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Volume 2. Number 5 Nay-Dec. 1996

Olympiad Corner~ BiJ ~ Ii aq II fiJ (-)

25th United States of AmericaMathematical Olympiad: ~ S mPart I (9am-noon, May 2, 1996)Pr bl IPr th th fth :W:~~~~:fj::;or\"m.m~fiiJ ' ~:5J,~

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Problem 2. For any nonempty set S of fiiJ~WJ. 0

real numbers, let a(S) denote the sum ofthe elements of S. Given a set A of npositive integers, consider the collection -.of all distinct sums a(S) as S ranges overthe nonempty subsets of A. Prove tbatthis collection of sums can be partitionedinto n classes so that in each class, theratio of the largest sum to the smallestsum does not exceed 2. B

Problem 3. Let ABC be a ttiangle.Prove that there is a line 1 (in the planeof triangle ABq such that theintersection of the interior oftriangle ABC and the interior of itsreflection A' B' C' in I h~ area more than2/3 the area of triangle ABC.

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Area of MCPArea of ABCP

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Area of MCPArea of MAP (continued on page 2)

Mathematical Excalibur. Vol. 2. No. Nav-Dec. 96 Pa~e 2

~gffl~~~~fiiJ (-)(continued from page 1) Error Correcting Codes (Part I)

Tsz-Mei Ko.l:t~ (*):fa (**)"DJ~AF AG-=-

Suppose one would like to transmit amessage, say "HELLO.. .", from onecomputer to another. One possible wayis to use a table to encode the messageinto binary digits. Then the receiverwould be able to decode the messagewith a similar table. One such table isthe American Standard Code forInformation Interchange (ASCII) shownin Figure 1. The letter H would beencoded as 1001000, the letter E wouldbe encoded as 1000101, etc. ~igure 2).

A 1000001 S 1010011 1100001 11100111000010 1010100 1100010 11101001000011 10101011100011 11101011000100 1010110 1100100 11101101000101 1010111 110010111101111000110 1011000 11qO110 l1l10001000111 1011001 1100111 l1l10011001000 1011010 1101000 l1l10101001001 '0110000 1101001 01011101001010 0110001 11101010 01011001001011 0110010 Ic 1101011 0ll11111001100 0110011 1 1101100 0101001

M 1001101 0110100 '~1101101 l1l1011N 1001110 0110101 1101110 010l1l1J 100l1l1, 0110110 110l1l1 0100110'" 1010000 I 011011111110000 I

I 01010111010001 0111000 c:r 1110001 01011011~10010 0111001 r 1110010 0l1l101

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ij, '--I-.E,'...,."Figure 4. Even parity code

Is there an encoding method so that thereceiver would be able to correcttransmission errors? Figure 5 shows onesuch method by arranging the bitsequence (e.g., 1001) into a rectangularblock and add parity bits to both rowsand columns. For the example shown,1001 would be encoded as 10011111 (byfIrst appending the row parities and thenthe column parities). If there is an errorduring transmission, say at position 2,the receiver can similarly arrange thereceived sequence 11011111 into arectangular block and detect that there isan error in row 1 and column 2.

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A

Figure 1. ASCII code

B cD

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Figure 2. Two computers talking

The receiver will be able to decode themessage correctly if there is no errorduring transmission. However, if thereare transmission errors, the receiver maydecode the message incorrectly. Forexample, the letter H (1001000) wouldbe received as J (1001010) if there is anerror at position 6.

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I I(f~.)Figure 5. A code that can correct 1 error.

Figure 3. Error at position 6.

One possible way to detect transmissionerrors is to add redundant bits, i.e.,append extra bits to the original

The above method can be used to correctone error but rather costly. For everyfour bits, one would need to transmit anextra four redundant bits. Is there abetter way to do the encoding? In 1950,Hamming found an ingenious method to

(continued on page 4)

message. For an even parity code, a 0 or1 is appended so that the total number of1's is an even number. The letters HandE would be represented by 10010000and 10001011 respectively. With aneven parity code, the receiver can detectone transmission error, but unable tocorrect it. For example, if 10010000 (forthe letter H) is received as 10010100, thereceiver knows that there is at least oneerror during transmission since thereceived bit sequence has an odd parity,i.e., the total number of l' s is an oddnumber.

Mathematical Excalibur, Vol. 2, No. Nov-Dec. 96 P~e3

Problem CornerWe welcome readers to submit solutionsto the problems posed below forpublication consideration. Solutionsshould be preceded by the solver's name,address, school affiliation and gradelevel. Please send submissions to Dr.Kin-Yin li, Dept of Mathematics, HongKong University of Science andTechnology, Clear Water Bay, Kowloon.The deadline for submitting solutions isIan 31, 1997.

Problem 46. For what integer a doesX2 -x +a divide Xl3 + x + 90?

Problem 47. If x, y, z are real numberssuch that X2 + y2 + Z2 = 2, then show thatx + y + z ~ xyz + 2.

Problem 48. Squares ABDE and BCFGare drawn outside of triangle ABC.Prove that triangle ABC is isosceles ifDG is parallel to AC.

Problem 49. Let UI, U2, U3, ...be a

sequence of integers such that UI = 29,U2 = 45 and Un+2 = Un+? -Un for n = 1, 2,

3, Show that 1996 divides infinitely

many terms of this sequence. (Source:

1986 Canadian Mathematical Olympiadwith modification)

Problem 42. What are the possible

values of oJ X2 + x + 1 -.J;2 ~ as xranges over all real numbers?

Solution: William CHEUNG Pot-man(STFA Leung KauKui College, Form 6).

Problem 50. Four integers are markedon a circle. On each step wesimultaneously replace each number bythe difference between this number andnext number on the circle in a givendirection (that is, the numbers a, b, c, dare replaced by a -b, b -c, c -d, d -a). Is it possible after 1996 such steps tohave numbers a, b, c, d such that thenumbers Ibc -adl, lac -bdl, lab-cd I are primes? (Source: unusedproblem in the 1996 IMO.)

Problem 44. For an acute triangle ABC,let H be the foot of the perpendicularfrom A to BC. Let M, N be the feet ofthe perpendiculars from H to AB, AC,respectively. Define LA to be the linethrough A perpendicular to MN andsimilarly define LB and Lc. Show thatLA, LB and Lc pass through a commonpoint O. (This was an unused problemproposed by Iceland in a past IMO.)

Let A=(x,O), B=(-t,~), c=(t,~).

The expression.J X2 + x + 1 -j;2=-;1

is just AB -AC. As x ranges over allreal numbers, A moves along the realaxis and the triangle inequality yields

-1 =-BC<AB-,.AC< BC= 1.

All numbers on the intergal (-1,1) arepossible.

Other commended solvers: CHAN MingChiu (La Salle College, Form 6),CHENG Wing Kin (S.K.H. Lam WooMemorial Secondary School, FOrm 5),LIU Wai Kwong (Poi Tak CanossianCollege), POON Wing Chi (La SalleCollege, FOrm 7), YU Chon Ling(HKU) and YUNG Fai (CUHK).

Solution: William CHEUNG Pok-man(STFA Leung KauKui College, F0ffil6).*****************

Solutions***************** Let LA interSect the circumcircle of

MBC at A and E. Since LAMH = 900 =LANH, A, M, H, N are concyclii::. SoLMAH = LMNH = 90° -LANM =LNAE = LCBE. Now LABE = LCBE+ LABC= LMAH + LABC = 90°. SoAE is a dianleter of the circumcircle and

Problem 41. Find all nonnegativeintegers x, y satisfying (xy -7Y = x2- +

2-y. Problem 43. How many 3-elemen'tsubse~ of the setX= {I, 2, 3, ..., 20} arethere such that the product of the 3numbers in the subset is diVisible by 4?

Solution: Gary NG Ka Wing (STFALeung Kau Kui College, Form 4). (continued on page 4)

Suppose x, y are nonnegative integerssuch that (xy -7)2 = r + y2. Then (xy -6)2 + 13 = (x + y)2 by algebra. So

13 = [(x+y) + (xy-6)][(x+y) -(xy-6)].

Since 13 is prime, the factors on theright side can only be :tl or :t13. Thereare four possibilities yielding (x,y) =(0,7), (7,0), (3,4), (4,3).

Other commended solvers: CHAN MingChin (La Salle College. Form 6),CHENG Wing Kin (S.K.H. Lam WooMemorial Secondary School, Form 5),William CHEUNG Pok-man (S.T.F.A.Leung Kau Kui College, Form 6), YvesCHEUNG Yui Ho(S.T.F.A. Leung KauKui College, Form 5), CHING WaiHung (S.T.F.A. Leung Kau Kui College,Form 5), CHUI Yuk Man (QueenElizabeth School, Form 7), LIU WaiKwong (pui Tak Canossian College),POON Wing Chi (La Salle College,Form 7), TING Kwong Chi & DavidGIGGS (SKH Lam Woo MemorialSecondary School, Form 5), YU ChonLing (HKU) and YUNG Fai (CUHK).

Solution: CHAN Ming ChiD (La SalleCollege, Form 6), CHAN Wing Sum(HKUST), CHENG Wing Kin (S.K.H.Lam Woo Memorial Secondary School,Form 5), CHEUNG Cheuk Lun(S.T.F.A. Leung Kau Kui College, Form6), William CHEUNG Pok-man(S.T.F.A. Leung Kau Kui College, Form6), Yves CHEUNG Yui Ho (S.T.F.A.Leung Kau Kui College, Form 5), CHUIYuk Man (Queen Elizabeth School,Form 7), FUNG Tak K wan (La SalleCollege, Form 7), LEUNG Wing Lon(STFA Leung Kau Kui College, Form6), LIU Wai Kwong (Poi TakCanossian College), Henry NG KaMan (STFA Leung Kau Kui College,Form 6), Gary NG Ka Wing (STFALeung Kau Kui College, Form 4),POON Wing Chi (La Salle College,Form 7), TSANG Sai Wing (ValtortaCollege, Form 6), YU Chon Ling(HKU), YUEN Chu Ming (Kiangsu-Chekiang College (Shatin), Form 6) andYUNG Fai (CUHK).

There are C;O = 1140 3-element subsets

of X. For a 3-element subset whose 3numbers have product not divisible by 4,the numbers are either all odd (there arecjO = 120 such subsets) or two odd and

one even, but the even one is notdivisible by 4 (therc are C~O x5=225

such subsets). So the answer to theproblem is 1140 -120 -225 = 795.

Mathematical Excalibur. Vol. 2, No. 5, Nov-Dec. 96 PMe4

point P such that LPAB = 10°, LPBA =20°, LPCA = 30°, LPAC = 40°. Provethat triangle ABC is isosceles.

Problem Corner(continued from page 3)

LA passes through the circumcenter O.Similarly, LB and Lc will passthrough O.

: ~0 , , Co6

)(

Problem 6. Determine (with proof)whether there is a subset X of theintegers with the following property: forany integer n there is exactly onesolution of a + 2b = n with a, b E X.

~'""DC><::5~--

Error Correcting Codes (part I)(continued from page 2)

add the redundancy. To encode a four-bit. sequence PIP1P3P4 (say 1001), onewould first draw three intersectingcircles A, B, C and put the informationbits Pl, Pl, PJ, P4 into the four overlappingregions AnB, AnC, BnC and AnBnC(Figure 6). Then three parity bits Ps, P6,P1 are generated so that the total numberof l's in each circle is an even number.For the example shown, 1001 would beencoded as 1001001.

Other commended solvers: CalvinCHEUNG Cheuk Lon (STFA LeungKau Kui College, FOrin 5), LIU WaiKwong (Pui Tak Canossian College),POON Wing Chi (La Salle College,Foml 7) and YU Chon Ling (HKU).

Figure 6. Hamming code

Problem 45. Let a, b, c> 0 and abc=lShow that

ab bc ca""l+ +~

as+bs+ab bs+cs+bc cs+as+ca

(This was an unused problem inIMO96.)

Solution: YUNG Fai (CUHK) ,

Expanding (a3 -b3)(a2 -b1 ~ 0, we geta5 + b5 ~ a2b2(a+b). So using this andabc = 1, we get

ab ab C2x-2

C

c.ti:;t: J:.tt3)as + b5 + ab

()f')

~

,~~

t> A~'~"" j-a

'\~~t> ~i~

a+b+cAdding 3 such inequalities, we get thedesired inequality. In fact, equality canoccur if and only if a = b = c = I.

-....

Other commended solvers: POON WingChi (La Salle College, Form 7) and YUChon Ling (HKU).

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Olympiad Corner(continued from page 1)

~Part n (lpm-4pm, May 2, 1996)

Problem 4. An n-term sequence (Xl' X2,..., xn) in which each term is either 0 or1 is called a binary sequence of length n.Let an be the number of binary sequencesof length n containing no threeconsecutive terms equal to 0, 1, 0 in thatorder. Let bn be the number of binarysequences of length n that contain nofour consecutive terms equal to 0, 0, 1, 1or 1, 1, 0, 0 in that order. Prove thatbn+l =2an for all positive integers n.

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Problem 5. Triangle ABC has thefollowing property: there is an interior

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point

If there is one error during transmission,say 1001001 received as 1011001, thereceiver can check the parities of thethree circles to find that the error is incircles B and C but not in A. This (7,4)Hamming code (the notation (7,4) meansthat every 4 information bits are encodedas a 7 bit sequence) can be generalized.For example, one may draw 4intersecting spheres in a three-dimensional space to obtain a (15,11)Hamming code. Hamming has alsoproved that his coding method isoptimum for single error correction.

(... to be continued)

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