PROBLEMS IN PLANE AND SOLID GEOMETRY v.1 Plane Geometry. Viktor Prasolov translated and edited by Dimitry Leites

Transcription

1 PROBLEMS IN PLANE AND SOLID GEOMETRY v.1 Plane Geometry Viktor Prasolov translated and edited by Dimitry Leites

2 Abstract. This book has no equal. The priceless treasures of elementary geometry are nowhere else exposed in so complete and at the same time transparent form. The short solutions take barely 1.5 times more space than the formulations, while still remaining complete, with no gaps whatsoever, although many of the problems are quite difficult. Only this enabled the author to squeeze about 000 problems on plane geometry in the book of volume of ca 600 pages thus embracing practically all the known problems and theorems of elementary geometry. The book contains non-standard geometric problems of a level higher than that of the problems usually offered at high school. The collection consists of two parts. It is based on three Russian editions of Prasolov s books on plane geometry. The text is considerably modified for the English edition. Many new problems are added and detailed structuring in accordance with the methods of solution is adopted. The book is addressed to high school students, teachers of mathematics, mathematical clubs, and college students.

3 Contents Editor s preface 11 From the Author s preface 1 Chapter 1. SIMILAR TRIANGLES 15 Background 15 Introductory problems Line segments intercepted by parallel lines 15. The ratio of sides of similar triangles The ratio of the areas of similar triangles Auxiliary equal triangles 18 * * * The triangle determined by the bases of the heights Similar figures 0 Problems for independent study 0 Solutions 1 CHAPTER. INSCRIBED ANGLES 33 Background 33 Introductory problems Angles that subtend equal arcs 34. The value of an angle between two chords The angle between a tangent and a chord Relations between the values of an angle and the lengths of the arc and chord associated with the angle Four points on one circle The inscribed angle and similar triangles The bisector divides an arc in halves An inscribed quadrilateral with perpendicular diagonals Three circumscribed circles intersect at one point Michel s point Miscellaneous problems 40 Problems for independent study 41 Solutions 41 CHAPTER 3. CIRCLES 57 Background 57 Introductory problems The tangents to circles 58. The product of the lengths of a chord s segments Tangent circles Three circles of the same radius Two tangents drawn from one point 61 3

4 4 CONTENTS Application of the theorem on triangle s heights Areas of curvilinear figures 6 8. Circles inscribed in a disc segment 6 9. Miscellaneous problems The radical axis 63 Problems for independent study 65 Solutions 65 CHAPTER 4. AREA 79 Background 79 Introductory problems A median divides the triangle into triangles of equal areas 79. Calculation of areas The areas of the triangles into which a quadrilateral is divided The areas of the parts into which a quadrilateral is divided Miscellaneous problems 8 * * * 8 6. Lines and curves that divide figures into parts of equal area Formulas for the area of a quadrilateral An auxiliary area Regrouping areas 85 Problems for independent study 86 Solutions 86 CHAPTER 5. TRIANGLES 99 Background 99 Introductory problems The inscribed and the circumscribed circles 100 * * * 100 * * * 100. Right triangles The equilateral triangles 101 * * * Triangles with angles of 60 and Integer triangles Miscellaneous problems Menelaus s theorem 104 * * * Ceva s theorem Simson s line The pedal triangle Euler s line and the circle of nine points Brokar s points Lemoine s point 111

5 CONTENTS 5 * * * 111 Problems for independent study 11 Solutions 11 Chapter 6. POLYGONS 137 Background 137 Introductory problems The inscribed and circumscribed quadrilaterals 137 * * * 138 * * * 138. Quadrilaterals Ptolemy s theorem Pentagons Hexagons Regular polygons 14 * * * 14 * * * The inscribed and circumscribed polygons 144 * * * Arbitrary convex polygons Pascal s theorem 145 Problems for independent study 145 Solutions 146 Chapter 7. LOCI 169 Background 169 Introductory problems The locus is a line or a segment of a line 169 * * * 170. The locus is a circle or an arc of a circle 170 * * * The inscribed angle Auxiliary equal triangles The homothety A method of loci The locus with a nonzero area Carnot s theorem Fermat-Apollonius s circle 173 Problems for independent study 173 Solutions 174 Chapter 8. CONSTRUCTIONS The method of loci 183. The inscribed angle Similar triangles and a homothety Construction of triangles from various elements Construction of triangles given various points Triangles Quadrilaterals Circles 185

6 6 CONTENTS 9. Apollonius circle Miscellaneous problems Unusual constructions Construction with a ruler only Constructions with the help of a two-sided ruler Constructions using a right angle 188 Problems for independent study 188 Solutions 189 Chapter 9. GEOMETRIC INEQUALITIES 05 Background 05 Introductory problems A median of a triangle 05. Algebraic problems on the triangle inequality The sum of the lengths of quadrilateral s diagonals Miscellaneous problems on the triangle inequality 07 * * * The area of a triangle does not exceed a half product of two sides Inequalities of areas Area. One figure lies inside another 09 * * * Broken lines inside a square The quadrilateral Polygons 10 * * * Miscellaneous problems 11 * * * 11 Problems for independent study 1 Supplement. Certain inequalities 1 Solutions 13 Chapter 10. INEQUALITIES BETWEEN THE ELEMENTS OF A TRIANGLE Medians 35. Heights The bisectors The lengths of sides The radii of the circumscribed, inscribed and escribed circles Symmetric inequalities between the angles of a triangle Inequalities between the angles of a triangle Inequalities for the area of a triangle 37 * * * The greater angle subtends the longer side Any segment inside a triangle is shorter than the longest side Inequalities for right triangles Inequalities for acute triangles Inequalities in triangles 39 Problems for independent study 40 Solutions 40 Chapter 11. PROBLEMS ON MAXIMUM AND MINIMUM 55

7 CONTENTS 7 Background 55 Introductory problems The triangle 55 * * * 56. Extremal points of a triangle The angle The quadrilateral Polygons Miscellaneous problems The extremal properties of regular polygons 58 Problems for independent study 58 Solutions 59 Chapter 1. CALCULATIONS AND METRIC RELATIONS 71 Introductory problems The law of sines 71. The law of cosines 7 3. The inscribed, the circumscribed and escribed circles; their radii 7 4. The lengths of the sides, heights, bisectors The sines and cosines of a triangle s angles The tangents and cotangents of a triangle s angles Calculation of angles 74 * * * The circles 75 * * * Miscellaneous problems The method of coordinates 76 Problems for independent study 77 Solutions 77 Chapter 13. VECTORS 89 Background 89 Introductory problems Vectors formed by polygons (?) sides 90. Inner product. Relations Inequalities Sums of vectors 9 5. Auxiliary projections 9 6. The method of averaging Pseudoinner product 93 Problems for independent study 94 Solutions 95 Chapter 14. THE CENTER OF MASS 307 Background Main properties of the center of mass 307. A theorem on mass regroupping The moment of inertia Miscellaneous problems The barycentric coordinates 310

8 8 CONTENTS Solutions 311 Chapter 15. PARALLEL TRANSLATIONS 319 Background 319 Introductory problems Solving problems with the aid of parallel translations 319. Problems on construction and loci 30 * * * 30 Problems for independent study 30 Solutions 30 Chapter 16. CENTRAL SYMMETRY 37 Background 37 Introductory problems Solving problems with the help of a symmetry 37. Properties of the symmetry Solving problems with the help of a symmetry. Constructions 38 Problems for independent study 39 Solutions 39 Chapter 17. THE SYMMETRY THROUGH A LINE 335 Background 335 Introductory problems Solving problems with the help of a symmetry 335. Constructions 336 * * * Inequalities and extremals Compositions of symmetries Properties of symmetries and axes of symmetries Chasles s theorem 337 Problems for independent study 338 Solutions 338 Chapter 18. ROTATIONS 345 Background 345 Introductory problems Rotation by Rotation by Rotations through arbitrary angles Compositions of rotations 347 * * * 348 * * * 348 Problems for independent study 348 Solutions 349 Chapter 19. HOMOTHETY AND ROTATIONAL HOMOTHETY 359 Background 359 Introductory problems Homothetic polygons 359. Homothetic circles Costructions and loci 360

9 CONTENTS 9 * * * Composition of homotheties Rotational homothety 361 * * * 36 * * * The center of a rotational homothety The similarity circle of three figures 363 Problems for independent study 364 Solutions 364 Chapter 0. THE PRINCIPLE OF AN EXTREMAL ELEMENT 375 Background The least and the greatest angles 375. The least and the greatest distances The least and the greatest areas The greatest triangle The convex hull and the base lines Miscellaneous problems 378 Solutions 378 Chapter 1. DIRICHLET S PRINCIPLE 385 Background The case when there are finitely many points, lines, etc Angles and lengths Area 387 Solutions 387 Chapter. CONVEX AND NONCONVEX POLYGONS 397 Background Convex polygons 397 * * * 397. Helly s theorem Non-convex polygons 398 Solutions 399 Chapter 3. DIVISIBILITY, INVARIANTS, COLORINGS 409 Background Even and odd 409. Divisibility Invariants Auxiliary colorings More auxiliary colorings 41 * * * Problems on colorings 41 * * * 413 Solutions 413 Chapter 4. INTEGER LATTICES Polygons with vertices in the nodes of a lattice 45. Miscellaneous problems 45 Solutions 46

10 10 CONTENTS Chapter 5. CUTTINGS Cuttings into parallelograms 431. How lines cut the plane 431 Solutions 43 Chapter 6. SYSTEMS OF POINTS AND SEGMENTS. EXAMPLES AND COUNTEREXAMPLES Systems of points 437. Systems of segments, lines and circles Examples and counterexamples 438 Solutions 438 Chapter 7. INDUCTION AND COMBINATORICS Induction 445. Combinatorics 445 Solutions 445 Chapter 8. INVERSION 449 Background Properties of inversions 449. Construction of circles Constructions with the help of a compass only Let us perform an inversion Points that lie on one circle and circles passing through one point Chains of circles 454 Solutions 455 Chapter 9. AFFINE TRANSFORMATIONS Affine transformations 465. How to solve problems with the help of affine transformations 466 Solutions 466 Chapter 30. PROJECTIVE TRANSFORMATIONS Projective transformations of the line 473. Projective transformations of the plane Let us transform the given line into the infinite one Application of projective maps that preserve a circle Application of projective transformations of the line Application of projective transformations of the line in problems on construction Impossibility of construction with the help of a ruler only 480 Solutions 480 Index 493

11 EDITOR S PREFACE 11 Editor s preface The enormous number of problems and theorems of elementary geometry was considered too wide to grasp in full even in the last century. Even nowadays the stream of new problems is still wide. (The majority of these problems, however, are either well-forgotten old ones or those recently pirated from a neighbouring country.) Any attempt to collect an encyclopedia of all the problems seems to be doomed to failure for many reasons. First of all, this is an impossible task because of the huge number of the problems, an enormity too vast to grasp. Second, even if this might have been possible, the book would be terribly overloaded, and therefore of no interest to anybody. However, in the book Problems in plane geometry followed by Problems in solid geometry this task is successfully perfomed. In the process of writing the book the author used the books and magazines published in the last century as well as modern ones. The reader can judge the completeness of the book by, for instance, the fact that American Mathematical Monthly yearly 1 publishes, as new, 1 problems already published in the Russian editions of this book. The book turned out to be of interest to a vast audience: about copies of the first edition of each of the Parts (Parts 1 and Plane and Part 3 Solid) were sold; the second edition, published 5 years later, had an even larger circulation, the total over copies. The 3rd edition of Problems in Plane Geometry was issued in 1996 and the latest one in 001. The readers interest is partly occasioned by a well-thought classification system. The collection consists of three parts. Part 1 covers classical subjects of plane geometry. It contains nearly 1000 problems with complete solutions and over 100 problems to be solved on one s own. Still more will be added for the English version of the book. Part includes more recent topics, geometric transformations and problems more suitable for contests and for use in mathematical clubs. The problems cover cuttings, colorings, the pigeonhole (or Dirichlet s) principle, induction, and so on. Part 3 is devoted to solid geometry. A rather detailed table of contents serves as a guide in the sea of geometric problems. It helps the experts to easily find what they need while the uninitiated can quickly learn what exactly is that they are interested in in geometry. Splitting the book into small sections (5 to 10 problems in each) made the book of interest to the readers of various levels. FOR THE ENGLISH VERSION of the book about 150 new problems are already added and several hundred more of elementary and intermideate level problems will be added to make the number of more elementary problems sufficient to use the book in the ordinary school: the Russian editions are best suited for coaching for a mathematical Olympiad than for a regular class work: the level of difficulty increases rather fast. Problems in each section are ordered difficulty-wise. The first problems of the sections are simple; they are a match for many. Here are some examples: 1 Here are a few samples: v. 96, n. 5, 1989, p (here the main idea of the solution is the right illustration precisely the picture from the back cover of the 1st Russian edition of Problems in Solid Geometry, Fig. to Problem 13.); v. 96, n. 6, p. 57, Probl. E319 corresponds to Problems 5.31 and 18.0 of Problems in Plane Geometry with their two absolutely different solutions, the one to Problem 5.31, unknown to AMM, is even more interesting.

12 1 CONTENTS Plane 1.1. The bases of a trapezoid are a and b. Find the length of the segment that the diagonals of the trapezoid intersept on the trapezoid s midline. Plane 1.5. Let AA 1 and BB 1 be the altitudes of ABC. Prove that A 1 B 1 C is similar to ABC. What is the similarity coefficient? Plane.1. A line segment connects vertex A of an acute ABC with the center O of the circumscribed circle. The altitude AH is dropped from A. Prove that BAH = OAC. Plane 6.1. Prove that if the center of the circle inscribed in a quadrilateral coincides with the intersection point of the quadrilateral s diagonals, then the quadrilateral is a rhombus. Solid 1. Arrange 6 match sticks to get 4 equilateral triangles with side length equal to the length of a stick. Solid 1.1. Consider the cube ABCDA 1 B 1 C 1 D 1 with side length a. Find the angle and the distance between the lines A 1 B and AC 1. Solid 6.1. Is it true that in every tetrahedron the heights meet at one point? The above problems are not difficult. The last problems in the sections are a challenge for the specialists in geometry. It is important that the passage from simple problems to complicated ones is not too long; there are no boring and dull long sequences of simple similar problems. (In the Russian edition these sequences are, perhaps, too short, so more problems are added.) The final problems of the sections are usually borrowed from scientific journals. Here are some examples: Plane Prove that l a +l b +m c 3p, where l a, l b are the lengths of the bisectors of the angles A and B of the triangle ABC, m c is the length of the median of the side AB, and p is the semiperimeter. Plane Let O be the center of the circle inscribed in ABC, K the Lemoine s point, P and Q Brocard s points. Prove that P and Q belong to the circle with diameter KO and that OP = OQ. Plane.9. The numbers α 1,..., α n, whose sum is equal to (n )π, satisfy inequalities 0 < α i < π. Prove that there exists an n-gon A 1...A n with the angles α 1,..., α n at the vertices A 1,..., A n, respectively. Plane 4.1. Prove that for any n there exists a circle on which there lie precisely n points with integer coordinates. Solid Consider several arcs of great circles on a sphere with the sum of their angle measures < π. Prove that there exists a plane that passes through the center of the sphere but does not intersect any of these arcs. Solid 14.. Prove that if the centers of the escribed spheres of a tetrahedron belong to the circumscribed sphere, then the tetrahedron s faces are equal. Solid In space, consider 4 points not in one plane. How many various parallelipipeds with vertices in these points are there? From the Author s preface The book underwent extensive revision. The solutions to many of the problems were rewritten and about 600 new problems were added, particularly those concerning the geometry of the triangle. I was greatly influenced in the process by the second edition of the book by I. F. Sharygin Problems on Geometry. Plane geometry, Nauka, Moscow,1986 and a wonderful and undeservedly forgotten book by D. Efremov New Geometry of the Triangle, Matezis, Odessa, 190. The present book can be used not only as a source of optional problems for students but also as a self-guide for those who wish (or have no other choice but) to study geometry

13 FROM THE AUTHOR S PREFACE 13 independently. Detailed headings are provided for the reader s convenience. Problems in the two parts of Plane are spread over 9 Chapters, each Chapter comprising 6 to 14 sections. The classification is based on the methods used to solve geometric problems. The purpose of the division is basically to help the reader find his/her bearings in this large array of problems. Otherwise the huge number of problems might be somewhat depressingly overwhelming. Advice and comments given by Academician A. V. Pogorelov, and Professors A. M. Abramov, A. Yu. Vaintrob, N. B. Vasiliev, N. P. Dolbilin, and S. Yu. Orevkov were a great help to me in preparing the first Soviet edition. I wish to express my sincere gratitude to all of them. To save space, sections with background only contain the material directly pertinent to the respective chapter. It is collected just to remind the reader of notations. Therefore, the basic elements of a triangle are only defined in chapter 5, while in chapter 1 we assume that their definition is known. For the reader s convenience, cross references in this translation are facilitated by a very detailed index.

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15 Chapter 1. SIMILAR TRIANGLES Background 1) Triangle ABC is said to be similar to triangle A 1 B 1 C 1 (we write ABC A 1 B 1 C 1 ) if and only if one of the following equivalent conditions is satisfied: a) AB : BC : CA = A 1 B 1 : B 1 C 1 : C 1 A 1 ; b) AB : BC = A 1 B 1 : B 1 C 1 and ABC = A 1 B 1 C 1 ; c) ABC = A 1 B 1 C 1 and BAC = B 1 A 1 C 1. ) Triangles AB 1 C 1 and AB C cut off from an angle with vertex A by parallel lines are similar and AB 1 : AB = AC 1 : AC (here points B 1 and B lie on one leg of the angle and C 1 and C on the other leg). 3) A midline of a triangle is the line connecting the midpoints of two of the triangle s sides. The midline is parallel to the third side and its length is equal to a half length of the third side. The midline of a trapezoid is the line connecting the midpoints of the trapezoid s sides. This line is parallel to the bases of the trapezoid and its length is equal to the halfsum of their lengths. 4) The ratio of the areas of similar triangles is equal to the square of the similarity coefficient, i.e., to the squared ratio of the lengths of respective sides. This follows, for example, from the formula S ABC = 1 AB AC sin A. 5) Polygons A 1 A...A n and B 1 B...B n are called similar if A 1 A : A A 3 : : A n A 1 = B 1 B : B B 3 : : B n B 1 and the angles at the vertices A 1,..., A n are equal to the angles at the vertices B 1,..., B n, respectively. The ratio of the respective diagonals of similar polygons is equal to the similarity coefficient. For the circumscribed similar polygons, the ratio of the radii of the inscribed circles is also equal to the similarity coefficient. Introductory problems 1. Consider heights AA 1 and BB 1 in acute triangle ABC. Prove that A 1 C BC = B 1 C AC.. Consider height CH in right triangle ABC with right angle C. Prove that AC = AB AH and CH = AH BH. 3. Prove that the medians of a triangle meet at one point and this point divides each median in the ratio of : 1 counting from the vertex. 4. On side BC of ABC point A 1 is taken so that BA 1 : A 1 C = : 1. What is the ratio in which median CC 1 divides segment AA 1? 5. Square PQRS is inscribed into ABC so that vertices P and Q lie on sides AB and AC and vertices R and S lie on BC. Express the length of the square s side through a and h a. 1. Line segments intercepted by parallel lines 1.1. Let the lengths of bases AD and BC of trapezoid ABCD be a and b (a > b). 15

16 16 CHAPTER 1. SIMILAR TRIANGLES a) Find the length of the segment that the diagonals intercept on the midline. b) Find the length of segment MN whose endpoints divide AB and CD in the ratio of AM : MB = DN : NC = p : q. 1.. Prove that the midpoints of the sides of an arbitrary quadrilateral are vertices of a parallelogram. For what quadrilaterals this parallelogram is a rectangle, a rhombus, a square? 1.3. Points A 1 and B 1 divide sides BC and AC of ABC in the ratios BA 1 : A 1 C = 1 : p and AB 1 : B 1 C = 1 : q, respectively. In what ratio is AA 1 divided by BB 1? 1.4. Straight lines AA 1 and BB 1 pass through point P of median CC 1 in ABC (A 1 and B 1 lie on sides BC and CA, respectively). Prove that A 1 B 1 AB The straight line which connects the intersection point P of the diagonals in quadrilateral ABCD with the intersection point Q of the lines AB and CD bisects side AD. Prove that it also bisects BC A point P is taken on side AD of parallelogram ABCD so that AP : AD = 1 : n; let Q be the intersection point of AC and BP. Prove that AQ : AC = 1 : (n + 1) The vertices of parallelogram A 1 B 1 C 1 D 1 lie on the sides of parallelogram ABCD (point A 1 lies on AB, B 1 on BC, etc.). Prove that the centers of the two parallelograms coincide Point K lies on diagonal BD of parallelogram ABCD. Straight line AK intersects lines BC and CD at points L and M, respectively. Prove that AK = LK KM One of the diagonals of a quadrilateral inscribed in a circle is a diameter of the circle. Prove that (the lengths of) the projections of the opposite sides of the quadrilateral on the other diagonal are equal Point E on base AD of trapezoid ABCD is such that AE = BC. Segments CA and CE intersect diagonal BD at O and P, respectively. Prove that if BO = PD, then AD = BC + AD BC On a circle centered at O, points A and B single out an arc of 60. Point M belongs to this arc. Prove that the straight line passing through the midpoints of MA and OB is perpendicular to that passing through the midpoints of MB and OA a) Points A, B, and C lie on one straight line; points A 1, B 1, and C 1 lie on another straight line. Prove that if AB 1 BA 1 and AC 1 CA 1, then BC 1 CB 1. b) Points A, B, and C lie on one straight line and A 1, B 1, and C 1 are such that AB 1 BA 1, AC 1 CA 1, and BC 1 CB 1. Prove that A 1, B 1 and C 1 lie on one line In ABC bisectors AA 1 and BB 1 are drawn. Prove that the distance from any point M of A 1 B 1 to line AB is equal to the sum of distances from M to AC and BC Let M and N be the midpoints of sides AD and BC in rectangle ABCD. Point P lies on the extension of DC beyond D; point Q is the intersection point of PM and AC. Prove that QNM = MNP Points K and L are taken on the extensions of the bases AD and BC of trapezoid ABCD beyond A and C, respectively. Line segment KL intersects sides AB and CD at M and N, respectively; KL intersects diagonals AC and BD at O and P, respectively. Prove that if KM = NL, then KO = PL Points P, Q, R, and S on sides AB, BC, CD and DA, respectively, of convex quadrilateral ABCD are such that BP : AB = CR : CD = α and AS : AD = BQ : BC = β. Prove that PR and QS are divided by their intersection point in the ratios β : (1 β) and α : (1 α), respectively.

17 . THE RATIO OF SIDES OF SIMILAR TRIANGLES 17. The ratio of sides of similar triangles a) In ABC bisector BD of the external or internal angle B is drawn. Prove that AD : DC = AB : BC. b) Prove that the center O of the circle inscribed in ABC divides the bisector AA 1 in the ratio of AO : OA 1 = (b + c) : a, where a, b and c are the lengths of the triangle s sides The lengths of two sides of a triangle are equal to a while the length of the third side is equal to b. Calculate the radius of the circumscribed circle A straight line passing through vertex A of square ABCD intersects side CD at 1 E and line BC at F. Prove that + 1 = 1. AE AF AB 1.0. Given points B and C on heights BB 1 and CC 1 of ABC such that AB C = AC B = 90, prove that AB = AC A circle is inscribed in trapezoid ABCD (BC AD). The circle is tangent to sides AB and CD at K and L, respectively, and to bases AD and BC at M and N, respectively. a) Let Q be the intersection point of BM and AN. Prove that KQ AD. b) Prove that AK KB = CL LD. 1.. Perpendiculars AM and AN are dropped to sides BC and CD of parallelogram ABCD (or to their extensions). Prove that MAN ABC Straight line l intersects sides AB and AD of parallelogram ABCD at E and F, respectively. Let G be the intersection point of l with diagonal AC. Prove that AB AC + AD = AE AF. AG 1.4. Let AC be the longer of the diagonals in parallelogram ABCD. Perpendiculars CE and CF are dropped from C to the extensions of sides AB and AD, respectively. Prove that AB AE + AD AF = AC Angles α and β of ABC are related as 3α + β = 180. Prove that a + bc = c The endpoints of segments AB and CD are gliding along the sides of a given angle, so that straight lines AB and CD are moving parallelly (i.e., each line moves parallelly to itself) and segments AB and CD intersect at a point, M. Prove that the value of AM BM is CM DM a constant Through an arbitrary point P on side AC of ABC straight lines are drawn parallelly to the triangle s medians AK and CL. The lines intersect BC and AB at E and F, respectively. Prove that AK and CL divide EF into three equal parts Point P lies on the bisector of an angle with vertex C. A line passing through P intercepts segments of lengths a and b on the angle s legs. Prove that the value of does a b not depend on the choice of the line A semicircle is constructed outwards on side BC of an equilateral triangle ABC as on the diameter. Given points K and L that divide the semicircle into three equal arcs, prove that lines AK and AL divide BC into three equal parts Point O is the center of the circle inscribed in ABC. On sides AC and BC points M and K, respectively, are selected so that BK AB = BO and AM AB = AO. Prove that M, O and K lie on one straight line Equally oriented similar triangles AMN, NBM and MNC are constructed on segment MN (Fig. 1). Prove that ABC is similar to all these triangles and the center of its curcumscribed circle is equidistant from M and N Line segment BE divides ABC into two similar triangles, their similarity ratio being equal to 3. Find the angles of ABC.

18 18 CHAPTER 1. SIMILAR TRIANGLES Figure 1 (1.31) 3. The ratio of the areas of similar triangles A point E is taken on side AC of ABC. Through E pass straight lines DE and EF parallel to sides BC and AB, respectively; D and E are points on AB and BC, respectively. Prove that S BDEF = S ADE S EFG Points M and N are taken on sides AB and CD, respectively, of trapezoid ABCD so that segment MN is parallel to the bases and divides the area of the trapezoid in halves. Find the length of MN if BC = a and AD = b Let Q be a point inside ABC. Three straight lines are pass through Q parallelly to the sides of the triangle. The lines divide the triangle into six parts, three of which are triangles of areas S 1, S and S 3. Prove that the area of ABC is equal to ( S1 + S + S 3 ) Prove that the area of a triangle whose sides are equal to the medians of a triangle of area S is equal to 3 4 S a) Prove that the area of the quadrilateral formed by the midpoints of the sides of convex quadrilateral ABCD is half that of ABCD. b) Prove that if the diagonals of a convex quadrilateral are equal, then its area is the product of the lengths of the segments which connect the midpoints of its opposite sides Point O lying inside a convex quadrilateral of area S is reflected symmetrically through the midpoints of its sides. Find the area of the quadrilateral with its vertices in the images of O under the reflections. 4. Auxiliary equal triangles In right triangle ABC with right angle C, points D and E divide leg BC of into three equal parts. Prove that if BC = 3AC, then AEC + ADC + ABC = Let K be the midpoint of side AB of square ABCD and let point L divide diagonal AC in the ratio of AL : LC = 3 : 1. Prove that KLD is a right angle In square ABCD straight lines l 1 and l pass through vertex A. The lines intersect the square s sides. Perpendiculars BB 1, BB, DD 1, and DD are dropped to these lines. Prove that segments B 1 B and D 1 D are equal and perpendicular to each other Consider an isosceles right triangle ABC with CD = CE and points D and E on sides CA and CB, respectively. Extensions of perpendiculars dropped from D and C to AE intersect the hypotenuse AB at K and L. Prove that KL = LB Consider an inscribed quadrilateral ABCD. The lengths of sides AB, BC, CD, and DA are a, b, c, and d, respectively. Rectangles are constructed outwards on the sides of

19 5. THE TRIANGLE DETERMINED BY THE BASES OF THE HEIGHTS 19 the quadrilateral; the sizes of the rectangles are a c, b d, c a and d b, respectively. Prove that the centers of the rectangles are vertices of a rectangle Hexagon ABCDEF is inscribed in a circle of radius R centered at O; let AB = CD = EF = R. Prove that the intersection points, other than O, of the pairs of circles circumscribed about BOC, DOE and FOA are the vertices of an equilateral triangle with side R. * * * Equilateral triangles BCK and DCL are constructed outwards on sides BC and CD of parallelogram ABCD. Prove that AKL is an equilateral triangle Squares are constructed outwards on the sides of a parallelogram. Prove that their centers form a square Isosceles triangles with angles α, β and γ at vertices A, B and C are constructed outwards on the sides of triangle ABC; let α +β +γ = 180. Prove that the angles of A B C are equal to α, β and γ On the sides of ABC as on bases, isosceles similar triangles AB 1 C and AC 1 B are constructed outwards and an isosceles triangle BA 1 C is constructed inwards. Prove that AB 1 A 1 C 1 is a parallelogram a) On sides AB and AC of ABC equilateral triangles ABC 1 and AB 1 C are constructed outwards; let C 1 = B 1 = 90, ABC 1 = ACB 1 = ϕ; let M be the midpoint of BC. Prove that MB 1 = MC 1 and B 1 MC 1 = ϕ. b) Equilateral triangles are constructed outwards on the sides of ABC. Prove that the centers of the triangles constructed form an equilateral triangle whose center coincides with the intersection point of the medians of ABC Isosceles triangles AC 1 B and AB 1 C with an angle ϕ at the vertex are constructed outwards on the unequal sides AB and AC of a scalene triangle ABC. a) Let M be a point on median AA 1 (or on its extension), let M be equidistant from B 1 and C 1. Prove that B 1 MC 1 = ϕ. b) Let O be a point of the midperpendicular to segment BC, let O be equidistant from B 1 and C 1. Prove that B 1 OC = 180 ϕ Similar rhombuses are constructed outwards on the sides of a convex rectangle ABCD, so that their acute angles (equal to α) are adjacent to vertices A and C. Prove that the segments which connect the centers of opposite rhombuses are equal and the angle between them is equal to α. 5. The triangle determined by the bases of the heights 1.5. Let AA 1 and BB 1 be heights of ABC. Prove that A 1 B 1 C ABC. What is the similarity coefficient? Height CH is dropped from vertex C of acute triangle ABC and perpendiculars HM and HN are dropped to sides BC and AC, respectively. Prove that MNC ABC In ABC heights BB 1 and CC 1 are drawn. a) Prove that the tangent at A to the circumscribed circle is parallel to B 1 C 1. b) Prove that B 1 C 1 OA, where O is the center of the circumscribed circle Points A 1, B 1 and C 1 are taken on the sides of an acute triangle ABC so that segments AA 1, BB 1 and CC 1 meet at H. Prove that AH A 1 H = BH B 1 H = CH C 1 H if and only if H is the intersection point of the heights of ABC a) Prove that heights AA 1, BB 1 and CC 1 of acute triangle ABC bisect the angles of A 1 B 1 C 1.

20 0 CHAPTER 1. SIMILAR TRIANGLES b) Points C 1, A 1 and B 1 are taken on sides AB, BC and CA, respectively, of acute triangle ABC. Prove that if B 1 A 1 C = BA 1 C 1, A 1 B 1 C = AB 1 C 1 and A 1 C 1 B = AC 1 B 1, then points A 1, B 1 and C 1 are the bases of the heights of ABC Heights AA 1, BB 1 and CC 1 are drawn in acute triangle ABC. Prove that the point symmetric to A 1 through AC lies on B 1 C In acute triangle ABC, heights AA 1, BB 1 and CC 1 are drawn. Prove that if A 1 B 1 AB and B 1 C 1 BC, then A 1 C 1 AC Let p be the semiperimeter of acute triangle ABC and q the semiperimeter of the triangle formed by the bases of the heights of ABC. Prove that p : q = R : r, where R and r are the radii of the circumscribed and the inscribed circles, respectively, of ABC. 6. Similar figures A circle of radius r is inscribed in a triangle. The straight lines tangent to the circle and parallel to the sides of the triangle are drawn; the lines cut three small triangles off the triangle. Let r 1, r and r 3 be the radii of the circles inscribed in the small triangles. Prove that r 1 + r + r 3 = r Given ABC, draw two straight lines x and y such that the sum of lengths of the segments MX M and MY M drawn parallel to x and y from a point M on AC to their intersections with sides AB and BC is equal to 1 for any M In an isosceles triangle ABC perpendicular HE is dropped from the midpoint of base BC to side AC. Let O be the midpoint of HE. Prove that lines AO and BE are perpendicular to each other Prove that projections of the base of a triangle s height to the sides between which it lies and on the other two heights lie on the same straight line Point B lies on segment AC; semicircles S 1, S, and S 3 are constructed on one side of AC, as on diameter. Let D be a point on S 3 such that BD AC. A common tangent line to S 1 and S touches these semicircles at F and E, respectively. a) Prove that EF is parallel to the tangent to S 3 passing through D. b) Prove that BFDE is a rectangle Perpendiculars MQ and MP are dropped from an arbitrary point M of the circle circumscribed about rectangle ABCD to the rectangle s two opposite sides; the perpendiculars MR and MT are dropped to the extensions of the other two sides. Prove that lines PR QT and the intersection point of PR and QT belongs to a diagonal of ABCD Two circles enclose non-intersecting areas. Common tangent lines to the two circles, one external and one internal, are drawn. Consider two straight lines each of which passes through the tangent points on one of the circles. Prove that the intersection point of the lines lies on the straight line that connects the centers of the circles. Problems for independent study The (length of the) base of an isosceles triangle is a quarter of its perimeter. From an arbitrary point on the base straight lines are drawn parallel to the sides of the triangle. How many times is the perimeter of the triangle greater than that of the parallelogram? The diagonals of a trapezoid are mutually perpendicular. The intersection point divides the diagonals into segments. Prove that the product of the lengths of the trapezoid s bases is equal to the sum of the products of the lengths of the segments of one diagonal and those of another diagonal A straight line is drawn through the center of a unit square. Calculate the sum of the squared distances between the four vertices of the square and the line.

21 SOLUTIONS Points A 1, B 1 and C 1 are symmetric to the center of the circumscribed circle of ABC through the triangle s sides. Prove that ABC = A 1 B 1 C Prove that if BAC = ABC, then BC = (AC + AB)AC Consider points A, B, C and D on a line l. Through A, B and through C, D parallel straight lines are drawn. Prove that the diagonals of the parallelograms thus formed (or their extensions) intersect l at two points that do not depend on parallel lines but depend on points A, B, C, D only In ABC bisector AD and midline A 1 C 1 are drawn. They intersect at K. Prove that A 1 K = b c Points M and N are taken on sides AD and CD of parallelogram ABCD such that MN AC. Prove that S ABM = S CBN On diagonal AC of parallelogram ABCD points P and Q are taken so that AP = CQ. Let M be such that PM AD and QM AB. Prove that M lies on diagonal BD Consider a trapezoid with bases AD and BC. Extensions of the sides of ABCD meet at point O. Segment EF is parallel to the bases and passes through the intersection point of the diagonals. The endpoints of EF lie on AB and CD. Prove that AE : CF = AO : CO Three straight lines parallel to the sides of the given triangle cut three triangles off it leaving an equilateral hexagon. Find the length of the side of the hexagon if the lengths of the triangle s sides are a, b and c Three straight lines parallel to the sides of a triangle meet at one point, the sides of the triangle cutting off the line segments of length x each. Find x if the lengths of the triangle s sides are a, b and c Point P lies inside ABC and ABP = ACP. On straight lines AB and AC, points C 1 and B 1 are taken so that BC 1 : CB 1 = CP : BP. Prove that one of the diagonals of the parallelogram whose two sides lie on lines BP and CP and two other sides (or their extensions) pass through B 1 and C 1 is parallel to BC. Solutions 1.1. a) Let P and Q be the midpoints of AB and CD; let K and L be the intersection points of PQ with the diagonals AC and BD, respectively. Then PL = a and PK = 1b and so KL = PL PK = 1 (a b). b) Take point F on AD such that BF CD. Let E be the intersection point of MN with BF. Then MN = ME + EN = q AF p + q + b = q(a b) + (p + q)b p + q = qa + pb p + q. 1.. Consider quadrilateral ABCD. Let K, L, M and N be the midpoints of sides AB, BC, CD and DA, respectively. Then KL = MN = 1 AC and KL MN, that is KLMN is a parallelogram. It becomes clear now that KLM N is a rectangle if the diagonals AC and BD are perpendicular, a rhombus if AC = BD, and a square if AC and BD are of equal length and perpendicular to each other Denote the intersection point of AA 1 with BB 1 by O. In B 1 BC draw segment A 1 A so that A 1 A BB 1. Then B 1C B 1 A = 1 + p and so AO : OA 1 = AB 1 : B 1 A = B 1 C : qb 1 A = (1 + p) : q.

22 CHAPTER 1. SIMILAR TRIANGLES 1.4. Let A be the midpoint of A 1 B. Then CA 1 : A 1 A = CP : PC 1 and A 1 A : A 1 B = 1 :. So CA 1 : A 1 B = CP : PC 1. Similarly, CB 1 : B 1 A = CP : PC 1 = CA 1 : A 1 B Point P lies on the median QM of AQD (or on its extension). It is easy to verify that the solution of Problem 1.4 remains correct also for the case when P lies on the extension of the median. Consequently, BC AD We have AQ : QC = AP : BC = 1 : n because AQP CQB. So AC = AQ + QC = (n + 1)AQ The center of A 1 B 1 C 1 D 1 being the midpoint of B 1 D 1 belongs to the line segment which connects the midpoints of AB and CD. Similarly, it belongs to the segment which connects the midpoints of BC and AD. The intersection point of the segments is the center of ABCD Clearly, AK : KM = BK : KD = LK : AK, that is AK = LK KM Let AC be the diameter of the circle circumscribed about ABCD. Drop perpendiculars AA 1 and CC 1 to BD (Fig. ). Figure (Sol. 1.9) We must prove that BA 1 = DC 1. Drop perpendicular OP from the center O of the circumscribed circle to BD. Clearly, P is the midpoint of BD. Lines AA 1, OP and CC 1 are parallel to each other and AO = OC. So A 1 P = PC 1 and, since P is the midpoint of BD, it follows that BA 1 = DC We see that BO : OD = DP : PB = k, because BO =PD. Let BC = 1. Then AD = k and ED = 1 k. So k = AD = AE + ED = k, that is k = 1 + k. Finally, observe that k = AD and 1 + k = BC + BC AD Let C, D, E and F be the midpoints of sides AO, OB, BM and MA, respectively, of quadrilateral AOMB. Since AB = MO = R, where R is the radius of the given circle, CDEF is a rhombus by Problem 1.. Hence, CE DF a) If the lines containing the given points are parallel, then the assertion of the problem is obviously true. We assume that the lines meet at O. Then OA : OB = OB 1 : OA 1 and OC : OA = OA 1 : OC 1. Hence, OC : OB = OB 1 : OC 1 and so BC 1 CB 1 (the ratios of the segment should be assumed to be oriented). b) Let AB 1 and CA 1 meet at D, let CB 1 and AC 1 meet at E. Then CA 1 : A 1 D = CB : BA = EC 1 : C 1 A. Since CB 1 D EB 1 A, points A 1, B 1 and C 1 lie on the same line A point that lies on the bisector of an angle is equidistant from the angle s legs. Let a be the distance from point A 1 to lines AC and AB, let b be the distance from point B 1 to lines AB and BC. Further, let A 1 M : B 1 M = p : q, where p + q = 1. Then the distances

23 SOLUTIONS 3 from point M to lines AC and BC are equal to qa and pb, respectively. On the other hand, by Problem 1.1 b) the distance from point M to line AB is equal to qa + pb Let the line that passes through the center O of the given rectangle parallel to BC intersect line segment QN at point K (Fig. 3). Figure 3 (Sol. 1.14) Since MO PC, it follows that QM : MP = QO : OC and, since KO BC, it follows that QO : OC = QK : KN. Therefore, QM : MP = QK : KN, i.e., KM NP. Hence, MNP = KMO = QNM Let us draw through point M line EF so that EF CD (points E and F lie on lines BC and AD). Then PL : PK = BL : KD and OK : OL = KA : CL = KA : KF = BL : EL. Since KD = EL, we have PL : PK = OK : OL and, therefore, PL = OK Consider parallelogram ABCD 1. We may assume that points D and D 1 do not coincide (otherwise the statement of the problem is obvious). On sides AD 1 and CD 1 take points S 1 and R 1, respectively, so that SS 1 DD 1 and RR 1 DD 1. Let segments PR 1 and QS 1 meet at N; let N 1 and N be the intersection points of the line that passes through N parallel to DD 1 with segments PR and QS, respectively. Then N 1 N = β RR 1 = αβ DD 1 and N N = α SS 1 = αβ DD 1. Hence, segments PR and QS meet at N 1 = N. Clearly, PN 1 : PR = PN : PR 1 = β and QN : QS = α. Remark. If α = β, there is a simpler solution. Since BP : BA = BQ : BC = α, it follows that PQ AC and PQ : AC = α. Similarly, RS AC and RS : AC = 1 α. Therefore, segments PR and QS are divided by their intersection point in the ratio of α : (1 α) a) From vertices A and C drop perpendiculars AK and CL to line BD. Since CBL = ABK and CDL = KDA, we see that BLC BKA and CLD AKD. Therefore, AD : DC = AK : CL = AB : BC. b) Taking into account that BA 1 : A 1 C = BA : AC and BA 1 + A 1 C = BC we get BA 1 = ac. Since BO is the bisector of triangle ABA b+c 1, it follows that AO : OA 1 = AB : BA 1 = (b + c) : a Let O be the center of the circumscribed circle of isosceles triangle ABC, let B 1 be the midpoint of base AC and A 1 the midpoint of the lateral side BC. Since BOA 1 BCB 1, it follows that BO : BA 1 = BC : BB 1 and, therefore, R = BO = a 4a. b

24 4 CHAPTER 1. SIMILAR TRIANGLES If EAD = ϕ, then AE = AD cos ϕ = AB cos ϕ and AF = AB sin ϕ. Therefore, 1 AE + 1 AF = cos ϕ + sin ϕ = 1 AB AB It is easy to verify that AB = AB 1 AC = AC 1 AB = AC a) Since BQ : QM = BN : AM = BK : AK, we have: KQ AM. b) Let O be the center of the inscribed circle. Since CBA + BAD = 180, it follows that ABO + BAO = 90. Therefore, AKO OKB, i.e., AK : KO = OK : KB. Consequently, AK KB = KO = R, where R is the radius of the inscribed circle. Similarly, CL LD = R. 1.. If angle ABC is obtuse (resp. acute), then angle MAN is also obtuse (resp. acute). Moreover, the legs of these angles are mutually perpendicular. Therefore, ABC = MAN. Right triangles ABM and ADN have equal angles ABM = ADN, therefore, AM : AN = AB : AD = AB : CB, i.e., ABC MAN On diagonal AC, take points D and B such that BB l and DD l. Then AB : AE = AB : AG and AD : AF = AD : AG. Since the sides of triangles ABB and CDD are pairwise parallel and AB = CD, these triangles are equal and AB = CD. Therefore, AB AE + AD AF = AB AG + AD AG = CD + AD = AC AG AG Let us drop from vertex B perpendicular BG to AC (Fig. 4). Figure 4 (Sol. 1.4) Since triangles ABG and ACE are similar, AC AG = AE AB. Lines AF and CB are parallel, consequently, GCB = CAF. We also infer that right triangles CBG and ACF are similar and, therefore, AC CG = AF BC. Summing the equalities obtained we get AC (AG + CG) = AE AB + AF BC. Since AG + CG = AC, we get the equality desired Since α + β = 90 1α, it follows that γ = 180 α β = α. Therefore, it is possible to find point D on side AB so that ACD = 90 1 α, i.e., AC = AD. Then ABC CBD and, therefore, BC : BD = AB : CB, i.e., a = c(c b) As segments AB and CD move, triangle AMC is being replaced by another triangle similar to the initial one. Therefore, the quantity AM BM remains a constant. Analogously, CM DM remains a constant Let medians meet at O; denote the intersection points of median AK with lines FP and FE by Q and M, respectively; denote the intersection points of median CL with lines EP and FE by R and N, respectively (Fig. 5). Clearly, FM : FE = FQ : FP = LO : LC = 1 : 3, i.e., FM = 1 FE. Similarly, 3 EN = 1FE. 3

25 SOLUTIONS 5 Figure 5 (Sol. 1.7) 1.8. Let A and B be the intersection points of the given line with the angle s legs. On segments AC and BC, take points K and L, respectively, so that PK BC and PL AC. Since AKP PLB, it follows that AK : KP = PL : LB and, therefore, (a p)(b p) = p, where p = PK = PL. Hence, 1 a + 1 b = 1 p Denote the midpoint of side BC by O and the intersection points of AK and AL with side BC by P and Q, respectively. We may assume that BP < BQ. Triangle LCO is an equilateral one and LC AB. Therefore, ABQ LCQ, i.e., BQ : QC = AB : LC = : 1. Hence, BC = BQ + QC = 3QC. Similarly, BC = 3BP Since BK : BO = BO : AB and KBO = ABO, it follows that KOB OAB. Hence, KOB = OAB. Similarly, AOM = ABO. Therefore, KOM = KOB + BOA + AOM = OAB + BOA + ABO = 180, i.e., points K, O and M lie on one line Since AMN = MNC and BMN = MNA, we see that AMB = ANC. Moreover, AM : AN = NB : NM = BM : CN. Hence, AMB ANC and, therefore, MAB = NAC. Consequently, BAC = MAN. For the other angles the proof is similar. Let points B 1 and C 1 be symmetric to B and C, respectively, through the midperpendicular to segment MN. Since AM : NB = MN : BM = MC : NC, it follows that MA MC 1 = AM NC = NB MC = MB 1 MC. Therefore, point A lies on the circle circumscribed about trapezoid BB 1 CC Since AEB+ BEC = 180, angles AEB and BEC cannot be different angles of similar triangles ABE and BEC, i.e., the angles are equal and BE is a perpendicular. Two cases are possible: either ABE = CBE or ABE = BCE. The first case should be discarded because in this case ABE = CBE. In the second case we have ABC = ABE + CBE = ABE + BAE = 90. In right triangle ABC the ratio of the legs lengths is equal to 1 : 3; hence, the angles of triangle ABC are equal to 90, 60, We have S BDEF S ADE = S BDE S ADE = DB = EF = S EFC AD AD S ADE. Hence, S BDEF = S ADE S EFC Let MN = x; let E be the intersection point of lines AB and CD. Triangles EBC, EMN and EAD are similar, hence, S EBC : S EMN : S EAD = a : x : b. Since S EMN S EBC = S MBCN = S MADN = S EAD S EMN, it follows that x a = b x, i.e., x = 1 (a + b ).

26 6 CHAPTER 1. SIMILAR TRIANGLES Through point Q inside triangle ABC draw lines DE, FG and HI parallel to BC, CA and AB, respectively, so that points F and H would lie on side BC, points E and I on side AC, points D and G on side AB (Fig. 6). Figure 6 (Sol. 1.35) Set S = S ABC, S 1 = S GDQ, S = S IEQ, S 3 = S HFQ. Then S1 S + S S + S3 S = GQ AC + IE AC + FQ AC = AI + IE + EC AC = 1, i.e., S = ( S 1 + S + S 3 ) Let M be the intersection point of the medians of triangle ABC; let point A 1 be symmetric to M through the midpoint of segment BC. The ratio of the lengths of sides of triangle CMA 1 to the lengths of the corresponding medians of triangle ABC is to : 3. Therefore, the area to be found is equal to 9S 4 CMA 1. Clearly, S CMA1 = 1 S (cf. the solution 3 of Problem 4.1) Let E, F, G and H be the midpoints of sides AB, BC, CD and DA, respectively. a) Clearly, S AEH + S CFG = 1S 4 ABD + 1S 4 CBD = 1S 4 ABCD. Analogously, S BEF + S DGH = 1 S 4 ABCD; hence, S EFGH = S ABCD 1S 4 ABCD 1S 4 ABCD = 1S ABCD. b) Since AC = BD, it follows that EFGH is a rhombus (Problem 1.). By heading a) we have S ABCD = S EFGH = EG FH Let E, F, G and H be the midpoints of sides of quadrilateral ABCD; let points E 1, F 1, G 1 and H 1 be symmetric to point O through these points, respectively. Since EF is the midline of triangle E 1 OF 1, we see that S E1 OF 1 = 4S EOF. Similarly, S F1 OG 1 = 4S FOG, S G1 OH 1 = 4S GOH, S H1 OE 1 = 4S HOE. Hence, S E1 F 1 G 1 H 1 = 4S EFGH. By Problem 1.37 a) S ABCD = S EFGH. Hence, S E1 F 1 G 1 H 1 = S ABCD = S First solution. Let us consider square BCMN and divide its side MN by points P and Q into three equal parts (Fig. 7). Then ABC = P DQ and ACD = P M A. Hence, triangle P AD is an isosceles right triangle and ABC + ADC = PDQ + ADC = 45. Second solution. Since DE = 1, EA =, EB =, AD = 5 and BA = 10, it follows that DE : AE = EA : EB = AD : BA and DEA AEB. Therefore, ABC = EAD. Moreover, AEC = CAE = 45. Hence, ABC + ADC + AEC = ( EAD + CAE) + ADC = CAD + ADC = 90.