2022年阿里巴巴全球数学竞赛预选赛试题及参考答案.pdf

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1、2022 Cpnn?mKKK555AAA r%N ABCD A1B1C1D1(b?AB=1)UXe 12 1:1)?6?AC,AB1,AD1,C1B,C1D,C1A1;2)N%:,6?9 12 c?n?/;3)18 n?/rN?12,zoN,zoNkcN?c;4)zoN=LN?coN(.1样玩具/G2.:3玩kU?/G,:m(3m)l?图1:磁性几何魔方图2:其它形状的例子11p7+42131+222KKKmmm*,F,/?|/k|?L+*.d/wL?%P:K.?+A1,A2,.,An,.,Ul A1m?Sg*?P1,P2,.,Pn,.,?Iven.1:An*?K?l?u 10=?n,KPn

2、10 2:An*?cz?lI?u 1=?n 2?1 m n 1,PmPn 1.3:3v1 2?cJeAnJ K?C?:?1*=F KPn?U?.XJv1 2?KPn?U?Pno AnJ?:.Xu A1,J vk2?J K?%10?C?:K?lT 10 u A2,F P2 K?lT 10=P23 C.du C kN:P1?l?u 1 J?:.(1)e=?(3?c1,c2,?n,A1,A2,.,An,NoJ k c1 KPn c2 3?c1,c2,?n,A1,A2,.,An,NoJ k c1n KPn c2n 3?c1,c2,?n,A1,A2,.,An,NoJ k c1n KPn c2n 3?c1

3、,c2,?n,A1,A2,.,An,NoJ k c1n2 KPn c2n2.(2)dukN?XJ,3,*?NC1?1?4.?i,j,XJ Pi?%16?KPj?o Aj?Ai?4?w?L?m.e=?(?k 60+*kw?L?m?k 60+*3kw?L?m?U5?k 800+*kw?L?m?k 800+*3kw?L?m?U5?k 10000+*kw?L?m?k 10000+*3kw?L?m?U5.3KKK/mmmmmm)%000_S!m/c0Ei?#S84zENRke?/m0/)0/%0?YXe8/m0/)0/%0=/mm)%0?4?0,3k X?b1,b2,.,bm?t X,3,bi(t)v|

4、f(x,t)f(x,bi(t)|0,3k X?a1,a2,.,an?t X,3,ai(t)v|f(t,x)f(ai(t),x)|?,x X,K f?y?dum5yyyKKK?n?,V=Rn n m,k|ei=(0,.,0|z i1,1,0,.,0|z ni)(1 i n)v(ei,ej)=i,j,i,j=?1 if i=j0 if i 6=jKronecker,(,)V?S.V?v,5C sv:V V sv(u)=u 2(u,v)(v,v)v,u V.u0u 0 n m?k,P Grk(V)V?k fm?8.u V?k fm W,P W Grk(V)?A?.?W?|?5 v1,.,vk(=,(

5、vi,vj)=i,j),sW:V V sW=sv1svk.(1)y sW6u?5 v1,.,vk?.(2)y s2W=id.(3)W0 Grk(V),tW(W0)=sW(W0),sW(W0)W03 sWe?”.Grk(V)?f8 X“k?8”etW(W0)=W0,W,W0 X.?Grk(V)k?8?,y.6)KKKaaa?444?4?3?:I(n,0)?4?4K3?:(0,0)?d?4?z 1?i3e?”?n?u 1(1)-P1,nT4?3r?T bn1.5c l4?lun2?Vylimn+P1,n=1(2)-P2,nT4?3c bn1.5c SQ?L4?Vylimn+P2,n=0(3)-P3

6、,nT4?c 2nSQ?L4?Vylimn+P3,n=17)KKK(1)y3vXe?S?a1,a2,a3,.,z 1,?kn,supNN?NXn=1ane2in?+.(2)y3z 1?S?a1,a2,a3,.v?kn supNN?NXn=1ane2in?2022.(3)3z 1?S?a1,a2,a3,.vk 1k 1?Q Z,supNN?NXn=1ane2in?+.8)KKKmmm444?!888?OOOb?cm4?EnKI?OY?n53m4?!8kY4+d?37?/G?|/wXv?PEV(t,v)(0)5x?|/?/?v RL?ou?m t,v1 v2Zv2v1(t,v)dvL?N0u v1

7、 v2m?Vdu 0.u(t)?-(1)?No 0u1 0?N(t)L 1 o N(t)3zLk.u?y?(?B?9?3|v|+?z(2)3B?q3?3w13?|/!uIu?p(t,x,v)(0)p x 0,2L 0.du 0.X?n?/;3)18 n?/rN?12,zoN,zoNkcN?c;4)zoN=LN?coN(.1图1:5A图2:/G?f:3玩具kU?/G,:m(3m)l?玩具/G2.(A)11(B)p7+42(C)13(D)1+22(E)1 YYYDABCD A1B1C1D1IPN?l:,?m?:AC1.o,:Ou6?oN,?B,C,D,D1,A1,B18:(NoC/)?1?8/(U

8、z).5?d12oN(l?/)?,d?7?oN?:.3oN,o:3a:3NAuAC1,111aaa:;:(l?m?c)u(m)8/BCDD1A1B1,111?aaa:;:AuN?%,?oN?,n:?l32,111nnnaaa:.?:mccc?l?,?eL:l1a:1?a:1na:1a:1+222+21+2+321?a:2+232+321na:1+2+322+322+3l?:m?lL1+22.?,Xe:21112,3KKKKKKmmm*,F,/?|/k|?L+*.d/wL?%P:K.?+A1,A2,.,An,.,Ul A1m?Sg*?P1,P2,.,Pn,.,?Iven.1:An*?K?l?u

9、10=?n,KPn 10 2:An*?cz?lI?u 1=?n 2?1 m n 1,PmPn 1.3:3v1 2?cJeAnJ K?C?:?1*=F KPn?U?.XJv1 2?KPn?U?Pno AnJ?:.Xu A1,J vk2?J K?%10?C?:K?lT 10 u A2,F P2 K?lT 10=P23 C.du C kN:P1?l?u 1 J?:.(1)e=?(A)3?c1,c2,?n,A1,A2,.,An,NoJ k c1 KPn c2(B)3?c1,c2,?n,A1,A2,.,An,NoJ k c1n KPn c2n (C)3?c1,c2,?n,A1,A2,.,An,NoJ k

10、 c1n KPn c2n (D)3?c1,c2,?n,A1,A2,.,An,NoJ k c1n2 KPn c2n2.(2)dukN?XJ,3,*?NC1?1?4.?i,j,XJ Pi?%16?KPj?o Aj?Ai?4?w?L?m.e=?(A)?k 60+*kw?L?m(B)?k 60+*3kw?L?m?U5?k 800+*kw?L?m(C)?k 800+*3kw?L?m?U5?k 10000+*kw?L?m(D)?k 10000+*3kw?L?m?U5.32 YYYB?KPn?dn.P1,P2,.,Pn1?%1?d Pn?2 C?S7,CX?K?%dn?d d2n(n 1)12+102,?d

11、nn+99 100n=10n.,P1,P2,.,Pn?%12?du P1,P2,.,Pnml?u 1?*d?.,?dK KP1,KP2,.,KPnAL dnek 1 m n 1?KPm?L?dnK AmJlC?Pn:gd?3 K?%dn+12?Sd(dn+12)2 n (12)2,?dnn212.5?n=1 d1=10?n 2,12n22n5=n10.dn10 dn 10n,B?(.3 YYYB?k 60 110sin6=120,?20sin60 1,S?u C?60/?u 1.d?P1,P2,.,P60T?60/?k:K8.?i,j,:Pi?KPj?l?u 10sin30d sin3015s

12、in6=110?l?u 1?kUw?L?m.,?k 800 KPi,o Aj?Ai?4?.b?vk-?kyXJ3?PiKPjv PiKPj112?1KPi+1KPj?l?o,Ai,Ajk?,?4?.”5b?KPi KPj,d KPi,KPj 10 PiKPj160,?b?.d Pi?KPj?Rv u KPj?S?Pi?KPj?lKPisinPiKPj 2.d3?PiKPjv PiKPj112?1KPi+1KPj?7,kw?L?m.nB?(.51114,5KKKKKK/mmmmmm)%000_S!m/c0Ei?#S84zENRke?/m0/)0/%0?YXe8/m0/)0/%0=/mm)%0?4

13、?tdt=Z0(1 PT t)dt=Z0(1 PTi t,1 i n)dt.Ym?PoissonL?thinning5ME?5?OC?Xd?r8zYiwpi?nPoissonLukET=Z0 1 nYi=1PTi t!dt.PTi t?|Poisson?LPTi t=1 ki1Xk=0epit(pit)kk!.7?LEN=Z0 1 nYi=1 1 ki1Xk=0epit(pit)kk!dt.3?K8n=38I8(k1,k2,k3)=(2,1,1)O?)?LEN=1+p1+?2p1+1p2+1p3?3Xi=111 pip1(p1+p2)2p1(p1+p3)2.3(p1,p2,p3)=(1/3,1

14、/3,1/3)?713YB5 YYYCO?YA?713Dw,?Y8?/%0iY?8?YA%3u?YBCYB3p*kV?|)2?E=1+p+?2p+1q+1r?1p+q+1p+r+1q+r?p(p+q)2p(p+r)2.B,CA?O7118,6223245YCZ?C81116KKKyyyKKK8 X,e f:X X 0,1 v?0,3k X?b1,b2,.,bm?t X,3,bi(t)v|f(x,t)f(x,bi(t)|0,3k X?a1,a2,.,an?t X,3,ai(t)v|f(t,x)f(ai(t),x)|0,dm?X k?b1,b2,.,bm.N?h:X 0,1m,h(x)=(f(x,

15、b1),f(x,b2),.,f(x,bm).d 0,1m?;53k c1,c2,.,cn?x X,3,cix?|h(x)h(cix)|?,l?|f(x,bi)f(cix,bi)|?,i 1,2,.,m.(1)dm|f(x,t)f(x,bi(t)|?(x X),|f(cix,t)f(cix,bi(t)|?,(1),?t X,|f(x,t)f(cix,t)|f(x,t)f(x,bi(t)|+|f(x,bi(t)f(cix,bii)|+|f(cix,bi(t)f(cix,t)|3?.?f?91117KKKyyyKKK?n?,V=Rn n m,k|ei=(0,.,0|z i1,1,0,.,0|z ni

16、)(1 i n)v(ei,ej)=i,j,i,j=?1 if i=j0 if i 6=jKronecker,(,)V?S.V?v,5C sv:V V sv(u)=u 2(u,v)(v,v)v,u V.u0u 0 n m?k,P Grk(V)V?k fm?8.u V?k fm W,P W Grk(V)?A?.?W?|?5 v1,.,vk(=,(vi,vj)=i,j),sW:V V sW=sv1svk.(1)y sW6u?5 v1,.,vk?.(2)y s2W=id.(3)W0 Grk(V),tW(W0)=sW(W0),sW(W0)W03 sWe?”.Grk(V)?f8 X“k?8”etW(W0)

17、=W0,W,W0 X.?Grk(V)k?8?,y.107 YYY(1)P W W 3 V?.KsW|W=1 sW|W=1.ddx?sWy?W?5 v1,.,vk?.(2)dusW|W=1sW|W=1,s2W=id.(3)(i)ytW(W0)=W0 W0=(W0 W)(W0 W).5y.75:b?tW(W0)=W0.K?u W0,k sW(u)W0.P u=u1+u2,u1 W u2 W.K u1+u2=sW(u)W0.,u1 W0u2 W0.d,W0(W0 W)(W0 W).w,(W0 W)(W0 W)W0.?,W0=(W0 W)(W0 W).(ii)PX=spanRei1,.,eik:1 i

18、1 ik n.d(i)?OOK,X“k?8”.8 X?nk?.(iii)n 8B,y:Grk(V)k?8 X?L?nk?.?n=1,w,?.b?d(3 n m.?n=m.?k=0 m,(w,.?1 k m 1.?W X.z?i(0 i k),PXi=W0 X:dimW0 W=k i.zW0 Xi,PYi=W0 W:W0 Xi Grki(W)Zi=W0 W:W0 Xi Gri(W).d(i)?OOK,Yi Grki(W)?k?8 Zi Gri(W)?k?8.du dimW=k m dimW=m k m,d8Bb?|Yi|?kk i?|Zi|?m ki?.?,|X|=X0ik|Xi|X0ik|Yi

19、|Zi|X0ik?kk i?m ki?=?mk?.ddy?(.111118KKK)KKKaaa?444?4?3?:I(n,0)?4d?4?z 1?i?4K3?:(0,0)?3e?”?n?u 1(1)-P1,nT4?3r?T bn1.5c l4?lun2?Vylimn+P1,n=1(2)-P2,nT4?3c bn1.5c SQ?L4?Vylimn+P2,n=0(3)-P3,nT4?c 2nSQ?L4?Vylimn+P3,n=1128 YYY(1)T4?3r?T bn1.5c l4?lV?u?u12?dn?k7,ue AA=4?d?:?lu?u12?n?XJ A u)K4?3?I?l?n4”?T

20、y-T4?1 k 3 y-?l?we?C?Yk=X1+X2+Xk P(Xi=0)=12,P(Xi=1)=14?E(Xi)=0,var(Xi)=12d?kP(A)2P?Ybn1.5c?n4?2 n1.52n2/16=16n0.5 1(2)dn?k7,ue AA=4?3cbn1.5cS?L?:?lu?un?/XJ A u)K4?3?I?L?n2?l=P(A)2P?maxkn1.5|Yk|n2?5?Yk?Doob?P?maxkn1.5|Yk|n2?Eh?Y2bn1.5c?i(n/2)2n1.52(n/2)2=2n0.5 2n)P(n,0)(b 2n)+P(n,0)(b 013u?n(C5kmin|X

21、0|2n/4PX0(|S2n/2|2n/4)P0(|S2n/2|2 2n/4)P(|B1|2)c2c2e4?2nU 2n/2?XJT4?3 2n ul(0,0):2n/4?SK7L3 2n/2,2 2n/2,2n/22n/2l(0,0):2n/4?S2di?5kP(n,0)(b 2n)h1 min|X0|2n/4PX0(|S2n/2|2n/4)i2n/2?1 c2?2n/2 1?u1?p2=P(n,0)(b b dp2=a(n,0)Ea(Sb)|b 5?d13 a(n,0)?log(n)Ea(Sb)|b?ndp2 1 y.1H.Kesten,Hitting probabilities of r

22、andom walks on Zd,Stochastic Process.Appl.25(1987)165184.141119KKK)KKK(1)y3vXe?S?a1,a2,a3,.,z 1,?kn,supNN?NXn=1ane2in?+.(2)y3z 1?S?a1,a2,a3,.v?kn supNN?NXn=1ane2in?2022.(3)3z 1?S?a1,a2,a3,.vk 1k 1?Q Z,supNN?NXn=1ane2in?+.159 YYY(1)Pz:=e2i.yb?3d N?d1?S?(an)vsupNN?NXn=1anzn?20222)g(3)(,?S?v?)?d5kww=1/

23、39N=3M?ai?supMN?M1Xn=0a3n+1!1+M1Xn=0a3n+2!e2i/3+M1Xn=0a3n+3!e4i/3?+Au?n1,e2i/3e4i/3,?O?1+e2i/3+e4i/3=0,1e2i/3e4i/3Figure 3:=1/3?.?4nk?4-i=1,i=29i=3S?(a3n+i)n111?aqup?3?5ATku1?17k?E?(an),1?Xe:n1234567891011an+111+1+1+1+1+1+111,+1y1!g,?X1y2!g,?+1y3!=6gXdUYe?eyTS?vk 0,-f(k)=0!+1!+k!.Ik NLvf(k)n 1.K?n?X

24、e/NXn=1ane2in=Xk0(1)kXnIknNe2inp/q.eT6uN?.Sk:=PnIknNe2inp/q(k 0).(i)ek q,d?|Sk|f(k+1)f(k)=(k+1)!(ii)ek q,|S?Ey?0.(a)ek qf(k+1)N,KSk=0.?Sk=f(k+1)1Xn=f(k)e2inp/q=?e2ip/q?f(k+1)?e2ip/q?f(k)e2ip/q 1.k q,(k+1)!U?q?,f0?e2ip/q?f(k)+(k+1)!?e2ip/q?f(k)=?e2ip/q?f(k)?e2ip(k+1)!/q 1?=0.(b)ek q?N Ik(=?),kSk=NXn=

25、f(k)e2inp/q=?e2ip/q?N+1?e2ip/q?f(k)e2ip/q 1.d?k|Sk|?e2ip/q?N?+?e2ip/q?f(k)?e2ip/q 1?=2?e2ip/q 1?.(c)ek qN f(k),KSk=0ST18d?NXn=1ane2in?q1Xk=0|Sk|+Xkq|Sk|f(q)+2?e2ip/q 1?.=supNN?NXn=1ane2in?+.1911110KKK)KKKmmm444?!888?OOOb?cm4?EnKI?OY?n53m4?!8kY4+d?37?/G?|/wXv?PEV(t,v)(0)5x?|/?/?v RL?ou?m t,v1 v2Zv2v1

26、(t,v)dvL?N0u v1 v2m?Vdu 0.u(t)?-(1)?No 0u1 0?N(t)L 1 o N(t)3zLk.u?y?(?B?9?3|v|+?z(2)3B?q3?3w13?|/!uIu?p(t,x,v)(0)p x 0,2L 0.du 0.X 0.K?t +k M(t)+?N(t)M(t)N(t)u?(2)du x?.?)?XeFp“?/p(t,x,v)=12+Xk=pk(t,v)eikx.pk(t,v)=Z20p(t,x,v)eikxdx.oIy?k 6=0 pk(t,v)Pk|Fp“?5?pkvtpk+ikvpk=v?(u(t)v)pk?v+vvpk,t 0,k Z,v

27、R.?)eu v?Fp“C-pk(t,)=12Z+pk(t,v)eivdv21LO?pk(t,)ve?t pk+(k)pk=?iu(t)+2?pk.?e5A?)dXeA?ddtk(t)=k(t)k.)k(t)k=ets(k(s)k).XA?pk(t,k(t)=pk(0,k(0)eRt0(k(s)2+iu(s)k(s)ds2|A?)?pk(t,)=pk(0,(k)et+k)eHt,k(k),Ht,k(z)=1 e2t2z2+2k(1 et)z+iZt0u(s)e(ts)dsz+k2t+ikZt0u(s)ds.,?LFp“_C?pk(t,v)?Lpk(t,v)=eikv(pt,k Gt,k)(v)

28、.pt,k(y)?Xept,k(y):=etp0,k(ety),p0,k(v):=eikvZ20pinit(x,v)eikxdx.?Gt,k(z)E?pdLXeGt,k(z)=12exp?(z )222?exp(ik)exp?k2D?.p(t)=Zt0e(ts)u(s)ds,(t)=p1 e2t,(t,z)=2(z (t)1 etZt0u(s)ds,D(t)=t 21 et1+et.5?t 0 pd Gt,kduPf exp?k2D?3PPu t.?|?kpk(t,v)kL1(R)=keikv(pt,k Gt,k)(v)kL1(R)kpt,k(g)kL1(R)kGt,kkL1(R)=kp0,k

29、(g)kL1(R)ek2D(t).y?k 6=0 pk(t,v)P?p(t,x,v)?C1p0(t,v)A?m?!222022 Alibaba Global Mathematics CompetitionSingle-Choice Problem:Magic Magnetic CubeDivide a solid cube ABCD A1B1C1D1(with AB=1)into 12 pieces(Figure 1)as follows:1)Take 6 diagonals of its surfaces AC,AB1,AD1,C1B,C1D,C1A1;2)Consider all tria

30、ngles with the center of the cube as a vertex,and one of the above 6diagonals and 12 edges as the opposite side;3)These 18 triangles cut the cube into 12 tetrahedra,and each tetrahedron has two edgesthat are cube edges;4)Each tetrahedron is connected to other tetrahedra only by its two cube edges.Fi

31、gure 1:Magic magnetic cubeSuch a toy can take on a variety of shapes(Figure 2).Figure 2:Examples of various shapesQuestion:Of all the possible shapes of this toy,what is the maximum distance(in space)between two points on it?111p7+42131+22None of the above2Single-Choice Problem:Onlook With DistanceO

32、ne day,there is a Street Art Show at somewhere,and there are some spectators around.We consider this place as an Euclidean plane.Let K be the center of the show.And namethe spectators by A1,A2,.,An,.They pick their positionsP1,P2,.,Pn,.,one by one.The positions need to satisfy the following three co

33、nditions simultaneously.(i)The distance between K and Anis no less than 10 meters,that is,KPn 10m holdsfor any positive integer n.(ii)The distance between Anand any previous spectator is no less than 1 meter,that is,PmPn 1m holds for any n 2 and any 1 m n 1.(iii)Analways choose the position closest

34、to K that satisfies(i)and(ii),that is,KPnreaches its minimum possible value.If there are more than one point that satisfy(i)and(ii)and have the minimum distance to K,Anmay choose any one of them.For example,A1is not restricted by(ii),so he may choose any point on the circle C whichis centered at K w

35、ith radius 10 meters.For A2,since there are lots of points on C which areat least 1 meter apart from P1,he may choose anyone of them.(1)Which of the following statement is true?There exist positive real numbers c1,c2such that for any positive integer n,no matter how A1,A2,.,Anchoose their positions,

36、c1 KPn c2alwayshold(unit:meter);There exist positive real numbers c1,c2such that for any positive integer n,no matter how A1,A2,.,Anchoose their positions,c1n KPn c2nalways hold(unit:meter);There exist positive real numbers c1,c2such that for any positive integer n,no matter how A1,A2,.,Anchoose the

37、ir positions,c1n KPn c2nalways hold(unit:meter);There exist positive real numbers c1,c2such that for any positive integer n,no matter how A1,A2,.,Anchoose their positions,c1n2 KPn c2n2always hold(unit:meter).(2)Since human bodies are 3-dimensional,if one spectators position is near another spec-tato

38、rs path of view,then the second ones sight will be blocked by the first one.Supposethat for different i,j,if the circle centered at Piwith radius16meter intersects with segmentKPj,then Ajs sight will be blocked by Ai,and Ajcould not see the entire show.Which of the following statement is true?If the

39、re were 60 spectators,then some of them could not see the entire show;If there were 60 spectators,then it is possible that all spectators could see theentire show,but if there were 800 spectators,then some of them could not seethe entire show;3If there were 800 spectators,then it is possible that al

40、l spectators could see theentire show,but if there were 10000 spectators,then some of them could not seethe entire show;If there were 10000 spectators,then it is possible that all spectators could seethe entire show.4Single-Choice Problem:Tiger Mystery BoxBrave NiuNiu(a milk drink company)organizes

41、a promotion during the Chinese New Year:one gets a red packet when buying a carton of milk of their brand,and there is one of thefollowing characters in the red packet“m”(Tiger),“)”(Gain),“%”(Strength).If one collects two“m”,one“)”and one“%”,then they form a Chinese phrases“mm)%”(Pronunciation:hu hu

42、 sheng wei),which means“Have the courage and strength of thetiger”.This is a nice blessing because the Chinese zodiac sign for the year 2022 is tiger.Soon,the product of Brave NiuNiu becomes quite popular and people hope to get a collection of“mm)%”.Suppose that the characters in every packet are in

43、dependently random,andeach character has probability13.(1)What is the expectation of cartons of milk to collect“mm)%”(i.e.one collects at least2 copies of“m”,1 copy of“)”,1 copy of“%”)?613713813913None of the above(2)In a weekly meeting of Brave NiuNiu,its market team notices that one often has to c

44、ollecttoo many“)”and“%”,before getting a collection of“mm)%”.Thus an improved planis needed for the proportion of characters.Suppose that the probability distribution of“m”,“)”and“%”is(p,q,r),then which of the following plans has the smallest expectation(among the 4)for a collection of“mm)%”?(p,q,r)

45、=(13,13,13)(p,q,r)=(12,14,14)(p,q,r)=(25,310,310)(p,q,r)=(34,18,18)5Given a set X and a function f:X X 0,1,we say f is right uniform if for any?0,there exist finitely many elements b1,b2,.,bmin X such that for any t X,the followingholds for some bi(t):|f(x,t)f(x,bi(t)|0,there exist finitely many ele

46、mentsa1,a2,.,anin X such that for any t X,the following holds for some ai(t):|f(t,x)f(ai(t),x)|?,x X.Prove that f is right uniform if and only if its left uniform.6Let n be a positive integer and V=Rnbe an n-dimensional Euclidean space with a basisei=(0,.,0|z i1,1,0,.,0|z ni)(1 i n)and with an inner

47、 product(,)defined by(ei,ej)=i,jwherei,j=?1 if i=j0 if i 6=jis Kroneckers symbol.For a nonzero vector v V,define sv:V V bysv(u)=u 2(u,v)(v,v)v,u V.For an integer k between 0 and n,write Grk(V)for the set of k-dimensional subspaces of V.For a k-dimensional subspace W of V,write W for the correspondin

48、g element of Grk(V).Choose an orthonomal basis v1,.,vk of W,define sW:V V bysW=sv1svk.(1)Prove that sWis independent of the choice of an orthonomal basis v1,.,vk.(2)Prove that s2W=id.(3)For another element W0 Grk(V),definetW(W0)=sW(W0),where sW(W0)is the image of W0under sW.We call a subset X of Grk

49、(V)a“niceset”iftW(W0)=W0,W,W0 X.Find the maximal cardinality of a“nice set”in Grk(V)and prove it.7A Busy CourierA courier picks up a package at coordinate(n,0)in the two-dimensional lattice,while hisstation locates at the origin(0,0).The courier then does a discrete time simple random walkon Z2.In t

50、he rest of this question,you may without loss of generality assume n is sufficientlylarge.(1)Let P1,nbe the probability that at his bn1.5cth step,the distance between this courierand his station is greater thann2.Prove thatlimn+P1,n=1(2)Let P2,nbe the probability that the courier has ever reached th