2005 AMC 12A Problems and Solution

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2005 AMC 10A Problems

2005 AMC 10A ProblemsProblem 1While eating out, Mike and Joe each tipped their server 2 dollars. Mike tipped 10% of his bill and Joe tipped 20% of his bill. What was the difference, in dollars between their bills?（A ）2 （B ）4 （C ）5 （D ）10 （E ）20Problem 2For each pair of real numbers b a ≠, define the operation as.What is the value of? （A ）32- （B ）51- （C ）0 （D ）21 （E ）This value is not defined. Problem 3The equations 372=+x and 210-=-bx have the same solution. What is the value of ?（A ）-8 （B ）-4 （C ）2 （D ）4 （E ）8Problem 4A rectangle with a diagonal of length is twice as long as it is wide. What is the area ofthe rectangle?（A ）241x （B ）252x （C ）221x （D ）2x （E ）223x Problem 5A store normally sells windows at <dollar/>100 each. This week the store is offering one free window for each purchase of four. Dave needs seven windows and Doug needs eight windows. How many dollars will they save if they purchase the windows together rather than separately?（A ）100 （B ）200 （C ）300 （D ）400 （E ）500Problem 6The average (mean) of 20 numbers is 30, and the average of 30 other numbers is 20. What is the average of all 50 numbers?（A ）23 （B ）24 （C ）25 （D ）26 （E ）27Problem 7Josh and Mike live 13 miles apart. Yesterday Josh started to ride his bicycle toward Mike's house. A little later Mike started to ride his bicycle toward Josh's house. When they met, Josh had ridden for twice the length of time as Mike and at four-fifths of Mike's rate. How many miles had Mike ridden when they met?（A ）4 （B ）5 （C ）6 （D ）7 （E ）8Problem 8In the figure, the length of side AB of square ABCD is50 and BE =1. What is the areaof the inner square EFGH ?（A ）25 （B ）32 （C ）36 （D ）40 （E ）42Problem 9Three tiles are marked X and two other tiles are marked O . The five tiles are randomly arranged in a row. What is the probability that the arrangement reads XOXOX ?（A ）121 （B ）101 （C ）61 （D ）41 （E ）31 Problem 10 There are two values of for which the equation 09842=+++x ax x has only onesolution for . What is the sum of those values of ?（A ）-16 （B ）-8 （C ）0 （D ）8 （E ）20Problem 11A wooden cube units on a side is painted red on all six faces and then cut into 3n unitcubes. Exactly one-fourth of the total number of faces of the unit cubes are red. What is ?（A ）3 （B ）4 （C ）5 （D ）6 （E ）7 Problem 12The figure shown is called a trefoil and is constructed by drawing circular sectors about the sides of the congruent equilateral triangles. What is the area of a trefoil whose horizontal base has length 2?（A ）2331+π （B ）π32 （C ）4332+π （D ）3332+π （E ）2332+π Problem 13How many positive integers satisfy the following condition: 200100502)130(>>n n ?（A ）0 （B ）7 （C ）12 （D ）65 （E ）125Problem 14How many three-digit numbers satisfy the property that the middle digit is the average of the first and the last digits?（A ）41 （B ）42 （C ）43 （D ）44 （E ）45Problem 15How many positive cubes divide 3!·5!·7! ?（A ）2 （B ）3 （C ）4 （D ）5 （E ）6Problem 16The sum of the digits of a two-digit number is subtracted from the number. The units digit of the result is 6. How many two-digit numbers have this property?（A ）5 （B ）7 （C ）9 （D ）10 （E ）19Problem 17In the five-sided star shown, the letters A, B, C, D, and E are replaced by the numbers 3, 5, 6, 7 and 9, although not necessarily in this order. The sums of the numbers at the ends of the line segments AB, BC, CD, DE, and EA form an arithmetic sequence, although not necessarily in this order. What is the middle term of the sequence?（A ）9 （B ）10 （C ）11 （D ）12 （E ）13Problem 18Team A and team B play a series. The first team to win three games wins the series. Each team is equally likely to win each game, there are no ties, and the outcomes of the individual games are independent. If team B wins the second game and team A wins the series, what is the probability that team B wins the first game?（A ）51 （B ）41 （C ）31 （D ）21 （E ）32 Problem 19Three one-inch squares are placed with their bases on a line. The center square is lifted out and rotated 45 degrees, as shown. Then it is centered and lowered into its original location until it touches both of the adjoining squares. How many inches is the point from the line on which the bases of the original squares were placed?（A ）1 （B ）2 （C ）23 （D ）212 （E ）2 Problem 20 An equiangular octagon has four sides of length 1 and four sides of length 22, arrangedso that no two consecutive sides have the same length. What is the area of the octagon?（A ）27 （B ）227 （C ）2245+ （D ）2254+ （E ）7 Problem 21For how many positive integers does evenly divide ?（A ）3 （B ）5 （C ）7 （D ）9 （E ）11 Problem 22Let S be the set of the 2005 smallest positive multiples of 4, and let T be the set of the 2005 smallest positive multiples of 6. How many elements are common to S and T ?（A ）166 （B ）333 （C ）500 （D ）668 （E ）1001 Problem 23Let AB be a diameter of a circle and let C be a point on AB with 2·AC=BC . Let D and E be points on the circle such that DC ⊥AB and DE is a second diameter. What is the ratio of the area of △DCE to the area of △ABD ?（A ）61 （B ）41 （C ）31 （D ）21 （E ）32 Problem 24For each positive integer 1>m , let )(m P denote the greatest prime factor of . Forhow many positive integers is it true that both n n P =)( and48)48(+=+n n P ?（A ）0 （B ）1 （C ）3 （D ）4 （E ）5 Problem 25In ABC we have AB =25, BC =39, and AC =42. Points D and E are on AB and AC respectively, with AD =19 and AE =14. What is the ratio of the area of triangle ADE to the area of the quadrilateral BCED ? （A ）1521266 （B ）7519 （C ）31 （D ）5619 （E ）12005 AMC 10A SolutionsProblem 1Let be Mike's bill and be Joe's bill. 210010=m , so m =20; 210020=j ,so j =10So the desired difference is 101020=-=-j mProblem 2Problem 32372-=⇒=+x x , 42102-=⇒-=--b b , Problem 4Let the width of the rectangle be w .Then the length is 2w . Using the Pythagorean Theorem :5)2(222x w w w x =⇒+=, So the area of the rectangle is 2522x w w =∙ Problem 5The store's offer means that every 5th window is free.Dave would get free window. Doug would get free window.This is a total of 2 free windows. T ogether, they would getfree windows.So they get 3-2=1 additional window if they purchase the windows together.Therefore they save 1·100=100 Problem 6Since the average of the first 20 numbers is 30, their sum is 20*30=600.Since the average of 30 other numbers is 20, their sum is 30*20=600.So the sum of all 50 numbers is 600+600=1200Therefore, the average of all 50 numbers is 1200/50=24 Problem 7Let be the distance in miles that Mike rode.Since Josh rode for twice the length of time as Mike and at four-fifths of Mike's rate, he rode m m 58542=∙∙ miles.Since their combined distance was 13 miles, 51358=⇒=+m m m Problem 8(C) We see that side BE , which we know is 1, is also the shorter leg of one of the four right triangles (which are congruent, I'll not prove this). So, AH=1. Then HB=HE+BE=HE +1, and HE is one of the sides of the square whose area we want to find. So:650)1(1222=⇒=++HE HE , So, the area of the square is 3662=.Problem 9There are10!3!2!5= distinct arrangements of three 's and two 's. There is only 1 distinct arrangement that reads xoxoxTherefore the desired probability is 1/10 Problem 10A quadratic equation has exactly one root if and only if it is a perfect square . So set22)(984n mx x ax x +=+++, 22222984n mnx x m x ax x ++=+++Two polynomials are equal only if their coefficients are equal, so we must have242±=⇒=m m 392±=⇒=n n1232228±=∙∙±==+mn a , a =4 or a =-20So the desired sum is (4)+(-20)=-16Alternatively, note that whatever the two values of are, they must lead to equations ofthe form 02=++r qx px and 02=+-r qx px . So the two choices of must makeq a =+81 and q a -=+82 so 160162121-=+⇒=++a a a a Alternate SolutionSince this quadratic must have a double root, the discriminant of the quadratic formula for this quadratic must be 0. Therefore, we must have 144160)9)(4(4)8(22-+⇒=-+a a a . We can use the quadratic formula to solve for its roots (we can ignore the things in the radical sign as they will cancel out due to the sign when added). So we must have216216something something --++- Therefore, we have (-16)(2)/2=-16 Problem 11Since there are 2n little faces on each face of the big wooden cube , there are 26n little faces painted red. Since each unit cube has 6 faces, there are 36n little faces total.Since one-fourth of the little faces are painted red, 3266n n =441=⇒n Problem 12 The area of the trefoil is equal to the area of a small equilateral triangle plus the area of four 60ºsectors with a radius of 2/2=1 minus the area of a small equilateral triangle. This is equivalent to the area of four 60ºsectors with a radius of 1.So the answer is: ππ3213606042=∙∙∙ Problem 13We're given 200100502)130(>>n n , so 502005010050502)130(>>n n (because all termsare positive) and thus 422130>>n n . Solving each part seperatly:4162>⇒>n n , n n n >⇒>1301302, so 1304<<n . Therefore the answer is thenumber of positive integers over the interval (4, 130) which is 125. Problem 14Solution 1If the middle digit is the average of the first and last digits, twice the middle digit must be equal to the sum of the first and last digits.Doing some casework :If the middle digit is 1, possible numbers range from 111 to 210. So there are 2 numbers in this case.If the middle digit is 2, possible numbers range from 123 to 420. So there are 4 numbers in this case.If the middle digit is 3, possible numbers range from 135 to 630. So there are 6 numbers in this case.If the middle digit is 4, possible numbers range from 147 to 840. So there are 8 numbers in this case.If the middle digit is 5, possible numbers range from 159 to 951. So there are 9 numbers in this case.If the middle digit is 6, possible numbers range from 369 to 963. So there are 7 numbers in this case.If the middle digit is 7, possible numbers range from 579 to 975. So there are 5 numbers in this case.If the middle digit is 8, possible numbers range from 789 to 987. So there are 3 numbers in this case.If the middle digit is 9, the only possible number is 999. So there is number in this case.So the total number of three-digit numbers that satisfy the property is2+4+6+8+9+7+5+3+1=45 Solution 2 (much faster and slicker)Alternatively, we could note that the middle digit is uniquely defined by the first and third digits, since it is half of their sum. This also means that the sum of the first and third digits must be even. Since even numbers are formed either by adding two odd numbers or two even numbers, we can split our problem into 2 cases:If both the first digit and the last digit are odd, then we have 1, 3, 5, 7, or 9 as choices for each of these digits, and there are 5·5=25 numbers in this case.If both the first and last digits are even, then we have 2, 4, 6, 8 as our choices for the first digit and 0, 2, 4, 6, 8 for the third digit. There are 4·5=20 numbers here.The total number, then, is 20+25=45Problem 15Solution 13!·5!·7!=(3·2·1) ·(5·4·3·2·1) ·(7·6·5·4·3·2·1)=12487532∙∙∙ Therefore, a perfect cube that divides 3!·5!·7! must be in the form d c b a 7532∙∙∙where c b a ,,, and are nonnegativemultiples of 3 that are less than or equal to 8, 4, 2 and 1, respectively. So:( posibilities)( posibilities)( posibility) ( posibility)So the number of perfect cubes that divide 3!·5!·7! is 3·2·1·1=6 Solution 2If you factor 3!·5!·7! You get 247532∙∙. There are 3 ways for the first factor of acube:302,2, and 62. And the second ways are: 303,3. Answer : Problem 16Let the number be b a +10 where and are the tens and units digits of the number.So a b a b a 9)()10(=+-+ must have a units digit of 6This is only possible if 369=a , so 4=a is the only way this can be true.So the numbers that have this property are 40, 41, 42, 43, 44, 45, 46, 47, 48, 49.Therefore the answer is 10. Problem 17Each corner (a,b,c,d,e) goes to two sides/numbers. (A goes to AE and AB, D goes to DC and DE). The sum of every term is equal to 2(3+5+6+7+9)=60Since the middle term in an arithmetic sequence is the average of all the terms in the sequence, the middle number is 60/5=12 DProblem 18There are at most 5 games played.If team B won the first two games, team A would need to win the next three games. So the only possible order of wins is BBAAA.If team A won the first game, and team B won the second game, the possible order of wins are: ABBAA, ABABA, and ABAAX, where X denotes that the 5th game wasn't played.Since ABAAX is dependent on the outcome of games instead of 5, it is twice as likely to occur and can be treated as two possibilities.Since there is 1 possibility where team B wins the first game and 5 total possibilities, thedesired probability is 1/5.Problem 19Consider the rotated middle square shown in the figure. It will drop until lengthis 1 inch. Then, because DEC is a triangle, 22=EC , and 21=FC . Weknow that BC =2, so the distance from to the line is 2121+=+-FC BCProblem 20The area of the octagon can be divided up into 5 squares with side 22 and 4 right triangles, which are half the area of each of the squares. Therefore, the area of the octagon is equal to the area of 7)21(45=+squares. The area of each squareis 21)22(2=, so the area of 7 squares is 27 .Problem 21If evenly divides , then nn +++...216 is an integer . Since 2)1(...21+=+++n n n we may substitute the RHS in the above fraction . So the problem asks us for how many positive integers is 11262)1(+=+n n n n an integer, or equivalently when 12)1(=+n k for a positive integer .112+n is an integer when n +1 is a factor of 12. The factors of 12 are 1, 2, 3, 4, 6 and 12, so the possible values of are 0, 1, 2, 3, 5,and 11.But 0 isn't a positive integer, so only 1, 2, 3, 5, and 11 are possible values of . Thereforethe number of possible values of is 5Problem 22Since the least common multiple 12)6,4(1=cm , the elements that are common to S and T must be multiples of 12.Since 4·2005=8020 and 6·2005=12030, several multiples of 12 that are in T won't be in S , but all multiples of 12 that are in S will be in T . So we just need to find the number of multiples of 12 that are in S . Since 4·3=12 every 3rd element of S will be a multiple of 12Therefore the answer isProblem 23Solution 1Let us assume that the diameter is of length 1. AC is 31 of diameter and CO is 613121=-. OD is the radius of the circle, so using the Pythagorean theorem height CD of △AOCIs 32)61()21(22=-. This is also the height of the .Area of theis 362326121=∙∙. The height of can be found using the area ofand DO as base.Hence the height of is922121362=∙ . The diameter is the base for both thetrianglesand .Hence, the ratio of the area of to the area of is 313292=Solution 2Since and share a base, the ratio of their areas is the ratio of theiraltitudes. Draw the altitude from C to DE .r OC r OD 31,==. Since ︒=∠=∠90DFC m DCO m ,then . So the ratio of the two altitudes is31==DO OC DC CFProblem 24 If ,)(n n P =then 21p n =, where is a prime number .If 48)48(+=+n n P , then 2248p n =+ , where is a different prime number.So: 4822+=n p , ,21n p = 48))((4812122122=-+⇒=-p p p p p pSince )()(:011121p p p p p ->+> Looking at pairs of divisors of 48, we have several possibilities to solve for and :48)(12=+p p 1)(12=-p p 2/471=p 2/492=p24)(12=+p p 2)(12=-p p 111=p 132=p16)(12=+p p 3)(12=-p p 2/131=p 2/192=p12)(12=+p p 4)(12=-p p 41=p 82=p8)(12=+p p 6)(12=-p p 11=p 72=pThe only solution ),(21p p where both numbers are primes is (11,13).Therefore the number of positive integers that satisfy both statements is 1. Problem 25The area of a triangle is A bc sin 21. Using this formula: A A ADE sin 133sin 141921][=∙∙∙= A A ABC sin 525sin 422521][=∙∙∙= Since the area of BCED is equal to the area of ABC minus the area of ADE , A A A BCED sin 392sin 133sin 525][=-=Therefore, thedesired ratio is sin 392sin 133=A A 2005 AMC 10A Answer Key1. D2. C3. B4. B5. A6.B7.B8.C 8. B 10. A11.B 12. B 13.E 14.E 15. E 16.D 17. D 18.A 19.D 20.A21.B 22.D 23.C 24.B 25.D。

2005年12月A级试题及答案解析

2005年12月A级试题及答案解析2005年12月高等学校英语应用能力A级试题Part I Listening Comprehension (15 minutes)Directions: This part is to test your listening ability. It consists of 3 sections.Section ADirections: This section is to test your ability to understand short dialogues. There are 5 recorded dialogues in it.After each dialogue, there is a recorded question. Both the dialogues and questions will be spokenonly once. When you hear a question, you should decide on the correct answer from the 4 choicesmarked A, B, C and D given in your test paper. Then you should mark the corresponding letter onthe Answer Sheet with a single line through the center.1. A. In a post office. B. In a bank. C. In a big store. D. In a hotel.2. A. He has a lot of experience. B. He appears to be a bit too quiet.C. He is nice-looking.D. He lacks experience.3. A. Prepare a price list. B. Go to the Brown Company.C. Send a parcel by post.D. Mail a price list.4. A. She should repair the telephone herself. B. She can turn to Mike for help.C. He knows nothing about telephone.D. He will call the telephone company.5. A. Seeing a film. B. Attending an evening party.C. Having another try.D. Finding a better theater.Section BDirections: This section is to test your ability to understand short conversations. There are 2 recorded conversations in it. After each conversation, there are some recorded questions. Both theconversations and questions will be spoken two times. When you hear a question, you should decideon the correct answer from the 4 choices marked A, B, C and D given in your test paper. Then youshould mark the corresponding letter on the Answer Sheet with a single line through the center. Conversation 16. A. In the office. B. In a factory.C. At the airport.D. At the railway station.7. A. She was busy working. B. She was having a holiday.C. She wanted to get relaxed.D. She wanted to shop for Christmas.Conversation 28. A. To make an apology. B. To place an order.C. To ask about delivery.D. To cancel an appointment.9. A. Trucks B. Printers C. Phones D. Cameras10. A. For the late payment. B. For the damaged products.C. For the lost goods.D. For the delayed delivery.Section CDirections: This section is to test your ability to comprehend short passages. Y ou will hear a recorded passage.After that you will hear five questions. Both the passage and the questions will be read two times.When you hear a question, you should complete the answer to it with a word or a short phrase (in notmore than 3 words). The questions and incomplete answers are printed in your test paper. Y ou shouldwrite your answer on the Answer Sheet correspondingly. Now the passage will begin.11. When did the family car become common in America?In the early .12. Why do more Americans have to drive to work?Because they have moved out of the to the suburbs.13. Why do people like smaller cars now?Because the cost of gas has .14. How popular are Japanese and German cars in the U.S.?They sell .15. What does the speaker mainly talk about?in the U.S.Part II Structure (15 minutes)Directions: This part is to test your ability to construct grammatically correct sentences.It consists of 2 sections. Section ADirections: There are 10 incomplete statements here. Y ou are required to complete each statement by choosing the appropriate answer from the 4 choices marked A, B, C and D. Y ou should mark the corresponding letter on the Answer Sheet with a single line through the center.16. My boss said that he was badly need of my assistance.A. atB. inD. with17. She didn’t know to express her ideas clearly when she was invited to speak at a meeting.A. whereB. whyC. whatD. how18. The professor, as a splendid speaker, was warmly received by the students.A. knownB. knowingC. to be knownD. having known19. tired of Tom’s all-talk-no-action attitude, Julia decided to do the job all by herself.A. To getB. To have gotC. GettingD. Have got20. He in this company since he graduated from Andong Technical College ten years ago.A. workedB. has been workingC. had workedD. was working21. you have any questions or needs, please contact the manager after 5:00 p.m. on weekdays.A. BecauseB. WhereD. Though22. It cost her a lot of money, but she doesn’t regret a year traveling around the world.A. to have spentB. to spendC. spentD. spending23. After for the job, you will be required to take a language test.A. being interviewedB. being interviewingC. interviewingD. having interviewed24. He is used to flying by air and on no occasion frightened.A. he has ever feltB. he ever feelsC. ever does he feelD. has he ever felt25. It is most desirable that he for the information by himself with a few clicks online.A. searchB. searchedC. has searchedD. will searchSection BDirections: There are also 10 incomplete statements here. Y ou should fill in each blank with the proper form of the word given in the brackets. Write the word or words in the corresponding space on the Answer26. Follow the (settle) of the strike, the train service is now back to normal.27. By the end of this year the factory (produce) 20,000 cell phones.28. Those who are working in the Human Resources Department are (humorous ) than we expected.29. We should read more and see more in order to (wide) our horizons.30. All the members of the club were present when the Chairman (elect) last week.31. When she was very young, my sister already knew where (put) her toys and dolls.32. Some experts suggest that we slow down the (economy) growth in the country.33. He might have been killed in that car accident yesterday if he (take) part in that activity with theteam.34. Doing a second job to earn more money also means you have to pay (addition)income tax.35. (watch) by a crowd of people, Charles felt embarrassed and couldn’t say a word.Part III Reading Comprehension (40 minutes)Directions: This part is to test your reading ability. There are 5 tasks for you to fulfill. Y ou should read the material carefully and do the tasks as you are instructed.T ask 1Directions: After reading the following passage, you will find 5 questions or unfinished statements, marked 36 to40. For each question or statement there are 4 choices marked A, B, C and D. Y ou should make thecorrect choice and mark the corresponding letter on the Answer Sheet with a single line through thecenter.The holiday shopping season got off to a flying start over the Thanksgiving (感恩节) weekend in the U.S. Retailers (零售商) attracted discount-hungry consumers with specials (特价商品) of televisions, DVD players and other gifts. Discounters like Wal-Mart and Target had the greatest success, offering customers early bird specials, while major department stores and clothing chains only met modest sales goals. Online sales were also strong.“It was as expected. It wasn’t a source of great prosperity, but it was a good start,”said Nevill, a man in a US reta il consulting firm. He said, “One thing we often heard was that only things marked down (降价) were selling. Consumers want a bargain, especially around the holiday.”Shoppers did seem to like the many markdowns. A lady in the State of Connecticut said, “I’ve never seen so many sales. I feel like a kid in a candy store.”Nevill estimated that discounters had a 4% gain over last year’s Thanksgiving figure, while sales at department stores and mall based clothing chains were up 2% for the weekend. However, the Thanksgiving weekend is not necessarily a good way to judge how retailers will be successful for the season. Over the past few years, the weekend accounted for less than 10% of sales.Online merchants also had a good weekend. Sales were up 61% last Friday, compared to the day after Thanksgiving a year ago.36. Over the Thanksgiving weekend, people bought a lot more than usual because .A. Wal-Mart and T arget offered birds as giftsB. shopping was the major weekend activityC. people had time to go shopping onlineD. they were allowed special discounts37. Retailers have the experience that consumers seem to .A. feel eager to buy discounted goodsB. be hungry for special kinds of goodsC. like to buy new DVD players as giftsD. be interested in major department stores38. From the context we know that Wal-Mart and Target are the names of two .A. food storesB. large retailersC. online servicesD. clothing chains39. By saying “I feel like a kid in a candy store”, the lady in Connecticut probably means that .A. she has difficulty choosing form such a wide variety of goodsB. she cannot decide whether she should buy some candyC. she is quite interested in buying candyD. she recalls her happy childhood days40. Which of the following statements is TRUE according to the passage?A. Online sales were 61% higher than retailers’.B. Online merchants had an average weekend sale.C. Retailers lost much profit because of the discounts they offered.D. Retailers had an increase in sales in the Thanksgiving weekend.T ask 2Directions: This task is the same as Task 1. The 5 questions or unfinished statements are numbered 41 to 45.The Future of General Motors (GM) Is Y oursWe have everything college students need to know about GM’s Cooperative Education and Inter n (实习) Programs in our Student Center. Each fall, GM recruiters (招聘人员) visit the campuses of many of the nation’s top engineering and business colleges and universities to recruit students. These students are considered for interesting assignments throughout our U.S. operations. Check out our Recruiting Calendar to see if GM will be visiting your campus. If your campus is not listed, please apply online.Will GM start your career moving? Fasten your seat belt!For full-time college students, General Motors offers both a Cooperative Education and an Intern Program. Participants in these real-business-world educational programs gain valuable degree-related experience, develop an insider’s understanding of how GM works and earn competitive wages. These programs are designed to provide GM with a source of highly talented candidates while giving students an opportunity for hands-on experience in their chosen field. As a result, participants in these programs are given serious consideration for full-time positions with GM when they graduate. Candidates for these programs must successfully complete an online assessment and possess qualifications that match the business needs of the organization.41. GM is likely to recruit college students for its programs who .A. do a full-time college programB. major in engineering and businessC. have had some practical experienceD. have gathered information about GM42. The Recruiting Calendar (Line 4, Para.1) gives the information about .A. the kinds of people GM needs to trainB. the nation’s top colleges and universitiesC. GM recruiters’ visits to colleges and universities.D. The interesting tasks GM expects the students to fulfill43. Those students whose university is not listed on the Recruiting Calendar may .A. apply to GM online.B. be given interesting tasksC. come to the GM’s training offices directlyD. invite recruiters to visit their universities.44. The programs that GM offers to full-time college students will help them to .A. gain information about business in generalB. get business experience and good wagesC. develop their talents fullyD. obtain a higher degree45. GM offers the Cooperative Education and Intern Program in order to .A. make its business needs known to the publicB. perform successful online assessmentsC. advertise its newly-designed productsD. find out highly talented candidatesT ask 3Directions:The following is an Introduction to Montserrat Publishing. After reading it, you are required to complete the outline below it (No.46 to No.50).Y ou should write your answerbriefly (in not morethan 3 words) on the Answer Sheet correspondingly.We at Montserrat Publishing have published two books devoted to improving the everyday vocabulary of intermediate and advanced English language students.Y ou probably find that even though you have been studying English for many years, you still have problems following a social conversation between English speaking people.Maybe you need English for business and you have a wide knowledge of the vocabulary related to your specialty (专业), but you find it difficult to conduct meetings in English. This is because you cannot understand the practical, everyday expressions being employed. Y ou may also need to understand journals and trade magazines written in English where a dictionary is not enough to convey the real meaning of what is being said.The Practical Everyday English series consists of a collection of words and expressions used every day by all classes of society. The idea is that after being presented with a word or expression the reader can see how it is used by reading two or more clear examples. Each example contains at least one word or even two or three which have been studied on earlier pages.Information on Newly-published BooksName of the series: (46)Number of the books: (47)Intended readers:(48) students studying EnglishPurposes: 1. to enlarge the students’ everyday (49) of English2. to help understand the meaning of practical, everyday expressionsContent: a collection of (50) used every day in varioussituationsT ask 4Directions: The following is a list of terms from a book on WTO. After reading it, you are required to find the items equivalent to (与……等同) those given in Chinese in the table below.Then you should put thecorresponding letters in brackets on the Answer sheet, numbered 51 through 55.A—dispute settlement body B—balance of international paymentsC— world trade organization D— risk managementE—investment in non-productive project F—grant the national treatmentG— appeal body H— common agriculture policyI— customs values J— export performanceK—food security L—free riderM— grey area measures N— import licensingO— market access P— market boardsQ—presence of national parson R—north american free trade areaS— international settlement T— peace clauseU—least-developed countries V—most-favored-nation treatmentExample:(F)实行国民待遇(P)营销机构51. ( )上诉机构( )最惠国待遇52. ( )和平条款( )进口许可53. ( )市场准入( )非生产性投资54. ( )食品安全保障( )共同农业政策55. ( )风险管理( )争端解决机构T ask 5Directions: There is an advertisement below. After reading it, you are required to complete the statements that follow the questions (No.56 to No.60). Y ou should write your answers (in not more than 3 words) onthe Answer Sheet correspondingly.Douglass Company (hereafter〈以下〉referred to as the Buyer) as one party and Anhui Import and Export Corporation (hereafter referred to as the Seller) as the other party agree to sign by their authorized representatives, as a result of friendly negotiation, the present Contract under the following terms and conditions:1) The Seller will arrange and sell 260 sets of drilling machines in the following two fiscal years (财政年度);2) The Seller should take care of the quality, quantity, packing, and shipping;3) The Buyer should pay the Seller by irrevocable L/C and take care of the insurance;4) Claim shall be made within 10 days of goods’ arrival;5) Neither party shall break the contract without mutual agreement;6) The present contract is written in English, valid for two years, after which it may be extended, amended ordiscontinued;7) The contract was signed on April 14, 2005 in Hefei by representatives of the two parties.(Signature) For the Anhui Mechanic Import and Export Co. of People’s Republic of China.(Signature) For the Douglass Company of New Y ork, N.Y, U.S.A.56. What is the contract product?.57. Which party is responsible for the insurance?.58. How should the Buyer pay the seller?By .59. How long is the contract valid for?.60. Who signed the contract?The of the two parties.Part IV T ranslation — English into Chinese (25 minutes)Directions: This part is to test your ability to translate English into Chinese. Each of the four sentences (No.61 to No.64) is followed by four choices of suggested Chinese translation marked A, B, C and D. Markthe best choice and write the corresponding letter on the Answer Sheet. Write your translation of theparagraph (No.65) in the corresponding space on the Translation / Composition Sheet.61. Y ou can buy a three-year-old car for only 60% of the price for a new vehicle and still have several years oftrouble-free driving.A. 你可以买一辆已使用3年的二手车，价格只是新车的60％，但你仍然可以开几年而不出故障。

美国数学竞赛AMC8 -- 2005年真题解析(英文解析+中文解析)

美国数学竞赛AMC8 – 2005年真题解析(英文解析+中文解析)Problem 1Answer: BSolution:If x is the number, then 2x=60 and x=30. Dividing the number by 2 yields 15.中文解析：按照Connie的计算，这个数乘以2是60，可知这个数是30. 应该做的计算是30除以2，因而正确答案应该是15. 答案是B。

Problem 2Answer: CSolution:Karl paid 5*2.5=$12.5. 20% of this cost that he saved is 12.5*0.2=$2.5.中文解析：Karl按原价买了5个文件夹，支付的费用是：2.5*5=12.5. 折扣价是：1.25*0.8=10。

如果Karl 等一天，可以省2.5元。

答案是C.Problem 3Answer: DSolution:Rotating square ABCD counterclockwise 45° so that the line of symmetry BD is a vertical line makes it easier to see that 4 squares need to be colored to match its corresponding square.中文解析：如上图所示，以BD为对称轴，标蓝色的方块需要涂黑。

共4块，答案是D。

Problem 4Answer: CSolution:The perimeter of the triangle is 6.1+8.2+9.7=24cm. A square's perimeter is four times its side length, since all its side lengths are equal. If the square's perimeter is 24, the side length is24/4=6, and the area is 6*6=36.中文解析：三角形的周长是：6.1+8.2+9.7=24. 正方形的周长和三角形相等，也是24，则其边长是24/4=6. 其面积是：6*6=36. 答案是C。

2005AMC12A

1.Two is 10%of x and 20%of y .What is x −y ?(A)1(B)2(C)5(D)10(E)202.The equations 2x +7=3and bx −10=−2have the same solution for x .What is the value of b ?(A)−8(B)−4(C)−2(D)4(E)83.A rectangle with a diagonal of length x is twice as long as it is wide.What is the area of the rectangle?(A)14x 2(B)25x 2(C)12x 2(D)x 2(E)32x 24.A store normally sells windows at $100each.This week the store is oﬀering one free window for each purchase of four.Dave needs seven windows and Doug needs eight windows.How many dollars will they save if they purchase the windows together rather than separately?(A)100(B)200(C)300(D)400(E)5005.The average (mean)of 20numbers is 30,and the average of 30other numbers is 20.What is the average of all 50numbers?(A)23(B)24(C)25(D)26(E)276.Josh and Mike live 13miles apart.Yesterday Josh started to ride his bicycle toward Mike’s house.A little later Mike started to ride his bicycle toward Josh’s house.When they met,Josh had ridden for twice the length of time as Mike and at four-ﬁfths of Mike’s rate.How many miles had Mike ridden when they met?(A)4(B)5(C)6(D)7(E)87.Square EF GH is inside square ABCD so that each side of EF GH can be extended to pass through a vertex of ABCD .Square ABCD has side length √50,E is between B and H ,and BE =1.What is the area of the inner square EF GH ?A BC DGHF E(A)25(B)32(C)36(D)40(E)428.Let A,M,and C be digits with (100A +10M +C )(A +M +C )=2005.What is A ?(A)1(B)2(C)3(D)4(E)59.There are two values of a for which the equation 4x 2+ax +8x +9=0has only one solution for x .What is the sum of those values of a ?(A)−16(B)−8(C)0(D)8(E)2010.A wooden cube n units on a side is painted red on all six faces and then cutinto n 3unit cubes.Exactly one-fourth of the total number of faces of the unit cubes are red.What is n ?(A)3(B)4(C)5(D)6(E)711.How many three-digit numbers satisfy the property that the middle digit is theaverage of the ﬁrst and the last digits?(A)41(B)42(C)43(D)44(E)4512.A line passes through A (1,1)and B (100,1000).How many other points withinteger coordinates are on the line and strictly between A and B ?(A)0(B)2(C)3(D)8(E)913.In the ﬁve-sided star shown,the letters A ,B ,C ,D and E are replaced by thenumbers 3,5,6,7and 9,although not necessarily in that order.The sums of the numbers at the ends of the line segments AB ,BC ,CD ,DE and EA form an arithmetic sequence,although not necessarily in that order.What is the middle term of the arithmetic sequence?AB C DE(A)9(B)10(C)11(D)12(E)1314.On a standard die one of the dots is removed at random with each dot equallylikely to be chosen.The die is then rolled.What is the probability that the top face has an odd number of dots?(A)511(B)1021(C)12(D)1121(E)61115.Let AB be a diameter of a circle and C be a point on AB with2·AC=BC.Let D and E be points on the circle such that DC⊥AB and DE is a second diameter.What is the ratio of the area of DCE to the area of ABD?B(A)16(B)14(C)13(D)12(E)2316.Three circles of radius s are drawn in theﬁrst quadrant of the xy-plane.Theﬁrst circle is tangent to both axes,the second is tangent to theﬁrst circle and the x-axis,and the third is tangent to theﬁrst circle and the y-axis.A circle of radius r>s is tangent to both axes and to the second and third circles.What is r/s?(A)5(B)6(C)8(D)9(E)1017.A unit cube is cut twice to form three triangular prisms,two of which arecongruent,as shown in Figure1.The cube is then cut in the same manner along the dashed lines shown in Figure2.This creates nine pieces.What is the volume of the piece that contains vertex W?(A)112(B)19(C)18(D)16(E)1418.Call a number“prime-looking”if it is composite but not divisible by2,3,or5.The three smallest prime-looking numbers are49,77,and91.There are168prime numbers less than1000.How many prime-looking numbers are there less than1000?(A)100(B)102(C)104(D)106(E)10819.A faulty car odometer proceeds from digit3to digit5,always skipping the digit4,regardless of position.For example,after traveling one mile the odometer changed from000039to000050.If the odometer now reads002005,how many miles has the car actually traveled?(A)1404(B)1462(C)1604(D)1605(E)180420.For each x in[0,1],deﬁnef(x)=2x,if0≤x≤12 2−2x,if12<x≤1.Let f[2](x)=f(f(x)),and f[n+1](x)=f[n](f(x))for each integer n≥2.For how many values of x in[0,1]is f[2005](x)=1/2?(A)0(B)2005(C)4010(D)20052(E)2200521.How many ordered triples of integers(a,b,c),with a≥2,b≥1,and c≥0,satisfy both log a b=c2005and a+b+c=2005?(A)0(B)1(C)2(D)3(E)422.A rectangular box P is inscribed in a sphere of radius r.The surface area of Pis384,and the sum of the lengths of its12edges is112.What is r?(A)8(B)10(C)12(D)14(E)1623.Two distinct numbers a and b are chosen randomly from the set{2,22,23,...,225}.What is the probability that log a b is an integer?(A)225(B)31300(C)13100(D)750(E)1224.Let P(x)=(x−1)(x−2)(x−3).For how many polynomials Q(x)does thereexist a polynomial R(x)of degree3such that P(Q(x))=P(x)·R(x)?(A)19(B)22(C)24(D)27(E)3225.Let S be the set of all points with coordinates(x,y,z),where x,y,and z areeach chosen from the set{0,1,2}.How many equilateral triangles have all their vertices in S?(A)72(B)76(C)80(D)84(E)88。

2003AMC12A(美国数学竞赛)

2003 AMC 12A ProblemsProblem 1What is the difference between the sum of the first even counting numbers and the sum of the first odd counting numbers?SolutionProblem 2Members of the Rockham Soccer League buy socks and T-shirts. Socks cost $4 per pair and each T-shirt costs $5 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 shirt for away games. If the total cost is $2366, how many members are in the League?SolutionProblem 3A 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?SolutionProblem 4It takes Mary minutes to walk uphill km from her home to school, but it takes her only minutes to walk from school to her home along the same route. What is her average speed, in km/hr, for the round trip?SolutionProblem 5The sum of the two 5-digit numbers and is . Whatis ?SolutionProblem 6Define to be for all real numbers and . Which of the following statements is not true?for all andfor all andfor allfor allifSolutionProblem 7How many non-congruent triangles with perimeter have integer side lengths?SolutionProblem 8What is the probability that a randomly drawn positive factor of is less than ?SolutionProblem 9A set of points in the -plane is symmetric about the orgin, both coordinate axes, and the line . If is in , what is the smallest number of points in ?SolutionProblem 10Al, Bert, and Carl are the winners of a school drawing for a pile of Halloween candy, which they are to divide in a ratio of , respectively. Due to some confusion they come at different times to claim their prizes, and each assumes he is the first to arrive. If each takes what he believes to be the correct share of candy, what fraction of the candy goes unclaimed?SolutionProblem 11A square and an equilateral triangle have the same perimeter. Let be the area of the circle circumscribed about the square and the area of the circle circumscribed around the triangle. Find .SolutionProblem 12Sally has five red cards numbered through and four blue cardsnumbered through . She stacks the cards so that the colors alternate and so that the number 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?SolutionProblem 13The polygon enclosed by the solid lines in the figure consists of 4 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 missing?SolutionProblem 14Points and lie in the plane of the square suchthat , , , and are equilateral triangles. If has an area of 16, find the area of .SolutionProblem 15A 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 calleda lune. Determine the area of this lune.SolutionProblem 16A point P is chosen at random in the interior of equilateral triangle . What is the probability that has a greater area than each of and ?SolutionProblem 17Square has sides of length , and is the midpoint of . A circle with radius and center intersects a circle with radius and center at points and . What is the distance from to ?SolutionProblem 18Let be a -digit number, and let and be the quotient and the remainder, respectively, when is divided by . For how many values of is divisibleby ?SolutionProblem 19A parabola with equation is reflected about the -axis. The parabola and its reflection are translated horizontally five units in opposite directions to become the graphs of and , respectively. Which of the following describes the graph of ?SolutionProblem 20How many -letter arrangements of A's, B's, and C's have no A's in thefirst letters, no B's in the next letters, and no C's in the last letters?SolutionProblem 21The graph of the polynomialhas five distinct -intercepts, one of which is at . Which of the following coefficients cannot be zero?SolutionProblem 22Objects and move simultaneously in the coordinate plane via a sequence of steps, each of length one. Object starts at and each of its steps is either right or up, both equally likely. Object starts at and each of its steps is either to the left or down, both equally likely. Which of the following is closest to the probability that the objects meet?SolutionProblem 23How many perfect squares are divisors of the product ?SolutionProblem 24If what is the largest possible value ofSolutionProblem 25Let . For how many real values of is there at least one positive value of for which the domain of and the range are the same set?Solution答案：SolutionSolution 1The first even counting numbers are .The first odd counting numbers are .Thus, the problem is asking for the valueof .Solution 2Using the sum of an arithmetic progression formula, we can write thisas .Solution 3The formula for the sum of the first even numbers, is , (E standing for even).Sum of first odd numbers, is , (O standing for odd).Knowing this, plug for ,.Problem 2SolutionSince 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, and was the cost per member, the number of members in the League isProblem 3SolutionThe volume of the original box isThe volume of each cube that is removed isSince there are corners on the box, cubes are removed.So the total volume removed is .Therefore, the desired percentage isProblem 4SolutionSolution 1Since she walked km to school and km back home, her total distanceis km.Since she spent minutes walking to school and minutes walking back home, her total time is minutes = hours.Therefore her average speed in km/hr is .Solution 2The average speed of two speeds that travel the same distance is the harmonic mean ofthe speeds, or (for speeds and ). Mary's speed going to schoolis , and her speed coming back is . Plugging the numbers in, we getthat the average speed is .Problem 5SolutionSince , , and are digits, , , .Therefore, .Problem 6SolutionExamining statement C:when , but statement C says that it does for all .Therefore the statement that is not true isProblem 7SolutionBy the triangle inequality, no side may have a length greater than the semiperimeter, which is .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 remaining two sides have a combined length of . So, the remaining sides must be either and or and . Therefore, the number of triangles is .Problem 8SolutionSolution 1For a positive number which is not a perfect square, exactly half of the positive factors will be less than .Since is not a perfect square, half of the positive factors of will be lessthan .Clearly, there are no positive factors of between and .Therefore half of the positive factors will be less than .So the answer is .Solution 2Testing all numbers less than , numbers , and divide . The prime factorization of is . Using the formula for the number of divisors, the total number of divisors of is . Therefore, our desired probabilityisProblem 9SolutionIf is in , then is also, and quickly we see that every point of theform or must be in . Now note that these points satisfy all of the symmetry conditions. Thus the answer is .Problem 10SolutionBecause the ratios are , Al, Bert, and Carl believe that they need to take , , and of the pile when they each arrive, respectively. After each personcomes, , , and of the pile's size (just before each came) remains. The pile starts at , and at the end of the original pile goes unclaimed. (Notethat because of the properties of multiplication, it does not matter what order the threecome in.) Hence the answer is.Problem 11SolutionSuppose that the common perimeter is Then, the side lengths of the square and triangle, respectively, are and The circle circumscribed about the square has a diameter equal to the diagonal of the square, which is Therefore, the radiusis and the area of the circle isNow consider the circle circumscribed around the equilateral triangle. Due to symmetry, the circle must share a center with the equilateral triangle. The radius of the circle is simply the distance from the center of the triangle to a vertex. This distance is of an altitude. By right triangle properties, the altitude is where s is the side. So, the radius is The area of the circleis So,Problem 12Let 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 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 one end of the 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 .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 tobottom:Therefore, the sum of the numbers on the middle three cards isProblem 13Solution 1Let 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" square that is attatched 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 folded.Thus the answer is .Solution 2Another way to think of it is that a cube missing one face has of its faces. Since the shape has faces already, we need another face. The only way to add another face is ifthe added square does not overlap any of the others. ,, and overlap, while squares to do not. The answer isProblem 14SolutionSolution 1Since the area of square ABCD is 16, the side length must be 4. Thus, the side length of triangle AKB is 4, and the height of AKB, and thus DMC, is .The diagonal of the square KNMC will then be . From here there are 2 ways to proceed:First: Since the diagonal is , the side length is , and the area isthus .Solution 2Since a square is a rhombus, the area of the square is , where and are the diagonals of the rhombus. Since the diagonal is , the areais .Problem 15SolutionLet denote the area of region in the figure above.The shaded area is equal 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 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 areaisProblem 16SolutionSolution 1After we pick point , we realize that is symmetric for this purpose, and so the probability that is the greatest area, or or , are all the same. Sincethey add to , the probability that has the greatest area isSolution 2We will use an approach of geometric probability to solve this problem. Let us take point P, and draw the perpendiculars to AB, BC, and AC, and call the feet of these perpendiculars D, E, and F respectively. The area of triangle ACP is simply 1/2 * AC * PF. Similarly we can find the area of triangles BCP and ABP. If we add these up and realize that it equals the area of the entire triangle, we see that no matter where we choose P, PD + PE + PF = the height of the triangle. Setting the area of triangle ACP greater than ABP and BCP, we want PF to be the largest of PF, PD, and PE. We then realize that PF = PD = PE when P is the orthocenter of ABC. Let us call the orthocenter of the triangle Q. If we want PF to be the largest of the three, by testing points we realize that P must be in the interior of quadrilateral QFCE. So our probability (using geometric probability) is the area of QFCE divided by the area of ABC. We will now show that the three quadrilaterals, QFCE, QEBD, and QDAF are congruent. As the definition of point Q yields, QF = QD = QE. Since ABC is equilateral, Q is also the circumcenter of ABC, so QA = QB = QC. Using the Pythagorean theorem, BD = DA = AF = FC = CE = EB. Also, angles BDQ, BEQ, CEQ, CFQ, AFQ, and ADQ are all equal to 90 degrees by the definition of analtitude. Also, angles DBE, FCE, DAF are all equal to 60 degrees as equilateral triangles are also equiangular. It is now clear that QFCE, QFAD, QEBD are all congruent. Summing up these areas gives us the area of ABC. QFCE contributes to a third of that area, as they are all congruent, so the ratio of the areas of QFCE to ABC is 1/3 (C). Problem 17Solution 1Let be the origin. is the point and is the point . We are given the radius of the quarter circle and semicircle as and , respectively, so their equations, respectively, are:Algebraically manipulating the second equation gives:Substituting this back into the first equation:Solving each factor for 0 yields . The first value of is obviously referring to the x-coordinate of the point where the circles intersect at the origin, , so the second value must be referring to the x coordinate of . Since is the y-axis, the distance to it from is the same as the x-value of the coordinate of , so the distancefrom to isSolution 2Note that is merely a reflection of over . Call the intersectionof and . Drop perpendiculars from and to , and denote their respective points of intersection by and . We then have , with a scale factor of 2. Thus, we can find and double it to get our answer. With some analytical geometry, we find that , implying that .Solution 3As in Solution 2, draw in and and denote their intersection point . Next, drop a perpendicular from to and denote the foot as . as they areboth radii and similarly so is a kite and by awell-known theorem.Pythagorean theorem gives us .Clearly by angle-angleand by Hypotenuse Leg. Manipulating similar triangles gives usProblem 18SolutionWhen a -digit number is divided by , the first digits become the quotient, , and the last digits become the remainder, .Therefore, can be any integer from to inclusive, and can be any integer from to inclusive.For each of the possible values of , there are atleast possible values of such that .Since there is "extra" possible value of that is congruent to , each of the values of that are congruent to have more possible value of such that .Therefore, the number of possible values of suchthat isProblem 19SolutionIf we take the parabola and reflect it over the x - axis, we have the parabola . Without loss of generality, let us say that the parabola is translated 5 units to the left, and the reflection to the right. Then:Adding them upproduces:This is a line with slope . Since cannot be (because would be a line) we end up withProblem 20SolutionThe answer is .Note that the first five letters must be B's or C's, the next five letters must be C's or A's, and the last five letters must be A's or B's. If there are B's in the first five letters, then there must be C's in the first five letters, so there must be C's and A's in the next five letters, and A's and B's in the last five letters. Therefore the number of each letter in each group of five is determined completely by the number of B's in the first 5 letters, and the number of ways to arrange these 15 letters with thisrestriction is (since there are ways to arrange B's and C's).Therefore the answer is .Problem 21SolutionSolution 1Let the roots be . According to Vieta's formulas, wehave . The first four terms contain and are therefore zero, thus . This is a product of four non-zero numbers, therefore must be non-zero .Solution 2Clearly, since is an intercept, must be . But if was , would divide the polynomial, which means it would have a double root at , which is impossible, since all five roots are distinct.Problem 22SolutionIf and meet, their paths connect and There are such paths, so the probability isProblem 23SolutionWe want to find the number of perfect square factors in the product of all the factorials of numbers from . We can write this out and take out the factorials, and then find a prime factorization of the entire product. We can also find this prime factorization by finding the number of times each factor is repeated in each factorial. This comes out to be equal to . To find the amount of perfect square factors, we realize that each exponent in the prime factorization must be even: . To find the total number of possibilities, we add to each exponent and multiply them all together. This gives us .Problem 24SolutionUsing logarithmic rules, we see thatSince and are both positive, using AM-GM gives that the term in parentheses must be at least , so the largest possible values isProblem 25SolutionThe function has a codomain of all non-negative numbers,or . Since the domain and the range of are the same, it follows that the domain of also satisfies .The function has two zeroes at , which must be part of the domain. Since the domain and the range are the same set, it follows that is in the codomain of ,or . This implies that one (but not both) of is non-positive. If is positive, then , which implies that a negative number falls in the domain of , contradiction. Thus must be non-positive, is non-negative, and the domain of the function occurs when, orCompleting the square, by the Trivial Inequality (remember that ). Since is continuous and assumes this maximal value at , it follows that the range of isAs the domain and the range are the same, we havethat (we can divide through by since it is given that is positive). Hence , which both we can verify work, and the answer is .。

amc12真题及答案

amc12真题及答案Problem 1What is the value of ?SolutionProblem 2For what value of does ?SolutionProblem 3The remainder can be defined for all real numbers and with bywhere denotes the greatest integer less than or equal to . What is the value of ?SolutionProblem 4The mean, median, and mode of the data values are all equal to . What is the value of ?SolutionProblem 5Goldbach's conjecture states that every even integer greater than 2 can be written as the sum of two prime numbers (for example, ). So far, no one has been able to prove that the conjecture is true, and no one has found a counterexample to show that the conjecture is false. What would a counterexample consist of?SolutionProblem 6A triangular array of coins has coin in the first row, coins in the second row, coins in the third row, and so on up to coins in the th row. What is the sum of the digits of ?SolutionProblem 7Which of these describes the graph of ?SolutionProblem 8What is the area of the shaded region of the given rectangle?SolutionProblem 9The five small shaded squares inside this unit square are congruent and have disjoint interiors. The midpoint of each side of the middle square coincides with one of the vertices of the other four small squares as shown. The common side length is , where and are positive integers. What is ?SolutionProblem 10Five friends sat in a movie theater in a row containing seats, numbered to from left to right. (The directions "left" and "right" are from the point of view of the people as they sit in the seats.) During the movie Ada went to the lobby to get some popcorn. When she returned, she found that Bea had moved two seatsto the right, Ceci had moved one seat to the left, and Dee and Edie had switched seats, leaving an end seat for Ada. In which seat had Ada been sitting before she got up?SolutionProblem 11Each of the students in a certain summer camp can either sing, dance, or act. Some students have more than one talent, but no student has all three talents. Thereare students who cannot sing, students who cannot dance, and students who cannot act. How many students have two ofthese talents?SolutionProblem 12In , , , and . Point lies on , and bisects . Point lies on ,and bisects . The bisectors intersect at . What is the ratio : ?SolutionProblem 13Let be a positive multiple of . One red ball and green balls are arranged in a line in random order. Let be the probability that at least of the green balls are on the same side of the red ball. Observe that andthat approaches as grows large. What is the sum of the digits of the least value of such that ?SolutionProblem 14Each vertex of a cube is to be labeled with an integer from through , with each integer being used once, in such a way that the sum of the four numbers on the verticesof a face is the same for each face. Arrangements that can be obtained from each other through rotations of the cube are considered to be the same. How many different arrangements are possible?SolutionProblem 15Circles with centers and , having radii and , respectively, lie on the same side of line and are tangent to at and , respectively, with between and . The circle with center is externally tangent to each of the other two circles. What is the area of triangle ?SolutionProblem 16The graphs of and are plotted on the same set of axes. How many points in the plane with positive -coordinates lie on two or more of the graphs?SolutionProblem 17Let be a square. Let and be the centers, respectively, of equilateral triangles with bases and each exterior to the square. What is the ratio of the area of square to the area of square ?SolutionProblem 18For some positive integer the number has positive integer divisors, including and the number How many positive integer divisors does thenumber have?SolutionProblem 19Jerry starts at on the real number line. He tosses a fair coin times. When he gets heads, he moves unit in the positive direction; when he gets tails, he moves unit in the negative direction. The probability that he reaches at some time during this process is where and are relatively prime positive integers. What is (For example, he succeeds if his sequence of tosses is ) SolutionProblem 20A binary operation has the properties that and that for all nonzero real numbers and (Here the dot represents the usual multiplication operation.) The solution to the equation can be written as where and are relatively prime positive integers. What isSolutionProblem 21A quadrilateral is inscribed in a circle of radius Three of the sides of this quadrilateral have length What is the length of its fourth side?SolutionProblem 22How many ordered triples of positive integers satisfy and ?SolutionProblem 23Three numbers in the interval are chosen independently and at random. What is the probability that the chosen numbers are the side lengths of a triangle with positive area?SolutionProblem 24There is a smallest positive real number such that there exists a positive real number such that all the roots of the polynomial are real. In fact, for this value of the value of is unique. What is the value ofSolutionProblem 25Let be a positive integer. Bernardo and Silvia take turns writing and erasing numbers on a blackboard as follows: Bernardo starts by writing the smallest perfect square with digits. Every time Bernardo writes a number, Silvia erases the last digits of it. Bernardo then writes the next perfect square, Silvia erases the last digits of it, and this process continues until the last two numbers that remain on the board differ by at least 2. Let be the smallest positive integer not written on the board. For example, if , then the numbers that Bernardo writes are , and the numbersshowing on the board after Silvia erases are and , and thus . What is the sum of the digits of ?2016 AMC 12A Answer Key1 B2 C3 B4 D5 E6 D7 D8 D9 E10 B11 E12 C13 A14 C15 D16 D17 B18 D19 B20 A21 E22 A23 C24 B25 E。

数学竞赛AMC12词汇

Aabbreviation 简写符号；简写absolute error 绝对误差absolute value 绝对值accuracy 准确度acute angle 锐角acute-angled triangle 锐角三角形add 加addition 加法addition formula 加法公式addition law 加法定律addition law(of probability)（概率）加法定律additive property 可加性adjacent angle 邻角adjacent side 邻边algebra 代数algebraic 代数的algebraic equation 代数方程algebraic expression 代数式algebraic fraction 代数分式；代数分数式algebraic inequality 代数不等式algebraic operation 代数运算alternate angle （交）错角alternate segment 交错弓形altitude 高；高度；顶垂线；高线ambiguous case 两义情况；二义情况amount 本利和；总数analysis 分析；解析analytic geometry 解析几何angle 角angle at the centre 圆心角angle at the circumference 圆周角angle between a line and a plane 直与平面的交角angle between two planes 两平面的交角angle bisection 角平分angle bisector 角平分线；分角线angle in the alternate segment 交错弓形的圆周角angle in the same segment 同弓形内的圆周角angle of depression 俯角angle of elevation xxangle of greatest slope 最大斜率的角angle of inclination 倾斜角angle of intersection 相交角；交角angle of rotation 旋转角angle of the sector 扇形角angle sum of a triangle 三角形内角和angles at a point 同顶角annum(X% per annum) 年（年利率X%）anti-clockwise direction 逆时针方向；返时针方向anti-logarithm 逆对数；反对数anti-symmetric 反对称apex 顶点approach 接近；趋近approximate value 近似值approximation 近似；略计；逼近Arabic system xx数字系统arbitrary 任意arbitrary constant 任意常数arc 弧arc length 弧长arc-cosine function 反xx函数arc-sin function 反正弦函数arc-tangent function 反正切函数area 面积arithmetic 算术arithmetic mean 算术平均；等差xx；算术xxarithmetic progression 算术级数；等差级数arithmetic sequence 等差序列arithmetic series 等差级数arm 边arrow 前号ascending order 递升序ascending powers of X X 的升幂associative law 结合律assumed mean 假定平均数assumption 假定；假设average 平均；平均数；平均值average speed 平均速率axiom 公理axis 轴axis of parabola 拋物线的轴axis of symmetry 对称轴Bback substitution 回代bar chart 棒形图；条线图；条形图；线条图base （1）底；（2）基；基数base angle 底角base area 底面base line 底线base number 底数；基数base of logarithm 对数的底bearing 方位（角）；角方向（角）bell-shaped curve 钟形图bias 偏差；偏倚billion 十亿binary number 二进数binary operation 二元运算binary scale 二进法binary system 二进制binomial 二项式binomial expression 二项式bisect 平分；等分bisection method 分半法；分半方法bisector 等分线；平分线boundary condition 边界条件boundary line 界（线）；边界bounded 有界的bounded above 有上界的；上有界的bounded below 有下界的；下有界的bounded function 有界函数brace 大括号bracket 括号breadth 阔度broken line graph 折线图Ccalculation 计算calculator 计算器；计算器cancel 消法；相消canellation law 消去律capacity 容量Cartesian coordinates xx坐标Cartesian plane xx平面category 类型；xxcentral line 中线central tendency 集中趋centre 中心；心centre of a circle 圆心centroid 形心；距心certain event 必然事件chance 机会change of base 基的变换change of subject 主项变换change of variable 换元；变量的换chart 图；图表checking 验算chord 弦chord of contact 切点弦circle 圆circular 圆形；圆的circular function 圆函数；三角函数circular measure 弧度法circumcentre 外心；外接圆心circumcircle 外接圆circumference 圆周circumradius 外接圆半径circumscribed circle 外接圆class 区；组；类class boundary 组界class interval 组区间；组距class limit 组限；区限class mark 组中点；区中点classification 分类clnometer 测斜仪clockwise dirction 顺时针方向closed convex region 闭凸区域closed interval 闭区间coefficient 系数coincide 迭合；重合collection of terms 并项collinear 共线collinear planes 共线面column (1)列；纵行；(2) 柱combination 组合common chord 公弦common denominator 同分母；公分母common difference 公差common divisor 公约数；公约common factor 公因子；公因子common logarithm 常用对数common multiple 公位数；公倍common ratio 公比common tangetn公切commutative law 交换律comparable 可比较的compass xxcompass bearing xx方位角compasses 圆规compasses construction 圆规作图complement 余；补余complementary angle 余角complementary event 互补事件complementary probability 互补概率completing the square 配方complex number 复数complex root 复数根composite number 复合数；合成数compound bar chart 综合棒形图compound discount 复折扣compound interest 复利；复利息computation 计算computer 计算机；电子计算器concave 凹concave downward 凹向下的Ddata 数据decagon 十边形decay 衰变decay factor 衰变因子decimal 小数decimal place 小数位decimal point 小数点decimal system 十进制decrease 递减decreasing function 递减函数；下降函数decreasing sequence 递减序列；下降序列decreasing series 递减级数；下降级数decrement 减量deduce 演绎deduction 推论deductive reasoning 演绎推理definite 确定的；定的distance 距离distance formula 距离公式distinct roots 相异根distincr solution 相异解distribution 公布distrivutive law 分配律divide 除dividend (1)被除数；(2)股息divisible 可整除division 除法division algorithm 除法算式divisor 除数；除式；因子divisor of zero 零因子dodecagon 十二边形dot 点double root 二重根due east/ south/ west /north 向xx/ xx/ 西/ xxdefiniton 定义degree (1)度；(2)次degree of a polynomial 多项式的次数degree of accuracy 准确度degree of precision 精确度delete 删除；删去denary number 十进数denary scale 十进法denary system 十进制denominator 分母dependence (1)相关；(2)应变dependent event(s) 相关事件；xx事件；从属事件dependent variable 应变量；应变数depreciation 折旧descending order 递降序descending powers of X X的降序detached coefficients 分离系数(法)deviation 偏差；变差deviation from the mean 离均差diagonal 对角diagram 图；图表diameter 直径difference 差digit 数字dimension 量；量网；维(数)direct proportion 正比例direct tax, direct taxation 直接税direct variation 正变（分）directed angle 有向角directed number 有向数direction 方向；方位discontinuous 间断(的)；非连续(的)；不连续(的)discount 折扣discount per cent 折扣百分率discrete 分立；离散discrete data 离散数据；间断数据discriminant 判别式dispersion 离差displacement 位移disprove 反证Eedge 棱；边elimination 消法elimination method 消去法；消元法elongation 伸张；展empirical data 实验数据empirical formula 实验公式empirical probability 实验概率；经验概率enclosure 界限end point 端点entire surd 整方根equal 相等equal ratios theorem 等比定理equal roots 等根equality 等(式)equality sign 等号equation 方程equation in one unknown 一元方程equation in two unknowns (variables) 二元方程equation of a straight line 直线方程equation of locus 轨迹方程equiangular 等角(的)extreme value 极值equidistant 等距(的)equilaeral 等边(的)equilateral polygon 等边多边形equilateral triangle 等边三角形equivalent 等价(的)error 误差escribed circle 旁切圆estimate 估计；估计量Euler's formula xx公式；xx公式evaluate 计值even function 偶函数even number 偶数evenly distributed 均匀分布的event 事件exact 真确exact solution 准确解；精确解；真确解exact value 法确解；精确解；真确解example 例excentre 外心exception 例外excess 起exclusive 不包含exclusive events 互斥事件exercise 练习expand 展开expand form xxexpansion 展式expectation 期望expectation value, expected value 期望值；预期值experiment 实验；试验experimental 试验的experimental probability 实验概率exponent 指数express…in terms of….. 以………表达expression 式；数式extension 外延；延长；扩张；扩充exterior angle 外角external angle bisector外分角external point of division 外分点extreme point 极值点Fface 面factor 因子；因式；商factor method 因式分解法factor theorem 因子定理；因式定理factorial 阶乘factorization 因子分解；因式分解factorization of polynomial 多项式因式分解FALSE 假(的)feasible solution 可行解；容许解Fermat’s last theorem xx最后定理Fibonacci number 斐波那契数；黄金分割数Fibonacci sequence 斐波xx序列fictitious mean 假定平均数figure (1)图(形)；(2)数字finite 有限finite population 有限总体finite sequence 有限序列finite series 有限级数first quartile 第一四分位数first term 首项fixed deposit 定期存款fixed point 定点flow chart 流程图foot of perpendicular 垂足for all X 对所有Xfor each /every X 对每一Xform 形式；型formal proof 形式化的证明format 格式；规格formula(formulae) 公式four rules 四则four-figure table 四位数表fourth root 四次方根fraction 分数；分式fraction in lowest term 最简分数fractional equation 分式方程fractional index 分数指数fractional inequality 分式不等式free fall 自由下坠frequency 频数；频率frequency distribution 频数分布；频率分布frequency distribution table 频数分布表frequency polygon 频数多边形；频率多边形frustum 平截头体function 函数function of function 复合函数；迭函数functional notation 函数记号Ggain 增益；赚；盈利gain per cent 赚率；增益率；盈利百分率game （1）对策；（2）博奕general form 一般式；通式general solution 通解；一般解general term 通项geoborad 几何板geometric mean 几何平均数；等比中项geometric progression 几何级数；等比级数geometric sequence 等比序列geometric series 等比级数geometry 几何；几何学given 给定；已知golden section 黄金分割grade 等级gradient (1)斜率；倾斜率；(2)梯度grand total 总计graph 图像；图形；图表graph paper 图表纸graphical method 图解法graphical representation 图示；以图样表达graphical solution 图解greatest term 最大项greatest value 最大值grid lines 网网格线group 组；grouped data 分组数据；分类数据grouping terms 并项；集项growth 增长growth factor 增长因子Hhalf closed interval 半闭区间half open interval 半开区间head 正面（钱币）height 高（度）hemisphere 半球体；半球heptagon 七边形Heron's formula 希罗公式hexagon 六边形higher order derivative xx导数highest common factor(H.C.F) 最大公因子；最高公因式；最高公因子Hindu-Arabic numeral xx数字histogram 组织图；直方图；矩形图horizontal 水平的；水平horizontal line 横线；水平线hyperbola 双曲线hypotenuse 斜边Iidentical 全等；恒等identity 等（式）identity relation 恒等关系式if and only if/iff 当且仅当；若且仅若if…., then 若….则；如果…..则illustration 例证；说明image 像点；像imaginary circle 虚圆imaginary number 虚数imaginary root 虚根implication 蕴涵式；蕴含式imply 蕴涵；蕴含impossible event 不可能事件improper fraction 假分数inclination 倾角；斜角inclined plane 斜面included angle 夹角included side 夹边inclusive 包含的；可兼的inconsistent 不相的(的)；不一致(的)increase 递增；增加increasing function 递增函数interior angles on the same side of the transversal 同旁内角interior opposite angle 内对角internal bisector内分角internal division 内分割internal point of division 内分点inter-quartile range 四分位数间距intersect 相交intersection (1)交集；(2)相交；(3)交点interval 区间intuition 直观invariance 不变性invariant (1)不变的；(2)不变量；不变式inverse 反的；逆的inverse circular function 反三角函数inverse cosine function 反xx函数inverse function 反函数；逆函数inverse problem 逆算问题inverse proportion 反比例；逆比例inverse sine function 反正弦函数inverse tangent function 反正切函数inverse variation 反变(分)；逆变(分)irrational equation 无理方程irrational number 无理数irreducibility 不可约性irregular 不规则isosceles triangle 等腰三角形increasing sequence 递增序列increasing series 递增级数increment 增量independence 独立；自变independent event 独立事件independent variable 自变量；独立变量indeterminate (1)不定的；(2)不定元；未定元indeterminate coefficient 不定系数；未定系数indeterminate form 待定型；不定型index,indices 指数；指index notation 指数记数法inequality 不等式；不等inequality sign 不等号infinite 无限；无穷infinite population 无限总体infinite sequence 无限序列；无穷序列infinite series 无限级数；无穷级数infinitely many 无穷多infinitesimal 无限小；无穷小infinity 无限(大)；无穷(大) initial point 始点；起点initial side 始边initial value 初值；始值input 输入input box 输入inscribed circle xx insertion 插入insertion of brackets 加括号instantaneous 瞬时的integer 整数integral index 整数指数integral solution 整数解integral value 整数值intercept 截距；截段intercept form 截距式intercept theorem 截线定理interchange 互换interest 利息interest rate 利率interest tax 利息税interior angle 内角Jjoint variation 联变(分)；连变(分) Kknown 己知LL.H.S. 末项law xx；定xxlaw of indices 指数xx；指数定xxlaw of trichotomy 三分xxleading coefficient 首项系数least common multiple, lowest common multiple (L.C.M) 最小公倍数；最低公倍式least value 最小值lemma 引理length 长(度)letter 文字；字母like surd 同类根式like terms 同类项limit 极限line 线；行line of best-fit 最佳拟合line of greatest slope 最大斜率的直；最大斜率line of intersection 交线line segment 线段linear 线性；一次linear equation 线性方程；一次方程linear equation in two unknowns 二元一次方程；二元线性方程linear inequality 一次不等式；线性不等式linear programming 线性规划literal coefficient 文字系数literal equation 文字方程load 负荷loaded coin 不xx钱币loaded die 不xx骰子locus, loci 轨迹logarithm 对数logarithmic equation 对数方程logarithmic function 对数函数logic 逻辑logical deduction 逻辑推论；逻辑推理logical step 逻辑步骤long division method 长除法loss 赔本；亏蚀loss per cent 赔率；亏蚀百分率lower bound 下界lower limit 下限lower quartile 下四分位数lowest common multiple(L.C.M) 最小公倍数Mmagnitude 量；数量；xx；大小major arc 优弧；大弧major axis 长轴major sector 优扇形；大扇形major segment 优弓形；大弓形mantissa 尾数mantissa of logarithm 对数的尾数；对数的定值部many-sided figure 多边形marked price 标价mathematical induction 数学xxmathematical sentence 数句mathematics 数学maximize 极大maximum absolute error 最大绝对误差maximum point 极大点maximum value 极大值mean 平均(值)；平均数；中数mean deviation 中均差；平均偏差measure of dispersion 离差的量度measurement 量度median (1)中位数；(2)中线meet 相交；相遇mensuration 计量；求积法method 方法method of completing square 配方法method of substitution 代换法；换元法metric unit 十进制单位mid-point 中点mid-point formula 中点公式mid-point theorem 中点定理million 百万minimize 极小minimum point 极小点minimum value 极小值minor (1)子行列式；(2)劣；较小的minor arc 劣弧；小弧minor axis 短轴minor sector 劣扇形；小扇形minor segment 劣弓形；小弓形minus 减minute 分mixed number(fraction) 带分数modal class 众数组mode 众数model 模型monomial 单项式multinomial 多项式multiple 倍数multiple root 多重根multiplicand 被乘数multiplication 乘法multiplication law (of probability) (概率)乘法定律multiplicative property 可乘性multiplier 乘数；乘式multiply 乘mutually exclusive events 互斥事件mutually independent 独立; 互相独立mutually perpendicular lines 互相垂直Nn factorial n阶乘n th root n次根；n次方根natural number 自然数negative 负negative angle 负角negative index 负指数negative integer 负整数negative number 负数neighborhood 邻域net 净(值)n-gon n边形nonagon xx形non-collinear不共线non-linear 非线性non-linear equation 非线性方程non-negative 非负的non-trivial 非平凡的non-zero 非零normal (1)垂直的；正交的；法线的 (2)正态的 (3)正常的；正规的normal curve 正态分记伲怀１分记伲徽媲伲徽忧normal distribution 正态分布，常态分布normal form 法线式notation 记法；记号number 数number line 数线number pair 数偶number pattern 数型number plane 数平面number system 数系numeral 数字；数码numeral system 记数系统numerator 分子numerical 数值的；数字的numerical expression 数字式numerical method 计算方法；数值法Ooblique 斜的oblique cone 斜圆锥oblique triangle 斜三角形obtuse angle 钝角obtuse-angled triangle 钝角三角形octagon 八边形octahedron 八面体odd function 奇函数odd number 奇数one-one correspondence 一一对应open interval 开区间open sentence 开句operation 运算opposite angle 对角opposite interior angle 内对角opposite side 对边optimal solution 最优解order (1)序；次序；(2)阶；级ordered pair 序偶origin 原点outcome 结果output 输出overlap 交迭；相交Pparabola拋物线parallel 平行(的)parallel lines 平行(直线)parallelogram 平行四边形parameter 参数；参变量partial fraction 部分分数；分项分式polar coordinate system 极坐标系统polar coordinates 极坐标pole 极polygon 多边形polyhedron 多面体polynomial 多项式polynomial equation 多项式方程positive 正positive index 正指数positive integer 正整数positive number 正数power (1)幂；乘方；(2)功率；(3)检定力precise 精密precision 精确度prime 素prime factor 质因子；质因素prime number 素数；质数primitive (1)本原的；原始的；(2)原函数principal (1)主要的；(2)本金prism 梭柱(体)；角柱(体)prismoid 平截防xx短?probability 概率problem 应用题produce 延长product 乘积；积product rule 积法则profit 盈利profit per cent 盈利百分率profits tax 利得税progression 级数proof 证(题)；证明proper fraction 真分数property 性质property tax 物业税proportion 比例proportional 成比例protractor 量角器pyramid 棱锥(体)；角锥(体) Pythagoras’ Theorem 勾股定理Pythagorean triplet xx三元数组partial sum 部分和partial variation 部分变(分)particular solution 特解Pascal’s triangle xxxx三角形pattern 模型；规律pegboard 有孔版pentadecagon 十五边形pentagon 五边形per cent 百分率percentage 百分法；百分数percentage decrease 百分减少percentage error 百分误差percentage increase 百分增加percentile 百分位数perfect number 完全数perfecr square 完全平方perimeter 周长；周界period 周期periodic function 周期函数permutation 排列perpendicular 垂线；垂直(于) perpendicular bisector 垂直平分线；中垂线perpendicular line 垂直线pictogram 象形图pie chart 饼图；圆瓣图pinboard 钉板place holder 补位数字place value 位值plan (1)平面图；(2)计划plane 平面plane figure 平面图形plot 绘图plus 加point 点point circle 点圆point of contact 切点point of division 分点point of intersection 交点point-slope form 点斜式polar axis 极轴polar coordinate plane 极坐标平面polar coordinate 极坐标系统Qquadrant 象限quadratic equation 二次方程(式) quadratic formula 二次公式quardratic function 二次函数quadratic inequality 二次不等式quadratic polynomial 四边形quantity 数量quartile 四分位数quotient 商；商式RR.H.S 右radian 弧度radian measure 弧度法radical 根式；根号；根数radius, radii 半径random 随机random experiment 随机试验random number 随机数range 值域；区域；范围；极差；分布域rate 率；利率ratio 比 ; 比率rational expression 有理式；有理数式rational function 有理函数rational index 有理数指数rational number 有理数rationalization 有理化raw data 原始数据raw score 原始分（数）real axis 实轴real number 实数real root 实根reason 理由reciprocal 倒数rectangle 长方形；矩形rectangular block 长方体rectangular coordinate plane 直角坐标平面rectangular coordinates 直角坐rectilinear figure 直线图形recurrent 循环的recurring decimal 循环小数reduce 简化reducible 可约的；可化简的reference angle 参考角reflex angle 优角；反角region 区域regular 正；规则regular polygon 正多边形reject 舍去；否定relation 关系；关系式relative error 相对误差remainder 余数；余式；剩余remainder term 余项remainder theorem 余式定理removal of brackets 撤括号；去括号repeated trials 重复试验resolve 分解revolution 旋转；xxrhombus 菱形right angle 直角right circular cone 直立圆锥（体）right circular cylinder 直立圆柱（体）right prism 直立棱柱；直立角柱(体)right pyramid 直立棱锥；直立角锥(体)right-angled triangle 直角二角形root 根rotation 旋转round angle 周角rounded number 舍数rounding(off) 舍入；四舍五入row 行；棋行rule 规则；法(则)ruler 直尺Ssalaries tax xx税sample 抽样；样本sample space 样本空间satisfy 满足；适合scale 比例尺；标度；图尺scalene triangle 不等边三角形；不规则三角形scientific notation 科学记数法solution of triangle 三角形解法solve 解special angle 特殊角；特别角speed 速率sphere 球形；球面square (1)平方；(2)正方形square bracket 方括号square number 正方形数；平方数square root xx；二次根standard deviation 标准差；标准偏离secant 割second 秒second quartile 第二四分位数(1)截面；截线；(2)截点section (1)截面；截线；(2)截点section formula 截点公式sector 扇式segment 段；节segment of a circle 弓形selling price 售价semi-circle 半圆semi-vertical angle 半顶角sentence 句；语句sequence 序列series 级数set square 三角尺；三角板shaded portion 有阴影部分shape 形状side 边；侧sign 符号；记号signed number 有符号数significant figure 有效数字similar 相似similar figures 相似图形similar triangles 相似三角形similarity 相似(性)simple equation 简易方程simple interest 单利；单利息simplify 简化simultaneous equations 联立方程simultaneous inequalities 联立不等式simultaneous linear equations in two unknowns 联合二次线性方程式sine 正弦sine formula 正弦公式slant edge 斜棱slant height 斜高slope 斜率；斜度；倾斜；坡度slope-intercept form 斜率截距式；斜截式solid 立体；固体soild with uniform corss-section 有均匀横切面的立体solution 解；解法solution of equation 方程解Uuniform 一致(的)；均匀(的)uniform cross-section 均匀横切面uniform speed 匀速率uniformly distributed 均匀分布unique solution 唯一解uniqueness 唯一性unit 单位unit area 单位面积unit circle 单位圆unit volume 单位体积unknown 未知数；未知量unlike 异类项upper bound 上界upper limit 上限upper quartile 上四分位数Vvalue 值variable 变项；变量；元；变元；变数variable speed 可变速率variance 方差variation 变数；变分verify 证明；验证vertex, vertices 顶(点)；极点vertical 铅垂；垂直vertical angle 顶角vertical line 纵线；铅垂vertically opposite angles 对顶角volume 体积Wweight (1)重量；(2)权weighted average, weighted mean 加权平均数whole number 整数；完整数width 阔度without loss of generality 不失一般性Xx-axis x轴x-coordinate x坐标x-intercept x轴截距Yy-axis y轴y-coordinate y坐标y-intercept y轴截距Zzero 零zero factor 零因子zeros of a function 函数零值统计学population 母体sample 样本census 普查sampling 抽样quantitative 量的qualitative/categorical 质的discrete 离散的continuous 连续的population parameters 母体参数sample statistics 样本统计量descriptive statistics 叙述统计学inferential/inductive statistics 推论/归纳统计学levels of measurement 衡量尺度nominal scale 名目尺度ordinal scale 顺序尺度interval scale 区间尺度ratio scale 比例尺度frequency distribution 次数分配relative frequency 相对次数range 全距class midpoint 组中点class limits 组限class boundaries 组界class width 组距cumulative frequency (以下) 累加次数decumulative frequency 以上累加次数histogram 直方图pie chart 饼图ogive 肩形图frequency polygon 多边形图cumulative frequency polygon 累加次数多边形图box plot 盒须图stem and leaf plot 枝叶图measures of central tendency 中央趋势量数mean 平均数median 中位数mode 众数location measures 位置量数percentile 百分位数quartile 四分位数decile 十分位数dispersion measures 分散量数range 全距interquartile-range IQR 四分位距mean absolute deviation 平均绝对离差variance 变异数standard deviation 标准差coefficient of variation 变异系数left-skewed 左偏negative-skewed 负偏right-skewed 右偏positive-skewed 正偏contingency table xx表sampling distribution (of a statistic)(某个统计量的) 抽样分布point estimate 点估计值point estimator 点估计式unbiased estimator 不偏点估计式efficient estimator 有效点估计式consistent estimator 一致点估计式confidence level 信赖水准confidence interval 信赖区间null hypothesis 虚无假设alternative hypothesis 对立假设left-tailed test 左尾检定right-tailed test 右尾检定two-tailed test 双尾检定test statistic 检定统计量critical value 临界值。

2005AMC12B解答

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15. (D) The sum of the digits 1 through 9 is 45, so the sum of the eight digits is between 36 and 44, inclusive. The sum of the four units digits is between 1 + 2 + 3 + 4 = 10 and 6 + 7 + 8 + 9 = 30, inclusive, and also ends in 1. Therefore the sum of the units digits is either 11 or 21. If the sum of the units digits is 11, then the sum of the tens digits is 21, so the sum of all eight digits is 32, an impossibility. If the sum of the units digits is 21, then the sum of the tens digits is 20, so the sum of all eight digits is 41. Thus the missing digit is 45 − 41 = 4. Note that the numbers 13, 25, 86, and 97 sum to 221.
OR
Each of the two-digit numbers leaves the same remainder when divided by 9 as does the sum of its digits. Therefore the sum of the four two-digit numbers leaves the same remainder when divided by 9 as the sum of all eight digits. Let d be the missing digit. Because 221 when divided by 9 leaves a remainder of 5, and the sum of the digits from 1 through 9 is 45, the number (45 − d) must leave a remainder of 5 when divided by 9. Thus d = 4.

2005年美国数学竞赛-AMC_12

2005A 1Two is 10%of x and 20%of y .What is x −y ?A.1B.2C.5D.10E.202The equations 2x +7=3and bx −10=−2have the same solution.What is the value of b ?A.-8B.-4C.-2D.4E.83A rectangle with a diagonal length of x is twice as long as it is wide.What is the area of the rectangle?A.14x 2B.25x 2C.12x 2D.x 2E.32x 24A store normally sells windows at $100each.This week the store is oﬀering one free window for each purchase of four.Dave needs seven windows and Doug needs eight windows.How much will they save if they purchase the windows together than rather separately?A.100B.200C.300D.400E.5005The average (mean)of 20numbers is 30,and the average of 30other numbers is 20.What is the average of all 50numbers?A.23B.24C.25D.26E.276Josh and Mike live 13miles apart.Yesterday,Josh started to ride his bicycle toward Mike’s house.A little later Mike started to ride his bicycle toward Josh’s house.When they met,Josh had ridden for twice the length of time as Mike and at four-ﬁfths of Mike’s rate.How many miles had Mike ridden when they met?A.4B.5C.6D.7E.87Square EF GH is inside the square ABCD so that each side of EF GH can be extended to pass through a vertex of ABCD .Square ABCD has side length √50and BE =1.What is the area of the inner square EF GH ?A.25B.32C.36D.40E.428Let A,M ,and C be digits with(100A +10M +C )(A +M +C )=2005What is A ?A.1B.2C.3D.4E.5This ﬁle was downloaded from the AoPS −MathLinks Math Olympiad Resources Page Page 1http://www.mathlinks.ro/20059There are two values of a for which the equation 4x 2+ax +8x +9=0has only one solution for x .What is the sum of these values of a ?A.-16B.-8C.0D.8E.2010A wooden cube n units on a side is painted red on all six faces and then cut into n 3unitcubes.Exactly one-fourth of the total number of faces of the unit cubes are red.What is n ?A.3B.4C.5D.6E.711How many three-digit numbers satisfy the property that the middle digit is the average ofthe ﬁrst and the last digits?A.41B.42C.43D.44E.4512A line passes through A(1,1)and B(100,1000).How many other points with integer coordi-nates are on the line and strictly between A and B?A.0B.2C.3D.8E.913The regular 5-point star ABCDE is drawn and in each vertex,there is a number.EachA,B,C,D,and E are chosen such that all 5of them came from set 3,5,6,7,9.Each letter is a diﬀerent number (so one possible ways is A=3,B=5,C=6,D=7,E=9).Let AB be the sum of the numbers in A and B.If AB,BC,CD,DE,and EA form an arithmetic sequence (not necessarily this order),ﬁnd the value of CD.A.9B.10C.11D.12E.1314On a standard die one of the dots is removed at random with each dot equally likely to bechosen.The die is then rolled.What is the probability that the top face has an odd number of dots?(A)511(B)1021(C)12(D)1121(E)61115Let AB be a diameter of a circle and C be a point on AB with 2·AC =BC .Let D and Ebe points on the circle such that DC ⊥AB and DE is a second diameter.What is the ratio of the area of DCE to the area of ABD ?(A)16(B)14(C)13(D)12(E)2316Three circles of radius s are drawn in the ﬁrst quadrant of the xy -plane.The ﬁrst circle istangent to both axes,the second is tangent to the ﬁrst circle and the x -axis,and the third is tangent to the ﬁrst circle and the y -axis.A circle of radius r >s is tangent to both axes and to the second and third circles.What is r/s ?200517A unit cube is cut twice to form three triangular prisms,two of which are congruent,as shownin Figure 1.The cube is then cut in the same manner along the dashed lines shown in Figure2.This creates nine pieces.What is the volume of the piece that contains vertex W ?[img]/Forum/album p ic.php ?pic i d =320[/img ]A)112B)19C)18D)16E)1418Call a number ”prime-looking”if it is composite but not divisible by 2,3,or 5.The three smallestprime-looking numbers are 49,77,and 91.There are 168prime numbers less than 1000.How many prime-looking numbers are there less than 1000?(A)100(B)102(C)104(D)106(E)10819A faulty car odometer proceeds from digit 3to digit 5,always skipping the digit 4,regardless ofposition.If the odometer now reads 002005,how many miles has the car actually traveled?(A)1404(B)1462(C)1604(D)1605(E)180420For each x in [0,1],deﬁnef (x )=2x,if 0≤x ≤12;f (x )=2−2x,if 12<x ≤1.Let f [2](x )=f (f (x )),and f [n +1](x )=f [n ](f (x ))for each integer n ≥2.For how many values of x in [0,1]is f [2005](x )=12?(A)0(B)2005(C)4010(D)20052(E)2200521How many ordered triples of integers (a,b,c ),with a ≥2,b ≥1,and c ≥0,satisfy both log a b =c 2005and a +b +c =2005?(A)0(B)1(C)2(D)3(E)422A rectangular box P is inscribed in a sphere of radius r .The surface area of P is 384,and the sumof the lengths of its 12edges is 112.What is r ?(A)8(B)10(C)12(D)14(E)16200523Two distinct numbers a and b are chosen randomly from the set {2,22,23,...,225}.What is theprobability that log a b is an integer?(A)225(B)31300(C)13100(D)750(E)1224Let P (x )=(x −1)(x −2)(x −3).For how many polynomials Q (x )does there exist a polynomialR (x )of degree 3such that P (Q (x ))=P (x )·R (x )?(A)19(B)22(C)24(D)27(E)3225Let S be the set of all points with coordinates (x,y,z ),where x,y,and z are each chosen from theset {0,1,2}.How many equilateral triangles have all their vertices in S ?(A)72(B)76(C)80(D)84(E)882005B 1A scout troop buys 1000candy bars at a price of ﬁve for $2.They sell all the candy bars at a price of two for $1.What was their proﬁt,in dollars?A.100B.200C.300D.400E.5002A positive number x has the property that x %of x is 4.What is x ?A.2B.4C.10D.20E.403Brianna is using part of the money she earned on her weekend job to buy several equally-priced CDs.She used one ﬁfth of her money to buy one third of the CDs.What fraction of her money will she left after she buys all the CDs?A.15B.13C.25D.23E.454At the beginning of the school year,Lisa’s goal was to earn an A on at least 80%of her 50quizzes for the year.She earned an A on 22of the ﬁrst 30quizzes.If she is to achieve her goal,on at most how many of the remaining quizzes can she earn a grade lower than an A?A.1B.2C.3D.4E.55An 8-foot by 10-foot ﬂoor is tiled with square tiles of size 1foot by 1foot.Each tile has a pattern consisting of four white quarter circles of radius 1/2foot centered at each corner of the tile.The remaining portion of the tile is shaded.How many square feet of the ﬂoor are shaded?A.80−20πB.60−10πC.80−10πD.60+10πE.80+10π6In ABC ,we have AC =BC =7and AB =2.Suppose that D is a point on line AB such that B lies between A and D and CD =8.What is BD ?A.3B.2√3C.4D.5E.4√220057What is the area enclosed by the graph of |3x |+|4y |=12?A.6B.12C.16D.24E.258For how many values of a is it true that the line y =x +a passes through the vertex of the parbola y =x 2+a 2?A.0B.1C.2D.10E.Inﬁnitely many 9On a certain math exam,10%of the students got 70points,25%got 80points,20%got 85points,15%got 90points,and the rest got 95points.What is the diﬀerence between the mean and the median score on this exam?A.0B.1C.2D.4E.510The ﬁrst term of a sequence is 2005.Each succeeding term is the sum of the cubes of thedigits of the previous terms.What is the 2005th term of the sequence?A.29B.55C.85D.133E.25011An envelope contains eight bills:2ones,2ﬁves,2tens,and 2twenties.Two bills are drawnat random without replacement.What is the probability that their sum is $20or more?A.14B.27C.37D.12E.2312The quadratic equation x 2+mx +n =0has roots that are twice those of x 2+px +m =0,and none of m,n,and p is zero.What is the value of n/p ?A.1B.2C.4D.8E.1613Suppose that 4x 1=5,5x 2=6,6x 3=7,...,127x 124=128.What is x 1x 2···x 124?A.2B.52C.3D.72E.414A circle having center (0,k ),with k >6,is tangent to the lines y =x,y =−x and y =6.What is the radius of this circle?A.6√2−6B.6C.6√2D.12E.6+6√215The sum of four two-digit numbers is 221.None of the eight digits is 0and no two of themare same.Which of the following is not included among the eight digits?A.1B.2C.3D.4E.516Eight spheres of radius 1,one per octant,are each tangent to the coordinate planes.What isthe radius of the smallest sphere,centered at the origin,thta contains these eight spheres?A.√2B.√3C.1+√2D.1+√3E.3200517How many distinct four-tuples (a,b,c,d )of rational numbers are there witha log 102+b log 103+c log 105+d log 107=2005?A.0B.1C.17D.2004E.inﬁnitely many 18AMC 122005B 18Let A (2,2)and B (7,7)be points in the plane.Deﬁne R as the region inthe ﬁrst quadrant consisting of those points C such that ABC is an actue triangle.What is the closest integer to the area of the region R(A )25(B )39(C )51(D )60(E )8019Let x and y be two-digit integers such that y is obtained by reversing the digits of x .Theintegers x and y satisfy x 2−y 2=m 2for some positive integer m .What is x +y +m ?A.88B.112C.116D.144E.15420Let a,b,c,d,e,f,g and h be distinct elements in the set{−7,−5,−3,−2,2,4,6,13}.What is the minimum possible value of(a +b +c +d )2+(e +f +g +h )2A.30B.32C.34D.40E.5021A positive integer n has 60divisors and 7n has 80divisors.What is the greatest integer ksuch that 7k divides n ?A.0B.1C.2D.3E.422A sequence of complex numbers z 0,z 1,z 2,....is deﬁned by the rulez n +1=iz nz n where z n is the complex conjugate of z n and i 2=−1.Suppose that |z 0|=1and z 2005=1.How many possible values are there for z 0?A.1B.2C.4D.2005E.2200523Let S be the set of ordered triples (x,y,z )of real numbers for whichlog 10(x +y )=z and log 10(x 2+y 2)=z +1.There are real numbers a and b such that for all ordered triples (x,y,z )in S we have x 3+y 3=a ·103z +b ·102z .What is the value of a +b ?2005A.152B.292C.15D.392E.2424All three vertices of an equilateral triangle are on the parabola y=x2,and one of its sides has a slope of2.The x-coordinates of the three vertices have a sum of m/n,where m and n are relatively prime positive integers.What is the value of m+n?A.14B.15C.16D.17E.1825Six ants simultaneously stand on the six vertices of a regular octahedron,with each ant at a diﬀerent vertex.Simultaneously and independently,each ant moves from its vertex to one of the four adjacent vertices,each with equal probability.What is the probability that no two ants arrive at the same vertex?A.5 256B.21 1024C.11 512D.23 1024E.3 128。

2005 AMC 10B Problems

2005 AMC 10B ProblemsProblem 1A scout troop buys 1000 candy bars at a price of five for $2. They sell all the candy bars at a price of two for $1. What was the profit, in dollars?（A ）100 （B ）200 （C ）300 （D ）400 （E ）500Problem 2A positive number has the property that x % of is . What is ?（A ）2 （B ）4 （C ）10 （D ）20 （E ）40Problem 3A gallon of paint is used to paint a room. One third of the paint is used on the first day. On the second day, one third of the remaining paint is used. What fraction of the original amount of paint is available to use on the third day?（A ）101 （B ）91 （C ）31 （D ）94 （E ）95 Problem 4For real numbers and , define . What is the value of?（A ）0 （B ）217 （C ）13 （D ）213 （E ）26 Problem 5Brianna is using part of the money she earned on her weekend job to buy several equally-priced CDs. She used one fifth of her money to buy one third of the CDs. What fraction of her money will she have left after she buys all the CDs?（A ）51 （B ）31 （C ）52 （D ）32 （E ）54 Problem 6At the beginning of the school year, Lisa's goal was to earn an A on at leastof her quizzes for the year. She earned an A on of the first quizzes. If she is toachieve her goal, on at most how many of the remaining quizzes can she earn a grade lower than an A?（A ）1 （B ）2 （C ）3 （D ）4 （E ）5Problem 7A circle is inscribed in a square, then a square is inscribed in this circle, and finally, a circle is inscribed in this square. What is the ratio of the area of the smaller circle to the area of the larger square?（A ）16π （B ）8π （C ）163π （D ）4π （E ）2πProblem 8An 8-foot by 10-foot ﬂoor is tiled with square tiles of size foot by foot. Each tile has a pattern consisting of four white quarter circles of radius 1/2 foot centered at each corner of the tile. The remaining portion of the tile is shaded. How many square feet of the ﬂoor are shaded?（A ）80-20π （B ）60-10π （C ）80-10π （D ）60+10π （E ）80+10π Problem 9One fair die has faces 1, 1, 2, 2, 3, 3 and another has faces 4, 4, 5, 5, 6, 6. The dice are rolled and the numbers on the top faces are added. What is the probability that the sum will be odd?（A ）31 （B ）94 （C ）21 （D ）95 （E ）32 Problem 10In △ABC , we have AC=BC=7 and AB =2. Suppose that D is a point on line AB such that B lies between A and D and CD =8. What is BD ?（A ）3 （B ）32 （C ）4 （D ）5 （E ）24Problem 11The first term of a sequence is 2005. Each succeeding term is the sum of the cubes of the digits of the previous term. What is the 2005th term of the sequence?（A ）29 （B ）55 （C ）85 （D ）133 （E ）250Problem 12Twelve fair dice are rolled. What is the probability that the product of the numbers on the top faces is prime?（A ）12)121( （B ）12)61( （C ）211)61( （D ）11)61(25 （E ）10)61( Problem 13How many numbers between 1 and 2005 are integer multiples of 3 or 4 but not 12?（A ）501 （B ）668 （C ）835 （D ）1002 （E ）1169Problem 14Equilateral △ABC has side length 2, M is the midpoint of AC , and C is the midpoint of BD . What is the area of △CDM ?（A ）22 （B ）43 （C ）23 （D ）1 （E ）2 Problem 15An envelope contains eight bills: 2 ones, 2 fives, 2 tens, and 2 twenties. Two bills are drawn at random without replacement. What is the probability that their sum is $20 or more?（A ）41 （B ）72 （C ）73 （D ）21 （E ）32 Problem 16 The quadratic equation 02=++n mx x has roots that are twice those of02=++m px x , and none of , , and is zero. What is the value of n/p ?（A ）1 （B ）2 （C ）4 （D ）8 （E ）16Problem 17Suppose that 76,65,54===c b a and 87=d . What is ?（A ）1 （B ）23 （C ）2 （D ）25 （E ）3 Problem 18All of David's telephone numbers have the form 555-abc-defg , where a, b, c, d, e, f , and g are distinct digits and in increasing order, and none is either 0 or 1. How many different telephone numbers can David have?（A ）1 （B ）2 （C ）7 （D ）8 （E ）9Problem 19On a certain math exam, 10% of the students got 70% points, 25% got 80 points, 20% got 85 points, 15% got 90 points, and the rest got 95 points. What is the difference between the mean and the median score on this exam?（A ）0 （B ）1 （C ）2 （D ）4 （E ）5Problem 20What is the average (mean) of all -digit numbers that can be formed by using each of thedigits 1, 3, 5, 7, and 8 exactly once?（A ）48000 （B ）49999.5 （C ）53332.8 （D ）55555 （E ）56432.8 Problem 21Forty slips are placed into a hat, each bearing a number 1, 2, 3, 4, 5, 6, 7, 8, 9, or 10, with each number entered on four slips. Four slips are drawn from the hat at random and without replacement. Let be the probability that all four slips bear the same number.Let be the probability that two of the slips bear a number and the other two bear anumber a b ≠. What is the value of q/p ?（A ）162 （B ）180 （C ）324 （D ）360 （E ）720Problem 22For how many positive integers less than or equal to 24 is n! evenly divisible by1+2+…+n ?（A ）8 （B ）12 （C ）16 （D ）17 （E ）21Problem 23In trapezoid ABCD we have AB parallel to DC , E as the midpoint of BC , and F as the midpoint of DA . The area of ABEF is twice the area of FECD . What is AB/DC ?（A ）2 （B ）3 （C ）5 （D ）6 （E ）8Problem 24Let and be two-digit integers such that is obtained by reversing the digits of . Theintegers and satisfy 222m y x =- for some positive integer . What is m y x ++?（A ）88 （B ）112 （C ）116 （D ）144 （E ）154 Problem 25A subset of the set of integers from to 100, inclusive, has the property that no twoelements of sum to 125. What is the maximum possible number of elements in?（A ）50 （B ）51 （C ）62 （D ）65 （E ）682005 AMC 12B SolutionsProblem 1 Expenses=400521000=∙ Revenue=500211000=∙Profit=Revenue-Expenses=100 Problem 2Since x % means 0.01x , the statement "x % of x is 4" can be rewritten as " 0.01x ·x =4":X =20 Problem 3After the first day, there is )131(1⨯- gallons left, or32 gallons. After the second day, there is a total of 94)3231(32=⨯-. Therefore, the fraction of the original amount of paintthat is left is 94194= Problem 4Problem 5Let Brianna's money. We have )(53)(3151CDs m CDs m =⇒=. Thus, the moneyleft over is m m m 5253=- , so the answer is Problem 6Lisa's goal was to get an A on 80%·50=40 quizzes. She already has A's on 22 quizzes, so she needs to get A's on 40-22=18 more. There are 50-30=20 quizzes left, so she canafford to get less than an A on 20-18=2 of them.Problem 7Let the side of the largest square be . It follows that the diameter of the inscribed circle is also . Therefore, the diagonal of the square inscribed inscribed in the circle is . The side length of the smaller square is 222x x =. Similarly, the diameter of the smallerinscribed circle is 22x . Hence, its radius is 42x . The area of this circle is 8)42(22x x ππ=, and the area of the largest squareis . The ratio of the areas is 8282ππ=x x.Problem 8There are 80 tiles. Each tile hasshaded. Thus: shaded area=ππ280))21(4141(802-=∙∙∙- Problem 9In order to obtain an odd sum, exactly one out of the two dice must have an odd number. We can easily find the total probability using casework. Case 1: The first die is odd and the second die is even. The probability of this happening is 946464=⨯ Case 2: The first die is even and the second die is odd. The probability of this happening is 916262=⨯ Adding these two probabilities will give us our final answer. 959194=+Problem10Draw height CH . We have that BH =1. From the Pythagorean Theorem, CH =48. Since CD =8, HD = 44882=-, andBD=HD -1, so BD=3 Problem 11Performing this operation several times yields the results of 133 for the second term, 55 for the third term, and 250 for the fourth term. The sum of the cubes of the digits of 250 equal 133, a complete cycle. The cycle is... excluding the first term, the 2nd , 3rd , and 4th terms will equal 133, 55, and 250, following the fourth term. Any term number that is equivalent to 1(mod3) will produce a result of 250. It just so happens that 2005≡1 (mod3), which leads us to the answer of 250.Problem 12 In order for the product of the numbers to be prime, of the dice have to be a , and the other die has to be a prime number. There are prime numbers (, , and ), and there is only one , and there are ways to choose which die will have the prime number, sothe probability is 101111)61(12)61(21112)61(63=⨯⨯=⎪⎪⎭⎫ ⎝⎛⨯⨯We can use the Principle of Inclusion-Exclusion to solve the problem as follows: We can count the number of multiples of 3 that are less than 2005, add the number of multiples of 4 that are less than 2005, and subtract the number of multiples of 12 twice that are less than 2005 (since those are counted twice in each of the 3 and 4 cases). Calculating, weget (where denotes the floor function). Problem 14Solution 1 The area of a triangle can be given by C ab sin 21. MC =1 because it is the midpoint of a side, and CD =2 because it is the same length as BC . Each angle of an equilateral triangleis 60º so ∠MCD =120º. The area is 23120sin )2)(1(21=︒ Solution 2In order to calculate the area of △CDM, we can use the formula bh A 21=, where CD is the base. We already know that CD =2, so the formula now becomes A =h . We can drop verticals down from A and M to points E and F , respectively. We can see that △AEC ∽△MFC . Now, we establish the relationship that MCAC MF AE =. We are given that 2=AC, and is the midpoint of AC , so 1=MC . Because △AEB is a 30—60—90 triangle and the ratio of the sides opposite the angles are 231-- AE is 3. Plugging those numbers in, we have 23=MF. Cross-multiplying, we see that23132=⇒⨯=⨯MF MF Since MF is the height △CDM , the area is 2/3Problem 15The only way to get a total of $ or more is if you pick a twenty and another bill, or if you pick both tens. There are a total of ways to choose bills out of . There are ways to choose a twenty and some other non-twenty bill. There is way to choose both twenties, and also way to choose both tens. Adding these up, we find that there are a total of ways to attain a sum of or greater, so there is a total probabilityof 14/28=0.5 .Let 02=++m px x have roots and . Thenab x b a x b x a x m px x ++-=--=++)())((22so )(b a p +-= and ab m =. Also, 02=++n mx x has roots 2a and 2b , so,4)(2)2)(2(22ab x b a x b x a x n mx x ++-=--=++ and )(2b a m +-=And ab n 4=. Thus 84)(42==+-=m m b a ab p n . Indeed, consider the quadratics 06416,016822=++=++x x x x .Problem 178=d 7 d c )6(8= dc b ))5((8=d c b a )))4(((8= d c b a ∙∙∙=48 d c b a ∙∙∙∙=2322 2323=∙∙∙∙=∙∙∙d c b a d c b a Solution using logarithmsWe can write a as 5log 4, b as 6log 5, c as 7log 6, and d as 8log 7. We know thata b log can be rewritten as ba log log , so232log 22log 34log 8log 7log 8log 6log 7log 5log 6log 4log 5log ===∙∙∙=⨯⨯⨯d c b a Problem 18 The only digits available to use in the phone number are 2, 3, 4, 5, 6, 7, 8, and 9. There are only 7 spots left among the 8 numbers, so we need to find the number of ways to choose 7 numbers from 8. The answer is justAlternatively, we could just choose 1 out of the 8 numbers not to be used. There are obviously 8 ways to do so.Problem 19To begin, we see that the remaining 30% of the students got 95 points. Assume that there are 20 students; we see that 2students got 70 points, 5 students got 80 points, 4 students got 85 points, 3 students got 90 points, and 6 students got 95 points. The median is 85, since the 10th and 11th terms are both 85. The mean is8620172020)6(95)3(90)4(85)5(80)2(70==++++. The difference between the meanand median, therefore, is 1 .Problem 20 We first look at how many times each number will appear in each slot. If we fix a number in a slot, then there are ways to arrange the other numbers, so each numberappears in each spot 24 times. Therefore, the sum of all such numbers isSincethereare such numbers, we divide to get 53332.8Solution (where the order of drawing slips matters)There are 10 ways to determine which number to pick. There are way to then draw those four slips with that number, and 40·39·38·37 total ways to draw four slips. Thus 37383940!410∙∙∙∙=p . There are 45210=⎪⎪⎭⎫ ⎝⎛ ways to determine which two numbers to pick for the second probability. There are ⎪⎪⎭⎫ ⎝⎛24 ways to arrange the order which we draw the non-equal slips, and in each order there are 3434⨯⨯⨯ ways to pick the slips, so 373839403464522∙∙∙∙∙∙=qHence, the answer is 162!4103464522=∙∙∙∙=p q Problem 22Since 1+2+…+n =2)1(+n n , the condition is equivalent to having an integer value for . This reduces, when, to having an integer value for . This fraction is an integer unless is an odd prime. There are 8 odd primes less than or equal to 25, so there are 24-8=16 numbers less than or equal to 24 that satisfy thecondition. Problem 23Since the height of both trapezoids are equal, and the area of ABEF is twice the area ofFECD , )2(22EF CD EF AB +=+, so . EF is exactly halfway between AB and CD , so 2CD AB EF +=. CD AB CD CD AB AB ++=++22, so AB/CD =5 Problem 24Let a b y b a x +=+=10,10, without loss of generality with b a >. Then222))((99)1111)(99())((m b a b a b a b a y x y x y x =+-=+-=+-=-It follows that , but so . Then we have 22)(33m b a =-. Thusis a perfect square. Also, because and have the same parity, is a one-digit odd perfect square, namely or 9. The latter case gives (a,b)=(10,1), which does not work. The former case gives (a,b )=(6,5) , whichworks, and we have 154335665=++=++m y x .Solution 2The first steps are the same as above. Let a b y b a x +=+=10,10, where we know that a and b are digits (whole numbers less than 10). Like above, we end up getting 2))((99)1111)(99(m b a b a b a b a =+-=+-. This is where the solution diverges.We know that the left side of the equation is a perfect square because m is an integer. If we factor 99 into its prime factors, we get 1132⨯. In order to get a perfect square on the left side, ))((b a b a +- must make both prime exponents even. Because the a and b are digits, a simple guess would be that )(b a + (the bigger number) equals 11 while )(b a -is a factor of nine (1 or 9). The correct guesses are 5,6==b a causing56,65==y x and m =33. The sum of the numbers is 154. Problem 25The question asks for the maximum possible number of elements. The integers from 1 to 24 can be included because you cannot make 125 with integers from 1 to 24 without the other number being greater than 100. The integers from 25 to 100 are left. They can be paired so the sum is 125: 25+100, 26+99, 27+98, …, 62+63. That is 38 pairs, and at most one number from each pair can be included in thein the set there are only the numbers 25 to 100 to consider. As 62+63 gives 125, the numbers 25 to 62 can be put in subset B without having two numbers add up to 125. Inthis way, subset B will have the numbers 1 to 62, and so2005 AMC 10B Answer Key1. A2. D3. D4. D5. C6.B7.B8.A 8. D 10. A11.E 12. E 13.C 14.C 15. D 16.D 17. B 18.D 19.B 20.C21.A 22.C 23.C 24.E 25.C。

2005年美国大学生数学建模大赛赛题

2005 MCM ProblemsPROBLEM A: Flood PlanningLake Murray in central South Carolina is formed by a large earthen dam, which was completed in 1930 for power production. Model the flooding downstream in the event there is a catastrophic earthquake that breaches the dam.Two particular questions:Rawls Creek is a year-round stream that flows into the Saluda River a short distance downriver from the dam. How much flooding will occur in Rawls Creek from a dam failure, and how far back will it extend?Could the flood be so massive downstream that water would reach up to the S.C. State Capitol Building, which is on a hill overlooking the Congaree River?PROBLEM B: TollboothsHeavily-traveled toll roads such as the Garden State Parkway , Interstate 95, and so forth, are multi-lane divided highways that are interrupted at intervals by toll plazas. Because collecting tolls is usually unpopular, it is desirable to minimize motorist annoyance by limiting the amount of traffic disruption caused by the toll plazas. Commonly, a much larger number of tollbooths is provided than the number of travel lanes entering the toll plaza. Upon entering the toll plaza, the flow of vehicles fans out to the larger number of tollbooths, and when leaving the toll plaza, the flow of vehicles is required to squeeze back down to a number of travel lanes equal to the number of travel lanes before the toll plaza. Consequently, when traffic is heavy, congestion increases upon departure from the toll plaza. When traffic is very heavy, congestion also builds at the entry to the toll plaza because of the time required for each vehicle to pay the toll.Make a model to help you determine the optimal number of tollbooths to deploy in a barrier-toll plaza. Explicitly consider the scenario where there is exactly one tollbooth per incoming travel lane. Under what conditions is this more or less effective than the current practice? Note that the definition of "optimal" is up to you to determine.2005 ICM ProblemPROBLEM C: Nonrenewable ResourcesSelect a vital nonrenewable or exhaustible resource (water, mineral, energy, food, etc.) for which your team can find appropriate world-wide historic data on its endowment, discovery, annual consumption, and price.The modeling tasks are:ing the endowment, discoveries, and consumption data, model thedepletion or degradation of the commodity over a long horizon using resource modeling principles.2.Adjust the model to account for future economic, demographic, politicaland environmental factors. Be sure to reveal the details of your model, provide visualizations of the model’s output, and explain limitations of the model.3.Create a fair, practical "harvesting/management" policy that may includeeconomic incentives or disincentives, which sustain the usage over a long period of time while avoiding severe disruption of consumption, degradation or rapid exhaustion of the resource.4.Develop a "security" policy that protects the resource against theft,misuse, disruption, and unnecessary degradation or destruction of the resource. Other issues that may need to be addressed are political and security management alternatives associated with these policies.5.Develop policies to control any short- or long-term "environmentaleffects" of the harvesting. Be sure to include issues such as pollutants, increased susceptibility to natural disasters, waste handling and storage, and other factors you deem appropriate.pare this resource with any other alternatives for its purpose. Whatnew science or technologies could be developed to mitigate the use and potential exhaustion of this resource? Develop a research policy to advance these new areas.。

2005AMC10-B

Wednesday, FEBRUARY 16, 2005Contest BThe MATHEMATICAL ASSOCIATION OF AMERICAAmerican Mathematics Competitions6th Annual American Mathematics Contest 10AMC 101. DO NOT OPEN THIS BOOKLET UNTIL YOUR PROCTOR GIVES THE SIGNAL TO BEGIN.2. This is a 25-question, multiple choice test. Each question is followed by answers marked A, B, C, D and E. Only one of these is correct.3. Mark your answer to each problem on the AMC 10 Answer Form with a #2 pencil. Check the blackened circles for accuracy and erase errors and stray marks completely. Only answers properly marked on the answer form will be graded.4. SCORING: You will receive 6 points for each correct answer, 2.5 points for each problem left unanswered, and 0 points for each incorrect answer.5. No aids are permitted other than scratch paper, graph paper, ruler, compass, protractor, erasers and calculators that are accepted for use on the SAT. No problems on the test will require the use of a calculator.6. Figures are not necessarily drawn to scale.7. Before beginning the test, your proctor will ask you to record certain information on the answer form. When your proctor gives the signal, begin working the problems. You will have 75 MINUTES to complete the test.8. When you finish the exam, sign your name in the space provided on the Answer Form.Stud ents who score 120 or above or finish in the top 1% on this AMC 10 will be invited to take the 23rd annual American Invitational Mathematics Examination (AIME) on Tuesday, March 8, 2005 or Tuesday, March 22, 2005. More details about the AIME and other information are on the back page of this test booklet.The Committee on the American Mathematics Competitions (CAMC) reserves the right to re-examine students before deciding whether to grant official status to their scores. The CAMC also reserves the right to disqualify all scores from a school if it is determined that the required security procedures were not followed.The publication, reproduction or communication of the problems or solutions of the AMC 10 during the period when students are eligible to participate seriously jeopardizes the integrity of the results. Dissemination at any time via copier, telephone, email, World Wide Web or media of any type is a violation of the competition rules. Copyright © 2005, The Mathematical Association of America1.A scout troop buys 1000candy bars at a price of ﬁve for $2.They thecandy bars at a price of two for $1.What was their proﬁt,in dollars?(A)100(B)200(C)300(D)4005002.A positive number x has the property that x %of x is 4.What is x ?(A)2(B)4(C)10(D)20403.A gallon of paint is used to paint a room.One third of the paint is used on the ﬁrst day.On the second day,one third of the remaining paint is used.What fraction of the original amount of paint is available to use on the third(A)110(B)19(C)13(D)49(E)94.For real numbers a and b ,deﬁne a b =√a 2+b 2.What is the value of(5 12) ((−12) (−5))?(A)0(B)172(C)13(D)13√2(E)265.Brianna is using part of the money she earned on her weekend job to buy several equally-priced CDs.She used one ﬁfth of her money to buy one third ofCDs.What fraction of her money will she have left after she buys all the (A)15(B)13(C)25(D)23(E)56.At the beginning of the school year,Lisa’s goal was to earn an A on at least 80%of her 50quizzes for the year.She earned an A on 22of the ﬁrst 30quizzes.If she is to achieve her goal,on at many of the remaining quizzes canshe earn a grade lower than an A?(A)1(B)23(D)4(E)57.A circle is inscribed in a square,then a square is inscribed in this circle,and ﬁnally,a circle is inscribed in this square.the ratio of the area of thesmaller circle to the area of the larger (A)π16(B)π8(C)3π16(D)π4(E)π28.An 8-foot by 10-foot ﬂoor is tiled with square tiles of size 1foot by 1foot.Each tile has a pattern consisting of four white quarter circles of radius 1/2foot centered at each corner of the tile.The of the tile is shaded.How many square feet of the(A)80−20π(B)60−10π(C)80−10π(D)60+10π(E)80+10π9.One fair die has faces 1,1,2,2,3,3and another has faces 4,4,5,5,6,6.The dice are rolled and the numberstopfaces areadded.What istheprobability thatthe sumwill be odd?(A)13(B)49(C)2(D)59(E)2310.In ABC ,we have AC =BC =7and AB =2.Suppose that D is a line AB such that B lies between A and D and CD =8.What is BD ?(A)3(B)2√3(C)4(D)5√211.The ﬁrst term of a sequence is 2005.Each succeeding term is the sum of thecubes of the digits of the previous term.What is the 2005th term of the se-quence?(A)29(B)55(C)85(D)133(E)25012.Twelve fair dice are rolled.What probability that the product of thenumbers on the top faces is prime?(A) 112 12(B) 16 12(C)216 11(D)52 16 11(E) 161013.numbers between 1and 2005are integer multiples of 3or 4but not12?(B)668(C)835(D)1002(E)116914.Equilateral ABC has side length 2,M is of AC ,and C is themidpoint of BD .What is the area of CDM (A)√22(B)34(C)√32(D)1(E)√215.An envelope contains eight bills:2ones,2ﬁves,2tens,and 2twenties.Twobills are drawn at replacement.What is the probability that their sum is $20or more?(A)14(B)25(C)37(D)12(E)2316.The quadratic equation x 2+mx +n =0has roots that are twice those ofx 2+px +m =0,and none of m ,n ,and p is zero.What is the value of n/p (A)1(B)2(C)4(D)8(E)17.Suppose that 4a =5,5b =6,6c =7,and 7d =8.What is a ·b ·c ·d ?(A)1(B)32(C)2(D)52(E)318.All of David’s telephone numbers have the form 555–abc –defg ,where a,b,c,d,e,f,and g are distinct digits and in increasing order,is either 0or 1.How many diﬀerent telephone numbers can David have?(A)1(B)2(C)7(E)919.On a certain math exam,10%of the students got 70points,25%got 80points,20%got 85points,15%got 90points,and the rest got 95is thediﬀerence between the mean and the median score on this exam?(A)0(B)1(C)2(D)4(E)520.What is the average (mean)of all 5-digit can be formed by usingeach of the digits 1,3,5,7,and 8exactly once?(A)48000(B)49999.5(C)53332.855555(E)56432.821.Forty slips are placed into a hat,each bearing a number 1,2,3,4,5,6,7,8,9,or 10,with each number entered on four slips.Four slips are drawn from the hat at random and without replacement.Let p be the probability that all four slips bear the same number.Let q be the probability that two of the slips bear a number a and the other two bear a number b =a .What is the value of q/p ?(A)162(B)180(C)324(D)360(E)72022.For how many positive integers n less than or equal to 24is n !evenly divisibleby 1+2+···+n ?(A)8(B)12(C)16(D)17(E)2123.In trapezoid ABCD we have AB parallel to DC ,E as the midpoint of BC ,and F as the midpoint of DA .The area of ABEF is twice the area of F ECD .What is AB/DC ?(A)2(B)3(C)5(D)6(E)824.Let x and y be two-digit integers such that y is obtained by reversing the digitsof x .The integers x and y satisfy x 2−y 2=m 2for some positive integer m .What is x +y +m ?(A)88(B)112(C)116(D)144(E)15425.A subset B of the set of integers from 1to 100,inclusive,has the property thatno two sum to 125.What is the maximum possible number ofelements in B ?(A)50(B)51(C)62(D)65(E)68Bob 数字签名者：BobDN ：cn=Bob,o=Shuren International School, ou=Student, email=shengyi.jiang @, c=CN日期：2010.11.1822:01:14 +08'00'WRITE TO US!Correspondence about the problems and solutions for this AMC 10 and ordersfor any of the publications listed below should be addressed to:American Mathematics CompetitionsUniversity of Nebraska, P.O. Box 81606Lincoln, NE 68501-1606Phone: 402-472-2257; Fax: 402-472-6087; email: amcinfo@ The problems and solutions for this AMC 10 were prepared by the MAA’s Committee on the AMC 10 and AMC 12 under the direction of AMC 10 Subcommittee Chair:Prof. Douglas Faires, Department of MathematicsYoungstown State University, Youngstown, OH 44555-00012005 AIMEThe AIME will be held on T uesday, March 8, 2005 with the alternate on March 22, 2005. It is a 15-question, 3-hour, integer-answer exam. You will be invited to participate only if you score 120 or above, or finish in the top 1% of the AMC 10, or if you score 100 or above or finish in the top 5% of the AMC 12. T op-scoring students on the AMC 10/12/AIME will be selected to take the USA Mathematical Olympiad (USAMO) on April 19 and 20, 2005. The best way to prepare for the AIME and USAMO is to study previous exams. 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All information (Rules and Instructions) needed to administer this exam iscontained in the TEACHERS’ MANUAL, which is outside of this package.PLEASE READ THE MANUAL BEFORE FEBRUARY 16.Nothing is needed from inside this package until February 16.2. Your PRINCIPAL or VICE PRINCIPAL must sign the Certification Formfound in the Teachers’ Manual.3. The Answer Forms must be mailed by First Class mail to the AMC no laterthan 24 hours following the examination.4. The publication, reproduction or communication of the problems or solu-tions of this test during the period when students are eligible to participate seriously jeopardizes the integrity of the results. Dissemination at any time via copier, telephone, email, World Wide Web or media of any type is a violation of the competition rules.Sponsored byThe MATHEMATICAL ASSOCIATION OF AMERICAUniversity of Nebraska – LincolnContributorsAkamai Foundation American Mathematical Association of T wo-Year Colleges American Mathematical Society American Society of Pension Actuaries American Statistical Association Art of Problem Solving Canada/USA Mathcamp Canada/USA Mathpath Casualty Actuarial Society Clay Mathematics Institute Institute for Operations Research and the Management Sciences L. G. Balfour & Company Mu Alpha Theta National Council of Teachers of Mathematics Pedagoguery Software Inc. Pi Mu Epsilon Society of Actuaries USA Math Talent Search W. H. Freeman & Company。

2010 AMC 12A Problems and Solution

2010 AMC 12A Problems and Solution Problem 1What is ?Solution.Problem 2A ferry boat shuttles tourists to an island every hour starting at 10 AM until its last trip, which starts at 3 PM. One day the boat captain notes that on the 10 AM trip there were 100 tourists on the ferry boat, and that on each successive trip, the number of tourists was 1 fewer than on the previous trip. How many tourists did the ferry take to the island that day?SolutionIt is easy to see that the ferry boat takes trips total. The total number of peopletaken to the island isProblem 3Rectangle , pictured below, shares of its area with square .Square shares of its area with rectangle . What is ?SolutionSolution 1Let , let , and let .Solution 2The answer does not change if we shift to coincide with , and add newhorizontal lines to divide into five equal parts:This helps us to see that and , where . Hence.Problem 4If , then which of the following must be positive?Solutionis negative, so we can just place a negative value into each expression and find the one that is positive. Suppose we use .Obviously only is positive.Problem 5Halfway through a 100-shot archery tournament, Chelsea leads by 50 points. For each shot a bullseye scores 10 points, with other possible scores being 8, 4, 2, and 0 points. Chelsea always scores at least 4 points on each shot. If Chelsea's next shots are bullseyes she will be guaranteed victory. What is the minimum value for ?SolutionLet be the number of points Chelsea currently has. In order to guarantee victory,we must consider the possibility that the opponent scores the maximum amount of points by getting only bullseyes.The lowest integer value that satisfies the inequality is .Problem 6A , such as 83438, 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 ?Solutionis at most , so is at most . The minimum value of is .However, the only palindrome between and is , which means thatmust be .It follows that is , so the sum of the digits is .Problem 7Logan is constructing a scaled model of his town. The city's water tower stands 40 meters high, and the top portion is a sphere that holds 100,000 liters of water. Logan's miniature water tower holds 0.1 liters. How tall, in meters, should Logan make his tower?SolutionThe water tower holds times more water than Logan'sminiature. Therefore, Logan should make his tower times shorterthan the actual tower. This is meters high, or choice .Also, the fact that doesn't matter since only the ratios are important.Problem 8Triangle has . Let and be on and , respectively,such that . Let be the intersection of segments and ,and suppose that is equilateral. What is ?SolutionLet .Since , triangle is a triangle, soProblem 9A solid cube has side length inches. A -inch by -inch square hole is cut into thecenter 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?SolutionSolution 1Imagine making the cuts one at a time. The first cut removes a box . Thesecond cut removes two boxes, each of dimensions , and the third cutdoes 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 2We can use Principle of Inclusion-Exclusion to find the final volume of the cube.There are 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 theirvolumes, as the central cube is included in each of these three cuts. Toget the correct result, we can take the sum of the volumes of the three cuts, and subtract the 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 3We 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 seen as a box, and each corner can be seen as abox..Solution 4First, you can find the volume, which is 27. Now, imagine there are three prisms of dimensions 2 x 2 x 3. Now subtract the prism volumes from 27. We have -9. From here we add two times 2^3, because we over-removed (LOL). This is 16 - 9 = 7 (A).Problem 10The first four terms of an arithmetic sequence are , , , and . Whatis the term of this sequence?Solutionand are consecutive terms, so the common difference is.The common difference is . The first term is and the term isProblem 11The solution of the equation can be expressed in the form . What is ?SolutionThis problem is quickly solved with knowledge of the laws of exponents and logarithms.Since we are looking for the base of the logarithm, our answer is .Problem 12In a magical swamp there are two species of talking amphibians: toads, whose statements 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, at least two are toads."How many of these amphibians are frogs?SolutionSolution 1We 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 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 exactly frog out of the two of them.Now suppose Mike is a toad. Then what he says is true because we already havetoads. However, if Brian is 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. Since either Chris or LeRoy is already a toad, Brain must bea 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 havefrogs total.Solution 2Start with Brian. If he is a toad, he tells the truth, hence Mike is a 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 toad.Hence we must have one toad and three frogs.Problem 13For how many integer values of do the graphs of and notintersect?SolutionThe image below shows the two curves for . The blue curve is ,which is clearly a circle with radius , and the red curve is a part of the curve.In the special case the blue curve is just the point , and as ,this point is on the red curve as well, hence they intersect.The case is symmetric to : the blue curve remains the same and the redcurve is flipped according to the axis. Hence we just need to focus on .Clearly, on the red curve there will always be points arbitrarily far from the origin: for example, as approaches 0, approaches . Hence the red curve intersects the blue one if and only if it contains a point whose distance from the origin is at most .At this point we can guess that on the red curve the point where is always closest to the origin, and skip the rest of this solution.For an exact solution, fix and consider any point on the red curve. Itsdistance from the origin is . To minimize this distance, it is enough tominimize . By the Arithmetic Mean-Geometric Mean Inequality we getthat this value is at least , and that equality holds whenever , i.e.,.Now recall that the red curve intersects the blue one if and only if its closest point is at most from the origin. We just computed that the distance between the originand the closest point on the red curve is . Therefore, we want to find all positiveintegers such that .Clearly the only such integer is , hence the two curves are only disjoint forand . This is a total of values.Solution 2:From the graph shown above, we see that there is a specific point closest to the center of the circle. Using some logic, we realize that as long as said furthest point is not inside or on the graph of the circle. This should be enough to conclude that the hyperbola does not intersect the circle.Therefore, for each value of k, we only need to check said value to determineintersection. Let said point, closest to the circle have coordinates derivedfrom the equation. Then, all coordinates that satisfyintersect the circle. Squaring, we findAfter multiplying through by and rearranging, wefind . We see this is a quadratic in and consider taking the determinant, which tells us that solutions are real when, after factoring:We plot this inequality on the number line to find it issatisfied for all values except: . We then eliminate 0 because it isextraneous as both and are points which coincide. Therefore,there are a total of values.Problem 14Nondegenerate has integer side lengths, is an angle bisector, ,and . What is the smallest possible value of the perimeter?SolutionBy the Angle Bisector Theorem, we know that . If we use the lowest possible integer values for AB and BC (the measures of AD and DC, respectively), then , contradicting the Triangle Inequality. If weuse the next lowest values (and ), the Triangle Inequality issatisfied. Therefore, our answer is , or choice .Problem 15A coin is altered so that the probability that it lands on heads is less than and when the coin is flipped four times, the probaiblity of an equal number of heads and tailsis . What is the probability that the coin lands on heads?SolutionLet be the probability of flipping heads. It follows that the probability of flipping tails is .The probability of flipping heads and tails is equal to the number of ways to flip ittimes the product of the probability of flipping each coin.As for the desired probability both and are nonnegative, we only need toconsider the positive root, henceApplying the quadratic formula we get that the roots of this equation are .As the probability of heads is less than , we get that the answer is . Problem 16Bernardo randomly picks 3 distinct numbers from the setand arranges them in descending order to form a 3-digit number. Silvia randomlypicks 3 distinct numbers from the set and also arranges them in descending order to form a 3-digit number. What is the probability that Bernardo's number is larger than Silvia's number?SolutionWe can solve this by breaking the problem down into cases and adding up theprobabilities.Case : Bernardo picks . If Bernardo picks a then it is guaranteed that hisnumber will be larger than Silvia's. The probability that he will pick a is .Case : Bernardo does not pick . Since the chance of Bernardo picking is , theprobability of not picking is .If Bernardo does not pick 9, then he can pick any number from to . SinceBernardo is picking from the same set of numbers as Silvia, the probability that Bernardo's number is larger is equal to the probability that Silvia's number is larger.Ignoring the for now, the probability that they will pick the same number is thenumber of ways to pick Bernardo's 3 numbers divided by the number of ways to pick any 3 numbers.We get this probability to beProbability of Bernardo's number being greater isFactoring the fact that Bernardo could've picked a but didn't:Adding up the two cases we getProblem 17Equiangular hexagon has side lengths and. The area of is of the area of the hexagon.What is the sum of all possible values of ?SolutionIt is clear that is an equilateral triangle. From the Law of Cosines, we getthat . Therefore, the area of is.If we extend , and so that and meet at , and meetat , and and meet at , we find that hexagon is formed bytaking equilateral triangle of side length and removing three equilateraltriangles, , and , of side length . The area of istherefore.Based on the initial conditions,Simplifying this gives us . By Vieta's Formulas we know that thesum of the possible value of is .Problem 18A 16-step path is to go from to with each step increasing either the -coordinate or the -coordinate by 1. How many such paths stay outside or on theboundary of the square , at each step?SolutionBrute Force SolutionThe number of ways to reach any point on the grid is equal to the number ofways to reach plus the number of ways to reach . Using this recursion, we can draw the diagram and label each point with the number of ways to reach it and go up until we reach the end. Luckily, the figure is not so big that this is too time-consuming or difficult to do.For example:etc.We soon reachCombinatorial Solution 1By symmetry we only need to count the paths that go through the second quadrant(, ).For each of these paths, let be the first point when it reaches . Clearlyand the previous point on such path has to be .Fix the value of . There are ways how the path can go from to, and then there are ways how the path can go from to .Hence for we get paths, for we getpaths, and for we getpaths. This gives us paths through the second quadrant, hence the total number of paths is .Combinatorial Solution 2Each path that goes through the second quadrant must pass through exactly one ofthe points , , and .There is exactly path of the first kind, paths of the second kind, andpaths of the third type. The conclusion remains the same. Problem 19Each 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 boxand successively draws a single marble at random from each box, in order. Shestops when she first draws a red marble. Let be the probability that Isabella stops after drawing exactly marbles. What is the smallest value of for which?SolutionThe probability of drawing a white marble from box is . The probability ofdrawing a red marble from box is .The probability of drawing a red marble at box is thereforeIt is then easy to see that the lowest integer value of that satisfies the inequalityis .Problem 20Arithmetic sequences and have integer terms withand for some . What is the largest possible value of ?SolutionSolution 1Since and have integer terms with , we can write the terms of each sequence aswhere and () are the common differences of each, respectively.Sinceit is easy to see that.Hence, we have to find the largest such that and are both integers. The prime factorization of is . We list out all the possible pairs thathave a product ofand soon find that the largest value is for the pair , and so thelargest value is .Solution 2As above, let and for some .Now we get , hence. Therefore divides . Andas the second term is greater than the first one, we only have to consider theoptions .For we easily see that for the right side is less than and for anyother it is way too large.For we are looking for such that . Notethat must be divisible by . We can start looking for the solution by trying thepossible values for , and we easily discover that for we get, which has a suitable solution .Hence is the largest possible . (There is no need to check anymore.)Problem 21The graph of lies above the lineexcept at three values of , where the graph and the line intersect. What is the largest of these values?SolutionThe values in which intersect atare the same as the zeros of .Since there are zeros and the function is never negative, all zeros must bedouble roots because the function's degree is .Suppose we let , , and be the roots of this function, and letbe the cubic polynomial with roots , , and .In order to find we must first expandout the terms of .[Quick note: Since we don't know , , and , we really don't even need the last 3terms of the expansion.]All that's left is to find the largest root of .Problem 22What is the minimum value of ?SolutionSolution 1If we graph each term separately, we will notice that all of the zeros occur at , where is any integer from to , inclusive.The minimum value occurs where the sum of the slopes is at a minimum, since it is easy to see that the value will be increasing on either side. That means theminimum must happen at some .The sum of the slope at isNow we want to minimize . The zeros occur at and , whichmeans the slope is where .We can now verify that both and yield .Solution 2Rewrite the given expression as follows:Imagine the real line. For eachimagine that there are boys standing at the coordinate . We now need to place a girl on the real line in such a way that the sum of her distances from all the boys is minimal, and we need to compute this sum.Note that there are boys in total. Let'slabel them from 1 (the only boy placed at ) to (the last boy placed at .Clearly, the minimum sum is achieved if the girl's coordinate is the median of the boys' coordinates. To prove this, place the girl at the median coordinate. If you nowmove her in any direction by any amount , there will be boys such that shemoves away from this boy. For each of the remaining boys, she moves at mostcloser, hence the total sum of distances does not decrease.Hence the optimal solution is to place the girl at the median coordinate. Or, more precisely, as is even, we can place her anywhere on the segment formed by boyand boy : by extending the previous argument, anywhere on thissegment the sum of distances is the same.By trial and error, or by solving the quadratic equation we getthat boy number is the last boy placed at and the next boy is the one placedat . Hence the given expression is minimized for any . Common part of both solutionsTo find the minimum, we want to balance the expression so that it is neither top norbottom heavy. .Now that we know that the sum of the first 84 's is equivalent to the sum of 's 85to 119, we can plug either or to find the minimum.Note that the terms to are negative, and the terms toare positive. Hence we get: andHence the total sum of distances is.Solution 3Given that the minimum exists, we know that we want all s to cancel out. Thus, we want to find some such that. It thereforefollows that . This gives , from which it follows that .There are therefore positive ones and , or negative ones, so.Problem 23The number obtained from the last two nonzero digits of is equal to . What is?SolutionWe will use the fact that for any integer ,First, we find that the number of factors of in is equal to. Let . The we want is therefore the lasttwo digits of , or . If instead we find , we know that, what we are looking for, could be , ,, or . Only one of these numbers will be a multiple of four, and whichever one that is will be the answer, becausehas to be a multiple of 4.If we divide by by taking out all the factors of in , we can write aswhere where every multiple of 5 is replaced by the number with all its factors of 5 removed. Specifically, every number in the form is replaced by , and every number in the form is replaced by .The number can be grouped as follows:Where the first line is composed of the numbers in that aren't multiples of five,the second line is the multiples of five and not 25 after they have been divided by five, and the third line is multiples of 25 after they have been divided by 25.Using the identity at the beginning of the solution, we can reduce toUsing the fact that (or simply the fact thatif you have your powers of 2 memorized), we can deduce that. Therefore .Finally, combining with the fact that yields . Problem 24Let . The intersection ofthe domain of with the interval is a union of disjoint open intervals. What is ?SolutionThe question asks for the number of disjoint open intervals, which means we need to find the number of disjoint intervals such that the function is defined within them.We note that since all of the factors are inside a logarithm, the function isundefined where the inside of the logarithm is equal to or less than .First, let us find the number of zeros of the inside of the logarithm.After counting up the number of zeros for each factor and eliminating the excess cases we get zeros and intervals.In order to find which intervals are negative, we must first realize that at every zero of each factor, the sign changes. We also have to be careful, as some zeros are doubled, or even tripled, quadrupled, etc.The first interval is obviously positive. This means the next interval is negative. Continuing the pattern and accounting for doubled roots (which do not flipsign), we realize that there are negative intervals from to . Since the function issymmetric, we know that there are also negative intervals from to .And so, the total number of disjoint open intervals isProblem 25Two quadrilaterals are considered the same if one can be obtained from the other by a rotation and a translation. How many different convex cyclic quadrilaterals are there with integer sides and perimeter equal to 32?Solution 1It should first be noted that given any quadrilateral of fixed side lengths, the angles can be manipulated so that the quadrilateral becomes cyclic.Denote , , , and as the integer side lengths of the quadrilateral. Without lossof generality, let .Since , the Triangle Inequality implies that .We will now split into cases.Case : (side lengths are equal)Clearly there is only way to select the side lengths , and no matter howthe sides are rearranged only unique quadrilateral can be formed.Case : or (side lengths are equal)If side lengths are equal, then each of those side lengths can only be integers fromto except for (because that is counted in the first case). Obviously there is stillonly unique quadrilateral that can be formed from one set of side lengths, resultingin a total of quadrilaterals.Case : (pairs of side lengths are equal)and can be any integer from to , and likewise and can be any integer fromto . However, a single set of side lengths can form different cyclic quadrilaterals(a rectangle and a kite), so the total number of quadrilaterals for this case is.Case : or or (side lengths areequal)If the equal side lengths are each , then the other sides must each be , whichwe have already counted in an earlier case. If the equal side lengths are each ,there is possible set of side lengths. Likewise, for side lengths of there are sets.Continuing this pattern, we find a total ofsets of side lengths. (Be VERY careful when adding up the total for this case!) For each set of side lengths, there are possible quadrilaterals that can be formed, so the total number ofquadrilaterals for this case is .Case : (no side lengths are equal) Using the same countingprinciples starting from and eventually reaching , we find that thetotal number of possible side lengths is . There are ways to arrange the sidelengths, but there is only unique quadrilateral for rotations, so the number ofquadrilaterals for each set of side lengths is . The total number of quadrilaterals is .And so, the total number of quadrilaterals that can be made is.Solution 2As with solution we would like to note that given any quadrilateral we can changeits angles to make a cyclic one.Let be the sides of the quadrilateral.There are ways to partition . However, some of these will not bequadrilaterals since they would have one side bigger than the sum of the other three.This occurs when . For , . There are ways topartition . Since could be any of the four sides, we have counteddegenerate quadrilaterals. Similarly, there are , for other values of . Thus, there arenon-degenerate partitions of by the hockey stick theorem. However, foror , each quadrilateral is counted times, for eachrotation. Also, for , each quadrilateral is counted twice. Since thereis quadrilateral for which , and for which , thereare quads for which or .Thus there are total quadrilaterals.。

2006 美国数学竞赛 AMC 12a solutions 完整答案

Phone: 402-472-2257; Fax: 402-472-6087; email: amcinfo@ The problems and solutions for this AMC 12 were prepared by the MAA’s Committee on the
AMC 10 and AMC 12 under the direction of AMC 12 Subcommittee Chair: Prof. David Wells, Department of Mathematics
6. (A) Let E represent the end of the cut on DC, and let F represent the end of the cut on AB. For a square to be formed, we must have
DE = y = F B and DE + y + F B = 18, so y = 6.
2. (C) By the deﬁnition we have
h ⊗ (h ⊗ h) = h ⊗ (h3 − h) = h3 − (h3 − h) = h.
3. (B) Mary is (3/5)(30) = 18 years old.
4. (E) The largest possible sum of the two digits representing the minutes is 5+9 = 14, occurring at 59 minutes past each hour. The largest possible single digit that can represent the hour is 9. This exceeds the largest possible sum of two digits that can represent the hour, which is 1 + 2 = 3. Therefore, the largest possible sum of all the digits is 14 + 9 = 23, occurring at 9:59.

美国数学竞赛AMC8 -- 2005年真题解析(英文解析+中文解析)

如果Karl 等一天，可以省2.5元。

共4块，答案是D。

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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2005 AMC 12A Problems and Solution Problem 1Two is of and of . What is ?Solution. Problem 2The equations and have the sam e solution. What is the value of ?SolutionProblem 3A rectangle with diagonal length is twice as long as it is wide. What is the area of the rectangle?SolutionLet be the width, so the length is . By the Pythagorean Theorem,. The area of the rectangle is.Problem 4A store norm ally sells windows at $each. This week the store is offering one free window for each purchase of four. Dave needs seven windows and Doug needs eight windows. How much will they save if they purchase the windows together rather than separately?SolutionFor windows, the store offers a discount of (floor function). Davereceives a discount of and Doug receives a disc ount of. These am ount to dollars in discounts. Together, they receivea discount of , so they save .Problem 5The average (m ean) of 20 numbers is 30, and the average of 30 other num bers is 20. What is the average of all 50 numbers?SolutionThe sum of the first 20 numbers is and the sum of the other 30 numbers is. Hence the overall average is .Problem 6Josh and Mike live 13 miles apart. Yesterday, Josh started to ride his bicycle toward Mike's house. A little later Mike started to ride his bicycle toward Josh's house. Whenthey m et, Josh had ridden for twice the length of tim e as Mike and at four-fifths of Mike's rate. How m any miles had Mike ridden when they m et?SolutionLet be the distances traveled by Josh and Mike, respectively, and let bethe tim e and rate of Mike. Using , we have that and. Then.Problem 7Square is inside the square so that each side of can beextended to pass through a vertex of . Square has side lengthand . What is the area of the inner square ?SolutionArguable the hardest part of this question is to visualize the diagram. Since each side of can be extended to pass through a vertex of , we realizethat must be tilted in such a fashion. Let a side of be .Notice the right triangle (in blue) with legs and hypotenuse. By the Pythagorean Theorem, we have. Thus,Problem 8Let , and be digits withWhat is ?SolutionClearly the two quantities are both integers, so we check the prime factorization of. It is easy to see now that works, so theanswer is .Problem 9There are two values of for which the equation has only one solution for . What is the sum of these values of ?SolutionSolution 1A quadratic equation always has two roots, unless it has a double root. That m eans we can write the quadratic as a square, and the coefficients 4 and 9 suggest this.Completing the square, , so. The sum of these is .Solution 2Another method would be to use the quadratic form ula, since our coefficient isgiven as 4, the coefficient is and the constant term is . Hence,Because we want only a single solution for, the determinant m ust equal 0. Therefore, we can write whichfactors to ; using Vieta's formulas we see that the sum of thesolutions for is the opposite of the coeffi cient of , or .Using the discriminant, the result m ust equal .Therefore, or , giving a sum of .Problem 10A wooden cube units on a side is painted red on all six faces and then cut into unit cubes. Exactly one-fourth of the total number of faces of the unit cubes are red. What is ?SolutionThere are sides total on the unit cubes, and are painted red.Problem 11How many three-digit numbers satisfy the property that the m iddle digit is the average of the first and the last digits?SolutionSolution 1Let the digits be so that . In order for this to be an integer,and have to have the sam e parity. There are possibilities for , and for .depends on the value of both and and is unique for each . Thus ouranswer is .Thus, the three digits form an arithmetic sequence.▪If the num bers are all the sam e, then there are possible three-digit numbers.▪If the numbers are different, then we count the number of strictly increasingarithmetic sequences between and and multiply by 2 for the decreasing ones:This gives us . However, the question asks for three-digit numbers, so we have to ignore the four sequences starting with . Thus our answeris .Problem 12A line passes through and . How m any other points withinteger coordinates are on the line and strictly between and ?SolutionFor convenience’s sake, we can transform to the origin and to (thisdoes not change the problem). The line has the equation.The coordinates are integers if , so the values of are , with a totalof coordinates.Problem 13The regular 5-point star is drawn and in each vertex, there is a number.Each and are chosen such that all 5 of them cam e from set. Each letter is a different number (so one possible way is). Let be the sum of the num bers onand , and so forth. If and form an arithmetic sequence(not necessarily in increasing order), find the value of .Solution. The sumwill always be , so the arithm etic sequence has a sum of . Since is the middle term, it must be theaverage of the five num bers, of .Problem 14On a standard die one of the dots is removed at random with each dot equally likely to be chosen. The die is then rolled. What is the probability that the top face has an odd number of dots?SolutionThere are dots total. Casework:▪The dot is rem oved from an even face. There is a chance ofthis happening. Then there are 4 odd faces, giving us a probability of .▪The dot is rem oved from an odd face. There is a chance ofthis happening. Then there are 2 odd faces, giving us a probability of .Thus the answer is .Problem 15Let be a diam eter of a circle and be a point on with . Letand be points on the circle such that and is a second diameter.What is the ratio of the area of to the area of ?SolutionSolution 1Notice that the bases of both triangles are diameters of the circle. Hence the ratio ofthe areas is just the ratio of the heights of the triangles, or (is the foot of the perpendicular from to ).Call the radius . Then , . Using the PythagoreanTheorem in , we get .Now we have to find . Notice , so we can write theproportion:By the Pythagorean Theorem in , we have.Our answer is .Solution 2Let the centre of the circle be .Note that .is midpoint of .is midpoint of Area of Area of Area ofArea of .Problem 16Three circles of radius are drawn in the first quadrant of the -plane. The first circle is tangent to both axes, the second is tangent to the first circle and the -axis, and the third is tangent to the first circle and the -axis. A circle of radius istangent to both axes and to the second and third circles. What is ?SolutionWithout loss of generality, let . Draw the segm ent between the center of thethird circle and the large circle; this has length . We then draw the radius of the large circle that is perpendicular to the x-axis, and draw the perpendicular from this radius to the center of the third circle. This gives us a right triangle with legsand hypotenuse. The Pythagorean Theorem yields:Quite obviously , so and .Problem 17A unit cube is cut twice to form three triangular prisms, two of which are congruent, as shown in Figure 1. The cube is then cut in the sam e m anner along the dashed lines shown in Figure 2. This creates nine pieces. What is the volum e of the piece that contains vertex ?SolutionIt is a pyramid, so .Problem 18Call a number "prime-looking" if it is composite but not divisible by 2, 3, or 5. The three sm allest prime-looking numbers are 49, 77, and 91. There are 168 prime numbers less than 1000. How many prime-looking numbers are there less than 1000?SolutionThe given states that there are prime numbers less than , which is a fact we m ust som ehow utilize. Since there seem s to be no easy way to directly calculate the number of "prime-looking" numbers, we can apply com plementary counting.We can split the num bers from to into several groups:. Hence, the num ber of prime-looking numbers is(note that are primes).We can calculate using the Principle of Inclusion-Exclusion: (thevalues of and their intersections can be found quite easily)Substituting, we find that our answer is .Problem 19A faulty car odom eter proceeds from digit 3 to digit 5, always skipping the digit 4, regardless of position. If the odometer now reads 002005, how many miles has the car actually traveled?SolutionWe find the number of numbers with a and subtract from. Quick counting tells us that there are numbers with a 4 in the hundreds place, numberswith a 4 in the tens place, and numbers with a 4 in the units place (counting). Now we apply the Principle of Inclusion-Exclusion. There are numberswith a 4 in the hundreds and in the tens, and for both the other two intersections.The intersection of all three sets is just . So we get:Alternatively, consider that counting without the number is almost equivalent tocounting in base ; only, in base , the number is not counted. Since is skipped,the sym bol represents miles of travel, and we have traveled miles. Bybasic conversion, .Problem 20For each in , defineLet , and for each integer . Forhow m any values of in is ?SolutionFor the two functions and ,wecan see that as long as is between and , will be in the right domain.Therefore, we don't need to worry about the dom ain of . Also, every tim e wechange , the final equation will be in a different form and thus we will get adifferent value of x. Every tim e we have two choices for ) and altogether wehave to choose tim es. Thus, .Problem 21How many ordered triples of integers, with , , and ,satisfy both and ?SolutionCasework upon :▪: Then . Thus we get .▪: Then . Thus we get .▪: Then the exponent of becom es huge, and since there is noway we can satisfy the second condition. Hence we have two ordered triples . Problem 22A rectangular box is inscribed in a sphere of radius . The surface area of is 384,and the sum of the lengths of its 12 edges is 112. What is ?SolutionThe box P has dimensions , , and . Therefore,▪▪Now we m ake a formula for . Since the diam eter of the sphere is the spacediagonal of the box,▪We square :▪We get thatProblem 23Two distinct num bers and are chosen randomly from the set .What is the probability that is an integer?SolutionLet , so . Define , ; then , so .Here we can just make a table and count the num ber of values of per value of .The largest possible value of is 12, and we get.The total number of ways to pick two distinct numbers is ,so we get a probability of .Problem 24Let . For how m any polynomials does thereexist a polynomial of degree 3 such that ?SolutionSince has degree three, then has degree six. Thus, hasdegree six, so must have degree two, since has degree three.Hence, we conclude , , and must each be , , or . Since aquadratic is uniquely determined by three points, there can bedifferent quadratics after each of the values of , , and are chosen.However, we have included which are not quadratics. Nam ely,Clearly, we could not have included any other constant functions. For any linearfunction, we have . Again, it is pretty obvious that we have not included any other linear functions. Therefore, the desired answer is.Problem 25Let be the set of all points with coordinates , where and are eachchosen from the set . How many equilateral triangles have all their vertices in ?SolutionSolution 1 (non-rigorous)For this solution, we will just find as many solutions as possible, until it becom es intuitive that there are no m ore triangles left.Take an unit cube. We try to m ake three of its vertices form an equilateral triangle. This we find is possible by taking any vertex, and connecting the three adjacentvertices into a triangle. This triangle will have a side length of ; a quick further examination of this cube will show us that this is the only possible side length. Each of these triangles is determined by one vertex of the cube, so in one cube we have 8 equilateral triangles. We have 8 unit cubes, and then the entire cube, giving us 9cubes and equilateral triangles.It m ay be tem pting to connect the centers of the faces and to call that a cube, but a quick look at this tells us that that figure is actually an octahedron.Now, we look for any additional equilateral triangles. Since the space diagonals of the cube cannot m ake an equilateral triangle, we will assume symmetry in the cube.A bit m ore searching shows us that connecting the midpoints of three non-adjacent, non-parallel edges gives us m ore equilateral triangles.Notice that picking these three edges leaves two vertices alone, and that picking any two opposite vertices determine two equilateral triangles. Hence there areadditional equilateral triangles, for a total of .Solution 2 (rigorous)The three dim ensional distance form ula shows that the lengths of the equilateraltriangle must be , which yields the possible edge lengths ofSom e casework shows that are the only lengths that work, from which we can use the sam e counting argument as above.See Math Jam solution.。