Produced For
PreCollege Philomaths
1984 -by 1985
Professor Samuel L. Qreitzer 1906-1988
Volume 3
Number 1 · 5
Second Edition Original Copyright © 1984
Second Edition Copyright © 1997
The American Mathematics Competitions
Professor Samuel L. Qreltzer in Memoria During the interval of time between 1982-1987 Professor Samuel L. Greitzer served as the editor and author of essentially all of the 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 Greitzer emigrated to the United States from Odessa, Russia in 1906. He graduated from the City College of New York in 1927 and earned his Ph.D. degree at Yeshiva University. He had more than 27 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 Revisited with H.S.M. Coxeter. I was extremely pleased that Professor Greitzer agreed to write and edit the Arbelos, since I frequently receive requests for references to pUblications which are appropriate for superior students, and for material which will help students prepare for the USA Mathematical Olympiad. Professor Greitzer served as a coach of the summer Mathematical Olympiad training program from 1974 to 198.3. Consequently, many of the articles in the Arbelos are a reflection of his lectures and thus appropriate for talented and gifted students. Professors Greitzer and Murray 5. Klamkin accompanied the USA team to the International Mathematical Olympiad from 1974 (the first year the USA participated) to 198.3. 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 Greitzer to the development of students of mathematics and teachers from many nations will be lasting. We shall miss his humor, words of wisdom, mathematical insight and friendship.
Dr. Walter E. Mientka Executive Director American Mathematics Competitions University of Nebraska-Lincoin
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Preface To The Second Edition
1984 - 1985 During the five year period from 1982 through 1987 Professor Greitzer authored 25 regular issues of the ARBELOS. This Volume :3 contains the issues published from 1984 - 1985. The tradition of Professor Greitzer to present a geometry problem on the cover of each of the issues of the ARBELOS led to a challenge to the readers for their solution. I was extremely pleased that he agreed to prepare a solutions manuscript and they appear in Volume 6 of the ARBELOS series. The Second Edition essentially remains the same as the First. However, it does reflect the use of current computer technology versus the original edition which was completed on a manual typewriter. In addition, page headings are included as well as the inclusion of one Table of Contents. These changes were made in a noteworthy manner by Mrs. Amy Fisher, a member of the American Mathematics Competitions Lincoln office staff. Your comments and suggestions are welcomed. Dr. Walter E. Mientka
Executive Director American Mathematics Competitions 1740 Vine Street University of Nebraska @ Lincoln Lincoln, NE USA 68588-0658
e-mail:
[email protected] -i
ICONTENTS Of THE ARBELOS, VOLUME Volume 3, Chapter 1* Cover Problem Preface to The First Edition Echoes of Summer The Binomial Theorem~Etc The Gamma Function Many Cheerful Facts Undetermined Coefficients Conjugate Coordinate Geometry Olympiad Problems
31 1-27 1 2 3 6 11 13 16 19 27
Volume 3, Chapter 2· * Cover Problem Preface to The First Edition Cayley v.s. Hamilton Week-Enders A Look at Inequalities Beware of the Obvious Olympiad Problems Partitions Kurschak Korner
29 59 29 30 31 39 40 45 52 53 59
Volume 3, Chapter 3 *. * Cover Problem Preface to The First Edition Transformations in Geometry Kurschak Korner Factoring is Fun The Cubic Equation Many Cheerful Facts Olympiad Problems
61-88 61 62 63 69 70 77 85 88
•Originally published as the ArbeJas Volume 3, Issue 1, September, 1984 "Originally published as the ArbeJas Volume 3, Issue 2, November, 1984 •••Originally published as the ArbeJas Volume 3, Issue 3, Januroy, 1985 -ii
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Volume 3, Chapter 4 * Cover Problem Preface to The First Edition From Our Readers The Parabola and the Right Angle ManyO Cheerful Facts Concurrence ~ Discovery and Invention A Combinatoric Formula Problems for Beginners Olympiad Type Problems Dimensional Analysis Kurschak Korner
Volume 3, Chapter 5** Cover Problem Preface to The First Edition Trigonometry Geometric Algebra Euler's Formula Term-Enders Math is Fun Kurschak Korner Olympiad Problems
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*Originally published as the Arbelos Volume 3, Issue 4, March, 1985 * *Originally published as the Arbelos Volume 3, Issue 5, May, 1985
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VOUJMII3, CllAl'1U 1
arbelos
Produced For PreCollege Philomaths
P
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A
Btl.
tl 2
C
A chain of circles is inscribed in the Arbelos, each tangent to the Arb~los and to the preceding circle, From the center of each circle, let a perpendicular 0iHj be constructed to line AC. Let d j be the diameter of circle
OJ' Prove that perpendicular OjH j is i x dj'
(For example, 02H2
1984-1985:
=
2 PQ).
No. 1
September, 1984
-1
ARBELOS
VOLliI'm:5, CHAI'mR 1
PREFACE It seems incredible that ARBELOS is starting its third year. We hope to continue our present policies on the basis of the numerous letters we have received from readers. First there were those who felt that no changes were needed - they were satisfied. There were some who felt that perhaps some applications could also he included. Oddly enough, there were a number who were against the inclusion of calculus. Now we have not done so yet but we do believe that if any problem can be most easily solved by calculus, then we should use it. At any rate, we plan to continue as before, if you - the readers - wish it. We will sneak in one or two applications and use a spot of calculus if that makes for a simple solution. At the June meeting of ARML, we were asked about the origin and meaning of ARBELOS. As the diagram on the cover shows, the word is Greek for a shoemaker's knife. Just leave out the diameter ABC. It was a favorite diagram for Archimedes, who worked out a number of properties of the diagram, which have appeared in previous issues. The cover illustrates another ARBELOS property.
Dr. Samuel L. Greitzer Mathematics Department Rutgers University New Brunswick, NJ 08903 -2
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ECHOES Of' SUMMER
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Usually, beginning in June, we get problems of varying age and difficulty presented to us. For once, this year, we did not get our usual angle trisection solution. This normally takes the form of an unholy mess of lines and arcs, accompanied by the claim, "I have trisected an angle ! Show me where I'm wrong." However, we did get a few "golden oldies", which we think might be of interest to our readers. The first came to us from Professor Leon Bankoff. He is too good a mathematician to have been stuck by this one, but his question puzzled me, too. Consider the diagram at the right. In the circle, M is the midpoint of chord PQ. AB and CD) are chords that intersect PQ at X and Y. Prove that PX = QY. This is well known as the Butterfly Problem. If you have not come on it yet, try to prove it on your own. Professor Bankoff merely wanted to know when it had acquired the name, "Butterfly Problem" . I didn't know either.
D
There are many references to this problem in the literature. It is mentioned in Johnson's "Geometry" on p. 78. You will also find it in Eves', "Survey of
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ARBELOS
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Geometry", Vol. L p. 171. We also put it into our "Geometry Revisited", pp. 45-46. The simplest proof we have uses projective geometry. The second "oldie" came to us from Dimitrios Vathis, of Chalkis, Greece. We like best the name "Propeller Problem". In the diagram to the right, we have the hexagon ABCDEF inscribed in a circle. Chords BC, DE, FA have lengths equal to the radius of the B circle. P, Q, Rare the midpoints of the three segments AB, CD, EF. Prove that the triangle '. C with vertices P, Q, F' R is equilateral. Again, if you have not met up with this r . one, you should try to derive a proof on E your own. Of the many proofs, we again recommend Eves", "Survey of Geometry", Vol. II, p.184. Also, we have before us an article by Bankoff, Erdos, and Klamkin, entitled "The Asymmetric Propeller", which extends the theorem. Of course, Mr. Vathis has included his own proof. For the extended proof, consult "Mathematics Magazine", for November 1973, pp. 270-272. Incidentally, I suggest that Mr. Vathis consult The proof found in Eves' book.
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The third oldie came to us from Professor Meisters, of the University of Nebraska. In the diagram to the right ABC is an Isosceles triangle with vertex angle A = 20 . Cevian BP divides angle B into two angles of 60 and 20 as shown, and cevian CQ divides the angle at C into two angles of 50 and 30 as shown. Problem: Find the number of degrees in angle BPQ. We have included this in our "Geometry Revisited on p. 26. A very ingenious solution is to be found in, "Mathematical Gems II", by Honsberger, pp. 16-18. Again, try to solve this for yourself, if you have not met with it before.
A
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, p
/
X':,>\
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! \;y. Then we may rewrite it in the form xn(l + 1.)n . x
We can then
expand the parenthetic term and then multiply by x n since 1. < 1. x Before continuing with the subject, we should note that we can use the Binomial Theorem with fractional exponent to compute any root of a number. The reader might be amused to compute the cube root of 123. -8
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Let us now examine, say, (x+y+z)n, where n is a natural number. We can "group" Y+z and expand (x + y + z)n. However, for (x+Y+z+ ... +w)n this would obviously be quite complicated. Hence we examine (x+y+z)(x+y+zl(x+y+z) ... (x+Y+z), where we have n factors. Now to obtain a term in the expansion, we select one object from each factor. Here, we are into combinatorics. For the term xaybzc, for example, we select ax's, by's and c z's. This can be done in nl al bl cl ways, where a + b + c = n. Thus, given (x+y+z)5: if we want the coefficient of the term x 2 yz 2, this will equal
51 =30, and the 211!21
term is 30x 2 yz 2. We have not been able to find any references to expansion of powers by the Multinomial Theorem except for the little we have presented here. Yet such expansions do exist, although they must be quite complicated. We can do something with a few simpler cases. For example, given (a+bx+cx 2 )6. Let us develop in terms of a, b, c. Then the term involving, say, a 3 b 1 c 2 , would have coefficient -2L 3!l!2!
60 ·108· x 5 .
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In the development above, we have found one term involving x 5 . To find all such terms will take a bit longer. Let us try. Suppose we want the coefficient of x 5 in (l + 2x + 3x 2 )6.
Rewrite this as (a + bx + cx 2 )6. The general term of the expansion of this in terms of a.b,c is
where r+s+t=6 and s+2t=5. We may have s= 1. t=2, r=3; or
s=3. t= 1. r=2; or
s=5, t=O. r= 1. Therefore the coefficient of x 5 will be
It might have been easier to use the Binomial Theorem, but we wanted to exhibit this result. If one wanted to find the coefficient of x 5 in (1 + 2x3x2 )1/2, the use of the Binomial Theorem would be strongly recommended. In a short article, we have had no opportunity to examine either the Binomial or Multinomial Theorems for complex exponents. We apologize.
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This article can be considered as being just a continuation of our article on binomial coefficients, but it so interesting on its own as to merit a separate treatment. It seems that it struck Euler (of whom we have written before) that it might be possible (and fun) to construct an expression that would yield values for all factorials, not merely the positive integers. He found two - one an integral which we will not examine, and one as follows:
r(x)
j.
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I
=
lim
nln x - I (n x(x+l)(x+2)...(x+n+l)
~
00).
How he derived the two expressions makes a fascinating story, which can be found by referring to the English translation of his works. We will work with the limit form, where we shall assume the limit (n ~ 00) understood for ease in typing. First, let us replace x by x+ 1 in the expression above. Then r(x + 1) = lim
I
x
n.n (x+l)(x+2)...(x+n)
=
lim
I x-I n.n nx x(x+l)(x+2)...(x+n-l)(x+n)
=
lim
n.n . ....K... x(x+l)(x+2)...(x+n-l) ~+1
I x-I
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=
Now, lim
we nl
2·3·4... n
x r(x). can
easily
find
that
= I. Hence r(2) = Ir(I) = 1, -11
r(l)
equals
ARBELOS
VOWM!l3, CtfAPTIIR 1
r(3) = 2r(2) = 2!, r(4) = 3r(3) = 3!, etc.
Certainly, Euler's expression, called the Gamma
Function, does what he set out to have it do, for
any positive integer, r(n) = (n -I)!. We now manipulate the expression a bit. Let us replace x by 1 ~x. Then we have i(1-x) = lim
I
-x
n.n (1- x)(2 - x)(3 - x) ... (n - x)
•
• •
Multiply,
and we find that r(x)f'(1- x) is equal to
~n~!.!..!.n~!n~-_I
lim
_
2 x(1- x 2 )(4 - x 2 )(9 - x 2 )... (-n_-1 - x 2 )(n - x)
If we distribute the (n!)2 in the denominator, we get lim
._1_ 1 x2 x2 x 2) ... 1+ x
x(1- -)(1-)(1n
12 22 32
At this point, we recall that, in the Arbelos of January, 1984, on page 11, we found that sin 1tX = 1tx(1- £)(1- £)(1- £) ....
2
1
22
3
2
Finally, in this last expression, let x= 1.. Then
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2
r(1.) r(1.)= 1t, or r(1.) = 2 2 2
~.
That is, we have the
value equivalent to factorial 1., and it involves 1t. 2
Note:
The interested reader can now evaluate the area under the normal probability curve. -12
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MANY CtlEERFUL fACTS We take the liberty of presenting three theorems which may be known to some readers, which have been used in solving problems in Olympiads and in the Putnam Exam, and which therefore merit presentation
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The first (in alphabetic order) is known as Beatty's Theorem. Simply stated, it asserts: Let a and ~ be two irrational numbers such that
l + 1 = 1. Now consider the sets Urn a)} and Un ~)},
ex ~ where m and n are natural numbers taking on the values 1,2,3.... Then every natural number will occur in just one of the two sets, with no repetition. We first met this theorem as a problem in Niven and Zuckerman's, Introduction to the Theory of Numbers, p.84, problem 23. It also appears in the USSR Olympiad Problem Book, p.26, problem 108, and there is a full discussion in Honsberger's Ingenuity in Mathematics, Volume 23 of the "New Mathematical Library", p.93 etc. The last two have proofs of the theorem. We suggest that the reader try to construct his or her own proof. Also, we suggest that the reader refer to one or more of the references.
It may be mentioned that Problem No.3 of the IMO for 1978 was easily solvable using Beatty's Theorem.
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The second theorem is known as Nelly's Theorem. We assume that the reader knows what a convex figure on the plane is . Let us call such a figure an oval. Then the theorem states: if each three ovals of a finite (or infinite) family of ovals have a common point, then all the ovals of the family have a common point. We first met this theorem in, Combinatorial Geometry in the Plane, by Nadwiger, Debrunner, and Klee. It is also in, Euclidean Geometry and Convexity, by Benson, p. 53, and there is a very fine exposition in Mathematical Time Exposures, by Schoenberg, Chapter 5. The first reference contains a number of allied theorems, including Sylvester's conjecture, If a finite set of points on a line is such that on the line determined by two of the points, there is always a third point of the set, then all the points lie on a line. It should be noted that this last dualizes to: If a finite set of lines is such that through the intersection point of any two of the lines, there passes a third line of the set, then all the lines are concurrent.
The reader will find several instances where Nelly's theorem has been used to solve problems on the Putnam exam. One such is Problem 3 of the Ninth Putnam Competition.
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The third theorem should be well-known, since it dates from 1899. Due to George Pick, it states: The area of a simple polygon whose vertices are lattice points is equal to the number of lattice points in the interior of the polygon plus half the number on the boundary minus 1. Or, K = i + lb -I. 2
We first met the theorem in, Mathematical Snapshots, by Steinhaus, p.94. An excellent development appears in Honsberger's, Ingenuity in Mathematics, pp. 27 - 30, and a short, snappy proof followed by examples in Coxeter's, Introduction to Geometry, p. 29. We recommend the proof, on p.211, of the theorem: If cevians be drawn to the trisection points of a triangle so as to enclose a triangle, the area of this triangle is one-seventh the .area of the entire triangle.
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UNDETERMINED COEffICIENTS The enterprising reader has undoubtedly met with undetermined coefficients, possibly in connection with partial fractions. We met them first at the start of the century, where they were used to develop series expressions for various functions. For example, we assume that: e X =A+Bx+Cx 2 +Dx 3 +Ex 4 + .... Let x=O, and we find that A= 1. Now differentiate the series, and we have: e X =B+2Cx+3Dx 2 +4Ex 3 .... Let x=O, and we find that B= 1. Differentiate the last series and we find: e X =2C+3x2Dx+4x3Ex 2 + .... Let x=O, and we find that C=.l!. 2
Similarly, D= 1-! 3
E= 1-!, etc. This gives us the usual 4
eX
2
.3
4
2!
3!
4!
= 1+~+1L+1L+1L+ ... 1
If you want to use calculus, there is an interesting development of Arctan x that you might try. Another application is to deriving Simpson's Rule. Suppose we are given a sequence of ordinates, Yj in terms of which we wish to develop a formula for the area under the curve joining the upper ends of this set of ordinates. We assume a rule in the form K=AYO+BYI+CY2 and we try to evaluate A, B, C. If there is such a formula, it should work when the curve is Y= 1 - that is, when Yo = YI = Y2 = 1. In this case, let the distance between ordinates be h. -16-
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Then the area is 2h=A+B+C. (see figure a)
y=l 1
Again, the formula should work when y=x. In this case, (see figure b) the area is 2h 2 =O+Bh+2Ch.
h
h (0)
Finally, the formula should work when y=x 2 . In this case (see Figure c), Yo=O, YI =h 2 'Y2=4h 2 . Hence we get 8h 3
3
y=x
= 0 + Bh 2 4ch 2 . h
Note: we remind you that we have already shown that the area
h
h
(b)
under the parabola is l. 3
that of the circumscribing rectangle. Solving the three equations A+B+C=2h Bh+2Ch=2h 2 Bh 2 +4Ch 2 =
8h 3 3
-
lh h
(c)
we arrive at A= h, B= 4h , C= h. That is, our formula 333
is K = h(yo + 4YI + Y2)' 3
This is Simpson's Rule. To
use it, divide the x-axis Into an even number of equal pieces, and find the area piece by piece. -17
ARBELOS
VOLUMl\ 3, ClIAI"mR 1
For a third application, consider a chemical reaction. We wish to write a balanced equation for this reaction. We know the valences of the elements involved (we must know something I), and will use the deep postulate: Whatever you put in will come out! We start with Aluminum (AI) and hydrochloric acid (HCl). We note that hydrogen comes off (H 2 ) and that aluminum chloride is left (A 1C 13 ), Write A(Al)+B(HCl)=C(AICI 3 )+0(H 2 ). We try to find A, B, C, O. Now for each element, we find: A=A for AI: for H:
A = C} B = 20
B=3A so that
for CI: B = 3C
C=A 0= 3A 2
Let A=2, and we have A=2, B=6, C=2, 0=3.
The
balanced equation is: 2AI+6HCI=2AICI 3 +3H 2 Let us try to balance the following reaction: ACu+BHN03=CCu(N03)2+0H20+ENO. Going by elements, we have (Cu) A=C; (H) B=20 (N) B=2C+E (0) 3B=6C+0+E. Solving in terms of E, we find A= 3E 2
B=4E
C= 3E
0=2E
2
E=E.
Now let E=2, to get integer coefficients, and we have: 3Cu + 8HN0 3 = 3 Cu(N0 3 )2 + 4H 20 + 2 NO. This will balance any equation without charges. -18-
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CONJUGATE COORDINATE GEOMETRY
Weare all acquainted with p the diagram (due to Wessel) im g showing the complex number z=x + iy graphically. It is also
usual to use the facts that
r y x=r cos 8 and y=r sin 8 to write z=r(cos 8+i sin 8). It is also usual to write z = x - iy . "-_8 ---'-_r:-:e::..;:a=1 Then z= r(cos 8-i sin 8). 0 x It struck several early mathematicians, among them Gauss, that one could express x and y in terms of z and In that case, we would have Y = z-z .
2 2i One could then express a relation in x and y in the form of another relation in terms of z and z. In a number of cases, this procedure makes it easier to derive results in coordinate geometry. We might note, for example, that zz= r 2 . Let us start with the linear equation y=mx+b. We x
= z+z
substitute and find z-z 2i
,. ,.i.i.
•
we arrive at
= m z-z + b.
On solving this for
2
z=
I-mi z - 2bi . First, the coefficient of l+mi l+mi z is not a slope. If we note that, in the diagram above, m=tan 8, we find I-mi l+mi
= l-itanS = cosS-isinS l+itanS
cosS+isinS
Now cos 8+i sin 8 = e i8 , cos 8-i sin 8 = e i8 . Hence, in the
linear equation above, the coefficient of z is e-2i8. In some texts, this is called a clinant. Let us represent it by
M. -19
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If we consider the segment or to represent a vector, which is often done, then, given two points z 1 and zz' then z 1+zz is the diagonal of the parallelogram of which z 1 and Zz are adjoining sides, and zZ - zl is the line segment joining the other vertices of that parallelogram. Moreover, every point on the line on z 1 and Zz may be written in the form z = zl + t(zz - zl) or z number.
= (1- t)zl + zz.
Here, t is a real
However, consider the determinant
This vanishes when Z=Z 1 and when z=zz' and is linear. Therefore this determinant, equated to zero, is also the equation of the line on zl and zz. By dint of a little algebraic manipulation, we obtain -
z = (Zt-Z2 )z + (Zt-Z2 - Z2Zt ) Zt- Z2 Zt- Z2 Comparing the two forms we have for the equation of a line, we may write this last z=Mz+B. Again, M is not a slope. When we solve M = (l-mi) for m we find m = 1. I-M. . (l+mi) , i l+M Now suppose we have intersecting lines with slopes m 1 and m z . Let the angle between the lines be (J). Then tan
(J)
m = (m2- t). (1+mtm2)
We will not show the
intermediate algebra, but just -20
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show the result:
Vowm 3,
tan co = i
CtfAPTER
2
1
(MI-M ). (Ml+M 2)
From this, it
follows that the two lines are parallel if M1 = M2 , and will be perpendicular if MI
= -M2 .
We can always proceed as we have thus far -start with an equation in x and y, replace with z and and thus arrive at a rule in Gaussian coordinates. Unless we can find something new and useful, it does not seem practical to do so. We will try to proceed further.
z,
r--
~b
First, we detach ourselves B from the origin. Given a line segment with endpoints a and b, suppose we select a segment with end point on the origin and at (b-a). Then 0 u======P~_ this line and the line with endpoints a and b form a parallelogram. Hence the line or is equal and parallel to the line AB in the diagram. Next, consider two triangles whose Guassian coordinates are zl' z2' z3' and u l , u 2 ' u 3 ·
If we refer back to the u J - u 1
triangles with one vertex at the origin, we get the configuration shown in the diagram at the right. Now assume that these two triangles are similar. Then
is z2 - ZI = Ile I; U2 - ul
= kqe i9 2;
z3 - zl
= qe i(8 1+a)
u3 - ul
= kqe i(9 2+ tan 45 = I, and we are done. (of course, 2
the sharp reader will have noticed that this is an example of the inequality (J! + .Q.) > 2, shown on the first page of this article.) b a Jensen's Inequality (see diagram) states that, if f(x) is convex, then f(a)+f(b) ~ f( a+b). 2 2 How to attack a given inequality depends on the form of the inequality and the skill of the solver. To show ~ -t- ~"+ ... + an-l + ~ > n, one may use the Arithmetic a2 a3 an al Geometric means, and note that the product of the terms of the (.!l+ a2 +...+.!n.) series is 1. Then a2 a3 al > n
prove that the series 1+ multiply by
,J;;
~, and we have it. However, to
-k + }; + ··· + ~ > ,J;;
I
all one does is
and note that each fraction (except the last)
exceeds unity. And to show that J... + _1_ + _1_ + ... + _1_ < 1., one 23
33
43
n3
4
could use mathematical induction. We could continue in this manner to show that there are many and varied ways of solving a problem. We will close with finding the minimum value of f
= x2 -
12x + 40 . -42
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We could use calculus, of course, but, instead note that f=x 2 -12x+36+4=(x-6)2+ 4 , which is obviously a minimum when x=6. We know we have a minimum without getting a second derivative. Finally, let us apply what we have developed to the problem of finding maximum and minimum values for f=xy+yz+zx-2xyz. (x+y+z=l). (The original problem included the limits, to which we objected.) First, let us change f to a homogeneous function by multiplication by x+y+z=l. We have f = (x + Y + z)(xy + yz + zx - 2xyz, or f = xyz + x 2 (y + z) + y2(z + x) + z2(x + y). Now we were told that x, y, Z were non~negative. Therefore each term in the above equals or exceeds 0, and the lower limit is f> O. The limit is attained when x= I, y=z=O, for example. Now let us examine (l-x)(l-y)(l-z)=g. On multiplying, we find that g = (xy + yz + zx) - xyz. Hence g = f+xyz. Now the sum of the factors of g equals 2, a constant. Therefore g is a maximum when the three factors are equaL that is, when x = y = z = 1.. 3
But then we get .J!... ~ f + _1 , or f ~.L. Equality 27 27 27 To summarize, .L ~ f ~ O. 3 27 This problem was No.1 on the 1984 IMO. It should not have been. occurs when x = y
= z < 1..
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Let us try a problem involving conjugate coordinate geometry (see September, 1984 issue.) We are given f(z) == azz + bz + bz + c, where a, care real numbers and b is complex. We wish to show that, if f(z) ~ 0, then ac ~ bb. We can work with g(z) == zz + bz + bz + c since this means division by a. We shall show c ~ bb. Consider (z + b)(z + b) = O. This is a circle with center at b and radius 0 - actually a point. This yields zz + bz + bZ + bb = 0 or zz + bz + bz = -bb . Therefore g(z) = -bb + c ~ 0, whence
c ~ bb. When
we return to f(z), we have ac ~ bb. The suggested solution has one return to complex numbers, i.e. z=x+iy, b=m+in, etc. It is much longer than the solution above. We recommend that those interested in inequalities will do well to read Introduction to Inequalities, by Beckenbach and Bellman, Volume .3 of the New Math Library, and "Geometric Inequalities" by Kazarinoff, Volume 4 of the New Math Library. I
Problem: 4 Show that (a + b + c)4 < 27(a + b 4 where a, b,c are real numbers.
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BEWARE Of' THE OBVIOUSI
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Imagination is most important in problem solving, but there are occasions when one must beware of the obvious. The history of mathematics has many cases where fine mathematicians jumped to hasty conclusions \:Vhich were later proved erroneous. We wish at this point to exhibit two cases from official examinations involving thousand of secondary school students that came to our attention.
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The first (in time) involved the following question: The discriminant of a quadratic equation with integer coefficients equals 23. What is the nature of the roots? We know the expected answer; the roots are real, irrational an unequal. But wait! The discriminant has the form b 2 -4ac. now 4ac is divisible by 4 with no remainder. Also, if b is even and equal to 2n, b 2 =4n 2 which is divisible by 4 with no remainder. In this case, therefore the discriminant is divisible by 4. Also, if b is odd and equal to 2m+ 1, then b 2 equals 4m 2 +4m+ 1, which leaves one when divided by 4. In every case, therefore, either the discriminant is exactly divisible by 4 or leaves a remainder of 1 when divided by 4. However when 23 is divided by 4, the remainder is 3. It follows that a quadratic equation with integer coefficients can not have 23 as ~ discriminant. We hope this surprised some readers, at least. The second example is geometrical in content: ·45·
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A
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In the diagram, A-BCDE is a square pyramid whose lateral faces are equilateral triangles. K-FGH is a regular tetrahedron each of whose faces are congruent with the lateral faces of the square pyramid. If face FGH is made to coincide with face AED, how many faces has the resulting solid? Again the answer is obvious. Both solids h~ve a total of five plus four faces, two faces disappear when merged. This leaves seven faces for the solid that is produced. Right? Wrong!
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Here is the solid we have created. Let each edge be denoted by s. Let M be the midpoint of edge AD. Then lines CM, EM and KM, being altitudes of their triangles, are perpendicular to AD at M,
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E
C
~.J3
M
2
K
~.J3 2
Now all lines perpendicular to a line at the same point lie in a plane perpendicular to that line. Thus, MC, ME, MK lie in a plane. Using the Law of Cosines in triangles MCE and MKE, 2s 2 s2
= .J.4 s 2 + .J. s 2 - .J. s 2 cos 42
= .J. s 2 + .J. s 2 -
x.
.J. s 2 cos y.
442
Subtraction yields s2 = 3s 2 (cosy-cosx). 2
Addition yields 0 = cos y + cos x. But cos(x + y) = cos x cos y -sin x sin y. From cosy +cosx = 0 cosy - cos x = 2 3
we find
cos y=.1
and
3
sin y= 2..]2
3
cos x= _.1 and sinx= 2..]2 3 3 When we SUbstitute in the formula for cos (x+y), we find cos(x + y)= _1. -.§. 9
9
= -1,
which means
that x+y=180. CMK is a straight line lying in face ACDK, which is not two faces, but one! In the same manner, we find that ABEK is not two faces, but one. Finally, the solid really has five faces, not seven. -47
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H
G
Once one has a solution to a problem, one often sees more ways to look at it. In this case, suppose we start with a cube (see diagram above) and join the centers of the six faces properly. One derives a regular octahedron, the dual of the cube. This is easy to show. For example, AH is half diagonal HAF, HD is half of equal diagonal HG, AD is half of diagonal FG, etc. So that HADE is a regular tetrahedron. Similarly, A-BCDE is a square pyramid with lateral faces equilateral triangles. Now, since A, C, D, H lie on plane FGH, face ACDH is planar, and so on. Lest the reader assume that this sort of lapse in intuition versus correctness does not occur in the case of professional mathematicians we show two additional examples where intuition has been at fault even for experts. Our first involves the Japanese mathematician Kakeya. In 1917, he proposed the following problem: Let AB be a unit segment in a plane. It is required to move AB in the plane so that A and B exchange places and so that segment AB sweeps out the minimum area. -48
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There seems to be a minimal area involved. We can rotate AB through a semicircle about A and then slide it down. The area covered by AB is then
.n.. Or we can rotate AB 2
about its midpoint through an angle of 180 and slide it to the desired position. The area covered by AB is then .K
4
A
Now Kakeya noted that the hypocycloid of three cusps,
produced when a circle with radius L rolls without :3
slipping inside a circle with radius r (see diagram) has
the property that a tangent to one part of the curve and
cut off by the other two arcs is constant. Therefore a line
one unit long, slid along this hypocycloid , will end up
with endpoints reversed after one complete pass.
Moreover, we know (calculus) that the area of the
hypocycloid is equal to .K. 8
Since the tangent always
remains constant while being moved, Kakeya conjectured that this was the minimum sought. Therefore, there was astonishment and surprise when Besicovitch proved that there was no minimum area that is, whatever area the line sweeps out, it is possible to sweep a lesser area while reversing the direction of the line. For more on this problem, consult, Mathematical Time Exposures, by I.J. Schoenberg, published by the Mathematical Association of America. There is also a film showing how this can be done, prepared by the M.A.A.
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Back in 1803, the Italian mathematician Malfatti proposed the problem which, intuitively, reduced to constructing three circles in a given triangle so that the sum of the areas of the circles would be a maximum. Malfatti thought he had proved that all one had to do was to construct three triangles as in the diagram, each circle being tangent to two sides of tl1e triangle and also tangent to the other two circles (see the figure above).
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Then Steiner, probably the best geometer of the time, gave a construction in 1826, without proof. Hall supplied elegant proofs in 1856. In 1929 (note the time spreads I), Lob and Richmond showed that, in this case, intuition was wrong ..
Even for the equilateral triangle, computation showed that the inscribed circle, with two small circles squeezed in, as shown at the right, above, yielded a greater area than three equal circles, as shown at the left, above. -50
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Moreover, it is easy to see (now that our suspicions have been aroused) that, in an isosceles triangle with a very short base and long sides, three circles placed one above the other must yield a far greater area that the Malfatti construction could yield.
Thus, so far, the problem of Malfatti has not yet been solved. The solution seems to depend on the shape of the triangle involved. A similar problem involving placing four spheres within a tetrahedron so that the cut out a maximum volume is also unsolved. Since an affine projection produces an equilateral triangle, it might be fun to compare the areas in the cases of the two situations shown in the diagram on the previous page. A very nice description of Malfatti's problem may be found in, A Survey of Geometry, Vol. II, by Howard Eves, In fact, this is an excellent text on geometry.
Moral:
Don't trust on intuition, all the time!
Shortie: We know that a right triangle cannot have integer sides, but still be isosceles. However, a 3-4-5 right triangle is nearly isosceles, and so is a 20-21-29 right triangle. Find the next largest nearly isosceles right triangle.
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OLYMPIAD PROBLEMS
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1.
Prove that for every natural number n, the number ?n) divides the least common multiple of the numbers
n
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1.2,... ,(2n -1),2n. Note: ?n)is a binomial coefficient.
n (Bulgaria)
2.
Suppose that al,a2,a3 that
(X-al)(x....:.a2)
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(X-a2n)+(-1)n-l(nf)2 =0 has
tit
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(Canada) Determine all pairs of positive real numbers (a, b) with a
* 1 such that
loga b < loga+l(b + 1) . (U.S.A.)
4.
5.
e
,a2nare distinct integers such
an integer solution r. Show that r = a l +a 2 +...+a 2n
3.
e e e
The circles C1(i = 1,2,3) are externally tangent to each other. The centers of the circles are at the vertices of triangle ABC. Find the centers of all circles C which intersect each of the circles Ci at endpoints of a diameter of Ci . (Finland) Prove that the product of five consecutive positive integers cannot be a perfect square. (Great Britain)
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PARTITIONS A partition of a positive integer n is a separation of n into positive integers whose sum is n. Thus, 1+2+3 is a partition of 6. In some cases, we require partitions to be distinct. Thus, 2+1+3 and 1+2+3 are not considered distinct and, in our study, one would be discarded.
Incidentally, in addition, we are given a set of integers and
seek to find their sum. In a partition, we are given the sum and seek Integers whose sum equals n. Thus, we consider partition to be the inverse of addition - not subtraction. First, if we allow non-distinct solutions, the number of partitions of the number n is easy to obtain. Assume that we . have produced a line of n dots. There are (n-l) spaces between them. We can either draw a separating line in any space or not. Therefore, the number of possible partitions is 2n-1. Let p(n) be defined as the number of ways in which the positive integer n can be written as the sum of positive integers. The first few p(n) are easy to evaluate. First, assume that p(O)= 1. Now p(I)=1. Next, 2=1+1 or 2, so p(2)= 2. Then, 3=1+1+1 or 1+2 or 3, so p(3)=3. By merely listing in this way, we find p(4)=S and p(S)=7, etc. To get a more efficient and useful way of evaluating p(n), we use generating functions (see Arbelos No.2, 1983-1984.) Note that (1- x)-l = 1 + x (1- x 2 )-1
+ x 2 + x.3 + x 4 + x 5 + ...
= 1+ x 2 + x 4 + x 6 + ...
(1- x.3)-1 = 1+ x.3
+ x 6 + ...
(1-
x 4 )-1 = 1+ x 4 + .
(1-
x 5 )-1 = 1+ x 5 + .. and so on. -53
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Then the coefficient of x n in the product of all these series equals p(n). Using the expressions given on the previous page, we multiply to get the first few terms of the product and find 2
4
5
II = I + x + 2x + 3x 3 + Sx + 7x ... from which we get p(O)=I, p(I)=I, p(2)=2, p(3)=3, p(4)=S, p(S)=7. We exhibit the partitions of 5 below: l I l l I , 1112, 122, 131,23, 14,5. Leave it to Euler (whom we have mentioned before) to derive a recursion formula for p(n). It looks formidable, but appearances are deceiving. p(n)
= ~(_I)i+lp(n -
.1 (3i 2 + i)) +
2 ~(_I)i+lp(n - .1 (3i 2 - i)). 2
From this formula, for the first 20 or so values of p(n), we have p(n)=p(n-I )+p(n-2)-p(n-5)-p(n-7)+ p(n-12) + p(n-IS)-.... Some theorems involving partitions are easy to derive. For example, let us denote a partition into unequal parts by U(n), a partition such that every part is odd by O(n), and one in which every part is even by E(n). Then
= (I + n)(1 + n 2 )(1 + n 3 ) ... O(n) = (1- n)-I(I_ n 3 )-I(I_ n 5 )-1 ...
U(n)
Now U(n)(I- n)(I- n 2 )... = (1- n 2 )(I- n 4 )(I- n 6 )... On dividing by (1- n)(I- n 2 )(I_ n 3 ) ... , we note that the factors containing even powers of n in both numerator and denominator cancel leaving the denominator equal to O(n). So we have: The number of partitions with no parts equal equals the number of partitions with all parts odd.
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For example, U(7)=7, 6+ 1, 5+2, 4+3, 1+2+4. 0(7)=7, 5+1+1, 3+3+1, 3+1+1+1, 1+1+1+1+1+1+1, and U(7)=O(7)=5. A partition can be represented in the form of an array of dots called a Ferrers graph. For example, we can represent the partition 1+2+4 by
• • • • • • • By reflecting this graph about the diagonal on the upper left hand, we obtain a second graph,
• • • • • • • From a partition with three parts (4,2,1) we have arrived at a partition in which the greatest part is three (3,2,1, 1). The theorem shown is thus: The number of partitions of n into m parts is equal to the number of partitions in which the greatest part is
m. The use of the Ferrers graph appears to be an example of "backsliding" to the days when figurate numbers were fashionable. Recall that the ancients arranged dots into triangles, squares, etc., to get the triangular numbers
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and square numbers
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or I, 4, 9, 16. Moreover, they did quite a lot with these figurate numbers. Merely by drawing a line, for example, they proved that the sum of two successive triangular numbers is a square number. -55
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Also, by adding a row and column (at the right and at the top) of every square number, one gets the next higher square number. The part added was called a "gnomon". Now, starting from just one dot and noting that the number of dots in each gnomon added is an odd number, one derives the fact that the sum of the odd numbers from one on equals a square, or 1+3=4, 1+3+5=9, 1+3+5+ 7= 16, etc. The rythagoreans did quite a lot of mathematics with these dot figures. . To continue with partitions: We define a perfect partition as being one such that, for a given n, every partition less than or equal to n can be derived in just one way. Thus, for p(7), we find that 1,1,1, 1, 1,1, 1 1, 1, 1,4 1,2,4 and 1,2,2,2 are perfect partitions. Note that, in each case, one can derive subsets whose sums are all the values from 1 through 7. One application of this result is that one has a set of values, for each partition, which yield all the values from 1 through 7. This solves the problem of finding weights with which one can find the weight of pieces weighing from one unit to seven units. Of these, the one using the least number of weights is the set 1, 2, 4. The theory of partitions has been applied in various ways. Among these are graph theory, the study of mathematical "trees", networks, as well as applications to chemistry and to physics.
It has been proved that p(5n + 4) =O(mod 5) and that p(7n + 5) =O(mod 7). Find a similar congruence for modulus 35. -56
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Another application of partitions is to the determination of the number of solutions of the linear Diophantine equation alxl + a2 x 2 + a3 x 3 + ... amx m = n. This involves finding the number of partitions of n into parts al,a2' ... a m . Sylvester called this the ~~denumerant" and wrote it it the form: n
D(n;al,a2, ... a m ) = ID(n;al,a2' ... am)t . The generating function for this would then be: (1- tal)-I(I_ t a2)-I(I_ t a3)-1 .... In this, we would find the coefficient of t n for a given n. Unfortunately, using this form involves a lot of careful counting, and it is easy for mistakes to occur. Euler used a recursion process, but this is still usually too complicated and subject to error. For example, let n=20, a l =2, a 2 =5. Then D(20 ;2,5)= (1- t 2 )-I(I_ t 5 )-I. When we expand we find that the coefficient of t 2 o=3. Hence there are three solutions. However, it seems more desirable to obtain these solutions, rather than merely the number. Suppose we wish to find out the number of ways of changing a dollar into dimes and quarters. We seek solutions, therefore, to the equation IOx+25y 100, or 2x+5y=20. Again, we are at D(20:2,5). In this case, we prefer to use the methods for solving Diophantine equations described in Arbelos, No.4, March, 1983. We find that x=5k-40 and y=20-2k. Positive solutions exist when 8 ~ k ~ 10. Not only do we know that there are three solutions, for k= 8, 9 and 10, but we know what these are. -57
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These are four quarters and no dimes, two quarters and five dimes, and no quarters and ten dimes. Next, how about changing a dollar using nickels, dimes and quarters? Now we have Sx+ 1Oy+2Sz= 100 to solve. Of course, we divide by S to get x+2y+Sz=20. It is simplest to consider five cases - z=O or z= 1, or z=2, or z=3, or Z=4.
When z=O, we have x + 2y = 20. Here, we can, by inspection, find eleven solutions. These are (20,0), (18, 1), (1 6,2), (14,3), (12,4), (1 0,S), (8 ,6 ), (6,7), (~, ,8 ), (2,9) and (0,10). When z= 1, we solve x+2y= IS, we easily find that there are eight solutions. These are (1S,O), (13,1), (11,2), (9,3), (7,4), (S,S), (3,6) and (1,7). There are eight solutions. When z=2, we must solve x+2y= 1O. There are six solutions. These are (10,0), (8,1),(6,2), (4,3), (2,4) and (O,S). When z=3, we solve x+2y=S. There are three solutions, namely (1,2), (3,1) and (S,O). Finally, when z=4, there is the one solution, namely (0,0). Counting, we find that there 29 solutions, and what is more, we know what these solutions are. There are variations, which can be taken care of individually. For example, in the above problem, suppose we are constrained to use all three kinds of coin ? What if we have only eight nickels? As a problem, using either the generating function or the method given above, show that the number of ways one may change a dollar into coins, using cents, nickels, dimes, quarters and half-dollars, is 292. Note:
For an excellent reference, consult Riordan's Combinatorial Analysis. ·58·
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KURSCliAK KORNER This column features those problems of the famous Hungarian Kurschak Contest which have not yet appeared in English translation. For the problems of 1894-1928, the reader should consult Hungarian Problem Books I &. II in the MAA's New Mathematical Library series (available from MAA hdqrs); the problems (with brief summaries of their solutions) covering the years from 1966-1981 were translated into English by Professor Csarmaz of Hungary in 1982. Student readers are uninterrupted 4-hour well-written solutions submit their work to critical evaluation. 1/1939
also invited to set aside an period to compose complete, to the problems below, and to the address given below for a
Let aI' a z , b I' b z , c I and C z be real numbers for which alaz > 0, alcl ~ bIz, azcz ~ b z z Prove that (al+az)(cI+cz)~(bl+bz)z.
2/1939
What is the highest power of 2 which divides 2 n ! evenly?
3/1939
Assume that MBC is a given acute triangle and that C I,A I and B I are points on the semicircles constructed outwardly on the sides AB, BC and CA, respectively. Show that these points can be chosen so that ABI=AC I , BAI=BC I AND CAI=CB I . Dr. George Berzsenyi Rose-Hulman Institute of Technology Terre Haute, IN 47803 -59
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On one side of line segment XV, similar triangles XYA, XYB, XYC, XYD, XYE, and XYF are constructed.
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Prove that points, A, B, C, 0, E and F line on the same cirlce.
1984-1985:
January, 1985
No. :3
-61
ARBeLOS
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PREfACE We begin this issue with our first article to be submitted by a reader. We hope it will be followed by more such articles. We did have problems sent in by Dimitrios Vathis, of Greece. Now we have an article by Professor Szeckeres, of Australia.
Weare happy to have the Kurschak Problems. They show what sort of problems Hungarian students are expected to be able to do. They are very good practice for the International Mathematical Olympiad. Our Olympiad Problems come from the proposed problems sent in by various nations for possible inclusion in the official Olympiad, but which have not been selected. After all, each year, there are many problems proposed, and these are quite good in their own right.
Again, because the preparation of students varies across the country, we may include subject matter that some of you find too easy, too hard, or too familiar.
Dr. Samuel L. Greitzer Mathematics Department Rutgers University New Brunswick, NJ 08903
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TRANSFORMATIONS IN GEOMETRY Many geometrical problems lend themselves to be solved by a transformation. We can often achieve by a transformation that a difficult problem becomes easy or an unwieldy solution is replaced by an elegant one. There are many well-known geometrical transformations: we shall discuss a few simple ones. a) Parallel translation: The following problem will illuminate the method: P
Figure 1
In a given circle two chords, AB and CD are given. Find a point P on the circle such that if PA, PB cut CD in X respectively and Y, then XY =d, a given distance. Solution: Move the distance XY until X coincides with A, and Y is in a new position. Then (Fig. 1) yy 1 is parallel to AP, so LY 1YB = LAPB, which we know, being the angle subtended by the chord AB in the given circle, So Y is a point on a circular arc which subtends the angle APB to the chord Y 1B. This arc will cut CD in 2, 1 or 0 points, giving the required solution. b) Reflection: We are given a straight line, L called the axis of reflection. We replace every point, P, by its mirror image in I. Every configuration -63
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goes into a congruent- configuration, but is described in an opposite sense: clockwise changes into counterclockwise direction. We say the figures are indirectly or oppositely congruent. Problem: Given three concurrent lines, 11' 1 2 , 1 3 and a point A on 11' Find B on 1 2 and C on 1 3 such that the three given lines should be the internal angle bisectors of the MBC. Solution: If 1 2 bisects the LABC then the line BC is the reflection on 1 2 of the line AB. That means that if we construct A', the mirror image of A on 1 2 (Fig.2), then A' is a point on BC or its extension. Similarly, we construct A", the mirror image of A in the line 1 3 , Then A' A" is the line BC, and cuts 1 2 in B,I 3 inC.
A
C A'
B
£1 Figure 2
c) Rotation: A figure is rotated around a given point by a given angle, thereby remaining congruent to itself. Problem: Given three parallel lines, 11' 1 2 , 1 3 , and a point A on 11' Find points B on 1 2 and C on 1 3 such that the triangle ABC is equilateral. -64
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C
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B
A Figure 3
Solution: Let B i run along 1 2 and imagine the points C i such that BiAC i is an equilateral triangle. Then it is easily seen that C i describes a line 1, some perpendicular distance from A as 1 2 , but rotated around A by 60° (Fig. 3). Then 1 intersects 1 3 in C and AC rotated by 60° meets 1 2 in B. There are two solutions, a clockwise and a counter clockwise equilateral triangle. d) Similarity Transformation: All linear dimensions of a figure are changed by the same proportion. We distinguish between central similarity and spiral similarity. Central similarity means that the lines connecting corresponding points in two similar figures all pass through a single point, which is called the center of similarity. It is not difficult to prove that if any two similar figures are drawn so that corresponding sides are parallel to each other, then it is a case of central similarity. Problem: Let ABCD be a quadrilateral with AB parallel to CD but BC intersecting AD in P. Show that the circumcircles of ~PCD and of ~PBA are tangent to each other (Fig.4). -65
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P
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Solution: Let Q be the circimcenter of ~ PBA and Q' the circumcenter of ~ PCD. These two triangles are centrally symmetric with the center P; therefore the line joining the corresponding points Q and Q' also passes through P. But P is a common point of these two circumcircles, if P is on the central line, there cannot be any other common point Le., the circles are tangent to each other. Spiral similarity. If two similar figures do not have their corresponding sides parallel, then there exists a point in the plane which can serve as the center of some rotation so as to bring one of the figures into a position of central similarity to the other.
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Problem: Given two circles, with centers 1and 02 intersecting at points A and B. Construct a line through A, cutting the circles in P, resp. Q, such that PA = AQ.
°
Solution: Draw the triangles 102B and PQB (Fig. 5). Then LAPB=L0 1 0 2 B and LAQB=L0 1 0 2 B. Therefore, the two triangles are similar for every position of the line PAQ. -66
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B Figure 5
We want the position where BA is a median. Let x
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be the midpoint of 1 02; then the L 1BP should be made equal to L XBA to achieve the required position. In conclusion, here are a few problems to be solved, using some of the above transformations: 1) Given two circles C l' C 2 and a straight line 1, with a distance AB marked on it; Find a point X on C 1 and Y on C 2 such that XY is parallel to -
and XY
= AB.
2) Given a river with parallel banks, distance d
apart, and points A and B on opposite sides of the river. Show where to build a bridge XY, perpendicular to the riverbank, so that the route AXYB should be the shortest possible. 3) We construct a square on each side of a given parallelogram. Show that the centers of these squares themselves form a square.
G
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4) C I and C2 are two circles, tangent externally at point T. Through T we draw two lines, meet the circles at AI' A 2 , respectively at B I , B 2 . Let the incenter ~AI TB I be II that of +A2 TB 2 be 12 , Show that the line I 112 passes through T. Prof. Esther Szekeres 94 Warragal Road Turramurra, NSW 2074 Australia 1te versus e 1t A fairly well-known problem is to compare 1t e and e 1t and determine which is greater. We offer three solutions: a)
Arithmetic: 10g1te =elog1t=2.718x.4972=1.3514. 10ge 1t = 1t loge = 3. 142x.4243 = 1.3646.
e X >I+x.
Let
x = (1t -1).
Then
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Algebra:
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Calculus: Let f = x x and take logs. Let y=ln f. Then
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I
In= 1, or x=e. Hence e ~ > 1t;, whence e 1t > 1t e . There are other ways of solving this inequality
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•• I, for example, we are, in effect enlarging the graph magnification, so to speak. When O(m). If this holds for any column it holds for all the terms in that column. We may discard the other columns, if we like. Also, in any column, there are (n) terms relatively prime to n. This holds for all columns. If we count, we find that there are (m) (n) terms left, and this must equal (mn). . a(l) a(2) a(i) Now, given the number N = PI P2 ... Pi where we have different primes, then the (N) will equal the product of the individual (pt). From this it is easy to derive Euler's formula (N) = N(I- .l)(I_~)... (1- .l). PI P2 Pi Incidentally, we can also show that, if a and mare relatively prime, then a(m) == I(modm). This is usually simpler to use than is Fermat's Theorem, -86
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(p prime).
Incidentally, if N = pap b ... p t, then the sum of the 1 2 i divisors of N may be found by using the generating function
a b t a(N) = (l + PI + ... + P )(1 + P2 + '" + P )... (1 + Pi + ... + P ) 1 2 i The number of these divisors is seen to equal 't(N) = (l + a)(1 + b) ... (1 + t). Of course, we could have used our special notation to define a(n) 't(n)
= L din d
and
= L din 1.
A function like (m), where (mn)= (m) (n)when m and n are relatively prime is called a multiplicative function. Under the same restriction, it happens that (n) and 't(n) are also multiplicative functions. The reader might try to prove that, when m and n are relatively prime, then (mn)= (m) (n) and that 't(mn)= 't(m) 't(n). There are some pretty problems that are readily solved by using these functions. For example: (a) Find all solutions (if any) of (n)=IO. (b) Compute (985) (c) If 't(n) is odd, then n is a perfect square. In a future issue, we hope to introduce the Mobius function Il(m) and apply it to multiplicative functions. ·87·
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OLYMPIAD PROBLEMS I) Determine all real values of the parameter a for
which the equation
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16x - ax + (2a + 17)x - ax + 16 = has exactly four distinct real roots which form a geometric progression. (Bulgaria)
2) Find the point inside the triangle ABC for which BC + CA + AB PD PE PF is a minimum, where PD, PE, PF are the perpendiculars from P to BC, CA, AB respectively. (Gt. Britain)
3) Prove that if a, b, C are integers and a+b+c=O, then 2(a 4 +b 4 +c 4 ) is the square of an integer. (USSR)
4) Prove: In any parallelepiped , the sum of the lengths of the edges is less than or equal to twice the sum of the lengths of the four diagonals. (Netherlands)
5) Let the set {I, 2, ... , 49} be divided into three subsets. Prove that at least one of these subsets contains three distinct integers a, b, c such that a+b=c. (Mongolia) -88
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B Given Triangle ABC: Construct line segment DE so that BD=DE=EC.
1984-1985:
March, 1985
No.4
-89
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PREFACE It is our policy to include, in each issue of Arbelos, some mathematical morsels that are new to our readers. Considering the great variety in the preparation of readers, this is not an easy task. In addition to the Arbelos, we take every opportunity to visit schools to see what is being done and what might possibly make for an article. In one school we visited, we found that the students had been taught the theorem of Menelaus, but not Ceva's Theorem. The teachers had made their decision of what would be more useful. We disagreed and presented the theorem, with proof, to the class. In another class in another schooL the students had no idea that the binomial theorem was correct for any real exponent. All they knew was the Pascal Triangle. Moreover, they had no idea that there might be a multinomial theorem. We found two trends that were somewhat disturbing. One was the rush toward introducing calculus to. students unprepared for it, as a result making the whole development a matter of rote. The other was what we consider an undue dependence on the use of hand calculators and computers, at the expense of understanding. We once attended a lecture on teaching algebra by hand calculator at Columbia. The lecturer was most enthusiastic about the results. One of the stories he told about the results was that about one student who got all the right answers. When asked how he happened to be so good, he answered, "It's only buttons man - only buttons". And there was the boy on TV who was at a computer. He told the interviewer that he had studied the computer for six years, and that he was four and a half years old. So much for his understanding of numbers.
To go into more advanced study of any kind, first prepare
yourself by learning the preliminaries first.
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fROM OUR READERS The letters we receive from some of our readers are enlightening and informative. We would like to get more. Mark Kantrowitz has been a good correspondent to date. We can see the improvement in his problem solving ability. His solutions to the problems from the November issue of Arbelos were quite good . We wish him success on the Westinghouse. We have received, for the first tine, solutions to some of the problems that are worth mentioning. C. Kenneth Fan submitted two good solutions. David Stephenson did nicely with his solutions, and
Japheth Wood also did well. We hope to hear further
from them and others as well. We read (in the Mathematical Digest for October, 1984), an analysis by Dimitris Vathis of a paragraph from von Daniken's "Signs of the Gods." He clearly pointed out where von Daniken had confused names and dates, and ascribed abilities that did not exist in some of the people he named. There are probably still some people who believe in von Daniken just as there those who accept the "explanations" that apply to the Bermuda Triangle, for example. They prefer to accept what is asserted instead of testing for accuracy. Just one note - those who wish to improve their knowledge in any field should do a lot of reading, examine what they read critically, and draw their own conclusions from what they read. We ourselves like to read old texts in mathematics, for example. For instance, we have an old (1890) arithmetic book used in the eighth grade then, containing problems that would glaze most people's eyes. It's fun! -91 ..
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THE PARABOLA AND THE RIGHT ANGLE by Isaac Ji KUO, 7th grade MCKinley Middle Magnet School Baton Rouge, LA
One day I looked at a corner. Hmmm. I looked at it again. I really began to wonder: how would it look if those walls were mirrors? Pretty funny, I guessed. Well, not really. I began painting a picture in my mind. This picture is shown in Fig. 1.
Fig. 2
Fig. 1
"That reminds me about a parabola!" I said right aloud. (Good thing no one heard me). OK, so I had a pretty wild imagination. But I still had somewhat of a point. Think of it this way: a ray of light, which is perpendicular to directrix, hits a parabolic mirror and goes right through the focus. It hits the mirror again and goes away parallel to its original path. "So what ?", you may be thinking. Well, Fig. 2 shows the resultant picture in my head. Amazing, I thought in my mind, when a chord passes through the focus in a parabola. the tangents of its endpoints are -92
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perpendicular!! I was quite excited about L~ discovery. So I went to Dad. I asked him if he COUlL prove it. After almost no time, he proved it. He said he used CALCULUS. I did not know any calculus at the time. When I gave him my proof, he was quite surprised. In fact, so surprised that he asked me to write this article. And now, for anyone wondering how I prove.d it all, here it is:
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2x+A+2y+B= 360°
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A+B= 180°
3.
Hence 2x+2y= 180° and so x+y= 90° and so C= 90°
Note: The article above was sent to us by Professor Hui-Hsiung Kuo, who has every reason to be proud of it. The boy's teacher had suggested having the Arbelos publish it! This makes us proud! We have attained the status of a "prestigious" journal! If any reader has an idea of his own that he or she is willing to submit for publication, we will do so after due consultation with a referee, as do all other IJprestigious" journals. -93
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MANY CHEERFUL FACTS Educationists are always at work. Recently, one has suggested that factoring of quadratic expressions no longer be taught. We happen to believe that factoring is fun, and that it would be wrong to keep students from the enjoyment it offers. The history of mathematics has many examples of great mathematicians relaxing with factoring. Thus, Gauss relates that he used to take time out to calculate a "chiliad" of prime numbers. Someone once sent Fermat a huge number for him to factor, if possible, and, by return maiL he sent back the factors of the number. And Euler showed that 2 32 _1 was composite. Factoring offers a lot. First, let us start with numbers and their factors. We all know that a number is divisible by 2 when the units digit is divisible by 2. Most of us know that a number is divisible by 4 when the number formed by the last two digits is divisible by 4. Similarly, a number is divisible by 8 when the number formed by the last three digits is divisible by 8, (etc.). We do not have to mention when a number is divisible by 5 or 10. Also, we know that a number is divisible by .3 when the sum of its digits is divisible by .3, and by 9 when the sum of the digits is divisible by 9. By combining these results, we are able to find out when a number is divisible by 2, .3, 4, 5, 6, 8, 9 and 10. Continuing, a number is divisible by 11 when the difference between the sum of its digits in the odd places and those in the even places is divisible by 11. We offer without proof a method for determining when a number is divisible by 7. Let that number be lOt + u. Subtract 2u from t. If the result is divisible by 7, so is the original number. One can practice this sort of factoring by trying to factor license numbers on passing cars for speed and accuracy. -94
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As something less elementary, consider the number 100 1. This is (obviously) divisible by 11, and (less obviously) divisible by 7 and 13. If this number be subtracted from (or added to) a given number, and if the result is divisible by 7 or 11 or 13, so is the original number. As an example, take 1183 we find that 1183-1001=182, and that 18-4=14. Hence 1183 is divisible by 7. It happens that 182 is also
divisible by 13, so that 1183 is divisible by 13. We
can now find factors for all numbers whose factors
include any number from 2 through 16.
a) Factor 1, 111 , 111 .
b) Invent a rule for finding when 19 is a factor of a
given number. If the reader thumbs through previous issues of Arbelos, he or she will find, in articles involving number theory and congruencies, more methods for factoring numbers. We continue with quadratic forms and their factors. Suppose we use x 2 -6x-247. When we subtract 4x247=988 from 1001, we see that the remainder is 13. This gives us 247= 13x 19, so x 2 -6x-247=(x - 19)(x + 13). We hope you will agree that just guessing at the factors of 247 might be somewhat time-consuming. We might try removing the second term of the expression. The work looks like this: 1 1 -
6
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x 2 -256=(x-16)(x+16).
whence we have On adding back what was subtracted, we get (x+13)(x-19). -95
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x 2 - 6x - 247
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= x2 -
6x + 9 - 9 - 247
or
(x-3?-256, or (x-3+16)(x-3-16), or, finally (x+13)(x-19). On the other hand, given x 2 - 6x - 616, one can alter the expression by dividing the second term by 2 and the constant term by 4. This gives us x 2 - 3x - 154, which is easy to factor. We get (x-14)(x+ll), so that, originally, we must have had (x-28)(x+22). Once again using what has been presented in other issues of Arbelos, suppose we start with 5x 2 +26x+24. We "multiply the roots" by 5, to get x 2 +26x+ 120. This is easy to factor (or one can manipulate the expression above further) . In either case we get (x+6)(x+20). Now we "divide the roots" and get (x + 6)(x + 20),
5
5
which simplifies to (5x+6)(x+4). We hope enough has been provided to make solving a problem involving just numbers or quadratics easy. The fun occurs when you can use some original method of your own, instead of merely mechanical means. When it comes to expressions of degree higher than the second, problems may become more difficult, but opportunities for originality also increase. There are some forms whose factors it is useful to remember. For example, if one remembers that the cube roots of unity are 1, ro, ro 2, and that 1+ ro + ro 2=0, then x 3 + y3 = (x + y)(x + roy)(x + ro 2 y) as well as (x+y)(x 2-xy+y2). In fact, x n - yn has the factor (x-y) for all n, and xn+yn has the factor (x-y) for all odd values of n. -96
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xn+yn is an example of a symmetric function. If one exchanges x and y, the function is unchanged. If y=kx makes the function take on the value 0, then y-kx will be a factor. In the example above, y=-x makes the expression zero, so x+Y is a factor. Again, consider (a-b)3+(b-c)3+(c-a)3. This is a symmetric function, and, as one finds by checking, becomes zero when b=a, c=b, a=c. Therefore the expression equals M(a-b)(b-c)(c-a) where
M is a constant. One merely compares terms to find OLlt
that the constant equals -1 so that (a-b)3 +(b-C)3 +(c-a)3 =-(a-b)(b-c)(c-a). c) Factor (a3+b3)(a-b)+(b3+c3)(b-c)+(c3+a3)(c-a) d) Factor a 3 +b 3 +c3-3abc. (Remember this one.) Sometimes, one gets extra mileage from factoring. We will try to simplify and solve the quadratic (x-a)(x-b) + (x-b)(x-c) + (x-c)(x-a) (c-a)(c-b) (a-b)(a-c) (b-c)(b-a)
= 1.
e e
This appears to be symmetric. Let x=a, and the expression on the left equals 1, so one root is a. Let x=b, and the expression on the left equals 1, so another root is b. Let x=c, and the expression on the left equals 1, so another root is c. However, a quadratic equation cannot have more that two roots hence we have been trying to solve an identity. This way, we have avoided much useless manipulation.
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As another example, let us attempt to factor f=(x+y)5- x 5_y 5. This equals zero when x=o and when y=O. Hence two factors are x and y. Next, let y=-x, and we find that f=O. Therefore (x+Y) is another factor. The remaining factor is quadratic, at most, so f=xy(x+y) (Ax 2 +Bxy+Ay 2)
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The coefficient A appears twice in the third factor because f is symmetric. Let x= 1, y= 1, and we have 30=2(2A+B). Let x=2, y=-l, and we have -30 = -2(5A -2B). When we solve these for A and B, we find A=B=5.
Therefore, f=5xy(x+y)(x 2 +xy+y2). e) The following problem appeared on the fMO for 1984: Find one pair of positive integers a and b such that: (i) ab(a+b) is not divisible by 7; (ii)(a+b)7- a 7_b 7 is divisible by 7 7 . Justify your answer.
HOW TO CATCH A LION WE OBSERVE THAT A
LION
HAS AT LEAST THE
CONNECTIVITY OF THE TORUS. WE TRANSPORT THE DESERT INTO FOUR-SPACE. POSSIBLE TO THAT THE
CARRY OUT SUCH LION
CAN
THREE-SPACE IN A
DEFORMATION
BE RETURNED INTO
KNOTTED
HE IS THEN
A
IT IS THEN
CONDITION.
HELPLESS.
ASPACE FilLER Suppose one can find a point P in the interior of the quadrilateral ABCD such that the four triangles PAB, PBC, PCD and PDA have the same area. Show that P is on one of the diagonals AC or BD. Let a, band c, be positive numbers. Prove that abc ~ (a + b-c)(a+c-b)(b+c-a) (These are from a Swedish Mathematical Olympiad.) -98
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CONCURRENCE There are many occasions when one is required to show that two or more lines are concurrent - that is, meet at the same point. We wish to present five ways in which this may be done. These range from the well-known to the (possibly) unknown. First is the well-known theorem of Ceva. In the diagram at the right, the necessary and A sufficient condition that the three "Cevians" AP, BQ, CR concur is that AR RB
x
BP PC
x CQ = 1
QA
With this theorem, it is very easy to show that the medians of a triangle
B
p
C
are concurrent. Remembering that, if the three Cevians are angle bisectors, so that BPjPC=cja, one can also show that the three angle bisectors are concurrent. Now, as an exercise, suppose that P is halfway around tIle perimeter of the triangle from A, Q is halfway around the perimeter from B and R is halfway around the perimeter from C, one can show that the lines AP, BQ, CR will concur. One can show in a similar manner that, if P, Q, R are the points where the incircle is tangent to the three sides of the triangle, then °AP, BQ and CR will also be concurrent. In the same manner, the three Cevians will be concurrent when the points P, Q, R are the points of tangency of the three escribed circles of the triangle. A second (and fundamental) theorem is Desargues' Theorem. As most readers know, it states that if two triangles (in two-space or three-space) are such that corresponding sides, produced if necessary, meet in three points that are collinear, then the three sets of corresponding vertices will lie in three concurrent lines. ..99..
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In this diagram, which shows two triangles in a plane, sides AB and A'B', BC and B'C', CA and C' A' meet on a line at P, Q and R. It follows that the lines on A and A', B and B' and C and C' will meet at the point O. For a diagram
C' .............
---Q .......
.......
showing Desargues' Theorem in three dimensions, see the cover of the May, 1983 issue of Arbelos. The diagram can be considered as R consisting of ten points and ten lines, with three points on every line and three lines on every point, which makes it a self-dual figure, technically called a "configuration".
If we recall that we have referred to the "line at infinity" when discussing projective geometry, and have agreed that lines that intersect on this line at infinity are parallel, one can easily show that if two triangles have their corresponding sides parallel, then their corresponding vertices lie on concurrent lines. A good exercise would be for the reader to try to show how, given two lines on a sheet of paper that would intersect somewhere off the paper, and given a point P, say, one can construct a line on P that, if extended, would pass through the intersection of these two lines. We have also presented a third method for showing that lines are concurrent by using Brianchon's Theorem. This is the dual of Pascal's Theorem and may be stated thus: If a hexagon is circumscribed about a conic, the lines joining opposite vertices are concurrent. The conic may be a circle, an ellipse, a parabola, an hyperbola, or even two lines. The diagram at the top of the next page shows some examples. In each diagram, we have AD, BE and CF concurrent at O. -100
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A
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B--==----=-~F
B F
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One can have some fun by trying to create special problems. For one example, what does Brianchon's Theorem look like when the hexagon 11as been squeezed down to form a triangle?
We come now to one theorem that is not too well
known -the dual of Pappus' Theorem. First we state Pappus' Theorem: Let A, B, C be points on line L, and A', B', C' be points on line L'. Then the A
L
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intersections of AB' and A'B, AC' and A'C, BC' and B'C are collinear. There may be some readers who do not know that we can prove Desargues" Theorem by means of Pappus " Theorem and vice versa - provided one assumes the concept of commutativity. Otherwise Desargues Theorem cannot be proved for triangles on a plane. There are non-Desarguesan geometries. When one tries to prove either theorem by using coordinates and linear algebra, one must beware of commutativity sneaking in. However, let us dualize the figure above. l
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f
c'
--=t-=:-bf------->,,------~
f'
a'
a
b We have the lines a, b, c On point f, and the lines a'l b', c' on point f'. Then rom the diagram, the lines (ab') (a'b), (be') (b'c), (ac') (a'c) meet on the point o. Note that, although Desargues' Theorem is self dual, (as is the dual of a triangle), Pappus' Theorem is not self-dual. The two diagrams look entirely different. (The same is really true of a complete quadrangle - four points and the six lines on them, and the complete quadrilateral - four lines and the six points on them. But that is another story). Just for fun, examine the following problem:
D
H
C
Ek:::--\"-'+---I F AL.--~-----=~
ABCD is a square and EF is parallel to AB, GH parallel to AD, as in the diagram. Prove that the lines on BE, DG and CK are concurrent. Can you recognize the upper diagram as equivalent to the lower one? Will there be concurrence when ABCD is a parallelogram? One can construct many hard-looking problems by slightly altering the figure. -} 02
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We have been trying to present theorems that are progressively less well known to the reader. We think this last one is least known. Examine the figure.
e
e
Let PX be perpendicular to BC, Py perpendicular to CA and PZ perpendicular to AB. Then
e
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(BX)2 - (XC)2 + (Cy)2 - (YA)2 + (AZ)2 - (ZB)2 = 0 The proof is extremely simple - one uses the Pythagorean
e
Theorem six times. (For example, (BX)2 so on)
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It
e
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e
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= (BP)2 -
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The converse is also true, and is equally easy to prove. We leave these proofs to the reader. Now see how easy it is to prove that the altitudes of a triangle are concurrent. Also, see how easy it is to show that the perpendicular bisectors of the sides of a triangle are concurrent (at the circumcenter). And that the lines perpendicular to the sides of a triangle at the points of tangency of the inscribed circle are concurrent (at the incenter). This theorem can be used to prove otherwise difficult problems. For example, the reader might try it on the case when the point P lies on the circumcircle, in which case, X, Y, Z are collinear. This line is the well-known Simson line. Should a problem ever arise where the reader is asked to prove lines concurrent (as in a test, a contest, or the USAMO or IMO, perhaps the reader will remember one of these methods for sl10wing concurrence and be able to use it. ~103~
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DISCOVERY AND INVENTION It is our belief that, in the sciences, mathematics, and when new products are invented, new ideas arise when the world is ready for them. We cannot come up with any thing in mathematics, for example, that has not been developed by two or more people. The story of the calculus, with its question of priority between Newton and Leibnitz comes to mind, as does the development of non-Euclidean geometry by Lobachevski and Bolyai simultaneously. It is interesting to note the simultaneous invention of logarithms by Napier, Briggs and Burgi. Fermat and Descartes are given simultaneous credit for analytic geometry, etc.
Nor have we been able to locate any invention that has not been simultaneously invented by at least two persons. Edison's life was full of lawsuits by others who claimed to have invented the same thing he had but earlier. Morse had the same trouble with the telegraph and Bell with the telephone. One of the reasons for reading up on the history of mathematics is the surprise aroused by the discovery of such simultaneous events. Another is, of course, reading about the lives of famous mathematicians, some of which were just as remarkable as the lives of, say, famous generals. All this is preliminary to a look at arithmetical computations before the invention of logarithms (and calculators). It is generally assumed that one of the difficulties ancient astronomers, arithmeticians, and accountants was with ordinary arithmetic. It is true that such men as Tycho Brahe and Johannes Kepler may have done a lot of long, complicated and boring arithmetic, but we take leave to doubt this. They had methods of their own to ease the burden, if it was a burden. I have already referred to Gauss who used to calculate a "chiliad" of prime numbers in his spare time, just for fun. -104
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Of course, the problem was usually multiplication. The arithmeticians naturally tried to convert such products to sums. This is what logarithms do.
One method used before logarithms was trigonometry. It was known that the following identity held: cos(A+B)+cos(A-B)=2 cos A cos B. We are told that calculators used this formula. Let us see how this could have been done by finding the product of, say, 652x.374. let cos A=.652
cos B= ..374
Using a table of the cosines of angles, we find that
(A + B) = 117°20', (A - B) = 18°44'. cos(A+B)=-.45917 cos(A-B)=.94702. cos(A+B)+cos(A-B)=.48785, and finally, dividing by 2 and placing the decimal correctly, 652x.374=24.3920 (approximately). The result is approximate because we used four place tables of cosines. The medieval calculators had the ten-place tables of Rhaeticus and the fifteen-place tables of Pitiscus. They also had lots of practice in arithmetic.
Then
We do not know why these calculators did not use the very simple rule (A+B)2_(A-B)2=4AB. All we would need in this case is a good table of the squares of numbers. We use this formula to calculate 652x.374. Let A=652 and B=.374. Then (A+B)=l026 (A-B)=278.
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(A+B)2= 1052676 (A-B)2=77284 4AB=975392 and AB=243848. This result is exact. We checked it on our little hand calculator. We suddenly have the urge to explain a method for multiplying numbers mentally that we happen to like. For two three-digit numbers, consider 2 ax + bx + mx
2
+nx +
When these are multiplied, the product containing no x-term is cpo To find the term in x, proceed as shown below: ax
2
+
mx 2 + to derive (bp+cn)x. For the term in x 2 , proceed as shown below:
~
~
to derive (ap+bn+cm)x 2 . For the term in x 3 , proceed as shown below:
~c ~+p
to get (an+bn)x 3 Finally, the term x 4 is obviously amx 4 . Now assume that the (x) has been replaced by 10, and we have a nice fast method for calculating a product. What's more, it looks pretty.
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Applying this method to finding the product of 652 and
374, the work locks like this:
3 6 2
4
3
8
4
8
6
(The column at the right is a record of what is to "carried."
This is what is done in multiplication, anyway.)
If a product is to be found in which the two numbers
have three digits or less each, this method can, with a
little practice, be faster than even a hand calculator.
We have seen a colleague "race" a hand calculator and
win every time. Of course, when it comes to multiplying
two two-digit numbers, there is almost "no contest." Finally, a method for multiplying two numbers which are alike except that the sum of their units digits equals ten, might be in place now. The two numbers are written as (lOa+b)(I0a+d) (b+d= 10).
If we multiply the two expressions and simplify, we find that the prodUct can be written as 1OOa(a+ 1) + bd. For two numbers like 57x53, which fulfil the conditions, we get the product 100(5)(6)+21 =3021. Of course, the same method will work on a product like, say, 257x253. Here we can use the device twice. Thus the product is 100(25)(26)+21=65021. What we have done is to use algebra (and trigonometry) in the service of arithmetic. Perhaps you might be able to invent some short-cut for finding the product of, say, numbers of the form (x+y) and (x-y). We have found this to be fun as well as a challenge. Perhaps you might agree. -107
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A COMBINATORIC fORMULA It happens that we encounter problems that sound quite different but that yield to the same process for their solution. We have met with several such problems recently. First we are asked to determine the number of solutions to the linear equation
xI + X2 + X3 + ... + Xr = n. If we write n ones and then insert r-I spacers between them, this will give us the r numbers wanted. We have n+r I objects, and can choose the spacers in r-I ways. Hence, . b·momIa . I coe ff·· usmg IClen t s, we h ave
(n +r-Ir- I) soIu t·Ions.
Actually, this formula gives us the number of nonnegative solutions. Some of the xj may be zero. Incidentally, if we seek solutions in positive integers only, we can use the same method and make sure no two spacers are adjacent. In this case, we will find that the number of · . soIu t Ions IS
(nr-I-I) .
What struck us was the frequent appearance of the first binomial coefficient. The same function showed up again in finding the number of arrangements, with repetition, of r objects selected from among n objects. Here we were reading our Euler, and found his solution very clever (as usual). Take a combination of r elements selected from the full set of n objects. Written in ascending order, let these be al,aZ,ay ... ,ar · It is understood that these a j are not necessarily distinct. Note: We assume that the reader understands the meaning of
t>. r -108
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Construct the associated set of elements a 1 + 0, a2 + I,
.a3 +2, ... ,a r + r-l. These elements are all distinct. Therefore, the number of ways or selecting them will equal the nLlmber of arrangements without repetition of elements from 1 to n+r-I. This will equal t + r - I ).
r We have already mentioned the fact that the Binomial Theorem holds for any exponent - not just positive integers. Applying this to a negative integer, we have (1- x)-n = en + (-n) x + (-n)(-n-I) x 2 + ... in which the
1
coefficient of
xr
Ix2 takes the form
(-n)(-n-I)(-n-2)...(-n-r+l) = (-I/t + r-I), and, once r! r
again, the same form occurs.
Incidentally, we have shown that the generating function that
generates these forms is (I-x)-n. The reader might like to
-n
show, by easy algebra, that (
) yields the same form as above.
r
It should be noted that we have a solution for the following problem: Find the number of ways of putting n like objects · · (n + r - I) . Into r d·f'& Ilerent b oxes. T h e answer IS r
If no box is to be empty, the answer is
tr-I - I). We seem to
have entered the realm of pigeonhole problems. We next examine the following problem: In n trials, some of which are successes and some failures, when will the r-th success tum up? -109
We shall assume that there are equal probabilities of success as of failure. According to elementary probability theory, we take p=q=1/2. There will be r+k-l trials, assuming k failures. At trial r+k, there is success. The probability sought will therefore be (r + k - 1)pr-1q k x p = {+ k -l)prqk Again, we have come to the k k same form as before. Before we come to our final problem, there are one or two theoretical matters to be presented. First, note that Newton and Gauss proved that: (
x r -1
x x+1 . )+ ( ) = ( ) where x may be any number and r an mteger. r
r
This is far more general than what Pascal was able to show. Next, consider an expression of the form (a+b+c+d)n. the expansion of this leads to the multinomial theorem. We state without proof the rule for expansion: A term of the form aibickct' (i+j+k+l=n) will have a coefficient of the form
(----.nL). i!j!k!l!
It is easy to show that the multinomial coefficient above is equal to the product of binomials
