AMC12B试题及解答Word文档下载推荐.docx

1、 drago nflySince the harm onic mea n isJ times their product divided by their sum, we get the equati on2 x 1 x 20161+2016which is the n4032（A） 22017which is fin ally closest toProblem 3Let X - 2016. What is the value of IM 力I 一 I刎兽?First of all, lets plug in all of the s into the equati on.| - 2016|

2、 一 （-2016）1 一 | - 20161 -（-2016）Then we simplify to get|2016 + 20161- 2016 12016which simplifies into2016 +2016（D） 4032and fin ally we get Problem 4The ratio of the measures of two acute an gles is 5:1, and the compleme ntof one of these two an gles is twice as large as the compleme nt of the other.

3、 What is the sum of the degree measures of the two an gles?（A） 75 （B） 90 （C） 135 （D） 150 （E） 270We set up equations to find each angle. The larger angle will bereprese nted as x and the larger an gle will we represe nted as /in degrees.This implies that4x = 5yand2 x （90 j:） = 90 y（C） 135since the la

4、rger the original angle, the smaller the complement.We the n find that 一 . and 彳 沉丄，and their sum isProblem 5The War of 1S12 started with a declarati on of war on Thursday,June 1812. The peace treaty to end the war was sig ned 019 days later,on December 24, 1814. On what day of the week was the trea

5、ty sig ned?To find what day of the week it is in 919 days, we have to divide 919 by 7 to see the rema in der, and the n add the rema in der to the curre nt day. We get919that 7 has a remain der of 2, so we in crease the curre nt day by 2 to（B）SatiLrdayget IProblem 6All three vertices of AAJ3C lie on

6、 the parabola defined by 9 = T , with A at the origin and BC parallel to the 工-axis. The area of the triangleis 64. What is the length of DC?（A） 4 （B） 6 （C） 8 （D） 10 （E） 16 Albert471Plotting points 门 and : on the graph shows that they are at （上、止）and ,r ”），which is isosceles. By setting up the trian

7、gle areaMak ing x=4, and the len gth（c）s64 = * 2t * rr2 = G4 formula you get: 2 of DC is so the answer isProblem 7Josh writes the numbers 1 2, 3,、99,100. He marks out 1, skips the next nu mber （2） ,marks out 3, and con ti nues skipp ing and marking out the next number to the end of the list. Then he

8、 goes back to the start of his list,marks out the first remaining number ,skips the next number ,marksout 6, skips 8, marks out 10, and so on to the end. Josh continues in this（D）64manner un til only one nu mber rema ins. What is that nu mber?（A） 13 （B） 32 （C） 56 （D） 64By Albert471Follow ing the pat

9、ter n, you are cross ing out.（E） 96Time 1: Every non-multiple ofTime 2:4Time 3:8Following this pattern, you are left with every multiple of G4 which isonlyProblem 8A thin piece of wood of uniform den sity in the shape of an equilateraltrian gle with side len gth 3 in ches weighs 12o un ces. A sec on

10、d piece of the same type of wood, with the same thick ness, also in the shape of an equilateral tria ngle, has side len gth of 5 in ches. Which of the follow ing isclosest to the weight, i n oun ces, of the sec ond piece?（A） 14,0 （B） 1G.0 （C） 20.0 （D） 33.3 （E） 55.6We can solve this problem by using

11、similar triangles, since two equilateraltriangles are always similar. We can then use100；which is closestAno ther approach to this problem, very similar to the previous one but perhaps expla ined more thoroughly, is to use proporti ons. First, since thethickness and density are the same, we can set

12、up a proportion based on the principle that V, thus dV = m.However, since den sity and thick ness are the same25,and 4 oc b3 （recognizing that the area of an equilateral triangle is we can say that m b 0. The area of PQRS is 16 .Whatis + b?By distanee formula we have 1 1 1 I ? SImplifying we get ）（a

13、 + b） = 8. Thus U + ft and a b have to be a- and rema in in tegers isfactor of 8. The only way for them to be factors ofif : I ； : and :丄 So the answer isSolution by l_Do nt_Do_MathSolution 2Soluti on by e_power_pi_times_iBy the Shoelace Theorem, the area of the quadrilateral is 2ti2 2bso a2 b2 8. S

14、i nee a and & are integers, u = 3 and b = 1,a + b = （A） 4 Iso .Problem 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 boun ded by the line 站=齐工，the line _0.1 and the line 龙= 5.1?（A） 30 （B） 41 （C） 45 （D

15、） 50 （E） 57（Note: diagram is n eeded）If we draw a picture show ing the tria ngle, we see that it would be easier to count the squares vertically and not horiz on tally. The upper boundis 1G squares （卩=5木汗），and the limit for the 工-value is 5 squares. Firstwe count the 1*1 squares. In the back row, th

16、ere are 12 squares with len gth 1 because y = 4 床 u gen erates squares from ,0） to （蠶仆）,and continuing on we have 9, 6, and 3 for values for 1, 2, and 3 in the equati on V 兀卫.So there are 12 + 9 + 6+ 3 = 30 squares with len gth I in the figure. For _ 一 squares, each square takes up _ un left and 2 u

17、n up. Squares can also overlap. For 2*2 squares, the back rowstretches from （N to（ *兀），so there are 8 squares with length 2 in a 2 by 0 box. Repeating the process, the next row stretches from （2,0）to （2,2打）,so there are 5 squares. Con ti nuing and addi ng up inthe end, there are& ” :, 乞一 、：:squares

18、with length - in the figure.Squares with len gth 3in the back row start at and end at （2*277）, sothere are 4 such squares in the back row. As the front row startsat （10 and ends at （1） there areI 1 squares with len gth . Assquares with len gth J would not fit in the tria ngle, the an sweris ； I 丨 wh

19、ich isD） 5（）Problem 12All the numbers 1, N 3,4,5,6, 7* & 0 are written in a 3x3 array of squares,one nu mber in each square, in such a way that if two nu mbers arecon secutive the n they occupy squares that share an edge. The nu mbers in the four corners add up to 18. What is the number in the cente

20、r?（A） 5 （B） 6 （C） 7 （D） 8 （E） 9Soluti on by Mlux: Draw a 3x3 matrix. Notice that no adjace nt nu mberscould be in the corners since two con secutive nu mbers must share anedge. Now find 4 noncon secutive nu mbers that add up to 18.Tryi ng 1 + 3 + 5+ 9 18 works. Place each odd nu mber in the corner i

21、na clockwise order. The n fill in the spaces. There has to be a 2 in betwee nthe 1 and 3. There is a 4 between 3 and 5. The final grid should similar to this.1,2.38,7 A9.6,5（C）7is in the middle.If we color the square like a chessboard, since the numbers altrenate betwee n eve n and odd, and there ar

22、e five odd nu mbers and four eve n numbers, the odd numbers must be in the corners/center, while the even nu mbers 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

23、apart. One day Alice looks due n orth from her house and sees an airpla ne. At the same time Bob looks due west from his house and sees the same airpla ne. The an gle of elevati on of the airpla ne is 30n from Alices positi on and 60 from Bobs positi on. Which of the following is closest to the airp

24、lanes altitude, in miles?（A） 3.5 （B）4 （C） 4*5 （D） 5 （E） 5.5Lets set the altitude = z, distanee from Alice to airplanes ground position （point right below airpla ne）二y and dista nee from Bob to airpla nes groundpositi on=xFrom Alices point of view,y. cos 30 品costan（0） = - GO = S1D 男=亞 工二From BobT. CU

25、S GO . So, V O、l = about 5.5.We know that * + / =丄Solv ing the equati on （by pluggi ng in x and y）, we get z=So, an swer issoluti on by sudeep naralaNon -trig soluti on by e_power_pi_times_iSet the distanee from Alices and Bobs position to the point directly below the airplane to be x and 甘,respectively. From the Pythagorean3 or . Solving theTheorem, ? +=100. As both are 30 60 90 tria ngles, the altitude of the airpla ne can be expressed aseq