2023年阿里巴巴全球数学竞赛预选赛赛题及参考答案.pdf

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1、2023 Cpnn?mGGG“?(R?kXe?!/(?G?5uy?G?Czv(t)vXe?v=ar+r3 r5.pr(t)LG?t mCvkG=r(0)=0A/kv(0)=0?a R?50/s?(?(k,He0/svm?jHe0/vXJHe?r(t)?mJp?u10/x?(k8U?8IG4?L224?”0/?(d?O?oY(s?YX0w?e?uy1?YkA?a=2,He?GL22?a=12B?a=3,He?GL22?a=13C?a=4,He?GL22?a=14D?a=5,He?GL22?a=151?lNO1,O2?zn?/,O13O2?SSS.PO1(O2)?c1(2).?O1/2,U?e=(

2、?)?K0.6411.441.9642A B?1/0iZiZmA kn?B kn+1n A?,/0kiZ5Ki)V?Ol?A kl B?ii)e,?gC?,Kiii)?k?b?zgl?Ve?n=n A?n=31n=32n=999n=1000k?nA?3,k10?10H?u100.?l,fuLzTg?u:.3Lzm=I?1I?1=I?2.?3?om?USKS 5050 S 9090 S 100100 S nn4.5r,|r|LrC?l|r|=min|r n|:n Z.1.3s,vlimn|(2+1)ns|=0?2.3s,vlimn|(2+3)ns|=0?6,i-?kN?b?N?kT?U-”?U*

3、?UU”?Xe-S1.-”?USU*?X”?1m Pk?&Ecm?N m?U2.z?”?I?=/uoffer3.XJ”?,uoffero Vp?V1 p u(c)k?XJT?offero”?2UY?e?XJToffero”?UYe 4.XJ”?,uofferoUYe U2?cL?5.ETS?k?offerXJvk?To”?k?N duN?S?dU?3N!?U5!”?k?&E?cL?”?3XS?cJe?N U?VzKXe(a)Xe”?kcm 1 +UXuofferl1m m?w?U3L?=uofferXKUY?e?c1yyu?N3m=mN?(kN )?V3kU?V?(b)b?p=1?N +mNN

4、?4(c)?p (0,1)?N +mNN?41/?c0?c?3k?L?()?uoffer?“?(R?kXe?!/(?G?5uy?G?Czv(t)vXe?v=ar+r3 r5.pr(t)LG?t mCvkG=r(0)=0A/kv(0)=0?a R?50/s?(?(k,He0/svm?jHe0/vXJHe?r(t)?mJp?u10/x?(k8U?8IG4?L224?”0/?(d?O?oY(s?YX0w?e?uy1?Yk(A).?a=2,He?GL22?a=12(B).?a=3,He?GL22?a=13(C).?a=4,He?GL22?a=14(D).?a=5,He?GL22?a=151 YYY(B

5、)PCzv=f(r;a).XJv 0 Kr mOXJv 0?kKr1=0 r5 0N?y?r (0,r5)v 0?r (r5,+)v 0?XJHeUG?O?r5?Lr5?L2I-r52I-a 2?(A)?14 a 0 r5 0AO/r5 1?r (r5,+)v U?”?(D)?a=14kKr1=0 r5=12aqXJdr=2?r=r5?ur5dGU?”?(C)?a 0 v?”(B)?(n?21112KKK?lNO1,O2?zn?/,O13O2?SSS.PO1(O2)?c1(2).?O1/2,U?e=(?)?K(A).0.64(B).1(C).1.44(D).1.96(E).42 YYY(A)(

6、B)(C)(D)360-70c“?)c?kL?K/oNV1 uoNV2S,yV1?c?uV2?c?43?0.p?/3uXJ?n?/u,n?/S,o?n?/=?un?/?,Xd.?3n/,NL?X?c?X?.?K?“?”AT=u1962cuL?Holszty nski,W.and Kuperberg,W.,O pewnej wlasn osci czworo scian ow,Wiadomo sci Matem-atyczne 6(1962),14-16.=?g,vko?,?1977c?=Holszty nski,W.and Kuperberg,W.,On a Property of Tetra

7、hedra,Alabama J.Math.1(1977),40-42.?1986c,Alabama?Carl Linderholmr(J2?pm?/:nnn:uRn?m/ST(c?u?S),?1 6 r 6 m.3Bm,r,?S?kr?LT?kr?Bm,r?.pBm,r?NOXe:?m+1=(r+1)q+s(),KBm,r=qr+1s(q+1)sm+1 r.(CARL LINDERHOLM,AN INEQUALITY FOR SIMPLICES,Geometriae Dedicata(1986)21,67-73.)?K,p?(A)?,:o(B)!(C)(D)y?o(E)Uy?pI?k:3(A)

8、?A:zn?/?lN,?k38/2=12c,udEuler,:6.(B):XJk:5c,o?7k,:5c,lN?:(5,5,4,4,3,3).d?,U?/z:4c(X?lN).(C):A:?lN,lN?:ml?73,:my.XJlN?z:4c,l3?:ABmy,o,o:?,lN?c?4(3,o:?AB?l?C),?u?lN5,b?z:4c,4n:CuA,nCuB,cCu6.?k?u1.5?y.(k(A),(B),(C)XJlN?:ml33?:my?,olN?c?3(3,o:?AB?l?C),?u?lN5,Eb?z:4c,4n:CuA,nCuB,cCu6.?k?u2?y.?XJd?lNlN?(?

9、,5?:?C,d,4:?C,oJp?8/3.?a?,XJlN?:ml23a?:b?:my?(+?),olN?cumin(a,b),?lN?cw,L12,(E)Uy?.41113KKKA B?1/0iZiZmA kn?B kn+1n A?,/0kiZ5Ki)V?Ol?A kl B?ii)e,?gC?,Kiii)?k?b?zgl?Ve?n=n A?(A).n=31(B).n=32(C).n=999(D).n=1000(E).k?nA?3 YYY(B)PAnA?anKa1=12+1212a1,(1)?ka1=23?a2=23+1313a2,(2)?ka2=34?4an=nn+1an2+1n+11n+

10、1an+1n+1nn+1pn,n1,(3)m1A?BoU?3BAkn 2Bkn 1Akm1?AB?m1nAB?BvA?pn,n1Akk?nBk?n 1A?kpn,n1=1 an2,(4)A?=U?dCAk?n 1Bk?n 2Bk?dB?an2?A?1 an25dan=nn+1an2+1n+11n+1an+n(n+1)2n(n+1)2an2(5)?an=nn+2an2+1n+2=nn+2(n 2nan4+1n)+1n+2=.(6)end4?an=n+32(n+2),(7)end4?an=n+42(n+2).(8)d a31=1733 a32=917 a999=5011001 a1000=2515

11、01Y(B)=Ak32A?61114KKK,k10?10H?u100.?l,fuLzTg?u:.3Lzm=I?1I?1=I?2.?3?om?USK(A).S 50(B).50 S 90(C).90 S 100(D).100 S/k?S?U?ku?S?.d/S?100/?kS?98 180.5S?U90180270?90ka270kbo90a+270b+180(100ab)=98180?n?ab=4.XJ?3?1?o90S?Am=180S?A1270S?A=3?om(100ab)+2b=100(ab)=96(min)XJ?3_?1?o90S?A=180S?A1270S?Am=3?om(100ab

12、)+2a=100+(ab)=104(min).dS=96(C)?(.5XJ?:/:?=?mOo?om?2J=?:?S=94?KJ?.71115KKK?n 2?.n n?X=(ai,j)1i,jn(ai,j=0 1)?8.(1)y:3?X vdetX=n 1.(2)e2 n 4,ydetX n 1.(3)en 2023,y3X?detX nn4.5 YYY(1)eXk1?0k1?,KdetX=0;eXk1k1,K?z?(n 1)?/;eXk1?1,k1kn 11,K?z?k1k1?/,?z?(n 1)?/.eu),KX?1k?U5,?.(2)?X0=(ai,j)1i,jn,ai,j=1 i,j,

13、1 i,j n.KdetX0=(1)n1(n1).en,-X=X0.en,-XNX0?1?.KdetX=n 1.(3)?n=2k 1,-Y=?1111?k.KY?1?(n+1)(n+1)?,detY=(2k)2k=2k2k1.5Y?1n+1=(1,.,1|z n+1).Pti=1Y?1i1?(1 i n).-0i=12(tii n+1).?K0i?(?u0),?n1i.-X0=(1,.,n)t.Pt=Y1inti=1.KX0?0,1?n n?,detX0=t2(k2)2k1+1.8ek7,KNX0?1,?0,1?n n?,vdetX=2(k2)2k1+1.”?2k 1 n 2k+1 1.?2k

14、 1 n 2(k2)2k1.,nn4 nn4.?3 2k1 n 2(3k7)2k2.,nn4(k+1)2k1.?,detX nn4.91116KKKr,|r|LrC?l|r|=min|r n|:n Z.1.3s,vlimn|(2+1)ns|=0?2.3s,vlimn|(2+3)ns|=0?6 YYY1.3?s=1=?(2+1)n=xn+2yn,K(2+1)n=xn2yn.l?x2n 2y2n=(1)n.dd|xn+2yn 2xn|=|2yn xn|=|2y2nx2n|2yn+xn 0.2.3yb?sv(2+3)ns=mn+?n,limn?n=0.P=2+3,=2+3.?s1 x=Xn=0mnx

15、n+Xn=0?nxn.(1 x)(1 x)=1 6x+7x2,1 6x+7x2?s(1 x)=(1 6x+7x2)Xn=0mnxn+(1 6x+7x2)Xn=0?nxn.(9)?(1 6x+7x2)Pn=0mnxn=Pn=0pnxn,(1 6x+7x2)Pn=0?nxn=Pn=0nxn,KpnZ,limnn=0.(9)gl?mvpn+n=0,n 2.nn?7kpn=n=0,=(9)mdXn=0?nxn=G(x)1 6x+7x2.m?/XH(x)+A1 x+B1x.limn?n=0,?1?,u17LA=B=0.?n?n=0,l?(2+3)ns=mn Z,g101117KKK,i-?kN?b?N?

16、kT?U-”?U*?UU”?Xe-S1.-”?USU*?X”?1m Pk?&Ecm?N m?U2.z?”?I?=/uoffer3.XJ”?,uoffero Vp?V1 p u(c)k?XJT?offero”?2UY?e?XJToffero”?UYe 4.XJ”?,uofferoUYe U2?cL?5.ETS?k?offerXJvk?To”?k?N duN?S?dU?3N!?U5!”?k?&E?cL?”?3XS?cJe?N U?VzKXe(a)Xe”?kcm 1 +UXuofferl1m m?w?U3L?=uofferXKUY?e?c1yyu?N3m=mN?(kN )?V3kU?V?(b)b?p

17、=1?N +mNN?4(c)?p (0,1)?N +mNN?47 YYYu?1 k N-Zk”?Lck 1?l1k m?=N Up?T?VKkZk Zk+1.(a)XJ”?1k U3ck?o”?uoffer3e”?Up?VYk=pkN+(1 p)Zk+1.1/?c0?c?3k?L?()?uoffer?)?11dd”?1k uoffer?=?3ck?UppkN+(1 p)Zk+1 Zk+1,(10)=kN Zk+12.dukNk4O?Zkk4ZkNk+1N?(10)7,k N 1 ddLJ,m 5?=v?(10)?k?d?XJk=m v?(10)K?k mvdu1m?/?c0A?uoffer(b

18、)-pm”?(a)?Up?V?p=1?uoffer?UpA?Nk=mAk?8AkA?1k N?Up?Ak?VP(Ak)=1Nm 1k 1,1NA?Up?Vm1k1/?V=ck 1?Up3cm 1?kpm=m 1NNXk=m1k 1.pmkQ?dm=mNAvpm pm+1?m=vNXk=m+11k 1 1?m?N d?Cq%Clog(N/m)mNN1e.(c)?up (0,1)?pm?e?VNk=mAk,AkA?1k N?Up!?!?offer.kpm=pNNXk=mqk,2XJ?vKL1k l1k+1 m?K?Z?V12qkb1k N Up?/?VKkqk=?m 1m+1 pm?mm+1+1

19、pm+1?k 2k 1+1 pk 1?=(m)(k p)(k)(m p),;?d?XJl1m m”?(kN )Up?Vpm=pN(m)(m p)NXk=m(k p)(k),pmum kO?d?Cq9?%CO?N 4pmz?mN?vmNN p11p.?p=1 41/e.132023 Alibaba Global Mathematics CompetitionBall lighteningAs a chief officer of a secret mission,you had the following conversation with the leadingscientist.Scient

20、ist:”Chief,we have mastered the control law of ball lightning.We found that the rateof change of the radius of ball lightning in the laboratory v(t)satisfies the following equation.v=ar+r3 r5.Here r(t)represents the radius of ball lightning,and t is the time variable.At the initialmoment,there is no

21、 ball lightning,that is,r(0)=0.Accordingly,we also have v(0)=0.And a R can be artificially controlled.You can quickly change the value of a by pulling acontrol lever.We set its preset value to a=1.”You:”Well done,Doctor!Is a our only way of control?It doesnt seem to be able to startthe ball lightnin

22、g.”Scientist:”Youre right,Chief.We do have another way of control,which is to kick theinstrument.”You:”Doctor,are you kidding me?Kick it?”Scientist:”Yes,if you kick it,the value of r(t)will instantly increase by (is much smallerthan 1).”You:”I see.Thats helpful indeed.Our test goal today is to start

23、 the ball lightning,makeits radius strictly exceed2,and then let it gradually disappear completely.”Scientist:”Yes,Chief.We have designed four control schemes for this.What do you think of these schemes,Chief?”You looked at these options and found that the feasible schemes are().Set a=2,kick the ins

24、trument,wait for the ball lightning radius to strictly exceed2,then set a=12;Set a=3,kick the instrument,wait for the ball lightning radius to strictly exceed2,then set a=13;Set a=4,kick the instrument,wait for the ball lightning radius to strictly exceed2,then set a=14;Set a=5,kick the instrument,w

25、ait for the ball lightning radius to strictly exceed2,then set a=15.1Let O1,O2be two convex octahedron whose faces are all triangles,and O1is inside O2.Letthe sum of edge kengths of O1(resp.O2)be 1(resp.2).When we calculate 1/2,whichvalue(s)among the following can be obtained?(Multiple Choice)0.6411

26、.441.9642Two players,A and B,play a game called“draw the joker card”.In the beginning,Player Ahas n different cards Player B has n+1 cards,n of which are the same with the n cards inPlayer As hand,and the rest one is a Joker(different from all other n cards).The rules arei)Player A first draws a car

27、d from Player B,and then Player B draws a card from PlayerA,and then the two players take turns to draw a card from the other player.ii)if the card that one player drew from the other one coincides with one of the cards onhis/her own hand,then this player will need to take out these two identical ca

28、rds anddiscard them.iii)when there is only one card left(necessarily the Joker),the player who holds that cardloses the game.Assume for each draw,the probability of drawing any of the cards from the other player isthe same.Which n in the following maximises Player As chance of winning the game?n=31n

29、=32n=999n=1000For all choices of n,A has the same chance of winning3There are 10 horizontal roads and 10 vertical roads in a city,and they intersect at 100crossings.Bob drives from one crossing,passes every crossing exactly once,and return tothe original crossing.At every crossing,there is no wait t

30、o turn right,1 minute wait to gostraight,and 2 minutes wait to turn left.Let S be the minimum number of total minutes onwaiting at the crossings,thenS 50;50 S 90;90 S 100;100 S nn4.5For a real number r,set|r|=min|r n|:n Z,where|means the absolute value ofa real number.1.Is there a nonzero real numbe

31、r s,such that limn|(2+1)ns|=0?2.Is there a nonzero real number s,such that limn|(2+3)ns|=0?6A company has one open position available,and N candidates applied(N is known).Assumethe N candidates abilities for this position are all different from each other(in other words,there is a non-ambiguous rank

32、ing among the N candidates),and the hiring committee canobserve the full relative ranking of all the candidates they have interviewed,and their observedrankings are faithful with respect to the candidates true abilities.The hiring committeedecides the following rule to select one candidate from N:1.

33、The committee interviews the candidates one by one,at a completely random order.They observe information on candidates relative ranking regarding their abilities forthe position.The only information available to them after interviewing m candidatesis the relative ranking among these m people.2.After

34、 each interview,the committee decides whether to offer the candidate the positionor not.3.If they decide to offer the position to the candidate just interviewed,then the candi-date will accept the job with probability p,and decline the offer with probability 1p,independently with all other candidate

35、s.If the selected candidate accepts the offer,then he/she gets the job,and the committee stops interviewing the remaining candi-dates.If he/she declines the offer,then the committee proceed to interviewing the nextcandidate.4.If they decide not to offer the position to the candidate just interviewed

36、,then theyproceed to interviewing the next candidate,and they can not turn back to previouslyinterviewed candidates any more.5.The committee continues this process until a candidate is selected and accepts the job,or until they finish interviewing all N candidates if the position has not been filled

37、before,whichever comes first.Since the interview order of the candidates are completely random,each ranking has equalprobability among the N!possibilities.The committees mission is to maximise the proba-bility of getting the candidate with the highest ranking(among N candidates)for the jobconstraine

38、d to the above selection process.Here are the questions(a)Fix 1 m N,and consider the following strategy.The committee interviews thefirst m 1 candidates,and do not give offer to any of them regardless of their rel-ative rankings.Starting from the m-th candidate,the committee offers him/her thepositi

39、on whenever the candidates relative ranking is the highest among all previouslyinterviewed candidates.If he/she declines the offer,then the committee continues theinterview until the next relatively best candidate1,and then repeat the process whenapplicable.Show that for every N,there exists m=mNsuc

40、h that the above strategy maximisesthe probability of getting the best candidate among all possible strategies.1“Relatively best candidate”refers to the candidate with the highest ability among all candidates who havebeen interviewed(including those who are offered the position and declined).7(b)Sup

41、pose p=1.What is the limit ofmNNas N +?(c)For p (0,1),what is the limit ofmNNas N +?82023 Alibaba Global Mathematics Competition1 Ball lighteningAs a chief officer of a secret mission,you had the following conversation with the leadingscientist.Scientist:”Chief,we have mastered the control law of ba

42、ll lightning.We found that the rateof change of the radius of ball lightning in the laboratory v(t)satisfies the following equation.v=ar+r3 r5.Here r(t)represents the radius of ball lightning,and t is the time variable.At the initialmoment,there is no ball lightning,that is,r(0)=0.Accordingly,we als

43、o have v(0)=0.And a R can be artificially controlled.You can quickly change the value of a by pulling acontrol lever.We set its preset value to a=1.”You:”Well done,Doctor!Is a our only way of control?It doesnt seem to be able to startthe ball lightning.”Scientist:”Youre right,Chief.We do have anothe

44、r way of control,which is to kick theinstrument.”You:”Doctor,are you kidding me?Kick it?”Scientist:”Yes,if you kick it,the value of r(t)will instantly increase by (is much smallerthan 1).”You:”I see.Thats helpful indeed.Our test goal today is to start the ball lightning,makeits radius strictly excee

45、d2,and then let it gradually disappear completely.”Scientist:”Yes,Chief.We have designed four control schemes for this.What do you think of these schemes,Chief?”You looked at these options and found that the feasible schemes are().(A).Set a=2,kick the instrument,wait for the ball lightning radius to

46、 strictly exceed2,then set a=12;(B).Set a=3,kick the instrument,wait for the ball lightning radius to strictly exceed2,then set a=13;(C).Set a=4,kick the instrument,wait for the ball lightning radius to strictly exceed2,then set a=14;(D).Set a=5,kick the instrument,wait for the ball lightning radius

47、 to strictly exceed2,then set a=15.11 AnswerThe answer is(B).We introduce the following notation for the rate functionv=f(r;a).When v 0,r is increasing in time.When v 0,we have two nonnegative roots:r1=0 and r5 0.Clearly,when r (0,r5),v 0;and when r (r5,+),v 0 and if we kick the instrument,wecan sta

48、rt the ball lightening,and its radius will grow to r5(but it will not exceed r5).To make the radius exceed2,we need r52.This means in the starting phase,we needa 2,and thus Scheme(A)fails.When 14 a 0,we have three nonnegative roots,which satisfy 0=r1 r3 r5.Inparticular,we have r5 1 and when r (r5,+)

49、,v 0.This means,if we start withr=2,the radius is getting smaller,but it will not become smaller than r5.Therefore,theball lightening will not vanish completely.Hence,Scheme(D)fails.When a=14,similar to the previous case,the radius will not be smaller than r5=12,andthe ball lightening will not vanis

50、h completely.Hence,Scheme(C)fails.When a 0,we always havev 0,and thus the ball lightening will vanish completely.This means Scheme(B)works.22Let O1,O2be two convex octahedron whose faces are all triangles,and O1is inside O2.Let the sum of edge kengths of O1(resp.O2)be 1(resp.2).When we calculate 1/2