2009 AMC 10A 試題及答案解析

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1、2009 AMC 10A 1 、One can holds ounces of soda. What is the minimum number of cans needed to provide a gallon (128 ounces) of soda? Solution cans would hold ounces, but , so cans are required. Thus, the answer is . 2 、Four coins are picked out of a piggy bank that contains a collec

2、tion of pennies, nickels, dimes and quarters. Which of the following could not be the total value of the four coins, in cents? Solution As all five options are divisible by , we may not use any pennies. (This is because a penny is the only coin that is not divisible by , and if we u

3、sed between and pennies, the sum would not be divisible by .) Hence the smallest coin we can use is a nickel, and thus the smallest amount we can get is . Therefore the option that is not reachable is . We can verify that we can indeed get the other ones: § § § § 3 、Which of the following

4、is equal to ? Solution We compute: 4 、Eric plans to compete in a triathlon. He can average miles per hour in the -mile swim and miles per hour in the -mile run. His goal is to finish the triathlon in hours. To accomplish his goal what must his average speed in miles per hour, be

5、 for the -mile bicycle ride? Solution Since , Eric takes hours for the swim. Then, he takes hours for the run. So he needs to take hours for the mile run. This is 5 、What is the sum of the digits of the square of ? Solution Using the standard multiplication algorithm, whos

6、e digit sum is Or Add up all the ones(thus deriving the sum of the number) of gives us Thus, 6 、A circle of radius is inscribed in a semicircle, as shown. The area inside the semicircle but outside the circle is shaded. What fraction of the semicircle's area is shaded? Solution

7、 Area of the circle inscribed inside the semicircle Area of the larger circle (semicircle's area x 2) (4, or the diameter of the inscribed circle is the same thing as the radius of the semicircle). Thus, the area of the semicircle is Part of the semicircle that is unshaded is Therefore, the shad

8、ed part is Thus the answer is 7 、A carton contains milk that is % fat, an amount that is % less fat than the amount contained in a carton of whole milk. What is the percentage of fat in whole milk? Solution Rewording the question, basically we are being asked "2 is 40% less than

9、 what number?" If x represents the number we are looking for, then 40% less than the number would be represented by or . Thus ; solving for x, we get , 8 、Three Generations of the Wen family are going to the movies, two from each generation. The two members of the youngest generation receive a % d

10、iscount as children. The two members of the oldest generation receive a discount as senior citizens. The two members of the middle generation receive no discount. Grandfather Wen, whose senior ticket costs $, is paying for everyone. How many dollars must he pay? Solution A senior ti

11、cket costs $, so a regular ticket costs dollars. Therefore children's tickets cost half that, or $, so we have: So Grandfather Wen pays $, or . 9 、Positive integers , , and , with , form a geometric sequence with an integer ratio. What is ? Solution The prime factorization of

12、is . As , the ratio must be positive and larger than , hence there is only one possibility: the ratio must be , and then , and . 10 、Triangle has a right angle at . Point is the foot of the altitude from , , and . What is the area of ? Solution It is a well-known fact that in any

13、 right triangle with the right angle at and the foot of the altitude from onto we have . (See below for a proof.) Then , and the area of the triangle is . Proof: Consider the Pythagorean theorem for each of the triangles , , and . We get: 1. . 2. 3. Substituting equations 2 and 3 into the le

14、ft hand side of equation 1, we get . Alternatively, note that . 11 、One dimension of a cube is increased by , another is decreased by , and the third is left unchanged. The volume of the new rectangular solid is less than that of the cube. What was the volume of the cube? Solution

15、 Let the original cube have edge length . Then its volume is . The new box has dimensions , , and , hence its volume is . The difference between the two volumes is . As we are given that the difference is , we have , and the volume of the original cube was . 12 、In quadrilateral , , , , , and is

16、an integer. What is ? Solution By the triangle inequality we have , and also , hence . We got that , and as we know that is an integer, we must have . 13 、Suppose that and . Which of the following is equal to for every pair of integers ? Solution We have . 1

17、4 、Four congruent rectangles are placed as shown. The area of the outer square is times that of the inner square. What is the ratio of the length of the longer side of each rectangle to the length of its shorter side? Solution The area of the outer square is times that of the inne

18、r square. Therefore the side of the outer square is times that of the inner square. Then the shorter side of the rectangle is of the side of the outer square, and the longer side of the rectangle is of the side of the outer square, hence their ratio is . 15 、The figures , , , and shown are the f

19、irst in a sequence of figures. For , is constructed from by surrounding it with a square and placing one more diamond on each side of the new square than had on each side of its outside square. For example, figure has diamonds. How many diamonds are there in figure ? Solution Solu

20、tion 1 Color the diamond layers alternately blue and red, starting from the outside. You'll get the following pattern: In the figure , the blue diamonds form a square, and the red diamonds form a square. Hence the total number of diamonds in is . Solution 2 When constructing from , we add

21、new diamonds. Let be the number of diamonds in . We now know that and . Hence we get: 16 、Let , , , and be real numbers with , , and . What is the sum of all possible values of ? Solution Solution 1 From we get that Similarly, and . Substitution gives . This gives . There

22、are possibilities for the value of : , , , , , , , Therefore, the only possible values of are 9, 5, 3, and 1. Their sum is . Solution 2 If we add the same constant to all of , , , and , we will not change any of the differences. Hence we can assume that . From we get that , h

23、ence . If we multiply all four numbers by , we will not change any of the differences. Hence we can assume that . From we get that . From we get that . Hence , and the sum of possible values is . 17 、Rectangle has and . Segment is constructed through so that is perpendicular to , and and l

24、ie on and , respectively. What is ? Solution The situation is shown in the picture below. Obviously, from the Pythagorean theorem we have . Triangle is similar to , as they have the same angles. Hence , and therefore . Also triangle is similar to . Hence , and therefore .

25、We then have . 18 、At Jefferson Summer Camp, of the children play soccer, of the children swim, and of the soccer players swim. To the nearest whole percent, what percent of the non-swimmers play soccer? Solution Out of the soccer players, swim. As the soccer players are of the wh

26、ole, the swimming soccer players are of all children. The non-swimming soccer players then form of all the children. Out of all the children, swim. We know that of all the children swim and play soccer, hence of all the children swim and don't play soccer. Finally, we know that of all the chil

27、dren are non-swimmers. And as of all the children do not swim but play soccer, of all the children do not engage in any activity. A quick summary of what we found out: § : swimming yes, soccer yes § : swimming no, soccer yes § : swimming yes, soccer no § : swimming no, soccer no Now we c

28、an compute the answer. Out of all children, are non-swimmers, and again out of all children are non-swimmers that play soccer. Hence the part of non-swimmers that plays soccer is . 19 、Circle has radius . Circle has an integer radius and remains internally tangent to circle as it rolls once around

29、 the circumference of circle . The two circles have the same points of tangency at the beginning and end of circle 's trip. How many possible values can have? Solution The circumference of circle A is 200, and the circumference of circle B with radius is . Since circle B makes a com

30、plete revolution and ends up on the same point, the circumference of A must be a perfect factor of the circumference of B, therefore the quotient must be an integer. R must then be a factor of 100, excluding 100 (because then circle B would be the same size as circle A). . Therefore 100 has fact

31、ors*. But you need to subtract 1 from 9, in order to exclude 100. Therefore the answer is . *The number of factors of and so on, is . 20、Andrea and Lauren are kilometers apart. They bike toward one another with Andrea traveling three times as fast as Lauren, and the distance between them decreas

32、ing at a rate of kilometer per minute. After minutes, Andrea stops biking because of a flat tire and waits for Lauren. After how many minutes from the time they started to bike does Lauren reach Andrea? Solution Let their speeds in kilometers per hour be and . We know that and that

33、. (The second equation follows from the fact that .) This solves to and . As the distance decreases at a rate of kilometer per minute, after minutes the distance between them will be kilometers. From this point on, only Lauren will be riding her bike. As there are kilometers remaining and , she

34、will need exactly an hour to get to Andrea. Therefore the total time in minutes is . 21 、Many Gothic cathedrals have windows with portions containing a ring of congruent circles that are circumscribed by a larger circle, In the figure shown, the number of smaller circles is four. What is the ratio

35、 of the sum of the areas of the four smaller circles to the area of the larger circle? Solution Draw some of the radii of the small circles as in the picture below. Out of symmetry, the quadrilateral in the center must be a square. Its side is obviously , and therefore its diago

36、nal is . We can now compute the length of the vertical diameter of the large circle as . Hence , and thus . Then the area of the large circle is . The area of four small circles is . Hence their ratio is: 22、Two cubical dice each have removable numbers through . The twelve numbers on the two d

37、ice are removed, put into a bag, then drawn one at a time and randomly reattached to the faces of the cubes, one number to each face. The dice are then rolled and the numbers on the two top faces are added. What is the probability that the sum is ? Solution Solution 1 At the momen

38、t when the numbers are in the bag, imagine that each of them has a different color. Clearly the situation is symmetric at this moment. Hence after we draw them, attach them and throw the dice, the probability of getting some pair of colors is the same for any two colors. There are ways how to pick

39、 two of the colors. We now have to count the ways where the two chosen numbers will have sum . Sum can be obtained as , , or . Each number in the bag has two different colors, hence each of these three options corresponds to four pairs of colors. Out of the pairs of colors we can get when throwi

40、ng the dice, will give us the sum . Hence the probability that this will happen is . Solution 2 Ignoring the numbers that do not affect the probability of the desired outcome (the ones that are not on top of the dice), say that the number on top of the first die is . For the sum of the numbers t

41、o be , the second die must have the number on top. There are remaining numbers that could be on top of the second die, of which are (since in all cases). Thus, the probability of the sum of the numbers being is , so the answer is . 23 、Convex quadrilateral has and . Diagonals and intersect at ,

42、, and and have equal areas. What is ? Solution Solution 1 Let denote the area of triangle . , so . Since triangles and share a base, they also have the same height and thus and with a ratio of . . Solution 2 Using the sine area formula on triangles and , as , we see that

43、 Since , triangles and are similar. Their ratio is . Since , we must have , . 24 、Three distinct vertices of a cube are chosen at random. What is the probability that the plane determined by these three vertices contains points inside the cube? Solution Solution 1 We will try to

44、 use symmetry as much as possible. Pick the first vertex , its choice clearly does not influence anything. Pick the second vertex . With probability vertices and have a common edge, with probability they are in opposite corners of the same face, and with probability they are in opposite corners

45、of the cube. We will handle each of the cases separately. In the first case, there are faces that contain the edge . In each of these faces there are other vertices. If one of these vertices is the third vertex , the entire triangle will be on a face. On the other hand, if is one of the two remain

46、ing vertices, the triangle will contain points inside the cube. Hence in this case the probability of choosing a good is . In the second case, the triangle will not intersect the cube iff point is one of the two points on the side that contains . Hence the probability of intersecting the inside of

47、 the cube is . In the third case, already the diagonal contains points inside the cube, hence this case will be good regardless of the choice of . Summing up all cases, the resulting probability is: Solution 2 There are ways to pick three vertices from eight total vertices; this is our denom

48、inator. In order to have three points inside the cube, they cannot be on the surface. Thus, we can use complementary probability. There are four ways to choose three points from the vertices of a single face. Since there are six faces, . Thus, the probability of what we don't want is . Using com

49、plementary, 25 、For , let , where there are zeros between the and the . Let be the number of factors of in the prime factorization of . What is the maximum value of ? Solution The number can be written as . For we have . The first value in the parentheses is odd, the second on

50、e is even, hence their sum is odd and we have . For we have . For the value in the parentheses is odd, hence . This leaves the case . We have . The value is obviously even. And as , we have , and therefore . Hence the largest power of that divides is , and this gives us the desired maximum of the function : . 2009 AMC 10A Answer Key 1. E 2. A 3. C 4. A 5. E 6. A 7. C 8. B 9. B 10.B 11.D 12.C 13.E 14.A 15.E 16.D 17.C 18.D 19.B 20.D 21.C 22.D 23.E 24.C 25.B - 18 -