2023阿里巴巴全球數(shù)學競賽阿里巴巴全球數(shù)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(yīng)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(shè)值是a=?1?！蹦悖骸白龅钠?，博士！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(shè)置a2,踢一下儀器，等球狀閃電半徑嚴格超過√2，再設(shè)置a123設(shè)置a3,踢一下儀器，等球狀閃電半徑嚴格超過√2，再設(shè)置a134設(shè)置a4,踢一下儀器，等球狀閃電半徑嚴格超過√2，再設(shè)置a145設(shè)置a=5,踢一下儀器，等球狀閃電半徑嚴格超過√2，再設(shè)置a=15設(shè)兩個凸八面體O1O2的每個面都是三角形,且O1在O2的內(nèi)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(shè)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(jīng)過每個路口恰一次，最后回到出發(fā)點.在經(jīng)過每個路口時，向右轉(zhuǎn)不需要等待，直行需要等待1分鐘，向左轉(zhuǎn)需要等待2分鐘.設(shè)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(shè)n2是給定正整數(shù).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(shù)s,滿足limn||(√21)ns||=試問是否存在非零實數(shù)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(shè)N位報名者所具有該職位相關(guān)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(xiàn)。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(shè)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(yīng)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(shè)值是a=?1。”你：“做的漂亮，博士！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(shè)置a2,踢一下儀器，等球狀閃電半徑嚴格超過√2，再設(shè)置a123設(shè)置a3,踢一下儀器，等球狀閃電半徑嚴格超過√2，再設(shè)置a134設(shè)置a4,踢一下儀器，等球狀閃電半徑嚴格超過√2，再設(shè)置a145設(shè)置a5,踢一下儀器，等球狀閃電半徑嚴格超過√2，再設(shè)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(shù)的非負實根數(shù)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?nèi)菀昨炞C，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(shè)兩個凸八面體O1O2的每個面都是三角形,且O1在O2的內(nèi)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(shù)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(nèi)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(nèi)部),和任意正整數(shù)1:::r:::m.存在常數(shù)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(shù)值計算如下:設(shè)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(guān)鍵是Y-說明為什么(B)、(C)和(D)可以實現(xiàn)× ×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(shù)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(xiàn),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(shè)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(xiàn).(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(shù)為3的頂點之間實現(xiàn)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(shè)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(xiàn).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(shù)為a的頂點和一個度數(shù)為b的頂點之間實現(xiàn)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(xiàn)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(shè)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(jīng)過每個路口恰一次，最后回到出發(fā)點.在經(jīng)過每個路口時，向右轉(zhuǎn)不需Y-等待，直行需Y-等待1分鐘，向左轉(zhuǎn)需Y-等待2分鐘.設(shè)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(nèi)角可能是平角，也有大千平角的內(nèi)角).由多邊形內(nèi)角和公式知這個100邊形的所有內(nèi)角之和為98180?.注意內(nèi)角只能是90?，180?和27?，設(shè)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(nèi)角對應(yīng)右轉(zhuǎn)，180?內(nèi)角對應(yīng)直行，270?對應(yīng)左轉(zhuǎn)，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(nèi)角對應(yīng)左轉(zhuǎn)，180?內(nèi)角對應(yīng)直行，270?內(nèi)角對應(yīng)右轉(zhuǎn)，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(zhuǎn)彎時間不計，那么等待的總時間還可以減少2分鐘(個左轉(zhuǎn)的位置作為起點)，這樣S94，但不影響選擇的選項第5題設(shè)n2是給定正整數(shù).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(shù)-<XXI.若n是偶數(shù)-<X為調(diào)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(diào)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(shè)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(shù)s,滿足limn→∞||(√21)ns||=試問是否存在非零實數(shù)s,滿足limn→∞||(√23)ns||=n存在，取s=1即可。設(shè)(√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(shè)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(shè)(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(shè)N位報名者所具有該職位相關(guān)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(xiàn)。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(shè)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(yīng)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(yīng)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(yīng)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(jīng)典的Γ函數(shù)。據(jù)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(shù)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