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Volume 2. Number5
Nay-Dec. 996
Olympiad Corner
~ BiJ~ Ii aq II fiJ (-)
25th United States of America
Mathematicallympiad: ~ S m
Part I (9am-noon,May 2, 1996)
Pr bl IPr th th fth :W:~~~~:fj::;or\"m.m~fiiJ ' ~:5J,~
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num~s n sm n (n = 2, 4, 6, ...,180) IS ~P).ljf- 0 ~:(3 .1if~~Mir{j\l:i-
cot 1 .~m.~~~fiiJJEJJ.' i1JQ~~~~
Problem 2. For any nonempty set S of fiiJ~WJ. 0
real numbers, et a(S) denote he sum of
the elementsof S. Given a set A of n
positive integers, consider he collection -.
of all distinct sumsa(S) as S rangesover
the nonempty subsetsof A. Prove tbat
this collection of sumscan bepartitioned
into n classesso that in each class, the
ratio of the largest sum to the smallest
sumdoesnot exceed . B
Problem 3. Let ABC be a ttiangle.
Prove that there s a line 1 (in the plane
of triangle ABq such that the
intersection of the interior of
triangle ABC and the interior of its
reflectionA' B' C' in I h~ areamore than
2/3 the area of triangle ABC.
&
AF
FB
Area of MCP
Area of ABCP
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ftm
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Area of MBD BD
=-0
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=1
FB DC EA
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B D C
(1].::.)
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.:=. ~ (J{j~ ':tl ijJ ~
AreaofMBD BD AreaofMBD
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Area of MDC DC Area of MDC
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a c a c
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Area of MCP
Area of MAP
(continuedon page 2)
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Mathematical Excalibur. Vol.2. No.
Nav-Dec. 96
Pa~e 2
~gffl~~~~fiiJ (-)
(continued from page 1)
Error Correcting Codes (Part I)
Tsz-MeiKo
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Supposeone would like to transmit a
message, say "HELLO.. .", from one
computer o another. One possibleway
is to use a table to encode he message
into binary digits. Then the receiver
would be able to decode he message
with a similar table. One such table is
the American Standard Code for
Information Interchange ASCII) shown
in Figure 1. The letter H would be
encodedas 1001000, he letter E would
be encoded s 1000101, tc. ~igure 2).
A 1000001 S 1010011 1100001 1110011
1000010 1010100 1100010 1110100
1000011 10101011100011 1110101
1000100 1010110 1100100 1110110
1000101 1010111 11001011110111
1000110 1011000 11qO110 l1l1000
1000111 1011001 1100111 l1l1001
1001000 1011010 1101000 l1l1010
1001001 '0110000 1101001 0101110
1001010 0110001 11101010 0101100
1001011 0110010 Ic 1101011 0ll1111
1001100 0110011 1 1101100 0101001
M 1001101 0110100 '
~1101101 l1l1011
N 1001110 0110101 1101110 010l1l1
J 100l1l1, 0110110 110l1l1 0100110
'" 1010000
I 011011111110000 I
I 0101011
1010001 0111000 c:r1110001 0101101
1~10010 0111001 r 1110010 0l1l101
FB GB
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:gg-][:iE~fII\]~:iE~~{PJ.~~ .f.:$
.L~.t87E-~Wfll\]E-i'lm~'~...:
E-I=P *1 (medians) # I i' E-~
(altitudes) I i' E- $j- ~ *1 (angle-
bisectors) I i 0 ~ E- i'l1 i $j- 7Jl ;:g t
m'L' (centroid)' ~ 'L' (orthocentre)~D
rf'\JWIiI'L'(incentre) 0
-~.#l im -.w-ar PJ.~H~~i1 ?
~-gg-][::iE:JJ.J'iJ~::iE:JJ.tf.m'rRJ~frj
~~o **-arjfrJffl#jf=-~~
~OO~~'f*' ~m (::I:/L\re~..$t
JjX,pjij~~~2:1J'iJ.~)m-~: 6' 1.
Ja~=-~~
ij, --I-.
E,'...,."
Figure4. Even parity code
Is there an encodingmethod so that the
receiver would be able to correct
transmissionerrors? Figure 5 showsone
such method by arranging the bit
sequence e.g., 1001) nto a rectangular
block and add parity bits to both rows
and columns. For the example shown,
1001 would be encodedas 10011111 by
fIrst appending he row parities and then
the column parities). If there s an error
during transmission, say at position 2,
the receiver can similarly arrange the
received sequence 11011111 into a
rectangularblock and detect hat there s
an error in row 1 and column2.
A
Figure 1. ASCII code
B
c
(/I)~)
Figure2. Two computers alking
The receiver will be able to decode he
message correctly if there is no error
during transmission. However, f there
are transmissionerrors, the receivermay
decode the message incorrectly. For
example, the letter H (1001000) would
be receivedas J (1001010) f there s an
error at position6.
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D
AB
-=-0
DC AC
9J- lj ;f,m I i.:;l- ~ '/1 ijJ m 11}] J:t
'I)(I I ,/
I I
,Ix
*1
I 1
I I
(f~.)
Figure 5. A code that can correct1 error.
Figure 3. Error at position6.
One possible way to detect ransmission
errors is to add redundant bits, i.e.,
append extra bits to the original
The abovemethod can be used o correct
one error but rather costly. For every
four bits, one would need to transmit an
extra four redundant bits. Is there a
better way to do the encoding? n 1950,
Hamming ound an ngeniousmethod o
(continuedon page 4)
message.For an evenparity code,a 0 or
1 is appended o that the total number of
1's is an evennumber. The letters Hand
E would be representedby 10010000
and 10001011 respectively. With an
even parity code, he receiver can detect
one transmission error, but unable to
correct t. For example, f 10010000 for
the letter H) is receivedas 10010100, he
receiverknows that there is at least one
error during transmission since the
receivedbit sequence as an odd parity,
i.e., the total number of l' s is an odd
number.
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Mathematical Excalibur, Vol. 2, No.
Nov-Dec. 96
P~e3
Problem Corner
We welcome readers to submit solutions
to the problems posed below for
publication consideration. Solutions
should be preceded by the solver's name,
address, school affiliation and grade
level. Please send submissions to Dr.
Kin-Yin li, Dept of Mathematics, Hong
Kong University of Science and
Technology, Clear Water Bay, Kowloon.
The deadline for submitting solutions is
Ian 31, 1997.
Problem 46. For what integer a does
X2 x +a divide Xl3 + x + 90?
Problem 47. If x, y, z are real numbers
such that X2+ y2 + Z2= 2, then show that
x + y + z ~ xyz + 2.
Problem 48. Squares ABDE and BCFG
are drawn outside of triangle ABC.
Prove that triangle ABC is isosceles if
DG 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 Olympiad
with modification)
Problem 42. What are the possible
valuesof oJ 2+ x + 1 -.J;2 ~ as x
ranges ver all realnumbers?
Solution: William CHEUNG Pot-man
(STFALeung KauKui College,Form 6).
Problem 50. Four integers are marked
on a circle. On each step we
simultaneously eplace each number by
the difference between his number and
next number on the circle in a given
direction (that is, the numbersa, b, c, d
are replaced by a -b, b -c, c -d, d -
a). Is it possibleafter 1996 such steps o
have numbers a, b, c, d such that the
numbers Ibc -adl, lac -bdl, lab-
cd I are primes? (Source: unused
problem n the 1996 MO.)
Problem 44. For an acute riangle ABC,
let H be the foot of the perpendicular
from A to BC. Let M, N be the feet of
the perpendiculars rom H to AB, AC,
respectively. Define LA to be the line
through A perpendicular to MN and
similarly define LB and Lc. Show that
LA, LB and Lc pass through a common
point O. (This was an unusedproblem
proposedby Iceland n a past MO.)
Let A=(x,O),B=(-t,~), c=(t,~).
The expression.J 2 + x + 1 -j;2=-;1
is just AB -AC. As x ranges over all
real numbers, A moves along the real
axis and he triangle inequality yields
-1 =-BC
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Mathematical Excalibur. Vol. 2, No. 5, Nov-Dec.96
PMe4
point P such hat LPAB = 10, LPBA =
20, LPCA = 30, LPAC = 40. Prove
that triangle ABC is isosceles.
Problem Corner
(continued rom page 3)
LA passes hrough the circumcenter O.
Similarly, LB and Lc will pass
through O.
:
~
, , Co
6
)(
Problem 6. Determine (with proof)
whether there is a subset X of the
integers with the following property: for
any integer n there is exactly one
solution of a + 2b = n with a, b E X.
~'""DC>A~'~"" -a
'\~~
t> ~i~
a+b+c
Adding 3 such inequalities, we get the
desired nequality. In fact, equality can
occur f and only if a = b = c = I.
-
..
.
Other commended olvers: POON Wing
Chi (La Salle College, Form 7) and YU
Chon Ling (HKU).
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