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Straight

Lines

and

Curves

N. Vasilyev

V. Gutenmacher

Mir Publishers

Moscow

H. B. BaCIIJILeB

B. ~. ryTeHMaxep

Ilpasnae B RpBBLle

H3AaTeJIbCTBO «HaYI{a»

MOCKBa

N. B. Vasilyev

V. L. Gutenmacher

Straight Lines

and

Curves

Mir Publishers Moscow

Transla ted from the Russian by Anjari Kundu

Contents

Preface (7) INTRODUCTION (9) Introductory Problems (9). Copernicus' Theorem (13).

1. SET OF POINTS (17) A Family of Lines and Motion (23). Construction Pro- blems (25). Set of Problems (30).
2. THE ALPHABET (35) A Circle and a Pair of Arcs (38). Squares of Distances (42). Distances from Straight Lines (51). The Entire Alphabet (57).
3. LOGICAL COMBINATIONS (60) Through a Single Point (60). Intersection and Union (67). The "Cheese" Problem (74).
4. MAXIMUM AND MINIMUM (78) Where to Put the Point (82). The "Motor-Boat" Problem (84).
5. LEVEL CURVES (90) The "Bus"··Problem (90). Functions on a Plane (93). Level Curves (94). Graph of a Function (94). The Map of a Func- tion (100). Boundary Lines (101). Extrema of Functions (103).
6. QUADRATIC CURVES (108) Ellipses, Hyperbolas, Parabolas (108). Foci and Tan- gents (113). Focal Property of a Parabola (117). Curves as

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the Envelopes of Straight Lines (121). Equations of Cur- ves (124). The Elimination of the Radicals (129). The End of Our Alphabet (130). Algebraic Curves (138).

7. ROTATIONS AND TRAJECTORIES (140) The Cardioid (141). Addition of Rotations ·(142). A Theorem on Two Circles (153). Velocities and Tangents (157). Paramet- ric Equations (166). Conclusion (170).

ANSWERS, HINTS, SOLUTIONS (172)

APPENDIX I. Method of Coordinates (181)

APPENDIX II. A Few Facts from School Geometry (183)

APPENDIX III. A Dozen Assignments (187)

Nota tion (196)

where it is. The beginning and the end of solutions are marked with the symbol 0 while t means that the solution or the answer to the problem is given at the end of the book. The problems at the beginning of each section are not usually difficult or else are analysed in detail in the book. The rest of the problems do not have to be solved in succession. One can, while reading the book, choose those which seem more attractive. It is useful to verify much of what is discussed in the problems through experi- ment: it is best to draw a diagram or-even better-several, with the figures in different positions. This experimental approach not only helps one to guess the answer and for- mulate a hypothesis but also often leads one to a mathe- matical proof. In drawing the diagrams in the margins the authors were convinced that almost behind every pro- blem there is hidden an auxiliary problem of constructing the points or lines which are stated in the problem. The preliminary problem often appears to be more simple but it is no less interesting than the problem itself! The authors are deeply thankful to I. M. Gelfand whose advice helped the entire work on the book, to I. M Yaglom, V. G. Boltyansky and J. M. Rabbot, who read the manu- script, for their significant remarks. Since the publication of the first edition (1970) of this book, it has been used in the work of the Moscow University Correspondence mathemat- ics school. The experience which the teachers of this school shared with us -and also the experience of our friends and colleagues has been taken into consideration in the detailed revision undertaken for the second edition. We thought it necessary to furnish the book with an additional appendix, Appendix III. This will assist in 'systematic study of the book, and will help to reveal rela- tionships between different sections of the book which are not immediately apparent.

N. B. Vasilyev, V. L. Gutenmacher

Introduction

Introductory Problems

0.1. A ladder standing on a smooth floor against a wall slides down. Along what line does a cat sitting at the middle of the ladder move? Let us suppose our cat is calm and sits quietly. Then. we can see behind this picturesque formulation the fol- lowing mathematical problem. A right angle is given.. Find the midpoints of all the possible segments of given length d, which have their end-points lying on the sides of the given angle. Let us try to guess what sort of a set this^ is.^ Obviously,^ when^ the segment rotates with its end-points sliding along the sides of the angle, its centre describes a certain line. (This is obvious from the first pictur- A esque statement of the problem.) First of all, let us determine where the end-points of this line lie. They correspond to the extreme positions It JB of the segment when it is vertical or d/

9

arc All belongs to the unknown set.

]t is easy to do this. Through any

point M of the arc AB we may draw

a ray OM, mark off the segment

, MT I == I OM I along it, drop per-

pendiculars T Land T K from the point T to the sides of the angle and the required segment KL with its midpoint at M is constructed. 0 The second half of the proof might appear to be unnecessary: It is quite clear that the midpoint of the seg- ment KL describes a "continuous line" with end-points. A and B; it means that the point M passes through the

whole of the arc AB and not just

through parts of it. This analysis is perfectly convincing, but it is not easy to give it a strict mathematical form. Let us now consider the motion of the ladder (from problem O.t) from another point of view. Suppose that the segment KL (the "ladder") is fixed and the straight. lines KO and LO ("the wall" and "the floor") rotate correspondingly about the points K and L so that the angle between them is always a right angle. The fact that the distance from the centre of the segment to the vertex 0 of the right angle always remains the same, reduces

to a well-known theorem: if two

points K and L are given in a plane, then the set of points 0 for which the

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~. angle KOL equals 90 0 is a circle with diameter KL. This theorem and also its generalization, which will be giv- en in the proposition E of Sec. 2, will frequently help us in the solution of problems. Let us return to pro- blem 0.1 and put a more general question. 0.2. Along what line does the cat move if it does not sit at the middle of the ladder? In the figure a few points on one such line are plotted. It can be seen that it :s neither a straight line nor a circle, i.e, it is a new curve for us. The coordinate method will help us to find out what sort of curve it is. D We introduce a coordinate system regarding the sides of the angle as the axes Ox and Oy. Suppose the cat sits at some point M (x; y) at a distance a from the end-point K of the ladder and at a distance b from L (a + b =

= d). We shall find the equation

connecting the x and y coordinates of the point M. If the segment KL is inclined to the axis Ox at an angle cp, then y =

= b sin cp and x = a cos q>; hence, for

any arbitrary q> (0 ~ q> ~ ~)

x2 y ~+b2== 1. (1)

The set of points whose coordinates satisfy this equation is an ellipse ..

12

B

,....- ....... A

B

-------..... A

the theorem on the inscribed angle. D Suppose that the point of the moving circle, which occupies posi- tion A on the stationary circle at the initial instant, has come to the posi- tion K, and that T is the point of contact of the circle at the present moment of time. Since the lengths of .....-... ---... the arcs KT and AT are equal and the radius of the movable circle is half as large, the angular size of the

arc KT in degrees is double that of

.....-... the arc AT. Therefore, if 0 is the centre of the stationary circle, then according to the theorem on the

inscribed angle (see p. 24), AOT =

= KOT. Hence, the point K lies on

the radius AO. This argument holds until the mo- ment when the moving circle has rolled around one quarter of the big- ger circle (the circles then touch at the point B of the bigger circle, for which

BOA = 90° and K coincides with 0).

After ~this, the motion will be contin- ued in exactly the same way-the whole picture will be simply reflected symmetrically about the straight line A' A .BO and then, after the point K -....t......--~-___ reaches the opposite end A' of the diameter AA' , the circle will roll along the lower half of the stationary

14

B

B'

circlo and the point K will return to A. 0 Let us compare the results of pro- blems 0.1 and 0.3. They are attractive probably for the' following reason. J30th problems deal with the motion of figures (the first with the motion of a segment, the second with the motion of a circle). The motion is quite complicated, but the paths of certain points appear to be unexpectedly simple. These two problems turn out to be not only related in appearance, but the motions themselves, discus- sed in the problems also coincide wi th each other. Indeed, suppose a circle of radius d/2 rolls along the inside of another circle of radius d, and suppose KL is a diameter of the moving circle rigidly fixed to it. According to Co- pernicus' theorem the points K and A'I---~------ IJ move along stationary straight lines A (along the diameters AA' and BB' of the bigger circle, respectively). Thus, the diameter KL slides with its end- points along two mutually perpendic- ular straight lines, i.e. it moves just like the segment in the problem 0.1. One more interesting problem con- nected with the motion of the seg- ment KL: what set of points is covered by this segment, or what is the union of all· the possible positions of the segment KL during its motion? The curve bounding this set is called the

15

I Set

of

Points

In this paragraph we shall discuss and illustrate with a number of examples the basic statements of the problems which the book deals with and also provide an arsenal of concepts and methods used for solving them. The paragraph ends with a set of various geometric problems. We shall first discuss the term which is most often used in the book and which is at the head of the para- graph. The concept of a "set of points" is very general. A set of points could be any figure, one point or several, a line or a domain in a plane. In many of the problems of our book, it is required to find a set of points which satisfy a certain condi- tion. Answers to such problems are, as a rule, figures known from school geometry (straight lines, circles, some- times pieces into which these lines divide a plane, etc.). The main task

17 2-

c

• (^) •
• •

• (^) •
• (^) •

o

is to guess what sort of a figure the answer is. Thus, in problem 0.1 about the cat, we have guessed the answer- it was a circle, and in problem 0. the answer turned out to be a straight line. In solutions of some problems we have to carry out a thorough investi- gation. One has to establish the fol- • lowing: (a) all the points satisfying the given condition belong to the figure; (b) all the points of the figure satisfy • the given condition. M Sometimes both of these statements are obvious, the direct statement as • well as its converse, sometimes only • one of them. Sometimes it is even • difficult to guess the answer. Let us investigate a few typical problems. 1.1. A point 0 lies on a segment AC. Find the set of points M for which

MOC = 2MAC.

o Answer: The union of the circle

with centre 0 and radius I AO I

(omitting the point A) and the ray OC (omitting the point 0). Let us establish this. Suppose, the point M of the unknown set does not belong to the straight line AO. We

shall prove that the distance I MO I

from the point M to the point 0 is A

equal to I AO I. Let us construct the

triangle OAM. According to' the theo-

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