2019年美國(guó)數(shù)學(xué)競(jìng)賽(AMC10A)的試題與解答

廣東省華南師范大學(xué)數(shù)學(xué)科學(xué)學(xué)院(510631) 李湖南

(A)0 (B)1 (C)2 (D)3 (E)4

解原式=20+19=2,故(C)正確.

2.What is the hundreds digit of(20!-15!)?

(A)0 (B)1 (C)2 (D)4 (E)5

譯文:(20!-15!)的百位數(shù)是多少?

解20!的標(biāo)準(zhǔn)分解式中含有個(gè)5,15!的標(biāo)準(zhǔn)分解式中含有=3個(gè)5,即它們的最后三位數(shù)均為0,從而(20!-15!)的百位數(shù)是0,故(A)正確.

3.Ana and Bonita were born on the same date in different years,nyears apart.Last year Ana was 5 times as old as Bonita.This year Ana's age is the square of Bonita's age.What isn?

(A)3 (B)5 (C)9 (D)12 (E)15

譯文:安娜和博尼塔出生在相隔了n年的同一天.去年安娜的年齡是博尼塔的5倍,今年安娜的年齡是博尼塔年齡的平方.問(wèn)n是多少?

解設(shè)去年博尼塔x歲,則安娜5x歲,依題意有5x+1=(x+1)2,解得x=3,于是有n=5x-x=12,故(D)正確.

4.A box contains 28 red balls,20 green balls,19 yellow balls,13 blue balls,11 white balls,and 9 black balls.What is the minimum number of balls that must be drawn from the box without replacement to guarantee that at least 15 balls of a single color will be drawn?

(A)75 (B)76 (C)79 (D)84 (E)91

譯文:一個(gè)盒子里有28個(gè)紅球,20個(gè)綠球,19個(gè)黃球,13個(gè)藍(lán)球,11個(gè)白球和9個(gè)黑球.在不放回的情況下,必須至少?gòu)暮凶又腥〕龆嗌賯€(gè)球才能確保會(huì)取出15個(gè)同種顏色的球?

解依題意,紅球、綠球、黃球得取出14個(gè),藍(lán)球、白球、黑球全部取出,再加任意1個(gè)就符合條件,即至少需要取出14×3+13+11+9+1=76個(gè)球,故(B)正確.

5.What is the greatest number of consecutive integers whose sum is 45?

(A)9 (B)25 (C)45 (D)90 (E)120

譯文:最多有多少個(gè)連續(xù)整數(shù)的和等于45?

解設(shè)有x個(gè)連續(xù)整數(shù),從n開(kāi)始,即n+(n+1)+(n+2)+···+(n+x-1)=45,可得x(2n+x-1)=90,最大值解為x=90,此時(shí)n=-44,故(D)正確.

6.For how many of the following types of quadrilaterals does there exist a point in the plane of the quadrilateral that is equidistant from all four vertices of the quadrilateral?

—a square

—a rectangle that is not a square

—a rhombus that is not a square

—a parallelogram that is not a rectangle or a rhombus

—an isosceles trapezoid that is not a parallelogram

(A)1 (B)2 (C)3 (D)4 (E)5

譯文:下列四邊形中,四邊形所在平面內(nèi)存在一點(diǎn)與四邊形的四個(gè)頂點(diǎn)距離都相等的有多少種?

①正方形;②不是正方形的矩形;③不是正方形的菱形;④不是矩形或菱形的平行四邊形;⑤不是平行四邊形的等腰梯形.

解存在一點(diǎn)與四邊形的四個(gè)頂點(diǎn)距離都相等,即四邊形存在外接圓.矩形、正方形和等腰梯形都存在外接圓,但菱形和平行四邊形就未必存在,故共有3種,(C)正確.

圖1

解先寫(xiě)出這兩條直線方程:y-和y-2=2(x-2),然后分別與方程x+y=10聯(lián)立,求得交點(diǎn)為B(6,4)和C(4,6),設(shè)點(diǎn)(2,2)為A,則同理可得.如圖示,作AD⊥BC于D,則D為BC中點(diǎn),從而AD=于是故(C)正確.

8.The figure below shows linelwith a regular,infinite,recurring pattern of squares and line segments.How many of the following four kinds of rigid motion transformations of the plane in which this figure is drawn,other than the identity transformation,will transform this figure into itself?

—some rotation around a point of linel

—some translation in the direction parallel to linel

—the reflection across linel

—some reflection across a line perpendicular to linel

圖2

(A)0(B)1(C)2(D)3(E)4

譯文:下圖表示一條有規(guī)則的、無(wú)限的、循環(huán)出現(xiàn)的正方形和線段圖案的直線l.下列四種平面上的剛體變換中,恒等變換除外,有多少種會(huì)將該圖形變換為自身?

①繞直線l上的一個(gè)點(diǎn)的一些旋轉(zhuǎn);②關(guān)于平行于直線l的方向上的一些平移;③關(guān)于直線l的反射;④關(guān)于一條垂直于l的直線的反射.

解看圖可知,繞直線l上一點(diǎn)做180°旋轉(zhuǎn)可以變回自身;關(guān)于平行于直線l的方向上做若干個(gè)周期的平移也可以變回自身;而關(guān)于直線l的反射、或者垂直于l的直線的反射則會(huì)將與正方形相連的線段方向變反.所以只有變換①②符合要求,故(C)正確.

9.What is the greatest three-digit positive integernfor which the sum of the firstnpositive integers is not a divisor of the product of the firstnpositive integers?

(A)995 (B)996 (C)997 (D)998 (E)999

譯文:使得前n個(gè)正整數(shù)之和不是前n個(gè)正整數(shù)之積的因子的最大三位數(shù)正整數(shù)n是多少?

解要使得不整除n!,即要使n+1不整除n!,由于997是個(gè)素?cái)?shù),所以n=996符合條件,故(B)正確.

10.A rectangular floor that is 10 feet wide and 17 feet long is tiled with 170 one-foot square tiles.A bug walks from one corner to the opposite corner in a straight line.Including the first and the last tile,how many tiles does the bug visit?

(A)17 (B)25 (C)26 (D)27 (E)28

譯文:一片10英尺寬17英尺長(zhǎng)的長(zhǎng)方形地板用170塊一英尺的正方形瓷磚鋪好.一只蟲(chóng)子從一個(gè)角落沿著直線爬到對(duì)角,包括第一塊和最后一塊瓷磚,這只小蟲(chóng)爬過(guò)了多少塊瓷磚?

圖3

解如圖示,建立直角坐標(biāo)系,設(shè)小蟲(chóng)的初始點(diǎn)為O(0,0),目標(biāo)點(diǎn)為P(17,10),先求出小蟲(chóng)與17條縱線的交點(diǎn)分別為···,A17(17,10);再求出小蟲(chóng)與10條橫線的交點(diǎn)分別為由于10與17是互素的,所以共有26個(gè)不同的交點(diǎn),小蟲(chóng)每爬到一個(gè)交點(diǎn)處,意味著它爬過(guò)了一塊瓷磚,即小蟲(chóng)共爬過(guò)了26塊瓷磚,故(C)正確.

11.How many positive integer divisors of 2019are perfect squares or perfect cubes(or both)?

(A)32 (B)36 (C)37 (D)39 (E)41

譯文:有多少個(gè)2019的正因數(shù)是完全平方數(shù)或者立方數(shù)(或者都是)?

解2019=39×679,其平方數(shù)因數(shù)可以表示成32i×672j,其中i,j=0,1,2,3,4,共25個(gè);立方數(shù)因數(shù)可以表示成33i×673j,其中i,j=0,1,2,3,共16個(gè);重復(fù)計(jì)算的有4個(gè),分別是1,36,676,36×676.所以符合條件的因數(shù)有25+16-4=37個(gè),故(C)正確.

12.Melanie computes the meanμ,the medianM,and the modes of the 365 values that are the dates in the months of 2019.Thus her data consist of 12 1s,12 2s,…,12 28s,11 29s,11 30s,and 7 31s.Letdbe the median of the modes.Which of the following statements is true?

(A)μ＜d＜M(B)M＜d＜μ(C)d=M=μ(D)d＜M＜μ(E)d＜μ＜M

譯文:米蘭妮在計(jì)算2019年12個(gè)月中的所有365個(gè)日期的平均數(shù)μ,中位數(shù)M,和眾數(shù).她的數(shù)據(jù)包含12個(gè)1,12個(gè)2,…,12個(gè)28,11個(gè)29,11個(gè)30和7個(gè)31.令d為所有眾數(shù)的中位數(shù),則下列哪個(gè)說(shuō)法是正確的?

解平均數(shù)為,中位數(shù)為從小到大排在第183的數(shù),即M=16,眾數(shù)為1,2,···,28,則眾數(shù)的中位數(shù)為d=14.5,比較后可知,(E)正確.

13.Let△ABCbe an isosceles triangle withBC=ACand∠ACB=40°.Construct the circle with diameterand letDandEbe the other intersection points of the circle with the sidesand,respectively.LetFbe the intersection of the diagonals of the quadrilateralBCDE.What is the degree measure of∠BFC?

(A)90 (B)100 (C)105 (D)110 (E)120

譯文:設(shè)△ABC是等腰三角形,BC=AC,∠ACB=40°.以BC為直徑作一個(gè)圓,分別交邊AC,BC于點(diǎn)D,E,令F為四邊形BCDE對(duì)角線的交點(diǎn),則∠BFC的度數(shù)是多少?

解如圖示,由于BC是直徑,則∠BDC=∠BEC=90°,從而∠CBD=90°-∠BCD=50°,且CE是等腰三角形△ABC的高,根據(jù)三線合一可知,CE也是角平分線,即.故∠BFC=180°-∠BCF-∠CBF=110°,(D)正確.

圖4

14.For a set of four distinct lines in a plane,there are exactlyNdistinct points that lies on two or more of the lines.What is the sum of all possible values ofN?

(A)14 (B)16 (C)18 (D)19 (E)21

譯文:對(duì)于一個(gè)平面上的四條不同的直線,恰好有N個(gè)不同的交點(diǎn).問(wèn)所有可能的N值之和是多少?

解分情況如下:(1)四條直線平行,則N=0;(2)四條直線交于一點(diǎn),則N=1;(3)三條直線平行,與第四條直線均相交,則N=3;(4)兩條直線平行,另兩條直線也平行,但相互不平行,則N=4;(5)兩條直線平行,另兩條直線相交,且與平行直線均相交于不同交點(diǎn),則N=5;(6)四條直線均不平行,且交點(diǎn)互不相同,則N=6.于是所有可能的N值之和為19,故(D)正確.

15.A sequence of numbers is defined recursively bya1=andfor alln≥3.Thena2019can be written as,wherepandqare relatively prime positive integers.What isp+q?

(A)2020 (B)4039 (C)6057 (D)6061 (E)8078

譯文:定義一個(gè)遞歸數(shù)列如下:a1=1,a2=,,?n≥3.則a2019可以寫(xiě)成,其中p,q是互素的正整數(shù),問(wèn)p+q是多少?

解代入可得···,我們猜測(cè),可驗(yàn)證如下:.于 是,a2019=,即有p=3,q=8075,p+q=8078,故(E)正確.

16.The figure below shows 13 circles of radius 1 within a larger circle.All the intersections occur at points of tangency.What is the area of the region,shaded in the figure,inside the larger circle but outside all the circles of radius 1?

譯文:下圖表示一個(gè)較大的圓內(nèi)有13個(gè)半徑為1的圓.所有的交點(diǎn)都發(fā)生在切點(diǎn)上.圖中陰影為大圓內(nèi)但在所有半徑為1的圓外部分,則陰影部分的面積是多少?

圖5

解如圖示,分別取大圓的圓心O,和上方三個(gè)小圓的圓心A,B,C,連結(jié)OA,OB,OC,AB,AC,BC,可知△OAC,△ABC均為邊長(zhǎng)為 2的正三角形.于是,即大圓半徑為,.故所求面積為,(A)正確.

17.A child builds towers using identically shaped cubes of different color.How many different towers with a height 8 cubes can the child build with 2 red cubes,3 blue cubes,and 4 green cubes?(One cube will be left out.)

(A)24 (B)288 (C)312 (D)1260 (E)40320

譯文:一個(gè)小孩用形狀相同、顏色不同的方塊搭建塔樓.問(wèn)小孩能用2個(gè)紅色方塊、3個(gè)藍(lán)色方塊和4個(gè)綠色方塊建造多少個(gè)高度為8層(8個(gè)方塊)的塔樓?(將有1個(gè)方塊剩余)

解分情況如下:(1)剩余1個(gè)紅色方塊,即需要1紅3藍(lán)4綠,此時(shí)塔樓的排列有種;(2)剩余1個(gè)藍(lán)色方塊,即需要2紅2藍(lán)4綠,此時(shí)塔樓的排列有種;(3)剩余1個(gè)綠色方塊,即需要2紅3藍(lán)3綠,此時(shí)塔樓的排列有種.故共有1260種,(D)正確.

18.For some positive integerk,the repeating base-krepresentationofthe(base-ten)fractionis.What isk?

(A)13 (B)14 (C)15 (D)16 (E)17

譯文:對(duì)于某個(gè)正整數(shù)k,(十進(jìn)制)分?jǐn)?shù)的k進(jìn)制循環(huán)表示為.問(wèn)k是多少?

解由于兩邊同時(shí)乘以k2可得,,整理可得7k2-102k-160=0,解得k=16或(舍去),故(D)正確.

19.What is the least possible value of(x+1)(x+2)(x+3)(x+4)+2019 wherexis a real number?

(A)2017 (B)2018 (C)2019 (D)2020 (E)2021

譯文:設(shè)x是實(shí)數(shù),則(x+1)(x+2)(x+3)(x+4)+2019的最小值是多少?

解(x+1)(x+2)(x+3)(x+4)+2019=其中y=,于是上式,從而當(dāng),即時(shí),上式有最小值2018,故(B)正確.

20.The numbers 1,2,…,9 are randomly placed into the 9 squares of a 3×3 grid.Each square gets one number,and each of the numbers is used once.What is the probability that the sum of the numbers in each row and each column is odd?

譯文:數(shù)字1到9被隨機(jī)地放入到3×3的九宮格中,每個(gè)格子放一個(gè)數(shù),且每個(gè)數(shù)只使用一次.問(wèn)每行每列的數(shù)字之和都是奇數(shù)的概率是多少?

解1到9共有5個(gè)奇數(shù)4個(gè)偶數(shù).要使得一行(或列)的數(shù)字和為奇數(shù),該行(或列)只能有1個(gè)或3個(gè)奇數(shù),從而三行(或列)的奇數(shù)個(gè)數(shù)分配只能是311,131,或113.而5個(gè)奇數(shù)可以隨意排列,4個(gè)偶數(shù)也隨意排列,因此符合條件的排列有3×3×5!×4!種.而9個(gè)數(shù)的全排列是9!種,故所求概率為,(B)正確.

21.A sphere with centerOhas radius 6.A triangle with sides of length 15,15,and 24 is situated in space so that each of its sides is tangent to the sphere.What is the distance betweenOand the plane determined by the triangle?

譯文:設(shè)球O的半徑為6,一個(gè)三邊長(zhǎng)分別為15,15和24的空間三角形的每條邊都和球相切.則點(diǎn)O到三角形所在平面的距離是多少?

解如圖示,這是三角形所在的平面圖形,AB=AC=15,BC=24,其內(nèi)切圓圓心為P,三個(gè)切點(diǎn)分別為D,E,F,連結(jié)AD,PE,PF,則AD⊥BC,從而B(niǎo)D=DC=12,于是BE=BD=12,AE=AB-BE=3,AD=.設(shè)內(nèi)切圓半徑為r,則在Rt△PEA中,有AP2=AE2+EP2,即(9-r)2=32+r2,解得r=4.又在球內(nèi)△OPD中,OP⊥PD,所以點(diǎn)O到該平面的距離,故(D)正確.

圖6

22.Real numbers between 0 and 1,inclusive,are chosen in the following manner.A fair coin is flipped.If it lands heads,then it is flipped again and the chosen number is 0 if the second flip is heads and 1 if the second flip is tails.On the other hand,if the first coin flip is tails,then the number is chosen uniformly at random from the closed interval[0,1].Two random numbersxandyare chosen independently in this manner.What is the probability that?

譯文:用下列方式選出區(qū)間[0,1]上的實(shí)數(shù):擲一枚均勻的硬幣,如果正面朝上,就再擲一次,若第二次正面朝上則選數(shù)字0,若第二次反面朝上則選數(shù)字1;如果第一次就反面朝上,則在區(qū)間[0,1]上隨機(jī)選擇一個(gè)數(shù)字.x和y均是用這種方式隨機(jī)獨(dú)立地選出來(lái)的.問(wèn)的概率是多少?

圖7

解設(shè)甲、乙兩人擲硬幣分別得到數(shù)字x和y,要使得|x-y|＞,則分如下情況:(1)甲連續(xù)擲出正面、正面,乙連續(xù)擲出正面、反面,此時(shí)x=0,y=1,概率為;(2)甲連續(xù)擲出正面、反面,乙連續(xù)擲出正面、正面,此時(shí)x=1,y=0,概率為;(3)甲連續(xù)擲出正面、正面,乙擲出反面,此時(shí)x=0,y∈[0,1],但要才有效,概率為;(4)甲連續(xù)擲出正面、反面,乙擲出反面,此時(shí)x=1,y∈[0,1],但要才有效,概率為;(5)甲擲出反面,乙連續(xù)擲出正面、正面,此時(shí)x∈[0,1],y=0,但要才有效,概率為;(6)甲擲出反面,乙連續(xù)擲出正面、反面,此時(shí)x∈[0,1],y=1,但要才有效,概率為;(7)甲擲出反面,乙也擲出反面,此時(shí)x,y∈[0,1],要符合條件,則點(diǎn)(x,y)必須落在陰影部分里,如圖示,概率為;綜上可得,所求概率為,故(B)正確.

23.Travis has to babysit the terrible Thompson triplets.Knowing that they love big numbers,Travis devises a counting game for them.First Tadd will say the number 1,then Todd must saythe next two numbers(2and3),then Tucker must say the next three numbers(4,5,6),then Tadd must say the next four numbers(7,8,9,10),and the process continues to rotate through the three children in order,each saying one more number than the previous child did,until the number 10,000 is reached.What is the 2019th number said by Tadd?

(A)5743 (B)5885 (C)5979 (D)6001 (E)6011

譯文:特拉維斯必須照顧麻煩的湯普森三胞胎.特拉維斯知道他們喜歡大數(shù)字,就為他們?cè)O(shè)計(jì)了一個(gè)數(shù)數(shù)游戲.首先塔德數(shù)1,然后托德必須數(shù)接下來(lái)的兩個(gè)數(shù)字(2和3),然后塔克必須數(shù)接下來(lái)的三個(gè)數(shù)字(4,5,6),接著塔德必須數(shù)接下來(lái)的四個(gè)數(shù)字(7,8,9,10),并且這個(gè)過(guò)程一直在這三個(gè)孩子之間按順序循環(huán)進(jìn)行,每個(gè)孩子都比前一個(gè)孩子多數(shù)一個(gè)數(shù)字,直到數(shù)字10000出現(xiàn)為止.問(wèn)塔德數(shù)的第2019個(gè)數(shù)字是多少?

解塔德第一輪數(shù)1個(gè)數(shù),第二輪數(shù)了4個(gè)數(shù),第三輪數(shù)了7個(gè)數(shù),……,第n輪數(shù)了3n-2個(gè)數(shù),共數(shù)了個(gè)數(shù),先求得的最大整數(shù)n=36,即塔德數(shù)到第36輪時(shí),共數(shù)了個(gè)數(shù).接著第37輪,他要數(shù)3×37-2=109個(gè)數(shù),此時(shí)只需要數(shù)到第2019-1926=93個(gè)數(shù)即可.另外,36輪過(guò)后,三個(gè)人一起數(shù)到1+2+3+···+108=5886,接下來(lái)的第93個(gè)數(shù)就是5886+93=5979,故(C)正確.

24.Letp,qandrbe the distinct roots of the polynomialx3-22x2+80x-67.It is given that there exist real numbersA,BandCsuch thatfor alls∈/{p,q,r}.What is?

(A)243 (B)244 (C)245 (D)246 (E)247

譯文:令p,q,r是多項(xiàng)式x3-22x2+80x-67三個(gè)不同的根,假設(shè)存在實(shí)數(shù)A,B,C使得式子對(duì)?s∈/{p,q,r}均成立.則是多少?

解由題意可知s3-22s2+80s-67=(s-p)(s-q)(s-r),根據(jù)根與系數(shù)的關(guān)系,有且比較后可得將A=-B-C代入后兩個(gè)式子,解得.因此,(q-p)(r-p)+(p-q)(r-q)+(p-r)(q-r)=p2+q2+r2-pqqr-rp=(p+q+r)2-3(pq+qr+rp)=222-3×80=244,故(B)正確.

25.For how many integersnbetween 1 and 50,inclusive,isan integer?(Recall that 0!=1.)

(A)31 (B)32 (C)33 (D)34 (E)35

譯文:1到 50之間(包括)有多少個(gè)整數(shù)n使得是一個(gè)整數(shù)?(注意0!=1.)

解由于,若n為素?cái)?shù),則分子含有個(gè)因子n,而分母含有n個(gè)n,從而不是整數(shù);若n為合數(shù),先引用一個(gè)結(jié)論:一定是整數(shù).于是有,當(dāng)n≥6時(shí),也是整數(shù),從而是整數(shù);但當(dāng)n=4時(shí),不是整數(shù).因此,不符合條件的n值為2,3,4,5,7,11,13,17,19,23,29,31,37,41,43,47,共16個(gè),即剩下的34個(gè)均滿足條件,故(D)正確.