Professor Samuel L. Greitzer in Memoria During the interval ohime between 1982-1987 Professor Samuel L. Gre itzer 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 PhD degree at Yeshiva University. He had more than 26 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 Revi8ited 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 publica tons which are appropriate for superior students, and for material which will help students prepare for the USA Mathematical Olympiad. Pro fessor Greitzer served as a coach of the summer Mathematical Olympiad training program from 1974 to 1983. 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 S.Klamkin accompanied the USA team to the International Mathematical Olympiad from 1974 [the first year 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 30-36 participating countries]. The contributions of Professor Greitller to the development of students of mathematies 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-Lincoln
I
PREFACE TO THE SECOND EDITION
[]
From September 1982 through May 1987 Professor Samuel L. Gre itzer authored a journal called Arbelo", for students talented and gifted in mathematics, thereby continuing the work with this nation's top high school mathematics scholars which he had begun as founding chair of our USA Mathematical Olympiad and as coach of the USA team in the International Mathematical Olympiad [IMO] 1974-1980.
Arbelo", kept contact, five times per year, with top students who were polishing their problem-solving skills to become contenders in the Mathe matical Olympiads. In places Arbelo", reads like an open newsletter to these "pre-college philomaths." Many of the articles Professor Greitzer wrote for Arbelo", were taken from lectures he gave at the Mathematical Olympiad Training Session [MOTS] for the USA team in the IMO. Just as topics covered at one MOTS are repeated at the next (but from a different point of view), one sees recur ring themes as one studies the various volumes of Arbelo-,. That Professor Greitzer was a geometer can be seen on almost every page. Since our U.S. students need additional training in geometry to compete in international mathematics competitions with countries whose educational systems em phasize geometry much more than in ours, Professor Greitzer had the ideal credentials for conducting the MOTS and leading the USA teams at the IMO. This second edition of Volume 5 consists of the articles Professor Gre itzer wrote for the journal's fifth and last year. In his preface to the first edition of the May issue that year (See p. 112) he made an unheard plea for someone to continue the work. Perhaps because the highly select audience toward which Arbelo", was aimed was so small or because no one with the appropriate credentials retired that year in order to take on the full-time job writing a journal at this level can very easily become, this was the last regular issue. There is one final volume in this series. In answer to numer ous requests, he wrote a manuscript the following year concerned primarily with presenting solutions to the geometry problems featured on the covers of the 25 regular issues of Arbelo",. After is death on February 22, 1988, I worked with some of his friends to put that special geometry issue into print as Volume 6. The mathematics (especially geometry) that I learned putting Vol ume 6 together and the contrast in format between the computer- and
lJ
laser-printer-generated Volume 6 and the first five volumes, which Profes sor Greitzer produced with his manual typewriter, made me feel that I would be doing a service to future olympiad-level mathematics students if I produced second editions of the first five volumes. This volume is the first to enter a second edition. As I went through the volume in detail I noted some unevenness in exposition and some typographical errors which I have corrected. I have tried to keep the flavor of the first edition, and so I included the preface Professor Greitzer wrote for each issue, including his pleas to readers to write to him about their solutions and their interests. I did make a few changes: In each issue, he had a selection of problems proposed for the IMO; I collected them in Appendix A. In each issue he featured a column, Kii.r~chd/c Korner, written by Professor George Berzsenyi, giving problems from that famous Hungarian mathematics competition; I collected all these problems in Appendix B where I also included Professor Berzsenyi's current address. He is still very interested in corresponding with readers about these problems. I would like to thank Mr Gregg Patruno of First Boston Corporation who read an early draft of this entire second edition and suggested some excellent improvements, and my colleagues at John Carroll University who read and helped me with several individual sections. As I worked on this edition for the last few years, every time Ire-read his mention of the Greitzer Point in "Blind Alley," (See p. 193) I recall fond memories of my friend, Professor Samuel L Greitzer.
Dr. Leo J. Schneider John Carroll Univer,ity July, 1991
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J J J J
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ICONTENTS OF THE ARBELOS. VOLUME VOLUME 5, CHAPTER 1* Cover Problem Preface to the First Edition Curve Fitting Just for Fun A Miscellany of Methods for Problem Solving A Brief Diversion on Dissections Another Application of Dissections The Pythagorean Theorem via Dissections . A Theorem and a Paradox Surface Area Volume The Paradox The Greatest Integer Function Some Problems A Postscript to the Greatest Integer Function Thinking about Numbers VOLUME 5, CHAPTER 2** Cover Problem Preface to the First Edition Areas on Lattices Triangle Area The Butterfly Problem Tchebyshev's Problem Many Cheerful Facts Pythagorean Triples Just for Fun Descriptive Geometry
51 1-28 1 2 3 8 9 12 14 15 17 17 19 20 21 24 25 28 29-58 29 30 31 36 37 40 42 44 50 51
[J * Originally published as the Arbelo~ Volume 5, Issue 1, September, 1986. ** Originally published as the Arbelo~ Volume 5, Issue 2, November, 1986. iii
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VOLUME 5, CHAPTER 3*
Cover Problem Preface to the First Edition Specialize!
Specializing Ptolemy's Theorem An Aside about the Talmud Specializing Hexagons in Conics Specializing in Algebra Supernumbers Stewart's Theorem Promiscuous Examples Miscellaneous Problems Circum-In- and Ex-centers Many Cheerful Facts VOLUME 5, CHAPTER 4** Cover Problem Preface to the First Edition Iteration Blind Alleys Dice and Mathematics Derivatives Promiscuous Examples Using Physics in Mathematics Many Cheerful Facts
57-84
57
58
59
59
61
62
62
66
69
73
74
75
80
85-110
85
86
87
91
93
96
102
103
109
* Originally published as the Arbelo" Volume 5, Issue 3, January, 1987. ** Originally published as the Arbelo" Volume 5, Issue 4, March, 1987. iv
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VOLUME 5, CHAPTER 5* Cover Problem Ave atque Vale A Note on Infinite Products Just for Fun Some Trigonometry Just Plain Numbers and a Magic Number Geometric Reveries Haversines A Space Filler Arithmetic Functions " APPENDIX A: Olympiad Problems from around the world APPENDIX B: Kiirschak Korner
*
111-144
111
112
113
117
118
122
125
129
132
133
137
142
Originally published as the Arbelo6 Volume 5, Issue 5, May, 1987. v
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ARBELOS -
Volume 5. Chapter 1
COVER PROBLEM from the
SEPTEMBER, 1986 ARBELOS
A
1.:....,t.----7 C
[J
x
z
AX
II
BY
II
CZ.
AX meets C B at X.
BY meets AC at Y.
C Z meets AB at Z.
PROVE: The area of 6XY Z is twice the area of 6ABC.( t)
[J
(t) The solutions to all of the geometry cover problems are published in Volume 6, the special geometry issue of the Arbelo-,.
[J
chapter was published as the
Copyrisht
0
The Mathematical Association of America, 1991. In its first edition, this
Arbelo-"
Volume 5, Issue 1, in September, 1988.
1
ARBELOS -
Volume 5, Chapter 1
J
J
PREFACE TO THE FIRST EDITION At the start of our fifth year of production, we take a few moments to see where we have been and where we expect to be going. We have done a lot of traveling, picking up problems on the way. We have helped with the First Ibero-American Mathematical Olympiad in Colombia. We have been to Finland, and collected more problems there, some of which will appear in this and future issues of Arbelos. We hope to receive more help and articles from colleagues who have shown an interest in the ArbeloJ. We hope to be in Poland to collect more Olympiad problems. We also hope to help with the Second Ibero-American Olympiad in Uruguay. Obviously, we will be on the move. The number of communications from readers has increased, and have given us increased pleasure. Rahul Pandharipande has solved two of our Cover Problems and tried his hand, successfully, at the Olympiad problems. An swers to our Arithmetic Problems have been received from Regina Grieco, E.L. Wilmer and Japheth Wood (nearly all correct). Constant writer Ambati Kr ishna has submitted problems solved by projection, and David Bernstein has solved our problem about the sum of fourth powers of sines and cosines very well.
J .
For the new year, we value letters from readers. Without these, we are working in a vacuum, not knowing whether out endeavors are having any effect. We also value suitable articles. Finally, we have much more in the way of mathematical morsels to present. We even welcome critical letters! So write and help make out efforts seem more worthwhile.
Your Editor Dr. Samuel L. Greitzer
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ARBELOS -
Volume 5. Chapter 1
CURVE FITTING In our very first issue (Aftermath - Sept. 198t), we introduced some concepts of the theory of Finite Differences. Because of the lapse of time, a short review might not be amiss. We started with a sequence UO,Ull'" ence", thus: Uo
'Un'
then we found the differ
etc.
We let ~uo = Ul - Uo, ~2UO = U2 - 2Ul + Uo, etc., and then derived an expression for Un; namely, the symbolic form Un = (1 + ~)nuo. To find
Un,
we used the binomial theorem to expand to get
Note that, after expanding as usual, we "dropped" all the exponents, pro ducing the subscripts shown. This gives us a simple way to find the nth term in a sequence and, by a slight extension, to find the sum of n terms of a sequence. For example, given the sequence 2,6, 12,20 we get 2
4
2
6
o
6 12
2
8 20
3
J
ARBELOS -
Volume 5, Chapter 1
so Uo = 2, ~uo = 4, ~2UO = 2, and all other differences are equal to zero. sllhstitllting in our formula, we find n
Un = (1 +~) Uo = Uo
n
l~uo
+
+
...
n(n -1) 2 ~2UO'
or Un
= 2 + 4n + n(n -
1)
= n 2 + 3n + 2.
By substitution, we find Uo = 2, Ul = 6, U2 = 12 and U3 = 20. We may eztrapolate here to get U4 = 30, Us = 42, and so on. We might also interpolate to find that U1.5 = 8.75 (whatever that may mean!). Note, however, that, if we write our data as n:
f(n)
0 2
1 6
2
3
12
20
we will have found a function that passes through each of the points indi cated. This function is f(n) = n 2 + 3n + 2. This is our example of what we call (mathematical) curve fitting. We may be wrong (and if we are, let us know), but this differs from statistical curve fitting. In the statistical situation, we would plot our points, take a guess on a curve that might come close to these, and derive an equation that is statistically suitable. In statistics, the derived curve might not pass through even one point plotted. As long as certain criteria are satisfied, so is the statistician! In our case, all points given must lie on the curve we have derived.
... ... ... ,.
It is true that much has been done to make interpolation and curve fitting in statistics as accurate and as useful as possible. Any text on numerical analysis will show this. There are statistically useful rules due to Newton, Gauss, Bessel, Stirling and others.
We would like to present a method of curve fitting due to Lagrange. As usual, we work with a small set of data, but the reader will see that the method is applicable in general. So, let us start with the data: z :
f(z)
Zo Yo
Zl Yl
Z2 Y2
Z3 Y3
We are not assuming that the Zi form an arithmetic progression, only that they are distinct. We now write: (z - zt}(z - Z2)(Z - Z3) (z - zo)(z - Z2)(Z - Z3) + Yl..,..--'---:'-;-'----':-':----'-:: (zo - zt}(zo - Z2)(ZO - Z3) (Zl - ZO)(ZI - Z2)(Zl - Z3) (z - zo)(z - zI)(z - Z3) (z - zo)(z - zt}(z - Z2) +~ +~ (Z2 - ZO)(Z2 - Zt}(Z2 - Z3) (Z3 - ZO)(Z3 - Zt}(Z3 - Z2)
f( z ) =Yo
4
,
... 11
.. ,... 1 ...
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ARBELOS -
Volume 5, Chapter 1
If the reader substitutes in the above, it will be obvious that 1 is a third degree polynomial, and I(xo) = Yo, I(:r.d = YJ, l(x2) = Y2, and l(x3) =
Y3· Inspection will show how the various terms are constructed. We will examine this in due time. As an exercise, let us take the following:
[J
x:
0
1
3
Y:
2 6 20
and substitute in our formula. We get
Y
= 2(x -l)(x - 3) (0 - 1)(0 - 3)
which simplifies to Y = x 2
+
6(x - O)(x - 3) (1 - 0)(1 - 3)
+
20(x - O)(x -1) (3 - 0)(3 - 1)
+ 3x + 2.
We can interpolate and find 1(2) = 12, or we can extrapolate and find 1(4) = 30. Just for fun, let us derive the usual formula for the Lagrange function. We use the auxiliary function
When this is differentiated, we find that
F'(x) =(x - xt}(x (x - xo)(x (x - xo)(x (x - xo)(x F'(xo) =(xo - xt}(xo F'(xt} =(X1 - XO)(X1 F'(X2) =(X2 - XO)(X2 F'(X3) =(X3 - XO)(X3
X2)(X - X3)+ X2)(X - X3)+ xt}(x - X3)+ xt}(x - X2), and that - X2)(XO - X3), - X2)(X1 - X3), - Xt}(X2 - X3), and - Xt}(X3 - X2).
It is not difficult to combine all this information to find that
1x
_
( )-
[J
~
~ (x -
YiF(X) xi)F'(Xi)
is the formula for an nth degree polynomial through n+1 points. We will use this form only when absolutely convenient.
5
ARBELOS -
Volume 5. Chapter 1
There are still intelligence tests that include questions like: Given the 1,2,4,5" .. write the nezt two term~. If you don't write these as 7,8, your intelligence may suffer. However, if we use the table ~eqttence
0 1
z: y:
1 2
2 4
J
3 5
and the Lagrange formula (which we leave it to the reader to do, as an exercise), we get: 2 3 /(z) = 6 - z + 9z - 2z 6 from which /(0) = 1, /(1) = 2, /(2) = 4 and /(3) = 5. However, /(4) = 3 and /(5) = -4. So 3, -4 is just as correct as the answer the testmakers want! We shall now show that any number can be selected to follow those in the table. Let us write:
+ P(z). F(z),
g(z) = /(z)
where /(z) and F( z) have already been defined, and where P( z) is a polynomial that you can select at will. This will yield g( z) = /( z) for z = 0,1,2, ... ,n because F(z) equals zero for these values of z. For z = n+l, F(z) is not zero, and P(z) can be manipulated so as to make g(z) whatever you want.
J J ,.
Thus, suppose we are given the table of values 0
z:
/(z):
1
2
3
1 9 8 6
There is no difficulty in finding that
" /(z) = 8z
3 -
51z
2
+ 91z + 6.
6
Thus, /(0) = 1, /(1) = 9, /(2) = 8, and /(3) = 6. However, /(4) = 11, and, for some strange reason, we want g( 4) to equal 1986. In ourformula for g(z) let z = 4, and we arrive at g(4) = 11 + P(4). F(4) or g(4) = 11 + P(4) ·24. So we let P(4) = 1~~5 and we have g(z) = 1986. Note that in this case, we let P(z) be the constant 1~~5. We hope we have convinced the reader that we can, if we wish, derive an equation that will give us any numbers we wish, or will attach to a given sequence of numbers, any numbers we wish. There are other uses to which
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ARBELOS -
Volume 5, Chapter 1
we can apply the Lagrange formula with our added term. We now present ll, !I~cond ll,pplication. Suppose that we have been given the data: x:
0
1
2 3
/(x): 8 1 2 8 (which we have just "made up"). We want to fit a curve to the data and, in addition, have the slope of this curve at x = 3 equal to zero. We leave the reader the task of finding that the associated /( x) is
Unfortunately, differentiation gives the slope at (3, /(3)) as
g(x)
We use our formula g(x) = /(x) + P(x)· F(x). For x = 0,1,2,3, = 8,1,2,8, respectively. Now we differentiate, and we have
=
When x
F(3)
15. 2
3, we have g'(3)
=
1'(3)
+ P(3)
. F'(3). (Remember that
= 0.) Now
so
g'(3)
15
= 2" + P(3) . (108 -
162 + 66 - 6)
15
= 2" + P(3) . 6.
Since we may let P(x) be a constant, in this case we let P(x) = -5/4, so P(3) = -5/4 and g'(3) = 0 as desired. We have given only two situations in which the Lagrange formula can be used. There are others, and we hope to return to the subject soon.
00, 00 2 , OOY?
Carolyn Well"
7
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ARBELOS -
Volume 5. Chapter 1
J
JUST FOR FUN a) A pole is being transported along an endless belt. When a man walks
in the direction in which the pole is moving, he goes from one end of
the pole to the other in 140 steps. Walking in the opposite direction
he goes from one end to the other in 20 steps. If each of his steps is
one yard, how long is the pole?
b) A right triangle is given complete with nine-point circle. Four tangents are drawn from the ends of the hypotenuse to the circle. The sum of the squares of these four tangents is 25. Find the length of the hypotenuse. c) If a number less than one million is selected at random, what is the probability that it contains exactly five 7's among its digits? d) A father plans to retire when his youngest son reaches the age of 21,
at which time his age will be the sum of the ages of all his sons. The
product of the ages of all his sons is now 2431. At what age will the
father retire? e) Evaluate the sum
lVij + lJ2j + lV3j + ... + lJiOOj·t
t The notation lt J indicates the largest integer less than or equal to t. The article beginning on page 23 in this volume summarizes many properties of this function.
8
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ARBELOS -
Volume 5, Chapter 1
A MISCELLANY OF METHODS
FOR PROBLEM SOLVING
To the extent that the would-be problem solver has alternative meth ods of attacking and solving a given problem, that student has an advantage over those who do not have these. It may even be the case that the solver is not aware of these alternative methods, even though they have been in troduced. These thoughts arose out of a problem we encountered recently. The problem:
C
F
IFigure
l.!
In the planer hezagon ABCDEF 8hown in Figure 1, AB i8 parallel to DE, BC i8 parallel to EF, and CD i8 parallel to AF. Prove that 6ACE and 6BDF have equal area8.
Perhaps the reader might take a few minutes to try to solve this before reading further. We hope to present three methods. If you have a fourth method, we will be happy to publish it - if it isn't too long. Our first method uses vectors. For this, let us imagine vectors ema nating from A and going to B, C, D, E and F.
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ARBELOS -
Volume 5. Chapter 1
Now there are two types of vector products. One, the scalar product, is equal to the product of the lengths of the vectors multiplied by the cosine of the angle between them. We won't be using this. The vector product, or crou product of two vectors equals the product of the lengths of the vectors, multiplied by the sine of the included angle, this product being the length of a new vector perpendicular to the plane of the vectors being multiplied. The order of the vectors multiplied determines the direction of the vector product with x a = -a x
ii
Ii.
J
J
.. .~
There are two points to notice. When two vectors are parallel, their cross product equals zero. Also, the absolute value of a cross product equals the area of the parallelogram of which the vectors are adjacent sides, so half of this equals the area of the triangle whose sides are the vectors. Now to Figure 1. Recall that A is the origin of all our vectors. Note that BC and EF are parallel, so (C - B) x (E - F) = 0 (a)
= O. E) = O.
Next, CD and AF are parallel, so (D - C) x F
(b)
Also, AB and DE are parallel, so B x (D -
(c)
Now for the areas K 1 and K 2 of 6.ACE and 6.BDF, we have 2K1 =
and
IC x
J
EI
2K2 = I(D - B) x (F - D)I = ID x F - D x D - B x F =
ID x
F - B x F
+B
+B
x DI
x DI.
Also, C x E - C x F - B x E
+B
x F =
C x F
and
B x E
0
=D =B
from (a) above,
x F
from (b)
x D
from (c).
...
Hence CxE-DxF-BxD+BxF=[
But So 2K2
ID x F
+B
x D - B x FI = 2K2 •
= IC x EI = 2K1 , and we are done.
Our second method for solving the problem of Figure 1 uses analytic geometry. Most texts on analytic geometry either ignore or barely mention the fact that coordinates need not be rectangular. It is permissible, and
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ARBELOS -
Volume 5, Chapter 1
y
(c, d) ...,:::;-----~J---x
(0,0)
(a,O)
IFigure
2.1
sometimes desirable, to use oblique coordinates. The difference in systems involves the sine or cosine of the angle between the coordinates. With this in mind, let us select oblique axes as shown in Figure 2. The coordinates of the six vertices of the hexagon are then calculated. The reader should check the coordinates in Figure 2 to see that they are labeled correctly. We can compute the number u, because the upper and lower bases of the hexagon are parallel. The slope of the lower edge is zero, so the slope of the upper edge must be zero. That explains the use of e in both places. Also through parallelism, we find that e-b
d-O
u-O
c-a
or
ud = (e - b)(c - a).
Now we may use the determinantal form to find the areas of the two triangles. For the one with vertex at the origin, we have
o 0 u e
e = ud - ce = (e - b)(c - a) - ce = - ea - be d
For the triangle with the vertex at (c, e), we get
2K2 sin (J
[J
1 1 1
o
b
c a
e = -ae
0
Again, we are done. 11
+ ab -
bc
2K
1 = --. sin (J
+ abo
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ARBELOS -
Volume 5, Chapter 1
The point is that the problem solver should not forget the existence of ohliCJue coordinates. In this problem, the Recond solution is surely shorter and easier than the vector solution, and sometimes, in a contest, time is important.
J
A BRIEF DIVERSION ON DISSECTIONS We come now to a concept that may be new to many, the idea of cutting up a figure and then rearranging the pieces to form another figure. This process may not change areas or lengths. The Greeks of old used this a lot. We shall call this method the method of dilJ6ectioniJ.
a
(a+b)j2J ~ b
b
K
K= ah
-
K= ~(a+b)
= !bh
IFigure 3.1 It is said that an Arabian mathematician, Bhaskara, drew a diagram to prove the Pythagorean Theorem and just said "Behold!". We display the three diagrams in Figure 3 for areas, and just say, "Look!".
What is more, we have already used dissections to derive the Pris matoid Formula as far back as the March, 1983 issue of the Arbe1oiJ. We can also consider the possibility that Cavalieri's Principle is an example of dissection. - - END DIVERSION -
J J ]
..
F a
IFigure
4.1
-] ~
12
ARBELOS -
Volume 5, Chapter 1
Our third method of solving the problem of Figure 1 uses dissections. Examine first the diagram in Figure 4. From the hexagon ABCDEF we select triangle B D F and construct three parallograms using adjacent pairs of sides of the hexagon with the three sides of 6BDF as their diagonals.
[J
Note the little triangle formed. We shall call its area T. Let the areas of the two congruent triangles forming each of the parallograms be r, q, and p, and let the lengths of the sides of the hexagon be a, b, c, d, e and I, as indicated in Figure 4. Then the area of triangle BDF is equal to p + q + r + T and the area of the whole hexagon is equal to 2p + 2q + 2r + T.
[J
Moreover, we can find the sides of T. These are c- I, e-b and a-d. We ask the reader to check that these are correct. Now we go to Figure 5. This time we concentrate on triangle ACE and, similar to the construction in Figure 4, we construct parallelograms with diagonals AC, CE, and AE and respective areas 2u, 2v and 2w. Then the area of 6AC E equals u + v + w + T' and the area of the hexagon is equal to 2u + 2v + 2w + T', where T' is the little triangle at the center.
IFigure 5.1
Again, we can find the sides of T'. Since T and T' have the same lengths as sides, they are equal in area. From 2p + 2q + 2r + T = 2u + 2v + 2w + T' and T = T', we can now decide that p + q + r + T = u + v + w + T', so triangles ACE and BDF have equal areas.
13
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ARBELOS -
Volume 5. Chapter 1
ANOTHER APPLICATION OF DISSECTIONS
J
J
The constant reader will recall that we have said that mathematics is fun. We cannot give rules for constructing dissections to solve problems. However, that is where the fun comes in. When you discover a dissection of your own, the feeling of accomplishment is like nothing else. For example, in a recent issue, we had a thorough discussion of the following problem (Cevians are segments from a vertex to the opposite side. ):
CevianJ are drawn to the triJection pointJ of triangle
ABG. They interJect in pairJ at X, Y, Z. Prove that
the area of 6XY Z iJ the area of 6ABC.
t
Solution: We locate the midpoints of the sides of 6ABC. Then the mid point M c ' the point X and a trisection point of side AC are collinear. The same is true for Y, Me and a trisection point of side AB, and for M b , Z and a trisection point of side BG. [See Figure 6.] Now rotate 6AX M c around M c to the position B.X M c • Do the same with 6BY Me and 6G Z Mb.
The area of the figure is unchanged, but we have parallelograms Y X X B,
Y ZX X and, in fact, a lot of parallelograms, as an examination of Figure
6 will show. These produce seven triangles, each equal in area to 6XY Z, so 6XYZ is the area of 6ABG.
t
We think this is pretty!
A
...
... ...
,
J J ] ..
IFigure
14
6.1
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Volume 5. Chapter 1
THE PYTHAGOREAN THEOREM VIA DISSEOTIONS One may conclude that dissections are limited in their use. This is far from the truth. One can use them to justify the fact that the area of a circle equals 1rr 2 and the volume of a sphere is ~1rr3. We will use dissections to show the Pythagorean Theorem.
x~---~---~y
B
."
1
o
b IFigure
7.1
The configuration at the left in Figure 7 indicates a proof, said to be due to Socrates, of the Pythagorean Theorem for an isosceles right triangle. One cuts across the squares on the equal sides with line XY and then places the pieces as indicated in the square on the hypotenuse. The diagram at the right of Figure 7 is a proof that consists of only three pieces. Squares with sides a and b are next to one another (forming one piece). One cuts along AB and AO and then swings these pieces to the positions shown. We once had a model of this that we made of plastic with hinges at Band O. A flip of the wrist changed the two smaller squares into one large square.
15
ARBELOS -
Volume 5. Chapter 1
We have not been able to locate the source of the proof given in Figure 8, hut t.hink it may be found in Howard Eves' "Geometry". It has remained in our memory for a long time. Note its relation to Socrates's proof for the isosceles right triangle.
IFigure
8.1
We start with the right triangle ABC with squares on the sides and hypotenuse. We slice parallel to AB through C and arrange the pieces as shown. A little piece of piece 2 may stick out at the bottom. Cut it off, and fit it into the space at A next to piece 4.
J
J
.. .. J
A good exercise would be to find the length of each line segment and thus prove that the dissection really works. A good lesson to be learned is: keep your eyes open for possible applications of vectors, oblique coordinates or dissections.
J
16
-
ARBELOS -
Volume 5, Chapter 1
A THEOREM AND A PARADOX
Surface Area Before the advent of the "New Math", textbooks on solid geometry included a sequence of theorems leading to, among others, a formula for the area of a sphere. These did use (as we will) some concepts of limits.
~----+H
E
B ~~========:::::::::7 K IFigure
9.1
Consider a frustum of a cone, as in Figure 9. Its altitude is AK, line DH is parallel to the bases, and DE is perpendicular to edge AB at midpoint D, meeting the axis at E. If we use the Pappus-Guldin Theorem (Arbelo." Vol. 4, No.3) we find that the surface area of the frustum equals 271-AB . D H. However, since triangles D H E and AKB are similar, we have ~~ = ~fI from which the area of the frustum equals 27rDE . AKj that is, the area of the frustrum equals the area of the side of a cylinder whose altitude is the same as that of the frustrum and whose radius is DE.
[J
Next let the surface of a sphere be divided into zones [See Figure 10.] and inscribe a frustum in each zone. Then the sum of the areas of the frustums will equal the product of: 27r, the sum of the altitudes of the frustums, and the length of, say, EO, (if ABC· .. is a regular polygon).
17
ARBELOS -
Volume 5. Chapter 1
J
IFigure 10.1
Now, let each of the edges AB, Be, etc. become smaller (approach zero as a limit); then the length EO approaches R, the radius of the sphere, as a limit. The sum of the areas of all the zones (which is itself a zone) will then equal 21rRH, where H is the sum of the altitudes of the original zones. We have two observations to make at this point. First, if H equals the diameter of the sphere, then the area equals 21rR . (2R) = 41rR 2 ; and second, all the development holds when the zone becomes a "cap". The area of a cap will also equal 21rRHo Suppose we have a sphere inside a cylinder of the same radius and altitude. If the whole arrangement be cut by two planes parallel to the base of the cylinder, these planes will cut off a zone on the sphere and a small cylinder on the cylinder. [See Figure 11.] Note that the zone and the small cylinder have the same area! Archimedes was proud of this result, which he had proved. He asked that the diagram be engraved on his tomb, and it was, but the tomb and the diagram are no longer to be found . (Perhaps one could get some students to make a collection to restore it!)
-.. ." I
1
.. .
... IFigure 1t.i 18
...,
..... ......
ARBELOS -
Volume 5. Chapter 1
Volume Now we come to volumes. First, let us find the volume of a spherical cap with height H. Its area is 2'TrRH. Let this area be subdivided into small bits, (each equal to Llx . Lly in area, say) and each the base of a small, very thin, pyramid with altitude equal to R. Since the volume of each of these slim pyramids is a third of the area of its base times its altitude, R, it follows that the sum of the volumes of all these pyramids will equal 2'Tr RH = ~'Tr H R 2. Notice that if H = 2R, the cap becomes the entire sphere and the volume becomes ~'TrR3.
k.
Unfortunately, the volume we have found for the cap has a cone with vertex the center of the sphere hanging down below. We will have to sub tract the volume of this cone to get the volume of the cap only. This cone has an altitude of (R-H) and a radius of base equal to JR2 - (R - H)2 = ../2RH - H2. Therefore, its volume equals
Subtract this from ~'Tr R 2 H, and, after some simple manipulation, we have: If a cap with altitude H be cut from a ~phere, it~ ~ur face area (without ba~e) equal~ 2'Tr RH, and it~ volume i~
f(3RH 2
-
H 3).
To find the volume contained within a zone and its two circular parts (top and bottom), the simplest way would seem to be to treat the zone as the difference between two solid caps, find the volume enclosed by each, and subtract. We have done this for the reader, and have arrived at the following: Let a and b be the radii of the top circle and bottom circle of the zone, and let h be it~ height. Then
[J
We cannot resist presenting one problem. In Figure 12 two planes intersect a sphere of radius 12 inches in circles which are tangent at A. Segment AB is a diameter of the sphere, and the planes cut off a 30° arc, Be. What is the area of the odd-shaped lune? (SuggeJtion: you could try diuection~ - or not.)
19
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ARBELOS -
Volume 5. Chapter 1
. ..,
..J
...
...
...
IFigure 12.1
The Paradox We start with a sphere on which we construct a zone and a cap. The bases are parallel and cap and zone have equal heights h. [See the sketch at the left in Figure 13.1 Therefore, the zone and the cap have equal areas. The situation is shown in the diagram at the left of Figure 13.
-
Now let the zone and the cap shrink, always keeping their heights equal. The areas will diminish, but will still remain equal. We let this situation continue until the cap shrinks to a point P and the zone shrinks to the circle. The situation is shown in the figure at the right of Figure 13. Can we assume that the single point P is equal to the circumference of the circle?
P
1
Figure
13·1
A similar situation can be found in "Hi~tory of Mathematic~", by Howard Eves, on page 272. It is due to Gallileo, and is worth examining. In fact, we strongly recommend all of the "Hidory of Mathematic~", which contains entertaining anecdotes, interesting problems, and useful methods for problem solving.
20
-
,
'"1
ARBELOS -
Volume 5. Chapter 1
THE GREATEST INTEGER FUNCTION Although it occurs frequently in problems, it is still customary to define the expression l x J as "the greateJt integer not exceeding the real number x" and to give some examples, such as
l5.4J
=
5, leJ
=
2, l-2.5J
= -3, etc.
We think it is useful to examine this function more closely.
[J
Clearly, x = l x J + a, where 0 ~ a < 1. We can write the same expression in a variety of ways, each of which has its uses, thus:
(a) lxJ
lxJ + 1, < l x J ~ x,
~ x
(b) x-I
(c) 0 ~ x -
<
l x J < 1, and
(d) - x - l <
l-xJ
~ -x.
Form (d) is derived from form (b) by replacing x by -x.
We now derive a few useful facts.
+ n J = l x J + n. We reason as follows:
(1) From (b) we have x + n - 1 < lx + nJ ~ x + n. (2) Next, x-I < l x J ~ x, which also follows from (b). (3) Obtain x + n - 1 < lx J + n ~ x + n by adding n to (2). Finally, compare (1) and (3) to obtain l x + n J = l x J + n since both First, if n iJ an integer, l x
are integers. Again, from (4) x-I
[J
L
(5) -x -
< l x J ~ x and 1 < l-x J ~ -x we get
-2 < lxJ + l-xJ
~
o. 21
. ... ..J ARBELOS -
Volume 5, Chapter 1
That is, if :I: iJ an integer, l:l:J + l-:I:J = 0, and if :I: iJ not an integer, lxJ + l-:l:J =-1. AIJo, l:l:J+lyJ::; l:l:+yJ::; l:l:J+ lyJ+1. Toseethis,let:l:= l:l:J+a and y = lyJ + b, where a and b are less than unity, or 0 ::; a + b < 2. Then
6) :I: + y
= l:l:J + lyJ + a + b
7) from which we get l:l: + yJ ::; l:l:J + lyJ + la + bJ. Since 0::; a+b < 2, la + bJ equals 0 or 1. Hence, l:l:J + lyJ ::; l:l: + yJ ::; l:l:J + lyJ + 1.
J
-... 1
Of the numerous results we can derive similarly, we shall restrict our selves to just two more. We Jha/l try to Jhow that
.... Let :I: = m + a (0 ::; a integers).
< 1), where m =
nq + r (0 ::; r
< n, m,
n, q and r
8) Then :I: = nq + r + a. :I: Therefore, ;;
=q+
r+a and l:l:J ----;;;;
= q.
9) Also l:l:J = m = nq + r. Therefore ll:J
J=
lq +
... "
~J = q which proves the assertion.
I
To Jee that l:l: - yJ ::; l:l:J - lyJ ::; l:l: - yJ + 1
10) take :I:
=
11) Then -y
l:l:J + a, and y = lyJ + b as before.
= -lyJ
However, -1
- b and :I: - Y
= l:l:J
- lyJ + (a - b).
b < 1.
Hence l:l: - yJ ::; l:l:J - lyJ ::; l:l: - yJ + 1. And now for some applications, old and new, of what we have thus far presented. To begin, we seek the greatest power of a prime, p, that will divide into nL The simplest way is to imagine n! written out seriatim. Then p divides into every pth factor. The number of times this occurs will be l ~
J. When these are removed, in the remaining product, every (p2)
factor is still divisible by p, so there are l ; 22
J of these.
th
Similarly, there
-
ARBELOS -
Volume 5. Chapter 1
will still be more factors divisible by p3, and these are
l; j
in number. In
short, we arrive at the series
l~ J +
l; J l; J +
+ ....
For example, we find the highest power of 5 that will divide into WOO! from this sum
1000j l1000j l1000j l1000j. l 5 + 25 + 125 + 625 249
This sum equals 249, which shows that 5 is the highest power of 5 that divides into 10001. (We can use the next-to-Iast theorem to make the com putations easier, thus: = 200, l2~O = 40, S0 = 8, l~J = 1, the sum of these being 249, as before.)
llO:O J
J
l4 J
Another problem involves proving that
We note that this is usually presented in texts on number theory as a prob lem for the reader to solve, as if it were a simple consequence of previously stated theorems. We will try to develop a proof. Let a ~ ;r; < a + 1 where a = l;r; J. Divide the interval between a and a+l into n parts. Then ;r; lies in one subinterval; that is,
p p+l a + - < ;r; < a + - - for some 0 nn p+l 1 p+2 a+--<;r;+-
Then
a
+ (1- ~) ~;r; + n - ~ -1 < a + 1,
and, continuing this sequence, n-p 1 a+l~;r;+--
[J
a+
p+n-l n-1 p+n <;r;+--
23
L
~ p
< n.
ARBELOS -
Volume 5, Chapter 1
To translate this into simpler language, for the inequalities down to (*) the greatest integer function of the center terms is a and there are n-p of these. For the inequalities beginning with (**) the greatest integer function of the center terms is a+ 1 and there are p of these. Hence we have
X+IJ + ... + lx+(n-l)J lx J + -nn = a( n l
p)
J ,.~
.... ·1
+ (a + l)p = an + p.
Also, from our inequality for x alone, we get
an
+ p S nx < an + p + 1
or lnx J = an + p. This proves the assertion.
Now for some problems:
1) How many zeroes are there at the end of n
.. ..
= 81!?
2) Determine all natural numbers that do not belong to the set
J where n is a natural number.
3) How many distinct numbers are there in the set
4) How about problem (e) in our
"Ju~t
...
...
for Fun" [page 8]?
...... 24
.....
I
ARBELOS -
Volume 5, Chapter 1
A POSTSCRIPT TO THE
GREATEST INTEGER FUNCTION
The constant reader will have noted that we have, on several occa sions, introduced subject matter that does not bear directly on high school mathematics. For example, we did an article on the Gamma Function, just because we found it pretty and interesting. We are about to do the same thing again. Our hope is that some readers may be sufficiently attracted to it as to do independent work with it. It occurred to Joseph Fourier that, if a function, f( x), were continuous or piecewise continuous in an interval (0, 2L), then it could be written in the form
f( x) = A
~x
2~x
L
L
+ (al cos - + a2 cos .
~x
.
3~x
+ a3 cos -
2~x
L
+ ...)
3~x
+ (b 1 sm L + b2 sm L + b3 sin L + ...), valid within that interval (and repeating in successive equal intervals). The constants A,
ak,
and bk could be computed simply:
1 A = 2L
f2L
Jo
1
2L
ak
= L1
bk
= L Jo
1
f(x) dx,
0
t
f(x) cos
L
•
f(x)sm
k~x dx, L kn
L
dx.
These days, the computation of A, ak and bk have become routine, even for the student of elementary integral calculus. Now in a spirited correspondence involving Euler, D'Alembert and Daniel Bernoulli concerning the shape of a vibrating string, in the late 1600's, this idea had already come up. However, when Fourier presented it in 1803, it met with universal rejection. Even Lagrange considered it wrong. Further study has put the concept on a rigorous basis, and there are no more arguments about its correctness.
25
ARBELOS -
o
1
Volume 5. Chapter 1
3
2
4
5
.
".
....
....
j
IFigure 14·1
Consider the graph in Figure 14. It represents the graph of f( x) == x in the interval (0,1) and copies translated horizontally one unit repeatedly as indicated. Naturally, it is called the ~awtooth curve.
...
Now we have computed the expressions for A, ak, and bk for you (although you may check them out yourselves). We get
...
1
1 (sin 27T"X
2
7T"
f() x=---
1
) + sin 247T"X + sin 367T"x + ....
Now it happens that x - lxJ also represents the sawtooth curve. Hence we equate these and solve for lxJ and arrive at the (surprising) result: lxJ
1
1 (Sin 27T"X
2
7T"
=x--+-
1
) + sin 247T"X + sin 367T"x + ....
One must be cautious when using this form when x is an integer. First, there is a condition called the Gibb~ Phenomenon which affects the series when x is close to an integer. Note that the series gives lxJ = ~ when x = I! We know that, for any integer, x - lxJ = O. Let us agree not to use integer values for x when we apply the formula.
i
... J J ]
It is possible to use the Fourier Series to derive all the usual rules for handling lxJ, although this might occasionally be complicated. For example, let us look at lx + n J, where n is an integer. We get
lx
+ nJ = x + n
1
2 1 (sin(27T"X + 27T"n) sin(47T"x + 47T"n) sin(67T"x + 67T"n) ) +- + + + ... 7T" 1 2 3 -
1 1 =x+n--+-
2
=
lxJ
7T"
(sin( 27T"x ) 1
+ n. 26
) + sin(47T"x + sin( 367T"x ) + ... ) 2
..
ARBELOS -
Volume 5, Chapter 1
Continuing, we have 1 (sin(271":1:) 1
1
l xJ =x--+2 11" l -x J = -x so
lxJ + l-xJ
1
2
+ -1 11"
+ sin(471":1:) + 2
(sin( - 211"x) 1
+
sin(671":1:) 3
sin( -471":1:) 2
+ ... )
+ sin( -671":1:) + ... ) 3
=-1.
These have been simple consequences of our new expression for l x J. There are other results that are somewhat more difficult, requiring care with Trigonometric terms. We leave it to the student to show the following (which is usually given without proof in most texts). Show that
~ l:XJ .,=1
(m - 1)(n - 1) 2
when m and n are relatively prime integers. (This may be a good start toward proving the Quadratic Reciprocity theorem.) This Fourier expansion has uses beyond those involving the greatest integer function. We may be able to use the series expression for l x J to evaluate some other series. So, let 211"x = 8 where 0 ~ 8 < 211". Then
or
sin 8 1
sin 28 2
sin 38 3
8
11" -
- - + - - + - - + ... = - - . If, in this series, we let 8 =
2
f, we find that
11" 111 -=1--+---+··· 4 3 5 7
a result we had previously obtained by other means. The following example takes a bit of originality. In our original expres sion for l x J, let x = Then
t.
o = l~J
1 1 1 (sin =---+3
2
11"
120 1
0
27
+
sin 240 2
0
+
sin 360 0 3
+ ...
)
'
. "'"
.J ARBELOS so
Volume 5, Chapter 1
i= V;(I-~+~-~+i-~+"-)·
J .....
Finally,
1 1 1 1 1 11" 1 - - + - - - + - - - + ... = - . 2
4
5
7
8
3V3
The reader is invited to try to find the sum of
..
1 1 1 1 1 1--+---+---+···. 5 7 11 13 17 We also note that, by suitable selection of x, it is possible to invent esoteric series that, to the uninitiated, seem to be impossible to evaluate in closed form. These may well be suitable for publication.
.... I
'11
THINKING ABOUT NUMBERS We wonder if anyone else is annoyed by thinking about numbers, their properties, and their peculiarities. We are often kept awake by these thoughts. For example, we know that if the product of four consecutive positive integers is increased by unity, the result is a perfect square. However, we now notice that if the sum of three squares is multiplied by three, the result is also equal to the sum of four squares.
] .
.",
..J
J
Examples:
2.3.4.5+1=11 2 •
We would welcome any similar curiosities you may have encountered or invented.
28
J
ARBELOS -
Volume 5. Chapter 2
COVER PROBLEM
from the
NOVEMBER, 1986 ARBELOS
[J
A
B
C
In 6ABC, AE, BJ, and CO are altitudes.
If OE
= 13, OJ = 14, and JE = 15, find the area of 6ABC.(t)
(t) The solutions to all of the geometry cover problems are published in Volume 6, the special geometry issue of the A rbelo~. Copyrisht
0
The Mathematical Allociation of America, 1991. In ita flrat edition, thll
chapter wal publllhed al the Arbelo~, Volume 5, blue 2, in November, 1986.
29
ARBELOS -
Volume 5, Chapter 2
J
,
"
....J PREFACE TO THE FIRST EDITION What with meetings, conferences and contests, our summers seem to be our busiest part of the year. We have finally gone through our correspondence and are delighted with the results. Our May cover problem attracted more than usual interest. To date, we acknowledge solutions from Bruce Denning and Francie Hsu of Bloom field Hills, the Mathematics Department of Roseburg High School (two ways). David Cheuk, Philip Lau and Ambati Jaya Krishna. Of special interest was the letter from L. J. Upton and A. Walker from Canada, which accompanied their solution. There may be more solutions coming in. The problem of the sum of fourth powers of sines and cosines of angles in arithmetic progression brought answers from Dr. Richard Gibbs of Fort Lewis College, and Russell May, reporting on the solution by his teacher, Steve Landy. The cover problem from the March edition was solved by Eddie Gutierrez, of Lufkin, Texas.
... '.'
J
J
Our Olympiad problem number 24, was solved by Ed Mullins, Stewart Kachru and Marshall Whittlesey. All in all, a rewarding experience for an editor. We have had requests for solutions of problems, the latest coming from Robert K. Franz. For the solutions to the Kurschak Problems, write to Profes sor George Berzsenyi, whose address is given. As for the Olympiad problems, we believe a set of solutions is being prepared by Professor Cecil Rousseau for the MAA/CAMC. Keep an eye open for its appearance. The best and most heartwarming news to the Editor is the mighty effort by Ms. Regina A. Grieco, who has prepared a sheet of formulas and an index of subjects that have been presented in all our Arbelo8 issues! We hope to add it to our May issue. Such help makes our work more rewarding.
.
...
J
Your Editor Dr. Samuel L. Greitzer
30
...
...
]
ARBELOS -
Volume 5. Chapter 2
AREAS ON LATTICES We have suggested that a good exercise in the use of ingenuity could be the derivation of as many formulas for finding the area of a triangle as one can - based, of course, on the use of various data. This article is the result of correspondence that leads us to suspect that using determinants to find the area of a triangle has disappeared from the curriculum.
IFigure
15·1
Let us assume that the reader can handle determinants. Given a tri angle with vertices at lattice points of a grid, we enclosed the triangle in a rectangle as shown in Figure 15. Then we find the area of the triangle by subtracting from the area of this rectangle the areas of the three unwanted triangular bits. That is, the area K will equal
31
1
ARBELOS -
Volume 5, Chepter 2
J
• "lI
...J .....
y
....
IFigure 16·1 When we expand and simplify this expression, we get
1 K =:= 2(XlY2 - X2Yl
+ X2Y3
- X3Y2
+ X3Yl
- XlY')'
Recalling the definition and properties of determinants, we see that we can rewrite this as 1 Xl Yl 1 K = X2 Y2 1 2 X3 Y3 1 Even this can be useful. For example, if (xl,yd, (X2,Y2) and (X3,Y3) are collinear, K = 0, so that we have a condition for three points to be on a line. Notice also that the determinant X
Y
1
Xl
Yl
1
x2
Y2
1
=0
is the condition for the arbitrary point (x, y) to lie on the line determined by points (Xl, yd and (X2, Y2)' We have a nice equation for a line through two points. We confess that we "fudged" a little in deriving the original determi nant. The reader will remember that, when two rows (or columns) of a determinant are interchanged, the sign of the determinant will change. We selected our areas so as to get the correct result. In all future work, let us agree that vertices are to be taken in counterclockwise order. This will help us avoid difficulties.
32
....
.... ..
,
..J
. ..,.
.J
.... J
1
ARBELOS -
Volume 5, Chapter 2
Let us apply all this to finding the area of 6.ABC in Figure 16 in terms of the areas of triangles with one vertex at the origin. Thust K = [OAB] = [OAB]
+ [OBC] - [OAC] + [OBC] + lOCAl.
Using the determinant form just derived, we easily derive
o
K= ~:I: 2 1
0
1
1
0
0
Y1
+ 2"
:1:2
Y2
:1:2
Y2
1 1
:1:3
Y3
110
0
+ 2"
:1:3
Y3
:1:1
Y1
1 1
1 1 1
If we expand these determinants using the top row, we find that
We can invent notation to aid the memory by transposing each determinant and combining the results into one simple form as follows: 1 :1: ) Y1
•
We can extend this form to take care of a figure with any number of vertices. Our final form is: 1 :1: ) Y1
•
The extra column at the right (an umbral column) is put there to make computation easier. To evalute, we perform diagonal multiplications down ward and get :l:1Y2
+ :l:2Y3 + ... + :I:"Y1
= D
and then perform diagonal multiplication upward and get
The area sought will then equal (D - U)/2. Notice that the columns of our new "determinant" are the coordinates of our polygon in counterclockwise order.
[J
t We let [XY Z] denote the area of 6.XY Z when X, Y, Z are listed counterclockwise. 33
... .~
I
.
Volume 5, Chapter 2
ARBELOS -
~
....J
y
...
...
(2,6)
1\
, -
I \ I II I
(3,3)
r--.. ........ """
(7, 2) 'I'
(1,1)
~
x
0 1Figure 17 ·1
This rule can be applied to find the area of any polygon whose sides do not intersect, even when the vertices are not lattice points. However, we will confine ourselves to polygons with vertices at lattice points for simplicity. We will apply our result to some simple cases.
For the polygon in Figure 17
K =
~
2
(1
1
7 3 2 1)
236 1
H
We find that D = 43 and U = 25, and thus K = 43 - 25). Hence the area sought is 9. At this juncture, let us remind the reader of Pick's Theorem: For a polygon with the area i~ K =
...
...
...
vertice~
at
crou-point~
of a lattice,
~ (lattice points on boundary) + (interior points)
- 1.
..,
... ,t
.. ...
!
In Figure 17, we count and find that K = ·4 + 8 - 1 = 9. Remember, however, that the determinant form is more general. Nev ertheless, Pick's Theorem is nice to know. To find the area of the triangle in Figure 18, the student could use the Pythagorean Theorem to find that AC = .J4O, BC = VIO and AB = J5ij. He could then use Heron's Formula to find the area. (Whoever does this 34
]
-... .. ..
1
ARBElOS -
Volume 5. Chapter 2
y (1,4)
B~
,
\
A
~ Jl,- ",-
.-.-
",-
"""""'"
(8, 3)
(2,1)
x
0
IFigure
18·1
y (0,6)
c (-5,0)
(0, -2) IFigure 19.1
problem this way deserves it.) Or, a student might note that the slope of AC is and the slope of BC is -3. Knowing that the product of the two is -1, one might know that ABC is a right triangle and find the area to be t . AC· BC = 10. Using our algorithm, he could find that
t
K =
1(2 81 2)
2 1 3
4
1
Or, using Pick's Theorem, one could get K
35
= 10.
= t ·4+ 9 -
1
= 10.
ARBELOS -
Volume 5. Chapter 2 • 11
Just for exercise, we have drawn a fairly complicated polygon in Figure 19. The reader is invited to use the determinantal form and Pick's Theorem to find the area enclosed by the figure. Then, one might try to find the area in some other way. (We did this by finding the area of the figure without the deep cuts at the right and left and then subtracting the two triangular cuts from the result.) It should be noted that the figure is symmetric about the Y-axis, so we can also find the area of the part to the right of the Y-axis and double the result.
...J
....
.. ....,
...
-
-
...
ANOTHER TRIANGLE AREA FORMULA
We remind the reader of two results which we have previously derived; namely, if R is the radius of the circumcircle of a triangle, the Law of Sines states abc --=--=--=2R sin A sin B sin C and this is a formula for the area of a triangle: K = we get abc = 8R3 sin A sin B sin C .
:~.
= 2R 2 sin
...
J
From the first,
Using the second, we finally get K
..
A sin B sin C.
...
Can our readers concoct additional formulas for the area of a triangle?
36
....
ARBELOS -
Volume 5. Chapter 2
THE BUTTERFLY PROBLEM In Figure 20, M is the midpoint of chord PQ, and AB and CD are two chords drawn through point M. Chord AD cuts PQ at X and chord CB cuts it at Y. Prove that MX = MY.
D
IFigure 20.' This problem has been around for a long time. One of the earlier solu tions was presented about 100 years ago by the same Horner who developed Horner's method for solving algebraic equations. It is called the Butterfly problem because of a fancied resemblance in the figure to the wings of a butterfly. We were reminded of the problem by some correspondence which gave the impression that, even to a mathematics professor, the problem was unknown. The present state of geometry in our educational institutions reinforces this impression. It is even possible that some of our readers have not heard of it. Hence this short article. To those who know of the problem, it may be a pleasant reminder. To those who do not, we hope it will prove a challenge and a diversion. Of the many proofs of the Butterfly problem, we shall present two. The first is rather sophisticated since it uses concepts from projective geometry.
37
ARBELOS -
Volume 5, Chapter 2
... ''1!!
We refer the reader to the article on projective geometry in the Volume 3, issue .5 of the Arbelo~, or to a good text on projective geometry. We define a conic (our circle in this case) as a projective but non perspective relation between two pencils of lines. In Figure 20, the pencils are AP, AD, AB, AQ and CP, CD, CB, CQ. Line PQ cuts both pencils in the corresponding points P, X, M, Q and P, M, Y, Q. The cross-ratios determined by these points are equal. That is, PX·MQ PQ.MX
=
PM·YQ PQ.YM·
Disregarding signs and remembering that PM = Q M, we find PX YQ MX = MY' MX+XP But MY + YQ
MX PX MY = YQ'
or
MP
= MQ = 1.
Hence M X = MY.
The second proof comes from Mr. Steve Conrad, and is probably the simplest of all proofs. We recall that the areas of two triangles which have an angle of one equal to an angle of the other are to each other as the products of the adjacent sides. In Figure 20 we have labeled the equal angles. Now, [X AM] XA·AM [MCY] = MC· CY' [XDM] _XD.DM [MBYj - MB·BY'
[CMY] [DMXj [BMY] _ [AMXj-
CM·MY DM·MX BM . MY AM·MX·
Note how the ratios of the terms at the left have been selected. We now multiply the four proportions. The product of the terms on the left equals one. We selected them so. Therefore, the product of the terms on the right equals one. That is,
j
....
...
.. ,
...
..
t
XA·AM CM·MY XD·DM BM·MY MC.CY· DM.MX· MB.BY· AM.MX =1.
Now we cancel like terms from numerator and denominator. This gives us:
XA.XD.(My)2 Cy. BY. (MX)2 = 1
or
38
XA· XD _ (MX)2 CY·BY - (MY)2'
..
ARBELOS -
Volume 5, Chapter 2
However, the "power of the point" X is
XA· XD
= Xp· XQ = (PM -
MX)(PM
+ MX) = (PM?
= QY· PY = (PM -
MY)(PM
+ MY) = (PM?
- (MX)2
and
CY· BY
- (MY?
Hence,
(PM? - (MX)2 (MX)2 (P M)2 - (MY)2 = (MY)2'
[J
so (MX)2 = (My)2 and finally, MX simplest concepts.
=
MY. This derivation uses the
For additi~nal proofs, the reader might consult Johnson's "Geometry", page 78 (which has a proof of a more general theorem), or "Geometry ReviJited" by Coxeter and Greitzer, page 45. This will provide four proofs (which may be more than the reader desires). But, as a challenge, can any reader send in a different proof, using, say, vectors, analytic geometry, trigonometry or any other mathematical methods?
[J
39
ARBELOS -
Volume 5, Chapter 2
TCHEBYSHEV'S PROBLEM
Our attention has been called to an interesting problem in probability theory proposed and solved by the Russian mathematician Tchebyshev. The problem:
If one write" a proper fraction ~ with both A and B "e lected at random, what i" the probability that the fraction i" in lowe"t term" P Tchebyshev's solution went somewhat as follows: Let P2,PS,P5, ... ,Pn,· .. be the probabilities, respectively, that 2,3,5, ... ,n, ... are not common factors of A and B. Then the probability that no prime is a factor of both A and B is the product p = P2PSP5 ... Pn .... The probability that A and B do have the prime n as a common factor is thus qn = 1 - Pn· If A (or B) is divided by n, the remainder must be one of 0, 1, 2, 3, ... , n-l. Each of these residues has the same probability of occurrence. Therefore, the probability that n divides A will be ~. Similarly for B: the probability that n divides B is ~. Thus, the probability that n divides both A and B is equal to ;,. That is, qn = 1 - Pn = ;,. Therefore, Pn = 1 - ;,. For the various primes, q2 = ~, qs = ~, etc. Then
At this point, we find the reciprocal 1 P -
J;.
1
(1 -
1 2 ,)
(1 40
~)
(1 -
51,) ....
.
.
...
q
1
...
... ...
'"
ARBELOS -
Volume 5. Chapter 2
We notice now that each factor on the right is the sum of a geometric progression. That is,
At this point, we notice that if we do perform the indicated multipli cation,.we will arrive at
11111 P = 12 + 2 2 + 3 2 + 4 2
+ ... ,
the sum of the squares of the reciprocals of all the integers. In a previous issue, we derived this sum in connection with a discussion of "in Ie as a product. We found then that the series of inverse squares had the value _pI
= 71"2 • 6
This completes the proof, yielding P
If the series for
=
6
71" 2
,
1> is expanded up to a moderately large prime number,
p is seen to be not too far from 0.6. Aside from its interest as a nice problem, it could be made the subject of an experiment. Have a class of students select numbers at random, and check to see, for two students, whether the numbers they chose are relatively prime. Perhaps one could get a fairly good value for 71". Problem: If one eliminates even numbers and numbers ending in five or zero, what is the probability that numbers then selected are relatively prime?
41
ARBELOS -
...
Volume 5. Chapter 2
. ..
..J
MANY CHEERFUL FACTS Trigonometry offers the reader many opportunities to exhibit ingenu ity. We could, for example, start with DeMoivre's Formula, (cos A
+ isin
A)n
= cos nA + isin nA.
...
This holds for all values of n. Let n = 2 and equate the imaginary and real parts to get sin 2A = 2sin A cos A,
'j
and 2
2
2
..J
2
cos 2A = cos A - sin A = 1 - 2 sin A = 2 cos A - 1. From n = 3, we get cos 3A = cos 3 A - 3 cos Asin2 A = 4cos 3 A - 3 cos A, which we used previously to solve a cubic, and also sin 3A = 3cos 2 Asin A - sin 3 A = 3sin A - 4sin 3 A, which we used to find the values of the sine of special angles. From n we can derive
. Sin
A
2"
=
/1 - cos A V 2
and
cos
~= 2
J+ 1
= ~,
J
cos A 2'
These are worth remembering. From a2 = b2 + c2 - 2bc cos A, we add and subtract 2bc, and find
a 2 =(b+c)2-2bc(l+cosA) Letting 28 = a cos
or
.J
(b+c+a)(b+c-a)=2bc(l+cosA).
+ b + c, we finally arrive
~2 = V~ ~
..,.
and then
at
. _J(
sln
~
2 -
8 -
b)( 8 be
-
c)
.
J J
42 ,
'"
.J
ARBELOS -
Volume 5, Chapter 2
It should be easy now for the reader to derive Heron's formula for the area of a triangle with sides given. Sometimes, it is useful to use the formulas e i9 = cos 8 + i sin 8
Then sin 8
e i9
=
e- i9 = cos 8 - i sin 8.
and
e- i9 2i
_
and
cos 8
=
ei9
+ e- i9 2
Perhaps the reader might spend a few moments finding ii , y'i, or maybe the i th root of i. For a triangle, where A + B + 0 = 7T, there are naturally special results. For example, we have tan A
+ tan B + tan 0
= tan
A· tan B· tan O.
We have used this to solve a problem involving a cubic equation. Historically, the formula cos
A
cos
B
=
cos(A
+ B) + cos(A 2
B)
was used before the days of logarithms (and hand calculators) to change a product into a sum. You let one number equal cos A, the other cos B, look up the angles A and B, and substitute on the right. A very pretty identity is the following:
A 2
B 2
tan - . tan -
+ tan
BOO + tan- . tan -A = 1. 2 2 2 2
- . tan -
It may take a bit of proving. The reader is also invited to derive the formula:
'A 'O A sm +'smB + sm = 4 cos "2 . cosB "2 . O cos "2' Again, if R is the circumradius of a triangle, then acos A
+ bcos
B
+ ccos 0
= 4Rsin A· sin B· sin O.
These formulas are enough to enable one to solve all sorts of problems. Thus, for example, if two angles, A and B, of a triangle sum to 45°, then (1 + tan A)(1 + tan B) = 2. Prove this. Also, if we allow
~=
a + ~ + c, then the area of the triangle, K, will
equal K = ~(~ - a) tan ~. We leave the question of algebraic identities for another time. Perhaps you might try to show:
z(1 - y2)(1 - z2)
+ y(1 -
z2)(1 - z2) 43
+ z(1 _
z2)(1 _ y2) = 4zyz.
ARBELOS -
Volume 5, Chapter 2
J
,... ,
,.
PYTHAGOREAN TRIPLES We assume that the reader is familiar with Pythagorean triple,; that
is, positive integers a, b, c, such that a 2 + b2 = c2 • A primitive Pythagorean triple [PPT] is one such that a, b, c have no common divisor. Thus, 3,4,5 is a PPT, but 6,8,10 is not. We shall call it a PT. Development of PT and PPT theory involves us with concepts of parity, divisibility, algebra and geometry with a large dose of ingenuity. First, a, b, and c cannot all be even if we are to have a PPT. Next, a, b, and c cannot all be odd because the sum of two odd numbers is an even number. Also, we cannot have exactly two of a, b, c even, for then the third one would be even due to parity considerations. This tells us that one of a or b is even, the other odd, and that c is odd. Let us assume that a is odd and b even. Then
The reader should verify that c; a and c; a are relatively prime. Now, if the product of two relatively prime integers is a perfect square, then each factor must be a perfect square. This means that there exist integers m and n such that c +a 2 -2-=m
and
c- a 2
... ... f
. ., ...J
.
=n 2
...
from which we easily find that
Conversely, if a, b, and c are as given in (*), then it is an easy matter to show that a 2 + b2 = c 2 • We should note that m and n must be relatively prime for a, b, c to be a PPT. Obviously, m and n cannot both be odd or both be even. 44
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...
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ARBELOS -
of
II
Volume 5. Chapter 2
We include a short table of m, n, a, b, c and, since a, b, and c are sides right triangle, the area and perimeter of the triangle in each case.
m,n
a,b,c
area
perimeter
2,1 3,2 4,1 4,3 5,2 6,1
3,4,5 5,12,13 15,8,17 7,24,25 21,20,29 35,12,37
6 30 60 84 210 210
12 30 40 56 70 94
Even this short table raises some questions that may be interesting. For example, how many PT's are there with perimeter equal to area? How many PT's are there with equal areas? How many PT's are there with any two of a, b, c consecutive integers? How many PT's are there whose area is a perfect square? We are sure that the ingenious reader will be able to pose other interesting questions. Some of these questions are simple. For example, if the square :1: 2 2a + 1 is an odd number, then a 2 + :1: 2 = a 2 + 2a + 1 = (a + 1)2, and we have PT's with one side less by one from the hypotenuse. We leave to the reader the exercise of finding PPT's like those in rows one and five, where the two short sides are consecutive integers. Again, suppose that two such triangles have equal areas and perime ters. Then a 2 +b2 =c 2 and ab=2(a+b+c). When we eliminate c from these equations and solve for b, we arrive at
8 b=4.+ - - . a-4 This has integer solutions when a = 5,6,8,12. And the only PT represent ing triangles with equal areas and perimeters are for 5,12,13 and 6,8,10. Further inspection of the table seems to show that, for any PPT a, b, c, one is divisible by 3, one by 4 and one by 5. The same side may be divisible by any two of these divisors, as in the 7,24,25 PPT. We leave this to the reader to prove. We also leave to the reader the proof that, in a PT and/or PPT, the product abc is divisible by 60. A lot can be done by realizing that, in the derivation of the forms for constructing a PT, we may let m and n be just about any expressions. Thus, let
45
ARBELOS -
J
J
Volume 5. Chapter 2
Construct PT's with al
a2
=m 2 - n2 = m 2 - ii.2
= 2mn b2 = 2mii bi
Cl C2
=m 2 +n 2 = m 2 + ii 2
and you will find that the two triangles are equal in area.
al,bl,Cl
u,v
m,n,n
2,1
7,3,5
3,1
13,8,7
3,2
19,5,16
4,1
21,15,9
4,2
28,12,20
4,3
37,7,33
5,1
31,24,11
5,2
39,21,24
Area
a2, b2 , C2 a3, bs , Cs
40,42,58 24,70,74 15,112,113 105,208,233 120,182,218 56,390,394 336,190,386 105,608,617 80,798,802 216,630,666 360,378,522 135,1008,1017 640,672,928 384,1120,1184 240,1792,1808 1320,518,1418 280,2442,2458 231,2960,2969 385,1488,1537 840,682,1082 264,2170,2186 1080,1638,1962 945,1872,2097 504,3510,3546
840 840 840 10920 10920 10920 31920 31920 31920 68040 68040 68040 215040 215040 215040 341880 341880 341880 286440 286440 286440 884520 884520 884520
IFigure 21.1
In fact, when one constructs a PT with a3
= (n + ii)2 -
m2
b3 = 2m(n + ii)
46
Cs
= m 2 + (n + ii)2
J
".
]
J
.
Volume 5. Chapter 2
ARBELOS -
one arrives at a PT with area equal to each of the two above it. Appropriate cll.lcll1ations for several triples of PT which have the same area are shown in Figure 21. The following is a very difficult result to obtain. We are indebted for it to "Canterbury PuzzleJ ", by Henry Dudeney. We read that Diophantus himself posed the problem of finding a PT such that the bisector of one acute angle would be rational. We thought we'd try it. [See Figure 22.]
,~
A
~~ B c
IFigure 22.1
We recall Stewart 'J Theoremt first. This says that, for any Cevian AT,
We also recall that, for an angle bisector, the segments m and n are proportional to the sides of the triangle adjacent to them. That is, m = kb and n = kc. Since m + n = a, we easily find that
ab b+c
m=--
Thus
2
2
b+c
b+c
and
ac b+c
n=--.
c ab b ac - +- = t 2 a+ -ab- . -ac- . b+c b+c
We spare the reader the manipulations, and get:
2 2-b ((b+c?-a ) (b + c)2 .
t - c
t For a further discussion of Stewart's Theorem see page 69 in this volume. 47
. .., .,j ARBELOS -
Volume 5. Chapter 2
Now let a = 2mn, b = m 2 - n 2 and c = m 2 simplify, check us if you wish to, and we have
t2
= (m 2 + n 2 )(m 2 -
+ n2•
Substitute and
n 2 )2
m2
J J
Therefore, if we see to it that m 2 + n 2 is a perfect square, then t will be rational. Let m = 4, n = 3. Then t = 2~t and t = 345. Of course, if we multiply all dimensions by 4, these will all become integers. We have been curious about how Diophantus solved this problem, but have not been able to discover it.
2
We have found that a number of texts on the theory of numbers state that a PPT cannot have an area with magnitude a perfect square, but without proof, or even a hint on how to proceed. We shall sketch our proof, but ask whether some reader may be able to provide a shorter and simpler one. Let the sides of our PPT be represented by m 2 - n 2 , 2mn, m 2 + n 2 • Of the totality of such PPT's, there is one for which m + n is a minimum. The area of the corresponding right triangle is mn(m + n)(m - n)
= K 2• ~
Because we are dealing with a PPT, each factor at the left of this is relatively prime to all the other factors. Since the product is a perfect square, each factor must then itself be a perfect square. Thus,
From the latter two expressions, we easily find and
Now assume that :&
u +v = -2-'
Y
u- V = --. 2
Then:&
2
2 v2 +y2 = u + = r2, 2
u2 _ v2 82 :&y 82 and :&,y,r form a PT. Also :&y = 4 = "2 so the area 2 = 4' This area is a perfect square because 8 is even since u and v are odd. Thus, :&, y, r form a PT with area 8 a perfect square. However, :& + y = u < u 2 = m + n, and we have arrived at a situation where :& + Y is less than the minimum m + n.
48
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ARBELOS -
Volume 5. Chapter 2
In theory, we could continue thus, getting ever smaller such "minima". Howt'!vt'!r, we are dealing with positive integers, and we have reached a contradiction. This is an example of the method of infinite de~cent proof, invented by Fermat. The directions in which one can go are limited only by the imagination of the researcher. For example: a) We note that there are PT's whose side are terms of an arithmetic progression. The sets 3t, 4t, 5t are such PT's. Are there any other ~uch PT'~P
b) Are there triples of squares of integers in arithmetic progression? There are. One such triple of squares is the sequence 1,25,49. The common difference here is 24. In the general case, the common differ
ence is called a congruum.
One result on congruums is the following: If a, b, c form a PT, then
2ab is a congruum. For, from a 2 + b2 = c2, we get, on adding and subtracting, a 2 _ 2ab + b2 = c2 - 2ab, (a - b? = c 2 - 2ab,
2 a 2 + 2ab + b = c 2 (a
+ b)2
= c
2
+ 2ab.
+ 2ab.
If we take a = 12, b = 5, c = 13 and substitute, we get the set 49,169,289. The congruum is 120.
Enough has been presented to show that investigation of PT's is inter esting. The reader might solve some problems now, like (1) Find three PT'~ with equal area... (Montucla said that one can find any number of such triangles, so next try to find four.) (2) A quadrilateral inscribed in a circle has sides (in order) of 7, 15, 20, 24. Determine the area of the quadrilateral, the length~ of the two diagonal~, and the radiu~ of the circle.
49
ARBELOS -
Volume 5, Chapter 2
...
..J
... JU8T FOR FUN The following problems use methods that we developed in previous issues of the Arbelo~. If your memory is good, the problems should not be too difficult.
1) This series is convergent. Find its sum: 3
3 5
3 5
7
8=1+-+-·-+-·-·-+· .. . 4 4 8 4 8 12
... ...
2) Solve the simultaneous equations
x+y+z=3 2
+ y2 + z2
=
x3
+ y3 + z3
= 6
x
3
3) Each of m points on one line is joined to each of n points on another line by segments with endpoints on both lines. Find an expression for the maximum number of points of intersection of these segments.
..J •
4) If n is an integer greater than unity, then
n5
-
5n 3
+ 60n 2 -
56n == 0
(mod 120).
5) Solve for x, y, z in terms of a, b, c if
x y z xyz ;=b=~=x+y+z
...
... 50
-
ARBELOS -
Volume 5, Chapter 2
DESCRIPTIVE GEOMETRY A problem included in a letter from a reader has brought back mem ories of a course we took very long ago at Stuyvesant High school. It was called Mechanical Drawing then. There are some old texts around in which it is referred to as Descriptive Geometry. We still recall working problems that seem to have disappeared with the de-emphasis on geometry. Let us describe what it is about as simply as possible. We assume that we have a solid in three dimensions. Of course, we can exhibit this in the form of a photograph or carefully made drawing. For our purposes, however, we draw what we see when looked at from three directions, front, top and side. For any point of the figure, this amounts to locating it as the intersection of three mutually perpendicular planes, which locates it uniquely. For example, suppose we start with a cube. The views from front, top and side will all be squares.
LII
TOP
FRONT
"" ,
" "" "
""
[J-tJ
SIDE
IFigure
23·1
Figure 23 shows the three views of the cube in a planar form. For point P inside the square, suppose that P is as shown on the top view, and also on the front view. Then the position of P on the side view would be
51
. ...
ARBELOS -
Volume 5, Chapter 2
as located by the dotted lines. If one could cut out the first quadrant and fold along the axes, pasting the positive x-axis to the positive y-axis, one would get what one sees in each direction. There is a lot more to Descriptive Geometry than this. We remember finding intersections, shades and shadows, and a lot more. We even devel oped blueprints. However, here let us restrict ourselves to two questions: Can one describe a figure completely from the diagram, and, given two views, can one find the third?
J
..
.. tl,
.. !
D
.. ,I'
.. I I
IFigure 24·1 Let us consider the second question first. To this, the answer is NO. Figure 24 shows us an example of a solid, two views of which are the same as views of the cube. There are many other such examples. At this juncture, we bring up the question of the ability to picture the solid from the diagram. This is a skill in the realm of spacial relations. There is the possibility that the ability to picture figures in three dimensions is a native trait and not capable of development. However, we believe it can be improved. (We believe the same is true of problem solving ability.) If you can "see" the figure, good! If not, try! It might be useful to show the plans for some other figures. The one in Figure 25 is that of a cone. We resist the temptation to add an intersecting plane and constructing the conic section, although we had to do this in class. The plan in Figure 26 shows a pyramid, upside down. The dashed lines in the top view represent edges that are there but not visible from the top. We have also included dotted lines used to locate the vertex. We hope the reader can picture the pyramid standing on its vertex.
52
J
..
~ ,I,
...
<
t ARBELOS -
Volume 5, Chapter 2
0
""
"
IFigure 25·1
I'
/
" \}//
/
/
"
/K- -
1/
I
"
,
I
""
v
v
IFigure
26·1
The diagram in Figure 27 shows a rectangular block with a hole going halfway through it. Of course, all lengths are to scale so that one can find out all lengths, areas and the volume. Sometimes, this is not as easy as it sounds. If we wish to find the length of an edge of the pyramid from its plan, we would have to proceed with caution. None of these edges are parallel to the planes of the plan. Nevertheless, we can find the length. For your edification, we describe the method. First, we rotate a selected edge so as to make it parallel to one of the planes, front, top or bottom. Then we can measure its length.
53
ARBELOS -
Volume 5, Chapter 2
....
...
-
J
IFigure 27.1
We leave it to the reader to determine just how this can be done. It is actions like this that raise the subject above mechanical drawing and make it descriptive geometry.
@
J
..
I
.. A I
I I
1
1
I 1 I
1 I
( ] I 1
( ] I I
B
..•
f
IFigure 28·1 We hope that the reader can now recognize that Figure 28 represents a sphere with a hole through it. We call attention to the fact that in the front view, each circle has a portion at the top and bottom that is a straight line. Now for a problem: 54
...
...
~
I
ARBELOS -
Volume 5. Chapter 2
In the diagram of the sphere with the hole in it [See Figure 28.], if AB equals 4 inches, what is the volume of the figure?
c
.A-
A,e-...:.-.;.....:::""",_ _=B-.r
\1 1
" I I
"
I
HJ ,
,
"
,
~_
E
IFigure 29·1 Next, Figure 29 shows a cube which has been cut by a plane passing through the midpoints of AD, AB, B F, FG, GH, H D. What would the plan be for that half of the figure containing vertices A, E, F, H, the rest being discarded?
IFigure 30·1
Next, the diagram in Figure 30 shows the front view and top view of a solid. What does the side view look like? (There are two possible answers.)
55
ARBELOS -
Volume 5. Chapter 2
J
.
....
IFigure 3t.1 Finally, the diagram in Figure 31 shows the front view and top view of a solid. What does the side view look like?
.J
... ... . ...
t
~
1
... 56
ARBELOS -
Volume 5. Chapter 3
COVER PROBLEM from the
JANUARY, 1987 ARBELOS
KVANT
ABCD is a square. K is any point inside ABCD. MN is any line through K meeting AB at M and CD at N. Circles are constructed through points AM K and C N K intersecting at P. Prove that P lies on AC.
Copyright
0
The Mathematical Association of America, 1991. It Ita lint edition, thi.
chapter was published as the
Arbelo",
Volume 5, Issue 3, in January, 1987.
57
ARBELOS -
Volume 5, Chapter 3
J
.... I
PREFACE TO THE FIRST EDITION One of the advantages of editing the A rbelo.! is the pleasure of receiving and sometimes answering letters from readers. The situation being what it is, this keeps us very busy. It also gives us ideas on articles for future issues.
-
]
The cover problems have elicited the greatest number of responses. Ap parently, geometry problems still have great appeal. The Roseburg High School Math Club of Roseburg, Oregon sent in an excellent solution, beautifully writ ten and a pleasure on the eyes. L. J. Upton, of Mississauga, Canada, sent in an elegant solution to a cover problem. So did Ambati Jaya Krishna. Good solutions to problems came from Brad Threatt, of Rock Hill, S.c., and Sam Vandervelde.
]
A letter from Professor Donald J. Brown, chairman of the mathematics department at St. Albans School in Washington, D.C., gave a solution to a problem of summing sines and cosines very nicely. We appreciate the material he sent us and promise to use the method.
...
We have acknowledged only a few of the letters we have received. Our problem is that of space. Thus, we have not usually given solutions to the cover problems we have proposed. However, requests for solutions keep coming in. t Your Editor Dr. Samuel L. Greitzer
1>'" 'I
,.~
I
...
,.
t The solutions of many of the cover problems are now available in Vol ume 6, the special geometry issue of the Arbelo.!. 58
,..
ARBELOS -
Volume 5. Chapter 3
SPECIALIZE! There was a famous mathematician who believed that "man muu im mer umkehren!" - that is, invert. Another famous expert insists that one should always "generalize". As an example of the latter, we have already shown that lines drawn from the vertices of a triangle to the points of tangency of the inscribed circle are concurrent. A generalization would be, "If line~ are drawn from the vertice~ of a tetrahedron to the point~ of tangency of the in~cribed ~phere with the oppo~ite face~, will the~e line~ be concurrentP" (We leave it to the reader to investigate this.)
It appears to us that there are times when it is more useful to specialize. If nothing is said about a triangle in a problem, perhaps using a right triangle or an equilateral triangle may make it easier to solve the problem. There are cases where it is useful to collapse the triangle into a line. We would like to present some illustrative results.
Specializing Ptolemy's Theorem
A
IFigure 59
32.'
ARBELOS -
Volume 5. Chapter 3
The last time we presented Ptolemy's Theorem was in Volume 1, Num bP.T:t. Ht>nce it may be useful to show the proof once more. [See Figure 32.J Let ABC D be an inscribed quadrilateral. Construct AM so that LAMD = LABC. (Thus LAMB = LADC.) Since LADM and LACB intersect the same arc, and LABM and LACD intersect the same arc, we have
J
J
... 'r
6ABC,...., 6AMD b MD Ac=T bd = AC . M D ac + bd =AC . (M D
6ABM,...., 6ACD a AC MB=~ ac = AC . M B
+ M B) =
AC . BD.
We think this is a beautiful theorem. We have also found it useful. For example, we worked out several formulas in trigonometry by using it. However, let us specialize. We take the inscribed quadrilateral to be a rectangle [See Figure 33.J, with sides a and b, and diagonal e. We apply Ptolemy's theorem and get a2
+ b2 =
ac
+ bd =
AC . BD = (AC)2 = e 2.
Thus, we have a nice proof of the Pythagorean Theorem.
-I
.... I
•
...
J
J
IFigure 33·1
.
...
Next, specialize again. Let the rectangle be a square. Note that if a = b in Figure 33, then 2a 2 = e 2 , so = J2. This reminds us that a we can use approximations of diagonals of certain squares to obtain good approximations of y'2.
=
60
."!
ARBELOS -
Volume 5, Chapter 3
An Aside about the Talmud A student once showed us something in the Talmud: It was a square with side ten plus another square joining the midpoints, as shown in Fiaure
34.
f----I----~
10
1Fiaure 34·1
First, we note that the area of the outer square is 100. Therefore, the area of the inscribed square is 50. This is close to 49, so the length of a side of the inscribed square is about 7. Now one has two choices. Using one of the four small squares, V2 ::::: ~. Using the square of side 7, we have M2 ~ !Q v~~ 7' For a better approximation, let the side of the large square be 99. Then the area of the inscribed square is ~ .99 2 = 4900.5, so the side of the inscribed square is approximately 70. Thus V2 ::::: ~~, which yields a value correct to at least three places.
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ARBELOS -
Volume 5, Chapter 3
J
,... ...,
... l
Specializing Hexagons in Conics One can specialize other theorems. We remind the reader first of Pas cal's Theorem. Roughly, for a hezagon in~cribed in a conic, the inter,ec tion~ of the oppo,ite ,ide, lie on a line. One may let the conic be a circle and the hexagon a regular hexagon. Or one may specialize the hexagon. As an example, let the hexagon be ABCDEF, where A and B coincide, and D and E coincide. Then, where AM and DM are tangents, BC and E F intersect at L, and CD and FA intereet at N, we have L, M and N collinear. [See Figure 35.]
.... 'j
J J ...."' J J I
N
1Figure 35·1 .... Exercise: If tangents drawn at C and F intersect at P, then L, M, Nand P lie on one line. One can operate in the same manner with Brianchon's Theorem, etc., but we hope we have made clear what we mean by specializing in geometric situations.
Specializing in Algebra One of the important steps involved in solving equations of degree three and four consists of transforming the equation into another with the second term missing. We explained the method in Volume 3, Number 3.
62
ARBELOS -
Volume 5. Chapter 3
This method can also be used for solving quadratic equations. Thus, to solve x 2 - 8x - 13 = 0, we reduce each root by ~ = 4, using z = x - 4: x2
-
8x -13 = (x - 4? - 29 =
Z2 -
29 = 0
and obtain z = ±V29. thus x = 4 ± V29. We don't say this method for solving quadratics is easier, only that it is different. In our article "Factoring i3 Fun" we referred to this method of making quadratics easier to factor. Many times we can simplify by means of a reduction without com pletely eliminating the second term. For instance, suppose we wish to fac tor x 2 - 28x + 187. The coefficients are rather large. We therefore reduce by 10, using z = x - 10: x
2
-
28x
+ 187 =
(x -
10? - 8(x - 10)
+7=
Z2 -
8z + 7.
Since Z2 - 8z + 7 = (z - 7) (z - 1), we add back the 10, to see that we originally had (x - 17)(x - 11). The reader may also recall the method for multiplying or dividing roots by any number desired, [IfAx 2 + Bx + C = 0, then z2 + Bi + AC = 0 where z = Ax, and go on from there.] and use this as well. However, our main purpose here is different. First recall that in Volume 2, Issue 3 of the Ar6elo3 we introduced Newton's Formulas for the sum of the products of the roots of any equation. Suppose the roots of an equation, say a cubic, aox 3 + al x 2 + a2x + a3 = 0, are rl, r2, r3' Let 81t = r~ + r~ + r:. Then Newton's Formulas are: ao81
+ al
= 0
+ al81 + 2a2 = 0 ao83 + al82 + a281 + 3a3 ao82
= 0
We have a set of recursion formulas from which we can operate when needed. We used these formulas when solving a problem from the first USA Math Olympiad, namely, x+y+z=3 x2
+ y2 + Z2 = 3
x3
+ y3 + Z3
= 3.
In this case, we let x, y and z be the roots ofthe cubic aP 3+bp2+ cP+d = 0, and solve using this: a81
+b =
0,
a82
+ b81 + 2c = 63
0,
a83
+ b82 + C81 + 3d
"
...J ARBELOS -
Volume 5, Chapter]
with "1 = "2 = "3 = 3 to find aP 3 + bp 2 + cP + d = a( P - 1)3, so the only solution is to the given equations is ~ = y = z = 1.
J
Now there is no reason why we can't specialize and use the same method for solving simultaneous quadratic equations or mixed equations. Our first example is particularly easy. We wish to solve ~
+y = 28 ~y
= 187
We let ~ and y be roots of aP 2 + bP + c = o. . Immediately, we get a(p 2 -28P + 187) = O. We solve this (note that we have already done so), and get or ~=17,y=11 ~ = 11, y = 17. Compare this with the method usually given to solve such equations.
....
..
Next, suppose we are given
= 125
~2 +y2 ~
+y
= 15
In the associated quadratic with roots ~ and y, p 2 + bP + c, we have b = -"1 = -15. Substitute first P = ~ then P = y into the quadratic and add to obtain the result, 125 - 15 . 15 + 2c = 0, so c = 50. Hence our quadratic is p 2 - 15P + 50 = 0 which solves easily with P = 10 or P = 5. Hence or ~ = 10, y = 5 ~ = 5, y = 10. Suppose we are faced with 3~
+ 2y = ~y
I
= 10
+ (2y)
3~
=0
and 2y, multiply the second
60.
-
16P + 60
or
3~
or
~
becomes p 2
= 10, 2y = 6 10 ~ = -, y = 3 3
3~
= 16
(3~)(2y) =
Then p 2 + bP + c P = 6. That is,
....
16
We let the roots of the allied quadratic be equation by 6, and go about solving (3~)
64
...
= O.
That is, P
= 6, 2y = 10;
= 2, y = 5.
= 10
or
.....
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ARBELOS -
Volume 5. Chapter 3
The point of this article is that, even when there is a set method for solving II. prohlem, it is not necessary to restrict oneself to that method when there is another, simpler method. Consider, for example, the equations
1
1
7
-+-= 12 z y zy = 12.
Rather than solve for z, say, substituting to get a quadratic equation and solving that, we could do as follows: . z +y 7 From the first equatIOn, - - = - , so z zy 12
For p 2 - bP + c P = 4. Hence z
+y =
= 0, we have, at once, p 2 = 3,
y
=4
or
7P
7.
+ 12, so P = 3 or
z = 4, y = 3.
We conclude with an example and an exercise that may be of interest to the reader: In Figure 36, the secant is special. It passes through the center of the circle of radius b. Then, we may write
and we have another proof of the Pythagorean Theorem.
IFigure
36·1
Finally, an exercise: Solve Z3
c
+ y3 = 407 z + y = 11. 65
"
...J
ARBELOS -
Volume 5, Chapter 3
J
-
SUPERNUMBERS
...
t
This all began when we explained a problem that had appeared in A rbel06 some time ago. The problem: If the ten6 digit of the 6quare of a number i6 6even, what i6 the unit6 digitP We give the solution first. Let the number be N = lOt + u. Its square is N 2 = 100t 2 + 20tu + u 2 • Now 100t 2 will not affect the units digit of N 2 • Since 20tu is even, u 2 must contribute to the tens digit to make it odd. Hence u cannot equal 1, 2 or 3. These contribute nothing to carry. Also, u = 5, 7, 8 and 9 have squares whose tens digits are even numbers. This leaves only u = 4 or 6 as possibilities. Each of these leaves a units digit equal to 6. For example, 24 2 = 576 and 26 2 = 676.
-
This example leads one to ask when the square of a number has the same units digit as the number itself. We can deteqnine this mentally, for: If a number ends in 1, its square ends in 1; If a number ends in 5, its square ends in 5; If a number ends in 6, its square ends in 6.
Naturally, one then asks whether it is possible for a number and its square to have the same tens and units digits. If the number end6 in 5, then N = lOt + 5, and N 2 = 100t 2 + lOOt + 25. Now N 2 - N should be divisible by 100. This means, since N2
-
N
= 100t 2 + 90t + 20,
that 90t + 20 must be divisible by 100. This happens only for the digit t = 2. That is:
......
.J
If the unit6 digit of a number i6 5 and the number and it6 6quare have the 6ame ten6 and unit6 digit6, then the number mud end in 25. If a number end6 in 6, we may write it N = lOt + 6. Its square is 100t 2 + l20t + 36. Again, N 2 - N = HOt + 30 must be divisible by 100. Obviously, this will occur for the digit t = 7. Therefore:
66
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ARBELOS -
Volume 5. Chapter 3
If the unit" digit of a number i" 6 and the number and it6 6quare have the "ame ten" and unit" digib, then the number mu"t end in 76. (We leave it to the reader to find out why we didn't work with a number ending in 1.) Of course, we are hot on the trail of something! Can a number and its square end in the same three digits? Let's see. If a number ends in 25, we may write it as n = lOOt + 25, whence N 2 = 10 4 t 2 + 5000t + 625. So, N2 - N = 104 t2 + 100( 49t+6), and the sum in parentheses must be divisible by 10. This occurs only for the digit t = 6.
We now have the following:
If the la"t two digit" of a number are 25 and the number and it" "quare have the 6ame lad three digit6, then the number mU6t end in 625. Now suppose a number ends in 76. Then N = lOOt + 76 and
N2
-
N = (10 4 t 2 + 15200t + 5776) - (lOOt + 76) = 10 4 t + 100(15lt + 57).
As before, 15lt+57 must be divisible by 10, which will occur when t = 3. We can now state:
If the lalt two digit" of a number are 76 and the number and it6 "quare have the lame lalt three digitI, then the number mU6t end in 376. We hope it is clear that we can continue in the same way to find numbers whose squares have the same k digits as the number itself. (Note that it is critical that there be a value of t such that the expression of the form at + b derived in each step be divisible by 10. This will always be the case if a is relatively prime to 10. Can you prove that all a in 6uch ezpreuionl will be relatively prime to 10 f) Just for fun, we continued the process and arrived at the following:
67
ARBELOS -
IL!!" num.1?~r
Volume 5. Chapter 3
epds in
it.1I
8qllll.~e
ends in
5 25 625 0625 90625
5 25 625 0625 90625
6 76 376 9376 09376
6 76 376 9376 09376
It would appear that we can continue as far as we wish. Suppose we agree to call the numbers we "finally" derive ~upernumber" S5 and Sa. Then, if Si has some integer in, say, the kth position from the right, Sl will have the same integer in the same position. This seems to mean that S~ = Ss and S~ = Sa. Finally, it seems that the equation z2 - Z = 0 has four solutions, namely, z = 0,1, Ss, Sa. This gives us something to think about!
J
." ..J p"""
..J
-.... ...
J
... ... I'
,..
J N ate: We are reminded of "Tristram Shandy", by Sterne. He is sup posed to take one year to describe one day in the life of Shandy. As in the case of the supernumbers, Sterne can, in theory, write the complete biography of Shandy. We welcome any thoughts any reader may have on this topic!
-.,. J
"111
f
68
ARBELOS -
Volume 5, Chapter 3
STEWART'S THEOREM We first introduced Stewart's Theorem:
If the point P (AC)2 BP
i~
on
~ide
BC in triangle ABC, then
+ (AB? PC =
in Volume 2, Issue 5 of the attention we feel it deserves.
(AP)2 BC
Arbelo~.
+ BP· PC· BC
We did not, probably, give it the
A
B
C
Recall that any line from a vertex of a triangle to the opposite side is called a cevian of the triangle, so AP is a cevian of 6.ABC. Let us first give a short proof of the theorem. Name the lengths of the sides and angles as in Figure 37. We apply the Law of Cosines twice once to 6.APB and once to 6.APC. This gives us
c 2 = m 2 + t2 - 2mt cos 8 b2 = n 2 + t 2 + 2nt cos 8
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ARBELOS -
Volume 5. Chapter 3
because cos('7I"-8) = - cos 8. Now we eliminate the terms involving cosines and get b2 m + c2 n = m 2 n + n 2 m + t 2 (m + n). Now let m
+n =a
to obtain
J
...
If AP is an angle bisector, for example, recall that m and n. are in the same ratio as c and b. Thus, since m + n = a,
c·a b+c
b·a b+c
n=--.
and
m=--
Substituting in the formula (*), we get
J
With a little manipulation, we can arrive at
bc.9(.9 - 2a)
(b + C)2 where.9
... .... '" -I
= Ha + b + c). A
B
a/2
M
G
a/2
If the cevian is a median [See Figure 38.], formula (*) takes the form
b2 ~ 2
+ c2 ~
2
= t2 a 70
+ ~ . ~ .a 2
... i
IFigure 38·1
2
J
... ...
ARBELOS -
Volume 5. Chapter 3
A
IFigure 39·1 Given a quadrangle ABCD and it~ diagonal~ AC = P and BD = q. The ~egment M N join~ the midpoint~ of the two diagonal~. Find the length of ~egment M N In term~ of a, b, c d, p and q. which eventually whittles down to t 2
=
b2 + c 2 a2 --2- - 4'
The reader who wishes to apply Stewart's Theorem to find the length of an altitude is welcome to try it. We did, and found the algebra complicated. It is easier to find the altitude by using other ways, like those involving areas. A very pretty problem is illustrated in Figure 39. All this problem requires is a continued use of the formula for the median of a triangle, just derived: First, BM is a median of 6ABC. Therefore,
P P a 2P - + b2P-= (BM)2 p+-._.p 2 2 2 2 which simplifies to
Similarly, DM is a median of 6DAC. Therefore,
2P d2P P P c-+ -= (DM)2 p+-.-.p 2 2 2 2 71
ARBELOS -
Volume 5. Chapter 3
or c2
...
+ d2 =
~
2
2(DlIf)2
+ e_.
...-'"
2
i
Adding the two results, we get a 2 + b2 + c2
+ d2 =
2(BM?
+ 2(DM)2 + p2.
Now we note that MN is a median of 6.BMD. Hence
(BM)2tI
2
+ (DM?tI == (MN)2 q + tI. tI. q 2
or
(BM)2
+ (DM)2
2
2
2
= 2(MN?
+ L. 2
Now we can use this to eliminate BM and DM from our previous result, and finally arrive at
Incidentally, if M N = 0, we get a parallelogram, and we see that the sum of the squares of the sides of a parallelogram equals the sum of the squares of the diagonals. The nice thing is that the converse also holds.
.. ...
.. 0"" I
...
....
.. .... ..J
We leave the reader with a problem that seems to be going the rounds at present; namely, locate the point in' ide a triangle /rom which the ,um of the squares of the diJtance, to the vertice, i, a minimum.
...
.. t
...
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ARBELOS -
Volume 5, Chapter 3
PROAfiSCUOUSEXAMPLES 31) Show that the product of the diameters of the inscribed and circum . I ' Ie 'IS equ al to 2abc scn'b ed CIrC es 0f a tnang b a+ +c 32) If a and p are the roots of :z:2 + p:z: + q = 0 find, in terms of a and ·the equation whose roots are p and q.
p,
33) Find the sum of the infinite series
S = 1+
2
2·5
2·5·8
(; + 6 . 12 + 6 . 12 . 18 + ....
34) Solve the equations :z:2
+ y2
= 61
:z:3 _ y3 = 91.
35) If the sides a, b, c of a triangle are roots of a cubic
find m,n,p in terms of R,T,S. 36) In triangle ABC with incenter at I, excenters at I"" I b, Ie (in usual position) prove that
AI AI",
+
BI BIb
CI
+ Cle =
1.
37) Solve for :z: and y: a:z:+by=1 c:z: 2
+ dy 2 = 1.
Note: These problems are not too difficult, and depend on materials and methods that have been discussed in this and previous issues of Arbelos. We hope you will find them relaxing and interesting. 73
..
ARBELOS -
.... ,
Volume 5. Chapter 3
~
..... M~CELLANEOUSPROBLEMS
a) [See the diagram at the left in Figure 40.] Square ABCD lies in Quadrant I of a rectangular coordinate grid. Vertex D has coordinates (12,17). What are the coordinates of vertex C? b) If f(l) = V2, f(2) = expression for f (n).
V2 - V2,
f(3) =
V2 - V2 - V2, ..., find
c) If 3x 2 + 9kx + 3x + 1 = 0, where k is a real number such that k what is the probability that the equation has real roots?
an
< 2,
d) [See the diagram at the right in Figure 40.] In square ABCD, arcs are drawn with centers at Band D, and radii equal to the side of the square. These arcs cut diagonal BD at X and Y. If XY = 2 units, what is the area of square ABCD?
., ...J
. '" ....I
J
e) The equation x 2 - 97x + A = 0 has roots that are the fourth powers of
the roots of x 2 - x + B = o. What are the possible the values for A?
D
J
A
.... C B
IFigure 40·1
, ..-Ii
74
...
...
,
ARBELOS -
Volume 5. Chapter 3
CIRCUl\f- IN- and EX-CENTERS
For a given triangle, the circumcenter is the center of the circle passing through the vertices, the incenter is the center of the circle tangent to the three sides, and an ezcenter is the center of a circle tangent to one side externally and the other two sides extended. [See Figure 41.]
IFigure 41.1 The point 0 is the circumcenter of the circle passing through vertices
A, B, Cj I is the incenter, and I", h, and I c are the excenters. Remember that AI bisects the interior angle at A, Alb bisects the exterior angle at A, etc. This article arose from a. problem presented to us: Prove that the circumcircle bi3ect3 each of 3egment3 I I",
Ih, and IIc .
75
ARBELOS -
J
Volume 5, Chapter 3
The diagram is quite interesting, and we decided to study it more carefully and closely. We have removed some lines and added others, and are now staring at Figure 42. Triangle IBM attracts us. We can find all its angles, for example. First, LIMB = C, since both subtend the same arc, AB. Next, LIAB = and LIBA = ~ = LIBC. As an exterior angle, LMIB = Now fCBM = since it subtends the same arc as LCAI. Therefore LMBI = LCBM + LIBC = (A + B)/2. Thus, 6M BI is isosceles with MB = MI. This is alluring, at least to us.
Ai B.
4,
..
... 'I
4
...
...
...
J . ..,.
M
1Figure 42·1
.
We have drawn the diameter UV containing 0 and I. Let the length of segment 01 be d. Since OU = OV = R, we have
UI . IV
= (R -
d)( R
+ d) = R 2 -
d2 •
We have two chords intersecting in the circle. Thus,
R2
-
d2
= AI . 1M = AI . M B.
... ~
... ...... "
(*)
... ~ I
76
.
~
•J
~
ARBElOS -
Volume 5. Chapter 3
We included IT = r, perpendicular to side AB. Let us now add line BN, and we have two similar triangles, ATI and N BM. Thus
2R AI
MB r
AI ·MB = 2rR.
or
Combining this with the formula (*), we have found that
In the form d 2 = R(R - 2r), this is another one of Euler's productions. It tells us also that r 1 - <-. or R ~ 2r R-2
Now we know that the area of triangle ABC equals
abc
4R
rs
or
vs(s - a)(s - b)(s - c).
or
Therefore, r
4R . abc· s 1
8abe or
= s(s - a)(s - b)(s - c), ~
(s - aHs - b)(s - c),
abe>(b+e-a)(e+a-b)(a+b-e).
77
· '"
ARBELOS -
Volume 5, Chapter 3
J
·...,I ~
A
J · .,
.j B&--.....- -.......·C
BH Next, - r
get
= cot
B HC - and - 2 r
= cot
C -. [See Figure 43.] Adding, we 2
~ + __ cos Q =
aBC cos _ = cot + t _ = _ _ 2 r "2 co 2 sin ~
2
sin Q
2
Multiply by sin to arrive at
Using ---=-----A SIn
2
2
2
2
4 in numerator and denominator, and then multiply by 2, 2a
-;: = a
2
(~
sin + Q) cos A 2 2 _ 2
sin ~ sin Q - sin ~ sin Q'
sin A
sin A sin ~ sin 2
2
Q . 2
= 2R, we substitute and find
...
'J ...
· ..
... ... I
''1t
-
r
4R
,A,
B , C
= SIn - sIn - sIn
2
2
2'
i
and, finally, 1 ,A, B , C - Sill - Sill 82 2 2'
- > Sill which holds for all triangles.
78
...
... "
I
ARBELOS -
Volume 5. Chapter 3
However, we do have a prnhlcll1 to solve. In Figure 44 we recall that f:::,M B I is isosceles. Since LB M I = e, o
I..MBI = I..MIB = 90 -
e
2'
Note, however, that IB is perpendicular to Bla . Therefore, LMBla
= 90 o
- LMBI
This yields LMlaB =
e = 2'
Also
LIMB
~, which means that
= !..AlBI + LMlaB.
f:::,M Bla is isosceles. We now
have shown MIa = M B = M I = Me, so that a circle with center at M and radius MI will also pass through B, e and la' and we are done.
IFigure
44·1
We seem to have gone about all this in a roundabout way, which is true. We also agree that there are short proofs for our results. All this means that there are simple ways for obtaining complicated results, like Euler's Formula. The reader could construct similar circles on A, B, I, and on A, e, I, and examine this complicated figure. We especially call attention to the triangle whose vertices are at la' h, Ie. This has many interesting properties.
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ARBELOS -
Volume 5, Chapter 3
J
J .~
...1 MANY CHEERFUL FACTS There are numerous occasions that require the use of series expansions for some functions - either trigonometric or logarithmic. It is, therefore, useful for the interested student to know some of these expansions. Nowa days, such expansions are obtained by using calculus concepts. However, we do know that they were originally derived without such means, mainly in the seventeenth century. Among those who derived them was Euler, whose methods are always worth examining. The reader must know by now that your author does not hold for calculus in the secondary school, believing that the college is a better place for its study.
2f1T o
XA
!Figure 45.1
Historically, the trigonometric functions were first conceived of as being certain lines on a unit circle. The idea of having trigonometric functions as ratios came later. In Figure 45, segment XY is the sine of the angle at O. (You can find its value by measuring the length of XY if the circle has radius 1.) Segment OX is the cosine of the angle at O. Finally, segment T A would be the tangent of angle O. Since we have a unit circle, arc AY equals the measure of the angle at 0, in radians. Now, we see that
XY
< AY < T A. That is, sin 8 < 8 < tan 8.
Assuming that sin 8 is positive, we divide by it. Then
8 1
1
< sin 8 < cos 8. 80
....
... ...
"" I
...,
J .,..,
... r
ARBELOS -
Volume 5. Chapter 3
Now, let 8 "approach" zero as a limit. Then the term at the right ap proaches 1 from above. This gives liS the first of onr useful formulas:
. sin 8 11m -8-
8-0
= l.
Of course, we could get the same result by using LH6pital's Rule, but we are avoiding calculus. Next, suppose we investigate lim n-(X)
(1 + ~)n n
We do have the Binomial Theorem, so we expand and write
lim 1 +
n-oo
(n)1 ~n+ (n)2 n~ + (n)3 ~n + .... 3
2
When we evaluate the binomial coefficients, these terms become lim
n-oo
Since lim
n-+oo
1+ 1+ ~2! (1 - n~) + ~3! (1 - n~) (1 - ~)n + ....
~ = 0,
n
we finally arrive at
lim
n-oo
( + -l)n = 1
n
1 1+1+ -
2!
+ -3!1 + ... = e.
We think Euler used this derivation first. In the same way, •
bm
n-oo
( 1) 1+
-n
nz
X
2
X
3
= 1 + x + - + - + ... = 2!
3!
z
e .
Again, no calculus. However, we have used a simple limit concept. We now a.ssume that t.he r("aoer has s("en a proof of DeMoivre's The orem. The proof is by induction. We remind the reader that the theorem holds for all exponents; that is, (cos 8 + i sin 8)n
= cos n8 + i sin n8
for all possible n.
c
81
ARBELOS -
Volume 5, Chapter 3
Now we follow Euler to find a series for sin:z:. We expand the left-hand side of DeMoivre's Theorem. Of course, the terms have various powers of i. We collect those terms which have odd powers of i. Then sin n8 =
(7)
cos n - I 8 sin 8 - (;) cosn- 3 8 sin3 8 + ....
Now let n8 = :z:, where lim n = to Euler!) Then
00
and lim 8
= 0,
. 8 .38 . n-I sIn :z: 3 n-3 SIn sIn:z: = :z:cos 8 - - - -cos 8 -38 3! 8 By virtue of our first limit formula, when 8 :z:3 sin :z: = :z: - 3!
:z:5
but n8 is finite. (Leave it 5
+ -:z:5!
-+
cos
n-5
.58 sIn 8 -58
+ ... .
0 this reduces to
:z:7
+ -5! - -7! + ... .
If we collect all the real terms of our expansion of DeMoivre's Theorem, then, in the same manner as above, we obtain :z:2 cos :z: = 1 - 2!
:z:4
:z:6
+ -4! - -6! + ... .
We spare you the necessity of deriving this, but, if you like to do so, we have no objection. We now have series expansions for three useful functions. eO!, sin :z:, and cos :z:. By ordinary division, we can also derive tan:z: =:Z:
1 3 2 5 + 3:Z: + 15:z: + ....
Obviously, we can now find series expansions for all the trigonometric func tions. Now we turn to an expansion of e iz , which is e
iz
.:z:2 .:z:3 = 1 + I:Z: - - - 1 2! 3!
:z:4
.:z:5
+ -4! + 15!- + ... '
and when we compare this with our expansions for cos :z: and sin :z:, we have arrived at another useful expansion: e iz = cos :z:
+ i sin:z:.
Similarly,
e- iz = cos :z: - i sin :z:.
We can now express sin :z: and cos :z: in terms of eiO! and e- iz • However, we can always derive these, so we don't have to memorize them. 82
J
J J J
. ,.,
...
., .... I
~
''''II)
ARBELOS -
Volume 5, Chapter 3
Just for fun, try to calculate ii and show that is a real number! You have all necessary information in this article. Finally, one more series, that for In(1 + z). Since this is rather tricky, just remember the result. We begin with :z:2 :z:3 e'" = 1 + :z: + - + - + ... 2! 3! . Let e a = u, so that a = In u. Substitution gives us u'" = e
a
2
",
= 1 + :z:ln u +
~! (In
3
u? +
~! (In
u)3 + ....
Next, let u = 1 + z in the above. Then :z:2 :z:3 (1 + z)'" = 1 + :z:ln(1 + z) + ,(In(1 + Z))2 + , (In(1 + z))3 + .... 2. 3. Now, we use the Binomial Theorem to expand (1 + z)"': '" :z:(:z: - 1)Z2 :z:(:z: -1)(:z: - 2)z3 ( 1+z ) =1+:z:z+ 2 + 6 + .... In this last expansion of (1 + z)'" , the coefficient of :z: is: Z2
z3
z4
z--+---+ .. ·. 234
However, in the expansion of (1 + z)'" just above it, the coefficient of:z: is In(1 + z). Therefore, we have Z2 Z3 z4 In(1 + z) = z - - + - - - + .... 234
Now it happens that, even when we have a series that represents a function, it may not do so except for suitable values of z. This is so for this expansion. (Even for z = 1, the series is what is called conditionally convergent.) However, it sometimes happens that this series involving the In function may be of use. We suggest that it and the other series expansion for e'" are worth remembering. As a short exercise, show that, if z2
Z3
z4
y=z-2"+3-4+'" then
83
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ARBELOS -
Volume 5. Chapter 3
.
.. .~
i
....
,
,.
,.
.... !
84
.,
....
•
ARBELOS -
Volume 5. Chapter 4
COVER PROBLEM from the
MARCH, 1987 ARBELOS
H is the orthocenter of 6ABC. H' is the orthocenter of 6A ' BC. Prove: AHH' A' is a parallelogram.
Copyrisht
@
The Mathematical A••ociation of America, 111111. In ito first edition, thl.
chapter was publl.hed a. the
Arbelo""
Volume 5, b.ue 4, In March, 11181.
85
ARBELOS -
Volume 5. Chapter 4
PREFACE TO THE FIRST EDITION
J
."
... ,. ... ~
I t
The cover problem for issue 2 of volume 5 of this ArbeloJ aroused a good deal of interest and corresponding correspondence. We received an ingenious solution from Mr. Michael Moser, of Edmonton, Alberta, and a very neat solution from our constant correspondent, Ambata Jaya Krishna. Mr. l. J. Upton was one of many readers who asked about solutions to the problems we have proposed. The Olympiad problems, with solutions, have been presented in various places by members of Olympiad committees, with solutions. Perhaps one could write the mathematician in charge of Olympiad matters and ask for solutions. t The size of ArbeloJ makes it difficult to present many solutions. The reader who knows a foreign language may find solutions in publications prepared in these languages. There is one such in Russian (Kvant), another in Hungarian (Matematikai Lapok), and others as well. Mr. Upton also provided us with material on the Butterfly Problem. We believed this to be so well know as to be too easy for the ArbeloJ. Now we wonder just how much geometry has been discarded to make room for whatever is being presented in our classes now. Most other countries have orthodox algebra and geometry in secondary school. Perhaps they are all out of step. Right now, we are puzzling over the following: SuppoJe two cycliJtJ ten mileJ apart on a draight road running eaJt-wed and traveling at five mph toward each other. A bee JtartJ from the wheel of one cycliJt toward the wheel of the other, then turnJ and headJ for the wheel of the other, and JO on, at a rate of 30 mph. The bee iJ finally cruJhed between the wheelJ. The problem: How far did the bee travel, AND which way was he facing when he was crushed?
"
... ,
..."
.J
J
Your Editor Dr. Samuel L. Greitzer
t As of the printing of this second edition, each year's USA Mathematical
Olympiad International Mathematical Olympiad problems and solutions, 1976 present, are published in an Olympiad Pamphlet available from the American Mathematics Competitions, University of Nebraska, Lincoln, NE 68588 at $3 per year.
86
..
...
.. . '" j
ARBELOS -
Volume 5, Chapter 4
ITERATION In any class of superior students, there are always some students who try to confuse the teacher by means of some stunt the teacher does not ex pect. Our first example of such a situation arose when we were still teaching at Bronx Science. We presented the class with the following example: Solve for z:
z2 - 2z - 2 =
o.
One student gave the following solution: z2 - 2 z= - 2 - . He claimed (with justice) that he had done what we wanted. After some giggling, another student gave us the solution we wanted, using the quadratic formula, and still another cleverly added 3 to both sides to get
(z - 1)2 = 3
or
z=l±vi
The next day, we returned to the same student who had given the first solution. This time we pointed out that there were other solutions for Z; for instance, z = y'2z + 2. Considering this as being equivalent to Zi+I = y'2z i
+2
suggested that one could make a guess at, say, Zo, use this to find Zll then Z2, and so on until we reached a point at which Zi+I equalled Zi. Fortunately, we still have our Standard Mathematical Tables, which we called the Chemical Rubber Book. Of course, we had it then, new. (It is the 1961 edition.) We also had our slide rules. We started with Zo = 2 (since that value for Z made z2 - 2z - 2 not too far from zero), and substituted, getting, in order, Z1 = 2.449, Z2 = 2.626, Z3 = 2.693, Z4 = 2.718, Z5 = 2.727, Z6 = 2.730, Z7 = 2.731, Z6 = 2.732, Zll = 2.732, 87
ARBELOS -
Volume 5, Chepler 4
and we were done. We derived two useful results from this example. First, we turned the joker's trick to our advantage. Second, we illustrated another way of solving certain problems. The method of solving such problems by iteration is not original. We just do not meet with it in the average algebra class. However, it i, used to a good extent in Numerical Analysis. Our student with the joke wanted to know why we hadn't used iteration on his formula. We told him we liked our formula better. (Actually, his formula would not have given us an answer.) Our next problem for the class, which was interested, was: Solve .jZ = 2:& - 5.
J
... .. ...
.., I
... ,."'"
f
,.
....
We first wrote it as:
When we started with :&0 = 4 (since that value for :& made .jZ is not far from 2:& - 5 = 2 . 4 - 5 = 3), we found :&} = 3.5, :&2 = 3.435, :&3 = 3.427, :&4
= 3.426,
:&1)
= 3.425,
:&6
= 2 which = 3.425. •
All the rules we have presented previously were for solving polynomial equations. There are formulas for solving quadratics, cubics (Cardan), and biquadratics (Ferrari). The iteration method, however, works with equations that are not polynomials. Suppose, for example, we are asked to: Solve for :&:
j
•
2:& -log}o:& = 7.
We first "solve for :&", and get
Referring to our Chemical Rubber book, we easily find :&0
= 3,
:&}
= 3.79,
:&2
= 3.786,
:&3
= 3.789 = :&4'
Just for fun, we did the same problem, using our slide rule. The same prob lem can be done on a hand calculator, of course, provided it has logarithms to base 10. If the reader has one, try it. 88
,.
.. '1If
I
ARBELOS -
Volume 5, Chapter 4
Before we explain the process more fully, let us attempt a solution of
3x = 1 + sin x. We first "solve for x":
1 + sin
Xi
3 At this point, we have a small problem - x will have to be in radians, not degrees, unless there is a certain bit of transformation. Fortunately for us, the Chemical Rubber book has a table of values of trigonometric functions expressed in radians. We have no problem. (How about a hand calculator?) Starting with Xo = .5, we get Xl = .493, X2 = .491, Xs = .491, and we are done. One question naturally arises: How do we know what to start with? Our method is to draw a rough sketch of y = the expreuion, then superim pose y = X and see approximately where the two cross. In fact, if you are a good artist and can make very fine graphs of functions, you may even get the solution by inspection. Making a large graph helps a lot, but detracts from the iteration process as a method of solution of a problem. A poor guess, however, makes for a lot of intermediate steps, and may even lead away from a solution. Another problem comes from the Twelfth Putnam Competition (Prob lem 7). It goes: Given any real number No; if Nj+l = cos N j for all j > 0, prove that .lim N j ezidJ and iJ independent of ' .... 00
No.
If we graph cos x and x [See Figure 46.], we see that the graphs inter sect in a unique point, which we proceed to locate. Again, we get help from the Chemical Rubber book, since we have to deal with radian measure. First, let us use and
Xo
= 0.7.
We get, in order, the following: Xl X5
= .765, = .745,
X2
= .721,
Xs
X8
= .735,
X7
= .751, X4 = .731, = .739 = X8'
Figure 46 shows the various operations. We start at Xo on the X -axis, go up to the curve at Xl, then across to where y = X at (X2' X2) then to the
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ARBELOS -
Volume 5, Chapter 4
J
J
.. ..
IFigure 46·1 curve at y = cos X2, and so on. We can see that the process will end where the graphs intersect. Iteration gives us a more accurate value of this point. X2
Finally note that if Xo = No, any real number, then Xl E [-1,1], Xi approaches the same number as above.
E [cos 1,1], and
The reader has doubtlessly met with other examples of iteration. We might refer one to the Newton-Raphson method for solving equations. Our interest is in a way of finding square roots. To find ..IN, we divide N by some X, and find the average of this with X, thus:
Let us find
v'24 by iteration. Xl
We begin with
Xo
= 5. Then
X2
... .
= 2"1 ( 5 + 524) = 4.9.
= ~2
(4.9 + 24) = 4.899, Ilnd that's it. 4.9 Students interested in learning more about iterative methods can learn more by consulting texts on Numerical Analysis and modeling methods. Our favorite is: Nu.merical Mathematical Analy~i~ by J .B. Scarborough. If you can do so, you might get a copy of Standard Mathematical Table~ published by the Chemical Rubber Co. Next, we find
J
J
J J J
90
..
..J ,....,
.J
... j
ARBELOS -
Volume 5, Chapter 4
We close with a problem from the Putnam Contest No.7, Problem 1. It goes:
If {an} i" a "equence of number" "uch that, for n ~ 1,
(2 - a n )an +l = 1,
prove that lim an ezi"t" and i" equal to one.
n-+oo
BLIND ALLEYSP Our interest in various named theorems in geometry dates from the point when we first saw Varignon'" Theorem. This theorem is so simple that we are sure it has been known for centuries, but not considered worth the dignity of having a name attached to it. The theorem states:
The midpoint" of the "ide" of a quadrilateral are vertice" of a parallelogram. and Varignon's name lives with this theorem. Euler is famous as much for results he discovered as for others which he did not. For example, we recall that the theorem: "For a polyhedron, V - E + F = 2 ", is called Euler's Formula, although Fermat seems to have discovered it first. We also have reservations about another result. In the Nine-point Circlet, the midpoints of the line segments connecting the vertices with the orthocenter have been given the name, Euler point". We haven't a notion why these points should be dignified by a name, especially Euler's. There is a theorem due to Carnot that goes: t Recall that the so-called Nine-point circle is the circle which passes through the midpoints of the sides of a triangle, the feet of the altitudes, and the midpoints of the segments of the altitudes between the vertices of the triangle and its orthocenter. 91
ARBELOS -
Volume 5. Chapter 4
The "um of the di"tance" of the circumcenter of a triangle from the three "ide" i" equal to the circumradiu" increa"ed by the inradiu".
J
J
This is a nice problem to solve, but we don't see why it is so important as to be given the name Carnot'" Theorem. We haven't been able to do much, starting with this problem. Yet another problem we think worth solving, but which we would not dignify with a special name is the following:
The "iz perpendicular" from the feet of the altitude" of a triangle to the "ide" determine, on the"e "ide", "iz concllclic point". This circle has been given the title The Taylor Circle. We have seen no further results depending on this circle, nor have we been able to extend its properties so far. The properties of the Brocard Point" are important, to be sure, and many results in modern geometry have been derived from them. One such result is:
The perpendicular" from the vertice" of a triangle to the corre "ponding "ide" of the jir"t Brocard triangle are concurrent. The point of concurrency has been named the Tarry Point.
,.
... ,.
..., .
.J
We would like to get in on the act. We have the following "theorem" we would like to "push":
If triangle" be con"trueted ezternally on the "ide" of any triangle, and if the outer angle" of the"e triangle" have a "um equal to 180 degree", then the circumcircle" of the"e three outer triangle" are concurrent at a point.edskip We are not asking too much, but, how about calling the common point of the three circles the Greitzer Point? We are told that Archimedes had his theorem on the cylinder and sphere inscribed on his tomb. For awhile, he was satisfied, we think. However, the tomb is no longer in existence. Sic tran"it gloria mundi. Suggestion: Try your hand at proving the theorems cited above.
.
... .
92
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ARBELOS -
Volume 5. Chapter 4
DICE AND MATH The theory of probability, some parts of which had been developed earlier, had its real birth in the work of Pascal and Fermat. It seems that Chevalier de Mere, an inveterate gambler whose results did not match his opinions, asked for help. He was losing too much. Pascal and Fermat helped him and also started the theory of probability. We would like to start our article by examining the mathematics of craps. We are told (we never gamble) that in the dice game called "crap,", two dice are tossed (rolled) and the resulting sum of spots on the dice examined. The player wins if, on the first roll, this sum equals either a seven or an eleven. He loses if, on the first roll, the sum of the spots is 2, 3, or 12. If another sum shows up, the player continues to roll the dice until that sum shows up, and he wins, or a seven shows up and he loses. First, note that there are 36 possible results of rolling two dice: (1,1), (1,2), ... ,(1,6), (2, 1), ... , (6,6), and we assume that each possibility has the same probability of is our fundamental assumption.
1 3 8'
This
One can roll a seven in the following six ways: (1,6), (2, 5), (3, 4), (4,3), (5, 2), (6, 1).
i.
Hence the probability of winning with a seven on the first roll is Also, one can roll eleven as follows: (5,6), (6, 5). The probability of this is 118 , Now suppose the playeT Tolls a fOllT. Since there are three possibilities: (1,3), (2,2) and (3,1), the probability of this happening is 112 , To win, the player must now roll any number but seven until a four turns up a second time. That is, until he wins or loses, he must roll any sum except four and seven. The probability for a four is and that for seven is Add these and subtract from 1 to see that the possibility of continuing to play is the sum, 1 = ~. In theory, the player may roll forever. Therefore,
l2
t
93
i.
ARBELOS -
Volume 5, Chapter 4
the probability that he can go on until he gets four a second time may be written as follows: C12 ) 2 + C1 ) 2 2
J
J
(~) + C12 ) 2 (~) 2 + ....
We have an infinite geometric progression here, namely:
So the probability of winning when the first roll has been a four is
1 3 6'
Now a ten can appear as follows: (4,6),(5,5),(6,4). Except for the tallies on the dice, this situation is precisely the same as for winning with a four. We can say that the probability of winning when one starts with a . 1 t en IS 36' Suppose the player first rolls a five. The possibilities are: (1,4), (2,3), (3,2), (4,1). The probability in this case is 3~ = ;. If the player is to continue rolling until a five turns up again, he must, in the meantime, roll any number except a five or a seven. The probability of a seven is land that for a five is ~. The probability that either occurs is the sum, 158' and the probability that neither occurs is ~:. As before, for a succession of rolls leading to a win after rolling an initial five, we will have the series
J
J
J . ...1
...
Again, we have an infinite geometric progression,
(
(13)2 9"1)2 [ 1 + 13 18 + 18 + (13)3 18 + ...]
2 = 45'
That is, the probability of winning when one rolls a five first is equal to
;5'
As before, we note that a nine can occur in four ways (3,6), (4,5), (5,4), (6,3), as is the case for five. Therefore the probability of winning when one first rolls a nine equals 425' We shall shorten the rest of the development a bit, since it is just like the previous work. First, a six can occur as (1,5), (2, 4), (3, 3), (4,2), (5,1). The related probability is To get a six or seven, the probabilities are
;6'
94
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ARBElOS -
Volume 5, Chapter 4
and ~. The sum equals ~~. Not to get these has probability ~:. Our geometric progression now looks like: 5 36
5) ( 36
2 [
25
1 + 36
+
(25) 2 36
+
(25) 3 36
+ ...
]
25
= 396'
Let us prepare a table of probabilities for winning when any number (that can win) appears first:
1 36
2 45
25
1
25
2
1
1
396
6
396
45
36
18
The probability of winning at all is the sum of all these fractions. Our addition yields a probability for winning to be ~:: which is slightly less than or even money. In the long run (whatever that means) the dice roller stands to lose.
!'
Nevertheless, many gamblers do play craps, possibly because any game must perforce be finite, and our development involves an infinity of rolls. As far as we are concerned, our advice is
SAY NO TO CRAPS!
95
ARBELOS -
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Volume 5. Chapter 4
DERIVATIVES The limit concept is quite sophisticated. In our opinion, its place is more properly in the college rather than in the secondary school. Its history bears this out. We have referred to, and used, Cavalieri's Method to find areas and volumes. The essence of this method is the concept of an indivilJible. But just what is an indivisible? Newton and Leibnitz, about 1685, separately invented calculus. New ton's development depended on something called a /luzion. What is a fluxion? Leibnitz used "dz" and such, which became evanescent.
J ..
None of the mathematicians of the day had a clear concept of what they were doing and why. It was not until the late 1800's that Weierstrass and Dedekind cleared matters up. The reader will enjoy reading The AnalYlJt, by Bishop Berkeley, criti cizing the infinitesimal method. We particularly liked his question, "May we not call them the gholJtIJ of departed quantitielJ P" For information only, we present a definition of a limit. The Jtatement, lim f( z) Z-+Q
=L
meanlJ that, for every
f
J
>0
(no matter how IJmall), there ilJ a b > 0 (depending on frm) lJuch that, for all z lJuch that 0 < Iz - al < b we have If(z) - LI < f.
For future reference, we state some "rules":
..
(a) Division by 0 is not permitted;
,.
(b) ~ is undefined; (c)
00
i~ not a number;
(d) ~ =
oo.t
t Remember, division by zero is not permitted, and What is (d) saying? 96
00
is not a number.
.... I
ARBELOS -
Volume 5, Chapter 4
The limit concept has edged its way into algebra. Recall: S = a + ar + ar 2 + ... + ar n -
1
rS = ar + ar 2 + ... + ar n •
Subtract (crossways) and we get S - rS = a - ar n
a - ar n S=-- l-r
or
Now the limit concept sneaks in. Rewrite the above as a ar n S=-- l-r l-r
and then assume that, for
Irl < 1, n--+oo lim r n
= O.
As a mild challenge, can you prove this last? Why can't the limit be something else, less than r but greater than zero? We have read somewhere that Euler first found 1 + z + z 2 + z 3 + ... = -1-
l-z
for Iz I <.1.
and then let z = 2 and -2, with the result that 1+2+4+8+···=-1
and
1 - 2 + 4 - 8 + ... =
1
3'
There are detractors of Euler who assert that he saw nothing wrong in this. We don't believe this! In most cases, it is possible, though complicated, once given an f > 0, to find a 0 > 0 such that the conditions for the existence of a limit are fulfilled. As a second mild challenge, let f( z) = Z2 - 1. We seek lim [f(z) - 8].
z--+3
!.
f
:
Problems become more complicated when a limit takes the form ~ or The usual method used in classes is as follows (We give an example):
=
Find the corresponding
o.
Let
z2 -
9
Problem: Find lim - - - . z--+3 Z -
3
97
Solution:
ARBELOS -
Volume 5, Chapter 4
Sub~titution yield~ ~.
:c-3 We factor, and get (:c + 3)--. :c-3 We then cancel get :c+3 and, for :c = 3, the limit i~ 6. Surely, one must wonder at the logic, since we first assume :c f= 3 so as to be able to cancel, and then let :c = 3. Study of the definition for a limit may provide the answer: In that definition, 0 < l:c - 31 < 6. That is, we must use a deleted neighborhood of 3. If we are given the problem lim "'->00
J
3:c 2 - 5:c - 7 , 4:c 2 + :c + 7
J
we might divide all through by :z:2 and get
assume .:, = 0 (!) and find the limit to be ~. Let us now examine l'
h~
f(:c
+ h) -
f(:c)
h
.
This obviously takes the form~. Nevertheless, this is a very important limit indeed. It is called the derivative of f(:c). We shall write it D[f] or
/'.
..." I
P
h
R
f(:c)
. h :c
:c+h IFigure 47.1
98
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ARBELOS -
Volume 5. Chapter 4
In Figure 47, PR = h, QR = f(z + h) - f(z), and the quotient is the tangent of LQPR. As h gets smaller, PQ appears to get ever closer to the tangent at P, which is what makes this limit useful. Of course, the limit f' (z) may not exist at P, but this is not mentioned in most situations. We shall first apply the definition to find D [zn]. Then
D [znl = lim (z+h)n - zn h->O h zn+nhz n - 1+h 2 {(;)zn-2+(;)hz n - s + ... +(~)hn-2} - zn h->O h n-l h{ } n-l · 1m nz + . . . = nZ . 1
= lim
= h->O
Now we can use this as a formula to find the slope of a tangent to a curve (where such exists). For example,
and the slope where z = 2 is 11. In our previous issue, we developed series forms for some functions. For example, we found that sin z
=z
zS - 3!
z5
+ 5!
z1
-
7T + ... = f(z).
If we differentiate this term by term, using our formula, we get
If we apply our formula to this previously derived series for cos z, we get, D [cos
zl = -z +
zS Z5 3! - 5!
z1
+ 7T - ... =
- sin z.
We now have derivatives for two trigonometric functions: D [sin z] = cos z
and
D [cos zl = - sin z.
We also developed an expansion for e"'. We write Z2 zS z4 z5 e'" = 1 + z + - + - + - + - + .... 2! 3! 4! 5!
99
ARBELOS -
Volume 5, Chapter 4
Apply our formula, term by term, and we get D [e"']
Z2
Z3
Z4
= 0 + 1 + z + -2! + -3! + -4! + ... = e'"
That is, e'" is its own derivative! We think this is odd, at any rate.
J
J J
Just to complete things, recall that we derived
In(l
+ z) =
Z2
z- -
2
z3
z4
+ -3 - -4 + ....
Then
D [In(l
+ z)] = 1 - z + z2 _
z3
+ ... = _1_ l+z
...,.
by the rules for summing geometric progressions. The interested reader can now find derivatives of many different ex pressions - polynomial, trigonometric and logarithmic, even exponential. However, our aim is to proceed in another direction, connected with limits. Suppose we are asked to find lim
It ++ ~~.
We did find such limits z a couple of pages back. However, we proceed in another way. We assume that I(z) = 0, g(z) = 0 and g(z + h) '" O. Then h .....O 9
I(z+h)-/(z) _ I(z+h) '" Q g( z + h) - g( z) - g( z + h) 0. Divide numerator and denominator by h and one has
I(z
+ h) - I(z)
g(z
+ h) - g(z)
h
h
the limit of which is
~~~~:~J.
This formula is called L'Hopital', Rule.
Under a peculiar arrangement, M. l'Hopital published Jacob Bernoulli's notes, including this formula. (Looks to us as if he bought the formula!) Suppose we look for a familiar limit, lim sin z, which we know from ", ..... 0 z before, equals 1. Using L'Hopital's Rule, . sin z 11m - -
", ..... 0
Z
cos z = ",l'1m - - = 1. ..... 0 1
100
... "'"",,,
II
ARBELOS -
Volume 5. Chapter 4
We note that, since we can alter : Rille for sl1ch limits as well.
to ~, we can also use L'Hopital's
As for some examples, first, x2 - 9 2x lim - - - = lim - = 6, ",--+3 X 3 ",--+3 1
as before. Now suppose we want lim sin x2 - x. Applying L'Hopital's Rule, we x
",--+0
get lim cos; - 1. But this also has the form ",--+0
x
~, so we use L'Hopital's Rule
. -sin x again. Now we find hm - - - = O. ",--+0 2
Finally, let's try one slightly more complicated limit:
lim ",--+0
e'" - sin x -1 . 1 - cos x
L'Hopital's Rule gives us lim e '" ~ cos x. We repeat, and ",--+0 sIn x
. e'" I1m ",--+0
+ sin x = 1. cos X
We have not developed further rules for derivatives because this was not what we wished to do. Naturally, once one develops rules for finding derivatives of products, quotients, etc., one can find more difficult limits. We conclude with one more example; namely, lim "'--+00
e'" + 1 x3 + x2
=
lim "'--+00
e'"
3x 2
+ 2x =
e'" lim - - 6x + 2
"'--+00
=
e'" lim 6
"'--+00
= 00.
We have reached our goal, namely, a short introduction to the limit concept. There is much more that can be done, and we must postpone it to another issue at another time.
101
ARBELOS -
Volume 5, Chapter 4
....''''' I
J
... PROMISCUOUS EXAMPLES (a) If a quadrangle is inscribed in one circle and circumscribed about a second circle, its area equals K = abcd. (b) A rectangular garden 30 feet long and 20 feet wide is surrounded by a paved walk with width w, creating a larger rectangle. The area of the walk equals the area of the garden. How wide is the walk?
...
(c) A quadrangle with sides (in order) a, b, c, d is inscribed in a circle. Prove that, if the diagonals are perpendicular, then a2 + c2 = b2 + d2 • (d) Through the vertices of a parallelogram PQ RS lines APB, BQC, CRD, DSA are drawn forming a larger quadrangle. If AP = PB, BQ = QC, CR = RD, prove that DS = SA. (e) Lines CHand C M trisect angle C of triangle ABC. If one of these,
C H, is an altitude, and the other, CM, is a median of the triangle
ABC, show that the angle at C is a right angle.
(f) A rectangle of white squares, each one unit in area, is surrounded by
a single layer of unit squares, all black. Find the minimum number of white and black squares such that the number of white squares equals
the number of black squares.
Note: We have always admired Lewis Carroll for his "Pillow Problem~", problems that he solved before going to sleep, and thus mentally. We wonder how many of the problems above could be "pillow problems" for our readers.
".
J ..
t
..,
.. I
102
ARBELOS -
Volume 5, Chapter 4
USING PHYSICS IN MATHEMATICS The most famous of mathematicians have also been physicists. Archi medes' work in hydrostatics and in statics is well known. Newton developed his physics using mathematics. Gauss earned fame by his work in locating the asteroid Ceres. He also developed a form of telegraph before Morse did. In 1906, a palimpsest turned out to be a letter from Archimedes to Eratosthenes describing how he discovered some of his mathematical results by means of physics. Apparently, he used physics to derive some result, and then used strict logic to prove what he surmised from physics. We will try to show how this might be done by using an adaptation of his method.
IFigure
48·1
In Figure 48, on the left we are looking at a cylinder whose base is a circle with radius r and whose height equals 2r. It contains a sphere with radius r and two cones with circular bases of radius r and with height equal to r. We make models of half of the cylinder, one cone and the sphere (I used putty) and then try to balance them as shown in Figure 48 an the right. We should find that the objects balance as shown. Therefore the sum of the moments on the right of the fulcrum equals the moment at the left, so the volume of the sphere equals the sum of the volumes of the cone and the cylinder, or
Archimedes must have had very fine balances and weights, and lots of confidence! 103
ARBELOS -
Volume 5. Chapter 4
...
Now consider the diagram at the left of Figure 49. At all vertices, we have placed equal forces directed toward the other two vertices, as shown. Obviously, the sum of these six forces equals zero, and the resultant of the forces is zero. However, the resultant of the forces at A runs along the bisector of the angle at A, and the same is true for the resultants at vertices Band C. We have shown:
The bi8ector8
0/ the
angle8
0/ a
triangle are concurrent.
...
1
...
.. .. ...
,.
IFigure 49.'
Again, suppose we place six forces at the vertices of a triangle, each force going from a vertex to the midpoint of a side, as in the diagram at the right of Figure 49. First, the sum of these six forces equals zero, so the three resultants must be concurrent. In this case, each resultant lies on a median. (Draw the parallelogram of forces for any pair). Therefore, we have proved:
The median8
0/ a
."
...J
triangle are concurrent.
In the same manner, it is possible to prove other theorems by using forces and their resultants. One complicated result is Ceva's Theorem. We leave this to our readers. We assume that the reader knows about the parallelogram of forces; namely, the sum of two forces equals the diagonal of the parallelogram of which the forces are adjacent sides. To this we add: The 8um 0/ the moment8 0/ any coplanar /orce8 about any point in their plane equal8 the moment 0/ their re8ultant about that point. We have to define the moment
104
.,
...
ARBELOS -
...
.,...
Volume 5, Chapter 4
'
/
d/
F
/
t.
o IFigure
[J
50·1
of a force about a point, which is the product of the force and the perpen dicular distance of the point from the line of action of the force. In Figure 50, the moment of force F about the point 0 equals F· d = F· OAsin fJ. Occasionally it is helpful to go to three dimensions to learn more about a two dimensional problem. Thus, insight into certain properties of a tri angle ABC may be gained by considering a prism with face ABC. [See Figure 51.]
C
A
IFigure 51.1 Now imagine we have the prism filled with a gas which presses equally in all directions, the pressure being equal to p units per square inch. The pressure on the two triangular faces is equal and in opposite directions, so we neglect them. The forces exerted by the pressure on each of the three rectangular faces are equal to the areas of the faces times the pressure. Thus, where h is the height of the prism, these forces are:
AB·h·p,
AC . h . p and
BC· h . p.
We can consider these forces as if they were acting at the center of gravity of the face; that is, at the midpoint of that face of the prism. 105
ARBELOS -
Volume 5, Chapter 4
Now, first the three forces are coplanar. Next, the sum of the three by these forc~s is zero. (Otherwise, the prism would move and we would have a case of perpetual motion.) Finally, each force acts at the midpoint of the side and perpendicular to each face, so corresponding lines of force are the perpendicular bisectors of the sides. We have thus proved: mom~nts cr~ated
The perpendicular concurrent.
bi~ector~
of the
~ide~
of a triangle are
In Figure 51, assume that the angle at A is a right angle, and that the prism is free to rotate about an axis through B perpendicular to the plane of ABC. The moments of the three forces about the axis at Bare:
AB AB·h·p·
2 '
AC AC·h·p. 2
and
BC·h .p.
.....
BC
+ ( AC· h . p . 2AC)2 = ( BC· h . p . 2BC)2 or
...,
2'
Note that the forces on the faces containing AB and AC are counterclock wise while that on the face containing BC is clockwise. Since the pressure in the prism will not cause it to move,
AB)2 ( AB . h . p . 2
J
J
J
AB 2 + AC 2 = BC 2,
so we have another proof of the Pythagorean Theorem.
... ...'" ~
F2
A
.. .
""'!II
..J
E
IFigure 106
52·1
... ~
ARBELOS -
Volume 5. Chapter 4
For our next trick, we refer you to Figure 52. We are looking down at 11. rip;ht circular cylinder. From this cylinder, we construct a container with one curved wall following arc AC and two flat walls along segments AB and BC. As we did before, we fill this container with gas at pressure p per square unit. We concern ourselves with moments about the axis of the original cylinder. If the height of the container equals h, we easily find
F 1 = AB· h·p
and
F2 = BC ·h·p.
and
M 2 = BC . h . p . NQ.
The respective moments are M 1 = AB . h . p . M P
[J
These must be equal, since the curved part adds nothing to any moments around the axis, all its forces being directed away from the axis. Now we have to operate on M P and NQ. We have
AB MP=--MB
BC NQ= -+BN 2
2
AB-2MB
BC =-+(NC-BC)
2
(AM
+ MB)
2
- 2MB
= NC _ BC
AM-MB
= NE- BC
2
2
2 2NE-BC
2
MD-MB 2
2
BD
BE - 2
2
We can now rewrite the moments, using these values:
BD
M 1 = AB · h · p · 2
and
M2
BE
= BC·h·p·-. 2
and, since M 1 = M 2 , finally derive AB· BD = BC· BE (The Power of the Point formula.) Perhaps the reader might enjoy the allied problem: In Figure 53, PT is a tangent to the circle, and P AB is a secant. Prove: PT 2 = P A . P B. We have outlined the container with edges PT, PA and AT. Fill the container with gas and derive the result.
107
ARBELOS -
Volume 5, Chapter 4
J
J
.. ..
,. IFigure 53·1
.J
For more interesting uses of physical methods to solve problems, we recommend the article by A.K. Dewdney in the Scientific American for June, 1985. It includes a pretty way to solve a cubic equation, for example.
. ..,
..J
...., .....
.
108
... .~ I
1
ARBELOS -
Volume 5. Chapter 4
MANY CHEERFUL FACTS A general equation for a conic section is
Ax 2
+ Bxy + C y 2 + Dx + Ey + F
= O.
By varying values of the coefficients, the graph of this expression may be a circle, an ellipse, a parabola, an hyperbola, or degenerate forms of these. To translate the coordinate system so as to have the origin (0,0) take the position (h,k), use x
= x' + h
y = y'
+ k.
To rotate the axes of the rectangular coordinate system about the origin through an angle 8, use y = x' sin 8 + y' cos 8.
x = x' cos 8 - y' sin 8
Some of the forms the conic equation can take are: (circle) (parabola) (ellipse) (hyperbola) The expression B 2 - 4AC remains unaltered under a translation or a rotation. The expression A + C remains unaltered under a translation or a ro tation. If B 2 - 4AC < 0, the conic is an ellipse. If B 2 - 4AC = 0, the conic is a parabola. If B 2 - 4AC > 0, the conic is an hyperbola. Note: In all these cases, it may happen that the conic is degenerate. One can arrive at a point, a line, two lines, etc. For example, try and
109
ARBELOS -
Volume 5, Chapter 4
J
· ,
..J
.
... ~
I
..... I
· "" '11
.J
...
...
. ...
...
~...,
<1
.......
110
· ...
-
ARBELOS -
Volume 5, Chapter 5
COVER PROBLEM
from the
MAY, 1987 ARBELOS
A
B
Given: X, Y, Z are points of tangency. The area of 6.ABC is K. Prove: BC· AX· XX = 4rK, where r is the inradius.
Copyright
0
The Mathematical Association of America, 1991. In ito flrot edition, thi.
chapter was publi.hed as the Arbelo~, Volume 5, Issue 5, in May, 1981.
111
· .,
ARBELOS -
Volume 5. Chapter 5
.J
.. ..,
..
1
PREFACE TO THE FIRST EDITION AVE ATQUE VALE First, an answer to those who asked what an Arbelos is. In the time of Archimedes, the shape was that of a shoemaker's knife. Archimedes admired it and was able to derive many mathematical curiosities from it. We have a list of at least ten such problems. Next, we take this opportunity to acknowledge some of the letters we have received. Our cover page problems seem to have aroused lots of interest. We have received solutions from Barry Stipe and, of course, Ambati Krishna. Also from Erik Winfree and Ravi Vakil. We also have solutions to some of our other problems which were correct and showed imagination. We have a problem of our own - we got a nice solution to a cover problem from Rock Hill, SC, but couldn't read the name! However, it is with a good deal of emotion that we must announce that, with this issue, we will be giving up the editorship of the Arbelo,. We have been at it for five years now, and feel it is time for other hands to take up the work. We have notified MAA of our intention, and hope they will find a new editor. However, we have agreed to prepare a set of solutions to our cover problems. We will have the welcome help of Dr Stanley Rabinowitz with this. With 25 problems, this should be a fairly comprehensive work.
.. """"
J ·.
J
Professor Mientka has promised to keep an eye on future developments and notify everybody in case something promising comes up. The Arbelo, has been work, but it has been fun! Until something comes up and we can correspond again, we say - so long until then!
Your Editor Dr. Samuel L. Greitzer
112
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....
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ARBELOS -
Volume 5.\ Chapter 5
A NOTE ON INFINITE PRODUCTS The reader who has been exposed to calculus will usually have been introduced to sequences and series. Of the many tests for convergence, one may select one or two as favorites. Even before this, there are discussions about infinite geometric series. However, there is little or (usually) nothing about infinite products, and the serious student should be made aware of them. Let
{'Un}
be an arbitrary infinite sequence of numbers. Then 00
IT
'Un
=
'Uo • 'Ul . 'U2 ••••. 'Un •.••
n=O
denotes the limit of the sequence of partial products
There are some consequences that can be drawn at once. First, if some 'Ui
= 0, then all the terms in the sequence of partial products will be zero, 00
and the limit,
IT
'Un,
will be zero. Next, if the infinite product has a finite
n=O
value other than zero, then the factors approach 1 as n increases. That is, the partial products should not differ too much as the number of factors increases. Mathematically stated, this means Given any E > 0 (no matter how ~mall), there N (which depend~ on E) ~uch that, for n > N, l'Un+l • 'U n +2 ••••• 'Un+i •••• -
Because of this situation, we agree to let product takes the form
'Ui
ezi~t~
11 < E. = 1 + ai, and our infinite
00
IT
'Un
= (1
+ ao)(1 + aI)(I + a2)'"
n=O
113
an
(1
+ an) .. ••
ARBELOS -
Volume 5, Chapter 5
In this case, an must approach zero as n increases. Usually, only infinite prooucts of positive terms are considereo. Since 00
IT Un = (uo .
00
Ul ..•..
n--+oo
Uk-I) .
IT Un n=k
n=O
and lim Un
J
J
= 1, there is no loss of generality in doing so.
Thus, in all the
following, we will assume that all Un = 1 + an are positive.
J ....
Since
J we see that, if the product is bounded, then the sum must be bounded. This, then, is one way of checking if an infinite product might have a finite value; that is, if the sum is not bounded, then the product does not converge.
J
Furthermore, since
In ((I+ao)(I+at} ... (I+an )) = In(I+ao) + In(I+at} + ... + In(I+a n )
and thus
(1 + ao)(1 + at}"'(1 + an) < e( .. o+..'+ .. + .... >,
we see that if the sum is bounded, then the product must be bounded. Consequently, 00
The infinite product
IT (1 + an) converge" if and only if n=l
00
the "erie"
L an converge".
J J J . '.
n=l
Now, just as knowing that an infinite series has a finite sum does not tell you what that limit is, knowing t.hat. t.he infinite product has a finite value does not tell us what that value is. We note that it was a century after it was determined that the sum of the reciprocals of the squares was finite until it was determined what that limit is.
(*) Can you "how how thi" inequality follow" from the f-N definition of convergence f 114
J, ,.
.
.,.
.,j
ARBELOS -
Volume 5, Chapter 5
Let us try to evaluate
g(1First, examining
~(-
n(n 2+
2
n(n +
1))
1))'
for convergence leads easily to a tele
scoping series which does converge. Thus, the product has a finite value.
.
RewrIte the general product term as
(n
+(2)( n -)1).
We note that we did nn+1 not let n=O or n=1. We write the first few partial products as follows:
4·1 2·3'
4·1 5·2 1·5
2·3' 3·4 = 3·3'
4·1 5·2 6·3 1·6 2-3· 3·4· 4·5 = 3·4' 4·1 5·2 6·3 7·4 1·7 2·3' 3·4' 4·5' 5·6 = 3·5'
4·1 5·2 (n+2)(n-1) 2·3·3·4· .. · n(n -1)
=
1.(n+2) 3· n
so that the limit (that is, the value of the infinite product) is
i.
We offer the following problems for your pleasure:
OO( 1 -
(1) Show that!!
(2) Show that
IT (1+ ,,=2
n3 3
=
1 n(n +2)) =2.
-1 II -= n +1 00
(3) Show that
1) 12'
n2
2 -. 3
Of course, infinite products can have the 11." be functions of a complex number, z = ;z; + iy. This raises interesting questions, which we simply do not have room enough to pursue completely. A few remarks must suffice: Consider a polynomial function h(z) equal to ao
+ alz + a2z2 + '" + a"z". 115
ARBELOS -
Volume 5. Chapter 5
Then eh(z) cannot equal zero for any z, from the properties of eZ • The FundAmental Theorem of Algebra states that if two nth degree polynomial equations have the same n roots, then they differ only by a constant fac tor. Now suppose that we have a polynomial Go(z). It has roots. Then eh(z)Go(z) has precisely the same roots. Question: Go(z) =
has no zeros. Therefore, it can be written in 1 the form eh(z). What would h(z) look like in this case? e
IZ -
2 '11":1:
=
J'
;.in z
In a previous issue of the Arbelol, we derived sin
J J
'11":1:
2
2
(1 _ :1:12 ) (1 _ :1:22 ) (1 _ :1:32 )
•••
•
Can you use this infinite product to evaluate
224 4 6 6
A=
l' 3. 3 . :5 . :5 . '7 .... ?
B =
l' 3 . :5 . '7 . 9 . 11 . 13· .. ·?
2 2 6 6 10 10 14
For more on infinite products, you might consult an (old) text on Function Theory.
.. J J J J
J
.,.I
...
..J ,. 116
ARBELOS -
Volume 5. Chapter 5
JUST FOR FUN (1) Carp and perch are natural enemies and always fight. A perch can kill one or two carp, is held at bay by three carp, and can be killed by four carp in three minutes. How long will a battle between four perch and thirteen carp last? (2) A column of troops one mile long is on the march. A messenger goes from the tail end of the column to the head and back, arriving at the tail end just as the column has moved one mile. How far will the messenger have traveled? (3) The numbers 1, 25, 49 are three squares in arithmetic progression with a common difference of 24. Find other sequences of three perfect squares in arithmetic progression and note the common differences of the arithmetic progressions. (4) Three indigents shared a meal. One had five sandwiches, a second had three sandwiches, and the third had eight dimes (which he had found in various public telephones). They shared the sandwiches equally, after which the indigent with the dimes gave them to the others to share. How many dimes should each of the others get? (5) Three equal circles are tangent as shown in Figure 54. Their centers lie on the same line, PQ. The line PT is drawn from point P on the left circle, and tangent to the circle at the right at point T. How long is the chord XY of the circle in the middle?
T
P,---1===::::===::r----,Q
IFigure
117
[J
54·1
ARBELOS -
J
J
J
J
Volume 5. Chapter 5
SOME TRIGONOMETRY We have always distinguished between talented and gifted mathematics students. A talented student may have a prodigious memory for formulas, and, if one of these helps solve a problem in a manner which he has seen, he may be successful. A gifted student is one who sees a novel method for solving a problem. He may not, however, have a more than ordinary memory for formulas or rules. This article is for such a student.
J
We are not advocating that the gifted student try to memorize all the formulas possible. We are suggesting that he learn to derive whatever formulas might be needed for a given problem. Since trigonometry is loaded with many formulas, we shall look into this area for our examples. We assume a minimal knowledge. The student should know that sin(A
+ B) =
sin A cos B
+ cos Asin B,
and also that
cos(A + B)
= cos
Acos B - sin Asin B.
~
Now it is easy to get sin 2A
J J
= 2 sin A cos
A
and
cos 2A
= cos 2 A -
sin 2 A.
There may be students who get the same results by use of De Moivre's formula, which is a bit too complicated. Of course, we should use De Moivre to find sin 3A and cos 3A, although it can be done without. In either case, one need not have memorized sin 3A or cos 3A. The talented student might.
..
J ~ " ..J ..
We get the half-angle formulas from the two rules: cos 2 A 2
+ sin2 A
= 1
2
cos A - sin A = cos 2A which looks like a pair of simultaneous equations.
118
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Volume 5. Chapter 5
We don't have to take the time and space to show that
.
B = 2
SIn -
/1-
cos B
2
COS
and
+ cos
B = /1 2
B.
2
Now one can derive the formulas
B
1- cos B
tan - = 2
l-cosB sin B - 1 + cos B' sin B
1 + cos B
These are so simple to derive that one needn't take the trouble to memorize them. At this point, we confess that we have memorized ' A+'sm B sm
' C= 4 cos 2 A C + sm cosB 2 cos 2
because we think it is pretty. We have also memorized tan A
+ tan
B
+ tan C = tan A tan
B tan C.
Both formulas are for A + B + C = 180 And we have found the second formula useful in solving several cubic equations. You might try it on: 0
•
If :c
+y +z =
:cyz, show that
There are some formulas that may be unfamiliar to the reader, but have many uses in problem solving. We prefer the following way of determining them. Start with the Law of Cosines,
and "complete the square". Thus:
+ 2ab + b2 - 2ab - 2ab cos (a + b? - 2ab(1 + cos C).
c2 = a2 =
C
Then (a
+ b? -
c 2 = 2ab(1
+ cos C) =
119
4abcos
2
C
2'
ARBELOS Now let s =
a+b+c 2
J
J
Volume 5. Chapter 5
and factor the above to get
s(s - c) = abcos 2
a
2
and finally,
a _)
cos 2 -
s(s - c) ab'
A similar process (subtract and add 2ab) yields
. a
sm
2
=
)(s - a)(s - b) ab .
We have textbooks with other ways of finding these two formulas, all more complicated. Incidentally, we call attention to the following:
...
J
The area of triangle ABO is equal to
K =
~b sin 0= ~2sin ~ cos ~
= -/s(s - a)(s - b)(s - c).
Here we do not recommend this device for problem solving, but prefer to use the specialized expression for K, the area of an inscribed quadrilat eral, which we have already derived; namely,
K = -/(s - a)(s - b)(s - c)(s - d) and then setting d = O. There are some occasions when it is good to recall Euler's formulas derived from the expressions: e ie = cos 8 + i sin 8
J
e- ie = cos 8 - isin 8.
Using these like simultaneous equations, we have: sin 8 =
J J
eie _ e- ie
2i
It may well be that some problem could become easier if one uses these definitions for sin 8 and cos 8. Try them on some of the formulas we have already presented. Also, since we all know the series form for e"', we can
J J
J
120
J
ARBELOS -
Volume 5. Chapter 5
use these formulas to derive series expressions for sin we lea.ve it to the reader to do this. Also, show that: ..
Sin
Z
+ Sin Y =
z+y
2'
:I:
and COS:l:. Again,
z-y
SIn -2- cos -2
using the exponential forms. (Actually, this is more complicated than stick ing to the usual forms.) Problems involving trigonometric equations are not common, and we will not discuss them here. Inverse functions do sometimes appear in con tests and in problems. Let us first define them. If sin A
= m, then we say that
If cos A =
VI -
A
= arcsin m.
m 2 (as it should be), then A = arctan
m Vl-m 2 You can now find arccot A, and so on. We find that the best way to solve inverse function problems is to draw a diagram and reason from it. Unfortunately, this is as far as we can go. We just add the following problems for your delectation. (v) Let r,B,t be roots of z(z - 2)(3z - 7) = O. (a) Prove that r,B,t are not negative. (b) Evaluate arctan r
+ arctan B + arctan t.
(w) Find all triangles for which tan( A - B) + tan( B - C) + tan( C - A) = O. (x) Eliminate A from
z = 3 sin A - sin y = 3 cos
A
3A
+ cos 3A
(y) Determine the minimum value of cot
".-A
4
+ cot
sin A
".-B
+ sin
121
4
B
+ cot
+ sin C
".-c 4
ARBELOS -
Volume 5. Chapter 5
JUST PLAIN NUMBERS AND A MAGIC NUMBER Lest one think that our articles must deal with unfamiliar or esoteric topics, we have found that there are those who are unfamiliar with ordinary positive integers and their properties. We would like to examine just plain numbers. We consider Fermat's Theorem, a P == a (mod p) for prime p, esoteric, but useful sometimes, of course. Instead, we ask a few questions about divisibility.
J J
We know that a number is divisible by 2 if the units digit is either 0, 2,4, 6, or 8. Now, a number is divisible by 4 if the two digits at the end are divisible by 4. You should find the proof easy. Next, a number is divisible by 8 is the three digits at the end are divisible by 8. Again, the proof is easy. Again, if a number ends in 5 or 0, it is divisible by 5. We hope you find this acceptable. It is fun to try to invent a proof for the next two rules that will satisfy a child. These are:
J
. ...
A number is divisible by 3 if the sum of its digits is divisible by 3. A number is divisible by 9 if the sum of its digits is divisible by 9. Now it appears that we can tell whether a number is divisible by 2, 3, 4, 5, 6, 8, 9, 10, 12. Of course, a number is divisible by 6 if it is divisible by 2 and 3, and similarly for divisibility by 12. It isn't easy, but one can devise a proof that a number is divisible by 11 if the difference between the sum of t.he rligit.s in the odd places and the sum of the digits in the even places is divisible by II.
We can thus check divisibility by every number from 2 through 12, except 7. This makes it fun to reduce many fractions, factor license numbers on cars, and so on. A rule for divisibility by 7 would be nice to have. Some time ago, we 122
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ARBELOS -
Volume 5. Chapter 5
did devise a rule which we now present with an example, the number 1834: Double the unit$ digit and $ubtract from the re$t of the num ber. This yields 183 - 8 = 175. If nece$$ary, repeat the proce$$. Thus 17 - 10 = 7. If we reach a number divi$ible by 7, the original number i$ di vi$ible by 7. Thus, we can state that 1834 number is divisible by 7.
The trouble is that we have been unable to devise a proof that would satisfy an elementary school student. We have therefore appealed to the magic number 1001. It happens that 1001 = 7· 11 ·13. Try it! Now suppose we start with a number N and subtract 1001. We get N - 1001 = r (remainder r). If r is divisible by 7 (or 11 or 13) then, since 1001 is divisible by 7, (or 11, or 13), the original number must be divisible by 7 (or 11 or 13). We get three divisors for the price of one, so to speak. Let's try this on 1834. First, 1834 - 1001 = 833. This is not too helpful, so let us apply 1001 again. Now 1001- 833 = 168, which 7 divides. Note that this method can be used very effectively if one is aware of which numbers between -500 and 500 are divisible by 7, 11 and 13. We welcomed this chance to introduce the magic number 1001 to our readers. They should be encouraged to invent their own magic numbers. This is all right if one wants to reduce a fraction to lowest terms. However, we offer two other easy problems on natural numbers for your delectation:
(q) A $iz-digit number, multiplied by 2, 3, 4, 5 or 6, yield$ a number containing the $ame $iz digit$ (naturally in different order). Find the number.
(w) Find a three-digit number all of who$e integral power$ end in the $ame three digit$ a$ doe$ the original number.
Of course, if one works a lot with numbers, one can discover new and interesting results. After all, Ramanujan must have just stared at numbers to be able to develop most of his surprising results. For instance, suppose we are looking for knowledge about perfect squares. We might notice, for example, that a perfect square must end 123
ARBELOS -
Volume 5, Chapter 5
with 1, 4,9,6,5, or O. So, if a number ends with a 7, it can't be a perfect
J J
SCJ11 flTt>.
We can see more if we try, by inspecting the sum of the digits of a number, iterating if necessary. (Dudeney calls this the digital number of the number.) Just writing a few squares will show us that the digital sum for a square ultimately equals 1, 4, 9 or 7. This gives us two checks on whether a number is a square or not.
...
When it comes to perfect cubes, matters are not so simple. First, a cube can end with any digit. However, the digital sum for a perfect cube will be either 1,8 or 9. Perhaps the reader can discover more such results. We would be happy to know them. We must end this article with some problems that depend on what we have covered here plus, if needed, other properties.
(k) Given a number N = 523abc. Determine a, b, c diviJible by 7, 8 and 9.
JO
that N will be
(w) A three-digit number N haJ the form N = abc. Find N given that it iJ diviJible by a! + b! + c!. (Factorials). (z) Given any 52 integerJ, there are two of theJe whoJe Jum or dif ference iJ diviJible by 100.
J J J
. ."
...
1
124
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Volume 5. Chapter 5
GEOMETRIC REVERIES We have often remarked at the ability of great mathematicians to de velop their concepts, seemingly without previous information. Actually, we think it is possible to do so in geometry, but cannot see how it can be done in other areas of mathematics. In geometry, it is possible to draw a dia.gram and, by merely staring at it, determine relations that are not readily apparent. To illustrate, we thought we'd try this with Figure 55. Let us describe the situation first.
IFigure 55.1
We begin with any triangle, 6ABG, and 6ABO has been constructed with BOil BG, OA II GA and AB II AB. In addition, we have constructed the orthictriangle Ho.HbH c . The orthocenter (where the altitudes intersect) we will call H. Also, 0 is the circumcenter, and Mo., M b and Afc are midpoints. 125
ARBELOS -
Volume 5, Chapter 5
.
Now we merely stare at the diagram and let our thoughts rove freely. It.
Moreover, the sides of 6ABC and 6AiJe are in the ratio of 1:2. Therefore, corresponding lines of the two figures have the ratio 1:2. Perhaps one can now see that H is the circumcenter of 6AiJe, since it is the intersection of the perpendicular bisectors AH,., B H band C He of the sides of 6AiJe. Now OM,. corresponds to AH. Therefore,
(b)
AH
= 20M,..
Incidentally, we don't want to deviate from our plan to discover useful connections, but, if we were to draw the median AM,., the centroid G would be ~ of the way between A and M,.. Then H, G and 0 would be collinear, and the line HGO (the Euler line) might lead us to other results. However, we merely mention this diversion, and let it to the reader to go on from there if desired. We have other fish to fry. Note first that the circle about the orthic triangle is the Nine-point circle of 6ABC. However, the circumcircle about 6ABC is the Nine point circle of 6AiJe since it passes through the midpoints of its sides. Let us save this result.
(c)
The circumcircle of 6ABC i~ the Nine-point circle of 6AiJ6.
Since corresponding lines in the two triangles have the ratio 1:2, we can say that
(d)
The 6AiJ6.
circumradiu~
of 6ABC
equal~
twice the
circumradiu~
.J .....
..J
AA, BiJ and ce are concurrent.
(a)
~
of
Further staring at the diagram discloses that since there are right angles at He and H b, the quadrilateral AHeH H b is inscriptible. Moreover, the circumcircle of this quadrilateral has t.he
-... ,.... I
J
J J • it
..J
.. :J ~ ~
.J
We now recall the Law of Sines, and write
(e)
HbHe
= AH.
sin A
J
Also, in 6ABC, 126
ARBELOS -
Volume 5, Chapter 5 a
(f)
- - =2R. sin A
We can eliminate sin A from these two expressions. Note that this might well be the first time one has to do any writing. This gives us:
(g)
Now, using relation (b), we can write
(h)
Now, think of having done the same to get
Add all three relations and we get
(i)
We believe that if you had been given asked to prove
(j)
The
~emiperimeter
circumradiu~
of the orthic triangle multiplied by the of the triangle it~elf equal~ the area of the triangle.
you might have had some trouble. We have found ten results mostly by staring at a diagram. We merely mention one more to show that we have not yet squeezed the diagram dry. This is
(k)
The
altitude~
of 6ABC
bi~eet
the
angle~
of
it~
orthic triangle.
To anyone who says that we have selected a special figure, we reply by presenting a second diagram in Figure 56 for similar development.
127
,. ARBElOS -
Volume 5. Chapter 5
IFigure
J J J J J J
56·1
In Figure 56 we have 6ABC with its orthic triangle H .. H&Hc • Also, we have drawn the circumcircle. Now we have constructed tangents to this circle at A, Band C, thus forming a tangential triangle. You can now try to deduce relations from the diagram. We will give just two.
(Q)
(f3)
The tangential triangle and the orthic triangle are
J
J
~imilar.
The perpendicular~ from the vertice~ of the triangle to the of the orthic triangle pa~~ through the circumcenter and are concurrent.
~ide~
•. 19
.J
.J 128
ARBELOS -
Volume 5. Chapter 5
HAVERSINES We read that, in the early days of trigonometry, the various functions were looked on as actual lengths of lines. Thus, in Figure 57, which shows a unit circle, the length of segment AB was the sine of the LAOB. That is, the six trigonometric functions were
AB = sin 8 TC = tan 8 OC = sec 8
= cos 8 DE = cot 8
OA
OD
= csc 8
At least that is what one finds in most stories about the genesis of trigonom etry!
Et---=::::
-.,..-;IOD
o
A
IFigure
T
57·1
There were, however, other line segments that had special values and functions, but which are now forgotten. One of these was AT = 1 - cos 8, known as the ver~ine of 8 and abbreviated vers 8. (There was also a coversed sine of 8, which we ignore.) Before such things as LORAN, the versine was used in navigation. It ma.de it almost automatic for the captain of a ship to locate himself.
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ARBELOS -
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Volume 5, Chapter 5
In Figure 58, we have a half-sphere with North Pole at N, we are assuming tha.t A a.nd B are points whose la.titude and longitude are known.
Thus, we have a spherical triangle NAB. Now we have done a bit of work with such triangles. The reader might consult Volume 2, Issue 1 this
Arbelo", for example. In that article, we derived a spherical Law of Cosines.
In our diagram, this Law takes the form:
cos c = cos a . cos b + sin a . sin b . cos N.
N
J
J'
J
J
J "
..J IFigure 58·1 In Figure 58, suppose we know the latitude and longitude of point B. Then b = 90 0 minus that latitude. Similarly, we know side a. As for the angle at N, that equals the difference in longitudes of the two points. In short, we know two sides of the triangle and the included angle. Unfortunately, one has to be a bit of a mathematician to use the Law of Cosines and, we are sorry to say, many captains were not (except Captain Bly). Therefore, the formula was altered so as to make it available to our common, garden variety captains. This was (probably) done as follows: Start with (*). Add and subtract sin a sin b to get
J J J
cos c = (cos a . cos b + sin a . sin b) - sin a· sin b(l - cos N). Then cos c = cos(a - b) - sin a· sin b· (1 - cos N). Add one to both sides after changing signs; 1 - cos c
= (1 -
cos(a - b))
+ sin a· sin b· (1 -
130
cos N).
J
... ... 1
''It
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ARBELOS -
Volume 5. Chapter 5
You see the versines coming out of the formula. But divide each term above by 2, Ilnd one has haversine c = haversine( b - c)
+ sin a . sin b· haversine N.
We have used the vernacular, calling half the versed sine, a times abbreviated as hav in formulas.
haver~ine,
many
Suppose we apply this formula to an example. Let us find the distance between Greenwich, England, and New York City. Our World Almanac tells us that New York City has a latitude of 40° 45' and a longitude of 74°. As for Greenwich, it has a latitude of 51 °30' and a longitude of zero. That tells us that, in our formula,
We can then fill in a table such as that which might have been used by an old sea captain: log hav N: 9.5589 -10 log sin a :
9.8794 -10
log sin b:
9.7942 10
Total :
29.2325 - 30
Antilog Total : hav 10°45' :
.0082
= hav:c
.1790
Sum
.1708
Now distances in sailing and flying are usually measured in nautical miles. The circumference of the earth is 360 degrees or 21600 minutes of arc. One lets one minute of arc be one nautical mile. We also know that the circumference of the earth is approximately 25000 miles. Simple arithmetic tells us that a nautical mile is 1.15 statute miles. We now finish our calculations. First, 50° equals 3000 nautical miles. Multiplying by 1.15 gives us 3450 miles as the distance from New York City to Greenwich. We are quite close - our World Almanac says that the distance is actually 3469 miles. It may be that tectonic motion explains the difference. It also happens that our Chemical Rubber book contains a table of haversines with their logarithms. (We mentioned this book in our last issue and we recommended it very highly.) Our sea captain would have
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Volume 5, Chapter 5
two tables, say table A and table B. One would have haversines and one sines. The whole process is actually mechanical.
J
Problem: Perhaps the reader would like to find the distance between Washington, DC (Lat 38°55'N, Lon 77°5'W) and Paris (Lat 48°50'N, Lon 2°20'E). It is a pity that we seem to have gone off in a different direction and thus have missed a pretty field of mathematics, namely, spherical trigonometry. We have previously discussed the area of a spherical triangle. When the three sides are known, there is a fascinating formula for finding the area. Let the sides, in degrees, be arcs a, b, c, and let s = (a + b + c)/2. Then the area of the triangle is: E tan - = 4
J
s-a s-b s-c s tan - tan - - tan - - tan - 2 2 2 2
where E is, as before, the spherical excess. (Note that you can use this rule to find E if you know the sides of the triangle.)
J ....
J A SPACE FILLER A triangle with one side equal to 8" is inscribed in a circle whose radius is 6". The sine of the angle opposite the eight-inch side is a mean proportional between the sines of the other two angles. Find the area of the triangle.
• .,
•
•
132
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Volume 5, Chapter 5
ARITHMETIC FUNCTIONS The imagination of the true mathematician knows no bounds. We have already mentioned, for example, the invention by Gauss of the congruence notation. To constant readers of the Arbe/oJ, the imagination of Euler needs no further mention. However, we shall proceed with one of Euler's ideas anyway. For somereason, Euler introduced the "phi-junction", q,(n), the num ber of positive integers d ::; n which are relatively prime to n. Thus, q,(10) = 4 because of {1,3, 7,9} We are sure the reader has encountered the phi-function before, so we feel it is not necessary to be repetitive about this function. However, just a little bit will help. First, for a prime p, q,(p) = p -1. Next, for p2, the positive integers not prime to p2 are 1, p, 2p, ... , p . p so the number of integers relatively prime to p2 comes to p2 - P = p2 (1 -
~).
In general,
We are assuming that the reader has seen all this before. One more, and we will have introduced all we need about the phi-function. That is: if m and n are relatively prime, then q,(mn) = q,(m)q,(n) That is, q,(n) is a multiplicative function. Note that q,(12) = q,(3)q,(4), but not q,(2)q,(6). Now we introduce what may be new to the reader, the sum of function values of elements taken over the positive divisors of the number or num bers involved for a given function, say f. This is usually written f(d).
L
din
Let us try to find
L q,( d). One way of doing this is: (1) Write all frac din
tionJ with denominator nand numeratorJ from 1 through n. We illustrate for n = 12: 1 2 3 4 5 6 7 8 9 10 11 12 12'12'12'12'12'12'12'12'12'12' 12'12'
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ARBELOS -
Volume 5. Chapter 5
J
(2) Reduce each fraction to lowe"t term". For our example we have
1111517235111 12' 6' 4' 3' 12' 2' 12' 3' 4' 6' 12' l' (3) The denominator" are all of the divi"or" d of n. For each divi"or d, all reduced fraction" with denominator d will appear. Group them together. Note, for example, that for n = 12, we can group thus:
I} {1
' {I} 2 ' {12} 3' 3 ' {13} 4' 4 ' {15} 6' 6 '
{I 5 7 11} 12' 12' 12' 12 .
(4) Al"o, ¢(d) i", of cour"e, the number of fraction" in each group. In our example, ¢(1) = 1, ¢(2) = 1, ¢(3) = 2, ¢(4) = 2, ¢(6) ::;:; 2 and ¢(12) ::;:; 4.
J J J
(5) Since all fraction" are included in thi" arrangement, we have
L¢(d) = n. din
It is customary to call a function f defined on the positive integers an arithmetic function if f(ab) = f(a)f(b) whenever a and b are relatively prime. Euler's ¢ function is one example of an arithmetic function. We could use some more arithmetic functions, so we go back to
First, note that we can get all possible divisors of n from the product
The sum of all the divisors of n can thus be written:
11"( n) = "
LJ
d::;:;
p"'1+ 1 -1
din
1
PI - 1
p"'.+1 _ 1 •
2
P2 - 1
p"'.+1 - 1 ............" -
PII - 1
Note that this function is also arithmetic. While we are at it, we can find the number of divisors of n, which we represent as 7'( n). This is simply
7'( n) ::;:; (01
....,
... i
J J
... .
~
...J
+ 1) . (02 + 1) ... (all + 1) , 134
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...1
ARBELOS -
Volume 5. Chapter 5
which is another arithmetic function. Next, we define the expression /L{n) recursively as follows:
Let /L(1) = 1.
For n
> 1, define /L{ n) so
L /L{ d) = O.
din
For example, note that, since /L(1)+/L{2) = 0, /L(2) = -1. In fact, for any prime P, /L{p) = -1. Next, /L{p lc ) = 0 for k ~ 2 since
0= L/L{d) dip·
+ (/L{p 2 ) + ... + /L{p lc ))
1 + (/L{p 2 ) + ... + /L(p lc )) for k = 2,3, ... , k
= 1 + /L{p)
=1-
Finally, if we let n be a product of distinct primes, then we can easily compute We have another arithmetic function at our service. Finally, we define an operation
* on the arithmetic functions as follows:
let
(f*g){n) = L f (d).g
(~).
din
First, since d and J run through the same values, this product is commu tative. Let us call an expression of this type the convolution of f and g. We can use it to get results involving combinations of arithmetic functions. For example,
(
=
when P is a prime. Exercise: Show that (
> O.
Note: We first encountered an analogos form of convolution many years ago in the form of the Beta function; namely,
135
,
....
..
ARBELOS -
Volume 5. Chapter 5
,,
We liked it then since it equals the Gamma expression ( m.n. )' We still m+n. like it! For more on the Gamma function, see Volume 3, chapter 1 of the Arbelo~.
In one of his classes, Gauss asked the question: "If we are given
g(n)
=L
f(d),
din
can we find, [rom this, an expression [or f( n)?" Thi~ wa~ ~olved
by
hi~
pupil,
Mobiu~,
(~)
f(n) = LJL (d) 9
who arrived at the = (JL
re~ult
* g)(n).
din
With the results presented in this article, we can enter the domain of arithmetic functions, using convolutions, the elementary arithmetic func tions, and the Mobius formula. There is not enough room in a short article to do more than introduce some of the elements. t We present a few short problems for the reader to try: (a) For which n = pClqbrc with p, q and r distinct primes, is it true that
LJL(d)r(d) = -1?
(c) Derive (q,
derive
* r)(p2)
(d) Show that""'
!
L...J d din
= p2
q,(n) = L n
+ p + 1 for
din
...
p prime.
n
din
r(n) =
L l. din
.
..... ·
(e) Apply the Mobius inversion formula to and
· .,
JL(d). d
= u(n).
u(n) = Ld
J ..J
din
(b) From
.
For more on arithmetic functions, the reader may refer to a good book on number theory. We like "Theory of Numbers" by Niven, Zuckerman, & Montgomery published by John Wiley. 136
· ..
... I
APPENDIX A: Olympiad Problems from around the world t
From Canada: 1) Find the average of the quantity
taken over the permutations (aI, a2, ... , an) of 1,2,3, ...
,n.
From Mongolia:
2) For each P inside 6ABC, let A(P), B(P), and C(P) be the points of intersection of the lines AP, BP and CP with the sides opposite to A, Band C respectively. Determine P in such a way that the area of the triangle A(P)B(P)C(P) is as large as possible. From the
Netherland~:
3) Prove that there are infinitely many pairs (k, N) of positive integers such that
1 + 2 + ... + k = (k
+ 1) + (k + 2) + ... + N.
From Poland:
4) Let p be a prime. For which k can the set {1,2, ... , k} be partitioned into p subsets with equal sum of elements? From the USA:
5) Let CD be a diameter of circle K, and let AB be a chord which is parallel to CD. Line segment AE with E on K is parallel to CB. Point F is the point of intersection of line segments AB and DE. Line segment FG with G on DC extended is parallel to CB. Is GA tangent to K at point A?
tThese problems are some of the problems proposed by various countries for use in the 1986 International Mathematical Olympiad (problems 1-20) or the 1987 Ibero-American Mathematical Olympiad (problems 21-25). In the first edition of the Arbe10J they appeared five per issue.
137
APPENDIX A: Olympiad Problems from around the world
From the Netherland3: 6) Find four positive integers, each not exceeding 70,000 and each having more than 100 divisors. From Bulgaria: 7) Let f( n) be the least number of distinct points in the plane such that for each k = 1,2, ... ,n there exists a straight line containing k of these points. Find an explicit expression for f(n). From brael: 8) The circle inscribed in a triangle ABC touches the sides BC, C A, AB in D, E, F respectively, and X, Y, and Z are the midpoints of EF, F D, DE, respectively. Prove that the centers of the inscribed circle and of the circles XYZ and ABC are collinear. From Canada: 9) Prove that the sum of the face angles at each vertex of a tetrahedron is a straight angle if and only if the faces are congruent triangles.
J
... J J •
~
..J ,~
...J • .,
..J
...
From Ea3t Germany:
...
10) To each vertex of a regular pentagon an integer is assigned in such a way that the sum of all the five numbers is positive. If three consecutive vertices are assigned the numbers z, y, z, respectively, and y < 0, then the following operation is allowed: the numbers z, y, z are replaced by z+y, -y, z+y, respectively. Such an operation is performed repeatedly as long as at least one of the five numbers is negative. Determine whether this procedure necessarily comes to an end after a finite number of steps.
.....
,I
...
.'" I
NOTE: Number 10 actually appeared on the 1986 International Mathe matical Olympiad in Poland. Joseph Keane of the USA received a special prize for his solution, the only special prize awarded for that Olympiad. 138
.....
..J ....
... j
APPENDIX A: Olympiad Problems from around the world
From the USA: 11) Let A, Band C be three points on the edge of a circular pond such that B is due west of C and ABC is an equilateral triangle of side 86 meters. A boy swam from A directly toward B. After covering a distance of x meters, he turned and swam westward reaching the shore after covering a distance of y meters. If x and yare both positive integers, determine y. From Turkey: 12) Let ABCD be a tetrahedron having each sum of opposite edges equal to 1. Prove that
rA +rB +ro +rv::;
v'3 3"""
where r A, rB, ro and rv are the inradii of the faces, equality holding only if ABCD is regular. From the USSR: 13) The numbers Xl' X2,' •• ,X n belong to the interval [a, bl, where 0 b. Prove the inequality ( Xl
+ X2 + ... + X n )
1 (
1
Xl
1 )
+ X2 + ... + X n
< (a + b)2 2 4ab
n.
From Ea6t Germany: 14) Find the smallest natural number N numbers r, 6, t such that
> 0 for which there are natural N
5
5
=t.
From Yugo61avia: 15) Let five different positive real numbers be given. Prove that there are two among them for which neither the sum nor the absolute value of the difference between them is equal to any of the remaining three numbers.
139
] APPENDIX A: Olympiad Problems from around the world
J
From Hungary:
.....
...
I
Prove that 1 -
al b1
is a perfect square.
From the
Netherland~:
17) Determine the maximum value of m 2 + n 2 where m and n are integers satisfying m,n E {1,2, ... , lOO}
and
(n 2
-
mn - m 2 )2
J
= 1.
From Sweden:
J
18) Determine all positive integer solutions (:1:0,:1:1" .• , :l: n ) of the system:
+1 5:1:1 + 1
4:1:1 = 5:1:0 4:1:2 =
... I
." From the USSR: 19) Find all integers n
~
1 such that
From Finland: 20) Show that for all (sin :1:) . (1
:I:
E (0,11"),
...'" I
+ cos :1:) ~ sin (~ + ~) . (1 + cos (~ + ~)) . 140
...
APPENDIX A: Olympiad Problems from around the world
21) Prove that if m, nand r are positive integers and 1 + m+ nV3 = (2 + V3)2r-l, then m is a perfect square. 22) If ABC D is a quadrilateral and P and Q are points on AD and BC, respectively, such that
AP AB BQ
PD = DC = QC'
prove that PQ is equally inclined to AB and to DC. 23) Let the 100 roots of
P(z) = z100 - 600z DD
+ ... + alz + ao
= 0
be real and suppose P(7) > 1. Prove that P(z) = 0 has a root greater than 7. 24) Find a pair of integers (r, s) such that 0< s
< 200
and
45 r 59 61 > ; > 80·
Also, prove the pair (r, s) unique.
25) Prove that the sum of the squares of P A, P B and PC from a point P within a triangle ABC is least when P is the centroid of the triangle.
141
APPENDIX B: Kurschcik Korner t The problems below are from the famous Kurschak Mathematical Compe tition of Hungary. Not only is this the oldest national talent search in mathe matics, but it also served as the model for most other national and international scholastic competitions ever since its inception in 1894. It is held in October of each year (exceptions have been 1919-21, 1944-46 and 1956), and while it was originally intended for the high school graduates of that year, in recent years it also attracted most of the other outstanding high school students as well. Usually, the number of contestants is around 500, two-thirds of whom submit solutions to the problems. The contest consists of 3 carefully chosen problems; the contestants are given 4 hours to submit well-written, complete solutions to them. Readers of this column are also invited to set aside an uninterrupted 4-hour period to compose their own solutions for any year's competition, and to submit them to Dr. George Berzsenyi Division of Mathematics Rose-Hulman Institute of Technology Terre Haute, IN 47803 for a critical evaluation.
J
..., • 'It
I
..
.J
J
... •
'!t
i
...
I
'11
Thi6 i6 the 1937 Kur6chak competition:
(1) Assume that at, a2, ... ,an are positive integers whose sum is less than
k, where k is also a positive integer. Prove that
(2) Let three circles be given so that each of them is tangent to each of the others and the three points of tangency are distinct. Prove that the three circles are either in a plane or on the surface of a sphere. (Note: If two circles are not in the same plane, then they are said to be tangent to one another only if their tangent lines at their common point coincide.) (3) Assume that the points At, A 2 , ••• , An are not on the same line. Let P and Q be two distinct points so that
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t In the first edition of the Arbelo6, each of the competitions was pub lished in a separate issue. In this edition they are all presented together.
142
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APPENDIX B: Kurschlik Korner
Prove that there is a point R such that
ThiJ iJ the 1974 KurJchdk competition:
(1) Two blackboards are located in a library, one at the entrance and one at the exit. Each new arrival registers on the board at the entrance the number of people he finds in the library, and upon leaving regis ters on the board at the exit the number of people left behind in the library. Prove that at the end of the day the two boards will have the same numbers written on them, though possibly in a different order. (Assume that only one person arrives at or leaves the library at any given time.) (2) Given
1,~,~, ... ,!., ..., an infinite sequence of side lengths of squares,
2 3 n show that there exists a square into which all of the above squares will fit without overlap, and find the smallest such square.
(3) Prove that for all integers k > 1 and all real x,
1- x
x2
X S
.
xi
x 2k
+ -2! - -3! + ... + (-1)1 -j! + ... + ->0 (2k)! - .
ThiJ iJ the 1979 K urJchtik competition:
(1) The base of a convex pyramid is a polygon with an odd number of sides, its lateral edges (i.e., those meeting at the apex) are all of the same length, and the angles between its neighboring lateral faces are also equal. Prove that the base of the pyramid is a regular polygon. (2) The function / satisfies the following inequalities for every pair of real numbers x and y:
:s x, /(x + y) :s /(x) + /(y). /(x)
Show that /( x) = x for every real number x. (3) Letters are arranged in an n X n array so that no two rows of the array are identical. Prove that it is possible to delete one of the columns of the array so that the remaining rows will remain distinct.
143
] APPENDIX B: Kurschlik Korner Thi~ i~
the 1981
Kti.r~chak
competition:
(1) Let A, B, P, Q, and R be points in a plane. Prove that
AB+PQ+QR+RPSAP+AQ+AR+BP+BQ+BR
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where XY denotes the distance between X and Y. (2) Let n be an even integer greater than 2. Assume that the squares of n2 an n x n board are colored with "2 colors so that there are exactly 2 squares of each color. Show that one can position n rooks on the board so that they occupy squares of different colors, and that no two of them are in the same row or column. (3) Let r( n) denote the sum of the n+ 1 remainders obtained upon dividing the positive integer n by 1,2,3, ... , n. Prove that there are infinitely many positive integers k for which r( k) = r( k 1).
Thi~ i~
the 1983
Kti.r~chak
then x
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competition:
(1) Prove that if x, y and z are rational numbers such that
x' + 3y' + 3z' -
,
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=0
= y = z = o.
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..
(2) Assume that the coefficients aI, a2, . .. ,an-l of
are non-negative real numbers, and that the equation /(x) real roots. Prove that /(2) ~ 3n •
= 0 has n
(3) In a plane, n+1 points, PI' P2 , ... Pn , and Q, are given, no three of which on the same line. We know I,hat. for each pair of distinct points Pi, Pj one can find a point Plo such that Q is interior to the triangle PiPjPIo. Show that n must be odd.
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