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Ian Volume 2, Number 1 January-Feb~ary,1996 Olympiad Corner Solution by Linear Combination Fourth Mathematical Olympiad of Taiwan: First Day Taipei,April 13, 1995 Problem 1. Let P(x) = ao + alx + ...+ all-lx,-1 + a"x" be a polynomial with complex coefficients. Supposethe roots ofP(x) are ai, a2, ..., all with jaIl> I, la21> I, ..., lafi > I, and laj+)! ~ I, .", lalll ~ 1. Prove: fila;! ~ 1~o12 +iaJ2+...:'anf ;=1 Problem 2. l j- . I Given a sequence of integers Xl' X2,X], X4,XS' X6, X7, Xg, One constructs a second sequence 1x2-xII, Ix] -x21, 1x4-x]l, Ixs -x41, IX6...:xsl, 1x7-x61, IXg-x71, Ix I -Xgl.. Such a process is called a single op~ration. Find all the 8-tenns integral sequences having the following property: after finitely many single operations it becomes an integral sequence with all tenns equal. (continuedonpage 4) Kin-Yin Li In mathematics, often we are factor mj omitted. ConsideiPj, 2Pj, ..., interested in fmding a solution to m?j. Upon division by m;, no two of equations.Considerthe following two these will have the same remainder problems: becausethe difference of any two of them is not divisible by m;- So one of Problem 1. Given real numbers ml, m2, these,say c?j, has remainderI. Let Vj= ..., fnn (all distinct) and aI' a2, ..., an, c?j,thenv;= I (modmjandv;=O(Iriod fmd a polynomial v(x) such that mk)for k * i becausePj = 0 (mod mJ. v(mJ = aI, v(mJ = a2, ..., v(mJ = an. Finally to solve problem I or 2 in Problem 2. Given positive integers ml, general, we use the special case mz, ..., mn (pairwise relatively prime) solutions VI' V2,..., v" to form V = alVj + and integers ai, az, ..., an,fmd an integeJ a2v2+ ...+ a"v". It is now easyto check v such that v = al (mod mJ, v = az(mod that the expression V solves both mz), ..., v = an(mod mn). problems I and 2. Problem I comes up first in algebra and analysis (later in engineering and statistics). It is an interpolation problem, where we try to fit the values a; at m; (i.e., to fmd a polyno~ial whose graph passes through the points (mJoaJ, (mz,az), ..., (mn,an». Problem 2 comes up in number theory. It is a congruence problem, where we try to count objects by inspecting the remainders (i.e., to find a number which has the same remainder as a; upon division by mJ. For problem ), v(x)=a) ~(x) ~(ml) + ...+an Pn(x) Pn(mn) is called Lagrange's interpolation formula. For problem2, althoughthe c)s may be tedious to find, we know a solution v = alc1P1+ ...+ anc"Pn exists. This is the assertion of the Chinese remaindertheorem.Note also that if we addto v anymultiple of (x-m J(x-mJ ... (x-mJ in problem 1 or any multiple of mlm2...mnin problem 2, we get other Th~..~ ;c, ~ +a~I.~:~..a +I.n+ ---1."." ." a ""'1"11'1"'" ..la. "'all uc solutions. applied to both problems. The idea is to solve first the special cases, where exactly one of the a;,sis 1 and all others O. For problem 1, this is easily solved by defming (for i = 1, 2, ..., n) the polynomial PAx) to be (x-mJ(x-mJ... (x-mJ with the factor (x-m;) omitted, i.e., n l}(x)=il(x-mj), j=l j~i and v;(x)= P;(x)/P;(mJ.Then v,{mJ = 1 and v,{mk)= 0 for k * ibecauseP;(mJ = 0 (for k * i). The expression of v, involving a sum of multiples of VI' V2, ..., V"' is so common in similar problems that it is now come to be called a linear combination ofv\, V2, ..., v". In passing, note that the a/s are numbers. However, the Vi'Sare polynomials in problem I and numbers in problem 2. Like vectors expressed in coordinates, the v/s are objects that may take on different values at different positions. So functions corresponding to solutions of equations are often viewed as vectors (with infinitely many coordinates). Concepts like these are the foundation of Linear For problem 2, this is solved similarly by first defining (for i = 1,2, ..., n) the integer Pi to be m.m2 ...mn with the (continuedon page 2) Mathematical Excalibur, Vol. 2, No.1, Jan-Feb,96 Solution by Linear Combination: (continuedfrom page 1) Algebra,which studiesthe propertiesof solutionsof thesekind of problemsin an abstractmanner. Example 1. If .f{x) is a polynomial of degree at most n and .f{k) = (n+Ik)/(k+l) fork= 0, I, ..., n, find.f{n+l). Solution 1. Applying Lagrange's interpolation formula, we define Pix) = x(x-l) ...(x-n) omitted. Then f), Pit) J(n+I)= where of (1-1) with Pin+l) -I )"+I~ ( k~- Solution 2. The (x+llf{x) -(n+I-x) :{£rp m[¥j ~t~~~~ r..': (n+I)! (k+I)!(n-k)'- [tpm*,Jtii5E.~] .~~*~ff;; g(x) has degree (n+2lf{n+ I) = (-I)"(n+2), ./(n+ I) = (-1)1/. N =70X2+21 x3+ISX2-IOSx2=23 = E~3~1 ' ~:I!{~70 ; ~lIt ' 70X2 1!;t;;3ji2o~1J:J.tt!? '21~7.w3~~'ffl:Wi ' fflEI!;t;;Sjil 'ffrP),21 X31!;t;;S~3; ~~ IS X21!;t;;~3.wS' ~7~2 0~.=.I!i:~~ 0 tlD*mm~l!I:~ [f*Tr..'M] ~~ftJ: , R1r:IrnB at most and get g(-l) and .Therefore, g(n+l) = which *M~~~~)t:tX7.wS~/j}ffl:l!{'m :t£g§nl!l:¥~LfJ!im~ n+1.Wearegiventhatg(O)=g(I)='" = g(n) = O. So g(x) = Cx(x-l)oo-(x-n). TofindC, wesetx=-l C = = -(n+2) (-If(n+2)/(n+I)! = C(-I)n+I(n+I)! ~s~1i-OO~~ fflJJi.~.A~~~ , :t£tp~~ ra'~.mE{$' last step. '~-~1i~N .=.A~ff-t;+~ 1i~:f:t;tt-tt -f::..yIllIfiI1E*J'3 4-':fifi¥JJ1'~~I!I:'::=.::=.~zllf ='1i1ilfl:zJJi::=.,-t::;-t::;lfI:zJJi - -illr..'M~~~~~iOO.JJ;;J:: ( I)" expansion ' r~~~*M$*~~ [f*.y~] :i§:EI: =+.=- l ) k+1 polynomial ~A*~~r~~~ /1X;~1!l: ~*~$ (~T.*~) rp , ~-mT"'i!i;g~ ' ;gm [!J'jjJ::f5;[jIf!:] r~'~~ = ' rl:l"¥JJ~ffiJ? and ~~1 ~sed the binomial mthe w~f4fX*~~~* the factor (x-k) = (n+l)!/(n+l- = (-l)'l-kk!(n-k)! ( ¥B*-ifI N=2 N=3 (mod3) (modS) N=2 *~/J\a91E~I!I:N ~'~~7pH~f~f~; ; ; ~f~~5/J\IE~:Wi (mod?) 0 .-s~fiEe ~r&' ~~3 .wSW7~/i.';:ffl:l!{'~N~m~,fEffr~ 00 0 ~~~llipHfI~ ~o ~IRJ implies. ","'" -" 'C'.Mt:. Example 2. Prove that for eachpositive integer n there exist n consecutive positive integers,none of which is an integral power of a prime number. (Source:1989 IMO.) 75-i1.~~~~ - CA-M)).l:t ~ 1;:-\~ I-'" .A.m c:fiJ A.1. , -m ~ I'Y\.A ! .5 1W. ~-1ti.~~ I ~':.~ nt} ~IS ~~~ P2II-IP2n)'Since PIPz, P3P4' ..., P2n-IP2nare pairwise relatively prime, by the Chinese remainder theorem, there is a positive integer solution v. Then each of the n consecutive numbers v+ 1, v+2, ..., v+n is divisible by more than one prime number. So each is not a power of a prime number. = v':-r;q~ )~ 1--.8.:-1ft. I~ ~ fT ~.J...4=:!. ~~ :.~.:.!h .d..=! -(A 2. ~ ~(") v Z- "'" ~~ .1~'5I :~A<' ,o! -/) (;.) 'a. -(AoI} 2.. '2.., A-I ~~(~-'),A. ~ --:: z. + ~~A -I f r:;A::t1 !~ ~A.. -- ~ ~- ~-=ij:-- M=-~~y'::lF if1 ~ A~I ~e-; ~- A > t J ~'D" I -~- ~~Q.Bj ~~'Oj~~ 9'.:.F~ ~~(ft&; , ~ flJ'(J Solution. LetPl'pZ' ."'P2n be 2n distinct prime numbers and consider the congruence problem v = -1 (mod PIPJ, v = -2 (mod P3P4), ..., v = -n (mod I,f!)t, )it \ o ~ ~ A=/ '-" @I J( ,.A)6 ~~. --~'=. ~"- """ -¥- .If -(I-A)";' -, ~ A~' (r~~)- ~ We Mathemat~~1 Excalibur, Vol. 2, No.J!Jan-Feb, 96 3 Problem 22. An acute-angledtriangle ABC is given in the plane. The circle Solutions with diameterAB intersectsaltitude CE ***************** and its extensionat points M and N, and the circle with diameterAC intersects Problem 21. Showthat if a polynomial altitude BD andits extensionat P and Q. P(x) satisfies Prove that the pointsM, N, P, Q lie on a common circle. (Source: 1990 USA p(2X2 -1) = MathematicalOlympiad). 2 it mustbe constant. ***************** Problem Corner We welcomereadersto submitsolutions to the problems posed below for publication consideration. Solutions should be preceded by the solver's name, address,school affiliation and gradelevel. Pleasesendsubmissionsto Dr. Kin- Yin Li, Dept of Mathematics, Hong Kong University of Scienceand Technology, Clear Water Bay, Kowloon. The deadline for submitting Solution 1: Independent solution by LIU Wai Kwong (Pui Tak Canossian solutionsis February28, 1996. College)andYUNG Fai (CUHK). ~ Problem the equation 26. Show COS1tX that the solutions = are all t of Construct irrational J¥ = Un-t u" numbers. (Source: 1974 Putnam a sequence Exam.) + U,.+l P(un) < I D have 3 for for n ?; we get * This Therefore, implies I. ..., P(I) = = 0 = J3.) P(x)P'(x). = zero= P'(l) Solution: William CHEUNG Pok Man (S.T.F.A. Leung Kau Kui College). P(UU l:t equation P'(UU = is P'(uJ the that P(un) P(U".J, get = Note 4xP'(2r-I) is P(un) functional 4, we P'(x)0 P(x) and ?; but the P(l) so n 2 I (otherwise P(Un.U, Differentiating P, ?; = O. P'(UJ) polynomial= If 1\1;N; P, Q are concyclic, then A must be the center because it is the intersection of the perpendicular bisectorsof PQ andMN. So it sufficesto showAP = AM. constant. Consideringthe similar triangles ADP and APC, we get ADIAP = APIAC, i.e., Comments:This problemwas from the Ap2 = ADxAC. Similarly, AJ.1l = 1991USSR Math Winter Camp. Below AExAB. SinceLBEC =LBDC, points B, we will provide a solution without C, D, E are concyclic. Therefore, calculus. ADxAC = AExAB and so AP = AM. -I D A ",,~~~~==~~==:::7 > -. n n rational, and 1C 0 imply are Since C for -1 * would for I~ U2 = -land I ti or (P(u,.+u2/2) ! I, 2 Un < Problem 27. Let ABCD be a cyclic quadrilateraland let lA, IB, lc, ID be the incenters of ABCD, MCD, MBD, MBC, respectively. Show that lAIJclD is a rectangle. Ul= -B Problem 28. The positive integersare separated into two subsets with no common elements. Show that one of thesetwo subsetsmust contain a three term arithmeticprogression. Problem 29. Suppose P(x) is a non constant polynomial with integer coefficients and all coefficients are greater than or equal to -1. If P(2) = 0, show that P(l) :;tO. Solution 2: + an is ...+ Suppose degree n ~ I. ao(2x2 -1)n = P(x) such a = al)X" + alX'.1 polynomial with Then +al(2x2 _1)n-1 + ( aox n +alx n-1 ...+an ) 2 -1. + ...+an 2 Comparing the coefficients of x2Il, find a02n = ao2/2, so ao = 2'1+1. Suppose ai, ..., ak are known Comparing the left yields side involving yields rational aD, be coefficients ..., a number a of ~l-k-l, ako but form involving rational. aD, rational. rational of the number ak+1 is also to we right side aOak+l plus ao, ..., Hence the number the aD, aI, ak. a So ..., an + are an all rational. is rational.Then However, P(I) = ao + P(l) a1 + = (P(1)2/2) a Problem 30. For positive integer n > I, define ./(n) to be I plus the sum of all prime numbers dividing n multiplied by their exponents, e.g.,./(40) =./(23 X 51)= 1+ (2x3 + 5xl) = 12. Show that ifn > 6, contradiction. the sequence n,./(n),fiJ{n»,fij{/(n»), ... must eventually be repeating 8, 7, 8, 7, 8,7, Other commended solver: William CHEUNG Pok Man (S.T.F.A. Leung Kau Kui College). -1 forces Therefore P(l) = l:t P(x) J3, must Other commendedsolvers: HO Wing Yip (ClementiSecondarySchool), LIU Wai Kwong (Pui Tak Canossian College), Edmond MOK Tze Tao (Queen'sCollege), WONG Him Ting (HKU) and YU Chun Ling (Ying Wa College). be constant. Problem 23. Determine all sequences i {at,a2,...} suchthat al = 1 and Ian-ami ~ 2mn/(m2+ n2)for all positive integers m and n. (Source: PastIMO problem proposedby Finland). Solution: Independent solution by CHAN Wing Sum (HKUST), LIU Wai Kwong (Pui Tak Canossian College) and YUNG Fai (CUHK). For fixed m, . I 11m an -nl n~(X) A a 1 <-n~(X)m lim 2 +n 2 .l.mn=o (continuedonpage 4) Mathematical Excalibur, ~o/. 2, No. 1, Ja~eb, 96 Problem Corner (contmuedfrompage 3) (For this solution, log means loglo') 9000possiblechoices.In particular,one Observethat of theseis 1995. So for all m, anI = lim an , Page4 . (1) 1 0 < 1 og ak ~ 1 og a30000 < 1996 1995 fork= It follows that all teTDlsare equal (to al = 1.) 1,2, ...,30000; (continuedfrom page I) }::Ioga j > 15000(logal+ loga30000) > I. j=1 Note the distance between {logI995} and {logI996} is log(1996/1995). Now bo, bl, ..., b3oooo is increasing and b3oooo -bo > 1 (by (ii», but 0 < b"J -b. < log 1996 (by (i)). J') 1995 Solution: Independentsolution by HO Wing Yip (ClementiSecondarySchool) Sothereis a k ~ 30000 suchthat and YU Chun Ling (Ying Wa College). {log 1995} < {bk} < {log 1996} Let bl ;?:b~;?:...;?:bn be the heights of the Now boys andg} ;?:g2;?:...;?:g" be those of the logIO8!+ L~=IIOgaj = log(l08 + k)!~ 8k girls. Suppose for some k, Ibk-gkl ;?:10. In the case bk-&;?: 10,we have bi-~;?:10 implies for 1 ~ i ~ k and k ~j ~ n. Consider the boys of height bi (1 ~ i ~ k) and the girls {log1995} <{log(108+k)!} <{logI996}. ofheightgj(k~j:S n). By the pigeonhole Adding [log1995] = [log1996] =3, we principle, two of these n+ 1 must be have paired originally. However, bi -gj;?: 10 log1995 < log(108+k)! -m < logI.996 contradicts the hypothesis. (The case gk -bk;?: 10 is handled similarly.) So for m = [log (108+k)!] -3. Therefore, Ibk-gkl < 10 for all k. 1995x10'" < (108+k)! < 1996x10m. Comments: This was a problem from the 1984 Tournament of the Towns, a competition started in 1980 at Moscow and Kiev and is now participated by students in dozens of cities in different continents. Other commended solvers: CHAN Wing Sum (HKUST), William CHEUNG Pok Man (S.T.F.A. Leung Kau Kui College, KU Yuk Lun (HKUST), LIU Wai Kwong (Pui Tak CanossianCollege) and WONG Him Ting (HKU). Problem 25. Are there any positive integersn such that the first four digits from the left side of n! (in base 10 representation)is 1995? Solution 1: LIU Wai Kwong (Pui Tak CanossianCollege). Let [x] be the greatest integer not exceedingx and {x} =x -[x]. Also, let a.:J = I + J"x10'&, b 0 = lo g 1081 and b.= .J log 10&!+ (log a) + ...+ log aj)forj> O. Olympiad Corner 30000 (n) Problem 24. In a party, n boys and n girls are paired. It is observedthat in eachpair,the difference in heightis less than 10 cm. Showthat the differencein height of the k-th tallest boy andthe k-th tallestgirl is also lessthan 10 cm for k = 1,2, ..., n. - og- n-+«J Consequently, the begins with 1995. number (108+k)! Problem3. Supposen personsmeetin a meeting, every one among them is familiar with exactly8 otherparticipants of that meeting. Furthermoresuppose that each pair of two participantswho are familiar with each other have 4 acquaintances in common in that meeting, and each pair of two participantswho are not familiar with eachotherhave only 2 acquaintancesin common. What are the possiblevalues ofn? ***************** SecondDay Taipei, April 15, 1995 Probiem 4. Given n (where n ~ 2) distinct integers ml, m2,..., mil' Prove that there exist a polynomial .f{x) of degreen and with integral coefficients which satisfiesthe following conditions: (i) .f{m)=-l,foralll ~i~n. (ii).f{x) cannot be factorized into a product of two nonconstant polynomials with integral coefficients. Problem 5. Let P be a point on the circumscribedcircle of MJA~J' Let H be the orthocenterof MIA~J. Let B1 (B2, BJ respectively) be the point of intersectionof the perpendicularfrom P to A~J (AJAljAJA2respectively). It is Solution 2: Let m = 1000100000.If k < known thatthe threepointsB" Bz,BJare 99999and (m+k)! =abcd... (in base 10 colinear. Prove that the line BIB2BJ representation), then (m+k+1)! = passesthroughthe midpoint of the line abcd... x 10001... = efgh..', where efgh segmentPH. equals abcd or the first four digits of abcd+1. So, the first four digits of each Problem 6. Let a, b, c, d be integers suchthat ad -bc = k> 0, (a,b) = 1, and of(m+l)!, (m+2)!, ..., (m+99999)!must (c,d) = 1. Prove that there are exactly k be the same as or increase by 1 ordered pairs of real numbers (Xt,xJ compared with the previous factorial. satisfying 0 ~ Xl' X2< 1 and both axt + Also, becausethe fifth digit of m+k (k < bX2 and CXl + dx2are integers. 99999)is 1,the fifth digit of (m+k)! will be addedto the first digit of (m+k)! in computing (m+k+1)!. So, in any ten consecutivefactorials among (m+1)!, (m+2)!, ..., (m+99999)!,there mus~be an increaseby 1 in the first four digits. So the first four digits of (m+1)!, (m+2)!, ..., (m+99999)!musttake on all Comments: With 1995 replaced by 1993,this problemappearedin the 1993 German Mathematical Olympiad. Below we will provide the (modified) official solution.