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

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.

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 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 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 Cho bon dudng trdn C, (i = M j Ky hieu Uj Id ddi cua tiep tuyen hai dudng trdn Q vd Cj Khi bdn dudng trdn C, Cling tiep xuc vdi mot dudng trdn (hodc mot dudng thang) C vd chl tnhi ± ^13^42 ± ^14^23 = Chii y rang tiep tuyen ducfc chon cua hai dudng trdn Q, Cj Id tiep fuyen chung ngodi vd chi cd hai dudng trdn Q, Cj cung tiep •^wc (hodc ngoai) vdi C, Id tiep tuyen chung 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 Cdc phuang phdp gidi todn qua cdc ky thi Olympic 158 hai dudng trdn d, Cj c6 mot dudng trdn tiep xiic trong, mot dudng trdn tiep xuc ngoai vdi C Ddu cua Ujtki Id "+" vd chi cdc doan thing noi hai tiep diem cua Q vd Cj, Ck vd Ci khong cat nhau, ddu "-" vd chi ngugc Igi De dang nhan thay k h i bon diTctng tron tren suy bien dirdng tron diem thi dinh ly Casey trd dinh ly Ptolemy (xem [2]) K h i ba difdng tr6n suy bien difdng tr6n diem t h i dinh ly Casey trd dinh ly Purser (xem [3]) 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 Cho hai dudng trdn {Ou Ri) vd {O2, R2) khong chijta I la mot diem nam ngoai hodc nam cd hai dudng trdn Phep nghich ddo cUc I phuang tich R^ idn lUdt bien ( O i , i ? i ) , (O2, R2) (O'l, i ? ; ) , (0'2, i?'2) Ggi T12, T^^ Ian lugt la dai tiep tuyen chung ngoai (hodc neu co) cua cdc cap dUdng trdn ( O i , -Ri) T" 12 - ' I Vd ( O i , R[), (O2, R2) vd (0'2, R'2) Khi = Dinh ly Casey vd Ung dung 159 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 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) 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' A, B' , C C, D' D Do A\ D' e (O) nen A, B, C, D thang h^ng Phep nghich ddo bao to^n Idn gdc giila hai du'dng cong tai giao diem nen A, B va C, O2, D thang hang Tuf 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 r B'D' _ JD' _ JB' _ ^JB'.JD' BD ~ JB ~ 7D k ~ ^JB.JD " JB.JD' Thie't lap cac bieu thiJc tu'dng tu" ta suy 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' ~ A' A'B'.C'D'' k Bay gid, phep nghich dao jf bien (O) (O'), bien A' A", B' B", C C", D' D" K h i {0') la difdng tron tryc giao v d i ( O i ) va (O2) Tu'dng tu" nhuf tren, ta cung chiJug minh du'dc Vav Til 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 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 dai tiep tuyen chung cua hai diTdng tron ( O i , n - k), {Oj, Tj -k) ike (0,min{ri, r , } ) ) bang U, 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 Q[, Q2 Q'2, Q3 Q^, (O, r - r4) / va Q[, Q'^, Q3 cung tiep xuc mot phia vdi / Goi t[j la 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 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 = Ta se chiJug minh Qu Q2, Q3, (O4, 0) cung tiep xuc vdi mot dudng tron tizMi ta cung suy 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 cho ^13 —• ^23 = ^12Tap hcJp cac diem cho tiep tuyen ke tijf diem den (7i, q^) bang i'i3 la difdng tron tam h , bin kinh ^qf + tf^, tap hdp cac did'm cho tiep tuyen ke tiif diem den {I2, 92) bang ^33 la du'dng tron tam I2, ban kinh \/^|T^ Hai du'dng tron cat 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 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 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 dai tiep tuyen chung (trong hade ngoai) cua C\ C2 duac tinh bdi cong thiic h2 = ^ay suy 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 tnfcfng hdp C i va C2 cung tiep xuc 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 ddi cdc tiep tuyen chung ngoai cua cdc cap dudng trdn {Cb, Cc), {Cc, Ca), {Ca, Cb) Khi a + h+ c tarn giac OOxO^ va AOB ta c6 OxO\ OOl + OOi - 2OO1.OO2 COSO1OO2 vi C O S O O O ) TO = 2B?(l- = ^R-r,f = (n , ^ - 2{R - n){R + [R- - r2)' - - r2)(l - ^ ) ( r i - r ) + 2{R - r i ) ( i ? - - (n - ^2) r2)^ A T3 Vay i i = —^{R-r{){R-T2) Tifdng tif neu C\a C 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 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 tiep xuc khac phia v d i C, chang han C i tiep xiic C tiep xiic ngoai, f 12 la dai cua tiep tuyen chung 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 , C v d i C A p dung Bo de 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 + = Vay t u - h ^ + ^14.^23 - ii3-i24 = 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, ) , {B, ) , , ^ a , a b+ c ^ a + c [C, 0) ta CO ta.a = 2-^+2'^' ^ '~Y~' ^^^"^ ^ ~2~' _ a + b • /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, ) , {C, 0) ta c6 ta.tc =^^-7: + tac-b, tff d o "J t-.;'^ ac _ ^"-^^ ~ ' J _ tac^ - Lfng dung B a i t o a n Cho tarn gidc ABC noi tiep dudng tron Ca, Cb, Cc Idn luat Id cdc dudng trdn tiep xuc (O) vdi (O) Goi tai {b + c){a + b) ac ~ ^ _ 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 ( 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 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 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 dai cac canh tam gidc A B C Ian lUdt Id a, b, c Ap dung dinh ly Casey cho bon diTdng tron (/), ( ^ , 0), {Bi, 0), {Ci, 0) vdi dai cdc tiep tuyen chung \in m B1B2 = la i C i = \, CxAr c - a , C1C2 (I = ^, A,B, = |, A^A^ = ^ b Ta c6 t h ^ chon cdc dau " + " , " - " cho a\b - c\± b\c - a\±c\a- b\ nen ton tai dtfdng tr6n tiep xuc vdi (7), ( ^ , 0), {Bi, 0), (Ci, 0) hay dirdng tr6n Euler cua tam gidc ABC tiep xuc vdi (7) • Bai toan 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 ta cd the chon cdc dau "+", "-" cho FAi ± FBx ± FCi = 2^{R-r)R = .{AiA2±BiB2±CiC2) {\b - c\±\c - a\±\a - b • Bai toan (Difdng tron Hart) Cho ba dudng tron C i , C2, C cdt 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 ton tai mot dudng tron tiep xuc ngodi vdi {IA), ( I B ) , (7C) va tiep xuc 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^ dai tiep tuyen chung ngoai, la dai tiep tuyen chung 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 nen tBC-t'AI+ t'AB-tci - t'AC-tBI = — tAB-t'ci = TH ba dang thiJc tren suy tAC-i'BI ~ ^BC^'AI (lufu y dau " + " hay " - " cac dang thufc tren khong quan trong, cd the dao dau nhufng phai thod man 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 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 ( IMO (O) / Id tiep tuyen thdng doi xitng vdi tao 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 A'B'C Khi 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 qua AB nen ds/i = ds/u, k va I doi xi^ng 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 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 ^ = 90° + -CB'A' BA) + {BA, + {BA, BC) BC) BC) + + {BC, B'C) {l,BC) (mod TT) = 180° - ABC ,) v ^ v Suy / 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) va chi B'C.y^a ± A ' C ^ b ± A'B'.^c = 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 ncn _ ABi.smPBiA _ dA/i ~ ~ cosBAC' cosBAC r EC _ AW {DC.PB + BD.PC) ' i ,,,,, DC.=FB + BD KA^ MA' TD • ' • FA MD [{PD - PC).PB + {PB - PD).PC , Lai c6 :„ 169 Dinh ly Casey vd ling dun^ dA/i=PAsinBiPA = PAsmPCA = MA MD PA.—- P nen A A = MA ^ ^ Suy 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 B'C'.^a ± A'C'.^k ± A'B'.^c = 2R'.^^—(PA.BC±PB.AC±PC.AB) V2R.R^ Vay (A'B'C) = tiep xuc vdi (O) theo (1) • Bai toan (Bo de Thebault) Cho tarn gidc ABC ngoai tiep (/), noi tiep (O) Mot dudng trdn (E) tiep xuc vdi (O) vd tiep xuc vdi cdc canh AB, AC idn luat tai P, Q Khi I la trung diem PQ Lcfi giai Ta chiing minh bd de sau Bo (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 EF di qua M vd chl 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 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 + = 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 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 r =^ra + n + TC- = a a b+ c a + b+ c dung sau mot so bie'n doi Suy / 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 (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 vdi (O) Khi I nam tren dUdng thing EF Goi J la giao cua AI va /, E, F thing hang va chi Lcfi g i a i IA EJ FD ED' FA' I J b+ c IA BC E J = Theo dinh ly Menelaus, IA FA' — I J = — 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 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 Tfe TOdng tir, - = r hay r = + n Tc P —C p—a TO dd — = ^ „ ,- = ^ Suy 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- = + rc • B a i toan (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 cho (Oi), (O2) va CD nam khdc phia vdi AB Ky hieu tpOi la tiep tuyen ke til P tdi dudng tron (Oi) Khi 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 P la die'm chinh giifa cung AB TOdng tiT suy HK, IJ, (O) dong quy tai P Mat khdc, AHK = -HOiK = -KOP = KJP nen ixi gidc HKJI noi tiep TO dd suy P nam tren true dang phifdng cua hai dudng trdn (Oi) v^ (O2), suy 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 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 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, qua tarn cac duTdng tr6n wi, Wi- Dufa vao phep v i tu" de d^ng chiJng minh cAc bo 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 C va chi k h i /I / cos I -ABC + y/2Ri.2R2 = \/h^2- \2 Vay ton tai Bai L a i goi I la dai tiep tuyen chung ngoai cua wi, W2; h, k la 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 ton tai w^ t a p tijf • luy$n Bai 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 ( ) tai E Ke tiep tuyen DT tdi ((7) Chvtng minh rang DT = DA , Bai (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 (Kostas Vittas) Cho dudng trdn [O] dudng kinh AB P, Q la Theo dinh ly Casey, ton tai ^ 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 AB PC PD idn lugt la tiep tuyen ke tU P den dudng trdn dudng kinh AT, BT Chiing minh rdng PC + PD = PQ d6 Bai Cho tam gidc ABC ngoai tiep dudng trdn (/), noi tiep dUdng trdn ijj uja la dudng trdn tiep xdc 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 (Iran 2009) Hai dudng trdn C i vd C2 c6 ban kinh bang vd cdt 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 phumg Cdc 320 phdp gidi todn qua cac ky thi Olympic Do d6, /(/c, k) = / ( n + + fc, 2n + - fc) = / ( n + + /c, k) v d i m o i /c = 1, 2, 3, , n va nhiT the, b i n g each sur dung quy nap, ta c6 difdc: ; /(A;, A;) = / ( ( n + l ) z + fc, n + - fc) = / ( ( n + 2)z + k, k) V Ld'i guiL de ilu cliun dgi tuye'n Olympic todn quoc gia nam 2013 Suf dung gii thie't quy nap, ta tinh dufdc l - ( - l ) / ( j + l + 2(m + l ) , i + l ) = vdri z € Z fc = 1, 2, 3, 2n Hdn nffa, sir xac dinh eac gia tri dufdng eheo tiep theo (nam ve phia ph^i) hoan toan giong d d a i cdc diem (fc, k) va {{2n + 1)?; -\-k,2n+l-k) nen cdc gia t r i tren cung tu-dng iJng bang Say gid tri cua / c6 sir tuan hoan va dufdc minh hoa nhU hinh ben dirdi, tu-c la cac gia t r i ciia /(x, y) vdi x - y chan la su" lap l a i cac gia tri dUdc gan cho cac diem nguyen tam giac Q eo tpa dp cac dinh ( , I), (2n + l , n ) , ( n + l , 1) m+l ,.vi:.> ''M; V', + (~ir+ia^+,-+2-(-l)-+^+V+i Suy (**) cung diing v d i i = j + 1, A; = m + Nhu" the, cac gid tri dufdc gdn cho cdc diem tam gidc Cl deu cd dang 6ij ± a, ± Uj v d i 6ij € { , 1}, gid t r i 6ij cung nhu" dafu cua Ui, Qj xac dinh nhat theo i , j T i e p theo, ta chpn a2k-i = ^2fc=T' "2/= 32{;rK^^ thi ta difa ve: 2^2 ^ xy(xy + yz + zx) (x - y) V (z - x)(z - y) 4^/5 + + a + a^ - f ^ ^ 2a Ta difa ve chufng minh: hay 3^y^ = (a; + y + 2) [(a- _ y)2 ^f^^_ = (xy + yz + zx) z'^{x - y)^ + xy{z - x){z - y) Do X = + c > 2\/bc = ^ nen: \ + y^z^ 13{x^ + y^ + z^ - 3xyz) 4(x3 + y3 + ^3 ^ 13 (xy + yz + zx)(x^ + y^ + z^ + xyz) > (xy + 2z)(2 + z^ + xyz) > xy(2 + z^) + 2z(2 + z^ + xyz) Ta dUa bai todn ve chiJng minh: 4xy(2 + z3) + 8z(2 + z3 + xyz) ^ 13xyz2(z + 2) , ^ 8z(2 + z3) ^ xy(9z3 + 18z2 - 8) ^ 8z4 - 9z3 - 18z2 + I6z + + (1 - xy)(9z3 + 18z2 - 8) ^ 328 Cdc phuang phdp gi&i todn qua cdc ky thi > nen (1 - xy){'iz^ Chu y r^ng x y ^ Ldi + ISz^ - 8) ^ 0; tiep tuc dat = + i v d i i ^ va thay vao: 82^ - 9z^ - IBz^ + I z + = Olympic de thi chon dqi tuyen Olympic todn quo'c gia nam 2013 329 giffa va Ta khong c6 y dinh di t i m XQ ma thong tin chi diing de i , gidi CO dinh htfdng cho viec chpn cdc gid t r i cua x de thay vao bat dang thiJc: 4(x + 2)(2x3 + x2 + l ) U'^ + 23t^ + 3*^ - 15i + v d i t ^ 2x2 + x De dang chiJng minh difdc bieu thtfc khong am nen bat dang thiJc Ta thay cdc gid t r i ddc biet nam can chtfng m i n h tren la diing Tuf dd ta c6 dieu phSi chufng minh ^ k • Thay x = 1, ta diTdc k ^ 16 Ta thay r^ng viec xuf ly bat d i n g thufc (**) de la tai l a i nghl cdch chon h = c D i l m tinh te' khong qud kho nhifng van \, a - \ c6 diTdc k < 14 • Thay 3; = 1, ta dtfdc k ^ 15 cung la kh6 nhat cua bai todn chinh la d day Trong bat d i n g thiJc da cho, d i n g thuTc xdy r a k h i a = = c = ta dif doan la giSm so bien b^ng each dat a = 1, = x , c = • Thay X = l nen ^ thi thu diTdc k ^ (x + j + ) , tiep tuc cho x —> thi c6 A: ^ ta diTdc A; < 13.98 o 16 Tuy nhieu dieu k i e n chifa chat! Ta tiep tuc cho = c = ^, a = x^ thi • Thay x = - , ta diTde k ^ 14.2 thay vdo bat d i n g thtfe ban dau, ta diTdc D e n day, ta chi difdc k < 14 4(2x^ + 5x3 ^ 2x^ + X + 2) ^ ^ (2) Do /(x) CO dang khd cong kenh nen neu de nguyen nhiT vay ma 2x2 + X _ ( x + 2)(2x3 + x2 + " khao sat thi se kha phtfc tap Do ta nghi den viec bien doi /(x) l) 2x2 + X = /(x), ta nghl den viec tdch va chia da thiJc: v d i X > fix) (***) K h i d6, ta phdi c6 k 4, m i n / ( x ) Difdi day ta se phan tich m6t so each fix) (1) Ta CO / ' ( x ) = tiTdng difdng v d i mpt phUdng trinh bac 5: = 4x^ + Sx" + - 4x - = chinh xdc cho k Tuy nhien, bai toan yeu cau t i m k nguyen difdng Idn nha't, nen ta cung khong can di t i m gia tri m i n , ma chi danh gia no nh^m t i m mpt chSn tren cho k Chu y phifdng trinh ^(x) = c6 the vie't diTdi dang: -4 + -4- = c6 nhieu nha't mpt n g h i e m dUdng g (5) = - | < 0, g{l) + X + - - ^ 2x + = x+ 1- + (2x + X _ x ^ - l - _ = x^ x^ l)2J Chu y rang = 12 > nen nghiem XQ cua phu'dng trinh nam ( x - l ) ( x + x + l) _ ^ 1_| 4(x2 + x + l ) (2x+l)2 X' (2x + l ) ^ Nhu" vay, ta eo the phan tich dufdc nhan tuf: = 8(x2 + X + 1) fix) X'^ x-^ T r e n m i e n (0, + 0 ) thi ve trai la ham tdng, ve phdi la ham giam nen phifdng trinh ^(x) \ Ta can gidi phiTdng trinh /'(x) = Quan sdt mpt ehiit, ta phdt hien difde: Ta khong giai difdc phiTdng trinh nay, v i the' khong the t i m difdc ehSn tren 4x2 + 8x + = =A(X^ D e n day thi viec lafy dao ham da trd nen kha de dang, ta tinh difdc i>0 de tCf (* * *) c6 the suy < 14 g{x) ve dang ddn gidn hdn de de tinh dao ham va v i /(x) c6 dang phan thiJc nen X - • + (2x + l)^ ti3f ta diTa di/dc viec xet /'(x) = ve xet mot phuTdng trinh ddn gian hdn: X - = 4x3 + 4^2 _ _ I ^ x^ {2x + lf + Cdc phucfng phdp 330 gidi todn qua cdc ky thi Olympic Ldi gidi de thi chon dgi tuycn D e n day, ta gap phdi mot kh6 khSn kha Idn, la phUdng trinh 42:^ + \/3 + l 2:0 < — ^ - ta quan tam den hai viec: phifdng trinh d6 c6 bao nhieu nghiem va la nhffng gia t r i nao V i e c thuf nhat c6 the de dang thiTc hien b^ng each viet :.' - ^ NhiT vay 2013 ^ 331 < - ^ 16 V i cac danh gia kha sat nen ta manh dan chon x i = •^"'".1 de thay vao / ( x ) De y r i n g x i la nghiem ciia 8xf - x i - = 0, do: l a i phifdng trinh dudi dang: : todn quoc gia nam TO day ta t i m diTdc chan tren cho XQ la X Q < - 3x - = khong c6 "nghiem d e p " K h i khdo sat mot phifcfng trinh, /Jill Olympic - 4x2 + 4x - - - = X ™2 _ Xi — 4x1 + ' — = 8x1-4, XI — =4-4x1 2x1 + Suy V e trai ciia phifdng trinh tren I I mot ham lien tuc va dong bien vcti moi 4x1 + / ( x i ) = — - — + 2x1 + 2(8x1 - 4) - (4 - X > 0, dong thcfi bkng k i e m tra trifc tiep ta de thay phiTdng trinh c6 it nhat mot nghiem thuoc (0, 1) nen bang each ket hdp hai dieu l a i , ta c6 the k h i n g dinh phiTdng trinh / ' ( x ) = c6 nghiem nhat thupc (0, 1) 95 y 4x1) 45^-50 V i e c thu" hai thifc siT rat kho khan trifcfng hdp ciia bai toan R6 rang v d i viec chi b^ng tinh tay, ta rat kho tinh dUdc gia t r i chinh xac nghiem ciia phifdng trinh 4x^ + 4x2 - 3x - = V d i ke't qu5 nay, ta thu diTdc k < = 45 • ^ - 25 45 • 0.866 2 25 = 13.97 Tat nhien, cung c6 the lUa chon mot gia tri nao dep hdn cho X de thuan tien hdn nffa cho viec tinh toan nhufng d day ta nen than trpng v i chi can chon mot so x lech hdi xa so v d i xo thoi la c6 the diTa den viec k se bi lech di may ddn v i sang 14, 15 tham chi la 16 Tot nhat, ta van ciJ nen suf dung gia tri nao ma ta da biet ch^c rang np "sat" v d i Xo du 1^ mpt chut cung khpng sap, bu l a i ta se cd the yen tam hdn ve ket qu^ l a i , neu tinh diTdc thi ch^c ch^n cong thuTc nghiem se c6 can thufc ch^ng chit va k h i thay v i o / ( x ) , ta se rat kho de danh gia xap x i sang dang thap phan de tif diTa nhan dinh ve k Co \h nhu" ta dang di vao ngo c u t Tuy nhien, cac ban hay chu y rang ta dang can t i m gia t r i nguyen diTdng Idn nhat cua k, do ta hoan toan khong can phai tinh dung gia t r i nh6 nhat cua / ( x ) l a m gi Thay vao do, ta c6 the nghl den viec t i m mot gia tri "sat" v d i gia t r i nh6 nhat cung dufdc M u o n vay, ta can t i m mot gia t r i x sat v d i nghiem ciia phifdng trinh / ' ( x ) = de thay vao tinh toan Tif nhu cau m d i nay, ta nghi den vide danh gia chan mien cho nghiem ciia phiTdng trinh 4x^ + 4x2 _ 3^ - = G o i xo € (0, 1) la nghiem cua phufdng trinh Y tirdng cua ta la l a m khijf du'dc dang bac ba de c6 diTdc mot phufdng trinh bac thap c6 the giai difdc nhanh chong b^ng tinh tay D a u tien ta c6 f Sau k h i da t i m dufdc k < 13, ta cp the'-thil di chufng minh bat dang thiJc dung v d i k = 13 b i n g nhieu each nhu^g v i dp chat ciia bat dang thiJc {k la h i n g sp tot nhat) nen tat nhien ta se chpn g i ^ i phap nap an tpan m l hieu quS nhaft va phiTdng phap ma ta chpn day I I l a m giam sp bien , b i n g dpn bien de y r^ng v d i xo € (0, 1) thi 4xg < 4x§, d6: 8x§ - 3x0 - > Cupi cung, ta se chi mpt cdch de t i m hSng SP thiTc k rat gan v d i I h i n g SP tpt nha't ciia (*) Dat: + Axl - 3x0 - = + V4T Tif day, ta t i m diTdc chan dadi cho xo la X Q > — — — > , F{x) - 4(2x4 + 5x3 ^ 2x2 + ^ + 2) _ k{2x^ + 2) ^ 0, x G R+ T i e p theo, ta se t i m chan tren cho X Q B^ng chu y nho r^ng v d i X Q G (0, 1) thi cac luy thufa ciia no kha nh6 va "xap x i " v d i nen de khiJf bac ba, ta manh dan sijf dung bat d i n g thufc A M - G M nhU sau: = 4x§ + 4x^ - 3x0 - = (4xg + xo) + 4x^ - 4xo - > Axl + 4x§ - 4x0 - = 8xg - 4x0 - V i v d i h i n g s6' tPt nhat thi d i l m cifc tieu X Q ciia da thufc gan v d i | nen ta gia su" r i n g v d i k can t i m thi diem ciTc tieu X Q b i n g diing |, i\tc la dap h i m F'{x) = 32x3 ^ 60x2 + 16x + - A;(4x + 1) b i n g tai x = |, ta tinh 'diTdc A; = i § ? V d i gia t r i k thi F{x) dat difdc ciTc tieu tai | nen: Cdc phuang 332 phdp gidi todn qua cac ky ilii Olympic Do do, bat dang thiJc (*) dung v d i fc = ^ « 13.85859 Chii y r^ng v d i A; ^ fci = ^ « 13.96825 va v d i /c2 = 13.85859 t h i (*) cung diing nen hang so thyc tot nhat k th6a man k^ ^ 'k ^ ki (ta tinh diTdc k w 13.96764) M o t dieu thii v i d day la du ^ du 16n hdn 13 kha nhieu nhUng bat dang thuTc trUdng hdp tu"dng iJng l a i de chiJng minh hdn Nhan xet chung, bai toan c6 hlnh thiJc kha ddn gian nhmig l a i doi h6i nhieu xuT ly trung gian tinh te", nhat la phan tinh toan dieu kien th5i gian c6 gidi han va khong c6 may tinh ho trd Nhieu ban chu quan gap b a i n^y, danh gia k v p i vang de A: = hoac = 16 r o i tijf "ket l u a n " luon va de maft diem dang tiec M o t suy nghi thiTdng thafy la viec chuTng minh bat dang thiJc mot bien la chuyen ddn gian (cac bai don bien thUcfng diTa ve cac bat ddng thufc mot bien hien nhien dung) nhu'ng thiTc khong p h ^ i vay; cung nhu" cac phiTdng trinh dai so, bat dang thiJc mot bien cung c6 the kho va tham chi la rat kho neu nhu" qua trinh xijf l y , chiing ta khong thu du'dc nghiem dac biet nao Co hai bai toan c6 each giai kha giong v d i bai (phan chuTng minh), nhien ddn gidn hdn C6 le b a i toan difdc phat trien tijf cac b a i toan difdi day§: (1) Cho a, 6, c la cac so thtfc du'dng cho ahc = ChiJng minh rang (a + + c)(c + a) + ^ 5(a + + c) (2) Cho a, h, c la cac so thufc du'dng cho ahc = ChiJng minh r i n g h - a + h-\- c B a i Cho tarn gidc ABC cao AD, BE, thing ^ ah + hc + ca nhon khong can c6 gdc A bang 45° Cdc dudng CF dong quy tai true tarn H Dudng thdng E F cat dudng BC tai P Goi I Id trung diem cua BC\ thdng IF cat PH tai Q (1) Chiing minh rang IQH = AIE (2) Goi K Id true tarn cua tarn gidc AEF tarn gidc KPD thdng IG Dudng thdng CK vd ( J ) la dudng iron ngoai tiep cat dudng trdn ( J ) tai G, dudng cat dudng trdn ( J ) tai M Dudng thang JC cat dudng trdn dudng kinh BC tai N Chiing minh rang cdc diem G, M, N, C cung nam tren mot dudng trdn ^Hai bai toan 50 va 55 chUdng cuon sach Algebraic Inequalities cua Vasile Cirtoaje Lot gidi de thi chon doi tuyen Olympic todn quoc gia nam 2013 333 Lcfi giai (1) Khong mat tinh tdng quat, ta gia su" AB < AC, k h i B se n i m gii?a P, C Tru'dng hdp AB > AC du'dc chiJng minh hoan toan tu"dng tu Trirdc het, ta se chiJng minh rang PH ± AI That vay, goi U, V l l n lifdt la trung d i e m ciia AH, IH thi ta c6 UV \\ De thay (P, D, B, C) = - nen theo tinh chat ciia hang diem dieu hoa t h i : PB Ta cung c6 PE • PF • PC = PD • PI = PB • PC nen PE • PF = PD tren true dang phu^dng cua du'dng tron du'dng kinh AH (tam V) • PI (tam U) va UV D o PH AI V i RAC = 45° nen EIF suy = 90°, ^ ^ IQH CO IH Hdn nffa H cQng n i m tren true dang phufdng cua difdng tron nen PH vay ta hay P nam IQH = = 90°-AIF AIE = EIF - AIF = AIE Cdc phuang phdp gidi todn qua cdc kS> thi Olympic 334 IP Lor/ gidi de thi chon dgi tuyen Olympic todn qudc gia ndm 2013 335 (2) Ta thay r&ng EKF + EOF = EKF + EAF = 180° nen K thuoc diTcJng tr6n dudng kinh BC Do hang diem D, P, B, C dieu h6a nen ta CO ID • IP = IC^, ma IM • IG = ID • IP (ciing b^ng phUdng tich ciia / den (J)) nen IM • IG = IC^ hay AIMC ~ AICG (c-g-c) nen: , IMC = TCG = ICK = A5° (1) Goi T m trung diem PD tW theo he thiJc Maclaurin, ta^^co^B • CT = CD-CP = CK• CG hay ti3f gidc GTBK npi tiep va BKG = 90° nen cung c6 GTD = 90° hay GT PD Tam gidc JPD can tai J c6 T la trung diem PD nen J T P D Do d6, G, J, T thing hang va7'^'> i •, B a i t$p 27 Cho p Id so nguyin ^, , - n+i = ChUng minh rdng tat cd cdc so hang cua to le, ddt f { x ) =: {x + 1) • • • {x + p - I ) (a) Chiing minh rdng p f { x ) = {x + 1) • f { x + 1) - x • f { x ) (b) Gid sH f{x) =xP-^ + aixP-^ ddy ta quy udc OQ = (c) (Dmh ly 1, 2, , p - + ••• + ap_2X + a p _ i Tit ddng thiic C^'^^aj Lagrange) Chiing B a i t§p 22 Cho cdc so nguyen topx, P2, P3, • • •, Pfc doi mot khdc vd cho cdc s:' lU nhien bat ky nx, n2, n^, , Uk deu Idn hon Chiing minh rang so cdc cgp {x, y) khong cd thii tu, nguyen to cung vd thoa man ddng thiic sau: — pi {7p+l)':qvd {7q+l)\p O Cac bai So hoc +y 349 B a i t$p 24 rim tat cd cdc cdp so nguyen to p, q cho tren, hay suy rdng: pm = X IQC ngoai T'^Pj.Tx ciing di qua true tam cua TXT2T3 nam ngodi ( O ) Goi CS Mgt so bai todn chpn B a i t§p 25 Cho hdm so f{x) diem c6 dinh cdc dudng cao cua tam gidc luc vd dudng tron {Ox) ndm tam gidc AC Dudng tron (O2) qua B, C vd tiep xiic B a i tap 20 Cho tam gidc ABC Phu minh rdng chia hit cho p vdi mpi i = (d) Chiing minh rdng neu p > Id so nguyen to thi mdu so cua phdn sd'l + ^ + - - + ^ chia het cho p^ Pk khong vuat qud 2^^^ B a i t$p 23 Cho p Id so nguyen to vd a, b, c Id cdc so nguyen bat ky (e) Chiing minh rdng neu p > Id so nguyen to thi Cl~\ ( m o d p^) Chiing minh rdng ton tai cdc so nguyen x, y, z khong dong thdi chia het B a i t0p 28 Tim tat cd cdc cdp so nguyen ducfng x , y cho ( i ^ + 1): y cho p cho ax^ + by^ + cz^ chia het cho p vd (y2 + l)':x inv, , ... au Oil O201 42 O20143 '^ 120 13 024 025 O 220 13 020 144 O20145 ^20 1 420 13 022 020 141 20 13 Trong d6 = i + ku h = 1 -2- 3 -20 13 = 20 13!, Mi = 1 ,20 13 20 13 aj, = a(,_i)^ + kj, kj = W a(,_i)^, Vj = 2; 20T4,... = 20 12 (b) n = 20 13 neu: 198 Cdc phucfng phdp gidi todn qua cdc kp thi Olympic Mgt so phuang vvTdt qua n - TiTOng tU, t o n t a i so 62 cho 62 to m a u va 62 + = = 620 12 to m^u 020 12 cho 020 12. .. phucfng phdp gidi todn qua cdc ky thi Olympic 22 0 1) Cho a, 6, c la cac so thiTc diTcJng thoa man a + + c = Chufng minh bat dang thuTc 4 ^2 + 62 + c2 4c2 + a2 + 62 - 4 62 + c2 + a2 ' c h+ c c+ a a+