




























































































Study with the several resources on Docsity
Earn points by helping other students or get them with a premium plan
Prepare for your exams
Study with the several resources on Docsity
Earn points to download
Earn points by helping other students or get them with a premium plan
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 ...
Typology: Study notes
1 / 495
This page cannot be seen from the preview
Don't miss anything!





























































































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 − 2 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 2000 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.
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”, 1–2 problems already published in the Russian editions of this book. The book turned out to be of interest to a vast audience: about 400 000 copies of the first edition of each of the Parts (Parts 1 and 2 — Plane and Part 3 — Solid) were sold; the second edition, published 5 years later, had an even larger circulation, the total over 1 000 000 copies. The 3rd edition of Problems in Plane Geometry was issued in 1996 and the latest one in 2001. 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 2 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 ( 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. 429–431 (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 13.22); v. 96, n. 6, p. 527, Probl. E3192 corresponds to Problems 5.31 and 18.20 of Problems in Plane Geometry — with their two absolutely different solutions, the one to Problem 5.31, unknown to AMM, is even more interesting.
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 29 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.
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 M N whose endpoints divide AB and CD in the ratio of AM : M B = DN : N C = p : q. 1.2. 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. 1.5. The straight line which connects the intersection point P of the diagonals in quadri- lateral ABCD with the intersection point Q of the lines AB and CD bisects side AD. Prove that it also bisects BC. 1.6. 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). 1.7. 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. 1.8. 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^2 = LK · KM. 1.9. 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. 1.10. 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 = P D, then AD^2 = BC^2 + AD · BC. 1.11. 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 M A and OB is perpendicular to that passing through the midpoints of M B and OA. 1.12. 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. 1.13. 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. 1.14. 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 P M and AC. Prove that ∠QN M = ∠M N P. 1.15. 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 = N L, then KO = P L. 1.16. 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 P R and QS are divided by their intersection point in the ratios β : (1 − β) and α : (1 − α), respectively.
§2. THE RATIO OF SIDES OF SIMILAR TRIANGLES 17
§2. The ratio of sides of similar triangles 1.17. 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. 1.18. 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. 1.19. A straight line passing through vertex A of square ABCD intersects side CD at E and line BC at F. Prove that (^) AE^12 + (^) AF^1 2 = (^) AB^12. 1.20. Given points B 2 and C 2 on heights BB 1 and CC 1 of △ABC such that AB 2 C = AC 2 B = 90◦, prove that AB 2 = AC 2. 1.21. 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. 1.22. Perpendiculars AM and AN are dropped to sides BC and CD of parallelogram ABCD (or to their extensions). Prove that △M AN ∼ △ABC. 1.23. 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 AE + AD AF = AC AG. 1.24. 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^2. 1.25. Angles α and β of △ABC are related as 3α + 2β = 180◦. Prove that a^2 + bc = c^2. 1.26. 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 CM^ · ·BMDM is a constant. 1.27. 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. 1.28. 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 (^1) a + (^1) b does not depend on the choice of the line. 1.29. 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. 1.30. 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^2 and AM · AB = AO^2. Prove that M , O and K lie on one straight line. 1.31. Equally oriented similar triangles AM N , N BM and M N C are constructed on segment M N (Fig. 1). Prove that △ABC is similar to all these triangles and the center of its curcumscribed circle is equidistant from M and N. 1.32. Line segment BE divides △ABC into two similar triangles, their similarity ratio being equal to
Find the angles of △ABC.
§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. 1.44. 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 △F OA are the vertices of an equilateral triangle with side R.
1.45. Equilateral triangles BCK and DCL are constructed outwards on sides BC and CD of parallelogram ABCD. Prove that AKL is an equilateral triangle. 1.46. Squares are constructed outwards on the sides of a parallelogram. Prove that their centers form a square. 1.47. Isosceles triangles with angles 2α, 2β and 2γ at vertices A′, B′^ and C′^ are con- structed 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. 1.48. 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. 1.49. 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 M B 1 = M C 1 and ∠B 1 M C 1 = 2ϕ. 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. 1.50. 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 M C 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. 1.51. 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 1.52. Let AA 1 and BB 1 be heights of △ABC. Prove that △A 1 B 1 C ∼ △ABC. What is the similarity coefficient? 1.53. 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 △M N C ∼ △ABC. 1.54. 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. 1.55. 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. 1.56. 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 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. 1.57. 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 1. 1.58. 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. 1.59. 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 1.60. 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 2 and r 3 be the radii of the circles inscribed in the small triangles. Prove that r 1 + r 2 + r 3 = r. 1.61. Given △ABC, draw two straight lines x and y such that the sum of lengths of the segments M XM and M YM 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. 1.62. 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. 1.63. 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. 1.64. Point B lies on segment AC; semicircles S 1 , S 2 , 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 2 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 BF DE is a rectangle. 1.65. Perpendiculars M Q and M P are dropped from an arbitrary point M of the circle circumscribed about rectangle ABCD to the rectangle’s two opposite sides; the perpendic- ulars M R and M T are dropped to the extensions of the other two sides. Prove that lines P R ⊥ QT and the intersection point of P R and QT belongs to a diagonal of ABCD. 1.66. 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 1.67. 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? 1.68. 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. 1.69. 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.