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ANNUAL HIGH SCHOOL MATHEMATICS EXAMINATIONS 1973-1982

Compiled and with solutions byRALPH A. ARTINO, ANTHONY M. GAGLIONE and NIEL SHELL

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The Mathematical Association of AmericaNew Mathematical Library

C

THE CONTESTPROBLEM BOOK IV

Annual High School Examinations1973-1982

ofThe Mathematical Association of America

Society of ActuariesMu Alpha Theta

National Council of Teachers of MathematicsCasualty Actuarial Society

NEW MATHEMATICAL LIBRARYPUBLISHED BY

THE MATHEMATICAL ASSOCIATION OF AMERICA

Editorial Committee

Ivan Niven, Chairman (1981-83) Anneli Lax, EditorUniversity of Oregon New York University

W. G. Chinn (1980-82) City College of San FranciscoBasil Gordon (1980-82) University of California, Los Angeles

M. M. Schiffer (1979-8 1) Stanford University

The New Mathematical Library (NML) was begun in1961 by the School Mathematics Study Group to makeavailable to high school students short expository bookson various topics not usually covered in the high schoolsyllabus. In a decade the NML matured into a steadilygrowing series of some twenty titles of interest not only tothe originally intended audience, but to college studentsand teachers at all levels. Previously published by Ran-dom House and L. W. Singer, the NML became apublication series of the Mathematical Association ofAmerica (MAA) in 1975. Under the auspices of the MAAthe NML will continue to grow and will remain dedicatedto its original and expanded purposes.

THE CONTESTPROBLEM BOOK IV

Annual High School Examinations

1973-1982

compiled and with solutions by

Ralph A. ArtinoThe City College of New York

Anthony M. GaglioneThe U.S. NavalAcademy

and

Niel ShellThe City College of New York

29

THE MATHEMATICAL ASSOCIATION

OF AMERICA

Fifth Printing

All rights reserved under International and Pan-American Copyright Conventions.Published in Washington by the Mathematical Association of America

Library of Congress Catalog Card Number: 82-051076

Complete Set ISBN 0-88385-600-X

Vol. 29 0-88385-629-8

Manufactured in the United States of America

Contents

Preface ........................................ vii

Editors' Preface ................................. xi

List of Symbols .................................. xiii

I Problems ......1973 Examination1974 Examination1975 Examination1976 Examination1977 Examination1978 Examination1979 Examination1980 Examination1981 Examination1982 Examination

....... 1

....... 1................................ 8

............................... 14

............................... 20

............................... 27

............................... 34

............................... 40

............................... 46

............................... 52

............................... 58

II Answer Keys ................................... 65

II Solutions ...................................... 67i973 Examination ............................... 671974 Examination ............................... 791975 Examination ............................... 881976 Examination ............................... 981977 Examination .............................. 1071978 Examination .............................. 1171979 Examination .............................. 1261980 Examination .............................. 1381981 Examination .............................. 1531982 Examination .............................. 167

IV Classification of Problems ......................... 179

v

NEW MATHEMATICAL LIBRARY

1. Numbers: Rational and Irrational by Ivan Niven2. What is Calculus About? by W W Sawyer3. An Introduction to Inequalities by E. E Beckenbach and R. Bellman4. Geometric Inequalities by N D. Kazarinoff5. The Contest Problem Book I Annual High School Mathematics Examinations

1950-1960. Compiled and with solutions by Charles T Salkind6. The Lore of Large Numbers by P J. Davis7. Uses of Infinity by Leo Zippin8. Geometric Transformations I by L M. Yaglom, translated by A. Shields9. Continued Fractions by Carl D. Olds

10. Replaced by NML-3411. Hungarian Problem Books I and 11, Based on the Ebtvos Competitions12. J 1894-1905 and 1906-1928, translated by E. Rapaport13. Episodes from the Early History of Mathematics by A. Aaboe14. Groups and Their Graphs by E. Grossman and W Magnus15. The Mathematics of Choice by Ivan Niven16. From Pythagoras to Einstein by K 0. Friedrichs17. The Contest Problem Book II Annual High School Mathematics Examinations

1961-1965. Compiled and with solutions by Charles T Salkind18. First Concepts of Topology by W G. Chinn and N E. Steenrod19. Geometry Revisited by H. S. M. Coxeter and S. L. Greitzer20. Invitation to Number Theory by Oystein Ore21. Geometric Transformations II by L M. Yaglom, translated by A. Shields22. Elementary Cryptanalysis-A Mathematical Approach by A. Sinkov23. Ingenuity in Mathematics by Ross Honsberger24. Geometric Transformations III by L M Yaglom, translated by A. Shenitzer25. The Contest Problem Book III Annual High School Mathematics Examinations

1966-1972. Compiled and with solutions by C. T Salkind and J. M Earl26. Mathematical Methods in Science by George P6lya27. International Mathematical Olympiads 1959-1977. Compiled and with solutions

by S. L. Greitzer28. The Mathematics of Games and Gambling by Edward W Packel29. The Contest Problem Book IV Annual High School Mathematics Examinations

1973-1982. Compiled and with solutions by R. A. Artino, A. M Gaglione, and N.Shell

30. The Role of Mathematics in Science byM M Schiffer and L. Bowden31. International Mathematical Olympiads 1978-1985 and forty supplementary

problems. Compiled and with solutions by Murray S. Klamkin32. Riddles of the Sphinx by Martin Gardner33. U.S.A. Mathematical Olympiads 1972-1986. Compiled and with solutions by

Murray S. Klamkin34. Graphs and Their Uses by Oystein Ore. Revised and updated by Robin J. Wilson35. Exploring Mathematics with Your Computer by Arthur Engel

Other titles in preparation.

PrefaceThis volume contains the Annual High School Mathematics Ex-

aminations, given 1973 through 1982. It is a continuation of ContestProblem Books I, II, III, published as Volumes 5, 17, and 25 of theNew Mathematical Library series and which contain the first twenty-three annual examinations. The Annual High School MathematicsExaminations (AHSME), it is hoped, provide challenging problemswhich teach, stimulate and provide enjoyment for not only the par-ticipants, but also the readers of these volumes.

All high school students are eligible to participate in the AnnualHigh School Mathematics Examinations. In 1982, over 418,000 stu-dents in the United States, Canada, Puerto Rico, Colombia, Jamaica,Australia, Italy, England, Hungary, Ireland, Israel, Finland, Belgiumand Luxembourg participated in the examination. It was administeredalso in many APO/FPO and other schools abroad. Each year aSummary of Results and Awards is sent to all participating highschools (in the United States and Canada). The problems are designedso that they can be solved with a knowledge of only "pre-calculus"mathematics, with emphasis on intermediate algebra and plane geome-try. The subject classification at the end of the volume indicates whichquestions are related to which topics.

The problems on each examination become progressively moredifficult. Between 1973 and 1977, the participants were given eightyminutes to complete the examination, and in subsequent years theywere allowed ninety minutes. The 1973 examination consists of fourparts containing 10, 10, 10 and 5 questions respectively worth 3, 4, 5,and 6 points each; to correct for random guessing, one fourth of thenumber of points assigned to incorrectly answered problems wasdeducted from the number of points assigned to correctly answeredproblems. The 1974 through 1977 examinations consist of 30 questionsworth five points per question; one point was deducted for eachquestion answered incorrectly. Since 1978, each examination consistsof 30 questions and was scored by adding 30 points to four times the

vii

viii PREFACE

number of correct answers and then subtracting one point for eachincorrect answer.

Each year since 1972, approximately one hundred of the highestscoring students on the AHSME and a number of members ofprevious International Mathematical Olympiad training classes havebeen invited to participate in the U.S.A. Mathematical Olympiad,currently a three and one half hour essay type examination consistingof five questions. Since 1974, a team of students has been selected toparticipate in the International Mathematical Olympiad.* An Interna-tional Mathematical Olympiad training class of approximatelytwenty-four students receives an intensive problem solving courseprior to the International Olympiad.

It is a pleasure to acknowledge the contributions of the manyindividuals and organizations who have made the preparation andadministration of these examinations possible. We thank the membersof the Committee on High School Contests and its Advisory Panel forProposing problems and suggesting many improvements in the pre-liminary drafts of the examinations. We are grateful to ProfessorStephen B. Maurer, who succeeded us as Committee Chairman in1981, for his assistance in the preparation of this book. We express ourappreciation to the regional examination coordinators throughout theUnited States and Canada who do such an excellent job of administer-ing the examinations in their regions, and to the members of theOlympiad Subcommittee who administer all the Olympiad activities.Particular thanks are due to Professor James M. Earl, who was thechairman of the Contests Committee until his death, shortly after the1973 examination was printed; to Professor Henry M. Cox, who wasthe executive director of the Contests Committee from 1973 to 1976;to Professor Walter E. Mientka, who has been the executive director ofthe Contests Committee since September 1976; and to ProfessorSamuel L. Greitzer, who has been the chairman of the OlympiadSubcommittee since the inception of the subcommittee. We expressappreciation to our sponsors, the Mathematical Association ofAmerica, the Society of Actuaries, Mu Alpha Theta, the NationalCouncil of Teachers of Mathematics, and the Casualty ActuarialSociety for their financial support and guidance; we thank the CityCollege of New York, the University of Nebraska, Swarthmore Col-lege and Metropolitan Life Insurance Company for the support theyhave provided present and past chairmen and executive directors; andwe thank L. G. Balfour Company, W. H. Freeman and Company,Kuhn Associates, National Semiconductor, Pickett, Inc., MAA, Mu

*The International Olympiads from 1959 to 1977 have been published in volume 27 ofthe New Mathematical Library series.

THE MAA PROBLEM BOOK IV

Alpha Theta, NCTM and Random House for donating awards tohigh-scoring individuals and schools on the AHSME.

The members of the Committee on High School Contests areparticularly pleased to acknowledge financial support for the U.S.A.Mathematical Olympiad and the participation of the U.S. team in theInternational Mathematical Olympiad. We express our gratitude tothe International Business Machines Corporation for an annual grantto sponsor an awards ceremony in honor of the winners of the U.S.A.Mathematical Olympiad; we thank the hosts of training sessions:Rutgers University, United States Military Academy and UnitedStates Naval Academy; we gratefully acknowledge financial supportof the training sessions and travel to the International MathematicalOlympiad from the following: Army Research Office, Johnson andJohnson Foundation, Office of Naval Research, Minnesota Mining &Manufacturing Corporation, National Science Foundation, SpencerFoundation, Standard Oil Company of California and Xerox Corpo-ration.

A few minor changes in the statements of problems have been madein this collection for the sake of greater clarity.

Ralph A. Artino

Anthony M. Gaglione

Niel Shell

ix

Editors' Preface

The editors of the New Mathematical Library, wishing to encouragesignificant problem solving at an elementary level, have published avariety of problem collections. In addition to the Annual High SchoolMathematics Examinations (NML vols. 5, 17, 25 and 29) described indetail in the Preface on the preceding pages, the NML series containstranslations of the Hungarian Eotvos Competitions through 1928(NML vols. 11 and 12) and the International Mathematical Olympiads(NML vol. 27). Both are essay type competitions containing only afew questions which often require ingenious solutions.

The present volume is a sequel to NML vol. 25 published at therequest of the many readers who enjoyed the previous Contest Prob-lem Books.

The Mathematical Association of America, publisher of the NMLseries, is concerned primarily with mathematics at the undergraduatelevel in colleges and universities. It conducts the annual PutnamCompetitions for undergraduate students. All three journals of theMAA, the American Mathematical Monthly, Mathematics Magazineand the Two Year College Mathematics Journal, have sections devotedto problems and their solutions.

The editors of the New Mathematical Library are pleased to ac-knowledge the essential contributions of R. A. Artino, A. M. Gaglioneand N. Shell, the three men who compiled and wrote solutions for theproblems in the present volume. The hard work of Stephen B. Maurer,current chairman of the MAA Committee on High School Contests,and that of other Committee members in the final editing of thiscollection is greatly appreciated.

We suggest that readers attempt their own solutions before lookingat the ones offered. Their solutions may be quite different from, butjust as good or better than, those published here.

People taking the AHSME are told to avoid random guessing, sincethere is a penalty for incorrect answers; however, if a participant can

xi

xii EDITORS' PREFACE

use his mathematical knowledge to eliminate all but one of the listedchoices, he will improve his score. A few examples of this kind ofelimination are indicated in some of the Notes appended to solutions.

Basil Gordon

Anneli Lax

1982

List of Symbols

Symbol

(x: )

D

f(x)f(a)

lx

n!

(n)

ixI

(abed)"

&an

Akan

[xi

Name and/or Meaning

set of all x such that; e.g. (x: x is a positive integer lessthan 4) is the set with members 1, 2, 3contained in; A C B means each member of A is in Bcontains: A v B means B c Anot equal tofunction f of the variable xthe value that f assigns to the constant aidentically equal; e.g. f(x) I means f(x) = I for allvalues of the variable xin number theory, for integers a, b, m, a-- b(mod m)is read "a is congruent to b mod mi" and means thata - b is divisible by m.less thanless than or equal togreater thangreater than or equal toapproximately equal to

absolute value; {xI = { if X >j 0

n factorial; n! = I * 2 *...- n

combinations of n things taken k at a time;(n) n!k k!(n - k)!

n

summation sign; E ai means a, + a2 + ... + an

00

infinite sum; a, = a, + a2 +

base n representation; an3 + bn2 + cn + d

first difference; for a sequence a,, a2 ,..., Aan meansan+ I - an

kth difference; Nlan = &an, Ak+lan = A (nkan) fork > Ithe largest integer not bigger than x

xiii

xiv THE MAA PROBLEM BOOK IV

Symbol Name and/or Meaning

a b determinant of the matrix (a b) equal to ad - bc

imaginary unit in the set of complex numbers, satisfying*2= -1; also used as a constant or as a variable (orindex) taking integer values (e.g. in E ,1 a = al + a2

+ - + a.)

z- complex conjugate of z; if z = a + ib with a, b real,then f = a - bi.

AB either line segment connecting points A and B or itslength

AB the minor circular arc with endpoints A and B4ABC either angle ABC or its measure1 is perpendicular to

is parallel tois similar tois congruent to

D2 parallelogram

I

Problems

1973 Examination

Part 1

1. A chord which is the perpendicular bisector of a radius of length12 in a circle, has length

(A) 3j3 (B) 27 (C) 6Vi (D) 12V3 (E) none of these

2. One thousand unit cubes are fastened together to form a largecube with edge length 10 units; this is painted and then separatedinto the original cubes. The number of these unit cubes whichhave at least one face painted is

(A) 600 (B) 520 (C) 488 (D) 480 (E) 400

3. The stronger Goldbach conjecture states that any even integergreater than 7 can be written as the sum of two different primenumbers.t For such representations of the even number 126, thelargest possible difference between the two primes is

(A) 112 (B) 100 (C) 92 (D) 88 (E) 80

tThe regular Goldbach conjecture states that any even integer greater than 3 isexpressible as a sum of two primes. Neither this conjecture nor the stronger version hasbeen settled.

I


4. Two congruent 30° - 60° - 90° triangles are placed so that theyoverlap partly and their hypotenuses coincide. If the hypotenuseof each triangle is 12, the area common to both triangles is

(A) 6V3 (B) 8V3 (C) 9r3 (D) 12F3 (E) 24

5. Of the following five statements, I to V, about the binary opera-tion of averaging (arithmetic mean),

I. Averaging is associativeII. Averaging is commutative

III. Averaging distributes over additionIV. Addition distributes over averagingV. Averaging has an identity element

those which are always true are

(A) All (B) I and II only (C) II and III only(D) II and IV only (E) II and V only

6. If 554 is the base b representation of the square of the numberwhose base b representation is 24, then b, when written in base10, equals

(A) 6 (B) 8 (C) 12 (D) 14 (E) 16

7. The sum of all the integers between 50 and 350 which end in I is

(A) 5880 (B) 5539 (C) 5208 (D) 4877 (E) 4566

8. If I pint of paint is needed to paint a statue 6 ft. high, then thenumber of pints it will take to paint (to the same thickness) 540statues similar to the original but only I ft. high, is

(A) 90 (B) 72 (C) 45 (D) 30 (E) 15

9. In AABC with right angle at C, altitude CH and median CMtrisect the right angle. If the area of ACHM is K, then the areaof AABC is

(A) 6K (B) 4VYK (C) 3V3 K (D) 3K (E) 4K

10. If n is a real number, then the simultaneous nx + y = 1system to the right has no solution if and only on + =if n is equal to hy + z = 1

x +nz =(A) -I (B) 0 (C) I(D) 0 or I (E) I

2

PROBLEMS: 1973 EXAMINATION

Part 2

11. A circle with a circumscribed and an inscribed square centered atthe origin 0 of a rectangular coordinate system with positive xand y axes OX and 0 Y is shown in each figure I to IV below.

IVY

~; - X

The inequalities

IXI + IY 1 2(X 2 +y 2 ) < 2Max(0x1, IYI)

are represented geometrically by the figure numbered

(A) I (B) II (C) III (D) IV (E) none of these

12. The average (arithmetic mean) age of a group consisting ofdoctors and lawyers is 40. If the doctors average 35 and thelawyers 50 years old, then the ratio of the number of doctors tothe number of lawyers is

(A) 3:2 (B) 3:1 (C) 2:3 (D) 2:1 (E) 1:2

13. The fraction ( )is equal to3 2 +

(B) 1 (C) 23 (D) 4 (E) 16

tAn inequality of the form f(x, y) S g(x, y), for all x and y is represented geometri-cally by a figure showing the containment

(The set of points (x, y) such that g(x, y) 6 a)

c (The set of points (x, y) such that f(x, y) 6 a}

for a typical real number a.

(A) 22

3

T 11


14. Each valve A, B, and C, when open, releases water into a tankat its own constant rate. With all three valves open, the tank fillsin I hour, with only valves A and C open it takes 1.5 hour, andwith only valves B and C open it takes 2 hours. The number ofhours required with only valves A and B open is

(A) 1.1 (B) 1.15 (C) 1.2 (D) 1.25 (E) 1.75

15. A sector with acute central angle 0 is cut from a circle of radius 6.The radius of the circle circumscribed about the sector is

(A) 3 cos 0 (B) 3 sec 0 (C) 3 cos 8 (D) 3 sec O (E) 3

16. If the sum of all the angles except one of a convex polygon is21900, then the number of sides of the polygon must be

(A) 13 (B) 15 (C) 17 (D) 19 (E) 21

x-117. If 8 is an acute angle and sin 20 = I, then tan 8 equals

(A) x (B) -(C) I (D) (E) x2- 1

18. If p > 5 is a prime number, then 24 divides p2 - I without re-mainder

(A) never (B) sometimes only (C) always(D) only if p = 5 (E) none of these

19. Define na! for n and a positive to be

nfa! = n(n - a)(n - 2a)(n - 3a)... (n - ka),

where k is the greatest integer for which n > ka.Then the quotient 72,!/182! is equal to

(A) 45 (B) 46 (C) 48 (D) 49 (E) 412

20. A cowboy is 4 miles south of a stream which flows due east. He isalso 8 miles west and 7 miles north of his cabin. He wishes towater his horse at the stream and return home. The shortestdistance (in miles) he can travel and accomplish this is

(A) 4 + V18- (B) 16 (C) 17 (D) I18 (E) 0_ + m13-

4


Part 3

21. The number of sets of two or more consecutive positive integerswhose sum is 100 is

(A) I (B) 2 (C) 3 (D) 4 (E) 5

22. The set of all real solutions of the inequality

Ix - 11 + Ix + 21 <5

is

(A) {x: - 3 < x < 2) (B) (x: - I < x < 2)(C) {x: -2 < x < 1} (D) {x:- < X < } (E) 0 (empty)

23. There are two cards; one is red on both sides and the other is redon one side and blue on the other. The cards have the sameprobability (2') of being chosen, and one is chosen and placed onthe table. If the upper side of the card on the table is red, then theprobability that the under-side is also red is

(A) 4L (B) 3 (C) 2 (D) 2 (E) 4

24. The check for a luncheon of 3 sandwiches, 7 cups of coffee andone piece of pie came to $3.15. The check for a luncheon consist-ing of 4 sandwiches, 10 cups of coffee and one piece of pie cameto $4.20 at the same place. The cost of a luncheon consisting ofone sandwich, one cup of coffee and one piece of pie at the sameplace will come to

(A) $1.70 (B) $1.65 (C) $1.20 (D) $1.05 (E) $.95

25. A circular grass plot 12 feet in diameter is cut by a straight gravelpath 3 feet wide, one edge of which passes through the center ofthe plot. The number of square feet in the remaining grass area is

(A) 367T - 34 (B) 30sr - 15 (C) 36v - 33(D) 35r- 9V3 (E) 307r - 93

26. The number of terms in an A.P. (Arithmetic Progression) is even.The sums of the odd- and even-numbered terms are 24 and 30respectively. If the last term exceeds the first by 10.5, the numberof terms in the A.P. is

(A) 20 (B) 18 (C) 12 (D) 10

5

(E) 8


27. Cars A and B travel the same distance. Car A travels half thatdistance at u miles per hour and half at v miles per hour. Car Btravels half the time at u miles per hour and half at v miles perhour. The average speed of Car A is x miles per hour and that ofCar B is y miles per hour: Then we always have

(A) x <y (B) x By (C) x =y (D) x <y (E) x >y

28. If a, b, and c are in geometric progression (G.P.) with I < a <b < c and n > 1 is an integer, then logan,logbn,logcn form asequence

(A) which is a G.P.(B) which is an arithmetic progression (A.P.)(C) in which the reciprocals of the terms form an A.P.(D) in which the second and third terms are the n th powers of

the first and second respectively(E) none of these

29. Two boys start moving from the same point A on a circular trackbut in opposite directions. Their speeds are 5 ft. per sec. and 9 ft.per sec. If they start at the same time and finish when they firstmeet at the point A again, then the number of times they meet,excluding the start and finish, is

(A) 13 (B) 25 (C) 44 (D) infinity (E) none of these

30. Let [t] denote the greatest integer < t where t > 0 and S ={(x, y): (x - T)2 + y 2 6 T2 where T = t - [t]}. Then we have

(A) the point (0, 0) does not belong to S for any t(B) 0 6 Area S < or for all t(C) S is contained in the first quadrant for all t > 5(D) the center of S for any t is on the line y = x(E) none of the other statements is true

Part 4

31. In the following equation, each of the letters represents uniquely adifferent digit in base ten:

(YE)- (ME) = TTT

The sum E + M + T + Y equals

(A) 19 (B) 20 (C) 21 (D) 22 (E) 24

6


32. The volume of a pyramid whose base is an equilateral triangle ofside length 6 and whose other edges are each of length VIK is

9V4(A) 9 (B) 9/2 (C) 27/2 (D) 2-3 (E) none of these

33. When one ounce of water is added to a mixture of acid and water,the new mixture is 20% acid. When one ounce of acid is addedto the new mixture, the result is 331 % acid. The percentage of acidin the original mixture is

(A) 22% (B) 24% (C) 25% (D) 30% (E) 331%

34. A plane flew straight against a wind between two towns in 84minutes and returned with that wind in 9 minutes less than itwould take in still air. The number of minutes (2 answers) for thereturn trip was

(A) 54 or 18 (B) 60 or 15 (C) 63 or 12 (D) 72 or 36(E) 75 or 20

35. In the unit circle shown in thefigure to the right, chords PQand MN are parallel to the unitradius OR of the circle withcenter at 0. Chords MP, PQand NR are each s units longand chord MN is d units long.Of the three equations

I. d-s= 1, II. ds= 1,III. d2 - S2 = C5

those which are necessarily true

R

are

(A) I only (B) II only (C) III only(D) I and II only (E) I, II, and III

7


1974 Examination

1. If x $ 0 or 4 and

to

y $ 0 or 6, then - + - = I is equivalentx y 2

(A) 4x + 3y = xy (B) y =4x6 -(y

(D) Y =x

(C) x + = 22 3

(E) none of these

2. Let xI and x2 be such that xl $ x2 and 3X2 - hxi = b,i= 1,2. Then xI + x2 equals

(A) - h (B) 3 (C) b (D) 2b (E) -b

3. The coefficient of X7 in the polynomial expansion of

(I + 2x -x2)4

is

(A) -8 (B) 12 (C) 6 (D) - 12 (E) none of these

4. What is the remainder when x51 + 51 is divided by x + 1?

(A) O (B) I (C) 49 (D) 50 (E) 51

5. Given a quadrilateral ABCD inscribed in a circle with side ABextended beyond B to point E, if 4BAD = 920 and 4ADC=68°, find 4EBC.

(A) 660 (B) 680 (C) 700 (D) 880 (E) 920

6. For positive real numbers x and y define x * y = Y * thenx +y

(A) " *" is commutative but not associative(B) "*" is associative but not commutative(C) "*" is neither commutative nor associative(D) "*" is commutative and associative(E) none of these

8


7. A town's population increased by 1,200 people, and then this newpopulation decreased by 11%. The town now had 32 less peoplethan it did before the 1,200 increase. What is the original popula-tion?

(A) 1,200 (B) 11,200 (C) 9,968 (D) 10,000(E) none of these

8. What is the smallest prime number dividing the sum 311 + 5'3?

(A) 2 (B) 3 (C) 5 (D) 311 + 513 (E) none of these

9. The integers greater than one are arranged in five columns asfollows:

2 3 4 59 8 7 6

10 11 12 1317 16 15 14

(Four consecutive integers appear in each row; in the first, thirdand other odd numbered rows, the integers appear in the last fourcolumns and increase from left to right; in the second, fourth andother even numbered rows, the integers appear in the first fourcolumns and increase from right to left.)

In which column will the number 1,000 fall?

(A) first (B) second (C) third (D) fourth (E) fifth

10. What is the smallest integral value of k such that

2x(kx - 4) _ x2 + 6 = 0

has no real roots?

(A) -1 (B) 2 (C) 3 (D) 4 (E) 5

11. If (a, b) and (c, d) are two points on the line whose equation isy = mx + k, then the distance between (a, b) and (c, d), interms of a, c and m, is

(A) la - clV/ 1+r 2 (B) la + GcI 1 2 (C) la-cla + m2 (

(D) ja - clIl + M2) (E) la - cl Iml

9


12. If g(x) = 1 x 2 and f(g(x)) = - 2

f(l/2) equals

(A) 3/4 (B) I

when x + 0, then

(C) 3 (D) V2/2 (E) F2

13. Which of the following is equivalentfalse."?

(A)(B)(C)(D)(E)

to "If P is true then Q is

"P is true or Q is false.""If Q is false then P is true.""If P is false then Q is true.""If Q is true then P is false.""If Q is true then P is true."

14. Which statement is correct?

(A) If x < 0, then x2 > x. (B) If x2 > 0, then x > 0.

(C) If x2 > x, then x > 0. (D) If x2 > x, then x < 0.

(E) If x < 1, then x2 < x.

15. If x < -2 then I - II + xlI equals

(A) 2 + x (B) -2 - x (C) x (D) -x (E) -2

16. A circle of radius r is inscribed in a right isosceles triangle, and acircle of radius R is circumscribed about the triangle. Then R/requals

(A) I + V (B) 2 + r

(D) 2

(C) r2

(E) 2(2 - )

17. If i2 = - 1, then (I + i) 20

- (I - i) 20 equals

(A) -1024 (B) -1024i (C) 0 (D) 1024 (E) 1024i

10


18. If log8 3 = p and log3 5 = q, then, in terms of p and q, log,05equals

(A) pq (B) 3p5q

(E) p2 + q2

(C) I + 3pqp +q

(D) 3pq()I + 3pq

19. In the adjoining figure ABCDis a square and CMN is anequilateral triangle. If the areaof ABCD is one square inch,then the area of CMN in squareinches is

(A) 2V3- 3

(C) V3/4(E) 4 - 2C3

M

(B) I -/3

(D) V2/3-i

D C

4 NB

20. Let

T= - I + - I +3 - a 4- r5 r; -C 6- r5 =- 2';

then

(A) T < I (B) T= I (C) I < T < 2 (D) T> 2I

(E) I =(3 - 8)(8- C7)(F7 - F)(C6- r§)(F- 2)

21. In a geometric series of positive terms the difference between thefifth and fourth terms is 576, and the difference between thesecond and first terms is 9. What is the sum of the first five termsof this series?

(A) 1061 (B) 1023 (C) 1024 (D) 768 (E) none of these

22. The minimum value of sin - - vcos2 is attained when A is2 2

(A) - 1800 (B) 600 (C) 1200 (D) 00 (E) none of these

I1I


23. In the adjoining figure TP and T'Q T 4 Pare parallel tangents to a circle ofradius r, with T and T' the pointsof tangency. PT"Q is a third tangentwith T" as point of tangency. If TP= 4 and T'Q = 9 then r is{AX 9IC le /\ / /V\ -C /AktX) /.-/O by3) 0 k%) ZJ/Q T' 9 Q(D) a number other than 25/6, 6, 25/4 T(E) not determinable from the given information

24. A fair die is rolled six times. The probability of rolling at least afive at least five times is

(A) 13/729 (B) 12/729 (C) 2/729 (D) 3/729(E) none of these

25. In parallelogram ABCD of the accompanying diagram, line DPis drawn bisecting BC at N and meeting AB (extended) at P.From vertex C, line CQ is drawn bisecting side AD at M andmeeting AB (extended) at Q. Lines DP and CQ meet at 0. Ifthe area of parallelogram ABCD is k, then the area of triangleQPO is equal to

(A) k (B) 6k/5

p

A B

26. The number of distinct positive integral divisors of (30)4 excludingI and (30)4 is


27. If f(x) = 3x + 2 for all real x, then the statement:

"If(x) + 41 < a whenever Ix + 21 < b and a > 0 and b > 0"is true when

(A) b < a/3 (B) b > a/3 (C) a < b/3 (D) a > b/3

(E) The statement is never true.

12

t-t


28. Which of the following is satisfied by all numbers x of the form

x a, + a2 + + a253 32 325'

where a, is 0 or 2, a2 is 0 or 2,..., a25 is 0 or 2?

(A) 0 < x < 1/3 (B) 1/3 < x < 2/3 (C) 2/3 < x < I

(D) 0 x < 1/3 or 2/3 < x < I (E) 1/2 < x < 3/4

29. For p = 1,2, . .. , 10 let S, be the sum of the first 40 terms of thearithmetic progression whose first term is p and whose commondifference is 2p - 1; then S, + S2 + * * + S10 is

(A) 80,000 (B) 80,200 (C) 80,400 (D) 80,600(E) 80,800

30. A line segment is divided so that the lesser part is to the greaterpart as the greater part is to the whole. If R is the ratio of thelesser part to the greater part, then the value of

R[R(R2+R-' 'R-1R + R -

is

(C) R -' (D)2+R 2 + R-

13

(A) 2 (B) 2 R (E) 2 + R


1975 Examination

1. The value of is2-

2 -2- 1

(A) 3/4 (B) 4/5 (C) 5/6 (D) 6/7 (E) 6/5

2. For which real values of m are the simultaneous equations

y = mx + 3

y = (2m - I)x + 4

satisfied by at least one pair of real numbers (x, y)?

(A) all m (B) all m $ O (C) all m $ 1/2 (D) all m $ I(E) no values of m

3. Which of the following inequalities are satisfied for all real num-bers a, b, c, x, y, z which satisfy the conditions x < a, y < b,and z < c?

I. xy + yz + zx < ab + bc + caII. X2 + y2 + Z2 < a2 + b2 + C2III. xyz < abc

(A) None are satisfied. (B) I only (C) II only(D) III only (E) All are satisfied.

4. If the side of one square is the diagonal of a second square, whatis the ratio of the area of the first square to the area of the second?

(A) 2 (B) V2 (C) 1/2 (D) 2C2 (E) 4

5. The polynomial (x + y)9 is expanded in decreasing powers of x.The second and third terms have equal values when evaluated atx = p and y = q, where p and q are positive numbers whosesum is one. What is the value of p?

(A) 1/5 (B) 4/5 (C) 1/4 (D) 3/4 (E) 8/9

6. The sum of the first eighty positive odd integers subtracted fromthe sum of the first eighty positive even integers is

(A) 0 (B) 20 (C) 40

14

(D) 60 (E) 80


7. For which non-zero real numbers x is | - a positive in-x

teger?

(A) for negative x only (B) for positive x only(C) only for x an even integer(D) for all non-zero real numbers x(E) for no non-zero real numbers x

8. If the statement "All shirts in this store are on sale." is false, thenwhich of the following statements must be true?

I. All shirts in this store are at non-sale prices.OII. There is some shirt in this store not on sale.

III. No shirt in this store is on sale.IV. Not all shirts in this store are on sale.

(A) II only (B) IV only (C) I and III only(D) II and IV only (E) I, II and IV only

9. Let a,, a 2 . - and b1, b2, ... be arithmetic progressions such thata, = 25, b, = 75 and alo + bloo = 100. Find the sum of thefirst one hundred terms of the progression a, + b , a2 + b2 .

(A) 0 (B) 100 (C) 10,000 (D) 505,000(E) not enough information given to solve the problem

10. The sum of the digits in base ten of (104n2+8 + 1)2, where n is apositive integer, is

(A) 4 (B) 4n (C) 2 + 2n (D) 4n2 (E) n2 + n + 2

11. Let P be an interior point of circle K other than the center of K.Form all chords of K which pass through P, and determine theirmidpoints. The locus of these midpoints is

(A) a circle with one point deleted(B) a circle if the distance from P to the center of K is less than

one half the radius of K; otherwise a circular arc of less than3600

(C) a semicircle with one point deleted(D) a semicircle (E) a circle

tOriginally, statement I read: All shirts in this store are not on sale.

15


12. If a + b, a3 - = 19x3 and a - b = x, which of the followingconclusions is correct?

(A) a = 3x (B) a = 3x or a =-2x(C) a = -3x or a = 2x (D) a = 3x or a = 2x (E) a = 2x

13. The equation x6 - 3x5 - 6x3 - x + 8 = 0 has

(A) no real roots(B) exactly two distinct negative roots(C) exactly one negative root(D) no negative roots, but at least one positive root(E) none of these

14. If the whatsis is so when the whosis is is and the so and so is is so,what is the whosis whatsis when the whosis is so, the so and so isso so, and the is is two (whatsis, whosis, is and so are variablestaking positive values)?

(A) whosis * is * so (B) whosis (C) is (D) so(E) so and so

15. In the sequence of numbers 1, 3, 2, . .. each term after the first twois equal to the term preceding it minus the term preceding that.The sum of the first one hundred terms of the sequence is

(A) 5 (B) 4 (C) 2 (D) 1 (E) -1

16. If the first term of an infinite geometric series is a positive integer,the common ratio is the reciprocal of a positive integer, and thesum of the series is 3, then the sum of the first two terms of theseries is

(A) 1/3 (B) 2/3 (C) 8/3 (D) 2 (E) 9/2

17. A man can commute either by train or by bus. If he goes to workon the train in the morning, he comes home on the bus in theafternoon; and if he comes home in the afternoon on the train, hetook the bus in the morning. During a total of x working days,the man took the bus to work in the morning 8 times, came homeby bus in the afternoon 15 times, and commuted by train (eithermorning or afternoon) 9 times. Find x.

(A) 19 (B) 18 (C) 17 (D) 16(E) not enough information given to solve the problem

16


18. A positive integer N with three digits in its base ten representa-tion is chosen at random, with each three digit number having anequal chance of being chosen. The probability that log2N is aninteger is

(A) 0 (B) 3/899 (C) 1/225 (D) 1/300 (E) 1/450

19. Which positive numbers x satisfy the equation (log3x)(log,5) =

log 3 5?

(A) 3 and 5 only (B) 3, 5 and 15 only(C) only numbers of the form 5 n * 3m, where n and m are posi-

tive integers(D) all positive x $ I (E) none of these

20. In the adjoining figure triangle ABC is such that AB = 4 andAC = 8. If M is the midpoint of BC and AM = 3, what is thelength of BC?

(A) 2V47 (B) 2T A(C) 9 (D) 4 + 2VP(E) not enough information

given to solve the problem B M C

21. Suppose f(x) is defined for all real numbers x; f(x) > 0 for allx; and f(a)f(b) = f(a + b) for all a and b. Which of the fol-lowing statements are true?

I. f(O) = 1 11. f(-a) = I/f(a) for all a

111. f(a) = 3f(3a) for all a IV. f(b) > f(a) if b > a

(A) III and IV only (B) 1, III and IV only(C) 1, 11 and IV only (D) I, II and III only (E) All are true.

22. If p and q are primes and x2 - px + q = 0 has distinct posi-

tive integral roots, then which of the following statements aretrue?

I. The difference of the roots is odd.11. At least one root is prime.

III. p2 - q is prime.IV. p + q is prime.

(A) I only (B) II only (C) II and III only(D) I, 11 and IV only (E) All are true.

17


23. In the adjoining figure AB and BC are D Cadjacent sides of square ABCD; M isthe midpoint of AB; N is the midpointof BC; and AN and CM intersect at0. The ratio of the area of A OCD tothe area of ABCD is

(A) 5/6 (B) 3/4 (C) 2/3

(D) J3/2 (E) (Vi - 1)/2

24. In triangle ABC, 4C = 0 and 4B = 20, where 0° < 0 < 600.

The circle with center A and radius AB intersects AC at D andintersects BC, extended if necessary, at B and at E (E maycoincide with B). Then EC = AD

(A) for no values of 0 (B) only if 0 = 450(C) only if 00 <8 < 450 (D) only if 450 < 0 < 600(E) for all 8 such that 00 < 0 < 600

25. A woman, her brother, her son and her daughter are chess players(all relations by birth). The worst player's twin (who is one of thefour players) and the best player are of opposite sex. The worstplayer and the best player are the same age. Who is the worstplayer?

(A) the woman (B) her son(C) her brother (D) her daughter(E) No solution is consistent with the given information.

26. In acute triangle ABC the bisector of gA meets side BC at D.The circle with center B and radius BD intersects side AB atM; and the circle with center C and radius CD intersects sideAC at N. Then it is always true that

(A) 4CND + 4BMD - 4DAC = 1200(B) AMDN is a trapezoid(C) BC is parallel to MN

(D) AM- AN = 3(DB - DC)2

(E) AB-AC = 3(DB DC)

27. If p, q and r are distinct roots of x 3 -X 2 + X -2 = 0, thenp3

+ q3 + r3 equals

(A) -1 (B) 1 (C) 3 (D) 5

18

(E) none of these

PROBLEMS: 1975 EXAMINATION 19

28. In triangle ABC shown in the adjoining figure, M is the mid-point of side BC, AB = 12 and AC = 16. Points E and F aretaken on AC and AB, respectively, and lines EF and AM in-tersect at G. If AE = 2AF then EG/GF equals

rC

(A) 3/2 (B) 4/3(C) 5/4 (D) 6/5(E) not enough information

given to solve the problem

29. What is the smallest integer larger than (VT + T)6?(A) 972 (B) 971 (C) 970 (D) 969 (E) 968

30. Let x = cos 360 - cos 720. Then x equals

(A) 1/3 (B) 1/2 (C) 3t-s (D) 2e-3(E) none of these


1976 Examination

1. If one minus the reciprocal of (1 - x) equals the reciprocal of(1 - x), then x equals

(A) -2 (B) -1 (C) 1/2 (D) 2 (E) 3

2. For how many real numbers x is - (x + 1)2 a real number?

(A) none (B) one (C) two(D) a finite number greater than two (E) infinitely many

3. The sum of the distances from one vertex of a square with sides oflength two to the midpoints of each of the sides of the square is

(A) 2r (B) 2 + C (C) 2 + 2V (D) 2 + r(E) 2 + 2r5

4. Let a geometric progression with n terms have first term one,common ratio r and sum s, where r and s are not zero. Thesum of the geometric progression formed by replacing each termof the original progression by its reciprocal is

(A) - (B) In (C) ___ (D) r E r'-S rTs r n S S

5. How many integers greater than ten and less than one hundred,written in base ten notation, are increased by nine when theirdigits are reversed?

(A) O (B) l (C) 8 (D) 9 (E) lo

6. If c is a real number and the negative of one of the solutions ofx2 - 3x + c = O is a solution of x2 + 3x - c = 0, then thesolutions of x2 - 3x + c = 0 are

(A) 1,2 (B) -1, -2 (C) 0,3 (D) 0, -3 (E) 2' 3

7. If x is a real number, then the quantity (1 - Ixj)(1 + x) is posi-tive if and only if

(A) {xl < I (B) x < I (C) Ixj > I (D) x <-1(E) x < -1 or -1 < x < 1

20


8. A point in the plane, both of whose rectangular coordinates areintegers with absolute value less than or equal to four, is chosen atrandom, with all such points having an equal probability of beingchosen. What is the probability that the distance from the point tothe origin is at most two units?

13 15 13 ()j(A) 8 (B) 8 (C) 64 (D) 6

(E) the square of a rational number

9. In triangle ABC, D is the midpoint of AB; E is the midpoint ofDB; and F is the midpoint of BC. If the area of AABC is 96,then the area of &AEF is

(A) 16 (B) 24 (C) 32 (D) 36 (E) 48

10. If m, n, p and q are real numbers and f(x) = mx + n andg(x) = px + q, then the equation f(g(x)) = g(f(x)) has a solu-tion

(A) for all choices of m, n, p and q(B) if and only if m = p and n = q(C) if and only if mq - np = 0(D) if and only if n(l - p) - q(l - m) = 0(E) if and only if (I-n)(l- p)-(I-q)(l -m) = 0

11. Which of the following statements is (are) equivalent to thestatement "If the pink elephant on planet alpha has purple eyes,then the wild pig on planet beta does not have a long nose"?

I. "If the wild pig on planet beta has a long nose, then the pinkelephant on planet alpha has purple eyes."

II. "If the pink elephant on planet alpha does not have purpleeyes, then the wild pig on planet beta does not have a longnose."

III. "If the wild pig on planet beta has a long nose, then the pinkelephant on planet alpha does not have purple eyes."

IV. "The pink elephant on planet alpha does not have purpleeyes, ort the wild pig on planet beta does not have a longnose."'

(A) I and III only (B) III and IV only(C) II and IV only (D) II and III only (E) III only

tThe word "or" is used here in the inclusive sense (as is customary in mathematicalwriting).

21


12. A supermarket has 128 crates of apples. Each crate contains atleast 120 apples and at most 144 apples. What is the largestinteger n such that there must be at least n crates containing thesame number of apples?

(A) 4 (B) 5 (C) 6 (D) 24 (E) 25

13. If x cows give x + I cans of milk in x + 2 days, how manydays will it take x + 3 cows to give x + 5 cans of milk?

(A) x(x + 2)(x + 5)(x + 1)(x + 3)

(C) (x + 1)(x + 3)(x + 5)x(x + 2)

(B) x(x + l)(x + 5)(x + 2)(x + 3)

(D) (x + l)(x + 3)x(x + 2)(x + 5)

(E) none of these

14. The measures of the interior angles of a convex polygon are inarithmetic progression. If the smallest angle is 1000 and the largestangle is 1400, then the number of sides the polygon has is

(A) 6 (B) 8 (C) 10 (D) 11 (E) 12

15. If r is the remainder when each of the numbers 1059, 1417 and2312 is divided by d, where d is an integer greater than one, thend - r equals

(A) I (B) 15 (C) 179 (D) d- 15 (E) d- 1

16. In triangles ABC and DEF, lengths AC, BC, DF and EF areall equal. Length AB is twice the length of the altitude of ADEFfrom F to DE. Which of the following statements is (are) true?

I. 4ACB and 4DFEII. 4 ACB and $DFE

III. The area of AABCIV. The area of AABC

must be complementary.must be supplementary.must equal the area of ADEF.must equal twice the area of ADEF.

(A) II only (B) III only (C) IV only(D) I and III only (E) II and III only

17. If 0 is an acute angle and sin 20 = a, then sinG + cosG equals

(C) a-+- a2 -a

(D) a+I+ 2 -ala 2 -a

22

(A) v G -+I (B) (C2 - I)a + I

(E) vra -+I + a2 - a


18. In the adjoining figure, AB is tangent at A to the circle withcenter 0; point D is interior to the circle; and DB intersects thecircle at C. If BC = DC = 3, OD = 2 and AB = 6, then theradius of the circle is

(A) 3 + V3

(B) 15/77

(C) 9/2

(D) 2V6

(E) 22

19. A polynomial p(x) has remainder three when divided by x - Iand remainder five when divided by x - 3. The remainder whenp(x) is divided by (x - 1)(x - 3) is

(A) x-2 (B) x+2 (C) 2 (D) 8 (E) 15

20. Let a, b and x be positive real numbers distinct from one. Then

4(o0gaX)2 + 3(logbX)2

= 8(logax)(logbX)

(A) for all values of a, b and x(C) if and only if b = a2

(E) none of these

(B) if and only if a = b2(D) if and only if x = ab

21. What is the smallest positive odd integer n such that the product

21/723/7 ... 2(2n+ 1)/7

is greater than 1000? (In the product the denominators of theexponents are all sevens, and the numerators are the successiveodd integers from I to 2n + 1.)

(A) 7 (B) 9 (C) 11 (D) 17 (E) 19

23

B


22. Given an equilateral triangle with side of length s, consider thelocus of all points P in the plane of the triangle such that the sumof the squares of the distances from P to the vertices of thetriangle is a fixed number a. This locus

(A) is a circle if a > 2(B) contains only three points if a = 2S2 and is a circle if

a > 2s 2

(C) is a circle with positive radius only if s2 < a < 2s2

(D) contains only a finite number of points for any value of a(E) is none of these

23. For integers k and n such that I < k < n, let(n) _ n!k} k!(n-k)!

Then ( -k -I )() is an integer

(A) for all k and n(B) for all even values of k and n, but not for all k and n(C) for all odd values of k and n, but not for all k and n(D) if k = 1 or n - 1, but not for all odd values of k and n(E) if n is divisible by k, but not for all even values of k and n

24. In the adjoining figure, circle K has diameter AB; circle L istangent to circle K and to AB at the center of circle K; andcircle M is tangent to circle K, to circle L and to AB. The ratioof the area of circle K to the area of circle M is

(A) 12

(B) 14

(C) 16

(D) 18 A

(E) not an integer

25. For a sequence U, u2 ,..-., define A'(un) = U+- un and, forall integers k > lAk(un) = AI(Ak-I(Un)) If un = n3 + n, thenAk(u ) = 0 for all n

(A) if k = I (B) if k = 2, but not if k = I(C) if k = 3, but not if k = 2 (D) if k = 4, but not if k = 3(E) for no value of k

24


26. In the adjoining figure, every point of circle O' is exterior tocircle 0. Let P and Q be the points of intersection of an inter-nal common tangent with the two external common tangents.Then the length of PQ is

(A) the average of thelengths of the internaland external commontangents

(B) equal to the length ofan external commontangent if and only ifcircles 0 and O'have equal radii

(C) always equal to theICnI1LIr1 VI dcli za

common tangent

(D) greater than the length of an external common tangent

(E) the geometric mean of the lengths of the internal and externalcommon tangents

27. If

Vi + 2 + Hi- - 2N = -F3 2r2

then N equals

C5 F5(A) I (B) 2V - I (C) 2 (D) 4

(E) none of these

28. Lines L1 , L2 ... , L 100 are distinct. All lines L4 n, n a positive in-teger, are parallel to each other. All lines L4,-3, n a positiveinteger, pass through a given point A. The maximum number ofpoints of intersection of pairs of lines from the complete set(LI, L 2 ,.. ., L 1 o) is

(A) 4350 (B) 4351 (C) 4900 (D) 4901 (E) 9851

25

)

26 THE MAA PROBLEM BOOK IV

29. Ann and Barbara were comparing their ages and found thatBarbara is as old as Ann was when Barbara was as old as Ann hadbeen when Barbara was half as old as Ann is. If the sum of theirpresent ages is 44 years, then Ann's age is

(A) 22 (B) 24 (C) 25 (D) 26 (E) 28

30. How many distinct ordered triples (x, y, z) satisfy the equations

x + 2y + 4z = 12xy + 4yz + 2xz = 22

xyz = 6 ?

(A) none (B) I (C) 2 (D) 4 (E) 6

PROBLEMS: 1977 EXAMINATIONS

1977 Examination

1. If y = 2x and z = 2y, then x + y + z equals

(A) x (B) 3x (C) 5x (D) 7x (E) 9x

2. Which one of the following statements is false? All equilateraltriangles are

(A) equiangular (B) isosceles (C) regular polygons(D) congruent to each other (E) similar to each other

3. A man has $2.73 in pennies, nickels, dimes, quarters and halfdollars. If he has an equal number of coins of each kind, then thetotal number of coins he has is

(A) 3 (B) 5 (C) 9 (D) 10 (E) 15

4. In triangle ABC, AB = ACand 4A = 800. If pointsD, E and F lie on sidesBC, AC and AB, respec-tively, and CE = CD an(BF = BD, then 4EDF equal

(A) 300 (B) 400(C) 50° (D) 65°fR I - -- - - trI c-anU) 1JLU11 VI tUlbr

5. The set of all points P such that the sum of the (undirected)distances from P to two fixed points A and B equals the dis-tance between A and B is

(A) the line segment from A to B(B) the line passing through A and B(C) the perpendicular bisector of the line segment from A to B(D) an ellipse having positive area(E) a parabola

27


6. If x, y and 2x + Y are not zero, then2

(2x + 2)-' [(2x)-1 + (2)

equals

(A) 1 (B) xy-' (C) x -'y (D) (xy)-(E) none of these

I7. If t= 4, then t equals

I - F

(A) (I- V-)(2 - r2)

(C) (I + 442)(l - r2)

(E) -(I + 71)(l F2)

(B) (I -472)(1

(D) (I + 4_)(l

8. For every triple (a, b, c) of non-zero realnumber

numbers, form the

a + b + c + abclal Jbi Icl labcl

The set of all numbers formed is

(A) (0) (B) (-4,0,4) (C) (-4, -2,0,2,4)(D) (-4, -2,2,4) (E) none of these

9. In the adjoining figure ,E = 400 and arc AB, arc BC and arcCD all have equal length. Find the measure of 4A CD.

(A) 100 (B) 150 .

(C) 200 (D) (2 )

(E) 300

B

A

E

10. If (3x - 1)7 = a7 x 7 + a 6 x 6 + * * * + ao, then a 7 + a6 + * * + ao

equals

(A) 0 (B) I (C) 64 (D) -64 (E) 128

I]

+

28

PROBLEMS: 1977 EXAMINATIONS

11. For each real number x, let [xl be the largest integer not exceed-ing x (i.e., the integer n such that n < x < n + 1). Which of thefollowing statements is (are) true?

I. Ix + 11 = lxi + I for all xII. [x + yl = [xl + lyJ for all x and y

III. Ixyl = Ixilyl for all x and y

(A) none (B) I only (C) I and II only(D) III only (E) all

12. Al's age is 16 more than the sum of Bob's age and Carl's age, andthe square of Al's age is 1632 more than the square of the sum ofBob's age and Carl's age. The sum of the ages of Al, Bob and Carlis

(A) 64 (B) 94 (C) 96 (D) 102 (E) 140

13. If a,, a2, a 3 ,... is a sequence of positive numbers such thatan+2 = anan+1 for all positive integers n, then the sequencea,, a2 , a 3 ,... is a geometric progression

(A) for all positive values of a, and a2(B) if and only if a, = a2 (C) if and only if a, = I(D) if and only if a 2 = 1 (E) if and only if a, = a2 = 1

14. How many pairs (m, n) of integers satisfy the equation m + n =

mn?

(A) I (B) 2 (C) 3 (D) 4 (E) more than 4

15. Each of the three circles in the adjoining figure is externallytangent to the other two, and each side of the triangle is tangent totwo of the circles. If each circle has radius three, then theperimeter of the triangle is

(A) 36 + 9V (B) 36 + 63

(C) 36 + 93 (D) 18 + 18V

(E) 45

29


16. If i2 = - 1, then the sum

cos450+ icos 1350 + *-- + ilcos(45+ 90n)0+ + i~cos3645o

equals

(A) -2 (B) - 10irV (C) 21

(D) r2 (21 - 20i) (E) r2 (21 + 20i)

17. Three fair dice are tossed (all faces have the same probability ofcoming up). What is the probability that the three numbers turnedup can be arranged to form an arithmetic progression with com-mon difference one?

~~927 54 3(A ' B'-9 (C)-2 (D)-5 (E) 3

18. If y = (1og 23)(10g34) ** (log,[n + 1]) ... (1og 3132) then

(A) 4 < y < 5 (B) y = 5 (C) 5 < y < 6(D) y = 6 (E) 6 <y < 7

19. Let E be the point of intersection of the diagonals of convexquadrilateral ABCD, and let P, Q, R and S be the centers ofthe circles circumscribing triangles ABE, BCE, CDE and ADE,respectively. Then

(A) PQRS is a parallelogram(B) PQRS is a parallelogram if and only if ABCD is a rhombus(C) PQRS is a parallelogram if and only if ABCD is a rectangle(D) PQRS is a parallelogram if and only if ABCD is a parallelo-

gram(E) none of the above are true

20. For how many paths consisting of a sequence of horizontaland/or vertical line segments, with each segment connecting apair of adjacent letters in the diagram below, is the word CON-TEST spelled out as the path is traversed from beginning to end?

(A) 63 (B) 128 C(C) 129 (D) 255 COC(E) none of these CONOC

CONTNOCCONTETNOC

CONTESETNOCCONTESTSETNOC

30

PROBLEMS: 1977 EXAMINATIONS 31

21. For how many values of the coefficient a do the equations

X2 +ax+ I =0,x2 - x - a = 0

have a common real solution?

(A) 0 (B) I (C) 2 (D) 3 (E) infinitely many

22. If f(x) is a real valued function of the real variable x, and f(x)is not identically zero, and for all a and b

f(a + b) +f(a - b) = 2f(a) + 2f(b),

then for all x and y

(A) f(O) = 1 (B) f(- x) = -f (x) (C) f (-x) = f (x)

(D) f (x + y) = f(x) + f(y) (E) there is a positive number Tsuch that f(x + T) = f (x)

23. If the solutions of the equation x2 + px + q = 0 are the cubes ofthe solutions of the equation x2 + mx + n = 0, then

(A) p = m 3 + 3mn (B) p = m 3 - 3mn (C) p + q= m3

(D) (im) - (E) none of thesen q

24. Find the sum

13 + 3-5 (2n - 1)(2n + 1) 255 - 257

127 128 1128 129(A) 2 (B) 2 (C) 2 (D) 2 (E) 25

25. Determine the largest positive integer n such that 1005! is divisi-ble by 10'.



26. Let a, b, c and d be the lengths of sides MN, NP, PQ and QM,respectively, of quadrilateral MNPQ. If A is the area of MNPQ,then

(A) A = (a )(b 2) if and only if MNPQ is convex

(B) A = (a 2 )( ) if and only if MNPQ is a rectangle

(C) A a + )( d) if and only if MNPQ is a rectangle

()A ( 2 )( 2 )gram

A ( 2 )( 2 )gram

if and only if MNPQ is a parallelo-

if and only if MNPQ is a parallelo-

27. There are two spherical balls of different sizes lying in two cornersof a rectangular room, each touching two walls and the floor. Ifthere is a point on each ball which is 5 inches from each wallwhich that ball touches and 10 inches from the floor, then the sumof the diameters of the balls is

(A) 20 inches(D) 60 inches

(B) 30 inches (C) 40 inches(E) not determined by the given information

28. Let g(x) = x5 + X4 + X3 + X2 + X + 1. What is the remainderwhen the polynomial g(x'2 ) is divided by the polynomial g(x)?

(A) 6 (B) 5 - x (C) 4-x+ x 2 (D) 3 - X + x2-X 3

(E) 2 - X + X2 - X3 + X4

29. Find the smallest integer n such that

(X 2 + y 2 + Z2) 2 < n(x 4 + y4 + Z4 )

for all real numbers x, y, and z.

(A) 2 (B) 3 (C) 4 (D) 6 (E) There is no such integer n.

32

PROBLEMS: 1977 EXAMINATIONS 33

30. If a, b and d are the lengths of a side, a shortest diagonal and alongest diagonal, respectively, of a regular nonagon (see adjoiningfigure), then

(A) d= a + b

(B) d2 = a2 + b2

(C) d2 = a2 + ab + b2

(D) b = a + d

(F) h2 = ad


1978 Examination

4 4 21. If I - -+-= 0, then - equals

x X2 x

(A) -I (B) I (C) 2 (D) -I or 2 (E) -I or - 2

2. If four times the reciprocal of the circumference of a circle equalsthe diameter of the circle, then the area of the circle is

(A) I17

(B) -IT

(C) I (D) ir (E) s72

3. For all non-zero numbers x and y such that x = l/y,

(x- )(Y+ y)

equals

(A) 2x2 (B) 2y2 (C) X2 + y2 (D) X2 - y2 (E) y2 - X2

4. If a = 1, b = 10, c = 100 and d = 1000, then

(a + b + c - d) + (a + b - c + d)

+ (a - b + c + d) + (-a + b + c + d)

is equal to

(A) 1111 (B) 2222 (C) 3333 (D) 1212 (E) 4242

5. Four boys bought a boat for $60. The first boy paid one half ofthe sum of the amounts paid by the other boys; the second boypaid one third of the sum of the amounts paid by the other boys;and the third boy paid one fourth of the sum of the amounts paidby the other boys. How much did the fourth boy pay?

(A) $10 (B) $12 (C) $13 (D) $14 (E) $15

34


6. The number of distinct pairs (x, y) of real numbers satisfyingboth of the following equations:

X = X2 +Y2

y = 2xy

is

(A) 0 (B) I (C) 2 (D) 3 (E) 4

7. Opposite sides of a regular hexagon are 12 inches apart. Thelength of each side, in inches, is

(A) 7.5 (B) 6r2 (C) 5Vr (D) 2 43 (E) 4V3

8. If x $ y and the sequences x, a,, a2 , y and x, bI, b2, b3, y eachare in arithmetic progression, then (a 2 - a1 )/(b 2 - bl) equals

(A) 2 (B) 3 (C) I (D) 4 (E) 3

9. If x < 0, then Ix - (x- 1)2 equals

(A) I (B) I-2x (C) -2x- I (D) I + 2x (E) 2x- I

10. If B is a point on circle C with center P, then the set of allpoints A in the plane of circle C such that the distance betweenA and B is less than or equal to the distance between A and anyother point on circle C is

(A) the line segment from P to B(B) the ray beginning at P and passing through B(C) a ray beginning at B(D) a circle whose center is P(E) a circle whose center is B

I1. If r is positive and the line whose equation is x + y = r is tan-gent to the circle whose equation is x + y2 = r, then r equals

(A) I (B) I (C) 2 (D) r (E) 22

35


12. In AADE,A!ADE = 1400 points B and C lie on sides AD andAE, respectively, and points A, B, C, D, E are distinct.t Iflengths AB, BC, CD and DE are all equal, then the measure of4EAD is

(A) 50 (B) 60 (C) 7.50 (D) 80 (E) 100

13. If a, b, c, and d are non-zero numbers such that c and d arethe solutions of x2 + ax + b = 0 and a and b are the solutionsof x2 + cx + d = 0, then a + b + c + d equals

(A) 0 (B) -2 (C) 2 (D) 4 (E) (-I + )/2

14. If an integer n, greater than 8, is a solution of the equationx2 - ax + b = 0 and the representation of a in the base nnumeration system is 18, then the base n representation of b is

(A) 18 (B) 28 (C) 80 (D) 81 (E) 280

15. If sinx + cosx = 1/5 and 0 < x < iT, then tanx is

(A) _ 4 (B3) - 3 (C) 3 (D) 44 3 34 4

(E) not completely determined by the given information

16. In a room containing N people, N > 3, at least one person hasnot shaken hands with everyone else in the room. What is themaximum number of people in the room that could have shakenhands with everyone else?

(A) 0 (B) I (C) N- I (D) N (E) none of these

17. If k is a positive number and f is a function such that, for everypositive number x,

[f(X2 + 1)] = k;

then, for every positive number y,

[f( +y2)]

is equal to

(A) Fk (B) 2k (C) kAk (D) k2 (E) yrk

The specification that points A, B, C, D, E be distinct was not included in the originalstatement of the problem. If B - D, then C - E and LEAD = 200.

36


18. What is the smallest positive integer n such that Cn - n -- 1< .01?

(A) 2499 (B) 2500 (C) 2501 (D) 10,000(E) There is no such integer.

19. A positive integer n not exceeding 100 is chosen in such a waythat if n < 50, then the probability of choosing n is p, and ifn > 50, then the probability of choosing n is 3p. The probabil-ity that a perfect square is chosen is

(A) .05 (B) .065 (C) .08 (D) .09 (E) .1

20. If a, b, c are non-zero real numbers such that

a + b - c a-b + c -a + b + cc b a

and

x (a + b)(b + c)(c + a)abc

and x < 0, then x equals

(A) -I (B) -2 (C) -4 (D) -6 (E) -8

21. For all positive numbers x distinct from 1,

1 1 1log3 x log4 x log5 x

equals

(A) ()(C)log6 0 x logx60 (1og3 x)(log4 x)(logsx)

(D) 12(log3 x) + (log4 x) + (log5 x)

(E) log2x + log3x + log5 x(log3x)(log1x) (1og2x)(log5x) (log2x)(log3x)


22. The following four statements, and only these, are found on acard:

(Assume each statement on the card is either true or false.) Amongthem the number of false statements is exactly

(A) O (B) I (C) 2 (D) 3 (E) 4

23. Vertex E of equilateral triangle ABE is in the interior of squareABCD, and F is the point of intersection of diagonal BD andline segment AE. If length AB is 1 + Vi then the area ofA&ABF is

F2 ,/ D C(A) 1 (l) - (C) 2

2 1 2)

A B

24. If the distinct non-zero numbers x(y - z), y(z - x), z(x - y)form a geometric progression with common ratio r, then r satis-fies the equation

(A) r 2 r+ I = 0 (B) r 2 -r + I = 0 (C) r4 + r2-I = 0(D) (r + 1) 4 + r = O (E) (r- l) 4 + r = 0

25. Let a be a positive number. Consider the set S of all pointswhose rectangular coordinates (x, y) satisfy all of the followingconditions:

ai < x < 2 a (i)a < y < 2a (iii) x + y > a

(iv) x + a > y (v) y + a > x

The boundary of set S is a polygon with

(A) 3 sides (B) 4 sides (C) 5 sides (D) 6 sides(E) 7 sides

On this card exactly one statement is false.On this card exactly two statements are false.On this card exactly three statements are false.On this card exactly four statements are false.

38


26. In AABC,AB= 10, AC= 8 and BBC = 6. Circle P is the circle withsmallest radius which passesthrough C and is tangent to AB.Let Q and R be the points of in-tersection, distinct from C, ofcircle P with sides AC and BC,respectively. The length of segmentQR is -A

(A) 4.75 (B) 4.8 (C) 5 (D) 4F2 (E) 3V3

27. There is more than one integer greater than I which, when dividedby any integer k such that 2 < k < 11, has a remainder of 1.What is the difference between the two smallest such integers?

(A) 2310 (B) 2311 (C) 27,720 (D) 27,721(E) none of these

28. If AAA 2A3 is equilateral and A3A, 3 is the midpoint of line seg-ment AAn+I for all positive in-tegers n, then the measure ofCA44A 45 A43 equals

(A) 300 (B) 450 (C) 600(D) 900 (E) 1200

Al A4 A2

29. Sides AB, BC, CD and DA, respectively, of convex quadrilateralABCD are extended past B, C, D and A to points B', C', D'and A'. Also, AB = BB' = 6, BC = CC' = 7, CD = DD' = 8and DA = AA' =9; and the area of ABCD is 10. The area ofA'B'C'D' is

(A) 20 (B) 40 (C) 45 (D) 50 (E) 60

30. In a tennis tournament, n women and 2n men play, and eachplayer plays exactly one match with every other player. If thereare no ties and the ratio of the number of matches won by womento the number of matches won by men is 7/5, then n equals

(C) 6 (D) 7 (E) none of these

39

(A) 2 (B) 4


1979 Examination

1. If rectangle ABCD has area 72 square meters and E and G arethe midpoints of sides AD and CD, respectively, then the area ofrectangle DEFG in square meters is

(A) 8 (B) 9

(D) 18 (E) 24

(C) 12

2. For all non-zero real numbers x1 1--- equals

(A) - (B) IX x - y

and y such that x - y = xy,

(C) 0 (D) -1 (E) y - x

3. In the adjoining figure, ABCD is a square, ABE is an equilateraltriangle and point E is outside square ABCD. What is the mea-sure of 4 AED in degrees?

(A) 10

(B) 12.5

(C) 15

(D) 20

(E) 25 C B

4. For all real numbers x, x[x{x(2 - x) - 4) + 10] + I =

(A)(B)(C)(D)(E)

-X 4 + 2x 3 + 4x2 +- X4- 2x3 + 4x 2 +-x 4 - 2x 3 -4x2 +-X 4 -2x 3 -4x 2 -- X

4 + 2x 3- 4x2 +

lox + Ilox + Ilox + Ilox + Ilox+ I

5. Find the sum of the digits of the largest even three digit number(in base ten representation) which is not changed when its unitsand hundreds digits are interchanged.

40

v

(A) 22 (B) 23 (C) 24 (D) 25 (E) 26


3 5 9 17 33 656. 2 + 4 + 8 + 16 + 32 + 64

(A) - 4 (1) - 6 (C) 0 (D) I64 16 16

7. The square of an integer is called a perfect square. Ifsquare, the next larger perfect square is

(A) x + I (B) x 2 + I (C) x 2 + 2x + I

(E) x + 2rx + 1

8. Find the area of the smallest region bounded byy = jxj and x2 + y2 = 4.

(C) X (D) 3

(E) 1

x is a perfect

(D) x 2 + x

the graphs of

(E) 2ff

9. The product of /4 and 4 equals

(A) g-1 (B) 2-12/ (C) 1372 (D) 1~32 ( 21~2m

10. If P1P2 P3 P4 P5 P6 is a regular hexagon whose apothem (distancefrom the center to the midpoint of a side) is 2, and Qj is themidpoint of side PjPj+I for i = 1, 2, 3, 4, then the area ofquadrilateral Q1Q2 Q3 Q4 is

(A) 6 (B) 2F6 (C) 8 (D) 3V3 (E) 4C3

11. Find a positive integral solution to the equation

I + 3 + 5 + - - + (2n - 1) 115

2 + 4 + 6 + - + 2n 116

(A) 110 (B) 115 (C) 116 (D) 231(E) The equation has no positive integral solutions.

41

397(A) ' (B)

4 T


12. In the adjoining figure, CD is the diameter of a semi-circle withcenter 0. Point A lies on the extension of DC past C; point Elies on the semi-circle, and B is the point of intersection (distinctfrom E) of line segment AE with the semi-circle. If length ABequals length OD, and the measure of ZEOD is 450, then themeasure of 4BAO is

(A) 100 (B) 150

(C) 200 (D) 250

(E) 30°

A

13. The inequality y - x < v is satisfied if and only if

(A) y < 0 or y < 2x (or both inequalities hold)(B) y > 0 or y < 2x (or both inequalities hold)(C) y2 < 2xy (D) y < 0 (E) x > 0 and y < 2x

14. In a certain sequence of numbers, the first number is 1, and, forall n > 2, the product of the first n numbers in the sequence isn2. The sum of the third and the fifth numbers in the sequence is

(A) 5 (B) 31 (C) 51 (D) 576 (E) 34

15. Two identical jars are filled with alcohol solutions, the ratio of thevolume of alcohol to the volume of water being p: 1 in one jarand q: 1 in the other jar. If the entire contents of the two jars aremixed together, the ratio of the volume of alcohol to the volume ofwater in the mixture is

(A) P 2 4 (B) p I q (C) pq

(D) 2(p 2 + pq + qq) (E) p + q + 2pq3(p +q) p +q +2

16. A circle with area A, is contained in the interior of a larger circlewith area A, + A2 . If the radius of the larger circle is 3, and ifA1, A2 , A, + A2 is an arithmetic progression, then the radius ofthe smaller circle is

(A) C (B) I (C) 2 (D) 3 (E) C2

42


17. Points A, B, C and D are distinct and lie, in the given order, ona straight line. Line segments AB, AC and AD have lengths x, yand z, respectively. If line segments AB and CD may be rotatedabout points B and C, respectively, so that points A and Dcoincide, to form a triangle with positive area, then which of thefollowing three inequalities must be satisfied?

I. x< 2 Z2

II.y < x +2

<z2 A B C D

(A) I only (B) II only (C) I and II only(D) II and III only (E) I, II and III

18. To the nearest thousandth, log, 02 is .301 and log103 is .477.Which of the following is the best approximation of log510?

8 9 10 7 12

19. Find the sum of the squares of all real numbers satisfying theequation

X2 5 6 - 25632 = 0.

(A) 8 (B) 128 (C) 512 (D) 65,536 (E) 2(25632)

20. If a = l and (a + 1)(b + 1) = 2, then the radian measure ofArctan a + Arctan b equals

(A) 2 (B) 3 (C) (D) 5 (E)6

21. The length of the hypotenuse of a right triangle is h, and theradius of the inscribed circle is r. The ratio of the area of thecircle to the area of the triangle is

(A) - (B) - (C) (D) irrh + 2r h + r 2h + r h2 + r'

(E) none of these

22. Find the number of pairs (m, n) of integers which satisfy theequation m3 + 6m2 + 5m = 27n3 + 9n2 + 9n + 1.

(A) 0 (B) I (C) 3 (D) 9 (E) infinitely many


23. The edges of a regular tetrahedron with vertices A, B, C and Deach have length one. Find the least possible distance between apair of points P and Q, where P is on edge AB and Q is onedge CD.

(A) I (B) 3

(C) -2 (D) -y

(E) 3A

24. Sides AB, BC and CD of simplee) quadrilateral ABCD havelengths 4, 5 and 20, respectively. If vertex angles B and C areobtuse and sin C =-cos B = 5, then side AD has length

(A) 24 (B) 24.5 (C) 24.6 (D) 24.8 (E) 25

25. If ql(x) and r1 are the quotient and remainder, respectively,when the polynomial x8 is divided by x + ', and if q2 (x) andr2 are the quotient and remainder, respectively, when ql(x) isdivided by x + 4, then r2 equals

(A) 256 (B) - (C) I (D) -16 (E) 256

26. The function f satisfies the functional equation

f(x) +f(y) =f(x +y) -xy- 1for every pair x, y of real numbers. If f(l) = 1, then the numberof integers n + I for which f(n) = n is

(A) 0 (B) 1 (C) 2 (D) 3 (E) infinite

27. An ordered pair (b, c) of integers, each of which has absolutevalue less than or equal to five, is chosen at random, with eachsuch ordered pair having an equal likelihood of being chosen.What is the probability that the equation x2 + bx + c = 0 willnot have distinct positive real roots?

(A) 106 (B) 108 (C) 10- (D) 12- (E) none of these

tA polygon is called " simple" if it is not self intersecting.

44


28. Circles with centers A, B and C each have radius r, where1 < r < 2. The distance between each pair of centers is 2. If B' isthe point of intersection of circle A and circle C which is outsidecircle B, and if C' is the point of intersection of circle A andcircle B which is outside circle C, then length B'C' equals

(A) 3r - 2

(B) r2

(C) r + F3(r-1)

(D) I + r3(r2 - 1)

(E) none of these

29. For each positive number x, let

+x 1) - (X6 + )

A x) =

( X ) ( 3 )

The minimum value of f (x) is

(A) I (B) 2 (C) 3 (D) 4 (E) 6

30. In AABC, E is the midpoint of side BC and D is on side AC.If the length of AC is 1 and 2BAC = 600, 4ABC = 100°,4ACB = 200and .DEC = 80°, then the area of AABC plustwice the area of ACDE equals

(A) I-cos 10° (B) 1000 E1 1

(C) I cos 400 (D) I cos 500 680 200A D C

1(E) -

45

)


1980 Examination

1. The largest whole number such that seven times the number is lessthan 100 is

(A) 12 (B) 13 (C) 14 (D) 15 (E) 16

2. The degree of (X 2 + 1)4 (x 3

+ 1)3 as a polynomial in x is

(A) 5 (B) 7 (C) 12 (D) 17 (E) 72

3. If the ratio of 2x - y to x + y is 3 what is the ratio of

x toy?

(A) I (B)5 (C) I (D) 5 (E) 4

4. In the adjoining figure, CDE is an equilateral triangle and ABCDand DEFG are squares. The measure of 4GDA is

(A) 900

(B) 1050

(C) 1200

(D) 1350

(E) 1500

B F

C E

5. If AB and CD are perpendicular diameters of circle Q, andQPC = 600, then the length of PQ divided by the length of A Q

is

(A) -y

(C) r2

(B) F3

(D) 2

C

A B

(E) 2D

A


6. A positive number x satisfies the inequality rx < 2x if andonly if

(A) X > 4 (B) x > 2 (C) x > 4 (D) X < I (E) X < 41 4

7. Sides AB, BC, CD and DA of convex quadrilateral ABCD havelengths 3, 4, 12, and 13, respectively; and 4CBA is a right angle.The area of the quadrilateral is C

(A) 32 (B) 36 12(C) 39 (D) 42 B(E) 48 D 13 A

8. How many pairs (a, b) of non-zero real numbers satisfy theequation

1 1 1 9a b a+b '

(A) none (B) I (C) 2 (D) one pair for each b $ 0(E) two pairs for each b $ 0

9. A man walks x miles due west, turns 1500 to his left and walks 3miles in the new direction. If he finishes at a point r1 miles fromhis starting point, then x is

(A) V3 (B) 2r3 (C) 3 (D) 3

(E) not uniquely determined by the given information

10. The number of teeth in three meshed circular gears A, B, C arex, y, z, respectively. (The teeth on all gears are the same size andregularly spaced as in the figure.) The angular speeds, in revolu-tions per minute, of A, B, C are in the proportion

(A) x:y:z (B) z:y:x (C) y:z:x (D) yz:xz: xy

(E) xz:yx:zy

A ~BC

47


11. If the sum of the first 10 terms and the sum of the first 100 termsof a given arithmetic progression are 100 and 10, respectively, thenthe sum of the first 110 terms is

(A) 90 (B) -90 (C) 110 (D) -110 (E) -100

12. The equations of L, and L2 are y = mx and y = nx, respec-tively. Suppose LI makes twice as large an angle with the hori-zontal (measured counterclockwise from the positive x-axis) asdoes L2 , and that L, has 4 times the slope of L2. If L1 is nothorizontal, then mn is

(A) C2 (B) - 2 (C) 2 (D) -2


13. A bug (of negligible size) starts at the origin on the co-ordinateplane. First it moves 1 unit right to (1,0). Then it makes a 900turn counterclockwise and travels 2 a unit to (1, -). If it con-tinues in this fashion, each time making a 90° turn counterclock-wise and traveling half as far as in the previous move, to which ofthe following points will it come closest?

( 3 3 ) ()(5 5 ) ()(3 5 ) ()(3 3)

(E) 2 4)

14. If the function f defined by

cx 3f (X) = 2x + 3 x + - 2 X c a constant,

satisfies f(f(x)) x for all real numbers x except - 2, thenC is

(A) - 3 (B) 3 (C) 3 (D) 33 3i


15. A store prices an item in dollars and cents so that when 4% salestax is added no rounding is necessary because the result is exactlyn dollars, where n is a positive integer. The smallest value of n is

(A) 1 (B) 13 (C) 25 (D) 26 (E) 100

48


16. Four of the eight vertices of a cube are vertices of a regulartetrahedron. Find the ratio of the surface area of the cube to thesurface area of the tetrahedron.

(A) fly (B) r

(C) 4(E) 2

(D) f2

17. Given thatteger?

i2 = - 1, for how many integers n is (n + i)4 an in-

(A) none (B) I (C) 2 (D) 3 (E) 4

18. If b > 1, sin x > 0, cos x > 0 and logbsin x = a, thenlogh cos x equals

(A) 2lOgb(l -ba/ 2 )

(D) I logb(l - b2 a)

(B) -a 2 (C) ba2

(E) none of these

19. Let Cl, C2 and C3 be three parallel chords of a circle on thesame side of the center. The distance between C, and C2 is thesame as the distance between C2 and C3. The lengths ofthe chords are 20, 16 and 8. The radius of the circle is

(A) 12 (B) 4V7 (C) 5/6 (D) 'I3 2


20. A box contains 2 pennies, 4 nickels and 6 dimes. Six coins aredrawn without replacement, with each coin having an equalprobability of being chosen. What is the probability that the valueof the coins drawn is at least 50 cents?

(A) 37 (B) 91 (C) 927 (D) 924

(E) none of these

49


21. In triangle ABC, 4CBA = 72°, E is the midpoint of side AC,and D is a point on side BC such that 2BD = DC; AD andBE intersect at F. The ratio of the area of A BDF to the area ofquadrilateral FDCE is

(A) I (B) 41 2

(C) I (D) 2

(E) none of these

B

22. For each real number x, let f(x) be the minimum of the num-bers 4x + 1, x + 2, and -2x + 4. Then the maximum value off(X) is

(A) I (B) I (C) 2 (D) 2 (E) 3

23. Line segments drawn from the vertex opposite the hypotenuse of aright triangle to the points trisecting the hypotenuse have lengthssin x and cos x, where x is a real number such that 0 < x < -

The length of the hypotenuse is

(A) 4 (B) 3 (C) 5 (D) 2


24. For some real number r, the polynomial 8x3 - 4X2 - 42x + 45is divisible by (x - r)2 . Which of the following numbers is closestto r?

(A) 1.22 (B) 1.32 (C) 1.42 (D) 1.52 (E) 1.62

25. In the non-decreasing sequence of odd integers (a,, a2 , a3 ,... } =(1,3,3,3,5,5,5,5,5.... ) each positive odd integer k appears ktimes. It is a fact that there are integers b, c and d such that, forall positive integers n,

an = b[ n c] + d,

where Ix] denotes the largest integer not exceeding x. The sumb + c + d equals

(A) O (B) I (C) 2 (D) 3 (E) 4

50


26. Four balls of radius 1 are mutually tangent, three resting on thefloor and the fourth resting on the others. A tetrahedron, each ofwhose edges has length s, is circumscribed around the balls. Thens equals

(A) 4r2 (B) 4V3 (C) 2C6 (D) I + 2C6 (E) 2 + 2C6

3 327. The sum T5 +2Vi + 15 -2Vy equals

3 __- )+V3

(A) 2 (B) 4 (C) 2 (D V

(E) none of these

28. The polynomial X2, + I + (x + 1)2n is not divisible byx2 +x + I if n equals

(A) 17 (B) 20 (C) 21 (D) 64 (E) 65

29. How many ordered triples (x, y, z) of integers satisfy the systemof equations below?

x2- 3xy +2y2 -Z2 = 31,-x2 +6yz +2Z2 = 44,

X2+ xy +8z 2 =100.

(A) 0 (B) I (C) 2 (D) a finite number greater than two(E) infinitely many

30. A six digit number (base 10) is squarish if it satisfies the followingconditions:

(i) none of its digits is zero;(ii) it is a perfect square; and

(iii) the first two digits, the middle two digits and the last twodigits of the number are all perfect squares when consideredas two digit numbers.

How many squarish numbers are there?

(C) 3 (D) 8 (E) 9

51

(A) 0 (B) 2


1981 Examination

1. If x±2= 2 then (x + 2)2 equals

(A) W (B) 2 (C) 4 (D) 8 (E) 16

2. Point E is on side AB of square ABCD. If EB has length oneand EC has length two, then the area of the square is

(A) A (B) V D, C

(C) 3 (D) 21i

(E) 5

3. For x * 0, 1 +1 + equalsx 2x 3x

1 1 C 5 11I(A) 2x (B) I1 ()6 (D) 6x (E) 3

4. If three times the larger of two numbers is four times the smallerand the difference between the numbers is 8, then the larger ofthe two numbers is

(A) 16 (B) 24 (C) 32 (D) 44 (E) 52

5. In trapezoid ABCD, sides AB and CD are parallel, and diago-nal BD and side AD have equal length. If 4DCB = 1100 and2CBD = 300, then 4ADB =

(A) 800 (B) 900

(C) 1000 (D) 1100

(E) 1200

D C

A B

X - = y 2 + 2y -I6IfX -1Iy+2 2,t tennxequals

(A) y2 + 2y- I

(D) y2 + 2y + I

(B) y2 + 2y - 2

(E) - y2- 2y + 1(C) y2 + 2y + 2

52


7. How many of the first one hundred positive integers are divisibleby all of the numbers 2, 3, 4, 5?

(A) 0 (B) I (C) 2 (D) 3 (E) 4

8. For all positive numbers x, y, z, the product

(x+y+z) (x '+y + z-')(xy+yz+zx) '[(xy) +(yz) +±(zx) I]

equals

(A) x- 2y- 2z-2

(D) Ixyz

(B) X-2 + y-2 + Z-2

(E) Ixy + yz + zx

(C) (X + y + Z)-2

Q9. In the adjoining figure, PQ is adiagonal of the cube. If PQ haslength a, then the surface area ofthe cube is

(A) 2a2

(C) 2V3a 2

(E) 6a 2

(B) 2V2a 2

(D) 3V3a 2

P

10. The lines L and K are symmetric to each other with respect tothe line y = x. If the equation of line L is y = ax + b witha + 0 and b + 0, then the equation of K is y =

(A) -x + ba (B) - l-x+ba (C) 1 ba a

(D) -x +- (E) -x--a a a a

11. The three sides of a right triangle have integral lengths which forman arithmetic progression. One of the sides could have length

(A) 22 (B) 58 (C) 81 (D) 91 (E) 361

12. If p, q and M are positive numbers and q < 100, then the num-ber obtained by increasing M by p% and decreasing the resultby q% exceeds M if and only if

(A) p > q (B) p > 1O0q

(D) p > 100q

(C) p > q1 - q

(E) p > 1 Oq100 - q

53


13. Suppose that at the end of any year, a unit of money has lost 10%of the value it had at the beginning of that year. Find the smallestinteger n such that after n years the unit of money will have lostat least 90% of its value. (To the nearest thousandth log103 is.477.)

(A) 14 (B) 16 (C) 18 (D) 20 (E) 22

14. In a geometric sequence of real numbers, the sum of the first twoterms is 7, and the sum of the first six terms is 91. The sum ofthe first four terms is

(A) 28 (B) 32 (C) 35 (D) 49 (E) 84

15. If b > 1, x > 0 and (2x)10gb2 - (3x)1 ogb3

= 0, then x is

(A) 26 (B) 6 (C) I (D) 6

(E) not uniquely determined

16. The base three representation of x is

12112211122211112222.The first digit (on the left) of the base nine representation of x is

(A) I (B) 2 (C) 3 (D) 4 (E) 5

17. The function f is not defined for x = 0, but, for all non-zero realnumbers x,

f (x) + 2f (-) = 3x. The equation f (x) = f (-x) is satisfied by

(A) exactly one real number(B) exactly two real numbers(C) no real numbers(D) infinitely many, but not all, non-zero real numbers(E) all non-zero real numbers

18. The number of real solutions to the equationx-0 = sin x

100 sn

is

(A) 61 (B) 62 (C) 63 (D) 64 (E) 65

54


19. In AABC, M is the midpoint of side BC, AN bisects 4BAC,BN I AN and 0 is the measure of $BAC. If sides AB and AChave lengths 14 and 19, respectively, then length MN equals

(A) 2 (B) 2

(C) 2 - sin G

(D) 2 - 2 sin O

(E) 5 1 i()2 2 (2)a

20. A ray of light originates from point A and travels in a plane,being reflected n times between lines AD and CD, before strik-ing a point B (which may be on AD or CD) perpendicularlyand retracing its path to A. (At each point of reflection the lightmakes two equal angles as indicated in the adjoining figure. Thefigure shows the light path for n = 3.) If 4CDA = 8°, what isthe largest value n can have?

(A) 6 (B) 10 (C) 38 (D) 98(E) There is no largest value.

21. In a triangle with sides of lengths a, b and c,

(a + b + c)(a + b - c) = 3ab.

The measure of the angle opposite the side of length c is

(A) 150 (B) 300 (C) 450 (D) 600 (E) 1500

22. How many lines in a three-dimensional rectangular coordinatesystem pass through four distinct points of the form (i, j, k),where i, j and k are positive integers not exceeding four?

(A) 60 (B) 64 (C) 72 (D) 76 (E) I 00


23. Equilateral AABC is inscribed in a circle. A second circle istangent internally to the circumcircle at T and tangent to sidesAB and AC at points P and Q. If side BC has length 12, thensegment PQ has length A

(A) 6

(B) 6r3(C) 8

(D) 8V3(E) 9

B C

T

24. If 0 is a constant such that 0 < O < 7T and x +

then for each positive integer n, x' + I equals

(A) 2 cos e(D) 2cosn6

(B) 2'cos 0(E) 2n cos,9

I- = 2 cos 0,x

(C) 2 cos'O

25. In triangle ABC in the adjoining figure, AD and AE trisect4BAC. The lengths of BD, DE and EC are 2, 3, and 6, re-spectively. The length of the shortest side of AABC is

(A) 2vi0

(B) 1 I

(C) 6C6

(D) 6

(E) not uniquely determinedby the given information

A

CB D E

26. Alice, Bob and Carol repeatedly take turns tossing a die. Alicebegins; Bob always follows Alice; Carol always follows Bob; andAlice always follows Carol. Find the probability that Carol will bethe first one to toss a six. (The probability of obtaining a six onany toss is 6, independent of the outcome of any other toss.)

( 2 (C) 5 )25 36(A) I (B) ~ C ~ (D)-j (E)

56


27. In the adjoining figure triangle ABC is inscribed in a circle. PointD lies on AC with D = 300, and point G lies on BAwith BG > GA. Side AB and side AC each have length equal tothe length of chord DG, and 2CAB = 300. Chord DG inter-sects sides AC and AB at E and F, respectively. The ratio ofthe area of AAFE to the area of AABC is

(A) 2 -

(C) 7Vi - 12

(B) 2/3 - 3

(D) 3V-5

A _.

9- 5V(E) 9 C

'-

28. Consider the set of all equations x3 + a2x2 + ax + a( = 0,where a2 , al, a0 are real constants and laxI < 2 for i = 0,1,2.Let r be the largest positive real number which satisfies at leastone of these equations. Then

(A) I < r < 32

(B) 3 < r < 2 (C) 2 <52

(D) 5 < r < 3 (E) 3 < r < 7

29. If a > 1, then the sum of the real solutions of

a- a+-x = x

is equal to

(A) V-I (B) 2

(D) a -I

(C) -- I

(E) - 3 -I() 2

30. If a, b, c, d are the solutions of the equationthen an equation whose solutions are

a+b+c a+b+d a+c+dd2 C2 b2

is

(A) 3x4 + bx + I = 0(C) 3x4 + bx3 - 1 = 0(E) none of these

x4 - bx - 3 = 0,

b+c+d

a2

(B) 3x 4 - bx + 1 = 0

(D) 3x 4 - bx 3 -I = 0

57


1982 Examination

1. When the polynomial X3 - 2 is divided by the polynomialx2- 2, the remainder is

(A) 2 (B) -2 (C) -2x-2 (D) 2x + 2 (E) 2x - 2

2. If a number eight times as large as x is increased by two, thenone fourth of the result equals

(B) x + 2

(E) 2x + 16

3. Evaluate (xx)(x ) at x = 2.

(A) 16 (B) 64 (C) 256

(C) 2x + 2 (D) 2x + 4

(D) 1024 (E) 65,536

4. The perimeter of a semicircular region, measured in centimeters, isnumerically equal to its area, measured in square centimeters. Theradius of the semicircle, measured in centimeters, is

(A) or (B)2- (C) I (D)2 (E)-+ 2

5. Two positive numbers x and y are in the ratio a: b, where 0 <a < b. If x + y = c, then the smaller of x and y is

(A) ba (B) b

(E) ac()b - a

(C) ac (D) bca +b a +b

6. The sum of all but one of the interior angles of a convex polygonequals 25700. The remaining angle is

(A) 90° (B) 1050 (C) 1200 (D) 130° (E) 1440

7. If the operation x * y is defined by x * y = (x + 1)(y + 1) - 1,then which one of the following is false?

(A) x * y = y * x for all real x and y.(B) x *(y + z) = (x * y) + (x * z) for all real x, y, and z.(C) (x - l)*(x + I) = (x * x) - I for all real x.(D) x * 0 = x for all real x.(E) x *(y * z) = (x * y)* z for all real x, y, and z.

(A) 2x + 1

58


8. By definition r! = r(r - 1) 1 and

= k!(j - k)!'

where r, j, k are positive integers and k <j. If (7)(2), (3)form an arithmetic progression with n > 3, then n equ s

(A) 5 (B) 7 (C) 9 (D) 11 (E) 12

9. A vertical line divides the triangle with vertices (0,0), (1, 1) and(9, 1) in the xy-plane into two regions of equal area. The equationof the line is x =

(A) 2.5 (B) 3.0 (C) 3.5 (D) 4.0 (E) 4.5

10. In the adjoining diagram, BO bisects 4CBA, CO bisects 4ACB,and MN is parallel to BC. If AB = 12, BC = 24, and AC = 18,then the perimeter of ALAMN is

(A) 30 (B) 33

(C) 36 (D) 39

(E) 42

A

N

11. How many integers with four different digits are there between1,000 and 9,999 such that the absolute value of the differencebetween the first digit and the last digit is 2?

(A) 672 (B) 784 (C) 840 (D) 896 (E) 1,008

12. Let f(x) = ax7 + bx3 + cx - 5, where a, b

stants. If f( -7) = 7, then f(7) equals

(A) - 17 (B) -7 (C) 14 (D) 21(E) not uniquely determined

and c are con-

13. If a > 1, b > I and p = ,gbl(gba) then aP equalslogba

(A) 1 (B) b (C) logb (D) logba

59

(E) alogba


14. In the adjoining figure, points B and C lie on line segment AD,and AB, BC and CD are diameters of circles 0, N and P, re-spectively. Circles 0, N and P all have radius 15, and the lineAG is tangent to circle P at G. If AG intersects circle N atpoints E and F, then chord EF has length

(A) 20 (B) 15C2(C) 24 (D) 25(E) none of these G

A

15. Let [z I denote the greatest integer not exceeding z. Let x and ysatisfy the simultaneous equations

y = 21x1 + 3

y = 31x - 21 + 5.

If x is not an integer, then x + y is

(A) an integer(C) between -4 and 4(E) 16.5

(B) between 4 and 5(D) between 15 and 16

16. In the adjoining figure, a wooden cube has edges of length 3meters. Square holes of side one meter, centered in each face, arecut through to the opposite face. The edges of the holes areparallel to the edges of the cube. The entire surface area includingthe inside, in square meters, is

(A) 54 (IS) 12

(C) 76 (D) 84

(E) 86

17. How many real numbers x satisfy the equation

32x+2 - 3x+3 - 3x + 3 = 0 ?

(C) 2 (D) 3 (E) 4

60

(A) 0 (B) I


18. In the adjoining figure of a rectangular solid, 4DHG = 450 andFHB = 600. Find the cosine of BHD.

(A) C3 (B) C2 (E) A- C(C) A (D) C

A

H F

19. Let f(x) = Ix -21 + Ix -41- 12x -61, for 2 < x < 8. The sumof the largest and smallest values of f(x) is

(A) I (B) 2 (C) 4 (D) 6 (E) none of these

20. The number of pairs of positive integers (x, y) which satisfy theequation X2 +y2 = X 3 is

(A) 0 (B) 1 (C) 2 (D) not finite (E) none of these

21. In the adjoining figure, the triangle ABC is a right triangle with4BCA = 90°. Median CM is perpendicular to median BN, andside BC = s. The length of BN is

(A) sC2 (B) 2sC (C) 2sC2 (D) 2 (E) sC

C

SN

B AM

61


22. In a narrow alley of width w a ladder of length a is placed withits foot at a point P between the walls. Resting against one wallat Q, a distance k above the ground, the ladder makes a 450angle with the ground. Resting against the other wall at R, adistance h above the ground, the ladder makes a 750 angle withthe ground. The width w is equal to

(A) a

(C) k

(E) h

(B) RQ

(D) h + k

R

h

w

23. The lengths of the sides of a triangle are consecutive integers, andthe largest angle is twice the smallest angle. The cosine of thesmallest angle is

(A) (B) (D) 94 (B) 3 1(C (E) none of these

24. In the adjoining figure, the circle meets the sides of an equilateraltriangle at six points. If AG = 2, GF = 13, FC = I and HJ = 7,then DE equals

(A) 2V2 (B) 7V3

(C) 9

(E) 13

(D) 10

25. The adjoining figure is a map of part of a city: the smallrectangles are blocks and the spaces in between are streets. Eachmorning a student walks from intersection A to intersection B,always walking along streets shown, always going east or south.For variety, at each intersection where he has a choice, he chooseswith probability 2 (independent of all other choices) whether togo east or south. Find the probability that, on any given morning,he walks through intersection C.

62


(A) 32 (B) I a l I L

42 21 ~(C) 4 (D) 21 L N

(E) 3 [ill [WLI E 1__

I I ED

,IS

26. If the base 8 representation of a perfect square is ab3c, wherea + 0, then c is

(A) (0) (B) I (C) 3 (D) 4 (E) not uniquely determined

27. Suppose z = a + bi is a solution of the polynomial equation

C4 Z4 + iC3 Z3 + C2 Z2 + iCI Z + Co = 0,

where cO, c1, C2 , C3 , C4 , a and b are real constants and i2 = - 1.Which one of the following must also be a solution?

(A) - a - bi (B) a - bi (C) -a + bi (D) b + ai(E) none of these

28. A set of consecutive positive integers beginning with I is writtenon a blackboard. One number is erased. The average (arithmeticmean) of the remaining numbers is 35 7. What number waserased?

(A) 6 (B) 7 (C) 8 (D) 9 (E) can not be determined

29. Let x, y and z be three positive real numbers whose sum is 1. Ifno one of these numbers is more than twice any other, then theminimum possible value of the product xyz is

(A) (B) (D) 1 (E) none of these(A) 3B 6 C 125 127

30. Find the units digit in the decimal expansion of

(15 + 2)'9 + (15 + l22)82.

(C) 5 (D) 9

63

(A) 0 (B) 2 (E) none of these

II

Answer Keys

1973 Answers

I . D 8. E 15. D 22.A 29. A2. C 9. E 16. B 23.D 30. B3. B 10. A 17. E 24.1D 31. C4. D 11. B 18. C 25.E 32. A5. D 12. D 19. D 26.E 33. C6. C 13. D 20. C 27.A 34. C7. A 14. C 21. B 28.C 35. E

1974 Answers 1975 Answers

1.D 7. D 13.D 19. A 25.C 1.B 7. E 13.D 19. D 25. B2.B 8. A 14.A 20. D 26.C 2.D 8. D 14.E 20. B 26. C3.A 9. B 1S.B 21. B 27.A 3.A 9. C 15.A 21. D 27. E4.D 10.B 16.A 22. E 28.D 4.A 10. A 16.C 22. E 28. A5.B 11.A 17.C 23. B 29.B 5.B 11. E 17.D 23. C 29. C6.D 12.B 18.D 24. A 30.A 6.E 12. B 18.D 24. E 30. B

1976 Answers 1977 Answers

1.B 7. E 13.A 19. B 25.D 1.D 7. E 13.E 19. A 25. E2.B 8. A 14.A 20. E 26.C 2.D 8. B 14.B 20. E 26. B3.E 9. D 15.B 21. B 27.A 3.E 9. B 15.D 21. B 27. C4.C 10.D 16.E 22. A 28.B 4.C 10. E 16.D 22. C 28. A5.C 11.B 17.A 23. A 29.B 5.A 11. B 17.B 23. B 29. B6.C 12.C 18.E 24. C 30.E 6.D 12. D 18.B 24. D 30. A

65


19. C20. A21. A22. D23. C24. A

1.

2.3.4.5.6.

1 .2.3.4.5.6.

25. D26. B27. C28. E29. D30. E

BCDBCE

CDECBA

7. B8. B9. B

10. AII. C12. A

13.14.15.16.17.18.

DCDBCD

1978 Answers

EDBBCE

13.14.15.16.17.18.

1979 Answers

BCAEDC

1.

2.3.4.5.6.

7.8.9.

10.II.12.

7.8.9.

10.II.12.

25. C 1.26. E 2.27. E 3.28. C 4.29. A 5.30. B 6.

1982 Answers

DCAECD

ECEDBB

13. A14. C15. E16. E17. C18. C

1980 Answers

19. A20. C21. B22. A23. C24. E

BAEDDC

13.14.15.16.17.18.

1981 Answers

BABBDD

19.20.21.22.23.24.

25.26.27.28.29.30.

25.26.27.28.29.30.

BBEDEB

ADCDED

DDCEDA

ECDCCA

BDEEAA

7.8.9.

10.11.12.

7.8.9.

10.11.12.

25.26.27.28.29.30.

BAAECE

13. E14. A15. B16. E17. B18. C

19.20.21.22.23.24.

BBDDCD

1. E2. A3. C4. E5. C6. D

19.20.21.22.23.24.

DBCBAD

66

III

Solutionst

1973 Solutions

Part 1

1. (D) Let 0 denote the center of the circle, and let OR and ABbe the radius and the chordwhich are perpendicular bisec-tors of each other at M.Applying the Pythagoreantheorem to right triangleOMA yields

(AM)' = (oA) 2 - (oM)2

- i,2 -2 -n= 1L -- = Itm,

AM = 6F3.

Thus the required chord has length 12Vi.

2. (C) The unpainted cubes form the 8 x 8 x 8 cube of interiorcubes. Therefore, 103 - 83 = 488 cubes have at least one facepainted.

t The letter following the problem number refers to the correct choice of the five listed inthe examination.

67


3. (B) Thirteen is the smallest prime p such that 126 - p is alsoprime. Thus the largest difference is 113 - 13 100.

4. (D) In the adjoining figure MV isan altitude of AABV. Since VAAMV is a 300 - 60° - 90°triangle, MV has length 2rY.The required area is, therefore,area AABV = '(AB)(MV) A B

= 2(12)2C3 = 12r;6.

5. (D) Let a * b denote the average, '(a + b), of a and b. Then

II a* b = '(a + b) = {(b + a) = b* a,

and

IV a + (b*c) =a + (b + c) = '(a + b + a + c)

= (a + b)*(a + c),

but

I (a *b)* c = '[!(a + b) + c] = 'a + lb + 2c,

while

a*(b*c) = I[a + (b + c)] = 'a + -b + 'c;

and

III a*(b + c) = I(a+ b + c),

while

(a*b) + (a*c) = -(a + b) + '(a + c) $ a*(b + c).

To see that V is false, suppose that e * a = a for some eand all a. Then 1(e + a) = a, so e = a. Clearly this can-not hold for more than one value of a. Thus only II and IVare true.

6. (C) Let b > 5 be the base. Since (24b )2 = (2b + 4)2 =4b2 + 16b + 16 and 5 5 4 b = 5b2 + 5b + 4, it follows that

5b2 + 5b + 4 = 4b2 + 16b + 16,

so

b2- llb - 12 = (b + I)(b - 12) = 0,

b = -I or b= 12.

68

SOLUTIONS: 1973 EXAMINATION

7. (A) The numbers to be added form an arithmetic progressionwith first term a = 51, last term I = 341 and common dif-ference d = 10. We use the formula 1 = a + (n - I)d todetermine the number n of terms to be summed and S= 2n(a + 1) to find their sum. Thus 341 = 51 + (n - 1)10yields n = 30, so S = 15(392) = 5880.

8. (E) The number of pints P of paint needed is proportional to thesurface area (hence to the square of the height h) of a statueand also to the number n of statues to be painted. ThusP = knh2; substituting n = 1, h = 6 and P = 1 into thisequation, we obtain the constant of proportionality k = 1/36,so that

1 2P= 36nh

Now when n = 540 and h = 1, P = (1/36)(540)(12) = 15.

9. (E) Right triangles CHM andCHB are congruent since theirangles at C are equal. There-fore the base MH of ACMHis 4 of the base AB ofA ABC, while their altitudes---- -- 1 TT__ _ - ____ _ Aare equal. nence we area OI M HAABC is 4K.

10. (A) If n - 1, then elementary operations on the system yieldthe solution x = y = z = 1/(n + 1). If n = - 1, adding thethree equations yields 0 = 3, which shows the system ofequations has no solution.

OR

We use the theorem: A system of three linear equations inthree unknowns has a unique solution if and only if thedeterminant of its coefficient matrix does not vanish. In ourexample, this determinant is

n 1 0D= 0 n I =n 3 +1.

1 0 nThe only real value of n for which D vanishes is - 1. Wesaw above that the system has no solution in this case.

69

'R


Part 2

11. (B) The given inequalities are represented geometrically as

(the set of points (x, y) such that 2max(ixl,lyl) < a)

C (the set of points (x, y) such that +2(x2 +y 2 ) < a}C (the set of points (x, y) such that lxI + IyI < a).

If the side length of the inner square in the figures is denotedby a, then the three sets in the above inclusions are the setsof points inside (or on) the inscribed square, the circle, andthe circumscribed square, respectively, in Figure II.

12. (D) Let m and n denote the number of doctors and lawyers,respectively. Then

35m + 50n= 40,

m + n

so 40(m + n)= 35m + 50n, 5m = 10n, and -= 2.n

13. (D) The square of the given fraction is

4(2 + 2v + 6) 4(8 + 41) 16(2 + v) 16

9(2±+ ) 9(2 + V) 9(2+VC) 9

Hence the fraction is equal to 1_ =6 4V 9 -3

OR

Multiplying top and bottom of the given fraction by C2, weget

2(2 + VC)_ 2(2 + I~)_ 4(1 + V) 4

3F2 + j 3A4 + 2F3 3 (I + C3)2 3

14. (C) Let x, y and z denote the number of tankfuls of water de-livered by valves A, B and C, respectively, in one hour. Then

1 1x + y + z=; x + z =-5; Y + z 2=

Subtracting the sum of the last two equations from twice thefirst yields x + y = 6, so that 5(x + y) = I tankful will bedelivered by valves A and B in 6 = 1.2 hours.

SOLUTIONS: 1973 EXAMINATION 71

15. (D) The center of the circle whichcircumscribes sector POQ is at C,the intersection of the perpendicularbisectors SC and RC. ConsideringAORC, we see that

sec 2 = 0 or OC = 3sec 2.2 - -.

Q

Note: All the answers except (D) can be eliminated by consider-ing the limiting case B = 7/2. For in this case (A) is 0, (B)is infinite, (C) is less than 3, and (E) is 3; all absurd.

16. (B) Let n denote the number of sides of the given convex poly-gon and x the number of degrees in the excepted angle. Then180(n - 2) = 2190 + x, so that

2190 x180 180

Since the polygon is convex, 0 < x < 180; it follows that

2190 2190180 180

i.e. 121 < n - 2 < 136. Since n is an integer, this forcesn - 2 = 13, so n = 15. (Incidentally, (13)(180) = 2340 =

2190 + 150, so the excepted angle has measure 1500.)

17. (E) Using the formula for the cosine of twice the angle Io, wehave

cos0 = cos2( 2) = I - 2sin2 I -2 x- = x

[Note that since 0 < 0 < 900, 0 <- < 1, so x > 1.] Nowx

tan 2 0= sec 2 0 - I 2= - I = X2cos

20

so

tan0= vX2 - I.


18. (C) Since the factors p - I and p + I of p2 - 1 are consecu-

tive even integers, both are divisible by 2 and one of them by4, so that their product is divisible by 8. Again (p - 1), pand (p + 1) are three consecutive integers, so that one ofthem (but not the prime p) is divisible by 3. Therefore theproduct (p - l)(p + 1) = p2 - 1 is always divisible by both3 and 8, and hence by 24.

19. (D) 728! = 72 X 64 x 56 X 48 x 40 x 32 x 24 x 16 X 8= 89 X (9!);

182! = 18 x 16 x 14 x 12 x 10 x 8 x 6 x 4 x 2= 29 x (9!).

The quotient (728!)/(182!) = 89/29 = 49.

20. (C) In the adjoining figure, S denotes an Darbitrary point on the stream SE, andC, H and D denote the position ofthe cowboy, his cabin and the point 8miles north of C, respectively. Thedistance CS + SH equals the distanceDS + SH, which is least when DSHis a straight line, and then

re - n crI- . 2 1 c2wan)IZ = J3 = Vet U - -

= 89 = 17 miles.

Part 3

21. (B) The number of consecutive integers in a set is either odd oreven. If odd, let their number be 2n + 1 and their average bex, the middle integer. Then (2n + I)x = 100 and x =100/(2n + 1), so that 2n + I can only be 5 or 25. If 2n + 1= 5, then x = 20 and n = 2, so that the integers are18,19,20,21,22. If 2n + 1 = 25, x = 4 and n = 12, whichis impossible because the integers must be positive.

If the number of consecutive integers is an even number2n, let the average of the integers (half way between themiddle pair) be denoted by x. Then 2nx = 100, x = 50/n.Since x is a half integer, 2x = 100/n is an integer, but x =50/n is not an integer, so that n is 4, 20 or 100. For n = 4,x = 12- and the integers are 9 through 16; n = 20 and 100are impossible since the integers are positive. Hence there areexactly two sets of positive integers whose sum is 100.

72

-E


22. (A) Interpreting Ix - 11 + Ix + 21 as

I $ I I t 11 1x

-3 -2 -I 0 1 2 3

the sum of the distances from x to 1 and from x to - 2, wesee that this sum is 3 if -2 < x < I and 3 + 2u if x is adistance u from the interval from - 2 to 1. Thus, the solutionset is (x: - 3 < x < 2).

23. (D) Let the sides of the first card (both red) be numbered I and 2.Let the red and blue sides of the second card be numbered 3and 4, respectively. On the draw any of the sides 1, 2 or 3 hasequal likelihood of being face up on the table. Of these three,two undersides are red and one blue, so that the probability ofa red underside is 2/3.

24. (D) Let s, c and p denote the cost in dollars of one sandwich,one cup of coffee and one piece of pie, respectively. Then3s + 7c + p = 3.15 and 4s + 10c + p = 4.20. Subtractingtwice the second of these equations from three times the firstyields s + c + p = 1.05 so that $1.05 is the required cost.

25. (E) Let 0 denote the center of theplot and M the midpoint of theside of the walk not passingthrough 0. If AB denotes onearc connecting opposite sides ofthe walk, then OMBA consistsof the 30° sector OAB and the30 - 600 - 900 triangle OMB.Thus the area of the walk is

2[ T62 + 2(3)3r I = 6] r + 93,Ln12 2

and the required area is

r62 - (67r + 9V3) = 30T - 91i.

A

73


26. (E) Let d and 2n denote the common difference and the evennumber of terms. Let SO and Se denote the sums of all nodd- and all n even-numbered terms respectively; then Se -SO = nd since each even-numbered term exceeds its odd-num-bered predecessor by d. Moreover the last term clearly ex-ceeds the first by (2n - 1)d. Hence in the present case wehave

nd= 30 - 24 = 6 and (2n - l)d= 10.5.

Therefore d = 2nd - 10.5 = 12 - 10.5 = 1.5, and

n = 6/1.5 =4, so 2n = 8.

Note: We note that in solving this problem we did not need toknow the values of SO and Se, but only their differenceSe - S.o

27. (A) Let s denote the distance. The total time t that car A travelsis

S St= 2 + v

and the average speed is

s s 2uv- = = = X.t S + S u+v

2u 2v

For car B, the average speed is clearly ' (u + v) = y. Now0 < (U- v) 2; 2uv _ u2 + V2; 4uv (U + v)2 ; and di-viding both sides of the last inequality by 2(u + v) yields

2uv u + vu + v < 2 =Y.

Note 1: The solution (A) can also be obtained without anycalculation, since car A clearly spends more than half the timedriving at the slower of the two speeds u and v.

Note 2: Observe that car A's average speed x is the reciprocal ofthe average of the reciprocals of u and v; this is called theharmonic mean (H.M.) of u and v:

H.M (M., v) = I = 2_uv1(1 +v

2 Tou v

The calculations above show that the harmonic mean of two


positive numbers never exceeds their arithmetic mean. Forcomparisons of the harmonic, geometric and arithmetic meansof two positive numbers, see An Introduction to Inequalities byE. Beckenbach and R. Bellman, NML vol. 3, Exercise 4 on p.62 and its solution on p. 120.

28. (C) Let r > I denote the common ratio in the geometric progres-sion a, b, c:

b = ar; lognb = logna + lognr

c = ar2 ; lognc = logna + 21ognr,

so that loga, logb, lognc is an arithmetic progression. Now

it follows from the identity log ,v = ' , that the recipro-

cals of logan, logbn, logan form an arithmetic progression.

29. (A) The boys meet for the first time when the faster has covered9/14 and the slower 5/14 of the track. They meet for the nthtime after the faster has travelled (9/14)n laps and the slower(5/14)n laps. Both of these are first whole numbers of lapswhen n = 14; there are thirteen meetings, excluding the startand finish.

30. (B) For any fixed t > 0, 0 < T < 1. Hence S, the interior ofthe circle with center (T, 0) and radius T, has an area be-tween 0 and v.

Part 4

31. (C) The integer TTT = T(lll) = T 3 - 37, with T equal to thefinal digit of E2 . Now 37 must divide one of ME and YE,say ME. Therefore ME = 37 or 74. The latter possibility isruled out since it would imply (ME) - (YE) > 74- 14 =1036, while clearly TTT < 999. Thus ME = 37, T = 9,

TTT 999and YE= -= 27.ME 37

tSee footnote on p. 82.

75


1'71 /A'% TIEU_ -rL-rs of - --l A ;Q CJZ_. gox) 11Ic; UvIUolVum V a pyialluu lb

equal to one third the area ofthe base times the altitude. Thebase of the given pyramid isan equilateral triangle withsides of length 6; hence it hasarea 9V3. Altitude h of thegiven pyramid may be foundby applving the Pvthagoreantheorem to right AABC in theadjoining diagram. Here B is the center of the base, so thatAB = 2(3C3) = 2V3, h2 = (vi5)2 - (2VY) 2, and h = rTherefore, the volume of the pyramid is 3(9i)(3) = 9.

33. (C) Let x and y denote the number of ounces of water and ofacid, respectively, in the original solution. After the additionof one ounce of water, there are y ounces of acid andx + y + I ounces of solution; after adding one ounce of acid,there are y + I ounces of acid and x + y + 2 ounces ofsolution. Therefore,

Y a 1d Ix+ y+ l 5 x + y+ 2 3

Solving these equations yields x = 3 and y = 1, from whichit follows that the original solution contained

x y =25% acid.

34. (C) Let d, v and w denote the distance between the towns, thespeed against the wind and the speed with the wind, respec-tively; then the plane's speed in still air is '(v + w). We aregiven that

()d d d( -) v = 84 and that (2) - = _ _ 9

V W -L(w+ V)

Set x = d/w, the required return time. Then by (2),

x = 2 - 9 2

d d x 84

Simplifying, we obtain X2 - 75x + 756 = 0. The solutionsof this equation are x = 63 and x = 12.

4


35. (E) We shall show that I, II, and III are all true by givinggeometric arguments for I and II and then showing algebrai-cally that III is a consequence of I and II.

In the adjoining fig-ure, chords QN andKM have length s. Thefive equal chords oflength s in the semicirclewith diameter KR sub-tend central angles of1800/5 = 36° each. Thefive isosceles triangles,each with base s andopposite vertex at thecenter 0, have base an-gles of measure

__ L2(180 -360) = 72°.

Now rotate the entire configuration clockwise through 720about 0. Then chord PQ, parallel to diameter KR, goes in-to chord NR, parallel to diameter PL. Denote by T the in-tersection of MN and PL. In parallelogram ORNT, TN =OR = 1, and TO = NR = s. We saw that 4MPO = 720;now 2MTP = 720 also, because MTIIKO. So APMT isisosceles with MT = MP = s. Therefore d = MT + TN= S + 1, so

I d-s= 1.

The segments PT, TL and MT, TN of intersecting chordsMN and PL satisfy (PT)(TL) = (MT)(TN). SincePT = OP - OT = I - s, and TL = OL + OT= 1 + s, thisequation takes the form

(I - s)(I + s) =s 1 or -S2 =s.

Multiplying equation (I) by s, we get ds - s2 = s, so

II ds = s 2 + s = 1.

The equation s2 + S = I is equivalent to S2 + s-= 0,

whose positive root is s= '(r- 1). Therefore

d=s+ I = 2 +

and

III d 2 - s 2 = (d+ s)(d- s) = d+ s = V5.

77


OR

Each chord of length s subtends an angle 360 at the center0, and MN of length d subtends an angle 3 36° = 1080.Therefore

s = 2 sin 18° and d = 2 sin 54°.It follows from this and trigonometric identities that

d = 2sin54' = 2cos36' = 2(1 - 2sin2180) = 2 - s2

and

s = 2 sin 18° = 2 cos72' = 2(cos236' - I) = d2 - 2.

Adding d= 2 - S2 and s = d2 - 2, we obtain d+ s=d2 - S2 = (d + s)(d - s), so that I. d - s = 1. Substitutings + l for d in d = 2 - s2, we find that s2 + S-l = 0,

which has one positive root, s = 2( 5 - ). Then

d=s + I= 2 ds = 1, d + s = F

and

d2- s2= A

78


1974 Solutions

1. (D) Multiplying both sides of the given equation by the leastcommon denominator 2xy yields 4y + 6x = xy or, equiva-lently, 4y = xy - 6x. Factoring x from the right side of thelast equation gives 4y = x(y - 6). Since y +# 6 we can di-vide both sides of this equation by y - 6 to obtain (D).

2. (B) Since x, and x2 are the two roots of the quadratic equation

3x2 - hx - b = 0, the sum of the roots is .3.

3. (A) The coefficient of x7 in (1 + 2x - x2 )4 is the coefficient ofthe sum of four identical terms 2x(-x 2 )3 , which sum is- 8x7 .

4. (D) By the remainder theorem x51 + 51 divided by x + 1 leaves

a remainder of (- 1)51 + 51 = 50. This can also be seenquite easily by long division.

5. (B) 4EBC = 4ADC since both angles are supplements ofYABC. Note the fact that 4BAD = 920 is not needed in thesolution of the problem.

6. (D) x * y = Y *x

andxyz

(x*y)*z = XY _Z +Y xyzx + y XY + z xy + xz + yz

x +y

Similarly x * (y * z)= xYZ so "* "is both com-xy + xz + yz'

mutative and associative.

Note: The result can also be derived from the identity

1 1 1x*y x y

Commutativity and associativity of * now follows from thefact that they hold for +.

79


7. (D) If x is the original population, then

x + 1,200 - 0.1l(x + 1,200) = x - 32.

Solving for x gives (D).

8. (A) Since 31" and 513 are both odd, their sum is even.

9. (B) All multiples of 8, including 1,000, fall in the second column.

10. (B) Putting the quadratic in its standard form:

(2k - l)x 2 - 8x + 6 = 0,

we see that the discriminant D is 64 - 4(2k - 1)6 =88 - 48k = 8(11 - 6k). A quadratic equation has no realroots if and only if its discriminant is negative. D is negativeif 11 - 6k < 0, that is, when k > 11/6; the smallest in-tegral value of k for which the equation has no real roots is2.

11. (A) Since (a, b) and (c, d) are on the same line, y = mx + k,they satisfy the same equation. Therefore,

b = ma + k d= mc + k.Now the distance between (a, b) and (c, d) is

(a - C)2

+ (b - d)2 .

From the first two equations we obtain (b - d) = m(a - c),so that

(a - c)2 +(b-d) 2 = V(a _ c) 2 + m2(a c) 2

= la - cll+m 2

Note we are using the fact that V = Ix for all real x.

ORLet 9 be the angle between the line and the x-axis; then

(tan) 2 = m =

so

(b - I

(a -

4 )2 (c, d)

2 ' Ib-d

(a,b)(b - d)2 = m2(a - c) 2 la - cl

and the square of the desired distance is

(a- )2 + (b -d)2 = (a - C)2 + m 2 (a - c)

= (a - C)2

(l + M2

).

80


12. (B) Since g(x) = 1/2 is satisfied by x = 1/2,

f(1/2) =f(g(J72))= 1.

OR

Since g(x) = I -x 2 for x 0, we have x2 1 -g(x)for x + 0; so

- X2 g(x) = (g(x)) for x -x l-g(x) )

and

A(i2) 2 11-2

13. (D) Statement (D) is the contrapositive of the given one and theonly one of the statements (A) through (E) equivalent to thegiven statement.

14. (A) Since x2 > 0 for all x + 0, x2 > 0 > x is true if x < 0.

Counterexamples to the other statements are easy to con-struct.

15. (B) By definition Ial = -a when a > 0 . If x < -2, then

1 + x < 0 and 11 + xi =-(1 + x) and 11 - I + xI= 1+ + 1+ xI = 12 + xi. Again if x < -2, then 2 + x < 0and 12 + xl = -2 - x.

16. (A) In the adjoining figure, AABC isa right isosceles triangle, with.4BAC = 900 and AB = AC,inscribed in a circle with centerO and radius R. The line seg-ment A 0 has length R and bi-sects line segment BC and2BAC. A circle with center O'lying on A 0 and radius r is in-

crr-h i- A AD" 'Th. -A-X ADbLI.AUVU III LIADZI. - aII lUub AiD

and A C are tangent to the inscribed circle with points oftangency T and T', respectively. Since ATO' has angles45 -45 -900 and O'T= r, we have AT= r and O'A= R - r = rC2. Hence R = r + rr2 and R/r = 1 + V2.

81


17. (C) Since i2 = - 1, (I + i) 2 = 2i and (I - i)2

=-2i. Writing

(1 + )20_ (1 )20= ((1 + j)2)° i((1-i)2)1O

we have

(1 + i)20 - (1 - i)2' = (2i)10 - (-2i)o = 0.

ORObserve that i(l - i) = i - i2 = I + i, and i4

= (-1)2 = 1.

Therefore

(I + i)4= i4(l- _)4 = (I -j)4

and

(1 + i)4n - (1- i)4

n = 0 for n = 1,2,3,....

OR

(1 + j)20 = lF 20[cos 900° + i sin 9000] = - 1024,

(1 - j)20 = V20[cos( 9goo) + isin(-9000)] = -1024,

and the difference is 0.

18. (D) By hypothesis we have

3 = 8P = 23 P, 5 = 3q; so 5 = (2 3p)q= 23pq.

We are asked to find log, 05 = x, i.e. x such that 10' = 5.Since 5 = 23, we have 10 X = 2 3pq; but also l0x= 2x 5= . 23pqx = 2 x(l0+ 3Pq) It follows that

x(l + 3pq) =3pq and x= l±3pq

OR

We recall the change of base formula

) 1ogb = logb and its corollary logab = logba

tTo verify the change of base formula (*) for positive numbers a, b, c, a s 1, c Iwe set

logab = U, so aU- b,logcb = V, so cv= b,logca = W, so cW= a.

Thus cV= b = au = (cW)u= cuW and UW= V, so U = V/W as claimed. Takingb = c, we also obtain the important corollary

1a lga =


and write

p = 10g83 =1 I 1 _ 1

1og38 1og323 31og32 '

so

1og 32 = I3 p,9

and

oI log3 10q

1og 32 + 1og 35q = 3pq

I + q I + 3pq'1 l3p

OR

Since we are asked for an answer in base 10, we convert allthe given information to that base, using (*). From

p = 1og83 = og 108

one obtains

log103 = p log10 8,

q = 1og35 = log103

log, 05 = qlogl,3,

log105 = pqlog 108 = pqlog 10)

= 3pq(l - log, 0 5).

Solving this for log105 gives choice (D).

Note: Conversion of all logarithms to a common base provides asystematic (if not always shortest) approach to problems like18, and this approach will be used again in later problems,with a reference back to (*). In the last solution to thisproblem, we converted to base 10 only because the answerwas requested in that base. The traditional reason for usingbase 10 - ease of doing numerical computations - has beenmade largely obsolete by computers.

83

thus


19. (A) Let DM = NB = x; then AM = AN = I - x. Now

area ACMN = area EABCD - area AANM-area A NBC - area A CDM

I- (I x)2 - 2-- X = l (I - x 2 ).

2 2 2 1

Denote the length of each side of the equilateral triangleCMN by y; using the Pythagorean theorem we see that

x2 + 1 2 = y 2 and (I-X) 2 + (I _ X) 2 = y2.

Substituting the first equation into the second we get

2(1 - x) = x2 + I or x2 - 4x + I = 0.

The roots of this equation are 2 - Vi and 2 + V3. Since2 + Vi > I we must choose x = 2 - ri and obtain areaACMN = 2Vi - 3.

20. (D) By rationalizing the denominator of each fraction, we see

T= (3 + C)- (r + v) + (vi + r)

-(rI + A) + (v + 2)=3 + 2 = 5.

21. (B) The sum of the first five terms of the geometric series withinitial term a and common ratio r is

S 5 = a + ar + ar2 + ar3 + ar4 = a(l-r5)1-r

By hypothesis ar 4 - ar3 = 576 and ar - a = 9. Dividing

the first equation by the last yields

r4__3_ = - r 64,

so r3 = 64, and r= 4. Since ar-a = 9 and r = 4, wehave a = 3 and therefore

S5= 3(1 = 1023.

22. (E) We recall that asinO + bcosO, if a2 + b2 +6 0, can be ex-pressed in the form

a2 ± b2 sin(G + qp),

84


wherea b

cos qP= i , sinq = - 'va2 + b 2 va2 2

and that the minimum of sin a is - I and occurs whena = 270° + (360m) 0, m = 0, + 1, +2,....

Applying this principle to the given function where

a 1 1 b v

1+3 2' a 2 +b 2 2

we find qp = - 60° and write

.AA I .A V4 A]sin -A _ cosA = 2I-sinA - 2-cos-

2 2 [ 2 2 2 2

= 2sin(A _ 600)

This expression is minimized when 'A - 600 = 2700 +(360m)0 , that is when A = 660° + (720m) 0 , m = 0,+ 1, + 2, .... None of (A) through (D)are angles of this form.

23. (B) Since TP = T"P, OT = OT" = r, and 4PT"O = 4PTO= 900, we have AOTP = AOT"P. Similarly AOT'Q-AOT"Q. Letting x = 4TOP = .POT" and y = 4T"OQ= 4QOT' we obtain 2x + 2 y = 1800. But this implies that2POQ = x + y = 900. Therefore APOQ is a right trianglewith altitude OT". Since the altitude drawn to the hypo-tenuse of a right triangle is the mean proportion of thesegments it cuts, we have

4 r-= or r= 6.

r 9

24. (A) Let A be the event of rolling at least a five; then the proba-bility of A is 6 = 3. In six rolls of a die the probability ofevent A happening six times is (3 )6. The probability of get-ting exactly five successes and one failure of A in a specificorder is (I)5 . 2. Since there are six ways to do this, theprobability of getting five successes and one failure of A inany order is

6l 5 2 126 3 * 3 729

The probability of getting all successes or five successes andone failure in any order is thus

12 + 1 = 13729 3 J 729


25. (C) Since DM = AM, 4QMA = 4DMC and 4CDM = AQAM,we have AQAM AMCD. Similarly ABPN= ADNC.Now,

area AQPO = area L27ABCD + area ADOC

and

area A DOC= 4(-2 area tABCD) =

so that

k 9karea AQPO = k + 8 = 8

26. (C) Writing 30 as a product of prime factors, 30 = 2 3 5, weobtain

(3°)4 = 24. 34 54.

The divisors of (30)4 are exactly the numbers of the form2' 3. 5 k, where i, j, k are non-negative integers betweenzero and four inclusively, so there are (5)3 = 125 distinctdivisors of (30)4; excluding I and (30)4 there are 123 divisors.

27. (A) Consider 1I(x) + 41 = 13x + 2 + 41 = 31x + 21. Nowwhenever Ix + 21 < a/3, then 1f(x) + 41 < a. Con-sequently whenever Ix + 21 < b and b < a/3, we haveIf(x)+41 <a.

28. (D) Using the formula for the sum of a geometric series, we have

2

0 < x < E -" < E, 3= l = 1.n=1 n=l I -

3

If a, = 0, then

2X 2 9 I

0 < x < E 3n 3

If a, = 2, then

2oe + t + o + *< x <

So either< x < ' or 2 X< 1.<


OR

If the number x is represented in base 3, then it takes theform .ala 2 a3 . . a25 where each digit is either 0 or 2. Thuswe see that either

x < (0100 ... )3 = or x > (0.200...) 3 = .

29. (B) For each p = 1,..., 10,

Sp=p+p+(2p-l)+p+ 2 ( 2 p-l)+.± +p+ 3 9 (2 p-I)

= 40p + (40)(39) (2p - 1)2

= (40 + 40 39)p - 20 39

= (40)2 p - 20 39.

Therefore,10 10

E Sp= (40)2 _ p - (10)(20)(39)P=I p=l

= (40)2 ( )( ) - (10)(20)(39)

= 80,200.

30. (A) Consider line segment AB cut by a point D with AD x,DB = y, y < x and

y x D Yx X + Y A | B

Since y/x = R we can choose x = 1 and therefore y R;

thus R = I and R2 + R-I = 0. We can therefore

write R-' = R + I so that

R2 + R-' = R2 + R + I = (R2 + R -1) + 2 = 2,

and

R[R(R2+R ')+R-'] + R- = R[R 2 +R I] + R-'

= R2 + R-1= 2.

87


1975 Solutions

IIl. (B) I

2- 1 2- 22- 2 3

2

1 4

2- 34

2. (D) The equations have a solution unless their graphs are parallellines. This will be the case only if their slopes are equal, i.e. ifm = 2m - 1 or m = 1. (The lines are not coincident sincethey have distinct y-intercepts for all values of m.)

3. (A) None of the inequalities are satisfied if a, b, c, x, y, z arechosen to be 1, 1, - 1, 0, 0, - 10, respectively.

4. (A) In the adjoining figure, if s is thelength of a side of the first square,then s/ l2 is the length of a sideof the second. Thus the ratio ofthe areas is s2 /(S/I r) 2 = 2.

5. (B) (x + y)9 = x9 + 9x8 y + 36x7 y 2 + . . + y9 ;

9p8q = 36p7 q2 and p + q = 1. Dividing the second equationby 9p7q, we obtain p = 4 q, and substituting 1 - p for qwe obtain p = 4(1 - p), so p = 4/5.

6. (E) Grouping the terms of the difference as

(2 - 1) + (4 - 3) + * + (160 - 159),

one obtains a sum of 80 terms, each equal to 1.

7. (E) I - x| is - 2 if x is negative and 0 if x is positive.x

8. (D) II and IV are the only negations, of the given statement.

88

/ l/


9. (C) Let d denote the common difference of the progressiona, + bl, a2 + b2,. Then

99d = (alo + bloo) - (a, + bl) = 0.

Thus d = 0, and 100(al + bl) = 10,000 is the desired sum.

10. (A) Let k be any positive integer. Then

(}ok + 1)2 = 1 0 2k + 2 * 1ok + 1.

The sum of the digits is therefore I + 2 + I = 4.

11. (E) Suppose P is the given point, 0 is the center of circle K,and M is the midpoint of a chord AB passing through P.Since 4OMP is 90°, M lies on a circle C having OP for itsdiameter. Conversely, if M is any point on the circle C, thenthe chord of circle K passing through P and M (the chordof K tangent to C at P if M = P) is perpendicular to OM.Hence M is the midpoint of this chord and therefore belongsto the locus.

12. (B) If a + b, a3 - b3 = 19x3 and a - b = x, then

a3 - b3 = (a-b)(a2 + ab + b2 ) = x(a2 + ab + b2 )

= 19x3

Dividing the last equality above by x and substitutingb = a - x, we obtain

18X2 + 3ax - 3a2 = O,

- 3(a - 3x)(a + 2x) = 0.

So a = 3x or a = -2x.

13. (D) For x < 0, the polynomial x6 - 3x5 - 6x3 - x + 8 is posi-tive, since then all terms are positive; so it has no negativezeros. At x = 1, the polynomial is negative and hence has atleast one positive zero (between 0 and 1).

14. (E) Let W, H, I and S denote whatsis, whosis, is and so, respec-tively. Then H = I and IS = 2S imply W= S, or equiva-lently, since S > 0, H = I= 2 implies W= S. Now if H= S, 2S =S 2 and I = 2, or equivalently H = I = S = 2,then W= S, so that HW= 4 = S + S.

89


15. (A) The first eight terms of the sequence are 1, 3, 2, - 1, - 3, -2,1, 3. Since the seventh and eighth terms are the same as thefirst and second, the ninth term will be the same as the third,etc.; i.e., the sequence repeats every six terms. Moreover, thesum of each six term period is 0. Hence the sum of the first 96terms is zero, and the sum of the first one hundred terms isthe sum of the last four terms 1 + 3 + 2 - I = 5.

16. (C) Denote the first term of the series by a and the commonratio by I/n; then the sum of the series is

a-(l/n) =3, and a = 3 - (3/n).

Since a and n are positive integers, 0 < 1/n < 1, n = 3 anda = 2. The sum of the first two terms is 2 + 2(1/3) = 8/3.

Note: The desired sum clearly lies between I and 3, so (A), (B)and (E) are impossible. If (D) were the answer, the first twoterms would both be 1, also impossible. This leaves only (C).

17. (D) Since the commuter makes two trips each work day, the totalnumber of trips is 2x; thus 2x = 9 + (8 + 15) = 32, andx = 16.

Note: The given information was deliberately redundant. "If hecomes home on the train, he took the bus in the morning" islogically equivalent to "If he takes the train in the morning,he comes home by bus in the afternoon."

18. (D) There are 900 three digit numbers, and three of them (128,256 and 512) have logarithms to base two which are integral.So 3/900 = 1/300 is the desired probability.

19. (D) For any fixed positive value of x distinct from one, let a =

log3 x, b = log,5 and c = 1og 35. Then x = 3a, 5 = xb and5 = 3C. These last equalities imply 3ab = 3c or ab = c. Notethat log,5 is not defined for x = 1.

OR

Converting all these logs to some fixed but arbitrary base d(see footnote on p. 82), we obtain

=logdX lOgd5 _ lod5(log3 x)(logx5) logd3 logdx logd3

for all x + 1.

90


20. (B) In the adjoining figure let h be the length of altitude ANdrawn to BC, let x = BM and let y = NM. Then

h2 + (x + y) = 64,

h2 + Y2 = 9,

h2 + (x - y) = 16.

A

B x-y Ny M x C

Subtracting twice the second equation from the sum of thefirst and third equations yields 2x2 = 62. Thus x = VfTand BC = 20iT.

OR

Recall that the sum of the squares of the sides of a parallelo-gram is equal to the sum of the squares of its diagonals.Applying this to the parallelogram having AB and AC asadjacent sides yields 2(42 + 82) = 62 + (2x)2 , x = vi3T.

21. (D) Letting a = 0 in the equation f(a)f(b) = f(a + b) (called afunctional equation) yields f (O)f(b) = f(b), or f(O) = 1; let-ting b = -a in the functional equation yields f(a)f( - a) =

f(0), or f (-a)= I/f(a); and3

f (a)f (a)f (a)= f (a)f (2a) = f (3a), or f (a) = f(3a).

The function f(x) =1 satisfies the functional equation, butdoes not satisfy condition IV.

22. (E) Since the product of the positive integral roots is the primeinteger q, q must be positive and the roots must be 1 and q.Since p = 1 + q is also prime, q = 2 and p = 3. Hence allfour statements are true.

91


23. (C) In the adjoining figure diagonals D CAC and DB are drawn. Since 0is the intersection of the mediansof A ABC, the altitude of AAOBfrom 0 is I the altitude ofAABC from C; i.e. I the sidelength, s, of the square. Hence

N

area AAOB = I (area AABC) A M B

= I(IS2) = IS2.

Similarly, area ACOB = 's 2 . The area of A OCD is ob-tained by subtracting the areas of triangles A OB and COBfrom that of the square, so area AOCD = 52 - 352 = 35

2.

ORIntroduce coordinates with respect to which AB is the unitinterval on the positive x-axis and AD is the unit interval onthe positive y-axis. Now

area AOCD = area AACD + area AAOC

1 112 1 22 21 l 31

The rows of the determinant are the coordinates of 0 and C.Those of 0 are (3, 3) because 0 is the intersection ofmedians of A ABC.

Note: Since the area of AABN is 4 of the area of the square, it isclear that the desired ratio r satisfies l < r < 3. Only (C)fulfills this condition.

24. (E) If 00 < 0 < 45°, then (see Figure 1) an application of atheorem on exterior angles of triangles to AEAC yields20 = EAC + 0. Therefore 4EAC = 0 and A EAC is iso-sceles. Hence EC = AE = AD.

A

180-2

r/

II E U; D tI-

Figure I Figure 2


If 0 = 450, then AABC is a 450-450-900E = B. Then EC = BC = AB = AD.

If 450 < 0 < 600, then (see Figure 2)

triangle and

,4EAC = 1800 - 4AEC - 4C

= 1800 - (1800 - 20) - 0

= 0.

Thus AEAC is isosceles and EC = EA = AD.

25. (B) If the son is the worst player, the daughter must be his twin.The best player must then be the brother. This is consistentwith the given information, since the brother and the soncould be the same age. The assumption that any of the otherplayers is worst leads to a contradiction:

If the woman is the worst player, her brother must be hertwin and her daughter must be the best player. But thewoman and her daughter cannot be the same age.

If the brother is the worst player, the woman must be histwin. The best player is then the son. But the woman and herson cannot be the same age, and hence the woman's twin, herbrother, cannot be the same age as the son.

If the daughter is the worst player, the son must be thedaughter's twin. The best player must then be the woman. Butthe woman and her daughter cannot be the same age.

26. (C) In the adjoining figure

BDICD = ABIAC,

since an angle bisector of a N Dtriangle divides the oppositeside into segments which areproportional to the two adjac- A M Bent sides. Since CN = CDand BM = BD, we have BM/CN = AB/AC, whichimplies that MN is parallel to BC. Since only one choice iscorrect, it must therefore be (C). Actually, it is easy to verifythat (A), (B), (D) and (E) are false if 4A = 900 - 8, 4B =600 and zC = 300 + 8, where 8 is any sufficiently smallpositive angle.

93

C


27. (E) If p, q, r are roots, then the polynomial can be factored asfollows:

x 3 -x 2 + x-2 = (x-p)(x-q)(x-r)

= x3 - (p + q + r)x2

+ (pq + pr + qr)x - pqr.

Equating coefficients of like powers of x, we findp + q + r = 1, pq + pr + qr = 1, pqr = 2.

In looking for the sum of the cubes of the roots of a cubicequation, let us use the fact that each root satisfies theequation:

p3 -p 2 + p - 2 =O

q3 -q 2 + q - 2 =O

r3- r2 + r - 2 = 0.Adding these, we obtain

(*) p 3 + q3 + r3 - (p 2 + q2 + r 2 ) + (p + q + r) - 6 = 0.

We saw that p + q + r = 1 and shall determine the sum ofthe squares of the roots by squaring this relation:

(p + q + r)2 = p2 + q2 + r2 + 2(pq + pr + qr) = I

p 2 + q2 + r2 + 2(l) =1,

P 2 + q 2 +r =-

Substituting this into (*), we obtain

p3 + q3 + r3 = -I - I + 6 = 4.

28. (A) Construct line CP parallel to EF and intersecting AB at P.ThenAP AFAC AE'that is,

AP AF16 2AF '

CA

AP = 8.

Let a, x, y, a, P3, 8 and 0 be as shown in the adjoining dia-gram. The desired ratio EG/GF is the same as y/x which

94


we now determine. By the law of sines,

a = 12sin a sin G '

a = 16 = 16sin i sin(1800 - 9) sinOG

Hence

sin _B 3sin a 4

Moreover

x 8sn= n8'sin a sin 8 '

y 16 16sinf, sin(180° - 8) sin 8

Hence

Y = 2 in = 3x sin a 2

OR

Join GB and GC. Triangle ABC issmaller triangles whose areas aredenoted by a, b, c, d, e, f, as in-dicated in the diagram. TrianglesAEG and AFG have the com-mon vertex A, so their areas arein the ratio EG to GF. Thus

EG = dGF a' A

subdivided into six

CE -

F B

and this we now calculate.Triangles ACM and ABM have equal areas, so

d+e+f =a+b+c.

Similarly f = c, and hence d + e = a + b by subtraction.Let x be the length AF, so that AE = 2x, FB = 12 -x,

EC = 16 - 2x. Then

b FB 12 - x e EC 16 - 2xa FA x and EA 2x

Hence

12 - x 12a + b=a + a = a-

x x

and

d+e= d 16 2xd d22x Ux

95


Thus a + b = d + e becomes

12 16 d 3a-= d 2, or 3a = 2d, so-=a 2

OR

Extend BC and FE until they Hintersect in a point Iadjoining figure. Thepoints A, G, M lie ontheir extensions) BFEof AFBH respectively.lie on extensionsCE, EH, HC of A Lmay therefore apply Itheoremd and find

A F B

HG FA BM HG EA CMFG BA HM ' EG CA HM

Since CM = BM and EA = 2FA, division of the first equa-tion by the second yields

EG BA 12 3- = 2- =2 6- =-FG CA 16 2

29. (C) Instead of trying to compute (V3 + r2 )6 directly, we com-pute something slightly larger and easier to compute, becausemany terms cancel; namely we compute

(ri±+V ) 6 +(v _-f2) 6 .

When (a + b) 2 k and (a - b)2k are expanded by the bi-nomial theorem, their even-powered terms are identical, andtheir odd-powered terms differ only in sign; so their sum is

2[a2k + ( 2k )a2k 2b2 + + b2k].

tMenelaus's theorem: If points X, Y, Z on the sides BC, CA, AB (suitably extended)of &ABC are collinear, then

BX CY AZ- - = 1.

CX A Y BZConversely, if this equation holds for points X, Y, Z on the three sides, then these threepoints are collinear.

For a proof, see e.g. H.S.M. Coxeter and S.L. Greitzer Geometry Revisited, vol. 19, p.66 in this NML series.

96


This principle, applied with a = V, b = r2, k = 3 yields

(VY + C)6 + (V3 - r2 = 2[27 + 15(18 + 12) + 8] = 970.

Since 0 < r- C < 1, 970 is the smallest integer largerthan (r3 + )6

30. (B) Let w = cos 360 and let y = cos 72°. Applying the identities

cos 28 = 2 cos2O - I and cos 2 8 1 -2 sin2 O,

with B = 36° in the first identity and 6 = 180 in the secondyields

y = 2w2 -1 and w = - 2y2 .

Adding these last two equations yields

w + y = 2(w 2 - y 2 ) = 2(w - y)(w + y)

and division by w + y yields 2(w - y) = 1, so

X = W -y 2-


1976 Solutions

1. (B) 1- 1 = 1- I 1-x-1 = 1; x = -1.

2. (B) If x + 1 + 0 then -(x + 1)2 < 0 and -(x+ 1)2 is not

real; if x + I = O then - (x±+ 1)2 = o. Thus x =-1 isthe only value of x for which the given expression is real.

3. (E) The distance to each of the two closer midpoints is one; thedistance to each of the other midpoints is 112 + 22.

4. (C) The sum of the terms in the new progression is

I I rn-1 + rn-2 + + I xI + 1 + -* * ++=l=

r r" ' r"-

Note: If r = 1, then s = n and the sum of the reciprocal pro-gression is also n. This eliminates all choices except (C).

5. (C) Let t and u be the tens' digit and units' digit, respectively, ofa number which is increased by nine when its digits arereversed. Then 9 = (lOu + t) - (lOt + u) = 9(u - t) andu = t + 1. The eight solutions are (12,23,..., 89).

6. (C) Let r be a solution of x 2- 3x + c = 0 such that -r is asolution of x2 + 3x - c = 0. Then

r2 3r + c = 0,r2- 3r - c = 0.

which implies 2c = 0. The solutions of x2 - 3x = 0 are 0and 3.

Note: The restriction to real c was not needed.

7. (E) The quantity (I - IxD(l + x) is positive if and only if eitherboth factors are positive or both factors are negative. Bothfactors are positive if and only if - 1 < x < 1, while bothfactors are negative if and only if x < - 1.

SOLUTIONS:1976 EXAMINATION

8. (A) The points whose coor-dinates are integers withabsolute value less thanor equal to four form a9X 9 array, and 13 ofthese points are at dis-tance less than or equalto two units from theorigin. See figure.

9. (D) Since F is the midpoint ofBC, the altitude of AAEFfrom P to A 1i (pIt~-bA,1

if necessary) is one half thealtitude of AABC fromCto AB (extended if nec-essary). Base AE of AAEFis 3/4 of base AB ofA ABC. Therefore, the areaof AAEF is A

(1/2)(3/4)(96) = 36.

10. (D) For the given functions, f(g(x)) = g(f(x)) is equivalent to

m(px + q) + n = p(mx + n) + q,

which reduces to

mq + n = pn + q or n(l-p) - q(l-m) = O.

If this last equation holds, then f(g(x)) = g(f(x)) is anidentity, i.e. true for every value of x. If on the other handn(I - p) + q(l - m), then f(g(x)) = g(f(x)) has no solu-tion in x.

Note: If the composites fIg(x)] and g[f(x)] of such functions fand g have the same output for some input x, then f and gcommute under composition.

11. (B) The statements "P implies Q," "Not Q implies not P"and "Not P or Q" are equivalent. The given statement,statement III and statement IV are of these forms, respec-tively.

F

B

99


12. (C) There are 25 different possibilities for the number of apples acrate can contain. If there were no more than five cratescontaining any given number of apples, there could be at most25(5) = 125 crates. Since there are 128 crates, n > 6. Weconclude that n = 6 by observing that it is quite possible thatthere are exactly six crates containing k apples in each of thecases k = 120, 121, 122, and exactly five crates containing kapples in each of the cases k = 123, 124,125,. .. , 144.

13. (A) If a cows give b cans in c days and A cows give B cans inC days, then we make the basic assumption

B bAC ac'

i.e. that the number of cans per cow-day is always the same.In the precent case a = x, b = x + 1, c = x + 2, and Ax + 3, B x + 5. Thus

x+ 5 x+ IC x(x + 2)(x + 5)

(x + 3)C x(x + 2) and (x + 1)(x + 3)

Note: As a partial check, we verify that our answer has theappropriate units:

xioews(x + 2)days(x + 5)c8I= Cdays.(x + 3)c~(x + I)car

14. (A) Let n be the number of sides of the polygon. The sum ofthe interior angles of a convex polygon with n sides is(n - 2)1800, and the sum of n terms of an arithmetic pro-gression is n/2 times the sum of the first and last terms.Therefore

(n - 2)180 =- (100 + 140).2

Solving this equation for n yields n = 6.

15. (B) Since each of the given numbers, when divided by d, has thesame remainder, d divides the differences 2312 - 1417 =895 = 5 * 179 and 1417 - 1059 = 378 = 2 * 179; and since179 is prime, d = 179. Now 1059 = 5 * 179 + 164, r = 164,and d - r = 179 - 164 = 15.


16. (E) Let G and H be the points at which the altitudes from Cand F intersect sides AB and DE, respectively. Right trian-gles A GC and FHD are congruent, since side AG and sideFH have the same length, and hypotenuse AC and hy-potenuse DF have the same length. Therefore, 4GAC=4DFH,

$ACG + 4GAC = ZACG + $DFH = 900,

so

$ACB + 4DFE = 2$ACG + 2$DFH = 1800,

and

area AABC = 2(area AACG) = 2(area ADFH) = area ADEF.

C

F

A G B D H E

17. (A) Using trigonometric identities, we obtain

(sinG + cos8) 2 = sin2O + cos 2G + 2sin cosG

= I + sin20= 1 + a.

Since 0 is acute, sin 0 + cos 0 > 0 and sin 8 + cos 0=+a.

18. (E) In the adjoining figure, E is the point of intersection of thecircle and the extension of DB, and FG is the diameter pass-ing through D. Let r denote the radius of the circle.Then

(BC)(BE) = (AB) 2,

3(DE + 6) = 36,

DE= 6. E

Also

(DE)(DC) = (DF)(DG),

6 . 3 = (r - 2)(r + 2),

18 = r2 - 4,

101


19. (B) Let r(x) be the desired remainder. Since its degree is lessthan the degree of the divisor (x - 1)(x - 3), r(x) is of theform ax + b. Thus

p(x) = (x - 1)(x - 3)q(x) + ax + b.

The given information says that p(l) = 3 and p(3) = 5. Set-ting x = I and then x = 3, we obtain

p(l) = a + b 3, p(3) = 3a + b = 5,

so a = 1, b = 2, and r(x) = x + 2.

20. (E) The given equation may be written in the form

4(logax ) 2 8(1ogaX)(logbX) + 3(logbx)2 = 0;

(210gax -logbX)(2logax - 3logbX) = 0;

logax 2 = logbx or log"x 2 = logbx.

Let r = logax2 . Then

ar = x2 and br = x, or ar = x2 and b r = x3

ar = or a 3 r = b2 r

Since x $ I we have r $ 0; and

a = b2 or a3 =b 2

21. (B) We recall that I + 3 + 5 + + 2n + I = (n + 1) 2 andwrite the product

2/7 3/7 . 2 (2n+1)/

7 = 2 I+3+-+2n+l]/

7 =(n+1)2/7

Since 210 = 1024, we consider values of n for which(n + 1)2/7 is approximately 10:

2(7+1)2/7 = 29 +(/7) < 29. 21/2 = (512)(1.41 ... )

< 1000 < 1024 = 210 < 2(9+1)2/7

and n = 9.

22. (A) Let point P have coordinates (x, y) in the coordinate sys-tem in which the vertices of the equilateral triangle are(0,0), (s, 0) and (s/2, sr/2). Then P belongs to the locus

102


if and only if

a=X2 + y2 + (X _ )2 + y +(

or, equivalently, if and only if

a = (3X2 - 3sx) + (3y2 - SVay) + 2S2 ,

a 2s = (x - s/2)2 + (y -sr/6 - S2/3,3

a - = (x - s/2)2 + (y - sli/6)23

Thus the locus is the empty set if a < s2; the locus is a singlepoint if a = s2 ; and the locus is a circle if a > s2

,

23. (A) Since all binomial coefficients ( n) are integers, the quantity

n -2k -I ln (n + 1) -2(k + 1) no

k + I Vk k + I Vk

n+I n! InIn + I n! 2(n-)!

(n + 1)! ( )(k + 1)!(n - k)! k

(k + I) Ik)

is always an integer.

24. (C) In the adjoining figure, MF is parallel to AB and intersectsKL at F. Let r, s(= r/2) and t be the radii of the circleswith centers K, L and M, respectively. Applying thePythagorean theorem to A FLM and A FKM yields

(MF)2 = (r + t) _-( 2 - t ,

(MF)2 = (r -)2 - t

2 .

Equating the right sides Aof these equalities yieldsr/t = 4. Therefore thedesired ratio is 16.

B


25. (D) For all positive integers n,

A'(u,) = (n + 1+ (n + 1) - 3 - n = 3n2 + 3n + 2

A 2 (U") = 3(n + 1)2 + 3(n + 1) + 2 - 3n2 - 3n - 2 = 6n + 6

z0(un) = 6

A4

(un) = O.

Note. If Un is a polynomial in n of degree r, then AlUn is apolynomial of degree r - 1, A2 u, is a polynomial of degreer - 2, etc. From this we see that ArUn is a non-zero constantsequence, and Ar± 'un = 0. In the present example r = 3, soA 3un is a non-zero constant, while A4u" = 0.

Readers familiar with calculus will note an analogy with thefact that if f(x) is a polynomial of degree r, then the rthderivative Drf(x) is a non-zero constant, while Dr+' f(x) =

0. The finite difference operator A is known as the forwarddifference operator and plays an important role in numericalanalysis.

26. (C) In the adjoining figure,X, Y, V and W are thepoints of tangency of theexternal common tan-gents; and R and S arethe points of tangency ofthe internal common tan-gent. From the fact thatthe tangents to a circlefrom an external pointare equal, we obtain:

I

PR = PX, PS = PY,QS= QW, QR= QV.

So

PR + PS + QS + QR = PX + PY + QW + QV,

and thus

2PQ = XY + VW.

Since XY= VW, PQ = XY= VW.

104

7


27. (A) A direct calculation shows that

( V+2 + J -2 2

and

3- 2C2 = 2- 2V2+ I = (V2 - 1)2.

Since a radical sign denotes the positive square root,

N = V - (C2 - l) = 1.

28. (B) One hundred lines intersect at most at 1 )( 2 2

4950 points. But the 25 lines L4, L 8 ,..., L 100 are parallel;

hence (25) = 300 intersections are lost. Also, the 25 lines

Li, L5,..., L97 intersect only at point A, so that (25) -

= 299 more intersections are lost. The maximum number ofpoints of intersection is 4950 - 300 - 299 = 4351.

29. (B) The table below shows the ages of Ann and Barbara atvarious times referred to in the problem. The first columnindicates their present ages. The second column shows theirages when Barbara was half as old as Ann is now-that was

y - (x/2) years ago, hence Ann's age was x - (Y - I) or

2 - y. The third column refers to the time when Barbara

was as old as Ann had been when Barbara was half as old as

Ann is-that was y - - Y) or 2y - years ago;

hence Ann's age then was x - (2y - 2 ) or2 - 2y.

3x 5xAnn x 2 Y 2 - 2y

2

x 3xBarbara Y x 2 Y

By the conditions stated in the problem, x + y = 44

and y = - 2y; the simultaneous solution of these yields

x = 24.

105


30. (E) We observe that we can find a system of symmetric equationsby the change of variables

() x = 2u, y = v, z =w.

This substitution yields the transformed system

u + v + w = 6,

(2) uV + vw + uw= 11,

uvw = 6.

Consider the polynomial p(t) = (t - u)(t - v)(t - w),where (u, v, w) is a solution of the system (2). Then

(3) p(t) = t3 - 6t 2 + lt - 6,

and u, v, w are the solutions of p(t) = 0. Conversely, if theroots of p(t) = 0 are listed as a triple in any order, this tripleis a solution to system (2).

It is not hard to see that p(t) = 0 has three distinct solu-tions; in fact, p(t) = (t - 1)(t - 2)(t - 3). So the triple(1, 2, 3) and each of its permutations satisfies the system (2).Since the change of variables (1) is one-to-one, the originalsystem has 6 distinct solutions (x, y, z): (2, 3, 1), (2,2, 3),(4, 1, 2), (4, 3, 1), (6, 1, 1) or (6, 2, 2

Note: There are methods for determining all solutions of a systemof linear equations in n unknowns, on the one hand, and of asingle n-th degree polynomial in one variable, on the other.No such simple methods are generally applicable to hybridsystems of the type presented in Problem 30. The problemshows that some very special systems can be transformed intoa system of n equations (n = 3 in the present case) involv-ing the elementary symmetric functions of n variables, thuspermitting solutions via a single polynomial equation of de-gree n in one variable.

106


1977 Solutions

1. (D) x + y + z = x + 2x + 2 y = x + 2x + 4 x = 7x.

2. (D) If three equal sides of one equilateral triangle have length s,and those of another have length t, then the triangles arecongruent if and only if s = t.

3. (E) Let n be the number of coins the man has of each type; theirtotal value, in cents, is

I * n + 5 * n + 10 * n + 25 * n + 50 * n = 91n = 273,and n = 3; three each of five types of coins is 15 coins.

4. (C) Since the base angles of an isosceles triangle are equal,4B = 2C = 500,

2EDC = 4CED = 65° and 4BDF= 4DFB = 650.It follows that

.4FDE = 1800 - 2(650) = 500.

Note: The measure of 4EDF is 500 even if AB + AC; see theFigure below.

4 FDE = 1800 - 4 EDC - 4BDF A

= 1800 - '(18O0

- 2(1800

= I'(B + ,C)

= '(180o - 4A)

= 'SOO I

5. (A) If P is on line segment AB, then AP + PB AB; other-wise AP + PB > AB.

6. (D) (2x + +[() () ] ( 4 x+y) [x + ]224x2 2_ 12 4 +xy = y = (xy)-

Note: The function f(x, y) = (2x + 2 ) [(2x) + (2) ] is

homogeneous of degree -2, which means that f(tx, ty) =t- 2f(x,y). Only choice D displays a function with thisproperty.

107


l 1 1_+ __ 1+ V I+y7. (E) = - i+V4

=-(I + W_)(l + Af2)-

8. (B) If a, b and c are all positive (negative), then 4 (respectively,- 4) is formed; otherwise 0 is formed.

9. (B) Let AB = x0 and AD = y'. Then

3x + y = 360,

2 (X - y) = 40.

Solving this pair of equations for y, we obtain y = 300, andhence 4C = ' = 15°.

10. (E) The sum of the coefficients of a polynomial p(x) is equal top(l). For the given polynomial this is (3 I - 1)7 = 128.

11. (B) If n < x < n + 1, then n + I < x + I < n + 2. HenceIx + 11 = lxI + 1. Choosing x = y = 2.5 shows that II andIII are false.

12. (D) Let a, b, and c denote the ages of Al, Bob and Carl, respec-tively. Then

a =16 + (b + c) and a2 = 1632 + (b + C)2;

so a2 = 1632 + (a - 16)2 which yields

1632 - 2 - 16a + 162 = 0 and a =59.Then b + c = 59 - 16 = 43 and a + b + c= 102.

OR

Since we are interested in the sum a + b + c, we try to writeit in terms of the information supplied by the problem:

la + b + c][a - (b + c)] a2 - (b + C)2

a+b+c a - (b + c) a - (b + c)

1632=1 =102.


13. (E) The second through the fifth terms of (an) are2 2 3a2 , ala2 , a1a2 ,a, a2

If these terms are in geometric progression, then the ratios ofsuccessive terms must be equal: a, = a2 = ala 2 . Since a,and a2 are positive, it is necessary that a, = a2 = 1. Con-versely, if a, = a2 = 1, then (an) is the geometric progres-sion 1,1,1,....

14. (B) If m + n = mn, then

m + n - mn = 0, m + n(l - m) = . n=

for m $ 1. There are no solutions for which m = 1. The

solutions (m, -l ) are pairs of integers only if m is 0

or 2.

OR

If mn = m + n, then (m - l)(n - 1) = 1. Hence eitherm - I = n - I = I or m - I = n - I =-1. Thus (m, n)= (2, 2) or (0, 0).

15. (D) In the adjoining figure, PB andOC are radii drawn to common

tangent AD of circle P andcircle Q. Since 4PAB = £QDC= 300, we have

AB = CD = 3V3.

Moreover BC = PQ = 6, andhence AD = 6 + 6F3. There-fore the perimeter is 18 + 18V3.

16. (D) Since

in 2cos(45 + 90(n + 2))0 = -in(-cos(45 + 90n)0)

= iP(cos(45 + 90n)0),

every other term has the same value. The first is r2/2, andthere are 21 terms with this value (n = 0,2,4,..., 40). Thesecond term is i cos 1350 = - iV/2, and there are 20 termswith this value (n = 1, 3,..., 39). Thus the sum is

C2 (21 - 20i).

109

Y


17. (B) The successful outcomes of the toss are the permutationsof (1,2,3), of (2, 3,4), of (3,4,5) and of (4,5,6). The proba-

bility that one of these outcomes will occur is 64 = 9

18. (B) Let y be the desired product. By definition of logarithms,2152 3 = 3. Raising both sides of this equation to the 1og34power yields 2(10523)(10o34) = 4. Continuing in this fashion, oneobtains 2Y = 32; thus y = 5.

OR

Express each logarithm in the problem in terms of some fixedbase, say 2 (see footnote on p. 82). Then we have

1og 2 3 1og2 4 1og 2 5 1og 2 31 1og 2 32

1 log2 2 1og 2 3 10g2 4 1og 2 30 1og2 31

which telescopes to

log232 51 1og 2 2 1

19. (A) The center of acircle circumscrib- R

O . .l In1no a tniannot, I.,

the point of inter-section of theperpendicular bi-sectors of the sidesof the triangle.Therefore, P, Q, Rand S are the in-tersections of theperpendicular bi-sectors of line seg-ments AE, BE, CEand DE. Since linesegments per-pendicular to thesame line are paral-1S>1 P/ID0 i- a

C,

jul, Ir V b aparallelogram.

110


20. (E) All admissible paths end at the center"T" in the bottom row of the diagram.Our count is easier if we go from the endto the beginning of each path; that is, ifwe spell TSETNOC, starting at the bot-tom center and traversing sequences ofhorizontally and/or vertically upwarddirected segments. The count becomeseasier still if we take advantage of thesymmetry of the figure and distinguishthose paths whose horizontal segmentsare directed to the left (see figure)

C

Co

CONCONT

CONTECONTES

CONTEST

from those whose horizontal segments are directed to theright. These two sets have the central vertical column incommon and contain an equal number of paths. Starting atthe bottom corner " T " in our figure, we have at each stage ofthe spelling the two choices of taking the next letter fromabove or from the left. Since there are 6 steps, this leads to 26paths in this configuration. We get 26 paths also in thesymmetric configuration. Since we have counted the centralcolumn twice, there are altogether 2 * 26 - I = 127 distinctpaths.

21. (B) Subtracting the second given equation from the first yieldsax + x + (I + a) = 0

or, equivalently,(a + l)(x + 1) = 0.

Hence, a= -I or x = -1. If a= -1, then the givenequations are identical and have (two complex but) no realsolutions; x = - I is a common solution to the given equa-tions if and only if a = 2. Therefore, 2 is the only value of afor which the given equations have a common real solution.

22. (C) Choosing a = b = 0 yields

2f(0) = 4a(0),

f(0) = 0.Choosing a = 0 and b = x yields

f(x) +f(-x) = 21(0) + 2f(x),

Af-x) =f(x).Note: It can be shown that a continuous function f satisfies the

given functional equation if and only if f(x) = cx 2, where cis some constant.

III


23. (B) Let a and b be the solutions of x2 + mx + n = 0; then

-m = a + b, -p=a3 +ib,

n = ab, q = a3 b3

Since

(a + b)3 = a3 + 3a2 b + 3ab2 + b3 = a3 + b3 + 3ab(a + b),

we obtain

- = 3 p + 3n(-m) or p = M3 - 3mn.

1 1 224. (D) We use the identity 2 2) to write the

given sum in the form

21 [ I + + 25-5 257]

which telescopes to

I I1 1 256 1282 257 2 257 257'

OR

Use the identity

1 1 2

n(n - k) n(n + k) (n - k)(n + k)

to group successive pairs of terms in the sum. Thus lettingk = 2, write

3 1 + 35 + 7.5 + 7 9

2( 1 + 5 +±)

then let k = 4 and write

S=2 2( 1 9+9 17 +-

This process will lead to 27 1 = 281 257 =257

25. (E) Let 2k315 m *-- be the factorization of 1005! into powers ofdistinct primes; then n is the minimum of k and m. Now201 of the integers between 5 and 1005 are divisible by 5;forty of these 201 integers are divisible by 52; eight of theseforty integers are divisible by 53; and one of these eightintegers is divisible by 54. Since 502 of the numbers between


2 and 1005 are even, k > 502; so

n = m = 201 + 40 + 8 + I = 250.

26. (B) If MNPQ is convex, as in Figure 1, then A is the sum ofareas of the triangles into which MNPQ is divided by diago-nal MP, so that

A = 'ab sin N + 'cdsin Q.

Similarly, dividing MNPQ with diagonal NQ yields

A = 'ad sin M + 'bc sin P.

We show below that these two equations for A hold also ifMNPQ is not convex. Therefore, in any case,

a+ c b+ dA < '(ab + cd + ad + bc) = 2 2

The inequality is an equality if and only if

sin M = sin N = sin P = sin Q = 1,

i.e. it and only it MA/PQ is a

PC

Q A A

d > /- " I

-S . I

M a I M a N

Figure I Figure 2

If MNPQ is not convex, for example if interior anglet Qof quadrilateral MNPQ is greater than 1800 as shown inFigure 2, then A is the difference

area of A MNP - area of A MQPso that

A absin N- Icdsin2CPQM22

= Iab sin N - cdsin(360' - 4MQP)2

= -ab sin N + -cdsin MQP.2

tTo distinguish interior angle Q of quadrilateral MNPQ from interior angle Q ofA PMQ in Figure 2, we label the angles in the counter-clockwise direction: 2 MQP isthe angle through which line segment MQ must be rotated counter-clockwise aboutpoint Q to coincide with the line through P and Q. Note that

MQP + 4PQM = 3600.

113


27. (C) In the corner of a room two walls meet in a vertical line, andeach wall meets the floor in a horizontal line. Consider thesethree mutually orthogonal rays as the positive axes of anx, y, z-coordinate system. A sphere with radius a, tangent toall three coordinate planes, has an equation of the form

(x - a)2 + (y - a)2 + (z - a)2 = a2

If the point (5, 5, 10) lies on such a sphere, then

(5 - a)2 + (5 - a)2 + (10 - a)2 = a2

2a 2 - 40a + 150 = 0,

a2 - 20a + 75 = 0.The two solutions of this quadratic equation give the radii ofspheres satisfying the given conditions. The sum of the solu-tions (the negative of the coefficient of a) is 20, so the sum ofthe diameters of the spheres is 40. (Since the last equation isequivalent to (a - 15)(a - 5) = 0 we see that the sphereshave radii 15 and 5.)

28. (A) We shall use the identity

(x -l)(X + X' + *+ X + 1) = X" 1

Thus, for example, (x - I)g(x) = x- 1. By definition ofthe function g, we have

g(x'2) = (X12)5 + (X12)

4 + (X12) 3

+ (X12)2

+ X12 +

= (X6)'O + (x6)8 +(X6)6

+ (X6)4

+ (X6)2

+ 1

Subtracting 1 from each term on the right yields the equation

g(x' 2 ) -6 = [(x6)IO- 1] + [(x6)8 -1] + . . . + [(x6)2 -1].

Each expression on the right is divisible by x6 - 1. We may

therefore write

g(x 1 2 ) - 6 = (x 6 - I)P(x),

where P(x) is a polynomial in x6. Expressing x6 - I in

terms of g(x), we arrive at

g(x'2 ) = (x - l)g(x)P(x) + 6.

When this is divided by g(x), the remainder is 6.

OR

Write g(x 12 ) = g(x)Q(x) + R(x), where Q(x) is a poly-nomial and R(x) is the remainder we are seeking. Since the

114


degree of the remainder is less than that of the divisor, weknow that the degree of R(x) is at most 4.

Since g(x)(x - 1) = x6 - 1, the five zeros of g(x) are- 1 and the other four (complex) sixth roots of unity; so if ais a zero of g(x), then a 6

= 1. Therefore

g(al2 ) = g[(a6)2] = g(1) = 6.

On the other hand,

g(a"2 ) = g(a)Q(a) + R(a),

6 = R(a),

and this holds for five distinct values of a. But the poly-nomial R(x) - 6 of degree less than 5 can vanish at 5 placesonly if R(x) - 6 = 0 for all x, i.e. if R(x) = 6 for all x.

29. (B) Let a = X2, b =y 2 and c = z 2. Noting that (a - b)2 > 0implies a2 + b2 > 2ab, we see that

(a + b + C)2 = a2 + b2 + c2 + 2ab + 2ac + 2bc

< a2 + b2 + c2 + (a 2 + b2 )

+ (a 2 + C2 ) + (b 2 + C2 )

= 3(a2 + b2 + C2 ).

Therefore n < 3. Choosing a = b = c > 0 shows n is notless than three.

30. (A) In the adjoining figure. sides PO Rand TS of the regular nonagonhave been extended to meet at Rand the circumscribed circle hasbeen drawn. Each side of thenonagon subtends an arc of3600/9 = 400; therefore

4TPQ = 4STP

1-. 3 -4O00=600.2

T

Since QSIIPT, it follows that both APRT and AQRS areequilateral. Hence

d = PT= PR = PQ + QR = PQ + QS = a + b.

115


OR

Inscribe the nonagon in a circle of radius r. Since the chordsof lengths a, b, d subtend central angles of 400,800, 1600respectively, we have

a = 2rsin200 , b = 2rsin400, d = 2rsin 800.

By means of the identity

sin x + sin y = 2 sin 2 Y cos 2Y2 2

with

x = 40°, y = 20,

we obtain sin 400 + sin 20° = 2 sin 30° cos 10°. Sincesin 300 = 2, this yields

sin 40 ° + sin 20 ° = cos 100 = sin 80 °,

and therefore b + a = d.


1978 Solutions

1. (B) I- 2=+ = (I- 2 ) = °; 2= 1.

2. (C) Let r be the radius of the circle. Then its diameter, cir-cumference, and area are 2r, 27Tr and 1Tr

2 , respectively. The4 =2

given information reads 2 = 2r. Therefore 4 = 47rand the area is 7Tr 2

= 1.

( I ) ( + y = (X-_Y)(X + y) = X2 _ y2.

4. (B) (a + b + c - d) + (a + b - c + d) + (a - b + c + d)+(-a + b + c + d) = 2(a + b + c + d) = 2222.

5. (C) Let w, x, y and z be the amounts paid by the first, second,third and fourth boy, respectively. Then since

w + x + y + z = 60,

w = I(x + Y + Z) = 2(60 - w), w = 20;1 1

x = I(W +y + z) = 1(60- x), x = 15;

1 3

y = I(w + x + z) = 1(60 - y), y = 12.

Any of these equations now yields z = 13.

6. (E) If y + 0, the second equation implies x = 2, and the firstequation then implies y = ± 2. If y = 0, the first equationimplies x = 0 or 1. Thus we have the four distinct solutionpairs (2, 0) 2 29(,), (1, 0).


7. (E) In the accompanying figure A3vertices A1 and A3 of the hexa-gon lie on parallel sides twelveinches apart, and M is the mid-point of A1 A3 . Thus AA1A2Mis a 30°-60°-900 triangle, and

A1A2 2 so A 1A2 = 4C.6 r

A,

8. (D) In the first sequence, it takes three equal steps to get from xto y; in the second, it takes 4. Hence the ratio of these stepsizes is

a 2 - a, (y - x)/3 4b2 - b= (y - x)/4 3

9. (B) IxA(x -1 | Ix - Ix - II I = Ix -(I -X)I= I - 1 + 2x1 = 1 - 2x.

10. (B) For each point A other than P, the point of intersection ofcircle C with the ray beginning at P and passing through Ais the point on circle C closest to A. Therefore the ray be-ginning at P and passing through B is the set of all points Asuch that B is the point on circle C which is closest topoint A.

11. (C) If the line with equation x + y = r is tangent to the circlewith equation X

2 + y2 = r, then the distance between the

point of tangency ( ') and the origin is Vr. Therefore,(r )2 + ( r )2 rand r 2.

118


12. (E) Let x = 4BAC = 4BCA; y = 4CBD = 4CDB and z =,(DCE = C DEC. Applying the theorem on exterior angles toA ABC and A A CD and the theorem on the sum of the inter-ior angles of a triangle to AADE yields

y = 2x,

z = x + y = 3x,

x + LADE + z = 1800,

1400 + 4x = 180°,

x= 10°. D

A C E

13. (B) Since the constant term of a monict quadratic equation is theproduct of its roots,

b = cd, d = ab.

Since the coefficient of x in a monic quadratic equation isthe negative of the sum of its roots,

-a = c + d, -c = a + b;

thus a + c + d = 0 = a + b + c, and b = d. But theequations b = cd and d = ab imply, since b = d $ 0,that I = a = c. Therefore, b = d = - 2, and

a+b+c+d= -2.

14. (C) Since n2 - an + b = 0, and a = (18)n = I n + 8, we have

b = an - n2 = (n + 8)n - n2

= 8n = (80)n.

OR

The sum of the roots of x2 - ax + b = 0 is a = (18)n. Since

one root is n = (10),, the other is (18),, - (10)n = (8)n = 8.Hence the product b of the roots is (8)n * (10)n = (80)n,

tA monic polynomial is one in which the coefficient of the highest power of the variableis 1.

119


15. (A) If sin x + cosx = 5 then cosx = 5 - sinx and

5' 2

cos2 x = sin2x = - sin x);

so

25 sin 2x - 5sinx - 12 = 0.

The solutions of 25s 2 - 5s - 12 = 0 are s = 4/5 and s =-3/5. Since 0 < x < f, sin x > 0, so sin x =4/5 andcosx = (1/5) - sinx = -3/5. Hence tanx = -4/3.

16. (E) Label the people A1, A 2 ,..., A N in such a way that Al andA2 are a pair that did not shake hands with each other.Possibly every other pair of people shook hands, so that onlyAl and A2 did not shake with everyone else. Therefore, atmost N - 2 people shook hands with everyone else.

17. (D) Applying the given relation with x = 3/y yields

9 y2

)23y = [(()2 + )j] =2

18. (C) We seek the smallest integer n such that C4 - n - I

< . Since, for positive a, a < c if and only if ->-,10O0 a c

we must find the smallest integer such that

I = + l > 100.

vA-Since 2500 + 2499 < 100 and 2501 + 2500 > 100,the least such integer is 2501.

19. (C) Since 50p + 50(3p) = 1, p = .005. Since there are seven per-fect squares not exceeding 50 and three greater than 50 butnot exceeding 100, the probability of choosing a perfectsquare is 7p + 3(3p) = .08.

120


20. (A) Observe that the given relations imply

a + b-c +2 a-b + c 2 -a+ b+c 2c b a

that is,

a+b+c a a+b+c a a+b+c= ac b a

These equalities are satisfied if a + b + c = 0, in which casex = -1. If a + b + c + 0, then dividing each member ofthe second set of equalities by a + b + c and taking thereciprocals of the results yields a = b = c. In that case x = 8.

21. (A) Since log, a = (see footnote on p. 82), we see thatlogab

± + = log,3 + log,4 + log,5log 3 x log4 x log5 x

= log60 og 0x

22. (D) Since each pair of statements on the card is contradictory, atmost one of them is true. The assumption that none of thestatements is true implies the fourth statement is true. Hence,there must be exactly one true statement on the card. In fact,it is easy to verify that the third statement is true.

23. (C) Let FG be an altitude of A AFB, and let x denote the lengthof AG. From the adjoining figure it may be seen that

D C

/l + V3 = AB = x(l + ),

I+ V3 = x2 (l + r,/)2

A x G x,/3 B

The area of AABF is

'(AB)(FG) = 2x2(1 + J = iFi.

121


24. (A) Let a = x(y - z) and observe that the identity

x(y - z) + y(z - x) + z(x -y) = 0

implies

a + ar + ar2 = 0;

I + r + r2 = 0

25. (D) The boundary of the set of points satisfying the conditions isshown in the adjoining figure.

y , Y=x+a

2a -

x+y=a \a,

a/2 --

(~n r.

I /Y=2a

/ / y=x-a

i y = a/2

I \, I WaX_nl n On

tuV) U/a U /U

26. (B) In the first figure, K is any circle tangent to AB and passingthrough C. Let T be the point of tangency and NT the di-ameter through point T. Let CH be the altitude of AABCfrom C. Then NT> CT> CH with NT = CH if and onlyif N= C, and T = H. There is such a circle with diameterCH, so it is the (unique) circle P of the problem, shown inthe second figure. By the converse of the Pythagorean theo-rem, .RCQ = 900; thus QR is also a diameter of P. Since

ACBH - AABC, CH = lo so QR = CH = 4.8.

A

Figure 2Figure I


27. (C) A positive integer has a remainder of 1 when divided by anyof the integers from 2 through 11 if and only if the integer isof the form mt + 1, where t is a nonnegative integer andm = 23 - 32 . 5 * 7 11 = 27,720 is the least common multi-ple of 2,3,..., 11. Therefore, consecutive integers with thedesired property differ by 27,720.

28. (E) Triangle A 2 A3 A4 has vertexangles 60°,300,90°, respectively. A3Since 4A,A 2 A 3 = 600, andA2A4 and A2 A5 have the samelength, AA2 A4 A5 is equilateral.Therefore, AA3 A4A. has vertexangles 30, 300, 1200, respective-ly. Then AA 4 A 5 A 6 has vertex

Ilr no _,no nno +soseagles u ,ouv-,u-, [espMLIvey. Al A 4 A 2Finally, since $A 4 A 5A6 = 600and A5A6 and AA7 have the same length, AA 5A 6 A. isagain equilateral. The next cycle yields four triangles, eachsimilar to the corresponding triangle in the previous cycle.Therefore AAnAn+i A n+2 - &A,+ 4 An+5An+ 6 with An andA,+4 as corresponding vertices. Thus

4A44A4 5A43 = $A4 A 5A3 = 1200.

29. (D) Since the length of base AA' of /AAA'B is the same as thelength of base AD of AABD, and the corresponding altitudeof AAA'B' has twice the length of the corresponding altitudeof AABD,

areaAAA'B' = 2areaAADB,

see figure on next page. (Alternatively, we could let B be themeasure of 2CDAB and observe

areaAAA'B' = I2 (AD)(2AB) sin (180- 0)

= 2[ (AD)(AB) sin9] = 2areaAABD.)

Similarly

areaABB'C' = 2areaABAC,

areaACC'D' = 2area ACBD,areaADD'A' = 2areaADCA.

123


Therefore

area A'B'C'D'

= (area /AAA'B' + area ABB'C')

+ (area ICC'D' + area A DD'A')

+ area ABCD= 2(area&ABD + areaABAC)

+2(areaACBD + areaADCA,

+ area ABCD= 5 area ABCD = 50.

30. (E) Let 7m, 5m be the total number of matches won by womenand men, respectively. Now there are

n(n - 1) matches between women, hence won by women.2

There are

2n(2n - 1) = n(2n - 1) matches between men, hence2

won by men. Finally, there are

2n n = 2n2 mixed matches, of which

k = 7m - (2 ) are won by women, and

2n2 - k = 5m - n(2n - 1) by men.

We note for later use that k < 2n2 . Then

7m - in(n - 1) + k I 1n(n - 1) + 2k5m n(2n-1) + 2n2 - k 2 4n2 - n-k

5n(n - 1) + l0k = 14(4n2 - n -k),

51n 2 - 9n - 24k = 3(17n2 - 3n - 8k) = 0,

17n2 -3n = 8k,

17n2 - 3n8

124


Since k < 2n2 , it follows that 17n 2 - 3n < 16n2, son(n - 3) < 0, n < 3. Since k is an integer, n +# 1 or 2;hence n = 3.

OR

The total number of games played was

(3n) 3n(3n-1)

Suppose that 7m games were won by women, and 5m gameswere won by men. Then

7m + 5m = 12m = 3n(3n- 1) and 3n(3n - 1) = 24m.2

Since m is an integer, 3n(3n - 1) is a multiple of 24. Butthis is not true for n = 2,4,6, 7. (It is true for n = 3.)


1979 Solutions

1. (D) Rectangle DEFG has area 18, one quarter of the area ofrectangle ABCD.

2. (D) l - l = -x = -x -y xyx y xy xy xy

3. (C) Since 4DAE = 900 + 60° = 1500 and DA = AB = AE,

4AED = (180 150)0 = 150

4. (E) x[x(x(2 - x) - 4) + 10] + I= x[x(2x _ - 4) + 10] + I = x[2x 2 - X3- 4x + 10] + I= -x 4 + 2x 3 - 4X2 + 10x + 1.

5. (D) The units and hundreds digits of the desired number mustclearly be equal. The largest such even three digit number is898. The sum of its digits is 25.

6. (A) Add and subtract directly or note that

3 5 652 4 64

-+-I+ ... +-- 1I -2 4 64 4

7. (E) Let x = n2, n > 0. Then

(n + 1) 2 = n2 + 2n + I = x + 21 + 1.

O AAr AA -£ 11A l-r. (C) Ine area 01 tUe sIallestregion bounded by y =xl and X2 + y2 = 4 isshown in the adjoiningfigure. Its area is

I (,7T22) =-.

-y= IxI


9. (E) -44 = (22)'/3(23)1/4 = 22/323/4 = 217/12 = 21+(5/12)

- 25/12 ~212= 2 1 20( L b = 2 1P2__P

10. (D) Let C be the center of P3- Q2 P2the hexagon; then thearea of Q1Q2Q3Q4 isthe sum of the areasof the three equilateraltriangles AAQ 1Q 2 C,

AQ2 Q3 C, AQ3 Q4 C,each of whose sideshave length 2. The areaof an equilateral trian-gle of side s is Is2r3.Therefore, P5 P6

22 C3area Q1Q2Q3Q4 = 3 = 3V-

4

11. (B) Summing the arithmetic progressions yields

115 1 + 3 + + (2n-1)116 2 + 4 + * + 2n

n2 n

n(n + 1) n+ 1

hence n = 115.

12. (B) Draw line segment BO, and let x and y denote the mea-sures of 4EOD and 2CBAO, respectively. Observe that AB= OD = OE = OB, and apply the theorem on exterior an-gles of triangles to &ABO and AAEO to obtain

4EBO = 4BEO= 2y

and

x = 3 y.

Thus

450 = 3y,

15° = y.A

0

127

I

C


OR

Since the measure of an angle formed by two secants is halfthe difference of the intercepted arcs,

Y I (x 1 y = x = 450 = 15

13. (A) Consider two cases:

Case 1: If x > 0 then y-x < V if and only ify - x < x, or equivalently, y < 2x.

Case 2: If x < 0 then y - x < Vix if and only ify - x < -x, or equivalently, y < 0.

The pairs (x, v)satisfying the condi-tions described ineither Case 1 or Case2 are exactly the pairs(x, y) such that y <0 or y < 2x (see ad-joining diagram forgeometric interpreta-tion).

x

Note 1: Since V72 xlI, the given inequality can be written

Y < x + lxi = ( 2x forx > 00 for x< 0

and the graph of x + [xl is the boundary of the shadedregion in the figure.

Note 2: The given inequality holds if x = 1, y = 0, and also ifx = y = - 1. This eliminates all choices except (A).

14. (C) Let an denote the nth number in the sequence; then

a~a2 ... an n2_ _

a1a2 ... an 1 (n- 1)2

for n > 2. Thus, the first five numbers in the sequence are

1,4,- 2 , and the desired sum is9+ = 1

128


15. (E) If each jar contains a total of x liters of solution, then onejar contains P+ liters of alcohol and +- liters of

water, and the other jar contains qx liters of alcohol andq + 1

q+ liters of water. The ratio of the volume of alcohol to

the volume of water in the mixture is then

(P+ q )+ l p(q+ l)+q(p+ I)

I + I )x (q + ) +(p 1)

p + q + 2pqp + q + 2

16. (E) Since A1, A2 , AI + A2 are in arithmetic progression,

A 2 -Al = (Al + A 2 )-A 2 ; 2A, = A2 .

If r is the radius of the smaller circle, then

9ir = Al + A2 = 3A, = 37r 2 ; r = V .

17. (C) Since the length of any side of a triangle is less than the sumof the lengths of the other sides,

x < y-x + z-y = z-x, which implies x < 2;2'

zy-x < x + z - y, which implies y < x + -;

2'z

z-y < x + y - x, which implies Y > 2

Therefore, only statements I and II are true.

X B C

,Y X X

A B C D

129


18. (C) Since logb a = lOI b (see footnote on p. 82),

105 1 1 10log10 5 log1 l0 10- logl 0 2 .699 7

(The value of log10 3 was not used.)

19. (A) Since 25632 = (28)32 = 2256, the given equation is equivalentto

1 x 256- = 1.

Among the 256 256th roots of 1, only I and - I are real.Thus x = 2 or x = -2 and 22 + (-2)2 = 8.

20. (C) Let x = Arctan a and y = Arctan b:

(a + 1)(b + 1)= 2,

(tan x + 1)(tan y + 1) =2,

tanax ± tany tan x tan y,

tan x + tany =1.I - tanxtany

The left hand side is tan(x + y), so

tan(x + y) = 1, x +y = = Arctan a + Arctan b.

ORSubstituting a = 2 in the equation (a + 1)(b + 1)= 2 andsolving for b, we obtain b = 3. Then

tan x + tan y a + b l + 3

tan(x + A) = = =1I - tan x tan y - -ab I-, 6

ITso x+y= .

Note: x + y cannot be (7r/ 4 ) + nir for a non-zero integer n be-cause 0 < x, y < 77/ 2.

21. (B) In the adjoining figure, 0 is the center of the circle and x, yare the lengths of the legs of the triangle. So

h = (y - r) + (x - r)

= x +y - 2r,


and

x +y = h + 2r.

lx

The area of z ABC is the sum of the areas of triangles A OB,BOC and A OC, whose altitudes have length r. Thus

area AABC = I(xr + yr + hr) = (x + y + h)

= 2 (h + 2r + h) = r(h + r).2

Thus the desired ratio is =hr + r2 h + r

Note: Alternatively, the area of AABC is

I 1I (X +y) 2 (X2 +y2)

2 2 2

- [(h + 2r)2 - h2] = hr + r2 .

22. (A) The left member of the given equation can be factored intom(m + 1)(m + 5) and rewritten in the form

m(m + 1)(m + 2 + 3) = m(m + 1)(m + 2) + 3m(m + 1).

For all integers m the first term is the product of threeconsecutive integers, hence divisible by 3, and the secondterm is obviously divisible by 3. So for all integers m, the leftside is divisible by 3.

The right side, 3[9n3 + 3n2 + 3n] + I has remainder Iwhen divided by 3. Therefore there are no integer solutions ofthe given equation.

131


23. (C) Let M and N be the Dmidpoints of AB andCD, respectively. Weclaim that M and Nare the uniquechoices for P and Qwhich minimize thedistance PQ. Toshow this we considerthe -.Pt .S of al nointsequidistant from Aand B. S is the plane perpendicular to AB through M.Since C and D are equidistant from A and B, they lie inS, and so does the line through C and D. Now M is thefoot of the perpendicular to AB from any point Q on CD.Therefore, if P is any point on AB,

MQ < PQ unless P = M.

Similarly, the plane through N perpendicular to CD con-tains AB. In particular, MN I CD; thus

MN < MQ unless Q = N.

By transitivity, MN < PQ unless P = M and Q = N. Thisproves the claim.

To compute the length of MN, we note that MN isthe altitude of isosceles ADMC with sides of lengths V3/2,V3/2, 1. The Pythagorean theorem now yields

V(MC2 _ Arc 3 _1MN= (MC) 2 -(NC) 2 = I

= C2 = minimal distance PQ.

24. (E) Let E be the intersection of lines AB and CD, and let /3and 8 be the measures of 4EBC and 2ECB, respectively;see figure. Since

cos A= -cos B= sinC = sinO,

/3 + 0 = 90°, so YBEC is a right angle, and

BE=BCsinO=3; CE=BCsin/3=4.

Therefore, AE = 7, DE = 24 and AD, which is the hypot-enuse of right triangle ADE, is V72 + 242 = 25.

132


5

C 20 D

25. (B) Let a =- . Applying the remainder theorem yields r=a8, and solving the equality x8 = (x - a)q,(x) + r, forq1(x) yields

q1 (x) = X a = x7 + ax 6 + +a 7

[or, by factoring a difference of squares three times,

q1(x) = (x 4 + a )(x 2 + a 2 )(x + a)].

Applying the remainder theorem to determine the remainderwhen q1(x) is divided by x - a yields

r2 = q, (a) = 8a7 = - 1

Note: For solvers familiar with calculus, there is anotherway to find qu(a). We differentiate the identity x8

-

(x - a)q,(x) + r, with respect to x, obtaining

8x7 = ql(x) + (x - a)q,(x).

Setting x = a yields 8a7 = qu(a).

26. (B) Substitute x = I into the functional equation and solve forthe first term on the right side to obtain

f(y + I) =f(y) +y + 2.

Since f(l) = 1, one sees by successively substituting y2, 3,4,... that f(y) > 0 for every positive integer. There-fore, for y a positive integer, f(y + 1) > y + 2 > y + 1,and f(n) = n has no solutions for integers n > 1. Solvingthe above equation for f(y) yields

f(y) = f(y + 1) - (y + 2).

Successively substituting y = 0, - 1, - 2,... into this equa-tion yields f(O) = -1, f(- 1) = - 2, fi( - 2) = -2,f(-3) = -1, f(-4) = 1. Now f(-4) > 0 and, for y <-4,-(y + 2) > 0. Thus, for y < -4, f(y) > 0. Therefore,

E

133

-


f(n) + n for n < -4; and the solutions n = 1, -2 are theonly ones.

OR

We write the functional equation in the form

f(x+y)-f(y)=f(x)+xy+ 1.

Setting x = I and using the given value f(l) = 1, we find

f(y + 1) -f(y) =y + 2.We now set y = 0, 1, 2,. . . n successively, then y =

0, - 1, - 2,. .. - m successively, and obtainf(l) -f(O) = 2 (so f(O)= -1) f(O f( l) = I

f(2) -f(l) = 3 f(- ) -f(-2) = O

f(n)-f(n I)=n + I f(- (- I))- f(-n)- (m-2).

Adding the set of equations in the left column and cancellinglike terms with opposite signs yields

n+I

f(n) -f(O) = 2 + 3 + - + n + 1=-1+ i.

Recalling that the sum of the first k positive integers is'k(k + 1) and using the value f(0) = -1, we obtain

f(n)+= (n + l)(n + 2)2

(I) f(n) = (n2 + 3n - 2)

for each non-negative integer n.The same procedure applied to the column of equations on

the right above shows that equation (1) is valid also fornegative integers; thus (1) holds for all integers.

The equation f(n) = n is equivalent to

n2 + 3n - 2 = 2n,n2 + n -2 = (n + 2)(n - 1) = 0,

n= I or n=-2;so there is only one integer solution other than n = 1.

Note 1: The computations involving the equations in the rightcolumn above can be avoided by setting y = -x in theoriginal equation to obtain

f(X)±( ( _ x) =f(O) + X2 1-

or since -f(O) = - 1,

(2) f(x) = -f(-x) + x2-2

for all x. Let x be negative and substitute (1) in (2) to get

134


A~X) = -l [(-I ) + 3(- x) -2] +x-2 = 2 (2+ 3x -2).

Thus negative integers as well as positive satisfy (1). In fact, itis easy to check that the function

x2 + 3x - 2f( ) = 2

satisfies the given functional equation for all real x and y.

Note 2: The technique used in the second solution is frequentlyapplied to solve many commonly occurring functional equa-tions involving the expression Af(x) = f(x + 1) - f(x);Af(x) is called the "first difference" of f(x) and behaves,in many ways, like the first derivative f'(x) of f(x). Forexample

f'(x)= x +2 implies f~) - + 2x +c;2

Af(x) = x + 2 implies f(x) = x(x - 1) + 2x + c,2

where c is a constant. The study of functions by means ofAf(x) is called the theory of finite differences.t

27. (E) These six statements are equivalent for integers b, c withabsolute value at most 5:

(1) the equation x2 + bx + c = 0 has positive roots;(2) the equation x2 + bx + c = 0 has real roots, the smaller

of which is positive;(3) b2 - 4c is real and -b - Vb2 - 4c > 0;(4) 0 b2 - 4c < b2 and b < 0;

(5) 0 < c s T and b < 0;

(6) b -2 and c = 1; or b = -3 and c = I or 2; orb =-4 and c = 1, 2, 3 or 4; or b =-5 and c = 1, 2,3, 4 or 5.

The roots corresponding to the pairs (b, c) described in (6)will be distinct unless b2 = 4c. Thus, deleting (b, c) =(-2, 1) and (-4,4) from the list in (6) yields the ten pairsresulting in distinct positive roots.

The desired probability is then I - Q = 11112 =121.

tSome references to this theory or to the theory of more general functional equationsare: Finite Differences by S. Goldberg; "Functional Equations in Secondary Mathe-matics", by S. B. Maurer, an article in The Mathematics Teacher, April 1974.

135


28. (D) Denote by A' the analogous intersection point of the circleswith centers at B and C, so that, by symmetry, A'B'C' andABC are both equilateral triangles. Again, by symmetry,AABC and A A'B'C' have a common centroid; call it K. LetM be the midpoint of the line segment BC. From the trian-gle A'BC we see that the length AkM = - I. Since cor-responding lengths in similar triangles are proportional,

B'C' A'KBC AK-

Since equilateral AABC has sides of length 2, we find that

B'C' = 2 A- and also that altitude AM has length V.AK

Consequently

AK = 2AM = 2-v/3-, MK = I AM= -;3 3 3 3

and

A'K =AM+MK= Vr2 l + (J /3).

Thus

+ +(r/3)B'C' = 2 2- /3) = 3(r2 - 1) + 1.

136


29. (E) By observing

that

that X3 + ] = X6 + 2 + one sees

+ })3]2 [X3 + l ]2

(X + -) +X3 + 3

= (X+ -) -(X3 + 13 )=3 3x

- I 2 =

+x).

x +-- 2, we have 2 < x +-,x x

and f(x) = 3(x +-) has a minimum value of 6, which is

taken on at x = 1.

30. (B) Let F be the point on theextension of side AB pastB for which AF = 1. SinceAF = AC and jFAC =60°, &ACF is equilateral.Let G be the point on linesegment BF for which4BCG = 200. Then ABCGis similar to ADCE andBC= 2(EC). Also AFGCis congruent to AABC.Therefore,

F

'\\

\ \'' \

A D C/u EgsH

/ o

area AACF = (area AABC + area AGCF) + area &BCG,

3= 2 area AABC + 4 area zCDE,

V8= area AABC + 2 area &CDE.

Since

0 (

137


1980 Solutions

100 21. (C) Since 7 14 + 7,the number is 14.

2. (D) The highest powers of x in the factors (x2 + 1)4 and

(X3

+ 1)3 are 8 and 9, respectively. Hence the highestpower of x in the product is 8 + 9 = 17.

3()f2x -y 23. (E) If + = , then

6x - 3y = 2x + 2y,

4x = 5y,

x 5y 4

4. (C) The measures of angles 4ADC, ZCDE and 4EDG are 900,60° and 90°, respectively. Hence the measure of 4GDA is

3600 - (900 + 600 + 900) = 1200.

5. (B) Triangle PQC is a 30°-600-90° right triangle. SinceAQ= CQ

PQ PQ I jAQ CQ =yy 3

6. (A) Since x is positive the following are equivalent:

Cx < 2x, x < 4x2, I < 4x, 1 < x, x > 1.4 41

7 (B) By the Pythagorean theorem, diagonal AC has length 5.Since 52 + 122 = 132, ADAC is a right triangle by the con-verse of the Pythagorean theorem. The area of ABCD is

(2)(3)(4) + (42)(5)(12) = 36.

138


8. (A) The given equation implies each of these equations:

a+b Iab a + b'

(a + b)2= ab,

a2 + ab + b2 = 0.Since the last equation is satisfied by the pairs (a, b) suchthat a = l [ - b ± i ,- 3 1, and since the only real pair amongthese is (0,0), there are no pairs of real numbers satisfyingthe original equation.

9. (E) As shown in the adjoining figure, there are two possiblestarting points; therefore, x is not uniquely determined.

Possible starting points.

3 -~---j An/

10. (D) Since the teeth are all the same size, equally spaced and aremeshed, they all move with the same absolute speed v (v isthe distance a point on the circumference moves per unit oftime). Let a, /3, y be the angular speeds of A, B, C, respec-tively. If a, b, c represent the lengths of the circumferences ofA, B, C, respectively, then

I' V Va = a' A= - Y ca b c'

Therefore, aa = 8ib = yc or, equivalently,

a / y-T= - = I'

a b c

Thus the angular speeds are in the proportion

a b C

Since a, b, c, are proportional to x, y, z, respectively, theangular speeds are in the proportion

x y z

Multiplying each term by xyz yields the proportionyz: xz: xy.


11. (D) The formula for the sum Sn of n terms of an arithmeticprogression, whose first term is a and whose common dif-ference is d, is 2Sn = n(2a + (n - I)d). Therefore,

200 = 10(2a + 9d),

20 = 100(2a + 99d),

2S lo = 110(2a + 109d).

Subtracting the first equation from the second and dividingby 90 yields 2a + 109d= -2. Hence 2SI10 = 110(-2), soS 110 = - 110.

12. (C) In the adjoining figure, LIand L2 intersect the linex = I at B and A, respec-tively; C is the intersectionof the line x = I with thex-axis. Since OC= 1, ACis the slope of L2 and BCis the slope of LI. There-fore, AC = n, BC = m,and AB = 3n. Since OAis an angle bisector

OC ACx

OB AB

This yields

I_ = 3nand OB= 3.

By the Pythagorean theorem I + (4n)2 = 9, 0 n = 2Since m = 4n, mn = 4n2 = 2.

OR

Let 01 and 02 be the angles of inclination of lines LI andL2, respectively. Then m = tan 0, and n = tan 02. Since 01= 202 and m = 4n,

4n = m = tan 0, = tan282 = 2 tan 02 2nI - tan2O2 1 n2

Thus

4n 1n 2 and 4n(1 - n2) = 2n.

Since n * 0, 2n2 = 1, and mn, which equals 4n2 , is 2.


13. (B) If the bug travels indefinitely, the algebraic sum of thehorizontal components of its moves approaches 4, the limitof the geometric series

1 1 1 1I - I I- .4 16

4J)

Similarly, the algebraic sum of the vertical components of itsmoves approaches

2 11 12-= I---+-...

5 2 8 32

Therefore, the bug will get arbitrarily close to (4, 5).

OR

The line segments may be regarded -as a complex geometricsequence with a, = 1 and r = i/2. Its sum is

aX a, 2 4 + 2i 4 2.ai = = - 5 = - + -'-

=1 - r 2-i 5 5 5

In coordinate language, the limit is the point ( , 5)

Note: The figure shows that the 4

limiting position (x, y) of 8

the bug satisfies x > 3/4, 2

y > 3/8. These inequalitiesalone prove that (B) is thecorrect choice.

314. (A) For all x - 2

X =f(f(x)) = , + 3 2cx + 6x + 9'

which implies (2c + 6)x + (9 -c2 ) = 0. Therefore, 2c + 6

= 0 and 9 -_c 2 = 0. Thus, c =-3.

OR

The condition f[f(x)] = x says that the function f(x) is itsown inverse. Thus, if

cx

Y 2x + 3

is solved for x as a function of y, the result will be x = f(y).


Indeed, we obtain

2xy + 3y - cx = 0, x(2y-c) =-3y,

and

-3yx=2y - ~c~

which implies c = -3.

Note: In order to form f [f(x)], we assumed that if x $ - 3/2,then f(x) $ - 3/2. This assumption can be justified ex postfacto now that we have found c = -3; indeed if

-3x - 3t~) 2x + 3 2'

then 6x = 6x + 9, a contradiction.

15. (B) Let m bethepriceoftheitemincents.Then (1.04)m = lOOn.Thus (8)(13)m = (100) 2n, so m = (2)(5)4 n . Thus m is an

13integer if and only if 13 divides n.

16. (B) The edges of the tetrahedron are face diagonals of the cube.Therefore, if s is the length of an edge of the cube, the areaof each face of the tetrahedron is

(r 4)2r3 52r

4 2'

and the desired ratio is

6s 2

4( 2 )

17. (D) Since i2 - 1,

(n + i)4 = n 4- 6n2 + 1 + i(4n3 - 4n).

This is real if and only if 4n3 - 4n = 0. Since 4n(n2 - 1) =

0 if and only if n = 0, 1, - 1, there are only three values ofn for which (n + i)4 is real; (n + i)4 is an integer in allthree cases.

18. (D) logbsin x = a; sin x = ba; sin2x = b2 a; cos x = (I -b2a)/2;

logbcos X = 2 logb(l - b2 ,).

142


19. (D) The adjoining figure is drawn and labelled according to thegiven data. We let r be the radius, x the distance from thecenter of the circle to the closest chord, and y the commondistance between the chords. The Pythagorean theorem pro-vides three equations in r, x, and y:

r2 =X2 + 102,

r2 = (X + y)2 + 82,

r2 =(x + 2y)2 + 42

Subtracting the first equationfrom the second yields

0 = 2xy + y2 - 36,

and suhtracting the secondequation from the third yields

0 = 2xy + 3y2 - 48.

This last pair of equations yields 2y 2 - 12 = 0, thus

y = l&, and x = 15/l&. Finally

r = rx2 + 102 = 5V2/2.

20. (C) The number of ways of choosing 6 coins from 12 is

(12) = 924.6

[The symbol ( n ) denotes the number of ways k things maybe selected from a set of n distinct objects.] "Having at least50 cents" will occur if one of the following cases occurs:

(1) Six dimes are drawn.(2) Five dimes and any other coin are drawn.(3) Four dimes and two nickels are drawn.

The numbers of ways (1), (2) and (3) can occur are

(6), (6)( 6) and (6)( 4), respectively. The desired proba-bility is, therefore,

(6) + ()(4) + (6)() 127924 924

143


21. (A) In the adjoining figure the line segment from E to G, themidpoint of DC, is drawn. Then

area ISEBG = (2 )(area AEBC),

area ABDF = ( 4 )(area AEBG) = ( 6)(areaAEBC)

(Note that since EG connects the midpoints of sides ACand DC in AACD, EG is parallel to AD.) Therefore,

area FDCE A

= (- 6)(areaAEBf

and

area tLBDF Iarea FlCPF 5 * I

The measure of 4 CBA was not needed.

C

22. (E) In the adjoining figure, the graphs of y = 4x + 1, y = x + 2and y = - 2x + 4 are drawn. The solid line represents thegraph of the function f. Its maximum occurs at the intersec-tion of the lines y = x + 2 and y = -2x + 4. Thus

x2x= -3

and

f(2) 833 3

1+

* 4

x

23. (C) Applying the Pythagorean theorem to ACDF and ACEGin the adjoining figure yields

4a 2 + b2 = sin2X,

a 2 + 4b2= cos2 x.

.


Adding these equations, we obtain

5(a 2+ b 2

) = sin2 x + cos2x = I.

Hence

AB = 3 a2 + b 2

= 3-

5 3 .- 5

a

145

24. (D) Denote the given polynomial by P(x). Since P(x) is divisi-ble by (x - r)2 , we have, for some polynomial L(x),

P(x) = (x -r)2L(x),

and since P(x) is of degree 3, L(x) is of degree 1, sayL(x) = ax + b. So we have the identity

8x 3 - 4X2 - 4 2x + 45 = (x2 - 2rx + r2 )(ax + b).

In the right member the coefficient of x3 is a, and theconstant term is br2 . Hence a = 8, and br2 = 45, or b =

45/r2 , and the identity becomes

8x3 - 4x2 - 42x + 45 = (x2 - 2rx + r2)(8x + 2

Equating coefficients of x2 and x, we obtain

45-16r +-= -4,

r2

2 - 90= -42.r

Multiplying the first equation by 2r and adding it to thesecond, we have

- 24r2 = - 8r - 42or

12r2 - 4r - 21 = 0 = (2r - 3)(6r + 7),

so that r = 3/2 or r = --7/6. Now P(3/2) = 0, butP(-7/6) + 0. We conclude that r = 3/2 = 1.5.

B

ID

! sinx : E_-A'F'5,_

I

I . IIb F b G b A

a


OR

For solvers familiar with calculus we include the follow-ing alternative. By hypothesis 8x3 - 4X2 - 42x + 45 =8(x - r)(x - s). Differentiating both sides, we obtain

24X2 -8x - 42 = 16(x - r)(x - s) + 8(x - r)2 .

Setting x = r anddividingby2, weget 12r2 - 4r - 21 = 0.From here on the solution proceeds as above.

25. (C) The given information implies that an > an-I if and only ifn + c is a perfect square. Since a 2 > a, and a5 > a4 , it fol-lows that 2 + c and 5 + c are both squares. The onlysquares differing by 3 are I and 4; hence 2 + c = 1, soc= - 1. Now

a2 = 3 = b[2+c] + d = b[] + d = b + d.

Hence b + c + d = 3 - 1 = 2. (Although we only needed tofind b + d here, it is easy to see by setting n = 1 that d = 1,and hence b = 2.)

Note: The last member of the k-th string of equal terms occupiesthe position I + 3 + 5 + *-- + 2k- I = k2 in the se-quence. Its successor is ak2+1 = ak2 + 2. Therefore

[ k2 + 1 + - -[ k 2 + = =

so c =-1 and b = 2.

26. (E) A smaller regular tetrahedron circumscribing just one of theballs may be formed by introducing a plane parallel to thehorizontal face as shown in Figure 1. Let the edge of thistetrahedron be t. Next construct the quadrilateral CIBIB 2C2 ,where C, and C2 are centers of two of the balls. By thesymmetry of the ball tetrahedron configuration the sidesC1 B1 and C2 B2 are parallel and equal. It follows that

BIB2 = CIC2 = 1 + 1 = 2.

Thus, s = t + 2, and our problem reduces to that of de-termining t.

146


Figure 2

Figure I

In AABCI, in Figure 2, AC, has a length of one radiusless than the length b of altitudes DB and AE of the smallertetrahedron; and AB has length 2 (V3/2)t, since B is thecenter of a face. Applying the Pythagorean theorem toAABC, yields

(b - 1)2 = 12 + [3 ( 2 t '

and applying the Pythagorean theorem to AADB yields

t= b2 + [(Ž2 t)j 2.

Solving the second equation for b, substituting this value forb in the first equation, and solving the resulting equation forthe positive value of t yields t = 2A6. Thus s = 2 + t=2 + 2A6.

OR

We first consider the regular tetrahedron T whose verticesare the centers of the four unit balls and calculate the distanced from the center of T to each face. We then magnify T sothat the distance from the center to each face is increased by1, the radius of the balls. We thus obtain the tetrahedron T'whose edge length e' is to be determined. Since T and T'are similar, and since the edges of T have length 2, we have

147


e' d+ 1(1) 2 d

We carry out this plan by placing the center of T at theorigin of a three dimensional coordinate system and denotethe vectors from the origin to the vertices by cl, c 2, C 3, C 4,and their common length by c. Since cl + C2 + C3 + C4 = 0,

(2) - = C2 + C3 + C4.

Taking the scalar product of both sides with c, yields

-C] q cl = -C 2 = C) * C2 + C, * C3 + Cl * C4 = 3c 2 cos,

where 0 is the angle between ci and cj, i $ j. Thus

Cos 0 = -3 .

The square of the length of an edge of T is

IC1 - c212 = 4 = 2c2 - 2c2 cos8 = c2(2 + 2) = 8c2,

so

(3) c = A3/.

Now the line from the origin perpendicular to a face of Tgoes through the centroid of the triangular face. So thedistance d is the length of the vector I (C2 + C3 + C4 ); by (2)we have,

d=I3(C2 + C3 + C4)= 3C.

Substituting for c from (3), we obtain

d = (1/3)3/2 = 1 .

Finally, from (1),

e' =2 = 2 2+ 2_ rd I/Vr

OR

Some solvers may be familiar with the fact that the altitudesof a regular tetrahedron intersect at a point 3/4 of the wayfrom any vertex to the center of the opposite face.t Given this

tThis can be proved by finding the center of mass M of four unit masses situated at thevertices. In finding M, the three masses on the bottom face can be replaced by a singlemass of 3 units located at the center of the bottom face. Hence M is 3/4 of the wayfrom the top vertex to this latter point. The result now follows by symmetry.

148


fact, it follows that the altitude DB of the small tetrahedronis 4. From AADB we find

2 /2V+ C 2 t2

t = 4 16 + +3

Hence 2 t2 = 16, or t = 4 = 2A.

27. (E) Let a= V5 +21fT,b= 15 -2Vfi and x= a+ b. Then

x3 =a3 + 3a 2 b + 3ab2 + b3,

x3 =a 3 + b3 + 3ab(a + b),

x3 =10 + 33 27x.

The last equation is equivalent to x3 + 9x - 10 = 0, or(x - 1)(x 2 + x + 10) = 0, whose only real solution is x = 1.

28. (C) Let f(x) = x2 + x + 1 and g,(x) = X2n + 1 + (x + 1)2n.Note that (x - 1)f(x) = x3 - 1, so that the zeros of f(x)are the complex cube roots of 1:

= - + -23i = cos 1200 + isin 1200 = e"l3 I t2 2

' = - I - C3 = cos240' + isin24 0 ' = e-i2i/3 22 2

Note that

W,3 = ()3 1

Now gn(x) is divisible by f(x) if and only if g&(w) =

gn(w') = 0. Since w and w' are complex conjugates, itsuffices to determine those n for which

gn(w) = W 2n + (W + 1)2n + 1 = 0.

We note that

+ I= + 2i = ei-f/6 , so (W + 1) 2 = eir32 2

tThe polar representation of the complex number a + ib is r(cos 6 + i sin 6), whereb

r = Va 2 + b2 and 6 = arctan-. For brevity, we set cos 6 + i sin 6 = e'0 , 6 in

radians. Note that ei2k4l = I for all integers k. This exponential notation is not arbi-

trary. Complex powers of e are defined as e = e' (cos 6 + isin0) for all real aand 9.

149


Thus, whenever n is a multiple of 3, say n = 3k, we have

g3 k(W) = (06k + (Co + 1 )6k + 1 = 1 + 1 + 1 = 3 $ 0.

Suppose n is not a multiple of 3.If n = 3k + 1,

W2n = 6k+2 =2,2 (W + ) = n = W3k+ =I ,

and

93k+ I()= W2

+ W + 1 = -I + 1 = 0.

If n = 3k + 2,

,2

n =6k+4 =W (w + ) =3k+2 = a2

and

93k+2(W) = W + W2 + 1 = 0.

Thus gn(W) + 0 if and only if n is a multiple of 3, andgn(x) fails to be divisible by f(x) only in that case.

OR

Both of the given polynomials have integer coefficients, andx2 + x + I has leading coefficient 1. Hence if

(1) X2n + I + (x + 1)2 = Q(x)(x2 + x + 1),

then Q(x) has integer coefficients. Setting x = 2 in (1), weobtain

22n + I + 322n = Q(2) 7,

which shows that 7 divides 22n + I + 32n, However if n isdivisible by 3, both 22n and 32n leave remainders of 1upon division by 7, since 26k = 64- = (7 * 9 + I)k, and3 `k (7 2 9 )k = (7 104 + l)k; so 22n + I + 322n leaves aremainder of 3 upon division by 7.

Note: The second solution, unlike the first, does not tell us whathappens when n is not divisible by 3, but it does enable usto answer the question. We know that only one listed answeris correct, and (C) is a correct answer since 3 divides 21.

150


29. (A) If the last equation is multiplied by 3 and added to the firstequation, we obtain

4x2 + 2y 2 + 23Z2 = 331.

Clearly z 2 is odd and less than 25, so Z2= 1 or 9. Thisleads to the two equations

2X2 + y2 = 154

and

2X2 + y 2 = 62,

both of which are quickly found to have no solutions. Notethat we made no use here of the second of the originalequations.

OR

Adding the given equations and rearranging the terms of theresulting equation yields

(x 2 - 2xy + y 2 ) + (y 2 + 6yz + 9z 2 ) = 175

or

(x -y) 2 + (y + 3z) 2 = 175.

The square of an even integer is divisible by 4; the square ofan odd integer, (2n + 1)2, has remainder 1 when divided by4. So the sum of two perfect squares can only have 0, 1 or 2as a remainder when divided by 4. But 175 has remainder 3upon division by 4, and hence the left and right sides of theequation above cannot be equal. Thus there are no integralsolutions.

30. (B) That N is squarish may be expressed algebraically as follows:there are single digit integers A, B, C, a, b, c such that

N = 104A2 + 102B2 + C2 = (102 a + 10b + C)2,

where each of A, B, C exceeds 3, and so a and c are posi-tive. Since 102B2 + C2 < 104 we can write

104A2 < (102a + 10b + c)2 < 104A2 + 104 < 104 (A + 1)2.

Taking square roots we obtain

lOOA < 100a + 10b + c < lOOA + 100,

151


from which it follows that A = a. Hence a > 4. Now con-sider

M = N - 104A2 = (102 a + lOb + c)2 - 104a2

= 103 (2ab) + 102 (b2 + 2ac) + 10(2bc) + c2 .

Since M has only four digits, 2ab < 10, which implies thatab < 4. Thus either (i) b = 0, or (ii) a = 4 and b 1.In case (ii),

N = (410 + C)2 = 168100 + 820c + c2 .

If c = I or 2, the middle two digits of N form a numberexceeding 81, hence not a square. If c > 3, then the leftmosttwo digits of N are 17. Therefore case (i) must hold, and wehave

N = (102

a + C)2 = 104

a2

+ 102 (2ac) + C

2 .

Thus a > 4, c > 4 and 2ac is an even two-digit perfectsquare. It is now easy to check that either a = 8, c = 4,N = 646416, or a = 4, c = 8, N = 166464.

152


1981 Solutions

1. (E) x + 2 = 4; (x + 2)2 = 16.

2. (C) The Pythagorean theorem, applied to A EBC,= 22 - 12 = 3. This is the area of the square.

yields (BC)2

1 1 1 6 3 2 1 13. (D) -+ I + l= 6- + - + 2 11

* x 2x 3x 6x 6 6x 6

4. (C) Let x be the larger number. Then x - 8 is the smaller num-ber and 3x = 4(x - 8), so that x = 32.

5. (C) In /ABDC, BDC = 40°. Since DC is parallel to AB,

4DBA = 40°.

Also, 4BAD = 40° since base angles of an isosceles triangleare equal. Therefore CADB = 1000.

6. (A) (y 2 + 2y - 2)x = (y 2 + 2y - I)x - (y2 + 2y - 1),

[(y2 + 2y - 2) - (y 2 + 2y - 1)]x = - (y2 + 2y - 1),

X =y 2 + 2y - 1.

OR

Rewrite the right member of the given equality as

(y 2 + 2y - 1)

(y2 + 2y - 1) - I

and note by inspection that x = y2 + 2y - 1.

7. (B) The least common multiple of 2, 3, 4 and 5 is 60. The num-bers divisible by 2, 3, 4 and 5 are integer multiples of 60.

153


8. (A) The given expression equals

I-- (I + I + ,8 1 Ar I + I +' I A

- I(xy+yz + zx+f I + )x +y+ Z)

+ + zVxyzz xyz

I = 2y-2-2

(xyz)2

Note: The given expression is homogeneous of degree -6; i.e. ifx, y, z are multiplied by t, the expression is multiplied byt - 6. Only choice (A) has this property.

9. (A) Let s be the length of an edge of the Qcube, and let R and T be vertices ofthe cube as shown in the adjoiningfigure. Then applying the Pythagoreantheorem to APQR and APRT yields

a2 - 52 = (pR) 2 = S2 + S2 ,

a2 = 3s2 .

The siurface area is 6s 2 = 2n22

10. (E) If (p, q) is a point on line L, then by symmetry (q, p) mustbe a point on K. Therefore, the points on K satisfy

x = ay + b.

Solving for y yields

x ba a

11. (C) Let the sides of the triangle have lengths s - d, s, s + d.Then by the Pythagorean theorem

(s - d) 2 + S2 = (s + d)2 .

Squaring and rearranging the terms yields

s(s - 4d) = 0.Since s must be positive, s = 4d. Thus the sides have lengths3d, 4d, 5d. Since the sides must have lengths divisible by 3, 4,or 5, only choice (C) could be the length of a side.

154


Note: The familiar 3-4-5 right triangle has sides which arein arithmetic progression. It is perhaps fairly natural to thinkof multiplying the sides by 27, thus getting 81, 108, 135.Since there is only one correct choice, it must be (C).

12. (E) The number obtained by increasing M by p% and decreas-ing the result by q% is M(l + j )(I _ and ex-ceeds M if and only if the following equivalent inequalitieshold:

M(l + p )(I- q )> M,

10 100

I + P. I > 100

100 1- q 100 - q'100

p 100 -q100 100- q 100 - q'

lOOqP 100-q

13. (E) If A denotes the value of the unit of money at a given time,then .9A denotes its value a year later and (.9)"A denotes itsvalue n years later. We seek the smallest integer n such thatn satisfies these equivalent inequalities:

(.9)'A 6 IAA

/ 9 )" I

logl 9 106 log 1 I,

n(2log1 03 - 1) 6 - 1,

I- 1 21.7.I -2og1 03

155


14. (A) Let a and r be the first term and the common ratio ofsuccessive terms in the geometric sequence, respectively. Then

a + ar = 7,

a(r 6 - 1)

r-lI

The result of dividing the second equation by the first is

a(r 6 - 1) = 91 =13(r - I)a(r + 1) 7

The left member reduces to

(r 2-2 )(r

4 + r2

+ I) = r + r2 + 1,

r- 1so

r4 + r2 - 12 = 0,

(r2 + 4)(r2 - 3) = 0.

Thus r2 = 3 and

a + ar + ar2 + ar3 = (a + ar)(l + r2 )

= 7(4) = 28.

15. (B) For this solution write log for logb. The given equation isequivalent to

(2x) = (3 x)Ig3

Equating the logarithms of the two sides, one obtains

log2(log 2x) = log 3(log 3x),

log2(log2 + logx) = log 3(log3 + log x),

(log 2)2 + log 2(logx) = (log 3)2 + log 3(log x),

(log 2)2 - (log 3)2 = (log 3 - log 2)log x,

-(log2 + log3) = logx,

log 6 = log x,

6 = x.

156


16. (E) Grouping the base three digits of x in pairs yields

x = (I 319 + 2 318) + (I. 317 + I 316)

+ -- + (2 *3 + 2)

= (I 3 + 2)(32)9 + (I .3 + 1)(32)8

+ * * * + (2 3 + 2).

Therefore, the first base nine digit of x is I 3 + 2 = 5.

17. (B) Replacing x by - in the given equationx

f (x) ±f ( 3x

yields

I )+ 2f(x) 3 -

Eliminating f (-) from the two equations yields

2-X 2

A X) = 2Xx

Then f(x) = f(-x) if and only if

2-x 2 2-(-X)2

x -x

or x2 = 2. Thus x = - are the only solutions.

18. (C) We have x = sin x if and only if x = sin( - x); thus,100 0

the given equation has an equal number of positive andnegative solutions. Also x = 0 is a solution. Furthermore, allpositive solutions are less than or equal to 100, since

lxi = lOOlsinxl < 100.

Since 15.5 < 100/(27,) < 16, thegraphsof x/100 and sin xare as shown in the figure on p. 158. Thus there is onesolution to the given equation between 0 and so and twosolutions in each of the intervals from (2k - I)ir to(2k + 1)ir, 1 < k < 15. The total number of solutions is,therefore,

I + 2(1 + 2 .15) = 63.

157


V

X/ X0 - -

---- --- --- --- 30,r 31# 10

19. (B) In the adjoining fig- Aure, BN is extendedpast N and meets ACat E. Triangle BNA iscongruent to AENA,since ZBAN= SEAN.AN = AN and 4ANB = KANE. Therefore N is the mid-point of BE, and AB = AE = 14. Thus EC = 5. SinceMN is the line joining the midpoints of sides BC and BE ofACBE, its length is 2(EC) = 2-

20. (B) Let 4DARI = 8 and let 8i be the (acute) angle the lightbeam and the reflecting line form at the ith point of reflec-tion. Applying the theorem on exterior angles of trianglesto AARID, then successively to the triangles Ri-,RiD,2 6 i 6 n, and finally to ARBD yields

#I = 0 + 80,

82 = 01 + 80 = 0 + 160,

83 = 82 + 80 = 8 + 240,---------------

en = - + 80 = 0 + (8n)0,

90° = 8,, + 80 = B + (8n + 8)0.

But 0 must be positive. Therefore,

0 < 8 = 900 - (8n + 8)0,

82n 8 <11.

The maximum value of n, 10, occurs when 8 = 20.

158


AK2 B

21. (D) Let 0 be the angle opposite the side of length c. Now

(a + b + c)(a + b - c) = 3ab,

(a + b)2 - C2 = 3ab,

a2 + b2 - ab = C2.

But

a2 + b2 - 2abcos0 = C2,

so that ab = 2ab cos 0, cos0 = 2 and 0 = 600.

22. (D) Consider the smallest cube containing all the lattice points(i, j, k), 1 6 i, j, k < 4, in a three dimensional Cartesiancoordinate system. There are 4 main diagonals. There are 24diagonal lines parallel to a coordinate plane: 2 in each offour planes parallel to each of the three coordinate planes.There are 48 lines parallel to a coordinate axis: 16 in eachof the three directions. Therefore, there are 4 + 24 + 48 = 76lines.

OR

Let S be the set of lattice points (i, j, k) with I < i, j, k <4, and let T be the set of lattice points (i, j, k) with 0 6i, j, k < 5. Every line segment containing four points of Scan be extended at both ends so as to contain six points of T.

v)

* Points of So Points of T - S

Io 0 o 0 0 00

-- O

I 0

, * * .Ox

159


(The figure on p. 159 shows two of these extended lines in thexy-plane.) Every point of the "border" T - S is contained inexactly one of these lines. Hence the number of lines is halfthe number of points in the border, namely half the numberof points in T minus half the number of points in S

1 (63 - 43) = 76.2

Note: The same reasoning shows that the number of ways ofmaking tic-tac-toe on an n-dimensional "board" of s' latticepoints is 2[(s + 2)' - s'].

23. (C) Let 0 and H be the points atwhich PQ and BC, respectively,intersect diameter AT. Sides ABand AC form a portion of theequilateral triangle circumscrib-ing the smaller circle and tangent Bto the smaller circle at T. There-fore, APQT is an equilateral tri-angle. Since AAPQ is an equi-lateral triangle with a side incommon with APQT,AAPQ t-PQT. Thus A0 is the center of the larger circle. This implies

A

T

0 = OT and

AO = 2(AH), so that PQ= 2 (BC)=8.

124. (D) Write x + -= 2 cos O as x2- 2x cos O + 1 = 0. Then

x

x = cos 0 -cos 2 o -1 =cos a + i sin 0 (= e ).

By De Moivre's theorem

xn =cosnO ± isinnO(= e± MO),

x" cosnO+isinnO =cosnO + isinnO(= e'@).

Thus

n + I- = 2cosnO.

Note 1: Squaring both sides of the given equation, we obtain

x + 2 + I = 4cos2O.x

2

160


Hence

x2 + I = 4cos 2O - 2 = 2(2cos2O -1) = 2cos20.

This eliminates all choices except (D).

Note 2: Several contestants wrote to the Contests Committee tosay that this problem had no solution because, for 0 < 0 < 1,

as given, 2cos 0 < 2, yet x +- 2 for all x. This claim

is true for real x, but nowhere in the problem did it say thatx must be real. When such a restriction is intended, it isalways stated.

25. (A) In the adjoining figure let 4BAC = 3a, c = AB, y = AD,z = AE and b = AC. Then by the angle bisector theorem

(1) C = 2 and Yb =I;z 3 b 2'

sob 2

z= 23c, b=2y.B 2 D 3 E 6

Using the law of cosines in AADB, AAED and AACE, re-spectively, yields the following expressions for cos a:

+Y 4 9C2 + y2 - 9 9 c2 + 4y2 - 36C 4 4 4

2cy 3cy 6cy

The equality of the first and second expressions implies

3c 2 - 2y 2 = 12.

tThe angle bisector theorem states that each angle bisector of a triangle divides theopposite side into segments proportional in length to the adjacent sides, i.e.

a a e() c b

in the adjoining figure. This is a consequence of thelaw of sines applied to each subtriangle and the factthat the supplementary angles at D have equalsines. B d D e C

161


The equality of the first and third expressions implies

3C2 - 4y2 = -96.

Solving these two equations for c2 and y 2 yields

c2 = 40, y 2 = 54.

Thus the sides of the triangle are

AB = c = 2h - 6.3,

AC = b = 2y = 24 = 66 14.7,

BC= 11.

OR

Using the angle bisector formulat and (1) above, we have

y 2 + 6 = cz = 2z2 z 2 + 18 =yb = 2Y2.3

Solving these equations for y 2 and z2 yields y2 = 54,z2 = 90. Therefore

AB = c = 2 X0 = 2V , AC = b = 2V = 6V.

tThe angle bisector formula states that the square of the angle bisector plus the productof the segments of the opposite side is equal to the product of the adjacent sides:

() k2

+ de= bc.

This is a consequence of the angle bisector theoremand the law of cosines:

c2+ k 2- 2ckcosa =d2

b 2 + k2 - 2bkcos a = e2

. B d D e C

Multiplying the first equation by b, the second by c and subtracting, we obtain

bc2

- b2

c - k2

(c - b) = d2

b - e2

c.

By (*) (see footnote on p. 161) the right member can be written dec - edb; thus

(c - b)[bc - k2 1 = (c - b)de.

Dividing by (c - b) and adding k2 yields formula (**). In the isosceles case, c - b =0; but the angle bisector is the altitude so that (**) becomes a consequence of thePythagorean theorem.

162


26. (D) The probability that the first 6 is tossed on the k-th toss isthe product( probability that never a 6 was (probability that a 6 is

tossed in the previous k-I tosses) tossed on the k' toss)

= (5 /6 )k l (1/6).

The probability that Carol will toss the first 6 is the sum ofthe probabilities that she will toss the first 6 on her first turn(3 rd toss of the game), on her second turn (6th toss of thegame), on her third turn, etc. This sum is

(5 )2 1 + ( 5 )5 1 + ** + ( 5 )3n1 16

an infinite geometric series with first term a = ) 6 and

common ratio r = (-)3. This sum is

a 52/63 52 25I-r I-(53/63) 63-53 91

27. (C) In the figure, line seg-ment DC is drawn.Since AC_= 1500, AD= AC - DC = 1500 -30° = 120°. Hence4ACD = 600. Since AC= DG GA = CD - AXD= AC - 1200 = 300.Therefore CG = 1800and 4CDG = 90°. SoADEC is a 30°-60°-90°triangle.

Since we are looking,0-An - -urn re; Wr I--o -----1or1 Lit; I dLIU UI LI a es C d-- - Blet us assume without lossof generality that AC = AB = DG = 1. Since AC and DGare chords of equal length in a circle, we have AE = DE. Letx be their common length. Then CE = I - x = 2x/ 4.Solving for x yields AE = x = 2r3 - 3. Since BA andDG are chords of equal length in a circle, we have FG = FA,and since AFAE is isosceles, EF = FA. Thus

EF = FG = '(1 - x).

163


Therefore

area AAFE= I(AE)(AF)sin300

1 I -x I x - x 2 7V3- 12=-r - .- = =I

2 2 2 8 4

area AABC= I(AB)(AC)sin3O=

Hence

area AAFE r-12area IA ABC =

28. (D) Let g(x) = X3 + a 2 x 2 + ax + a0 be an arbitrary cubic withconstants of the specified form. Because X3 dominates theother terms for large enough x, g(x) > 0 for all x greaterthan the largest real root of g. Thus we seek a particular g inwhich the terms a2 x2 + ax + ao "hold down" g(x) asmuch as possible, so that the value of the largest real root is aslarge as possible. This suggests that the answer to the problemis the largest root of f(x) = X - 2x 2 - 2x - 2. Call thisroot r0 . Since f(O) = -2, r. is certainly positive. To verifythis conjecture, note that for x > 0,

-2X 2 < a2 x2 , -2x < alx, and -2 < ao.

Summing these inequalities and adding X3 to both sidesyields f(x) < g(x) for all x > 0. Thus for all x > ro,o < f(x) < g(x). That is, no g has a root larger than ro,so ro is the r of the problem.

A sketch of f shows that it has a typical cubic shape,with largest root a little less than 3. In fact, f(2) = -6 andf(3) = 1. To be absolutely sure the answer is (D), not (C),compute f(' ) to see if it is negative. Indeed,

(5) - 31/2= 8

29. (E) Since x is the principal square root of some quantity, x > 0.For x > 0, the given equation is equivalent to

a- +x= x 2

or

a = a+x X 2 .

The left member is a constant, the right member is an

164


increasing function of x, and hence the equation has exactlyone solution. We write

aI= a- x2,

-a ±x + x = (a + x) -x,

= (Va + x)(± a ± x - x).

Since a +x + x > 0, we may divide by it to obtain

1= +x- x or x + = a+x,

so

x2 + 2x + I = a + x,and

x2 + x + 1 - a = 0.

Therefore x= I 3, and the positive root is x

2 3, the only solution of the original equa-

tion. Therefore, this is also the sum of the real solutions.

OR

As above, we derive a - a+x= x 2 , and hencea-x2 = a + x. Squaring both sides, we find that

a 2 - 2x 2 a + X4 = a + x.

This is a quartic equation in x, and therefore not easy tosolve; but it is only quadratic in a, namely

a - (2x 2 + I)a + x - x = 0.

Solving this by the quadratic formula, we find that

a = l2[2X2 + 1 + V4X4 + 4X2 + 1 - 4X4 + 4x]

- X2 + X + 1.

[We took the positive square root since a > x2 ; indeeda - x2 = la +x.]

Now we have a quadratic equation for x, namely

x2 + x +I -a=0,which we solve as in the previous solution.

Note: One might notice that when a = 3, the solution of theoriginal equation is x = 1. This eliminates all choices except(E).


30. (D) Since the coefficient of x3 in the polynomial function f(x)= X4 - bx - 3 is zero, the sum of the roots of f(x) is zero,and therefore

a + b + c a + b + c + d-d -1d2 d 2 d

Similarly,

a + c + d -I a + b + d -I b + c + d -I

b2 b' c2 C' a2 a

Hence the equation f I) = 0 has the specified solutions:

I + b -3=0,X 4 x

I + bx 3 - 3x4 = 0,

3x 4 - 1X3 _ I 0.


1982 Solutions

1. (E) xx 2 +Ox-2 x3 + OX2 +Ox -2

x3 + OX2 - 2x2x -2 = remainder.

ORWe have

x3 - 2 X 3 -2x + 2x-2 2x-2x2-2 x2-2 x2 -2'

so the remainder is 2x - 2.

2. (A) The answer is 84 2x + 2

3. (C) For x = 2, the expression equals

(22)(22) = 44 = 4 4 4* 4 = 256.

4. (E) Let r be the radius of the semicircle. The perimeter of asemicircular region is irr + 2r. The area of the region is

1 221Tr . Therefore

irr + 2r = 2 r2'

27r + 4 = ir,

2 +-= r.1T

b b5. (C) Since y = -x, - > 1, and x > 0, it follows that x is thea a

smaller number. Also, x + y = c. Thus

bX +-X = C,

a

ax + bx = ac,ac

Xa +b -


6. (D) The sum of the angles in a convex polygon of n sides is(n - 2)1800. Therefore, if x is the unknown angle,

(n - 2)180° = 25700 + x, withO0 < x < 1800.

If n = 17, then (n - 2)180° = 15(1800)= 27000 and x =130°. Smaller values of n would yield negative values of x,and larger values of n would yield values of x greater than1800.

7. (B) x*(y + z) = (x + l)(y + z + 1)- 1,

(x * y) + (x * z) = [(x + l)(y + 1) - 1]

+ [(X + l)(Z + 1) - 1]

= (x + l)(y + z + 2) - 2.

Therefore, x * (y + z) +# (x * y) + (x * z). The remainingchoices can easily be shown to be true.

8. (B) Since

(2) _ (1) = (3) -(2)

we have

n(n -1) n n(n - 1)(n -2) n(n - 1)2 6 2

Thus

n- 9n2 + 14n = 0,n(n - 2)(n - 7) = 0.

Since n > 3, n = 7 is the solution.

(The answer may also be obtained by evaluating the sequence

(I), (2), (n3) for the values of n listed as choices.)

9. (B) In the adjoining Y x = afigure, ABC is thegiven triangle andx = a is the divid-ing line. Since areaAABC = 2(1)(8)= 4, the two re-gions must each B(O,have area 2. Since Ithe portion of I

)


AABC to the left of the vertical line through vertex A hasarea less than area AABF = 2-, the line x = a is indeed rightof A as shown. Since the equation of line BC is y = x/9,the vertical line x = a intersects BC at a point E:(a, a/9).Thus

area &DEC = 2= 1( - 9)(9 -a),

or

(9 - a)2 = 36.

Then 9-a = ±6, and a = 15 or 3. Since the line x = amust intersect AABC, x = 3.

10. (A) Since MN is parallel to BC, A

4MOB = 2CBO = 2OBM,

and12 18

4CON= 4OCB = NCO. MN

Therefore MB = MOand ON = NC. Hence B C

AM + MO + ON + AN = (AM + MB) + (AN + NC)

= AB + AC = 12 + 18 = 30.

Note that the given value BC = 24 was not needed; in fact,only the sum of the lengths of the other two sides was used.

11. (C) From the set (0, 1,..., 9) there are sixteen pairs of numbers{(0, 2), (2,0), (1, 3), (3, 1),... } whose difference is ± 2. Allbut (0, 2) can be used as the first and last digit, respectively,of the required number. For each of the 15 ordered pairsthere are 8 * 7 = 56 ways to fill the remaining middle twodigits. Thus there are 15 * 56 = 840 numbers of the requiredform.

12. (A) Since f(x) = ax7 + bx3 + cx - 5,

f(-x) = a(-x)7 + b(-x)3 + c(-x) - 5.

Therefore, f(x) + f(-x) = -10 and f(7) + f(-7) = -10.Hence, since f ( - 7) = 7, f(7) = - 17.

169


13. (D) We have

p (logb a) = logb (logb a),logb(aP) = logb(logb a),

aP = logb a.

14. (C) In the adjoining figure, MN is perpendicular to AG at M,and NF and PG are radii. Since AAMN- LXAGP, it fol-

lows t MN GP or MN = 15. Thus MN= 9. Ap-

plying the Pythagorean theorem to triangle MNF yields(MF)2

= (15)2 - 92 = 144, so MF = 12. Therefore EF=24. G

A

15. (D) We have

21x] + 3 = 31x - 21 + 5,

21xJ + 3 = 3(IxJ - 2) + 5,

Jxl = 4.

Therefore, 4 < x < 5, and y = 21x1 + 3 = 11. Hence,15 < x + y < 16. (Alternatively, if one draws the graphs ofy = 21x1 + 3 and y = 3[x - 21 + 5, one can see that theyoverlap when 4 < x < 5).

16. (B) Each exterior unit square which is removed exposes 4 interiorunit squares, so the entire surface area in square meters is

6- 32 - 6 + 24 = 72.

17. (C) Let y = 3x; then 32x+2 = 9y 2 and 3x+3 = 27y. The givenequation now becomes

9y 2 - 28y + 3 = (9y - 1)(y - 3) = 0

and has the solutions y = 3 and y = 1. Hence the originalequation has exactly two solutions, namely x = I and x=-2.

170


18. (D) Without loss of ieneral- Aity, let HF= 1 in theadjoining figure. ThenBH = 2, BF = = DG=GH, and DH=F6.Since DC = HC = F3, VADCB - AHCB andDB = HB = 2. Sincez DBH is isosceles,

2 4

B

F3

H I F

19. (B) When 2 < x . 3, f(x) = (x - 2)-(x - 4) + (2x - 6) =-4 + 2x. Similar algebra shows that when 3 < x < 4, f(x)= 8-2x; and when 4 < x < 8, f(x)= 0. The graph off(x) in the adjoin- ving figure showsthat the maximum 2and minimum of0,f(x) are 2 and 0,respectively. 0 O x

Note: Since a linear 1 2 3 4 5 8function reaches its extreme values at the endpoints of aninterval, and since the given function is linear in each subin-terval, it suffices to calculate f(2) = 0, f(3) = 2, f (4) = f (8)= 0.

20. (D) Since

X2 +y 2 = x 3,

y = x 2 (x- 1).

Therefore, if k is an integer satisfying x - I = k2 , i.e.,x = I + k2 , then there is a y satisfying X2 + y 2 = X 3 .

Hence there are infinitely many solutions.

21. (E) Since the medians of a triangle intersect at a point two thirdsthe distance from the vertex and one third the distance fromthe side to which they are drawn, we can let x = DN and

171


2x = BD. Right triangles BCN and BDC are similar, so 2x= BD. Right triangles BCN and BDC are similar, so

s _ 2x C3x s

Thus s2 = 6x2, or x = ,

and BN= 3x = = sV'7 )VD

22. (E) In the adjoining figure RX isperpendicular to QB at X.Since CQPR = 600, ARPQ isequilateral and RQ = a. Also4ARP = 4QRX = 150.Therefore, &RXQ =- RAP.Thus w = h.

A P B

OR W

w = AP + PB = acos75' + acos45'.

From the identity

cos A + cos B = 2 cos cos22it follows that

750 + 450 750 - 450w = a2cos 2 cos 2

= a2cos60°coslS0 = acosl5' = asin75' =h.

23. (A) In the adjoining figure, n denotes the length of the shortestside, and 6 denotes the measure of the smallest angle. Usingthe law of sines and writing 2 sin e cos 9 for sin 29, we obtain

sin = 2 sin 0 cos 0n n+2

n n + 2cos 0= 2

Equating 2n to the expression of cos 0 obtained from

I


the law of cosines yields

n + 2 (n + 1)2 + (n + 2)2 - n22n 2(n + l)(n + 2)

(n + l)(n + 5) n_+_ 5

2(n + 4)(n + 2) 2(n + 2)

Thus n = 4 and cos 0 =4 4(2) = 3

24. (A) In the adjoining figure, letAH = y, BD = a, DE = x andEC = b. We are given AG = 2,GF= 13, HJ= 7 and FC= 1.Thus the length of the side of theequilateral triangle is 16. Also,using the theorem about sec-ants drawn to a circle froman external point. we havey(y + 7) = 2(2 + 13), or

0 = y2 + 7y - 30 = (y - 3)(y + 10).

Hence y = 3 and BJ = 6. Using the same theorem we haveb(b + x) = 1(1 + 13) = 14 and a(a + x) = 6(6 + 7) = 78.Also, a + b + x = 16.There are many ways of solving the system

(1) a2 + ax = 78,

(2) b2 + bx = 14,

(3) a+b+x= 16.

The one given below allows us to find x without first havingto find a and b.

Subtract the second equation from the first, factor out(a - b), and use (3):

a2 - b2 + (a - b)x = (a - b)(a + b + x)

= (a - b)16 = 78 - 14 = 64.

Therefore

a - b = 4.

Adding this to (3) we obtain 2a + x = 20, whence

a = 10 - (x/2).

C


This, substituted into (1), yields

(0 2 )[1° 2 +x] = (10 - x)(10 + X2)

= 100 - = 78;4

x2

= 22, x = 2V22

25. (D) The probability that the - Jstudent passes throughC is the sum from i = 0to 3 of the probabilitiesthat he enters intersec-tion Ci in the adjoiningfigure and goes east. Thenumber of paths from Ato C, is (2 + i) be- -cause each such path has2 eastward block seg- Dments and they can oc- 1 Icur in any order. Theprobability of taking anyone of these paths to C, and then going east is (2)3+ be-cause there are 3 + i intersections along the way (includingA and C,) where an independent choice with probability 2

is made. So the answer is

3 (2 + i) I 83 I 3 6 10 212) \2 ) 8 + 6 + 32 + 64 32

OR

One may construct a tree-diagram of the respective probabil-ities, obtaining the values step-by-step as shown in the schemeto the right (the final 1 1 I - 'also serves as a check on 2 8 8the computations). I I 3 5

It is important to re- 2 2 S 16cognize that not all I 3 3

twenty of the thirty-five 4 8 8 2

paths leading from A to 4 4 4 4I 5~..J ~21B through C are 8 4 16 32

equally likely; hence 4 4 4 4answer (C) is incorrect! 16 1-6 32


26. (B) If n2 = (ab3c)8 , let n = (de)8 . Then n2 = (8d + e)2

=

64d 2 + 8(2de) + e2 . Thus, the 3 in ab3c is the first digit (inbase 8) of the sum of the eights digit of e2 (in base 8) and theunits digit of (2de) (in base 8). The latter is even, so theformer is odd. The entire table of base 8 representations ofsquares of base 8 digits appears below.

|e l 2 |3 4 |5 6 7e 2

1 4 1 20 31 1 44 61

The eights digit of e2 is odd only if e is 3 or 5; in either casec, which is the units digit of e2 , is 1. (In fact, there arethree choices for n: (33)8, (73)8 and (45),. The squares are(1331)8, (6631). and (2531)8, respectively.)

ORWe are given

n2 = (ab3c)8 = 83a + 82b + 8 3 + c.

If n is even, n2 is divisible by 4, and its remainder upondivision by 8 is 0 or 4. If n is odd, say n = 2k + 1, thenn2 = 4(k2 + k) + 1, and since k 2 + k = k(k + 1) is al-ways even, n2 has remainder I upon division by 8. Thus inall cases, the only possible values of c are 0, 1 or 4. If c = 0,then n2 = 8(8K + 3), an impossibility since 8 is not a square.If c = 4, then n2 = 4(8L + 7) another impossibility since noodd squares have the form 8L + 7. Thus c = 1.

27. (C) We recall the theorem that complex roots of polynomials withreal coefficients come in conjugate pairs. Though not applica-ble to the given polynomial, that theorem is proved by atechnique which we can use to work this problem too. Namely,conjugate both sides of the original equation

0 = C4 Z4 + ic3z3 + C2Z2 + iCJZ + Co,

obtaining

0 = C4Z4 -ic 3 Z3 + C2 Z2 -icf + co

= C4(-Z) + iC3(_z) + C2(-z) + ici(-_) + co.

That is, -z = -a + ib is also a solution of the originalequation. (One may check by example that neither -a - binor a - bi nor b + ai need be a solution.) For instance, con-sider the equation Z 4

- iz3 = 0 and the solution z = i. Herea = 0, b = 1. Neither - i nor I is a solution. [Alternatively,the substitution z = iw into the given equation makes thecoefficients real and the above quoted theorem applicable.]

175


28. (B) Let n be the last number on the board. Now the largestaverage possible is attained if I is erased; the average is then

(n + I)n - 12+3+ -+n 2 n+2

n - n - 1 2

The smallest average possible is attained when n is erased;the average is then

n(n- 1) n

2(n- 1) 2

Thusn 7 n+22 317 -< -2

n < 70-1 < n + 2,17

68 l < n < 70 -41 7 17

Hence n = 69 or 70. Since 35 7 is the average of (n - 1)integers, (35 -7)(n - 1) must be an integer and n is 69. If xis the number erased, then

2 (69)(70) - x 768 17

So69 35 - x = (35#1)68

= 35 68 + 28,35 - x = 28,

x = 7.

29. (A) Let m = x0y0 z0 be the minimum value, and label the num-bers so that x0 < yo S z0 . In fact z0 = 2xo, for if z0 < 2xo,then by decreasing x0 slightly, increasing z0 by the sameamount, and keeping y0 fixed, we would get new valueswhich still meet the constraints but which have a smallerproduct-contradiction! To show this contradiction formally,let xl = x0 - h and z, = z0 + h, where h > 0 is so smallthat z1 < 2x, also. Then xl, yo, z1 also meet all the originalconstraints, and

x1yoz1 = (xo - h)yO(zo + h)

= x 0 yozo + yO[h(xo - zo)-h2] < xoyoZ.


So zo = 2xo, y 0 = I- x - z0 = I - 3x,, and

m = 2x(1 - 3x 0 ).

Also, x0 < I - 3x0 < 2xo, or equivalently, x0 < 4.

Thus m may be viewed as a value of the function

f(x) = 2x 2 (l - 3x)

on the domain D = (x I x < 4}. In fact, m is the small-est value of f on D, because minimizing f on D is just arestricted version of the original problem: for each x E D,setting y = I - 3x and z = 2x gives x, y, z meeting theoriginal constraints, and makes f(x) = xyz.

To minimize f on D, firstsketch f for all real x. (See Fig- yure.) Since f has a relativeminimum at x = 0 (f(x) hasthe same sign as x2 for x < 4),and cubics have at most one rela- Y =Xx)tive minimum, the minimum of fon D must be at one of the end- xpoints. In fact, 3

4= 32 5= 125

30. (D) Let d, = a + vb and d2 = a - rb, where a = 15 and b= 220. Then using the binomial theorem, we may obtain

dn+dn=2[an'+ (n)an -2b+ (n)an -4b2 +...

where n is any positive integer. Since fractional powers of bhave been eliminated in this way, and since a and b areboth divisible by 5, we may conclude that dn + d n is divisi-ble by 10.

We now apply the above result twice, taking n = 19 andn = 82. In this way we obtain

d' 9 + d' 9 = 10k1 and d 2 + d22 = 10k2 ,

where k, and k2 are positive integers. Adding and rearrang-ing these results gives

d19 + d 2 = 10k - (d19 + d 82

where k = k, + k2 . But

d2=15- 2- 5_ < I15+ 220 3

Therefore, d19 + d82 < 1. It follows that the units digit of10k - (d29 + d 2 ) is 9.

177

IV

Classification of Problems

To classify these problems is not a simple task; their content is sovaried and their solution-possibilities so diverse that it is difficult topigeonhole them into a few categories. Moreover, no matter whichheadings are selected, there are borderline cases that need cross-index-ing. Nevertheless, the following may be helpful to the reader whowishes to select a particular category of problems.

The number preceding the semicolon refers to the last two digits ofthe examination year, and the numbers following the semicolon referto the problems in that examination. For example, 82; 5 meansProblem 5 in the 1982 examination.

Algebra

Absolute Value

Addition of Signed Numbers

Binary Operations

Binomial Expansions(coefficients)

Complex Numbers

73; 22 74; 11, 15, 27 75; 777; 8 78; 9 79; 13 82; 19

78;4 79;4

73; 5 74; 6 82; 7

74; 3 75; 5 76; 23 77; 1082; 8, 30

74; 10, 17 76; 2 77; 16, 2180; 17 82; 27

179


De Moivre's Theorem

EquationsCubicExponentialFractionalFunctional (see functions)IntegerLinearQuadratic

81; 24

75;79;74;

73;73;73;76;80;

73;76;80;73;

Systems ofLinearNon-linear

Trigonometric

Exponents

Factoring

Fractions

Identities

Inequalities

Logarithms

Optimization

Polynomials

ProgressionsArithmetic

12, 27 81; 2819 82; 171 79;11 80;3,8 81;6

31 77; 14 79; 11 82; 2016 74; 734 74; 2, 10 75; 12, 206 77; 21, 23 78; 1, 138 81; 29

10, 14, 24 75; 2 77; 130 77; 21 78; 619, 29 81; 1817 81; 18, 24

73; 19 74; 8 75; 10 76; 20,2179; 19 81; 13, 15 82; 3, 30

75; 29 78; 1, 4

78;2,20 79;29 81;3,8

74; 1,30 77; 6 78; 3, 2079; 2

73;77;81;

73;76;80;

73;77;82;

74;76;78;

73;77;81;

22, 30 74; 14, 27 75; 329 78; 18 79; 13 80; 612, 13, 28 82; 14, 28

28 74; 18 75; 18, 1920 77; 18 78; 21 79; 1818 81; 13, 15 82; 13

3, 20 74; 22 76; 12, 2825, 29 80; 22 81; 13,2819, 29

2, 3, 4 75; 5, 13, 276, 19,27 77; 10,21,23,2813 79; 19 80; 2, 28 81; 30

7, 26 74; 29 75; 9 76; 1417 78; 8 79; 16 80; 1111

180

CLASSIFICATION OF PROBLEMS

Geometric

Proportions

Radicals

Reciprocals

Remainder Theorem

Sequences (also seeProgressions)

Word ProblemsAge ProblemsDistance, Rate, TimeGeneral

MixtureMoney

Approximation

Arithmetic Mean

Fractions

Percent

Rationalization

73; 28 74; 21, 28 75; 1676; 4 77; 13 78; 24 80; 1381; 14

73; 8, 12, 29 74; 30 80; 1082; 5

75; 29 76; 29 79; 9 80; 2781; 1, 29 82; 30

76; 1 78; 2 81; 8

74; 4 77; 28 79; 2580; 24. 28 82; 1

75; 15 79; 14, 7

76; 29 77; 1273; 27, 3475; 14, 17, 25 76; 13 78; 3080; 9 81; 4 82; 273; 33 79; 1577; 3 78; 5

Arithmetic

75; 29

73; 2

73; 1380; 1

73; 33

74; 20

76; 21 78; 18 79; 18

75; 1 77; 24 79; 6

74; 7 81; 12, 13

77; 7

Geometry

Analytic GeometryCircles

Lines

73; 11, 30 78; 11 79; 881; 27 82; 475; 1, 2 80; 12, 22 81; 1082; 9

181


Spheres

Angles

Area

Circles

External TangentsSecant From External Point

Congruent Triangles

Coordinate Geometry

Inequalities

Locus

Medians

Menelaus' Theorem

Parallel Lines

PolygonsConvexHexagonsNonagonsParallelogramsQuadrilaterals

RectanglesSquares

Proportion (Similarity)

Pythagorean Theorem

Similarity (see Proportion)

Solid GeometryCubes (and rectangular solids)Pyramids

77; 27

73; 9, 16 74; 5 77; 4 78; 1279; 12 80; 4 81; 20, 25

73; 4, 9, 25, 30 74; 19, 2575; 4, 23 76; 9, 16, 24 77; 2678; 1, 2, 23, 29 79; 1, 8, 21, 30

73; 1, 11, 15, 25, 35 74; 5, 16, 2375; 11 76; 18, 24 77; 1578; 2, 10, 11, 26 79; 12, 21, 2881; 23 82; 2476; 26 78; 2677; 9 82; 24

73; 4 76; 16 79; 30 81; 19

73; 11,30 74; 11 75; 23 76; 878; 25 79; 8 81; 22 82; 9

73; 11, 27 79; 17

76; 22 77; 5 78; 10

82; 21

75; 28

73; 35 75; 26 82; 10

73;78;77;74;75;79;79;73;

73;78;

73;79;82;

1673025424

76; 1479; 10

82; 6

76; 3 77; 19, 26 78; 29

111 75; 23 78; 23 79; 30

9, 35 74; 23 75; 2626 79; 30 82; 14

1, 20 75; 20 76; 3, 2428 80; 7, 19 81; 2, 9, 1114

73; 2 80; 16 81; 9 82; 16, 1873; 32

182

CLASSIFICATION OF PROBLEMS

SpheresTetrahedronsVolumes

TrianglesGeneral

Equilateral

IsoscelesRight300-600-900450-450-900

77;79;73;

75;82;73;78;82;74;74;73;75;

272332

20

80; 2680; 16, 26

78; 12 80; 212123 74; 1923, 28 79;22, 2416 75; 2423 79; 2125 77; 152, 24

81; 19

76; 22 77; 2, 153, 30 81; 23

81;5,27 82;1880; 2380; 5 81; 27

Basics

Double Angle Cosine

Double Angle Sine

Identities

Inverse Trigonometric Functions

Law of Cosines

Law of Sines

Radians

Trigonometry

73; 15 79; 24

73; 17 75; 30

76; 17

74; 22 76; 17 78; 15

79; 20

81; 21, 25 82; 23

75; 28

79; 20

Miscellaneous

Combinatorics 76; 28 77; 2082; 11, 25

78; 16 80; 20

FunctionsAbsolute Value (see Absolute Value)Composite 74; 12Greatest Integer Function 73; 30Functional Equations 75; 21

80; 14

74; 13 75; 8 76; 11 78; 22

76; 1077; 1177; 2281; 17

78; 1780; 2578; 1782; 12

82; 1479; 26

183

Logic


Number TheoryDiophantineDivisibility

Factorial ExpressionsGreatest Integer Function(see functions)Least Common MultipleNumber Bases

Perfect SquaresPrime FactorizationPrime NumbersRecursive DefinitionSums of Integers

Probability

78; 3073; 18,31 74; 8, 9, 26 76; 15, 1978; 27 80; 15, 29 81; 773; 19 76; 23

78; 27 81; 773; 6 75; 10 76; 5 78; 14 79; 581; 16 82; 2679; 7 80; 3077; 2573; 3, 18 74; 8, 26 75; 2276; 25 80; 2573; 21 75; 6

73; 2377; 1781; 26

74; 2478; 1982; 25

75; 18 76; 879; 27 80; 20

184

The Annual High School Mathematics Exami-nation (AHSME) began as a local contest in NewYork City in 1950. By 1960, 150,000 studentsthroughout the United States and Canada took theAHSME. The 1982 Examination was administeredto 418,000 participants in the United States andCanada and to 20,000 students in various coun-tries of other continents. In the United States andCanada, one use of AHSME is to select approxi-mately one hundred participants in the U.S.A.Mathematical Olympiad, and the Olympiads areused in the selection of a student team to repre-sent the United States in the International Mathe-matical Olympiad. Since the difficulty of problemsappearing in the AHSME varies over a wide range,they are a valuable teaching aid for all high schoolstudents interested in mathematics.

This volume contains the 1973-1982 Examina-tions and solutions. The 1950-1960, 1961-1965and 1966-1972 Examinations and solutions werepublished in NML Volumes 5, 17, and 25. Theproblems and solutions are prepared by the Com-mittee on High School Contests, an MM Commit-tee whose members are representatives of the fiveorganizations that jointly sponsor the AHSME: theMathematical Association of America, Society ofActuaries, Mu Alpha Theta, Casualty ActuarialSociety, and National Council of Teachers ofMathematics.

The questions for the 1973-1982 AHSME inthis volume were compiled by Professors R. A.Artino, A. M. Gaglione and N. Shell who co-chaired the Committee from 1973 to 1977; N.Shell served as chairman from 1977-1981. Profes-sor S. B. Maurer (Committee Chairman since 1981)and R.A. Artino, A. M. Gaglione and N. Shell workedjointly with the editors of the New MathematicalLibrary series in the preparation of this volume.They made some changes in the originally pub-lished solutions and added some alternate solu-tions and some explanatory notes.

NEW MATHEMATICAL LIBRARY Front cover ca graphy by Nanc y Foldfi.

[Ralph Artino, Anthony Gaglione and Niel Shell (Co(BookSee.org)

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