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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.<. <. I
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I. I
2 V
=
1
288 Ie
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p2
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p2
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No. 1
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2
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c
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Sept. 1982
Copyright ~ The Mathematical Association
of America, 1982
APOLOGIA Fran experience with the USA Mathematical Olympiad and the International Mathematical Olympiad, we have cane to believe that there are a large number of students in our schools who are interested in mathematics and who may be well above average in the subject, but who have no opportunity of developing whatever abilities they have. The purpose of this small (as yet) effort is to cater to the tastes of those gifted in mathematics who would like to develop their abilities. Each issue will have one or more short articles on a topic not usually studied in school, together with several problems which require above-average imagination and ability to solve. Most of these problems cane from other nations, where concern for the better student is considered important. We have not been able to find any publication that addresseS itself to the mathematically gifted student. Hence our hope is to provide such a publication. The future of this effort depends on your support. If we get sufficient support, this publication will grow. We also hope that it will be an outlet for students who have ideas they would like to express to others.
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For one way to express your opinion, see the back page.
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I.• ••,.•• i.I.J. i.i.,. 1.i.,. 1.,. ,.i.,.'. ,.,. i.,.I.'••. i.J. •• •,. i. ,.••• l
•• ••
RECURSIONS Finite differences and recursion equations are beginning to occupy a greater place in the problems presented for use in mathell&tical contests, including the International Math Olympiad. It seems worthwhile, therefore, to take some time and space to present at least an introduction to this area of mathematics. As a start, consider the table below, in (i = 0,1,2, ••• ) are successive i terms of a sequence that follow a definite but
which the
u
unknown rule of formation: U
u
o
u
1
u
u
2
u - u 1 - u0 2 1 u - 2u + U u 2 o 3 1 u - JU + JU 2 1 J • • • In any row except the first,
u
J 2u
J
...
u 2
2 + u
1
u0 each term is the
difference between the two terms immediately above it in the row above. Thus the ele.ents in the second row are called "first differences", those in the third row "sec ond differences", etc. It is usual to write
u
- U aa AUO' 1 o u - u as AU , and so on for the second row • 2 1 1 As for the third row, note that ~ - 2u 1 + Uo equals A{AU - AU ) which we write in the form O 1 2 6 u • I n the same manner we are able to rewrite O J ~ - JU2 + Ju1 - Uo as A u O•
We may thus rewrite the first table in the form u
U o AU
O
u 2
1
Au 2 A U o
1
AJll
0
2 L:J;. u
AU
2 1 •••
~ ••• ••• etc •
•••
Now, since li.uO= u - u o ' u = Uo + li.U O • 1 1 We take the liberty of detaching the operational
=
symbol A and writing
u (1 + li.) u • If we o 1 2 substitute this into A U = u - 2u + u ' we 2 o O 1 2 find that u = U + 2Au + A U ' On detaching o O o 2 the symbols of operation, we have u = (1+li.)2 uO ' 2 In fact, it is easy to show by induction that un c (1 + A)nUo ' The expression
(1 + A)n
is
an operator that happens to behave just like an ordinary algebraic expression. The formula
un
= (1
+ li.)nuO is remarkably useful. For example, it can be used. to find the general term in a sequence. Let us exaaine the sequence
6
2
6
4 2
10
8 2
30
20
12
2
•••
•••
o
Here,
•••
0 2 U = 2, li.u = 4, A u = 2. FrCII the formula, O O o
(~)A + (~)li.2 + ••• )uo = 2 U + (~)li.Uo + (~)li.2Uo = n + 3h + 2. o
un = (1 +
•• •• •• •• •• •• •• •• • •• •• ••• •• •• •• ••• ••
I.I.,. i. <. i.,i.. • 'i.• i
,e ,e
Ie ie
'!be formula can also be used for summing a series. If we add another row above the first in the first ta'b1e, we have
o
Ie
U
sequence, and we apply the formula to the augmented table. Let us apply this process to the sequence used in our first example. The augment,d. table is 0
8
2
20
6
2
20
6
8
2
30
2 o •
= 0, ~uo = 2, ~2uO = 4, ~3uo = 2,
Here
U
that
(1 + ~)nuo
simplifies to
=0
70
10
2 0
o
40
12
4
e
',.'.. '.••• ••• ,.,.• •
•••
sum of the first n terms of the original
.e ,e
•• •
~
u2
The n-th term of the new row is equal to the
.e
• • •••
u1
o
+ (~)2 + (~)4 + (~)2
and we which
n(n +3 1 )(n + 2) •
If we represent a summation by using the symbol E, we have found that
~
i(i + 1)
1
We already know that
~ i 1
= n(n+1)(n+2). 3
= n(n+1). 2
n
Similarly, we can show that E i(i +1)(i + 2)(i + 3) equals
n(n+1)£n+2) (n+3)
and 1so on.
We can use these expressions to find many sums of series. For example, to find the sum of the squares of n natural numbers, we note n n n that E i 2 = E i(i + 1) - E i which equals 111 n(n+1) (n+2) n(n+1). 3 2 n As an exercise, find E i 3 • 1
Finally, let us agree to write ux+ as Eux • 1 2 Then ux+2 = EU + = E Ux • In fact, we easily x 1 k see that uX~I\. oU_ = E u • When we compare this new x notation, using the operator E , with the notation using A, we find that (1 + A)nUo = un = EnU we may write
• It appears that o (1 +A) =E and (1 +A)n =E n •
We have now introduced those operators from the calculus of Finite Differences that we will find useful in working with recursion formulas. These are A, E = 1 + A, and E. They behave to a large extent like ordinary algebraic sYJIbo1s.
••• •• •
•• •• •• •• •• •• •• •
.•
•• ••
•
.•
••
•• •
•
••• •
,.!. ~.
• • >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
•• •• •
••
!.ie
,.i.•• ,.i. •• . I
,e
7 Now suppose we are ~ven f(E)u f(E)
,. ,. i. ,.,.I.•• i.•• •• .••<. ;.• ,
X., O.
= X, where
Things become
much more difficult.
There are two forms that
X can take for
which solutions can be found. x
First, suppose that
,
••• ••<. •• <.• • •••• •• •••• •
is a polynomial, but
x
a
X = aX. Then E(a )
=
x+1 _ x 2( x) _ 2 x n x - a.a ,E a - a a , ••• ,E n( a x) =aa.
Fran this we conclude that F(E)ax = F(a)ax • Fran this, we have
aX/F(E) = aX/F(a) •
To illustrate, let
x ux +2 - 7ux+1 + 12ux = 5 •
Using the canp1eaent&ry equation, we first find the canp1ementary solution
U
x
= A(J)x + B(4)X.
For the particular solution corresponding to x X = 5 , we proceed as follows I (E 2 _ 7E + 12)u = 5X x
u
x
=
1
E2 _ 7E + 12
and, letting E = 5, solution is
u
x
U
x
.5x
= 5x/2.
The cCllp1ete
= A(J)X + B(4)x + 5x/2 •
The same procedure will work when
X
involves
trigonometric or hyperbolic expressions if we rewrite these expressions in exponential form •
X is a polynomial function of degree
If
m,
we can assume a particular solution of the farm U
m
x
= aOx
m-1 + a x + • • • + am ' substitute in 1
the recursion equation am. equate coefficients. Let us work on First, let
ux +1
information as
ux+2 + 4u + + 4 x 1
= vx'
=x.
••
am. we may rewrite the
v x+1 + 4vx
=x
•
- 4.
The complementary solution is obviously
v U
x
x
= A( -4)x. For the particular solution, = ax + b. Substitution gives us
let
ax + a + b + 4ax + 4b = x - 4.
Equating coefficients, we fim. The complete solution is v However, we want to find by
x-1
am. arrive at
a
x
x
= A(-4)
•
.."
b
u, so we replace U
••
=!' =- ~~.
= A(-4)x + (5x-21)/25.
x
•
•
•• •• •• ••
x
+
x
5x - 26 25
Unfortunately, there are recursions that do
•• •• •
not have the forms discussed above. For these, it becomes necessary to make some special substitution or use some inspiration to attack the equation.
•• ••• ••
I. I.i. ,.I.. ,.:. • • ••<. •• •• •• •• I
9
For exatnp1e,
•
Divide by
and. we have
v
v
x
= ux+1
+ u • x
xx+1.ux (assuming neither is zero),
1/ux +1 + 1/ux
= A( _l)X +
=1
x+1 + v x
1/2,
so
= 1.
U
Now let 1/u x
= vx ,
to solve. We find
x is the reciprocal of
this • Again, we are given
ux + = 2ux 1
form sugges.ts that we let
u
x
2
- 1.
The
= cos v • This x
gives us
••,e
or
ie
and., u
i
ux + ·u 1 x
and we have
,
ie
let
•
I.-. I.i.,.'. ,.'. •• i.•• ••• •
••
v x+1 x
=2
The solution is
vx •
= cos(2xA)
v
x
= A(2)X,
•
For a more complete discussion of recursions, texts on Numerical analysis frequently have at least a chapter on the subject. If one can get to the t'Treatise on the Calculus of Finite Differences", by Boo1e, one will find an excellent development. It is also instructive to eX8JIine the solutions of related problems in the various mathematical contests. For example, see Problem
6 of the International Mathematical Olympiad for 1967.
/0 MANY CHEERFUL FACTS We assume that the ('
Law of Sines is known to all our readers. However, from the diagram at the left, note that LA'
A
am
sin' A'
where
2R
=LA
= a/2R, is the
diameter of the circumcircle. Therefore, we
sin A = a/2R. In the same manner,
may write
sin B = b/2R
we have
sin C = c/2R.
am
Therefore, we have an extended Law of Sines
a b = --:--= sin A sin B
--:-=-~
That
2R
= sinc
C
= 2R.
is very useful in many situations.
For example, we know that the area of AAB:: is
K =
~
ab sin C. However, sin C = c/2R, and
we have a new formula for the area of a triangle, K - abc
- mr·
We also know that, if K
= rs.
r
is the inradius,
Therefore, we have
4Rrs
= abc.
••• •• • • •• .•• •• ."
•• •• •• • •
We can also derive other formulas, such as 2 K 2R sin A.sin B.sin C,
=
and see what results when
b, c
are replaced
by their values fran the Law of Sines in the
Law of cosiness a
2
= b
2
+ c
2
- 2bc cos A.
•• •• ••
,.I.,.• ;.',. ,.I.'<.. ••;.>. ;.;.I. ,. I. i.. I.,.,.•• i.•• ,.i.I. ,.••;. i.;.• • ,.• ,
,e
1e ,
"
ANSWERS AND QUESTIONS
n n
=
x ) First, show that A x n! Next, since n Anx (E - 1)n we have, on expansion,
=
Anx n • (En _ (~)En-1 + (~)En-2 _ ••• )x n
= (x+n)n
_ (~)(x+n_1)n + (~)(x+n_2)n _ •••
Finally, let n!
x
=0
in lines
1
and
4, and
= nn
• ••
How could you use this result to prove Wilson's Theorem - namely, if prime number, then
( p-1 ) ! + 1
p
is a is d i visible
by p? y) We pro'tably have heard about the number 1729, which Ramanujan told Hardy was the smallest integer which could be expressed as the sum of two cubes in two ways. In fact, it is easy to see that
13 + 123
=
r
+ 103.
What is the next larger number that can be expressed as a sum of two cubes in two ways? z) Any power of a number ending in 1, 5 or 6 will also end in a 1, 5, or 6. For which nWlbers ending with two digits will every power of the number end in the same two digits in the same order?
.•
ASSORTED PROBIEMS 1) Determine all real values of
a
for which
the equation 4 2 16x - ax3 + (2a+17)x - ax + 16
=0
has four distinct real roots which form a (Bulgaria)
progression.
~ometric
2) A sequence and
a , a , a , ••• satisfies a 1 1 2 3 mn a m+n = 4 a ma n for all m and n.
Determine the smallest value of which
n
=2
for
a
(written in decimal notation) n has at least 3000 digits. (Netherlands) 3) Given k
•• •• ., • •• • •
•
•
a fixed non-negative integer and
P(2x)
= 2k - 1 (
p(x) + p(x +
t»
Prove that
k 1 p(3x) = 3 - ( p(x) + p(x
+~) + p(x + 3» (Gt. Britain)
•• •••
•••
••• •• •
,.'. i.. •• ,.'. .
••I.
••• I.••• •
,.'.
,.,.••,.
,..•• ••i. ;.'i.. i. ,. ,.
,
,e <e '
Ie '
i
•
<
•
,
~.
CON'lENTS
Cover •••••Voluae of Tetrahedron Page l ••••Recursions Page 10 •••Many Cheerful Facts Page 11 •••Answers & Questions
••• • •
Page 12 •••Assorted Problems
• •• ••• •
,.•• . ;.I.,.••
.•• • ,.,. '. • 'i..•
';.!. i.i.. ,. '. ,.I.,.j.
i.'••. ,i.i. .
,.j.
I
,
arbelos
,
i
PRODUCED FOR
PRECOLLEGE
PHI LOMATHS
•
No.2
November, 1982
Copyright ~ The Mathematical Association of America, 1982
.• •., . '
1
PREFACE Mathematics possesses a "cold and austere beauty" which fascinates good matheMticians, and which makes the study of mathematics for its own sake worthwhile for them. Just as an artist may see a painting as being beautiful, a mathematician may see a theorem as beautiful. It 1s a fact, however, that many of the greatest mathematicians had no hesitation in using other disciplines when developing mathematical ideas. For exuple, a letter from Archimedes to Eratosthenes ( resuscitated in 1906 by Dr. O. Neugebauer from a palimpsest) contains an explanation of how Archimedes used physics to derive mathematical results which he then proved mathematically. We know that Gauss and Newton worked in science as well as mathematics. Hence we feel that it is not unreasonable for us to present problems that involve science as well as mathematics.
•
In fact, we recommend that students of mathematics be conversant with sciences as applications of mathematics and as aids in deriving mathematical results. Therefore. we have included one or two problems reqUiring knowledge other than pure mathematics in their solution. We hope you will like them. We will gladly accept others of the same type or similar for inclusion in this publication.
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2
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THE ARBELOS
~
4
4 4
4 -4
The Arbel06 (see cover) consists of the three
. '.." .. .. .•
that of a shoemaker's knife ar arbelos (Greek) •
. 4
•• • •• >.
'.
•• •• •• •• •• •• •• •
points A, B, C which are collinear,
to~ther
with the semicircles ATB, AXC, CYB, as shown. It was so namen because of 1ts fanciful shape
It engaged the attention of no less a mathe matician than Archimedes, who was certainly
the greatest mathematician of antiquity, and
possibly the greatest of mathematicians. He played with this figure for fun, which is an excellent reason far
doin~
mathematics •
In the figure, we have added line CT, which is tan~ent
to the slllall semicircles, and also
lines AXT and BYT. We have also added l1ne XY, intersectin~
CT at O.
Now, just far fun and re laxati on, try the followi~
(a)
problems:
XY is tangent to the small semicirclesl
(b) XY and CT bisect each other; (c) The area of the arbe105 is equal to the area of the circle on CT as diameter. (d) The circles inscribed in segments ACT and BeT have equal rad 11. Archimedes derived these results. Can you ?
FUNCTIONAL EQUATIONS
A functional equation is an equation in which an unknown function is expressed in terms of known functions, the object being to determine the nature of the unknown function. Examples aret f(mx) = a f(x) f(x) + f(a-x) = c 2 f(a+x).f(a-x) = a 2 _ x2 Sanetimes a functional equation may be solved by making use of the concept of "periodic" functions. If F(x) is of such a nature that when it is performed twice on a quantity, the result is the quantity itself, we call F(x) periodic of order two. Examples of such periodic functions
- xx. ( 1 - x2)t • etc. aret a/ x , a - x • 1 1 + Examples of periodic functions of order three are 1 ,1 ( - x-2)+ , etc. 1 - x There are an infinite number of periodic functions of all orders. As an example, take (f(x)}2.f(~ : ~) =c 2x •
••
We note that ~:~ is periodic of order two.
Therefore we replace
- x
x 'by 11 + x in the
•• ••• •
•
•
•
•
•
•
•
•
•
• •• ••
•
•••• •
•
•
•
•
•
•
-. • •
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
4
given equation and find that () - x}2 [ f ( 11 + x • f x
= c 2 • (11
- x) + x •
Divide the square of the given equation by this secom. equation, and we fim.
[ f ( x )} 3 -_ c 2x 2 '11 +- xx from which we have
x
f(x)
immediately.
Again, suppose
n f(x) + af( -x) = x • Replace
by
fe-x) + af(x)
-x
Eliminate
to get fe-x)
= (_x)n.
from these two equations, and
we fim. that f(x)
= xn
- a(_x)n 2 1 - a '!be method of differences (see issue No.1) can sometimes be used to solve a functional equation. The following method is due to Laplace. Suppose we are given x
= ut
=vt
and
We let f(x)
am.
feu) ax
f(mx)
=a
= u t +1
= v t +1 •
f(x) •
• Then we let
Then, first,
,ut + - aUt = 0 , from which we have 1 Next,
=0,
v t +1 - a v t
Now we elt-inate
t
fram which
u vt
t
=A • t •
=B a t •
from these two equations.
5
From the first equation, we find that t
=
so that
log u
- log A t log m
v
Now replace
_ B (log u - log A)/log m t a •
t
u
by
t
x
and
v
t
f(x),
by
and we have
= B a(log
f(x)
x - log A)/log m.
We can simplify' this a bit further to get f(x) = C a log x/log m
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
•
C is not necessarily a constant. It can be any function It should. be pointed out that x
of
which does not change when
replaced by
x
is
1lX.
As another application of the Laplace method, take ux+1
f(x) + f(a - x)
=a
- x,
V
x
=c2•
=f(x),
Since
u
Fran
v
Eliminate
Let
v X+1
x x
u
= f(a
x
= x, - x).
= A( _l)X
+.! • 2
= B(_l)x
+ c 2/2 •
•
•
•
•
•
•
•
(-1?, and. we finally arrive at f(x)
= m(x
- ~) + ~
2
•
•
•
•
•
•
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6
Let us next examine the equation f(a + x).f(a - x) = a 2 _ x2 • We could use the same method to solve this. However, it is easier to note that fCa + x) fCa - x) a + x· a - x or F(x).F(-x)
= 1,
or
= 1,
which is obviously satisfied x b,y F(x) = eX. That is, f(a + x) = (a + x).e • Finally, f(x) = x.C x- a • Frequet1y, therefore, one can use special methods to solve functional equations, the method depending on the ingenuity of the solver. Consider the equation (f(x)}2 _ (f(_x)}2
= 4x.
Here it is best to factor. We obtain (f(x) + f(-x»(f(x) - fe-x»~ = 4x and ass\Dlle that f(x) is a po1ynania1 of farm 2 n f(x) =a O + a x + a 2x + • • • + anx • Then 1 _ 2 n f ( -x ) - a O - a x + a 2x - • •• ~ anx • 1 Substituting and simplifying, we find (a + a x 2 + ••• )(a + a x2 + ••• ) = 1. o 1 2 J Then a a = 1 and all other coefficients are O1 zero. Thus, f(x) = a O + x/a o.
?
In some cases, it is possible to solve a functional equation by using eleaentary calculus. Let
r(x + y) + f(x - y)
= f(x).r(y).
Differentiate tWice, one time with respect to x
and the other tiae with respec t to y. Then f"(x + y) + f"(x - y)
= f"(x).r(y)
f"(x + y) + f"(x - y)
= f(x).f"(y).
Therefore,
~= ~ fW f(Y)
, which means that
either f'raction is equal to a constant. Now let
f('j)
=.! n2 and solve. We have
f x
f(x) or
= Ae nx
f(x) = A cos nx
+ Be- nx + B sin nx.
Occasionally, one can insert special values ror
x
and/or y
to determine a pattern of
behavior of the function, and then use induction to prove your guess. You are now in a position to try your hand on the Olympiad problem elsewhere in this
•• • •• •• •• •• •• •• •
•
.."
•
•
•• •• ••
issue. Or, consider Problem 5 of Olympiad 10. f(x + a)
= 1/2
+ (f(x) - (f(x»2)
t .
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8 MANY CHEERFUL FACTS
Identities appear in all branches of mathe matics, but are apparently most entertaining in trigonanetry. For example, starting with the formula for the cosine or a double angle, one can easily derive sin ~ = + ~ - cos A 2
-
+ cos A
2
2
These can be used to derive
A
tan '2 • However,
c...
A~~~L.,.o<:;.._ _~17
,$-a..
from the
dia~am
above, one can see that
tan
~ =s
: a
etc •
Using the formulas for functions of multiple angles and of sub-multiple angles, try these. (a) Find an expression for
sin,6
o
;
(b) For any triangle ABC, prove that tan A + tan B + tan C = tan A. tan B.Tan C. A ta ~B + ta ~. B ~_..Q. + ta C ta A 1 ( c ) ta ~. 1.OoJ~ ~. ~ = •
9 PASCAL AND BRIANCKON In 1639, Blaise Pascal discovered the theorem "If a hexagon is inscribed in a conic, then the intersections of the opposite sides of the hexagon are collinear." We know that one of the Bernoulli brothers saw the proof and canplimented Pascal highly on it. Unfortunately, the proof has disappeared.
., J
I I
, , ,,R. ,•
•• •• ••• ••• •• •• •••
I
I I
, I
I
,Q Now it does not matter how the vertices of the
hexagon are labeled - the theorem holds in all
cases. Question. how many Pascal lines are possible for a given hexagon ? This is a theorem in Projective Geaaetry, and does not involve lengths, areas, angles, or any metric properties of the figure.
••
•• •• •• •
••
•• ••• •
•• •• • • •• •• • •• ••• •• •• ••>.• >. ••• >.• ••• •• •• • • ••
:.'. ,.
'.
'.
10 If, in the diagram, it happens that vertices A and F coincide, side AF becanes a tangent to the conic. Nevertheless, Pascal's Theorem still holds.
Questiona In the diagram above, ABeD is inscribed
in a conic. Can you show that P, Q. R are
collinear ?
One of the properties of figures in project
ive geanetry is that they can be dualized. That is, i f one describes a figure in terms of points and lines, and then replaces the ward point by the word line and the word line by the ward point where these arise, we arrIve at a dual figure. Thus, we can say that Ita triangle is a figure consisting of three points not on a line. together with the lines on them. It The dual is itA figure consisting of three lines not on a point. together with the points on them It •
11
It is easy to see that a triangle is a self dual f,igure. However, a quadrilateral is not. By definition, "a quadrilateral consists of four lines, no three
bein~
on the same point, together
with the (six) points on these lines in pairs." The dual is therefore, "a figure consisting of f our points, no three on the same line, together with the (siX) lines on these points in pairs." This figure is called a quadrangle, and the difference between quadrangle and quadrilateral is evident once both are drawn. Next, the graph of a function may be considered as consisting of points, in which case we call it a locus. Its dual may be considered as consisting of lines, in which case we call it an envelope. Consider the midpoints of all equal chords (not diameters) drawn in a circle. These midpoints fara a circle. The chords themselves "envelope" the same circle. Questiona A ladder slides along a wall and the ground so that its ends always tOltch the wall or the ground. What is the locus of the midpoint of the ladder? What is the envelope determined by the ladder ?
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'.•••
•• •• ••
•• •• • •• ••• • 1.• •• •• •• ••• •
,.,. ,.'. ,'. '. ,.
12 A few remarks are in order here. First, if one can dualize a figure in a problem, it may be easier to solve the dual problem. Once this has been done, the original problem is solved. Second, it may be that the conic in Pascal's Theorem "degenerates" to form two lines. The theorem is still true. However, this case was stated by Pappus (circa 300 AD). Questions Can you draw the figure and state the theorem for the dual of Pappus' Theorem ? Third, since, in Projective GeOllletry, two distinct lines always intersect, there is no parallelism. If one wishes to do so, one can select a line, call it the "line at infinity", and consider two lines that would intersect on this special line "parallel".
Problem: In the diagram below, ABeD is a para11t \
ogram. Point P is selected on side CD and lines
PA, PB drawn. Point Q is selected on side AB and lines QD, QC drawn. If AP, DQ intersect at M, and BP, QC intersect at N, and line MN intersects AB at Y and CD at X, prove that DX
= BY.
1)
It 1s a fact that the dual of a locus is an envelope. Points become 11nes and v1ce versa. Therefore the Pascal configurat1on has a dual. Vert1ces become tangents, coll1near po1nts become concurrent 11nes, etc. However, 1t was not unt11 1806 that Br1anchon dual1zed Pascal's Theorem. Br1anchon's Theorem states, "1f a hexagon is circumscr1bed about a con1c, then the 11nes join1ng oppos1te vert1ces are con current. "
p
Once again, 1f the f1gure becomes a pentagon, then one of the vertices becomes a tangent po1nt, but the theorem st111 holds. It 1s 1nterest1ng to try var10us configurat1ons and see what one gets by apply1ng Br1anchon's Theorem.
••• ••• •• •• •• •• •• •• •
• •
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•• •• •• •
•• •• •• ••. e • ••• >
14
A A
•
•• ••• •• •• •• • • '.
:.
,.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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;.
••• • • ,e ••• ••• • • .e • ,e •• •• •
:. ;.'.• ••
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•
,.1·
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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,.'.• •• •• '.•••
•• ••
e
1.j. ,e ,e
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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1
INEQUALITIES The subject of inequalities has assumed much importance in mathematics below the collep;e level of late. Problems involvin~ inequalities have appeared in under~aduate contests and are bec ominp; part of the sec omary school curriculum. One has only to examine recent International Mathematical Olympiads to see this. for example. Hence this short intr
2
fOUM by addinp; the squares of the scores p;1ven. dividinp; by the number of scores. am. then finHnp; the square root of the result. That is. (RMS) L(Ea 2 )/nJ1/ 2 •
=
Now. the diagram on the cover page shows a trapezoid ABeD. In this diagram. a) line MN bisects sides AD am. MC. Find MN in terms of x and y. Which average is it? b) line PQ is parallel to AB am passes through the intersection 0 of diagonals AC am BD. Fim PQ in terms of x am y. Which averap;e is it ? c) line RS is parallel to AB am bisects the trapezoid - that is. divides it ato two parts of equal area. Fim RS in terms of x am y. Which average is it? d) Line UV is ~allel to AB am divides the trapezoid into two similar trapezoids. Fim UV in terms of x and y. Which averap;e is it ? Now it is a fact that
(AM) > (GM).
All
s~ujents interested in a beautiful. proof· should
consult any text on inequalities for Cauchy's proof. In it. he used "'backward" imuction for the first time. For two terms. x am y. we have
(-v'X -
x + Y~ 2
-v;)2 ~ 0 • from which we get
,JXY •
fraa which
(AM) ~ (GM).
In fact. for two terms. we can do more. From 2/(HM) (HM) so that
=
l/x + l/y • we get
= x+y 2xy = (x+y 2 )(xy) = (GM)2/(AM) • (GM)2 = (AM)(HM). Now we can say that
(AM)
~ (GM) ~ (HM).
This inequali tyrelatlon holds
(A) in general.
• •
••
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•
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••
.•"
•• ••• •••
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',.•.• •
• ••• •
••• •• • • • •• •• •• •• ••• •• • I
•
,.,. '.
J Now far some possibly less familiar but equally useful inequalities. We return to the (AM)-(CM) inequality - labeled (A). We shall use m+n terms, of which m are equal to ~ am n equal to E. On substitution, we have (a + • eo + a) + (b + ••• + b) > m..-nVambn m+ n -
Let m+n
til
m+n = p, n = q.
p1 + qi_-
Then
,
1, am we
am we may substitute, deriving
~p
+
E q
If we let a l / P
>
-
=x
i!. q
a 1/ Pb 1/ q
am >
(B)
•
b1/ q
= y,
we get (c)
xy •
These results have wide applications am are warth remembering. Next, we know that
2
r(aix + b i ) ~ 0 •
Expa.min~,
(ra 2 )x 2 + 2(rab)x + (rb2 ) > 0 • The ~aph of this quadratic equation does not cross the X-ax is (it may be tan~ent to it). Hence the discriminant cannot be positive. That is, 4(tab)7. - 4(ra 2 )(Lb2 ) < 0 , ar (ra 2 )(rb2 ) ->
(rab)2. .
(n)
This inequality has many titles. It has been named after Cauchy, Schwarz, Buniakowski am also Lagrange •
4
Finally. let us return to inequality b = y (Eyq)-l/q Let a = x (ExP)-l/p i
i
=1
l/p + l/q p
1 xi P (Exp)
(n). where
• p > 1 • q > 1 • Substituting. q
1 Yi x i Yi + Ii (Eyq ) ~ -(Ex-P)-':'l"T/P--.-(-E-yq-)~lf"r"q
•
Add all such inequalities far i = 1 ••••• n , am we have (Exy) • 1 (Ex P) + 1 (Eyq) > q P (ExP) q (~yq) - (ExP)l!p(Eyq)l/ On simplifying this, we finally arrive at (Ex P)l/P(Eyq)l/q > (Exy).
(E).
This is known as Hl:Jlder's Inequal1 ty. am is one of about five essential inequalities. Incidentally. far P = q = 2. we ~t (Ex 2 )( Ey2) ~ (Exy)2 am we have Cauchy again ! Now perhaps we might try the followings n+l n ~ n2 for n > 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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:.,. •• :. •• ••
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,e
>. •• •• •• •• ••
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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'i..
7
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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.
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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
••• •• •• •• •• •• •• •• •• •• •••
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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arbelos
PRODUCED FOR
PRECOLlEGE
PHILOMATHS
No.4
Copyright
March, ~ The
1983
Mathematical Association of America, 1983
i
PREFACE I t is pleasant to report that the number of subscribers has continued to ~ow. It is more pleasant to report that some readers are writing to give their opinions of the articles in the Arbelos. Happily, nearly all are compU:mentary. However, one correspondent wrote to say that he had seen something we had written about before. Now one reason for the Arbelos is to present subject matter that is not normally touched on in classes in mathematics. Naturally, we do not know what is beinp; taup;ht everywhere. Therefore, it must happen that the reader will sometimes see somethinp; he has seen before. We can only hope that what we present is unfamiliar to a majority of you. This situation can work in reverse. I still remember when I presented a problem based on an example I found in "Theory of Numbers", by Niven and Zuckerman (page 84, No.23) and heard Randy Dougherty say, "That's Beatty's Theorem". I had never heard of Beatty's Theorem, and had to search about until I found it. So it is quite possible that you know somethinp; the editor does 't. So - if you find that we have written some thing that is familiar to you, forp;ive us. If you have discovered somethinp; new, write us. We will p;ladly print it and give you credit. Finally, if you would like to see some topic appear in Arbelos. let us know. We're happy to oblip;e. S. Greitzer
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1
PRISMATOID AND PRISMOID So much new mathematics has been developed in the past few years that some of it has gotten into the secondary school curriculum, thus displacing other topics that are, nevertheless, still valuable. We will examine a few of these. Let us first examine a prisma toid. This is defined as being a solid figure whose upper and lower bases are parallel polygons ( not necessar ily with the same number of edges) and whose lateral faces are planes. These are therefore triangles or quadran~les.
c
The diagram above shows a prtSUlatold ~ri to upper base U, lower base L, and 'mid-section M. First, all quadrangular faces are cut by dia gonals to farm triangles. Next, fro. any point 0 in tho mid-section, lines are drawn to all the vertices of the prislllatoid as well as to all the intersection points of the faces and the mid-section. Lots of triangular pYramids are formed. We examine one such pyramid, O-ABC • (It is shown at the left.)
2
Since PQ bisects AB and AC, MBC has
area four t1.lles that of MPQ. Hence pyramid
O-ABC has a volUlle four time that of pyramid O-APQ. However, pyramid C-APQ = pyramid A-OPQ which is equal to, say, H x H/6 in volume. 1 Therefor the volUllle of pyramid C-ABC is 4M 1 xH/6. Going room the solid, the sum of the volwtes of all the pyruids whose bases are on lateral faces equals 4M x H/6. We must inchde two mare WX'8Jll1ds - with bases U and L. These have volUllles equal to U x H/6 am L x H/6. Adding these, we find the volUllle of the whole prislllatoid to equal
(n)
V
*
=
(U + L + 4M) •
We come next to Bonaventura Cavalieri (1598 1647). To paraphrase his theorem, called Cavalieri's Principle) If two plane figures are on the same line as tase and if the lengths of intersections of these figures bylines parallel to the base are always in the constant ratio t , then the areas of the figures are in the same ratio. (If the umerllned wards are replaced, plane by solid, line by plane, areas by volUllles, we obtain the other half of Cavalieri's Theorem. (n) I have been unable to find any source far this formula among the ancient Greek geometers. The earliest reference I fim attributes the fOI'Jlula to Johannes Kepler (1571-16)0). Doesn't that make this a . "modern" theorem?
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•• •• •• r ••.e • i ••• • •• •• •• •• ••• •• •• ••• •• •• • ,e
I. ''..
3
Finally, let us look At the rip;ht, we have a section, at a distance is a quadratic function 8 = Ax 2 + Bx
at the dla~am above. solid whose cross x units from the top, of that distance. Then + C.
As for the so1ids pictured. at the left, the section of the pyramid x units from the top 1s 8 = Ax 2 • The section of the triangular 1 prism x un! ts from the top is 8 Bx, am 2 the section of the rectangular prism x units
=
from the top is 8 = C. Therefore, the solid a t the right and t~e sum of the three solids at the left have equal volumes, by Cavalieri's Principle. However, the volume of each of the three solids at the left can be foum by usinp; the Prismatoid Formula, which means that we can find the volume of the solid at the rip;ht by using the same formula. Hence, If a solid has the property that every section parallel to the base is a quadratic function of 1ts distance from the base, then the volume of the solid is given by the formula
V =
*
(U + L + 4M) •
4 Such a fip;ure is called a Prismoid. Now we have enou~ information to determine areas and volumes for many figures, which are usually found by calculus. For example, since any cross-section of a sphere is a quadratic function of its distance from a pole, the Pris~oid Formula holds, so the volume is V
=
¥ x(
2 0 + 0 + 4nr )
=~
nr J •
Of course, there is the (relatively) well known method for finding the volume of a sphere, attributed to Archimedes. In the diagram above, we have, on the same plane as base, a hemisphere and a cylinder, radius of base r am height r, from which a cone has been removed, as shown. Sections of both solids are taken x units up.
In the hemisphere. the area of the section is 2 2 n(r - x ). In the cylinder, the area of the 2 2 ring is also n(r - x ) • Hence the hemisphere and the cylindrical
fi~e
The cylinder has volume
have equal volumes.
Tfl'J and the cone has
Tfl'J IJ. Therefore the hemisphere has a volume of 2nrJ IJ and the whole sphere 4Tfl'J IJ.
volume
On the basis of this proof and the assertion that this was the way Archimedes did it, I still cannot imagine that Archimedes did not know Cavalieri's principle and the Prismatoid formula.
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,e
j
5
Again, in the dia~am above, we have a sphere with radius r and a triangular pyramid in which edges AB, BC and CD are equal in len~th to 2r and are mutually perpendicular. Let a plane x units above the canJIlon plane on which both sit. Then it is a simple matter to show that the section by this p~ne of the pyramId has an area equal to (2rx - x ) while the section of the
2
'
sphere has an area equal to w(2rx - x ). Hence the volumes have the ratio
l/tr.
Now the volume
of the pyramid is 4r J lJ. Therefore the volume of the sphere is 4wr J /). These three methods for finding the volume of a sphere have been presented just to show the variety of ways the Prismoid, Prismatoid and Cavalipri's Theorems can be used.
We present another application. In the
fiKUre above, we start with a parabolic arc,
the equation of which will be y2 = h 2x/B.
6
This arc is enclosed in the rectangle shown. which has base B and altitude h. Next to this parabolic arc. we have erected a pyramid with base B and altitude h. inverted. Both fi~es are on the same base and both are cut by a plane parallel to that base. For the parabolic arc. the section 51 is related to B • h and x/B
= y2/h2.
y by the formula
For the section 52 of the pyramid.
5 /B = y2/h2. Hence 51 = 52 and so 2 the measures of the two figures are equal. Now
we have
the volume of the pyramid is equal to Therefore the area of
K
Bh/3.
• •• •• •• •• •• ••
is also equal to Bh/3.
(If the thought of comparing an area with a length worries one. just make the figure at the left into a solid by giving a thickness of 1 to it.) We have thus shown that the area under the Parabola is 2/3 of the area of the rectangle that encloses it. We must regretfully end this article. but we shall use it to solve just one more problem. We know about the curve produced when a point on a circle moves while the circle itself rolls alone a strai~t line. It is called a cycloid. Parametric equations for the cycloid are easy to derive. (One ~ts from A directly to P alonp; the path x + iy. or from A to B to 0 to C to p. or ae + ia - a cos(e - 90 ) + i sinCe - 90 ).) We find x = aCe - sin e) y = a(1 - cos e)
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,e
-. -.
'.
7
(trQ.,J.tO.)
CO,JaH.-------=:;--'"'"""::"':'W-1
In the diagram above. the circle has rolled along AX. and the point on the circumference has traveled from A to P. It should be noted that
arc BP
= segment
drawn the curve
AB
= ae.
We have also
x = a9. y = a(1 - cos 9). which
1s called the "companion" to the cycloid. It is actually a cosine curve. passes through the point Q. am. bisects the !:!!! of the rectangle ! Now the distance from point P to the Y-axis is equal to (a9 - a sin 9 • The distance frOll point Q to the Y-axis is equal to a9. This makes PQ equal to a sin e. which happens to be the section of the semicircle. Hence the area between the cycloid am. its cOlipanion is equal to the area of the half-circle. Let
K be the area um.er the cycloid between 2 A am. na. The area um.er the companion is Tfa 2 2 2 Therefore. K - Tfa = TTa. 12 • K = Jna. I? •
am. the area um.er one complete arch of the
2 cycloid is equal to 3Tfa • It may be of interest to students interested in the Putnam Competition to remark that the first problem in the first Competition in 1938
8
involves proving that the Prismoid Formula works when the section parallel to the base is a cubic function of the distance frOll the base, and that Problem 2 of the Eighth Competition in 1948 uses Cavalieri's Principle. Serious stooents might be interest in trying to solve the following problems, using the ideas presented in this article. a)
Find the area enclosed between the parabolas y2 = 2px and x2 = 2py.
b) Find the volume of a paraboloid of revolution and cOllpare this with the volume of the enclosing cylinder. c) The hyperbola
x2 _ y2 = a 2
is rotated
about the X-axis, am a cap of height
a
is then cut frOil it by a plane perpendicular to the X-axis. F1nd the volume of this cap. d) Derive a formula for the voluae of a frustum of a circular cone whose bases have radii H am r. and whose alt1tooe 1s h. e) Two cylinders with equal radius a intersect so that their axes are at right angles and intersect. Fim the voluae cOlllllon to both cyl1mers. (Avery caamon problem. Sorry about that !) f) A right circular cylimer has a circular base with radius r units and altltooe a. Throu~ a diameter of the upper base. two planes are drawn touching the lower base on opposite sides (see diagram on cover). What is the volume of that part of the cylinder between the planes? (Our text used triple integrals to get V = (IT - 4/J)ar 2 .)
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9 ANSWE~S
AND QUESTIONS
1. When the Easter vacation arrived, Hyde am Zeke decided to go to Florida. Hyde left in a transport of joy, but Zeke took the trip more calmly, on a higher plane, so to speak.
If the plane traveled nine times as fast as the transport, am Zeke ~ot 36 hours more than Hyde at the beach, how long did each take travelin~ ? 2. The area of a plane section of a circular cylimer is equal to 16,.,. The area of a plane section of the same cylimer, which is perpemicular to the plane of the first section, is equal to 9rr. What is the radius of the cy11mer ?
3. If a yardstick marked
off into 1/4 inches am a meter stick marked. off into centi meters are placed with their initial points together, which two marks on the two sticks will most nearly coincIde ? (our almanac tells us that one meter is exactly 39.37 inches long.)
4.
The equation x 2 - 9?x + A = 0 has roots equal to the fourth powers of the roots of x 2 _ x + B = O. What is the value of A?
5. A 5-digit number 1s a multiple of 41. If the highest order digit be removed am placed to the right of the units digit, the new number is a perfect cube. What was the original number?
10
DIOHfANTUS IN SPRINr.TIME
There are a great many mathematicians in this country, and sprin~time - especially April First seems to brinp; them out of hibernation. We have already received. our first anp;le trisection, a friend assures us that he has a method of con structing a re~lar heptap;on, and we expect our ftrst circle-squaring soon. We also ~et our share of old prob1ems,~which repeat annually. This time, it is the following:
.-.
When a man cashed a check at the bank, the
teller, by mistake (what else ??) gave him
as many dollars as the check called for cents
and as many cents as the check called for
dollars. After spendin~ 68 cents, the man
found he still had double the amount of the
orip;ina1 check. What was the amount of the
original check ?
This naturally set us Equations.
thinkin~
of Diophantine
First, a definition: A Diophantine equation is an equation that has to be satisfied in integers. With this in mind, suppose that the original check was for 100x + y cents. Then 100y + x - 68
= 2(
100x + y) , or
98y - 199x = 68. There are lots of ways to SOlove a linear Diophantine equation, provided it has a solution. Let us illustrate a few of these methods, using the equation
5x + ?y for simpl1city.
= 41
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11
(a) We can ~ess! For this equation, y can be only 1, 2, J, 4, 5. Trial gives us x = 4 and y = J. ~ess a solution for ax + by = 1. Then a(cx) + b(cy) = c, and we have a solution. In this example,
(b) We can
50) But
I
5(123) + 7(-82) 5(x) + 7(y)
= 41 = 41
5(123 - x) + 7(-82 - y) = o. Rewrite this as a proportion, and we have
Hence
'.
'. '. '. ,.
+ 7(-2) = 1
123 - x = 82 + y
1 5'
We may now assume that
= 123 - 7t Only t = 17 will x
so x = 4,
(c)
y
= 3.
y = 5t - 82 • yield positive x and y,
Euler had his own method. Start with 5x + 7y Then x + This means Therefore, Now divide Hence
= 41 8 = (1
and divide by
5.
y - 2y)/5. that (1 - 2Y)/5 is an inte~er• we may write 2y - 1 = 5u, say. this equation by 2, and we have
y - 2u = (u + 1)/2. (u + 1) /2 is an integer, and we
arrive at u = 2t - 1. Substitute back to find y, and we have y = 5t - 2. Finally, substitute in the original equation, and we have x = 11 - 7t.
•• •• •• ••• ••• ••
12 Here only t = 1 yields non-negative
results, so x = 4, y = J.
(d)
One can use Farey sequences. This looks a little like the Pascal triangle. The n-th line in a Farey sequence consists of all the fractions, in ascending order, whose denominators do not exceed n. The first few lines there go as follows:
-01
1 1 2 1
0
1 0
1 1
! -1 2 J
1 0
1
1
7i
1
3
1 1
2
J
1 2
2
J
~
1
etc.
"1
There are two properties of Farey sequences that interest us. First, if the series contains three successive fractions, thus,
-ham - k
then
b n
~ + m=~ + n b
• We say that alb
is the mediant between h/k and min. This allows us to write out a line of the Farey sequence easily. The sec ond property is, that given alb between h/k and min, then hb - ka
= -1
and
an - bm
This gives us two solutions for For our problem, 517
= -1.
ax + by'
is in line
7
= 1.
of
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•• •
,. •• ;.• • •• •••
,. '. '.•••
''••.. • '.••• • ,.'.• •
•
•
•
the Farey sequence. In fact, we find that
follow each other, and we have
5(3) - 7(2) = 1 and 7(3) - 5(4) = 1.
We can now proceed as in Method (b), thus:
5( -4) + 7(J )
=1
=41
5(-164) + 7(123) 5x + 7y
= 41
5(-164 x) + 7(123 - y)
=0
164 + x = 1 123 - y 5 x = 7t - 164 y 123 - 5t Only t = 24 yields positive values for
=
x, y,
so x
= 4,
y
= 3.
Note: Using the other expression obtained
from the Farey sequence would give us
precisely the same result as Method (b).
(e) We come now to Continued Fractions. A continued fraction has the form E=a q
b
0
+-1 a
1
b2 +-
b
..::1
a2 + a
3
+
where each fraction is attached to the denominator of the previous fraction. An example would be 1 + 1
2+,6
J +
t +4 "5
14
To save space, we ap;ree to lower the signs, so p/q = 1 +
~
+
j
+
~
+
~
, for example.
For our purposes, we can let all the bi equal 1 and the a be positive integers. In i this Cl'lse, we agree to write the continued fraction in the form (aO:a1,a2,a3,···,an)· Such fractions are called simple continued fractions. Now any fraction can be rewritten as a simple continued fraction. For example, 7/5
=1
=1
+ 2/5
Consider now .E q
+ 1/(5/2)
=a0
+ 1
a
=1
+
~
+
~
•
1 1
+ a
1 + ••• + a n 2
If we drop all terms after a given one, we get what is called a partial convergent, written Pi/qi' Thus, polqo = aol1 ,P1/q 1 = (a Oa 1 + 1) /a 1 ' P2/q 2 = (a Oa 1a 2 + a O + a 2 )/(a1a 2 + 1) , etc. Here, PO P2
= a O'
= a Oa 1a 2
P1
= a Oa 1
+ 1, q1
+ a O + a 2 ' q2
= ~1a2
+ 1 , etc.
qo
=1
Someone noticed that
P2
= a 2P1
+ Po
q2
= a 2q 1
+ qo
=a 1
and, using mathematical induction, discovered Pi qi
= a i Pi -1 = a i q i-1
+ Pi - 2
+ qi-2
I
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15
ConsIderIng these as a pair of sImultaneous equatIons, let us elimInate the terms contaInIng a • Then l PI q 1-1 = a l P l - 1q l-1 + P I - 2q l-1
= a l P l - 1q l-1
ql P l - 1 Subtractln~,
we have
(P l q l-1 - Ql P l-1)
=-
(P I -1QI-2 - Ql-1 PI-2)
=• • •
(-1)2(P I _2Ql_J - QI-2PI-J) to (-1)1-1(P Qo 1 that Is,
+ P l - 1q l-2
untIl we get
= (-1)1-1(a Oa 1
- Q1 PO)
=
+ 1 - a a ) O1
= ± 1.
P Q QP 1 1-1 - 1 1-1
If we start wIth the fractIon a/b, then, when we wrIte thIs In contInued fractIon form, the last or ultImate convergent Is a/b. Therefore, the next-to-last or penultImate convergent, with a/b wIll, from the formula above, yIeld aql-1 - bP1 _ 1 =11. That Is, Ql-1' Pi-l are a solutIon of the equatIon ax + by
=1
,
due care bel~ taken wIth sIgns. For example, for our sample equatIon, we can wrIte
a
p
9
1
2 2 am. we see that 50) + 7(-2) = 1 • Now we can contInue as In method (b). We shall return to the matter of contInued fractIons In due time.
16
(f) We come at last to a method using congruence relations. It is a lot like Method (c). We have 5x + 7y = 41. Divide by 5, and. we get so
2y.: 1(mod 5)
=)(mod
y
5)
or
or
y
2y:= 6{nod 5),
= 5t
+ ). Now we
continue as in method (c). Now we can get 'tack to our check problem. We have
98y - 199x = 68.
Methods (a) am (b) are out - we can 't ~ess an answer (at least I can't). Method (c) is possible. Divide the equation by 98, am we have -
y - 2x -
)x +
68 98
)x + 68 = 98u.
from which
Divide this by),
x + 22 + ~ = )2u + ~
am we fim
)
fran which
2u - 2
am we have
u - 1
)
= )v,
say. Divide this by 2,
=v
'2v '
+
from which
v
= 2t.
Substituting successively in the previous equa tions, we finally get x y
= 98t + 10 = 199t + 21
Any value of t other thah t = 0, leads to an impossible number of cents. Therefore, the original check was for $10.21. We won't even consider method (d) • Getting the requisite terms of the Farey sequence would be too harrowing. Using method (e), we easily find that 199/98
= (21
)2, 1, 2). We find the\(onvergent~.
••• •• •• ••• •• ••• •• •• • ••• •• ••• •
••• •
•
•
•• ••
•• e
•• ••<.• ••• ••
•••
•
ie
,. •
•
•••••
•• <. •• ••• •• •• • •••
,.
•• •
17
a 2 32 1
2
am see that
p 2
q 1
65
32
67 \ 199
X
/33 98
=1
199(33) - 98(67)
The rest is just heavy arithmetic, using method (b).
Finally, we try method (f). We divide by am have
98,
-3x: 68(mod 98)
-3x ~ -30(mod 98)
x
=10(mod
98)
= 98t
x
+ 10.
Then we substitute in the original equation, am
98y - 98x199t - 1990
= 68,
The original check was for
or
y
= 199t
+ 21.
$10.21.
We em this with a simple problem, which can be done all six ways. On a certain planet, there are two inimical forms of life. The Septicapita have seven heads but only two lep;s each, while the Pentapods have only two heads but do have five lep;s each. One day an odd lot of Septicapita encountered an odd lot of Pentapods, am a wild melee ensued. Heads am legs were flying allover - one observer counted 180 of both together. How many of each type were involved in the fracas ? Note: Unfortunately, we must leave further work on continued fractions to another issue. We would like to recommend the reader to "Continued Fractions", by C.D.Olds. This is Volume No. 9 of the New Mathematical Llbrary, pu blished by the M. A. A.
18
MANY CilEERFUL FACTS It is sometimes useful, when given an equation, to derive another equation whose roots will be the sguares of the roots of the original one. Here is an easy method for doing this. We illustrate with a cubic equation, but the method is general. Let the cubic equation have the form: f(x) : ax) + bx 2 + cx + d : a(x-r )(x-r )(x-r)) : O. 1 2 Then fe-x) : ax J _bx 2 + cx - d : 0 will have rOots which are the negatives of the roots of the of the orig;inal equation. For, in that case, we have a(x + rt)(x + r 2 )(x + r)) : o. Now we find the product f(x).f(-x) : O. Writing; f(x) : a(x fe-x): a(x f(x).f(-x) Now merely
- rt)(x - r 2 )(x + r~)(x + r 2 )(x + : a (x 2 - r 12 )(x 2 replace x2 with
r)) : 0 r)) , we find - r 22 )(x 2 - r)2) : O. y, and we have
g(y) : a 2 (y - r 2 )(y - r 2 )(y - r)2) : O. t 2 Finding the product f(x) .f( -x) is easiest done by using detached coefficients. We illustrate with the cubic equation ) 2 . f(x) : x - 6x + tlx - 6 : 0 (roots 1,2,)). Then
and
~~f--t: ~~
~ - t4y2 + 491 - )6 : 0 has roots 1,4,9.
This process us used in Graeffe's Method of finding approximations to the roots of equations in NUlIlerical Analysis. Consult any text for more.
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19 ASSORTED PROBLEMS 10)
Show that, for all x € (0, n), the inequality rr+x) si~ n+x ( 1 + cos x ) sin x ~ ( 1 + co~ Finlam
11) {~} (k = 1, ••• ~) is a convex sequence orreal numbers, i. e •
~+ ~+2 ~. 2 ~+1
for (k=1,2, ••• ).
Prove that a
1
+ a
J
+ ••• + a
+ 2n 1
n + 1
Sweden 12) Prove that for every natural number k,
there exists a natural number ~ suck that ~ 2 + J Xk + 5 is divisible k by 15 • Bul~ria
Note:
Thus far, we have received solutions to our simple problems. There have been no solutions to the Olympiad Type problems. We would like to see solutions to these in time to acknowledge them in our May issue. Sem them to
20
..
,
KURSCHAK CORNER
Students who enjoyed Hungarian Problem Books I & II covering the famous Kurscha~ 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. 1/1954. Assume that AB + BD $ AC + CD in a convex quadrangle ABCD. Prove that AB < AC.
••• •• •• •• •• •• •• •• •
•
2/1954. Prove that if every planar section of a three- • dimensional solid is a circle, then the solid is a sphere.
4t
3/1954. Prove that in a round robin tournament (i.e. a • tournament in which each contestant is matched against • every other contestant) without ties, there must be a • contestant who will list all of his opponents when he 1is~ the ones whom he beat as well as the ones beaten by those. whom he beat. Dr. George Berzsenyi 2040 Chevy Chase Beaumont, Texas 77706
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21 VOLUME OF PYRAMID
Since we havo had occasion to make use of the rule for find ing the volume of a pyramid, am in order to make our discussion more nearly canplete, we have taken the liberty of addi~ the note below as one method for obtainin~ it. Below are two patterns. Usi~ the dimensions ~iven (or multiples of these), construct three patterns - one of the dia~ at the right am two of the diap;ram at the left. Fold the patterns thus made alon~ the dotted lines, am you have three pyramids. Two are co~uent am therefore have equal volumes. The third is symmtric to the others, am, fran Cavalieri's Theorem, has a volume equal to each of the two others. These three pyramids, however, can be assembled to form a trian~lar prism, whose volume is equal to its base multiplied by its altitude. Hence anyone of the pyramids will have its volume equal to one-third of the product of its base by its altitude Assembling the pyramids to form the prism is not really as difficult as solving Rubik's Cube.
/.
II ' \ \ I , I.e \
-I"''' I I
I
4.~
&~,e G.
I
•
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•
I
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\ \
I
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_~:/
CONTENTS Preface. • • • • • • • • • • • i
Prismatoid &: Prismoid. • • • • 1
• 9
Answers &: Questions. • • Diophantus in
Sprin~1me
••
10
Many Cheerful Facts. • • ••
18
Assorted Problems. • • • ••
19
Kurschak Corner • • • • • ••
20
Volume of Pyramid. • • • • •
21
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arbelos PRODUCED FOR PRECOLIJErrE PHILOMATHS
o
I.
No.5
May, 1983 Copyright ~ The Mathematical Association of America, 1983
i
EPILOCUE It was with a great deal of trepidation that we launched this small (as yet) effort almost a year ago. We were not certain we would have sufficient readership and support, and the start of the project was without publicity of any sort. We just selected the na.Iqes of a few students who, in our opinion, mi~ht be willin~ to assist us, and hoped for moderate success. Actually, the reaction was very p.;ood indeed! We have received solutions to problems, letters from readers, and requests for more. Therefore, we have decided to continue the production far a second year, subject to action on your part to our request far subscriptions far the year. We have done our best to provide articles on topics not usually studied in school, and we hope we have been successful. If any reader has an idea about some topic he or she would like to have us cover, please let us know. We admit we have had an ulterior motive in mind. Spring is the season far various contests, both local and regional. We hope these issues will help a contestant improve his ar her achievement in whatever contest comes up. We also hope that the student will be able to do better in class. We also welcome contributions from our readers. Have a pleasant Summer and do come 'tack Dr. Samuel L. Greitzer Mathematics DePartment Rut~rs University New Brunswick, NJ 08903
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e
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I. ;.i. I.
I.
i.I.
1
LOCI AND ENVELOPES We are accustomed by habit to see things in certain ways, ways which may inhibit our ability to solve problems. For example, consider the equation ax = b. Through custom, most of us wpu1d consider a and b to be constants, and determine that the equation locates a point x on the X-axis and equal to b/~. Suppose, however, that a and x are variables and b constant. Then the equation has a graph which is a hyperbola. And if all three letters represent variables, the ~aph of the equation is a quadric surface. With this in mind, let us consider the equation ax + by +
C
=0
a,b,c constants
c/o.
In this case, the graph consists of all the points on a line. The graph can be constructed in the usual way. The intercepts of the line will be
(-cIa, -c/b).
So far, there is nothing new.
Let us now divide the linear equation by and rewrite it as u
ux + vy + 1
= ale
a",d
v
=0 , = b/c .
c,
where
Then, if we allow x and y to be constants,
and u, v to be variables, we also have a linear
equation. Now, for every pair of values of u, v,
we get a line, and all these lines lie on a point. For example, take
1u + 2v + 1
= O.
Substitutin~ as usual far u and finding v, we have the following table of values:
~ vi 1/2 I -1 I -2 I
•• •• •• ••
2
If we prepare a the intercepts
~aph,
remembering that
(-cia, -c/b) = [-1/u. -1/v] ,
we see that the lines all lie on the point (1,2). That is, the equation
ux + vy + 1
=0
,
looked on as an equation in x, y, yields a locus of points
~
.! line. Looked on as an
equation in u, v, it yields an envelope of lines .Q!L.! point. There is an important and useful duality here, which we can use. The equation
x
222 + y =r , when graphed,
yields a locus of points on a circle. Let us see what happens whel) we change to u-v coord
inates.
We start from
solve tbis for
ux + vy + 1
= 0,
x, am substitute. Then
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••• •• ••• •• •• ••• •• .,• •• •• ••
••
•• • •• •• •• •• •• •• •• • •• •• •
3
,.
'. ',.. ',.,..•
•• •• •• •• •
x
-1 - vy =~----u
The two roots of this determine the two points in which the line ux + vy + 1 = 0 intersects the circle x2 + y2 = r 2 • Since we seek the tangent line, the discriminant of the quadratic must be zero. On computing this, we find that 2( 2 2 + 2 2 u r v r - 1) -- 0 , and since ufO, u 2 + v2 = 1/r2 • This is U
~phed
by
findin~
lines that satisfy the last
equation, and the result is the same circle as that determined by x2 + y2 = r 2 , but in the form of an envelope of lines. 2 2 For the circle u + v = 1/4 , we prepare the following table of values& u
.! 1/3
v
.!.373
First, remember to use ~phing,
.! 1/4
ne~ative
.! 1/6
reciprocals in
am second, note that each pair of
values gives us f2!:!! lines to ~aph, and we get the envelope shown on the next page.
4
One can see the circle determined by the lines. Point-and-line cooroinates can be used to derive many theorems in analytic geometry, affine ~ometry and projective ~eometry. We shall limit ourselves to two possible uses, however. One of these involves finding envelopes of lines and curves. One example, taken from our old Calculus, by Granville (circa 1911) goes as follows I Find the rectangular equation of the envelope of the straight line y=rnx+.E m
where the slope is the variable parameter. We first rewrite the equation in the form
m2
m ---p y + 1 = O. p x
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'.
'.
5 2
2 , and u = v p. This is the equation in line coordinates. Now we use the
=~ p'
=- ~p
v
Then
u
form
ux + vy + 1
=0
equation. This yields
to eliminate u from the v 2px + vy + 1 = O. And to
have a tangent, the discriminant must be zero, so y2 _ 4px
=0
is the envelope in point coordinates.
The reader might try his hand on the fo110wingl A line of constant length
a
moves so that its
endpoints are on the coordinate axes. Find the rectangular equation of the envelope. (The x2/ 3 + y2/3 = a 2/ 3 .) answer is A second application of the theory depends on the fact that we have dual relations between the locus and the envelope form of a curve. Thus, if two curves considered as loci intersect and have a point in common, then the curves considered as envelopes have one line: in common, or have a canmon tangent.
As an example, let us find the area of the
quadrilateral formed by the common tangents to
the circle x2 + y2 = 9 and the ellipse x2 + 8y2 = 16.
-
'.
I-
We have already determined that the circle x2 + y2 = 9 transforms into 9u 2 + 9v2 = 1. As for the general ellipse, b2x2 + a 2y2 = a 2 b2 , we eliminate
u
between this and
ux + vy + 1
= O.
6
As before,
u = -1 - vy , and. substituting, u
b 2 (1 + yy)2 + a 2 y 2u 2 = a2b2u2 • This yields. after sane al~ebra, the quadratic y2(b2y 2 + a 2u 2 ) + Y(2b2y) + (b2 _ a 2 b2u 2 ) = A~in.
o.
tangency means that the discriminant of
this equation must be zero. That meansl 2 a 2 b2u 2( a 2u 2 + b y 2 - 1) -- 0 • a nd since neither a, b. nor u can be zero, the line 2 2 2 2 equation for the ellipse is a u + b y = 1. We will be worki~ with the ellipse x 2 + 8y2= 16 or x 2 /16 + y2/2 = 1. Therefore a 2 = 16, b2 = 2. The dual equation is 16u2 + 2y2 = 1. Solving 16u2 + 2y2 = 1 2 2 9u + 9v = 1 we fim that !U = four tangent
.Ix .!
:!Y = 3,f2 • Thus there are
l1~es,
y =
3-{2
which form a square enclosing the two The area of this square equals graph appears as followsl
36
fi~es.
units. The
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7 We should note that there are restrictions in this development - that UfO , v f 0, for example. This restriction is removed by assumin~ a line "at infinity", and usinp; the equation xu + yv + zw = 0 Now we have homogeneous coordinates, and can do much more. However, we would be encroachinp; on the domain of projective p;eometry. We add here some problems that are usually by means of calculus.
~olved
1)
Find, in point coordinates, the equation of the line 4u
joinin~
5v
+ 2
the points
=0 =0
5u + 6v 1 2) Determine the envelope of all ellipses whose axes lie on the coordinate axes, whose centers are at the orip;in, and which have equal areas. (Area of ellipse b2x 2 + a 2y 2
= a 2b2
is
nab.)
2) Find the equations, in point coordinates, of the canmon tanp;ents to the circle 2x 2+2y2=1 and the parabola
y2 = 4x.
But he opened out the h1~6
Pushed and pulled the jo1nts and h1~s
T111 1t lo~ed all squares and oblon~
L1ke a c.-pl1cated t1~
In the Second B~ of Euclid. C.L.D.
8
DERANGEMENTS There is a puzzle that has been p;oi~ round for sane 270 years, in various forms. It was first proposed by Montmort in 1713 am called for selecti~ numbered balls or ticke-ts so that the ball numbered r should not be the r-th ball selected. The reader has undoubtedly seen variants of this puzzle - for example, placinp; enclosures in envelopes so that no envelope will contain the correct enclosure, or havinp; a hatcheck p;1rl distribute n hats so that no person gets the right hat. Your writer does not know hoW' Montmort solved the puzzle, but would like to present some ways of doing so. A monumental work titled "Canbinatory Analysis" by P.A. MacMahon has the following solution: let + a + ••• + an am fim the coefficient 2 1 of a a •••a n in (X - a )(x - a )· •• (X - an). 2 1 2 1 Note that the element from the first parenthesis X= a
can't be
a
can't be
a
1
, that from the secom parenthesis
, etc. In the expansion, the r-th 2 term contributes (n) expressions equal to r
a a ••• a n ' so the solution is 1 2 ~n) _ (n) + (n) _ + (_1)n(n) o 1 2· n • A secom method makes use of the Principle of Inclusion am Exclusion. This principle occurs in logic, probability am combinatorics. Briefly, it goes somewhat like this: the number of total derangements will equal the number of arrangements minus the number of those in which one element is in ita' place plus the number in which two are in place minus the number in which three are in place, am so 00.
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9
Symbolllcally, if
N(~)
arranp;ements in which N(aj~)
~
is the number of is in its proper place,
is the number of arran~ments in which
a j am ~ are in place, am if Dn equals the total number of derangements, then we have Dn = N(a a 2 ···an ) - (Na 1 + Na2 + ••• + N~n) + 1 (Na a 2 + N ala) + ••• + Na n_1a n ) 1 (Na 1a 2a) + Na 1a 2a 4 + ••• + Nan_2an_lan) + etc. am on calculating each of these, we get Dn = (~) - (~) + (~) - (~) -
which comes to
t! - ... ).
Dn = n!(l - 1 +~! - 1! + Incidentally, the series in parentheses above is the start of the series for l/e (e=2.71828 ••• ) so that, as n increases, the probability of having D derangements is approximately l/e, and n more independent of the value of n. is more am Naturally, Euler had his own method. for solving the puzzle. His reasoning was sanewhat as follows I Suppose that element a is in spot one. Then i the number of derangements among the rest of the elements is Dn- 2' and because i can take on n-l values, then, for this case, the partial total is (n - 1)D n_2 ,provided a is in spot i. 1 However, if a is not in spot i, then we have 1 (n - 1) elements a 1 , a 2 , ""~i-l' a i +1 , ••• , an to deal with, so that, in this case, the partial total will be
(n - l)D
nderangements is therefore
l ' The total number of D = (n-l)(D 1 + D 2)' n nn
10
Now it happens that Euler's form far the solution can be rewritten aSI (D n - nD n- 1) + (D n- 1 - (n-1)D n-~ ~) = o. Therefare,by reduci~ the subscripts again and again, one finally reaches the simple form I D = nO n
n-
1 + (_1)n.
This is a recursion formula, it is true, am. one can get to the solution by continually depressing subscripts, as befare. However, we can solve it by using a subterfuge (trick ?) Let
D = n! v n
n! v
n
n
am. substitute. Then
= n! v _ + (_1)n n 1
Division yields
v n - v n-1 =
Now reduce subscripts by unity until one gets to v - v ' say. We have a telescoping series, am o 1 we fim that v=11+1_1+ n - I! 2! )! ••• am. finally, Dn = n!(1 - 1 +!! - 1!+ • • .) Note I We cannot help but womer how Archimedes would have solved the problem! After all, he did. solve a very complicated Pellian Equation am he did not even have our number system to wark with.
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11 QUESTIONS AND ANSWERS 1) A man walkinp; across a railroad bridge had reached a point three-eighths of the way along when he heard the Orient Express behind him and coming at him at sixty miles per hour. As a devoted reader of Arbe1os, he immediately figured that he could save himself by runni~ to either end of the bridge. How fast can he run? 2)
How many times does 2 appear as a factor in the first positive integer that is greater than
(4 + 2~)100 ?
J)
A locker room in the form of a rectanp;u1ar parallelepiped is 15 feet lonp;, 12 feet wide and 9 feet hig;h. There are two 1ip;hts set flush with the cei1inp; and an athlete is standinp; under each light. One of the two is three inches taller than the other and casts a shadow 6 feet lonp;. The other man's shadow is 7 feet long;. How far apart are the two lights? What's the name of the ~e ?
4)
In our fip;ure of the Arbe10s (see cover of issue No.2) the common tangents to the small semicircles, XY and CT, intersect at O. If the sum of the diameters of the small circles remains 10 centimeters, but the radii of the small circles are permitted to vary, what is the locus of the point 0 ? Whenever I see a number, I usually do some arithmetic with it (like factoring). This is a common addiction. I have just seen a four d igit number such that (a) all the d igits are different, and (b) if the first two dip;its be added to the last two digits and the sum squared, the result is the original number. What is the original number ?
12 NEWTON'S POLYGON There are problems for the solution of which it is useful to have a graph of a function bein~ considered. Of course, one can assume the functlQn to be polynomial in form, substitute values for x, find corresponding values for y, and thus p;et a graph as a locus. Sometimes, one can find the graph as an envelope.However, this sometimes turns out to be complicated and time7consuminP-, especially if all one needs is a sketch. There is a method of obtaininp; a sketch in many cases. Historically, it p;oes back to Newton and his method of fluxions, which he used in the development of calculus his way. Considerations of space make it impossible to present a complete explanation, and we hope that our presentation will be fairly acceptable. Consider the circle at the right, with diameter AB, tangent AD and secant ~D. As C "flows" along arc CA toward A, the ratio BC/AB "flows" 5 toward 1. From similarity considerations, so does the ratio CA/AD. Now both CA and AD become progressively smaller. Nevertheless, we say that they are infinitesimals of the same order, unity. Now CD/AD = AC/AB becomes small D as C "flows" toward A, so we say that CD is an infinitesimal of the second order with respect to AB. We can find infinitesimals of increasing order - for example, by d.rawing a perpendicular from C to AD. The point is that there exist infinitesimals of different orders. Moreover, the ratio between an infinitesimal of order m and one of order n equals zero, a constant or infinity according as m is less than, equal to, or greater than n.
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'<.••. • ,.•• ,.•• •• •• •• •• •• •• •• •• • ••• ••• ••
'.
Still historically, Leibnitz used infinitesi mals in his version of calculus. The differences were in notation, in the tmplied use of a ttme factor and, most tmportant, in the fact that where Leibnitz, and modern calculus, retains all infinitestmals until the end of the process of findinK a derivative, Newtonians discarded all infinitesimals that would not affect the result as they arose. With this distinction in mind, one might write out the equation which we wish to sketch, divide each term by x and see whether y/x can equal a constan then divide by x again to see whether y/x can equal a constant, etc. However, there is a very stmple (and surprising) method that accomplishes. the same result.
z'
Graph the exponents of each term of the equation on graph paper, join these points so as to form a convex-polygon, and then examine those terms that lie on the sides of the polygon. These terms will yield approximations to the shape of the graph. What is more, if a side of this polygon faces the origin, the terms will yield an approx tmation to the shape of the graph at the origin. If a side faces the "point at infinity", the terms :will yie1d an approxtmation to the shape of the ~aph "near infinity". This polygon is called the "Newton Polygon", naturally. Let us take, as an example, y5 - 4xy2 + x5 = o. Note that, when x and yare replaced by -y and -x, the equation does not change. This means that the graph has central symmetry.
14
We now ~aph the exponents. Point A corresponds to the tem ;S. Point B corresponds to to
4xy2. Point C corresponds x5 • We ignore signs and
c oeffic ients •
~--'----,.-
Fran side AB, we now get y5 _ 4~J2 = 0 as an approximation at or near the origin. We get
y2= 0 and
? - 4x
= 0 as
approximations. The first is rather rough, the second is a cubical parabola, lying in the first and third quadrants. Let'c call this I Fran side Be, we find x5 = 4xy2 , or x = 0 and x4 = 4y2. This last represents two parabolas symmetric with the Y-axis, one lyin~ in the first and second quadrants, the other in the third and fourth quadrants. We will call these II. Fran side AC, we have x5 + y5 = 0, which can give us only x + y = 0 , as the shape of the graph near infinity. We call this III. We put these parts all together in the graph belowl ---...------~--
v
- -I
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''..
15 The line
x + y
=0
is an asymptote.
Let us now examine a well-known curve, the Folium of Descartes. This has the equation x3 + ~ = 3axy (a positive).
A corresponds to the term the term
3axy, and
C
Correspoming to side This gives us y y2
= ax
=0
y3,
B corresponds to to the term x 3 •
AB, we have
y3 = 3axy.
as an approximation, but
as a better one. This is a parabola
symmetric with the X-axis and lying to the right of the Y-axis. Side Be gives us x3 = 3axy, so 2 that x = 0 i& an approximation, but x = 3ay a better one. This is & parabola lying above the X-axis and symmetric with the Y-axis • Finally, side AC yields x 3 + y3 = 0, and since this equals (x + y)(x 2 - xy + y2) = 0, a first approximation would be the line
x+y
= O.
We can, however, get a better second. approximation.
16 Let us rewrite the ori~inal equation as: + _ 3axy x y - 2 2 x - xy + y am use our first approximation,
= -x,
y
to
see what happens. We p;et
x +
so a
~tter
y ~
2
-3ax
2
_
2 - -a,
2
x + x + x approximation near infinity would be x +
Y+
a
= O.
Now we can sketch our curve.
~~
V
~~ I
"\ .........
/ J'
"
h
1/
V
~ 1\
~
~
Next, we examine the Strophoid, whose equation is
or
ay? - Xy? = ax? + x 3 •
We note that the ~ph
will be symmetric with the X-axis. Next, we graph exponents.
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17
.!
From side X-axis at
x
CD, we find ulat the graph cuts the
=0
(twice) and x:: -a. From side 2 2 AI', we note that ax = ay or.! x :: ! y. Side AB gives us gives us
x:: a as an asymptote, and side BC x(x 2 + y2) :: 0, for nothing useful. The
graph is shown above. Of course, it makes sense to use any other data that might help in sketching a curve. For example, look for symmetry relations. If it is easy, one mi11;ht differentiate to find local maxima and minima. Look for re11;ions where the ~aph cannot occur (we used to call them sign lines). The aim is to sketch a curve so as to enable one to use it to solve a problem. Finally, you might try your hand at, i) (x-a)(x-b)y2 - a 2x 2 = 0: a > b > 0 i1) y4 _ x4 _ 96y2 + 100 x 2 = O. (This is Problem No.4 of the Eleventh Putnam Exam).
18
MANY CHEERFUL FACTS
Stewart's Theorem I In a triangle, a line segment like AP, going from a vertex to a point on the opposite side, is called a Cevian. The theorem states I If a Cevian ~ divides side Be = a into segments of lenp;th m, n (as shown above), then b2m + c 2n = t 2a + mna
C
A simple proof using the Law of Cosines iSI 2 2 c = m2 + t - 2mt cos e b2 = n2 + t 2 + 2nt cos e and eliminate the tems containin~ the cosine tem. One can see how hardy this theorem would be for findin~ the lenp;th of a median, an~le bisector, altitude, etc. Ceva's Theoreml I~ too well known to necessitate a proof, of which there are many. It statesl If Cevians AP, BQ, CR are concurrent at a ......._-....,;;:.. c
point 0, then AR BP Qg RB x 'pC x QA = +1 , am conversely. We can see how Ceva's Theorem could be used to show that the medians, altitudes, or anp;le bisectors in a triangle are concurrent. Remember that the segments are all directed line segments. If one remembers this, one can use the the~rem for any position of the point O. ~_
,
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,.
,.'.
19 Menelaus' Theorem: This is as well-known as is Ceva, am there are many proofs far this as well. It says: If the sides of a triangle are cut by a transversal
(see diagram), then
AR x BP x .Qg
= -1
QA • Again, the segments are directed line se"nents, and. one m'.lst take care of signs. RB
PC
Desargues' Theorem: This states, If correspondin~ vertices of two trian~les lie on three concurrent lines, then correspoming sides of the trian~les meet in three collinear points, am conversely. For a proof, please refer to the cover dia~am: AB, A'B' lie on plane OA'B'. Hence they intersect .t P. Also, P lies on planes ABC and. A'B'C'. Be, B'C' lie on plane OB'C'. Hence they intersect at Q. Also, Q lies on planes ABC and. A'B'C'. AC, A'C' lie on plane OA'C'. Hence they intersect at R. Also R lies on planes ABC am A'B 'C' .. Since two planes intersect in a line, P, Q, R are collinear. This is a theorem in Projective Geometry, where all lines intersect. Moreover, it is not necess arily true on the plane - not unless one adds an axiom of continuity ar admits commutativity. This makes many of the "proofs" we hav€ seen suspect especially those involving analytic geometry. Note also that this thearem is self-dual. If one restates it, replacing point by line, line by point, concurrency by collinearity am collinear ity by concurrence, one gets the same theorem.
,
20
ASSORTED PROBLEMS 13) Find all integers
n < 1 for which n In + 4 + • • • + (n+l)n + (n+2)n = (n+J)n.
(U.S.S.R.) 14) A finite set of unit circles is ~iven on a plane, such that the area of their union U is
S. Prove that there exists
a subset of mutually disjoint circles such that the area of their union is
~ ~a t er th an 9 S .
(Yu~oslavia)
15)
2 Let f(x) = ax + bx + c and 2 g(x) = cx + bx + a. One knows that If(O)1 ::: 1,
If(l)1 ~
Prove that for
Ixl
i) lr(x)1
~
t
•
1 , 'f(-l)' < 1 •
< 1 • 11)
1~(x)1 ~
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2 •
(Vietnam) Congratulations to Mark Kantrowitz and to David Moews for their solutions to various probl~ms. We are especially interested in the "Assorted Problems" which are actually among those proposed by the countries indicated. Our apolo~ies for the error in Asserted Problem No.7. What was wanted was a 5-digit number.
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21 "
,
KURSCHAK CORNER
· .Students who enjoyed Hungarian Problem Books I & II , Ikovering the famous KUrschdk Contest (formerly named .fter Eotvos) for the years 1894-1928, should welcome
ttthe challenge provided by this competition in more
If.ecent years. They should set aside an uninterrupted
, tt4-hour period to compose complete well-written solutions, extensions, generalizations, etc. to at least some of · ~hese problems, and submit their work (enclosing a ,~tamped, self-addressed envelope) to the address given · ~elow. Every solution will be thoroughly evaluated and , .ach respondent will receive a set of instructive solu tfions to the problems posed. I
~/1961. Consider the six distances determined by four
~oints in a plane. Prove that the quotient of the largest , ~f these distances to the smallest of them can not be less · .han 12 .
1
•
Prove that if a,b, and c are positive numbers, each less than 1, then the products (l-a)b, (l-b)c, and , 4t(1-c)a can not all be greater than 1/4.
iill961.
•
· ./1961. Given two circles, exterior to one another, along ~ith a common inner and a common outer tangent. The re tfulting points of tangency define a chord in each of the '.ircles. Prove that the point of intersection of these ~wo chords (or extensions thereof) is collinear with the ttenters of the given circles.
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Dr. George Berzsenyi 2040 Chevy Chase Beaumont, Texas 77706
CONTENTS
Frontispiece - Desarp;ues' Theorem i
Epllop;ue
1.
Loci am Envelopes
8•
Deran~ments
11.
Questions am Answers
12.
Newton •s Polygon
18.
Many Cheerful Facts
20.
Assorted Problems
21.
Kurschak Corner
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