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1、2010 AMC 10A problems and solutions.The test was held on February 8, 2010. The first link contains the full set of test problems. The rest contain each individual problem and its solution.Problem 1 Marys top book shelf holds five books with the following widths, in centimeters: , , , , and . What is
2、 the average book width, in centimeters? SolutionTo find the average, we add up the widths , , , , and , to get a total sum of . Since there are books, the average book width is The answer is . Problem 2 Four identical squares and one rectangle are placed together to form one large square as shown.
3、The length of the rectangle is how many times as large as its width? SolutionLet the length of the small square be , intuitively, the length of the big square is . It can be seen that the width of the rectangle is . Thus, the length of the rectangle is times large as the width. The answer is . Probl
4、em 3 Tyrone had marbles and Eric had marbles. Tyrone then gave some of his marbles to Eric so that Tyrone ended with twice as many marbles as Eric. How many marbles did Tyrone give to Eric? SolutionLet be the number of marbles Tyrone gave to Eric. Then, . Solving for yields and . The answer is . Pro
5、blem 4 A book that is to be recorded onto compact discs takes minutes to read aloud. Each disc can hold up to minutes of reading. Assume that the smallest possible number of discs is used and that each disc contains the same length of reading. How many minutes of reading will each disc contain? Solu
6、tionAssuming that there were fractions of compact discs, it would take CDs to have equal reading time. However, since the number of discs can only be a whole number, there are at least 8 CDs, in which case it would have minutes on each of the 8 discs. The answer is . Problem 5 The area of a circle w
7、hose circumference is is . What is the value of ? SolutionIf the circumference of a circle is , the radius would be . Since the area of a circle is , the area is . The answer is . Problem 6 For positive numbers and the operation is defined as What is ? Solution. Then, is The answer is Problem 7 Crys
8、tal has a running course marked out for her daily run. She starts this run by heading due north for one mile. She then runs northeast for one mile, then southeast for one mile. The last portion of her run takes her on a straight line back to where she started. How far, in miles is this last portion
9、of her run? SolutionCrystal first runs North for one mile. Changing directions, she runs Northeast for another mile. The angle difference between North and Northeast is 45 degrees. She then switches directions to Southeast, meaning a 90 degree angle change. The distance now from travelling North for
10、 one mile, and her current destination is miles, because it is the hypotenuse of a 45-45-90 triangle with side length one (mile). Therefore, Crystals distance from her starting position, x, is equal to , which is equal to . The answer is Problem 8 Tony works hours a day and is paid $ per hour for ea
11、ch full year of his age. During a six month period Tony worked days and earned $. How old was Tony at the end of the six month period? SolutionTony worked hours a day and is paid dollars per hour for each full year of his age. This basically says that he gets a dollar for each year of his age. So if
12、 he is years old, he gets dollars a day. We also know that he worked days and earned dollars. If he was years old at the beginning of his working period, he would have earned dollars. If he was years old at the beginning of his working period, he would have earned dollars. Because he earned dollars,
13、 we know that he was for some period of time, but not the whole time, because then the money earned would be greater than or equal to . This is why he was when he began, but turned sometime in the middle and earned dollars in total. So the answer is .The answer is . We could find out for how long he
14、 was and . . Then is and we know that he was for days, and for days. Thus, the answer is . Problem 9A palindrome, such as , is a number that remains the same when its digits are reversed. The numbers and are three-digit and four-digit palindromes, respectively. What is the sum of the digits of ? Sol
15、ution is at most , so is at most . The minimum value of is . However, the only palindrome between and is , which means that must be . It follows that is , so the sum of the digits is . Problem 10 Marvin had a birthday on Tuesday, May 27 in the leap year . In what year will his birthday next fall on
16、a Saturday? Solution(E) 2017 There are 365 days in a non-leap year. There are 7 days in a week. Since 365 = 52 * 7 + 1 (or 365 is congruent to 1 mod 7), the same date (after February) moves forward one day in the subsequent year, if that year is not a leap year. For example: 5/27/08 Tue 5/27/09 Wed
17、However, a leap year has 366 days, and 366 = 52 * 7 + 2. So the same date (after February) moves forward two days in the subsequent year, if that year is a leap year. For example: 5/27/11 Fri 5/27/12 Sun You can keep count forward to find that the first time this date falls on a Saturday is in 2017:
18、 5/27/13 Mon 5/27/14 Tue 5/27/15 Wed 5/27/16 Fri 5/27/17 Sat Problem 11 The length of the interval of solutions of the inequality is . What is ? SolutionSince we are given the range of the solutions, we must re-write the inequalities so that we have in terms of and . Subtract from all of the quantit
19、ies: Divide all of the quantities by . Since we have the range of the solutions, we can make them equal to . Multiply both sides by 2. Re-write without using parentheses. Simplify. We need to find for the problem, so the answer is Problem 12 Logan is constructing a scaled model of his town. The city
20、s water tower stands 40 meters high, and the top portion is a sphere that holds 100,000 liters of water. Logans miniature water tower holds 0.1 liters. How tall, in meters, should Logan make his tower? Solution The water tower holds times more water than Logans miniature. Therefore, Logan should mak
21、e his tower times shorter than the actual tower. This is meters high, or choice . Problem 13 Angelina drove at an average rate of kph and then stopped minutes for gas. After the stop, she drove at an average rate of kph. Altogether she drove km in a total trip time of hours including the stop. Which
22、 equation could be used to solve for the time in hours that she drove before her stop? SolutionThe answer is （ ）because she drove at kmh for hours (the amount of time before the stop), and 100 kmh for because she wasnt driving for minutes, or hours. Multiplying by gives the total distance, which is
23、kms. Therefore, the answer is Problem 14 Triangle has . Let and be on and , respectively, such that . Let be the intersection of segments and , and suppose that is equilateral. What is ? Solution Let . Since , Problem 15 In a magical swamp there are two species of talking amphibians: toads, whose st
24、atements are always true, and frogs, whose statements are always false. Four amphibians, Brian, Chris, LeRoy, and Mike live together in this swamp, and they make the following statements. Brian: Mike and I are different species. Chris: LeRoy is a frog. LeRoy: Chris is a frog. Mike: Of the four of us
25、, at least two are toads. How many of these amphibians are frogs? Solution Solution 1 We can begin by first looking at Chris and LeRoy. Suppose Chris and LeRoy are the same species. If Chris is a toad, then what he says is true, so LeRoy is a frog. However, if LeRoy is a frog, then he is lying, but
26、clearly Chris is not a frog, and we have a contradiction. The same applies if Chris is a frog. Clearly, Chris and LeRoy are different species, and so we have at least frog out of the two of them. Now suppose Mike is a toad. Then what he says is true because we already have toads. However, if Brian i
27、s a frog, then he is lying, yet his statement is true, a contradiction. If Brian is a toad, then what he says is true, but once again it conflicts with his statement, resulting in contradiction. Therefore, Mike must be a frog. His statement must be false, which means that there is at most toad. Sinc
28、e either Chris or LeRoy is already a toad, Brain must be a frog. We can also verify that his statement is indeed false. Both Mike and Brian are frogs, and one of either Chris or LeRoy is a frog, so we have frogs total. Solution 2 Start with Brian. If he is a toad, he tells the truth, hence Mike is a
29、 frog. If Brian is a frog, he lies, hence Mike is a frog, too. Thus Mike must be a frog. As Mike is a frog, his statement is false, hence there is at most one toad. As there is at most one toad, at least one of Chris and LeRoy is a frog. But then the other one tells the truth, and therefore is a toa
30、d. Hence we must have one toad and three frogs. Problem 16 Nondegenerate has integer side lengths, is an angle bisector, , and . What is the smallest possible value of the perimeter? Solution By the Angle Bisector Theorem, we know that . If we use the lowest possible integer values for AB and BC (th
31、e measures of AD and DC, respectively), then , contradicting the Triangle Inequality. If we use the next lowest values ( and ), the Triangle Inequality is satisfied. Therefore, our answer is , or choice . Problem 17 A solid cube has side length inches. A -inch by -inch square hole is cut into the ce
32、nter of each face. The edges of each cut are parallel to the edges of the cube, and each hole goes all the way through the cube. What is the volume, in cubic inches, of the remaining solid? Solution Solution 1 Imagine making the cuts one at a time. The first cut removes a box . The second cut remove
33、s two boxes, each of dimensions , and the third cut does the same as the second cut, on the last two faces. Hence the total volume of all cuts is . Therefore the volume of the rest of the cube is . Solution 2 We can use Principle of Inclusion-Exclusion to find the final volume of the cube. There are
34、 3 cuts through the cube that go from one end to the other. Each of these cuts has cubic inches. However, we can not just sum their volumes, as the central cube is included in each of these three cuts. To get the correct result, we can take the sum of the volumes of the three cuts, and subtract the
35、volume of the central cube twice. Hence the total volume of the cuts is . Therefore the volume of the rest of the cube is . Solution 3 We can visualize the final figure and see a cubic frame. We can find the volume of the figure by adding up the volumes of the edges and corners. Each edge can be see
36、n as a box, and each corner can be seen as a box. . Problem 18 Bernardo randomly picks 3 distinct numbers from the set and arranges them in descending order to form a 3-digit number. Silvia randomly picks 3 distinct numbers from the set and also arranges them in descending order to form a 3-digit nu
37、mber. What is the probability that Bernardos number is larger than Silvias number? Solution We can solve this by breaking the problem down into cases and adding up the probabilities. Case : Bernardo picks . If Bernardo picks a then it is guaranteed that his number will be larger than Silvias. The pr
38、obability that he will pick a is . Case : Bernardo does not pick . Since the chance of Bernardo picking is , the probability of not picking is . If Bernardo does not pick 9, then he can pick any number from to . Since Bernardo is picking from the same set of numbers as Silvia, the probability that B
39、ernardos number is larger is equal to the probability that Silvias number is larger. Ignoring the for now, the probability that they will pick the same number is the number of ways to pick Bernardos 3 numbers divided by the number of ways to pick any 3 numbers. We get this probability to be Probabil
40、ity of Bernardos number being greater is Factoring the fact that Bernardo couldve picked a but didnt: Adding up the two cases we get Problem 19 Equiangular hexagon has side lengths and . The area of is of the area of the hexagon. What is the sum of all possible values of ? Solution Solution 1It is c
41、lear that is an equilateral triangle. From the Law of Cosines, we get that . Therefore, the area of is . If we extend , and so that and meet at , and meet at , and and meet at , we find that hexagon is formed by taking equilateral triangle of side length and removing three equilateral triangles, , a
42、nd , of side length . The area of is therefore . Based on the initial conditions, Simplifying this gives us . By Vietas Formulas we know that the sum of the possible value of is . Solution 2As above, we find that the area of is . We also find by the sine triangle area formula that , and thus This si
43、mplifies to . Problem 20 A fly trapped inside a cubical box with side length meter decides to relieve its boredom by visiting each corner of the box. It will begin and end in the same corner and visit each of the other corners exactly once. To get from a corner to any other corner, it will either fl
44、y or crawl in a straight line. What is the maximum possible length, in meters, of its path? SolutionThe distance of an interior diagonal in this cube is and the distance of a diagonal on one of the square faces is . It would not make sense if the fly traveled an interior diagonal twice in a row, as
45、it would return to the point it just came from, so at most the final sum can only have 4 as the coefficient of . The other 4 paths taken can be across a diagonal on one of the faces, so the maximum distance traveled is . Problem 21 The polynomial has three positive integer zeros. What is the smalles
46、t possible value of ? SolutionBy Vietas Formulas, we know that is the sum of the three roots of the polynomial . Also, 2010 factors into . But, since there are only three roots to the polynomial, two of the four prime factors must be multiplied so that we are left with three roots. To minimize , and
47、 should be multiplied, which means will be and the answer is . Problem 22 Eight points are chosen on a circle, and chords are drawn connecting every pair of points. No three chords intersect in a single point inside the circle. How many triangles with all three vertices in the interior of the circle
48、 are created? SolutionTo choose a chord, we know that two points must be chosen. This implies that for three chords to create a triangle and not intersect at a single point, six points need to be chosen. Therefore, the answer is which is equivalent to 28, Problem 23 Each of 2010 boxes in a line contains a single red marble, and for , the box in the position also contains white marbles. Isabella begins at the first box and successively draws a single marble at random f