AMC美國數(shù)學(xué)競賽2003AMC10A試題及答案解析

1、如果您需要使用本文檔，請(qǐng)點(diǎn)擊下載按鈕下載!2003 amc 10a1、what is the difference between the sum of the first even counting numbers and the sum of the first odd counting numbers? solution the first even counting numbers are . the first odd counting numbers are . thus, the problem is asking for the value of . . alternative

2、ly, using the sum of an arithmetic progression formula, we can write . 2、members of the rockham soccer league buy socks and t-shirts. socks cost per pair and each t-shirt costs more than a pair of socks. each member needs one pair of socks and a shirt for home games and another pair of socks and a s

3、hirt for away games. if the total cost is , how many members are in the league? solution since t-shirts cost dollars more than a pair of socks, t-shirts cost dollars. since each member needs pairs of socks and t-shirts, the total cost for member is dollars. since dollars was the cost for the club, a

4、nd was the cost per member, the number of members in the league is . - 1 - / 23- 1 -如果您需要使用本文檔，請(qǐng)點(diǎn)擊下載按鈕下載!3、a solid box is cm by cm by cm. a new solid is formed by removing a cube cm on a side from each corner of this box. what percent of the original volume is removed? solution the volume of the ori

5、ginal box is the volume of each cube that is removed is since there are corners on the box, cubes are removed. so the total volume removed is . therefore, the desired percentage is . 4、it takes mary minutes to walk uphill km from her home to school, but it takes her only minutes to walk from school

6、to home along the same route. what is her average speed, in km/hr, for the round trip? solution since she walked km to school and km back home, her total distance is km. since she spent minutes walking to school and minutes walking back home, her total time is minutes = hours. therefore her average

7、speed in km/hr is 5、let and denote the solutions of . what is the value of ? solution using factoring: - 2 - / 23- 2 -如果您需要使用本文檔，請(qǐng)點(diǎn)擊下載按鈕下載!or so and are and . therefore the answer is or we can use sum and product. 6、define to be for all real numbers and . which of the following statements is not tru

8、e? solution examining statement c: when , but statement c says that it does for all . therefore the statement that is not true is " for all " alternatively, consider that the given "heart function" is actually the definition of the distance between two points. examining all of th

9、e statements, only c is not necessarily true; if c is negative, the distance between and is the absolute value of , not itself, because distance is always nonnegative. 7、how many non-congruent triangles with perimeter have integer side lengths? solution by the triangle inequality, no one side may ha

10、ve a length greater than half the perimeter, which is - 3 - / 23- 3 -如果您需要使用本文檔，請(qǐng)點(diǎn)擊下載按鈕下載!since all sides must be integers, the largest possible length of a side is therefore, all such triangles must have all sides of length , , or . since , at least one side must have a length of thus, the remainin

11、g two sides have a combined length of . so, the remaining sides must be either and or and . therefore, the number of triangles is . 8、what is the probability that a randomly drawn positive factor of is less than ? solution for a positive number which is not a perfect square, exactly half of the posi

12、tive factors will be less than . since is not a perfect square, half of the positive factors of will be less than . clearly, there are no positive factors of between and . therefore half of the positive factors will be less than . so the answer is . 9、simplify solution . therefore: - 4 - / 23- 4 -如果

13、您需要使用本文檔，請(qǐng)點(diǎn)擊下載按鈕下載!10、the polygon enclosed by the solid lines in the gure consists of congruent squares joined edge-to-edge. one more congruent square is attached to an edge at one of the nine positions indicated. how many of the nine resulting polygons can be folded to form a cube with one face mis

14、sing? solution let the squares be labeled , , , and . when the polygon is folded, the "right" edge of square becomes adjacent to the "bottom edge" of square , and the "bottom" edge of square becomes adjacent to the "bottom" edge of square . so, any "new&q

15、uot; square that is attached to those edges will prevent the polygon from becoming a cube with one face missing. therefore, squares , , and will prevent the polygon from becoming a cube with one face missing. squares , , , , , and will allow the polygon to become a cube with one face missing when fo

16、lded. thus the answer is . another way to think of it is that a cube missing one face has of it's faces. since the shape has faces already, we need another face. the only way to add anopther face is if the added square does not overlap any of the others. - 6 - / 23- 6 -如果您需要使用本文檔，請(qǐng)點(diǎn)擊下載按鈕下載!, and

17、 overlap, while 9 do not. the answer is 11、the sum of the two -digit numbers and is . what is ? solution since , , and are digits, , , . therefore, . 12、a point is randomly picked from inside the rectangle with vertices , , , and . what is the probability that ? solution the rectangle has a width of

18、 and a height of . the area of this rectangle is . the line intersects the rectangle at and . the area which is the right isosceles triangle with side length that has vertices at , , and . the area of this triangle is therefore, the probability that is - 6 - / 23- 6 -如果您需要使用本文檔，請(qǐng)點(diǎn)擊下載按鈕下載!13、the sum

19、of three numbers is . the rst is times the sum of the other two. the second is seven times the third. what is the product of all three? solution solution 1 let the numbers be , , and in that order. the given tells us that therefore, the product of all three numbers is . solution 2 alternatively, we

20、can set up the system in matrix form: or, in matrix form to solve this matrix equation, we can rearrange it thus: solving this matrix equation by using inverse matrices and matrix multiplication yields - 7 - / 23- 7 -如果您需要使用本文檔，請(qǐng)點(diǎn)擊下載按鈕下載!which means that , , and . therefore, 14、let be the largest in

21、teger that is the product of exactly distinct prime numbers, , , and , where and are single digits. what is the sum of the digits of ? solution since is a single digit prime number, the set of possible values of is . since is a single digit prime number and is the units digit of the prime number , t

22、he set of possible values of is . using these values for and , the set of possible values of is out of this set, the prime values are therefore the possible values of are: the largest possible value of is . so, the sum of the digits of is 15、what is the probability that an integer in the set is divi

23、sible by and not divisible by ? solution there are integers in the set. - 8 - / 23- 8 -如果您需要使用本文檔，請(qǐng)點(diǎn)擊下載按鈕下載!since every 2nd integer is divisible by , there are integers divisible by in the set. to be divisible by both and , a number must be divisible by . since every 6th integer is divisible by , th

24、ere are integers divisible by both and in the set. so there are integers in this set that are divisible by and not divisible by . therefore, the desired probability is 16、what is the units digit of ? solution since : therefore, the units digit is 17、the number of inches in the perimeter of an equila

25、teral triangle equals the number of square inches in the area of its circumscribed circle. what is the radius, in inches, of the circle? solution let be the length of a side of the equilateral triangle and let be the radius of the circle. in a circle with a radius the side of an inscribed equilatera

26、l triangle is . so . the perimeter of the triangle is the area of the circle is so: - 9 - / 23- 9 -如果您需要使用本文檔，請(qǐng)點(diǎn)擊下載按鈕下載!18、what is the sum of the reciprocals of the roots of the equation solution multiplying both sides by : let the roots be and . the problem is asking for by vieta's formulas: so

27、 the answer is . 19、a semicircle of diameter sits at the top of a semicircle of diameter , as shown. the shaded area inside the smaller semicircle and outside the larger semicircle is called a lune. determine the area of this lune. - 10 - / 23- 10 -如果您需要使用本文檔，請(qǐng)點(diǎn)擊下載按鈕下載!solution the shaded area is eq

28、ual to the area of the smaller semicircle minus the area of a sector of the larger circle plus the area of a triangle formed by two radii of the larger semicircle and the diameter of the smaller semicircle. the area of the smaller semicircle is . since the radius of the larger semicircle is equal to

29、 the diameter of the smaller semicircle, the triangle is an equilateral triangle and the sector measures . the area of the sector of the larger semicircle is . the area of the triangle is so the shaded area is 20、a base- three-digit number is selected at random. which of the following is closest to

30、the probability that the base- representation and the base- representation of are both three-digit numerals? solution to be a three digit number in base-10: thus there are three-digit numbers in base-10 - 11 - / 23- 11 -如果您需要使用本文檔，請(qǐng)點(diǎn)擊下載按鈕下載!to be a three-digit number in base-9: to be a three-digit n

31、umber in base-11: so, thus, there are base-10 three-digit numbers that are three digit numbers in base-9 and base-11. therefore the desired probability is . 21、pat is to select six cookies from a tray containing only chocolate chip, oatmeal, and peanut butter cookies. there are at least six of each

32、of these three kinds of cookies on the tray. how many different assortments of six cookies can be selected? solution solution 1 let the ordered triplet represent the assortment of chocolate chip cookies, oatmeal cookies, and peanut butter cookies. using casework: pat selects chocolate chip cookies:

33、pat needs to select more cookies that are either oatmeal or peanut butter. the assortments are: assortments. pat selects chocolate chip cookie: - 12 - / 23- 12 -如果您需要使用本文檔，請(qǐng)點(diǎn)擊下載按鈕下載!pat needs to select more cookies that are either oatmeal or peanut butter. the assortments are: assortments. pat selec

34、ts chocolate chip cookies: pat needs to select more cookies that are either oatmeal or peanut butter. the assortments are: assortments. pat selects chocolate chip cookies: pat needs to select more cookies that are either oatmeal or peanut butter. the assortments are: assortments. pat selects chocola

35、te chip cookies: pat needs to select more cookies that are either oatmeal or peanut butter. the assortments are: assortments. pat selects chocolate chip cookies: pat needs to select more cookies that are either oatmeal or peanut butter. the assortments are: assortments. pat selects chocolate chip co

36、okies: pat needs to select more cookies that are either oatmeal or peanut butter. the only assortment is: assortment. the total number of assortments of cookies that can be collected is solution 2 it is given that it is possible to select at least 6 of each. therefore, we can make a bijection to the

37、 number of ways to divide the six choices into three categories, since it is assumed that their order is unimportant. using the ball and urns formula, the number of ways to do this is - 14 - / 23- 14 -如果您需要使用本文檔，請(qǐng)點(diǎn)擊下載按鈕下載!22、in rectangle , we have , , is on with , is on with , line intersects line a

38、t , and is on line with . find the length . solution s olution 1 (opposite angles are equal). (both are 90 degrees). (alt. interior angles are congruent). therefore and are similar. and are also similar. is 9, therefore must equal 5. similarly, must equal 3. because and are similar, the ratio of and

39、 , must also hold true for and . , so is of . by pythagorean theorem, . - 14 - / 23- 14 -如果您需要使用本文檔，請(qǐng)點(diǎn)擊下載按鈕下載!. so . . therefore . solution 2 since is a rectangle, . since is a rectangle and , . since is a rectangle, . so, is a transversal, and . this is sufficient to prove that and . using ratios:

40、since can't have 2 different lengths, both expressions for must be equal. - 15 - / 23- 15 -如果您需要使用本文檔，請(qǐng)點(diǎn)擊下載按鈕下載!solution 3 since is a rectangle, , , and . from the pythagorean theorem, . lemma statement: proof: , obviously. since two angles of the triangles are equal, the third angles must equal

41、 each other. therefore, the triangles are similar. let . also, , therefore we can multiply both sides by to get that is twice of 10, or - 16 - / 23- 16 -如果您需要使用本文檔，請(qǐng)點(diǎn)擊下載按鈕下載!solution 4 we extend bc such that it intersects gf at x. since abcd is a rectangle, it follows that cd=8, therefore, xf=8. let

42、 gx=y. from the similarity of triangles gch and gea, we have the ratio 3:5 (as ch=9-6=3, and ea=9-4=5). gx and gf are the altitudes of gch and gea, respectively. thus, y:y+8 = 3:5, from which we have y=12, thus gf=y+8=12+8=20. b. 23、a large equilateral triangle is constructed by using toothpicks to

43、create rows of small equilateral triangles. for example, in the gure we have rows of small congruent equilateral triangles, with small triangles in the base row. how many toothpicks would be needed to construct a large equilateral triangle if the base row of the triangle consists of small equilatera

44、l triangles? solution solution 1there are small equilateral triangles. each small equilateral triangle needs toothpicks to make it. but, each toothpick that isn't one of the toothpicks on the outside of the large equilateral triangle is a side for small equilateral triangles. so, the number of t

45、oothpicks on the inside of the large equilateral triangle is therefore the total number of toothpicks is solution 2- 17 - / 23- 17 -如果您需要使用本文檔，請(qǐng)點(diǎn)擊下載按鈕下載!we see that the bottom row of small triangles is formed from downward-facing triangles and upward-facing triangles. since each downward-facing tria

46、ngle uses three distinct toothpicks, and since the total number of downward-facing triangles is , we have that the total number of toothpicks is 24、sally has ve red cards numbered through and four blue cards numbered through . she stacks the cards so that the colors alternate and so that the number

47、on each red card divides evenly into the number on each neighboring blue card. what is the sum of the numbers on the middle three cards? solution let and designate the red card numbered and the blue card numbered , respectively. is the only blue card that evenly divides, so must be at one end of the

48、 stack and must be the card next to it. is the only other red card that evenly divides , so must be the other card next to . is the only blue card that evenly divides, so must be at the other end of the stack and must be the card next to it. is the only other red card that evenly divides , so must b

49、e the other card next to . doesn't evenly divide , so must be next to , must be next to , and must be in the middle. this yields the following arrangement from top to bottom: therefore, the sum of the numbers on the middle three cards is . 25、let be a -digit number, and let and be the quotient and remainder, respectively, when is divided by . for how many values of is divisible by ? solution - 18 - / 23- 18 -如果您需要使用本文檔，請(qǐng)點(diǎn)擊下載按鈕下載!solution 1 when a -digit number is divided by , the first digits become the quotient, , and the last digits become the remainder, . therefore, can be any i