problems in plane geometry (bài tập hình học phẳng) bởi i. f. sharygin

Science for Everyone 0.«1» maplllrHB 3a~aqH no reOMeTpHH TIJIaHHMeTpHH HSJUlTeJlbCTBO tHayKa~, MOCKB8 I.F Sharygin Problems in Plane Geometry Mir Publishers Moscow Translated from Russian by Leonid Levant First published 1988 Revised from the 1986 Russian edition m17020'0000-304 056(01)-88 21-88 ' •• Ha ane uucIWM Jl.awlCe Printed In the Union of Soviet Soetalis; Republic @ HSAateJIbCTBO 4lHayxat I'nasaaa peAaxD;HJI t!»B8HKo-MaTeM8TB1fOOKOii JlHTepaTypw, t 986 @ English translation, Mir Publishers, 1988 ISBN 5-03-000180-8 Contents Preface to the English Edition Section t Fundamental Geometrical Facts and Theorems Computational Problems Section Selected Problems and Theorems of Plane Geometry 65 Carnot's Theorem 65 Ceva's and Menelaus' Theorems Affine Problems 70 Loci of Points 82 Triangles A Triangle and a Circle 89 Quadrilaterals t t Circles aod Tangents Feuerbach's Theorem t29 Combina- tions of Figures Displacements in the Plane Polygons t37 Geometrical Inequalities Problems on Extrema 147 Answers, Hint!, Solutions Section t Section too too 214 Appendix Inversion 392 Preface to the English Edition This is a translation from the revised edition of the Russian book which was issued in 1982 It is- actually the first in a two-volume work on solving problems in geometry, the second volume "Problems in Solid Geometry" having been published in English first by Mir Publishers in ~986 Both volumes are designed for schoolchildren and teachers This volume contains over 600 problems in plane geometry and" consists of two parts The first part contains rather simple problems to be solved in classes and at home The second part also contains hints and detailed solutions Over ~OO new problems have been added to the 1982 edition, the simpler problems in the first addition having been eliminated, and a number of new sections- (circles and tangents, polygons, combinations of figures, etc.) having been introduced, The general structure of the book has been changed somewhat to accord with the new, more detailed, classification of the problems As a result, all the problems in this volume have been rearranged Although the problems in this collection vary in "age" (some of them can be found in old books and journals, others were offered at mathematical olympiads or published in the journal "Quant" (Moscow», I still hope that some of the problems in Preface this collection will be of interest to experienced geometers Almost every geometrical problem is nonstandard (as compared with routine exercises on solving equations, inequalities, etc.): one has to think of what additional constructions must be made, or which formulas and theorems must be used Therefore, this collection cannot be regarded as a problem-book in geometry; it is rather a collection of geometrical puzzles aimed at demonstrating the elegance of elementary geometrical techniques of proof and methods of computation (without using vector algebra and with a minimal use of the method of coordinates, geometrical transformstions, though a somewhat wider use of trigonometry) In conclusion, I should like to thank A.Z Bershtein who assisted me in preparing the first section of the book for print I am also grateful to A.A Yagubiants who let me know several elegant geometrical facts The Author /tppendix: Inversion 395 intersection points of the circles 001 and 002 , ~ and li the tangents to WI and 6)2' respectively, passing through A Let us also assume that the centre of inversion does not lie on the straight lines and In the inversion with centre 0, the circles (a)t and Cl)2 go into w~ and co~, respectively, and the lines '1 and 12 into the circles I; and 'i touching ro; and ro~ at the point A' of their intersection (Property 7), that is, the angle between 1; and 12 is equal to the angle between ~~ and m;, and since the angle between 1; and ,; is equal to the angle between 11 and It (Property 8), the angle between (J) ~ and CO is equal to the angle between (0\ and (Ot' to If the circles ex and CI) are orthogonal, that is, the angle between them is equal to 90°, then In io,:ersion with respect to ex the circle CI) goes into itself And conversely, if in inversion with respect to the circle ex the circle CI) not coinciding with ex goes into itself, then a and (a) are orthogonal Obviously, the last property is symmetric with respect to a and CI) The radii of the circles a and 6) are, respectively, equal to the tangents drawn from the centre of one circle to the other circle On the basis of Property 10, the inversion can be defined in the following way All the points of the circle a go into themselves If A does not belong to a and does not coincide with its centre, then the image of the point -i is represented by the point A' which is the second point of inter.section of any two circles orthogonal to a and passing through A, Now, the sense of the synonymic name for inversion-symmetry with respect to a circle-becomes clearer From this definition and the property of inversion to preserve the "angle between two Intersecting circles, it follows that: t t , For any circle CIl and two points A and B going into each other in the inversion with respect to (0 their images in the inversion with respect to the circle a whose centre does not belong to (a) are represented by the circle ro' and polnta A' 't 26* '2' Problems in Plane Geometry 396 B' which go into each other in the inversion with respect to (0' If the centre of a, lies on 0), then CJ) goes into the straight line l, and the points A and B into the points A' and B', symmetric with respect to I 80d The Radical Axis of Two Circles Solve the following problem Given two non-concentric circles 6)1 and (lls Find the locus of points M for which the tangents drawn to the circles (1)1 - and (llt are equal Solution Let 01 and denote the centres of the circles Ctll and (1)2' Tl and TI their radii, Al and A I the points of tangency, respectively We have IM01 12 - I MO" 12 = (I MAl 12 rl) (I MA" T~) = rf - rl Thus, all the points belong to one and the same straight line perpendicular to 101 - This line is called the radical axis of the circles 6)1 and (I)" To complete the solution of the problem, it remains to determine which points of the found line satisfy its conditions I t is possible to show that if the circles not intersect, then all the points of the radical axis are suitable If (1)1 and (1)2 intersect, then the radical axis contains their common chord; but all the points of the common chord are not contained in the required locus of points Therefore, if (1)1 and (1),1 touch each other, then the point of tangency is' -excluded Consider the circle a with centre M on the radical axis of the circles 0)1 and 0), and radius equal to the length of the tangent drawn from M to (01 or (1)1 (M is assumed to be located outside 0)1 and (1),-) The circle a is orthogonal to the circles (01 and (01 Thus, the points of the radical axis situated outside the circles which intersect or touch each other constitute the locus of centres of the circles orthogonal simultaneously to 0)1 and (0" and there is an inversion that carries each ot them into itself + + Appendix: Inversion 397 Now, let us prove one more property of the inversion 12 If the circles WI and 002 Dot Jntersect, then there is an inventon carrying them into eoncentric circles " Let us take a circle a orthogonal to"~l and ro2 with centre on the straight line I containing the centres of rot and IDs Since the circle~l and (Ill not intersect, such a circle a is ~tent Let be one of the intersection pointsref the circle a and the line I In the-inversion with centre 0, the line t goes into itself, and the circle a into the straight line p _ The lines I and p intersect and are orthogonal to the circles co; and roi whieh are the images of 0)1 and COt in the inversion with respect to a Hence it fol ows that the centres of Q); and ro~ coincide with the point of intersection of the lines I and p, that is, CIlI and ro' are concentric circles, (Prove that if a straight fine is orthogonal to a circle, then the former passes through its centre.) Here, we should like to note that any circle orthogonal to the concentric circles 0)' and 00; is a straight line, that is, a circle of infinite radius Hence, in the inversion with respect to the circle a all the circles, orthogonal to the circles (1)1 and CJ)t must go into straight lines Consequently, all the circles orthogonal to rol and IDs intersect the line l at two fixed points 13 For aoy two circles (&)1 and (1)2' there exists at least one inversion which carries them into each other The circle defining this inversion is called the middle circle of WI and CJ)I_ Theorem 13 should be formulated more exactly in the Iollowing way If 6)1 and w, intersect, then there exist exactly two inversions in which 001 goes into W t and vice versa If W1 and CJ)B touch each other or do", not intersect, then there is only one such inversion Let us first consider the case of intersecting circles CDt and CIl~ Apply an inversion I with centre 398 Problems in Plane Geometry in one of the points of their intersection; a8 a result, (01 and IDs go into intersecting straight lines 11 and ll' The lines 11 and have two bisectors with respect to which 11 and " are symmetric, Consequently (Property t 1), in the inversion I those bisectors go into two circles with respect to which (1)1 and 6) t are symmetric, If ml and ID not intersect, then there is an inversion I (Property i2) carrying them into concentric circles 00; and 6)g Let denote the centre of m; and m;, and rl and r their radii Inversion with respect to the circle a,' with centre at and radius './ TIT, carries 6>i and 6)2 into each other, In the inversion I applied, the circle a' goes into the required circle ex with respect to which 6>1 and (01 are symmetric To conclude this section, let us give the definition of the radical centre of three circles Consider three circles Cl)t, CI)~H and 6>3 whose centres not lie on a straight line It is possible to prove that three radical axes corresponding to three pairs of those circles intersect at a point M This point is called the radical centre of the circles COt, CI) 2' and Cila, The tangents drawn from M to the circles - lAC 1• I BD ~ + 15 In a triangle ABC, the side AC is the greatest Prove that for any point M the following inequality holds: I AM I I CM I:> I BM I· ~ 16 Prove that all the circles passing through a 'given fo.nt A and intersecting a circle r We apply the inversion whose centre is at a distance from the common centre of the circles al and at Let, for definiteness, a > R The circles at and