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&~~a~ti~~~ .?;1t~(~t~~e'1ea~~at ~ Volume 2. Number5 1996 Nay-Dec. Olympiad Corner ~ BiJ~ Ii aq II fiJ (-) 25th United States of America Olympiad: Mathematical ~ S m Part I (9am-noon,May 2, 1996) Pr 0 bl em IPr ..o:e th at th e average 0 fth .e :W:~~~~:fj::;or\"m.m~fiiJ ~ffij:1fif~#;~~fiiJ'l1.:W~f&~~~ ' ~:5J,~ & num~s n sm n (n = 2, 4, 6, ...,180) IS ~P).ljf- 0 ~:(£3!.1if~~Mir{j\l:i- AF Area of MCP cot 1 .~m.~~~fiiJJEJJ.' !i1JQ~~~~ FB Area of ABCP Problem 2. For any nonempty set S of fiiJ~WJ. 0 real numbers,let a(S) denotethe sum of ~ * ~O ~ .l:.jzI!;'=-~.1:\ 1JI.!J;f§ ~ ~ CJM the elementsof S. Given a set A of n .:fJtfrlj:$'r.~~:PJ~tt'g-flM¥[fij":m: ftm positive integers, considerthe collection ~tt'g'l1.W~~ : -. AF BD CE =1 0 of all distinct sumsa(S) as S rangesover FB DC EA the nonempty subsetsof A. Prove tbat this collection of sumscan bepartitioned ~ ~ ~ JJ. ~ * iflJA "@"][ (Giovanni into n classesso that in eachclass, the Ceva)1£+-t;-tlt*C~a~fI'i1' ffrPJf& ratio of the largest sum to the smallest ~ A fi Z ~ r"@"][ (Ceva) JJ. j 0 "@" sumdoesnot exceed 2. B c ][~JJ."CJm~lPJlJ~m~)! , f.§.-.t~ Problem 3. Let ABC be a ttiangle. (II-) m~):5~:tiP.ifIJm T m"III~OOtRffi Prove that there is a line 1 (in the plane ~fI'i1:=f:.~' f~ff**mm 0 of triangle ABq such that the !il-~moo':='~~:;FJ/;}#~AD' intersection of the interior of ~irJ~~~~ ' ~!l:t g§][~:@.~~$Q~~{t?~Vi?m triangle ABC and the interior of its m~$Q,tt.~re~$Q~~*.ff Area of MBD BD reflectionA' B' C' in I h~ areamore than =-0 ~°tt.g§][~:@.*~'~~~$Q 2/3 the area of triangle ABC. Areaof MDC DC ~: (continuedonpage 4) A ~.=.~ ~ABC~g§][~ [~ (Cevians) rn~:¥.Jj;iJ;~~~]AD" BE~CF~ ~ Jf:. fIt- F ;~.. AF BD CE =1 (*) /..- .::.t: :~ FB DC EA B D C RU!l:t=-g§"K~#1!i 0 (1].::.) ffl 1l:t~~:@.al;Jm$~ I!i ~ DJff§: ~ ~ 111=, iii ~ (J{j ~ AD ' BE~D :@.: W;Jlt ~ B~=-~][.f!:'f'-~#I5' fH CRf§ ~ ~ fii]-~ p 0 ~IJffll:.))I!; #.iI ;tt~~*g§][.f!£\~#15 0 ~irlJDJ .:=. ~ (J{j ':tl ijJ ~ ~ ,~ ~ADfOBE:3(~P (III=-) ~p{lp~ ][.f!CG, ~U AreaofMBD -- BD AreaofMBD Area of MDC DC Area of MDC ~.~.~=1 (**) GR DC EA mffl$}.~'~ : A ;E=-=- , ~U-=-= a-c a c a c b d b d b-d E CJlf BD Area of MBP B D DC Area of MCP CI)-=-) IRJ~ CE Area of MCP EA Area of MAP (continuedonpage 2) Mathematical Excalibur. Vol.2. No. Nav-Dec. 96 Pa~e 2 ~gffl~~~~fiiJ (-) (continued from page 1) Error Correcting Codes (Part I) .l:t~ (*):fa (**)"DJ~ Tsz-MeiKo AF AG -=- Supposeone would like to transmit a FB GB message.For an evenparity code,a 0 or message, say "HELLO.. .", from one so 1 is appended that the total number of ~Jl:tFWG~ABJ::~IffJ-1!i 0 computerto another. One possibleway 1's is an evennumber. The letters Hand is to use a table to encodethe message E would be representedby 10010000 :gg-][:iE~fII\]~:iE~~{PJ.~~ .f.§:$ into binary digits. Then the receiver and 10001011 respectively. With an .L~.t87E-~Wfll\]E-i'lm~'~...: would be able to decode the message even parity code, the receiver can detect E-I=P *1 (medians) # I!i' E-~ with a similar table. One such table is one transmission error, but unable to # (altitudes) I!i' E- $j- ~ *1 (angle- the American Standard Code for correctit. For example,if 10010000(for ~ tl bisectors) I!i 0 ~ E- i'l1!i $j- 7J ;:g t£ Information Interchange(ASCII) shown the letter H) is receivedas 10010100, the m'L' (centroid)' ~ 'L' (orthocentre)~Din Figure 1. The letter H would be receiverknows that there is at least one rf'\JWIiI'L'(incentre) 0 encodedas 1001000,the letter E would error during transmission since the as etc. be encoded 1000101, ~igure 2). has receivedbit sequence an odd parity, =-~.#l!im -.w-ar PJ.~H~~i1!? i.e., the total number of l' s is an odd A 1000001 S 1010011 1100001 1110011 ~-gg-][::iE:JJ.£J'iJ~::iE:JJ.tf.m'rRJ~frj 1000010 1010100 1100010 1110100 number. m~~o **£-arjfrJffl#jf=-~~ 1000011 10101011100011 1110101 ~OO~~'f*'~m (::I:/L\re~..$t 1000100 1000101 1010110 1010111 1100100 1110110 11001011110111 m-~:!6'!1. JjX,pjij~~~2:1£J'iJ.~) 1000110 1011000 11qO110 l1l1000 ft"o 1000111 1011001 1100111 l1l1001 1001000 1011010 1101000 l1l1010 1001001 '0110000 1101001 0101110 ij,'--I-. mJa~ =-~#1!i1l:t'~'1fijJ;fIJ =-~~ 1001010 0110001 11101010 0101100 E,'...,." 3Ift~=-~~"'I;~~~.m'{tl. 1001011 1001100 0110010 Ic 1101011 0110011 1 1101100 0ll1111 0101001 Figure4. Even parity code f3J~rI1J~irJ~a~~o M 1001101 ~ 0110100 ' 1101101 l1l1011 N 1001110 0110101 1101110 010l1l1 Is there an encodingmethod so that the =-~~(l{j5j-~~::fif-@M.ffijm~ J 100l1l1, '" 1010000 0110110 110l1l1 I 011011111110000 I 0100110 receiver would be able to correct ~.~~: 1010001 0111000 c:r1110001 I 0101011 0101101 transmissionerrors? Figure 5 showsone A 1~10010 0111001 r 1110010 0l1l101 such method by arranging the bit Figure 1. ASCII code sequence (e.g., 1001) into a rectangular block and add parity bits to both rows and columns. For the example shown, 1001 would be encodedas 10011111(by B D c fIrst appendingthe row parities and then (/I)~) the columnparities). If there is an error during transmission, say at position 2, the receiver can similarly arrange the Figure2. Two computers talking received sequence 11011111 into a rectangularblock and detectthat there is The receiver will be able to decodethe an error in row 1 and column2. messagecorrectly if there is no error ~1I:t. during transmission. However, if there are transmissionerrors, the receivermay BD -=-0 AB decode the message incorrectly. For ~??) -=-9J-flj ;f,m I!i.:;l- .J:\~ilio # DC AC ~ '/1 ijJ m 11}]JJ:t ~ (f~.) example, the letter H (1001000)would be receivedas J (1001010)if there is an error at position6. 0 °1 1 I * I I 0 I I ,/ I I 'I)( ,Ix Figure 5. A code that can correct1 error. The abovemethod can be usedto correct one error but rather costly. For every Figure3. Error at position6. four bits, one would need to transmit an extra four redundant bits. Is there a One possibleway to detect transmission better way to do the encoding?In 1950, errors is to add redundant bits, i.e., Hammingfound an ingeniousmethodto ~ append extra bits to the original (continuedon page 4) Mathematical Excalibur, Vol. 2, No. Nov-Dec. 96 P~e3 Supposex, y are nonnegative integers Solution: CHAN Ming ChiD (La Salle Problem Corner suchthat (xy -7)2 = r + y2. Then (xy - College, Form 6), CHAN Wing Sum We welcome readers to submit solutions 6)2+ 13 = (x + y)2 by algebra. So (HKUST), CHENG Wing Kin (S.K.H. to the problems posed below for 13 = [(x+y) + (xy-6)][(x+y) -(xy-6)]. Lam Woo Memorial SecondarySchool, publication consideration. Solutions Form 5), CHEUNG Cheuk Lun should be preceded by the solver's name, Since 13 is prime, the factors on the (S.T.F.A. Leung Kau Kui College, Form address, school affiliation and grade right side can only be :tl or :t13. There 6), William CHEUNG Pok-man level. Please send submissions to Dr. are four possibilities yielding (x,y) = (S.T.F.A. Leung Kau Kui College, Form Kin-Yin li, Dept of Mathematics, Hong (0,7), (7,0), (3,4), (4,3). 6), Yves CHEUNG Yui Ho (S.T.F.A. Kong University of Science and Leung Kau Kui College, Form 5), CHUI Technology, Clear Water Bay, Kowloon. Other commended solvers: CHAN Ming Yuk Man (Queen Elizabeth School, The deadline for submitting solutions is Chin (La Salle College. Form 6), Form 7), FUNG Tak K wan (La Salle Ian 31, 1997. CHENG Wing Kin (S.K.H. Lam Woo College, Form 7), LEUNG Wing Lon Memorial SecondarySchool, Form 5), (STFA Leung Kau Kui College, Form Problem 46. For what integer a does William CHEUNG Pok-man (S.T.F.A. 6), LIU Wai Kwong (Poi Tak X2-x +a divide Xl3 + x + 90? Leung Kau Kui College, Form 6), Yves Canossian College), Henry NG Ka CHEUNG Yui Ho(S.T.F.A. Leung Kau Man (STFA Leung Kau Kui College, Problem 47. If x, y, z are real numbers Kui College, Form 5), CHING Wai Form 6), Gary NG Ka Wing (STFA such that X2+ y2 + Z2= 2, then show that Hung (S.T.F.A. Leung Kau Kui College, Leung Kau Kui College, Form 4), x + y + z ~ xyz + 2. Form 5), CHUI Yuk Man (Queen POON Wing Chi (La Salle College, Elizabeth School, Form 7), LIU Wai Form 7), TSANG Sai Wing (Valtorta Problem 48. Squares ABDE and BCFG Kwong (pui Tak Canossian College), College, Form 6), YU Chon Ling are drawn outside of triangle ABC. POON Wing Chi (La Salle College, (HKU), YUEN Chu Ming (Kiangsu- Prove that triangle ABC is isosceles if Form 7), TING Kwong Chi & David ChekiangCollege (Shatin), Form 6) and DG is parallel to AC. GIGGS (SKH Lam Woo Memorial YUNG Fai (CUHK). SecondarySchool, Form 5), YU Chon Problem 49. Let UI, U2, U3, ...be a Ling (HKU) and YUNG Fai (CUHK). There are C;O = 1140 3-elementsubsets sequence of integers such that UI = 29, of X. For a 3-element subsetwhose 3 = U2= 45 and Un+2 Un+? -Un for n = 1, 2, Problem 42. What are the possible numbershave product not divisible by 4, 3, Show that 1996 divides infinitely the numbersare either all odd (there are X2 valuesof oJ + x + 1 -.J;2 ~ as x or cjO = 120 such subsets) two odd and many terms of this sequence. (Source: overall realnumbers? ranges 1986 CanadianMathematical Olympiad one even, but the even one is not with modification) Solution: William CHEUNG Pot-man divisible by 4 (therc are C~Ox5=225 Form 6). (STFALeungKauKui College, such subsets). So the answer to the Problem 50. Four integers are marked problemis 1140 -120 -225 = 795. on a circle. On each step we Let A=(x,O),B=(-t,~), c=(t,~). simultaneouslyreplace each number by Problem 44. For an acutetriangle ABC, the difference betweenthis number and X2 The expression.J + x + 1 -j;2=-;1 let H be the foot of the perpendicular next number on the circle in a given is just AB -AC. As x ranges over all from A to BC. Let M, N be the feet of direction (that is, the numbersa, b, c, d real numbers, A moves along the real the perpendicularsfrom H to AB, AC, are replacedby a -b, b -c, c -d, d - axis and the triangle inequality yields respectively. Define LA to be the line a). Is it possibleafter 1996 suchstepsto through A perpendicular to MN and have numbers a, b, c, d such that the -1 =-BC<AB-,.AC< BC= 1. similarly define LB and Lc. Show that numbers Ibc -adl, lac -bdl, lab- All numbers on the intergal (-1,1) are LA, LB and Lc pass through a common cd I are primes? (Source: unused possible. point O. (This was an unusedproblem problem in the 1996 IMO.) proposedby Iceland in a past IMO.) Othercommended solvers: CHAN Ming Chiu (La Salle College, Form 6), Solution: William CHEUNG Pok-man ***************** CHENG Wing Kin (S.K.H. Lam Woo (STFALeungKauKui College, F0ffil6). Memorial SecondarySchool, FOrm 5), Solutions LIU Wai Kwong (Poi Tak Canossian Let LA interSect the circumcircle of ***************** College), POON Wing Chi (La Salle MBC at A and E. Since LAMH = 900 = College, FOrm 7), YU Chon Ling LANH, A, M, H, N are concyclii::. So Problem 41. Find all nonnegative (HKU) and YUNG Fai (CUHK). LMAH = LMNH = 90° -LANM = integers x, y satisfying (xy -7Y = x2-+ LNAE = LCBE. Now LABE = LCBE 2- y. Problem 43. How many 3-elemen't + LABC= LMAH + LABC = 90°. So AE is a dianleter of the circumcircle and Solution: Gary NG Ka Wing (STFA subse~of the setX= {I, 2, 3, ..., 20} are there such that the product of the 3 Leung Kau Kui College,Form 4). numbersin the subset diVisible by 4? is (continuedon page 4) Mathematical Excalibur. Vol. 2, No. 5, Nov-Dec.96 PMe4 Problem Corner point P such that LPAB = 10°, LPBA = (continuedfrom page 3) 20°, LPCA = 30°, LPAC = 40°. Prove LA passesthrough the circumcenter O. Similarly, LB and Lc will pass through O. that triangle ABC is isosceles. Problem 6. Determine (with proof) whether there is a subset X of the : ~ 6 0 , )( , Co integers with the following property: for Other commended solvers: Calvin code Figure6. Hamming any integer n there is exactly one CHEUNG Cheuk Lon (STFA Leung solution of a + 2b = n with a, b E X. Kau Kui College, FOrin 5), LIU Wai If there is one error during transmission, Kwong (Pui Tak Canossian College), say 1001001 received as 1011001, the POON Wing Chi (La Salle College, ~'""DC><::5~-- receiver can check the parities of the Foml 7) and YU Chon Ling (HKU). three circles to find that the error is in Error Correcting Codes(part I) circles B and C but not in A. This (7,4) Problem 45. Let a, b, c> 0 and abc=l (continued from page 2) Hammingcode(the notation (7,4) means Showthat that every4 information bits are encoded add the redundancy. To encode a four- as a 7 bit sequence) can be generalized. ab bc ca"" bit. sequence PIP1P3P4(say 1001), one For example, one may draw 4 + +~ l as+bs+ab bs+cs+bc cs+as+ca would first draw three intersecting intersecting spheres in a three- (This was an unused problem in circles A, B, C and put the information dimensional space to obtain a (15,11) bits Pl, Pl, PJ,P4 into the four overlapping Hamming code. Hamming has also IMO96.) regions AnB, AnC, BnC and AnBnC proved that his coding method is Solution: YUNG Fai (CUHK) , (Figure 6). Then three parity bits Ps, P6, optimum for single error correction. P1 are generated so that the total number of l's in each circle is an even number. (... to be continued) Expanding (a3-b3)(a2 -b1 ~ 0, we get For the example shown, 1001 would be a5 + b5 ~ a2b2(a+b). So using this and abc = 1, we get encoded as 1001001. R~h ~ IJ'1-e-!&\ ab ab C2 x- 2 J:.tt3) c.ti:;t: as + b5 + ab C ()f') a+b+c Adding 3 such inequalities, we get the desiredinequality. In fact, equality can t>A~'~""j-a '\~~ -~ occur if and only if a = b = c = I. t> ~i~ Other commended solvers: POON Wing .. .. Chi (La Salle College, Form 7) and YU Chon Ling (HKU). ~~, /jl}.~t' .-" ~t yotl.~ A Xyz .i ?? Igo. 1~~lv1'~ ~A8c.'; --~""c:>C><:5" ~ ~ /r ". Y. z ~:fJ f3c, Olympiad Corner C~.R8 "~1'D'~,. Y L). )C z. 1m j 1£4t £1/fe-c.-'I@~ (continuedfrom page 1) A )( Y z. -iff) h :JtJ"~~ ~ FtBc.1~ ~ ColA'" Co;'" c.~nt,.t. 0 Part n (lpm-4pm,May 2, 1996) ABc. ~'~1'f1 G ~ 1/':", ~ £). Problem 4. An n-term sequence X2, (Xl' ,go ..&iJ.-:1ji." -, r}f";" o. f-r.H ..., xn) in which each term is either 0 or {iiJ--1li,~~l,~A o~::~H = /=2. 1 is called a binary sequenceof lengthn. Let anbe the numberof binary sequences o!.! ! iJ.' 9~A'f ~ ..), ~~ .A X Yz.!j: I~~ of length n containing no three 50 Yo ,:tt ISOD, l<~ fJ L). XYz. p~~ consecutive terms equal to 0, 1, 0 in that Cir-c.umC£l:Itr~ N , ~ GN : 0& : GrH order. Let bn be the number of binary --= 1 ~ 2.. "" sequencesof length n that contain no four consecutive terms equal to 0, 0, 1, 1 or 1, 1, 0, 0 in that order. Prove that ~).. ~ "of/v' N . '1'.'Z. bn+l=2an for all positive integersn. "!1~A. ~'f!1 ,.V. ~ -:s: ~~ S. T (HA. 1-18. 11c. Problem 5. Triangle ABC has the I N t..t: '04.~ AASc. ~'ft. ".."e following property: there is an interior point ,~ ~