2018年美國數(shù)學競賽(AMC12A)的試題與詳細解答

廣東省華南師范大學數(shù)學科學學院(510631) 李湖南

1.A large urn contains 100 balls,of which 36%are red and the rest are blue.How many of the blue balls must be removed so that the percentage of red balls in the urn will be 72%?(No red balls are to be removed.)

(A)28 (B)32 (C)36 (D)50 (E)64

譯文 一個大缸里有100個球,其中36%是紅球,其余的是藍球.問要拿走多少個藍球才能使大缸里紅球的比例是72%(不拿走紅球)?

解紅球有100×36%=36個,拿走藍球數(shù)為100?36÷72%=50個,故(D)正確.

2.While exploring a cave,Carl comes across a collection of 5-pound rocks worth$14 each,4-pound rocks worth$11 each,and 1-pound rocks worth$2 each.There are at least 20 of each size.He can carry at most 18 pounds.What is the maximum value,in dollars,of the rocks he can carry out of the cave?

(A)48 (B)49 (C)50 (D)51 (E)52

譯文 一個巖洞爆破過后,卡爾發(fā)現(xiàn)一大堆巖石,每塊5磅的巖石價值14美元,每塊4磅的巖石價值11美元,每塊1磅的巖石價值2美元.每種型號的巖石都至少有20塊.卡爾至多能帶走18磅巖石.問他能帶走最多價值多少美元的巖石離開巖洞?

解很明顯,為了價值最大,卡爾要帶走盡可能多的4磅或5磅的巖石.實際上,他帶走2塊4磅的和2塊5磅的巖石,價值最大,為2×14+2×11=50美元,故(C)正確.

3.How many ways can a student schedule 3 mathematics courses—algebra,geometry,and number theory—in a 6-period dayifnotwomathematicscoursescanbetakeninconsecutiveperiods?(What courses the student takes during the other 3 periods is of no concern here.)

(A)3 (B)6 (C)12 (D)18 (E)24

譯文 如果一個學生一天上6節(jié)課,任意兩節(jié)數(shù)學課不相連,問一個學生有多少種方法安排他的3節(jié)數(shù)學課—代數(shù)、幾何和數(shù)論?(其它3節(jié)課是什么課與此無關(guān))

解要使得任意兩節(jié)數(shù)學課不相連,只能將3節(jié)數(shù)學課安排在第 1、3、5 節(jié),或第 1、3、6 節(jié),或第 1、4、6 節(jié),或第 2、4、6節(jié),共4種可能,而每一種可能,都有3!=6種方法,因此共有24種方法,故(E)正確.

4.Alice,Bob and Charlie were on a hike and were wondering how far away the nearest town was.When Alice said,“We are at least 6 miles away,”Bob replied,“We are at most 5 miles away.”Charlie then remarked,“Actually the nearest town is at most 4 miles away.”It turned out that none of the three statements was true.Letdbe the distance in miles to the nearest town.Which of the following intervals is the set of all possible values ofd?

(A)(0,4) (B)(4,5) (C)(4,6) (D)(5,6) (E)(5,∞)

譯文 艾莉絲、鮑勃和查理在徒步旅行,他們想知道他們離最近的鎮(zhèn)有多遠.艾莉絲說:“我們至少有6英里遠.”鮑勃說:“我們至多有5英里遠.”查理說:“實際上最近的鎮(zhèn)至多還有4英里遠.”事實證明三個人說的都不對.設(shè)是他們到最近的鎮(zhèn)的距離,下列哪個區(qū)間是d可能的取值?

解艾莉絲說的不對說明d不到6英里,鮑勃和查理說的也不對說明d超過5英里,即d應該在5英里到6英里之間,故(D)正確.

5.What is the sum of all possible values ofkfor which the polynomialsx2?3x+2andx2?5x+khavearootincommon?

(A)3 (B)4 (C)5 (D)6 (E)10

譯文 多項式x2?3x+2和x2?5x+k有一個公共根,則k的所有可能取值的和是多少?

解多項式x2?3x+2的根為1和2,代入x2?5x+k得k=4或k=6,故(E)正確.

6.For positive integersmandnsuch thatm+10＜n+1,both the mean and the median of the set{m,m+4,m+10,n+1,n+2,2n}are equal ton.What ism+n?

(A)20 (B)21 (C)22 (D)23 (E)24

譯文 已知m和n是正整數(shù),且m+10＜n+1,集合{m,m+4,m+10,n+1,n+2,2n}的平均數(shù)和中位數(shù)均等于n,問m+n是多少?

解中位數(shù)為平均數(shù)為解得于是m+n=21,故(B)正確.

7.For how many(not necessarily positive)integer values ofnis the value of 4000an integer?

(A)3 (B)4 (C)6 (D)8 (E)9

譯文 有多少個整數(shù)n(不必是正的)可以使得的值是一個整數(shù)?

解當n≥0時,是整數(shù),則n=0,1,2,3;當n＜0時,是整數(shù),則n=?1,?2,?3,?4,?5.因此n共有 9個取值,故(E)正確.

8.All of the triangles in the diagram below are similar to isosceles triangleABC,in whichAB=AC.Eachofthe7smallesttriangleshasarea 1,and△ABChas area 40. What is the area of trapezoidDBCE?

圖1

(A)16 (B)18 (C)20 (D)22 (E)24

譯文 如圖1,所有的三角形都相似于等腰三角形ABC,其中AB=AC,7個小三角形的面積均為1,三角形ABC的面積為40,則梯形DBCE的面積是多少?

解分別延長小三角形的腰線,可得圖2.易得所有小三角形均全等,面積均為1,因此三角形ADE的面積為1+3+5+7=16,梯形DBCE的面積=40?16=24,故(E)正確.

圖2

9.Whichofthefollowingdescribesthelargestsubsetofvalues ofywithin the closed interval[0,π]for which sin(x+y)≤sinx+sinyfor everyxbetween 0 andπ,inclusive?

譯文 下列哪個區(qū)間是y值在閉區(qū)間[0,π]上的最大子集,使得式子sin(x+y)≤sinx+siny對所有的0≤x≤π成立?

解當0≤x,y≤π時,sinx≥0,siny≥0,于是有sin(x+y)=sinxcosy+cosxsiny≤sinx+siny,區(qū)間內(nèi)的一切y值均滿足,故(E)正確.

10.How many ordered pairs of real numbers(x,y)satisfy the following system of equations

(A)1 (B)2 (C)3 (D)4 (E)8

譯文 有多少對有序?qū)崝?shù)對(x,y)滿足以下方程組

解先去絕對值符號,||x|?|y||可得四種結(jié)果:x?y,x+y,?x?y,?x+y,代入后分別與x+3y=3聯(lián)立方程組,解得三組解:經(jīng)檢驗,它們均為原方程組的解,故(C)正確.

11.A paper triangle with sides of lengths 3,4 and 5 inches,as shown,is folded so that point A falls on point B.What is the length in inches of the crease?

譯文 一張三角形紙的三邊長分別為3,4,5英寸,如圖3,現(xiàn)將紙折疊使得點A落到點B上,問折痕的長度是多少英寸?

圖3

解如圖3,設(shè)折痕為DE,則AD=DB=2.5,令EC=x,則AE=BE=4?x,由勾股定理可得:(4?x)2=x2+32,解得故(D)正確.

12.LetSbe a set of 6 integers taken from{1,2,…,12}with the property that ifaandbare elements ofSwitha＜b,thenbis not a multiple ofa.What is the least possible value of an element ofS?

(A)2 (B)3 (C)4 (D)5 (E)7

譯文 設(shè)S是一個從{1,2,…,12}中取出的6個元素的集合,具有以下性質(zhì):若a和b均是S中的元素,且a＜b,則b不是a的倍數(shù).問S中元素的最小可能取值是多少?

解這個性質(zhì)說明:集合S中的元素不存在倍數(shù)關(guān)系.顯然,S中最小元素不能是1;若S中最小元素為2,則S={2,3,5,7,9,11},元素3和9不符合性質(zhì),矛盾!若S中最小元素為3,則S?{3,4,5,7,8,10,11},元素4和8、5和10均不符,矛盾!因此,S中可能的最小元素是4,此時S={4,5,6,7,9,11}或S={4,6,7,9,10,11}均滿足要求,故(C)正確.

13.How many nonnegative integers can be written in the forma7·37+a6·36+a5·35+a4·34+a3·33+a2·32+a1·31+a0·30,whereai∈{?1,0,1}for 0≤i≤ 7?

(A)512 (B)729 (C)1094 (D)3281 (E)59,048

譯文 有多少個非負整數(shù)可以寫成如下形式:a7·37+a6·36+a5·35+a4·34+a3·33+a2·32+a1·31+a0·30,其中ai∈{?1,0,1},0≤i≤7?

解由于 3n＞3n?1+3n?2+···+31+30,n是正整數(shù),所以當以上3的多項式中首項系數(shù)an,0≤n≤7取 1的時候,其它的系數(shù)ai,i=0,1,···,n?1可以任意取,顯然首項系數(shù)不能取?1,因此所有可能的組合有37+36+35+34+33+32+31+30+1=3281種,故(D)正確.

14.The solutions to the equation log3x4=log2x8,wherexis a positive real number other thancan be written aswherepandqare relatively prime positive integers.What isp+q?

(A)5 (B)13 (C)17 (D)31 (E)35

譯文 設(shè)x是不等于的正實數(shù),方程log3x4=log2x8的解可以寫成其中p和q是互素的正整數(shù),問p+q是多少?

解由換底公式可得化簡成解得故(D)正確.

15.A scanning code consists of a 7×7 grid of squares,with some of its squares colored black and the rest colored white.There must be at least one square of each color in this grid of 49 squares.A scanning code is called symmetric if its look does not change when the entire square is rotated by a multiple of 90°counterclockwisearounditscenter,norwhenitisreflectedacross a line joining opposite corners or a line joining midpoints of opposite sides.What is the total number of possible symmetric scanning codes?

(A)510 (B)1022 (C)8190 (D)8192 (E)65,534

譯文 一個掃描碼包含一個7×7正方形方格,其中一些方格是黑色的,而其余的是白色.在這49個方格中,每種顏色必須至少有一格.如果一個掃描碼圍繞著它的中心做90°的整數(shù)倍的逆時針旋轉(zhuǎn)還是一樣的,或者沿著對角線或?qū)呏悬c連線的翻轉(zhuǎn)也不變,則稱這個掃描碼是對稱的.問對稱的掃描碼共有多少個?

解由于對稱的掃描碼繞著中心90°旋轉(zhuǎn)是一樣的,從而只需要考慮左上角的4×4正方形方格即可;又由于該4×4的正方形方格關(guān)于對角線對稱,所以只需考慮圖4中標有a,b,c,d,e,f,g,h,i,j的10個方格即可,每個方格的顏色有黑色和白色2種選擇,故有210=1024種,再減去全黑和全白的2種,共有1022種,故(B)正確.

圖4

16.Which of the following describes the set of values ofafor which the curvesx2+y2=a2andy=x2?ain the realxy-plane intersect at exactly 3 points?

譯文a取何值時,曲線x2+y2=a2與y=x2?a在實平面上恰好有3個交點?

解代入可得x2+(x2?a)2=a2,化簡得x2(x2?2a+1)=0,要使得該方程有3個不同的解,則x2?2a+1=0必須有2個解,即有x2=2a?1＞0,于是故(E)正確.

圖5

17. Farmer Pythagoras has a field in the shape of a right triangle.The right triangle’s legs have lengths of 3 and 4 units.In the corner where those sides meet at a right angle,he leaves a small unplanted squareSso that from the air it looks like the right angle symbol.The rest of the field is planted.The shortest distance fromSto the hypotenuse is 2 units.What fraction of the field is planted?

譯文 農(nóng)夫畢達哥拉斯有一塊直角三角形的地,直角邊長分別為3個單位和4個單位.在構(gòu)成直角的那個角落里,他留了一小塊未種植的正方形地S,從空中看上去就像一個直角符號.其余的地均進行了種植.S到斜邊的最短距離是2個單位.問種植的地占多少比例?

圖6

解如圖6,分別延長ED、FD交AC于點L、K,作LH⊥BC于H,設(shè)正方形邊長為x個單位,則由于△KDL～=△ABC,可以設(shè)KD=3k,DL=4k,于是解得因為解得則S占地比例為因此所求的占地比例為故(D)正確.

18.TriangleABCwithAB=50 andAC=10 has area 120.LetDbe the midpoint ofAB,and letEbe the midpoint ofAC.The angle bisector of∠BACintersectsDEandBCatFandG,respectively.What is the area of quadrilateralFDBG?

(A)60 (B)65 (C)70 (D)75 (E)80

譯文 三角形ABC的面積為120,AB=50,AC=10,D是AB的中點,E是AC的中點,∠BAC的平分線分別交DE和BC于F和G.求四邊形FDBG的面積是多少?

圖7

解如圖7示,DE是△ABC的中位線,根據(jù)角平分線定理,從而可得而于是SFDBG=100?25=75,故(D)正確.

19.LetAbe the set of positive integers that have no prime factors other than 2,3,or 5.The infinite sumof the reciprocals of the elements ofAcan be expressed aswheremandnare relatively prime positive integers.What ism+n?

(A)16 (B)17 (C)19 (D)23 (E)36

譯文 設(shè)A是不含除2,3或5外的素因子的正整數(shù)集合,A中元素的倒數(shù)之和可以表示成其中m和n是互素的正整數(shù),問m+n是多少?

解由題目條件,可得

m+n=19,故(C)正確.

20.TriangleABCis an isosceles right triangle withAB=AC=3.Let M be the midpoint of hypotenuseBC.Points I and E lie on sidesACandAB,respectively,so thatAI＞AEandAIMEis a cyclic quadrilateral.Given that triangleEMIhas area 2,the length CI can be written aswherea,bandcare positive integers andbis not divisible by the square of any prime.What is the value ofa+b+c?

(A)9 (B)10 (C)11 (D)12 (E)13

譯文 三角形ABC是一個等腰直角三角形,AB=AC=3.M是斜邊BC上的中點.點I和E分別在邊AC和AB上,使得AI＞AE,且AIME是一個圓內(nèi)接四邊形.設(shè)三角形EMI的面積為2,CI的長度可以寫成其中a,b,c是正整數(shù),且b不含素因子的平方,問a+b+c是多少?

圖8

解如圖8所示,連結(jié)AM,由四點共圓,可得∠EMI=90°,∠EIM=∠EAM=45°,從而△EMI也是等腰直角三角形,得MI=2,而根據(jù)余弦定理,MI2=MC2+CI2?2MC·CIcos∠C,代入 后得解得于是a+b+c=12,故(D)正確.

21.Which is the following polynomials has the greatest real root?

(A)x19+2018x11+1 (B)x17+2018x11+1

(C)x19+2018x13+1 (D)x17+2018x13+1

(E)2019x+2018

譯文 下列哪個多項式的實根最大?

解容易分析這些多項式均是單調(diào)遞增的.對于多項式f(x)=x2m+1+2018x2n+1+1,m,n為自然數(shù),當x≤?1時,有f(x)＜0;當x≥0時,有f(x)＞0,故f(x)的根一定在區(qū)間(?1,0)內(nèi).當x∈(?1,0)時,有x11＜x13＜x17＜x19,這表明在前四個選項中,x17+2018x11+1的值最小,x19+2018x13+1的值最大.因此,x17+2018x11+1的根最大.另外,2019x+2018的根為而可知x17+2018x11+1的根大于x0,故(B)正確.

22.The solutions to the equationsandform the vertices of a parallelogram in the complex plane.The area of this parallelogram can be written in the formwherep,q,randsare positive integers and neitherqnorsis divisible by the square of any prime number.What isp+q+r+s?

(A)20 (B)21 (C)22 (D)23 (E)24

解解方程得解方程得z=則平行四邊形的頂點坐標分別為O為兩條對角線交點,可得OA= 4,OB= 2,由余弦定理,進而于是故(A)正確.

圖9

23. In△PAT,∠P=36°,∠A=56°,andPA=10.PointsUandGlie on sidesTPandTA,respectively,so thatPU=AG=1.LetMandNbe the midpoints of segmentsPAandUG,respectively.What is the degree measure of the acute angle formed by linesMNandPA?

圖10

(A)76 (B)77 (C)78 (D)79 (E)80

譯文 在△PAT中,∠P=36°,∠A=56°,PA=10.點U和G分別在邊TP和TA上,使得PU=AG=1.設(shè)M和N分別是PA和UG的中點.問直線MN和PA所夾的銳角是多少度?

解如圖所示,以P為原點,PA為x軸建立直角坐標系,則有點坐標M(5,0),U(cos36°,sin36°),G(10?cos56°,sin56°),從而

所求的銳角為∠NMA,于是

因此∠NMA=80°,故(E)正確.

24.Alice,Bob and Carol play a game in which each of them chooses a real number between 0 and 1.The winner of the game is the one whose number is between the numbers chosen by the other two players.Alice announces that she will choose her number uniformly at random from all the numbers between 0 and 1,and Bob announces that he will choose his number uniformly at random from all the numbers betweenArmed with this information,what number should Carol choose to maximize her chance of winning?

譯文 艾莉絲、鮑勃和卡羅爾在玩一種游戲,每個人在0和1之間選一個實數(shù),誰的數(shù)在其他兩人之間誰就贏了.艾莉絲宣布她會在0和1之間隨機選一個數(shù),鮑勃宣布他會在之間隨機選一個數(shù).有了這些信息,卡羅爾應該選什么數(shù)才能使他贏的機會最大?

解設(shè)艾莉絲選的數(shù)為x,x∈[0,1],鮑勃選的數(shù)為y,卡羅爾選的數(shù)為z,z∈[0,1],于是P(x＜z)=z,P(z＜x)=1?z,卡羅爾贏的兩種情況為:(1)當x＜z＜y時:概率為當y＜z＜x時:概率為因此卡羅爾贏的概率為時,概率最大為故(B)正確.

25.For a positive integernand nonzero digitsa,bandc,letAnbe then-digit integer each of whose digits is equal toa;letBnbe then-digit integer each of whose digits is equal tob;and letCnbe the 2n-digit(notn-digit)integer each of whose digits is equal toc.What is the greatest possible value ofa+b+cfor whichthereareatleasttwovaluesofnsuchthatCn?Bn=A2n?

(A)12 (B)14 (C)16 (D)18 (E)20

譯文 已知n是正整數(shù),a,b,c是非零數(shù)字,設(shè)An是數(shù)字均為a的n位數(shù),Bn是數(shù)字均為b的n位數(shù),Cn是數(shù)字均為c的2n位數(shù)(不是n位).求a+b+c可能取到的最大值,使得至少存在兩個n值能讓等式Cn?Bn=A2n成立?

即c·(10n+1)=b+a2·進一步可得

由于至少存在兩個n值使該式成立,所以解得.因此a+b+c的最大值等于18,故(D)正確.