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97

T H E O L Y M P I A D C O R N E R

No, 134

R . E . W O O D R O W

All communications about this column should be sent to Professor R.E. Woodrow ,

Department of Mathematics and Statistics, The U niversity of Calgary, Calgary, Alberta,

Canada, T2N IN4.

The first problems we present are from the American Invitational Mathematics

Exa m ination (A.I.M .E.) wr itten April 2, 1992. Th e tim e allowed was three hours. Th e

problems are copyrighted by the Committee on the American Mathematics Competi t ions

of the Mathematical Association of America and may not be reproduced without per

mission. The num erical solutions only will be published next mo nth. Full solutions,

and additional copies of the prob lem s, may be obta ined for a nom inal fee from Professor

Walter E. Mientka, C.A.M.C. Executive Director, 917 Oldfather Hall, University of

Nebraska, Lincoln, NE, U.S.A., 68588-0322.

19 92 A M E R I C A N I N V I T A T I O N A L M A T H E M A T I C S E X A M I N A T I O N

1 , Find the sum of all positive rationa l num bers tha t are less tha n 10 and tha t

have denominator 30 when written in lowest terms.

2* A positive integer is called "ascen ding" if, in its decim al rep res enta tion, the re

are at least two digits and each digit is less than any digit to its right. How many ascending

positive integers are there?

3* A tennis player com putes her "win ratio " by dividing the num ber of ma tches

she has won by the tota l num ber of ma tches she has played. At the sta rt of a weekend,

her win ratio is exactly .500. During the weekend she plays four matches, winning three

and losing one. At the end of the weekend her win ratio is greater than .503. What is the

largest number of matches that she could have won before the weekend began?

4* In Pascal 's triangle, each entry is the sum of the two entries above it . The first

few rows of the triangle are shown below.

Row

Row 1

Row 2

Row 3

Row 4

Row 5

Row 6

1

1 5 10 10 5 1

1 6 15 20 15 6 1

In which row of Pascal's triangle do three consecutive entries occur that are in the ratio

3 : 4 : 5 ?

r V Y \ Printed on recycled paper

y - i A / Impnme sur papier recycle

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98

5 . Let

S

be the set of all rational numbers r, 0 <

r <

1, that have a repeating

decimal expansion of the form

O.ahcabcabc... = O.a&c,

where the digits a, 6, c are not necessarily distinct. To write the elements of

S

as fractions

in lowest terms, how many different numerators are required?

6 . For how many pairs of consecutive integers in { 100 0,1 001 ,10 02,. . . ,

2000}

is no

carrying required when the two integers are added?

7. Faces ABC and BCD of tetrahedron ABCD me et at an angle of 30°. The area

of face

ABC

is 120, th e area of face

BCD

is 80, and

BC

= 10. Find the volume of the

tetrahedron.

8 . For any sequence of real num bers

A

= ( a i , a

2

,a3,.. .)> define

A A

to be the

sequence (a

2

— a i , a s — a

2

,a4 — a

3

, . . . ) , whose nth te rm is a

n+

i — a

n

. Suppo se that all of

the terms of the sequence A (AA) are 1, and th at a i

9

= a

92

= 0. Find a\.

9. Trapezoid

ABCD

has sides

AB

= 92,

BC

= 50,

CD

=

19,

^n d

AD

= 70, with

AB parallel to CD. A circle with center P on AB is drawn tangen t to BC and AD. Given

tha t

AP

= ra/n, where

m

and

n

are relatively prime positive integers, find

m

+

n.

1 0 . Consider the region

A

in the complex plane that consists of all points

z

such

tha t bo th z /40 an d 40 /z have real and imagina ry parts between 0 and 1 inclusive. W hat is

the integer that is nearest the area of A? (If

z

=

x

+

iy

with

x

and

y

real, then ~z

= x

—

iy

is the conjugate of z.)

1 1 . Lines

£\

and f

2

both pass through the origin and make first-quadrant angles of

~ and ~ radian s, respectively, with the positive x-axis. For any line £, the transform ation

R(£)

produces another line as follows:

£

is reflected in

£i,

and the resulting line is then

reflected in

l

2

.

Let

R^(£)

=

R(£),

and for integer

n >

2 define

R^(£) =

i ? ^

7 1

"

1

^ ) ) .

Given tha t

£

is th e line

y

= | | x, find th e smallest positive integer m for which

R(

m

\£) = L

1 2 . In a gam e of Chomp, two players alternately take

a

bites" from a 5-by-7 grid

of unit sq uares. To take a bite, the player chooses one of the rem aining squares, th en

removes ("eats") all squares found in the quadrant defined by the left edge (extended up

ward) an d th e lower edge (extended rightward) of the chosen square. For exam ple, the

bite determined by the shaded square in the diagram would remove the shaded square and

the four squares marked by x . (Th e squares with two or mo re dotted edges have been

removed from the original board in previous

moves.) The object of the game is to make one's

opponent take the last bite. The diagram shows

one of the many subsets of the set of 35 unit

squares that can occur during games of Chomp.

How many different subsets are there in all?

Include the full board and the empty board in your

count.

V J I '

i ih

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99

1 3 . Triangle ABC has AB = 9 and BC : C A = 40 : 41. W hat is the largest area

that this triangle can have?

1 4 . In triangle ABC, A

1

, B', and C are on sides B~C,

~AC,

and

~AB,

respectively.

Given that AA', BB', and CC are concurrent at the point 0, and that

AO BO CO

+ +

find the value of

OA' OB' OC

AO BO CO

OA' ' OB' ' OC

7 = 92,

1 5 • Define a positive integer

n

to be a "factorial tail" if there is some positive

integer

m

such tha t the base-ten representation of m ends with exactly

n

zeros. How

many positive Integers less than 1992 are no t factorial tails?

*

The Olympiad problems we give this month are those of the

21st Austrian Math

ematical Olympiad, 1990, Many thanks to Walther Janous, Ursulinengymnasium, Inns-

bruck

9

Austria, for translating the problems. He points out that the distinction between

^natural

5

and 'intege r' numbe r in 2nd Round # 1 and F inal Round # 4 w as the cause of

confusion and quite a few erroneous 'solutions

5

,

2 1s t A U S T R I A N M A T H E M A T I C A L O L Y M P I A D

2nd Round (May 3

9

1990): 4 hours

1 • Prove: There exists no natural number n such that the total number of integer

factors of

n

equals 1990 and the sum of the inverses 1/6 of all natural number factors

b

of

n

equals 2,

2 ,

Solve (in R ) the equation

^2x

-

7 + ^3x

- 3 =

^T^~8 + tyix

- 2.

3*

Let

ABC

be a triangle with

E

and

D

the feet of the altitudes to sides

h

and a,

respectively. Let

M

be the point on

AD

such that

AD

=

DM.

(a) Show there exists no acute-angled triangle

ABC

such tha t (7, D ,

E, M

lie on

a circle.

(b) Determine all triangles

ABC

such that

CDEM

do He on a circle.

4 . For natura l numbers &,n > 2, determine the sum

S(k,n)

=

[2

n + 1

+ l"

2*-

1

+ 1

+

[3

n + i

+ r

I

n

~

+

+ ...+

k*-

1

+

1

where [x] denotes the greatest integer

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100

Final Round

1st Day — May 30, 1990: 4.5 hours

1 . Determine the num ber of al l natur al numbers

n

such that 1 <

n < N

= 1990

1990

with n

2

—

1 and

N

relatively prime.

2 . Show that for all na tura l num bers

n > 2

fH-

...

tfn + ' ( - ) =

( i

- ^ : r

9 )

^

for all

x ^ 0,

—3. Furthermore, for each fixed natural number

n

determine all integers

x

such that

f(x)

is an intege r.

5 . De term ine all rationa l numb ers r such th at all solutions of rx

2

+(r+l)x+(r—1) =

0 are integers.

6 . Th e convex pentagon ABODE has a circumcircle. Th e perpen dicular distances

of

A

from the lines through

B

and C,

C

and D, and

D

and

E

are a,

h

and c, respectively.

Determine (as a function of a, b and c) the perpendicular distance of A from the diagonal

BE.

We next move to solutions of problems posed in the 1991 volume of

Crux.

2 .

[1991:

1]

1980 Celebration of Chinese New Year Contest.

Let

n

be a positive integer. Is the gre atest integer less than (3 + \ /7 )

n

odd or even?

Solution by Seung-Jin Bang, Seoul, Republic of Korea.

If (3 + \ / 7 )

2 n

= a

n

+ b

n

y/1 (a

n

and b

n

are integers), from the binomial formula

(3 - V7 )

2 n

= a

n

-

b

n

yjl. It follows th at (3 + v ^ )

2 n

+ (3 - V7)

2n

= 2a

n

and (3 + v ^ )

2

^

1

+

(3 - v /7 )

2 n + 1

= 2(3o

n

+ 76„). Since 0 < 3 -

y/7 <

1, we have

[(3 + v^)

2

"] = 2

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101

4 . [1991: 1]

1980 Celebration of Chinese New Year Contest

Denote by

a

n

the Integer closest to

y/n.

Determine

1 1 1

—

+

—

+

»_

1

(0) - 2 ) ^ ( 0 ) = 2

•

4

2

" + 4P:_

X

(0)

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102

and

P (0)

= 2. Solving the recurrence relation, we have

The coefficient of

x

2

in case the expansion results from

n

applications is thus

=^

4

n

+ § 4

2 n

.

2 , [1991: 2]

1981 Celebration of Chinese New Year Contest.

Prove that 1980

19811982

+ 1982

198ll98

° is divisible by

1981

1981

.


Let

a

= 1981 = 7

•

283. From the binomial formula

(«-i)

a

"

+1

+(«+ir

1

= E «

r

{(

a

7) ( -

1

)

r

+ '

f l

+ a

a

L

r J ' \ r

for some integer L. It is well known that if p is a prime and p

3

divides r\ , then

5 <

+

+ •••<

. p - 1

Since 7 and 283 are primes and [r/6] < r

—

1, [r/282] < r

—

1, (r > 1), we have

r °

+

J

a

f

=

a

a

M

a nd r "

J

a

r

=

a

a

N

for some integers

M

an d iV for r > 1. It follows th a t

(a —

l )

a a + 1

+ (a +

I )

0

"

- 1

is divisible

b y a

a

.

3 . [1991:

2] 1981 Celebration of Chinese New Year Contest.

Let

f(x) = x" + x

98

+ x

97

+ h x

2

+ x

+ 1. Determine the remainder when

f(x

10 0

)

is divided by

f(x).


Note that

/ ( * ) =

c

10 0

- 1

x-1

'

If

xo

is a (complex) zero of

f(z)

then

XQ

is a simple zero of

f(z)

because

100xl

9

(x

o

- 1) -

(x l

00

- 1) _ 100

f *o)

=

# 0 .

(XQ — I )

2

# o ( # 0 - 1 )

Let x\

9

X 2

?

. . . , 99 be all the distinct zeroes of /(#), and let

r(x) = a

9 8

x

9 8

+ a

97

x

97

H h ai# + a

0

be the remainder when

f(x

10 0

)

is divide d by

f(x).

From / (# J

0 0

) = 100 =

r(x{)

for

t =

1,2,...,

99

we

have the following system of equations for ao,.. . ,

a^g:

a

0

+ aixi+ ••• + a

9 8

x

9 8

= 100

a

Q

+ aix

2

+ - • • + a

9 8

;r

9 8

= 100

o>o

+ ai#99+

+^ 98^99 = 100 .

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103

Note that the determinant of the coefficient matrix is

A =

1 xi ... xf*

1 X

2

• • • % 2

8

1

Xg g

. . • X

9 9

3

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104

1 . Let ax , . . . , a

n

, 6 1 , . . . , b

n

, c j , . . . , c

n

be positive real numbers. Show that

(t a

k

b

k

cX

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105

where £?= i

q

{

= 1,

q

{

>

0, unless either

(an, a,-

2

,...,

a,-

n

) ( i = 1 , . . . , m) are all proport ional ,

o r ( a , i , . . . , a

f

-

n

) = ( 0 , . . . , 0) for some

i

= 1 , . . . , ro . (Let ^ =

q

2

= g

3

= 1/3,

a

k

\

= a| ,

&&2 = bl, a

k3

= c |). See Hardy, Littlewood, & Polya, Inequalities, 2nd Edition (1952),

pp .

21-24.]

2« Elach point of the pla ne ( R

2

) is coloured by one of the two colours A and B.

Show that there exists an equilateral triangle with monochromatic vertices.

Solutions by Seung-Jin Bang, Seoul, Republic of Korea; Margherita Barile, student,

Genova, Italy; and John Morvay, Sp ringfield, Missouri. We give Barile's solution.

Let the two colours be black and white and let ABC be an equilateral triangle. We

may assume, without loss of generality, that the vertices

A

and

B

are black. Suppose

C

is

white and consider the figure where

ADGEFC

is a regular hexagon (with

B

the centre) .

Now proceed as follows:

TJ

•

ADB

is equilateral,

A, B

are

black; so suppose

D

is white.

• CDB is equilateral; C , D are

white; so suppose E is black.

F

•

BEF

is equ ilater al; J3,

E

are

black; so suppose

F

is white.

•

CFH

is equ ilatera l; (7,

F

are

white; so suppose H is black.

E

•

BHI

is equilateral; J3,

H

are

black; so suppose

I

is white.

G

Now 7JDF is equilateral and 7, 7}, F are white, completing the

proof.

3*

Determine all natura l numbe rs

N

(in decimal representation) satisfying the

following properties:

(1)

N

= (aa66)io, where (aaft)io and (a66)i

0

are primes.

(2) N = P i • P2 * P35 where P& (1 < fc < 3) is a prime consisting of k (decimal) digits.

Solutions by Seung-Jin Bang, Seoul, Republic of Korea; Margherita Barile, student,

Genova, Italy; and Stewart Metchette, Culver City, C alifornia. We give Bang's solution.

Since

(aab)io

= 110a +

b

is a prim e, we obta in 6 = 1,3,7 or 9. F rom (1) we have

N = l l (100a + b) . In a table of primes we obtain the pairs ((aab)

10 j

(aab)io) satisfying (1)

as follows:

(223,233), (227,277), (331,311), (443,433),

(449,499), (557,577), (773,733), (881,811),

(887,877), (991,911), (997,977).

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106

Corresponding to 100a +

h

are

203 = 7 x 29, 207 = 9 x 23 , 301 = 7 x 43 , 403 = 13 x 31 ,

409 = prim e , 507 = 3 x 13

2

, 703 = 19 x 37, 801 = 3

2

x 89,

807 = 3 x 269, 901 = 17 x 53, 907 = pri m e

Answer:

N

= (8877)i

0

= 3 x 11 x 269.

5» Let A be a vertex of a cube u; circumscribed about a sphere

&

of radius 1. We

consider lines

g

through

A

containing at least one point of

K.

Let

P

be the point of

g

fl K

having minimal distance from

A.

Furthermore ,

g

fl u? is

AQ ,

Determine the maximum

value of

AP

•

AQ

and characterize the lines

g

yielding the maximum

e


Let the vertices of the given cube

u>

be (0 ,0 ,0) , (2 ,0 ,0) , (0 ,2 ,0) , (0 ,0 ,2) , (2 ,2 ,0) ,

(2,0,2), (0,2,2), (2,2,2) and let the equation of the given sphere

K

be

( * - l )

2

+ ( y - l )

2

+ ( 2 - l )

2

= L

We may assume that

A

= (0,0,0), and the equation of the l ine

g

is

t(a

}

6,

c) where

t

is real

and

a

2

+ b

2

+ c

2

= 1, a,6,c > 0. From

(ta

- I )

2

+

(tb-

I )

2

+ ( t c - I)

2

= 1 we have

p = (

a

+

6

+

c

_

y/{a +

b

+ c)

2

- 2 ) ( a , 6 , c ) .

Let

d

= max(a ,

6,

c ). Since gC\u>

~ AQ,

we obtain

td

= 2 and

2

Q =

„ / „ * x ( g > M .

max(a ,

6,

c)

Note that

AP-AQ = 2

a + b + c-^J(a + b + c)

2

-2

max(a ,

6,

c)

4

max(a , 6, c)(a + b + c + y (a + 6 + c)

2

— 2 )

We may assume that a = max(a, 6,

c ).

Then

A ( 6 ,c ) = A P . A Q = /

a(a + b+c+ yj(a + b + c)

2

- 2 )

and b

2

+ c

2

= 1 — a

2

. Note that /

t t

is strictly decreasing as a function of b + c and

\ / l

—

a

2

<

6

+ c holds for 6 = 0 or c = 0. So we ma y tak e 6 = 0 and

c

=

y/ \ — a

2

.

From

a

= max(a ,

6,

c) we then have

a > y/l —

a

2

, so a > l/\/2 . It follows that

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107

The last inequality holds because

^ / , n o\ y l — a

2

— a

-

(a +

VT^)=

V i

_

o 2

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108

Let

C = J2{p

€ N : c„ is a prime or the square of a prime for every

n < p}.

If

C

is finite,

then

C = A +

1 and

A

is finite. We prove that

C

is finite.

For

n -

1 (5) yields Pf

1

= 26

x

+ 1. Suppose next tha t

A > n

0

=

2b

x

+

2. By (5),

P«

0

"o = 2ft + 46a + 3 = 66

x

+ 3 = 3 P ?

1

.

This implies that 3|P„

0

, i.e. P

n o

= 3, since P

no

is a prime.

Since

P?

1

>

1,

it

follows

tha t

a

0

= 2, so tha t 6Pi + 3 = 3

2

= 9, implying 6

:

= 1. Thus by (3) 6

2

= 2 and a

2

= b\ = 4, so

that, by (6)

C l

= 4 - 1 =

c

2

= 5

c

3

= 7

c

4

= 9 = 3

2

c

5

= l l

Ce= 13

3 prime

prime

prime

square of a prime

prime

prime

Since c

+

2 = 15, which is neither a prim e nor the sq uare of a prim e, (7 = 6 and therefore

A

= 7. In fact

a\

= 1, a

2

= 4, a

3

= a

2

+ c

2

= 4 + 5 = 9, a

4

= a

3

+ c

3

= 9 + 7 = 16,

a

5

= a

4

+ c

4

= 16 + 9 = 25, a

6

= a

5

+ c

5

= 25 + 11 = 36 and

a

7

= a

6

+ c

6

=

36 + 13 = 49. We

now prove th at this is the sequence of max ima l length A by showing that the corresponding

sequence {c

n

: n > 1} is of maximal length C. If C > 3 then one of the numbers

cu c

2

= C i + 2 , c

3

= c i + 4

is divisible by 3, and so is equal to 3 or 9. By (5), since

b\ >

0, we have

c\ >

3. Therefore,

in the first case

c\

= 3, which yields the above example. Otherwise

c

n

= 9 for a certain

n

€ {1 ,2 ,3} . But then c

n

+

3

=

c

n

+ 3

•

2 = 15 , so tha t C = n + 2 < 5 < 6 . Th i s completes

the

proof.

* * *

That's all the space we have this issue. We'll continue with solutions for the Team

Competit ion of the 12th Austrian-Polish Mathematics Competit ion next issue. The alert

reader will note we did not give a solution to number 4. Meanwhile send me your national

and regional Olympiads, as well as your nice solutions.

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109

P R O B L E M S

Problem proposals and solutions should be sent to B. Sands, Department of

Mathematics and Statistics, University of Calgary, Calgary, Alberta, Canada T2N IN4.

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 m ay also

be acceptable provided they are not too w ell known and references are given as to their

proven ance. Ordina rily, if the originator of a problem can be

located,

it should not be

submitted by somebody else without permission.

To facilitate their consideration, yo ur solutions, typewr itten o r neatly handw ritten

on

signed,

separate sheets, should preferably be mailed to the editor before

N o v e m b e r 1 ,

1992

?

although solutions received after that d ate will also be considere d until the time when

a solution is published.

1 7 3 1 * .

Proposed by Vaclav Konecny, Ferris State University, Big Rapids,

Michigan.

Let

P

be a point within or on an isosceles right triangle and let ci,c

2

, c

3

be the

lengths of the three concurrent cevians through P. Prove or disprove tha t ci ,c

2

, c

3

form

the sides of a nonobtuse triangle. [This problem was inspired by Mu rray K lam kin's problem

1631,

solution this issue.]

1732.

Proposed by Walther Janous, Ursulinengymnasium, Innsbruck, Austria.

Let

pal(n)

be the nth pal indromic number ( i .e.

pal (I)

== 1, . . . ,

pal(9)

= 9,

pal(10)

= 11,

pal(ll)

= 22, etc .). Determ ine the set of all exponen ts

a

such that

00 ^

5

&»'(»)]

converges.

1733.

Proposed by Toshio Seimiya, Kawasaki, Japan.

ABC

is a triangle with circumcen ter

O

and such that

IA

> 90° and

AB < AC.

Let

M

and

N

be the midpoints of

BC

and

AOj

and let

D

be the intersection of

MN

with

side

AC.

Suppose that

AD

=

(AB +

AC)/2. Find

LA.

1734. Proposed by Murray S. Klamkin, University of Alberta.

Determine the m inimum value of

y/^T^)

2

+

(ay)

2

+ (azf

+

y/(l

-

by Y

+

(bzf

+

(bxf +

yftT-

czf + (cxY + (cyY

for all real values of a,

6,

c,

x,

t/,

z.

1735. Proposed by P. Penning, Delft, The Netherlands.

In a conic (ellipse or hyperbola) with centre 0 , chords AB have the property that

all triangles OAB have the same area. Find the locus of the midp oint of AB.

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I l l

Solution by Walther Janous, U rsulinengymnasium, Innsbruck, Austria.

This is a particular case of a famous problem from probability theory, sometimes

called the "collector's problem" (see, e.g., pp. 174-175 of W. Feller, Introduction to Prob

ability Theory and its Applications,

Vol. I, Joh n Wiley, New York-London , 1950), with the

result for an n-sided die:

Expe cted num ber of throws = n ( — H h -\ h i ) ;

\n n —

1 2 /

i.e., for n = 6, the expected number of throws is 14.7.

Also solved by MAR GHE RITA BARILE , student, Universitd degli Studi di Genova,

Italy; HANS ENGELH AUPT, Fmnz-Ludwig-Gymnasium, Bamberg, Germany; RICHARD

I. HESS, Rancho Palos Verdes, California; FRIEN D H. KIERST EAD JR., Cuyahoga

Falls, Ohio; MU RRAY S. KLAM KIN, University of Alberta; MA RGIN E. KUC ZMA ,

Warszawa,

Poland;

TOM LEINSTER, New College,

Oxford;

CHRIS WILDHAGE N,

Rotterdam, The Netherlands (two solutions); and the proposer.

Klamkin also referred to Feller's book, but p. 211 of the 1961 edition. He also

mentions the paper

a

The double dixie cup problem" by D.J. Newm an and L. Shepp, in

the

American Math. Monthly 67 (1960),

pp .

58-61,

which considers the problem if each

number is to be obtained m times. For a more recent reference, see

u

Majorization and the

birthday inequality" by M.L. Clevenson and W . W atkins, in Mathematics Magazine 64

(1991), pp .

183-188,

especially p.

187.

5fc $ sfe * >fc

1 6 2 8 * . [1991: 79]

Proposed by Remy van de Ven, student, University of Sydney,

Sydney, Australia.

Prove that

where k is a positive integer.

Solution by G.P. Henderson, Campbe llcroft, Ontario.

It is not necessary that k be an integer. The relation is true provided k is not zero

or a negative integer.

The coefficient of

r

n

on the left side is

(1)

i = 0

s=0

n - 1

= £

=Q

n - 1 n-t—l

= E E C - ^ u n

n

_

s

j(k + ty

r^zweri::-

1

)

(k + ty

8/17/2019 Crux v18n04 Apr

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112

We have

f

k\ (k + n - s - l \ k{k + n - s - l)(fc + n - s - 2 ) . . . (k - s + 1)

n

— 5

s (ra

—

s )

(fc

—

s)(n

—

s) (k + n

—

3)5

fc + n - a - l ) . . . ( f c - 3 + l )

s (n — 5)

"( . )( n K - l ) ( » J'

Therefore in (1), the inner sum is

s = l

[

(

-

v

("7>)(*+»

B

—')-

(

-ir*(;:l)(*+;-)

= ( - I ) " " ' "

1

n - 1 \ /

A;

+

n

J

and

n - 1 \ (k + t\ 1

fc + n - T

n

L j j is a polynomial of degree n — 1 in

k

and its constant term is (—I)

71

""

1

/™. Hence

where P is a polynomial of degree n — 2 if

n >

2 and is zero if

n =

1. Then

( _ 1 ) H " 1 1

, n - l

P(k + t) +

n k + t

(-l)

n

~

l

1

A

n

~

1

P(k) + ^-^- A

n

-

X

T

[Here

A / ( * ) = / ( * + 1)

- / ( * ) ,

A

m

/ ( » ) = ACA—V C*)), m = 2 , 3 , . . . ,

are the

forward differences.

—

Ed.]

The first part of this is zero because

P

is either zero or

a polynomial of degree n

—

2. The second part is

1

which is the coefficient of

r

n

on the right side.

The proposer found the identity (used without proof) in an old paper "The negative

binom ial distribution", by R.A. Fisher, in

The Annals of Eugenics 11 (1941),

pp.

182-187.

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113

1 6 2 9 .

[1991: 79]

Proposed by Rossen Ivanov, student, St. Klirnent O hridsky

University, Sofia, Bulgaria.

In a tetrahedron

x

and v,

y

and u,

z

and

t

are pairs of opposite edges, and the

distances between the midpoints of each pair are respectively l^rn.n. The tetrahedron has

surface area 5, circumradius

R,

an d inradius r. Prove that , for any real num ber

A

with

0 < A < 1,

x

2X

v

2X

l

2

+ y

2X

u

2X

rn

2

+ z

2X

t

2X

n

2

> (^) ( 2 S )

1 + A

( / 2 r ) \

Solution by the proposer.

We use the inequality of Neuberg-Pedoe:

J2 a

2

(-a

a

+ b

f2

+ c

n

)>16FF

,

J

where a, 6, c and a', &',

d

are the sides of triangles with areas

F

and

F\

respectively, and

the sums here and below are cyclic (e.g., see p. 355 of [1]). If g

1?

5*2? 5s and

S

4

are the

areas of the faces of the tetrahedron determined by edges v,y,z; x,u,z; x,y,t; and v,t/ , t

respectively, then we have four inequalities:

£ a

2

( - z

2

+ u

2

+ z

2

) > 16 $ ,^ ,

£ a

2

( - *

2

+ y

2

+ f

2

) > 1 6 S

3 J

F ,

5 3 a

2

( - u

2

+

u

2

+ *

2

) > 1 65

4

F.

If we sum up these four inequalities, we get

2 £

a

2

{-x

2

-v

2

+ y

2

+ u

2

+ z

2

+ t

2

) > 16SF.

But the bimedian connecting th e midpoints of edges with lengths

x

and

v

has length /, so

4 / 2

=

_

x

2 _

v

2

+ y

2

+ u

2

+ z

2

+

^

e t c <

(e.g., see the remark after 1.3 on p. 547 of [1]). Therefore

8 X >

2

/

2

> 1 6 S F .

Now we will use Oppenheim's result: if a, 6,

c

are the sides of a triangle with area

F,

and

if 0 1 6 £

h@\

F

\

8/17/2019 Crux v18n04 Apr

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114

Furthermore, according to Crelle's theorem (item 1.6, p. 549 and (4), p. 555 of [1]), we can

let

a

=

xv, h

=

yu ,

c =

zt, F

=

2SRr,

from which the proposed inequality follows.

Reference:

[1] D.S. Mitrinovic, J.E. Pe caric, and V. Volenec,

Recent Advances in Geometric Inequal

ities, Kluwer Academic Publishers, Dordrecht, 1989.

1 6 3 0 .

[1991: 79]

Proposed by Isao Ashiba, Tokyo, Japan.

Maximize

over all permutations a

1?

a

2

, . . . ,

a

2n

o f the s e t {1 ,2 , . . . ,

2n}.

I.

Solution by Pavlos Maragoudakis, student, University of Athens, Greece.

Since

1 / 2n

n

\

a ia

2

+

a

3

a

4

+ h

a

2n

^

x

a

2n

= r ( 5 3

a

t

?

"" ] L (

a2

*

~~

a

2»-i )

2

I

-I / 2n n \

=

2

E

j 2

- E K -

f l

« - i )

2

L

the above sum is maximized when

Y^l-\

{p>2%

—

02t- i )

2

i

s

minimized. But

n

2 (

a

2i - a

2

, - i )

2

> n,

with equality when \a

2

% — a

2

*-x| = 1? * = 1,2, . . . , n , for example when a

t

- =

i

for all

i.

Therefore we obtain the maximization of ax2n-ia>2n

is maximized over all permutations of S by pairing the largest two numbers, then the

remaining largest two numbers, and so on down to the smallest two numbers. For consider

any four elements

a, a + A

x

, a + Ax + A

2

,

a +

Ax + A

2

+ A

3

of 5, where Ax, A

2

, A3 > 0, and let

M

=

(a +

Ax + A

2

+ A

3

) a + Ax + A

2

) +

a(a

+ Ax),

Li = a + Ax + A

2

+ A

3

) a + A

x

) +

a(a

+ A

x

+

A

2

,

L

2

= (a +

Ax + A

2

+ A

3

)a + a + Ax + A

2

) a +

A

x

.

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115

Then

M-Li = A

2

(Ai + A

2

+ A

3

) > 0 and M - L

2

= (A

a

+ A

2

) (A

2

+ A

3

) > 0,

so

M

is the best to pick. A permutation claiming to be maximizing in a way other than

above could be improved by finding the largest a

t

- deviating from the rule and switching

the permutation to obey the rule. [So if a,- were not paired off with its mate a,j according

to th e above rule , suppose a,- were paired with a& and dj with ai

}

where a

f

- > aj > a^ai\

then aiQk +

a$a\

is increased by switching to a{Uj +

ctkat.

—

Ed.]

Also solved by H.L.

ABBOTT,

University of Alberta; MAR GHE RITA BARILE,

student, Universitd degli Studi di Geno va, Italy; ILIYA

BLUSKOV,

Technical

University, Gabrovo, Bulgaria; HANS ENG ELHAU PT, Franz-Ludwig-Gymnasium,

Bamberg, Germany; C. FESTRA ETS-HAM OIR, Brussels, Belgium; PETER HURT HIG,

Columbia College, Burnaby, B.C.; W ALTHE R J ANOU S, Ursulinengymnasium, Innsbruck,

Austria; FRIEND H. KIERSTE AD JR., Cuyahoga Falls, Ohio; HURR AYS. KLAM KIN,

University of Alberta; MAR GIN E. KUC ZMA, Warszawa,

Poland;

TOM LEINSTER,

New College,

Oxford;

AND Y LIU, University of Alberta; JEAN-M ARIE MO NIER, Lyon,

France; P. PEN NING, Delft, The Netherlands; DAVID G. POO LE, Trent University,

Peterborough, Ontario; EDW ARD T.H. WAN G, Wilfrid Laurier University, Waterloo,

Ontario; CHR IS WILDH AGEN , Rotterdam, The Netherlands; and the proposer.

The solutions of Janous, Leinster, and the proposer wer e similar to Solution

I.

Several solvers obtained the more general result in Solution

II .

Several more noted that the

minimum of

ai

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117

Hence (4) is a necessary condition for (1) to hold.

We will show that it is also sufficient so that

A

= A

0

is actually the threshold value

for the problem in question. Write for convenience p = A

0

/2 and denote the right side of

(3) with

A

= A

0

by

F(x,y).

So we claim

F(x, y) := f(xY + f(yf - f(xyY > 0 for x, y > 1. (5)

This is a continuous function, with boundary values

F(x

J

l)

= F ( l , y ) =

f(l)

p

= (3 /4)

p

;

its limit behavior at infinity can be estimated from (2):

3 \

p

liminf F(x,y)>(-)

Since F vanishes at {XQ,X

0

), we see tha t F{x,y) mu st atta in its minim um value at some

point (x^y) with 1 1 . Then 2t + 2 = (» + l)

2

/*> so

« > 1

X

1 * *

2t + 1

* / ' ( * ) =

(z +

l )

2

2t + 2 2 t +

'

ar ( a j - l ) / * \

3

/ 1 \ _ 1 V ^ ^ T

OHH

x + i )

3

\{x + iyj \ xj 2 (t + iy

and hence by (6)

_ (a

i r

1

^ - )

1 7

'

_.

h(t)

9{x)

- — » { t + iy+*— -

m

Now

sign

h\t)

= sign [2(p - 1)(* +

l)(t

2

- 1) - (p + 1)(2* + 1)(<

2

- 1) +

8/17/2019 Crux v18n04 Apr

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118

left of

J,

such that

g

increases in / and decreases in

J.

Consequently, (7) forces

x = y

6 J,

x

2

€ J, and

g(x)

=

g(x

2

).

The last equation can have at most one solution in ( l , o o ) ,

in view of the mon otonicity relations just established. So it will be enough to check that

x =

XQ

is

a solution.

For x

0

we have f(xl)/f(x

0

) = 9/8 and (9/8)

p

= 2. Hence by (6) (and by f'(x) =

(x - l)(x + I)"

3

)

9(4)

_. . /7(*3)Y"

1

(*o + 1 )

4

_ ,„ , . / * . 16 . (3 +

Vz)

4

= 1?

^)-

X0

{JM) R T I F -

( 2 +

^

3 )

y(*o)

\f(xo)J (x l + lf

v

' 9 (8 + 4 V3 )

3

as needed.

All this taken into account, we infer that

F(x,y)

is minimized at

(X

Q I

X Q ) . SO

its

m inim um value is 0 and (5) is settl ed. Therefore (1) holds for A = A

0

, hence for every

positive A < Ao. Th e lengths of th e thr ee cevian s, raised to pow er A, satisfy t he triang le

inequality for every P if and only if A < Ao, wh ere AQ is the con stant defined in (4).

Also solved by VACLAV KONE&NY, Ferris State University, B ig Rapids,

Michigan; and P. PEN NING , Delft, The Netherlands.

As both Kuczm a and Penning observe, the points giving solutions to problems 1621

an d 1631 coincide Is there a simple proof that this nice "coincidence" had to be?

Penning also calculates that the minimum value of A satisfying the problem is

A = - l n 4 / l n ( 4 / 3 ) « - 4 . 8 2 .

Konecny's work on this problem led him to propose problem 1731, this issue.

1 6 3 2 . [1991:

113]

Proposed by Stanley Rabinow itz, Westford, Massachusetts.

Find all

x

and

y

which are rational multiples of

n

(with 0 <

x < y <

7r/2) such

tha t tan x + t an y = 2.

Solution by Diane and Roy Dowling, University of Manitoba, Winnipeg.

The only solut ion is x =

T T / 1 2 ,

y =

TT/3 .

A resul t of Go rd an [1] says th at fo r u ,

v

a n d

w

ra t iona l mul t ip l es of ?r with 0 <

u <

v < w <

7r, t h e

equat ion

cos u + cos v + cos w + 1 = 0

has only

t h e t w o

solut ions

2n 2T T 4TT

_ 7r

2T T 2n

u

= — ,

v

= — ,

w

= — and

u

=

—

,

v

= -— ,

w

= — .

5 ' 3 ' 5 2

5

3 ' 3

It follows that, when

0

and are rational multiples of

TT

between 0 and 7r,

the equation

cos 0 + 2 cos + 1 = 0 has only one solution: 0 = 7r/2 a n d = 2w/3.

Note that if tan x + tan y = 2 then

sin(y + x) = 2 cos y cos x =

cos(j/

+ x) + cos(y — #),

8/17/2019 Crux v18n04 Apr

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119

so

2 - 2sin(2y +

2x) =

2[sin(y +

x) - cos{y + x)}

2

=

2cos

2

(y -

x)

= 1 + cos(2j/ - 2a;),

so

cos(ir -

2y + 2x) + 2

cos

(

y

-2y-2x\ +1 = - cos(2y - 2x) - 2

sin(2y +

2x )

+ 1

= 0. (1)

Suppose now that

x

and y satisfy the conditions of the problem; then 0 <

y

—

x <

7r/2, SO

0

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120

Solution by Dag Jonsson, Uppsala, Sweden (slightly altered by the editor).

Let

LA

= 2a ,

LB

= 2/2,

LG

= 27,

and let the bisectors intersect at

O.

We will

show that

sin

2

x + sin

2

y

1.

Case 1. 30° < a < 45 °. Th en th e

circumcircle of

AEOD

intersects the exten

sion of the bisector AO at a point F beyond

A, since x = LEDO = LEFO and similarly

y

=

LDFO

so th at

Z F = x + y = 0 + 7 = 90° - a (90° - a ) / 2 . Now for given s, 0

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121

Case

2. 45° <

a <

90°. Then 0 <

x

+

y

< 45°, giving (since the function sin

2

x

is

increasing and convex on 0 <

x <

90°)

sin

2

x + si n

2

y

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122

towards the centre. Each tray is inscribed in a circular sector with central angle 2ir/n. Let

$

= 7r/n. Then from the diagrams it is obvious that

Subtraction yields

Then

n > r

2

4=>

cot

2

0>3

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123

Solution by the proposer .

Consider the direct similarity

S

x

so

th a t Si(BiCi) = B

2

C

2

. Its centre O

x

is

the intersection (other than

E)

of the circles

EB

X

B

2

and

EC

X

C

2

,

where

E

=

B

1

C

1

nB

2

C

2

.

[Because in the figure

LEC\0\ = LEC

2

0\

and

LEB

x

O

x

= LO

x

B

2

C

2

.— Ed.)

Analo-

gously 5*2 denotes the similarity ^ ( J ^ C ^ ) =

B3C3

of centre 0

2

« Since

A1B1C1

is simi-

lar to ^2^(72, we wil l have

A2

= *5i(yli)

and analogously

A3

=

S2(A

2

).

Conv ersely, if

Si(Ai) = A

2

and

5

2

A

2

)

= A

3j

A1B1C1 is

similar to A2B2C2 &nd i4

2

52C

2

is similar to

A3B3C3.

Therefore the problem reduces to

constructing an equilateral triangle

A1A2A3

such that

Si(Ai)

=

A

2

an d ^ ( A O = A3.

It holds that

O1A1A2

is similar to

O1B1B2

(and to any O1X1X2, where

X

2

=

5 i ( X i ) )

?

and O2A2A3 is similar to O2B2B3. Let A ^^ A g be any equilateral tr iangle.

Construct

0{

and

0

f

2

such that

0'

l

A

l

A

2

is similar to

OiBiB

2

,

and

0'

2

A

2

A

z

to

O2B2B3.

Then triangles O1O2A1 and O1O2A2 are similar to

0\0

2

N

X

an d O J O ^ ^ r esp ec tive ly .

Knowing 0i and 02? we can construct Ai and A

2

>

Also solved by TOSHIO SE IMIYA, Kawasaki, Japan.

* *

1 6 3 6 * .

[1991:

114] Proposed by Walther Janous, Ursulinengymnasium, Innsbruck,

Austria.

Determine the set of all real exponents r such that

d

r

(x,y)

=

{x +

y

y

satisfies the triangle inequality

d

r

(x, y) + d

r

(y

1

z) > d

r

(xj z)

for all

x,

y,

z >

0

(and thus induces a metric on ft

4

"—see

Crux

1449, esp. [1990: 224]).

Solution by Murray S, Klamkin, University of Alberta.

We will show that the only valid exponents are r £ [0,1] and r = — 1 .

Assume that x > y > z and consider all three permuted versions of the triangle

inequality. If

r

= — 1, then rf

r

(x

?

t/)

5

d

r

(j /

5

z), d

r

(x, z)

become

x

2

— j

2

5

y

2

— z

2

^

x

2

-

z

2

which

obviously satisfy the condition since the first two sum to the third.

Suppose next that 0 < r

(x + z)

r

(x + y)

=

d

r

(x,y).

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8/17/2019 Crux v18n04 Apr

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125

References:

[1] M.S. Klamk in and A. Meir, Ptole m y's inequality, chorda m etric, multiplicative me tric,

Pacific J. Math.

10 (1982) 389-392.

[2] Doris J. Schattschneider, A multiplicative metric, Mathematics Magazine 49 (1976)

203-205.

Also solved by MARGIN B. KUCZM A, Warszawa,

Poland.

A further solution, due to H. HAM ETNE R, appears in the article

u

Eine Metrik auf

R

+

^

;

Wissenschaftliche Nachrichten 85 (Jdnner 1991J 31-32. The problem had appeared

in the same journal (Vol

83,

April

1990

;

pp .

29-30,)

in an earlier article (of the same

title) by the proposer.

The proposer would like to know whether the function

d

'

r{x,y)

=

IRTw

'

*>y*

R

-w>

satisfies the triangle inequality for any r € [0,1] U { —1}. A similar problem, with x and y

in

R

2

—

{ (0 ,0)}

;

appeared in Hametner's article.

1 6 3 7 .

[1991:

114]

Proposed by George Tsintsifas, Thessaloniki, Greece.

Prove that

^ sin

B +

sin

C

12

where the sum is cyclic over the angles

A,B^C

(measu red in radians) of a nonob tuse

triangle.

Comm ent by the Editor.

A solution to this problem, by Ian Goldberg, has already been published as part of

his solution to Crux 1611 [1992: 62]. Other solvers of the present problem are listed below.

Two other readers suggest that the bound in this problem could be improved to 9\/3/7r,

with equality holding when

ABC

is equ ilater al (one of the m also guesses th at th e result

should be true for all triangles). However, the editor feels that the only proof received of

this (probably true) claim is incomplete, and so prefers to wait and see whether a better

proof will be forthcoming.

Also solved by gEFK BT ARSLA NAG IC, Trebinje, Yugoslavia; WAL THER

JANO US, Ursulinengymnasium, Innsbruck, Austria; MAR GIN E. KUC ZMA , Warszawa,

Poland;

PAV LOS MA RAG OUD AKIS, student, University of Athens, Greece; VED ULA N.

MURTY,

Penn State University at Harrisburg; BOB PRIELIP P, University of Wisconsin-

Oshkosh; and the proposer.

* *

*

* *

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126

1 6 3 8 .

[1991: 114]

Proposed by Juan C. Candeal, Universidad de Zaragoza, and

Esteban Indurain, Universidad Publica de Navarra, Pamplona, Spain.

Fin d all continuo us functions / : (0, oo) —> (0,o o) satisfying th e following two

conditions:

(i) / is not one-to-one;

(ii) if

f(x)

=

f(y)

then

f(tx)

=

f(ty)

for every * > 0.

I.

Solution by Robert B. Israel, University of British Columbia.

All such functions are constant.

Consider any two points u^ v with 0

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128

we get

AB

2

+ BC

2

+ CD

2

+ DA

2

=

2(BM

2

+ AM

2

) +

2(DM

2

+ AM

2

)

=

2(BM

2

+ DM

2

) +

4AM

2

= 4(BN

2

+ M N

2

) +

4AM

2

= AC

2

+ BD

2

+

4MN

2

.

(1)

Therefore we have

AB

2

+ BC

2

+ CD

2

+ DA

2

> AC

2

+ BD

2

. (2)

In a general quadrilateral, it is well known that

AB

•

CD + AD

•

BC > AC

•

£ D (3)

(e.g., p. 63 of R.A. Johnson,

Advanced Euclidean Geometry).

From (2) and (3) we obtain

(A B

+

CD)

2

+ (AD

+

BC)

2

- (AC + B D)

2

=

(A B

2

+ BC

2

+ CD

2

+

DA

2

- A C

2

-

BD

2

)

+ 2(AB -CD + AD-BC-AC- BD) >

0.

Hence we have the required inequality. As shown in the

proof,

the condit ion that ABCD

be cyclic is unnecessary.

Also solved by SEFKET ARSLANA GIC, Trebinje, Yugoslavia; SEUNG-JIN BANG,

Seoul, Republic of Korea; ILIYA BLUSKOV, Technical University, G abrovo, Bulgaria;

JORD IDOU , Barcelona, Spain; IAN GOL DBE RG, student, University of Toronto Schools;

JEFF HIGHAM, student, University of Toronto; WA LTHE R JAN OUS, Ursulinen-

gymnasium, Innsbruck, Austria; MU RRA Y S, KLA MK IN, University of Alberta; M ARG IN

E. KUCZM A, Warszawa,

Poland;

KEE-WA ILAU, Hong Kong; ANDY LIU, University of

Alberta; R.S. OD ONK OR, International Secondary School, Agege, Nigeria; P. PEN NING ,

Delft, The Netherlands; D.J. SMEE NK , Zaltbommel, The Netherlands; and the proposer.

Several readers observe that equality holds if and only if ABC D is a rectangle, as

can be seen from the above proof

Janous and Klamkin also observe that the result holds for arbitrary qua drilaterals.

Janous points out that (1) was known to Euler (e.g., see p. 547", remark following item 1.3,

of Mitrinovic et al, Recent Advances in Geometric Inequalities^). Klamkin remarks that

the inequality of the problem is a special case of the following: if

a, a

1

; 6,

b

f

; and

c,

d are the

lengths of the three pairs of opposite edges of an arbitrary tetrahedron, then a + a', b + b

f

,

c + c

f

are the sides of an acute triangle. See [1979: 130-131] or pp. 69-70 of Klamkin's

USA Mathematical Olympiads 1972-1986, MAA, 1988.

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* • • •

BOUND VOLUMES

* • * *

J • • •

VOLUMES RELIES

• • • |

THE FOLLOWING BOUND VOLUMES OF/ CHACUN DES VOLUM ES SUIVANTS DE

CRUX MA THEM A

T CORUM

ARE A VAILABLE AT 10.00 PER

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PLEASE SEND CHEQUES MADE PAYABLE TO

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577 King Edward

Ottawa, Ontario

C anada K1N 6N5

Vo lum es: x 10 .00 =

Shipping and handling charges:

C anadian addresse s: 1.0 0 per book ; US and foreign addresses: 1.2 5 per book

(Canadian residents, please add GST / Residents canadiens, veuillez ajouter la TPS)

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POSTALE I

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PUBLI CA Tl O N S

The Canadian Mathematical Society

577 King Edward, Ottawa, Ontario K1N 6N5

is pleased to annou nce the availability of the following publications:

1001 Problems in High School Mathematics

(C ollected and edited by E J . Barbeau, M.S.

Klamkin and W.O.J. Moser.)

Book 1

Book II

Book III

Book IV

1 Book V

Problems 1-100 and Solutions 1-50, 58 pages

Problems 51-200 and Solut ions 51-150, 85 pages



Problems 3 51 -50 0 and Solutions 3 51 -45 0, 86 pages

( 7 .50,

( 7 .50,

( 7 .50,

OUT OF PRINT

CMS Members: 6 .00)

OUT OF PRINT

CMS Members: 6 .00)

CMS Members : 6 .0 0 )J

The Canadian Mathematical Olympiads

( 1 9 6 9 - 1 9 7 8 ) a n d ( 1 9 7 9 - 1 9 8 5 )

Prob lems se t in the f i r s t ten O lymp iads (19 69 -19 78 ) tog e the r w i th sugge s ted

so lu t ion . Ed i ted by E .J . Barbeau and W .O .J . Mose r . 89 page s .

( 7 . 5 0 , C M S M em ber s : 6 . 00 )

P r ob lem s s e t in t he O l y m p i ads ( 197 9 - 1 985 ) t oge t he r w i t h s ugges t ed s o l u t i on .

Ed i ted by C M . Reis and S .Z . D i to r . 84 pag es .

( 7 . 5 0 , C M S M e m b e r s : 6 . 0 0 )

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