2023阿里巴巴全球数学竞赛阿里巴巴全球数学竞赛预选赛赛题及参考答案

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2023)v=ar+r3−这里r(t)表示球状闪电的半径，而t是时间变量。初始时刻，没有球状闪电，即r(0)=0。相应地，我们也有v(0)=0。而a∈R可以被人为控制，您可以通过拉动一个控制杆来迅速的改变a的值。我们给它的预设值是a=−1。”你：“做的漂亮，博士！a是我们的唯一控制方式吗？这似乎并不能把球状闪电启动起和学家：“没错，如果踢一下的话，r(t)的值就会瞬间提高ε(ε远小千1)2A设置a2,踢一下仪器，等球状闪电半径严格超过√2，再设置a123设置a3,踢一下仪器，等球状闪电半径严格超过√2，再设置a134设置a4,踢一下仪器，等球状闪电半径严格超过√2，再设置a145设置a=5,踢一下仪器，等球状闪电半径严格超过√2，再设置a=15设两个凸八面体O1O2的每个面都是三角形,且O1在O2的内部.记O1(O2)的棱长之和为当我们计算f1/f2时,可能得到以下哪个(些)值?14A与B二人进行“抽鬼牌”游戏。游戏开始时，A手中有n张两两不同的牌。B手上有n1张牌，其中n张牌与A手中的牌相同，另一张为“鬼牌”，与其他所有牌都不同。游双方交替从对方手中抽取一张牌，A先从B假设每一次抽牌从对方手上抽到任一张牌的概率都相同，请问下列n中哪个n使An=n=n=n=对所有的n，A某个城市有10条东西向的公路和10条南北向的公路，共交千100个路口.小明从某个路口驾车出发，经过每个路口恰一次，最后回到出发点.在经过每个路口时，向右转不需要等待，直行需要等待1分钟，向左转需要等待2分钟.设小明在路口等待总时间的最小可能值是S分S<50≤S<90≤S<100≤S<S≥设n2是给定正整数.考虑nn矩阵X=(ai,j)1≤i,j≤n(ai,j=0或者1)的集合证明:存在这样的X满足detX=n若2n4,证明detXnn若n2023,证明存在X使得detX>n4n,=:试问是否存在非零实数s,满足limn||(√21)ns||=试问是否存在非零实数s,满足limn||(√23)ns||=某公司要招聘一名员工，有N人报名面试。假设N位报名者所具有该职位相关的能力值两招聘委员会按随机顺序逐个面试候选人，且他们能观察到当时所见候选人的相对排名。比如委员会面试到第m位候选人时，他们拥有的信息是前m位面试者的相对排名，但不知后N−m位候选人的能力情况。−如果委员会决定给某位候选者发offer，那么这位候选者以概率p接受，以概率1p拒绝，且独立千(之前)所有其他面试者的决定。如果该候选人接受offer，那么委员会将−如果委员会决定不给某位面试者发offer反复该面试程序，直到有候选者接受offer面试完所有的N由千N位面试者的顺序是完全随机的，因此他们能力的排名在N!的可能性中是均匀分布。上程序的前提下，找到一个策略，使得招到N位候选者中能力最优者的概率最大化。−考虑如下策略。委员会先面试前m1位候选者，不管其能力排名如何，都不发工作offer。从第m位开始，一旦看到能力在所面试过候选人中的最优者，即发工作offer。如对方拒绝，则继续面试直到下一位当前最优者1出现。试证明：对千任意的N，都存在一个m=mN，使得依靠上述策略找到(所有N位候选人中)最优者的概率值，在所有可能的策略所给出的概率值中是最大的。−N假设p1。当N+∞，求N

N对一般的p(01)，当N+∞，求N

2023第1 球状闪的变化率v(t)v=ar+r3−这里r(t)表示球状闪电的半径，而t是时间变量。初始时刻，没有球状闪电，即r(0)=0。相应地，我们也有v(0)=0。而a∈R可以被人为控制，您可以通过拉动一个控制杆来迅速的改变a的值。我们给它的预设值是a=−1。”你：“做的漂亮，博士！a是我们的唯一控制方式吗？这似乎并不能把球状闪电启动起和学家：“没错，如果踢一下的话，r(t)的值就会瞬间提高ε(ε远小千1)2设置a2,踢一下仪器，等球状闪电半径严格超过√2，再设置a123设置a3,踢一下仪器，等球状闪电半径严格超过√2，再设置a134设置a4,踢一下仪器，等球状闪电半径严格超过√2，再设置a145设置a5,踢一下仪器，等球状闪电半径严格超过√2，再设置a15答案选(B)v=f(r;如果v0则r随时间增长;如果v0则r随时间下降;如果v0则r我们首先注意到f(0a0，即r0永远是一个根。但是变化率函数的非负实根数量受a取值影响。事实上，我们可以算出来f(ra)=0r1= r2=

√4a+ r3

√4a+ r4=

1+√4a+✓ r5✓

1+√4a+✓ ✓下面我们分类讨论，当a0的时候，我们有两个非负实根r1=0和r50。我们容易验证，当r∈(0r5)时v>0，但r∈(r5+∞)时v<0。千是当a>0的时候，如果我们踢一下为了使得半径严格超过√2，我们rTTIY--<r5√2。所以启动时，我们rTTIY--<a2了选项(A)当1<a<0的时候，我们有三个非负实根，从小到大依次是r1=0,r3>0和r5>0别地，r5<1且当r∈(r5+∞)时，v<0半径缩小。如果此刻r=√42到r=r5，但不会小千r5。所以此时，球状闪电不能完全消失。这样，排除了选项(D)4242径会逐步缩小直到r=r5，但不会小千r5了选项(C)a< r1= r>a< r1= r> v<4全消失。选项(B)第2题设两个凸八面体O1O2的每个面都是三角形,且O1在O2的内部.记O1(O2)和为f1(f2).当我们计算f1/f2时,可能得到以下哪个(些)值?14答案选(A)(B)(C)(D)说明：在60-70年代全苏中学生数学奥林匹克中，有过这样一个题：“四面体V14,

倍”.3维平面上一个三角形位千另一个三角形内部,那么小三角形不仅面积是严格小千大三角形的周长也是如此.而在三维情形,Holszy´ski,wlasnosciczworoscianow,Holszy´ski,定理:对千Rn中的两个m维单形S和T(前者完全位千后者的内部),和任意正整数1:::r:::m.存在常数Bm,r,使得S的所有r维面的面积之和不超过T的所有r维面的面积之和的Bm,r倍.这里Bm,r的具体数值计算如下:设m1=(r1)qs(带余除法),则qr+1−s(q+Bm,r

m+1−21,67-73.)回到本题,这里的选项(A)是平凡的,关键是Y-说明为什么(B)、(C)和(D)可以实现× ×一点点招论:如果有一个顶点引出5条棱,那么简单讨论可知必有另一个顶点也引出5条棱这个八面体的各顶点度数为(554433).除此之外,唯一可能的情形就是每个顶一点点凸几何知识:因为我们考虑的都是凸八面体,•如果大八面体的每个顶点都引出4条棱,且最大距离f在两个不相邻顶点A和B之间实现,那么因为另四个顶点与这两个顶点均相邻,所以大八面体的棱长之和至少是4f(且在另四个顶点到直线AB的距离充分小的时候可以充分接近),而对千小八面体来说,假设也是每个顶点引出4条棱,让三个顶点趋近千A,另三个趋近千B,其棱长之和会趋近千6f.这样所有小千1.5的比例均可实现.(所以有选手会选(A),(B),(C))••如果大八面体的两点间最大距离是在两个度数为3的顶点之间实现的,那么大八面体的棱长之和至少是3f(且在另四个顶点到直线AB的距离充分小的时候可以充分接近),而对千小八面体来说,仍假设每个顶点引出4条棱,让三个顶点趋近千A,另三个趋近千B,其棱长之和会趋近千6f.这样所有小千2的比例均可实现.而如果此时小八面体与大八面••作简单的分类讨论可知,如果大八面体的两顶点间最大距离f2是在一个度数为a的顶点和一个度数为b的顶点之间实现的(不管它们是否相邻),那么大八面体的各棱长度之和大千min(a,b)f,而小八面体的棱长之和显然不超过12f,所以(E)是不可能实现的.•第3题A与B二人进行“抽鬼牌”游戏。游戏开始时，A手中有n张两两不同的牌。B手双方交替从对方手中抽取一张牌，A先从B假设每一次抽牌从对方手上抽到任一张牌的概率都相同，请问下列n中哪个n使An=n=n=n=答案选(B)

3故有a1=23

a1=2+2·2 a2=3+3·3 4故有a2=34 an=n+1an−2+n+1n+1an+n+1n+1 − 其中右端第一项为A未抽中鬼牌的情况，这时B无论抽中什么都能成功配对(鬼牌在B手上)，这时A手上有n2张牌，B手上有n1张牌且A先手。右端第二项为A，B均抽中对方手上的鬼牌的情况，右端第三项为A抽中B手上的鬼牌而B没抽中手上的鬼牌的情况，而pn,n−1为A先手，手上有包含鬼牌的n张牌，B手上有不包含鬼牌的n1张牌时− pn,n−1=1−2 这是因为A无论抽到哪一张均能配对，此时变为A手上有包含鬼牌的n−1张牌，B包含鬼牌的n2张牌且为B先手，故此时B的胜率为2而A的胜率为12

an=n+1an−2+n+1n+1an+(n+1)2−(n+1)2 an=n+2an−2+n+ (n−2

)

n+=

n+

n+

an=2(n+2) n+an=2(n+2) a31=a32= a1000=第4题某个城市有10条东西向的公路和10条南北向的公路，共交千100个路口.小明从某个路口驾车出发，经过每个路口恰一次，最后回到出发点.在经过每个路口时，向右转不需Y-等待，直行需Y-等待1分钟，向左转需Y-等待2分钟.设小明在路口等待总时间的最小可能值是S分钟，则S<50≤S<90≤S<100≤S<S≥答案选(C)− − ×− 由题意知小明行驶的路线是一条不自交的闭折线.将每个路口看作一个顶点，那么他行驶的路线可以看成是一个100边形(有的内角可能是平角，也有大千平角的内角).由多边形内角和公式知这个100边形的所有内角之和为98180◦.注意内角只能是90◦，180◦和27◦，设90◦有a个，27◦有b个，那么90a+270b+180(100ab)=98180，整理得ab=4.如果小明在这条路上是顺时针行驶的，那么90◦内角对应右转，180◦内角对应直行，270◦对应左转，他在路口等待的总时间是(100ab)+b=100(ab)=96(min);如果小明在这条路上是逆时针行驶的，那么90◦内角对应左转，180◦内角对应直行，270◦内角对应右转，他在路口等待的总时间是(100ab)+2a=100+(ab)=− − ×− 注：如果小明的起点/终点处的转弯时间不计，那么等待的总时间还可以减少2分钟(个左转的位置作为起点)，这样S94，但不影响选择的选项第5题设n2是给定正整数.考虑nn矩阵Xjj(ai,j0或者1)的集合证明:存在这样的X满足detX=n若2n4,证明detXnn若n2023,证明存在X使得detXn4n−−−若X有一行全为0或者有两行相等则detX0;若X有一行只有一个1,则可约化到(n1)阶矩阵的情形;若X有一行全为1,还有一行有n1个1,则可约化到有一行只有一个1的情形,进一步约化到(n1)阶矩阵的情形.若以上都不发生,则X的各行有很少的可能性,我们可以−−−取XI=jjnai,j=1− 1≤i,j≤则detXI−1)n−1(n1).若n是奇数-<XXI.若n是偶数-<X为调换XI得矩阵.则detX=n (3)当n2k1时 − −Y 则Y是元素为±1的(n1)(n1)矩阵,detY=(√2k)2k=2k2k−1、比,,注意Y的最后一行为αn+1=(11).记ti=±1为Y的第i行的最后一个元素(1i、比,, βi=2(tiαi−.XI=(β1,...,记则XI是元素为0,1的nn矩阵,

t

ti=detXI=k11×若有必Y-,则调换XI的最后两行,可以得到一个元素为0,1的n n矩阵,满足detX=k11×不妨设2k1n2k+11.当2k1n3·2k−1且n2023时则k11.存在元素为0,1的n×n矩阵X,满足

detX≥kk1>kk k+1 k

n4< = n这样detXn4n

(k−2)2k−1≥3(k+1)·当3·2k−1n2k+11且n2023时则k10.存在元素为0,1的nn矩阵X,detX≥kk11kk2>kk

k+1 k

n4< = n这样detXn4n

(3k−7)2k−2>(k+1)·试问是否存在非零实数s,满足limn→∞||(√21)ns||=试问是否存在非零实数s,满足limn→∞||(√23)ns||=n存在，取s=1即可。设(√2+1)n=xn+√2yn,则(−√2+1)n=xn−√2yn.nnn√n而x22y2−1)n.由此nn√n

+

|=

−x|=|2y2−x2

→nnnnnn不存在。反证法，假设s满足(√23)ns=mn+n其中limn→∞En=0.记α=√23¯=nn−√23.

=∞m区n1− 区n

xn+∞区n区n

(1αx)(1¯x)16x7x2,上式两边乘以16x7x2s(1−¯x)=(1−6x+7x2)

mxn+(1−6x+7x2)nn

设(16x7x2)=0mnxn==0pnxn(16x7x2)=0Enxn==0ηnxn,则pn∈Zlimnηn=0.因为(9)左边是一次式，从而右边满足pnηn=0n≥2.n充分大时ηn很小，所以必有pn=η ∞

xn= .n1−6x+n右边写成部分分式形如H(x)+A+B.因为limn→∞En=0, 为1，而α¯均大千1，所以必须A=B=0.这样当n充分大时，En=0,从而(23)smnZ,第7题某公司Y-招聘一名员工，有N人报名面试。假设N位报名者所具有该职位相关的招聘委员会按随机顺序逐个面试候选人，且他们能观察到当时所见候选人的相对排名。比如委员会面试到第m位候选人时，他们拥有的信息是前m位面试者的相对排名，但不知后N−m位候选人的能力情况。−如果委员会决定给某位候选者发offer，那么这位候选者以概率p接受，以概率1p拒绝，且独立千(之前)所有其他面试者的决定。如果该候选人接受offer，那么委员会将−如果委员会决定不给某位面试者发offer反复该面试程序，直到有候选者接受offer面试完所有的N由千N位面试者的顺序是完全随机的，因此他们能力的排名在N!的可能性中是均匀分布。上程序的前提下，找到一个策略，使得招到N位候选者中能力最优者的概率最大化。−考虑如下策略。委员会先面试前m1位候选者，不管其能力排名如何，都不发工作offer。从第m位开始，一旦看到能力在所面试过候选人中的最优者，即发工作offer。如对方拒绝，则继续面试直到下一位当前最优者1出现。试证明：对千任意的N，都存在一个m=mN，使得依靠上述策略找到(所有N位候选人中)最优者的概率值，在所有可能的策略所给出的概率值中是最大的。−N假设p=1。当N→+∞，求N

N对一般的p(01)，当N+∞，求N

N，我们-<Zk为委员会完全略过前k Zk≥Zk+1如果委员会面试了第k位候选人，且其能力在前kYk=N+(1−p)Zk+1N+(1−p)Zk+1≥Zk+1 NNN即k≥Zk+12.由千{k}k递增，而{Zk}k递减，且Zk≤N−k+1，易见不等式(10)NNN 对某一个 1成立。由此可知，最优策略可以通过选择某个m ≥足不等式(10)的k中的最小值。此外，如果k=m满足不等式(10)，则任意的 ≥-<pm为委员会采取(a)中的策略，且找到能力最高者的概率。当p=1时，被发NN的不相交并集，其中Ak对应的事件为第k位候选人是N了。这样的话，事件Ak1P(Ak)=

m−1·k−·其中

对应的是这位候选者是能力最高者的概率，

是他/ 前k前NmNk− =m−1mNk−易见pm先曾后减，因此其最优值m∗=mNpm≥NN区区≤k=m+1k−的最小m。当N很大时，由左端的近似逼近为log(N/m可知，N

→e11NAk其中Ak对应的事件为第k位候选人是N人中的能力最高者、被面试了、并且接受ppmN

区qk区2如果这个不等式不满足，则略过第k个人，且从第k1其中qk为假定第k位候选人为N人中能力最高者后，他/q=(m−1+1−p)(m+1−p)···(k−2+1−p)=Γ(m)Γ(k−p)

m+

m+

k−

k−

Γ(k)Γ(m−其中Γ为经典的Γ函数。据此，我们得到如果从第m位开始，委员会找到(所有NNmN =p·Γ(m)NmNpm对千m先增后减。由Γ函数的近似及积分对求和的逼近，我们计算得，当N大时，让pm最大化的mN→ p11−p.→N当p1时，极限为PAGEPAGE12023AlibabaGlobalMathematicsBallAsachiefofficerofasecretmission,youhadthefollowingconversationwiththeleadingScientist:”Chief,wehavemasteredthecontrollawofballlightning.Wefoundthattheratev=ar+r3−Herer(t)representstheradiusofballlightning,andtisthetimevariable.Attheinitialmoment,thereisnoballlightning,thatis,r(0)=0.Accordingly,wealsohavev(0)=0.Anda∈Rcanbeartificiallycontrolled.Youcanquicklychangethevalueofabypullingacontrollever.Wesetitspresetvaluetoa=−1.”You:”Welldone,Doctor!Isaouronlywayofcontrol?Itdoesntseemtobeabletostarttheballlightning.”Scientist:”You’reright,Chief.Wedohaveanotherwayofcontrol,whichistokicktheYou:”Doctor,areyoukiddingme?KickScientist:”Yes,ifyoukickit,thevalueofr(t)willinstantlyincreasebyε(εismuchsmallerthan1).”You:”Isee.That’shelpulindeed.Ourtestgoaltodayistostarttheballlightning,itsradiusstrictly 2,andthenletitgraduallydisappearWhatdoyouthinkoftheseschemes,Chief?”Youlookedattheseoptionsandfoundthatthefeasibleschemesare(2Seta=2,kicktheinstrument,waitfortheballlightningradiustostrictlyexceed√2,thenseta=−1;2Seta=3,kicktheinstrument,waitfortheballlightningradiustostrictlyexceed3thenseta=−13Seta=4,kicktheinstrument,waitfortheballlightningradiustostrictlyexceed4thenseta=−14Seta=5,kicktheinstrument,waitfortheballlightningradiustostrictlyexceed5thenseta=−15LetO1,O2betwoconvexoctahedronwhosefacesarealltriangles,andO1isinsideO2.LetthesumofedgekengthsofO1(resp.O2)be€1(resp.€2).Whenwecalculate€1/€2,whichvalue(s)amongthefollowingcanbeobtained?(MultipleChoice)14Twoplayers,AandB,playagamecalled“drawthejokercard”.Inthebeginning,PlayerAhasndifferentcardsPlayerBhasn+1cards,nofwhicharethesamewiththencardsinPlayerA’shand,andtherestoneisaJoker(differentfromallotherncards).TherulesPlayerAfirstdrawsacardfromPlayerB,andthenPlayerBdrawsacardfromPlayerA,andthenthetwoplayerstaketurnstodrawacardfromtheotherplayer.ifthecardthatoneplayerdrewfromtheotheronecoincideswithoneofthecardsonhis/herownhand,thenthisplayerwillneedtotakeoutthesetwoidenticalcardsanddiscardthem.whenthereisonlyonecardleft(necessarilytheJoker),theplayerwhoholdsthatcardlosesthegame.Assumeforeachdraw,theprobabilityofdrawinganyofthecardsfromtheotherplayeristhesame.WhichninthefollowingmaximisesPlayerA’schanceofwinningthegame?n=n=n=n=Forallchoicesofn,AhasthesamechanceofThereare10horizontalroadsand10verticalroadsinacity,andtheyintersectat100crossings.Bobdrivesfromonecrossing,passeseverycrossingexactlyonce,andreturntotheoriginalcrossing.Ateverycrossing,thereisnowaittoturnright,1minutewaittogostraight,and2minuteswaittoturnleft.LetSbetheminimumnumberoftotalminutesonwaitingatthecrossings,thenS<50≤S<90≤S<100≤S<S≥Letn≥2beagivenpositiveinteger.Considerthesetofn×nmatricesX=(ai,j)1≤i,j≤nwithentries0and1.showthat:thereexistssuchanXwithdetX=n−when2≤n≤4,showthatdetX≤n−nWhenn≥2023,showthatthereexistsanXwithdetX>n4Forarealnumberr,set||r||=min{|r−n|:n∈Z},where|·|meanstheabsolutevalueofarealnumber.Isthereanonzerorealnumbers,suchthatlimn→∞||(√2+1)ns||=Isthereanonzerorealnumbers,suchthatlimn→∞||(√2+3)ns||=Acompanyhasoneopenpositionavailable,andNcandidatesapplied(Nisknown).AssumetheNcandidates’abilitiesforthispositionarealldifferentfromeachother(inotherwords,thereisanon-ambiguousrankingamongtheNcandidates),andthehiringcommitteecanrankingsarefaithfulwithrespecttothecandidates’trueabilities.ThehiringcommitteedecidesthefollowingruletoselectonecandidatefromN:Thecommitteeinterviewsthecandidatesonebyone,atacompletelyrandomorder.Theyobserveinformationoncandidates’relativerankingregardingtheirabilitiesfortheposition.Theonlyinformationavailabletothemafterinterviewingmcandidatesistherelativerankingamongthesempeople.ornot.Iftheydecidetoofferthepositiontothecandidatejustinterviewed,thenthecandi-datewillacceptthejobwithprobabilityp,anddeclinetheofferwithprobability1−p,independentlywithallothercandidates.Iftheselectedcandidateacceptstheoffer,thenhe/shegetsthejob,andthecommitteestopsinterviewingtheremainingcandi-dates.Ifhe/shedeclinestheoffer,thenthecommitteeproceedtointerviewingthenextIftheydecidenottoofferthepositiontothecandidatejustinterviewed,thentheyproceedtointerviewingthenextcandidate,andtheycannotturnbacktopreviouslyinterviewedcandidatesanymore.Thecommitteecontinuesthisprocessuntilacandidateisselectedandacceptsthejob,oruntiltheyfinishinterviewingallNcandidatesifthepositionhasnotbeenfilledbefore,whichevercomesfirst.Sincetheintervieworderofthecandidatesarecompletelyrandom,eachrankinghasequalprobabilityamongtheN!possibilities.Thecommittee’smissionistomaximisetheproba-bilityofgettingthecandidatewiththehighestranking(amongNcandidates)forthejobconstrainedtotheaboveselectionprocess.HerearetheFix1≤m≤N,andconsiderthefollowingstrategy.Thecommitteeinterviewsthefirstm−1candidates,anddonotgiveoffertoanyofthemregardlessoftheirrel-ativerankings.Startingfromthem-thcandidate,thecommitteeoffershim/herthepositionwheneverthecandidate’srelativerankingisthehighestamongallpreviouslyinterviewedcandidates.Ifhe/shedeclinestheoffer,thenthecommitteecontinuestheinterviewuntilthenextrelativelybestcandidate1,andthenrepeattheprocesswhenShowthatforeveryN,thereexistsm=mNsuchthattheabovestrategymaximisestheprobabilityofgettingthebestcandidateamongallpossiblestrategies.1“Relativelybestcandidate”referstothecandidatewiththehighestabilityamongallcandidateswhohavebeeninterviewed(includingthosewhoareofferedthepositionanddeclined).NSupposep=1.WhatisthelimitofN

asN→NForp∈(0,1),whatisthelimitofN

asN→2023AlibabaGlobalMathematics1BallAsachiefofficerofasecretmission,youhadthefollowingconversationwiththeleadingScientist:”Chief,wehavemasteredthecontrollawofballlightning.Wefoundthattheratev=ar+r3−Herer(t)representstheradiusofballlightning,andtisthetimevariable.Attheinitialmoment,thereisnoballlightning,thatis,r(0)=0.Accordingly,wealsohavev(0)=0.Anda∈Rcanbeartificiallycontrolled.Youcanquicklychangethevalueofabypullingacontrollever.Wesetitspresetvaluetoa=−1.”You:”Welldone,Doctor!Isaouronlywayofcontrol?Itdoesntseemtobeabletostarttheballlightning.”Scientist:”You’reright,Chief.Wedohaveanotherwayofcontrol,whichistokicktheYou:”Doctor,areyoukiddingme?KickScientist:”Yes,ifyoukickit,thevalueofr(t)willinstantlyincreasebyε(εismuchsmallerthan1).”You:”Isee.That’shelpulindeed.Ourtestgoaltodayistostarttheballlightning,itsradiusstrictly 2,andthenletitgraduallydisappearWhatdoyouthinkoftheseschemes,Chief?”Youlookedattheseoptionsandfoundthatthefeasibleschemesare(2Seta=2,kicktheinstrument,waitfortheballlightningradiustostrictlyexceed√2,thenseta=−1;2Seta=3,kicktheinstrument,waitfortheballlightningradiustostrictlyexceed3thenseta=−13Seta=4,kicktheinstrument,waitfortheballlightningradiustostrictlyexceed4thenseta=−14Seta=5,kicktheinstrument,waitfortheballlightningradiustostrictlyexceed5thenseta=−15AnswerTheanswerisv=f(r;Whenv>0,risincreasingintime.Whenv<0,risdecreasingintime.Whenv=0,Wecanfindalltherootsoff(r,a)=0,whichwelistinther1= r2=

r3

r4=

1+√4a+✓ r5✓

1+√4a+✓ ✓Whena>0,wehavetwononnegativeroots:r1=0andr5>0.Clearly,whenr∈(0,r5),v>0;andwhenr∈(r5,+∞),v<0.Thus,whena>0andifwekicktheinstrument,wecanstarttheballlightening,anditsradiuswillgrowtor5(butitwillnotexceedr5).Tomaketheradiusexceed√2,weneedr5>√2.Thismeansinthestartingphase,wea>2,andthusScheme(A)4When−1<a<0,wehavethreenonnegativeroots,whichsatisfy0=r1<r3<r5.4particular,wehaver5<1andwhenr∈(r5,+∞),v<0.Thismeans,ifwestartr 2,theradiusisgettingsmaller,butitwillnotbecomesmallerthanr5.Therefore,Whena=−1,similartothepreviouscase,theradiuswillnotbesmallerthanr5=√1, 4Whena<−1,wehaveonlyonenonnegativerootr1=0.Whenr>0,wealways4v<0,andthustheballlighteningwillvanishcompletely.ThismeansScheme(B)LetO1,O2betwoconvexoctahedronwhosefacesarealltriangles,andO1isinsideO2.LetthesumofedgekengthsofO1(resp.O2)be1£2).Whenwecalculate12whichvalue(s)amongthefollowingcanbeobtained?(MultipleChoice)142AnswerTheansweris(A)(B)(C)CommentsInthe60’s-70’s,thefollowingquestionappearedinAll-UnionMathOlympiadofUSSR:AtetradehronV1sitsinsideanothertetrahedronV2,provethatthesumof4lengthsofV1doesnotexceed3timesthatofV2.Whatisanti-intuitiveisthat,onaifatrianglesitsinsideanothertriangle,thennotonlytheareaofthefirsttriangleissmallerthanthatofthesecondone,buttheperimeteralsois.Nowinathreedimensionalsituation,thoughthe“order”ofvolumeandsurfaceisstillkept,itisnotthecaseforthesumofedgelengths.The“origine”oftheproblemislikelythefollowingpaperinHolszy´ski,wlasnosciczwoscianow,atyczne6(1962),14-16.Holszy´ski,4042.Thenin1986,CarlLinderholmoftheUniversityofAlabamageneralizedtheaboveresulttohigherdimensionalEuclideanspaces:second,and1�r�m.ThethereexistsconstantsBm,r,suchthatthesumofalldimensionalfacesofSdoesnotexceedBm,rtimesthatofT.HereBm,riscalculatedasfollows:Letm+1=(r+1)q+s(Euclideandivision),thenqr+1−s(q+Bm,r

m+1−(CARLLINDERHOLM,ANINEQUALITYFORSIMPLICES,GeometriaeDedicata(1986)21,67-73.)Nowbacktothecurrentproblem,theChoice(A)istrivial,sowefocuswhy(B),(C)and(D)canbewhy(E)Themathematicsthatweneedhereedges,sobyEuler’sFormula,thenumberofverticesis6.abitofgraphtheory:ifonevertexhasdegree5,thenbyaveryeasyargumentonehasanothervertexwithdegree5also,andthedegreesoftheverticesare(5,5,4,4,3,3).Theonlyotherpossibilityisthateveryvertexhasdegree4(likethatofaregularalittlebitofconvexgeometry:asweconsiderconvexoctahedron,sothemaximumdistanceoftwopointsonitmustbeattainedbetweentwovertices.Ifeveryvertexofthebigoctahedronisofdegree4,andthemaximumdistance£isrealizedbetweentwoverticesAandBthatareNOTadjacent,thenastheotherfourverticesarealladjacenttothem,so£2isatleast4£2(andcanbearbitrarilyclosetothatvalurwhentheotherfourverticesarecloseenoughtolineAB),andforthesmalloctahedron,ifeveryvertexisofdegree4,wecanmakethreeverticesveryclosetoA,whiletheotherthreeveryclosetoB,so£1wouldbeverycloseto6£2.Henceanyratiolessthan1.5isrealizable.(sotheChoices(A),(B)and(C))Ifthemaximumdistance£isrealizedbetweentwoverticesofdegree3inthebigoctahedron,then£2isatleast3£2(andcanbearbitrarilyclosetothatvalurwhentheotherfourverticesarecloseenoughtolineAB),whileforthesmalloctahedron,wecanstilltakeeachvertextobeofdegree4,andthreeofthemveryclosetoA,whiletheotherthreeveryclosetoB,so£1wouldbeverycloseto6£2.Henceanyratiolessthan2isrealizable.(sotheChoice(D))Actually,ifthesmalloctahedronhasthesometopologicalconfigurationasthatofthebigone,andthetwoverticesofdegree5areveryclosetoeachother,whiletheotherfourverticesareveryclosetogether,thentheratiocanactuallyapproach8/3.Aftersomeeasycasebycasediscussion,weconcludethat,ifthemaximum£isrealizedbetweenavertexofdegreeaandavertexofdegreeb(whetherthey12£2,So(E)isimpossible.3Twoplayers,AandB,playagamecalled“drawthejokercard”.Inthebeginning,PlayerAhasndifferentcards.PlayerBhasn+1cards,nofwhicharethesamewiththencardsinPlayerA’shand,andtherestoneisaJoker(differentfromallotherncards).TherulesPlayerAfirstdrawsacardfromPlayerB,andthenPlayerBdrawsacardfromPlayerA,andthenthetwoplayerstaketurnstodrawacardfromtheotherplayer.ifthecardthatoneplayerdrewfromtheotheronecoincideswithoneofthecardsonhis/herownhand,thenthisplayerwillneedtotakeoutthesetwoidenticalcardsanddiscardthem.whenthereisonlyonecardleft(necessarilytheJoker),theplayerwhoholdsthatcardlosesthegame.Assumeforeachdraw,theprobabilityofdrawinganyofthecardsfromtheotherplayeristhesame.WhichninthefollowingmaximisesPlayerA’schanceofwinningthegame?n=n=n=n=Forallchoicesofn,Ahasthesamechanceof3AnswerTheanswerisSowehave a1=2+2·2 3Therefore,a1=2.Inaddition,we3 4soweconcludethata2=34

a2=3+3·3 Actually,wecanobtainthefollowinginduction an=n+1an−2+n+1n+1an+n+1n+1 wherethefirsttermontheRHSisthescenariowhenAdoesnotdrawthejokercardfromB.Inthiscase,nomatterwhichcardBdrawsfromA,thiscardwouldmatchoneofthecardsthatBhasinhishand(becauseBholdsthejokercard).ThenAwillhaven−2cardsandBhasn−1cards,withAdrawingfromBfirstandBholdingthejokercard.ThetermontheRHSisthescenariowhenAfirstdrawsthejokercardfromB,andthenBdrawsthejokercardfromA.ThethirdtermontheRHSisthescenariowhenAdrawsthejokercardfromBbutBdoesnotdrawthejokercardfromA,andpn,n−1istheprobabilityforAtowinthegamewhenAdrawsfirstwithncardsincludingajokercard,Bdrawsnextwithn−1cardsthatdonotincludethejokercard.Wehavepn,n−1=1−2 becausenomatterwhichcardAdrawsfromB,AwouldhaveonecardinhandthatmatchthisdrawncardfromB(becausethejokercardisinA’shand).Therefore,afterA’sdrawing,Awillhaven−1cardsincludingthejokercard,Bwillhaven−2cardswithoutthejokercard,andBdrawsfirst.Inthiscase,theprobabilityforBtowinwillbe2sowepn,n−1=1− an=n+1an−2+n+1n+1an+(n+1)2−(n+1)2an−2, andwecansimplifytheaboveequationto an=n+2an−2+n+ (n−2

)

n+ =

n+Byinduction,ifnisanoddnumber,n+an=2(n+2) Ontheotherhand,ifnisanevennumber,thenbyinductionwen+an=2(n+2) a31=a32= a1000=501Sothecorrectansweris(B),andn=32initialcardswillgiveAthebiggestchanceof4Thereare10horizontalroadsand10verticalroadsinacity,andtheyintersectat100crossings.Bobdrivesfromonecrossing,passeseverycrossingexactlyonce,andreturntotheoriginalcrossing.Ateverycrossing,thereisnowaittoturnright,1minutewaittogostraight,and2minuteswaittoturnleft.LetSbetheminimumnumberoftotalminutesonwaitingatthecrossings,thenS<50≤S<90≤S<100≤S<S≥4AnswerTheanswerisasavertex,thentherouteisregardedasa100-gon.Aninterioranglemaybegreaterthanorequaltoastraightangle..Bytheformulaofthesumoftheanglesofthepolygon,thesumofallinterioranglesis98×180◦.Notethattheinterioranglecanonlybe90◦,180◦98×180,soa−b=4.IfBobdrivesclockwise,then90◦,180◦and270◦corrspondstoturnright,gostraightandturnleft,respectively.Thetotaltimeonwaitingatthecrossingsiscorrspondstoturnleft,gostraightandturnright,respectively.Thetotaltimeonwaitingatthecrossingsis(100−a−b)+2a=100+(a−b)=104(min).Therefore,S=96,and(C)iscorrect.Note:Ifweignorethewaitingtimeonthebeginning/endingcrossing,thetotaltimeonwehavethatS=94,butdonotaffectthecorrectchoice.5Letn≥2beagivenpositiveinteger.Considerthesetofn×nmatricesX=jjwithentries0and1.showthat:thereexistssuchanXwithdetX=n−when2≤n≤4,showthatdetX≤n−nWhenn≥2023,showthatthereexistsanXwithdetX>n45IfXhasazerorowortwoequalrows,thendetX=0;ifXhasarowwithonlyonenonzeroentry,itreducesto(n−1)matrixcase;ifXhasarowwithnnonzeroentriesandarowwith(n−1)nonzeroentries,itreducestothecasethatXhasarowwithonlyonenonzeroentryandfurtherreducesto(n−1)matrixcase.Whentheaboveallnothappen,thenrowsofXhavefewpossibilitiesandonecouldtakeacasebycaseverification.takeX,=jjai,j=1− 1≤i,j≤Then,detX,=(−1)n−1(n−1).Ifnisodd,letX=X,.Ifniseven,getXbyswitchingthefirsttworowsofX,.Then,detX=n−1. whenn=2k−1, − −Y Then,Yisan(n+1)×(n+1)matrixwithentries±1anddetY=(√2k)2k=2k2k−1 ThelastrowofYisequalto =(1,...,1).Writet 1forthelastentryof 1rowαiofY(1≤i≤n).

,1,βi=2(tiαi−Removingthelastentryofβi,(whichis0),wegetannrowvectorβi.X,=(β1,...,and

t

ti=Then,X,isann×nmatrixwithentries0and1andwedetX,=k11×SwitchingthefirsttworowsofX,ifnecessary,wegetann suchthatdetX=2(k−2)2k−1+1.×Assumethat2k−1≤n<2k+1−1.When2k−1≤n<3·2k−1andn≥2023,wek≥11.Thereexistsann×nmatrixXwithentries0,1suchdetX≥kk1>kkWe

k+1 kDuetok≥11,we

n4< = nThen,detX>n4

(k−2)2k−1≥3(k+1)·When3·2k−1≤n<2k+1−1andn≥2023,wehavek≥10.Thereexistsann×nXwithentries0,1suchdetX≥kk11kk2>kkWe

k+1 kDuetok≥10,we

n4< = nThen,detX>n4

(3k−7)2k−2>(k+1)·6Forarealnumberr,set||r||=min{|r−n|:n∈Z},where|·|meanstheabsolutevalueofarealnumber.Isthereanonzerorealnumbers,suchthatlimn→∞||(√2+1)ns||=Isthereanonzerorealnumbers,suchthatlimn→∞||(√2+3)ns||=6tattheproper. xn,yn∈Z.Then(−2+1)n=xn 2ynandx2−2y2=(−1)n.Itfollowsthat|xn |2y2−x22yn−2xn|=|2yn−xn|=√ n→No.number=that(2+3)ns=mn+En,wherelimn→∞En=0.Denoteα 2+3,¯=−2+3.

=∞1−

mxn+nn

Since(1−αx)(1−¯x)=1−6x+7x2,multiplyingbothsidesoftheaboveequationby1−6x+7x2wegets(1−¯x)=(1−6x+7x2)

mxn+(1−6x+7x2)nn

=0=0Denote(16x7x2)=0mnxn==0pnxn(16x7x2)=0Enxn==0ηnxn,wherepn∈Zlimnηn=0.Becausethelefthand =0=0∞

Exn= .n1−6x+nWritetherighthandsideasH(x)+A+B,whereH(x)isapolynomialandA,B

n→∞En=0,theradiusofconvergenceofthepowerseriesintheleftsideisatleast1.Whileαand¯arelargerthan1AandBmustbezero.HenceEn=0largen.Itfollowsthat(2+3)ns=mn∈Zforlargen.It’sa7Acompanyhasoneopenpositionavailable,andNcandidatesapplied(Nisknown).AssumetheNcandidates’abilitiesforthispositionarealldifferentfromeachother(inotherwords,thereisanon-ambiguousrankingamongtheNcandidates),andthehiringcommitteecanobservethefullrelativerankingofallthecandidatestheyhaveinterviewed,andtheirobservedrankingsarefaithfulwithrespecttothecandidates’trueabilities.ThehiringcommitteedecidesthefollowingruletoselectonecandidatefromN:Thecommitteeinterviewsthecandidatesonebyone,atacompletelyrandomorder.Theyobserveinformationoncandidates’relativerankingregardingtheirabilitiesfortheposition.Theonlyinformationavailabletothemafterinterviewingmcandidatesistherelativerankingamongthesempeople.ornot.Iftheydecidetoofferthepositiontothecandidatejustinterviewed,thenthec