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ARBELOS
PRODUCED FOR
PRECOLLEGE
PHILOMATHS'
'.
.•...
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ARBELOS
PRODUCED FOR
PRECOLLEGE
PHILOMATHS'
'.
.••••
-
.1982 - 1983
Copyright © 1982
Committee on the American Mathema~ics Competitions
Mathematical Association of America.
-, -'.'-' .,
. t
_I
Professor Samuel L. Greitaer in Memoria During the interval of time betwe~n 1982-1987 Prof~ssor Samuel L. Gre itzerserved as the editor and author ofessentially allofthe articles which appear in the Arbelos. His untimely death, on February 22, 1988, was indeed a sad day for all those who knew him. Professor Greiber emigrated to the United States from Odessa, Russia in 1906. He graduated from the City College of New York in 1927 and earned his PhD degree at Yeshiva University. He had more than 25 years experience as a junior and senior high school teacher. He taught at Yeshiva University, the Polytechnic Institute of Brooklyn, Teachers College and the School of General Studies of Columbia University. His last academic teaching position was at Rutgers University. He was the author or co-author of several books including Geometry Ret1i,ited with H.S.M. Coxeter. I was extremely pleased that Professor Greiber agreed to write and edit the Arbelos, since I frequently receive requests for references to publica tons which are appropriate for superior students, and for material which will help students prepare for the USA Mathematical Olympiad. Pro fessor Greiber served as a coach of the summer Mathematical Olympiad training program Crom 1974 to 1983. Consequently, many oUhe articles in the Arbelos are a reflection oC his ledures and thus appropriate for talented and gifted students. ProCessors Greiber and Murray S. Klamen accompanied the USA team to the International Mathematical Olympiad Crom 1974 [the first yeat the USA participated] to 1983. Their success in coaching the team is indicated by the fact that it usually placed among the top three [out of 30-35 participating countries]. The contributions of Professor Greiber to the development oC students oC mathematics and teachers Crom many nations will be lasting. We shall miss his humor, words of wisdom, mathematical insight and Criendship. Dr. Walter E. Mientka Executive Director American Mathematics Competitions University oC Nebraska-Lincoln
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I.I.I. i.I. la f termathl ,. ••• •• .•• • • I.I.e ••••
An equation involving x, u , and differences x is known as a Difference Equation. We may write it as F'(x, u , AU , x x
A simple example
••>. • •••• ••• >. •• >. >.• •• • ••• ••• ••
rewrite a difference equation in the form
;. .e
I
•
I.1. ,.,. I.
By virtue of the relation
1 +
A =
E, we may
F(x, ux ' Eux ' ••• ,Enux ) = O•
Finally, since Ekux = uX+k ' even this last equation may be rewritten in the form
=
F(x, ux ' ux+1 ' ux+2 ' • • • , uX+n) 0 which we shall call a recursion equation. The simplest recursion equations are those with constant coefficients. An example will show how such equations may be solved. Let 2
uX+2 - 7ux+1 + 12ux = 0, or (E -?E + 12)ux Let ux = aX and. substitute. Then a x+2 7ax+1 + 12ax = a X(a2 - 7a + 12) = O. Since aX
r 0,
a 2 _ 7a + 12
= O.
it follows that the roots of
=0
and only those, satisfy the
equation. These roots are a = 3, a = 4. Hence the solution is u =A(3)x + B(4)x. A and B x are constants.
Except for occasional difficulties that may arise due to oddities in the form. any recursion equation f(u. Eu •••••Enu ) = 0 which is a x x x polynomial form can be solved. by fiMing the roots of the auxiliary equation first. Thus. for
uX + - 6ux +2 + l1ux +l - 6ux
3 (E 3 _ 6E2 + llE - 6)u = 0
=0
we have
x
(E - l)(E - 2)(E - 3)u
x
aM
u
x
Given
=A(l)x
=0
+ B(2)x + C(3)x •
ux +3 - 3ux+l - 2ux = 0 (E 3_ 3E - 2)u 0
x
we fiM
=
(E + 1)2(E - 2)u
x
=0
aM we have one oddity - a repeated. root. In
•• •• •• •• •• •• ••• •• •• -. ••
this case. we fiM the solution to be u
x
= (A
+ Bx)(_l)x + C(2)x •
A famous recursion is that involving the Fibonacci sequence. where ux +2 = uX +1 + U x • (E 2 - E - 1)u = O. The roots Wri te this as x
•• •• ••
of the related. (or complementary) equation are 1
u
;.,f5 x
= A(1
aM
1
2 -15
• Therefore
- ,15)~
+ ='5)X+ B(l 2 2
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••
!.ie
,.i.•• ,.i. •• . I
,e
7 Now suppose we are ~ven f(E)u f(E)
,. ,. i. ,.,.I.•• i.•• •• .••
14
A A
•
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:.
,.i. >.• •• ,.•• i.'•. ,e
o
v (A)
(B)
(c)
The diagrams show the appearance of the Brianchon configuration for special cases. In (A), V is taken as a vertex in addition to A, B, C, D and E. In Figure (B), Vi and V2 are taken
as vertices, and in Figure (C), we use Vi' V2' VJ. Incidentally, in case (C) we have labeled point G because it is called tbe Gergonne Point. Now you might try to solve Problem 2 of the XXXIII International Mathematical Olympiad, held in Budapest in July, 1982. A non-isosceles triangle A1AzAJ is given with sides ai' a 2 , a (a i is the side opposite J Ai). For all i = 1, 2, J, Mi is the midpoint of side ai' Ti is the point where the incircle touches side ai' ans the reflection of Ti in the interior bisector of Ai yields the point Si. Prove that the lines M1S , M2S2 , and MJS 1 J are concurrent.
15 WEEK-ENDERS 1) What is the greatest integer that will divide into
3999, 5585 and 6378 and leave
the same remainder ?
2) Find the roots of
x
that the roots are If
e~~
2
a
+ ax + b
and
= 0,
given
b.
are removed frail a basket two,
three, four, five, six
and seven at a time,
there remain respectively one, two, three, four, five and six
eg~.
What is the least
number of eggs in the basket ?
4) Not only is Ann four tlmes as old as Mary was when Ann was as old as Mary is now, but Ann is twice as old as Mary was when
Ann was six years older than Mary is now.
How old is Ann?
5) In right triangle ABC with right angle C, lines CP and CQ divide the hypotenuse into three eq ual CP
= 7,
se~ents.
and CQ = 9, how
Given that
lon~
is hypotenuse
AB ?
Notes Lewis Carroll used to call problems like these Pillow Problems - to be done to help him go to sleep. For example, try to do No. 2 and 3 mentally !
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16 OLYMPIAD PROBIEMS
4) Determine the smallest integer p > n
that for every integer dissect a square into
p
(a) n the recursion
such
one can
squares (not
necessarily congruent).
5) A sequence
n
(France)
is defined by means of
16a n+ = 1 + 4a n + -,/1 + 24an • 1 Find an explicit formula for
a =1. 1
a. n
(W. Germany)
6)
Let
M be the set of all functions
f
with the follOWing properties. (a) f
is defined for all integers and
takes on only real values. (b) For all integers
x,y
= f(x+y) + f(x-y). f(O) I 0 , f(l) = 5/2.
f(x).f(y)
(c) Find
fen).
(E. Germany)
1 ? APPLICATIONS? When Moses inflicted the plague of darkness on the Egyptians, the Egyptian god Set decided to use the three days to protect himself frOll his enemy, Osiris. When the sun set Set set out to build a wall. He ma.de it 3 layers thick and 12 layers high, while working throughout the dark ness. When light a~in appeared, the cemont Set set set, and the wall was finished. If Set Get 20 bricks per minute, and each brick was 1 1/4 Kabs long, how long was the wall ? (Notes the kab was an ancient Ep;J~tl;:.•: You can't get a kab these days.)
ll,'aS"i'e.
While besieging a tower in M~occo 180 feet high, a Knight at the foot of the tower hurled a stone at a Dey at the top 01' t.he tower at the same instant tha.t the Dey dropped t', rock on ~h(,( Knight. The stone and the rock paf''31'd: each other at a point halfway up the tower. Who p' q struck first, and how much sooner? (Notel Use g
= )2
2
ft/sec }.
Evariste Galois had. a sUDl1ler job 0'"' ~ farJll. Asked to weigh four sacks of potatoos, he wet~ed. thell two at a time (don It ask why !) and found the wei~ings to be 92 lb., 93 lb. I 95 lb., 9? lb., 99 lb. I and 100 lb. What was the weight of each individual sack ? (Notes Galois was sacked.)
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CON'mNTS i.
Preface
1.
Applications?
2.
The Arbel08
3.
Functional Equations
8.
Many Cheerful Facts
9.
Pascal and Brlanchon
15.
Week-enders
16.
Olympiad Problems
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I.I. . I.
arbelos
,
No.1
JanuRrv, 1981
Copyright ~ The Mathematical Association of America, 1983
i
PREFACE We are (aoderate1y) pleased at the increase in the nuaber Of subscribers. Although this nuaber has aore than doubled, it is still SJla11 We urge sttdents who are interested to support this effort. Our confidence in the value of the pamphlet will be increased exponentially. We are also happy to add a "departaent" to the paaph1et - namely, the KURSCHAK CORNER. This will be presided over by Professor Geor~e Berzs.ny1. We ur~e all subscribers to join in. The cc.ments and solutions alone make it worth while. We thank the subscribers who were kind enough to send in their cOlUlendatory reaarks. These make the whole effort Ilore worth-while. Solutions are also beginning to come to me. I promise to read these, list the successes, and. present the neatest solutions, beginning with the March issue.
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•• •• •• ••• 3 . (a) Show that (2) 2 (b) Show that (1:&)(E:) ~ n • (c)
Prove
(d) If
(x 2 + y2 + z2)1/2 _> )x + 4y + 12z
x +y + z
13
= 1.
show that
(1 + 1)(1 + 1)(1 + 1) > 64 • x y z
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5 PI'OLEMY'S THEOREM
Possibly because Ptolemy's Theorem first appeared about 100 AD and. because Euclid was the source of ~OIletry for all until FeI'1llat and Descartes, it does not usually form part of the usual high school curriculum. Howe~er, it is often useful and always fun. We first prove the theorem geometrically. Given a quadrilateral inscribed in a circle, as shown in the figure at the right. Then ac + bi = ACxBD. Construct angle ADP equal to angle BDC. Then bAPD and. 6 BCD are similar, so that bd = (AP)xn. Also, 6 DAB and 6 DPC are similar, so that ac = (PD)xn. Adding, ac + 'td = mn. The converse is also true. Incidentally, if the inscribed quadrilateral happens to be a rectan~le with two sides equal to a arrl the other two sides equal to b, am the dia~onals each equal to c, Ptolemy's theorem yields a short arrl snappy proof of the Pythagorean Theorem. ~ Next, suppose we have the diagram at the right, where A B , 4 8 is a diameter. Then using ac + bd = mn, m = 2R, n = 2Rsin(a+a) p and we have 2 2Rcosa.2Rsina + 2Rsina.2Rcosa = 4R sin(o+a), and we have derived the formula sin(a+a) = slna.cosa + c08a.sina.
6
Can you use Ptolemy's Theorem to derive sin(a - a) ? How about cos(a ~ ~) ? Specializing the fi~e often produces special results. For example. in the figure at the right. ABC is an equilateral trian~le. If P is any point on the arc Be. prove that PA
= PB
to
A harrler problem is proVf~. frOll the d ia~am,
+ PC.
t/(PD)
= t/(PH)
+
t/(pc).
This last formula is frequently used in Nomography. far example. in problems involving resistances. optics, etc. Again, suppose that, in the figure below. ABCD 1s a parallelogram. Go Prove that 2 (AB)(AP) ~ (AD)(AQ) = AC • Findinr, the area of an inscribed quadrilateral 1s complicated" but easy. If we lise the firs t d ia~am • we note that angles A and C are supplementary, so that Rin C = sin A but cos C - - cos A. Let the Lr84 soup;ht be K. Then K:o!ad: s1n A + ire s1n C so that
sin A:
(ad
2K +
be) •
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Next,
?
m m2 2
= a 2 + = b2 +
m .- b
or
2
+
d2 c2 c2
2ad cos A
2bc cos r. +
2bc cos A.
Subtract the third line from the first, am we derive cos A = (a 2 _b?'_c 2 +d 2 )/(2ad + 2bc). We now elimlnate the an~le A, because 2 sin A + cos 2A = 1, am, after $OGle complicated but easy algebra, derive the results 2 K = (s - a)(s - b)(s - c)(s - d) where
s
= (a+b+c+d)/2
•
Notice that, if we let d equal zero, we have Hero's Formula for the area of a trlan~le. For some reason, the Arabian mathematicians spent some time on a special quadrilateral one whose dIagonals were perpendicular to each other. Let us call this D a cyc Hc orthaUagonal .--- quadrilateral-see the d ia~am. For such a fi~re, show that a 2 + b2 + c 2 + d 2 = OR?. BrahmaKupta showed that the altitude on side c of the riKht triangle 0 wi th hypotenuse CD would, i f extended, bisect side a of the quadrilateral. This can be the start of a number of interesting properties of this figure. For example, join the mid points of sides a, b. c, d. What figure Is formed, and what is its area ? Wha t happens when perpendiculars are drawn from the inter section of the diagonals to all four sides?
8 ANSWERS AND QUESTIONS 1. When the roots, real am complex, of z3 - 1
=0
are plotted, they determine
an equilateral roots of
tria~le.
z4 - 1
=0
Similarly, the
determine a square.
(a) What fl~e do the roots of the 2 equation z4 + 41z 3 - 6z - 4iz -
1 =0
determine? (b) For what relation among the coefficients 2 of a z 3 + a z + a z + a = 0 do the 2 o 1 3 roots determine an equilateral trlangl~?
2. When f(x)
is divided by (x-a) , the
remaimer is
f(a) - Remaimer Theorem.
(c) What is the remaimer when divided by 2
3. If ax +
f(x)
is
(x-a)(x-b)?
bx + C
=0
has roots
r , r , 2 1 then the discriminant is (r - r )2. Far, 1 2 if r , r are real, then (r - . r )2 is 2 2 1 1 positive; if r r , the discriminant is 2 1 zero; if the roots are complex, then they
=
have the form is
p+qi, p-qi, their difference
2qi, the square of which is
ne~t1ve.
- r )2 = (r + r )2 - 4r r 1 1 2 2 2 -_ b2/a 2 - 4e / a -_ ( b2 - 4ac )/a2. Since a 2 is always positive, we have b2 - 4ac Now
(r
1
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9
as the quadratic discriminant, which we already know • (d) What is the discr1.Jlinant of the cubic xJ + JHx + G = 0 in terms of the coefficients ?
4. De Moivre's 'nlearelll (of wai«!h mare later) states that (cos e + i sin e)n
=
cos n9 + i s1n nee In add1tion, 1 + i
= ~(cos
H 1 s1n H). +
(e) Show that (1+1)n
= 2n/2(cos ~
+ i s1n
~).
(f) What do we get 1f we expand. and. then eq ua te real parts 1n the a hove ?
(g) What do we get if we expand and then eq uate 1Jlag1nary parts ? We wish to take this oppartun1ty of congratulat ing Douglas Davidson far being the first one to sem in solutions. He did No. 1 and No.2 of the assarted problems am No. z of the Answers am questions. I hope to hear from others, so we are not
~iving
his solutions yet. S. Gre1tzer
10 MANY CHEERFUL FACTS
Given the polynomial equation
n n-l
f ( x ) = aOx + a l x + • • • + an = 0, the relations between roots and coefficients are well-known - namely I r l + r 2 + • ~ • + r n = -a l /aO r l r 2 + r l r J + • • • + r n_ l r n
= +&2/a O
... What is not as we11 known, however, are the relations involving powers of the roots, and known as Newton's Identities. These are often useful in problem solving. We define
~
=r l
= -a l /a O '
Then sl
k
k
+ r2
or
aOs l + a l
Next, (r l +r2+ ••• +rn )2 2 _
k + •• • + r n •
,
= s2
=o
•
+ 2a2/aO' or 2,
1 - s2 + 2a2,aO• Since 51 = -a1s 1,aO we rewrite this aOs 2 + a 18 1 + 2a2 = 0 • 8
Now we restrict ourselves to a cubic. Then
a r ) + a r 2 + a r + a) = 0 2 l t 1 O1 ) + 2 + a Or 2 alr 2 a 2r 2 + a) = 0 a Or)3 + a l r)2 + a 2r) + a) = 0 and, adding, a 8 + a l s 2 + a s 1 + )a = o • 0 3 2 3
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11
The easiest way to derive this is by calculus. It is also possible to derive
s_k. Thus,
consider the cubic forms at the bottom of the last page. Provided no root is
~ro,
a Or 2 + a r + a + a) I r 1 2 1
1 1 a Or 2 2 + a 1r 2 + a 2 + a) I r 2
ge t
a Or)2 + a l r) + a 2 + a)/r)
we may
=0 = 0
=0
and, when we add these, we have aOs 2 + a l s 1 + )a2 + a)s_1 x
x, y
Then b
2
y =a 2 + 1 = J4
2 to be the roots of X + aX + b = O.
s1 = -a = ""8, and
= 15,
•
x +
Now consider
Assume
=0
J4 -
and our quadratic is
fA, + 2b = 0,
2 X -
ax
+ 15 = O.
The roots are ) and
5, so, for our siau1taneous
equations, we have
(x,y) = (),5) or (5,)).
Now you might have fun solving theses x + y + z
Mathesis
(1904)
x+y+z=) 222 x + y + z =)
x5 +
r
+ z5 z 3 First USA Olympiad
=)
22 x2 + y + z =) x) + y3 + z) 6 Math Tripos
=
-IX + ...;y = 6 l/x + lJy = 5/16
12
ASSORTED PROBLEMS 7)
Find a J-digit number with all its digits distinct and different from zero such that the sum of all the numbers that can be formed with three of the digits (repetition not permitted) is equal to the original number.
8)
(CulB) x (0 < x < "'), the
Show that for all inequality
sin x (1 + cos x) < (1 + cos(~))(sin(~) holds,
9)
(Finland)
Determine all positive integer solutions
(x O' xl' ••• ,x n )
of the system
4x l
= 5xO
+ 1
4x2
= 5x1
+ 1
4xn
....
= 5xn- 1 +
1
(Sweden)
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KURSCHAK CORNER Students who enjoyed Hungarian Problem Books I & I I covering the famous Kurschak Contest (formerly named after Eotvos) for the years 1894-1928, should welcome the challenge provided by this competition in more recent years. They should set aside an uninterrupted 4-hour period to compose complete well-written solu tions, extensions, generalizations, etc. to at least some of these problems, and submit their work (enclos ing a stamped, self-addressed envelope) to the address given below. Every solution will be thoroughly eval uated and each respondent will receive a set of in structive solutions to the problems posed. 1/1970. What is the maximum number of acute interior angles of an n-sided planar polygon which does not cross itself? 2/1970. Five distinct numbers are chosen from the set {10,11,12, ••• ,99}. What is the probability that there are at least two among the chosen numbers whose diff erence is I? 3/1970. Assume that n points are given so that no three of them lie on a straight line. Some of the segments connecting them are colored red, while some are cc10red blue so that from any given point one can get to any other point along a colored segment in one and only one way. Prove that it is possible to color blue or red the remaining segments connecting the points so that all of the triangles defined by the n points will have an odd number of red sides.
Dr. George Berzsenyi 2040 Chevy Chase Beaumont, Texas 77706
CONTENTS
Covert Quadrilateral and Averages. i
Preface
1
Inequalities
5
Ptoleay's Theore.
8
Answers and Questions
10
Many Cheerful Facts
12
Assorted Proble.s
1)
Kurschak Corner
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Yet lIhat an aU sue _leUes to . . ~. IIh_ thou'lhtS an rull or lnllc.. ani surds? x2 • 1x • S3
•
u/).
C.L.D.
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