Tài liệu Cac phuong phap giai toán qua cac thoi ky olympic

156 Cdc phUcfng phdp gidi todn qua cdc ky thi Olympic 0|NH LY CASEY VA L/NG DUNG i T a i li$u tham khao [1] Ha V u Anh, Dudng doi trung, Chuyen de Bao cao tai Hoi thao Toan sd cap nam 2010 tai B a V i , m Noi. Nguyen v a n Linh+ ; ' [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 con du-cfc gpi la djnh ly Ptolemy md rong (xem [1]), du-dc phat bieu nhu" sau Nauka, 1996. [5] Cae nguon tai heu tuf Internet: : ' www.mathscope.org; www.diendantoemhoc.net; www.mathlinks.org; www.imo.org.yu. / I'l, Dinh ly 1. Cho bon dudng trdn C, (i = M j . Ky hieu Uj Id do ddi cua tiep tuyen hai dudng trdn Q vd Cj. Khi do bdn dudng trdn C, Cling tiep xuc vdi mot dudng trdn (hodc mot dudng thang) C khi vd chl khi 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 • w trong (hodc ngoai) vdi C, Id tiep tuyen chung trong khi vd chl ^ c * 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 Cdc phuang phdp gidi todn qua cdc ky thi Olympic 158 khi trong hai dudng trdn d, Cj c6 mot dudng trdn tiep xiic trong, mot dudng trdn tiep xuc ngoai vdi C. Ddu cua Ujtki Id "+" khi vd 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. De dang nhan thay k h i bon diTctng tron tren suy bien thanh dirdng tron diem thi dinh ly Casey trd thanh dinh ly Ptolemy (xem [2]). K h i ba difdng tr6n suy bien thanh difdng tr6n diem t h i dinh ly Casey trd thanh dinh ly Purser (xem [3]). 2. Chufng m i n h d i n h ly L d i giai sau d\ia theo [4]. Ta se phat bieu dl. chiJng minh mot bd ' B o de 1. Cho hai dudng trdn {Ou Ri) vd {O2, R2) khong chijta nhau. I la mot diem nam ngoai hodc nam trong cd hai dudng trdn. Phep nghich ddo cUc I phuang tich R^ idn lUdt bien ( O i , i ? i ) , (O2, R2) 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) 2 T" 12 - ' I Vd ( O i , R[), (O2, R2) vd (0'2, R'2). Khi do = Dinh ly Casey vd Ung dung 19 5 Chiang m i n h . Ta chi c h t o g minh cho tnrdng hcJp tiep tuyen chung ngoai, tnrdng hcJp tiep tuyen chung trong chiJng minh ttfcJng tu". 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 tren true dang phiTcfng cua ( d ) va (02)- Goi k la phiTdng tich ti^ J den hai diTdng tron ( O i ) va (O2). Phep nghich dao bien A' thanh A, B' thanh 5 , C thanh C, D' thanh D. Do A\ D' e (O) nen A, B, C, D thang h^ng. Phep 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. 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 ) va (O2) ( M G ( O i ) , G (O2)). G o i P la hinh chieu vuong goc cua Oi tren O2N. Ta c6 = 0,0l - O2P' = 0\0\ {R, - R2)' = 0,P' MN' = {O1O2 + Ri- i?2)(Oi02 -Ri + R2) = BD.CA. Ngoai ra, ta cung c6 B'D' _ JD' _ JB' _ ^JB'.JD' BD r ~ JB ~ 7D k ~ ^JB.JD " JB.JD' Thie't lap cac bieu thiJc tu'dng tu" ta suy ra B'D'.JB'.JD' TI2 _ BD.CA R1R2 _ AB.CD C'A'.JC'.JA' • A'B'.JA'.JB' k ' _ B'D'.a C'D'.JC'.JD' 1 ~ A' A'B'.C'D'' 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 Vav B"D".C"A" R'^R!2 A"B".C"D" -'12 = -'12 R1R2 R'^R'2' B'D'.a A' A'B'.C'D' • Cdc phU(fni; 160 phdp s^idi todn qua cdc ky thi Olympic Trd lai b^i todn. TriTdc tien ta chiJng minh chieu thuan cua dinh ly Casey cho trufdng hcfp ca bon diTdng tron d deu tiep xuc trong vdi C, cac tnrdng hcJp c6n lai chiJng minh tiTcJng tif. * Ky hieu Oi Ian lifdt la ban kinh va tam cua cac dvTdng tron Ci\r,0 m ban kinh va tam cua dtfdng tron C. Khong mat tinh tong quat, gid s\i = m m { r i } . Ta nhan thay cdc dUcJng tron ( O i , r i i=l,4 n), (O2, r2 - u), (O3, rz - r4), (O4, 0) cung tiep xiic vdi dufdng tron (O, r - r^) va do dai tiep tuyen chung cua hai diTdng tron ( 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 tron ( d , r i - u), (O2, - r^), (O3, rs - u), (O4, 0) (ki hieu la Qi, Q2, Qs, Qi v6i ban kinh Ian liTdt la Ri, R2R3, 0). Xet phep nghich dao ag' bien Qi thanh Q[, Q2 thanh Q'2, Q3 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 mat tong quat gia suf tiep diem cua Q'^ va / nam tren doan thang ndi hai tiep diem cua Q[ va Q'^ vdi /. Ta c6 ^'12 + ^23 - ^13 = 0Ap dung Bo' d l 1, =4 - V i , j e {1,2,3}. Ngoai ra, ta cung c6 Ri do vay hi = R-^^- t^i t\ 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 = 0. Ta se chiJug minh Qu Q2, Q3, (O4, 0) cung tiep xuc vdi mot dudng tron. tizMi ta cung suy ra t'^^ +1'23 - i'13 = 0- 161 Khong mat tong quat gia su" i?^ = minji?;, R'2, R's}. Cung giam bdn kinh ba diTdng trdn Q[, Q'2, Q'3 mot doan i?3 ta duTdc ba du'dng tron (/i, qi), {I2, 92), {h, 0). Ta se tim quy tich cac diem I3 sao cho ^13 —• ^23 = ^12Tap hcJp cac diem sao cho tiep tuyen ke tijf diem do den (7i, q^) bang i'i3 la difdng tron tam h , bin kinh ^qf + tf^, tap hdp cac did'm sao cho tiep tuyen ke tiif diem do den {I2, 92) bang ^33 la du'dng tron tam I2, ban kinh \/^|T^. Hai du'dng tron nay cat nhau tai hai diem nam tren tiep tuyen chung ngoai cua (/i, qi) va {I2, 92)NhiT the' 73 nam tren tiep tuyen chung ngoai cua (/i, gi) va (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. Vay ton tai mot difdng tron tiep xuc vdi bon diTdng tron Q va dinh ly Casey du'dc chiJug minh. NhSn xet. Chieu thuan ciia hhi toan c6 the chiJng minh theo hufdng sau day khong suf dung ph6p nghich dao, diTcJc coi la he qua cua dinh ly Ptolemy. 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 hade ngoai) cua C\ C2 duac tinh bdi cong thiic h2 = ^ay suy ra Suf dung ph6p nghich dao Dinh ly Casey va itng dung -^\/{R±ri){R±r2). Cdc phucfng phdp gidi todn qua cdc ky thi 162 Olympic ChuTng m i n h . Ta chtfng minh bo de trong tnfcfng hdp C i va C2 cung tiep xuc trong v d i C. De thay tl^ = 0\0l - ( r i - r^)^. A p dung dinh ly ham so cosine cho Dinh ly Casey vd ling dung 163 diem chinh giOta cung nhd BC, CA, AB vd idn luat tie'p xuc vdi cdc 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 a + h+ c tarn giac OOxO^ va AOB ta c6 OxO\ OOl + OOi - 2OO1.OO2. COSO1OO2 vi 4 C O S O 1 O O 2 ) . TO do = 2B?(l- = ^R-r,f = (n , ^ - 2{R - n){R + [R- - r2)' - - r2)(l - ^ ) ( r i - r 2 ) 2 + 2{R - r i ) ( i ? - - (n - ^2) 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 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 cong thuTc ti2 = -pr-\/{R-ri){R + r2). • Trd l a i bai toan. G o i A, B, C, D Ian lUdt la tiep diem cua C i , C2, C 3 , C 4 v d i C. A p dung Bo de 2 ta c6 ^12^34 + tu.t23 AB.CD + AD.BC - AC.BD i?2 Do tur gidc ABCD AD.BC - AC.BD - ii3-^24 = V(^-n)(i?-r-2)(i?-r'3)(i?-r4). npi tiep nen theo dinh ly Ptolemy, AB.CD + = 0. Vay t u - h ^ + ^14.^23 - ii3-i24 = 0. Chieu thuan cua dinh ly Casey difdc chufng minh. 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 giffa cac c u n g nho BC, x u c v d i BC, CA, AB CA, AB n e n ba diTdng t r o n I a n Iffcft t i e p t a i trung d i e m m o i canh. 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 ) , , ^ a , a b+ c ^ a + c [C, 0) ta CO ta.a = 2-^+2'^' ^ '~Y~' ^^^"^ ^ ~2~' _ a + b 2 • /r^ 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 ) , {C, 0) ta c6 ta.tc =^^-7: + tac-b, tff d o "J t-.;'^ ac _ ^"-^^ ~ ' J _ tac^ - 3. Lfng dung B a i t o a n 1. Cho tarn gidc ABC noi tiep dudng tron Ca, Cb, Cc Idn luat Id cdc dudng trdn tiep xuc trong (O). vdi (O) Goi tai {b + c){a + b) ac 4 ~ ^ _ a + b+ c ^ ^ . TffOng t i r suy r a d i e u p h a i chiJng m i n h . ' ' • Cdc phucmg phdp gidi todn qua cdc ky thi Olympic 164 Bai toan 2 ( Dinh ly Feuerbach). Chiing minh rdng dudng tron Euler cua tam gidc tiep xiic vdi cdc dudng tron noi tiep vd bang tiep tarn gidc do. 165 Dinh ly Casey vd Ung dung Chrfng minh. Goi R, r Ian Ixidt la ban kinh dudng tron Euler va (7). Ap dung Bo de 2 cho dUdng tron (7) va {Bi, 0) ta c6 BiF R = y/iRCxF = AxF -y=^^=.BiB2, r)R R = ViR- r)R R ,/{R - r)R •C1C2, .A1A2. Do dd AiF ± BiF ± CiF = 7? y/{R-r)R R Chiing minh. Ta chtfng minh b ^ i todn cho tnTcfng hdp dufdng tron noi tiep. Cac tnTcfng hcJp khac chiJng minh tufdng ttf. Gid sur ta c6 tam gidc A B C vdi Ax, B i , Ci Ian liTcft la trung diem cdc canh B C , CA, A B ; A2, B2, C2 Ian lufcJt la tiep diem cua dirdng tron n6i tiep (7) vdi B C , CA, A B . Dat do dai cac canh tam gidc A B C Ian lUdt Id a, b, c. Ap dung dinh ly Casey cho bon diTdng tron (/), ( ^ 1 , 0), {Bi, 0), {Ci, 0) vdi do dai cdc tiep tuyen chung \in m B1B2 = la 5 i C i = \, CxAr c - a , C1C2 (I = ^, A,B, = |, A^A^ = ^ b 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. 2^{R-r)R = 0. .{AiA2±BiB2±CiC2) .{\b - c\±\c - a\±\a - b • Bai toan 4 (Difdng tron Hart). Cho ba dudng tron C i , C2, C 3 cdt nhau Idn lugt tai cdc cap diem {A, A ) , {B, B'), (C, C ) . Goi (I), {IA), (7B), (7C) idn lugt Id dudng tron noi tiep tam gidc cong ABC, ABC, AB'C, ABC. Khi do ton tai mot dudng tron tiep xuc ngodi vdi {IA), ( I B ) , (7C) va tiep xuc trong vdi (7), ggi Id dudng trdn Hart. Cdc phucmg phdp gidi todn qua cdc ky thi 166 Olympic 167 Dinh ly Casey vd vCng dung 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. — tAB-t'ci = . 0 TH ba dang thiJc tren suy ra tAC-i'BI ~ ^BC^'AI (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 ca " + " va " — " ) . Theo dinh ly Casey dang dao, ton tai mot du'dng trdn tiep xuc ngoai v d i ba difdng trdn {IA), {IB), {IC) va tiep xiic trong v d i ( / ) . • 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]). B a i toan 5 ( IMO (O). / Id tiep tuyen thdng doi xitng vdi tao thanh tarn gidc A'B'C tiep xuc vdi 2011). Cho tarn gidc ABC noi tiep dudng tron bat ky cua (O). Ggi la, h, Ic l^n lugt Id dudng I qua ba canh BC, CA. AB. la, h, Ic cat nhau A'B'C. Khi do dudng trdn ngoai tiep tarn gidc (O). Lcfi giai. K y hieu dx/i 1^ khodng cdch tir X den diTdng t h i n g I. I cat BC, CA, AB Ian liTdt tai Ai, Bu Ci. P Ik tiep diem cua / va (O). Do Ic -vh I doi xiJng nhau qua AB nen ds/i = ds/u, k va I doi 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'. TOdng tir suy ra AA', BB', tarn gi^c A'B'C. {B'A!, B'C) dong quy tai tam noi tiep / cua M a t khdc, = {B'A', = {BAJ) = 2{BA, Do dd CIA' CC 1 ^ = 90° + -CB'A' BA) + {BA, + {BA, BC) BC) BC) + + {BC, B'C) {l,BC) (mod TT). = 180° - ABC. ,) v ^ v Suy ra / G (O). G o i Pa, Pb, Pc Ian liTdt \h phufdng tich tir A', B', C t d i {0). dung dinh ly Casey cho bon difdng tron {A', 0), {B', 0), {C, ( O ) , ta cd {A'B'C) Ap 0) va tiep xuc v d i (O) khi va chi khi B'C.y^a ± A ' C . ^ b ± A'B'.^c = 0. Efat bdn kinh du'dng tron ngoai tiep tam giac ABC (1) va A'B'C liTdt la R va i?', bin kinh dUdng trdn noi tiep tam giac A'B'C B'C r. Theo dinh ly ham so' sine, , . = 2B! nen sinB'A'C B'C = 2R'. sin(180° - 2BAC) Ngoai ra, ta cung cd = 2R'. sm{2BAC). Ian la Cdc phuang phdp gidi todn qua cdc ky thi Olympic 168 suy ra ncn _ ABi.smPBiA _ dA/i ~ ~ EC _ AW {DC.PB + BD.PC) ' i ,,,,, DC.=FB + BD. KA^ MA' TD • ' • FA MD [{PD - PC).PB + {PB - PD).PC cosBAC' cosBAC r. , Lai c6 :„ 169 Dinh ly Casey vd ling dun^ dA/i=PAsinBiPA = PAsmPCA = MA MD PA.—-. P nen A A = MA ^ ^. Suy ra 2Rco^BAC B'C'.\/AA'.A'I = 2R'. PD PD.{PB-PC) MD BC. PD Chilu dio cua bd de chuTng minh tifdng tU di^a theo chieu dSo cua dinh ly Menelaus. Bo de difdc chiJng minh. • sm{2BAC).PA. = PA. sin BACAR'.^ = V2R cos BAG y/2R = PA.BC.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) V2R.R^ Vay (A'B'C) = 0. 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 tren cdc canh BC, CA, AB cua AABC vd M e AD. Khi do EF di qua M khi vd chl khi DC.= FB FA + BD.= EC EA = BC. MD MA Chtfng minh. Gia sur M e EF. Goi {P} = EF nBC. Ap dung dinh ly Menelaus cho ADAB Ung vdi dxicfng thang PMF va PEM ta CO FA MA PD EA PD MA Trd lai bai todn. Ap dung dinh ly Casey cho bdn dUdng tron {A, 0), {B, 0), (C, 0), (E) ta c6 AP.BC - AB.CQ + AC.BP hay c{b - AP) + b{c - AP) 2bc AP = — ^ -, suy ra AP = r o ^ a+b+c \S Goi {D} = Ain BC, ta c6 DI " ,CD AI b+ c ac . Ap dung Bd de 3, ta can chiJng minh b+ c PA AQ Ar = ^ , B D 6 + = 170 Cdc phuang phdp gidi todn qua cdc ky thi Olympic 111 Dinh ly Casey vd ling dung i hay 1^ c— ah b+ c 26c a + b+ c 2bc + ac b- 2bc a + b+ c 2bc b+ c a + b+ c 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 trdn tiep xuc vdi cdc tia AB, AC vd tiep xuc ngodi vdi u ^ . Tuang tu ta xdc dinh a;^, ^ c r, Va, f t , TC idn luat la bdn kinh cdc dudng trdn (/), u)'^, u[, u'^. Khi do r =^ra + n + TC- = a. a b+ c a + b+ c 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. • B a i t o a n 7 (Bo de Sawayama). Cho tam gidc ABC noi tiep dudng A 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 /, E, F thing hang khi va chi khi Lcfi g i a i . EJ ED' IA FD FA' I J b+ c IA BC. E J = 1 Theo dinh ly Menelaus, IA FA' I — . —J = 1. nen ta can chu-ng minh a.FA ={b + c)EJ. (1) Ap Ma I J a dung dinh ly Casey cho bon du-^ng tron {A, 0), (B, 0), (C, 0), a; ta c6 AF.BC + AB.CE b.BE ^ a.AF + ac= ac = AC.BE, hay BE{b + c). la a.AF + c.{BC - BE) = 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 / nam tren difdng thing EF. • Ke tiep tuyen qua T cua dtfdng tron (/) va song song vdi BC, cat AB, AC Ian lifdt tai B', C . Ta chiTng minh dUdng tron noi tiep tam giac AB'C la u'^. Lcfi g i a i . Goi Ai, Bi, Ci la tiep diem cua (7) vdi BC, CA, AB; A2, B^, C2 la tiep diem cua diTdng tron noi tiep (/„) cua tam giac AB'C vdi B'C, CA, AB'. Ki hieu p, p' la nu:a chu vi tam giac ABC, AB'C Ta c6 p' = ABi = p - a nen hai tam giac AB'C va ABC dong dang theo ti so' suy ra B'C = TiT day ta difdc P P BC2 = C-AC2 = p'-B'C = c- (p-a)- (p — a)a {P P a? 172 Cdc phucfng phdp gidi todn qua cdc ky thi Olympic Dinh IS Casey vd ling dung (v - a)'^ TiTdng W, CB2 = b- A2T = B'C'-2B'A2 . Ngo^i r a , v d i gia thiet b>c P = p —a ~ P {P - a)b P = 173 H- thi (b-c). p_~_a P Ap dung dinh ly Casey cho bon dUdng trdn ( B , 0), (C, 0), (/), (/„) ta c6 BA1.CB2 + BC.A2T = - BC2.CA1 jp - g)^ p {p-b) c— {p - ay p + a{b-c). P ip-c)=0, ntn (la) tiep xuc ngo^i vdi Ua hay {Q = ^ p —h Te f TOdng tir, - = r hay r = ra + n Tc P —C p—a . TO dd — = ^ „ ,- = ^ . Suy ra p r p i~a + n + rc 3p - a - b - c P Ap dung dinh ly Casey cho bon dufdng tron (C, 0), (P, 0), (Oi), " (O2) ta cd CP. HI = tcoi-tpOi + *co2-tpoi • Lai dp dung dinh ly Casey cho bon dUdng tron (D, 0), (P, 0), ( d ) , (O2) ta cd f: . . DP.HI = tooi-tpOi + toOi-ipoi- = 1 + rc. • B a i toan 9 (Juan Carlos). Goi AB va CD la hai day cung song 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) va CD nam khdc phia vdi AB. Ky hieu tpOi la tiep tuyen ke til P tdi dudng tron (Oi). Khi do tco, + tco^ = too, + too^L d i giai. Gpi /, J, K Ian Itfdt la tiep diem cua (O2), (Oi) vdi AB, (O) v^ P la giao diem cua HK va (O). Do K la tam vi tU cua (O) va (Oi) nen OP \\ hay OP ± AB, suy ra P la die'm chinh giifa cung AB. TOdng tiT suy ra HK, IJ, (O) dong quy tai P. 1 Mat khdc, AHK = -HOiK = -KOP = KJP nen ixi gidc HKJI noi tiep. TO dd suy ra P nam tren true dang phifdng cua hai dudng trdn (Oi) v^ (O2), suy ra tpo^ = tpo^. (1) Do CD AB va C la trung diem cung AB nen CP = DP, tir dd tcOi-tpOi + tc02-tP0i = tD0i-tp02 + tD02-iP0i- Ket hdp vdi (1) suy ra tcoi + tco2 = tooi + ^002- • B a i toan 10 (Iran TST 2012). Cho hlnh binh hdnh ABCD. Goi wi, W2 Idn luat la hai dudng trdn tiep xuc vdi cdc cap doan thing AB vd AD, BC vd CD. Gid sA ton tai mot dUdng tron tiep xuc vdi dudng thdng AD vd DC vd tiep xuc ngodi vdi Wi vd W2. Khi dd ton tai mot dudng trdn tiep xuc vdi dudng thdng AB vd BC vd tiep xuc ngodi vdi wi vd W2Lcri giai. Goi 103 la dUdng tron tiep xiic vdi wi, W2, AD, DC; Ri, R2, R3 Ian iudt la ban kinh cua wu W2, W3; hi, h2 la 2 dUdng cao cua hinh binh hanh ABCD uTng vdi cdc canh AB, AD; M, N, P, 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 vdi Q, 5 qua tarn cac duTdng tr6n wi, Wi- Dufa vao phep v i tu" de d^ng chiJng minh cAc bo 3 diem (P, M , U), (Q', M , T), (T, iV,' R), {U, N, S') thang hang. '! 175 Dinh ly Casey vd vCng dung D i l u n^y dan den /. 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 Vay ton tai 4. Bai 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 TQ = y/TM.TQ', US = h = VuM.up, k = VUN.US' VTN.TR. W3 J k h i wh. chi khi ton tai w^. t a p tijf • luy$n Bai 1. Cho hinh vuong ABCD noi tiep dudng trdn (O). E Id mot diem nam tren cung AC chvta B. Dudng trdn {O) tiep xiic vdi AC vd tiep xiic vdi ( 0 ) tai E. Ke tiep tuyen DT tdi ((7). Chvtng minh rang DT = DA. , 1 ... Bai 2 (Hongkong 2009). Cho tarn gidc ABC vuong tai C, dudng cao CD. Dudng trdn u tiep xiic vdi cdc cqnh AC, AB idn lugt tai N, M vd tiep xdc ngodi vdi dudng trdn dudng kinh BC. Chiing minh rdng BD.CN + BC.DM = CD.BM vd BM = BC. Bai 3 (Kostas Vittas). Cho dudng trdn [O] dudng kinh AB. P, Q la Theo dinh ly Casey, ton tai ^ 3 di qua U, T va tiep xiic v d i i ^ i , W2 nen l.TU + TQ.US = h.l2, hai diem bat ky tren (0) vd khdc phia vdi AB. Ke QT 1 AB. PC PD idn lugt la tiep tuyen ke tU P den dudng trdn dudng kinh AT, BT. Chiing minh rdng PC + PD = PQ. do d6 Bai 4. Cho tam gidc ABC ngoai tiep dudng trdn (/), noi tiep dUdng trdn ijj. uja la dudng trdn tiep xdc trong vdi u vd tiep xiic vdi cdc cqnh AB, AC. AI giao u idn thd hai tai S. Ke tiep tuyen ST tdi ST \b - c\ u!a- Chiing minh rdng SA b+c l.TU + yjTM.TQ'.UN.US' = VUA4.UP.TN.TR. A p dung dinh ly ham so' sine, ta c6 /.2i?3.sin 1- 90-^ADC \ J +2R3\fuS'.sinNUS.TQ'.sinMTQ UP. sin MUS.TR. sin NTQ. Bai 5 (Iran 2009). Hai dudng trdn C i vd C2 c6 ban kinh bang nhau vd cdt nhau tqi hai diem. Mot dUdng thang I idn luat cat Ci, C2 tqi cdc diem theo thvc tu A, B, C, D (A, C e C2, B, D e C J . Dung 176 Cdc phucfng phdp gidi todn qua cdc ky thi Olympic 111 Dinh ly Casey vd ling di^ng hai dudng trdn Wi vd sao cho hai dudng trdn cung tiep xiic ngodi vdi Ci, tiep xiic trong vdi Ci vd cung tiep xiic khdc phia vdi I. Gid sit uji vd u)2 tiep xiic vdi nhau. CMng minh rang AB = CD. trdn ndi tiep tam gidc ABC khi vd chi khi AAi, BBi, CCi dong Bai 6. Cho hai dudng trdn OJ\ U2 tiep xiic ngodi vdi nhau tai I vd cung tiep xiic trong vdi u. Tiep tuye'n chung ngodi cua toi vd giao u tai B, C. Tiep tuye'n chung tai I cat uj tai A sao cho Avd I nhm cung mpt phia doi vdi BC. Chiing minh rhng I la tdm ndi tiep tarn gidc ABC. Tai li$u tham khao Bai 7. Cho dudng trdn (O). ( O i ) vd (O2) ciing tiep xiic trong vdi iP) vd tiep xiic ngodi nhau tai X. Tiep tuye'n chung tai X cua hai dudng trdn cat (O) tai A vd B. Ki hieu ti2 Id do ddi tiep tuye'n i 1 2 chung ngodi cua (Oi) vd (O2). Chvtng minh rang —— + -— = —. XA XB t\2 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. Dudng trdn u>i bdn kinh r i tiep xiic vdi cdc tia DA, DB vd tiep I quy. [1] Casey's theorem, from Wolfram Mathworld http://mathworld.wolfram.com/CaseysTheorem.html [2] Ptolemy's theorem, from Wolfram Mathworld http://mathworld.wolfram.com/PtolemysTheorem.html [3] Purser's theorem, from Wolfram Mathworld http://mathworld.wolfram.com/PursersTheorem.html , [4] Roger A. Johnson, Advanced Euclidean Geometry, Dover Publications, New York, 1965. [5] Hart circle, from Wolfram Mathworld http://mathworld.wolfram.com/HartCircle.html xiic trong vdi ( O ) , dUdng trdn u>2 bdn kinh r2 tiep xiic vdi cdc tia DA, DC vd tiep xiic trong vdi (O). Dat ADB = a. Chvtng minh rang T\. COS^ 7: -\-'r2- sm = ^- [6] Art of Problem Solving Forum http://www.artofproblemsolving.com/Forum/portal.php [7] Lev Emelyanov, A Feuerbach type theorem on six circles. Fo- Bai 9 (Luis Gonzalez), Cho tam gidc deu ABC canh a. Goi (/) vd {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 rum Geometricorum Vol. 1, 2001. Mot so phuong phap giai bai toan ton tai trong to M O T S O PHUdNG PHAP GIAI BAI TOAN TON TAI TRONG TO HCJP • • hap 179 1.2. Nguyen li Dirichle mof rpng 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 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 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 phap giai cac dang toan nay rat phong phii va da dang. Nham giiip cac em hoc sinh c6 du'dc he thong tu' duy de tim hMdng giai cho 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 toan nay. cho Yl thuoc > k. \S\. K h i do ton tai mot phan tu" x G 5 sao cho x It nhat trong + 1 tap cua ho { 8 1 , 8 2 , • • •, 8m} • gr.'id 1.4. Nguyen li Dirichle trong hinh hoc Cho mot hinh pbang {H) va {Hi),i = l , n la cac hinh phang nam trong [H). K i hieu 5, Si la dien tich cua cac hinh phang (H) va (H,). K h i do, neu 8 < Si thi ton tai hai hinh phang (Hi), i=i CO giao khac rong v d i i, j E {1, 2 , . . . , n } . {H^) 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 Nguyen l i Dirichle (hay la nguyen l i chuong tho) du'dc phat bieu het siJc ddn gian nhu'ng lai c6 nhieu iJng dung trong toan hoc va dac biet nguyen l i Dirichle la mot cong cu manh de chufng minh 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 1.5. Cac vi du minh hpa V i du 1. Trong mot tam gidc deu canh bang 3 cho 2012 diem phan biet. Chiing minh rang ton tai mot tam gidc deu canh bang 1 chi^a trong no it nhat 224 diem trong 2012 diem dd cho. 180 Cdc phucmg phdp gidi todn qua cdc ky thi Olympic Ldi gi^i. Ta c6: 2 0 1 2 = 8 nen ta se tao ra 9 tarn gidc deu va moi 224 tarn gi^c deu c6 canh b^ng 1 nam trong tarn giac c6 canh bang 3. Ta thifc hien ph6p chia nhif sau Chia tarn giac da cho th^nh 9 tarn gidc deu c6 canh b^ng 1. K h i do CO mot tam giac chuTa it nhat 224 diem trong 2012 diem da cho. • Mot so phuang phdp gidi Inn ux'in ton tai trong to hap 181 V i du 3. Cho da giac deu AxA2-->li98i noi tiep ( O ) . Chvtng minh rdng trong so 64 dinh bat ki ciia da giac ludn cd 4 dinh Id cdc dinh cua mot hinh thang. L t / i g i a i . Ta c6 nhan xet: Neu c6 2 day (diTcJc tao thanh tijf 1981 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 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". .4. 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. Ldi giai. V i can chiJng minh di/dng tron chiJa n + l diem trong 2n + 1 diem nen ta can chia 2n + 1 diem nay thanh 2 nhom. X e t A m mot trong 2n + 1 diem. X e t dUdng tron (5) = {A, 1). • Neu (5) chuTa 2n diem con l a i thi ta c6 dieu phai chtfng minh. • N e u B i ( 5 ) , ta xet difdng tron {S') = {B, 1). K h i do v d i diem C bat k i khdc A va B, ta c6: AC < 1 BC < 1 Ce{S) C e {S') Do d6 trong 2n - 1 diem con lai (khdc A \k B) hoSc thupc (5) hoac thuoc {S') nen trong hai dirdng tron d6 chtfa it nhat n diem. Hay dudng tron d6 chtfa it nhat n + l da cho. diem trong 2n + 1 diem • Xet do dai cac day cung AiAigsu AiA-i = AiAiQSQ, AiA^, >liAiggi AiAj.,..., • • • , ^ i ^ o i = ^ 1 ^ 9 2 va se c6 cac A1A2 dp dai = nay doi mot khac nhau. Vay c6 990 dp dai cac day cung c6 mot dinh la A i va do cung la tat ca cac dp dai cua cdc day cung. Trong 64 dinh se c6 = 2016 day cung suy ra c6 it nhat 3 day cung c6 ciing dp dai. Neu cac day cung nay deu doi mot c6 dinh chung thi se tao thanh mot tam giac deu (vi chi c6 dung 2 day cung chung dinh c6 cung dp dai) nhu" hinh ve: Khi do dudng tron se dUpc chia ra thanh 3 cung bang nhau suy 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. Vay trong 3 day cung co cung dp dai nay c6 it nhat hai day cung khong c6 chung dinh. Ta c6 dieu phai chUng minh. • V i du 4. Cho da giac loi 2013 canh cd cdc dinh deu ed toa do Cdc phucfng phdp gidi todn qua cdc ky thi Olympic 182 nguyen. Chiing minh rang trong da gidc c6 it nhat 402 dii'm c6 toa do nguyen. Lcfi giai. X6t 5 dinh lien tiep cua da giac A, B, C, D, E tao thanh mot ngii gidc (nhiT hinh ve). LoTi giai. X ^ t bdng sau 2 1 4 3 5 au Oil O20142 '^12013 024 025 O22013 020144 O20145 ^20142013 022 020141 2013 O20143 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. K i hieu A, B, C, D, E Ian lifcJt la cdc goc cua ngu giac. V i (A + 5 ) + ( 5 + C) + (C + D) + (D + + (E + ^ ) = 2(A + 5 + C + D + £;) = 6 • 180°. Suy ra trong 5 dinh A, B, C, D , E luon ton tai hai dinh chung canh ma tong cua hai goc do Idn hdn 180". Gia suf hai gdc do la A + B > 180°. Mat khac: A + B^C^^Ei= 360° A + Ex> 180° B + Ci> 180° Gia suf B + Ci > 180°. Difng hinh binh hanh ABCF, trong tuf giac ABCE. suy ra F nam V i ABCF la hinh binh hanh va A, B, C c6 toa do nguyen nen dan tdi F cung c6 toa do nguyen. Ap dung tifdng tiT cho cac bo 5 diem Hen tiep rdi nhau con lai, ta CO them [ ^ ] = 402 diem nguyen nifa. TiT do ta c6 ket luan bai toan. • V I du 5. Cho A la mot tap con cua tap cdc so tu nhien duang. Biet rang trong 2013 so tU nhien lien tiep bat ki luon ton tai mot so thuoc A. Chiang minh rang trong A luon ton tai hai so sao cho so nay chia hit cho 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 bang CO 2013 cot nen trong 2014 so' d6 c6 it nhat hai so thuoc ciing mot cot. Nhuf vay trong hai so do luon c6 mot so' la boi cua so kia, ta CO ket luan bai toan. • Vi du 6. Trong bang 4 x 7 (4 hang, 7 cot) ngudi ta to cdc 6 vuong bdi hai man: Den vd trang, moi 6 mot mdu. ChvCng minh rdng vdi 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. Lcfi giai 1. Ta x^t bang 3 x 7 (3 hang, 7 cot). Ta x6t cac trUdng hdp sau • Trtfcfng ii^p 1: Trong bang ton tai it nhat m6t cot du'dc to duy nhat mot mau. Chang han toan bo cot do drfdc to mau den. Ta xet 6 cot con lai. o Neu ton tai it nhat 1 cot c6 2 6 diTcJc to mau den thi bai toan diTdc giai quye't. o Neu 6 cot con lai, mSi cot c6 it nhat 2 6 difdc to mau trang. V i c6 4 trang thai to mau cho 6 cot con lai la TD-T, D-T-T, T-T-D, T-T-T nen theo nguyen l i Dirichle, trong 6 cot do c6 it nha't 2 cot c6 trang thai to mau giong nhau. Chon 2 cot do ta c6 difdc cdc to thoa yeu cau bai toan. Cdc phuang phdp gidi todn qua cdc kp thi 184 Olympic • TrufoTng hgTp 2: Khong c6 cot n^o difdc t6 1 mhu, nen c6 6 trang thai to mau cho 7 cot la T - D - T , D - T - T , T - T - D , D - D T , D - T - D , T - D - D . Suy ra se c6 2 cot c6 cung trang thai to mau. Chon hai hang do ta c6 dtfdc each to thoa y e u cau bai todn. • Chu-ng m i n h ho^n tat. Lcfi giai 2. V i bang da cho c6 28 6 va dxXdc to bdi 2 mau nen c6 it nhat 14 6 diTcJc to cung mau. G o i a, la so 6 difOc to mau den tren 185 Mgt so phucfng phdp gidi bai todn ton tai trong to hap Lcfi giai. K i hieu cdc so Idn thtf ba la ag < ag < • • < ao. K h i d6 so phan tuf Idn hon ao nhieu nhat la 20 (nhieu nhat la 2 phan i\i d m o i hang). Suy ra ao ^ 80. TiTdng t u a i ^ 78. M a t khac , 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 - ' • ' * > ao + a i + • • • + ag ^ 80 + 78 + (ag + 1) + • • • + (ag + 7) = Sag + 180. X e t hang chiJa ag. Tong cac so cua dong chuTa ag 1^ 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 ^ ao + a i 7 H h ag. Ta CO dieu phai chiJng minh. i=i cSp 6 cung mau den Ta chieu bang T r e n cot thiir i c6 xuong mot difdng thang d song song v d i cot cua bang. M 6 i 6 cua 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: > ai{ai - Cho tap S = {1,2 ,..., 999} vd A la mot tap con bat ki cua S ma \A\ 835. Chiing minh rang ludn ton tai 4 phan tA a, 6, c, d thuoc A sao cho a + 2b + 3c = d. Lcfi giai. Dat A = { a i , 0 2 , . . . , agas} v d i a i < aa < • • • < agas- X e t 'hieu 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 ) , . . . , ~ i=i y 2-165; 165), = 7 > 6. - ai i=i • {d- ( d - 2 - 1 6 6 ; 166). V i c6 164 phan tuf S khong thuoc tap A, nen trong cac cap tren ton tai it nhat mot cap {x; y) v d i y ^ ax / 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 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 tong cdc so trong hang do nhd hem tong cdc chQ so Idn thii ba ducfe chon. A; = d - 3c = (d - c) - 2c ^ 834 - 2 • 165 > 3 • 166. X6t cac cdp so {k - 2i; z) v d i z = 1 , 2 , . . . , 166 va i c thi do A; > 3 • 166 nen ta thay cd tat cd 166 - 1 = 165 cap (do loai mot cap chti-a c) nhiT the v d i 330 so doi mot khdc nhau va khdc c. Cdc phucfng 186 phdp gidi todn qua cdc ky thi Olympic Mot so'phucfng phdp 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 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 . = 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 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 difPc b p so t h o a m a n de b a i . Vi du 9. Mdi dinh to mot trong tam giac dinh duac cua mot cHu giac hai mau xanh c6 dinh to ciing • Id dinh hoac deu (da giac do. cua citu giac, ChvCng minh dong dang to hop 187 v d i AiAj < AjA^ ^ A ^ A - G p i a^^ l a s6' AiAj k h o n g chiJa A^, T a t h a y m o i t a m g i a c AiAjAk du'Pc d i n h n g h i a ajk,aki tu'dng i^ng v d i m p t b p ajk, aki) t h o a 1 ^ aij ^ ajk ^ aki ^ 7 va a^j + ajk + aki = 9. V I {aij, d u t a m g i a c ^3^5^19 tufdng iJng v d i b p (2; 3; 4 ) . 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 dugc g i a c d o n g d a n g t h u o c T. ton tai hai n g h i e m c u a phtTdng t r i n h deu 9 canh) rang iun loc'in ton tai trong X e t t a m g i a c A,AjAk 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 D a t k-2b gidi vdi nhau vd cdc NhU" v a y m 6 i Ti tu'dng ufng v d i m p t b p a + 6+ c = 9 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. thanh 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 ChvCng minh rang mot so vd hgn cdc thuoc Cling mot tap thi x-y tap cdc so nguyen tap vd hgn sao cho: vd z -w Neu thuoc x, cd the chia ?/, z, w deu tap do thi | = f;- +00 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 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 cUng duang 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 . Ak = { ( 2 / c - l ) - 2 ' | / C G N * , 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 dang v d i nhau. D a t cufu g i d c d o l a A1A2 ten). 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• • • ^ 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 " ' . Cdc phU(/ng 188 phdp i^idi todn qua cdc ky thi Olympic Do d6 189 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 each la: , , . y w a:-y = {2kz-w Mot SO phuang phdp giai bai loan ton tai trong to hop = {2k- n - l p + m-l+ 1)2^' (2^'"^' - l ) , ^ (i-l)=p + m - l + i=m+2 1)2"" (2^'-^' - 1) . i i=m+l nin — 1) mim — 1) Suy ra x-y e Ak z z — w ^ Ak w Bai toan dufdc chiJng minh. • ChiJng minh hoan taft. n Mng Vi du 11. Cho neN* vd songuyen h thoa l2i ^ h ^ ililLii). minh rang ton tai n diem phdn biet trong mat phdng sao cho tun dugc dung h khodng cdch phdn biet trong chiing. Vi du 12. Tren bang kich thudc 2" x n (2" hang vd n cot) ta dien moi so mot so 1 hodc —1 sao cho khong c6 hai hang nao giong nhau. Sau do ta thay mot vdi so d mot vdi 6 bdi so 0. Chiing minh rang ta c6 the chon dugc mot so hang sao cho tong cdc so tren moi cot tinh tren cdc 6 thuoc cdc hang dugc chon deu bang 0. 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 Lcfi giai. Ta xet cac triTdng hdp sau _ ^ / 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, nen trong n diem nay c6 dung h khoang each phan biet. n-1 cho ^ h ^ • khi do, ton tai duy nhat so nguyen m sao - 1 tai cac vi tri ma x = I . 1 tai cac vi tri ma a; G { - 1 , 0 } . -•). i-'f- Ta xay diTng day h = /(ai)) n(n-l) m(m-l) n(n - 1) (m - 2)(m - 1) 2 2 2 2 Ta lay n diem thuoc mot du'dng thang nhu" sau. Chon m diem dau tien la 1,2,3,..., m. Diem tiep theo la m + p, vdi p = h - {m - 1) + m{m + 1) n(n — 1) b2 = h + f{g{h)) = 6 2 n _ l + / ( ^ ( 6 2 n _ l ) ) ' • '''^ Ta c6: ai = ( 1 , 1 , . . . , 1). Suy ra bi=f{a,) = (1,1,...,0,...) = ( a i , . . . , a 2 " ) ; a i G {0,1} Ta CO ^(60 l^p , = h- n(n-l) (m-l)(m-2) ^ ^ ' + ^+m^m. Ta CO khoang each giifa m + l diem dau la = Suy ra f{g{hi)) ta suy ra dufdc (A,...,/32^) vdi = (^ , I - 1 khi Qfi = 1. + 61 G { 0 , 1 } =^ 6 2 G {0,1} . Tiif do, bang quy nap 61,62,•••,^2" {1,2,3,...,m + p - 1}. A Ta xet cac trtfdng hdp sau e {0,1}. i lom ' Cdc 190 phuctng phdp gidi todn qua cdc ky thi Neu t6n tai z de ft, = (0, 0,..., 0). Mgt Olympic • o N e u z = 1 => / ( a i ) = (0,0,..., 0) ta chon hang 1. f{g{k^x)) + f{9{b.-2)) Ta chon hang 1, gibi),..., • N e u ton t a i i j ma k = bj + --- + ^(fei-i) thoa y e u cau bai to^n. (**) ta chon i , j : \i- j \o C i = a2 = 1, 03 = 04 = 2 , . . . , a i 9 = + f{g{bj-i)) Suy ra f{g{k)) + ^bi + f{g{k)) • + fig{b,-i)) Suy ra g{bi),... + ••• + = 0,6^ ^ .,g{bj-i) la phan biet. L a p luan tiTcfng tif, ta cung thay rang cac so 61, 62, 6 3 , . . . , 620 gia t r i 1, 2, 3,..., 10. Ta l a i cd hai trUdng hdp: thoa y e u cau bai toan. o N e u ton t a i i, j md 6 2 1 - 1 ^ 62J-1 v d i i < j thi k h i dd de • ky thi chon ddi tuyen hoc sinh gidi tlnh CO 20 em tham gia. Moi hoc sinh phdi dUcfc goi la mot bdi thi. Diem thay todn cua mot thi 2 vdng, moi 0-21-1 = 0'2i = i < 3 = a2j-i vdng cua mdi bai thi dugc cho Id mot so ta nhien tit 1 den 10. PhUorng thvCc chon ddi tuyen Id so sdnh ke't qua diem cua tCcng bdi thi tuang ling (vdng ^ cung di/dc chia thanh 10 cap va m o i cap nhan mot trong cac B a i toan diCdc chuTng minh. V i du 13. Trong = 10, va 62i-i < 6 2 i V d i z = l , 2, 3,..., 10. f{g{bj_i)). phan biet. Ta chon hang g{bi),.. 191 • N e u khong ton tai ba so nhiT the thi ro rang cdc so nay se du'cfc chia thanh 10 cap, m d i cap la hai so' bang nhau va lay dung 1 gia t r i khong vifdt qua 10. Khong mat tinh td'ng quat, ta gia sur f{a,). nhat thoa ( * * ) b, = gidi bai todn ton t 62^-1 v d i m o i Thi sinh A goi Id so sdnh duac vdi thi sinh B neu diem mdi bdi thi \ ^ z < J ^ 10. Do dd, ta suy ra dUdc rang 619 = ciia A khdng nho hon diem I Biet rang khdng cd hai thi sinh ndo cd cung cap diem so tuang i(ng. Chiing minh rdng mdi bdi thi tuang ling cua thi sinh B. cd the chon duac ba thi sinh A, B, C sao cho A so sdnh duac vdi B vd B so sdnh duac vdi C. Lcfi giai. D a t Xi = (oi, bi) dai dien cho dic'-i-i thi cua hoc sinh thuf i v d i dieu kien i = 1, 2, 3,..., 20 va 1 ^ a^, 6 ^ 10. ^ Do khong c6 hai thi sinh nao c6 ciang cap diem so nen Xi 7^ Xj vdi m o i i 7^ j . Ta quy Udc goi Xi > Xj neu ^ a^, 6, ^ bj. 1, 6 1 7 - 2 , 6 1 5 = 3,..., 61 = 10. Nhir the' ta da cd A^ = (1,10), ^ = (2,9), As = (3,8),..., A,g = (10,1) va cdc cap nay d o i mot khong the so sanh dUdc v d i nhau. Ngoai ra, c6n 10 cap niTa la A2 = (1,62), A4 = (2,64), AQ = ( 3 , 6 6 ) , . . . , ^20 = (10,610) v d i 62, 64, 6 6 , . . . , 620 la mot hoan v i cua 10, 9, 8 , . . . , 1 ma khong cd diem ba't dong, ttfc la v d i m o i ^ = 1, 2, 3,..., 10 thi 62i 7^ 11 — z (do cdc diem cua thi sinh A2i-i va A2i khdc nhau). 192 ' Cdc phuang phdp gidi todn qua cdc ky thi Olympic 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 m o i triTcJng hcJp, ta luon c6 dieu phai chufng minh. B a i toan diTdc giai quyet hoan toan. • V i du 14. Cho hai tap hap A, B la tap hap cdc so nguyen duang thoa man \A\ \B\ n (vdi n Id so nguyen duang) va c6 tong cdc phan tvc bdng nhau. Xet bang 6 vuong nxn. Chiing minh rang ta c6 the dien vdo moi 6 vuong cua bdng mot so nguyen khong dm thoa man dong thdi cdc dieu kien: (ii) Tap hap tdng cdc so d moi cot Id tap B. tit chung cua Avd + k so 0 trong bdng vdi k id sd cdc phan B. 0 X1 0 2 0 0 X 0 0 0 0 0 193 Ta se chufng minh bai todn n^y bang quy nap. ^ ^ Gpi T la tap hdp cac dieu kien ( i ) , {ii), (Hi) nh\i tren (dieu kien {Hi) tiXdng Ung v d i trUdng hdp xet sd nguyen dtfcfng n). >i 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 se chufng minh rang v d i hai tap A, B c6 n phan tuf, ta cung c6 the xay diTng mot bang nxn thoa man dieu kien T. That vay, ta xet hai tap -4 = { a i , a 2 , . . . , a „ } , B = 62, • • •, &n} , va hai tap n^y ,^ ^„ Gia suf ai < bi. V i tdng cac phan tu" bang nhau nen tdn tai chi so' i sao cho Qi > bi > hi - ai =^ ai - (61 - ai) > 0. Lcfi giai. Trirdc het, ta thay rang neu mot gia tri k sao cho ton tai 2 phan tuf bang nhau 6 m o i tap la a/t = 6fc = ^ thi ta dien so t vao 6 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 va cot nay thoa man de bai va khong anh hu-dng den cac hang 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 hang va cot theo each tiTcJng tiT nhiT tren), tiJc la sd phan tuf chung cua hai tap la = 0. 0 phdp gun hai todn ton tai trong td hap trong 66 ai < a2 < • • • < an, 61 < 62 < • • • < khong CO phan tuf chung. (i) Tap hap tdng cdc so d mSi hang la tap A. (Hi) Co it nhdt {n - if Mot sdphuang 0 X T - 1 1 2 n-\ X e t tap A', B' nhvt sau A' = { a 2 , a s , . . . , a i _ i , - 61 + a i , . . . , a „ } , B' = { ^ 2 , •. •, • 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 kien T (trong bang nay c6 it nhat (n - 2)^ sd 0). Ta them vao ben trai bang mot cot va ben tren bang mot hang nffa nhu" hinh ve. 6 6 goc ben trai va phia tren, ta dien sd a i , 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. 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 dufdc khong nho hdn (n - 2)^ + 2(n - 1) - 1 = (n - 1)^ va do do no thoa man dieu kien T. Do do, bai toan cung dung v d i m o i tap hPp c6 n phan tu*. Theo nguyen l i quy tap, bai toan nay dung v d i moi sd nguyen dufcfng n . V a y ta c6 dieu phai chiJng minh. • Cdc phuang phdp gidi todn qua cdc ky thi Olympic 194 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 |AiUA2U---UA„| = - ^ fc=l Hem I so hoc sinh dqt diem gidi d mon Todn cung dong thdi dqt diem gidi d mon Vat Ly. \Air\Aj\-\---- l^iKj^n + {-ir-'\A,nA2n---nAn\. De chiJng minh ton tai mot phan tuf thuoc dong thcJi tat ca cac tap Ai,A2,..., An, ta se chiJng minh \Ai n A2n • • • n An\ 0. Vi du 15. Bay da gidc deu c6 dien tick bang 1 nam trong mot hinh vuong CO do dai canh bang 2. Chiing minh rang c6 it nhdt hai da gidc cat nhau c6 dien tich phan chung khong nhd hem ^. L m giai. K i hieu Hi, H2,.. •, H7 \a 7 6a giac deu c6 dien tich bang 1 nam trong hinh vuong H c6 canh bang 2 va S{A) la dien tich cua hinh A. Gia sijf yeu cau bai toan khong thoa man. Suy ra SiHiDHj) vdi moi 1 < 2 < j ^ 7. Khi do S{Hi U H2) - S{Hx) + S{H2)-S{HrnH2)>l S{Hi UH2U +l - ^ = < \ . li Hon I so hoc sinh dqt diem gidi d mon Vat Ly cung dong thdi dqt diem gidi d mon Van. Hon I so hoc sinh dqt diem gidi d mon Van cung dong thdi dqt diem gidi d mon Lich Sut. Hari I so hoc sinh dqt diem gidi d mon Lich Sit cung dong thdi dqt diem gidi d mon Todn. 'I.'' • Chang minh rang cd it nhdt 1 hoc sinh dqt diem gidi cd 4 mon Todn, Vat Ly, Van vd Lich SH. Lcri giai. K i hieu T, L, V, S Ian Itfdt la tap cdc hoc sinh c6 diem gioi d mon Toan, Ly, Van, SuT. X = TnL, 2-]^. H3) m>lm, ivi>lm. = SiHi U H2) + S{H,) - S{Hi n Hi) - S{H2 n H^) /I > 2 - - + i n 7 + 7 xnYnz\ \xnY\ TiTcJng tijf, ta suy ra S > 7- ]JH \i=i y 6\ - + - + V7 7 = 4. / m>H|v|. \z\~\{xnY)uz\ X\ \Y\ \Z\-\X[JY\-\{Xn Y) u z\. Ma Tuy nhien 7 0 , Khong mat tinh tong qudt, ta gia sijf |T| ^ \L\ \V\ \S\. Theo nguyen l i bu truf, ta c6 = ^ - 7 + 7 / 7 : ' Theo de b^i, ta c6 n i ^ ) U (i?2 n ifa))"''' + s{Hi n i/2 n Hs) z = vns. Y = Lnv, De chu-ng minh yeu cau b^i toan, ta chtfug minh X nY n Z hay \X nY nZ\> 0. = S{Hi U H2) + S{Hs) - S{{Hi U H2) n i/3) = S{Hi UH2) + S{Hs) - • \ C//=^|J//iC//:^5|Jifi ^ S{H) = 4. \i=l / 1=1 X u r = (T n L) u (L n y ) c L =^ |x u r | ^ \L {xr\Y)uz dV ^\{xnY)i}Z\^\v\.