2016AMC12B精彩试题及解答

1、2016 AMC 12BProblem 12犷】+丐1 a =What is the value ofawhe n2?(A) 1(B) 2(C) |(D) 10(E) 20Soluti onBy: Drago nflyIWe find that is the same as 丄,since a number to the power of . is just the reciprocal of that nu mber. We the n get the equati on to be2x2 + |1(D) 102We can then simplify the equation to get

2、Problem 2The harm onic mean of two nu mbers can be calculated as twice their product divided by their sum. The harm onic mean of 1 an d201G is closest to which in teger?(A) 2(B) 45(C) 504(D) 1008(E) 2015Soluti onBy: drago nflySince the harm onic mean is 2 times their product divided by their sum, we

3、 get the equati on2 x 1 x 20161+2016which is the n40322017which is fin ally closest toProblem 3Let X = 2016. What is the value of lll 一 1尤Soluti onBy: drago nflyFirst of all, lets plug in all of the h's into the equati on.| - 2016| 一 (-2016)1 - | - 20161 -(-2016)Then we simplify to get12016 + 20

4、16| 2016 +2016which simplifies into2016 I+201G(D) 4032and fin ally we get Problem 4The ratio of the measures of two acute angles is 5 : 1, and the complement of one of these two an gles is twice as large as the compleme nt of the other. What is the sum of the degree measures of the two an gles?(A) 7

5、5(B) 90(C) 135(D) 150(E) 270Soluti onBy: drago nflyWe set up equations to find each angle. The larger angle will be represented as and the larger an gle will we represe nted as, in degrees. This implies that 5yand2 x (90 - x) = 90 - ysince the larger the original angle, the smaller the complement.We

6、 then find thatand *'J , and their sum isProblem 5(C) 135The War of 1812 started with a declarati on of war on Thursday, June18,1812.The peace treaty to end the war was sig ned919 days later, onDecember 24, 1814. On what day of the week was the treaty signed?(A) Friday (B) Saturday (C) Sunday (D

7、) Monday(E) TYiesdaySoluti onBy: drago nflyTo find what day of the week it is in 919 days, we have to divide 919 by 7 to the rema in der, and the n add the rema in der to the curre nt day. We get919seethat 7 has a rema in der of 2, so we in crease the curre nt day by-toget(B) SaturdayProblem 6All th

8、ree vertices of AUG lie on the parabola defined by 射=工，with A at the origin and UC parallel to the x-axis. The area of the triangle is 64. What is the length of UC?(A) 4(B) 6(C) 8(D) 10(E) 16Soluti onBy: Albert471,which is isosceles. By setting up the triangle areaG4 =* 2rr * a;2 = G4 = x3formula yo

9、u get:2Making x=4, and the len gth(C)SPlotting points D and O on the graph shows that they are at (一叫 *) and (") of BC is 2h, so the an swer isProblem 7Josh writes the numbers 1,2, 3, 99,100. He marks out 1, skips the next nu mber , marks out 3, and continues skipping and marking out the next n

10、u mber to the end of the list. Then he goes back to the start of his list, marks out the first remai ning nu mber ,skips the n ext nu mber ,marks out 6, skips 8, marks out 10, and so on to the end. Josh con ti nues in this manner un til only one nu mber rema ins. What is that nu mber?(A) 13(B) 32(C)

11、 56(D) 64(E) 96Soluti onBy Albert471Follow ing the patter n, you are cross ing out.Time 1: Every non-multiple of 2Time 2: Every non-m ultiple of 4Time 3: Every non-m ultiple of 8.1(0)64Follow ing this patter n, you are left with every multiple of bl which is on ly Problem 8A thin piece of wood of un

12、iform density in the shape of an equilateral triangle with side length 3 inches weighs 12ounces. A second piece of the same type of wood, with the same thickness, also in the shape of an equilateral triangle, has side length of 5 inches. Which of the following is closest to the weight, in ounces, of

13、 the sec ond piece?(A) 14,0(B) 16.0(C) 20.0(D) 33.3(E) 55.6Solution 1By: drago nflyWe can solve this problem by using similar triangles, since two equilateral triangles are always similar. We can then use100We can the n solve the equati on to getSolution 2； which is closest to(D) 33,3Ano ther approa

14、ch to this problem, very similar to the previous one but perhaps expla ined more thoroughly, is to use proporti ons. First, since the thick ness and density are the same, we can set up a proportion based on the principlethatV, thus dV 二 m.However, since density and thickness are the same and=4 <x

15、 s2 (recognizing thatthe area of an equilateral triangle is 4), we can say that :x .Then, by in creas ing s by a factor of5；,-is in creased by a factor of25",thus25"i" orProblem 9Carl decided to fence in his rectan gular garde n. He bought20 fence posts,placed one on each of the four

16、corn ers, and spaced out the rest eve nly along the edges of the garde n, leav ing exactly 4 yards betwee n n eighbori ng posts. The lon ger side of his garde n, in clud ing the corn ers, has twice as many posts as the shorter side, in clud ing the cor ners. What is t he area, in square yards, of Ca

17、rl garde n?(A) 256(B) 336(C) 384(D) 448(E) 512Soluti onBy Albert471To start, use algebra to determine the number of posts on each side. You have(the long sides count for 2 because there are twice as many) G；!； = 20 + 4 (each corner is double coun ted so you must add 1) Making the shorter end have 1,

18、 and the longer end have 8. (8 一 1)术 4) * (4 一 1) * 4)二 28 来 12 = 336.(B) 336Therefore, the an swer is Problem 10A quadrilateral has vertices 卩(為 b), Q(b,o), 71(a. b), ando),where a and 6 are integers with a > t? > 0. The area of PQRS is 1G. What is a + b?(A) 4(B) 5(C) 6(D) 12(E) 13Solution 1B

19、y dista nee formula we have(a 6)2 + (6 a)2 * 2 * (a + &)2 = 256Slmplifying we get W b)(n + b) = 8. Thus u -b and a b have to be a(A) 4factor of 8. The only way for them to be factors of 8 and rema in in tegers isif !i and 乙 V So the answer isSolution by l_Do nt_Do_MathSolution 22a2 - 21)-J二 Si n

20、eeand ' are integers,飞3 and ： ；I,Soluti on by e_power_pi_times_iBy the Shoelace Theorem, the area of the quadrilateral is so asoa + b = (A) 4Problem 11How many squares whose sides are parallel to the axes and whose vertices have coord in ates that are in tegers lie en tirely within the regi on b

21、oun ded by the line y = T, the line U 一°a nd the line X 5,1?(A) 30(B) 41(C) 45(D) 50(E) 57Soluti onSoluti on by e_power_pi_times_i (Note: diagram is n eeded)If we draw a picture showing the triangle, we see that it would be easier to count the squares vertically and not horiz on tally. The uppe

22、r boundis 16 squares (y = 5l*7T, and the limit for the 龙-value is 5 squares. First wecount the 1*1 squares. In the back row, there are 12 squares withlen gth 1 because / = 4 * 7T gen erates squares from (4,0) to (£4tt) ,and continuing on we have 9, 6, and 3 for x-values for 1,2, and 3 in theequ

23、atio n V 兀.So there are 12 + 9 + 6 + 3 = 30 squares with len gth J. in the figure. For 2*2 squares, each square takes up 2 un left and 2 un up.Squares can also overlap. For 2*2 squares, the back row stretchesfrom )忆&3tt),so there are 8 squares with length 2 in a 2 by 9 box.are 5 squares. Continu

24、ing and add ing up in the end, there areRepeating the process, the next row stretches from - " to - -, so there汽-1 I 匸一 I squares with length in the figure. Squares with length . in the back row start at 0) and end at 2tt), so there are 4 such squares in theback row. As the front row starts at

25、0,0)and ends at (】")thereD) 50are 4+1 = 5 squares with len gth 3. As squares with len gth 4 would not fit in the trian gle, the an swer is 30 +15 + 5 which isProblem 12All the numbers 23,4,5,6,7, 8,9 are written in a 3x3 array of squares, one nu mber in each square, in such a way that if two nu

26、 mbers are con secutive the n they occupy squares that share an edge. The nu mbers in the four cor ners add up to 18. What is the nu mber in the cen ter?(B) 6(C) 7(D) 8(A) 5(B) 6(C) 7(D) 8(E) 9Soluti onSoluti on by Mlux: Draw a 3x3 matrix. Notice that no adjace nt nu mbers could be in the corners si

27、nce two consecutive numbers must share an edge. Now find 4 noncon secutive nu mbers that add up to IS. Trying 1 + 3 + 5+ 9 = 18 works. Place each odd nu mber in the corner in a clockwise order. Then fill in the spaces. There has to be a 2 in betwee n the 1 and 3. There is a 1 betwee n 3 and 5. The f

28、inal grid should similar to this.1,2,38,7 A9S 6.5(°)了 is in the middle.Solution 2If we color the square like a chessboard, si nee the nu mbers altre nate betwee n eve n and odd, and there are five odd nu mbers and four eve n nu mbers, the odd numbers must be in the eorners/eenter, while the eve

29、n numbers must be on the edges. Since the odd nu mbers add up to 25, and the nu mbers in the corners add up to 18, the nu mber in the cen ter must be 25-18=7Problem 13Alice and Bob live 10 miles apart. One day Alice looks due n orth from her house and sees an airpla ne. At the same time Bob looks du

30、e west from his house and sees the same airpla ne. The an gle of elevati on of the airpla ne is 30° from Alice's position and 60° from Bob's position. Which of the following is closest to the airpla ne's altitude, in miles?(A) 3.5(U4(C) 4*5(D) 5(E) 5.5Soluti onLet's set the

31、 altitude = z, distance from Alice to airplane's ground position (point right below airplane)=y and distance from Bob to airplane's ground position二xt辿(&) =三 30 =畔=士From Alice's point of view,:;.v .zsin GO l蛊From Bob's point of view,怙叩 2 -.曲 G0= = V. So, r 二不We know that 工'+

32、话=102Solv ing the equati on (by pluggi ng in x and y), we get z= 730 =about 5.5.So, answer is 匕)55soluti on by sudeep naralaSolution 2Non -trig soluti on by e_power_pi_times_iSet the distanee from Alice's and Bob's position to the point directly below the airplane to be x and V, respectively

33、. From the PythagoreanTheorem, * +=10(). As both are 30 60 90 triangles, the altitude ofx/3the airpla ne can be expressed as 3 or 目Solv ing the/q=Jv_ Qequatio n * + "" = 100, weequation3', we get 心勿.Plugging this into theget11 or : 一： ( cannot ben egative), so the altitude isx/aTw =阿 w

34、hich is closest toE) 5.5Problem 14The sum of an infinite geometric series is a positive number S, and the second term in the series is I. What is the smallest possible value of S?(A)(B) 2(C) /5(D) 3(E) 4Soluti onThe sec ond term in a geometric series is= Q * 广，where r is the com mon ratiofor the ser

35、ies and a is the first term of the series. So we know that * r = 1 and we wish to find the minimum value of the infinite sum of the series. We knowa1a ,we get1that:弘that1 : and substituting in丄11-一卜I r r(l r) rI 一 f.From here, you can either usecalculus or AM-GM.Calculus: Let/(X)= = -then 畑=一佃一护严 *

36、(1- 2訟).since /(°)and X1)areun defi ned1. This means that we on ly n eed to find where the derivativee1 = 0=>x = - r =-equals , meaning_. So _, meaning thatS % 1 _ (1)22 2 <« +AM-GM For 2 positive real numbers 】and ',/I |11 r V 1 r1,1 珂在、(t y 兰Let . and I . Then:-gThis implies th

37、at _'. or "" l. :.Rearranging : 、-、 二一- -. Thus, the smallestvalue is - .Solution 2y01% = a =-A simple approach is to initially recognize that1 r and r . Weknow that丨，si nee the series must con verge. We can start by observ ing ther- > 3greatest an swer choice, 4. Therefore,3, becau

38、se that would make rwhich would make the series exceed 4. In order to minimize both the initial term1and the rest of the series, we can recog nize thatthe an swer is(E) 4-is the opitimal ratio, thusProblem 15All the numbers 4,6, 7 are assigned to the six faces of a cube, onenu mber to each face. For

39、 each of the eight vertices of the cube, a product of three nu mbers is computed, where the three nu mbers are the nu mbers assig ned to the three faces that in clude that vertex. What is the greatest possible value of the sum of these eight products?(A) 312(B) 343(C) G25(D) 729(E) 1680Soluti onFirs

40、t assign each face the letters ® U C. rf, c, f. The sum of the product of the faces is 血 + 血忖 + 侬仏 + atb + fbc + fcd+ fde + /eft. We can factornu mbers 1-1: or(D) 729Problem 16In how many ways can 345 be written as the sum of an increasing sequenee of two or more con secutive positive in tegers

41、?(A) I (B) 3(C) 5(D) 6(E) 7Soluti onWe proceed with this problem by con sideri ng two cases, whe n: 1) There are an odd nu mber of con secutive nu mbers, 2) There are an eve n nu mber of con secutive nu mbers.because the n our sum will just be. We now haveFor the first case, we can cleverly choose t

42、he convenient form of our seque nee to be a 一口 + 1丫匕 + 驛345and a will have a solution when 2fi +1 is an integer, namely when+ 1 is a divisor of 345. We check that 2" + 1 = 3,5,23work, and no more, because 2n + 1 = 1 does not satisfy the requireme nts of two or morecon secutive in tegers, and wh

43、e n2n + 1 equals the n ext biggest factor, 69, theremust be n egative in tegers in the seque nee. Our soluti onsare 口4,115,11G. f67?，71, 16,，3(), 4, 26For the eve n cases, we choose our seque nee to be of the form: 一 (n - 1),+ 1,+ r】so the sum(E)7is.In this case, we find our soluti ons tobe Id 173,

44、55,，60, 30,39We have found all 7 soluti ons and our an swer isSolution 2The sum from a to b where a and b are integers and a > 6 is(a b + l)(a + b)“ 2345= b + l)(a +b)J _223523 = (a-¥+l)(a+bjLet C = a + 1 and J = u + &2-3 *5-23 =If we factor ；:.：i in to all of its factor groups ；.：、二；.:

45、'广 -； !' we will have several ordered pairs 二:where -The number of possible values for c is half the number of factors of G90 which is 冷 2 However, we have one extra neous case of(1,690)because here,wehave the sum of one con secutive nu mber which is not allowed by the questi on.Thus the ans

46、wer is 81 = 7(eTFProblem 178, CA 9, and AH is an respectively, soat QIn AUG shown in the figure, .47? = 7, BC = altitude. Points D and E lie on sides AC a nd AB, that Hd and CE are an gle bisectors, in tersect ing respectively. What is PQ?Soluti onGet the area of the triangle by heron's formula:

47、Jss-a)(s -b)s - c) = y(12)(3)(4)(5)二 12v/5uUse the area to findArea = 12/5 = bli = 2-AH2 =the height AH with known base BC:AH = 3卯 H = y/AB./. f /' i. J ' Apply angle bisector theorem on丄,；/ J '1' and、J ： 1" ', respectively. From now, you cansimply use the an swer choices be

48、cause only choice D has in it and we knowtriangle . ' ' and triangle U .：, wegetthat r '： I the segme nts on it all have in tegral len gths so that 'will remai n there. Howeverratio： AH : AP : PH = 45 : 27 : 18 and AQ : QH = 45 : 35 : 10. wePQ = 3/5 * = get AH : PQ = 45 ： (18 一 10) =

49、15 : 8.佃Problem 18(D)nWhat is the area of the regi on en closed by the graph of the22equatio n 龙 + V创 + if/i?(A) 7T /2(B)?r + 2(C) 7T + 2/2(D) 2?r + 辺 (E)Soluti on+ 22Tom, Dick, and Harry are playing a game. Starting at the same time, each of them flips a fair coin repeatedly un til he gets his firs

50、t head, at which point hestops. What is the probability that all three flip their coins the same number of times?(A) -(B) -(C) -(D) -(E)-Solution 1By: drago nflyWe can solve this problem by listing it as an infinite geometric equation. We get1.Wethat to have the same amount of tosses, they have a 8

51、cha nee of gett ing all heads. Then the next probability is all of them getting tails and then on the sec ond try, they all get heads. The probability of that happe ning isthen get the geometric equationAnd the n we find that 力 equals to(li) zbecause of the formula of the sum fori 181.=-*= .877an in

52、fin ite series,sSolution 2Call it a "win "if the boys all flip their coins the same nu mber of times, and the1 probability that they win is P. The probability that they win on their first flip is 8 . If they don't win on their first flip, that mea ns they all flipped tails (which also1

53、happe ns with probability 8) and that their cha nces of winning have retur ned to what they were at the beg inning. This covers all possible seque nces of winning flips. So we haveSolving for P gives II.Problem 20A set of teams held a roun d-rob in tour name nt in which every team played every other

54、 team exactly once. Every team won 丨1 'games and lost I: games; there were no ties. How many sets of three teams 仇 D.C were there inwhich 一 ； beat /.-, /. beat : , and beat 一 ；(A) 385(B) 665(C) 915(D) 1140(E) 1330Soluti onWe use compleme ntary coun ti ng. Firstly, because each team played 20 oth

55、er teams, there are 21 teams total. All sets that do not have A beat D, D beat and (? beat # have one team that beats both the other teams. Thus we must count the nu mber of sets of three teams such that one team beats the two other teams and subtract that nu mber from the total nu mber of ways to c

56、hoose three teams.There are 21 ways to choose the team that beat the two other teams,45to choose two teams that the first team both beat. This(3is :- 一 讣！二 sets. There are二 1330sets of three teams total.Subtract ing, we obta in1330 - 945 = A )385as our an swer.Problem 21Let 丄 ,/ be a unit square. Let ：丨 be the midpoint of rFor J -二 let y be the in tersect ion of 曲 and TTd ,and let Qt+i be the8ilrca of £DQiPi ? foot of the perpe ndicular from “ to CD .What is t=i(A) £(B)扌 (C) *(D)(E) 1Soluti on(By Qwertazertl)DQAWe are tasked with finding the