CruxPublished by the Canadian Mathematical Society.

http://crux.math.ca/

The Back FilesThe CMS is pleased to offer free access to its back file of all issues of Crux as a service for the greater mathematical community in Canada and beyond.

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

C R U X P i A T H E M A T I C O R U M

Vol. 11, No. 4

April 1985 ISSN 0705 - 0348

Sponsored by Carleton-Ottawa Mathematics Association Mathematique dsOttawa~Carietoa

Publie par le College Algonquin, Ottawa

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

CRUX MATHEMAT1CORUM is a problem-solving journal at the senior secondary and miiverslty undergraduate levels for those who practise or teach mathematics. Its purpose is primarily educational, but it serves also those who read it for professional, cultural, or recreational reasons*

It is published monthly (except July and August) . The yearly subscription rate for ten issues is $25 in Canada, $28 Cor US$21} elsewhere,, Back issues: each $2«50 in Canada, $3*00 Cor US$2„Sa) elsewhere. Bound volumes with indext Vols* l£2 (combined) and each of Vols* 3-10, $19 in Canada* $20 (or US$15) elsewhere. Cheques and money orders* payable to CRUX MATHEMAT1C0RUM# should be sent to the Managing Editor.

All communications about the content (articles, problems, solutions, etcJ should be sent to the Editor* All changes of address and inquiries about st±>scriptions and back issues should be sent t© the Managing Editor.

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

Managing Editors Kenneth S. Williams, Algonquin College, 203 Lees Ave., Ottawa, Ontario, Canada K1S 0C5o

Assistant to the Editors: Mrs. Siu Kuen Lam,

Second Class Mail Registration No, 5432. Return Postage Guaranteed.

* * *

CONTENTS

The Olympiad Corner: 64 . . . . . . . , . . . . . . . . . . . . M.S. Klamkin 102

Mathematical Swifties , . . . . . . . . . . . . . . . , . . . . M.S. Klamkin 120

Problems - Problfimes: 1031-1040 . . , , . . , , . . . . , . . . . . . . . . . 121

Mathematical Clerihew . . . . . . , . . . . . . . . . . . . . . . Alan Wayne 122

Solutions: 897, 9049 912~9149 918-921 , . . , . t , . . . . . . . . . . . . . 123

- 101 -

- 102 -

THE OLYMPIAD CORNER: m

M.S. KLAMKIN

I begin with another set of 25 problems which further extends the list of pro­

blems submitted by various participating countries (but unused) in past International

Mathematics Olympiads. I solicit from all readers elegant solutions to these pro­

blems* if possible with extensions or generalizations. Readers submitting solutions

should clearly identify the problems by giving their numbers* as well as the year

and page number of the issue where they appear8

511 Proposed by Australia.

Let P be a regular convex 2n-gon9 Show that there is a 2n-gon Q with the

same vertices as P (but in a different order) such that Q has exactly one pair of

parallel sides,

52 8 Proposed hy Belgium.

If n > 2 is an integer and [>] denotes the greatest integer < x9 show that

'n(n-f-l)' 4n-2

n+ll

. * J' 5 3 , Proposed by Belgium.

If P% Q, and R are polynomials with complex coef f ic ien ts , not a l l of them

constant polynomials5 prove that the ident i ty Pn+Qn+Rn = o9 where n is a natural

number, implies that n < 3.

{Note hy M.S.K* As s ta ted , the implication is inval id . Just consider Q = -p

and R = 0; then P3+§3+i?3 = o. Prove or disprove the implication when the condition llP*Q9R are not al l constant polynomials18 is changed to UP,Q9R are nonproportional

polynomials18,]

5 1, Proposed by Bulgaria.

The incircle of triangle AXA2A3 touches the sides A ^ ^ A ^ A ^ at the

points Tl9T29T3f respectively. If MlsM2,M3 are the midpoints of the sides A2A3,A3Ai,

A1A2s respectively, prove that the perpendiculars through the points Ml5M2,M3 to

the lines T2T3,1311*12129 respectivley, are concurrent.

55 1 Proposed by Bulgaria.

Prove that the volume of a tetrahedron inscribed in a closed right circular

cylinder (capped by two disks) of volume 1 does not exceed 2/3TT.

56 , Proposed hy Canada.

Given that ai*a2*« « » *a are distinct integers such that the equation

- 103 -

(x~a\ )(x™a2)^ . (x-a ) + ( - ! ) ( n ! ) 2 = 0

has an integer solution r , show that r = (a1+a2+« « « +a )/2«.

57 @ Proposed by Canada.

Let m and n be nonzero integers. Prove that %?n ~ m - n can be a square

infinitely often, but that this is never a square if either m or n is positive.

58 s Proposed by Finland.

Determine all sequences {a l 9a 2^^} such that aY = l and

\a - a < -r~r

for all positive integers 777 and na

59 s "Proposed by Finland*

A s t r i c t l y increasing function f defined on [0,1] s a t i s f i e s f(o) = o9

/ ( l ) = 1, and

1 < ti£^y£isi < 2

2 - f(x)-fix-y) ~

for all x32/ such that 0 < #±z/ < 1. Show that /(1/3) < 76/135.

60 1 Proposed by France.

ABC is an isosceles right triangle with right angle at A, Determine the

minimum value of

BP + CP - /3APS

where P is any point in the plane of the triangle.

61, Proposed by the Federal Republic of Germany.

You start with a white balls and b black balls in a container and proceed

as follows:

Step 1. You draw one ball at random from the container (each ball being equally

likely). If the ball is white, then stop.

Step 2t If the drawn ball is black, then add two black balls to the balls re­

maining in the container and repeat Step 1.

Let s denote the number of draws until stop. For the cases a = b = 1 and

a - b = 2 only, determine

a = Pro = n)9 b = Pr(s < n), 1 im b 9 n n n-*00 n

and the expectation E(s) = \ na . n>l n

- 104 -

52, Proposed by the Federal Republic of Germany.

A 2x2xi2 box is to be filled with twenty-four lxix2 bricks. In how many

different ways can this be done if the bricks are indistinguishable?

63fi Proposed by Great Britain.

Prove that the product of five consecutive integers cannot be a perfect

square.

6 % Proposed by Great Britain.

The function f{n) is defined for nonnegative integers n by fiO) = 09

fil) = l, and

for

fin) = fin - — ~ ~ ) - f ( — — ~ - n)

rnOn^l < n < wOs+ii m > 2 # 2 2

Determine the smallest integer n for which fin) = 5,

65 i Proposed by Morocco.

Determine all continuous functions / such that, for all real x and z/,

f(x+y)f(x-y) = (f(x)/(y)}2.

66, Proposed by the Netherlands.

Determine positive integers p*q3r such that the diagonal of a block con­

sisting of px-qxr unit cubes passes through exactly 1984 of the unit cubes, while

its length is a minimum, (The diagonal is said to pass through a unit cube if it

has more than one point in common with the unit cube.)

67 1 Proposed by Poland*

Let a^b%o be positive numbers such that

/3 Ja + S + S = rs

Prove that the system

\Jy-a + y^s-a = 1

v ^ i + v^Tt = 1

vx-e + ry-c = 1

has exactly one real solution ix9y9z)»

68 1 Proposed by Poland*

Let X be an arbitrary nonempty point set in a plane and let Al%A2>>.. .9A

and £l9£9,...,B be its images under translations (in the plane). If the sets 1 ^ n

- 105 -

41,^25...>A are pairwise disjoint and

AiuAoU . . 9uA C ^-JU^OU . . .u£ .

prove that m < n.

69 i Proposed by Rumania.

Let {a^} and {£ }9 n = l 9 2 9 3 9 a 9 0 , be two sequences of natural numbers

such tha t , for all n > l ,

a „ = wa + 1 and £> = nfc - 1. n+1 w n+1 n

Prove that the two sequences can have only a f in i t e number of terms in common,

70 i Proposed by Rumania. k k k

Let sv = x\ + x2 + . . . + x , where the x. are real numbers. If

prove that a:, e {0,1} for every i = l 5 2 5 , . . 5 n ,

711 Proposed by Spain*

Construct a nonisosceles t r iangle ABC such tha t

<z(tan B - tan C) = 2?(tan A - tan C),

where a and b are the side lengths opposite angles A and B3 respect ively .

72 . Proposed by Spain*

Let P be a convex n»gon with equal interior angles^ and let l\^li^ ..*l be

the lengths of its consecutive sides. Prove that a necessary and sufficient condi­

tion for P to be regular is that

T>\ l i ln

73 i Proposed by Sweden.

Let {a1,a29a3so..} be an infinite real sequence such that, for all positive

integers n and m9

a £ a < a + a .

Prove that a In has a l imit as n ->- oo

n 74, Proposed by the U.S.A.

A fair coin is tossed repeatedly until there is a run of an odd number of

heads followed by a tail. Determine the expected number of tosses.

75, Proposed by the U.S.A.

Inside triangle ABC, a circle of radius l is externally tangent to the

- 106 -

incircle and tangent to sides AB and AC, A circle of radius 4 is externally tangent

to the incircle and tangent to sides BA and BC. A circle of radius 9 is externally

tangent to the incircle and tangent to sides CA and CB. Determine the inradius of

the triangle,

*

I now present solutions and comments for various problems from past columns.

1( [1982: 133] From the 1982 Australian Mathematical Olympiad.

If A and Btossn + l and n fair coins, respectively, what is the probability

P that A gets more heads than B? n

Comment by M,S*K«

It is easy to show that l-p is the probability that A gets more tails than B.

By symmetry l-p = P , so P = 1/2. For a detailed solution and extensions, see

Problem 12-3 [1980: 149, 311].

3, [1982: 133] From the 1982 Australian Mathematical Olympiad,

Let ABC be a triangle, and let the internal bisector of the angle A meet the

circumcircle again at P. Define Q and R similarly. Prove that

AP + BQ + CR > BC + CA + AB. (1)

Solution by M.S>K9

Let 0,I,i?,r be the circumcenter, incenter, circumradius, and inradius, respec­tively, of the tr iangle. By the power of a point theorem,

Al -IP = E2 - 0I2 = 2Rp9

and so

IP = ^ = 2i?sin^9

With this and two similar resul ts , inequality (l) is found to be equivalent to

Z(rcsc ~ + 2i?sin | ) > 2j?Zsin A,

where the sums are cyclic over A,B,Ce, or, using

equivalent to

A B C A B+C r = 4i? sin - s i n - s i n - and s i n - = cos -r—, etc,

B -C C A A B cos — + cos ~ + cos ~ > sin A + sin B + sin C. (2)

Finally, (2) follows from

- 107 -

B-C B-C . B+C l, . n . rx cos — > cos — s i n — = -(sin B + sin C), etc. D

An inequality stronger than (2)9

B-C C-A A-B 2 , . . . n . rs cos -— + cos -T- + cos -T- > y=(sin A + sin B + sin C) 3

is proved in Problem 613 [1982: 559 138].

L\ 8 [1982: 133] From the 1982 Australian Mathematical Olympiad.

Find what real numbers d have the following property:

If fix) is a continuous function for 0 < x < 1 and if fio) = fil)9 there exists

t such that o < t < t + d < 1 and fit) = fit+d). Equivalently9 every continuous

graph from (0,0) to (1,0) has a horizontal chord of length d.

Comment by M.S.K.

The graph of every function fix) continuous for 0 < x < 1 such that fio) = fil)

cer ta in ly has a horizontal chord of length d = 1; and i t follows from the Parallel

Chord Theorem of P. Levy Cl] that for such a function there is a t such that

0 < t < t+d < l for which fit) = fit+d), and hence a horizontal chord of length d9

only if d = l /n for some n = 1 9 2,3 5 8 0 S .

Levy also gave the following example:

fix) = s i n 2 ^ - x s i n 2 J 9 0 < x < l. d

Here fio) = f(l) = 0 ands for every t such that 0 < t < t-td < 1, we have fit) = fit+d),

and hence there is a horizontal chord of length d9 if d = l/n for any n = 192939«.9 .

This follows from

d* fit+d) = /(£) - dsin2^.

REFERENCE

1. Ha Hadwiger and H* Debrunners Combinatorial Geometry in the Plane , Holt5 Rinehart & Winston, New York, 1964, p8 238

5„ [1982: 134] From the 1982 Australian Mathematical Olympiad.

The sequence p 1 $ p 2 3 ... is defined as follows:

p1 = 2; p = the largest prime divior of PiP2---P + 1' n - 2* Yl (LA.

Prove that 5 is not a member of this sequence.

- 108 -

Solution by Paul Wagner3 Chicago3 Illinois.

The definition of the sequence implies that p is odd for all n > 2, and so

PlP2.-Pw_1 E 2 (mod *0> n > 2. (1)

AT so 9 P2 = 3 and p > 3 for al l n > 2. Suppose now that p = 5 for some r > 2.

Then we must have

PiP2.--Py.-i_ + 1 = 5???

for some integer m > l . Hence

PiP2- e -P p„ 1 = $m - l E o (mod 4) ,

contradicting (l). Therefore 5 is not a member of the sequence. A

1, [1982: 134] From the 1982 British Mathematical Olympiad.

PQRS is a quadrilateral of area A. 0 is a point inside it. Prove that if

2A = OP2 + OQ2 + OR2 + OS2,

then PQRS is a square and 0 is its centre.

Solution by M.S.K.

We do not assume that the quadrilateral is convex, nor that 0 is an interior

point. From

(OP - OQ)2 + (OQ - OR)2 + (OR - OS)2 + (OS - OP)2 > 0, (1)

we obtain

Z4 = OP2 + OQ2 + OR2 + OS2

> 0P*0Q + 0Q®0R + 0R*0S + 0S«0P

> 2([0PQ] + [OQR] + [ORS] + [OSP])

> 2A9

where the square brackets denote area, so equality holds throughout here as well as

in (i). Hence

OP = OQ = OR = OS and ^POQ = /QQR = ROS = /SOP,

or, equivalently, PQRS is a square and 0 is its center. •

Note that one need not assume that 0 is an interior point of the quadrilateral,

since this follows from the above proof.

2, C1982: 134] From the 1982 British Mathematical Olympiad.

A multiple of 17 when written in the scale of 2 contains exactly three di-

- 109 -

gits 1. Prove that it contains at least six digits 0§ and that if it contains

exactly seven digits 0, then it is even.

From the official solutions.

In base 2, 17 = 10001* If the given multiple is 17m and m contains at most

four binary digits, say m = abed (perhaps with some initial zeros), then we have

17m = abedabed* The number of digits 1 in 17m is then even and so cannot equal

three* Therefore m must contain at least five binary digits and so 17m has at least

nine digits* Since exactly three of these are lss, there must be at least six o's.

If there are exactly three ifs and seven ofs, and if 17m is odd, then

17m = 2 9 + 2 + l, where i < x < 8.

Now for x = 1,2,...,8, we find that 2X = 2,4,8,16,15,13,9,1 (mod 17), respectively,

while 29+l = 3. Hence 2 9 + 2X + l t o (mod 17), and we have a contradiction. There­

fore 17m is even.

3, C1982: 134] From the 1982 British Mathematical Olympiad.

If

prove t h a t

l l l s = l + ~ + ; r + « * . + - * n > 2 ,

n 2 3 n

n ( n + l ) a - n < s < n - (w-l )n ,

where a and b a r e given in terms of n by an = 1 and b(n~l) = - 1 ,

Solution by A«C*L« Dyson3 Leeds Grammar Schools England.

By the A.M.-G.M. i n e q u a l i t y , for n > 2 ,

n+1 n 3 2 _—. + « _ . + + _ + _ n n -1 2 1 , , N l /n n

> (n+1)

Also, for n -1 > 2,

(1+-) + (1+—T) + *** + (1+ | ) + (1+D > n ( n + l ) 1 / n

n n -1 2.

n + s > n ( n + l ) n

n -1 n-2 2 1 I T + n ^ l + ••• + 3 + 2 ,1 l/("-D

n-1 > (.T)

= > ( 1 ~ ) + ( 1 - ~ T ) + . . . + ( 1 ~ ) + ^ 4 ) > (n-l)n~1/in~1) n n A. o z.

, „ N - l / ( n - i ) = > n - s > ( n - i ) n 7 \

- 110 -

Hence, for all n > 29

n(n+l) - n < s < n - (n-l)n , n

where a = l/n and b = -l/(w-l).

pt [1982: 135] From the 1982 British Mathematical Olympiad*

Prove that the number of sequences a\a^. ..a with each of their n terms

a. = 0 or 1 and containing exactly m occurrences of 01 is

v2m+r9

Solution by Brian Brunswick^ Great Britain

For each sequence a\ai

quence

ac\a\ao«»«a a A 9

a of the required type, consider the augmented se-

(1)

where a0 = 1 and a 1 = 0. It is easy to show by induction that each augmented

sequence exhibits exactly 2 +1 occurrences of 01 or 10, consisting of m occurrences

of 01 from the original sequence and m+l occurrences of 10 from the augmented se­

quence. Conversely, every sequence (1) with a0 = l, a = 0S and exactly 2m+l

occurrences of 01 or 10 gives a sequence aia2«**ci of the desired type when a0 and

a are deleted. The longer sequence is completely specified when the 2m+l places

for the occurrences 01 or 10 are identified. Since there are m l possible places

s the desired number is a .a, A t z+1

x2m+l;

l\ t [1983; 303] From the 1980 Leningrad High School Olympiad^ Third Round

Is it possible to arrange the natural numbers from l to 64 on

an 8x8 checkerboard in such a way that the sum of the numbers in any

figure of the form shown on the right is divisible by 5? (The fi­

gure can be placed on the board with any orientationa)

Solution by Gali Salvatore3 Perkinss Quebec*

We show that the arrangement is not possible even if we are

allowed to use 64 different natural numbers from 1 to 78,

Suppose the arrangement is possible, and let a9b9c9d,e be the

numbers used in any part of the checkerboard having the sh^pe of

Figure i. We then h$ye (congruences throughout are modulo 5)

1 b a

e

a

d\

Figure 1

I l l

and

Now

a+b+Q+e=a+d+c+e

b+a+d + e = b+c+d + e

and a

(1)

(2)

(3)

nil

llll 1

bm 1 1

"EH

mUL

• ~

r~

llll

ill!

III!

III!

zzz

—_

EEf

/III

ml

'llll

llll'

~

~

rEE

=£

———i

III!

ill!

HI!

M

~ |

~ "

^ E ~ 1

follow from (i) and (2), respectively. We conclude

from (3) that the numbers on the 16 squares with hori­

zontal shading in Figure 2 are all congruent to one

another modulo 5a None of them is a multiple of 5,

for if one of them is then all of them are, and there Figure 2 are only 15 multiples of 5 from 1 to 78* A similar

argument shows that no multiple of 5 appears in any of the 16 squares with vertical

shading in Figure 2. Therefore no multiple of 5 appears in any of the 32 shaded

squares (the black squares, say) of Figure 29 A similar argument shows that no

multiple of 5 appears in any of the 32 white squares* Therefore each of the 64

squares of the board is filled by a number that is not a multiple of 59 Since there

are only 63 such numbers from 1 to 78, not enough to fill the board, we have the

required contradiction.

P. 383« C1984: 76] From Kozepiskolai Matematikai Lapok 67 (1983) 80.

Does there exist a multiple of 5 1 0 0 which contains no zero in its

decimal representation?

Solution by Andy Liu3 University of Alberta,

More generally, we prove by induction that, for every k = ls29358e,9 there is a k fc-digit multiple m« of 5 which contains no zero in its decimal representation,, For

k = l, we can take 7771 = 5„ We now define

777-

777. fc + 1

10kt + mk = 5k{2kt + -|h 5

k k where t is one of 1,2,3,4,5, so chosen that 2 £ + 777,/5 is divisible by 5a Then

777. is a (fc+i)-d1git multiple of 5 which contains no zero in its decimal repre­

sentation. The first few terms of the sequence are

7771 = 5l = 5* ^5 = 1>5 5 =

7772 = 5 2 = 2 5 , 7??6 = 2 9 ® 5 6 =

77?3 = 5 3 = 1 2 5 ^ my

53125,

453125,

57.57 = 4453125,

7774 5 * 5 L 3125, Wft = 37®58 = 14453125,

- 112 -?s 384. [ 984: 76] From Kosepiskolai Matematikai Lapok 67 (1983) 80.

In a party of 25 members, whenever two members donst know each other,

they have common acquaintances among the others. Nobody knows all the others.

Prove that if we add the numbers of acquaintances of all members the sum is at

least 72. Can the sum be 92? (By acquaintance we mean mutual acquaintance.)

Solution by Andy Liu3 University of Alberta.

We first point out that no member has exactly one acquaintance9 because this

sole acquaintance would have to know all the others, and this has been ruled out.

To show that the sum, s9 of the numbers of acquaintances of all members is at least

72, we assume, on the contrary, that s < 70 (the sum must be even, so it cannot be

7i). By the pigeonhole principle, at least one member, say the President, has

exactly two acquaintances, say the Secretary and the Treasurer. We refer to the

other 22members as ordinary members. Each of them must know either the Secretary

or the Treasurer in order to have access to the President. Hence the sum of the

numbers of acquaintances of the Secretary and the Treasurer is at least 2+22 = 24.

The sum of the numbers of acquaintances of the ordinary members is at least 2*22 = 44.

Adding the 2 acquaintances of the President, the total count is 70, which must be

exact. Hence the Secretary and the Treasurer do not know each other. Moreover,

each ordinary member knows exactly one of the Secretary and the Treasurer, and

exactly one other ordinary member. Thus they can be classified as Undersecretaries

and Undertreasurers. By the pigeonhole principle again, we may assume that there

are at least ll Undersecretaries. In order for the Treasurer to have common ac­

quaintances with each of them, there must be ll Undertreasurers each knowing a

different Undersecretary. However, it is impossible for an Undersecretary and an

Undertreasurer who do not know each other to have common acquaintances. This is a

contradiction. So S > 72,

S = 92 is possible with a President and a Vice-President who do not know each

other, and 23 ordinary members each knowing only the President and the Vice-President.

By a more detailed analysis, Bollobis [1] showed that s z 4n-l0 for a party

of n members. When n = 25, the strangest party (i.e., one with the maximum number

of mutual strangers) may consist of a President, a Vice-President, a Secretary, a

Treasurer, and 21 ordinary members, where the President knows only the Vice-Presi­

dent and the Secretary, the Vice-President knows only the President and the Treas­

urer, the Secretary does not know the Treasurer, and each ordinary member knows only

the Secretary and the Treasurer. Here s = 90, in accordance with the Bollobcfs result.

REFERENCE

1, BeU Bollob^s, Extremal Graph Theory* Academic Press, New York, 1978, pp.

173-175.

- 113 -1, [1984: 182] From the 1984 West German Olympiad.

Given are 2n rcss in a row. Two players alternately change an x into one

of the digits l, 25 39 4, 5, and 69 The second player wins if and only if the re­

sulting 2n~digit number (in base ten) is divisible by 9e For which values of n is

there a winning strategy for the second player?

Solution by Gali Salvatore3 Perkins3 Quebec,

We show that there is a winning strategy for the first player if n i 0 (mod 9)

and a winning strategy for the second player if n = 0 (mod 9). We recall that a

positive integer is divisible by 9 if and only if the sum of its decimal digits is

divisible by 9.

Suppose n = i (mod 9), where 0 < i < 8* The first player selects his first

digit XQ3 as explained below, and thereafter he chooses 7-# for every x chosen by

the second player (except the last). After the second player has chosen his last

digit xi, the sum of the digits is

S - x0 + 7(w-l) + x\ = x0 + 7(i-l) + xi (mod 9).

The following table gives values of 0 which will assure the first player of a win,

because in each case there is no x\ for which 5 = 0 (mod 9).

XQ 2 4 6 1 1

If n =0 (mod 9), the second player can win by choosing l-x for each x chosen by

the first player. At the end of the game, the sum of the digits is

S = In = 0 (mod 9).

2 8 [1984: 182] From the 1984 West German Olympiad,

Determine the sum of the squares of all line segments joining two vertices

of a regular n~gon with circumradius 1.

Solution by M»S*K«

More generally, let AiA2...A be a cyclic polygon, with circumradius l, for

which the centroid coincides with the circumcenter 0 (as happens when the polygon

is regular), and let a. = o L , i = 1,2,...,n. Then the required sum is

n n A n n ^ ^

1=1 3=1 d %•=! 3=1

n n nl ' ( I «!•)'( I a . ) = n'

i=l % j=l 3

- 1 1 4 -n

since \ a./n = 0 when 0 is the centroid. i=1 ^ *

3, C1984: 182] FTOW the 2Sfi West German Olympiad* Prove that the product of the positive integers a and b is even if and only

if there exist positive integers c and d such that

a2 + £2 + e2 = ^ ( 1 )

Solution by Daniel Ropp3 student3 Stillman Valley High School3 Illinois.

Suppose ab is even. If one of a9b is even and the other is odd, then a2+fc2 is

odd, and equation (1) is satisfied by the positive integers

a2 + b2 - 1 . , a2 + b2 + 1 e = _ _ _ _ _ _ _ a n c j ^ - _ _ _ e

If a and Z> are both even, then a2+b2 = 0 (mod 4), and equation (l) is satisfied by

the positive integers

a2 + b2 - 4 . , a 2 + b2 + 4 e = _ _ _ _ _ and ^ = _ _ _ _ _ _

Conversely, suppose equation (l) has a solution in positive integers. If a

and b are both odd, then a2^b2 = 2 (mod 4). This implies that d2-e2 = 2 (mod 4).

However, the square of any integer is congruent to o or l modulo 4, and so d2-c2 4 2

(mod 4). This contradiction shows that at least one of a and b9 and hence their

product db9 is even.

28 [1984: 214, 316] Fiw?? the 1983 Dutch Olympiad,

Prove that if n is an odd positive integer, then the last two digits in base

ten of 2 (2^w - l) are 28.

II. Solution by K»S« Murray3 Brooklyn3 N@Ya

With n = 2&7+1, we must show that

2 ^ + 2 ( 2 ^ + 3 _ 1 } _ 2 8

is d iv i s ib le by 100 = 22®52 for m = 0 , 1 , 2 , . . . , or , equivalently, tha t

Now

2em+3 _ 2%* _ ? E Q ( m o d 2 5 K ( 1 )

2^ - 1 = 0 (mod 5) => ( 2 % - l ) 2 E 28m - 2 % ? + 1 + 1 E 0 (mod 25)

Thus (1) is equivalent to 2 3 (2 4 m + 1 - l ) - 24"1 - 7 = o (mod 25), or to

15(2^" - i ) = o (mod 25), and th i s is c lear ly t r u e .

- 115 -1, [1984: 214] From the 1983 Swedish Mathematical Contest.

The positive integers are summed in groups in the following way:

1, 2+3, 4+5+6, 7+8+9+10, ... .

Find the sum of the nth group.

Comment by Gali Salvatore3 Perkins3 Quebec*

This problem has appeared many times in the past, including in this journal.

See Problem 518 [1981: 63-64] where a simple solution is given, and the editors

comment following the solution for references to earlier occurrences of the problem.

2, [1984: 214] From the 1983 Swedish Mathematical Contest.

Prove t h a t , for a l l real numbers x and y%

c o s x 2 + cosy2 - cos xy < 3*

Solution by Aleksandar Zorovic3 Elizabethtown High School3 Eastview3 Kentucky.

Since |cos 0 I £ l9 we ce r ta in l y have cos x2 + cosy2 - cos xy < 3, so we need

only show that

c o s x 2 + cos y2 - cos xy * 3 (1)

for a l l real numbers x and y . Suppose equal i ty holds in ( l ) . Then

cos x2 = 1 , cosy2 = 1 , cosxy = -1

and

x = ±/2/ar, y = ±v^rr 9 xy = (2m+i)iT9

where k and I are nonnegative integers and m is an integer. It follows that

2m + 1 = ±2/kX,

and this is impossible since 2}/kZ is either irrational or an even integer. Hence

(l) is establ isheda

l\ t [1984: 214] From the 1983 Swedish Mathematical Contest.

Two concentric circles have radii r and R. A rectangle has two adjacent

vertices on one of the circles* The two other vertices are on the other circle.

Determine the length of the sides of the rectangle when its area is maximal.

Solution by K* Seymour3 Toronto3 Ontario.

It is clear that for maximal area the centre of the circles must not lie out­

side the rectangle. Referring to the figure3 we find that the area function is

A(x) = 2x(v^2-x2 + Vr2^2), 0 < x < r.

LU

Since Aix) > 0 for all x e [09r] and AiO) = 09

it is seen that if the function has a unique

critical point in the open interval iO,r), then

the corresponding value of the function will be

the absolute maximum over the closed interval

[09r], provided this value exceeds Air). This

is indeed what happens, for upon differentiating

we find that

- f(x)ivR2~x2®vr2-~x2 - x2)$

where fix) > 0 for all x e (09r), and the deri­

vative vanishes if and only if

x Rr

-p € (0,r). (1)

+!>'

Furthermore^ for t h i s x the corresponding value of the funct ion is

2Rr > 2rS2-r2 = Air).

The side lengths of the rectangle of maximal area can now be calculated from (l)

They are

2x 2Rr

M and

+r'

2Rr 2x

JR2 •\-x*

[It would be interesting to have a noncalculus solution for this problem,

(M.S.K.)]

F8 [1984: 214] From the 1983 Swedish Mathematical Contest*

Prove that (x9y) = (1,2) is the unique (real) solution of the system

( xix + y)2 = 9

xiy3 - x3) = 7,

Comment by M.S.K,

To within a permutation of x and y9 this is the same problem as J-10, which is

solved in [1981: 109].

1, [1984: 215] From the 1984 British Mathematical Olympiad.

Ps Q5 R are arbitrary points on the sides BC9 CA9 ABf respectively, of tri­

angle ABC. Prove that the triangle whose vertices are the centres of the circles

AQRS BRP5 CPQ is similar to triangle ABC.

- 117 -Comment by M.S.K.

This result is the first of several corollaries to the theorem [l]: All the

Miguel triangles of a given point P are directly similar^ and P is the center of

similitude or self-homologous point in every case*

REFERENCE

1. Roger A, Johnson, Advanced Euclidean Geometry (Modern GeometryJ, Dover,

New York, i960, p. 1349

5, [1984: 215] From the 1984 British Mathematical Olympiad.

(i) Prove that, for all positive integers m,

(2 - i)(2 - )(2 - -)...(2 - ) <ml, m m m m

(ii) Prove that if a5b^c^d$e are positive real numbers, then

£ C a e a a b c d e

Solution by M.S.K.

( i ) Our proof is inductive. The resu l t is true for 777 = 1. Suppose i t is

true for some m = k > l . Then

(2fc-l)(2fc-3).,.3-l < ^ ^ ! ,

and the resu l t will be valid for m = fc+i if

or , equivalemtly, if

(2fc+l)&*•&! < (&+l) f e+1(£+l)!,

1 k

2k + 1 < (1 + £) (fc + I ) 2 .

7/ To es tabl ish t h i s , we f i r s t note that (l + l/k) > 2 for all k > 1 by the binomial

theorem, and then 1 k

2k+± < 2(k+l)2 < ( l+i) (k+l) 2e

I t follows from our proof that the proposed inequali ty is always s t r i c t except when

m = 1, when there is equal i ty .

( i i ) The following resu l t appears in [ 1 ] : for a l l real nonzero a9b9o9

a2 fo2 Q2 ra b c b c a-t £2 e 2 a2 b c a a b c

We generalize th i s as follows. Let n be a posit ive integer and l e t ^ , 9 i = 0 9 l 9 9 e . 9 n 9

be n+l posi t ive real numbers (with x = ^ 0 ) e Th e n

118 -

XQ n x\ n w (_£) + _i) + . . . + (JL) > max

sX> ( i / T* A T* T "^

<—+—+. . .+—, —+—+ . . . + — \ . ( i ) \X j Xo XQ X A X I X I

The proposed inequality will then follow when n = 4„ It is clear that equality holds

in (1) for n = l. We now assume that n > l.

We first note that, for any m > 1 (all sums run from i = 0 to n),

x, m t x. \m x6

X . . P^X . . J ^X . .

w+1 I n+1 n+1

The first inequality follows from the power mean inequality, and the second from the

fact that Z(a

777 = n we get

fact that E(x./x. )/(n+l) > 1 by the A.M.-G.M* inequality. In particular, when

x» n x.

We now rewrite 5 in the form

5 = i/(5-(^)n) + (s-(x-±)n) + . . . + (5-(^)n)| . ?t I Xi X Q * n /

By the A.M.-G.M. inequality, for i = 0,1,...3n9

x. n

s - ( -M n x. *

and hence

5 > 1 - ^ . (3)

Finally, (1) follows from (2) and (3).

REFERENCE

1. M.S* Klamkin, "Asymmetric Triangle Inequalities11, Pttblikacije Elektro-

tehnickog Fakulteta Univerziteta U Beogradu (Serija: Matematika I Fizika), No. 357 -

No. 380 (1971) 42.

{J, [1984: 215] From the 1984 British Mathematical Olympiad.

Let ff be a positive integer. Determine, with proof, the number of solutions

of the equation

x2 - [x2l - (x - [xl)2

- 119 -

lying in the interval 1 < x < N« (The square brackets denote the greatest integer

function.)

Comment by MSS«K»

This problem appeared earlier in the 1982 Swedish Olympiad. For a solution,

see [1984: 46]s

5B [1984: 215] From the 1984 British Mathematical Olympiad*

A plane cuts a right circular cone with vertex V in an ellipse E and meets

the axis of the cone at C; A is an extremity of the major axis of E. Prove that the

area of the curved surface of the slant cone with V as vertex and E as base is

jp@(area of E).

Solution by M0S,K»

Let the equation of the cone be

z2 = (x2 + y2) tan2ou

If E% is the orthogonal projection of E

onto the #2/-plane9 then it is known that

LEll s cos a

where the square brackets denote area*

Let the plane of E be parallel to the iy-axis

and consider the traces of the configuration

on the s-plane, as shown in the figure. If

6 is the angle that segment AC makes with

the jc-axis, then LE*1 = [#]cos 6 and

LEI cos 6

J » i-™

cos a

3 *

LEI a/AC _ VA W V A ~ AC*

LEI.

'z = x tan a

5, [1984: 215] From the 1984 British Mathematical Olympiad*

Let a and m be positive integers* Prove that, if there exists an integer x

such that a2x - a is divisible by m, then there exists an integer y such that both

a2y - a and ay2 - y are divisible by m.

Solution by Stewart Metchette3 Culver City3 California.

Clearly, if m\a then y can be any multuple of m. So we assume that (a9m) = l.

Then ax = 1 (mod m). Since

1ZU -

ay' a a(ay - 1) and ay' y = y(ay - 1),

y - x + km will satisfy the desired conditions for any integer fc.

7S [1984: 215] From the 1984 British Mathematical Olympiad

ABCD is a quadrilateral which has an inscribed

circle. With the side AB is associated

uAB =p!Sin(^DAB) + p 2 sin (/ABC),

where p\ and p 2 are the perpendiculars from A and Bs

respectively* to the opposite side CD6 Define uRp9

urns ^DA "^'^wise^ using in each case perpendiculars

to the opposite side. Show that

um ~ UBC u CD u DA8

Solution by K*S» Murray9 Brooklyn^ N»Y«

Referring to the figure, we have

Hence

Pi = (a+d?)sinD and p2 = (b+c) sin C,

= (a+d) sin D sin A + (2>+c) sin C sin B. u AB

Also, air - cot (A/2), e t c . , and so

1 A D R C ~«uflD = (cot - + cot -) sin D sin A + (cot - + cot -) sin C sin B. r AB 2 2 2 2 '

1 p 9 WAB

This is eas i ly transformed into

sinD(l+cosA) + sinA(l+cosD) + sinC(l+cosB) + sinB(l+cosC)

= s inA + s inB + s inC + s inD + sin (A+D) + sin (B+C)

= sinA + sinB + sinC + sin Ds

since (A+D) + (B+C) = 360°. As th i s r e su l t is symmetric in A,B,C,D, we are done.

Editor8s 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.

ft ft ft

MATHEMATICAL SWIFTIES

"\z\ + Z£I |zi| + |Z2|," Tom maintained absolutely.

"Sin^x + 1000 has interesting properties/8 Tom repeated periodically.

"The decimal expansion of 1/7 is rather long/1 Tom stated indefinitely.

M.S. KLAMKIN

- 121 -

P R O B L E M S — P R O B L E M 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 problems may also be acceptable provided they are not too well known and references are given as to their provenance. Ordinarily, 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, 1985, although solutions received after that date will also be considered until the time when a solution is published.

10311 Proposed by Allan Wm« Johnson Jr«3 Washington^ D«C«

Independently solve the alphametics

MAN MAN WINS and WEDS, MAID WIFE

but FIND A MAN AND A WOMAN common to both alphametics.

1032 i Proposed by J«T* Groenman3 Arrihem3 The Netherlands.

The following formulas are given in R.A. Johnsonfs Advanced Euclidean

Geometry (Dover, New York, I960, p. 205):

IH2 = 2p2 - 2Rr and OH2 = i?2 - ^Br9

where, in Johnson's notation, 09I,H,/?,p,r are the circumcenter, incenter, ortho-

center, circumradius, inradius, and inradius of the orthic triangle, respectively,

of a given triangle. Johnson claims (at least tacitly) that these formulas both

hold for all triangles. Prove that neither formula holds for obtuse triangles,

1033 i Proposed by W«B„ Utz3 University of Missouri-Columbia.

Let D be any symmetric determinant of order n in which the elements

in the principal diagonal are all l's and all other elements are either 1's or -l's,

and let D be the determinant obtained from D by replacing the non-principal -n n _

diagonal elements by their negatives. It is easy to show that D2D2 = 0 for all D2

and D3D3 = 0 for all D3» For which n > 3 is it true that zyJ^ = o for all Z^?

1034, Proposed by Kenneth M« Wilke3 Topeka3 Kansas.

Find all positive integers n such that nh + 1 has a divisor of the form

dn - l, where d is a positive integer.

- 122 -1035 i From a Trinity College, Cambridge, examination paper dated December 63

1901.

If the equations

axy + bx + cy + d = 0,

ays + Z?2/ + cs + <i = 0S

asw + i>s + cw + d = 0,

awx + i>u + ex + d = 09

are sa t i s f ied by values of x9y9z,w which are al l d i f fe ren t , show that

b2 + a2 = 2ai*

1036 i Proposed by Gali Salvatore3 Perkins, Quebec.

Find sets of positive numbers {a,i,e,d,e,/} such that, simultaneously.

ae/ 9 a + £ + / 9 a £ / 9 a b c

or prove that there are none*

10371 Proposed by (the late) H« Kestelman3 University College, London, England.

If A and B are Hermitian matrices of the same order and A is positive

definite, show that AB is similar to a Hermitian matrix and that the latter is

positive definite if B is so. Show that if A is assumed only to be positive semi-

definite then it may happen that the matrix AB is not similar to any diagonal matrix.

1038 t Proposed by Jordi Dou3 Barcelona, Spain*

Given are two concentric circles and two lines through their centre.

Construct a tangent to the inner circle such that one of its points of intersection

with the outer circle is the midpoint of the segment of the tangent cut off by the

two given 1ines*

1039 8 Proposed by Kesiraju Satyanarayana3 Gagan Mahal Colony, Hyderabad, India.

Given are three collinear points 0, P, H (in that order) such that

OH < 30P. Construct a triangle ABC with circumcentre 0 and orthocentre H and such

that AP is the internal bisector of angle A. How many such triangles are possible?

1040 i Proposed by Clark Kirribevling, University of Evansville3 Indiana.

ABC being a given triangle, describe the locus of all points P such

that /ACP = /ABP. A A A

MATHEMATICAL CLERIHEW Hermann Amandus Schwarz , (As Philip J. Davis reports)

Indulged in reflections ALAN WAYNE, Holiday, Florida In complex directions.

- 123 -

0 L U T I 0 N S

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

8971 [1983: 313; 1985: 63] Proposed by Vedula NB Murty> Pennsylvania State

University3 Capitol Campus,

If X > y and a > b > c > Os prove that

7 2X 2y 2X 2y 2X.2y ^ ,- ,X+y ( NXry , x^A+y , . N b c K + c a + a fcH> (fee) M + (ca) + (a&) , U )

with equality just when a = b = e.

II. Extracted from a comment by M.S. Klamkin3 University of Alberta,

The published solution [1985: 63] is incomplete* In a couple of places the

solver tacitly assumes that A+y > 0, and this is not guaranteed by the proposal.

Still to be proved is that the proposed inequality also holds when A+y < 0,

In an editor's comment following the solution [1985: 64]s it is stated that !,A = 3/2 and y = 1/2 give

b2c(b-c) + c2a(c~a) + a2b(a-b) > 09 (2)

an inequality given at the 1983 International Mathematical Olympiad [1983: 207;

1984: 73],if It should be noted that the IeM«0. problem asked for a proof of (2)

subject to the condition that a3b3c are the sides of a triangle. Now (2) certainly

holds if a > b > c > 0S whether or not aj)9c are the sides of a triangle. But it

does not hold, for example, if a = l, b = 10, c = 12. Since we can have a < b < c

in a triangle, it follows that the I.M.O. problem is not a special case of (l).

III. Completion of proof extracted from the proposer9s solution.

Still to be proved is that (l) holds when

A > y, A + y < 0, a > b > c > 08 (3)

Suppose X,\x3a,b,c s a t i s f y (3) and l e t

- 1 - l ~ 1 Xi = ~y9 \x\ = -X, a\ = c , b\ = b 9 o\ - a .

Then

Xi > Hi , Xi + ]ii > 0, aY > bl > ci > 0* f4)

I t follows from (4) and the e a r l i e r s o l u t i o n t h a t (1) holds for X 1 , y 1 , a 1 , f c 1 , c 1 ;

and s ince

b2K2^ = a2Xb2V, e t c . and ( V i > X l + P l = (ab)X+]i, e t c . ,

i t follows t h a t ( l ) holds when X9y,as&9c? s a t i s f y ( 3 ) .

Klarnkin a l s o f i l l e d the gap in the e a r l i e r s o l u t i o n .

- 124 -904, [1984: 19 ; 1985: 90] Correction, The display in the penultimate line of

the published solution [1985: 91] should, of course, read

I u m u n9

not I n m n n5 as some typographical gremlins would have it. The statement will

appear corrected in any subsequent reprint of the issue concerned and in the 1985

bound volume.

912 i [1984: 53] Proposed by Shrnuel Avital3 Technion - Israel Institute of

Technology3 Eaifa3 IsraeI *

It is not difficult to verify that the nonlinear recursive relation

a Aa A r a + 1, n = 192939...9 n-l n+1 n

generates a periodic sequence of period length 5 for any given CLQ and a\. Show how

to construct nonlinear recursive relations which will generate periodic sequences of

any given period length k»

I. Comment by M.S. Klamkin3 University of Alberta.

According to Sawyer m , the proposer's recursive relation was first discovered

by R.C. Lyness. In [2]9 Lyness himself explains how he discovered the relation and

shows how to construct nonlinear recursive relations which generate periodic sequen­

ces of any given period length k when the first k-3 arbitrary nonzero starting values"

are given.

II. Comment by Stanley Rabinowitz3 Digital Equipment Corp«3 Nashua3 New

Hampshire.

For n = 1,2,3,..., the following nonlinear recursive relations generate periodic

sequences of the stated period length k« In each case, a§a\ * o and the value of r

is restricted only by the requirement that a * 0 for n < k-1.

k - 29 a a A = r; n n-1

& = 39 a (a ^ + P ) = -r2; n n~l

k = 4, a (a A + 2r) = ~2r2; n n-1

k = 59 a ^a A = r(a + r); n+1 n-l n

k = 69 a (a A + 3r) = -3r2; n n-l

k = 9, a „ + a = la I. n+1 n-l ! n1

The last example is due to Morton Brown [3].

- 125 -

III. Solution by Leroy F. Meyers3 The Ohio State University,

The proposerss sequence of period length 5 (see also Crux 191 rig?7: 77])

has repetend

l a ° 9 ai9 a0 9 aQa1 * aY

U

provided that none of the terms is 0. The sequence generated by the slightly simpler

recurrence relation

a Aa A = a 9 n = 192939. .., n-1 w+1 n

has period length 6 and repetend

/ a\ 1 1 ao\ {<2Q9 a \ 9 —3 — s — , — ; ,

^0 a0 al al

provided again that none of the terms is 0„ However, the period may be shorter for

certain initial values. In particular, the proposerfs sequence has period length i

if a 0 = a\ = (l±/5~)/2. The simpler sequence has period length 1 if a0 = a\ = l; it

has period length 2 if

-1 ± i/3" . 2 -1 + i/3" ag = ~~~rT—' al = ao = " ^ ~ ~ J

i t has period length 3 i f aj) = af = 1 but a 0 ai + 1 * a0 + a\.

A homogeneous l inear recurrence re la t ion

r

with initial values ^ 0 5c 1 ? e.. 9e and constant coefficients ai,a2»...,a , can be

converted Into a nonlinear relation

r a* a = TI a • 9 n > i %

e by setting a = & n for some suitable constant b« If the coefficients a. of the

linear recurrence relation are integers, then no restrictions need be placed on the

initial values (other than that they be nonzero). Hence we may first deal with linear recurrence relations, which have a long history*

In terms of vectors and matrices, the linear recurrence relation above can be

expressed as

v = Av A9 n > r9

n n-1 where (with r for transpose)

v ~ (c Q A ... a . ) 9 n > r~l9

n n n-1 n-r-t-1

- 126 -

and

i o

a . a\

0 0

0 0

The vector of initial values is v A. (Note that A is a companion matrix.) Since k r~1

v , ~ A v for k > 0 and n > r-l, the requirement that the positive integer k be a period length of the sequence is equivalent to the requirement that A be the r*v identity matrix I . Now for any companion matrix A3 a consequence of the Hamilton-

Cayley theorem is that

T

Ar - I a. Ar J = 0. 3=1 J

Hence for each k it is sufficient to choose r and the coefficients al9a29...5a so

that the polynomial p defined by

r-J p i t ) = t - I a « * .7=1 J

shall divide £ - l. The trivial choice r = k and p(t) = t - 1 yields the linear

recurrence relation

e = <? 7 s n > k9

n n-k Q

and exponentiation (a = 2? n ) yields the similar recurrence re la t ion

Since

a = a 7 , n > k. n n-k

tk - 1 = (t - DC^*"1 + . . . + * + i ) ,

another linear recurrence relation of period length fc is obtained by letting k-l

v = fe-l and p(t) = t + ... + t + l. This yields the relation

Cn " en-l °n~2 ~ #ae ^n-£+2 ~ n-k+1 n > k-l3

and exponentiation yields the re la t ion

a n a A a ^« * • a 7 ^ a 7 . n-1 n-2 n-k+2 n-k+1 9 n > k~l9

with repetend

(ao» #i 9 . . . 9 a k-23 a^a\..«av„n )

- 127 -k

Other factors of t - i may be used to generate recurrence relations, A general l inear re la t i on

v c% = Y o u e 1 „ + (3, n > r9

can also be converted into a nonlinear relation by exponentiation (a = b n ) , but it

can be obtained from a homogeneous linear relation by setting c% = c + v, where y ^ n n ] '

is chosen so that p

yd - I a.) = 3 .

The simpler recurrence relation

a .a . = a 9 n = l 9 2 9 3 f . 9 3 9 n-l n+1 w

with period length 69 comes from the factor t2 - t + 1 of £ 6 - i, and can be general­

ized to

a Aa A = Aa , n = l 9 2 9 3 , 9 e e , n-l n+1 n '

by setting 3 = y ar|d A = 2> . This new recurrence relation has repetend

/ Aai A2 A2 Xan% (a09 a l 9 —-A-, —-, — , — " - ) .

It is likely that there are many more periodic recurrence relations with simple

formulas*

Also solved by FRIEND He KIERSTEAD, JR., Cuyahoga Falls, Ohio. A comment was received from WALTHER JANOUS^ Ursulinengymnasium^ Innsbruck^ Austria.

REFERENCES

1. W.W. Sawyer, "Lyness8 periodic sequence", Mathematical Gazette^ 45 (1961) 207,

2. R.C. Lyness, "Cycles", Mathematical Gazette, 45 (1961) 207-209,

30 James F. Slifker, solution to Problem 6439 (proposed by Morton Brown),

American Mathematical Monthly3 92 (1985) 218.

913, [1984: 53] Proposed by Stanley Rabinowitz3 Digital Equipment Corp«5

Nashua3 New Hampshire*

Let n f \ n ^ n-l ^ n-2 , n-3 f A ,

f (x) = x + 2x + 3x + 4ic + .. . + nx + (n+1). J n

Prove or disprove that the discriminant of / (x) Is

f A,n(n~l)/2 -f .vW-L . ,n-2 (-1) •2(n+2) (n+1)

128 -

Solution by John McKay3 Concordia University^ Montreal3 Quebec k

Let r., be the roots of f and sv = Jr., k = 09l32 Alsos let F = (v..) be the square matrix of

the required discriminant is

disc (f ) = det (VV ) = det (w) 9

where w = ( w . . ) with w.. = s . . . for i , j ^J ^^ ^+J~2

For the funct ion

# + a i x + . . . + <z Ax + a 9 1 ' n - 1 n

Newtonfs identities give sg = n5

sk + aiSk-± + ••• + a f c - i s i +

and

sv + a l S , + a 2 s , 0 + e e 6 4 a s * = 0 , fe > n. fc-1 Jk-2

For our function f , where a« = £+1, Newton1s identities give Jn % ' y

order n defined by u V

r\ . Then

1 2 n

faz, = 0, 1 < k < n9

n k-n

SQ = n ; s . = - 2 , ^ = 1 ,2 , . Is

\n n+1

; s = 2 J t = n 2 + 3n, 72 + 1 ^

1 t = l The matrix W therefore has the fo l lowing appearance (the missing elements w i l l not

be needed):

n -2 -2 . . J -2 -2

-2 -2 -2 . . J -2 -2

-2 -2 -2 . . J -2 s

W = I -2 -2 -2 . . . | v „ n+1

k~2 - 2 s w+1

Subtracting the second row from all the oth

disc (f ) = det (W) = (n+2)*(

( - D l+(n-2)(n-3)/2

n+1

|ers gives

(n-2}

n-2 2 ' "2),(*n+l+2) (~ 1 }

®2(n+2) (n+1)

, .*n(n-l)/2 0/ ^n-1, «<n-2 = (-1) *2(w+2) (n+1)

since

i + i!L^!L±I = r L ^ - 2(n~2) = r^L±l (mod 2).

Also solved by GEORGE SZEKERES, University of New South Wales, Australia.

& * &

- 129 -Ol'l. M98U: b3l Proposed hy Vedula N. Muvty3 Pennsylvania State University3

Capitol Campus.

If a,b9c > 0, then the equation x3 - (a2-tb2+c2)x ~ 2abc = 0 has a unique positive

root ^0. Prove that

2 -Aa+b+o) < XQ < a-^-bi-a.

I . Solution by Andy Liu3 University of Alberta,

Let f(x) = x 3 - (a2+&2+c?2)x - 2aZ?c. By the c o n t i n u i t y of / , the des i red r e s u l t

wil l follow from

/(|(a+2>+<?)) < 0 (1)

and

f(a+bi-o) > 0e (2)

Now

y(|<a+]b+c)) = - ^ Z - ^ - J 9

where

X = (i>~c)2(i>+cO + (<—a)2(<?+a) + ( a - i ) 2 ( a+&)

and

J = a(a~b)(a-c) + b(b-c)(b~a) + c ( c - a ) ( c - t ) .

C lea r ly 5 J > 03 with e q u a l i t y i f and only i f a = 2? = e; and 7 > 0 follows by s e t t i n g

t = 1 in Schur ' s i n e q u a l i t y

t £ £ a ( a - £ ) ( a - c ) + i> (2>-c)(ib-a) + e (c-a)(<?-2>) > 05

which holds for any real number t when a9b9o > 09 with equality if and only if

a - b ~ e. Thus (1) is established, with equality if and only if a = b = c.

As for (2), it follows immediately from

f(a+b+c) = 2(i>+e)(e+a)(a-t-&).

II. Comment by M.S. Klamkin3 University of Alberta,

For a geometric interpretation of the positive root XQ9 s e e ^ e solution by

Howard Eves to Crux 787 rigs^: 571.

Also solved by W.J. BLUNDON, Memorial University of Newfoundland| J„Te GROENMAN, Arnhem, The Netherlands; RICHARD I. HESS, Rancho Palos Verdes, California; WALTHER JANOUS, Ursulinengymnasium, Innsbruck, Austria; FRIEND H9 KIERSTEAD, JR., Cuyahoga Falls, Ohio? M.S. KLAMKIN, University of Alberta; D.J. SMEENK, Zaltbommel, The Netherlands; and the proposer

- 130 -

918 i T1984: 54] Proposed by John P9 Hoyt3 Lancaster3 Pennsylvania; and

Leroy F, Meyers3 The Ohio State University.

Let k be a positive integer and g a polynomial of degree at most fc-2. Show

that

n_0 (n+l)(w+2). .. (n+fc)

converges to a rational number.

Solution by M9S« Klamkin3 University of Alberta.

If k = l, then the series converges to 09 since the zero polynomial is the only

one of degree at most -1. If k > 25 the result is not valid as stated. No doubt the

proposers tacitly assumed that g{n) is rational for all integral n. If this is so,

then g(n) has the representation

gin) = aQ + ai(n+l) + a2(n+l)(n+2) + ...-* a, _ (w+l)(w+2)... (n+fe-2),

where the a. are all rational (just let n = -l s-2 s e s a successively). The desired

result will follow if we show that

00

A(n+r)(n+r+l).. .(n+k)

is rational for r = l929oa.5k-le It is well known that S can be written in the form

CO

s = JL y { 1 „ i \ k-r L^in+r)in+r+±) e. .(n+fc-1) Xn+ri^(n+r+2 ) . . . (w+fc) *

and this sum telescopes down to

1 1 _ 1 ( r -1 ) ! S = b r ( r + l ) , . . ( M ) fc-r (fc-l)P

which is rational.

Also solved by J,Ta GROENMANf Arnhemf The Netherlands (partial solution); FRIEND H. KIERSTEAD, JRe, Cuyahoga Falls, Ohio; RICHARD PARRIS, Phillips Exeter Academy, New Hampshire; McAe SELBY, University of Windsor; JORDAN B. TABOV, Sofia, Bulgaria; and the proposers,, A comment was received from WALTHER JANOUS, Ursulinen-gymnasium, Innsbruck, Austria.

Editor fs comment*

The proposers noted that this is essentially Problem 106 on page 277 of Konrad

Knopp's Theorie und Anwendung der unendlichen Reihen3 5th edition (1964),

- 131 -

9]9 i f:i98i+: 54] Proposed by Jordi Dou3 Barcelona3 Spain.

Show how to construct a point P which is the centroid of triangle A'B'C,

where A\B\CS are the orthogonal projections of P upon three given lines a,b3c, respectively,

Solution by J.T. Groenrnan3 Arnhem3 The Netherlands; and Jordan /. Taloe, Sofia,

BuIgaria (independently).

We assume that the proposers unstated intention was that the lines a,b,c be coplanar and intersecting in pairs. Let

b n c - A^ c n a = B, a n 2> = C5

and let G be the centroid of triangle ABC. Then the required point P is the Lemoine

point K of the triangle, the isogonal conjugate of G. This is an immediate con­

sequence of the following known result ril: The symmedian point TLemoine point]

is the median point [centroid] of its own pedal triangle, and it is the only point

with that property.

Also solved by J*Ta GROENMANf Arnhemf The Netherlands (second solution); D.J. SMEENK^ Zaltbommelf The Netherlands; and the proposer.

REFERENCE

1. Roger A* Johnson9 Advanced "Euclidean Geometry3 Dover, New York, I960,

p. 217 (first theorem and corollary to second theorem). A A •*• rf£ ?\» *•*

9201 [1984: 54] Proposed by Basil C. Rennie3 James Cook University of North

Queensland^ Australia.

If a triangle of unit base and unit altitude is in the unit square9 show that

the base of the triangle must be one side of the square.

Solution by the proposer.

The desired result is an immediate consequence of the following

THEOREM, Let [T] be the area of a triangle T inside a rectangle R of area [R~].

Then r n < ri?1/2, with equality if and only if one si"^ of T coincides with a side

of R and the opposite vertex of T is on the opposite ?ie of R .

Proof. Let T be a triangle of maximal area in­

side j?. (The existence of such a triangle is a sim­

ple exercise in analysis, for the area is a continu­

ous function of the six coordinates of the vertices

on a compact set.) Each side of R must contain at

least one vertex of T3 for otherwise R could be made

M

- 132 -

smaller, still containing T9 and then the whole figure expanded back to the original

size of R to give a triangle larger than T in R. Therefore one vertex of T lies in two adjacent sides, and hence at a vertex9 of

R9 and the configuration is as shown in the figure^ with A at Ms B on NP5 and C on

PQ« If either B is at N or C is at Q3 the result is clear. If not, then [T] can be

increased by moving either B to N or C to Qf contradicting the fact that [Tl is maxi­mal .

Also solved by CLAYTON W. DODGE (Maine), JORDI DOU (Spain), J.T. GROENMAN (The Netherlands), RICHARD I. HESS (California), WALTHER JANOUS (Austria), FRIEND He

KIERSTEAD, JR. (Ohio), M.S. KLAMKIN (Alberta), ANDY LIU (Alberta), LEROY F6 MEYERS (Ohio), FLORENTIN SMARANDACHE (Maroc), D.J. SMEENK (The Netherlands), DAN SOKOLOWSKY (New York), ESTHER SZEKERES (Australia, two solutions), JORDON B. TABOV (Bulgaria), and KENNETH M. WILKE (Kansas).

* * *

921« 1~1984: 88] Proposed by Allan Wm. Johnson Jr.3 Washington,, B.C.

In U.S. liquid measure, a gill is 4 fluid ounces (fl oz) and a pint is

4 gills^ that is:

FLOZ GILL 4 H ±

GILL dna PINT*

Solve these alphametical multiplications independently without reusing the digit 4.

Solution by Charles W» Trigg3 San Diego3 California.

Clearly, in the first multiplication F = 1 or 2 and 4JLL. Thus LL = 00 or 88,

and so LOZ = 025, 075s 872, or 897. If F = 1, then all the LOZ candidates must be

rejected, either because of a letter duplication or because an unwanted 4 appears

in the product. If F = 2, the only possibilities for LOZ are 075 and 897, In the

latter case the product is too large, and the former gives the unique reconstruction

2075*4 = 8300.

In the second multiplication, there are only three values of LL which avoid a

4 or duplicated digits among L, N, and T. Thus:

4*55 = 209 4*77 E 089 4*88 = 52 (mod 100),

The f i r s t p o s s i b i l i t y leads to I = 6, G = 1 , and P = I ; the second leads to I = 9,

G = l , and L = P; and the t h i r d and only sa t is fac tory one leads to I = 9, G = l ,

and P = 7. The unique reconstruct ion is 1988-4 = 7952.

Also solved by CLAYTON W. DODGE (Maine), RICHARD I. HESS (California), JACK LESAGE (Ontario), J. A. McCALLUM (Alberta), STEWART METCHETTE (California), GLEN E. MILLS (Florida), STANLEY RABINOWITZ (New Hampshire), KENNETH Ma WILKE (Kansas), ANNELIESE ZIMMERMANN (West Germany), and the proposer.