2014美国数学竞赛AMC10A、10B试题及答案
AMC10B试题及答案解析
AMC10B试题及答案解析2010 AMC 10B1 . What is ?SolutionWe first expand the first term, simplify, and then compute to get an answer of .2 、Makarla attended two meetings during her -hour work day. The first meeting took minutes and the second meeting took twice as long. What percent of her work day was spent attending meetings?SolutionThe percentage of her time spent in meetings is the total amount of time spent in meetings divided by the length of her workday. The total time spent in meetings isTherefore, the percentage is3、A drawer contains red, green, blue, and white socks with at least 2 of each color. What is the minimum number of socks that must be pulled from the drawer to guarantee a matching pair?SolutionAfter you draw socks, you can have one of each color, so (according to the pigeonhole principle), if you pull then you will be guaranteed a matching pair.4 、For a real number , define to be the average of and . What is ?SolutionThe average of two numbers, and , is defined as . Thus the average of and would be . With that said, we need to find the sum when we plug, , and into that equation. So:.5 、A month with days has the same number of Mondays and Wednesdays.How many of the seven days of the week could be the first day of this month?Solution（B）. 36 、A circle is centered at , is a diameter and is a point on the circle with . What is the degree measure of ?SolutionAssuming the reader is not readily capable to understand how will always be right, the I will continue with an easily understandable solution. Since is the center, are all radii, they are congruent. Thus, and are isosceles triangles. Also, note that and are supplementary, then . Since is isosceles, then . They also sum to , so each angle is .7 、A triangle has side lengths , , and . A rectangle has width and area equal to the area of the rectangle. What is the perimeter of this rectangle?SolutionThe triangle is isosceles. The height of the triangle is therefore given byNow, the area of the triangle isWe have that the area of the rectangle is the same as the area of the triangle, namely 48. We also have the width of the rectangle: 4.The length of the rectangle therefor is:The perimeter of the rectangle then becomes:The answer is:8 、A ticket to a school play cost dollars, where is a whole number. A group of 9th graders buys tickets costing a total of $, and a group of 10th graders buys tickets costing a total of $. How many values for are possible?SolutionWe see how many common integer factors 48 and 64 share. Of the factors of 48 - 1, 2, 3, 4, 6, 8, 12, 16, 24, and 48; only 1, 2, 4, 8, and 16 are factors of 64. So there are possibilities for the ticket price.9 、Lucky Larry's teacher asked him to substitute numbers for , , , , and in the expression and evaluate the result. Larry ignored the parenthese but added and subtracted correctly andobtained the correct result by coincidence. The number Larry sustitued for , , , and were , , , and , respectively. What number did Larry substitude for ?SolutionSimplify the expression . I recommend to start with the innermost parenthesis and work your way out.So you get:Henry substituted with respectively.We have to find the value of , such that(the same expression without parenthesis).Substituting and simplifying we get:So Henry must have used the value for .Our answer is:10、Shelby drives her scooter at a speed of miles per hour if it is not raining, and miles per hour if it is raining. Today she drove in the sun in the morning and in the rain in the evening, for a total of miles in minutes. How many minutes did she drive in the rain?SolutionWe know thatSince we know that she drove both when it was raining and when it was not and that her total distance traveled is miles.We also know that she drove a total of minutes which is of an hour. We get the following system of equations, where is the time traveled when it was not raining and is the time traveled when it was raining:Solving the above equations by multiplying the second equation by 30 and subtracting the second equation from the first we get:We know now that the time traveled in rain was of an hour, which is minutesSo, our answer is:11 、A shopper plans to purchase an item that has a listed price greater than $and can use any one of the three coupns. CouponA givesoff the listed price, Coupon B gives $off the listed price, and Coupon C gives off the amount by which the listed price exceeds $. Let and be the smallest and largest prices, respectively, for which Coupon A saves at least as many dollars as Coupon B or C. What is ??SolutionLet the listed price be , whereCoupon A saves us:Coupon B saves us:Coupon C saves us:Now, the condition is that A has to be greater than or equal to either B or C which give us the following inequalities:We see here that the greatest possible value for p is and the smallest isThe difference between and isOur answer is:12 、At the beginning of the school year, of all students in Mr. Wells' math class answered "Yes" to the question "Do you love math", andanswered "No." At the end of the school year, answered "Yes" and answerws "No." Altogether, of the students gave a different answer at the beginning and end of the school year. What is the difference between the maximum and the minimum possible values of ?SolutionThe minimum possible value occurs when of the students who originally answered "No." answer "Yes." In this case,The maximum possible value occurs when of the students who originally answered "Yes." answer "No." and the of the students who originally answered "No." answer "Yes." In this case,Subtract to obtain an answer of13 、What is the sum of all the solutions of ?SolutionCase 1:Case 1a:Case 1b:Case 2:Case 2a:Case 2b:Since an absolute value cannot be negative, we exclude . The answer is14 、The average of the numbers and is . What is ?SolutionWe must find the average of the numbers from to and in terms of . The sum of all these terms is . We must divide this by the total number of terms, which is . We get: . This is equal to , as stated in the problem. We have: . We can now cross multiply. This gives:This gives us our answer.15 、On a -question multiple choice math contest, students receive points for a correct answer, points for an answer left blank, and point for an incorrect answer. Jesse’s total score on the contest was . What is the maximum number of questions that Jesse could have answered correctly?SolutionLet be the amount of questions Jesse answered correctly, be the amount of questions Jesse left blank, and be the amount ofquestions Jesse answered incorrectly. Since there were questions on the contest, . Since his total score was , . Also,. We can substitute this inequality into the previous equation to obtain another inequality:. Since is an integer, the maximum value for is .16 、A square of side length and a circle of radius share the same center. What is the area inside the circle, but outside the square?Solution(B)17 、Every high school in the city of Euclid sent a team of students to a math contest. Each participant in the contest received a different score. Andrea's score was the median among all students, and hers was the highest score on her team. Andrea's teammates Beth and Carla placed th and th, respectively. How many schools are in the city?SolutionLet the be the number of schools, be the number of contestants, and be Andrea's score. Since the number of participants divided by three isthe number of schools, . Andrea received a higher score than her teammates, so . Since is the maximum possible median, then is the maximum possible number ofparticipants. Therefore, . This yields thecompound inequality: . Since a set with an even number of elements has a median that is the average of the two middle terms, an occurrence that cannot happen in this situation, cannot be even.is the only other option.18 、Positive integers , , and are randomly and independently selected with replacement from the set . What is the probability that is divisible by ?Solution（E）13/2719 、A circle with center has area . Triangle is equilateral,is a chord on the circle, , and point is outside . What is the side length of ?Solution(B)620 、Two circles lie outside regular hexagon . The first is tangent to , and the second is tangent to . Both are tangent to lines and . What is the ratio of the area of the second circle to that of the first circle?SolutionA good diagram here is very helpful.The first circle is in red, the second in blue. With this diagram, we can see that the first circle is inscribed in equilateral triangle while the second circle is inscribed in . From this, it's evident that the ratio of the red area to the blue area is equal to the ratio of the areas of triangles toSince the ratio of areas is equal to the square of the ratio of lengths, weknow our final answer is From the diagram, we can see that this isThe letter answer is D21 、A palindrome between and is chosen at random. What is the probability that it is divisible by ?SolutionView the palindrome as some number with form (decimal representation): . But because the number is a palindrome, . Recombining this yields . 1001 is divisible by 7, which means that as long as , the palindrome will be divisible by 7. This yields 9 palindromes out of 90 () possibilities for palindromes. However, if , then this givesanother case in which the palindrome is divisible by 7. This adds another 9 palindromes to the list, bringing our total to22 、Seven distinct pieces of candy are to be distributed among three bags. The red bag and the blue bag must each receive at least one piece of candy; the white bag may remain empty. How many arrangements are possible?SolutionWe can count the total number of ways to distribute the candies (ignoring the restrictions), and then subtract the overcount to get the answer.Each candy has three choices; it can go in any of the three bags.Since there are seven candies, that makes the total distributionsTo find the overcount, we calculate the number of invalid distributions: the red or blue bag is empty.The number of distributions such that the red bag is empty is equal to , since it's equivalent to distributing the 7 candies into 2 bags.We know that the number of distributions with the blue bag is empty will be the same number because of the symmetry, so it's also .The case where both the red and the blue bags are empty (all 7 candies are in the white bag) are included in both of the above calculations, and this case has only distribution.The total overcount isThe final answer will beThat makes the letter choice C23 、The entries in a array include all the digits from through ,arranged so that the entries in every row and column are in increasing order. How many such arrays are there?Solution（D）6024 、A high school basketball game between the Raiders and Wildcatswas tied at the end of the first quarter. The number of points scored by the Raiders in each of the four quarters formed an increasing geometric sequence, and the number of points scored by the Wildcats in each of the four quarters formed an increasing arithmetic sequence. At the end of the fourth quarter, the Raiders had won by one point. Neither team scored more than points. What was the total number of points scored by the two teams in the first half?SolutionRepresent the teams' scores as: andWe have Manipulating this, we can get, orSince both are increasing sequences, . We can check cases up to because when , we get . WhenChecking each of these cases individually back into the equation, we see that only when a=5 and n=2, we get an integer value for m, which is 9. The original question asks for the first half scores summed, so we must find25 、Let , and let be a polynomial with integer coefficients such that, and.What is the smallest possible value of ?SolutionThere must be some polynomial such thatThen, plugging in values of we getThus, the least value of must be the . Solving, we receive , so our answer is .。
2014 AMC 10A Problems
2014 AMC 10A ProblemsProblem 1 What is 1)1015121(10-++∙ ? （A ）3 （B ）8 （C ）225 （D ）3170 （E ）170 Problem 2Roy's cat eats 1/3 of a can of cat food every morning and 1/4 of a can of cat food every evening. Before feeding his cat on Monday morning, Roy opened a box containing 6 cans of cat food. On what day of the week did the cat finish eating all the cat food in the box?（A ）Tuesday （B ）Wednesday （C ）Thursday （D ）Friday （E ）Saturday Problem 3Bridget bakes 48 loaves of bread for her bakery. She sells half of them in the morning for $2.50 each. In the afternoon she sells two thirds of what she has left, and because they are not fresh, she charges only half price. In the late afternoon she sells the remaining loaves at a dollar each. Each loaf costs $0.75 for her to make. In dollars, what is her profit for the day?（A ）24 （B ）36 （C ）44 （D ）48 （E ）52 Problem 4Walking down Jane Street, Ralph passed four houses in a row, each painted a different color. He passed the orange house before the red house, and he passed the blue house before the yellow house. The blue house was not next to the yellow house. How many orderings of the colored houses are possible?（A ）2 （B ）3 （C ）4 （D ）5 （E ）6 Problem 5On an algebra quiz, 10% of the students scored 70 points, 35% scored 80 points, 30% scored 90 points, and the rest scored 100 points. What is the difference between the mean and median score of the students' scores on this quiz?（A ）1 （B ）2 （C ）3 （D ）4 （E ）5 Problem 6Suppose that cows give gallons of milk in days. At this rate, how many gallons ofmilk will cows give in days?（A ）ac bde （B ）bde ac （C ）c abde （D ）abcde （E ）de abc Problem 7Nonzero real numbers ,,,a y x and b satisfy a x < and b y <. How many of the following inequalities must be true?（I ）b a y x +<+ （II ）b a y x -<- （II ）ab xy < （IV ）b a y x //<（A ）0 （B ）1 （C ）2 （D ）3 （E ）4Problem 8Which of the following numbers is a perfect square?（A ）2!15!14 （B ）2!16!15 （C ）2!17!16 （D ）2!18!17 （E ）2!19!18 Problem 9 The two legs of a right triangle, which are altitudes, have lengths 32 and 6. How long is the third altitude of the triangle?（A ）1 （B ）2 （C ）3 （D ）4 （E ）5 Problem 10Five positive consecutive integers starting with have average . What is the averageof 5 consecutive integers that start with ?（A ）3+a （B ）4+a （C ）5+a （D ）6+a （E ）7+a Problem 11A customer who intends to purchase an appliance has three coupons, only one of which may be used:Coupon 1: 10% off the listed price if the listed price is at least $50Coupon 2: $20 off the listed price if the listed price is at least $100Coupon 3: 18% off the amount by which the listed price exceeds $100For which of the following listed prices will coupon 1 offer a greater price reduction than either coupon 2 or coupon 3?（A ）$179.95 （B ）$199.95 （C ）$219.95 （D ）$239.95 （E ）$259.95 Problem 12A regular hexagon has side length 6. Congruent arcs with radius 3 are drawn with the center at each of the vertices, creating circular sectors as shown. The region inside the hexagon but outside the sectors is shaded as shown What is the area of the shaded region?（A ）π9327-（B ）π6327-（C ）π18354-（D ）π12354-（E ）π93108- Problem 13Equilateral △ABC has side length 1, and squares ABDE, BCHI, CAFG lie outside the triangle. What is the area of hexagon DEFGHI ?（A ）43312+ （B ）29 （C ）33+ （D ）2336+ （E ）6 Problem 14The y -intercepts, P and Q , of two perpendicular lines intersecting at thepoint A (6,8) have a sum of zero. What is the area of △APQ ?（A ）45 （B ）48 （C ）54 （D ）60 （E ）72Problem 15David drives from his home to the airport to catch a flight. He drives 35 miles in the first hour, but realizes that he will be 1 hour late if he continues at this speed. He increases his speed by 15 miles per hour for the rest of the way to the airport and arrives 30 minutes early. How many miles is the airport from his home?（A ）140 （B ）175 （C ）210 （D ）245 （E ）280 Problem 16In rectangle ABCD, AB =1, BC =2, and points E, F , and G are midpoints of CD BC ,, and AD , respectively. Point H is the midpoint of GE . What is the area of the shaded region?（A ）121 （B ）183 （C ）122 （D ）123 （E ）61 Problem 17Three fair six-sided dice are rolled. What is the probability that the values shown on two of the dice sum to the value shown on the remaining die?（A ）61（B ）7213 （C ）367 （D ）245 （E ）92 Problem 18A square in the coordinate plane has vertices whose -coordinates are 0, 1, 4, and 5.What is the area of the square?（A ）16 （B ）17 （C ）25 （D ）26 （E ）27Problem 19Four cubes with edge lengths 1, 2, 3, and 4 are stacked as shown. What is the length of the portion of XY contained in the cube with edge length 3?（A ）5333 （B ）32 （C ）3332 （D ）4 （E ）23Problem 20The product (8)(888…8), where the second factor has digits, is an integer whose digitshave a sum of 1000. What is ?（A ）901 （B ）911 （C ）919 （D ）991 （E ）999Problem 21Positive integers a and b are such that the graphs of 5+=ax y and b x y +=3 intersect the -axis at the same point. What is the sum of all possible-coordinates of these points of intersection?（A ）-20 （B ）-18 （C ）-15 （D ）-12 （E ）-8Problem 22In rectangle ABCD, AB =20 and BC =10. Let E be a point on CD such that ︒=∠15CBE . What is AE ?（A ）3320 （B ）310 （C ）18 （D ）311 （E ）20 Problem 23 A rectangular piece of paper whose length is 3 times the width has area A . The paper is divided into three equal sections along the opposite lengths, and then a dotted line is drawn from the first divider to the second divider on the opposite side as shown. The paper is then folded flat along this dotted line to create a new shape with area B . What is the ratio B:A ?（A ）1:2 （B ）3:5 （C ）2:3 （D ）3:4 （E ）4:5Problem 24A sequence of natural numbers is constructed by listing the first 4, then skipping one, listing the next 5, skipping 2, listing 6, skipping 3, and, on the th iteration, listing3+n and skipping . The sequence begins 1,2,3,4,6,7,8,9,10,13.What is the 500,000thnumber in the sequence?（A ）996,506 （B ）996507 （C ）996508 （D ）996509 （E ）996510 Problem 25The number 8675 is between 20132 and 20142. How many pairs of integers ),(n m are there such that 20121≤≤m and 125225++<<<n m m n ?（A ）278 （B ）279 （C ）280 （D ）281 （E ）2822014 AMC 10A SolutionsProblem 1 We have 1)1015121(10-++∙Making the denominators equal gives2254510)108(10)101102105(1011⇒∙⇒∙⇒++∙⇒-- Problem 2 Each day, the cat eats 1274131=+ of a can of cat food. Therefore, the cat food will last for 7726127= days, which is greater than 10 days but less than 11 days. Because the number of days is greater than 10 and less than 11, the cat will finish eatingin on the 11th day, which is equal to 10 Problem 3She first sells one-half of her 48 loaves, or 48/2=24 loaves. Each loaf sells for $2.50, so her total earnings in the morning is equal to 24·$2.50=$60.This leaves 24 loaves left, and Bridget will sell 162432=⨯ of them for a price of 25.1$250.2$=. Thus, her total earnings for the afternoon is 16·$1.25=$20. Finally, Bridget will sell the remaining 24-16=8 loaves for a dollar each. This is a total of $1·8=$8.The total amount of money she makes is equal to 60+20+8=$88.However, since Bridget spends $0.75 making each loaf of bread, the total cost to make the bread is equal to $0.75·48=$36.Her total profit is the amount of money she spent subtracted from the amount of money she made, which is ?????Problem 4The following problem is from both the 2014 AMC 12A #3 and 2014 AMC 10A #4Solution 1Attack this problem with very simple casework. The only possible locations for the yellow house (Y ) is the 3rd house and the last house.Case 1: Y is the 3rd house.The only possible arrangement is B-O-Y-RCase 2: Y is the last house.There are two possible ways: B-O-R-Y and O-B-R-YSolution 2There are 24 possible arrangements of the houses. The number of ways with the blue house next to the yellow house is 3!2!=12, as we can consider the arrangements of O, (RB), and Y . Thus there are 24-12 arrangements with the blue and yellow houses non-adjacent.Exactly half of these have the orange house before the red house by symmetry, and exactly half of those have the blue house before the yellow house (also by symmetry), so our answer is 3212112=∙∙. Problem 5The following problem is from both the 2014 AMC 12A #5 and 2014 AMC 10A #5Without loss of generality, let there be 20 students(the least whole number possible) who took the test. We have 2 students score 70 points, 7 students score 80 points, 6 students score 90 points and 5 students score 100 points.The median can be obtained by eliminating members from each group. The median is 90 points.The mean is equal to the total number of points divided by the number of people, which gives 87Thus, the difference between the median and the mean is equal to 90-87=3. Problem 6The following problem is from both the 2014 AMC 12A #4 and 2014 AMC 10A #6Solution 1We need to multiply by a d / for the new cows and c e / for the new time, so theanswer is acbde c e a d b =∙∙, or . Solution 2We plug in ,5,4,3,2====d c b a and 6=e . Hence the question becomes "2 cows give 3 gallons of milk in 4 days. How many gallons of milk do 5 cows give in 6 days?"If 2 cows give 3 gallons of milk in 4 days, then 2 cows give 3/4 gallons of milk in 1 day, so 1 cow gives 3/(4*2) gallons in 1 day. This means that 5 cows give (5*3)/(4*2) gallons of milk in 1 day. Finally, we see that 5 cows give (5*3*6)/(4*2) gallons of milk in 6 days.Substituting our values for the variables, this becomes ac dbe / , which is Solution 3We see that the the amount of cows is inversely proportional to the amount of days and directly proportional to the gallons of milk. So our constant is b ac /.Let be the answer to the question. We haveacbde g bde gac b ac g de =⇒=⇒=Problem 7SolutionFirst, we note that (I) must be true by adding our two original inequalities. b a y x b y a x +<+⇒<<,Though one may be inclined to think that (II) must also be true, it is not, for we cannot subtract inequalities.In order to prove that the other inequalities are false, we only need to provide one counterexample. Let's try substituting 1,1,2,3==-=-=b a y x(II) states that 0111)2(3<⇒-<---⇒-<-b a y x Since this is false, (II) must also be false.(III) states that 1611)2)(3(<⇒∙<--⇒<ab xy . This is also false, thus (III) is false. (IV) states that 15.12123<⇒<--⇒<b a y x . This is false, so (IV) is false.One of our four inequalities is true, hence, our answer is 1. Solution 2Also, with some intuition, we could have plugged ,3,1,0Y A X =-== and B =-2 and then plugged these values into the equations to see which ones held.Problem 8Note that for all positive , we have21)!(2)1()!(2)!1(!22+∙⇒+⇒+n n n n n n We must find a value of such that21)!(2+∙n n is a perfect square. Since 2)!(n is a perfect square, we must also have 21+n be a perfect square. In order for21+n to be a perfect square, 1+n must be twice a perfect square. From theanswer choices, 181=+n works, thus, 17=n and our desired answer is 2!18!17 Problem 9Solution 1 We find that the area of the triangle is 3636=⨯. By the Pythagorean Theorem , we have that the length of the hypotenuse is 346)32(22=+ . Dropping an altitude from the right angle to the hypotenuse, we can calculate the area in another way.Let h be the third height of the triangle. We have 336234=⇒⨯=h h Problem 10The following problem is from both the 2014 AMC 12A #9 and 2014 AMC 10A #10Solution 1Let 1=a . Our list is {1,2,3,4,5} with an average of 15/5=3. Our next set starting with 3 is {3,4,5,6,7}. Our average is 25/5=5.Therefore, we notice that 5=1+4 which means that the answer is 4+a . Solution 2We are given that 254321+=⇒++++++++=a b a a a a a b We are asked to find the average of the 5 consecutive integers starting from in termsof . By substitution, this is4565432+=+++++++++a a a a a aThus, the answer is 4+a . Problem 11The following problem is from both the 2014 AMC 12A #8 and 2014 AMC 10A #11Solution 1Let the listed price be . Since all the answer choices are above $100, we can assume100>x . Thus the discounts after the coupons are used will be as follows:Coupon 1: x x 1.0%10=∙Coupon 2: 20Coupon 3: 1818.0)100(%18-=-∙x xFor coupon 1 to give a greater price reduction than the other coupons, we must have 200201.0>⇒>x x and 2251818.01.0<⇒->x x x .From the first inequality, the listed price must be greater than $200, so answer choices(A) and (B) are eliminated.From the second inequality, the listed price must be less than $225, so answer choices(D) and (E) are eliminated.The only answer choice that remains is $219.95. Problem 12The area of the hexagon is equal to 35423)6(32= by the formula for the area of a hexagon. We note that each interior angle of the regular hexagon is 120º which means that each sector is 1/3 of the circle it belongs to. The area of each sector is ππ33/9=. The area of all six is ππ1836=⨯.sectors, which is equal to π18354- Problem 13Solution 1The area of the equilateral triangle is 4/3. The area of the three squares is 3*1=3. Since 360=∠C , 120609090360=---=∠GCH .Dropping an altitude from C to GH allows to create a 30-60-90 triangle since GCH ∆ is isosceles. This means that the height of GCH ∆ is 1/2 and half the length of GH is 2/3. Therefore, the area of each isosceles triangle is 432321=⨯. Multiplying by 3 yields 4/33 for all three isosceles triangles.Therefore, the total area is 33433433+=++ Solution 2As seen in the previous solution, segment GH is 3 . Think of the picture as one large equilateral triangle, JKL ∆ with the sides of 132+, by extending EF ,GH, and DI to points J, K, and L , respectively. This makes the area of JKL ∆ 431312)132(432+=+Triangles △DIJ , △EFK , and △GHL have sides of3, so their total area is439))3(43(32=.Now, you subtract their total area from the area of △JKL : 3439431312+=-+Problem 14Solution 1Note that if the -intercepts have a sum of 0, the distance from the origin to each of theintercepts must be the same. Call this distance . Since the ∠PAQ=90º, the length of themedian to the midpoint of the hypotenuse is equal to half the length of the hypotenuse. Since the median's length is 108622=+, this means 10=a , and the length of the hypotenuse is 202=a . Since the of A is the same as the altitude to thehypotenuse, []602/620=∙=APQ Solution 2We can let the two lines be b x my b mx y --=+=1,. This is because the lines are perpendicular, hence theandm1- , and the sum of the y-intercepts is equal to 0, hence the b b -,. Since both lines contain the point (6,8), we can plug this into the two equations to obtain b m +=68and b m--=168. Adding the two equations gives m m 6616-+=. Multiplying by m gives06166661622=--⇒-=m m m m . Factoring gives 0)3)(13(=-+m m .We can just let 3=m , since the two values of do not affect our solution - one is theslope of one line and the other is the slope of the other line.Plugging 3=m into one of our original equations, we obtain 10)3(68-=⇒+=b bSince △APQ has hypotenuse 202=b and the altitude to the hypotenuse is equal tothe x-coordinate of point A , or 6, the area of △APQ is equal to 20·6/2=60. Problem 15The following problem is from both the 2014 AMC 12A #11 and 2014 AMC 10A #15 Solution 1 (Algebra)Note that he drives at 50 miles per hour after the first hour and continues doing so until he arrives.Let be the distance still needed to travel after the first 1 hour. We have that 355.150d d =+, where the 1.5 comes from1 hour late decreased to 0.5 hours early.Simplifying gives ,105257d d =+ or 175=d .Now, we must add an extra 35 miles traveled in the first hour, giving a total of (C) 210 miles.Solution 2 (Answer Choices)Instead of spending time thinking about how one can set up an equation to solve the problem, one can simply start checking the answer choices. Quickly checking, we know that neither choice(A) or choice(B) work, but(C) does. We can verify as follows. After 1 hour at 35mph , David has 175 miles left. This then takes him 3.5 hours at 50mph . But 210/35=6 hours. Since 1+3.5=4.5 hours is 1.5 hours less than 6, our answer is 210.Problem 16Solution 1Denote D=(0,0). Then A=(0,2), F=(1/2,0), H=(1/2,1). Let the intersection of AF and DH be X , and the intersection of BF and CH be Y . Then we want to find the coordinates of X so we can find XY . From our points, the slope of AF is 4)2(21-=-, and its -intercept isjust 2. Thus the equation for AF is 24+-=x y . We can also quickly find that the equation of DH is x y 2=. Setting the equations equal, we have 3/1242=⇒+-=x x x . Becauseof symmetry, we can see that the distance from Y to BC is also 1/3, so 313121=∙-=XY . Now the area of the kite is simply the product of the two diagonals over 2. Since the lengthHF =1, our answer is 612131=∙ .Solution 2Let the area of the shaded region be . Let the other two vertices of the kitebe I and J with I closer to AD than J . Note that [][][][][]BCJ ADI x DCH ABF ABCD ++-+= . The area of ABF is 1 and the area of DCH is 1/2. We will solve for the areas of ADI and BCJ in terms of x by noting that the area of each triangle is the length of the perpendicularfrom I to AD and J to BC respectively. Because the area of IJ x *21=based on the areaof a kite formula, 2/ab for diagonals of length and , x IJ 2=. So each perpendicularis length 221x -. So taking our numbers and plugging them into [][][][][]BCJ ADI x DCH ABF ABCD ++-+= gives us x 3252-=. Solving thisequation for gives us 6/1=x .Solution 3From the diagram in Solution 1, let be the height of XHY and f be the height of XFY . Itis clear that their sum is1 as they are parallel to GD . Let be the ratio of the sides of thesimilar triangles XFY and AFB , which are similar because XY is parallel to AB and the triangles share angle F . Then 2/f k = , as 2 is the height of AFB . Since XHY and DHC are similar for the same reasons as XFY and AFB , the height of XHY will be equal to the base, like in DHC , making e XY =. However, XY is also the base of XFY , so AB e k /= where AB =1 so e k =. Subbing into 2/f k = gives a system of linear equations, 1=+f e and 2/f e =. Solving yields 3/1==XY e and 3/2=f , and since the area of the kite is simply the product of the two diagonals over 2 and HF =1, our answerIs 612131=∙. Solution 4Let the unmarked vertices of the shaded area be labeled I and J , with I being closer to GD than J . Noting that kite HJFI can be split into triangles HJI and JIF . Because HJI and JIF are similar to HDC and ABF , we know that the distance from line segment JI to H is half the distance from JI to F . Because kite HJFI is orthodiagonal, we multiply6/12/))3/1(*1(= Problem 17Solution 1 (Clean Counting)First, we note that there are 1,2,3,4 and 5 ways to get sums of 2,3,4,5,6 respectively--this is not too hard to see. With any specific sum, there is exactly one way to attain it on the other die. This means that the probability that two specific dice have the same sum as the other is 725)3654321(61=++++. Since there are ⎪⎪⎭⎫ ⎝⎛13 ways to choose which die will be theone with the sum of the other two, our answer is 2457253=∙ . Solution 2 (Casework)Since there are 6 possible values for the number on each dice, there are 21663= total possible rolls.The possible results of the 3 dice such that the sum of the values of two of the die is equal to the value of the third die are, without considering the order of the die, (1,1,2), (1,2,3), (1,3,4), (1,4,5), (1,5,6), (2,2,4), (2,3,5), (2,4,6), (3,3,6). There are 3!/2=3 ways to order the first, sixth, and ninth results, and there are 3!=6 ways to order the other results.Therefore, there are a total of 3*3+6*6=45 ways to roll the dice such that 2 of the dice sumto the other die, so our answer is 45/216=5/24. Problem 18Let the points be )5,(),1,(),0,(321x C x B x A ===, and )4,(4x D =Note that the difference in value of B and C is 4. By rotational symmetry of the square,the difference in value of A and B is also 4. Note that the difference in valueof A and B is 1. We now know that AB , the side length of the square, is equal to174122=+, so the area is 17. .Problem 19By Pythagorean Theorem in three dimensions, the distance XY is 3321044222=++ . Let the length of the segment XY that is inside the cube with side length 3 be . By similartriangles, 10/3323/=x , giving 5/333=x . Problem 20The following problem is from both the 2014 AMC 12A #16 and 2014 AMC 10A #20Note that for 2≥k , , which has a digit sum of k k +=++-+94027. Since we are given that said number has a digit sum of 1000,we have 99110009=⇒=+k k Problem 21Note that when 0=y , the values of the equations should be equal by the problemstatement. We have that a x ax /550-=⇒+=,3/30b x b x -=⇒+=.Which means that 153//5=⇒-=-ab b a .The only possible pairs ),(b a then are )1,15(),3,5(),5,3(),15,1(),(=b a . These pairs give respective x -values of 3/1,1,3/5,5----Problem 22Note that 3102010/15tan -=⇒=︒EC EC . (If you do not know the tangent half-angle formula, it is aa sin cos 1-. Therefore, we have 310=DE . Since ADE is a30-60-90 triangle, 201022=∙=∙=AD AE . Problem 23Solution 1I've no clue how to draw pictures on here, so I'll give instructions. Find the midpoint of the dotted line. Draw a line perpendicular to it. From the point this line intersects the top of the paper, draw lines to each endpoint of the dotted line. These two lines plus the dotted line form a triangle which is the double-layered portion of the folded paper. WLOG, assume the width of the paper is 1 and the length is 3 . The triangle we want to find has side lengths3321)33(,3322=+ , and 3321)33(2=+. It is an equilateral triangle with height 1333=∙, and area 3321332=∙. The area of the paper is 331=∙, and the folded paper has area 32333=-. The ratio of the area of the folded paper to that of theoriginal paper is thus 2:3. Solution 2 Our original paper can be divided like this:After the fold across the dotted line, our paper becomes:Since our original sheet of paper has six congruent 30-60-90 triangles and and our newone has four, the ratio of the area B:A is equal to 4:6=2:3 Problem 24Solution 1If we list the rows by iterations, then we get1,2,3,46,7,8,9,1013,14,15,16,17,18 etc.so that the 500,000th number is the 506th number on the 997th row.(4+5+6+7+…+999=499,494). The last number of the 996th row (when including the numbers skipped) is 499,494+(1+2+3+4+…+996)=996,000 , (we add the 1—996 becauseof the numbers we skip) so our answer is 996,000+506=996,506. Solution 2Let's start with natural numbers, with no skips in between. 1,2,3,4,5,…, 500000All we need to do is count how many numbers are skipped, , and "push" (add on to)500000 however many numbers are skipped.Clearly, 000,5002)1000(999≤ . This means that the number of skipped number "blocks" in the sequence is 999-3=996 because we started counting from 4.Therefore 506,4962)997(996==n Problem 25The following problem is from both the 2014 AMC 12A #22 and 2014 AMC 10A #25Between any two consecutive powers of 5 there are either 2 or 3 powers of 2 (because 312252<<). Consider the intervals ).5,5),...(5,5(),5,5(8678662110 We want the number of intervals with 3 powers of 2.From the given that 20148672013252<<, we know that these 867 intervals together have 2013 powers of 2. Let of them have 2 powers of 2 and of them have 3 powers of 2.Thus we have the system 201332,867=+=+y x y x from which we get 279=y , sothe answer is2014 AMC 10A Answer Key 1. C 2. C3.E4. B5. C6.A7.B8.D 8. C 10.B 11.C 12. C13.C 14.D 15. C 16.E 17. D 18.B 19.A 20.D 21.E 22.E23.C 24.A 25.B。
2014年美国“数学大联盟杯赛”(中国赛区)初赛七年级(初一)详解
(七年级) 一、 选择题 1. B. No even number can be written as the product of two odd integers. Since 11 is the product of 1 and 11, Skip may have run 11 kilometers. A) 10 B) 11 C) 12 D) 14
27. D. As shown, only choice D is not a product of a divisor of 24 and a divisor of 35.
A) 1 × 1 B) 6 × 7 C) 8 × 7 D) 6 × 11 28. B. The sum of the dimensions is 40 ÷2 = 20. Its dimensions are 16 × 4, and its area is 64. A) 100 B) 64 C) 40 D) 24 29. A. Her average score for 6 tests was 82. So her total was 6 × 82 = 492. Adding 2 × 98, her total for 8 tests was 688. Her average score was 86. A) 86 B) 88 C) 90 D) 94 30. D. From 1 to 99 there are 9; in every 100 #s after there are 19. Include 1000. A) 162 B) 171 C) 180 D) 181 二、填空题 31. 8. 32. 2257. 33. 21. 34. 37. 35. 120. 36. 7410. 37. 2. 38. 81. 39. 32451. 40. 671.
AMC10 2014最新版 (完整版)
6 Suppose that a cows give b gallons of milk in c days. At this rate, how many gallons of milk
will d cows give in e days?
(A) 24 (B) 36 (C) 44 (D) 48 (E) 52
4 Walking down Jane Street, Ralph passed four houses in a row, each painted a different color. He passed the orange house before the red house, and he passed the blue house before the yellow house. The blue house was not next to the yellow house. How many orderings of the colored houses are possible?
(A) Tuesday (B) Wednesday (C) Thursday (D) Friday (E) Saturday
3 Bridget bakes 48 loaves of bread for her bakery. She sells half of them in the morning for $2.50 each. In the afternoon she sells two thirds of what she has left, and because they are not fresh, she charges only half price. In the late afternoon she sells the remaining loaves at a dollar each. Each loaf costs $0.75 for her to make. In dollars, what is her profit for the day?
2014AMC10A试题及简答(繁体中文)
(D) 12
(E) 8
22. 在長方形 ABCD 中， AB 20 、 BC 10 。若點 E 在 CD 上使得 CBE 15 ，則 AE 長度
是多少？
(A)
20 3 3
(B) 10 3
(C) 18
(D) 11 3
(E) 20
23. 一張長方形紙片的長是寬的 3 倍，其面積為 A。將兩個長邊三等分，如圖所示；圖中的
19. 四個邊長分別為 1、2、3、4 的正立方體堆疊如圖所示。試問 XY 包含在那個邊長為 3 的
正立方體的截線段之長度為何？
(A)
3 33 5
(B) 2 3
(C)
2 33 3 X 1
(D) 4
(E) 3 2
2 3 4
Y
20. 考慮 8 8888 兩項的乘積，其中第二項是 k 位數，若乘積為一整數，其各位數字的和
16. 在長方形 ABCD 中， AB 1 、 BC 2 ，點 E、F、G 分別為 BC 、 CD 、 AD 的中點，點 H
為 GE 的中點。試問圖中陰影區域的面積為多少？
(A)
1 12
(B)
3 18
(C)
2 12
(D)
3 12
(E)
1 6
A
B
1
G
H
E
1
D 1 F 1 C 2 2
17. 投擲三顆公正的骰子，試問出現在其中兩顆骰子上數字的和等於剩下那顆骰子出現的數 字之機率為多少？ 1 13 7 5 2 (B) (C) (D) (E) 6 72 36 24 9 18. 若坐標平面上的某個正方形，它各頂點的 y 坐標分別為 0、1、4、5，則此正方形的面積 為多少？ (A) (A) 16 (B) 17 (C) 25 (D) 26 (E) 27
-年amc 10a和b竞赛真题及答案(英文版)
2010-2015年AMC 10A和B竞赛真题及答案（英文版）
你好，目前，只分享了2010-2013年AMC 10A和B竞赛真题及答案（英文版），2014-2015年的暂时还没有上传，等采纳后再私信我吧！需要的话，我可以将Word文档中的原图文件一同发给你，求采纳，毕竟我花了差不多一个下午才整理完，谢谢！(@_@)
2010年AMC 10Aห้องสมุดไป่ตู้赛真题及答案（英文版）
2010年AMC 10B竞赛真题及答案（英文版）
2011年AMC 10A竞赛真题及答案（英文版）
2011年AMC 10B竞赛真题及答案（英文版）
Problem 4
2012年AMC 10A竞赛真题及答案（英文版）
go去wentgone
2012年AMC 10B竞赛真题及答案（英文版）
get得到gotgot
sink下沉sank / sunksunk / sunken
give给gavegiven
hit打hithit
不规则动词表
2013年AMC 10A竞赛真题及答案（英文版）
mistake误认mistookmistaken
grow成长grewgrown
think思考thoughtthought
AMC10美国数学竞赛讲义全
AMC 中的数论问题1：Remember the prime between 1 to 100：2 3 5 7 11 13 17 19 23 29 31 37 41 43 47 53 59 61 67 71 73 79 83 89 91 2：Perfect number:Let is the prime number.if 21p - is also the prime number. then 1(21)2pp --is the perfectnumber. For example:6，28，496。
3: Let ,0n abc a =≠ is three digital integer 。
if 333n a b c =++Then the number n is called Daffodils number. There are only four numbers ： 153 370 371 407Let ,0n abcd a =≠ is four digital integer 。
if 4444d n a b c +=++ Then the number n is called Roses number. There are only three numbers: 1634 8208 94744:The Fundamental Theorem of ArithmeticEvery natural number can be written as a product of primes uniquely up to order.5：Suppose that a and b are integers with b =0. Then there exists unique integers q and r such that 0 ≤ r< ｜b ｜ and a = bq + r.6:（1)Greatest Common Divisor: Let gcd (a, b) = max {d ∈ Z： d | a and d | b}。
AMC10美国数学竞赛讲义全
AMC 中的数论问题1：Remember the prime between 1 to 100：2 3 5 7 11 13 17 19 23 29 31 37 41 43 47 53 59 61 67 71 73 79 83 89 91 2：Perfect number:Let is the prime number.if 21p - is also the prime number. then 1(21)2pp --is the perfectnumber. For example:6，28，496。
3: Let ,0n abc a =≠ is three digital integer 。
if 333n a b c =++Then the number n is called Daffodils number. There are only four numbers ： 153 370 371 407Let ,0n abcd a =≠ is four digital integer 。
if 4444d n a b c +=++ Then the number n is called Roses number. There are only three numbers: 1634 8208 94744:The Fundamental Theorem of ArithmeticEvery natural number can be written as a product of primes uniquely up to order.5：Suppose that a and b are integers with b =0. Then there exists unique integers q and r such that 0 ≤ r< ｜b ｜ and a = bq + r.6:（1)Greatest Common Divisor: Let gcd (a, b) = max {d ∈ Z： d | a and d | b}。
2014-2015年度美国”数学大联盟杯赛“(中国赛区)初赛-(十、十一、十二年级).doc
2014-2015年度美国”数学大联盟杯赛“（中国赛区）初赛（十、十一、十二年级）一、选择题（每小题10分，答对加10分，答错不扣分，共100分，请将正确答案A 、B 、C 或者D 写在每题后面的圆括号内。
）正确答案填写示例如下：=-⨯⨯20522 ？ （A ）A)5 B)15 C)25 D)301. Meg loves her megaphone! The large circular end has a circumference that is the reciprocal of its diameter. What is the area of the circle? ( )A)π14 B) π12 C) 14 D) 122. How many solutions does the equation x x +=233 have? ( )A)0 B)1 C)2 D)43. If y x =-1, which of the following is always true for any value of x ? ( )A) ()()x y -=-2211B) ()()x x y y -=-222211 C) ()()x x y y --=-222211 D) ()()()()x x y y -+=-+22221111 4. Lee the crow ate a grams of feed that was 1% seed, b grams of feed that was 2% seed, and c grams of feed that was 3% seed. If combined, all the feed he ate was 1.5% seed. What is a in terms of b and c ?( )A)b c +3B)b c +3 C)b c +23 D)b c +32 5. If <x 0 and <.x 2001, then x -1 must be ( )A)less than -10B)between-0.1 and 0 C)between 0and 0.1 D) greater than 106. At 9:00 A.M., the ratio of red to black cars in a parking lot was 1 to 5. An hour later the number of red cars had increased by 2, the number of black cars had decreased by 5, and the ratio of red to black cars was 1 to 4. How many black cars were in the lot at 10:00 A.M.? ( )A)13 B)15 C)60 D)657. If x ≠1and x ≠-1, then ()()()x x x x x --++-32241111=( ) A)x -21 B) x +21 C) x -241 D) x -341 8. The Camps are driving at a constant rate. At noon they had driven 300 km.At 3:30 P.M. they had driven 50% further than they had driven by 1:30 P.M.What is their constant rate in km/hr? ( )A)150 B)120 C)100 D)909. The letters in DIGITS can be arranged in how many orders without adjacent I ’s? ( )A)240 B)355 C)600 D)71510. Al, Bea, and Cal each paint at constant rates, and together they are painting a house. Al and Bea togethercould do the job in 12 hours; Al and Cal could to it in 15, and Bea and Cal could do it in 20. How many hours will it take all three working together to paint the house?( )A)8.5 B)9 C)10 D)10.5二、填空题（每小题10分，答对加10分，答错不扣分，共200分）11. What is the sum of the degree-measures of the angles at the outer points ,,,A B C D and E of a five-pointed star, as shown? Answer: . 12. What is the ordered pair of positive integers (,k b ), with the least value of k , which satisfiesk b ⋅⋅=34234?Answer: .13. A face-down stack of 8 playing cards consisted of 4 Aces (A ’s) and 4 Kings (K ’s).After I revealed and then removed the top card, I moved the new top card to thebottom of the stack without revealing the card. I repeated this procedure until thestack without revealing the card. I repeated this procedure until the stack was leftwith only 1 card, which I then revealed. The cards revealed were AKAKAKAK ,in that order. If my original stack of 8 cards had simply been revealed one card at atime, from top to bottom (without ever moving cards to the bottom of the stack),in what order would they have been revealed?Answer: .14. For what value of a is one root of ()x a x a -+++=222120 twice the other root?Answer: .15. Each time I withdrew $32 from my magical bank account, the account ’sremaining balance doubled. No other account activity was permitted. My fifth$32 withdrawal caused my account ’s balance to become $0. With how manydollars did I open that account?Answer: .16. In how many ways can I select six of the first 20 positive integers, disregarding the order in which these sixintegers are selected, so that no two of the selected integers are consecutive integers?Answer: .17. If, for all real ,()()xx f x f x =-21, what is the numerical value of f (3)?Answer: .18. How many pairs of positive integers (without regard to order) have a least common multiple of 540?Answer: .19. If the square of the smaller of consecutive positive integers is x , what is the square of the larger of thesetwo integers, in terms of x ?Answer: .20. A pair of salt and pepper shakers comes in two types: identical and fraternal.Identical pairs are always the same color. Fraternal pairs are the same colorhalf the time. The probability that a pair of shakers is fraternal is p andthat a pair is identical is .q p =-1 If a pair of shakers is of the same color, AE DCBword 格式-可编辑-感谢下载支持 determine, in terms of the variable q alone, the probability that the pair is identical. Answer: .21. As shown, one angle of a triangle is divided into four smaller congruentangles. If the lengths of the sides of this triangle are 84, 98, and 112, as shown,how long is the segment marked x ?Answer: .22. How long is the longer diagonal of a rhombus whose perimeter is 60, if threeof its vertices lie on a circle whose diameter is 25, as shown?Answer: .23. The 14 cabins of the Titanic Mail Boat are numbered consecutively from1 through 14, as are the 14 room keys. In how many different ways canthe 14 room keys be placed in the 14 rooms, 1 per room, so that, for everyroom, the sum of that room ’s number and the number of the key placed inthat room is a multiple of 3?Answer: .24. For some constant b , if the minimum value of ()x x b f x x x b -+=++2222is 12, what is the maximum value of ()f x ? Answer: .25. If the lengths of two sides of a triangle are 60cos A and 25sin A , what is the greatest possible integer-length of the third side?Answer: .26. {}n a is a geometric sequence in which each term is a positive number. If a a =5627, what is the value oflog log log ?a a a +++3132310Answer: . 27. What is the greatest possible value of ()=sin cos ?f x x x ++3412Answer: .28. Let C be a cube. Triangle T is formed by connecting the midpoints of three edges of cube C . What is the greatest possible measure of an angle of triangle T ?Answer: .29. Let a and b be two real numbers. ()sin f x a x b x =++34 and (lg log )f =3105. What is the value of (lg lg )f 3?Answer: .30. Mike likes to gamble. He always bets all his chips whenever the number of chips he has is <=5. He always bets n （10-）chips whenever the number of chips he has is greater than 5 and less than 10. He continues betting until either he has no chips or he has more than 9 chips. For every round, if he bets n chips. The probability that he wins or loses in each round is 50%. If Mike begins with 4 chips, what is the probability that he loses all his chips?Answer: .1129884xword格式-可编辑-感谢下载支持。
(参考资料)2014年美国“数学大联盟杯赛”(中国赛区)初赛五、六年级试卷
A) 100
B) 160
C) 200
D) 250
二、填空题(每小题 5 分，答对加 5 分，答错不扣分，共 50 分，答案请填涂在答题卡上)
31. The sum of the digits of 2014 is 2 + 0 + 1 + 4 = 7. Let n be a natural number.
m = n + 2014. The sum of the digits of m is half the sum of the digits of n.
What is the minimum value of n?
Answer: ______.
32. The sum of 5 different prime numbers is 200. Each of the 5 prime
1
2
26. If 5 of the 200 stripes on Frank’s giant shell are blue, 5 of the remaining
stripes are brown, and the rest are white, there are ? more white stripes
D) 110
22. The average of 2014 sixes is equal to the average of 4028 ? .
A) threes
B) sixes
C) nines
D) twelves
23. What is 0.625% of 8% of 500?
A) 0.25
B) 2.5
数字为三个连续的偶数，个位数字为三个连续的奇数。如果将四位数的
amc10 立体几何题目
amc10 立体几何题目AMC10立体几何题目立体几何是数学中的重要分支，它考察的是在三维空间中的物体的形状、大小和位置关系。
AMC10是美国数学竞赛中的一项考试，其中也包括了对立体几何的考察。
本文将给出一些AMC10中常见的立体几何题目，并进行解析。
题目一：已知正方体ABCDA1B1C1D1的棱长为1，点P在AC1边上，使得BP与平面A1B1C1垂直且长度为1/3，求AP的长度。
解析：首先，我们将正方体ABCDA1B1C1D1绘制出来，然后标记出点P 和线段BP。
由题目中的条件可知，BP与平面A1B1C1垂直且长度为1/3，那么BP的长度就为1/3，由此我们可以得到三角形ABP的边长比例。
三角形ABP与三角形ABC是相似的，所以我们可以得到以下比例关系：AP/AB = BP/BC。
由于AB的长度为1，BC的长度为√2，代入已知条件，可以得到AP的长度为1/√2。
答案为1/√2。
题目二：如图，四个球共面且两两相切，半径依次为3，2，1和r，若r为正实数，则球心四点共面且所在平面与第一个球的切点到第四个球心的直线垂直，求r的值。
解析：首先，我们可以通过观察来推测四点球心共面并且所在平面与第一个球的切点到第四个球心的直线垂直。
通过连接四个球心，我们可以得到一个四面体，由题目中的条件，四面体的底面是一个等腰三角形，而球心则位于其高的延长线上。
由此，我们可以知道该四面体是一个正四面体，也就是四个球心共面。
接下来，我们要确定r的值。
通过观察，我们可以发现正四面体的侧面中心与四个球心连线构成一个等腰三角形，这个三角形的底边长为3+1=4，高为r-1。
根据等腰三角形的性质，底边长的平方等于两腰的乘积的二倍。
那么，我们可以得到方程4^2 = 2(3^2 + (r-1)^2)，解方程可得r=3/2。
因此，r的值为3/2。
题目三：如图，正方体ABCDA1B1C1D1的棱长为2，X为AC边上的一点，X1为对点X关于平面A1B1B的镜像点，连结XX1，若交于棱AA1的点为P，则AP : XA = t^2 - at + b，其中a，b皆为正整数，t为正整数大于等于2。
AMC美国数学竞赛AMC B 试题及答案解析
2003A M C10 B 1、Which of the following is the same as2、Al gets the disease algebritis and must take one green pill and one pink pill each day for two weeks. A green pill costs more than a pink pill, and Al’s pills cost a total of for the two weeks. How much does one green pill cost?3、The sum of 5 consecutive even integers is less than the sum of the ?rst consecutive odd counting numbers. What is the smallest of the even integers?4、Rose fills each of the rectangular regions of her rectangular flower bed with a different type of flower. The lengths, in feet, of the rectangular regions in her flower bed are as shown in the ?gure. She plants one flower per square foot in each region. Asters cost 1 each, begonias each, cannas 2 each, dahlias each, and Easter lilies 3 each. What is the least possible cost, in dollars, for her garden?5、Moe uses a mower to cut his rectangular -foot by -foot lawn. The swath he cuts is inches wide, but he overlaps each cut by inches tomake sure that no grass is missed. He walks at the rate of feet per hour while pushing the mower. Which of the following is closest to the number of hours it will take Moe to mow his lawn?.6、Many television screens are rectangles that are measured by the length of their diagonals. The ratio of the horizontal length to the height in a standard television screen is . The horizontal length of a “-inch” television screen is closest, in inches, to which of the following?7、The symbolism denotes the largest integer not exceeding . For example. , and . Compute.8、The second and fourth terms of a geometric sequence are and . Which of the following is a possible first term?9、Find the value of that satisfies the equation10、Nebraska, the home of the AMC, changed its license plate scheme. Each old license plate consisted of a letter followed by four digits. Each new license plate consists of three letters followed by three digits. By how many times is the number of possible license plates increased?11、A line with slope intersects a line with slope at the point . What is the distance between the -intercepts of these two lines?12、Al, Betty, and Clare split among them to be invested in different ways. Each begins with a different amount. At the end of one year they have a total of . Betty and Clare have both doubled their money, whereas Al has managed to lose . What was Al’s origin al portion?.13、Let denote the sum of the digits of the positive integer . For example, and . For how many two-digit values of is ?14、Given that , where both and are positive integers, find the smallest possible value for .15、There are players in a singles tennis tournament. The tournament is single elimination, meaning that a player who loses a match is eliminated. In the first round, the strongest players are given a bye, and the remaining players are paired off to play. After each round, the remaining players play in the next round. The match continues until only one player remains unbeaten. The total number of matches played is16、A restaurant offers three desserts, and exactly twice as many appetizers as main courses. A dinner consists of an appetizer, a main course, and a dessert. What is the least number of main courses that the restaurant should offer so that a customer could have a different dinner each night in the year ?.17、An ice cream cone consists of a sphere of vanilla ice cream and a right circular cone that has the same diameter as the sphere. If the ice cream melts, it will exactly ?ll the cone. Assume that the melted ice cream occupies of the volume of the frozen ice cream. What is the ratio of the cone’s height to its radius?18、What is the largest integer that is a divisor offor all positive even integers ?19、Three semicircles of radius are constructed on diameter of a semicircle of radius . The centers of the small semicircles divide into four line segments of equal length, as shown. What is the area of the shaded region that lies within the large semicircle but outside the smaller semicircles?20、In rectangle , and . Points and are on so that and . Lines and intersect at . Find the area of.21、A bag contains two red beads and two green beads. You reach into the bag and pull out a bead, replacing it with a red bead regardless of the color you pulled out. What is the probability that all beads in the bag are red after three such replacements?22、A clock chimes once at minutes past each hour and chimes on the hour according to the hour. For example, at 1 PM there is one chime and at noon and midnight there are twelve chimes. Starting at 11:15 AM on February , , on what date will the chime occur?23、A regular octagon has an area of one square unit. What is the area of the rectangle ?24、The ?rst four terms in an arithmetic sequence are , , , and, in that order. What is the ?fth term?25、How many distinct four-digit numbers are divisible by and have as their last two digits?。
amc10数学竞赛题目
amc10数学竞赛题目
以下是一道AMC 10数学竞赛题目的例子：
问题：在平面上，有一个边长为10的正方形ABCD，点E是AB
的中点，点F是BC的中点。
连接线段AF和BE，交于点P。
求AP的长度。
A) 2
B) 4
C) 5
D) 6
E) 8
解析：
首先，我们可以观察到三角形APE与三角形FPC共有一个角A，因此它们是相似三角形。
根据相似三角形的性质，我们可以得出以下比例关系：AP/FP = AE/FC = 1/2。
另一方面，我们可以利用勾股定理来计算AE和FC的长度。
由于AE是AB的中点，而AB的长为10，所以AE的长度为10/2 = 5。
同样地，FC是BC的中点，而BC的长也是10，所以FC的长度也是10/2 = 5。
将上面的结果代入比例关系中，我们得到AP/FP = 5/5 = 1。
因此，AP和FP的长度是相等的。
最后，我们可以发现三角形APF是一个等边三角形，因此AP的长度等于FP的长度，即AP = FP = 5。
所以，答案是C) 5。