Problem and Solution
生活中的问题及解决方法英语作文初二
生活中的问题及解决方法英语作文初二全文共10篇示例，供读者参考篇1In our daily life, we may encounter various problems that bother us. But don't worry, here are some common problems and their solutions.1. Waking up early in the morningProblem: It's always difficult to wake up early in the morning, especially during weekdays when we have to go to school.Solution: Try setting an alarm clock far away from your bed so that you have to get up to turn it off. Also, try going to bed early to ensure you get enough sleep.2. ProcrastinationProblem: Many students have the habit of procrastinating, which affects their study and work efficiency.Solution: Make a to-do list and prioritize tasks. Break down large tasks into smaller ones to make them more manageable. Set a timer for work sessions and reward yourself after completing tasks.3. Feeling stressedProblem: Sometimes we feel overwhelmed with schoolwork, exams, or personal issues.Solution: Take breaks to rest and relax. Talk to someone you trust about your feelings. Practice mindfulness, meditation, or exercise to relieve stress.4. Unhealthy eating habitsProblem: Eating too much junk food or skipping meals can lead to health issues.Solution: Plan and prepare healthy meals and snacks in advance. Drink plenty of water and eat a variety of fruits and vegetables. Limit sugary drinks and snacks.5. Lack of exerciseProblem: Spending too much time on screens or sitting for long periods can be harmful to our health.Solution: Set aside time for physical activity each day. Go for a walk, ride a bike, dance, or do yoga. Find activities you enjoy and make them a regular part of your routine.Remember, it's normal to face problems in life, but there are always solutions to overcome them. Stay positive, be proactive,and seek help when needed. Life is full of challenges, but with the right attitude and strategies, we can conquer any problem that comes our way!篇2Title: Problems in Daily Life and How to Solve ThemHey guys! Today I want to talk about some of the problems we face in our daily lives and how we can solve them. Life can be tough sometimes, but with the right attitude and approach, we can overcome any obstacle that comes our way.One common problem we all face is time management. We often find ourselves running out of time to complete all our tasks and homework. The key to solving this problem is to make a schedule and stick to it. We should prioritize our tasks and allocate specific time slots for each of them. By staying organized and disciplined, we can make sure we have enough time to do everything we need to do.Another problem that many of us face is peer pressure. We often feel pressured to fit in with our friends or classmates, even if it means doing things that we are not comfortable with. The best way to deal with peer pressure is to stand up for ourselves and our values. We should not be afraid to say no if somethingdoesn't feel right to us. Surrounding ourselves with supportive and positive friends can also help us resist negative peer pressure.One more problem that can be challenging is stress. We all experience stress from time to time, whether it's from school, family, or other responsibilities. To cope with stress, we can try practicing relaxation techniques such as deep breathing, meditation, or yoga. It's also important to take breaks, exercise regularly, and get enough sleep to maintain our mental and physical well-being.Lastly, a common problem that many of us face is conflicts with others. Whether it's with our friends, family, or teachers, disagreements can arise that cause tension and stress. The key to resolving conflicts is communication. We should express our feelings and listen to the other person's perspective in a calm and respectful manner. By finding common ground and compromising, we can work together to find a solution that benefits everyone involved.In conclusion, life is full of challenges, but with the right mindset and approach, we can overcome any problem that comes our way. By managing our time effectively, standing up to peer pressure, coping with stress, and resolving conflicts throughcommunication, we can lead happier and more fulfilling lives. Remember, we are all capable of facing and solving life's problems – we just need to believe in ourselves and keep pushing forward!篇3Title: Problems in Daily Life and Their SolutionsHi everyone! Today I want to talk about some problems we might face in our daily lives and how we can solve them. Let's get started!Problem 1: Forgetting to do homeworkDo you ever forget to do your homework? It happens to all of us! One solution is to write down your homework in a planner or notebook so you can remember what you need to do. You can also set reminders on your phone or ask your parents to help you remember.Problem 2: Getting up early for schoolWaking up early can be tough, especially when you're tired. One solution is to go to bed early so you can get enough sleep. You can also set an alarm clock across the room so you have to get out of bed to turn it off.Problem 3: Fighting with friendsArguments with friends can be upsetting. One solution is to talk calmly and listen to each other's point of view. You can also take a break and cool off before discussing the issue rationally.Problem 4: Eating unhealthy snacksSometimes we can't resist unhealthy snacks, but it's important to eat nutritious foods. One solution is to keep healthy snacks on hand, like fruits and nuts. You can also limit the junk food you buy so you're not tempted to eat it.Problem 5: ProcrastinationPutting off tasks can lead to stress and anxiety. One solution is to break tasks into smaller steps and set deadlines for each step. You can also reward yourself for completing tasks on time.Remember, everyone faces problems in their daily lives. But with patience and determination, we can find solutions and make our lives happier and more productive. I hope these tips help you tackle any challenges you encounter. Good luck!篇4Title: Life Problems and SolutionsHey guys, have you ever faced any problems in your daily life and wondered how to solve them? Well, don't worry, I'm here to help you out! In this article, we will talk about some common problems we face and the solutions to overcome them.First of all, let's talk about the problem of time management.I know we all have a lot of things to do every day, like homework, extracurricular activities, and spending time with family and friends. It can be really hard to manage our time effectively. But don't worry, because there is a simple solution to this problem. One way to manage your time better is to make a schedule or a to-do list. By writing down what you need to do and when you need to do it, you can stay organized and make sure you have enough time for everything.Another common problem we face is peer pressure. Sometimes our friends might want us to do things that we are not comfortable with, like skipping class or trying drugs. It can be really hard to say no to our friends, but it's important to stand up for what we believe in. One way to deal with peer pressure is to be assertive and say no firmly. Remember, true friends will respect your decisions and won't pressure you into doing something you don't want to do.Lastly, let's talk about the problem of stress. As students, we have to deal with a lot of pressure from school, family, and friends. It's important to find healthy ways to cope with stress, like exercising, talking to someone you trust, or practicing mindfulness. Remember, it's okay to take a break and take care of yourself.So, guys, next time you face a problem in your life, don't worry! Just remember these simple solutions and you'll be able to overcome any challenge that comes your way. Life is full of ups and downs, but with the right attitude and mindset, you can conquer anything. Stay positive and keep smiling! Life is beautiful, and so are you!篇5Life is full of problems, big and small. Sometimes it's hard to find a solution, but there's always a way to tackle them. Let me share with you some common problems and their solutions that we encounter in our daily lives.Problem 1: Waking up early in the morningSolution: Set an alarm clock and place it far away from your bed so you have to get up to turn it off. Also, go to bed early to ensure you get enough sleep.Problem 2: Forgetting homework or important datesSolution: Use a planner or diary to write down important dates and deadlines. Check it every day to stay on top of your tasks.Problem 3: ProcrastinationSolution: Break down tasks into smaller, manageable parts and set deadlines for each part. Reward yourself after completing each task to stay motivated.Problem 4: Fighting with friendsSolution: Communication is key. Talk to your friend about the issue calmly and try to understand their perspective. Apologize if needed and work together to find a solution.Problem 5: Feeling stressed or overwhelmedSolution: Take breaks when needed, practice deep breathing or mindfulness, and talk to someone you trust about how you're feeling.Remember, it's important to face problems head-on and not avoid them. By being proactive and seeking solutions, you can overcome any challenge that comes your way. Life is full of upsand downs, but with the right mindset and approach, you can navigate through it successfully.篇6Living in this world, there are always many problems that we need to solve. Some problems are big and some problems are small, but no matter what, we need to find a way to solve them. Today, I want to talk about some problems that we might face in our daily lives and how we can solve them.One common problem that many people face is not having enough time to do all the things they need to do. This problem can be solved by making a schedule and planning out your day. By setting aside specific times for each task, you can make sure that you have enough time to get everything done.Another problem that people often face is feeling stressed or overwhelmed. This problem can be solved by taking breaks and practicing self-care. It's important to take time for yourself and do things that make you happy, whether it's reading a book, going for a walk, or spending time with loved ones.One more problem that many people face is not knowing how to communicate effectively with others. This problem can be solved by practicing good communication skills, such as activelistening and using "I" statements to express your feelings. By improving your communication skills, you can build better relationships with others and avoid misunderstandings.In conclusion, there are many problems that we might face in our daily lives, but with the right mindset and approach, we can find solutions to these problems. By making a plan, taking care of ourselves, and improving our communication skills, we can overcome any challenge that comes our way. Remember, it's okay to ask for help when you need it, and it's important to stay positive and believe in yourself.篇7Title: Everyday Problems and Solutions in My LifeHey everyone! Today, I want to talk about some of the problems I face in my everyday life and how I try to solve them. Life can be tough sometimes, but with a positive attitude and some creative thinking, we can overcome any challenge!One common problem I face is forgetting to do my homework. It's so easy to get caught up in playing games or watching TV that I forget all about my assignments. To solve this problem, I started using a planner to write down all my tasks anddeadlines. Now I can easily keep track of what I need to do and when it's due. It's been a huge help!Another issue I often encounter is feeling stressed about school exams. I used to get really anxious before a test, which made it hard for me to focus and study effectively. To combat this, I started setting aside time each day to review my notes and practice problem-solving. I also learned some relaxation techniques, like deep breathing and meditation, to calm my nerves. Now I feel more confident and prepared for exams.One more problem I face is getting up early in the morning. I'm not a morning person at all, and I struggle to wake up on time for school. To tackle this issue, I started going to bed earlier and setting multiple alarms to ensure I wake up on time. I also try to create a morning routine that I enjoy, like listening to music or taking a quick walk, to jumpstart my day with positivity.In conclusion, life is full of challenges, but we can always find ways to overcome them. By staying organized, staying focused, and staying positive, we can tackle any problem that comes our way. So don't give up when things get tough – keep pushing forward and you'll be amazed at what you can achieve!篇8Living in this big world is so hard sometimes. There are so many problems we face every day! But don't worry, because I'm here to tell you some solutions to these problems.One problem we face is forgetting our homework. It happens to the best of us, but it's important to remember our assignments so we can do well in school. One solution is to write down our homework in a notebook or on a piece of paper as soon as it's assigned. That way, we won't forget it by the time we get home! Another solution is to set a reminder on our phone or ask our parents to help us remember.Another problem we face is getting enough sleep. Sometimes it's hard to go to bed early, especially when we have so much fun stuff to do! But it's super important to get enough rest so we can focus in school and have energy throughout the day. One solution is to create a bedtime routine, like brushing our teeth, reading a book, or listening to music before bed. This can help us relax and get ready for sleep. Another solution is to turn off all screens, like phones and computers, at least an hour before bed. The bright lights can keep us awake and make it harder to fall asleep.One more problem we face is eating healthy. It's so tempting to reach for a snack or fast food, but it's important to eat fruits,vegetables, and other nutritious foods to keep our bodies strong. One solution is to pack healthy snacks, like an apple or carrots, to bring to school or when we go out. Another solution is to involve our parents in meal planning and cooking. They can help us make nutritious meals that taste good too!In conclusion, life is full of challenges, but with a little creativity and effort, we can overcome them. Remember to write down your homework, create a bedtime routine, and eat healthy foods. You've got this!篇9Life is full of problems, but don't worry, because there are always solutions to every problem! Let me tell you about some common problems in life and how to solve them.One problem that many people face is feeling stressed out. This can be caused by too much homework, too many activities, or just feeling overwhelmed. The solution to this problem is to take breaks and relax. You can go for a walk, listen to your favorite music, or talk to a friend. Remember, it's important to take care of yourself and not let stress get the best of you.Another problem that some people have is not getting enough sleep. This can make you feel tired and grumpy duringthe day. The solution to this problem is to establish a bedtime routine and stick to it. Try to go to bed at the same time every night and avoid using your phone or computer before bed. Also, make sure your bedroom is quiet and dark so you can get a good night's sleep.One more problem that many people face is feeling lonely. This can happen when you don't have anyone to talk to or hang out with. The solution to this problem is to reach out to others and make new friends. Join a club or sports team, volunteer in your community, or just start a conversation with someone new. Remember, everyone feels lonely sometimes, but there are always people who care about you and want to be your friend.In conclusion, life is full of problems, but with a positive attitude and some creativity, you can overcome any obstacle that comes your way. Remember, it's okay to ask for help when you need it, and don't be afraid to try new things. And most importantly, always believe in yourself and your ability to solve any problem that life throws at you.篇10Title: Problems in Daily Life and How to Solve ThemHey guys, have you ever encountered some problems in your daily life? Well, I have some too! But don't worry, here are some solutions that might help you out.Problem 1: Forgetting to bring homework to schoolSolution: Always pack your bag the night before and double-check in the morning.Problem 2: Fighting with friendsSolution: Talk it out calmly and try to understand each other's perspectives. Remember, friends are important!Problem 3: Getting bored during weekendsSolution: Plan some fun activities with your family or friends, like going to the park or watching a movie.Problem 4: Feeling stressed about examsSolution: Remember to study regularly and take breaks to relax. Don't forget to ask for help from teachers or parents if you need it.Problem 5: Wasting too much time on electronic devicesSolution: Limit your screen time and find other hobbies to keep yourself engaged, like reading books or playing sports.Problem 6: Not getting enough sleepSolution: Create a bedtime routine and try to go to bed at the same time every night. Avoid using electronic devices before sleeping.By following these solutions, you can make your daily life much smoother and happier. Remember, it's okay to have problems, but it's important to find ways to solve them. Stay positive, stay motivated, and keep smiling!。
2004 AMC 10 试卷
You and five friends need to raisedollars in donations for a charity, dividing the fundraising equally. How manydollars will each of you need to raise?Solution For any three real numbers , , and , with , the operation is defined by: What is?Solution Alicia earns 20 dollars per hour, of whichis deducted to pay local taxes. How many cents per hour of Alicia'swages are used to pay local taxes?Solution What is the value of if?SolutionA set of three points is randomly chosen from the grid shown. Each three point set has the same probability of beingchosen. What is the probability that the points lie on the same straight line?SolutionBertha has 6 daughters and no sons. Some of her daughters have 6 daughters, and the rest have none. Bertha has a total of 30 daughters and granddaughters, and no greatgranddaughters. How many of Bertha's daughters and granddaughters have no daughters?2004 AMC 10A ProblemsProblem 1Problem 2Problem 3Problem 4Problem 5Problem 6SolutionA grocer stacks oranges in a pyramidlike stack whose rectangular base is 5 oranges by 8 oranges. Each orange above the first level rests in a pocket formed by four oranges below. The stack is completed by a single row of oranges. Howmany oranges are in the stack?SolutionA game is played with tokens according to the following rule. In each round, the player with the most tokens gives one token to each of the other players and also places one token in the discard pile. The game ends when some player runs out of tokens. Players, , and start with 15, 14, and 13 tokens, respectively. How many rounds will there be in the game?SolutionIn the figure, and are right angles. , andand intersect at . What is the difference between the areas of and?SolutionCoin is flipped three times and coin is flipped four times. What is the probability that the number of heads obtained from flipping the two fair coins is the same?SolutionA company sells peanut butter in cylindrical jars. Marketing research suggests that using wider jars will increase sales. If the diameter of the jars is increased by without altering the volume, by what percent must the height be decreased?Problem 7Problem 8Problem 9Problem 10Problem 11SolutionHenry's Hamburger Heaven offers its hamburgers with the following condiments: ketchup, mustard, mayonnaise, tomato,lettuce, pickles, cheese, and onions. A customer can choose one, two, or three meat patties, and any collection ofcondiments. How many different kinds of hamburgers can be ordered?SolutionAt a party, each man danced with exactly three women and each woman danced with exactly two men. Twelve menattended the party. How many women attended the party?Solution The average value of all the pennies, nickels, dimes, and quarters in Paula's purse is cents. If she had one morequarter, the average would be cents. How many dimes does she have in her purse?Solution Given that and, what is the largest possible value of ?Solution The grid shown contains a collection of squares with sizes from to. How many of these squarescontain the black center square?SolutionProblem 12Problem 13Problem 14Problem 15Problem 16Problem 17Brenda and Sally run in opposite directions on a circular track, starting at diametrically opposite points. They first meet after Brenda has run 100 meters. They next meet after Sally has run 150 meters past their first meeting point. Each girlruns at a constant speed. What is the length of the track in meters?SolutionA sequence of three real numbers forms an arithmetic progression with a first term of 9. If 2 is added to the second term and 20 is added to the third term, the three resulting numbers form a geometric progression. What is the smallest possiblevalue for the third term of the geometric progression?SolutionA white cylindrical silo has a diameter of 30 feet and a height of 80 feet. A red stripe with a horizontal width of 3 feet ispainted on the silo, as shown, making two complete revolutions around it. What is the area of the stripe in square feet?Solution Points and are located on squareso that is equilateral. What is the ratio of the area ofto that of ?Problem 18Problem 19Problem 20Solution Two distinct lines pass through the center of three concentric circles of radii 3, 2, and 1. The area of the shaded region in the diagram is of the area of the unshaded region. What is the radian measure of the acute angle formed by the two lines? (Note:radians isdegrees.)Solution Square has side length . A semicircle with diameter is constructed inside the square, and the tangent to the semicircle from intersects side at . What is the length of ?Problem 21Problem 22Solution Circles , , and are externally tangent to each other and internally tangent to circle . Circles and arecongruent. Circle has radius and passes through the center of . What is the radius of circle ?Problem 23Solution Let, be a sequence with the following properties.(i), and (ii) for any positive integer .What is the value of?SolutionProblem 24Copyright © 2016 Art of Problem Solving Three pairwisetangent spheres of radius 1 rest on a horizontal plane. A sphere of radius 2 rests on them. What is thedistance from the plane to the top of the larger sphere?Solution2004 AMC 10A (Problems • Answer Key • Resources)Preceded by2003 AMC 10B Problems Followed by 2004 AMC 10B Problems1 •2 •3 •4 •5 •6 •7 •8 •9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25All AMC 10 Problems and SolutionsAMC 10AMC 10 Problems and SolutionsAMC Problems and SolutionsMathematics competition resourcesThe problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.Retrieved from "/wiki/index.php?title=2004_AMC_10A_Problems&oldid=71805"Problem 25See also。
2010 AMC 10A 试题及答案解析
2010 AMC 10A problems and solutions.The test was held on February 8, 2010. The first link contains the full set of test problems. The rest contain each individual problem and its solution.Problem 1Mary’s top book shelf holds five books with the follow ing widths, incentimeters: , , , , and . What is the average book width, in centimeters?SolutionTo find the average, we add up the widths , , , , and , to get a total sum of . Since there are books, the average book width isThe answer is .Problem 2Four identical squares and one rectangle are placed together to form one large square as shown. The length of the rectangle is how many times as large as its width?SolutionLet the length of the small square be , intuitively, the length of the big square is . It can be seen that the width of the rectangle is .Thus, the length of the rectangle is times large as the width. The answer is .Problem 3Tyrone had marbles and Eric had marbles. Tyrone then gave some of his marbles to Eric so that Tyrone ended with twice as many marbles as Eric. How many marbles did Tyrone give to Eric?SolutionLet be the number of marbles Tyrone gave to Eric. Then,. Solving for yields and . The answer is .Problem 4A book that is to be recorded onto compact discs takes minutes to read aloud. Each disc can hold up to minutes of reading. Assume that the smallest possible number of discs is used and that each disc contains the same length of reading. How many minutes of reading will each disc contain?SolutionAssuming that there were fractions of compact discs, it would take CDs to have equal reading time. However, since the number of discs can only be a whole number, there are at least 8 CDs, in which case it would have minutes on each of the 8 discs. The answer is .Problem 5The area of a circle whose circumference is is . What is the value of ?SolutionIf the circumference of a circle is , the radius would be . Since the area of a circle is , the area is . The answer is . Problem 6For positive numbers and the operation is defined asWhat is ?Solution. Then, is The answer isProblem 7Crystal has a running course marked out for her daily run. She starts this run by heading due north for one mile. She then runs northeast for one mile, then southeast for one mile. The last portion of her run takes her on a straight line back to where she started. How far, in miles is this last portion of her run?SolutionCrystal first runs North for one mile. Changing directions, she runs Northeast for another mile. The angle difference between North and Northeast is 45 degrees. She then switches directions to Southeast, meaning a 90 degree angle change. The distance now from travelling North for one mile, and her current destination is miles, because it is the hypotenuse of a 45-45-90 triangle with side length one (mile). Therefore, Crystal's distance from her starting position, x, is equal to , which is equal to . The answer isTony works hours a day and is paid $per hour for each full year of his age. During a six month period Tony worked days and earned $. How old was Tony at the end of the six month period?SolutionTony worked hours a day and is paid dollars per hour for each full year of his age. This basically says that he gets a dollar for each year of his age. So if he is years old, he gets dollars a day. We also know that he worked days and earned dollars. If he was years old at the beginning of his working period, he would have earned dollars. If he was years old at the beginning of his working period, he would have earned dollars. Because he earned dollars, we know that he was for some period of time, but not the whole time, because then the money earned would be greater than or equal to . This is why he was when he began, but turned sometime in the middle and earned dollars in total. So the answer is .The answer is . We could find out for how long he was and . . Then isand we know that he was for days, and for days. Thus, the answer is .Problem 9A palindrome, such as , is a number that remains the same when its digits are reversed. The numbers and are three-digit and four-digit palindromes, respectively. What is the sum of the digits of ?Solutionis at most , so is at most . The minimum value ofis . However, the only palindrome between and is , which means that must be .It follows that is , so the sum of the digits is .Marvin had a birthday on Tuesday, May 27 in the leap year . In what year will his birthday next fall on a Saturday?Solution(E) 2017There are 365 days in a non-leap year. There are 7 days in a week. Since 365 = 52 * 7 + 1 (or 365 is congruent to 1 mod 7), the same date (after February) moves "forward" one day in the subsequent year, if that year is not a leap year.For example: 5/27/08 Tue 5/27/09 WedHowever, a leap year has 366 days, and 366 = 52 * 7 + 2. So the same date (after February) moves "forward" two days in the subsequent year, if that year is a leap year.For example: 5/27/11 Fri 5/27/12 SunYou can keep count forward to find that the first time this date falls on a Saturday is in 2017:5/27/13 Mon 5/27/14 Tue 5/27/15 Wed 5/27/16 Fri 5/27/17 Sat Problem 11The length of the interval of solutions of the inequality is . What is ?SolutionSince we are given the range of the solutions, we must re-write the inequalities so that we have in terms of and .Subtract from all of the quantities:Divide all of the quantities by .Since we have the range of the solutions, we can make them equal to .Multiply both sides by 2.Re-write without using parentheses.Simplify.We need to find for the problem, so the answer isProblem 12Logan 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's miniature. Therefore, Logan should make his towertimes shorter than the actual tower. This ismeters high, or choice .Problem 13Angelina drove at an average rate of kph and then stopped minutes for gas. After the stop, she drove at an average rate of kph. Altogether she drove km in a total trip time of hours including the stop. Which equation could be used to solve for the time in hours that she drove before her stop?SolutionThe answer is （）because she drove at kmh for hours (the amount of time before the stop), and 100 kmh for because she wasn't driving for minutes, or hours. Multiplying by gives the total distance, which is kms. Therefore, the answer isProblem 14Triangle 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 ,Problem 15In 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 at least frog out of the two of them.Now suppose Mike is a toad. Then what he says is true because we already have toads. However, if Brian 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 be a frog. We can also verify that his statement is indeed false.Both Mike and Brian are frogs, and one of either Chris or LeRoy is a frog, so we have frogs total.Solution 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 16Nondegenerate 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 we use the next lowest values (and ), the Triangle Inequality is satisfied. Therefore, our answer is , or choice .Problem 17A solid cube has side length inches. A -inch by -inch square hole is cut into the center 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 . The second cut removes two boxes, each of dimensions, and the third cut does the same as the second cut, on the last two faces. Hence the total volume of all cuts is .Therefore the volume of the rest of the cube is.Solution 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 their volumes, as the central cube is included in each of these three cuts. To get the correct result, we can take the sum of the volumes of the three cuts, and subtract the 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 a box..Problem 18Bernardo randomly picks 3 distinct numbers from the setand arranges them in descending order to form a 3-digit number. Silvia randomly picks 3 distinct numbers from the set and also arranges them in descending order to form a 3-digit 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 the probabilities.Case : Bernardo picks . If Bernardo picks a then it is guaranteed that his number 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 , the probability of not picking is .If Bernardo does not pick 9, then he can pick any number from to . Since Bernardo is picking from the same set of numbers as Silvia, the probability that 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 the number 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 19Equiangular hexagon has side lengthsand . The area of is of the area of the hexagon. What is the sum of all possible values of ?SolutionSolution 1It is clear that is an equilateral triangle. From the Law of Cosines, we get that . Therefore, the area of is .If we extend , and so that and meet at , and meet at , and and meet at , we find that hexagon is formed by taking equilateral triangle of side length and removing three equilateral triangles, , and , of side length . The area of is therefore.Based on the initial conditions,Simplifying this gives us . By Vieta's Formulas we know that the sum of the possible value of is .Solution 2As above, we find that the area of is .We also find by the sine triangle area formula that, and thusThis simplifies to.Problem 20A fly trapped inside a cubical box with side length meter decides to relieve its boredom by visiting each corner of the box. It will begin and end in the same corner and visit each of the other corners exactly once. To get from a corner to any other corner, it will either fly or crawl in a straight line. What is the maximum possible length, in meters, of its path?SolutionThe distance of an interior diagonal in this cube is and the distance of a diagonal on one of the square faces is . It would not make sense if the fly traveled an interior diagonal twice in a row, as it would return to the point it just came from, so at most the final sum can only have 4 as the coefficient of . The other 4 paths taken can be across a diagonal on one of the faces, so the maximum distance traveled is.Problem 21The polynomial has three positive integer zeros. What is the smallest possible value of ?SolutionBy Vieta's Formulas, we know that is the sum of the three roots of the polynomial . Also, 2010 factors into. But, since there are only three roots to the polynomial, two of the four prime factors must be multiplied so that we are left with three roots. To minimize , and should be multiplied, which means will be and the answer is .Problem 22Eight points are chosen on a circle, and chords are drawn connecting every pair of points. No three chords intersect in a single point insidethe circle. How many triangles with all three vertices in the interior of the circle are created?SolutionTo choose a chord, we know that two points must be chosen. This implies that for three chords to create a triangle and not intersect at a single point, six points need to be chosen. Therefore, the answer is which is equivalent to 28,Problem 23Each of 2010 boxes in a line contains a single red marble, and for , the box in the position also contains white marbles. Isabella begins at the first box and successively draws a single marble at random from each box, in order. She stops when she first draws a red marble. Let be the probability that Isabella stops afterdrawing exactly marbles. What is the smallest value of for which ?SolutionThe probability of drawing a white marble from box is . Theprobability of drawing 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 inequality is .Problem 24The 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 last two digits of , or . Since there is clearly an excess of factors of 2, we know that , so it remains to find .If we divide by by taking out all the factors of in , we canwrite as where 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:Using the identity at the beginning of the solution, we can reducetoUsing the fact that (or simply the fact that if you have your powers of 2 memorized), we can deduce that . Therefore.Finally, combining with the fact that yields.Problem 25Jim starts with a positive integer and creates a sequence of numbers. Each successive number is obtained by subtracting the largest possible integer square less than or equal to the current number until zero is reached. For example, if Jim starts with , then his sequence contains numbers:Let be the smallest number for which Jim’s sequence has numbers. What is the units digit of ?SolutionWe can find the answer by working backwards. We begin with on the bottom row, then the goes to the right of the equal's sign in the row above. We find the smallest value for whichand , which is .We repeat the same procedure except with for the next row and for the row after that. However, at the fourth row, wesee that solving yields , in which case it would be incorrect since is not the greatest perfect square less than or equal to . So we make it a and solve . We continue on using this same method where we increase the perfect square until can be made bigger than it. When we repeat this until we have rows, we get:Hence the solution is the last digit of , which is .。
英语作文写问题和解决方案
英语作文写问题和解决方案Title: Writing Problems and Solutions in English CompositionIntroductionEnglish composition writing is an essential skill that students need to master in today's education system. However, many students often face various challenges when it comes to writing essays, reports, and other compositions in English. In this article, we will discuss some common problems encountered by students and provide practical solutions to help them improve their writing skills.Problems and Solutions1. Lack of VocabularyProblem: One of the most common challenges students face is a limited vocabulary, which can make it difficult for them to express their ideas clearly and accurately.Solution: To overcome this problem, students should make a conscious effort to expand their vocabulary by reading books, articles, and other written materials. They can also use online resources such as vocabulary building apps and websites to learn new words and their meanings.2. Poor Grammar and PunctuationProblem: Another common issue is poor grammar and punctuation, which can affect the clarity and coherence of a student's writing.Solution: To improve their grammar and punctuation skills, students should practice writing regularly and seek feedback from teachers or peers. They can also use grammar checking tools and online resources to identify and correct errors in their writing.3. Lack of OrganizationProblem: Many students struggle with organizing their ideas and structuring their writing in a logical and coherent manner.Solution: To address this problem, students should create outlines before writing to plan their main points and supporting details. They should also use transition words and phrases to connect ideas and ensure smooth flow between paragraphs.4. Difficulty in Developing IdeasProblem: Some students find it challenging to generate ideas and develop them effectively in their writing.Solution: Students can overcome this problem by brainstorming ideas, conducting research, and organizing their thoughts before writing. They can also use techniques such as mind mapping and freewriting to explore different angles and perspectives on a topic.5. Lack of ConfidenceProblem: Many students lack confidence in their writing abilities, which can hinder their creativity and expression.Solution: To boost their confidence, students should practice writing regularly, receive constructive feedback, and celebrate small victories. They can also seek support from teachers, tutors, or writing centers to improve their skills and gain confidence in their abilities.ConclusionIn conclusion, writing problems are common among students, but they can be overcome with practice, perseverance, and the right strategies. By identifying their weaknesses, seeking help when needed, and implementing effective solutions, students can improve their English composition writing skills and become confident and proficient writers. With dedication andeffort, students can enhance their writing abilities and achieve academic success in their studies.。
问题和解决方案英语作文50字
问题和解决方案英语作文50字英文回答：1. What is the problem?The problem is that the world is facing a climatecrisis. The Earth's average temperature is rising, and this is causing a number of serious problems, such as more extreme weather events, rising sea levels, and the loss of biodiversity.What is the solution?The solution is to reduce greenhouse gas emissions. Greenhouse gases are gases that trap heat in the atmosphere, and they are released into the atmosphere by humanactivities such as burning fossil fuels. We need totransition to a low-carbon economy by investing in renewable energy sources, such as solar and wind power, and by making our homes and businesses more energy-efficient.2. What is the problem?The problem is that there is a global food crisis. The world's population is growing, and this is putting a strain on the world's food supply. In addition, climate change is making it more difficult to grow food in many parts of the world.What is the solution?The solution is to increase food production. We need to invest in agricultural research and development to develop new and more productive crops. We also need to make sure that food is distributed fairly around the world.3. What is the problem?The problem is that there is a global water crisis. The world's population is growing, and this is putting a strain on the world's water supply. In addition, climate change is making it more difficult to access water in many parts ofthe world.What is the solution?The solution is to conserve water. We need to make sure that we are using water wisely and that we are not wasting it. We also need to invest in water infrastructure, such as dams and reservoirs, to ensure that we have enough waterfor the future.4. What is the problem?The problem is that there is a global energy crisis. The world's population is growing, and this is putting a strain on the world's energy supply. In addition, climate change is making it more difficult to access energy in many parts of the world.What is the solution?The solution is to transition to a renewable energy economy. We need to invest in renewable energy sources,such as solar and wind power, and we need to make our homes and businesses more energy-efficient.5. What is the problem?The problem is that there is a global poverty crisis. The world's population is growing, and this is putting a strain on the world's resources. In addition, climate change is making it more difficult for people to escape poverty in many parts of the world.What is the solution?The solution is to invest in education and healthcare. We need to make sure that everyone has access to quality education and healthcare, so that they can improve their lives and escape poverty.中文回答：问题 1：问题，世界正面临气候危机。
2017 AMC 12B Problem and Solution
2017 AMC 12B2017 AMC 12A problems and solutions. The test was held on February 7, 2017.Problem 1Kymbrea's comic book collection currently has comic books in it, and she is adding to her collection at the rate of comic books per month. LaShawn's collection currently has comic books in it, and he is adding to his collection at the rate of comic books per month. After how many months will LaShawn's collection have twice as many comic books as Kymbrea's?SolutionKymbrea has comic books initially and every month, she adds two. This can be represented as where x is the number of months elapsed. LaShawn's collection, similarly, is . To find when LaShawn will have twice the number of comic books asKymbrea, we solve for x with the equation and get .Problem2Real numbers , , and satisfy the inequalities , , and . Which of the following numbers is necessarily positive?SolutionNotice that must be positive because . Therefore the answer is . The other choices:As grows closer to , decreases and thus becomes less than .can be as small as possible (), so grows close to as approaches .For all , , and thus it is always negative.The same logic as above, but when this time.Problem3Supposed that and are nonzero real numbers such that . What is the value of ?SolutionsSolution 1Rearranging, we find , or . Substituting, we canconvert the second equation into .Solution 2Substituting each and with , we see that the given equation holds true,as . Thus,Problem4Samia set off on her bicycle to visit her friend, traveling at an average speed of kilometers per hour. When she had gone half the distance to her friend's house, a tire went flat, and she walked the rest of the way at kilometers per hour. In all it took her minutes to reach her friend's house. In kilometers rounded to the nearest tenth, how far did Samia walk?Solution 1Let's call the distance that Samia had to travel in total as , so that we can avoid fractions. We know that the length of the bike ride and how far she walked are equal, so they areboth , or .She bikes at a rate of kph, so she travels the distance she bikes in hours. She walks at a rate of kph, so she travels the distance she walks in hours.The total timeis . This is equal to of an hour. Solving for , we have:Since is the distance of how far Samia traveled by both walking and biking, and we want to know how far Samia walked to the nearest tenth, we have that Samia walked about .Problem 5The data set has median , firstquartile , and third quartile . An outlier in a data set is a value that is more than times the interquartile range below the first quartle () or more than times the interquartile range above the third quartile (), where the interquartile range is definedas . How many outliers does this data set have?SolutionThe interquartile range is defined as , which is . times this value is , so all values more than below = is an outlier. The only one that fits this is . All values more than above = are also outliers, of whichthere are none so there is onlyProblem 6The circle having and as the endpoints of a diameter intersects the -axis at a second point. What is the -coordinate of this point?SolutionBecause the two points are on a diameter, the center must be halfway between them at the point (4,3). The distance from (0,0) to (4,3) is 5 so the circle has radius 5. Thus, the equationof the circle is .To find the x-intercept, y must be 0, so ,so , , .Problem 7The functions and are periodic with least period . What is the least period of the function ?The function is not periodic. Solutionhas values at its peaks and x-intercepts. Increase them to . Then we plug theminto . and .So, isSolution by TheUltimate123 (Eric Shen)Solution IIStart by noting that . Then realize that under this function the negative sine values yield the same as their positive value, so take the absolute value of the sine function to get the new period. This has period , so the answer is surprisingly !!!!Problem 8The ratio of the short side of a certain rectangle to the long side is equal to the ratio of the long side to the diagonal. What is the square of the ratio of the short side to the long side of this rectangle?Solution 1: Cross-MultiplicationLet be the short side of the rectangle, and be the long side of the rectangle. The diagonal,therefore, is . We can get the equation . Cross-multiplying, weget . Squaring both sides of the equation, we get , which simplifies to . Solving for a quadratic in , using the quadratic formulawe get which gives us . We know that the square of the ratio must be positive (the square of any real number is positive), so the solutionis .Solution by: vedadehhcSolution 2: SubstitutionSolution by HydroQuantumLet the short side of the rectangle be and let the long side of the rectangle be . Then, the diagonal, according to the Pythagorean Theorem, is . Therefore, we can write the equation:.We are trying to find the square of the ratio of to . Let's let our answer, , be . Then, squaring the above equation,.Thus, .Multiplying each side of the equation by ,.Adding each side by ,.Solving for using the Quadratic Formula,.Since the ratio of lengths and diagonals of a rectangle cannot be negative, and ,the symbol can only take on the . Therefore,.Problem 9A circle has center and has radius . Another circle has center andradius . The line passing through the two points of intersection of the two circles has equation . What is ?Solution 1The equations of the two circlesare and . Rearrange themto and , respectively. Their intersection points are where these two equations gain equality. The two points lie on the line with the equation . We cansimplify this like thefollowing.. Thus, .Solution by TheUltimate123 (Eric Shen)Solution 2: Shortcut with right trianglesNote the specificity of the radii, and , and that specificity is often deliberately added tosimplify the solution to a problem.One may recognize as the hypotenuse of the right triangle and as thehypotenuse of the right triangle with legs and . We can suppose that the legs of these triangles connect the circles' centers to their intersection along the gridlines of the plane.If we suspect that one of the intersections lies units to the right of and units above the center of the first circle, we find the point , which is infact unit to the left of and units below the center of the second circle at .Plugging into gives us .A similar solution uses the other intersection point, .Problem10At Typico High School, of the students like dancing, and the rest dislike it. Of those who like dancing, say that they like it, and the rest say that they dislike it. Of those who dislike dancing, say that they dislike it, and the rest say that they like it. What fraction of students who say they dislike dancing actually like it?SolutionWLOG, let there be students. of them like dancing, and do not. Of those who like dancing, , or of them say they dislike dancing. Of those who dislike dancing, ,or of them say they dislike it. Thus,Problem11Call a positive integer if it is a one-digit number or its digits, when read from left to right, form either a strictly increasing or a strictly decreasing sequence. Forexample, , , and are monotonous, but , , and are not. How many monotonous positive integers are there?Solution 1Case 1: monotonous numbers with digits in ascending orderThere are ways to choose n digits from the digits 1 to 9. For each of these ways, we can generate exactly one monotonous number by ordering the chosen digits in ascending order. Note that 0 is not included since it will always be a leading digit and that is not allowed. Also, (the empty set) isn't included because it doesn't generate a number. The sum isequivalent toCase 2: monotonous numbers with digits in descending orderThere are ways to choose n digits from the digits 0 to 9. For each of these ways, we can generate exactly one monotonous number by ordering the chosen digits in descending order. Note that 0 is included since we are allowed to end numbers with zeros.However, (the empty set) still isn't included because it doesn't generate a number. The sumis equivalent to We discard the number 0 since it is not positive. Thus there are here.Since the 1-digit numbers 1 to 9 satisfy both case 1 and case 2, we have overcounted by 9. Thus there are monotonous numbers.Solution 2Like Solution 1, divide the problem into an increasing and decreasing case:Case 1: Monotonous numbers with digits in ascending order.Arrange the digits 1 through 9 in increasing order, and exclude 0 because a positive integer cannot begin with 0.To get a monotonous number, we can either include or exclude each of the remaining 9 digits, and there are ways to do this. However, we cannot exclude every digit at once, so we subtract 1 to get monotonous numbers for this case.Case 2: Monotonous numbers with digits in descending order.This time, we arrange all 10 digits in decreasing order and repeat the process tofind ways to include or exclude each digit. We cannot exclude every digit at once, and we cannot include only 0, so we subtract 2 to get monotonous numbers for this case.At this point, we have counted all of the single-digit monotonous numbers twice, so we must subtract 9 from our total.Thus our final answer is .Problem12What is the sum of the roots of that have a positive real part?Solution 1The root of any polynomial of the form will have all of it roots will havemagnitude and be the vertices of a regular -gon in the complex plane (This concept is known as the Roots of Unity). For the equation , it is easy tosee and as roots. Graphing these in the complex plane, we have four vertices of a regular dodecagon. Since the roots must be equally spaced, besides , there are fourmore roots with positive real parts lying in the first and fourth quadrants. We also know that the angle between these roots is . We only have to find the real parts of the roots lying in the first quadrant, because the imaginary parts would cancel out with those from the fourth quadrant. We have two triangles (the triangles formed by connecting the origin to the roots, and dropping a perpendicular line from each root to the real-axis), both withhypotenuse . This means that one has base and the other has base . Adding these and multiplying by two, we get the sum of the four roots as . However, wehave to add in the original solution of , so the answer is .Solution by vedadehhcSolution 2has a factor of , so we need to remember to multiply our solution below, usingthe Roots of Unity. We notice that the sum of the complex parts of all these roots is , because the points on the complex plane are symmetric. The rootswith are and by the Roots of Unity. Their real partsare and . Their sumis . But, remember to multiply by . Theanswer is .Solution by TheUltimate123 (Eric Shen)Problem 13In the figure below, of the disks are to be painted blue, are to be painted red, and is to be painted green. Two paintings that can be obtained from one another by a rotation or a reflection of the entire figure are considered the same. How many different paintings are possible?SolutionLooking at the answer choices, we see that the possibilities are indeed countable. Thus, we will utilize that approach in the form of two separate cases, as rotation and reflection take care of numerous possibilities. First, consider the case that the green disk is in a corner. This yields possible arrangements for the blue disks and red disks in the remaining available slots. Now, consider the case that the green disk is on an edge. This yields more possiblearrangements for the blue disks and red disks in the remaining available slots. Thus, our answer isProblem14An ice-cream novelty item consists of a cup in the shape of a 4-inch-tall frustum of a right circular cone, with a 2-inch-diameter base at the bottom and a 4-inch-diameter base at the top, packed solid with ice cream, together with a solid cone of ice cream of height 4 inches, whose base, at the bottom, is the top base of the frustum. What is the total volume of the ice cream, in cubic inches?SolutionSolution 1:The top cone has radius 2 and height 4 so it has volume .The frustum is made up by taking away a small cone of radius 1, height 4 from a large cone of radius 2, height 8, so it has volume .Adding, we get .Solution by: SilverLionSolution 2:Find the area of the cone with the method in Solution 1. The area of the frustrumisAdding, we getProblem 15Let be an equilateral triangle. Extend side beyond to a point sothat . Similarly, extend side beyond to a point so that , and extend side beyond to a point so that . What is the ratio of the area of to the area of ?Solution 1: Law of CosinesSolution by HydroQuantumLet .Recall The Law of Cosines. Letting ,Since both and are both equilateral triangles, they must be similar dueto similarity. This means that .Therefore, our answer is .Solution 2: InspectionNote that the height and base of are respectively 4 times and 3 times thatof . Therefore the area of is 12 times that of .By symmetry, . Adding the areas of these three triangles and for the total area of gives a ratio of ,or .Solution 3: CoordinatesFirst we note that due to symmetry. WLOG,let and Therefore, . Using the conditionthat , we get and . It is easy to checkthat . Since the area ratios of two similar figures is the square of the ratio of their lengths, the ratio isProblem 16The number has over positive integer divisors. One of them is chosen at random. What is the probability that it is odd?SolutionIf a factor of is odd, that means it contains no factors of . We can find the number of factors of two in by counting the number multiples of , , , and that are less than or equal to .After some quick counting we find that this number is . If the prime factorization of has factors of , there are choices for each divisor for how many factors of should be included (to inclusive). The probability that a randomlychosen factor is odd is the same as if the number of factors of is which is . Solution by: vedadehhcSolution 2We can write as its prime factorization:Each exponent of these prime numbers are one less than the number of factors at play here. This makes sense; is going to have factors: , and the other exponents will behave identically.In otherwords, has factors.We are looking for the probability that a randomly chosen factor of will be odd--numbers that do not contain multiples of as factors.From our earlier observation, the only factors of that are even are ones with at least one multiplier of , so our probability of finding an odd factor becomes the following:Problem 17A coin is biased in such a way that on each toss the probability of heads is and the probability of tails is . The outcomes of the tosses are independent. A player has the choiceof playing Game A or Game B. In Game A she tosses the coin three times and wins if all three outcomes are the same. In Game B she tosses the coin four times and wins if both the outcomes of the first and second tosses are the same and the outcomes of the third and fourth tosses are the same. How do the chances of winning Game A compare to the chances of winning Game B?The probability of winning Game A is less than the probability of winning Game B.The probability of winning Game A is less than the probability of winning Game B.The probabilities are the same.The probability of winning Game A is greater than the probability of winning Game B.The probability of winning Game A is greater than the probability of winning Game B. SolutionThe probability of winning Game A is the sum of the probabilities of getting three tails andgetting three heads which is . The probability of winning Game B is the sum of the probabilities of getting two heads and getting two tailssquared. This gives us . The probability of winningGame A is and the probability of winning Game B is , so the answer isProblem18The diameter of a circle of radius is extended to a point outside the circle sothat . Point is chosen so that and line is perpendicular to line . Segment intersects the circle at a point between and . What is the areaof ?Solution 1Let be the center of the circle. Notethat .However, by Power of apoint ,, so .Now. Since .Solution 2: Similar triangles with Pythagoreanis the diameter of the circle, so is a right angle, and therefore by AA similarity, .Because of this, , so . Likewise, , so .Thus the area of .Solution 3: Similar triangles without PythagoreanOr, use similar triangles all the way, dispense with Pythagorean, and go for minimal calculation:Draw with on . ... ( ratio applied twice).Problem19Let be the -digit number that is formed by writing the integers from to in order, one after the other. What is the remainder when is divided by ?SolutionWe will consider this number and . By looking at the last digit, it is obvious that the number is . To calculate the number , note thatso it is equivalent toLet be the remainder when this number is divided by . We knowthat and , so by the Chinese remainder theorem,since , ,or . So the answer isProblem 20Real numbers and are chosen independently and uniformly at random from theinterval . What is the probability that , where denotes the greatest integer less than or equal to the real number ?SolutionFirst let us take the case that . In this case, both and lie in the interval . The probability of this is . Similarly, in the casethat , and lie in the interval , and the probabilityis . It is easy to see that the probabilitiesfor for are the infinite geometric series that startsat and with common ratio . Using the formula for the sum of an infinite geometric series, we get that the probability is .Problem 21Last year, Isabella took 7 math tests and received 7 different scores, each an integer between 91 and 100, inclusive. After each test she noticed that the average of her test scores was an integer. Her score on the seventh test was 95. What was her score on the sixth test?Solution 1Let us simplify the problem. Since all of Isabella's test scores can be expressed as the sumof and an integer between and , we rewrite the problem into receiving scoresbetween and . Later, we can add to her score to obtain the real answer.From this point of view, the problem states that Isabella's score on the seventh test was . We note that Isabella received integer scores out of to . Since is already given as the seventh test score, the possible scores for Isabella on the other six testsare .The average score for the seven tests must be an integer. In other words, six distinct integers must be picked from set above, and their sum with must be a multiple of . The interval containing the possible sums of the six numbers in S arefrom to . We must now find multiples of from the interval to . There are fourpossibilities: , , , . However, we also note that the sum of the six numbers (besides ) must be a multiple of as well. Thus, is the only valid choice.(The six numbers sum to .)Thus the sum of the six numbers equals to . We apply the logic above in a similar way for the sum of the scores from the first test to the fifth test. The sum must be a multiple of . The possible interval is from to . Since the sum of the five scores must be less than , the only possibilities are and . However, we notice that does not work because the seventh score turns out to be from the calculation. Therefore, the sum of Isabella's scores from test to is . Therefore, her score on the sixthtest is . Our final answer is .Solution 2Let be Isabella's average after tests. , so . The only integer between and that satisfies this condition is . Let be Isabella'saverage after tests, and let be her sixth test score. ,so is a multiple of . Since is the only choice that is a multiple of , the answeris .Solution 3Let be the total sum of Isabella's first five test scores, and let be her score on the sixthtest. It follows that , , and , since at each step, her average score was an integer. Using the lastequivalence, , so we have a system of equivalences for . Solving this using the Chinese Remainder Theorem, weget .Now let's put a bound on . Using the given information that each test score was a distinct integer from to inclusive and that the seventh score was 95, weget . Since , we get . Therefore,The last preparation step will involve calculating all the possible test scores . Heretheyare:. This means that . Note that is not in the previous list because it corresponds to a score of , which we cannot have.We must have , and using the possible values we found for and , the only two that sum to are and . This corresponds to an value of , so the answer is .Problem 22Abby, Bernardo, Carl, and Debra play a game in which each of them starts with four coins. The game consists of four rounds. In each round, four balls are placed in an urn---one green, one red, and two white. The players each draw a ball at random without replacement. Whoever gets the green ball gives one coin to whoever gets the red ball. What is the probability that, at the end of the fourth round, each of the players has four coins?SolutionIt amounts to filling in a matrix. Columns are the random draws each round; rows are the coin changes of each player. Also, let be the number of nonzero elements in .WLOG, let . Parity demands that and must equal or .Case 1: and . There are ways to place 's in , so there are ways.Case 2: and . There are ways to place the in , ways to place the remaining in (just don't put it under the on top of it!), and ways for one of the other two players to draw the green ball. (We know it's green because Bernardo drew the red one.) We can just double to cover the case of , for a total of ways.Case 3: . There are three ways to place the in . Now, there are two cases as to what happens next.Sub-case 3.1: The in goes directly under the in . There's obviously way for that to happen. Then, there are ways to permute the two pairs of in and . (Either the comes first in or the comes first in .)Sub-case 3.2: The in doesn't go directly under the in . There are ways to place the , and ways to do the same permutation as in Sub-case 3.1. Hence, thereare ways for this case.There's a grand total of ways for this to happen, along with total cases. The probabilitywe're asking for is thusSolution 2 (Less Casework)We will proceed by taking cases based on how many people are taking part in this "transaction." We can have 2, 3, or 4 people all giving/receiving coins during the 4 turns. Basically, (like the previous solution), we are thinking this as filling out a 4x2 matrix of letters, where a letter on the left column represents this person gave, and a letter on the right column means this person recieved. We need to make sure that for each person that gave a certain amount, they received in total from other people that same amount, or in other words there are an equal number of A's, B's, C's, and D's on both columns of the matrix.Case 1: people. In this case, we can 4C2 ways to choose the two people, and 6 ways to get order them to get a count of 6 * 6 = 36 ways.Case 2: people. In this case, we have 4*(3C2)*4! = 288 ways to order 3 people.Case 3: people. In this case, we have 3*3*4! = 216 ways to order 4 people.So we have a total of 36+288+126=540 ways to order the four pairs of people. Now we divide this by the total number of ways - (4*3)^4 ( 4 times, 4 ways to choose giver, 3 to choose receiver). So the answer is 5/192.~ccx09 (NOTE: Due to the poor quality of this solution, please PM me and I will explain the numbers, I have some diagrams but I can't show it here)Problem 23The graph of , where is a polynomial of degree , containspoints , , and . Lines , , and intersect the graph again at points , , and , respectively, and the sum of the -coordinates of , , and is 24. What is ?SolutionFirst, we can define , which contains points , ,and . Now we find that lines , , and are defined by theequations , , and respectively. Since we want to find the -coordinates of the intersections of these lines and , we set each of them to ,and synthetically divide by the solutions we already know exist (eg. if we were looking atline , we would synthetically divide by the solutions and , because wealready know intersects the graph at and , which have -coordinates of and ). After completing this process on all three lines, we get that the -coordinates of , ,and are , , and respectively. Adding these together, weget which gives us . Substituting this back into the original equation,we get ,andProblem 24Quadrilateral has right angles at and , , and . There is a point in the interior of such that and the areaof is times the area of . What is ?Solution 1Let , , and . Note that . By the PythagoreanTheorem, . Since , the ratios of sidelengths must be equal. Since , and . Let F be a point on such that is an altitude of triangle . Notethat . Therefore, and . Since and form altitudes of triangles and , respectively, the areas of these triangles can be calculated. Additionally, the area of triangle can be calculated, as it is a right triangle. Solving for each of these yields:Therefore, the answerisSolution 2Draw line through , with on and on , . WLOGlet , , . By weighted average .Meanwhile, .. We obtain , namely .The rest is the same as Solution 1.Solution 3Let . Then from the similar triangles condition, we compute and . Hence, the -coordinate of is just . Since lies on the unit circle, we can compute the coordinate as . By Shoelace, we wantFactoring out denominators and expanding by minors, this is equivalent toThis factorsas , so and so the answer is .ProblemA set of people participate in an online video basketball tournament. Each person may be a member of any number of -player teams, but no two teams may have exactly thesame members. The site statistics show a curious fact: The average, over all subsets of size of the set of participants, of the number of complete teams whose members are among those people is equal to the reciprocal of the average, over all subsets of size of the set of participants, of the number of complete teams whose members are among those people. How many values , , can be the number of participants?SolutionSolution by Pieater314159Let there be teams. For each team, there are different subsets of players including that full team, so the total number of team-(group of 9) pairs isThus, the expected value of the number of full teams in a random set of players isSimilarly, the expected value of the number of full teams in a random set of players isThe condition is thus equivalent to the existence of a positive integer such that。
2007 AMC 12B Problems and Solution
2007 AMC 12B Problems and Solution Problem 1Isabella's house has 3 bedroom s. Each bedroom is 12 feet long, 10 feet wide, and 8 feet high. Isabella must paint the walls of all the bedrooms. Doorways and windows, which will not be painted, occupy 60 square feet in each bedroom. How many square feet of walls m ust be painted?SolutionThere are four walls in each bedroom, since she can't paint floors or ceilings. So we calculate the num ber of square feet of wall there is in one bedroom:We havethree bedrooms, so she must paint square feet of wall.Problem 2A college student drove his com pact car 120 miles home for the weekend and averaged 30 miles per gallon. On the return trip the student drove his parents' SUV and averaged only 20 miles per gallon. What was the average gas mileage, in miles per gallon, for the round trip?SolutionThe trip was miles long and took gallons. Therefore,the average m ileage wasThe point is the center of the circle circum scribed about triangle , withand , as shown. What is the degree m easure of ?SolutionProblem 4At Frank's Fruit Market, 3 bananas cost as m uch as 2 apples, and 6 apples cost as much as 4 oranges. How many oranges cost as m uch as 18 bananas?Solution18 bananas cost the sam e as 12 apples, and 12 apples cost the sam e as 8 oranges,so 18 bananas cost the sam e as oranges.The 2007 AMC 12 contests will be scored by awarding 6 points for each correct response, 0 points for each incorrect response, and 1.5 points for each problem left unanswered. After looking over the 25 problems, Sarah has decided to attem pt the first 22 and leave the last 3 unanswered. How many of the first 22 problems must she solve correctly in order to score at least 100 points?SolutionShe must get at least points, and that can only be possible byanswering at least questions correctly.Problem 6Triangle has side lengths , , and . Two bugs startsimultaneously from and crawl along the sides of the triangle in oppositedirections at the sam e speed. They m eet at point . What is ?SolutionOne bug goes to . The path that he takes is units long. The lengthof isProblem 7All sides of the convex pentagon are of equal length, and. What is the degree m easure of ?SolutionSince and are right angles, and equals , is a square, and is5. Since and are also 5, triangle is equilateral. Angle is thereforeProblem 8Tom's age is years, which is also the sum of the ages of his three children. His ageyears ago was twice the sum of their ages then. What is ?SolutionTom's age years ago was . The ages of his three children years ago wassince there are three people. If his age years ago was twice the sum ofthe children's ages then,Problem 9A function has the property that for all real numbers .What is ?SolutionProblem 10Som e boys and girls are having a car wash to raise m oney for a class trip to China.Initially % of the group are girls. Shortly thereafter two girls leave and two boys arrive, and then % of the group are girls. How many girls were initially in thegroup?SolutionIf we let be the num ber of people initially in the group, the is the number of girls. If two girls leave and two boys arrive, the number of people in the group is stillbut the num ber of girls is . Since only of the group are girls,The num ber of girls isProblem 11The angles of quadrilateral satisfy . What isthe degree m easure of , rounded to the nearest whole number?SolutionThe sum of the interior angles of any quadrilateral isProblem 12A teacher gave a test to a class in which of the students are juniors and areseniors. The average score on the test was . The juniors all received the sam e score, and the average score of the seniors was . What score did each of thejuniors receive on the test?SolutionWe can assum e there are people in the class. Then there will be junior andseniors. The sum of everyone's scores is Since the average score ofthe seniors was the sum of all the senior's scores is The only score that has not been added to that is the junior's score, which isProblem 13A traffic light runs repeatedly through the following cycle: green for seconds,then yellow for seconds, and then red for seconds. Leah picks a randomthree-second tim e interval to watch the light. What is the probability that the color changes while she is watching?SolutionThe traffic light runs through a second cycle.Letting reference the m oment it turns green, the light changes at threedifferent tim es: , , andThis m eans that the light will change if the beginning of Leah's interval lies in, orThis gives a total of seconds out ofProblem 14Point is inside equilateral . Points , , and are the feet of theperpendiculars from to , , and , respectively. Given that ,, and , what is ?SolutionDrawing , , and , is split into three sm aller triangles. Thealtitudes of these triangles are given in the problem as , , and .Summing the areas of each of these triangles and equating it to the area of the entire triangle, we get:where is the length of a sideProblem 15The geom etric series has a sum of , and the terms involving odd powers of have a sum of . What is ?SolutionSolution 1The sum of an infinite geom etric series is given by where is the first term and is the common ratio.In this series,The series with odd powers of is given asIt's sum can be given byDoing a little algebraSolution 2The given series can be decomposed as follows:Clearly . We obtain that, hence .Then from we get , and thus . Problem 16Each face of a regular tetrahedron is painted either red, white, or blue. Two colorings are considered indistinguishable if two congruent tetrahedra with those colorings can be rotated so that their appearances are identical. How many distinguishable colorings are possible?SolutionA tetrahedron has 4 sides. The ratio of the number of faces with each color must be one of the following:, , , orThe first ratio yields appearances, one of each color.The second ratio yields appearances, three choices for the first color, andtwo choices for the second.The third ratio yields appearances since the two colors areinterchangeable.The fourth ratio yields appearances. There are three choices for the first color, andsince the second two colors are interchangeable, there is only one distinguishable pair that fits them.The total is appearancesSolution 2Every colouring can be represented in the form, where is the num ber of white faces, is the number of red faces, and is the number of blue faces. Everydistinguishable colouring pattern can be represented like this in exactly one way, and every ordered whole number triple with a total sum of 4 represents exactly one colouring pattern (if two tetrahedra have rearranged colours on their faces, it is always possible to rotate one so that it m atches the other).Therefore, the number of colourings is equal to the num ber of ways 3 distinguishable nonnegative integers can add to 4. If you have 6 cockroaches in a row, this number is equal to the num ber of ways to pick two of the cockroaches to eat for dinner (because the rem aining cockroaches in between are separated in to three sections with a non-negative number of cockroaches each), which isProblem 17If is a nonzero integer and is a positive number such that , what isthe m edian of the set ?SolutionNote that if is positive, then, the equation will have no solutions for . Thisbecom es more obvious by noting that at , . The LHS quadraticfunction will increase faster than the RHS logarithmic function, so they will never intersect.This puts as the sm allest in the set since it m ust be negative.Checking the new equation:Near , but at ,This implies that the solution occurs som ewhere in between:This also implies thatThis m akes our set (ordered)The m edian isProblem 18Let , , and be digits with . The three-digit integer lies one third of theway from the square of a positive integer to the square of the next larger integer. The integer lies two thirds of the way between the sam e two squares. What is?SolutionThe difference between and is given byThe difference between the two squares is three tim es this amount orThe difference between two consecutive squares is always an odd number, therefore is odd. We will show that must be 1. Otherwise we would belooking for two consecutive squares that are at least 81 apart. But already theequation solves to , and has m ore than threedigits.The consecutive squares with common difference are and .One third of the way between them is and two thirds of the way is .This gives , , .Problem 19Rhombus , with side length , is rolled to form a cylinder of volum e bytaping to . What is ?SolutionWhere andProblem 20The parallelogram bounded by the lines , , , andhas area . The parallelogram bounded by the lines ,, , and has area . Given that , , , andare positive integers, what is the sm allest possible value of ?SolutionThis solution is incomplete. You can help us out by completing it.Plotting the parallelogram on the coordinate plane, the 4 corners are at. Because , wehave that or that , whichgives (consider a hom othety, or dilation, that carries the first parallelogram to the second parallelogram; because the area increases by , it follows that thestretch along the diagonal is ). The area of triangular half of the parallelogram onthe right side of the y-axis is given by , so substituting:Thus , and we verify that , will give us a minimum value for . Then.Solution 2This solution is incomplete. You can help us out by completing it.The key to this solution is that area is invariant under translation. By suitably shifting the plane, the problem is mapped to the linesand . Now, thearea of the parallelogram contained by is the former is equal to the area of arectangle with sides and , , and the area contained bythe latter is . Thus, and must be even if the form erquantity is to equal . so is a m ultiple of . Putting this alltogether, the minimal solution for , so the sum is . Problem 21The first positive integers are each written in base . How m any of thesebase-representations are palindromes? (A palindrome is a number that reads the sam e forward and backward.)SolutionAll numbers of six or less digits in base 3 have been written.The form of each palindrome is as follows1 digit -2 digits -3 digits -4 digits -5 digits -6 digits -Where are base 3 digitsSince , this gives a total ofpalindromes so far.7 digits - , but not all of the num bers are less thanCase:All of these numbers are less than giving more palindromesCase: ,All of these numbers are also small enough, giving more palindromesCase: ,It follows that , since any other would make the value too large. This leavesthe number as . Checking each value of d, all of the three are sm all enough, so that gives more palindromes.Summing our cases there areProblem 22Two particles m ove along the edges of equilateral in the directionstarting simultaneously and moving at the sam e speed. Onestarts at , and the other starts at the midpoint of . The midpoint of the linesegm ent joining the two particles traces out a path that encloses a region . Whatis the ratio of the area of to the area of ?SolutionFirst, notice that each of the midpoints of ,, and are on the locus.Suppose after som e time the particles have each been displaced by a short distance, to new positions and respectively. Consider and drop aperpendicular from to hit at . Then, and .From here, we can use properties of a triangle to determine thelengths and as m onomials in . Thus, the locus of the midpoint will be linear between each of the three special points m entioned above. It follows that the locus consists of the only triangle with those three points as vertices. Com paring inradii between this "midpoint" triangle and the original triangle, the area containedby must be of the total area.Problem 23How many non-congruent right triangles with positive integer leg lengths have areas that are num erically equal to tim es their perimeters?SolutionUsing Euclid's formula for generating primit ive triples: , ,where and are relatively prime positive integers, exactly one of which being even.Since we do not want to restrict ourselves to only primitives, we will add a factor ofk. , ,Now we do som e casework.Forwhich has solutions , , ,Removing the solutions that do not satisfy the conditions of Euclid's formula, theonly solutions are andForhas solutions , , both of which are valid.Forhas solutions , of which only is valid.Forhas solution , which is valid.This m eans that the solutions for aresolutionsProblem 24How many pairs of positive integers are there such that andis an integer?SolutionCombining the fraction, must be an integer.Since the denominator contains a factor of ,Rewriting as for some positive integer , we can rewrite the fraction asSince the denominator now contains a factor of , we get.But since , we must have , and thus .For the original fraction simplifies to .For that to be an integer, must divide , and therefore we must have. Each of these values does indeed yield an integer.Thus there are four solutions: , , , and the answer isProblem 25Points and are located in 3-dim ensional space withand .The plane of is parallel to . What is the area of ?SolutionLet , and . Since , we could let ,, and . Now to get back to we need another vertex. Now if we look at this configuration as if it was two dim ensions, we would see a square missing a side if we don't draw . Now we can bend thesethree sides into an equilateral triangle, and the coordinates change: ,, , , and . Checking for all the requirements, they are all satisfied. Now we find the area of triangle . Itis a triangle, which is an isosceles right triangle. Thus the area of it is.。
2010AMC10A试题及答案解析
2010 AMC 10A problems and solutions.The test was held on February 8, 2010. The first link contains the full set of test problems. The rest contain each individual problem and its solution.Problem 1Mary’s top book shelf holds five books with the follow ing widths, incentimeters: , , , , and . What is the average book width, incentimeters?SolutionTo find the average, we add up the widths , , , , and , to geta total sum of . Since there are books, the average book width isThe answer is .Problem 2Four identical squares and one rectangle are placed together to form one large square as shown. The length of the rectangle is how many times as large as its width?SolutionLet the length of the small square be , intuitively, the length of the big square is . It can be seen that the width of the rectangle is .Thus, the length of the rectangle is times large as thewidth. The answer is .Problem 3Tyrone had marbles and Eric had marbles. Tyrone then gavesome of his marbles to Eric so that Tyrone ended with twice as many marbles as Eric. How many marbles did Tyrone give to Eric?SolutionLet be the number of marbles Tyrone gave to Eric. Then,. Solving for yields and . Theanswer is .Problem 4A book that is to be recorded onto compact discs takes minutes toread aloud. Each disc can hold up to minutes of reading. Assumethat the smallest possible number of discs is used and that each disc contains the same length of reading. How many minutes of reading will each disc contain?SolutionAssuming that there were fractions of compact discs, it would takeCDs to have equal reading time. However, since the number of discs can only be a whole number, there are at least 8 CDs,in which case it would have minutes on each of the 8 discs.The answer is .Problem 5The area of a circle whose circumference is is . What is thevalue of ?SolutionIf the circumference of a circle is , the radius would be . Sincethe area of a circle is , the area is . The answer is .Problem 6For positive numbers and the operation is defined asWhat is ?Solution. Then, is The answer isProblem 7Crystal has a running course marked out for her daily run. She starts this run by heading due north for one mile. She then runs northeast for one mile, then southeast for one mile. The last portion of her run takes her on a straight line back to where she started. How far, in miles is this last portion of her run?SolutionCrystal first runs North for one mile. Changing directions, she runs Northeast for another mile. The angle difference between North and Northeast is 45 degrees. She then switches directions to Southeast, meaning a 90 degree angle change. The distance now from travellingNorth for one mile, and her current destination is miles, because it is the hypotenuse of a 45-45-90 triangle with side length one (mile). Therefore, Crystal's distance from her starting position, x, is equal to, which is equal to . The answer isTony works hours a day and is paid $per hour for each full yearof his age. During a six month period Tony worked days and earned$. How old was Tony at the end of the six month period?SolutionTony worked hours a day and is paid dollars per hour for eachfull year of his age. This basically says that he gets a dollar for each year of his age. So if he is years old, he gets dollars a day. Wealso know that he worked days and earned dollars. If he wasyears old at the beginning of his working period, he would have earned dollars. If he was years old at the beginningof his working period, he would have earned dollars.Because he earned dollars, we know that he was for someperiod of time, but not the whole time, because then the money earned would be greater than or equal to . This is why he waswhen he began, but turned sometime in the middle and earneddollars in total. So the answer is .The answer is . We could findout for how long he was and . . Then isand we know that he was for days, and for days. Thus,the answer is .Problem 9A palindrome, such as , is a number that remains the same whenits digits are reversed. The numbers and are three-digit andfour-digit palindromes, respectively. What is the sum of the digits ofSolutionis at most , so is at most . The minimum value ofis . However, the only palindrome between and is ,which means that must be .It follows that is , so the sum of the digits is .Marvin had a birthday on Tuesday, May 27 in the leap year . Inwhat year will his birthday next fall on a Saturday?Solution(E) 2017There are 365 days in a non-leap year. There are 7 days in a week. Since 365 = 52 * 7 + 1 (or 365 is congruent to 1 mod 7), the same date (after February) moves "forward" one day in the subsequent year, if that year is not a leap year.For example: 5/27/08 Tue 5/27/09 WedHowever, a leap year has 366 days, and 366 = 52 * 7 + 2. So the same date (after February) moves "forward" two days in the subsequent year, if that year is a leap year.For example: 5/27/11 Fri 5/27/12 SunYou can keep count forward to find that the first time this date falls on a Saturday is in 2017:5/27/13 Mon 5/27/14 Tue 5/27/15 Wed 5/27/16 Fri 5/27/17 Sat Problem 11The length of the interval of solutions of the inequality is. What is ?SolutionSince we are given the range of the solutions, we must re-write the inequalities so that we have in terms of and .Subtract from all of the quantities:Divide all of the quantities by .Since we have the range of the solutions, we can make them equal to .Multiply both sides by 2.Re-write without using parentheses.Simplify.We need to find for the problem, so the answer isProblem 12Logan 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's miniature. Therefore, Logan should make his towertimes shorter than the actual tower. This ismeters high, or choice .Problem 13Angelina drove at an average rate of kph and then stoppedminutes for gas. After the stop, she drove at an average rate ofkph. Altogether she drove km in a total trip time of hoursincluding the stop. Which equation could be used to solve for the time in hours that she drove before her stop?SolutionThe answer is （）because she drove at kmh for hours (theamount of time before the stop), and 100 kmh for because shewasn't driving for minutes, or hours. Multiplying by gives thetotal distance, which is kms. Therefore, the answer isProblem 14Triangle has . Let and be on and ,respectively, such that . Let be the intersection ofsegments and , and suppose that is equilateral. What isSolutionLet .Since ,Problem 15In 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 at least frog out of the two of them.Now suppose Mike is a toad. Then what he says is true because we already have toads. However, if Brian is a frog, then he is lying, yethis 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 isalready a toad, Brain must be a frog. We can also verify that his statement is indeed false.Both Mike and Brian are frogs, and one of either Chris or LeRoy is afrog, so we have frogs 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 16Nondegenerate has integer side lengths, is an anglebisector, , and . What is the smallest possible value ofthe 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 we use the next lowest values (and ), the Triangle Inequality is satisfied. Therefore,our answer is , or choice .Problem 17A solid cube has side length inches. A -inch by -inch square hole iscut into the center 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 . The second cut removes two boxes, each of dimensions, and the third cut does the same as the second cut, on thelast 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 cannot just sum their volumes, as the central cube is included ineach of these three cuts. To get 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 beseen as a box..Problem 18Bernardo randomly picks 3 distinct numbers from the setand arranges them in descending order to form a 3-digit number. Silvia randomly picks 3 distinct numbers from the setand 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 andadding up the probabilities.Case : Bernardo picks . If Bernardo picks a then it is guaranteedthat his number will be larger than Silvia's. The probability that he willpick a is .Case : Bernardo does not pick . Since the chance of Bernardopicking is , the probability of not picking is .If Bernardo does not pick 9, then he can pick any number from to .Since Bernardo is picking from the same set of numbers as Silvia, the probability that 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 samenumber is the number 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 19Equiangular hexagon has side lengthsand . The area of is of the area of thehexagon. What is the sum of all possible values of ?SolutionSolution 1It is clear that is an equilateral triangle. From the Law ofCosines, we get that . Therefore,the area of is .If we extend , and so that and meet at , andmeet at , and and meet at , we find that hexagonis formed by taking equilateral triangle of side lengthand removing three equilateral triangles, , and ,of side length . The area of is therefore.Based on the initial conditions,Simplifying this gives us . By Vieta's Formulas we knowthat the sum of the possible value of is .Solution 2As above, we find that the area of is .We also find by the sine triangle area formula that, and thusThis simplifies to.Problem 20A fly trapped inside a cubical box with side length meter decides torelieve its boredom by visiting each corner of the box. It will begin and end in the same corner and visit each of the other corners exactly once. To get from a corner to any other corner, it will either fly or crawl in a straight line. What is the maximum possible length, in meters, of its path?SolutionThe distance of an interior diagonal in this cube is and the distanceof a diagonal on one of the square faces is . It would not make sense if the fly traveled an interior diagonal twice in a row, as it would return to the point it just came from, so at most the final sum can onlyhave 4 as the coefficient of . The other 4 paths taken can be across a diagonal on one of the faces, so the maximum distance traveled is.Problem 21The polynomial has three positive integer zeros. What is the smallest possible value of ?SolutionBy Vieta's Formulas, we know that is the sum of the three roots ofthe polynomial . Also, 2010 factors into. But, since there are only three roots to the polynomial,two of the four prime factors must be multiplied so that we are left with three roots. To minimize , and should be multiplied, whichmeans will be and the answer is .Problem 22Eight points are chosen on a circle, and chords are drawn connecting every pair of points. No three chords intersect in a single point insidethe circle. How many triangles with all three vertices in the interior of the circle are created?SolutionTo choose a chord, we know that two points must be chosen. This implies that for three chords to create a triangle and not intersect at a single point, six points need to be chosen. Therefore, the answer iswhich is equivalent to 28,Problem 23Each of 2010 boxes in a line contains a single red marble, and for , the box in the position also contains white marbles.Isabella begins at the first box and successively draws a single marble at random from each box, in order. She stops when she first draws ared marble. Let be the probability that Isabella stops after drawing exactly marbles. What is the smallest value of for whichSolutionThe probability of drawing a white marble from box is . Theprobability of drawing 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 satisfiesthe inequality is .Problem 24The 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 thereforethe last two digits of , or . Since there is clearly anexcess of factors of 2, we know that , so it remains tofind .If we divide by by taking out all the factors of in , we canwrite as where whereevery 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:Using the identity at the beginning of the solution, we can reducetoUsing the fact that (or simply the fact thatif you have your powers of 2 memorized), we can deducethat . Therefore.Finally, combining with the fact that yields.Problem 25Jim starts with a positive integer and creates a sequence of numbers. Each successive number is obtained by subtracting the largest possible integer square less than or equal to the current number until zero is reached. For example, if Jim starts with , then hissequence contains numbers:Let be the smallest number for which J im’s sequence has numbers.What is the units digit of ?SolutionWe can find the answer by working backwards. We begin withon the bottom row, then the goes to the right of the equal'ssign in the row above. We find the smallest value for whichand , which is .We repeat the same procedure except with for the next rowand for the row after that. However, at the fourth row, wesee that solving yields , in which case it would beincorrect since is not the greatest perfect square less than orequal to . So we make it a and solve . We continue onusing this same method where we increase the perfect square until can be made bigger than it. When we repeat this until we have rows,we get:Hence the solution is the last digit of , which is .。
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In recent years, Chinese education system has become quite utilitarian. Study to get high marks in order to be accepted by a university and it seems that you then have a promised life. And with the change of policies to focus more on children’s balanced development nowadays, students are even burden with much more. As I am tutoring a girl recently, I know only too well about that.
The girl’s mother has me to teach her English, has someone else to teach her Math, drawing, playing the zither (古筝) and even playing basketball. Her mother herself helps her with dictations every night for about one hour. I was just shocked when the girl tell me about that, I think I was much happier that she is when I was at her age. Honestly, I have to say, that even she takes so many classes, she is not really good at study.
So I just wonder what we could do to change this kind of phenome non, that is what can we do to help children from being burdened with more than they can bear.
The long-term effective practice definitely should be the change of people’s opinion. I forgot to say one thing, that is when I ask the girl I tutor what she thinks of those varied classes and is she tired, to my surprise, she said it’s O kay and she don’t mind. I think she is somehow a little numb and she needs someone to tell her that she shouldn’t be force to burden so much. I really want to tell her mother abou t that, however, ironically, I was the one who earn money out of this kind of idea and how can I convince them. So I think maybe people who are more authoritative like headmasters or teachers of children like her should take most of the responsibility to tell the truth, helping not only the children but also the parents to realize that it’s not
true that you take more courses and you become more easily to succeed. But those headmasters and teachers’ opinion should have changed first and that need efforts to o.
Then comes another useful measurement. It is the conversion of the assessment system of students. Assess a student more depends on his ability of critical thinking, his imagination, his morality and some other good qualities a man should have will greatly reduce parents’ and schools’ attention on students points. But it’s hard to change a system in such a short time because it needs careful consideration.
So maybe the abolishment of the preferential policy that awards bonus points in the national college entrance exams (高考加分政策) may be more practical. In doing so, people may pay less attention on those items that can earn their child ren bonus points so children will not be rushed so hard to take so much courses that they don’t like at all.
At last, I want to say that to change a phenomenon always need long-term efforts, but be sure that we will finally make it if we are willing and start to do.。