in the whole pegboard, so there is only arrangement of the pegs. The answer is14Mrs. Walter gave an exam in a mathematics class of five students. She entered the scores in random order into a spreadsheet, which recalculated the class average after each score was entered. Mrs. Walter noticed that after each score was entered, the average was always an integer. The scores (listed in ascending order) were , , , , and . What was the last score Mrs. Walter entered? SolutionThe sum of the first scores must be even, so we must choose evens or the odds to be the first two scores.Let us look at the numbers in mod .If we choose the two odds, the next number must be a multiple of , of which there is none.Similarly, if we choose or , the next number must be a multiple of , of which there is none.So we choose first.The next number must be 1 in mod 3, of which only remains.The sum of the first three scores is . This is equivalent to in mod .Thus, we need to choose one number that is in mod . is the only one that works.Thus, is the last score entered.15Two non-zero real numbers, and , satisfy . Which of the following is a possible value of ?SolutionSubstituting , we get16The diagram shows lattice points, each one unit from its nearest neighbors. Segment meets segment at . Find the length of segment .SolutionSolution 1Let be the line containing and and let be the line containing and . If we set the bottom left point at , then , , , and .The line is given by the equation . The -intercept is , so . We are given two points on , hence we cancompute the slope, to be , so is the lineSimilarly, is given by . The slope in this case is , so . Plugging in the point gives us , so is the line .At , the intersection point, both of the equations must be true, soWe have the coordinates of and , so we can use the distance formula here:which is answer choiceSolution 2Draw the perpendiculars from and to , respectively. As it turns out, . Let be the point on for which ., and , so by AA similarity,By the Pythagorean Theorem, we have ,, and . Let , so , thenThis is answer choiceAlso, you could extend CD to the end of the box and create two similar triangles. Then use ratios and find that the distance is 5/9 of the diagonal AB. Thus, the answer is B.17Boris has an incredible coin changing machine. When he puts in a quarter, it returns five nickels; when he puts in a nickel, it returns five pennies; and when he puts in a penny, it returns five quarters. Boris starts with just one penny. Which of the following amounts could Boris have after using the machine repeatedly?SolutionConsider what happens each time he puts a coin in. If he puts in a quarter, he gets five nickels back, so the amount of money he has doesn't change. Similarly, if he puts a nickel in the machine, he gets five pennies back and the money value doesn't change. However, if he puts a penny in, he gets five quarters back, increasing the amount of money he has by cents.This implies that the only possible values, in cents, he can have are the ones one more than a multiple of . Of the choices given, the only one is18Charlyn walks completely around the boundary of a square whose sides are each km long. From any point on her path she can see exactly km horizontally in all directions. What is the area of the region consisting of all points Charlyn can see during her walk, expressed in square kilometers and rounded to the nearest whole number?SolutionThe area she sees looks at follows:The part inside the walk has area . The part outside the walk consists of four rectangles, and four arcs. Each of the rectangles has area . The four arcs together form a circle with radius . Therefore the total area she can see is, which rounded to the nearest integer is .19Through a point on the hypotenuse of a right triangle, lines are drawn parallel to the legs of the triangle so that the trangle is divided into a square and two smaller right triangles. The area of one of the two small right triangles is times the area of the square. The ratio of the area of the other small right triangle to the area of the square is SolutionLet the square have area , then it follows that the altitude of one of the triangles is . The area of the other triangle is .By similar triangles, we haveThis is choice(Note that this approach is enough to get the correct answer in the contest. However, if we wanted a completely correct solution, we should also note that scaling the given triangle times changes each of the areas times, and therefore it does not influence the ratio of any two areas. This is why we can pick the side of the square.)20Let , , and be nonnegative integers such that . What is the maximum value of ? SolutionThe trick is to realize that the sum is similar to the product .If we multiply , we get.We know that , therefore.Therefore the maximum value of is equal to the maximum value of . Now we will find this maximum.Suppose that some two of , , and differ by at least . Then this triple is surely not optimal.Proof: WLOG let . We can then increase the value ofby changing and .Therefore the maximum is achieved in the cases where is a rotation of . The value of in this case is . And thus the maximum of is.21If all alligators are ferocious creatures and some creepy crawlers are alligators, which statement(s) must be true?I. All alligators are creepy crawlers.II. Some ferocious creatures are creepy crawlers.III. Some alligators are not creepy crawlers.SolutionWe interpret the problem statement as a query about three abstract concepts denoted as "alligators", "creepy crawlers" and "ferocious creatures". In answering the question, we may NOT refer to reality -- for example to the fact that alligators do exist.To make more clear that we are not using anything outside the problem statement, let's rename the three concepts as , , and .We got the following information:▪If is an , then is an .▪There is some that is a and at the same time an .We CAN NOT conclude that the first statement is true. For example, the situation "Johnny and Freddy are s, but only Johnny is a "meets both conditions, but the first statement is false.We CAN conclude that the second statement is true. We know that there is some that is a and at the same time an . Pick one such and call it Bobby. Additionally, we know that if is an , then is an. Bobby is an , therefore Bobby is an . And this is enough to prove the second statement -- Bobby is an that is also a .We CAN NOT conclude that the third statement is true. For example, consider the situation when , and are equivalent (represent the same set of objects). In such case both conditions are satisfied, but the third statement is false.Therefore the answer is .22One morning each member of Angela's family drank an -ounce mixture of coffee with milk. The amounts of coffee and milk varied from cup to cup, but were never zero. Angela drank a quarter of the total amount of milk and a sixth of the total amount of coffee. How many people are in the family?SolutionThe exact value "8 ounces" is not important. We will only use the fact that each member of the family drank the same amount.Let be the total number of ounces of milk drank by the family and the total number of ounces of coffee. Thus the whole family drank a total of ounces of fluids.Let be the number of family members. Then each family member drank ounces of fluids.We know that Angela drank ounces of fluids.As Angela is a family member, we have .Multiply both sides by to get .If , we have .If , we have .Therefore the only remaining option is .23When the mean, median, and mode of the list are arranged in increasing order, they form a non-constant arithmetic progression. What is the sum of all possible real values of ? SolutionAs occurs three times and each of the three other values just once, regardless of what we choose the mode will always be .The sum of all numbers is , therefore the mean is .The six known values, in sorted order, are . From this sequence we conclude: If , the median will be . If , the median will be . Finally, if , the median will be .We will now examine each of these three cases separately.In the case , both the median and the mode are 2, therefore we can not get any non-constant arithmetic progression.In the case we have , because. Therefore our three values inorder are . We want this to be an arithmetic progression. From the first two terms the difference must be . Therefore thethird term must be .Solving we get the only solution for this case: . The case remains. Once again, we have ,therefore the order is . The only solution is when , i. e., .The sum of all solutions is therefore .24Let be a function for which . Find the sum of all values of for which .SolutionIn the definition of , let . We get: . As we have , we must have , in other words .One can now either explicitly compute the roots, or use Vieta's formulas. According to them, the sum of the roots ofis . In our case this is .(Note that for the above approach to be completely correct, we should additionally verify that there actually are two distinct real roots. This is, for example, obvious from the facts that and .)25In year , the day of the year is a Tuesday. In year , the day is also a Tuesday. On what day of the week did the of year occur?SolutionClearly, identifying what of these years may/must/may not be a leap year will be key in solving the problem.Let be the day of year , the day of year and the day of year .If year is not a leap year, the day will bedays after . As , that would be a Monday.Therefore year must be a leap year. (Then is days after .) As there can not be two leap years after each other, is not a leap year. Therefore day is days after . We have . Therefore is weekdays before , i.e., is a.(Note that the situation described by the problem statement indeed occurs in our calendar. For example, for we have=Tuesday, October 26th 2004, =Tuesday, July 19th, 2005 and =Thursday, April 10th 2003.)。
2023年AMC10美国数学竞赛A卷附中文翻译和答案
2023AMC10美国数学竞赛A卷1. A cell phone plan costs $20 each month, plus 5¢ per text message sent, plus 10¢ for each minute used over 30 hours. In January Michelle sent 100 text messages and talked for 30.5 hours. How much did she have to pay?(A) $24.00 (B) $24.50 (C) $25.50 (D) $28.00 (E) $30.002. A small bottle of shampoo can hold 35 milliliters of shampoo, Whereas a large bottle can hold 500 milliliters of shampoo. Jasmine wants to buy the minimum number of small bottles necessary to completely fill a large bottle. How many bottles must she buy?(A) 11 (B) 12 (C) 13 (D) 14 (E) 153. Suppose [a b] denotes the average of a and b, and {a b c} denotes the average of a, b, and c. What is {{1 1 0} [0 1] 0}?(A) (B)(C)(D) (E)4. Let X and Y be the following sums of arithmetic sequences:X= 10 + 12 + 14 + …+ 100.Y= 12 + 14 + 16 + …+ 102.What is the value of(A) 92 (B) 98 (C) 100 (D) 102 (E) 1125. At an elementary school, the students in third grade, fourth grade, and fifth grade run an average of 12, 15, and 10 minutes per day, respectively. There are twice as many third graders as fourth graders, and twice as many fourth graders as fifth graders. What is the average number of minutes run per day by these students?(A) 12 (B) (C) (D) 13 (E) 146. Set A has 20 elements, and set B has 15 elements. What is the smallest possible number of elements in A∪B, the union of A and B?(A) 5 (B) 15 (C) 20 (D) 35 (E) 3007. Which of the following equations does NOT have a solution?(A) (B) (C)(D) (E)8. Last summer 30% of the birds living on Town Lake were geese, 25% were swans, 10% were herons, and 35% were ducks. What percent of the birds that were not swans were geese?(A) 20 (B) 30 (C) 40 (D) 50 (E) 609. A rectangular region is bounded by the graphs of the equations y=a, y=-b, x=-c, and x=d, where a, b, c, and d are all positive numbers. Which of the following represents the area of this region?(A) ac + ad + bc + bd (B) ac – ad + bc – bd (C) ac + ad – bc – bd(D) –ac –ad + bc + bd (E) ac – ad – bc + bd10. A majority of the 20 students in Ms. Deameanor’s class bought pencils at the school bookstore. Each of these students bought the same number of pencils, and this number was greater than 1. The cost of a pencil in cents was greater than the number of pencils each student bought, and the total cost of all the pencils was $17.71. What was the cost of a pencil in cents?(A) 7 (B) 11 (C) 17 (D) 23 (E) 7711. Square EFGH has one vertex on each side of square ABCD. Point E is on AB with AE=7·EB. What is the ratio of the area of EFGH to the area of ABCD?(A) (B)(C)(D) (E)12. The players on a basketball team made some three-point shots, some two-point shots, some one-point free throws. They scored as many points with two-point shots as with three-point shots. Their number of successful free throws was one more than their number of successful two-point shots. The team’s total score was 61 points. How many free throws did they make?(A) 13 (B) 14 (C) 15 (D) 16 (E) 1713. How many even integers are there between 200 and 700 whose digits are alldifferent and come from the set {1, 2, 5, 7, 8, 9}?(A) 12 (B)20 (C)72 (D) 120 (E) 20014. A pair of standard 6-sided fair dice is rolled once. The sum of the numbers rolled determines the diameter of a circle. What is the probability that the numerical value of the area of the circle is less than the numerical value of the circle’s circumference?(A) (B)(C)(D) (E)15. Roy bought a new battery-gasoline hybrid car. On a trip the car ran exclusively on its battery for the first 40 miles, then ran exclusively on gasoline for the rest of the trip, using gasoline at a rate of 0.02 gallons per mile. On the whole trip he averaged55 miles per gallon. How long was the trip in miles?(A) 140 (B) 240 (C) 440 (D) 640 (E) 84016. Which of the following in equal to(A) (B) (C) (D) (E)17. In the eight-term sequence A, B, C, D, E, F, G, H, the value of C is 5 and the sum of any three consecutive terms is 30. What is A + H?(A) 17 (B) 18 (C) 25 (D) 26 (E) 4318. Circles A, B, and C each have radius 1. Circles A and B share one point of tangency. Circle C has a point of tangency with the midpoint of AB. What is the area inside Circle C but outside Circle A and Circle B?(B) (C) (D) (E)(A)19. In 1991 the population of a town was a perfect square. Ten years later, after an increase of 150 people, the population was 9 more than a perfect square. Now, in 2023, with an increase of another 150 people, the population is once again a perfect square. Which of the following is closest to the percent growth of the town’s popu lation during this twenty-year period?(A) 42 (B) 47 (C) 52 (D) 57 (E) 6220. Two points on the circumference of a circle of radius r are selected independently and at random. From each point a chord of length r is drawn in a clockwise direction. What is the probability that the two chords intersect?(A) (B) (C) (D) (E)21. Two counterfeit coins of equal weight are mixed with 8 identical genuine coins. The weight of each of the counterfeit coins is different from the weight of each of the genuine coins. A pair of coins is selected at random without replacement from the 10 coins. A second pair is selected at random without replacement from the remaining 8coins. The combined weight of the first pair is equal to the combined weight of the second pair. What is the probability that all 4 selected coins are genuine?(A) (B) (C) (D) (E)22. Each vertex of convex pentagon ABCDE is to be assigned a color. There are 6 colors to choose from, and the ends of each diagonal must have different colors. How many different colorings are possible?(A) 2500 (B) 2880 (C) 3120 (D) 3250 (E) 375023. Seven students count from 1 to 1000 as follows:·Alice says all the numbers, except she skips the middle number in each consecutive group of thre e numbers. That is Alice says 1, 3, 4, 6, 7, 9, …, 997, 999, 1000.·Barbara says all of the numbers that Alice doesn’t say, except she also skips the middle number in each consecutive grope of three numbers.·Candice says all of the numbers that neither Alice nor Barbara says, except she also skips the middle number in each consecutive group of three numbers. ·Debbie, Eliza, and Fatima say all of the numbers that none of the students with the first names beginning before theirs in the alphabet say, except each also skips the middle number in each of her consecutive groups of three numbers.·Finally, George says the only number that no one else says.What number does George say?(A) 37 (B) 242 (C) 365 (D) 728 (E) 99824. Two distinct regular tetrahedra have all their vertices among the vertices of the same unit cube. What is the volume of the region formed by the intersection of the tetrahedra?(A) (B) (C) (D) (E)an integer. A point X in the interior of R is25. Let R be a square region andcalled n-ray partitional if there are n rays emanating from X that divide R into N triangles of equal area. How many points are 100-ray partitional but not 60-ray partitional?(A) 1500 (B) 1560 (C) 2320 (D) 2480 (E) 25002023AMC10美国数学竞赛A卷1. 某通讯公司手机每月基本费为20美元, 每传送一则简讯收 5美分（一美元=100 美分）。
AMC8(美国数学竞赛)历年真题、答案及中英文解析
AMC8（美国数学竞赛）历年真题、答案及中英文解析艾蕾特教育的AMC8 美国数学竞赛考试历年真题、答案及中英文解析：AMC8-2020年：真题 --- 答案---解析（英文解析+中文解析）AMC8 - 2019年：真题----答案----解析（英文解析+中文解析）AMC8 - 2018年：真题----答案----解析（英文解析+中文解析）AMC8 - 2017年：真题----答案----解析（英文解析+中文解析）AMC8 - 2016年：真题----答案----解析（英文解析+中文解析）AMC8 - 2015年：真题----答案----解析（英文解析+中文解析）AMC8 - 2014年：真题----答案----解析（英文解析+中文解析）AMC8 - 2013年：真题----答案----解析（英文解析+中文解析）AMC8 - 2012年：真题----答案----解析（英文解析+中文解析）析）AMC8 - 2010年：真题----答案----解析（英文解析+中文解析）AMC8 - 2009年：真题----答案----解析（英文解析+中文解析）AMC8 - 2008年：真题----答案----解析（英文解析+中文解析）AMC8 - 2007年：真题----答案----解析（英文解析+中文解析）AMC8 - 2006年：真题----答案----解析（英文解析+中文解析）AMC8 - 2005年：真题----答案----解析（英文解析+中文解析）AMC8 - 2004年：真题----答案----解析（英文解析+中文解析）AMC8 - 2003年：真题----答案----解析（英文解析+中文解析）AMC8 - 2002年：真题----答案----解析（英文解析+中文解析）AMC8 - 2001年：真题----答案----解析（英文解析+中文解析）AMC8 - 2000年：真题----答案----解析（英文解析+中文解析）析）AMC8 - 1998年：真题----答案----解析（英文解析+中文解析）AMC8 - 1997年：真题----答案----解析（英文解析+中文解析）AMC8 - 1996年：真题----答案----解析（英文解析+中文解析）AMC8 - 1995年：真题----答案----解析（英文解析+中文解析）AMC8 - 1994年：真题----答案----解析（英文解析+中文解析）AMC8 - 1993年：真题----答案----解析（英文解析+中文解析）AMC8 - 1992年：真题----答案----解析（英文解析+中文解析）AMC8 - 1991年：真题----答案----解析（英文解析+中文解析）AMC8 - 1990年：真题----答案----解析（英文解析+中文解析）AMC8 - 1989年：真题----答案----解析（英文解析+中文解析）AMC8 - 1988年：真题----答案----解析（英文解析+中文解析）析）AMC8 - 1986年：真题----答案----解析（英文解析+中文解析）AMC8 - 1985年：真题----答案----解析（英文解析+中文解析）◆AMC介绍◆AMC（American Mathematics Competitions) 由美国数学协会（MAA）组织的数学竞赛，分为 AMC8 、 AMC10、 AMC12 。
AMC 美国数学竞赛 2001 AMC 10 试题及答案解析
USA AMC 10 20011The median of the listis . What is the mean?Solution2A number is more than the product of its reciprocal and its additive inverse. In which interval does the number lie?Solution3The sum of two numbers is . Suppose 3 is added to each number and then each of the resulting numbers is doubled. What is the sum of the final two numbers?Solution4What is the maximum number of possible points of intersection of a circle and a triangle?Solution5How many of the twelve pentominoes pictured below have at least one line of symettry?Solution6Let and denote the product and the sum, respectively, of thedigits of the integer . For example, and . Supposeis a two-digit number such that . What is the units digit of ?Solution7When the decimal point of a certain positive decimal number is moved four places to the right, the new number is four times the reciprocal of the original number. What is the original number?Solution8Wanda, Darren, Beatrice, and Chi are tutors in the school math lab. Their schedule is as follows: Darren works every third school day, Wanda works every fourth school day, Beatrice works every sixth school day, and Chi works every seventh school day. Today they are all working in the math lab. In how many school days from today willthey next be together tutoring in the lab?Solution9The state income tax where Kristin lives is levied at the rate of of the first of annual income plus of any amount above . Kristin noticed that the state income tax she paid amounted to of her annual income. What was her annual income?Solution10If , , and are positive with , , and , then isSolution11Consider the dark square in an array of unit squares, part of which is shown. The first ring of squ ares around this center square contains unit squares. The second ring contains unit squares. If we continue this process, the number of unit squares in the ring isSolution12Suppose that is the product of three consecutive integers and that is divisible by . Which of the following is not necessarily a divisor of Solution13A telephone number has the form , where each letter represents a different digit. The digits in each part of the numbers are in decreasing order; that is, , , and . Furthermore, , , and are consecutive even digits; , , , and are consecutive odd digits; and . Find .Solution14A charity sells 140 benefit tickets for a total of . Some tickets sell for full price (a whole dollar amount), and the rest sells for half price. How much money is raised by the full-price tickets?Solution15A street has parallel curbs feet apart. A crosswalk bounded by two parallel stripes crosses the street at an angle. The length of the curb between the stripes is feet and each stripe is feet long. Find the distance, in feet, between the stripes.Solution16The mean of three numbers is 10 more than the least of the numbers and 15 less than the greatest. The median of the three numbers is 5. What is their sum?Solution17Which of the cones listed below can be formed from a sector of a circle of radius by aligning the two straight sides?A cone with slant height of and radiusA cone with height of and radiusA cone with slant height of and radiusA cone with height of and radiusA cone with slant height of and radiusSolution18The plane is tiled by congruent squares and congruent pentagons as indicated. The percent of the plane that is enclosed by the pentagons is closest toSolution19Pat wants to buy four donuts from an ample supply of three types of donuts: glazed, chocolate, and powdered. How many different selections are possible?Solution20A regular octagon is formed by cutting an isosceles right triangle from each of the corners of a square with sides of length . What is the length of each side of the octagon?Solution21A right circular cylinder with its diameter equal to its height is inscribed in a right circular cone. The cone has diameter and altitude , and the axes of the cylinder and cone coincide. Find the radius of the cylinder.Solution22In the magic square shown, the sums of the numbers in each row, column, and diagonal are the same. Five of these numbers are represented by , , , , and . Find .Solution23A box contains exactly five chips, three red and two white. Chips are randomly removed one at a time without replacement until all the red chips are drawn or all the white chips are drawn. What is the probability that the last chip drawn is white?Solution24In trapezoid , and are perpendicular to , with, , and . What is ?Solution25How many positive integers not exceeding are multiples of or but not ?。
AMC 美国数学竞赛 2002 AMC 10A 试题及答案解析
2002 AMC 10A1、The ratio is closest to which of the following numbers?SolutionWe factor as . As , ouranswer is .2、For the nonzero numbers , , , define.Find .Solution. Ouranswer is then .Alternate solution for the lazy: Without computing the answer exactly,we see that , , and . The sumis , and as all the options are integers, the correct one is obviously .3、According to the standard convention for exponentiation,.If the order in which the exponentiations are performed is changed, how many other values are possible?SolutionThe best way to solve this problem is by simple brute force.It is convenient to drop the usual way how exponentiation is denoted,and to write the formula as , where denotes exponentiation. We are now examining all ways to add parentheses to this expression. There are 5 ways to do so:1.2.3.4.5.We can note that . Therefore options 1 and 2 are equal, and options 3 and 4 are equal. Option 1 is the one given in the problem statement. Thus we only need to evaluate options 3 and 5.Thus the only other result is , and our answer is .4、For how many positive integers does there exist at least one positive integer such that ?infinitely manySolutionSolution 1For any we can pick , we get , therefore theanswer is .Solution 2Another solution, slightly similar to this first one would be using Simon's Favorite Factoring Trick.Let , thenThis means that there are infinitely many numbers that can satisfythe inequality. So the answer is .5、Each of the small circles in the figure has radius one. The innermost circle is tangent to the six circles that surround it, and each of those circles is tangent to the large circle and to its small-circle neighbors. Find the area of the shaded region.SolutionThe outer circle has radius , and thus area . The littlecircles have area each; since there are 7, their total area is . Thus,our answer is .6、Cindy was asked by her teacher to subtract from a certain numberand then divide the result by . Instead, she subtracted and thendivided the result by , giving an answer of . What would heranswer have been had she worked the problem correctly?SolutionWe work backwards; the number that Cindy started with is. Now, the correct result is . Ouranswer is .7、If an arc of on circle has the same length as an arc of oncircle , then the ratio of the area of circle to the area of circle isSolutionLet and be the radii of circles A and B, respectively.It is well known that in a circle with radius r, a subtended arc oppositean angle of degrees has length .Using that here, the arc of circle A has length . The arcof circle B has length . We know that they are equal,so , so we multiply through and simplify to get . As all circles are similar to one another, the ratio of the areas is just thesquare of the ratios of the radii, so our answer is .8、Betsy designed a flag using blue triangles, small white squares, anda red center square, as shown. Let be the total area of the bluetriangles, the total area of the white squares, and the area of thered square. Which of the following is correct?SolutionThe blue that's touching the center red square makes up 8 triangles, or 4 squares. Each of the corners is 2 squares and each of the edges is 1, totaling 12 squares. There are 12 white squares, thus we have.9、There are 3 numbers A, B, and C, such that ,and . What is the average of A, B, and C?More than 1SolutionNotice that we don't need to find what A, B, and C actually are, just their average. In other words, if we can find A+B+C, we will be done.Adding up the equations gives soand the average is . Our answer is .10、Compute the sum of all the roots of.SolutionSolution 1We expand to get which isafter combining like terms. Using the quadratic partof Vieta's Formulas, we find the sum of the roots is . Solution 2Combine terms to get, hence the rootsare and , thus our answer is .11、Jamal wants to store computer files on floppy disks, each ofwhich has a capacity of megabytes (MB). Three of his files requireMB of memory each, more require MB each, and theremaining require MB each. No file can be split between floppydisks. What is the minimal number of floppy disks that will hold all the files?SolutionA 0.8 MB file can either be on its own disk, or share it with a 0.4 MB. Clearly it is not worse to pick the second possibility. Thus we will have 3 disks, each with one 0.8 MB file and one 0.4 MB file.We are left with 12 files of 0.7 MB each, and 12 files of 0.4 MB each.Their total size is MB. The total capacity of 9 disks is MB, hence we need at least 10 more disks. And wecan easily verify that 10 disks are indeed enough: six of them will carry two 0.7 MB files each, and four will carry three 0.4 MB files each.Thus our answer is .12、Mr. Earl E. Bird leaves his house for work at exactly 8:00 A.M. every morning. When he averages miles per hour, he arrives at hisworkplace three minutes late. When he averages miles per hour, hearrives three minutes early. At what average speed, in miles per hour, should Mr. Bird drive to arrive at his workplace precisely on time?SolutionSolution 1Let the time he needs to get there in be t and the distance he travelsbe d. From the given equations, we know that and. Setting the two equal, we have andwe find of an hour. Substituting t back in, we find . From, we find that r, and our answer, is .Solution 2Since either time he arrives at is 3 minutes from the desired time, the answer is merely the harmonic mean of 40 and 60. The harmonicmean of a and b is . In this case, a and b are 40 and 60,so our answer is , so .Solution 3A more general form of the argument in Solution 2, with proof:Let be the distance to work, and let be the correct average speed.Then the time needed to get to work is .We know that and . Summing these twoequations, we get: .Substituting and dividing both sides by , we get ,hence .(Note that this approach would work even if the time by which he is late was different from the time by which he is early in the other case - we would simply take a weighed sum in step two, and hence obtaina weighed harmonic mean in step three.)13、Give a triangle with side lengths 15, 20, and 25, find the triangle's smallest height.SolutionSolution 1This is a Pythagorean triple (a 3-4-5 actually) with legs 15 and 20. Thearea is then . Now, consider an altitude drawn to anyside. Since the area remains constant, the altitude and side to which it is drawn are inversely proportional. To get the smallest altitude, it must be drawn to the hypotenuse. Let the length be x; we have, so and x is 12. Our answer is then.Solution 2By Heron's formula, the area is , hence the shortest altitude'slength is .14、Both roots of the quadratic equation are prime numbers. The number of possible values of isSolutionConsider a general quadratic with the coefficient of being and theroots being and . It can be factored as which is just. Thus, the sum of the roots is the negative of the coefficient of and the product is the constant term. (In general, this leads to Vieta's Formulas).We now have that the sum of the two roots is while the product is. Since both roots are primes, one must be , otherwise the sumwould be even. That means the other root is and the product mustbe . Hence, our answer is .15、Using the digits 1, 2, 3, 4, 5, 6, 7, and 9, form 4 two-digit prime numbers, using each digit only once. What is the sum of the 4 prime numbers?SolutionOnly odd numbers can finish a two-digit prime number, and a two-digit number ending in 5 is divisible by 5 and thus composite,hence our answer is .(Note that we did not need to actually construct the primes. If we had to, one way to match the tens and ones digits to form four primes is , , , and .)16、Let . What is?SolutionLet . Since one ofthe sums involves a, b, c, and d, it makes sense to consider 4x. We have. Rearranging, we have , so .Thus, our answer is .17、Sarah pours four ounces of coffee into an eight-ounce cup and fourounces of cream into a second cup of the same size. She then transfers half the coffee from the first cup to the second and, after stirring thoroughly, transfers half the liquid in the second cup back to the first. What fraction of the liquid in the first cup is now cream?SolutionWe will simulate the process in steps.In the beginning, we have:▪ounces of coffee in cup▪ounces of cream in cupIn the first step we pour ounces of coffee from cup to cup ,getting:▪ounces of coffee in cup▪ounces of coffee and ounces of cream in cupIn the second step we pour ounce of coffee and ounces of cream from cup to cup , getting:▪ounces of coffee and ounces of cream in cup▪the rest in cupHence at the end we have ounces of liquid in cup , and outof these ounces is cream. Thus the answer is .18、A cube is formed by gluing together 27 standard cubicaldice. (On a standard die, the sum of the numbers on any pair of opposite faces is 7.) The smallest possible sum of all the numbers showing on the surface of the cube isSolutionIn a 3x3x3 cube, there are 8 cubes with three faces showing, 12 with two faces showing and 6 with one face showing. The smallest sum with three faces showing is 1+2+3=6, with two faces showing is 1+2=3, and with one face showing is 1. Hence, the smallest possiblesum is . Our answer is thus.19、Spot's doghouse has a regular hexagonal base that measures oneyard on each side. He is tethered to a vertex with a two-yard rope.What is the area, in square yards, of the region outside of the doghouse that Spot can reach?SolutionPart of what Spot can reach is of a circle with radius 2, whichgives him . He can also reach two parts of a unit circle, whichcombines to give . The total area is then , which gives .20、Points and lie, in that order, on , dividing it intofive segments, each of length 1. Point is not on line . Point lieson , and point lies on . The line segments andare parallel. Find .SolutionAs is parallel to , angles FHD and FGA are congruent. Also,angle F is clearly congruent to itself. From SSS similarity,; hence . Similarly, . Thus,.21、The mean, median, unique mode, and range of a collection of eight integers are all equal to 8. The largest integer that can be an element of this collection isSolutionAs the unique mode is , there are at least two s.As the range is and one of the numbers is , the largest one can beat most .If the largest one is , then the smallest one is , and thus the meanis strictly larger than , which is a contradiction.If the largest one is , then the smallest one is . This means that wealready know four of the values: , , , . Since the mean of all thenumbers is , their sum must be . Thus the sum of the missing fournumbers is . But if is the smallest number,then the sum of the missing numbers must be at least ,which is again a contradiction.If the largest number is , we can easily find the solution. Hence, our answer is .NoteThe solution for is, in fact, unique. As the median must be , thismeans that both the and the number, when ordered by size,must be s. This gives the partial solution . For themean to be each missing variable must be replaced by the smallestallowed value.22、A sit of tiles numbered 1 through 100 is modified repeatedly by the following operation: remove all tiles numbered with a perfect square, and renumber the remaining tiles consecutively starting with 1. How many times must the operation be performed to reduce the number of tiles in the set to one?SolutionSolution 1The pattern is quite simple to see after listing a couple of terms.Solution 2Given tiles, a step removes tiles, leaving tiles behind. Now,, so in the next step tilesare removed. This gives , another perfect square.Thus each two steps we cycle down a perfect square, and insteps, we are left with tile, hence our answer is.23、Points and lie on a line, in that order, with and. Point is not on the line, and . The perimeterof is twice the perimeter of . Find .SolutionFirst, we draw an altitude to BC from E.Let it intersect at M. As triangle BEC is isosceles, we immediately get MB=MC=6, so the altitude is 8. Now, let . Using the Pythagorean Theorem on triangleEMA, we find . From symmetry,as well. Now, we use the fact that the perimeter of is twice the perimeter of .We have so. Squaring both sides, we havewhich nicely rearranges into. Hence, AB is 9 so our answer is .24、Tina randomly selects two distinct numbers from the setand Sergio randomly selects a number from the set. The probability that Sergio's number is larger than the sum of the two numbers chosen by Tina isSolutionThis is not too bad using casework.Tina gets a sum of 3: This happens in only one way (1,2) and Sergio can choose a number from 4 to 10, inclusive. There are 7 ways that Sergio gets a desirable number here.Tina gets a sum of 4: This once again happens in only one way (1,3). Sergio can choose a number from 5 to 10, so 6 ways here.Tina gets a sum of 5: This can happen in two ways (1,4) and (2,3). Sergio can choose a number from 6 to 10, so 2*5=10 ways here.Tina gets a sum of 6: Two ways here (1,5) and (2,4). Sergio can choose a number from 7 to 10, so 2*4=8 here.Tina gets a sum of 7: Two ways here (2,5) and (3,4). Sergio can choose from 8 to 10, so 2*3=6 ways here.Tina gets a sum of 8: Only one way possible (3,5). Sergio chooses 9 or 10, so 2 ways here.Tina gets a sum of 9: Only one way (4,5). Sergio must choose 10, so 1 way.In all, there are ways. Tina chooses twodistinct numbers in ways while Sergio chooses a number inways, so there are ways in all. Since , ouranswer is .25、In trapezoid with bases and , we have ,, , and . The area of isSolutionSolution 1It shouldn't be hard to use trigonometry to bash this and find the height, but there is a much easier way. Extend and to meet atpoint :Since we have , with the ratio ofproportionality being . Thus So the sides of are , which we recognize to be aright triangle. Therefore (we could simplify some of the calculation using that the ratio of areas is equal to the ratio of the sides squared),Solution 2Draw altitudes from points and :Translate the triangle so that coincides with . We getthe following triangle:The length of in this triangle is equal to the length of the original, minus the length of . Thus .Therefore is a well-known right triangle. Its area is, and therefore its altitude is.Now the area of the original trapezoid is.。
2021年美国数学竞赛AMC12A 真题加答案解析
2021 AMC 12A Peeyush Pandaya et al.February 20211 Answer1.B2. D3. D4.D5.E Key6.C7.D8.C9.C10.E11.C12.A13.B14.E15. D16.C17.D18.E19.C20.B21.A22.D23.D24.D25.E2 Problems and SolutionsProblem 1. What is the value of21+2+3- (2¹+2²+23)?(A)0 (B)50 (C)52 (D)54 (E)57Solution.2⁶-(2¹+2²+23)=64-(2+4+8)=64-14=(B)50Problem 2.Under what conditionsisva²+62=a+btrue,whereaand bare real numbers?(A)It is never true(B) It is true if and only ifab=0(C) It is true if and only ifa+b≥0(D)It is true ifandonlyifab=0anda+b≥0(E)It is always trueSolution. It is clear that both sides of the equation must be nonnegative.Squaring,a²+b²=a²+2ab+b2→ab=0The answer is (D)Problem 3.The sum of two natural numbers is 17,402.One of the two numbers is divisible by 10.If the units digit of that number is erased, the other number is obtained. What is the difference of these two numbers?(A)10,272 (B)11,700 (C)13,362 (D)14,238 (E)15,426Solution. Let the first number mentioned be 10n;the second is n. Then10n+n=17,402,from which it follows thatProblem 4. Tom has a collection of 13 snakes,4 of which are purple and 5 of which are happy. He observes that●all of his happy snakes can add,·none of his purple snakes can subtract●all of his snakes that can't subtract also can't add.Which of these conclusions can be drawn about Tom's snakes?Solution. Together, the second and third conditions imply that none of Tom's purple snakes can add.Thus,(D) is correct: happy snakes are not purple.Problem 5.When a student multiplied the number 66 by the repeating decimal,where a and b are digits, he did not notice the notation and just multiplied 66 times 1.ab. Later he found that his answer is 0.5 less than the correct answer. What is the 2-digit integer gb? (A)15 (B)30 (C)45 (D)60 (E)75Solution. The student computed 66 ; the correct answer is 66 . Thus,Problem 6.A deck of cards has only red cards and black cards. The probability of a randomly chosen card being red is . When 4 black cards are added to the deck, the probability of choosing red becomes . How many cards were in the deck originally?(A)6 (B)9 (C) 12 (D)15 (E)18Solution. If the deck begins with x red cards and 3x cards in total, thenProblem 7.What is the least possible value of(ry- 1)²+(x+y)2 for real numbersa and y?Solution. We have(ry- 1)2+(x+y)2=(ry)2-2ry+1+x²+2ry+v²=x2v²+z²+y2+1= (r²+1)(y²+1),which achieves a minimum of (D)1 atx=y=0.D Do D ₁ D ₂D ₃ D ₁ D ₅D ₆ D ₇ D ₈ D, D1o Problem 8.A sequence of numbers isdefinedbyDo=0,D ₁=0,Dz=1,andDn=Dn- 1+Dn-3 forn≥3.What are the parities(evenness oroddness)of the triple of numbers(D2021,D2022,D2023), whereE denotes even and O denotes odd?Solution.0/10 01 D2+Do=1+0=1 D ₃+Di=1+0=1D ₄+Dz=1+1=0 Ds+D ₃=0+1=1D ₆+D4=1+1=0D-+Ds=0+0=0D ₈+D ₆=0+1=1Dg+D ₇=1+0=1We can see that the pattern repeats in cycles of length7.and as 2021=5 mod7,we have D2021= Ds,D2022=D6,D2023=D7→(C)(E,O,E) Problem 9.Which of the following is equivalent to(2+3)(2²+3²)(2⁴+34)(2⁸+3⁸)(216+316)(2³²+3³2)(264+364)?(A)3127+2127 (B)3127+2127+2.363+3.263 (C)3128-2128 (D)3128+2128 (E)5127Solution.(3-2)(2+3)(2²+3²)(2⁴+3⁴)(2⁸+3⁸)(216+316)(2³²+3³2)(264+364)=(3²-2²)(2²+3²)(2⁴+3⁴)(2⁸+3⁸)(216+316)(232+3³2)(264+364)=(3⁴-2⁴)(2⁸+3⁸)(216+316)(232+332)(264+364)=(C)3128-2129=(364-264)(264+364)Problem 10.Two right circular cones with vertices facing down as shown in the figure below contain the same amount of liquid. The radii of the tops of the liquid surfaces are 3cm and 6cm. Into each cone is dropped a spherical marble of radius lcm, which sinks to the bottom and is completely submerged without spilling any liquid. What is the ratio of the rise of the liquid level the narrow cone to the rise of the liquid level in the wide cone?(A)1:1 (B)47:43 (C) 2:1 (D)40:13 (E)4:1Solution. The two cones have equal volume, so the height of the first is times that of the second. Since the volumes increase by equal proportions, the heights increase by equal proportions. Thus, the ratio of the rise in liquid levels is (E)4:1Problem 11.A laser is placed at the point(3,5). The laser beam travels in a straight rry wants the beam to hit and bounce off the y-axis, then hit and bounce off the x-axis,then hit the point (7,5). What is the total distance the beam will travel along this path?(A)2√10 (B)5√2(C)10√2(D)15√2 (E)10√5Solution. Reflect about the y-axis then the z-axis. It is well-known that the image under the two reflections must be a straight line.The answer isv(3-(-7))²+(5-(-5)= (C) 10√2Problem 12.All the roots of polynomialz6- 10z ⁵+Az ⁴+Bz³+Cz²+Dz+16are positiveintegers, possibly repeated. What is the value of B?(A)-88 (B)-80 (C)-64 (D)-41 (E)-40Solution. By Vieta's,the sum of the 6 roots is 10 and the product is 16,hence they are all powers of 2. It is not hard to find that the only working unordered sextuple is(2,2,2,2,1,1). As(z-2)4=24-8z3+24z2-32z+16 and(z- 1)2=z2-2z+1.the z3 coefficient is -8.1+24 · (-2)+(-32) · 1= (B)-88Problem 13.Of the following complex numbers z, which has the property that z5 has the greatest real part?(A)-2 (B)-√3+i (C)-√2+√2i Solution. The magnitude of each complex number is the same, so it suffices to look at the argu- ment. The angles are π, ,and ,which after raising to the 5th power give π, and . We seek the angle that reaches farthest to the right(smallest argument),which is . Thus, our answer is (B)-√3+iProblem 14.What is the value of(A)21 (B)100logs3 (C)200log35 (D)2,200 (E)21,000Solution.And,Therefore, their product is210logs3.100log35=(E)21,000(D)- 1+√3i (E)2iProblem 15.A choir director must select a group of singers from among his 6 tenors and 8 basses. The only requirements are that the difference between the numbers of tenors and basses must be a multiple of 4,and the group must have at least one singer.Let be the number of groups that could be selected. What is the remainder when N is divided by 100?(A)47 (B)48 (C)83 (D)95 (E)96Solution. Suppose we mark down(1) the tenors that are in the group,and(2)the basses that aren't in the group. Then we necessarily mark down a number of people that is a multiple of 4. This is also sufficient; we mark down some people numbering a multiple of 4, then select the marked tenors and unmarked basses to form our choir. Clearly, we just mark down at least one person. The answer is thus(D)95Problem 16.In the following list of numbers, the integer n appears times in the list for l≤n≤200.1,2,2,3,3,3,4,4,4,4,.…,200,200,...200.What is the median of the numbers in this list?(A)100.5 (B)134 (C)142 (D)150.5 (E)167Solution.For general n, we have numbers. We want to approximate a such that is close to . Since the formula is a quadratic in n and we are halving this value,we can find that a is approximately .Plugging in n = 200,this is about 100√2,or 141.Of the answer choices, (C)142 is the closest, and indeed it is our answer.To verify, we can see that and ), so clearly 142 works.Problem 17.Trapezoid ABCD has ABICD,BC=CD=43,and AD1BD.Let O be the intersection of the diagonals AC and BD,and let P be the midpoint of BD.Given that OP=11, the length AD can be written in the form myn,where m and n are positive integers and n is not divisible by the square of any prime. What is m+n?(A)65 (B)132 (C)157 (D)194 (E)215Solution. Let M be the intersection of CPand AB.Since DCBMisakite,andCMIBD,we have MP1PB,and by considering the homothethy taking △MBD to △ABD with scale factor 2,we can see that M is the midpoint of AB.In particular,we haveSince AD 1BD,we have AD1DOandthusZADO=90°,andasCD=CB,wehaveCP1BD and ZCPD=2CPO=90°.Also,ZAOD=ZCOP,so △AOD~ △COP.Therefore,so DO=22.Thus,AD=√AB²-BD²=√86²-66²=4√ 190→m+n= (D)194The desired answer isProblem 18.Let f be afunction defined on the set of positive rational numbers with the property that f(a ·b)=f(a)+f(b)for all positive rational numbersa and b.Suppose that falso has the property that f(p)= pfor every prime numberp.For which of the following numbers zis f(x)<0?(A)整(B) (C) (D) (E) 51Solution. Note that f(a ·1)=f(a)+f(1)= f(1)=0,andIn particular, it follows by induction that f(p*)= kp for each k ∈Z.Thus,(A) f(2-5. 17)=-5·2+17=7(B) f(2-4. 11)=-4·2+11=3(C) f(3-2.7)=-2.3+7=1(D)f(2- 1.3- 1.7)=- 1.2+(- 1) ·3+7=2(E)f(52.11- 1)=2.5- 11=- 125The answer is (E)11Problem 19.How many solutions does the equation sinclosed interval [0,π]?(A)0 (B)1 (C) 2 (D) 3 (E)4Solution. Note on the interval , the left-hand side is negative while the right-hand side is positive.We thus restrict our attention to ]. The arguments cosx and sz are both between 0 and .ForsinA=cosBinA,B ∈[0,],we must have . This implies sinz+cosz=1,hence . There are (C) 2 solutions.Problem 20.Suppose that on a parabola with vertexV and focus F there exists a point Asuch that AF=20 and AV=21.What is the sum of all possible values of the length FV?(A)13 (B) 40 (C) (D)14 (E)Solution. Let the directrix be the x-axis,F=(0,2d),V=(0,d),A=(x,y),andB=(0,y)for some d>0.By the definition of a parabola,y=20.We compute x in two ways:x²=AF²-BF²=20²-|20-2d|2=AV²-BV²=21²-|20-d²Subtracting,O=20²-21²+(20-d)²-(20-2d)2=3d²-40d+41.The sum of all possible values ofProblem 21.The five solutions to the equation(z- 1)(z²+2z+4)(z²+4z+6)=0may be written in theformzk+ykiforl≤k≤5,where xk and yk are real.Let E be the unique ellipse that passes through the points(Ti,yi),(x2,32),(r3,Y3),(x4,y4)and(xs,ys).The eccentricity ofE can be written in the form ,where m and n are relatively prime positive integers. What is m+n?(Recall that the eccentrictiy of an ellipse E is the ratio,where 2a is the length of the major axis ofE and 2c is the distance between its two foci.)(A)7 (B)9 (C)11 (D)13 (E)15Solution. The roots of the polynomial arez=1,z=- 1±i√3,andz=-2±i√2,hence the five points onE are(1,0),( - 1,土√3),( - 2,±√2) .By symmetry through the x-axis, the ellipse is of the formE: a(x-r)²+by²=1.We then have the relationsa(1-r)²=1a(1+r)²+3b=1a(2+r)²+2b=1.Eliminating b from the latter two,1=3[a(2+r)²+2]-2[a(1+r)²+36]=a(r²+8r+10),henceit follows that and so the eccentricity is 1/√6. The requested sum is1+6= (A)7Problem 22.Suppose that the roots of the polynomialP(x)=x³+ax²+bx+care cos 2π,COS 47 and cos ,where angles are in radians. What is abc?(C) (D) (E)Solution. Recall1+e2m/7+ …+e12mi/7=0→e2mi/7+e4xi/7+e6mi/7=- 1/2,and in particular caNote are solutions to the equation cosb+cos28+cos30=- 1/2,so lettingx=co sθimplies thathas roots ,COS 4π7Thus, the polynomial in question is and the requested answer isRemark. Perhaps it is easier to motivate the solution as follows.Lett=e2mi/7andx=t+t- 1= 2cos2π/7.Thenx²-2=t²+t-2andx³-3x=t³+t-3.Moreover,t⁶+t⁵+ …+1=0impliest³+t²+t+t- 1+t-2+t-3=- 1,i.e.x+(x²-2)+(r³-3r) has root 2cos2π/7.It certainly seems logical that the Galois conjugates of 2cos would be 2cos and 2cos (especially given the phrasing of the problem),so simply replacex → to get the desired polynomial form.Let w = e2ix/7.Note thatLet these be r,s,t respectively. By Vieta's formulas, note that the desired quantity is(-rst)(rs+st+tr)(-r-s-t)=(r+s+t)(rs+st+tr)(rst).Note that 1+w+ …+w⁶=0.We haveThen,FinallyMultiplying yields the answer of (D)1 32Solution Manual 2021AMC12AProblem 23.Frieda the frog begins a sequence of hops on a 3×3 grid of squares,moving one square on each hop and choosing at random the direction of each hop up,down, left,or right. She does not hop diagonally. When the direction of a hop would take Frieda off the grid,she ”wraps around”and jumps to the opposite edge.For example if Frieda begins in the center square and makes two hops ”up”,the first hop would place her in the top row middle square, and the second hop would cause Frieda to jump to the opposite edge,landing in the bottom row middle square.Suppose Frieda starts from the center square, makes at most four hops at random,and stops hopping if she lands on a corner square. What is the probability that she reaches a corner square on one of the four hops?(A)G (B) (D)强(E)8Solution. We complementary count,and determine the probability we never reach a corner square. Denote by A the center,and B a square adjacent to the center. Then the first hop lands on B.● If the second hop lands on A(with probability ,then the third hop lands on B always,andthere is a chance the fourth hop lands on a non-corner square. The probability in this case is● If the second hop lands on B, then there is a chance the third hop lands on A,and achance the third hop lands on B.In the former subcase, the fourth hop always lands on a non-corner square, and in the latter subcase, there is a chance the fourth hop lands on a non-corner square. The probability in this case isThe requested probability isProblem 24.SemicircleT has diameter AB of length14.Circle Ωlies tangent to AB at a point P and intersectsI at pointsQandR.IfQR=3√3and ZQPR=60°,then the area of △PQR is ,where a and c are relatively prime positive integers, and b is a positive integer not divisible by the square of any prime. What isa+b+c?(A)110 (B)114 (C)118 (D)122 (E)126Solution. First, by Extended Law of Sines,we have that the radius of Q is .Let M be the midpoint of QR,O be the center ofT,and X be the center of 2.Since △OQR is isosceles, OM is perpendicular to QR. Thus we have that X lies on OM since it is the circumcenter of △puting lengths, we hav thagorean theorem on △ORM andfrom isosceles triangle XQR. Thus and OP=4 from Pythagorean theorem on △OXP.To find [PQR],we will find the height from P to QR.Let the foot of the perpendicular fromX to the P-altitude be D. Since PDⅡOM,we know that △XDP~△OPX.This means that . Now note that the bottom portion of the P-altitude after subtractingPD is equal to XM,so the height of the triangle is . The area is simply2021 AMC 12A11 Solution ManualProblem 25.Let d(n)denote the number of positive integers that divide n,including l and n. For example,d(1)=1,d(2)=2,and d(12)=6.(This function is known as the divisor function.) LetThere is a unique positive integer N such that f(N)> f(n)for all positive integers n≠N.What is the sum of the digits of N?(A)5 (B)6 (C)7 (D)8 (E) 9Solution. Letn=II;p',wherepi=2,Pz=3,etc.are the primes in increasing order and e; are nonnegative integers. ThenIt is equivalent to maximize Thus, it remains to find the optimal e; for each i. We go term-by-term, noting that it is only necessary to check until the expression first decreases, as exponentials increase more quickly than polynomials.ei 0 1 2 3 4 0 1 2 3 0 1 2 0 1 2 ((ez+1)3)/p⁸1³/20=1 23/2¹=433/2²=6.25 4³/2³=85³/2⁴<8 13/3⁰=1 23/31≈2.67 3³/32=3 43/3³<3 13/5⁰=1 2³/5¹=1.6 3³/5²=1.08 13/70=1 23/71≈1.14 3³/7²<1Note that we do not need to check p≥11,ase;=1yields which is suboptimal.Thus,the answer isN=23.32.5.7=2520,which has a Remark:It is sufficient to stop at p=3,for 3²|Nleaves N among the answer choices.digit sum of (E)9only one possibility for the digit sum ofi 1 1 1 1 1 2 2 2 2 3 3 3 4 4 4。
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Solution
Answer: (C)
Lemma) if , then
The tedious algebra is left to the reader. (it is not bad at all)
Well, let us consider the cases where each of those step is definite (is never evaluate).
So, we have ,
--- (exception -> case 2)
--- (exception -> case 3)
--- (exception -> case 4)
--- (exception -> case 5)
If it is not any of the above 5 cases, then
if (--- exception -> case 6), then , ,
Hence, it is possible maximum of and minimum is 1.
2 possible combination of are and . Verification is left upto the reader. Right now, (C) is the most possible answer out of those 5.
Case 2) , then
Case 3) = 1, which is in the range.
Case 5) , hence
Case 6) Since , ,
Case 4) , this is quite an annoying special case. In this case, ,
is not define.
In this case,
and
Hence, and . Once,
you work out this system, you will get no solution with .
Thus, answer is (C).
SOLUTION 2:
After a bit of tedius algebra (that isn't too bad, but a little lengthy) we obtain
where ,
, , and
. In order for , we must have , ,
and . The first implies or , the second implies ,
, or , and the third implies or .
Since , in order to satisfy all 3 conditions we must have either or
. In the first case . For the latter case note that
so that and hence
. On the other hand so that
. Thus . Hence in any case the maximum
value for is while the minimum is (which can be achieved for instance when
or respectively). Hence the answer is .\\ \
Solution
From , we know that .
From the first inequality, we get . Subtracting
from this gives us , and thus .
Since must be an integer, it follows that .
Similarly, from the second inequality, we get . Again
subtracting from this gives us , or
. It follows from this that .
We now have a system of three equations: , , and
. Solving gives us and from this we find that
Since , we find that
.。