đề thi tim kiếm tài năn tóan học quốc tế

8/8/2019 thi tim kim ti nn tan hc quc t

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KOREAGERMAN

International Mathematical Talent Search (IMTS)[Cuc thi Tm kim Ti nng Ton hc Quc t]Translated from English-progenitor by Hn Ngc c [KoreaGerman]

Ch : Cc ch c du (?) c th cha chnh xc! ngh bn c kim tra v gp vi ngi dch

Copyright 2004 by KoreaGerman & www. diendantoanhoc.net


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Tm kim Ti nng Ton hc Quc t diendantoanhoc.net

International Mathematical Talent Search - 1 - [email protected]

IMTS vng 1Bi ton 1/1. Vi mi s nguyn dng n, lp s n/s(n), s(n) l tng cc

ch s ca n trong h thp phn. Tnh gi tr nh nht ca n/s(n) trong mitrng hp sau:

(i) 10 n 99(ii) 100 n 999(iii) 1000 n 9999(iv) 10000 n 99999Bi ton 2/1. Tm tt c cc cp s nguyn (n, k), 2 < k < n, sao cho cc s ,, to thnh mt dy s tng.

1knC

knC

1knC+

Bi ton 3/1. Trn mt bng c 8 x 8 ngi ta t n qun c min, mi qunchim hai k nhau, sao cho khng th t thm qun min no vo cc cnli. Hi gi tr nh nht ca n l bao nhiu trng thi trn cn ng?Bi ton 4/1. Chng minh rng mt tam gic nhn tu c th b ct ra bi ccon thng thnh ba phn theo ba cch khc nhau sao cho mi phn c mt trci xng.Bi ton 5/1. Chng minh rng c th chia mt t din tu thnh 6 phn bicc mt phng hoc phn mt phng sao cho mi mt phn c mt mt phng ixng.


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IMTS vng 2Bi ton 1/2. S nguyn nh nht l bi s ca 9997, khc 9997, ch cha mt

ch s l, l bao nhiu?Bi ton 2/2. Chng minh rng mi tam gic c th chia thnh 9 hnh ng gicli khng suy bin.Bi ton 3/2. Chng minh rng nu x, y v z l cc s nguyn dng i mtnguyn t cng nhau, v nu

z

1

y

1

x

1=+ th x + y, x - z v y - z l cc s chnh

phng.Bi ton 4/2. Cho a, b, c v d l din tch ca cc mt tam gic ca mt t dinv ha, hb, hc, v hd l cc ng cao tng ng ca t din. K hiu V l th tch

ca t din, chng minh rng(a + b + c + d)(ha + hb + hc + hd) 48V.Bi ton 5/2. Chng minh rng c v s s nguyn dng n sao cho hnh hpc n n n khng th c ghp bi cc khi lp phng c 2 2 2 v 3 3 3.


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IMTS vng 3Bi ton 1/3. Ch rng nu tch ca hai phn t khc nhau ca tp {1, 16, 27}

tng thm 9 th kt qu l mt s chnh phng. Hy tm cc s nguyn dng nsao cho n + 9, 16n + 9, v 27n + 9 cng l cc s chnh phng.Bi ton 2/3. Ch rng 1990 c th "tr thnh mt s chnh phng" (turnedinto a square) bng cch thm mt ch s vo bn phi n v mt s ch s bntri n; chng hn 419904 = 6482. Chng minh rng 1991 khng th tr thnh mts chnh phng bng cch trn; c ngha l, khng tn ti cc ch s d, x, y, ...sao cho ...yx1991d l mt s chnh phng.Bi ton 3/3. Tm k nu P, Q, R v S l cc im trn cc cnh ca t gicABCD sao cho

kSADSRDCRQCBQPBAP ==== ,v din tch ca t gic PQRS bng ng 52% din tch ca t gic ABCD.

P

QS

CD

A B

Bi ton 4/3. Cho n im vi cc to nguyn trn mt phng to xy. Gitr nh nht ca n chc chn rng c 3 trong cc im trn l cc nh ca mttam gic vi din tch nguyn (chp nhn 0), l bao nhiu?Bi ton 5/3. Hai ngi, A v B chi tr chi vi mt c bi 32 l. A l ngibt u, v tip hai ngi chi xen k lun phin nhau. Mi ngi ly hocmt l bi hoc mt s nguyn t l bi. Cui cng tt c cc l bi c chn, vngi khng ly l bi cui cng l ngi thua. Ai s thng nu h u chi theochin lc ti u?


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IMTS vng 4Bi ton 1/4. Dng mi ch s 1, 2, 3, 4, 5, 6, 7, 8, 9 ng hai ln lp cc s

nguyn t khc nhau sao m tng cng nh cng tt. Tng ny nh nht phibng bao nhiu? (Ch : 5 s nguyn t nh nht l 2, 3, 5, 7, v 11).Bi ton 2/4. Tm s nguyn dng n nh nht sao m c th biu din thnhtng ca cc s nguyn dng phn bit a, b, v c sao cho a + b, a + c v b + c lcc s chnh phng?Bi ton 3/4. Chng minh rng mt s nguyn dng c th biu din didng 3x2 + y2 nu v ch nu n cng c th biu din di dng u2 + uv + v2, x, y, u, v v l cc s nguyn dng.Bi ton 4/4. Cho ABC tu , dng P, Q v R sao cho mi mt gc to thnhl 30

0

. Chng minh rng PQR l tam gic u.C

P

B

Q

A

Bi ton 5/4. S o cc cnh ca ABC l 11, 20 v 21 n v. Ta gp n dctheo PQ, QR v RP, P, Q, R l cc trung im ca cc cnh ca tam gic, chon khi A, B, v C trng nhau. Hi th tch ca t din to thnh l bao nhiu?


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IMTS vng 5Bi ton 1/5.Tp hp S gm 5 s nguyn. Nu mi cp phn t phn bit ca S

c cng vi nhau th 10 tng thu c l 1967, 1972, 1973, 1974, 1975, 1980,1983, 1984, 1989, 1991. Tm nhng phn t ca S?Bi ton 2/5.Cho cc s nguyn n 3 v k 2, v hnh thnh cc hiu s (saiphn tin - forward difference) ca cc phn t ca dy

1, n, n2, ..., nk - 1

v c th ly cc hiu s lin tip ca cc dng trn c dng di, nh chra trn bng di vi (n, k) = (3, 5). Chng minh rng cc kt qu (s dng cuicng) l khc nhau vi cc cp (n, k) khc nhau.

1 3 9 27 812 6 18 544 12 36

8 2416

Bi ton 3/5.Trong mt trn bng chy "kiu ton hc" (mathematical version),trng ti chn mt s nguyn dng m, mn, v bn phi on mt s nguyndng bit c m. Nu s bn on nh hn smca trng ti th ng ta gin l mt "bng" (ball); nu s ln hn hoc bng mth trng ti gi l mt

"c nh" (strike). "nh trng" (hit) n, bn phi tm c ng gi tr ca msau c nh th 3 hoc ln on th 6, tu theo ci no trc. Gi tr ln nht canl bao nhiu tn ti mt chin lc cho php bn nh (bat) c 1000 im,ngha l lun tm ng m? Hy nu chi tit chin lc .Bi ton 4/5.Chng minh rng nu f l mt hm thc khc hm hng sao chovi mi x ta c

f(x + 1) + f(x - 1) = 3 f(x).th f l hm tun hon. Tm s dng p nh nht sao cho f(x + p) = f(x) vi mi x?Bi ton 5/5.Trong ABC (hnh v), gi r l bn knh ng trn ni tip. GirA, rB, rC l cc bn knh ca cc ng trn tip xc vi ng trn ni tip v vicc cnh ca tam gic, tng ng vi cc nh A, B, C. Chng minh rng:

r rA + rB + rC.

rC

rA

rrB

C

AB


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IMTS vng 6Bi ton 1/6. Chn ng thng, cng song song vi mt cnh ca mt tam

gic v chia mi cnh cn li thnh 10 on bng nhau v chia din tch thnh 10phn khc nhau. Tm din tch ca tam gic ban u, nu din tch ca phn lnnht l 76.Bi ton 2/6.C bao nhiu cch biu din 1992 di dng tng ca mt hocnhiu cc s nguyn lin tip?Bi ton 3/6.Chng minh rng tn ti mt lc gic c cc gc bng nhau trnmt phng m di cc cnh ca n l 5, 8, 11, 14, 23 v 29 n v, theo mt tht no .Bi ton 4/6.Mt cng ty quc t c 250 nhn vin, mi ngi c th ni vingn ng. Trong mi cp nhn vin (A, B), c mt ngn ng

c A ni m Bkhng v c mt ngn ng c B ni m A khng. C t nht bao nhiu ngnng c ni trong cng ty?Bi ton 5/6. Mt bn c v hn (infinite checker-board) c chia bi mtng k nm ngang thnh na trn v na di nhhnh v. Mt s qun c c t vo bn c bn di ng k (trong cc ). Mt "bc i" (move) lmt qun c nhy dc hoc ngang qua mt qun c khc v n qun c .

Gi tr nh nht ca n c th t qun c cui cng vo dng th 4 phatrn ng k ngang sau n- 1 bc i l bao nhiu? Hy m t v tr ban u ca

cc qun c v tng bc i.


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IMTS vng 7Bi ton 1/7.Trong mt hnh thang ABCD, hai ng cho ct nhau ti E. Din

tch ca ABE l 72, v din tch ca CDE l 50. Din tch ca hnh thang ABCDl bao nhiu?Bi ton 2/7.Chng minh rng nu a, b v c l cc s nguyn dng sao cho c2= a2 + b2, th c c2 + ab v c2 - ab u c th biu din di dng tng ca hai schnh phng.Bi ton 3/7.Vi n l mt s nguyn dng, k hiu P(n) l tch ca tt c ccc s nguyn dng ca n. Tm n nh nht

P(P(P(n))) > 1012.Bi ton 4/7.Khi chp li trn bng mt dy 6 s nguyn dng l mt cp scng, mt sinh vin vit 5 s: 113, 137, 149, 155, 173, v b st mt s. Sau cuta nhn ra rng cng chp sai mt s trong chng. Bn hy gip cu ta v tmli dy ban u.Bi ton 5/7.Cho T = (a, b, c) l mt tam gic vi cc cnh a, b, v c v din tch. K hiu T' = (a', b', c') l tam gic vi cc cnh l ng cao ca T (ngha l a' =ha, b' = hb, c' = hc) v k hiu din tch ca n l '. Tng t k hiu T'' = (a'', b'', c'')l tam gic to bi cc ng cao ca T', v k hiu din tch ca n l ''. Cho ' =30 v '' = 20. Hy tm .


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IMTS vng 8Bi ton 1/8.Chng minh rng khng c tam gic no c di cc ng cao

l 4, 7 v 10 n v.Bi ton 2/8.Nh trn hnh v, c mt s thc x, 0 < x < 1, sao cho ta c thchia hnh vung n v thnh 7 tam gic ng dng. Khi x phi tho mn mta thc dng chun bc 5. Tm a thc . (a thc dng chun l a thc c hs ca bc cao nht ca x l 1.)Bi ton 3/8.(i) C th sp xp li cc s 1, 2, ..., 9 thnh a(1), a(2), ..., a(9) sao

cho cc s sau y i mt khc nhau hay khng? Chng minh khng nh ca

bn:

1

1

x

|a(1) - 1|, |a(2) - 2|, ..., |a(9) - 9|.(ii) C th sp xp li cc s 1, 2, ..., 10 thnh a(1), a(2), ..., a(10) sao cho cc s

sau y i mt khc nhau hay khng?. Chng minh khng nh ca bn:|a(1) - 1|, |a(2) - 2|, ..., |a(10) - 10|.Bi ton 4/8.Trong mt cuc chy 50 mt, Anita c th chp Bob nhiu nht 4

mt v ui kp cu ta ti vch ch. Trong mt cuc chy 200 mt, Bob c thchp Carol nhiu nht 15 mt v ui kp c ta ti vch ch. Gi s rng tt c 3ngi lun chy vi mt vn tc khng i. Hi Anita c th chp Carol nhiunht bao nhiu mt trong mt cuc chy ua 1000 mt m vn c th ui kp cta.Bi ton 5/8.Cho cc s thc a, b, x v y tho mn:

a + b = 23,ax + by = 79,ax2 + by2 = 217,ax3 + by3 = 691,Hy tnh ax4 + by4.


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IMTS vng 9Bi ton 1/9.Mt li m x n c t sao cho mt gc ca n ti (0, 0) v mt

gc ti (m, n). Mt bc i hp l c nh ngha l mt bc chuyn mt nv theo hng dng ca y hoc mt n v theo hng dng ca x. im (i, j)ca li, vi 0 i m v 0 j n, b b i v khng c ng i no qua n bngcc bc i n im (m, n).

C bao nhiu con ng c th i t (0, 0) n (m, n), bng cc bc i nhvy?Bi ton 2/9.Cho mt im P v hai on thng trn mt mnh giy hnh chnht sao m giao im Q ca cc ng thng cha chng khng nm trn mnhgiy. Lm th no dng ng thng PQ vi mt ci thc nu ch c php

v trong phm vi ca mnh giy? (*)Bi ton 3/9. Mt a gic li c 1993 nh c t mu sao cho hai nh knhau c mu khc nhau. Chng minh rng ngi ta c th chia a gic thnh cctam gic bng cc ng cho khng giao nhau m hai u mt c mu khcnhau.Bi ton 4/9.Mt tam gic c gi l Heronian nu s o cc cnh v dintch ca n l cc s nguyn. Xc nh tt c 5 tam gic Heronian c chu vi v dintch cng bng mt s nguyn.Bi ton 5/9.Mt b gm 5 "Hnh lp phng k o" (con xc sc - Trick MathCubes) c cho di dng khai trin phng nh hnh v. Mt "Php s"

(magician) yu cu bn gieo chng v cng 5 s mt trn li. ng ta cng nhmv vit ngay ra kt qu vo mt t giy trc khi bn cng xong! ng ta lmth no? Trnh by v gii thch tr mo ny.

179 564278 377 872 762 366 861

971 168773 780 960

483 186 285384

741 681 756147 543 840 855 558 459

642 954345 657


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IMTS vng 10Bi ton 1/10.Tm x2 + y2 + z2 nu x, y v z l cc s nguyn dng tho mn

7x2 - 3y2 + 4z2 = 8 v 16x2 - 7y2 + 9z2 = - 3.Bi ton 2/10.Tc lng n gin 1 < 3 < 2, hy suy ra rng 6 < 33 < 7.Bi ton 3/10.Vi mi s nguyn dng n, n 2, hy xc nh hm s:fn(x) = an + bnx + cn|x - dn|,

an, bn, cn, dn ch ph thuc vo n, sao cho:fn(k) = k + 1 vi k = 1, 2, ..., n - 1 v fn(n) = 1.Bi ton 4/10.Mt ti cha 1993 bng v 1993 bng en. Chng ta hy ly

i hai bng mt lc nhiu ln v(i) Loi b chng nu chng c cng mu.

(ii) Nu chng khc mu, loi b bng en v b li bng vo ti.Hi xc sut qu trnh s kt thc vi mt qu bng cn li trong ti lbao nhiu?Bi ton 5/10.Cho P l mt im trn vng trn ngoi tip ca ABC khngtrng vi A, B v C. Gi s rng BP ct AC ti X v CP ct AB ti Y. Cho Q l giaoim khc A ca hai vng trn ngoi tip ABC v AXY. Chng minh rng PQchia i on XY. (Cc giao im khc nhau c th xy ra vi phn ko di cacc on thng).


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IMTS vng 11

Bi ton 1/11. Hy biu din 9419

di dng n1

m

1+ , vi m v n l hai s

nguyn dng.Bi ton 2/11.Cho n l mt s nguyn ln hn 5. Chng minh rng c nhiunht 8 phn t ca tp {n + 1, n + 2, ..., n + 30} l s nguyn t.Bi ton 3/11.Mt 2n-gic li c gi l "Thoi gic" (rhombic) nu tt c cccnh ca n bng n v di v nu hai cnh i din l song song vi nhau. Mtth d minh ho c cho trn hnh v (vi n = 4), mt Thoi gic 2n-gic c thc chia thnh cc hnh thoi cnh 1 theo nhiu cch khc nhau. Vi gi tr noca n mt Thoi gic 2n-gic c th c chia thnh 666 hnh thoi?

Bi ton 4/11. Chng minh rng nu 3 trong 4 ng phn gic trong ca mtt din ng quy th c 4

ng phn gic y phi ng quy ti mt im.Bi ton 5/11. Cho f(x) = x4 + 17x3 + 80x2 + 203x + 125. Tm a thc bc nhnht g(x) sao cho f( 33 ) = g( 33 ) v f( 55 ) = g( 55 ).


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IMTS vng 12Bi ton 1/12.Thy gio vit mt s nguyn dng nh hn 50.000 ln bng.

Mt sinh vin xc nhn l mt bi ca 2; mt sinh vin th hai ni rng n lmt bi ca 3; v c nh vy... cho n khi sinh vin th 12 ni rng n mt bica 13. Thy gio quan st v thy rng tr hai sinh vin ra th tt c nhng ngikhc ni ng. V hai sinh vin pht biu sai ni cch nhau mt ngi khc.

Hi s c vit trn bng l s no?Bi ton 2/12.Mt hnh 12-gic u c ni tip trong mt hnh vung dintch bng 24 nhhnh v, 4 nh ca hnh 12-gic l trung im ca cc cnhhnh vung. Tm din tch ca hnh 12-gic u.

Bi ton 3/12.Cho S l tp hp gm 30 im trn mt phng, c tnh cht lkhong cch gia hai im phn bit bt k trong S nh nht bng 1. nh ngha Tl tp con ln nht ca S sao cho khong cch gia hai im phn bit bt k ca Tl nh nht bng 3 . C bao nhiu im trong tp T?

Bi ton 4/12. Chng minh rng nu 33 42 + l nghim ca mt a thc bcba vi cc h s nguyn, th n l nghim thc duy nht ca a thc ny.Bi ton 5/12. Trong hnh v, l1 v l2 l hai ng thng song song, AB l onvung gc vi chng, v P, Q, R, S l cc giao im ca l1 v l2 vi mt ng trnvi ng knh ln hn AB v tm C nm trn on thng AB. Chng minh rngtch PS.PR khng ph thuc vo v tr ca im C trn on thng AB.

l2

l1S

P

B

A

C

Q


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IMTS vng 13Bi ton 1/13. Milo l mt sinh vin trng Mindbender High. Sau mi bi

kim tra, cu ta tnh im trung bnh ca mnh v lun lm trn ti phn trmgn nht. (85,49 lm trn xung thnh 85, nhng 85,50 th lm trn ln thnh 86).Hm nay cu ta c hai bi kim tra. Bi u cu ta c 75 im mn ting Php,lm gim im trung bnh ca cu ta i 1 im. Sau cu ta c 83 im mnlch s, lm tng im trung bnh ca cu ta ln 2 im.

Hi im trung bnh ca Milo hin nay l bao nhiu?Bi ton 2/13. Erin ang ngh ra mt tr chi v mun chn 4 loi ng tin tcc loi $1, $2, $3, $5, $10, $20, $25 v $50 chi vi tin. Cu ta chn chngnh th no m mi gi tr t $1 n $120 c th nhn c bng cch dng

nhiu nht l 7 t bc.Bi ton 3/13. Vi s nguyn dng d no th c th t mu cc s nguyn mu hoc xanh sao cho khng c hai im no cch nhau d v khng c haiim xanh no cch nhau 1?Bi ton 4/13.Chng minh rng c v hn b ba s nguyn dng sp th t(x, y, z) sao cho

x3 + y5 = z7.Bi ton 5/13. Ch c trang b mt compa (khng c cnh thng) v haing trn ct nhau theo mt gc vung; tc l dng nhng ng trn lnnhau trong cng mt mt phng, trc giao (cc tip tuyn ti ln lt hai giao

im ca chng vung gc).


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IMTS vng 14Bi ton 1/14. Cho a, b, c, d l cc s dng sao cho a2 + b2 + (a - b)2 = c2 + d2 +

(c - d)2. Chng minh rng a4 + b4 + (a - b)4 = c4 + d4 + (c - d)4.Bi ton 2/14. Cc nhn gi trong mt ca hng bch ho ghi nh sau:$ 2.00 $ 5.50$ 0.75

Ch rng tng ca ba gi trn l 8,25 $ v tch ca ba s cng bng 8,25. Hytm 4 nhn gi c tng bng 8,25 $ v tch bn s cng bng 8,25.Bi ton 3/14.Trong mt nhm 8 nh ton hc, mi v thy rng c ng 3 vkhc c mt s thch chung. C th ghp h tng i mt sao cho trong mi cp

(trong 4 cp) , hai v khng c chung s thch no hay khng?Bi ton 4/14. Cho cc s nguyn dng a v b. K hiu a ~ b nu ab + 1 l mts chnh phng. Chng minh rng nu a ~ b, th tn ti mt s nguyn dng csao cho a ~ c v b ~ c.Bi ton 5/14. Cho tam gic ABC, ko di cc cnh dng thnh hai hnh lcgic nhhnh v. So snh din tch ca hai lc gic .


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IMTS vng 15Bi ton 1/15.C th ghp i cc s nguyn dng 1,2, 3, ..., 50 sao cho tng

ca mi cp s l cc s nguyn t khc nhau hay khng?Bi ton 2/15.Hy thay th cc ch ci khc nhau bi cc ch s khc nhautrong {0, 1, ..., 9} sao cho cc php cng sau ng. (Hai php ton c lp vinhau).

H A R R I E T D I A N AM A R R I E D A N D

+ H E R S A R A HD E N T I S T + A R E

R E B E L S

Bi ton 3/15.Hai hnh chp c chung y 7 cnh, vi cc nh k hiu l A1,A2, ..., A7. Chng c cc nh khc l B v C. Khng c ba im no trong 9 iml thng hng. Mi cnh trong 14 cnh BAi v CAi (i = 1, ..., 7), 14 ng cho cay, v on BC, c t mu xanh hoc . Chng minh rng c ba on trongchng c cng mu v l ba cnh ca mt tam gic.Bi ton 4/15. Gi thit cho cc s nguyn dng a, b, c v x, y, z tho mn a2 +b2 = c2 v x2 + y2 = z2. Chng minh rng:

(a + x)2 + (b + y)2 (c + z)2.V hy xc nh khi no du bng xy ra?

Bi ton 5/15. Cho C1 v C2 l hai ng trn ct nhau ti A v B. C0 l ngtrn qua A vi tm B. Hy xc nh iu kin cho dy cung chung ca C0 v C1tip xc vi C2?


17/45



IMTS vng 16Bi ton 1/16. Chng minh rng nu a + b + c = 0 th a3 + b3 + c3 = 3abc.Bi ton 2/16.Vi n l mt s nguyn dng, t P(n) l tch ca cc ch s

khc khng trong h thp phn ca n. Ta gi n l "prodigitious" nu P(n) chia htn. Chng minh rng khng th c mt dy 14 s nguyn dng lin tip m tt clprodigitious.Bi ton 3/16. Cc a c nh s t 1 n n v xp vo mt hng cc vung vi mt vung trng. Mtbc chuynbao gm ly i mt a v dichuyn n vo trng. Mc ch l sp xp li cc a vi s bc chuyn t nhtsao cho a 1 trong vung 1, a 2 trong vung 2, ..., a n trong vung n, v vung cui cng l trng. Chng hn, nu v tr ban u l

3 2 1 6 5 4 9 8 7 12 11 10th mt t nht 14 bc chuyn; c th l chng ta c th di chuyn cc a vo trng theo th t sau: 7, 10, 3, 1, 3, 6, 4, 6, 9, 8, 9, 12, 11, 12.

Hi v tr ban u i hi s ln nht cc bc chuyn l th no nu n = 1995?Ch r s bc chuyn cn thit.Bi ton 4/16. Cho ABCD l mt t gic li tu , vi E, F, G, H l nhng trungim ca cc cnh nhc ch ra trn hnh v. Chng minh rng ngi ta c thghp cc mnh tam gic AEH, BEF, CFG, DGH li vi nhau thnh mt hnh

bnh hnh tng ng (congruent) vi hnh bnh hnh EFGH.C

H

F

G

D

A E B

Bi ton 5/16. Mt hnh bt gic ng gic (cc gc bng nhau) ABCDEFGH ccc cnh vi di l 2, 2 2 , 4, 24 , 6, 7, 7, 8. Cho AB = 8, hy tm di EF.


18/45



IMTS vng 17Bi ton 1/17.S 154 ch s 19202122 ...939495 nhn c bng cch vit lin

nhau cc s nguyn t 19 n 95 theo chiu tng dn. Chng ta xo i 95 ch sca s c mt s ln nht c th c. Hi 19 ch s ca s c 59 ch sny l g?Bi ton 2/17. Tm tt c cc cp s nguyn dng (m, n) sao m m2 - n2 = 1995.Bi ton 3/17. Chng minh rng c th sp xp trn mt phng 8 im sao chokhng c 5 im no trong chng l cc nh ca mt ng gic li. (Mt ng gicl li nutt c cc gc trong ca n khng vt qu 180 ).Bi ton 4/17. Mt ngi n ng hn v ng ta 6 tui. ng ta cho bit 4 nmtrc ng ta ci v c ng mt na s tui ng ta. ng ta s bao nhiu

tui vo l k nim 50 nm ngy c

i nu vo 10 nm ti th b v tri qua 2/3cuc i ly ng ta? (*)Bi ton 5/17.S b nht cc hnh ch nht 3 x 5 l bao nhiu s ph kn mthnh vung 26 x 26? Cc hnh ch nht c th ln nhau v/hoc ln cc cnhca hnh vung. Bn nn chng minh kt lun ca bn bng mt v d c th.


19/45



IMTS vng 18Bi ton 1/18. Xc nh di nh nht ca khong [a, b] sao cho a x + y b

vi mi s thc x y 0 tho mn 19x + 95y = 1995.Bi ton 2/18.Vi s nguyn dng n 2, k hiu P(n) l tch ca cc c snguyn dng (bao gm 1 v n) ca n. Tm s n nh nht P(n) = n10.Bi ton 3/18. Cho th (graph) trn hnh v vi 10nh, 15 cnh v mi nh c bc 3 (ngha l c 3 cnhni vi mi nh). Mt s cnh c nh s 1, 2, 3, 4,5 nhhnh v. Chng minh rng c th nh s 6, 7, ...,15 cc cnh cn li sao cho ti mi nh tng ca cc strn cc cnh ni vi n l bng nhau.

Bi ton 4/18. Cho a, b, c, d l cc s thc khc nhausao cho a + b + c + d = 3 v a2 + b2 + c2 + d2 = 45.

Tm gi tr ca biu thc:

)db)(cb)(ab(b

)da)(ca)(ba(a 55

+

.

)cd)(bd)(ad(d

)dc)(bc)(ac(c 55

+

+

Bi ton 5/18. Cho a v b l hai ng thng trn mt phng, v C l mt imnh hnh v. Ch dng mt compa v mt thc thng khng chia vch, hydng mt tam gic vung cn ABC, sao cho A nm trn a, B nm trn b v AB lcnh huyn ca tam gic ABC.

b

a

C


20/45



IMTS vng 19Bi ton 1/19. C th thay th mi du di y thnh - hoc +

1 2 3 4 5 ... 96 = 1996.Nhiu nht bao nhiu du c th c thay bi du +?Bi ton 2/19.Ta ni (a, b, c) l mtprimitive Heronian triplenu a, b v c l

cc s nguyn dng khng c tha s chung (khc 1), v din tch ca tam gicm di cc cnh a, b, c cng l mt s nguyn. Chng minh rng nu a = 96 thb v c phi cng l l.Bi ton 3/19.Cc s trong hnh ch nht 7 x 8 trn hnh v c to thnhbng cch t 28 qun min khc nhau ca mt b c thng thng, ghi li ccs (s chm, t 0 n 6) trn mi mt ca cc qun min v xo i bin ca

chng. Xc nh li bin ban u ca cc qun min. (Ch : Mi qun mincha hai vung lin nhau bi mt cnh).

5 5 5 2 1 3 3 46 4 4 2 1 1 5 26 3 3 2 1 6 0 33 0 5 5 0 0 0 63 2 1 6 0 0 4 20 3 6 4 6 2 6 52 1 1 4 4 4 1 5

Bi ton 4/19. Gi s rng hm f tho mn phng trnh hm sau2f(x) + 3

+

2x29x2f = 100x + 80.

Hy tm f(3).Bi ton 5/19.Trong hnh v, hy xc nh din tch ca hnh bt gic t mtheo din tch ca hnh vung. Trong cc bin ca hnh bt gic u l ccng thng k t cc nh ca hnh vung n trung im ca cc cnh idin.


21/45



IMTS vng 20Bi ton 1/20. Hy xc nh s im (x, y) trn hyperbol

2xy - 5x + y = 55sao m c x v y l cc s nguyn.Bi ton 2/20. Tm gi tr nh nht ca n sao cho pht biu sau y l ng:Vi mi tp hp gm n s nguyn dng ngi ta lun lun chn ra c by snguyn dng c tng chia ht cho 7.Bi ton 3/20. Nhng ng chng ca 11 nh ton hc i theo cc b v ca hn d mt cuc meeting. Thnh thong nhng ngi chng i qua mt ngikhc trn i snh, nhng mt khi i no i qua nhau mt ln, th hkhng bao gi i qua nhau ln no na. Khi h qua i qua nhau, hoc ch mt

ng

i nhn thy ng

i kia, hoc h nhn ra nhau, hoc khng ai nhn ra ng

ikhc. Chng ta s quy cho s kin mt ng chng nhn ra ngi khc nh l mt"ci nhn" (sighting), v s kin h nhn ra nhau l mt "cuc tn gu" (chat), v lrng trong trng hp ny h ngng li tn gu. Ch rng mi cuc tn gung vi hai ci nhn.

Nu 61 ci nhn xy ra, chng minh rng mt trong nhng ng chng phi ct nht hai cuc tn gu.Bi ton 4/20.Gi s rng a v b l hai s nguyn dng sao cho cc phn sa/(b - 1) v a/b, khi lm trn (theo quy tc thng thng, ngha l gp ch s 5 vln hn th lm trn ln, cn ch s 4 v nh hn th lm trn xung) ti 3 ch s

thp phn sau du phy, th c hai s thp phn c gi tr l 0,333.Bi ton 5/20.Trong hnh v, cc tm ca nhng ng trn C0, C1, v C2 lthng hng, A v B l cc giao im ca C1 v C2, v C l giao im ca C0 vng thng i qua AB. Chng minh rng hai ng trn nh, tip xc vi C0,C1, BC, v C0, C2,BC tng ng, l bng nhau.


22/45



IMTS vng 21Bi ton 1/21. Hy xc nh cc s trong trng ca hnh vung

k o (magic square) trn hnh v sao cho tng ca ba s trn mihng, ct v trn mi ng cho chnh u bng mt hng s k. Sk l bao nhiu?

3331 28

Bi ton 2/21. Tm s nguyn dng nh nht xut hin trong mi cp s cngcho di, v chng minh rng c v hn s nguyn dng xut hin trong c bady ny.

5, 16, 27, 38, 49, 60, 71, ...7, 20, 33, 46, 59, 72, 85, ...8, 22, 36, 50, 64, 78, 92, ...

Bi ton 3/21. Sp xp li cc s nguyn 1, 2, 3, 4, ..., 96, 97 thnh mt dy a1, a2,a3, a4, ..., a96, a97 sao cho gi tr tuyt i ca hiu ca ai+1 v ai l 7 hoc 9 vi mi i= 1, 2, 3, ..., 96.Bi ton 4/21. Tha nhn qu trnh v hn nh trn hnh v bn tri (m ngita thng gi l lin phn s - ND), c li khi biu din mt s thc dng. Hyxc nh s thc ny.

...1

...51

...3

...151

...1

...53

...3

...11

5

3

...1

...51

...3

...15

3

...1

...53

...3

...11

1

1

1

+

++

+

++

+

+

++

+

++

+

+

+

++

+

++

+

+

++

+

++

+

+

+

E

B

D

A

CBi ton 5/21. Cho tam gic cn (trn hnh bn phi) ABC vi ABC = ACB =78

o

. Gi D v E l cc im t

ng ng trn AB v AC sao cho

BCD = 24o

vCBE = 51o. Hy xc nh, vi mt chng minh, gc BED.


23/45



IMTS vng 22Bi ton 1/22. Trong nm 1996 khng ngi no m vo ngy sinh nht ca h

th tui ca h bng tng cc ch s ca nm sinh ca h. Nm cui cng trc1996 c cng tnh cht l nm no?Bi ton 2/22.Hy xc nh s nguyn dng ln nht n c duy nht snguyn dng m sao cho m < n2 v mnmn ++ l mt s nguyn.Bi ton 3/22. Gi th rng hin c 120 triu my in thoi ang c sdng nc Anh. C th phn thnh cc s-in-thoi-10-ch-s (trong khong(0, 1, ..., 9) cho h sao cho bt k li n no khi quay s u c th pht hin vkhc phc hay khng? (V d, nu mt thu bao l 812-877-2917 v nu mt cucquay s sai thnh 812-872-2917, th khng mt s no khc m sai khc 812-872-

2917 mt ch s, l s ca mt thu bao no ).Bi ton 4/22. Nh trn hnh v u, mt khi g lp phng rng c mt gcb ca thnh mt t din ABCD. Hy xc nh di ca CD nu AD = 6, BD = 8v dt(ABC) = 74.Bi ton 5/22.Nh ch ra trn hnh th hai, trong mt hnh vung cnh 96 cmt ng trn bn knh r1, c hai ng trn bn knh l r2 v c ba ng trnbn knh l r3. Tt c cc ng trn tip xc vi ng thng v/hoc vi mtng khc nh ch ra, v ng trn nh nht i qua nh ca tam gic nhhnh v. Hy tm cc bn knh r1, r2, v r3.


24/45



IMTS vng 23Bi ton 1/23. Trong php cng trn hnh v, mi ch ci thay th cho mt chs khc nhau t 0 n 9. Hy xc nh chng sao cho tng s l ln nht c thc. Gi tr ca GB l bao nhiu vi cc ch s thay th?

A R L OB A R TB R A D

E L T O N+ R O G E RBi ton 2/23. Ta s gi s nguyn n l s may mnnu n c th biu din

d

i dng 3x2

+ 32y2

, trong x v y l cc s nguyn. Chng minh rng nu n ls may mn th 97n cng vy.Bi ton 3/23.Hy ch ra trn mt phng 19 ng thng sao cho chng ctnhau ti ng 97 im. Chp nhn rng c th c mt s im c hn hai ngthng giao nhau. Li gii ca bn nn km theo mt hnh v c chun b cnthn.Bi ton 4/23. Chng minh rng:

cot100.cot300.cot500.cot700 = 3.Bi ton 5/23. Tam gic cn ABC b ct ra thnh 13 tam gic nhn cn, nhc ch ra trn hai hnh bn di. Trong cc on c cng di c nh

du ging nhau, v hnh th hai ch ra chi tit cch chia tam gic EFG. Bit rnggc y ca tam gic ABC l mt s nguyn . Hy xc nh .


25/45



IMTS vng 24Bi ton 1/24.Li im ca gc phn t th nht c nh s nh ch ra

trn hnh v. Nh vy, chng hn, im li th 19 l im (2, 3), cn im lith 97 l im (8, 5). Hy xc nh, vi chng minh, im li th 1997 theo ssp xp trn.

o o o o o o o o

22 35 o o o o o o21 23 34 o o o o o11 20 24 33 o o o o10 12 19 25 32 o o o4 9 13 18 26 31 o o

3 5 8 14 17 27 30 o1 2 6 7 15 16 28 29Bi ton 2/24. Cho Nk = 131313... 131 l mt s c (2k + 1) ch s (trong h thp

phn), c to thnh t k + 1 ch s 1 v k ch s 3. Chng minh rng Nk khngchia ht cho 31 vi mi k = 1, 2, 3, ...Bi ton 3/24. Cho tam gic ABC vi AB = 52, BC = 64, CA = 70 v gi s rngP v Q ln lt l cc im c chn trn cc cnh AB v AC sao cho tam gicAPQ v t gic PBCQ c cng din tch v chu vi. Hy xc nh bnh phng

di ca on PQ.Bi ton 4/24. Hy xc nh cc s nguyn dng x < y < z sao cho9719

xyz1

xy1

x1

= .

Bi ton 5/24.Cho P l mt a gic phng li n nh. V t mi nh ca P tak cc ng vung gc vi n - 2 cnh (hoc vi cc ng ko di ca chng)ca P khng cha nh . Chng minh rng hoc mt trong cc ng vunggc nm hon ton trong P, hoc n l mt cnh ca P.


26/45



IMTS vng 25Bi ton 1/25. Gi s rng chng ta c 12 thanh, mi thanh di 13 n v.

Chng c ct thnh cc phn di 3, 4 v 5 n v. T cc phn c th cghp li thnh 13 tam gic vi cc cnh 3, 4 v 5. Cc thanh s c ct nh thno?Bi ton 2/25. Cho f(x) l mt a thc vi cc h s nguyn, v gi s f(0) = 0,f(1) = 2. Chng minh rng f(7) khng phi l mt s chnh phng.Bi ton 3/25.Ngi ta c th chng minh rng vi mi phng trnh bc hai(x - p )( x - q) = 0, tn ti cc hng s a, b v c vi c 0, sao cho phng trnh (x -a)(b - x) = c tng ng vi phng trnh , v lp lun khng hon ton:"Hoc x - a hoc b - x phi bng c" dn n cu tr li ng l "x = p hoc x = q".

Xc nh cc hng s a, b v c, vi c 0, sao cho ph

ng trnh (x - 19)(x - 97) = 0c th "c gii" theo cch .Bi ton 4/25.Cho ABC l mt tam gic khng cn vi AB l cnh di nht.Ko di AB n im D sao cho B nm gia A v B trn on AD v BD = BC.Chng minh rng ACD l gc t.Bi ton 5/25.Cho hnh chp PABCD (hnh v) vi y ABCD l hnh thoi cDAB = 60o. Gi s PC2 = PB2 + PD2. Chng minh rng PA = AB.


27/45



IMTS vng 26Bi ton 1/26. Cho x, y v z l cc s thc dng tho mn cc phng trnh:

x + y + xy = 8,y + z + yz = 15,z + x + zx = 35.

Tnh gi tr ca x + y + z + xyz.Bi ton 2/26. Xc nh s cc a gic u khng ng dng m mi s o gctrong ca chng l mt s nguyn .Bi ton 3/26.Thay th cc ch s khc nhau (0, 1, ..., 9) vo cc ch ci khcnhau trong php tnh sau, sao cho php tnh l ng, v gi tr ca kt quMONEY cng ln cng tt. Gi tr l bao nhiu?

S H O WM E

+ T H EM O N E Y

Bi ton 4/26. Chng minh rng nu a b c > 0, th( )

++++++

ac4

cb

ba

31cabcab

38c5b3a2

222

Bi ton 5/26.Cho ABCD l mt t gic li ni tip trong mt ng trn. GiM l giao im ca hai ng cho ca ABCD, v E, F, G, H l cc chn ca ccng vung gc h t M xung cc cnh ca ABCD, nh trn hnh v. Hy xcnh (vi chng minh) tm ca ng trn ni tip trong t gic EFGH.


28/45



IMTS vng 27Bi ton 1/27.C hay khng cc s nguyn M, N, K sao cho M + N = K v(i) Mi s trong chng cha mi ch s trong by ch s 1, 2, 3, ..., 7 ng mt

ln?(ii) Mi s trong chng cha mi ch s trong chn ch s 1, 2, 3, ..., 9 ng mt

ln?Bi ton 2/27. Gi s R(n) l s cc biu din ca s nguyn dng n didng tng ca cc bnh phng ca bn s nguyn khng m, trong chng taxem hai biu din l trng nhau nu chng ch khc nhau th t ca cc shng. (Chng hn, R(7) = 1 v 22 + 12 + 12 + 12 l biu din duy nht ca 7.)

Chng minh rng nu k l mt s nguyn dng, th R(2k) + R(2k + 1) = 3.

Bi ton 3/27.Gi thit rng f(1) = 0, v vi mi s nguyn m v n ta cf(m + n) = f(m) + f(n) + 3(4mn - 1).Xc nh f(19).Bi ton 4/27.Trong mt phng to vung gc, ABCD l mt hnh vung

v cc im (31, 27), (42, 43), (60, 27), v (46, 16) ln lt nm trn cc cnh AB,BC, CD, DA ca n. Xc nh din tch ca ABCD.Bi ton 5/27.C th dng trung im ca mt on thng cho trc ch sdng compa hay khng? (ngha l, khng s dng thc thng tr khi v onthng ban u).


29/45



IMTS vng 28Bi ton 1/28. Vi s nguyn b v c no th 9819x += l nghim ca

phng trnh x4 + bx2 + c = 0?Bi ton 2/28. Cc cnh ca tam gic c di l a, b, c, a, b, c l cc snguyn, a > b, v s o gc i din vi c l 60o. Chng minh rng a phi l mthp s.Bi ton 3/28. Hy xc nh, km theo mt chng minh ton hc, gi tr ca[x], ngha l s nguyn ln nht nh thua hoc bng x, vi:

10000001

...3

12

11x ++++= .

Bi ton 4/28. Cho n l mt s nguyn dng v gi thit rng vi mi snguyn k, 1 k n, ta c hai a

c nh s k. Yu cu sp xp 2n a trn mthng sao cho vi mi k, 1 k n, c k a nm gia hai a nh s k. Chngminh rng:

(i) nu n = 6, th khng s sp xp no l thch hp;(ii) nu n = 7, th c th sp xp cc a theo yu cu.? Bi ton 5/28.Gi S l tp hp tt c cc im ca mt hnh lp phng n

v (tc l hnh lp phng vi di cc cnh bng 1) m c khong cch ngnnht t bt k nh no ca hnh lp phng bng t tm ca hnh lp phng,ti n. Xc nh hnh dng v th tch ca S.


30/45



IMTS vng 29Bi ton 1/29. Mt vi cp s nguyn dng (m, n) tho mn phng trnh

19m + 90 + 8n = 1998. D thy trong (100, 1) l cp vi gi tr nh nht ca n.Hy xc nh cp vi gi tr nh nht ca m.

Bi ton 2/29.Tm s hu t nh nhtsr sao cho

n

1

m

1

k

1++

sr

, trong k, m,

n l cc s nguyn dng tho mn bt ng thcn

1

m

1

k

1++ < 1.

Bi ton 3/29.C th sp xp tm trong chn s2, 3, 4, 7, 10, 11, 12, 13, 15,

vo cc vung trng ca bng 3 hng 4 ct trn hnh v sao cho trung bnh cng

ca cc s trong mi hng v trong mi ct u bng mt s nguyn. Hy ch ramt cch sp xp v ch r s no trong 9 s phi b i khi hon thnh bng.

19 5

14Bi ton 4/29.Chng minh rng c th sp xp by im phn bit trn mt

phng sao cho vi bt k 3 im no trong 7 im , th hai trong ba im cchnhau mt khong bng n v. (Li gii ca bn nn bao gm mt phc ho cn

thn ca by im, cng vi tt c (*) cc on thng c di n v).Bi ton 5/29. S bn v hnh elip c phng trnh

O

R3 R4

R1R2

y

x

.199898

)98y(19

)19x( 22=

+

Gi R1, R2, R3, R4 ln lt l din tch pha trong hnh eliptrong cc gc phn t th I, th II, th III v th IV. Tnhgi tr ca

R1 - R2 + R3 - R4.


31/45



IMTS vng 30Bi ton 1/30. Xc nh cp s thc duy nht (x, y) tho mn phng trnh

(4x2 + 6x + 4)(4y2 - 12y + 25) = 28.Bi ton 2/30. Chng minh rng c v s b ba sp th t cc s nguyn dng(a, b, c) sao cho c chung ln nht ca a, b v c l 1; v tng a2b2 + b2c2 + c2a2 lbnh phng ca mt s nguyn.Bi ton 3/30. Chn tm th c th c nh s bng cch s dng cc s bn-nguyn dng (1/2, 1, 3/2, 2, 5/2, ...) sao cho tng ca cc s trong mt cp thc chn ngu nhin l mt s nguyn t 2 n 12. V tng xut hin vicng tn s ca s kin n xut hin khi ta gieo hai con xc sc v ly tng s ntca chng. Cc s trn chn tm th l s no v mi s xut hin bao nhiu ln

trn cc tm th?Bi ton 4/30. Hnh vung PQRS ni tip trong tam gic vung ABC, vung tiC, vi S v P ln lt trn cnh BC v CA, cn Q v R nm trn cnh AB. Chngminh rng AB 3QR v ch ra khi no c du ng thc.

Bi ton 5/30.Cho t gic li ABCD. K, L, M v N ln lt l cc trung imca cc cnh v PQRS l t gic c to thnh bi cc giao im ca AK, BL, CMv DN. Xc nh din tch ca t gic PQRS nu dt(ABCD) = 3000 (din tch cat gic ABCD); dt(AMQP) = 513; dt(CKSR) = 388.


32/45



IMTS vng 31Bi ton 1/31. Xc nh ba ch s ngoi cng bn tri ca s:

11 + 22 + 33 + ... + 999999 + 10001000.Bi ton 2/31. C v s cp sp th t s nguyn dng (m, n) cho tngm + (m + 1) + (m + 2) + ... + (n - 1 ) + n

bng vi tch mn. Bn cp vi gi tr nh nht ca m l (1, 1), (3, 6), (15, 35), v(85, 204). Hy tm thm ba cp (m, n) na.Bi ton 3/31. Cc s nguyn t 1 n 9 c th sp xp vo trong bng 3 3 saocho tng cc s trong mi hng, mi ct v mi ng cho l mt bi ca 9.

(a) Chng minh rng s tm ca bng phi l mt bi ca 3.(b) Hy ch ra mt v d vi s tm ca bng l 6.

Bi ton 4/31. Chng minh rng nu 0 < x < /2 thsec6 x + csc6 x + (sec6 x)(csc6 x) 80.Bi ton 5/31.Trn hnh v, O l tm ca ng trn, OK v OA l hai bnknh vung gc nhau, M l trung im OK, BN song song vi OK, v AMN =NMO.

Xc nh s o tnh theo ca ABN.


33/45



IMTS vng 32Bi ton 1/32. Hy ch ra mt s nguyn c 13 ch s N l bi nguyn ca 213

v ch bao gm cc ch s 8 v 9.Bi ton 2/32. Vi s nguyn khc 0 i, s m ca 2 trong php phn tch ratha s nguyn t ca ic k hiu l ord2(i). Chng hn, ord2(9) = 0 do 9 l sl, v ord2(28) = 2 do 28 = 2

2 7. Cc s 3n - 1 vi n = 1, 2, 3, ... u l s chn, do ord2(3

n - 1) 1 vi n > 0.Vi s nguyn dng n no t ha) ord2(3n - 1) = 1 ?b) ord2(3

n - 1) = 2 ?c) ord2(3n - 1) = 3 ?

Chng minh li gii ca bn.Bi ton 3/32. Cho f l a thc bc 98, sao cho f(k) = 1/k vi k = 1, 2, ..., 99. Hyxc nh f(100).Bi ton 4/32. Tp A gm 16 phn t ca tp hp {1, 2, 3, ..., 106}, sao chokhng c hai phn t no ca A chnh nhau (*) 6, 9, 12, 15, 18, hoc 21. Chngminh rng c hai phn t ca A chnh nhau 3.Bi ton 5/32.Trong tam gic ABC, gi D, E, v F l cc trung im cc cnhca tam gic, v gi P, Q, R ln lt l cc trung im ca cc trung tuyn tngng AD, BE, v CF (hnh v). Chng minh rng gi tr ca

222

222222

CABCAB

CQCPBRBPARAQ

++

+++++

khng ph thuc vo hnh dng ca tam gic ABC v hy tm gi tr .


34/45



IMTS vng 33Bi ton 1/33. Cc ch s ca cc s nguyn c ba ch s a, b v c l cc ch s

trong 9 ch s khc khng 1, 2, 3, ..., 9, mi ch s xut hin ng mt ln. Cho ts a : b : c bng 1 : 3 : 5, hy xc nh a, b v c.Bi ton 2/33. Cho N = 111...222...2, trong c 1999 ch s 1 v 1999 ch s 2.Hy phn tch N thnh tch ca bn s nguyn ln hn 1.Bi ton 3/33. Tam gic ABC c s o gc A l 30o, gc B l 60o v gc C l 90o.Hy trnh by bn cch chia tam gic ABC thnh bn tam gic, mi tam gic ngdng vi tam gic ABC v c din tch bng dt(ABC)/4. Chng minh rng cc gcv cnh ca tam gic nh hn l chnh xc (*).Bi ton 4/33. C 8436 bi thp, mi vin c bn knh 1cm, c xp chng vomt ng hnh t din. Vi 1 vin trn nh, 3 vin lp th hai, 6 vin lp thba, 10 vin lp th t, v c nh th... Hy xc nh chiu cao ca ng bi tnhbng centimet.Bi ton 5/33. Trong mt ng gic li ABCDE cc cnh c di 1, 2, 3, 4 v 5(khng nht thit theo th t ). Gi F, G, H, v I ln lt l trung im ca cccnh AB, BC, CD v DE. Gi X l trung im ca on FH, v Y l trung im caon GI. di ca on XY l mt s nguyn. Tm tt c cc gi tr c th ca di cnh AE.


35/45



IMTS vng 34Bi ton 1/34. S N gm 1999 ch s sao cho nu mi cp ch s lin nhau ca

N c xem nh l mt s c hai ch s , th s s hoc l mt bi ca 17, hocl mt bi ca 23. Tng cc ch s ca N l 9599. Hy xc nh 10 ch s ngoicng bn phi ca N.Bi ton 2/34.Cho C l tp hp cc s nguyn khng m m c th biu dindi dng 1999s+ 2000t, sv t cng l cc s nguyn khng m.

(a) Chng minh rng 3994001 khng thuc tp hp C.(b) Chng minh rng nu 0 n 3994001 v nl mt s nguyn khng thuc

C, th 3994001 - n thuc C.Bi ton 3/34. S bn l bn ca Squareville, mi thnh ph c cng c. Hai ng

i bn,Alexandra v Brianna, sng ti cc gc nh du lnlt l A v B. H bt u i b hng v nh ca mingi khc, khi hnh ng thi, i b vi cng mtvn tc, v chn mt cch c lp mt con ng nnh ngi kia phn phi u (*) cho tt c cc conng ngn nht c th (tt c cc con ng ngnnht c chn nh nhau). Hi vi xc sut l baonhiu h s gp nhau?

B

A

Bi ton 4/34. Tam gic PQR c PQ = 8, QR = 13 v RP = 15. Chng minh rngc mt im S nm trn on thng PR, khng trng vi hai u mt, sao cho PSv QS u l cc s nguyn.

C

A HM B

Bi ton 5/34.Cho tam gic ABC vi AC > BC, CM l trung tuyn v CH lng cao xut pht t C (hnh v). Hy xc nh s o ca MCH, nu mi gcACM v BCH c s o l 17o.


36/45



IMTS vng 35Bi ton 1/35. Chng ta gi s nhc li(repetition number) ca mt s nguyn

dng n l s cc ch s khc nhau ca nkhi vit trong h thp phn. Chngminh rng mi s nguyn dng c mt bi c s nhc li nh hn hoc bng 2.Bi ton 2/35. Cho a l mt s thc dng, n l mt s nguyn dng, v nhngha thp lu tha(power tower)a n mt cch quy nh sau: a 1 = a, a (i+ 1) = aa i vi i = 1, 2, ... Chng hn, ta c 4 3 =

444 = 4256, mt s c 155 ch s.Vi mi s nguyn dng k, gi xk l nghim thc dng duy nht ca phngtrnh xk= 10 (k+ 1). S no ln hn, x42hay x43?Bi ton 3/35. Gi s rng 32 my tnh trong mt mng my tnh no cnh s bi cc s nguyn 5-bit 00000, 00001, ..., 11111 (bit l vit tt ca ch s

nh phn - binary digit). Gi s rng c mt cch ni my tnh A vi my tnh Bnu v ch nu A v B chung 4 trong cc bit ca n vi bit cn li l 0 A v 1 B.(V d, 10101 c th nhn tin ti 11101 v ti 10111.) Chng ta ni rng mt mytnh l c cp ktrong mng nu n c ng kbit 1 trong nhn ca n (k = 0, 1, 2,3, 4, 5). Gi s tip l chng ta bit rng 12 my tnh, ba chic mi cp 1, 2, 3 v 4l ang gp s c, nhng ta khng bit ci no. Chng ta vn chc chn rng cth gi mt tin nhn t my 00000 ti my 11111 c hay khng?Bi ton 4/35.Chng ta gi mt tam gic trong mt phng to l nguynnu ba nh ca n c cc to nguyn v ba cnh ca n c di nguyn.

(a) Tm mt tam gic nguyn vi chu vi bng 42.

(b) C tam gic nguyn no vi chu vi bng 43 khng?Bi ton 5/35. Chng ta ni rng mt tp hu hn cc im l phn tn tt(well scattered) trn mt mt cu nu mi bn cu m (Mt na mt cu khngtnh ng bin) cha t nht mt im ca n. Trong khi {(1, 0, 0), (0, 1, 0), (0, 0,1)} khng phn tn tt trn mt cu n v (mt cu c bn knh 1, tm ti gc to), nhng nu bn thm ng mt im P, n tr thnh phn tn tt. Tm, vichng minh, tt c cc im P m c th to thnh tp phn tn tt.


37/45



IMTS vng 36Bi ton 1/36.Xc nh s nguyn c 9 ch s duy nht M c nhng tnh cht

sau: (1) Tt c cc ch s ca n phn bit v khc 0; v (2) vi mi s nguyndng m = 2, 3, 4, ..., 9, s nguyn to thnh bi m ch s ngoi cng bn tri caM chia ht cho m.Bi ton 2/36. Cc s Fibonacci c nh ngha bi: F1 = F2 = 1 v Fn = Fn - 1 +Fn - 2 vi n > 2. Mt iu quen thuc l tng ca 10 s Fibonacci lin tip bt k thchia ht cho 11. Hy tm s nguyn dng N nh nht sao cho tng ca N sFibonacci lin tip bt k th chia ht cho 12.Bi ton 3/36.Tnh gi tr ca

.

2000

1

1999

11...

3

1

2

11

2

1

1

11S 222222 +++++++++=

Bi ton 4/36.Chng ta s ni rng mt hnh bt gic l nguyn nu n c ccgc bng nhau, cc nh ca n l cc nt li (ngha l cc im vi cc to nguyn), v din tch ca n l mt s nguyn. Chng hn, s bn ch ra mtbt gic nguyn vi din tch 21. Xc nh, c chng minh, s nguyn dng nhnht K sao cho vi mi s nguyn dng k K th c mt bt gic nguyn vidin tch k.

Bi ton 5/36.Cho P l mt im nm bn trong hnh vung ABCD sao cho PA= a, PB = b, PC = c, v c2 = a2 + 2b2. Ch cho cc di a, b, c v ch dng compa vthc k thng, hy dng mt hnh vung c cng kch thc vi hnh vung

ABCD.


38/45



IMTS vng 37Bi ton 1/37.Tm s nguyn N nh nht c 5 ch s sao cho 2N cng l s c

5 ch s v tt c 10 ch s t 0 n 9 c dng N v 2N.Bi ton 2/37.Mi y ngi ta chng minh c rng khng phi ls nguyn t. Tm bn ch s ngoi cng bn phi ca s ny.

12242 +

Bi ton 3/37. Xc nh cc s nguyn a, b, c, d v e sao cho(x2 +ax + b)(x3 + cx2 + dx + e) = x5 - 9x - 27.Bi ton 4/37. Mt dy s thc s0, s1, s2, ... c tnh cht: sisj = si + j + si - j vi mi s

nguyn khng m i v j vi i j; si = si + 12 vi mi s nguyn khng m i; v s0 > s1> s2 > 0.

Tm ba s s0, s1, v s2.

Bi ton 5/37.Trong hnh bt din hnh v bn, mt y v mt nh l cctam gic u vi di cc cnh l 9 v 5 n v, v tt c cc cnh cn li c di l 6 n v. Hy xc nh chiu cao ca hnh bt din; tc l khong cch giamt y v mt nh.


39/45



IMTS vng 38Bi ton 1/38. Mt cch kim tra tnh chia ht cho 19 ca mt s l nh sau:

B i ch s cui cng ca s , cng hai ln ch s b i vo s lc bn, v lp li th tc ny cho ti khi thu c mt s nh hn 20. Khi , s banu chia ht cho 19 khi v ch khi s cui cng l 19. Phng php c minh hobi s . D dng kim tra rng ng l 67944 chia ht cho 19 trong khi 44976th khng.

Tm v chng minh mt php kim tra tng t cho tnh chia ht cho 29.6 7 9 4 4 4 4 9 7 6

8 1 26 8 0 2 4 5 0 9

4 1 86 8 4 4 6 88 1 6

7 6 6 21 2 41 9 1 0Bi ton 2/38.Tnh 17761492! (mod 2000), tc l phn d khi chia 17761492! cho

2000. (Du chm than k hiu cho giai tha.)Bi ton 3/38. Cho cp s cng cc s nguyn308, 973, 1638, 2303, 2968, 3633, 4298.

Xc nh cp s nhn cc s nguyn duy nht b1, b2, b3, b4, b5, b6, sao cho308 < b1 < 973 < b2 < 1638 < b3 < 2303 < b4 < 2968 < b5 < 3633 < b6 < 4298.Bi ton 4/38. Chng minh rng mi khi a din c hai nh m ti c

cng s cnh gp nhau.Bi ton 5/38.Trong tam gic ABC, cc on PQ, RS v TU tng ng songsong vi cc cnh AB, BC v CA, v giao nhau ti cc im X, Y v Z nhhnh v.

Xc nh din tch ca tam gic ABC nu mi on PQ, RS v TU chia i dintch ca tam gic ABC v din tch ca tam gic XYZ l 1 n v. p n ca bnnn di dng a + 2b , a v b l cc s nguyn dng. (*)


40/45



IMTS vng 39Bi ton 1/39.Tm s nguyn dng nh nht c tnh cht l n c cc c s

tn cng bi mi ch s thp phn; ngha l tn cng bi 0, 1, 2, ..., 9.Bi ton 2/39. Gi s rng cc phn s ti gin nm gia 0 v 1, vi mu s lnnht bng 99, c lit k theo th t tng dn. Tm hai phn s k vi 17/76trong dy ny.Bi ton 3/39. Cho p(x) = x5 + x2 + 1 c cc nghim r1, r2, r3, r4, r5. Gi q(x) = x2 -2. Hy tnh tch sau

q(r1)q(r2)q(r3)q(r4)q(r5).Bi ton 4/39.Gi s mi phn t ca dy =

1ii hoc l du + hoc l du -.Xc nh dy cc du + v - thch hp sao cho

...6662 321 =

V xc nh dy cc du no cn thit khi thay cc s 6 trong cc du cnbi cc s 7. Lit k tt c cc s nguyn v tr ca cc s 6 v cc dy nhng ducn thit vi chng. (*)Bi ton 5/39. Ba tam gic vung cn c dng ng t cnh ln ca mthnh ch nht v pha trong ca hnh ch nht nh hnh v, M l trung im cacnh . Nm ng trn ni tip tip xc vi mt s cnh v vi ng khcnhhnh v. Mt trong cc ng trn tip xc vi nh ca tam gic ln nht.

Tm t s gia cc bn knh ca nm ng trn.


41/45



IMTS vng 40Bi ton 1/40.Tm tt c cc s nguyn dng c tnh cht l chng ln hn

tng cc bnh phng cc ch s vit trong h thp phn ca chng 1 n v.Bi ton 2/40. Chng minh rng nu n l mt s nguyn dng l thN = 2269n + 1779n + 1730n - 1776n

l mt bi nguyn ca 2001.Bi ton 3/40. Hnh bn c th c chia thnh hai na thch hpm c lin quan vi nhau bi mt s phn x trt (a glidereflection), nh hnh di n. Mt s phn x trt phn x mthnh qua mt ng thng, ng thi cng di chuyn hnh phn xtheo hng song song vi ng thng . Vi mt hnh li

vung (square-grid figure), ch c nhng

ng thng m gi hnhphn chiu ca n (*) trn li l ng nm ngang, ng thngng, ng cho 45o v ng cho 135o. Trong hai hnh v di,hy chia mt hnh thnh hai na thch hp lin quan vi nhau bimt s phn x trt, v ch ra ti sao hnh kia khng th chia cnhvy.

Bi ton 4/40. Cho A v B l cc im nm trn mt ng trn m khng ixng qua tm, v gi C l trung im ca cung nh AB. Gi D, E, v F l cc imc xc nh bi giao im ca cc ng tip tuyn vi ng trn ti A, B, vC. Chng minh rng din tch ca tam gic DEF ln hn mt na din tch catam gic ABC.Bi ton 5/40. Bt gic RSTUVW c dng nh sau: bt u vi mt tam gicvung c di cc cnh gc vung l p v q, dng cc hnh vung hng rapha ngoi trn cc cnh ca tam gic ny, v sau ni cc nh pha ngoi ca

cc hnh vung li, nh hnh v. Cho p v q l cc s nguyn vi p > q v dintch ca RSTUVW l 1922. Tm p v q.


42/45



IMTS vng 41Bi ton 1/41. Tm cc s nguyn dng c hai ch s m v n duy nht cho

xp x m/n = 0.2328767 chnh xc n by ch s thp phn; ngha l 0.2328767 m/n < 0.2328768.Bi ton 2/41. Ta bit rng c v hn b ba cc s nguyn (a, b, c) c c chungln nht bng 1 v tho mn phng trnh a2 + b2 = c2.

Chng minh rng cng c v hn b ba cc s nguyn (r, s, t) c c chung lnnht bng 1 v tho mn phng trnh (rs)2 + (st)2 = (tr)2.

Bi ton 3/41. Gi thit31

xcosx3cos= vi gc x no thuc [0, /2]. Tnh

xsinx3sin

vi cng gi tr ca x.Bi ton 4/41. Mt phng chiu bc ba (The projective plane of order three)gm c 13 im v 13 ng thng. Nhng ng thng ny khng l ngthng Euclid, m chng l tp hp gm bn im c tnh cht mi cp ngthng c ng mt im chung v mi cp im c ng mt ng thng chac hai im . Gi s cc im c nh s t 1 n 13 v su trong s ccng thng l A = {1, 2, 4, 8}, B = {1, 3, 5, 9}, C = {2, 3, 6, 10}, D = {4, 5, 10, 11}, E ={4, 6, 9, 12}, v F = {5, 6, 8, 13}. ng thng no cha 7 v 8?Bi ton 5/41.Trong tam gic PQR, QR < PR < PQ sao cho ng phn gicgc ngoi qua P ct tia QR ti im S, v phn gic ngoi ti R ct tia PQ ti imT, nhhnh v. Cho PR = PS = RT, tnh s o gc PRQ.


43/45



IMTS vng 42Bi ton 1/42.C bao nhiu s nguyn dng c 5 ch s t 1 n 9 m trong

n mt ch s xut hin mt ln v hai ch s xut hin hai ln? Chng hn, s41174 l mt s nhvy, trong khi 75355 th khng phi.Bi ton 2/42.Xc nh s nguyn dng c bnh phng ng bng s

.)2i4(12001

1i

3=

+

Bi ton 3/42.Phn tch thnh nhn t biu thc30(a2 + b2 + c2 + d2) + 68ab - 75ac - 156ad - 61bc - 100bd + 87cd.Bi ton 4/42.Cho X = (x1, x2, ..., x9) l mt vc-t 9 chiu cc s nguyn. Tm X

nu by vc-t sau thu c t X bng cch xo i 3 trong s cc thnh phn ca

n Y1 = (0, 0, 0, 1, 0, 1), Y2 = (0, 0, 1, 1, 1, 0),Y3 = (0, 1, 0, 1, 0, 1), Y4 = (1, 0, 0, 0, 1, 1),Y5 = (1, 0, 1, 1, 1, 1), Y6 = (1, 1, 1, 1, 0, 1),

Y7 = (1, 1, 0, 1, 1, 0).Bi ton 5/42. Cho R v S tng ng l cc im nm trn cc cnh BC v ACca tam gic ABC, v gi P l giao im ca AR v BS. Tnh din tch ca tam gicABC nu cc din tch ca cc tam gic APS, APB, v BPR tng ng l 5, 6, v 7.


44/45



IMTS vng 43Bi ton 1/43. Chng ta s ni rng mt s sp xp li cc k t ca mt t l

khng c cc ch ci c nh(fixed letters) nu khi t t sp xp li ngay di t th khng ct no c cng mt ch ci c lp li. lm v d th cc ch cibn di ch ra ESARET l mt s sp xp li khng c ch ci c nh ca tTERESA, cn REASTE th khng phi.

T E R E S A T E R E S AE S A R E T R E A S T E

C bao nhiu s sp xp li khc nhau khng c ch ci c nh ca tTERESA? (Hai ch E c xem nhging nhau hon ton.Bi ton 2/43. Tm 5 b ba khc nhau cc s nguyn dng {k, m, n} sao cho k< m < n v

8419

n1

m1

k1

=++ .

Bi ton 3/43. Gi s p(x) = xn + an - 1xn - 1 + ... + a1x + a0 l mt a thc vi cc hs nguyn v gi s (p(x))2 l mt a thc c tt c cc h s khng m. C nhtthit rng tt c cc h s ca p(x) phi l khng m hay khng? Chng minh cutr li ca bn.Bi ton 4/43.Trong tam gic ACF, B l trung im ca AC, D v E chia cnhCF thnh ba phn bng nhau, cn G, H, v I chia cnh FA thnh bn phn bngnhau (hnh v).

Mi by on c v ni su im vi nhau v vi cc nh i din catam gic. Xc nh cc im bn trong tam gic ACF ti c ba on hoc nhiuhn trong s cc on thng trn i qua.

Bi ton 5/43. Hai mt phng vung gc ct mt mt cu theo hai ng trn.Cc ng trn ny ct nhau ti hai im cch nhau 14 n v, o dc theo ngthng ni chng (*). Nu cc bn knh ca cc ng trn l 18 v 25 n v thbn knh ca mt cu l bao nhiu?


45/45


IMTS vng 44Bi ton 1/44. Trong mt ngn ng k l ch c hai ch ci av b, v c quy

nh l ch ci al mt t. Hn na, tt c cc t thm vo c hnh thnh datheo cc quy tc sau y:

1. Cho t no , mt t mi c th c hnh thnh t n bng cch thm mtch ci bvo bn phi ca t .

2. Nu trong t no xut hin mt dy aaath mt t mi c th c hnhthnh bng cch thay thaaabi ch ci b.

3. Nu trong t no xut hin mt dy bbbth mt t mi c th c hnhthnh bng cch b i bbb.

4. Cho t no , mt t mi c th c hnh thnh bng cch vit ra dy to

thnh do vit t hai ln.Chng hn, do (4) th aal mt t, li bi (4) nn aaaal mt t. T (2) th balmt t, v do (1) nn babcng l mt t. Mt ln na do (1) th babbl mt t, vdo (4) nn babbbabbcng l mt t. Cui cng, do (3) chng ta thy rng baabbl mt t.

Chng minh rng trong ngn ng ny th baabaabaakhng phi l mt t.Bi ton 2/44.Cho f(x) = x.[x.[x.[x]]] vi mi s thc dng x, trong [y] khiu s nguyn ln nht khng vt qu y.

1. Tm x f(x) = 2001.2. Chng minh rng phng trnh f(x) = 2002 khng c nghim.

Bi ton 3/44.Cho f l mt hm s xc nh trn tp cc s nguyn, v gi srng n tho mn cc tnh cht sau:1. f(0) 0;2. f(1) = 3; v3. f(x)f(y) = f(x + y) + f(x - y) vi mi s nguyn x v y.Hy tnh f(7).? Bi ton 4/44.Mt cng ty no c mt h thng in thoi khng chnh

xc, thnh thong i ch cp ch s k nhau khi ai quay mt s my nhnh cba ch s. T mt cuc gi x318 s lm chung ti s my x318 hoc x138hoc x381, trong khi mt cuc nhn in ti s my x044 th c th l gi cho s

x404 hoc x044. Hn l thay th h thng , cng ty thm vo mt my tnh suy ra cuc quay s my nhnh no l b li v tr li nhng s chnh xc cachng. Chng phi loi b vi my nhnh c th cho vic ny. Hi s ln nhtcc s my nhnh c ba ch s m cng ty c th n nh vi k hoch ny l baonhiu?? Bi ton 5/44.Tm s nh nht cc hnh vung m ngi ta c th chia mthnh ch nht 11 13 v nhn c. Cc hnh vung khng cn khc c, cnh cachng l cc s nguyn v chng khng ln nhau.

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