Tuyển tập các phương pháp giải toán qua các kỳ thi Olympic: Phần 2

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Nối tiếp nội dung phần 1 tài liệu Các phương pháp giải toán qua các kỳ thi Olympic, phần 2 giới thiệu các chuyên đề: Định lý Casey và ứng dụng, một số phương pháp giải bài toán tồn tại trong tổ hợp, một cách đổi biến và ứng dụng trong chứng minh bất đẳng thức,... Mời các bạn cùng tham khảo nội dung chi tiết.

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Nội dung Text: Tuyển tập các phương pháp giải toán qua các kỳ thi Olympic: Phần 2

156 Cdc phUcfng phdp gidi todn qua cdc ky thi Olympic T a i li$u tham khao 0|NH LY CASEY VA L/NG DUNG i [1] Ha V u Anh, Dudng doi trung, Chuyen de Bao cao tai Hoi Nguyen v a n Linh+ ; ' thao Toan sd cap nam 2010 tai B a V i , m Noi. [2] Nguyin V a n Ban, Hoang Chung, Hinh hoc cua tarn gidc, Nha xuat ban Giao due, 1996. [3] Doan Quynh (chu bien), Tdi lieu gido khoa chuyen Todn, Nha , xuat ban Giao due, 2010. 1. Gidri thieu [4] I. F . Sharyghin, Cdc bdi todn hinh hoc phdng, Nha xuat ban Dinh ly Casey du-dc dat theo ten nha Toan hoc John Casey, hay Nauka, 1996. con du-cfc gpi la djnh ly Ptolemy md rong (xem [1]), du-dc phat bieu nhu" sau [5] Cae nguon tai heu tuf Internet: / Dinh ly 1. Cho bon dudng trdn C, (i = M j . Ky hieu Uj Id do ddi www.mathscope.org; cua tiep tuyen hai dudng trdn Q vd Cj. Khi do bdn dudng trdn C, :' www.diendantoemhoc.net; I'l, Cling tiep xuc vdi mot dudng trdn (hodc mot dudng thang) C khi vd www.mathlinks.org; chl khi www.imo.org.yu. tnhi ± ^13^42 ± ^14^23 = 0. Chii y rang tiep tuyen ducfc chon cua hai dudng trdn Q, Cj Id tiep fuyen chung ngodi khi vd chi khi cd hai dudng trdn Q, Cj cung tiep •^wc trong (hodc ngoai) vdi C, Id tiep tuyen chung trong khi vd chl * S i n h v i e n D a i h o c N g o a i thiTdng H a N o i . 157
158 Cdc phuang phdp gidi todn qua cdc ky thi Olympic Dinh ly Casey vd Ung dung 159 khi trong hai dudng trdn d, Cj c6 mot dudng trdn tiep xiic trong, Chiang m i n h . Ta chi c h t o g minh cho tnrdng hcJp tiep tuyen chung mot dudng trdn tiep xuc ngoai vdi C. Ddu cua Ujtki Id "+" khi vd ngoai, tnrdng hcJp tiep tuyen chung trong chiJng minh ttfcJng tu". chi khi cdc doan thing noi hai tiep diem cua Q vd Cj, Ck vd Ci khong cat nhau, ddu "-" khi vd chi khi ngugc Igi. Goi (O, R) la du-dng tron triTc giao v d i ( O i ) va (O2). (O) giao ( O i ) t a i A ' , B', giao (O2) tai C, D'. Lay J tren (O) sao cho J nam De dang nhan thay k h i bon diTctng tron tren suy bien thanh tren true dang phiTcfng cua ( d ) va (02)- Goi k la phiTdng tich ti^ J dirdng tron diem thi dinh ly Casey trd thanh dinh ly Ptolemy (xem den hai diTdng tron ( O i ) va (O2). Phep nghich dao bien A' thanh [2]). A, B' thanh 5 , C thanh C, D' thanh D. K h i ba difdng tr6n suy bien thanh difdng tr6n diem t h i dinh ly Do A\ D' e (O) nen A, B, C, D thang h^ng. Phep Casey trd thanh dinh ly Purser (xem [3]). nghich ddo bao to^n do Idn gdc giila hai du'dng cong tai giao diem nen A, B va C, O2, D thang hang. Tuf do A, B, C, D nam tren du'dng noi tam O1O2. 2. Chufng m i n h d i n h ly Khong mat tinh tdng quat gia su" A, B, C, D nam tren O1O2 theo thu" tif, Ri < R2. G o i MN la tiep tuyen chung ngoai cua ( O i ) L d i giai sau d\ia theo [4]. Ta se phat bieu chiJng minh mot bd va (O2) ( M G ( O i ) , G (O2)). G o i P la hinh chieu vuong goc dl. ' cua Oi tren O2N. Ta c6 B o de 1. Cho hai dudng trdn {Ou Ri) vd {O2, R2) khong chijta nhau. MN' = 0,P' = 0,0l - O2P' = 0\0\ {R, - R2)' I la mot diem nam ngoai hodc nam trong cd hai dudng trdn. Phep = {O1O2 + Ri- i?2)(Oi02 -Ri + R2) = BD.CA. nghich ddo cUc I phuang tich R^ idn lUdt bien ( O i , i ? i ) , (O2, R2) Ngoai ra, ta cung c6 thanh (O'l, i ? ; ) , (0'2, i?'2). Ggi T12, T^^ Ian lugt la do dai tiep tuyen chung ngoai (hodc trong neu co) cua cdc cap dUdng trdn ( O i , -Ri) B'D' _ JD' _ JB' _ ^JB'.JD' k r 2 12 T" - ' I BD ~ JB ~ 7D ~ ^JB.JD " JB.JD' Vd ( O i , R[), (O2, R2) vd (0'2, R'2). Khi do = Thie't lap cac bieu thiJc tu'dng tu" ta suy ra B'D'.JB'.JD' C'A'.JC'.JA' TI2 _ BD.CA _ 1 • _ B'D'.a A' R1R2 AB.CD A'B'.JA'.JB' C'D'.JC'.JD' ~ A'B'.C'D'' k ' k Bay gid, phep nghich dao jf bien (O) thanh (O'), bien A' thanh A", B' thanh B", C thanh C", D' thanh D". K h i do {0') la difdng tron tryc giao v d i ( O i ) va (O2). Tu'dng tu" nhuf tren, ta cung chiJug minh du'dc Til B"D".C"A" B'D'.a A' R'^R!2 A"B".C"D" A'B'.C'D' Vav -'12 = -'12 • R1R2 R'^R'2'
Cdc phU(fni; phdp s^idi todn qua cdc ky thi Olympic 160 Dinh ly Casey va itng dung 161 Trd lai b^i todn. TriTdc tien ta chiJng minh chieu thuan cua dinh Khong mat tong quat gia su" i?^ = minji?;, R'2, R's}. Cung giam ly Casey cho trufdng hcfp ca bon diTdng tron d deu tiep xuc trong bdn kinh ba diTdng trdn Q[, Q'2, Q'3 mot doan i?3 ta duTdc ba du'dng vdi C, cac tnrdng hcJp c6n lai chiJng minh tiTcJng tif. tron (/i, qi), {I2, 92), {h, 0). Ta se tim quy tich cac diem I3 sao * Ky hieu Oi Ian lifdt la ban kinh va tam cua cac dvTdng tron cho ^13 —• ^23 = ^12- Ci\r,0 m ban kinh va tam cua dtfdng tron C. Khong mat tinh tong Tap hcJp cac diem sao cho tiep tuyen ke tijf diem do den (7i, q^) quat, gid s\i = m m { r i } . Ta nhan thay cdc dUcJng tron ( O i , r i - bang i'i3 la difdng tron tam h , bin kinh ^qf + tf^, tap hdp cac did'm i=l,4 sao cho tiep tuyen ke tiif diem do den {I2, 92) bang ^33 la du'dng n), (O2, r2 - u), (O3, rz - r4), (O4, 0) cung tiep xiic vdi dufdng tron tam I2, ban kinh \/^|T^. Hai du'dng tron nay cat nhau tai tron (O, r - r^) va do dai tiep tuyen chung cua hai diTdng tron hai diem nam tren tiep tuyen chung ngoai cua (/i, qi) va {I2, 92)- ( O i , n - k), {Oj, Tj -k) ike (0,min{ri, r , } ) ) bang U,. Do do ta chi can chufng minh chieu thuan cua dinh ly Casey cho bon du'dng NhiT the' 73 nam tren tiep tuyen chung ngoai cua (/i, gi) va tron ( d , r i - u), (O2, - r^), (O3, rs - u), (O4, 0) (ki hieu la (I2, q2)- Suy ra Q'2, Q'3 cung tiep xiic vdi mot difdng thing. NghTa la ton tai mot du'dng tron di qua O4 va tiep xiic vdi Qi, Q2, Q3. Qi, Q2, Qs, Qi v6i ban kinh Ian liTdt la Ri, R2R3, 0). Vay ton tai mot difdng tron tiep xuc vdi bon diTdng tron Q va dinh Xet phep nghich dao ag' bien Qi thanh Q[, Q2 thanh Q'2, Q3 ly Casey du'dc chiJug minh. thanh Q^, (O, r - r4) thanh / va Q[, Q'^, Q3 cung tiep xuc mot phia vdi /. Goi t[j la do dai tiep tuyen chung ngoai cua va Q'j. Khong NhSn xet. Chieu thuan ciia hhi toan c6 the chiJng minh theo hufdng mat tong quat gia suf tiep diem cua Q'^ va / nam tren doan thang sau day khong suf dung ph6p nghich dao, diTcJc coi la he qua cua dinh ly ndi hai tiep diem cua Q[ va Q'^ vdi /. Ta c6 ^'12 + ^23 - ^13 = 0- Ptolemy. Ap dung Bo' d l 1, =4 - V i , j e {1,2,3}. Ngoai ra, B6' dl 2. Cho hai dudng tron Ci{0, ri) va C2(02, r2) cung tiep xiic vdi €{0, R) Idn luat tai hai diem A, B. Khi do do dai tiep tuyen chung (trong ta cung c6 hade ngoai) cua C\ C2 duac tinh bdi cong thiic Ri t^i t\ h2 = -^\/{R±ri){R±r2). do vay hi = R-^^- ^ay suy ra J Ri R2 R3 ni p/ n/ R!y.R!2-R!i ^• Chieu thuan dtfdc chiJng minh. Bay gid ta chiJng minh chieu dao. Gia suf ii2.^34 + hzMi - tizMi = 0. Ta se chiJug minh Qu Q2, Q3, (O4, 0) cung tiep xuc vdi mot dudng tron. Suf dung ph6p nghich dao ta cung suy ra t'^^ +1'23 - i'13 = 0-
162 Cdc phucfng phdp gidi todn qua cdc ky thi Olympic Dinh ly Casey vd ling dung 163 ChuTng m i n h . Ta chtfng minh bo de trong tnfcfng hdp C i va C2 cung tiep diem chinh giOta cung nhd BC, CA, AB vd idn luat tie'p xuc vdi cdc xuc trong v d i C. canh BC, CA, AB. Goi the, tea, tab la do ddi cdc tiep tuyen chung ngoai cua cdc cap dudng trdn {Cb, Cc), {Cc, Ca), {Ca, Cb). Khi do De thay tl^ = 0\0l - ( r i - r^)^. A p dung dinh ly ham so cosine cho a + h+ c tarn giac OOxO^ va AOB ta c6 OxO\ OOl + OOi - 2OO1.OO2. COSO1OO2 , ^ vi = 2B?(l- C O S O 1 O O 2 ) . TO do 4 = ^R-r,f + [R- - 2{R - n){R - r2)(l - ^ ) - (n - ^2) = (n - r2)' - ( r i - r 2 ) 2 + 2{R - r i ) ( i ? - r2)^ A T3 Vay i i 2 = —^{R-r{){R-T2). Tifdng tif neu C\a C 2 cung tiep xiic ngoai v d i C, ti2 = -^\/{R + n){R + r2), Lofi g i a i . G o i ta, tb, tc l a do d ^ i c^c t i e p t u y e n k6 tuf A, B, C t d i Ca, Cb, Cc. D o Ca, Cb, Cc t i e p x u c t r o n g v d i ( O ) t a i d i e m c h i n h neu C i va C 2 tiep xuc khac phia v d i C, chang han C i tiep xiic trong con C 2 tiep xiic ngoai, f 12 la do dai cua tiep tuyen chung trong difdc tlnh bdi giffa cac c u n g nho BC, CA, AB n e n ba diTdng t r o n I a n Iffcft t i e p x u c v d i BC, CA, AB t a i trung d i e m m o i canh. cong thuTc ti2 = -pr-\/{R-ri){R + r2). • A p d u n g d i n h l y Casey cho b o n diTdng t r o n Ca, {A, 0 ) , {B, 0 ) , Trd l a i bai toan. G o i A, B, C, D Ian lUdt la tiep diem cua C i , C2, C 3 , /r^ , ^ a , a b+ c ^ a + c C 4 v d i C. A p dung Bo de 2 ta c6 ^12^34 + tu.t23 - ii3-^24 = [C, 0) ta CO ta.a = 2-^+2'^' ^ '~Y~' ^^^"^ ^ ~2~' AB.CD + AD.BC - AC.BD _ a + b i?2 V(^-n)(i?-r-2)(i?-r'3)(i?-r4). 2 • Do tur gidc ABCD npi tiep nen theo dinh ly Ptolemy, AB.CD + L a i a p d u n g d i n h l y Casey cho b o n d i f d n g t r o n Ca, Cc, {A, 0 ) , AD.BC - AC.BD = 0. Vay t u - h ^ + ^14.^23 - ii3-i24 = 0. Chieu thuan cua dinh ly Casey difdc chufng minh. {C, 0) ta c6 ta.tc =^^-7: + tac-b, tff d o "J t-.;'^ ac {b + c){a + b) ac 3. Lfng dung _ ^"-^^ ~ ' J _ 4 ~ ^ _ a + b+ c ' ' tac- ^ - ^ - ^ . B a i t o a n 1. Cho tarn gidc ABC noi tiep dudng tron (O). Goi TffOng t i r suy r a d i e u p h a i chiJng m i n h . • Ca, Cb, Cc Idn luat Id cdc dudng trdn tiep xuc trong vdi (O) tai
164 Cdc phucmg phdp gidi todn qua cdc ky thi Olympic Dinh ly Casey vd Ung dung 165 Bai toan 2 ( Dinh ly Feuerbach). Chiing minh rdng dudng tron Euler Chrfng minh. Goi R, r Ian Ixidt la ban kinh dudng tron Euler va cua tam gidc tiep xiic vdi cdc dudng tron noi tiep vd bang tiep tarn (7). Ap dung Bo de 2 cho dUdng tron (7) va {Bi, 0) ta c6 gidc do. R BiF = -y=^^=.BiB2, y/iR- r)R R CxF = •C1C2, ViR- r)R R AxF = .A1A2. ,/{R - r)R Do dd 7? AiF ± BiF ± CiF = .{AiA2±BiB2±CiC2) y/{R-r)R R .{\b - c\±\c - a\±\a - b 2^{R-r)R Chiing minh. Ta chtfng minh b ^ i todn cho tnTcfng hdp dufdng tron = 0. • noi tiep. Cac tnTcfng hcJp khac chiJng minh tufdng ttf. Bai toan 4 (Difdng tron Hart). Cho ba dudng tron C i , C2, C 3 cdt Gid sur ta c6 tam gidc A B C vdi Ax, B i , Ci Ian liTcft la trung nhau Idn lugt tai cdc cap diem {A, A ) , {B, B'), (C, C ) . Goi (I), diem cdc canh B C , CA, A B ; A2, B2, C2 Ian lufcJt la tiep diem cua {IA), (7B), (7C) idn lugt Id dudng tron noi tiep tam gidc cong dirdng tron n6i tiep (7) vdi B C , CA, A B . Dat do dai cac canh tam ABC, ABC, AB'C, ABC. Khi do ton tai mot dudng tron tiep xuc gidc A B C Ian lUdt Id a, b, c. Ap dung dinh ly Casey cho bon diTdng ngodi vdi {IA), ( I B ) , (7C) va tiep xuc trong vdi (7), ggi Id dudng tron (/), ( ^ 1 , 0), {Bi, 0), {Ci, 0) vdi do dai cdc tiep tuyen chung trdn Hart. \in m la 5 i C i = \, CxAr = ^, A,B, = |, A^A^ = ^ (I b B1B2 = c - a , C1C2 Ta c6 t h ^ chon cdc dau " + " , " - " sao cho a\b - c\± b\c - a\±c\a- b\ nen ton tai dtfdng tr6n tiep xuc vdi (7), ( ^ 1 , 0), {Bi, 0), (Ci, 0) hay dirdng tr6n Euler cua tam gidc ABC tiep xuc vdi (7). • Bai toan 3. Cho tam gidc A B C ngoai tiep (7) c6 A i , B i , Ci Ian luat Id trung diem B C , CA, A B . F Id tiep diem cua (7) vd dudng tron Euler (E). Khi do ta cd the chon cdc dau "+", "-" sao cho FAi ± FBx ± FCi = 0.
166 Cdc phucmg phdp gidi todn qua cdc ky thi Olympic Dinh ly Casey vd vCng dung 167 Lcfi giai. K y hieu Uj 1^ do dai tiep tuyen chung ngoai, la do dai tiep tuyen chung trong cua hai du'dng tron (/j) ( / , ) . Do ( / A ) , ( / B ) , ( / C ) , (/) cung tiep xuc v d i Ci nen iAI-t'BC + ^AB-t'ci~t'AC-'tBI = Q- Tifdng tuf, bon duTdng tr6n cdng tiep xiic v d i C2 nen ' ' ' tAC-t'BI'^i'sC-^AI— t'AB-^CI = 'V Bon du'dng tron ciing tiep xiic v d i C 3 nen tBC-t'AI+ t'AB-tci - t'AC-tBI = 0. . TH ba dang thiJc tren suy ra tAC-i'BI ~ ^BC^'AI — tAB-t'ci = 0 TOdng tir suy ra AA', BB', CC dong quy tai tam noi tiep / cua tarn gi^c A'B'C. M a t khdc, (lufu y dau " + " hay " - " trong cac dang thufc tren khong quan trong, cd the dao dau nhufng phai thod man trong mot dang thiJc ton tai {B'A!, B'C) = {B'A', BA) + {BA, BC) + {BC, B'C) ca " + " va " — " ) . Theo dinh ly Casey dang dao, ton tai mot du'dng = {BAJ) + {BA, BC) + {l,BC) trdn tiep xuc ngoai v d i ba difdng trdn {IA), {IB), {IC) va tiep xiic = 2{BA, BC) (mod TT). ,) v ^ v trong v d i ( / ) . • 1 ^ Do dd CIA' = 90° + -CB'A' = 180° - ABC. Suy ra / G (O). Nh$n x6t. Chu y r^ng cd tS't ci tdm dUdng tr6n Hart iJng vdi ba dirdng tron cho trifdc (xem [4], [5]). G o i Pa, Pb, Pc Ian liTdt \h phufdng tich tir A', B', C t d i {0). Ap dung dinh ly Casey cho bon difdng tron {A', 0), {B', 0), {C, 0) va B a i toan 5 ( IMO 2011). Cho tarn gidc ABC noi tiep dudng tron ( O ) , ta cd {A'B'C) tiep xuc v d i (O) khi va chi khi (O). / Id tiep tuyen bat ky cua (O). Ggi la, h, Ic l^n lugt Id dudng thdng doi xitng vdi I qua ba canh BC, CA. AB. la, h, Ic cat nhau B'C.y^a ± A ' C . ^ b ± A'B'.^c = 0. (1) tao thanh tarn gidc A'B'C. Khi do dudng trdn ngoai tiep tarn gidc Efat bdn kinh du'dng tron ngoai tiep tam giac ABC va A'B'C Ian A'B'C tiep xuc vdi (O). liTdt la R va i?', bin kinh dUdng trdn noi tiep tam giac A'B'C la B'C Lcfi giai. K y hieu dx/i 1^ khodng cdch tir X den diTdng t h i n g I. I r. Theo dinh ly ham so' sine, , . = 2B! nen cat BC, CA, AB Ian liTdt tai Ai, Bu Ci. P Ik tiep diem cua / va sinB'A'C (O). B'C = 2R'. sin(180° - 2BAC) = 2R'. sm{2BAC). Do Ic -vh I doi xiJng nhau qua AB nen ds/i = ds/u, k va I doi Ngoai ra, ta cung cd xi^ng nhau qua BC nen ds/i = ds/u. TO dd ds/i^ = ds/i, hay B'B la phan gi^c gdc B'.
Cdc phuang phdp gidi todn qua cdc ky thi Olympic Dinh ly Casey vd ling dun^ 169 168 ncn suy ra _ ABi.smPBiA _ dA/i r. , ' i ,,,,, DC.=FB + BD. EC _ AW {DC.PB + BD.PC) ~ cosBAC ~ cosBAC' FA KA^ MA' TD • ' • Lai c6 MD [{PD - PC).PB + {PB - PD).PC MA PD :„ dA/i=PAsinBiPA = PAsmPCA = PA.—-. MD PD.{PB-PC) MD = BC. P MA PD nen A A = ^ ^. Suy ra 2Rco^BAC Chilu dio cua bd de chuTng minh tifdng tU di^a theo chieu dSo cua dinh ly Menelaus. Bo de difdc chiJng minh. • B'C'.\/AA'.A'I = 2R'. sm{2BAC).PA. V2R cos BAG = PA. sin BACAR'.^ = PA.BC.2R'.- ^ y/2R ' • ' V2R.R TiTcJng W va ap dung dinh ly Ptolemy, ta suy ra B'C'.^a ± A'C'.^k ± A'B'.^c = 2R'.^^—(PA.BC±PB.AC±PC.AB) = 0. V2R.R^ Vay (A'B'C) tiep xuc vdi (O) theo (1). • Bai toan 6 (Bo de Thebault). Cho tarn gidc ABC ngoai tiep (/), noi tiep (O). Mot dudng trdn (E) tiep xuc trong vdi (O) vd tiep xuc vdi cdc canh AB, AC idn luat tai P, Q. Khi do I la trung diem PQ. Lcfi giai. Ta chiing minh bd de sau Bo 3 (Dinh ly Cristea). Gpi D, E, F Ian luat la ba diem nam Trd lai bai todn. Ap dung dinh ly Casey cho bdn dUdng tron tren cdc canh BC, CA, AB cua AABC vd M e AD. Khi do EF {A, 0), {B, 0), (C, 0), (E) ta c6 AP.BC - AB.CQ + AC.BP hay di qua M khi vd chl khi c{b - AP) + b{c - AP) 2bc AP = — ^ -, suy ra AP = r o ^ a+b+c \S FB EC MD DC.= + BD.= = BC. FA EA MA DI Goi {D} = Ain BC, ta c6 " ,CD = ^ , B D = Chtfng minh. Gia sur M e EF. Goi {P} = EF nBC. Ap dung AI b+ c 6 + ac dinh ly Menelaus cho ADAB Ung vdi dxicfng thang PMF va PEM . Ap dung Bd de 3, ta can chiJng minh ta CO b+ c FA MA PD EA PD MA PA AQ Ar
170 Cdc phuang phdp gidi todn qua cdc ky thi Olympic Dinh ly Casey vd ling dung 111 i hay 1^ B a i t o a n 8. Cho tam gidc ABC ngoai tiep dudng trdn (/). Goi Ua la dudng trdn qua hai diem B, C vd tiep xiic vdi (/), la dudng c— 26c b- 2bc trdn tiep xuc vdi cdc tia AB, AC vd tiep xuc ngodi vdi u ^ . Tuang tu ah a + b+ c ac a + b+ c a b+ c 2bc + b+ c 2bc = a. b+ c ta xdc dinh a;^, ^ c - r, Va, f t , TC idn luat la bdn kinh cdc dudng a + b+ c a + b+ c trdn (/), u)'^, u[, u'^. Khi do r =^ra + n + TC- dung sau mot so bie'n doi. Suy ra / G PQ. Ma tam giac APQ can tai A CO phan gidc AI nen / la trung diem PQ. • A B a i t o a n 7 (Bo de Sawayama). Cho tam gidc ABC noi tiep dudng trdn (O), ngoqi tiep dudng trdn (/). D Id diem bat ky tren BC. Mot dudng trdn u) tiep xuc vdi cdc tia DC, DA idn luat tai E, F vd tiep xuc trong vdi (O). Khi do I nam tren dUdng thing EF. Goi J la giao cua AI va Lcfi g i a i . BC. Theo dinh ly Menelaus, Ke tiep tuyen qua T cua dtfdng tron (/) va song song vdi Lcfi g i a i . /, E, F thing hang khi va chi khi BC, cat AB, AC Ian lifdt tai B', C . Ta chiTng minh dUdng tron noi EJ FD IA E J IA tiep tam giac AB'C la u'^. ED' FA' I J = 1 — FA' . — I J = 1. IA b+ c Goi Ai, Bi, Ci la tiep diem cua (7) vdi BC, CA, AB; A2, B^, C2 Ma I J a nen ta can chu-ng minh a.FA ={b + c)EJ. (1) Ap la tiep diem cua diTdng tron noi tiep (/„) cua tam giac AB'C vdi dung dinh ly Casey cho bon du-^ng tron {A, 0), (B, 0), (C, 0), a; ta B'C, CA, AB'. Ki hieu p, p' la nu:a chu vi tam giac ABC, AB'C c6 AF.BC + AB.CE = AC.BE, hay la a.AF + c.{BC - BE) = Ta c6 p' = ABi = p - a nen hai tam giac AB'C va ABC dong b.BE ^ a.AF + ac= BE{b + c). dang theo ti so' suy ra B'C = TiT day ta difdc ac P P Do BJ = b + c nen a.AF + {b + c)BJ - BE{b + c), tuf d6 ' a.AF ={b + c){BE - BJ) = {b + c)EJ. Nhif vay (1) dung, tiJc la (p — a)a {P - a? / nam tren difdng thing EF. • BC2 = C-AC2 = p'-B'C = c- (p-a)- P
172 Cdc phucfng phdp gidi todn qua cdc ky thi Olympic 173 Dinh IS Casey vd ling dung (v - a)'^ H- TiTdng W, CB2 = b- P . Ngo^i r a , v d i gia thiet b>c thi {P - a)b A2T = B'C'-2B'A2 = p —a ~ = (b-c). p_~_a P P P Ap dung dinh ly Casey cho bon dUdng trdn ( B , 0), (C, 0), (/), (/„) ta c6 BA1.CB2 + BC.A2T - BC2.CA1 = {p-b) jp - g)^ p—a + a{b-c). p P {p - ay c— ip-c)=0, p Ap dung dinh ly Casey cho bon dufdng tron (C, 0), (P, 0), (Oi), ntn (la) tiep xuc ngo^i vdi Ua hay {Q = . TO dd — = ^ " P (O2) ta cd ^ Tfe p —h Tc P —C „ CP. HI = tcoi-tpOi + *co2-tpoi • TOdng tir, - = ,- = ^ . Suy ra r p r p Lai dp dung dinh ly Casey cho bon dUdng tron (D, 0), (P, 0), ( d ) , i~a + n + rc 3p - a - b - c (O2) ta cd f: . . = 1 DP.HI = tooi-tpOi + toOi-ipoi- hay r = ra + n + rc. • Do CD AB va C la trung diem cung AB nen CP = DP, tir dd B a i toan 9 (Juan Carlos). Goi AB va CD la hai day cung song tcOi-tpOi + tc02-tP0i = tD0i-tp02 + tD02-iP0i- song cua dudng tron [O). Hai dUdng tron (Oi) va (O2) cung tiep xiic ngodi vdi (O) va cd AB la tiep tuyen chung sao cho (Oi), (O2) Ket hdp vdi (1) suy ra tcoi + tco2 = tooi + ^002- • va CD nam khdc phia vdi AB. Ky hieu tpOi la tiep tuyen ke til P B a i toan 10 (Iran TST 2012). Cho hlnh binh hdnh ABCD. Goi tdi dudng tron (Oi). Khi do tco, + tco^ = too, + too^- wi, W2 Idn luat la hai dudng trdn tiep xuc vdi cdc cap doan thing L d i giai. Gpi /, J, K Ian Itfdt la tiep diem cua (O2), (Oi) vdi AB vd AD, BC vd CD. Gid sA ton tai mot dUdng tron tiep xuc vdi AB, (O) v^ P la giao diem cua HK va (O). Do K la tam vi tU dudng thdng AD vd DC vd tiep xuc ngodi vdi Wi vd W2. Khi dd ton cua (O) va (Oi) nen OP \\ hay OP ± AB, suy ra P la die'm tai mot dudng trdn tiep xuc vdi dudng thdng AB vd BC vd tiep xuc chinh giifa cung AB. TOdng tiT suy ra HK, IJ, (O) dong quy tai P. ngodi vdi wi vd W2- 1 Lcri giai. Goi 103 la dUdng tron tiep xiic vdi wi, W2, AD, DC; Ri, Mat khdc, AHK = -HOiK = -KOP = KJP nen ixi gidc R2, R3 Ian iudt la ban kinh cua wu W2, W3; hi, h2 la 2 dUdng cao HKJI noi tiep. TO dd suy ra P nam tren true dang phifdng cua cua hinh binh hanh ABCD uTng vdi cdc canh AB, AD; M, N, P, hai dudng trdn (Oi) v^ (O2), suy ra tpo^ = tpo^. (1) Q, R, S, T, U Ian liTdt la wi n W3, W2 n W3, AB n wi, AD n w i , BC n W2, CD n W2, AD n W3, CD n W3; Q', S' la diem doi xiJng
174 Cdc phuang phdp gidi todn qua cdc ky thi Olympic Dinh ly Casey vd vCng dung 175 vdi Q, 5 qua tarn cac duTdng tr6n wi, Wi- Dufa vao phep v i tu" de D i l u n^y dan den d^ng chiJng minh cAc bo 3 diem (P, M , U), (Q', M , T), (T, iV,' R), {U, N, S') thang hang. '! /. cos ^]:ADC\ V'2Ri.2R2 = y/JhM.' ' ' \ / M o t cdch tufdng tif, ton tai dUdng trdn W4 tiep xiic v d i w\, W2, AB. 5 C khi va chi k h i /I /. cos I -ABC 1 + y/2Ri.2R2 = \/h^2- \2 J Vay ton tai W3 k h i wh. chi khi ton tai w^. • 4. Bai t a p tijf luy$n Bai 1. Cho hinh vuong ABCD noi tiep dudng trdn (O). E Id mot diem nam tren cung AC chvta B. Dudng trdn {O) tiep xiic vdi AC vd tiep xiic vdi ( 0 ) tai E. Ke tiep tuyen DT tdi ((7). Chvtng minh rang DT = DA. , 1 ... L a i goi I la do dai tiep tuyen chung ngoai cua wi, W2; h, k la do d^i tiep tuyen tur U, T tdi cdc diTdng trdn wi, W2. Ta c6 Bai 2 (Hongkong 2009). Cho tarn gidc ABC vuong tai C, dudng cao CD. Dudng trdn u tiep xiic vdi cdc cqnh AC, AB idn lugt tai N, TQ = y/TM.TQ', US = VUN.US' M vd tiep xdc ngodi vdi dudng trdn dudng kinh BC. Chiing minh rdng BD.CN + BC.DM = CD.BM vd BM = BC. h = VuM.up, k = VTN.TR. Bai 3 (Kostas Vittas). Cho dudng trdn [O] dudng kinh AB. P, Q la Theo dinh ly Casey, ton tai ^ 3 di qua U, T va tiep xiic v d i i ^ i , W2 hai diem bat ky tren (0) vd khdc phia vdi AB. Ke QT 1 AB. PC nen PD idn lugt la tiep tuyen ke tU P den dudng trdn dudng kinh AT, l.TU + TQ.US = h.l2, BT. Chiing minh rdng PC + PD = PQ. do d6 Bai 4. Cho tam gidc ABC ngoai tiep dudng trdn (/), noi tiep dUdng trdn ijj. uja la dudng trdn tiep xdc trong vdi u vd tiep xiic vdi cdc l.TU + yjTM.TQ'.UN.US' = VUA4.UP.TN.TR. cqnh AB, AC. AI giao u idn thd hai tai S. Ke tiep tuyen ST tdi ST \b - c\ A p dung dinh ly ham so' sine, ta c6 u!a- Chiing minh rdng SA b+c 1- /.2i?3.sin 90-^ADC +2R3\fuS'.sinNUS.TQ'.sinMTQ Bai 5 (Iran 2009). Hai dudng trdn C i vd C2 c6 ban kinh bang nhau \ J vd cdt nhau tqi hai diem. Mot dUdng thang I idn luat cat Ci, C2 tqi UP. sin MUS.TR. sin NTQ. cdc diem theo thvc tu A, B, C, D (A, C e C2, B, D e C J . Dung
176 Cdc phucfng phdp gidi todn qua cdc ky thi Olympic Dinh ly Casey vd ling di^ng 111 hai dudng trdn Wi vd sao cho hai dudng trdn cung tiep xiic ngodi trdn ndi tiep tam gidc ABC khi vd chi khi AAi, BBi, CCi dong vdi Ci, tiep xiic trong vdi Ci vd cung tiep xiic khdc phia vdi I. Gid quy. I sit uji vd u)2 tiep xiic vdi nhau. CMng minh rang AB = CD. Bai 6. Cho hai dudng trdn OJ\ U2 tiep xiic ngodi vdi nhau tai I Tai li$u tham khao vd cung tiep xiic trong vdi u. Tiep tuye'n chung ngodi cua toi vd giao u tai B, C. Tiep tuye'n chung tai I cat uj tai A sao cho Avd I [1] Casey's theorem, from Wolfram Mathworld nhm cung mpt phia doi vdi BC. Chiing minh rhng I la tdm ndi tiep http://mathworld.wolfram.com/CaseysTheorem.html tarn gidc ABC. [2] Ptolemy's theorem, from Wolfram Mathworld Bai 7. Cho dudng trdn (O). ( O i ) vd (O2) ciing tiep xiic trong vdi http://mathworld.wolfram.com/PtolemysTheorem.html iP) vd tiep xiic ngodi nhau tai X. Tiep tuye'n chung tai X cua hai [3] Purser's theorem, from Wolfram Mathworld dudng trdn cat (O) tai A vd B. Ki hieu ti2 Id do ddi tiep tuye'n http://mathworld.wolfram.com/PursersTheorem.html , i 1 2 chung ngodi cua (Oi) vd (O2). Chvtng minh rang —— + -— = —. XA XB t\2 [4] Roger A. Johnson, Advanced Euclidean Geometry, Dover Pub- lications, New York, 1965. Bai 8 (Thebault). Cho tam gidc ABC ngoai tiep dudng trdn {I, r) vd ndi tiep dudng trdn (O). D Id mot diem nam tren canh BC. [5] Hart circle, from Wolfram Mathworld Dudng trdn u>i bdn kinh r i tiep xiic vdi cdc tia DA, DB vd tiep http://mathworld.wolfram.com/HartCircle.html xiic trong vdi ( O ) , dUdng trdn u>2 bdn kinh r2 tiep xiic vdi cdc tia [6] Art of Problem Solving Forum DA, DC vd tiep xiic trong vdi (O). Dat ADB = a. Chvtng minh http://www.artofproblemsolving.com/Forum/portal.php rang T\. COS^ 7: -\-'r2- sm = ^- [7] Lev Emelyanov, A Feuerbach type theorem on six circles. Fo- Bai 9 (Luis Gonzalez), Cho tam gidc deu ABC canh a. Goi (/) vd rum Geometricorum Vol. 1, 2001. {O) Idn luat Id dudng trdn ndi tiep vd ngoai tiep tam gidc ABC. P Id diem bat ky tren (/), XYZ Id tam gidc pedal cua P iCng vdi tam gidc ABC. Cdc dudng trdn Ci, C2, C 3 Idn luat tie'p xiic vdi canh BC, CA, AB tai X, Y, Z vd tiep xiic vdi cdc cung BC, CA, AB khong chiia dlnh ddi dien. Goi T12, T23, T31 idn luat Id do ddi cdc tie'p tuye'n chung ngodi cua Ci, C2, C3. Chvlng minh rang 35 Tn + ^23 + T31 = — a . 16 Bai 10 (Lev Emelyanov). Cho ba diem Ai, Bi, C i Idn luat thuoc ba canh BC, CA, AB cua tam gidc ABC. Dung cdc dudng trdn uja, Ub, ujc Idn luat tiep xiic vdi ba canh BC, CA, AB tai Ai, By, C\ tiep xiic vdi dudng trdn ngoai tiep (O) cua tam gidc ABC tai cdc diem ndm tren cung khdng chvta A, B, C. Chvlng minh rang ton tai mot dudng trdn tiep xiic ngodi vdi u>\, UI2, u>3 vd tie'p xiic vdi dudng
Mot so phuong phap giai bai toan ton tai trong to hap 179 M O T S O PHUdNG PHAP GIAI BAI 1.2. Nguyen li Dirichle mof rpng TOAN TON TAI TRONG TO HCJP N e u nhot n con tho vao m chuong ( m ^ n) thi c6 mot chuong chuTa [^^±^1 • • it nhat con. Nguyen Tat Thu^ 1.3. Nguyen li Dirichle cho tap hofp Cho S la tap hdp hffu han. ^ i , ^ 2 , . . . , -S^ la cac tap con cua S sao m Chu de to hdp thiTdng xuyen xua't hien trong cac ky thi chon cho Yl > k. \S\. K h i do ton tai mot phan tu" x G 5 sao cho x hoc sinh gioi va day la chu de du'Oc danh gia la kho nhaft trong de thi. Ve toan to hdp ta thufdng gap bai toan d dang yeu cau chung thuoc It nhat trong + 1 tap cua ho { 8 1 , 8 2 , • • •, 8m} • gr.'id ta chiJng minh ton tai (hoac khong ton tai) mot trang thai, mot cau hlnh to hop thoa mot tinh chat nao do. N o i dung cung nhu'phu'Ong 1.4. Nguyen li Dirichle trong hinh hoc phap giai cac dang toan nay rat phong phii va da dang. Nham giiip Cho mot hinh pbang {H) va {Hi),i = l , n la cac hinh phang nam cac em hoc sinh c6 du'dc he thong tu' duy de tim hMdng giai cho trong [H). K i hieu 5, Si la dien tich cua cac hinh phang (H) va dang toan nay, chiing toi he thong mot so phifdng phap giai bai toan ton tai, de qua do hoc sinh c6 du'dc cai nhin tdng the ve dang (H,). K h i do, neu 8 < Si thi ton tai hai hinh phang (Hi), {H^) toan nay. i=i CO giao khac rong v d i i, j E {1, 2 , . . . , n } . De suf dung nguyen l i Dirichle, ta can tao ra so chuong va so tho. 1. SuT dung nguyen li D i r i c h l e 1.5. Cac vi du minh hpa Nguyen l i Dirichle (hay la nguyen l i chuong tho) du'dc phat bieu V i du 1. Trong mot tam gidc deu canh bang 3 cho 2012 diem phan het siJc ddn gian nhu'ng lai c6 nhieu iJng dung trong toan hoc va biet. Chiing minh rang ton tai mot tam gidc deu canh bang 1 chi^a dac biet nguyen l i Dirichle la mot cong cu manh de chufng minh trong no it nhat 224 diem trong 2012 diem dd cho. bai toan ton tai. Sau day, chiing ta di x6t mot so ufng dung cua nguyen l i Dirichle cho bai toan ton tai. 1.1. Nguyen li Dirichle Neu nhot ??, + 1 con tho vao n chuong thi c6 mot chuong chiJa it nha't 2 con tho. Tru-dng T H P T Chuyen LiTcfng The Vinh, Dong Nai. 178
180 Cdc phucmg phdp gidi todn qua cdc ky thi Olympic Mot so phuang phdp gidi Inn ux'in ton tai trong to hap 181 Ldi gi^i. Ta c6: 2 0 1 2 = 8 nen ta se tao ra 9 tarn gidc deu va moi V i du 3. Cho da giac deu AxA2--- >li98i noi tiep ( O ) . Chvtng minh 224 tarn gi^c deu c6 canh b^ng 1 nam trong tarn giac c6 canh bang 3. rdng trong so 64 dinh bat ki ciia da giac ludn cd 4 dinh Id cdc dinh Ta thifc hien ph6p chia nhif sau cua mot hinh thang. Chia tarn giac da cho th^nh 9 tarn gidc deu c6 canh b^ng 1. K h i L t / i g i a i . Ta c6 nhan xet: Neu c6 2 day (diTcJc tao thanh tijf 1981 do CO mot tam giac chuTa it nhat 224 diem trong 2012 diem da cho. dinh cua da giac) c6 do dai bang nhau va khong c6 dinh chung thi • ta se CO mot hinh thang. NhSn x6t: Vdi cdch l^m nhiT tren, ta c6 the thiJc hiSn doi vdi tam .4. giac deu c6 canh b^ng n hoac vdi hinh vuong va c6 the thay doi each hoi de CO b^i loan mdi. Chdng han ta c6 bai toan sau: "Trong hinh vuong canh bling n cho Gn^ + 1 diem. Chtfng minh rang luon ton tai mot dUdng U-on c6 bdn kinh b^ng 1 chiJa it nhat 7 diem trong Qn? + 1 diem da cho". De gidi bai toan tren, ta chia hinh vuong da cho thanh v? hinh vuong dcfn vi. V i c6 Qv? + 1 diem va n? hinh vuong nen c6 it nhat mot hinh vuong chufa 7 trong 6n^ + 1 diem da cho. Difdng tron ngoai tiep hinh vuong ddn vi nay c6 bin kinh bing < 1 chiJa 7 diem noi tren. Do d6, yeu cau b^i toan dUdc chiJng minh. V i du 2. Trong mat phdng cho 2n + 1 diem sao cho vdi 3 diem bat ki ludn CO 2 diem ma khoang each givta chung nho han 1. Chitng minh rang ton tai mot dUdng tron c6 ban kinh bang 1 chvCa it nhdt n + l diem trong 2n + 1 diem da cho. Xet do dai cac day cung AiA^, AiAj.,..., >liAiggi se c6 A1A2 = AiAigsu AiA-i = AiAiQSQ, • • • , ^ i ^ o i = ^ 1 ^ 9 2 va cac dp dai nay Ldi giai. V i can chiJng minh di/dng tron chiJa n + l diem trong 2n + 1 doi mot khac nhau. Vay c6 990 dp dai cac day cung c6 mot dinh diem nen ta can chia 2n + 1 diem nay thanh 2 nhom. la A i va do cung la tat ca cac dp dai cua cdc day cung. X e t A m mot trong 2n + 1 diem. X e t dUdng tron (5) = {A, 1). Trong 64 dinh se c6 = 2016 day cung suy ra c6 it nhat 3 • Neu (5) chuTa 2n diem con l a i thi ta c6 dieu phai chtfng minh. day cung c6 ciing dp dai. Neu cac day cung nay deu doi mot c6 dinh chung thi se tao • N e u B i ( 5 ) , ta xet difdng tron {S') = {B, 1). thanh mot tam giac deu (vi chi c6 dung 2 day cung chung dinh c6 K h i do v d i diem C bat k i khdc A va B, ta c6: cung dp dai) nhu" hinh ve: AC < 1 Ce{S) Khi do dudng tron se dUpc chia ra thanh 3 cung bang nhau suy BC < 1 C e {S') ra so dinh cua da giac phai la so nguyen Ian cua 3. dieu nay la v6 li vi 1981 khong chia het cho 3. Do d6 trong 2n - 1 diem con lai (khdc A \k B) hoSc thupc (5) Vay trong 3 day cung co cung dp dai nay c6 it nhat hai day hoac thuoc {S') nen trong hai dirdng tron d6 chtfa it nhat n diem. cung khong c6 chung dinh. Ta c6 dieu phai chUng minh. • Hay dudng tron d6 chtfa it nhat n + l diem trong 2n + 1 diem da cho. • V i du 4. Cho da giac loi 2013 canh cd cdc dinh deu ed toa do
182 Cdc phucfng phdp gidi todn qua cdc ky thi Olympic nguyen. Chiing minh rang trong da gidc c6 it nhat 402 dii'm c6 toa LoTi giai. X ^ t bdng sau do nguyen. 1 2 3 4 5 2013 Lcfi giai. X6t 5 dinh lien tiep cua da giac A, B, C, D, E tao Oil au '^12013 thanh mot ngii gidc (nhiT hinh ve). 022 024 025 O22013 020141 O20142 O20143 020144 O20145 ^20142013 Trong d6 ai. = i + ku h = 1-2-3---2013 = 2013!, Mi = 1,2013. 2013 aj, = a(,_i)^ + kj, kj = W a(,_i)^, Vj = 2;20T4, i = T720T3 i=i Ta thay trong cac so cung mot cot luon ton tai hai so sao cho so nay la boi ciia so' kia. Theo de bai, trong cac so' cung mot hang luon c6 it nha't 1 so' thuoc A. Trong bang tren ta luon tim dUdc 2014 so thuoc A, ma K i hieu A, B, C, D, E Ian lifcJt la cdc goc cua ngu giac. V i bang CO 2013 cot nen trong 2014 so' d6 c6 it nhat hai so thuoc ciing (A + 5 ) + ( 5 + C) + (C + D) + (D + + (E + ^ ) = 2(A + 5 + mot cot. Nhuf vay trong hai so do luon c6 mot so' la boi cua so kia, C + D + £;) = 6 • 180°. Suy ra trong 5 dinh A, B, C, D , E luon ta CO ket luan bai toan. • ton tai hai dinh chung canh ma tong cua hai goc do Idn hdn 180". Vi du 6. Trong bang 4 x 7 (4 hang, 7 cot) ngudi ta to cdc 6 vuong Gia suf hai gdc do la A + B > 180°. bdi hai man: Den vd trang, moi 6 mot mdu. ChvCng minh rdng vdi Mat khac: bat ki cdch to nao ta luon tim duc/c mot hinh chit nhdt c6 cdc canh ndm tren cdc dudng ludi md 4 dinh d 4 6 cung mdu. A + Ex> 180° A + B^C^^Ei= 360° Lcfi giai 1. Ta x^t bang 3 x 7 (3 hang, 7 cot). Ta x6t cac trUdng B + Ci> 180° hdp sau Gia suf B + Ci > 180°. Difng hinh binh hanh ABCF, suy ra F nam • Trtfcfng ii^p 1: Trong bang ton tai it nhat m6t cot du'dc to trong tuf giac ABCE. duy nhat mot mau. Chang han toan bo cot do drfdc to mau V i ABCF la hinh binh hanh va A, B, C c6 toa do nguyen nen den. Ta xet 6 cot con lai. dan tdi F cung c6 toa do nguyen. o Neu ton tai it nhat 1 cot c6 2 6 diTcJc to mau den thi bai Ap dung tifdng tiT cho cac bo 5 diem Hen tiep rdi nhau con lai, toan diTdc giai quye't. ta CO them [ ^ ] = 402 diem nguyen nifa. TiT do ta c6 ket luan o Neu 6 cot con lai, mSi cot c6 it nhat 2 6 difdc to mau bai toan. • trang. V i c6 4 trang thai to mau cho 6 cot con lai la T- V I du 5. Cho A la mot tap con cua tap cdc so tu nhien duang. Biet D-T, D-T-T, T-T-D, T-T-T nen theo nguyen l i Dirichle, rang trong 2013 so tU nhien lien tiep bat ki luon ton tai mot so thuoc trong 6 cot do c6 it nha't 2 cot c6 trang thai to mau giong A. Chiang minh rang trong A luon ton tai hai so sao cho so nay chia nhau. Chon 2 cot do ta c6 difdc cdc to thoa yeu cau bai hit cho so kia. toan.
184 Cdc phuang phdp gidi todn qua cdc kp thi Olympic Mgt so phucfng phdp gidi bai todn ton tai trong to hap 185 • TrufoTng hgTp 2: Khong c6 cot n^o difdc t6 1 mhu, nen c6 6 Lcfi giai. K i hieu cdc so Idn thtf ba la ag < ag < • • < ao. K h i d6 trang thai to mau cho 7 cot la T - D - T , D - T - T , T - T - D , D - D - so phan tuf Idn hon ao nhieu nhat la 20 (nhieu nhat la 2 phan i\i d T , D - T - D , T - D - D . Suy ra se c6 2 cot c6 cung trang thai to m o i hang). Suy ra ao ^ 80. TiTdng t u a i ^ 78. M a t khac , mau. Chon hai hang do ta c6 dtfdc each to thoa y e u cau bai todn. as ^ ag + 1, ay ^ ag + 1 , . . . , a2 ^ ag + 7. '" - ' K e t hdp v d i cdc dieu tren l a i v d i nhau, ta diTdc •> ' * Chu-ng m i n h ho^n tat. • ao + a i + • • • + ag ^ 80 + 78 + (ag + 1) + • • • + (ag + 7) = Sag + 180. Lcfi giai 2. V i bang da cho c6 28 6 va dxXdc to bdi 2 mau nen c6 it X e t hang chiJa ag. Tong cac so cua dong chuTa ag 1^ nhat 14 6 diTcJc to cung mau. G o i a, la so 6 difOc to mau den tren cot i (1 ^ i ^ 7). 5 ( 0 9 ) ^ 100 + 99 + ag + ag 1 + • • • + ag - 7 = Sag + 171 Ta c6: ^ f l i ^ 14, va ta c6 the gidm so 6 den xuong de < Sag + 180 7 ^ ao + a i H h ag. Ta CO dieu phai chiJng minh. • i=i Cho tap S = {1,2 ,..., 999} vd A la mot tap con bat ki cua T r e n cot thiir i c6 cSp 6 cung mau den Ta chieu bang S ma \A\ 835. Chiing minh rang ludn ton tai 4 phan tA a, 6, c, d xuong mot difdng thang d song song v d i cot cua bang. M 6 i 6 cua thuoc A sao cho a + 2b + 3c = d. bdng bie'n thanh mot doan thang nam tren d va so doan thang tren d la: C4 = 6 doan. Mat khac, so cap d i e m den tren cot la: Lcfi giai. Dat A = { a i , 0 2 , . . . , agas} v d i a i < aa < • • • < agas- X e t 'hieu > ai{ai - 1) d - a835 - 3ai = 3(a835 - a i ) - 20335 ^ 3 • 834 - 2.999 = 504. t=i Do do 166 cap so sau la phan biet [d - 2; 1 ) , (d - 4; 2 ) , . . . , {d- 2-165; 165), ( d - 2 - 1 6 6 ; 166). V i c6 164 phan tuf S khong thuoc tap - ai = 7 > 6. A, nen trong cac cap tren ton tai it nhat mot cap {x; y) v d i y ^ ax i=i ~ i=i y / max,ye A. Gia suf cap do la [d - 2k; k) v6i k e {1, 2,... 166} . V i so cap chieu xuong nhieu hcJn so doan thang nen se c6 hai cap K h i do ta cd ngay: x + 2y + 3ai = agas, suy ra dieu phai chtfng ma hinh chieu cua chiing trung nhau. Bon 6 d hai cap d6 tao thanh minh. G o i c, d Ian lifdt Id phan tuf nhd nhat va Idn nhat cua tap S. mot hinh chi? nhat thoa yeu cau bai toan. • De thay d - c ^ 834, c ^ 165. Ta cd ••f A; = d - 3c = (d - c) - 2c ^ 834 - 2 • 165 > 3 • 166. Vi 7. Tren ban cd 10 x 10 ngudi ta viet cdc so tic 1 den 100. M6i hang chon ra so Idn thvC ba. Chiing minh rang ton tai mot hang c6 X6t cac cdp so {k - 2i; z) v d i z = 1 , 2 , . . . , 166 va i c thi do tong cdc so trong hang do nhd hem tong cdc chQ so Idn thii ba ducfe A; > 3 • 166 nen ta thay cd tat cd 166 - 1 = 165 cap (do loai mot chon. cap chti-a c) nhiT the v d i 330 so doi mot khdc nhau va khdc c.
186 Cdc phucfng phdp gidi todn qua cdc ky thi Olympic Mot so'phucfng phdp gidi iun loc'in ton tai trong to hop 187 D o t a p h d p A c 6 835 p h a n tur n e n c 6 d u n g 164 p h a n tur c u a S X e t t a m g i a c A,AjAk v d i AiAj < AjA^ ^ A ^ A - G p i a^^ l a s6' m a k h o n g t h u o c t a p A. S u y r a v d i 165 c a p da c h o t h i d d t h a y c 6 It " l a " n a m tren cung AiAj k h o n g chiJa A^, ajk,aki du'Pc d i n h n g h i a n h a t m o t c a p m a ca h a i so d e u t h u o c A, g i a suf l a {k - 2b, b). tifdng t y . T a t h a y m o i t a m g i a c AiAjAk tu'dng i^ng v d i m p t b p D a t k-2b = a t h i de t h a y b p so (a, b, c, d) t h o a m a n d a n g {aij, ajk, aki) t h o a 1 ^ aij ^ ajk ^ aki ^ 7 va a^j + ajk + aki = 9. V I thuTc a + 2b + Sc = d n e n t r o n g t a p h p p c6 835 p h a n tu", ta l u o n t i m d u t a m g i a c ^3^5^19 tufdng iJng v d i b p (2; 3; 4 ) . difPc b p so t h o a m a n de b a i . • C h i a T t h a n h cac tSp c p n Tj m a m o i t a p c o n Ti chiJa cac t a m Vi du 9. Mdi dinh cua mot cHu giac deu (da giac deu 9 canh) dugc g i a c d o n g d a n g t h u o c T. NhU" v a y m 6 i Ti tu'dng ufng v d i m p t b p to mot trong hai mau xanh hoac do. ChvCng minh rang ton tai hai n g h i e m c u a phtTdng t r i n h tam giac c6 dinh Id dinh cua citu giac, dong dang vdi nhau vd cdc a + 6+ c = 9 dinh duac to ciing mot mau. (*) v a ngu'dc l a i . . Lcfi giai. C h u n g t a g p i m o t t a m g i a c d o ( x a n h ) n e u t a t ca cac d i n h c u a t a m g i a c diTPc to m a u d o ( x a n h ) . Phifdng t r i n h (*) c 6 7 b p n g h i e m : ( 1 ; 1; 7 ) , ( 1 ; 2; 6 ) , ( 1 ; 3; 5 ) , ( 1 ; 4; 4 ) , (2; 2; 5 ) , (2; 3; 4 ) , (3; 3; 3 ) . D o do ta c d 7 t a p T,, m a t r o n g T c d 10 t a m g i a c n e n t h e o n g u y e n l i D i r i c h l e , t r o n g cac t a p Ti cd i t n h a t m p t t a p chiJa i t n h a t h a i t a m g i a c . B a i t o a n dUdc g i a i quye't. • 2. Phiftfng phap xay drfng T r o n g m p t sp t r t f d n g h d p de chiJng m i n h b a i t p a n t p n t a i , ta x a y d i f n g m p t c a u h i n h to h d p n a o do v a c a u h i n h d o thoa y e u c a u b a i t o a n . M p t so trUdng h d p , c a u h i n h to h d p diTdc x a y diTng dUa v a o phUdng phap q u y n a p . Vi du 10. ChvCng minh rang tap cdc so nguyen duang cd the chia thanh mot so vd hgn cdc tap vd hgn sao cho: Neu x, ?/, z, w deu thuoc Cling mot tap thi x-y vd z -w cUng thuoc tap do thi | = f;- V i 9 d i n h c i i a cufu g i a c dUPc t o b d i h a i m a u n e n theo n g u y e n l i D i r i c h l e , c 6 i t n h a t 5 d i n h diTPc t o m o t m a u . T a g i a su" 5 d i n h dUPc +00 to m a u d o . Suy r a c 6 i t n h a t C| = 10 t a m g i a c d o , k l h i e u T l a t a p Lcfi giai. T a p h a n h o a c h : N * = [J Ak. Trong dd g o m 10 t a m g i a c d o n a y . T a chuTng m i n h t r o n g 10 t a m g i a c d o n a y , c 6 h a i t a m g i a c d o n g Ak = { ( 2 / c - l ) - 2 ' | / C G N * , ten). dang v d i nhau. T a c d Ak l a t a p v 6 h a n v d i m p i A; = 1 , 2 , . . . X e t x, y, 2, w e Ak- D a t cufu g i d c d o l a A1A2 • • • ^ 9 v a dtfdng t r o n ( O ) l a d t f d n g t r o n Suy r a n g o a i t i e p cu'u g i a c . K h i d o cuTu g i a c se chia du'dng t r o n ( O ) t h a n h 9 cung nho bang nhau. T a g p i m 6 i cung nhp la m p t " l a " . x = {2k- 1) 2^\y = {2k - 1) 2^', z = {2k - 1) 2'\ = {2k - 1) 2 " ' .
188 Cdc phU(/ng phdp i^idi todn qua cdc ky thi Olympic Mot SO phuang phdp giai bai loan ton tai trong to hop 189 Do d6 Bay gid, n — m — 1 diem con lai ta chon sao cho khong c6 khoang each bat ki nao difdc tinh 2 Ian. Suy ra so khoang y each la: , , - . w n - l a:-y = {2k- 1)2^' (2^'"^' - l ) , p + m-l+ ^ (i-l)=p + m - l + i i=m+2 i=m+l z-w = {2k- 1)2"" (2^'-^' - 1) . nin — 1) mim — 1) Suy ra x-y e Ak z Bai toan dufdc chiJng minh. • z — w ^ Ak w Vi du 12. Tren bang kich thudc 2" x n (2" hang vd n cot) ta dien ChiJng minh hoan taft. • moi so mot so 1 hodc —1 sao cho khong c6 hai hang nao giong Vi du 11. Cho neN* vd songuyen h thoa l2in ^ h ^ ililLii). Mng nhau. Sau do ta thay mot vdi so d mot vdi 6 bdi so 0. Chiing minh minh rang ton tai n diem phdn biet trong mat phdng sao cho tun rang ta c6 the chon dugc mot so hang sao cho tong cdc so tren moi dugc dung h khodng cdch phdn biet trong chiing. cot tinh tren cdc 6 thuoc cdc hang dugc chon deu bang 0. Lcfi giai. Ta xet cac triTdng hdp sau Lofi giai. Ta danh so cac hang la 0 , 1 , 0 2 , . . . , 0 2 " . K i hieu / ( o i ) la hang Ui sau khi thay ddi va g{x) la hang duy nhat cua x thoa man _ ^ / i ^ n - 2, ta xet 2/i + 1 da giac deu. Do 2/i + 1 > n, nen ta chon n diem lien tiep cua da giac deu . V i n ^ / i - 2, - 1 tai cac vi tri ma x = I . nen trong n diem nay c6 dung h khoang each phan biet. 1 tai cac vi tri ma a; G { - 1 , 0 } . -•). i-'f- n-1 ^ h ^ khi do, ton tai duy nhat so nguyen m sao Ta xay diTng day cho h = /(ai)) n(n-l) m(m-l) n(n - 1) (m - 2)(m - 1) b2 = h + f{g{h)) 2 2 2 2 Ta lay n diem thuoc mot du'dng thang nhu" sau. Chon m diem = 6 2 n _ l + / ( ^ ( 6 2 n _ l ) ) ' • '''^ dau tien la 1,2,3,..., m. Diem tiep theo la m + p, vdi Ta c6: ai = ( 1 , 1 , . . . , 1). Suy ra m{m + 1) n(n — 1) p = h - {m - 1) + bi=f{a,) = (1,1,...,0,...) = ( a i , . . . , a 2 " ) ; a i G {0,1} Ta CO ^(60 = (A,...,/32^) vdi A = (^ , , n(n-l) (m-l)(m-2) I - 1 khi Qfi = 1. l^p = h- ^ ^ ' + ^+m^m. Suy ra f{g{hi)) + 61 G { 0 , 1 } =^ 6 2 G {0,1} . Tiif do, bang quy nap ta suy ra dufdc Ta CO khoang each giifa m + l diem dau la 61,62,•••,^2" e {0,1}. {1,2,3,...,m + p - 1}. Ta xet cac trtfdng hdp sau i lom '
190 Cdc phuctng phdp gidi todn qua cdc ky thi Olympic Mgt so'phuctng phdp gidi bai todn ton t Xj neu ^ a^, 6, ^ bj. va A2i khdc nhau).
Mot sdphuang phdp gun hai todn ton tai trong td hap 193 192 ' Cdc phuang phdp gidi todn qua cdc ky thi Olympic Ta se chufng minh bai todn n^y bang quy nap. ^^ Ta gia sii hi = 10, i > 1 thi b2 = k < 10 thi k h i do A2 = (l.k) < = (1,10) < A^i = {i, 10) la bo ba so sanh diTcJc. Do do, trong Gpi T la tap hdp cac dieu kien ( i ) , {ii), (Hi) nh\i tren (dieu m o i triTcJng hcJp, ta luon c6 dieu phai chufng minh. kien {Hi) tiXdng Ung v d i trUdng hdp xet sd nguyen dtfcfng n). >i B a i toan diTdc giai quyet hoan toan. • V d i n = 1, bai toan hien nhien diing. , Gia su" bai toan dung v d i m o i so" tap hdp c6 n - 1 phan tuf. Ta V i du 14. Cho hai tap hap A, B la tap hap cdc so nguyen duang se chufng minh rang v d i hai tap A, B c6 n phan tuf, ta cung c6 the thoa man \A\ \B\ n (vdi n Id so nguyen duang) va c6 tong cdc xay diTng mot bang nxn thoa man dieu kien T. phan tvc bdng nhau. Xet bang 6 vuong nxn. Chiing minh rang ta c6 That vay, ta xet hai tap the dien vdo moi 6 vuong cua bdng mot so nguyen khong dm thoa man dong thdi cdc dieu kien: -4 = { a i , a 2 , . . . , a „ } , B = 62, • • •, &n} , trong 66 ai < a2 < • • • < an, 61 < 62 < • • • < va hai tap n^y (i) Tap hap tdng cdc so d mSi hang la tap A. khong CO phan tuf chung. ,^ ^„ (ii) Tap hap tdng cdc so d moi cot Id tap B. Gia suf ai < bi. V i tdng cac phan tu" bang nhau nen tdn tai chi so' i sao cho (Hi) Co it nhdt {n - if + k so 0 trong bdng vdi k id sd cdc phan tit chung cua Avd B. Qi > bi > hi - ai =^ ai - (61 - ai) > 0. Lcfi giai. Trirdc het, ta thay rang neu mot gia tri k sao cho ton tai X e t tap A', B' nhvt sau 2 phan tuf bang nhau 6 m o i tap la a/t = 6fc = ^ thi ta dien so t vao 6 A' = { a 2 , a s , . . . , a i _ i , - 61 + a i , . . . , a „ } , B' = { ^ 2 , •. •, • vuong nam d hang thiJ k va cot thuT k, cac 6 con lai cua hang thu" k va cot thuf k deu dien vao so 0. Nhir the thi tdng cac sd d hang Hai tap hdp nay c6 cung sd phan tuf la n - 1 nen theo gia thiet quy nap, ton tai mot bang c6 kich thiTdc (n - 1) x (n - 1) thoa man dieu va cot nay thoa man de bai va khong anh hu-dng den cac hang kien T (trong bang nay c6 it nhat (n - 2)^ sd 0). va cot khac. Do do, khong mat tinh tdng quat, ta xet triTdng hdp AnB = 0(trircJng hdp c6 cac phan tuf chung thi dien them vao cac Ta them vao ben trai bang mot cot va ben tren bang mot hang hang va cot theo each tiTcJng tiT nhiT tren), tiJc la sd phan tuf chung nffa nhu" hinh ve. 6 6 goc ben trai va phia tren, ta dien sd a i , cua hai tap la = 0. d hang tM i cua bang ban dau (hang c6 tdng cac phan tu" bang a, - 61 + a i ) , ta dien so" bi - a i , con tat ca cac 6 con lai cua hang va cot vijfa them vao, ta dien vao cac sd 0. 0 0 X1 1 2 n-\ K h i do, bang nay c6 tdng cac phan tuf d moi hang la tap A va tdng cac phan tuf d m d i cot la tap B, sd cac sd 0 d bang vufa lap 0 2 dufdc khong nho hdn (n - 2)^ + 2(n - 1) - 1 = (n - 1)^ va do do 0 0 X 0 0 0 no thoa man dieu kien T. Do do, bai toan cung dung v d i m o i tap hPp c6 n phan tu*. 0 X Theo nguyen l i quy tap, bai toan nay dung v d i moi sd nguyen dufcfng n . 0 0 T - 1 V a y ta c6 dieu phai chiJng minh. •
194 Cdc phuang phdp gidi todn qua cdc ky thi Olympic Mot so phUcmg phdp gidi bdi todn ton tai trong to hop 195 3. Suf dung cong thuTc nguyen li bu trijf TO do suy ra mau thuan. Vay bai toan difcfc chiJug minh. • Cong thuTc nguyen l i bu tnjT: Vdi n tap Ai, A2,. •., An khi do, ta c6 V I du 16. Khi dieu tra mot Idp hoc, ngudi ta thdy n Hem I so hoc sinh dqt diem gidi d mon Todn cung dong thdi dqt |AiUA2U---UA„| = - ^ \Air\Aj\-\---- diem gidi d mon Vat Ly. fc=l l^iKj^n . li + {-ir-'\A,nA2n---nAn\. Hon I so hoc sinh dqt diem gidi d mon Vat Ly cung dong thdi dqt diem gidi d mon Van. De chiJng minh ton tai mot phan tuf thuoc dong thcJi tat ca cac tap Hon I so hoc sinh dqt diem gidi d mon Van cung dong thdi dqt Ai,A2,..., An, ta se chiJng minh \Ai n A2n • • • n An\ 0. diem gidi d mon Lich Sut. Vi du 15. Bay da gidc deu c6 dien tick bang 1 nam trong mot hinh Hari I so hoc sinh dqt diem gidi d mon Lich Sit cung dong thdi vuong CO do dai canh bang 2. Chiing minh rang c6 it nhdt hai da dqt diem gidi d mon Todn. gidc cat nhau c6 dien tich phan chung khong nhd hem ^. 'I.'' • Chang minh rang cd it nhdt 1 hoc sinh dqt diem gidi cd 4 mon L m giai. K i hieu Hi, H2,.. •, H7 \a 7 6a giac deu c6 dien tich bang Todn, Vat Ly, Van vd Lich SH. 1 nam trong hinh vuong H c6 canh bang 2 va S{A) la dien tich cua hinh A. Lcri giai. K i hieu T, L, V, S Ian Itfdt la tap cdc hoc sinh c6 diem Gia sijf yeu cau bai toan khong thoa man. Suy ra SiHiDHj) < \ gioi d mon Toan, Ly, Van, SuT. vdi moi 1 < 2 < j ^ 7. Khi do X = TnL, Y = Lnv, z = vns. : ' S{Hi U H2) - S{Hx) + S{H2)-S{HrnH2)>l +l - ^ = 2-]^. De chu-ng minh yeu cau b^i toan, ta chtfug minh X nY n Z 0 S{Hi UH2U H3) hay \X nY nZ\> 0. = S{Hi U H2) + S{Hs) - S{{Hi U H2) n i/3) Theo de b^i, ta c6 = S{Hi UH2) + S{Hs) - n i ^ ) U (i?2 n ifa))"''' = SiHi U H2) + S{H,) - S{Hi n Hi) - S{H2 n H^) m>lm, ivi>lm. m>H|v|. , + s{Hi n i/2 n Hs) Khong mat tinh tong qudt, ta gia sijf |T| ^ \L\ \V\ \S\. Theo /I n > 2 - - + i 7 + 7 = ^ - 7 + 7 nguyen l i bu truf, ta c6 TiTcJng tijf, ta suy ra xnYnz\ \xnY\ \z\~\{xnY)uz\ / 7 6\ X\ \Y\ \Z\-\X[JY\-\{Xn Y) u z\. S ]JH > 7- - + - + = 4. V7 7 Ma \i=i y Tuy nhien X u r = (T n L) u (L n y ) c L =^ |x u r | ^ \L 7 / \ C//=^|J//iC//:^5|Jifi ^ S{H) = 4. 1=1 \ i = l / {xr\Y)uz dV ^\{xnY)i}Z\^\v\.