2023阿里巴巴全球數學競賽阿里巴巴全球數學競賽預選賽賽題及參考答案

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2023)v=ar+r3?這里r(t)表示球狀閃電的半徑，而t是時間變量。初始時刻，沒有球狀閃電，即r(0)=0。相應地，我們也有v(0)=0。而a∈R可以被人為控制，您可以通過拉動一個控制桿來迅速的改變a的值。我們給它的預設值是a=?1?！蹦悖骸白龅钠粒┦?！a是我們的唯一控制方式嗎？這似乎并不能把球狀閃電啟動起和學家：“沒錯，如果踢一下的話，r(t)的值就會瞬間提高ε(ε遠小千1)2A設置a2,踢一下儀器，等球狀閃電半徑嚴格超過√2，再設置a123設置a3,踢一下儀器，等球狀閃電半徑嚴格超過√2，再設置a134設置a4,踢一下儀器，等球狀閃電半徑嚴格超過√2，再設置a145設置a=5,踢一下儀器，等球狀閃電半徑嚴格超過√2，再設置a=15設兩個凸八面體O1O2的每個面都是三角形,且O1在O2的內部.記O1(O2)的棱長之和為當我們計算f1/f2時,可能得到以下哪個(些)值?14A與B二人進行“抽鬼牌”游戲。游戲開始時，A手中有n張兩兩不同的牌。B手上有n1張牌，其中n張牌與A手中的牌相同，另一張為“鬼牌”，與其他所有牌都不同。游雙方交替從對方手中抽取一張牌，A先從B假設每一次抽牌從對方手上抽到任一張牌的概率都相同，請問下列n中哪個n使An=n=n=n=對所有的n，A某個城市有10條東西向的公路和10條南北向的公路，共交千100個路口.小明從某個路口駕車出發(fā)，經過每個路口恰一次，最后回到出發(fā)點.在經過每個路口時，向右轉不需要等待，直行需要等待1分鐘，向左轉需要等待2分鐘.設小明在路口等待總時間的最小可能值是S分S<50≤S<90≤S<100≤S<S≥設n2是給定正整數.考慮nn矩陣X=(ai,j)1≤i,j≤n(ai,j=0或者1)的集合證明:存在這樣的X滿足detX=n若2n4,證明detXnn若n2023,證明存在X使得detX>n4n,=:試問是否存在非零實數s,滿足limn||(√21)ns||=試問是否存在非零實數s,滿足limn||(√23)ns||=某公司要招聘一名員工，有N人報名面試。假設N位報名者所具有該職位相關的能力值兩招聘委員會按隨機順序逐個面試候選人，且他們能觀察到當時所見候選人的相對排名。比如委員會面試到第m位候選人時，他們擁有的信息是前m位面試者的相對排名，但不知后N?m位候選人的能力情況。?如果委員會決定給某位候選者發(fā)offer，那么這位候選者以概率p接受，以概率1p拒絕，且獨立千(之前)所有其他面試者的決定。如果該候選人接受offer，那么委員會將?如果委員會決定不給某位面試者發(fā)offer反復該面試程序，直到有候選者接受offer面試完所有的N由千N位面試者的順序是完全隨機的，因此他們能力的排名在N!的可能性中是均勻分布。上程序的前提下，找到一個策略，使得招到N位候選者中能力最優(yōu)者的概率最大化。?考慮如下策略。委員會先面試前m1位候選者，不管其能力排名如何，都不發(fā)工作offer。從第m位開始，一旦看到能力在所面試過候選人中的最優(yōu)者，即發(fā)工作offer。如對方拒絕，則繼續(xù)面試直到下一位當前最優(yōu)者1出現。試證明：對千任意的N，都存在一個m=mN，使得依靠上述策略找到(所有N位候選人中)最優(yōu)者的概率值，在所有可能的策略所給出的概率值中是最大的。?N假設p1。當N+∞，求N

N對一般的p(01)，當N+∞，求N

2023第1 球狀閃的變化率v(t)v=ar+r3?這里r(t)表示球狀閃電的半徑，而t是時間變量。初始時刻，沒有球狀閃電，即r(0)=0。相應地，我們也有v(0)=0。而a∈R可以被人為控制，您可以通過拉動一個控制桿來迅速的改變a的值。我們給它的預設值是a=?1?！蹦悖骸白龅钠粒┦?！a是我們的唯一控制方式嗎？這似乎并不能把球狀閃電啟動起和學家：“沒錯，如果踢一下的話，r(t)的值就會瞬間提高ε(ε遠小千1)2設置a2,踢一下儀器，等球狀閃電半徑嚴格超過√2，再設置a123設置a3,踢一下儀器，等球狀閃電半徑嚴格超過√2，再設置a134設置a4,踢一下儀器，等球狀閃電半徑嚴格超過√2，再設置a145設置a5,踢一下儀器，等球狀閃電半徑嚴格超過√2，再設置a15答案選(B)v=f(r;如果v0則r隨時間增長;如果v0則r隨時間下降;如果v0則r我們首先注意到f(0a0，即r0永遠是一個根。但是變化率函數的非負實根數量受a取值影響。事實上，我們可以算出來f(ra)=0r1= r2=

√4a+ r3

√4a+ r4=

1+√4a+? r5?

1+√4a+? ?下面我們分類討論，當a0的時候，我們有兩個非負實根r1=0和r50。我們容易驗證，當r∈(0r5)時v>0，但r∈(r5+∞)時v<0。千是當a>0的時候，如果我們踢一下為了使得半徑嚴格超過√2，我們rTTIY--<r5√2。所以啟動時，我們rTTIY--<a2了選項(A)當1<a<0的時候，我們有三個非負實根，從小到大依次是r1=0,r3>0和r5>0別地，r5<1且當r∈(r5+∞)時，v<0半徑縮小。如果此刻r=√42到r=r5，但不會小千r5。所以此時，球狀閃電不能完全消失。這樣，排除了選項(D)4242徑會逐步縮小直到r=r5，但不會小千r5了選項(C)a< r1= r>a< r1= r> v<4全消失。選項(B)第2題設兩個凸八面體O1O2的每個面都是三角形,且O1在O2的內部.記O1(O2)和為f1(f2).當我們計算f1/f2時,可能得到以下哪個(些)值?14答案選(A)(B)(C)(D)說明：在60-70年代全蘇中學生數學奧林匹克中，有過這樣一個題：“四面體V14,

倍”.3維平面上一個三角形位千另一個三角形內部,那么小三角形不僅面積是嚴格小千大三角形的周長也是如此.而在三維情形,Holszy′ski,wlasnosciczworoscianow,Holszy′ski,定理:對千Rn中的兩個m維單形S和T(前者完全位千后者的內部),和任意正整數1:::r:::m.存在常數Bm,r,使得S的所有r維面的面積之和不超過T的所有r維面的面積之和的Bm,r倍.這里Bm,r的具體數值計算如下:設m1=(r1)qs(帶余除法),則qr+1?s(q+Bm,r

m+1?21,67-73.)回到本題,這里的選項(A)是平凡的,關鍵是Y-說明為什么(B)、(C)和(D)可以實現× ×一點點招論:如果有一個頂點引出5條棱,那么簡單討論可知必有另一個頂點也引出5條棱這個八面體的各頂點度數為(554433).除此之外,唯一可能的情形就是每個頂一點點凸幾何知識:因為我們考慮的都是凸八面體,?如果大八面體的每個頂點都引出4條棱,且最大距離f在兩個不相鄰頂點A和B之間實現,那么因為另四個頂點與這兩個頂點均相鄰,所以大八面體的棱長之和至少是4f(且在另四個頂點到直線AB的距離充分小的時候可以充分接近),而對千小八面體來說,假設也是每個頂點引出4條棱,讓三個頂點趨近千A,另三個趨近千B,其棱長之和會趨近千6f.這樣所有小千1.5的比例均可實現.(所以有選手會選(A),(B),(C))??如果大八面體的兩點間最大距離是在兩個度數為3的頂點之間實現的,那么大八面體的棱長之和至少是3f(且在另四個頂點到直線AB的距離充分小的時候可以充分接近),而對千小八面體來說,仍假設每個頂點引出4條棱,讓三個頂點趨近千A,另三個趨近千B,其棱長之和會趨近千6f.這樣所有小千2的比例均可實現.而如果此時小八面體與大八面??作簡單的分類討論可知,如果大八面體的兩頂點間最大距離f2是在一個度數為a的頂點和一個度數為b的頂點之間實現的(不管它們是否相鄰),那么大八面體的各棱長度之和大千min(a,b)f,而小八面體的棱長之和顯然不超過12f,所以(E)是不可能實現的.?第3題A與B二人進行“抽鬼牌”游戲。游戲開始時，A手中有n張兩兩不同的牌。B手雙方交替從對方手中抽取一張牌，A先從B假設每一次抽牌從對方手上抽到任一張牌的概率都相同，請問下列n中哪個n使An=n=n=n=答案選(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 ? 其中右端第一項為A未抽中鬼牌的情況，這時B無論抽中什么都能成功配對(鬼牌在B手上)，這時A手上有n2張牌，B手上有n1張牌且A先手。右端第二項為A，B均抽中對方手上的鬼牌的情況，右端第三項為A抽中B手上的鬼牌而B沒抽中手上的鬼牌的情況，而pn,n?1為A先手，手上有包含鬼牌的n張牌，B手上有不包含鬼牌的n1張牌時? pn,n?1=1?2 這是因為A無論抽到哪一張均能配對，此時變?yōu)锳手上有包含鬼牌的n?1張牌，B包含鬼牌的n2張牌且為B先手，故此時B的勝率為2而A的勝率為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題某個城市有10條東西向的公路和10條南北向的公路，共交千100個路口.小明從某個路口駕車出發(fā)，經過每個路口恰一次，最后回到出發(fā)點.在經過每個路口時，向右轉不需Y-等待，直行需Y-等待1分鐘，向左轉需Y-等待2分鐘.設小明在路口等待總時間的最小可能值是S分鐘，則S<50≤S<90≤S<100≤S<S≥答案選(C)? ? ×? 由題意知小明行駛的路線是一條不自交的閉折線.將每個路口看作一個頂點，那么他行駛的路線可以看成是一個100邊形(有的內角可能是平角，也有大千平角的內角).由多邊形內角和公式知這個100邊形的所有內角之和為98180?.注意內角只能是90?，180?和27?，設90?有a個，27?有b個，那么90a+270b+180(100ab)=98180，整理得ab=4.如果小明在這條路上是順時針行駛的，那么90?內角對應右轉，180?內角對應直行，270?對應左轉，他在路口等待的總時間是(100ab)+b=100(ab)=96(min);如果小明在這條路上是逆時針行駛的，那么90?內角對應左轉，180?內角對應直行，270?內角對應右轉，他在路口等待的總時間是(100ab)+2a=100+(ab)=? ? ×? 注：如果小明的起點/終點處的轉彎時間不計，那么等待的總時間還可以減少2分鐘(個左轉的位置作為起點)，這樣S94，但不影響選擇的選項第5題設n2是給定正整數.考慮nn矩陣Xjj(ai,j0或者1)的集合證明:存在這樣的X滿足detX=n若2n4,證明detXnn若n2023,證明存在X使得detXn4n???若X有一行全為0或者有兩行相等則detX0;若X有一行只有一個1,則可約化到(n1)階矩陣的情形;若X有一行全為1,還有一行有n1個1,則可約化到有一行只有一個1的情形,進一步約化到(n1)階矩陣的情形.若以上都不發(fā)生,則X的各行有很少的可能性,我們可以???取XI=jjnai,j=1? 1≤i,j≤則detXI?1)n?1(n1).若n是奇數-<XXI.若n是偶數-<X為調換XI得矩陣.則detX=n (3)當n2k1時 ? ?Y 則Y是元素為±1的(n1)(n1)矩陣,detY=(√2k)2k=2k2k?1、比,,注意Y的最后一行為αn+1=(11).記ti=±1為Y的第i行的最后一個元素(1i、比,, βi=2(tiαi?.XI=(β1,...,記則XI是元素為0,1的nn矩陣,

t

ti=detXI=k11×若有必Y-,則調換XI的最后兩行,可以得到一個元素為0,1的n n矩陣,滿足detX=k11×不妨設2k1n2k+11.當2k1n3·2k?1且n2023時則k11.存在元素為0,1的n×n矩陣X,滿足

detX≥kk1>kk k+1 k

n4< = n這樣detXn4n

(k?2)2k?1≥3(k+1)·當3·2k?1n2k+11且n2023時則k10.存在元素為0,1的nn矩陣X,detX≥kk11kk2>kk

k+1 k

n4< = n這樣detXn4n

(3k?7)2k?2>(k+1)·試問是否存在非零實數s,滿足limn→∞||(√21)ns||=試問是否存在非零實數s,滿足limn→∞||(√23)ns||=n存在，取s=1即可。設(√2+1)n=xn+√2yn,則(?√2+1)n=xn?√2yn.nnn√n而x22y2?1)n.由此nn√n

+

|=

?x|=|2y2?x2

→nnnnnn不存在。反證法，假設s滿足(√23)ns=mn+n其中l(wèi)imn→∞En=0.記α=√23ˉ=nn?√23.

=∞m區(qū)n1? 區(qū)n

xn+∞區(qū)n區(qū)n

(1αx)(1ˉx)16x7x2,上式兩邊乘以16x7x2s(1?ˉx)=(1?6x+7x2)

mxn+(1?6x+7x2)nn

設(16x7x2)=0mnxn==0pnxn(16x7x2)=0Enxn==0ηnxn,則pn∈Zlimnηn=0.因為(9)左邊是一次式，從而右邊滿足pnηn=0n≥2.n充分大時ηn很小，所以必有pn=η ∞

xn= .n1?6x+n右邊寫成部分分式形如H(x)+A+B.因為limn→∞En=0, 為1，而αˉ均大千1，所以必須A=B=0.這樣當n充分大時，En=0,從而(23)smnZ,第7題某公司Y-招聘一名員工，有N人報名面試。假設N位報名者所具有該職位相關的招聘委員會按隨機順序逐個面試候選人，且他們能觀察到當時所見候選人的相對排名。比如委員會面試到第m位候選人時，他們擁有的信息是前m位面試者的相對排名，但不知后N?m位候選人的能力情況。?如果委員會決定給某位候選者發(fā)offer，那么這位候選者以概率p接受，以概率1p拒絕，且獨立千(之前)所有其他面試者的決定。如果該候選人接受offer，那么委員會將?如果委員會決定不給某位面試者發(fā)offer反復該面試程序，直到有候選者接受offer面試完所有的N由千N位面試者的順序是完全隨機的，因此他們能力的排名在N!的可能性中是均勻分布。上程序的前提下，找到一個策略，使得招到N位候選者中能力最優(yōu)者的概率最大化。?考慮如下策略。委員會先面試前m1位候選者，不管其能力排名如何，都不發(fā)工作offer。從第m位開始，一旦看到能力在所面試過候選人中的最優(yōu)者，即發(fā)工作offer。如對方拒絕，則繼續(xù)面試直到下一位當前最優(yōu)者1出現。試證明：對千任意的N，都存在一個m=mN，使得依靠上述策略找到(所有N位候選人中)最優(yōu)者的概率值，在所有可能的策略所給出的概率值中是最大的。?N假設p=1。當N→+∞，求N

N對一般的p(01)，當N+∞，求N

N，我們-<Zk為委員會完全略過前k Zk≥Zk+1如果委員會面試了第k位候選人，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遞增，而{Zk}k遞減，且Zk≤N?k+1，易見不等式(10)NNN 對某一個 1成立。由此可知，最優(yōu)策略可以通過選擇某個m ≥足不等式(10)的k中的最小值。此外，如果k=m滿足不等式(10)，則任意的 ≥-<pm為委員會采取(a)中的策略，且找到能力最高者的概率。當p=1時，被發(fā)NN的不相交并集，其中Ak對應的事件為第k位候選人是N了。這樣的話，事件Ak1P(Ak)=

m?1·k?·其中

對應的是這位候選者是能力最高者的概率，

是他/ 前k前NmNk? =m?1mNk?易見pm先曾后減，因此其最優(yōu)值m?=mNpm≥NN區(qū)區(qū)≤k=m+1k?的最小m。當N很大時，由左端的近似逼近為log(N/m可知，N

→e11NAk其中Ak對應的事件為第k位候選人是N人中的能力最高者、被面試了、并且接受ppmN

區(qū)qk區(qū)2如果這個不等式不滿足，則略過第k個人，且從第k1其中qk為假定第k位候選人為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為經典的Γ函數。據此，我們得到如果從第m位開始，委員會找到(所有NNmN =p·Γ(m)NmNpm對千m先增后減。由Γ函數的近似及積分對求和的逼近，我們計算得，當N大時，讓pm最大化的mN→ p11?p.→N當p1時，極限為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,ˉ=?2+3.

=∞1?

mxn+nn

Since(1?αx)(1?ˉx)=1?6x+7x2,multiplyingbothsidesoftheaboveequationby1?6x+7x2wegets(1?ˉx)=(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.Whileαandˉarelargerthan1AandBmustbezero.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