2014美国数学竞赛AMC10A、10B试题及答案

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 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。

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}。

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）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。