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Journal title history:

➢ The first 32 issues, from Vol. 1, No. 1 (March 1975) to Vol. 4, No.2 (February 1978) were published under the name EUREKA.

➢ Issues from Vol. 4, No. 3 (March 1978) to Vol. 22, No.8 (December 1996) were published under the name Crux Mathematicorum.

➢ Issues from Vol 23., No. 1 (February 1997) to Vol. 37, No. 8 (December 2011) were published under the name Crux Mathematicorum with Mathematical Mayhem.

➢ Issues since Vol. 38, No. 1 (January 2012) are published under the name Crux Mathematicorum.

Mat he m

at ico ru m

ISSN 0705 - 0348

C P !! X M A T H E-w A T I C 0 P !! P

Vol. 10, No. 4

April 1984

Sponsored by Carieton-Ottawa Matheaatics Association Mathematique d'Ottawa-Carleton

Publie par le College Algonquin, Ottawa

The assistance of the publisher and the support of the Canadian Mathematical Society, the Carleton University Department of Mathematics and Statistics, the University of Ottawa Department of Mathematics T and. the endorsement of the Ottawa Valley Education Liaison Council are gratefully acknowledged-

CRUX MATHEMATJCDRUM is a prdbiesa-solvi&g jouraal at the senior secondary and university undergraduate lewis for those who practise or teach mathematics. Its purpose is primarily educational, but it serros also those who read it for professional t cultural, or recreational reasons e

It is published monthly (ejscept July and August! . The yearly subscription rate for ten issues is $22 in Canada, §24.50 (or US$20) elsewhere.. Bade issues s each $2.25 in Canada, $2.50 (or US$2) elsewhere. Bound volumes with iadess Vols. 1&2 (eoefeimed) and each of Vols* 3-9, $17 in Canada, $18,40 (or US$15) elsewhere. Cheques and money orders, payable to CRUX MATHEMATICORUM, should be sent to the managing editor.

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Editors Leo Sauve, Algonquin College, 140 Main Street, Ottawa, Ontario, Canada K1S 1C2.

Managing Editor; F.G.B. Haskell, Algonquin College, ZOO Lees Ave. f Ottawa, Ontario, Canada K1S 0CS.

Typist-compositor: Kuen Huynh.

Second Class mil Registration No. 5432. Return Postage Guaranteed.

CONTENTS

Prime Aspects of 37 . . Charles W. Trigg 100

The Olympiad Corner: 54 . . . . . . . . M.S. Klamkin 107

Problems - ProblSmes . . . . 113

Advent of the Breakfast Serial ^ 5

Letter to Edith Orr . . . . . . . . . . . . . . . . . . . . . . . 115

Solutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 1 6

- 99 -

- 100 -

PRIME ASPECTS OF 37

CHARLES W. TRIGG

37, its two digits9 and its reverse are all primes in the usual sense. Thus

37 is a prime prime. That is9 it is one of the set of primes that remain prime

after successively dropping the right-hand digit [14], It is also one of the set

of primes that remain prime after successively droppina the leftmost digit. Hence

37 (and its reverse) is a right prime prime 9 a left prime prime, and a double prime

prime [ 1 ].

37 is an isolated prime [2], since it is not one of a twin prime pair (35 and

39 being composite).

Let „ 1 D _ 1 _ _ 5 _ 691 _ 37*208360028141 B\ = 6- 5 2 ~ 30* . . . . B5 - ~ 9 B6 = 273^> • • - B16 = 5 1 0 . . .

be Bernoul l i numbers, each expressed in i t s lowest terms. With p an odd pr ime, con­

s ider the set { B l 9 5 2 9 . . . , B m l > 9 where u - ( p - i ) / 2 . I f the numerators of none

of these f rac t ions are d i v i s i b l e by p5 then p is said to be a regular prime; o ther­

wise i t is said to be irregular [ 1 3 ] . Kummer in 1850 found tha t a l l the primes less

than loo are regular except 37, 59, and 67. Thus 37 is the smal lest i r r e g u l a r prime

37 re ta ins i t s prime s t a t u s , as does i t s reverse 73, even under S te in 's whim­

s ica l ex t rapo la t ion o f G u l l i v e r ' s account of mathematics as pract iced on the f l y i n g

is land of Laputa [ 7 ] . This defines the good numbers as 1 and a l l natural numbers

obtained from 1 by adding a mu l t i p l e of 3S A good natural number is a Lagado prime

(named a f t e r the earthbound metropol is of the Kingdom) i f i t has exact ly two good

d i v i s o r s . Hence

37 = !»(! + 12»3) and 73 = 1-(1 + 24»3)

are Laqado primes.

An absolute pseudoprime, n9 is a composite number such that for each integer a

the number an - a is divisible by n [6]. Such a number is n - 37.13-61.

37 is the norm of the Gaussian primes 1 + ei9 i - 6i9 and their associates. 73

is the norm of the Gaussian primes 3 + 8i9 3 - 8i9 and their associates.

Prime loops.

A prime loop i s formed when primes are placed around a c i r c l e i n such an order

tha t each pa i r of adjacent primes concatenated forms a prime. Thus (37, 67) is a

prime loop since 3767 and 6737 are both prime. Prime loops contain ing from two to

ten two -d i g i t primes e x i s t w i th 37 as an element. For example:

- 101 -

(37, 79) (37, 19, 13, 61, 97, 43)

(37, 97, 43) (37, 19, 13, 67, 61, 97, 43)

(37, 79, 19, 31) (37, 79, 19, 13, 67, 61, 97, 43)

(37, 61, 43, 97, 67) (37, 79, 19, 13, 61, 97, 43, 73, 31)

(37, 61, 97, 43, 73, 31, 19, 13, 67, 79)

The last loop is one of eleven prime loops each of which contains all ten two-

digit primes of the form 6k + i once each Cio].

37 expressed by primes less than 37,

37 = 2 + 3 + 13 +19 = 2 + 5 + 7 + 23 = 2 + 5 + 11 +19

= 2 + 5 + 1 3 + 1 7 = 2 + 7 + 1 1 + 1 7 = 3 + 5 + 2 9

= 3 + 11 + 23 = 5 + 13 + 19 = 7 + 11 + 19 = 7 + 13 + 17.

37 = 2«3 + 31 = 2-5 + 3 + 7 + 17 = 2-7 + 23 = 2-11 + 3 + 5 + 7

= 2-13 + 11 = 2-17 + 3 = 2-19 - 3 - 5 + 7

= 2*23 + 3 + 5 - 17 = 2-29 - 3 - 5 - 13 = 2-31 - 5 - 7 - 13.

37 = 3-5 + 2 + 7 + 13 = 3-7 + 2(19-11) = 3-11 - 1 9 + 2 3

= 3-13 + 17 - 19 = 3-17 - 2-7 = 3-19 - 7 - 13

= 3°23 - 13 - 19 = 3-29 - 19 - 31 = 3-31 - 7(3+5),

37 = 2 + 5-7 = 5-11 + 13 - 31 = 5*13 - 11 - 17

= 5-17 - 19 - 29 = 5*19 - 2*29 = 5*23 - 7 - 11 - 29 - 31

= 5*29 - 7-13 - 17 = 5-31 - 2-3-11 - 23 - 29.

For other representations of 37 by other primes, see [8],

37 in prime arithmetic progressions.

The only three arithmetic progressions of positive prime numbers containing

37 in which 37 is not the leading term are

31, 37, 43; 13, 37, 61; and 7, 37, 67, 97, 127, 157,

the last of which contains six primes. The progression 37, 73, 109 contains both

37 and its reverse.

In progressions of prime numbers greater than 39 all common differences, d,

are multiples of 6. For 30 < d < 2310, no progression of primes starting with 37

has more than five members. The progression for d = 2310 is

37, 2347, 4657, 6967, 9277, 11587.

Within the stated interval9 the thirty-seven three-prime progressions have d = 36, 60, 72, 102, 156, 186, 270, 312, 336, 396, 450, 486, 606, 696, 786, 792,

816, 996, 1026, 1116, 1176, 1242, 1260, 1410, 1506, 1512, 1572, 1590, 1716, 1740,

1746, 1752, 1836, 1992, 2046, 2202, and 2256.

- 102 -

The sixteen four-prime progressions in the interval have d = 1209 192, 360,

420, 540, 582, 5729 8829 10509 10929 1200, 1542, 1632s 16629 1710, and 1830.

The two five-prime progressions in the interval have d = 870 and 1290.

Z7 9 a prine-generated prime.

37 is the f i f t h term of the sequence of primes of the form 2p + 3, where p

is prime. That is9

79 13, 17, 29, 37, 41, 61, 89, ...,

wherein 7 + 13 + 17 = 37 and (13 + 6D/2 = 37. In the expression 2»37 + 3 = 7*n,

each integer is prime.

37 is the fifth term of the sequence of primes of the form 3p - 2, where p is

prime. That is,

7, 13, 19, 31, 37, 67, 109, ...,

wherein -7 + 13 + 31 = 37 = (7 + 67)/29 and 3-37 - 2 = 109.

37 is the third term of the sequence of primes of the form 3p + 4, where p is

prime. That i s ,

13, 19, 37, 43, 61, 73, 97, ...,

wherein (13 + 6D/2 = 37, and a pair of reversal primes, 37 and 73, appears.

37 is the second term of the sequence of primes of the form 6p + 7, where p is

prime. That is,

19, 37, 73, 109, 181, 193, 229, ...,

wherein 229 = 7 + 3*37*2, and again the reversal primes 37 and 73 appear, this time

in adjacent positions.,

37 and its reverse are the first two terms of the sequence of primes of the forms

9p + 10 and I2p + 13, where p is prime. That is,

37, 73, 109, 127, 163, 181, 271, ...

and

37, 73, 97, 241, 457, ... .

We observe tha t 9«37 + 10 = 7 3 and 12-37 + 13 = 457,

37 in other prime sequences.

37 and i t s reverse 73 occupy the f i f t h and ninth positions in the sequence of

primes of the form 4n + l , thus:

5, 13, 17, 29, 37, 41, 53, 61, 73, 89, 97, 101, ... .

They occupy the same positions in the sequence of primes of the form 6n + 1, thus:

7, 13, 19, 31, 37, 43, 61, 67, 73, 79, 97, 103, ... .

- 103 -

In the sequence of primes of tire form 9n + 1, the reverses are adjacent, thus:

19, 37, 73, 109, ... .

Sequences of primes of other forms in which 37s but not 73s appears are:

7, 17, 37, 47, 67, 97, 107, ..., 5n+2;

239 37s 79, 107, ..., 7n+2;

37, 59, 103, ..., lln+4;

37, 113, 151, .... 19n-l.

There are 19 primes less than 10^ of the form n2 + l, namely: 2, 5, 17, 37, ioi9

197, 257, 401, 577, 677, 1297, 1601, 2917, 3137, 4357, 5477, 7057, 8101, and 8837.

Therein9 (3+7)2 + l = 101 and (2«37)2 + l = 5477. These two primes have digital

roots of 2 and 5S which are members of the sequence. Furthermore,

(2*5 + 10D/37 = 3, While (2 + 257)/37 = 7.

37 = 3137 - 2917 - 1 9 7 + 1 7 - 5 + 2 ,

372 = 8837 - 7057 - 4 0 1 - 1 7 + 5 + 2 , and

373 = 5(8837 + 1297) - 17.

37 is the sequence starter in the prime positional sequence

37 157 919 7193 72727 919913

wherein f o r any term n the next term is the nth prime [ 1 6 ] .

Primitive roots of primes.

g is a p r i m i t i v e root of p i f <f~ = l (mod p) and gx i 1 (mod p) f o r o < x < p - l .

The twelve p r i m i t i v e roots of 37 are [ 5 ]

2 5 13 15 17 18 19 20 22 24 32 35.

Four of the roots are consecutive in tegers . Roots equ id is tan t from the middle of

the sequence sum to 37. Each pa i r of equ id is tan t roots except (159 22) contains one

prime. Six o f the roots are even and s i x are odd. There are 5 primes, four mul t ip les

of 53 and one 5th power. There are 5 roots r e l a t i v e l y prime to 3 7 - 1 , and 6 roots do

not exceed (37-1 ) /2 .

37 is a p r i m i t i v e roo t of the fo l low ing f i f t y - e i g h t primes less than 103 [ 5 ] :

59, 79, 97, 109, 113, 131, 167, 179, 193s 227, 239, 241, 251, 257, 283, 311, 313, 331,

347, 353, 383, 389, 401, 409, 4319 439, 463, 4675 479, 503, 523, 541, 547, 557, 563,

569, 587, 643, 647, 653, 661, 701, 709, 727, 757, 769, 797, 809, 827, 829, 857, 859,

907, 911, 919, 947, 967, 977, Eight of these primes are palindromes. Those of the

forms 4& - 1 and 4fc + i are in the r a t i o 33:25; those of the form 6k - i and 6k + l

are in the r a t i o 34:24.

- 101 -

Miscellanea.

In the sequence resulting from the application of the Collatz algorithm (if

it is odd, triple it and add l; if it is even9 divide it by 2) starting with 37,

all odd terms save the final 1 are prime. Thus: 37_, 112, 56, 28, 14, ]_, 22, n_,

34, 17_, 52s 26, 13 , 40, 20 10, 5_, 16, 8, 4, 2, l. Indeed, 5, 7, li, 13, and 17

are consecutive primes.

In the following sequence, the first 37 positive integers have been placed so

that the sum of each pair of adjacent numbers is a prime: 37 - 36 - 31 - 12 - 19 -

28 - 25 - 6 - 11 - 8 - 23 - 14 - 3 3 - 4 - 7 - 3 0 - 29 - 18 - 5 - 24 - 13 - 10 - 3 -

16 - 1 - 2 - 17 - 20 - 21 - 22 - 15 - 2 6 - 3 5 - 3 2 - 9 - 3 4 - 2 7 . In the following

addition of the th i r ty -s ix prime sums, the frequencies of occurrence (other than i )

of the prime sums precede the parentheses: (3) + (11) + (13) + 2(17) + 3(19) + 2(23)

+ (29) + 3(31) + 6(37) + 3(41) + 3(43) + 3(47) + (53) + (59) + 2(61) + 2(67) + (73)

= 1342, and (1+2) + (3+4) = 3 + 7, The largest prime sum is the reverse of 37. As

a sum, 37 occurs twice as often as the next most frequently occurring prime.

Euclid used the expression

Q(p) = 2»3-5«7-ll-13-.,.-p + 1,

where p is a prime, to demonstrate the infinitude of primes. A related expression

which can be used for the same purpose is

Rip) = 2*3-5®7«ll-13-...*p - 1.

It turns out that

5(37) = 181®60611»6764219

but no value of Qip) is divisible by 37. In contrast,

i?(23) = 37-131-46027.

The smallest Rip) that contains four d ist inct prime factors is [3]

i?(37) = 229-541*1549-38669.

Curiously enough, the sum of the digits of the four factors of i?(37) is 2-37.

There are exactly 37 palindromic primes less than 107 composed of prime digits

[9]. The twelve in which 37 is imbedded are: 373, 37273, 37_573, 73237_, 77377,

3337333, 37_22273, £732373, 3773773, 7327237_, 7352537_, and 7733377.

According to Samuel Yates [15], the prime divisors of the repunit #37 = (io37-i)/9

are

2028119-247629013-2212394296770203368013.

Every repunit consisting of 3n ones is divisible by 37, since ill = 3-37.

- 105 -

The smallest solutions of the equation x2 + 3y2 = p, where p is a prime, are:

(p, a% y) = (7, 2, 1), (13, 1, 2), (19, 4S 1), (31, 22 3), and (379 5, 2),

The smallest solutions of the equation x1 + 4w2 = p3 where p is a prime, are:

(p, x, y) = (5, 1, 1), (13, 39 1), (17, 1, 2), (29, 59 1), and (37, 1, 3).

The smallest solutions of the equation x3 - y3 = p% where p is a primes are:

(p, x9 z/) = (7, 29 1)9 (19, 39 2), and (379 49 3).

Curiosa. Each of the prime digits of 37 is generally considered to be a lucky number, 3 7 = 7 + 3 + 7 + 3 + 7 + 3 + 7 9 a smoothly undulating sequence including 3

threes amongst i ts 7 terms. (3 + 7)/2 = 5 involves the f irs t four primes. 37 + 3 + 7 = 47 and 37 - 3 + 7 = 4i„ Both of these sums are primes with the

prime digit sums n and 59 and have the prime digital roots 2 and 5 differing by 3 and summing to 7.

37 - 3 - 7 = 27 and 37 + 3 - 7 = 33, Both of these sums are composite, and have composite digital roots9 9 and 69 differing by 3,

The related primes 373 and 3773773 have the prime digit sums 13 and 37, re­spectively.

37 is the twelfth prime, its reverse 73 is the twenty-first prime9 and 12 is the reverse of 21. Furthermore,

H I H = 22 = 7 - 3 9 and 3*7 = 2 1 .

The only set of four distinct two-digit primes composed of a pair of twin primes and their reverses is 71, 739 17, and 37, Indeed, 17 and 37 belong to each other [ i l l in the sense that

(1 - 7)2 + 1 = 37 and (3 - 7)2 + 1 = 17.

Their respective digital roots are the cubes 1 and 89 which differ by 7 whereas their

cube roots sum to 3.

37 also belongs to the prime ioi, in the sense that (3 + 7) 2 + 1 = 101. [11]

The sum of the squares of the first 7 primes is

22 + 32 + 52 t 72 + ll2 f 132 + 172 = 2«37*32 = 666,

the number of the beast, Each of Enma Lehmer's twin primesf 9 « 2 2 1 1 ± 1 =

2961908203 1781708758 7894452860 3230908968 9245581404 4375402749 62432 ± 1,

- 106 -

contains 37 even and 28 odd digits [4]. The sum of the even digits is 2°73. Both

37 and 28 are imbedded in the primes .

Among the 37n, n < 37, there are ll powers with prime digit sums9 Z, namely:

(n, Z) = (2, 13), (3, 19), (12, 73), (14, 109), (15, 109), (18, 127), (21, 163),

(25, 127), (29, 199), (30, 181), and (37, 271). [12]

REFERENCES

1. Murray Berg and John E. Walstrom9 "Right and Left Primes", Journal of

Recreational Mathematics, 3 (3), July 1970, 179.

2. Richard L. Francis, "Isolated Primes"9 Journal of Recreational Mathematics ,

11 (1), 1978-79, 17.

3. Sidney Kravitz and David E. Penney, "An Extension of Triag's Table", Mathe­

matics Magazine, 48 (2), March 1975, 92-96.

4. Rudolf Ondrejka, "65»digit Twin Primes ", Journal of Recreational Mathematics,

3 (4), October 1970, 257.

5. Roger Osborn, Tables of All Primitive Roots of Odd Primes Less Than 1000 ,

University of Texas Presss Austin, 1961.

6. Wacjaw Sierpiflski, A Selection of Problems in the Theory of numbers, Macmillan

New York, 1964, pp. 51-52.

7. Sherman K. Stein, Mathematics^ A Man-Made Universe, W.H. Freeman, San

Francisco, 1963, pp. 32-33.

8. Charles W. Trigg* "A Close Look at 37", Journal of Recreational Mathematics,

2 (2), April 1969, 117-128. 9- » "Special Palindromic Primes", Journal of Recreational Mathe­

matics, 4 (3), July 1971, 169-170.

10. , "Loops of Two-Digit Primes", Journal of Recreational Mathe­

matics, 6 (4), Fall 1973, 302-305.

11. , "Some Related Prime Sequences", Journal of Recreational Mathe­

matics, 12 (2), 1979-80, 119-120.

12. , "The Powers of 37", Journal of Recreational Mathematics3 12

(3), 1979-80, 186-191.

13. H.S. Vandivers "Is There an Infinity of Regular Primes?", Scripta Mathematica,

21 (4), December 1955, 306-309.

14. John E. Walstrom and Murray Berg9 "Prime Primes", Mathematics Magazine3 42

(5), November 1969, 232.

15. Samuel Yates, Repunits and Repetends , published by the author at Delray

Beach, Florida, 1982, p. 143.

16. Jacques P. Sauve* and Charles W. Trigg, "Prime Positional Sequences", Crux

Mathematicorum, 9 (2), February 1983, 37-41.

2404 Loring Street, San Diego, California 92109.

- 107 -

THE OLYMPIAD CORNER: 54

M*S, KLAMKIN

I present two new problem sets this month. The f i r s t consists of the problems

se t a t the second round of the 1983 Chinese Mathematics Olympiad. I am grateful to

Genozhe Chang (University of Science and Technology of China a t Hefei) for supplying

these problems in English t rans la t ion . Next I give s through the courtesy of Walther

Janous (Ursulinengymnasium at Innsbruck), the problems of the 1982 Austrian-Polish

Mathematics Competition*

1983 CHINESE MATHEMATICS OLYMPIAD (Second Round)

October 16, 1983 - Time: 2 hours

1 , (8 points) Show that

Arcs in x + Arccosa: = | s - l < x < l .

2 6 (12 points) A function / defined on the interval [ o , i ] s a t i s f i e s

f(0) = / ( l ) and 0 < xi*x2 - 1 = = > l/C«2)-f Ca?i)l < \x2-xi\.

Prove that 0 < xi*x2 < 1 = > \f(x2)-f(#i)\ < | .

3 8 (16 points) For the quadrilateral ABCD of

the f igure s the following proportion holds:

CABD] : [BCD] : [ABC] = 3 : 4 : l ,

where the brackets denote area* If the points M e AC

and N e CD are such that AM : AC = CN : CD and B9M9N

are col l i nea r , prove that M and N are the midpoints

of AC and CD9 respect ively.

L\, (16 points) Determine the maximum volume of

a tetrahedron whose six edges have lengths

2S 39 39 49 5, and 50

5 , (18 points) Determine

min max |cos2a? + 2sin£rcosa? - sin2x + Ax + B\. A9B 0<a?<3*rr/2

it

AUSTRIAN-POLISH MATHEMATICS COMPETITION

1st days July 9, 1982 - Time: 41 hours

- 108 -

1, Determine all pairs of natural numbers (n,k) such that Ty "k + 3

g.c.d.((w+l) - n, (n+1) - n) > 1. 28 We are given a unit circle C with center N and a closed convex region i?

in the interior of c. From every point P of circle c9 there are two tangents to the boundary of i? that are inclined to each other at 60°. Prove that i? is a closed circular disk with center M and radius £.

38 Prove that9 for all natural numbers n > 29

.e t an{ f ( 1 + ji-)f= S cot | f ( l - - | i ) f . ^=l 3 -1 x-1 3 -1

2nd day: July 10, 1982 - Time: 41 hours

lf8 N being the set of natural numbers, for every n e N le t Pin) denote the product of all the digits of n (in base ten). Determine whether or not

the sequence k>%K where

x\ £ Ns x, = x. + P(x. ), fc = l ? 2 9 3 s 0 O . 9

can be unbounded ( i . e . , for every number M9 there exists an x. such that a?. > M) e

3 3 58 Let {ASB} be a partition of the unit interval J = [0,1] (so that AuB = I

and AnB = 0), and$ for any real number x3 let {^} + 4 = iy \ y - x+a s a e A) .

Prove that there is no real number x such that B - ix) + A .

6, Let a be a fixed natural number* Find all functions / defined on the set

D of natural numbers x > a and satisfying the functional equation

f(x+y) = f(x)°f(y)

for all x9y e D.

3rd day: J u l y 1 1 , 1982 - Time: 4 hours

(Team Competi t ion)

(Here each team, u s u a l l y c o n s i s t i n g of s i x s t u d e n t s , s o l v e s t h e t h r e e problems c o l l e c t i v e l y and submits a t most one s o l u t i o n p e r problem.)

1, Determine all tr iplets (x9y9z) of natural numbers, with z as small as possible, for which there exist natural numbers a9b9c9d satisfying

(i) oft - a = c with x > a > a, (i i) z = ab = ad 9

( i l l ) x t y = a + b.

2, Point X is in the interior of a given regular tetrahedron ABCD of edge

- 109 -

lenath 1. If <f(X,YZ) denotes the shortest distance from X to edge YZ, prove that

<?(X,AB) + d(X,AC) - <7(X,AD) + d(X,BC) - J(X,BD) + d(X,CD) > 4=,

and show that equality holds if and only if X is the center of the tetrahedron.

3, Let

Determine a real constant c such that, for all natural numbers n > 39

n < S < en* n

Remark* The smaller the number o determined, the greater w i l l be the number o f

points awarded f o r the s o l u t i o n .

I now present so lu t ions to some problems published e a r l i e r in th is column.

1 2 , [ 1981 : 74] From a 1979 Moscow Olympiad (for Grade 10).

The func t ion / i s defined on the in te rva l [ o , i ] and i s twice d i f f e r e n t i a b l e

at each po in t . The absolute value of the de r i va t i ve is never greater than 1 on the

i n t e r v a l , and / ( o ) = / ( i ) = o. What is the greatest possible maximum value that fix)

can have on the in te rva l?

Solution by M,S.K.

The given funct ion f i s d i f f e r e n t i a t e , , hence continuous, on the in te rva l [ o , i ] ,

so i t has an absolute maximum at some po in t x0,

and f (XQ) = o i f x0 i s an i n t e r i o r po in t o f

the i n t e r v a l * We compare / w i t h the funct ion

g defined by

i f o < x < | s

gix)

/x9 if 0 < x < i ,

i 1-x i f 1 < x < l ,

whose graph is shown in the figure. Since

f (0) = f(i) - o, no point of the graph of /

can lie above the graph of g. For otherwise, geometrically or by the mean value

theorems \f Or) I would exceed l at some point. Therefore fix$) £ g(i) = £.

We show that in fact f(x0) < £. Suppose on the contrary that f(xo) = i. Then

we must have x0 = i and f (!) = o. Now9 for o < h < J,

i = \ - ii-h) = I - gjj-h) < fjj) - fjj-h)

110 and taking the limit as h + o shows that f[(i) * l, contradictina /*(£) = oB There­

fore f(xQ ) < £.

So far we have assumed that / is only (once) different!able on the interval

fo,i]. Assuming that f is twice differentiable on the interval provides no new

information; nothing* in particular, about the supremum of f(x0) as / ranges over

all suitable functions. However, if we assume differentiability of a sufficiently

high order, it is possible to find a function /which otherwise satisfies the con­

ditions of the problem and for which f(i) is as close to i as we please. Considers

for example, the following function / (obtained by smoothina out function g in the

neighborhood of x = £):

,r (2a r

fix) 2r

(1-ar) J2(l-x)f

2r

if 0 < x < i,

if 1 < x < l.

This function satisf ies f(o) - f{i) = os \f%ix)\ < i, and i t is n times differentiable

on the interval r o , i ] i f r > n> I ts absolute maximum is

f(i) = 2rs

which can be made as close to i as we please by taking r large enough.

2 , [1982: 237] From a 1982 Bulgarian Olympiad,

In a plane there are n circles each of unit radius. Prove that at least

one of these circles contains an arc which does not intersect any of the other

circles and whose length is not less than 2ir/n9

Solution by George Tsintsifas3 Thessaloniki3 Greece.

Let Kl9 Ko, S8es K be the consecutive

vertices and a i , a29 . . . , a the corresponding

inter ior angles of the convex hull of the

centers of the n given c i rc les, so that

m < n. We assume that these have been

labeled so that min a. = on. As a conse» i

quence, we have (777-2 )TT

a ! < (1)

As shown in the f igure, the perpendiculars KXP and Kfi to the sides of angle a! form an angle e such that

- I l l -0 + a i = TT9 ( 2 )

and i t follows from (1) and (2) that

Q > 2T[> 27[ m n

Therefore KiP and KiQ intersect circle (Ka) in an arc AB whose length is not less than 2-n/n,

For i < i < n9 no circle (K.) has a point in common with arc AB„ For the

points of intersection of circles Ki and K.f if anyf lie on the perpendicular

bisector of segment KxK., and this perpendicular bisector cannot intersect the

interior of angle AKiB.

2 8 [1982: 269] From a 1963 Peking Mathematics Contest for Grade 12.

Nine points are randomly selected inside a square of side l. Show that

three of the points are the vertices of a triangle of area at most 1/8.

Solution by M.S9K,

Partition the square into four squares of side 1, By the pigeonhole principles

at least one of the four squares contains three of the given points, and they are

the vertices of a triangle of area at most 1/8.

Eider. Show that the maximum area of a triangle inscribed in a rectangle is

half the area of the rectangle. A

3, [1982: 269] From a 1963 Peking Mathematics Contest for Grade 12*

Given are 2n+3 points in the planes no three col linear and no four concyclic*

Is it possible to construct a circle passing through three of the points so that

exactly half of the remaining 2n points lie inside the circle? Justify your answer.

Comment by M.S.K.

The answer is yes. For two solutions, see Ross Honsbergers Mathematical Morsels, Mathematical Association of America, Washington, D9C,9 19789 pp8 48-51.

£}, [1982: 269] From a 1963 Peking Mathematics Contest for Grade 12.

A set of 2n objects is partitioned into a number of subsets* A move consists of transferring from one of the subsets to an equal or smaller subset a number of objects equal to the cardinality of the second subset. Prove that, irrespective of the initial partition., all the subsets can be combined into a single set by a finite number of moves.

Solution by Willie longs Singapore.

We use induction on n„ The conclusion is immediate if n = l . Suppose i t holds

112 nn+l f o r some n > i . In any p a r t i t i o n of a set of 2 ob jec ts , the number o f subsets

of odd c a r d i n a l i t y i s even. Any move invo lv ing two subsets o f odd c a r d i n a l i t y

resu l ts i n two subsets of even c a r d i n a l i t y (one being possibly the empty subset,

which is of even c a r d i n a l i t y ) . Thus, a f t e r a f i n i t e number of moves, i t i s pos­

s ib le to have a l l the subsets o f even c a r d i n a l i t y . We now imagine tha t i n every

subset the objects are glued together in pa i r s . As a r e s u l t , we now have a t o t a l

of 2n ob jec t s , and the desired conclusion fo l lows from the induct ion assumption.

1 B [198 2 : 270] From a 1964 Peking Mathematics Contest for Grade 12.

ABC is a t r i a n g l e in which angle A is nonacute. Let B^ECi be any inscr ibed

square of t r i a n g l e ABC wi th B1 on AB, Cx on AC, and DE on BC. Let B2DiEiC2 be any

inscr ibed square of t r i ang le AB1C1 wi th B2 on ABi , C2 on AC l9 and DxEx on BiC 1 • The

process is continued i n the same way f o r a f i n i t e number of s teps. Prove tha t the

sum of the areas of a l l the inscr ibed squares is s t r i c t l y less than ha l f the area

of t r i ang le ABC.

Solution by M»S,K»

We f i r s t construct square APRQ where

AP = h is the a l t i t u d e from A (see f i g u r e ) .

The squares B:DECi and APRQ are nomothetic,

wi th B the nomothetic center. Let BP = a l 9

PC = a 2 s so t ha t BC = a = a 1 + a2» and

l e t PE = 2/ i , ECj = xi. By s im i l a r t r i ­

angles, we have

By

A

!^\c>/^ —^

Xi

Q

h

xi If

a 1+2/1 <zi-y\

a 1 +h a2

from which fo l lows

^ 2 . z/i = —7- and **L a+h

ah

Since t r i ang le A B ^ i w i th inscr ibed square B2DiEiC2 o f side x2 is nomothetic to

t r i ang le ABC w i th inscr ibed square BiDECx o f side a?l9 we have

BC " B1C1 or X2 = *i

A simple induction now shows that, if the process in performed n times, then n

x xn " n-19

a n = 1,2,3,.

- 113 -

We now show that the desired conclusion holds even i f the process is performed

in f in i te ly often. The sum of the areas of a l l the inscribed squares is

00 2 0 „

r 2 ^1 - ha to

ni± Xn - —* - ^2ah < 2 >

Since 2h2a2< 2h2a2+ha3. a

Editor's note. All communications about this column should be sent to Professor M.S. Klamkin, Department of Mathematics, University of Alberta, Edmonton, Alberta, Canada T6G 2G1.

P R O B L E M S - ' P O O B L E ^ E S

Problem proposals and solutions should be sent to the editor, whose address appears on the front page of this issue. Proposals should? whenever possible, be accompanied by a solution, references, and other insights which are likely to be of help to the editor. An asterisk (*) after a number indicates a problem submit­ted without a solution.

Original problems are particularly sought. But other interesting piToblems may also be acceptable provided they are not too well known and references are given as to their provenance. Ordinarilyf if the originator of a problem can be located, it should not be submitted by somebody else without his permission.

To facilitate their consideration, your solutions, typewritten or neatly hand­written on signed, separate sheets, should preferably be mailed to the editor before November 1, 1984, although solutions received after that date will also be considered until the time when a solution is published.

931, Proposed by Allan Wm, Johnson Jr.3 Washington, D.C.

Replace the letters with digits to obtain a decimal addition:

PASTE STAPLE. fWETT

Then PASTE* STAPLE* or FASTEN somehow your solution to a postcard and send it along

to the editor.

932, Proposed by Kenneth S. Williams^ Carleton Universitys Ottawa,

Let p = 7 (mod 8) be a prime so there is an integer w such that

w2 ~ 2 (mod p).

Prove that the value of the Legendre symbol (-—-) is given by

2 ^ (+1. 1 f P ~ 1 5 ( m ° d 1 6 h

p ( -1 , if p = 7 (mod 16).

933 B Proposed by Jordan B. Tabovs Sofias Bulgaria.

- 1W -A truncated triangular pyramid with base ABC and lateral edges AAl9 BBl9 and

CCi is given. Let P be an arbitrary point in the plane of triangle AiBiCl9 and let Q be the common point of the planes as 3* and y passing respectively through BxCl9

CiAi, and AiBi, and parallel respectively to the planes PBC9 PCAS and PAB. Prove that the tetrahedron ABCQ and the pyramid ABBiAxC have equal volumes.

934, Proposed by Leon Barikoffs Los Angeles., California, As shown in the figure, the diameter AB9 a variable chord AJ9 and the

intercepted minor arc JB of a circle (0) form a mixtilinear triangle whose inscribed circle (W) touches arc JB in K and whose mixtilinear excircle (V) touches arc JB in L. The projections of W and V upon AB are C and D, respectively. As J moves along the circumference of circle (0), the ratio of the arcs KL and LB varies.

(a) When arcs KL and LB are equal, what are their values? (b) Show that BD is equal to the side of the inscribed square lying in the

right angle of triangle ADV.

9 3 5 j Proposed by J.T. Gvoenman3 Amhems The Netherlands.

It is easy to show that (o,o) and (9,9) are the only solutions (x9y) of the equation

x3 + z/3 = l&xy

in which x = y . Find a solution in integers (x9y) with x > y and show that it is unique.

936, Proposed by Stanley Rabinowitzs Digital Equipment Corp., Nashua, New

Hampshire*

- 115 -

Find a-1 V eight-digit palindromes in base ten that are also palindromes in at

least one of the bases two, three, ..., nine,

937 , Proposed by Jordi Dous Barcelonas Spain,

ABCD is a trapezoid inscribed in a circle <|>, with AB 1! DC, The midpoint

of AB is M, and the line DH meets the circle again in P, A line I through P meets

line BC in A!5 line CA in B

!s line AB in C

f, and the circle again in FF.

Prove that (A8B'? C'F1) is a harmonic range.

938. Proposed by Charles W. Trigg3 San Diego5 California.

[s there an i n f i n i t y of pronic numbers of the form a b in the decimal n n

system? (A pronic number is the product of two consecutive integers. The symbol

x indicates x repeated n times* For example, 5323=555222,)

939. Proposed by George Tsintsifass Thessaloniki3 Greece,

Trianale ABC is acute-angled at Bs and AB < AC, M being a po in t on the

a l t i t u d e AD, the l ines BM and CM in te rsec t AC and AB9 respec t i ve ly , i n B* and C \

Prove that BB! < CC! .

9^40, Proposed by Jack Garfunkels Flushings N.I.

Show that, for any triangle ABC9

s i n B s i n C + s i n C s i n A + s i n A s i n B < - + 4 s in ~ s in - sin - < ^ 4 2 2 2 4

ft it ft

ADVENT OF THE BREAKFAST SERIAL

I was j us t goina to say9 when I was i n t e r r u p t e d , that one of the many ways o f

c lass i f y ing minds is under the heads of ar i thmet ica l and algebraical i n t e l l e c t s .

A l l economical and p rac t i ca l wisdom is an extension or va r ia t i on of the fo l l ow ing

ar i thmet ica l formula: 2 ->- 2 = 4, Every phi losophical proposi t ion has the more

general character o f the expression a + b = a. We are mere opera t ives , empi r ics ,

and e g o t i s t s , u n t i l we learn to th ink in l e t t e r s instead of f i g u r e s .

The above is the opening paragraph in the first of a series of articles by Oliver Wendell Holmes (like A. Conan Doyle a medical doctor turned author, not to be confused with his equally famous son. Justice Oliver Wendell Holmes of the U.S. Supreme Court) in the early issues (1857) of a magazine that he himself christened The Atlantic Monthly, The articles were published under the running title "The Autocrat of the Breakfast-Table" and later collected in a book of the same name. A. Conan Doyle so admired the book that he named Sherlock Holmes after its American author.

5% * ft

LETTER TO EDITH ORR

Re your comment [1983s 211]: Better dat Mary shoulda spent a sheepless night.

G.S., Brooklyn, N.Y.

- 116 -

S O L U T I O N S

No problem is ever permanently closed. The editor will always be pleased to consider for publication new solutions or new insights on past problems.

8 0 1 1 [1983 : 21] Proposed hy Sidney Kravitz^ Dovers flew Jersey.

At this time (late January 1983), Canada's peripatetic Prime Minister,

the apostle of the North-South dialogue* is catching his breath at home* So le t

us lose no time in proposinq the following alphametic before he takes of f again

(or bows out):

PIERRE ELLIOTT. TRUDEAU

Solutions were received from CLAYTON W. DODGE, University of Maine at Orono; MEIR FEDER, Haifa,, Israel? J.T. GROENMAN, Arnhem, The Netherlands; RICHARD I4 HESS, Rancho Palos Verdes, California? J.A.H. HUNTER, Toronto, Ontario; ALLAN WM. JOHNSON JR., Washington, D.C.; J.A. McCALLUM, Medicine Hat, Alberta; STANLEY RABINOWITZ, Digital Equipment Corp., Nashua, New Hampshire; KENNETH M. WILKE, Topeka, Kansas; ANNELIESE ZIMMERMANN, Bonn, West Germany; and the proposer.

Editorfs comment*

The unique answer9 obtained by all sol vers, is

246116 6884577. 7130693

Reaching this answer required a lot of brute force on the part of the unaided solver,

so the problem was more suitable as a programming exercise. Most solvers submitted

only the answer: they apparently just twiddled their thumbs while they waited for

a computer to burp out the answer. Apart from the proposer, who was in honour bound

to prove the uniqueness of the answer (to get his problem published), only Kenneth

M. Wilke, of Topekaf Kansas, was unfazed by the difficulties and submitted a full­

blown solutions too long to be reproduced here (motto of Kansas: ad astra per aspera).

With the exception of a few months in 1980 when Joe Clark9 that hiccup of history,

was Prime Minister, Pierre-Elliott Trudeau has been Prime Minister of Canada for the

last 16 years. He had announced his retirement four years ago, then changed his mind.

Our prescient proposer had raised the possibility that Trudeau would bow out again

soon. On February 29, 1984s Trudeau announced his coming retirement. So this is the

Second Leaving. If Trudeau does not change his mind again9 his successor as head of

the Liberal Party and Prime Minister of Canada (and possible subject of a future alpha­

metic) will be chosen in mid-June of 1984 at a Liberal Party convention in Ottawa,

Ontario (not Ottawa, Kansas, the birthplace of Senator Gary Hart, another Kansan who

is reaching ad astra per aspera). We have enjoyed watching Trudeau pirouetting his

way across the world's stage, but now acta est fabula. Unless...

- 117 -

pP2 , [1983: 21] Proposed by Joseph Gillis3 Weizmann Institute of Science^

Rehovot3 Israel.

Points ?i and P2 move with equal angular velocities along circles Ci and C2s

respectively, in the plane. Prove that in general there is one point S such th&t

ISPJ 2 - |SP2I2 is constant (where | SPi denotes the distance between S and P), and

discuss the exceptional cases.

(This problem generalizes Problem 3 of the 1979 International Mathematical

Olympiad [1979: 194].)

I. Solution by Jan van de Craatss Leiden University3 The Netherlands,

We take C\ to be the unit circle in the complex plane and t the affix of the

moving point Px (so that tt = 1). We further assume that P2, with affix 2, moves on

a circle C2 with centre m9 and that z = a when t - 1. Two cases will be considered:

Case l: If Pl and P2 move in the same sense, then (see Figure 1)

z = m + t(a-rn). (1)

Case 2: If Pi and P2 move in opposite senses, then (see Figure 2)

z = m + t(a-rn), (2)

In both cases, we look for a point S9 with affix w, and a (real) constant y such that

(w-t)iw-t) = iw-z )(w-z ) + y (3)

for all t such that tt = 1, (Of course, if a = i9 then y = o. Case 1 is then the

original Olympiad problem.) We now examine the two cases separately.

0. Figure 1 Figure 3

- 118 -Case i. Upon subst i tu t ing (i) into (3), we get a quadratic polynomial in t

which vanishes for all t on the uni t c i r c l e , so i t s coefficients are all zero:

w - (w-m)ia-m) = 0, 0+)

w - (w-m) (a-m) = 09 (5)

-ww - 1 + (w-m) (w-m) + (a-m) (a-m) + y = 0, (6)

If a-m * l, then there is only one point w given by (4) or (5)9 and then y

follows from (6).

If a-m = 1, then only m = o gives solutions, and circles c^ and £2 then coincide.

But when the circles coincide there is one solution (w = 0) if a * 1 and infinitely

many solutions (all complex w) if a = 1. In both cases y = 0.

Case 2„ Substituting (2) into (3) and proceeding as in Case 1, we obtain

w - (w-m) (a-m) = 09 (7)

w - (w-m) (a-m) = 0S (8)

and9 again, equation (6). Substi tut ing w from (7) into (8), we get

w((a-m)(ja-m) - l ) = (a~m)(m + 777 (a-m)). (9)

If (a~m)(a-m) * 1, there is only one point w given by (9), and then y follows

again from (6).

If (a-m) (a-m) = l , then from (6) we get

(w-m) (w-m) = ww - y s (10)

while (9) yields

772 + m(a-m) = 0, (11)

If 777 = o9 then (11) i s an ident i ty and (10) gives y = o* Now9 since 2 - at from (2)

and aa = l , equation (3) is equivalent to

wF + wt ~ wat + &az?T,

or9 f ina l ly 9 to w = aw. This represents a Une on the origin^ and a is the ref lect ion

of 1 in this l i n e . If m * o3 then (11) i s equivalent to

a = 777 - ~ . ( 1 2 )

Unless (12) holds* the problem has no solution. If (12) holds* then, for any choice

of y9 (10) represents a line of solutions perpendicular to the line joining 0 and m.

In particular, if y = os then m and a are the reflections of 0 and 1, respectively,

in this line (see Figure 3).

- 119 -II, Comment by 0. Bottema, Delft, The Netherlands; and J.T. Groenman3 Arnhem3

The Netherlands.

Two solutions to this problem have appeared in Nieuv Tijdsehrift voor Wiskundei

one by E.C. Buissant des Amorie [67 (March issc^ 188-190] and one by 0. Bottema [68

(September 1980) g - n ] .

Also solved by the COPS of Ottawa; JORDX DOUr Barcelona, Spain; J«TB GROENMAN, Arnhem, The Netherlands; RICHARD I. HESS, Rancho Palos Verdes, California; and KES1RAJU SATYANARAYANA, Gagan Mahal Colony, Hyderabad, India.

* * ft

803 i [1983: 21] Proposed by Leroy F, Meyers, The Ohio State University,

Let / be the operation which takes a positive integer n to In if n is even*

and to 3n+i if n is odd. It is as yet unknown whether or not every positive integer

n can be reduced to i by successively applying f to it, (This is the substance of

Problem 133 [1976: 67s 1^-150, 221].) In comment I (Trigg, p. im), it is stated

that it is unnecessary to carry the sequence .v, fN9 ffN9 fffN9 ... for N beyond any

term less than N. In particular, it is not necessary to test N if N is even or is

0f tne forrn + i a

Is this argument justified?

Solution by Charles W, Triggs San Diego, California.

The argument is justified. The portion of my Comment I [1976: 1^-147] mentioned

in the proposal came right after I had stated that the process is known to converge

to i for all values of N < 10^°. What followed was clearly meant to imply that, in

testino numbers N > iok° in order of magnitude, i!it is not necessary to carry the

sequence for any N beyond a term < N." For if in the sequence of any N a term M < N

is encountered, then that sequence need not be carried further because from that

point on it will coincide with the sequence for M which has already been examined,

"Thus it is not necessary to test any even values [for fN = N/2 < Nl, nor [among

others] odd values of the form 4k+is 16&+3, or 128&+7,1! For if we denote the nth

iterate of / by /% then we have

fH^k+l) = 3k+l < 4fc+l, ^HekiS) = 9k+2 < 16k+39 fn(±28k+7) = Slk+5 < 128^+7.

Also so lved by CLAYTON W. DODGE, U n i v e r s i t y of Maine a t Orono; RICHARD I . HESS, Rancho Pa los Verdes , C a l i f o r n i a ; STANLEY RABXNOWITZ, D i g i t a l Equipment Corp*, Nashua, New Hampshire; and t h e p r o p o s e r .

Editor!s comment.

Much his tor ica l and up-to-date computational information about the !f3i?+i problem1'

(sometimes called the Collatz algorithm* althouah i t s actual origin is hard to pin

down) can be found in Hayes [1] and Gruenberger [ 2 ] . Hayes, in par t iculars wr i tes :

- 120 -

"...there" is no need to follow the path of a number a l l the way to 1; once the value

of N fa l l s below the i n i t i a l value the candidate can be dismissed/1

REFERENCES

1. Brian Hayes, "Computer Recreations", Scientific American, January 1984.

2. Fred Gruenbergers "Computer Recreations"s Scientific American, April 1984. & & &

304 , [1983; 22] Proposed by V.N. Murty, Pennsylvania State University,

Let P O ) be a polynomial of degree n > 2 with real coefficients, lead­

ing coefficient a * o5 and n real zeros a:, with

a? i < xo ^ . . . < a; . 1 *• n

I t is easily veri f ied that

fX2

) \ dx ~ -~G?2 _ ar i ) 3

and (more tediously) that

)| da; = ^{(ar 2 -a: i ) 3 (3a:3-Sa: . ) + (a? 3 -ar 2 ) 3 (^ -3a : 1 ) } ,

where the indicated sum is for i = is293.

Find a "nice" compact formula for

01 dr.

Editorfs comment.

No solution was received for this problem* which therefore remains open. Writing

the result for n = 2 in the form

fX2

Jx\ | P 2 ( x ) | da? = ~^(x2-xi)2(2x2~lxi)

and comparing with the result for n = 3 shows an emerging pattern9 but i t is not easy

to see how the pattern w i l l develop for n > 3. I t might help to know the result for n = 4 .

805, C1983: 22] Proposed by M.S. Klamkin, University of Alberta.

If xsy9z > 0, prove that

x + y + z > yz -t sx + xy 3* 3 vyljryz+z2 + vz2+zx+x2 + \/x2+xy+y2

with equality i f and only i f a? = y = s.

- 121 -

Solution by Vedula N. Murty_, Pennsylvania State University^ Capitol Campus;

and the proposer (independently).

The inequalities

I = Vy2+yz+z2 > - ( 1 / + 2 ) ,

m = i/z2+zx+x2 > ^y(z+x),

and n = vx2-txy+y2 > — Or H/)

are equivalent to (y-z)2 > o9 (z-x)2 > os and (x-y)2 > os respectively. Hence

Also

_L > _±^iiB (i) 3 l+m+n

(x+y+z )2

yZ+ZX+Xy (2) 3

is equivalent to (y-z)2 + (3-a?)2 + (a:-z/)2 > 0. Now, from (1) and (2),

x+y+z > (x-ty+z ) 2 yz+zx+xy 3/3 *" 3(Z+?rz+n) ~ l+m+n 9

which establishes and sharpens the proposed inequality. Equality clearly holds

throughout if and only if a; = y = 2 ,

Also solved by J.T. GROENMAN, Arnhem, The Netherlands? GEORGE TSINTSIFAS, Thessaloniki, Greece; and the proposer (second solution),

5% ;'; ii

806, [1983: 22] Proposed by Kesiraju Satyanarayana^ Gagan Mahal Colonys India.

Let LMN be the cevian triangle of the point S for the triangle ABC (i.e.,

the lines AS,BS,CS meet BC,CASAB in L,M,N, respectively). It is trivially true that

S is the centroid of AABC = > S is the centroid of ALMN.

Prove the converse.

Solution by Howard Evess University of Maine.

Denote by P the (real or ideal) point of intersection of MN and BC, and by Q

the point of intersection of AL and MN. In the complete quadrilateral BCMN-AP,

(NM,QP) and (BC.LP) are harmonic ranges. If S is the centroid of ALMN, then Q is

the midpoint of NM9 and P is an ideal point at infinity. It then follows that L

must be the midpoint of BC, and AL is a median of AABC. Similarly, BM and CN are

medians of AABC, and S is the centroid of AABC.

Also solved by J.T. GROENMAN, Arnhem, The Netherlands? STANLEY RABINOWITZ, Digital Equipment Corp., Nashua, New Hampshire? GEORGE TSINTSIFAS, Thessaloniki, Greece; and the proposer.

- 122 -

Editorfs comment.

Let a = A Q A ^ . A be a simplex in n-dimensional Euclidean space5 and let

a1 = AQ A i... AJ be the cevian simplex of the point S for the simplex a (i.e., for

i = o,i,...,r. the line A.S intersects the (n-i)-face opposite A. in A'.). Tsintsifas

used barycentric coordinates to prove that

S is the centroid of a => S is the centroid of a'.

He then simply added that "the converse is proved in the same way". * * *

8 0 7 , £1393: 22] Proposed by D.J. Smeeriks Zaltbommel3 The Netherlands*

Three balls are taken at random from an urn that contains w white balls

and r red bal ls . The probability that the three balls are white is p. I f the urn

had contained one more white b a l l , the probability of three white balls would have

been 4p/3. Find a l l possible values of the pair (w,r).

I . Solution by Richard Rhoads New Trier High School^ Winnetkay Illinois,

We assume that the three balls are taken without replacement and9 to eliminate

t r i v i a l solutions., that w > 3 (there are in f in i te ly many solutions i f w = o or l ,

and none i f w = 2). From the given data9 we have

_ £ _ = 2 e _ i — ( 1 ) ,w+r+l% 3 rW+r%

an equation that is soon found to be equivalent to

(ll-w)r = <tf-2)(w+l),

so solutions exist only i f w < 11. With this rest r ic t ion, the equation is equivalent

to r s ,w „ 10 + J£L9

11 -w

and we need test only w = 10, 9, 89 7, 5. Each of these is satisfactory, and we have the solutions

(w,2») = ( 1 0 5 8 8 ) s ( 9 , 3 5 ) , ( 8 , 1 8 ) , ( 7 , 1 0 ) , ( 5 , 3 ) .

I I . Comment by Vedula N, Marty3 Pennsylvania State University3 Capitol Campus.

The proposer wisely did not specify that the balls are taken without replacement.

He didn't have to5 because there is no solution with replacement. For then the equa­

tion corresponding to ( l ) is

W i » + l ' " 3 w r ' "

- 123 -

This is equivalent to

w+r+1* w 3

wi th l e f t side ra t iona l and r i a n t side i r r a t i o n a l .

Also solved by SAM BAETHGE, San Antonio, Texas? BARNARD H. BISSINGER, Pennsyl­vania State University, Capitol Campus (three solutions); W.J. BLUNDON, Memorial University of Newfoundland; RAIZAH BUJANG, Pennsylvania State University, Capitol Campus? S.C. CHAN, Singapore? the COPS of Ottawa? CLAYTON W. DODGE, University of Maine at Orono? JORDI DOU, Barcelona, Spain? J*T. GROENMAN, Arnhem, The Netherlands? THOMAS F. HALLEY, student, Pennsylvania State University, Capitol Campus; RICHARD I. HESS, Rancho Palos Verdes, California? J.A.H. HUNTER, Toronto, Ontario; FRIEND H. KIERSTEAD, JR*, Cuyahoga Falls, Ohio? J.A. McCALLUM, Medicine Hat, Alberta? VEDULA N. MURTY, Pennsylvania State University, Capitol Campus; STANLEY RABINOWITZ, Digital Equipment Corp., Nashua, New Hampshire? KENNETH M. WILKE, Topeka, Kansas? and the proposer.

808, [1983: 22] Pro-posed by Stanley Rabinowitzs Digital Equipment Corp,s Merri­

mack s New Hampshire,

Find the length of the largest circular arc contained within the right triangle

with sides a < b < a.

Comment by the proposer.

I do not have a d e f i n i t i v e so lu t ion to t h i s . Surp r i s ing ly , the i n c i r c l e i s not

necessari ly the la rges t c i r c u l a r arc. As the r a t i o b:a gets l a rge , a c i r c u l a r arc

l i k e the one shown in the f i gu re (tangent to one side of the r i g h t angle) gets to be

laraer than the i n c i r c l e . I f i t can be shown tha t these are the only two cases to

consider3 then the problem shouldn ' t be too hard.

One incorrect solution was received.

Editorrs comment.

The proposer's comment should be helpful in arriving at a solution to this open

problem. * * *

- 124 -809 i Cl9:r : 23] Proposed by G.C. Giris Midnapore College3 West Bengal, India.

Let £ and B be two angles of a tr iangle* with opposite sides a and b5

respectively. Drove that

y i (cos 2nA - cos2nB) = I n - . LA nx a

«=1

Solution zy S.C. Chcan9 Singapore; Friend Ha Kiersteads Jr*3 Cuyahoga Falls9

Ohio; and Stanley Rabinowitzs digital Equipment CorpBJ Nashuas New Hampshire

(independently,. The ser ies summation

l c o i M = - l n 2 s i n | , o < e < 2* 72 = 1

is given by J c ^ e y [ i ] , who states that a proof can be found in [ 2 ] . This r e s u l t

y ie lds

I -(cos 2rcA - cos 2nB) = - In 2 s in A + In 2 s in B «=1 n

= ln-s

a the last equality being a consequence of the law of sines.

Also solved by W.J. BLUNDON, Memorial University of Newfoundland; the COPS of Ottawa? J.T. GPJOENMAN, Arnhem, The Netherlands; RICHARD I. HESS, Rancho Palos Verdes, California; VEDULA N. MURTY, Pennsylvania State University, Capitol Campus; BOB PRIE-LIPP, University of Wisconsin-Oshkosh; KESIRAJU SATYANARAYANA, Gagan Mahal Colony, Hyderabad, India; KENNETH S. WILLIAMS, Carleton University, Ottawa; and the proposer.

REFERENCES

1. L.B.W. Jolley, Summation of Series, Dover, New York9 1961, p. 96, Series No*

503.

2. T.J. Bromwich, Introduction to the Theory of Infinite Series , Macmillan9 London, 1926, D. 356=

& * *

810, C135 3: 23] Proposed by Charles W. Trigg; San Diego3 California,

Place different integers on the vertices of a triangle, chosen so that the

sums of integers on the extremities of each side of the triangle will be squares.

I. Historical note by Howard Evess University of Maine*

Problem 5, Book III (using T.L. Heath's numbering) of Diophantus' Arithmetica (ca. A.D. 250?) asks: Find three numbers such that their sum is a square and the sum of any pair is a square. Diophantus' answer: 80, 320s 41.

- 125 -

II. Joint solution by Malcolm A. Smith and David H. Stone3 Georgia Southern

Gollege3 Statesboro3 Georgia (revised by the editor).

Suppose the triple (A3Bsc) of distinct integers is a solution. Then there

exists a triple (a9b9c) of integers (which we assume without loss of generality to

be nonnegative) such that

B + G = a2, C + A = b2, A + B = a2, (1)

and from this easily follows

b2 + a2 ~ a2 _, _ o2 + a2 - b2 a2 + b2 - e2 ... A = — 9 a - —— '—•—, o = — - — — — — — — . \2.)

2 2 2

We conclude from (2) that a$b$c are distinct and their sum is even. Conversely,

for any triple (a,b9c) of distinct nonnegative inteaers whose sum is even9 the triple

(A,B,C) defined by (2) consists of distinct integers that satisfy (l).

Let D be the set of all triples (a9b9c) of distinct nonnegative integers whose

sum is even, and let E be the set of all triples (A9B9C) of distinct integers that

are solutions to our problem. It follows from the preceding paragraph that there is

a bisection from D onto E, and that for every (a*b$e) e D the corresponding (A9B9C)

€ E is given by (2). We now show that E is the union of infinite sequences of solu­

tions

(A , B , G ), n = i52935 . .., (3) n n n

some of which may have interesting properties. Let F be the set of a l l t r ip les (x9y5z) of d ist inct integers such that

y+z + 2x * ~Wn9 z+x+2y * -4ft 9 x+y+2z * -4n, w = 1 , 2 , 3 , . , . , (4)

and9 for each (x9y9z) e F, le t the sequence (3) be defined by

(A = 2(n+x)2 + (z-x)ix-y) 9

\B = 2(ft+t/)2 + (x~y)(y-z)9 (5)

C = 2(n+z)2 + (y-z)iz-x) o n °

Now B - C = (ty-z )(4n+y+3+2a?)

and it follows from this and two similar results that A , B , C are distinct inteqers. n ft ft

And since also /B + C = (2ft + y + 2 ) 2 , ! ft ft ^ '

( ^ + 4 n = (2ft + z + x)29 (6)

\A + £ = (2ft + a; + y)2 , ft ft ^

- 126 -

i t follows that (A 9B 9C ) e E for n = 1,2,3,. . . . Every solution {ASB9C) e E is the nth term of some sequence (3). For let (A9B*C) be the solution correspond!no to (a,b,c) e D. If we s e t

b+o-a c+a-b a+b-e /ri. x - —£— - n 9 y = — - j — - n , JS = — j — - n9 (7)

then x9z/ss are distinct integers, they satisfy (4), and

2n+y+z = a , 2n+z+x = b9 2n+a?+z/ = o.

It now follows from (6) that (A9B9C) is the nth term of the sequence (3) defined by (5) for the triple (x9y9z) given in (7).

One interesting seauence is that resultinq from the choice (x,y9z) = (1,0,-1),

A = 2n2 + 4n, B - In2 + 1, C - 2n2 - 4n, n n n

in which the sums in pairs are consecutive squares:

B + C = (2n - 1 ) 29 <7 + 4 = ( 2 n ) 2

9 A + B = (2n + I ) 2 . n n n n n n

The choice (#9z/9£) = (3,1,-1) yields

4 = 2n2 + 12n + 10, B = 2n2 + 4n + 6, C = 2n2 - 4n- 6, n n n

and the sums in pairs are the squares of consecutive even integers:

B + C - ( 2 n ) 2 , C + A = (2n + 2 ) 29 4 + 5 = (2n + 4 ) 2 .

n n n n n n Other interesting sequences can be singled out at will.

III. Comment by Stanley Rabinowitzs Digital Equipment Corp.3 Nashua3 New

Hampshire.

There are many ways to place three integers in the desired manner. For example,

there are exactly 16 solutions in positive integers less than 100:

( 2 , 239 98) ( 5 , 20, 44) (12 , 52, 69) (26 , 74, 95)

( 2 , 34, 47) ( 6 , 19, 30) (14 , 35, 86) (29 , 52, 92)

( 3 , 22, 78) ( 6 , 75 , 94) ( 1 5 , 34, 66) (30 , 5 1 , 70)

( 4 , 2 1 , 60) ( 1 0 , 54, 90) (16 , 33 , 48) ( 4 8 , 7 3 , 96)

To make the problem more challenging9 additional restrictions should be imposed. For example, one can require that the three numbers be in arithmetic progression (see r i ] ) 9 that the sum of all three be a perfect square (see r2]), or that the differences between each two also be squares (see [3]).

The problem can be extended in several ways:

- 127 -

(a) There are many ways to put inteners on the vertices of a quadrilateral

so that the sums of the integers on each side will be squares. For example9

(1, 3, 335 48) and (49 5, 765 45).

It should be even easier for polygons with more sides.

(b) Suppose, instead, that we want to place integers at the vertices of a

tetrahedron so that the numbers on each face sum to a square. Again, there are

many ways to do this. If the numbers are all positive and less than 1005 there are

exactly three solutions:

(1, 22, 41, 58)s (9, 34, 57, 78)9 and (14, 41, 66, 89).

(c) Suppose we wish to place integers at the vertices of a tetrahedron so that

the numbers on each edge sum to a square. Call the numbers a3 b , e s and d. We pick

a number that is the sum of two squares in multiple ways. For example,

5525 = 552 + 502 = 622 + 412 = 702 + 252 .

Then we solve the simultaneous equations

a+b = 552, c+d = 502, a+c = 622, a+d = 702, b±d - 412, b+c = 252.

The system is consistent because the last two equations can be derived from the first

fours and solvinq the latter yields the solution

(a, 2>, c, d) = (3122, -97, 722, 1778).

Also solved by ELWYN ADAMS, Gainesville, Florida; SAM BAETHGE, San Antonio, Texas? HIPPOLYTE CHARLES, Waterloo, Quebec? CLAYTON W. DODGE, University of Maine at Orono; JORDI DOU, Barcelona, Spain? J.T. GROENMAN, Arnhem, The Netherlands? RICHARD I. HESS, Rancho Palos Verdes, California? FRIEND He KIERSTEAD, JR., Cuyahoga Falls, Ohio? BOB PRIELIPP, University of Wisconsin-Oshkosh? KESIRAJU SATYANARAYANA, Gagan Mahal Colony, Hyderabad, India? KENNETH M* WILKE, Topeka, Kansas? and the proposer.

REFERENCES

1. American Mathematical Monthly_, 1 (1894) 96,

2. Math. Quest, Educ. Times, 61 (1894) 115-116,

3. Leonard Eugene Dickson, History of the Theory of Numbers9 Che!sea9 New York,

1952, Vol. II9 p. 448, }*» ft

8 1 1 , C1983; 45] Proposed by J.A.H. Hunter3 Toronto3 Ontario, Some say that's how al l the trouble started. But i t must

have been very tempting9 for that Adam's APPLE was truly prime!

ADAM ATE

THAT RED

APPLE

- 128 -

Solution by Edwin M, Kleins University of Wisconsin-Whitewater.

It is clear that A = l , T > 7, E = 3, 7, or 9, and M+T+D = 10 or 20. If T = 7, then P = o and M+D = 13. Since E = 9 implies L = o = P9 we must

have E = 3, but then L = 4 and D+H+R = 18, and no three of the remaining digits sum to 18. Therefore T * 7,

If T = 8, then P = o and D+H+R = 89 an impossible sum with the remaining digits. So T * 8.

Therefore T = 9, P = 2, and M+D = n . If E = 7, then L = o and APPLE = 12207 is not prime. Hence E = 3, L = 69 and D+H+R =20. If D = 4, then H+R = 16, an impos­sible sum with the remaining digits. Therefore D = 7 and M = 4. Finally9 H+R = 13 and the digits {H5R} = {5,8} are interchangeable.

Thus there are two solutions with a unique prime APPLE:

1714 1714 193 193 9519 9819 837 md 537

12263 12263°

Also solved by HAYO AHLBURG, Benidorm, Alicante, Spain; SAM BAETHGE, San Antonio, Texas? CLAYTON w. DODGE, University of Maine at Orono; MEIR FEDER, Haifa, Israel? ALLAN WM. JOHNSON JR., Washington, D.C.; STANLEY RABINOWITZ, Digital Equipment Corp., Nashua, New Hampshire? CHARLES W. TRIGG, San Diego, California? KENNETH M. WILKE, Topeka, Kansas? ANNELIESE ZIMMERMANN, Bonn, West Germany? and the proposer. One incorrect solution was received.

Editor's comment*

Dodge noted that the solution is unique except for "the interchange of the H & R

Block", a useful reminder that this issue of Crux is that of the Income Tax month,

April 1984. & * *

812 i C1983: 45] Proposed by Dan Sokolowsky9 California State University at Los

Angeles.

Let c be a given circle* and let c9 i = 1,2,3,4, be circles such that (i) c. is tangent to C at A. for i = 1,2,3,4;

( i i) C. is tangent to C. for i = 1,2,3. Furthermore, let I be a line tangent to C at the other extremity of the diameter

of C through Al9 and, for i = 2,3,4, let k\k. intersect I in P.. Prove that, if C% c l s and Ci* are fixed, then the ratio of unsigned lengths

P2p3/p3pt+ is constant for all circles C2 and c 3 that satisfy (i) and ( i i ) .

Solution by Dan Pedoe, University of Minnesota.

Let A be the other extremity of the diameter of c through Alo Take the circle

wi th centre A! and radius AiA as the c i r c l e of invers ion 9 and i nve r t the given

con f igu ra t ion . Then the c i r c l e c inver ts in to the l i ne l\ the c i r c l e C\ inver ts

in to a l i n e pa ra l l e l to Z9 which we ca l l c\\ the c i r c l e C2 inver ts in to a c i r c l e

C2 touching l i n e c\ and touchino l a t AXA2 n I = P2; the c i r c l e C3 inver ts in to

a c i r c l e c\ touching c i r c l e C2 and touching I at AxA3 n I - P3; and f i n a l l y the

c i r c l e Ck i nver ts i n to a c i r c l e c i touching c i r c l e Cl3 and touching Z at A ^ n I = Pk.

Since l i n e c\ i s f i x e d 3 c i r c l e C2 has a f i xed rad ius , say i?. C i rc le ^4 also has

a f i xed rad ius , say r. Let p be the radius of c i r c l e c 3 . Since C2 and c3 touch

e x t e r n a l l y , we have

(P2P3)2 = (i?+p)2 ~ (B-Q)2 = 4/?p;

and since C3 and dj. touch e x t e r n a l l y , we also have

( P A ) 2 = (p+r)2 - ( P - P ) 2 = 4rp.

Hence P2P3/P3P1+ = /#7?s which is a constant.

Also solved by JORDI DOU, Barcelona, Spain; HOWARD EVES, University of Maine; KESIRAJU SATYANARAYANA, Gagan Mahal Colony, Hyderabad, India; and the proposer.

A A A

813, C1983: 45] Proposed by Charles W. Triggs San Diegos California. 31

The array on the riant is a "staircase" of primes of the form 331

3,1. When 3 is replaced by some other digit9 the furthest any staircase 3331

of Drimes goes is 666is since 66661 = 7*9523. 33331

How much further does the 3,1 staircase ao before a composite number 333331

appears? Subsequent to that, what is the next prime in the staircase? 3333331

Solutions were received from LEON BANKOFF, Los Angeles, California; MEIR FEDER, Haifa, Israel; J.T. GROENMAN, Arnhem, The Netherlands; FRIEND H. KIERSTEAD, JR., Cuyahoga Falls, Ohio; STANLEY RABINOWITZ, Digital Equipment Corp., Nashua, New Hamp­shire; and KENNETH M. WILKE, Topeka, Kansas,

Editor ?s corrwn.ent*

The following information was abstracted from the solutions received.

All factors appearing in the table (see next page) are primes.

The first part of the table, where complete factorizations are given, was taken

from KiersteacTs solution. All his results are confirmed by other solvers except

the last9 317i = primes which, says Kiersteads required 40 minutes on a fast computer.

No other solver gave complete results beyond 316i.

The second part of the table, where only the smallest prime factor is given, was

taken from Feder's solution. The question mark after 330i presumably indicates that

this number, if not prime, has no "small" prime factor. Our proposer, a Count of

130 -

3 i i = prime

3 2 i = prime

s 3 i = prime

3^1 = prime

3 5 i = prime

361 = prime

3 7 i = prime

381 = 17-19607843

3 9 1 = 6 7 3 * 4 9 5 2 9 4 7

3 1 0 1 = 3 0 7 * 1 0 8 5 7 7 6 3 3

3 n l = 1 9 - 8 3 - 2 1 1 3 7 1 8 0 3

3 1 2 1 = 5 2 3 - 3 0 4 9 - 2 0 9 0 3 5 3

3 1 3 1 = 6 0 7 « 1 5 1 1 - 1 9 9 7 « 1 8 1 9 9

3 1 L f l = 1 8 1 - 1 8 4 1 6 2 0 6 2 6 1 5 1

3 1 5 1 = 1 9 9 - 1 6 7 5 0 4 1 8 7 6 0 4 6 9

3 1 6 1 = 3 1 - 1 4 9 9 - 7 1 7 3 2 4 0 9 4 1 9 9

3 1 7 i = prime

D i g i t Del vers9 w i l l be happy to learn tha t 31 div ides 3 3 1 i 5 and tha t 19 and 29 take

in each o the r ' s washina,

Wilke showed tha t i n f i n i t e l y many of the numbers 3 7 i are composite, and in doing

3 i 8 l = 1009* . . .

3 1 9 1 = 2 9 - . . .

3 2 0 1 = 2 3 - . . .

3 2 1 1 = 1 7 7 9 4 3 * . ,

3 2 2 1 = 6 1 * . . .

3 2 3 1 = 3 1 2 9 2 9 * . ,

3 2 i t l = 1 7 * . . .

3 2 5 1 = 8 2 1 - . . .

3 2 6 1 = 3 5 3 * . . .

3 2 7 1 = 3 6 3 9 4 1 - . .

3 2 8 1 = 8 2 9 - . . .

3 2 9 1 = 1 9 - . . .

3301 = ?

3 3 1 1 = 3 1 - . . .

so he found much interesting information.

Since

He reasoned as follows:

V = l O ^ 1 - 7

k

r.k + 1

l92s39..

a prime p > 3 divides 3,1 if and only if lQA'"r"L = 7 (mod p ) . He then used the theory

of indices (and the table of indices in the CRC Standard Mathematical Tables) to show

that

8 + 16-£s i - 0 9 1 9 2 , . . . ,

11 + 18-£9

20 + 2 2 £ 9

19 + 28<£,

I + 30£ and 16 + 3 0 i 9

37 + 46<£9

43 + 5 8 i ,

22 + 6 0 i 9

I I + 8 2 i and 52 + 8 2 i 9

52 + 9 6 £ 9

171V 19!3^1

23 |3^1

291 3 ,1

3 1 | 3 - 1

47 |3^1

59| 3, 1

61 |3^1

83 | 3- 1

97 |3^1

for

for

for

for

for

for

for

for

for

for

k =

k =

k =

k =

k =

k =

k =

k =

k =

k =

- 131 -and that no other primes p < 100 divide any of the numbers 3,1. Our proposer may

be disappointed to learn that his pet prime 37 (see his article in this issue) does

not divide any of the numbers 3,1. Not a very useful pet for a numerologist to

have around the house. • A A

8 W , C1983: 45] Proposed by Leon Bankoff^ Los Angeles, California.

Let D denote the point on BC cut by the internal bisector of angle BAC

in the Heronian trianale whose sides are AB = 14, BC - 139 CA = 15. With D as

center9 describe the circle touching AC in L and cutting the extension of AD in J.

Show that AJ/AL = (/5+D/2, the Golden Ratio.

I. Solution by the proposer.

The location of the point D is a red herrina. Instead, let D be any point on

the line I which bisects angle BAC, With L and J as defined in the proposal, the

ratio AJ/AL is constant fors as D varies on Z, the circles (D) are all nomothetic

with nomothetic center A, In particular., when circle (D) is the incircle of tri­

angle ABC9 whose radius is

then AL = s-b = 8, AD = 4/5, AJ = 4(/5+i)9 and so

AJ /5 + 1 AL " 2 *

II. April 1 "solution" by Hayo Ahlburg3 Benidorm^ Alicante3 Spain,

Major premise: The Golden Ratio pops up in the most unexpected places.

Minor premise: The ratio AJ/AL in the Heronian 13-14-15 triangle is a most

unexpected place.

Conclusion: AJ/AL = the Golden Ratio.

What I really want to express is my wonder at the Golden Ratio showing up here

between 13, 14, and 159 and my admiration for Dr. Bankoff who discovered it there.

Congratulations, Dr. Bankoff!

Also solved by ELWYN ADAMS, Gainesville, Florida? SAM BAETHGE, San Antonio, Texas; W.J. BLUNDON, Memorial University of Newfoundland; the COPS of Ottawa; CLAYTON W. DODGE, University of Maine at Orono; JORDI DOU, Barcelona, Spain; HOWARD EVES, University of Maine? JACK GARFUNKEL, Flushing, N.Y.; J„T. GROENMAN, Arnhem, The Netherlands; WALTHER JANOUS, Ursulinengymnasium, Innsbruck, Austria; STANLEY RABINO-WITZ, Digital Equipment Corp*, Nashua, New Hampshire; KESIRAJU SATYANARAYANA, Gagan Mahal Colony, Hyderabad, India; MALCOLM A, SMITH, Georgia Southern College, States-boro, Georgia; et ROBERT TRANQUILLE, College de Maisonneuve, Montreal, Quebec.

Editorrs comment.

Only Dou gave evidence of being aware of the red herring. For the others9

April fool I

- 132 -

8 1 5 , [1983 : M-6] Proposed by J.T. Groenman3 Arnhems The Netherlands,

Let ABC be a t r i ang le w i th sides a 5£,c s in terna l anale b isectors t , t.3

t , and semi perimeter s. Prove tha t the fo l lowing inequa l i t i es ho ld , w i th equa l i t y

i f and only i f the t r i ang le is e q u i l a t e r a l :

a b a

at bt. et (b) 3 / 3 - ^ ^ ^ * 4 ^ " 'at + T t I + at ~ "¥(abc)'5'

Solution by Bob Prielipp^ University of Wisconsin-Oshkosh,

(a) We will use the followinq known results, in which T , T*9 T are the a b a

in terna l angle b isectors extended u n t i l they are chords of the c i rcumc i rc le 9 R is

the c i rcumradius, and the sums (here and l a t e r ) are cyc l i c over A9B,C and a9b9c: R f B-C 2

be = T t , e t c . ; T = 2i?cos 2-—-, e t c . ; Ecos - T - > -^Zs in A.

For the f i r s t r e s u l t , see [1976: 233] ; f o r the second9 see [1982: 115] ; the t h i r d

is Garfunkel 's i nequa l i t y [1982: 67s 138], From these resu l t s 9 we get

/ 3 I - T - = "T~^T = —r—ECOS -~~ > -r-~S2i?SinA = - r -Za = -r—. a t a£c a aba 2 a#<2 a&c aba

a As in GarfunkeVs inequality, equality holds if and only if ABC is equilateral.

(b) Here all we need is

obtained by setting X = l 1n Inequality 8,12 on page 76 of the Bottema Bible9 Geometric Inequalities. The inequality we seek here is obtained by dividing corresponding

members of (1) and the inequality of part (a). As in (l), equality holds if and only

if ABC is equilateral.

Also solved by JACK GAKFUNKELf Flushing, NJ.j VEDULA N. MURTY, Pennsylvania State University, Capitol Campus? KESIRAJU SATYANARAYANA, Gagan Mahal Colony, Hydera­bad, India; GEORGE TSINTSIFAS, Thessaloniki, Greece; and the proposer.

Editor's comment.

It is clear from the proposer's solution that he is having his little joke in part (b) . It is easy to manufacture, by multiplying together two or more simpler inequalities, ever more monstrous inequalities that will stump everyone but their creator. To be fair, proposed inequalities should be, or at least be thought to be, "prime". In this case, fortunately, nearly all solvers were able to find, in their veil-thumbed Bottema Bible, the second "factor" that led to (b).

April fool to you, Mr. Proposer.

END OF THE APRIL ISSUE