2014美国数学竞赛AMC10A10B试题

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。

2000 AMC 10 ProblemsProblem 1In the year 2001, the United States will host the International Mathematical Olympiad. Let , , and be distinct positive integers such that the product 2001=••O M I . What is the largest possible value of the sum O M I ++ ?（A ）23 （B ）55 （C ）99 （D ）111 （E ）671Problem 22000 （20002000）＝（A ）20012000 （B ）20004000 （C ）40002000 （D ）2000000,000,4 （E ）000,000,42000Problem 3Each day, Jenny ate 20% of the jellybeans that were in her jar at the beginning of that day. At the end of the second day, 32 remained. How many jellybeans were in the jar originally?（A ）40 （B ）50 （C ）55 （D ）60 （E ）75Problem 4Chandra pays an on-line service provider a fixed monthly fee plus an hourly charge for connect time. Her December bill was , but in January her bill was because she used twice as much connect time as in December. What is the fixed monthly fee?（A ）2.53 （B ）5.06 （C ）6.24 （D ）7.42 （E ）8.77Problem 5Points M and N are the midpoints of sides PA and PB of △PAB. As P moves along a line that is parallel to side AB, how many of the four quantities listed below change? (a) the length of the segment MN (b) the perimeter of △PAB(c) the area of △PAB (d) the area of trapezoid ABNM（A ）0 （B ）1 （C ）2 （D ）3 （E ）4Problem 6The Fibonacci sequence 1，1，2，3，5，8，13，21，…… starts with two s, and each term afterwards is the sum of its two predecessors. Which one of the ten digits is the last to appear in the units position of a number in the Fibonacci sequence?（A ）0 （B ）4 （C ）6 （D ）7 （E ）9Problem 7 In rectangle , , is on , and and trisect .What is the perimeter of ?（A ）333+ （B ）3342+ （C ）222+ （D ）2533+ （E ）3352+ Problem 8At Olympic High School, 52 of the freshmen and 54 of the sophomores took the AMC-10. Given that the number of freshmen and sophomore contestants was the same, which of the following must be true?（A ）There are five times as many sophomores as freshmen.（B ）There are twice as many sophomores as freshmen.（C ）There are as many freshmen as sophomores.（D ）There are twice as many freshmen as sophomores.（E ）There are five times as many freshmen as sophomores.Problem 9If , where , then（A ）－2 （B ）2 （C ）2－2p （D ）2p-2 （E ）｜2p-2｜Problem 10The sides of a triangle with positive area have lengths , , and . The sides of a second triangle with positive area have lengths , , and . What is the smallest positive number that is not a possible value of ｜x-y ｜?（A ）2 （B ）4 （C ）6 （D ）8 （E ）10Problem 11Two different prime numbers between 4 and 18 are chosen. When their sum is subtracted from their product, which of the following numbers could be obtained?（A ）21 （B ）60 （C ）119 （D ）180 （E ）231Problem 12Figures 0, 1, 2 and 3 consist of 1, 5, 13, and 25 nonoverlapping unit squares, respectively. If the pattern were continued, how many nonoverlapping unit squares would there be infigure 100?Figure 0 Figure 1 Figure 2 Figure 3（A ）10401 （B ）19801 （C ）20201 （D ）39801 （E ）40801 Problem 13There are 5 yellow pegs, 4 red pegs, 3 green pegs, 2 blue pegs, and 1 orange peg to be placed on a triangular peg board. In how many ways can the pegs be placed so that no (horizontal) row or (vertical) column contains two pegs of the same color?Problem 14Mrs. Walter gave an exam in a mathematics class of five students. She entered the scores in random order into a spreadsheet, which recalculated the class average after each score was entered. Mrs. Walter noticed that after each score was entered, the average was always an integer. The scores (listed in ascending order) were 71, 76, 80, 82, and 91. What was the last score Mrs. Walter entered?（A ）71 （B ）76 （C ）80 （D ）82 （E ）91Problem 15Two non-zero real numbers, and , satisfy. Find a possible value of ab ab b a -+ . （A ）-2 （B ）-21 （C ）31 （D ）21 （E ）2 Problem 16The diagram shows 28 lattice points, each one unit from its nearest neighbors. Segment AB meets segment CD at E. Find the length of segment AE.（A ）354 （B ）355 （C ）7512 （D ）52 （E ）9565Boris has an incredible coin changing machine. When he puts in a quarter, it returns five nickels; when he puts in a nickel, it returns five pennies; and when he puts in a penny, it returns five quarters. Boris starts with just one penny. Which of the following amounts could Boris have after using the machine repeatedly?（A ）<dollar/>3.63 （B ）<dollar/>5.13 （C ）<dollar/>6.30 （D ）<dollar/>7.45 （E ）<dollar/>9.07Problem 18Charlyn walks completely around the boundary of a square whose sides are each 5 km long. From any point on her path she can see exactly km horizontally in all directions. What is the area of the region consisting of all points Charlyn can see during her walk, expressed in square kilometers and rounded to the nearest whole number?（A ）24 （B ）27 （C ）39 （D ）40 （E ）42Problem 19Through a point on the hypotenuse of a right triangle, lines are drawn parallel to the legs of the triangle so that the triangle is divided into a square and two smaller right triangles. The area of one of the two small right triangles is times the area of the square. The ratio of the area of the other small right triangle to the area of the square is（A ）121 m （B ）m （C ）1-m （D ）m 41 （E ）281mProblem 20Let A, M, and C be nonnegative integers such that A+M+C=10. What is the maximum value of A ·M ·C + A ·M + M ·C +C ·A?（A ）49 （B ）59 （C ）69 （D ）79 （E ）89Problem 21If all alligators are ferocious creatures and some creepy crawlers are alligators, which statement(s) must be true?I. All alligators are creepy crawlers.II. Some ferocious creatures are creepy crawlers.III. Some alligators are not creepy crawlers.（A ）I only （B ）II only （C ）III only （D ）II and III only （E ）None must be true Problem 22One morning each member of Angela's family drank an 8-ounce mixture of coffee with milk. The amounts of coffee and milk varied from cup to cup, but were never zero. Angela drank a quarter of the total amount of milk and a sixth of the total amount of coffee. How many people are in the family?（A ）3 （B ）4 （C ）5 （D ）6 （E ）7When the mean, median, and mode of the list 10, 2, 5, 2, 4, 2, x are arranged in increasing order, they form a non-constant arithmetic progression. What is the sum of all possible real values of ? （A ）3 （B ）6 （C ）9 （D ）17 （E ）20Problem 24Let be a function for which. Find the sum of all values of for which. （A ）-31 （B ）-91 （C ）0 （D ）95 （E ）35 Problem 25In year , the day of the year is a Tuesday. In year, the day is also a Tuesday. On what day of the week did the day of year occur?2000 AMC 10 SolutionProblem 1 The following problem is from both the 2000 AMC 12 #1 and 2000 AMC 10 #1, so both problems redirect to this page.The sum is the highest if two factors are the lowest.So, 1·3·776=2001 and 1+3+667=671 (E)Problem 2 The following problem is from both the 2000 AMC 12 #2 and 2000 AMC 10 #2, so both problems redirect to this page.2000 （20002000）＝12000（20002000）=20012000 (A)Problem 3 The following problem is from both the 2000 AMC 12 #3 and 2000 AMC 10 #3, so both problems redirect to this pageSince Jenny eats 20% of her jelly beans per day, 80%=4/5 of her jelly beans remain after one day. Let be the number of jelly beans in the jar originally.325454=••x x=50 (B) Problem 4Let be the fixed fee, and be the amount she pays for the minutes she used in the first month.x+y=12.48 x+2y=17.54 y=5.06 x=7.42 We want the fixed fee, which is (D) Problem 5(a) Clearly AB does not change, and MN=0.5AB, so MN doesn't change either. (b) Obviously, the perimeter changes.(c) The area clearly doesn't change, as both the base AB and its corresponding height remain the same.(d) The bases AB and MN do not change, and neither does the height, so the area of the trapezoid remains the same.Only quantity changes, so the correct answer is .Problem 6 The following problem is from both the 2000 AMC 12 #4 and 2000 AMC 10 #6, so both problems redirect to this page.Note that any digits other than the units digit will not affect the answer. So to make computation quicker, we can just look at the Fibonacci sequence in :The last digit to appear in the units position of a number in the Fibonacci sequence is 6 (C).Problem 7AD=1 Since ∠ADC is trisected, ∠ADP=∠PDB=∠BDC=30º Thus, PD=332 BD=2 BP=332333=- Adding, 3342+Problem 8Let be the number of freshman and be the number of sophomores.s f 5452= f=2s There are twice as many freshmen as sophomores. Problem 9 The following problem is from both the 2000 AMC 12 #5 and 2000 AMC 10 #9, so both problems redirect to this page.When x<2, x-2 is negative so ∣x -2∣=2-x =p and x=2-pThus x-p=(2-p)-p=2-2p (C)Problem10From the triangle inequality, 2<x <10 and 2<y <10. The smallest positive number not possible is 10-2, which is . (D)Problem 11 The following problem is from both the 2000 AMC 12 #6 and 2000 AMC 10 #11, so both problems redirect to this page.All prime numbers between 4 and 18 have an odd product and an even sum. Any odd number minus an even number is an odd number, so we can eliminate B andD. Since the highest two prime numbers we can pick are 13 and 17, the highest number we can make is (13) (17) – (13+17) = 221-30=191. Thus, we can eliminateE. Similarly, the two lowest prime numbers we can pick are 5 and 7, so the lowest number we can make is (5) (7) – (5+7) =23. Therefore, A cannot be an answer. So, the answer must be .Problem 12 The following problem is from both the 2000 AMC 12 #8 and 2000 AMC 10 #12, so both problems redirect to this page.Solution 1We have a recursion: .I.E. we add increasing multiples of each time we go up a figure. So, to go from Figure 0 to 100, we addWe then add to the number of squares in Figure 0 to get, which ischoice Solution 2We can divide up figure to get the sum of the sum of the first odd numbers and the sum of the first odd numbers. If you do not see this, here is the example for :The sum of the first odd numbers is 2n , so for figure , thereare 22)1(n n ++ unit squares. We plug in n=100 to get 20201, which is choice Solution 3Using the recursion from solution 1, we see that the first differences of 4,8,12, … form an arithmetic progression, and consequently that the second differences are constant and all equal to . Thus, the original sequence can be generated from a quadratic function.If c bn an n f ++=2)(, and f(0)=1, f(1)=5, and f(2)=13, we get a system of three equations in three variables:f(0)=1 gives c=1; f(1)=5 gives a+b+c=5; f(2)=13 gives 4a+2b+c=13 Plugging in into the last two equations gives a+b=4 4a+2b=12 Dividing the second equation by 2 gives the system: a+b=4 2a+b=6Subtracting the first equation from the second gives , and hence . Thus, our quadratic function is: 122)(2++=n n n fCalculating the answer to our problem, f(100)=20201Problem 13In each column there must be one yellow peg. In particular, in the rightmost column, there is only one peg spot, therefore a yellow peg must go there.In the second column from the right, there are two spaces for pegs. One of them is in the same row as the corner peg, so there is only one remaining choice left forthe yellow peg in this column.By similar logic, we can fill in the yellow pegs as shown:After this we can proceed to fill in the whole pegboard, so there is only arrangement of the pegs. The answer is (B)Problem 14 The following problem is from both the 2000 AMC 12 #9 and 2000 AMC 10 #14, so both problems redirect to this page.Solution 1The first number is divisible by 1.The sum of the first two numbers is even.The sum of the first three numbers is divisible by 3.The sum of the first four numbers is divisible by 4.The sum of the first five numbers is 400.Since 400 is divisible by 4, the last score must also be divisible by 4. Therefore, the last score is either 76 or 80.Case 1: 76 is the last number entered.Since 400≡76≡1 (mod 3), the fourth number must be divisible by 3, but none of the scores are divisible by 3. Case 2: 80 is the last number entered. Since 80≡2 (mod 3), the fourth number must be 2 (mod 3). Thatnumber is 71 and only 71. The next number must be 91, since the sum of the first two numbers is even.So the only arrangement of the scores 76, 82, 91,71,80Solution 2We know the first sum of the first three numbers must be divisible by 3, so we write out all 5 numbers (mod 3), which gives 2,1,2,1,1, respectively. Clearly the only way to get a number divisible by 3 by adding three of these is by adding the three ones. So those must go first. Now we have an odd sum, and since the next average must be divisible by 4, 71 must be next. That leaves 80 for last, so the answer is .Problem 15 The following problem is from both the 2000 AMC 12 #11 and 2000 AMC 10 #15, so both problems redirect to this page.22)()()(22222=-=----+=--+=-+ba ab b a b a b a b a b a ab b a ab a b b a (E)Alternatively, we could test simple values, like )21,1(),(=b a , which would yield 2=-+ab ab b a Another way is to solve the equation for giving 1+=a a b , then substituting this into the expression and simplifying gives the answer ofProblem 16Solution 1Let be the line containing A and B and let be the line containing C and D. If we set the bottom left point at (0,0), then A=(0,3), B=(6,0), C=(4,2), and D=(2,0) . The line is given by the equation 11b x m y +=. The -intercept is A=(0,3), so 1b =3. We are given two points on , hence we can compute the slope, to be 210630-=-- , so is the line 321+-=x y Similarly, is given by 22b x m y +=. The slope in this case is 12402=--, so 2b x y +=. Plugging in the point (2,0) gives us 2b =-2, so is the line 2-=x y . At E, the intersection point, both of the equations must be true,2-=x y , 321+-=x y so 3212+-=-x x SO 310=x 34=y We have the coordinates of and , so we can use the distance formula here: 355)334()0310(22=-+- which is answer choiceSolution 2Draw the perpendiculars from andto , respectively. As it turns out, . Let be the point onfor which . , and, so by AA similarity,By the Pythagorean Theorem, wehave, ,。

2010-2014MCMProblems建模竞赛美赛题目重点2010 MCM ProblemsPROBLEM A: The Sweet SpotExplain the “sweet spot” on a baseball bat.Every hitter knows that there is a spot on the fat part of a baseball bat where maximum power is transferred to the ball when hit. Why isn’t this spot at the end of t he bat? A simple explanation based on torque might seem to identify the end of the bat as the sweet spot, but this is known to be empirically incorrect. Develop a model that helps explain this empirical finding.Some players believe that “corking” a bat (h ollowing out a cylinder in the head of the bat and filling it with cork or rubber, then replacing a wood cap enhances the “sweet spot” effect. Augment your model to confirm or deny this effect. Does this explain why Major League Baseball prohibits “corking”?Does the material out of which the bat is constructed matter? That is, does this model predict different behavior for wood (usually ash or metal (usually aluminum bats? Is this why Major League Baseball prohibits metal bats?MCM 2010 A题：解释棒球棒上的“最佳击球点”每一个棒球手都知道在棒球棒比较粗的部分有一个击球点，这里可以把打击球的力量最大程度地转移到球上。

AMC美国数学竞赛AMCB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 losesa 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?。