Collection of all geometric world champion P3 SOLUTIONS 201 8.77. On one of the given lines take segment AB and construct its midpoint, M (cf. Problem 8.74). Let A 1 and M 1 be the intersection points of lines P A and PM with the second of the given lines, Q the intersection point of lines BM 1 and MA 1 . It is easy to verify that line P Q is parallel to the given lines. 8.78. In the case when point P does not lie on line AB, we can make use of the solution of Problem 3.36. If point P lies on line AB, then we can first drop perpendiculars l 1 and l 2 from some other points and then in accordance with Problem 8.77 draw through point P the line parallel to lines l 1 and l 2 . 8.79. a) Let A be the given point, l the given line. First, let us consider the case when point O does not lie on line l. Let us draw through point O two arbitrary lines that intersect line l at points B and C. By Problem 8.78, in triangle OBC, heights to sides OB and OC can be dropped. Let H be their intersection point. Then we can draw line OH perpendicular to l. By Problem 8.78 we can drop the perpendicular from point A to OH. This is the line to be constructed that passes through A and is parallel to l. In order to drop the perpendicular from A to l we have to erect p erpendicular l ′ to OH at point O and then drop the perpendicular from A to l ′ . If point O lies on line l, then by Problem 8.78 we can immediately drop the perpendicular l ′ from point A to line l and then erect the perpendicular to line l ′ from the same point A. b) Let l be the given line, A the given point on it and BC the given segment. Let us draw through point O lines OD and OE parallel to lines l and BC, respectively (D and E are the intersection points of these lines with circle S). Let us draw through point C the line parallel to OB to its intersection with line OE at point F and through point F the line parallel to ED to its intersection with OD at point G and, finally, through point G the line parallel to OA to its intersection with l at point H. Then AH = OG = OF = BC, i.e., AH is the segment to be constructed. c) Let us take two arbitrary lines that intersect at point P. Let us mark on one of them segment P A = a and on the other one segments PB = b and P C = c. Let D be the intersection point of line P A with the line that passes through B and is parallel to AC. Clearly, P D = ab c . d) Let H be t he homothety (or the parallel translation) that sends the circle with center A and radius r to circle S (i.e., to the given circle with the marked center O). Since the radii of both circles are known, we can construct the image of any point X under the mapping H. For this we have to draw through point O the line parallel to line AX and mark on it a segment equal to r s ·AX r , where r s is the radius of circle S. We similarly construct the image of any point under the mapping H −1 . Hence, we can construct the line l ′ = H(l) and find its intersection points with circle S and then construct the images of these points under the map H −1 . e) Let A and B be the centers of the given circles, C one of the points to be constructed, CH the height of triangle ABC. From Pythagoras theorem for triangles ACH and BCH we deduce that AH = b 2 +c 2 −a 2 2c . The quantities a, b and c are known, hence, we can construct point H and the intersection points of line CH with one of the given circles. 8.80. a) Let us draw lines parallel to lines OA and OB, whose distance from the latter lines is equal to a and which intersect the legs of the angles. The intersection point of these lines lies on the bisector to be constructed. b) Let us draw the line parallel to OB, whose d istance from OB is equal to a and which intersects ray OA at a point M. Let us dr aw through points O and M another pair of parallel lines the distance between which is equal to a; the line that passes through point O contains the leg of the angle to be found. 202 CHAPTER 8. CONSTRUCTIONS 8.81. Let us draw through point A an arbitrary line and then draw lines l 1 and l 2 parallel to it and whose d istance from this line is equal to a; these lines intersect line l at points M 1 and M 2 . Let us draw through points A and M 1 one more pair of parallel lines, l a and l m , the distance between which is equal to a. The intersection point of lines l 2 and l m belongs to the perpendicular to be found. 8.82. Let us draw a line parallel to the given one at a distance of a. Now, we can make use of the results of Problems 8.77 and 8.74. 8.83. Let us draw through point P lines P A 1 and P B 1 so that PA 1  OA and PB 1  OB. Let line P M divide the angle between lines l and P A 1 in halves. The symmetry through line P M sends line P A 1 to line l and, therefore, line P B 1 turns under this symmetry into one of the lines to be constructed. 8.84. Let us complement triangle ABM to parallelogram ABMN. Through point N draw lines parallel to the b isectors of the angles between lines l and MN. The intersection points of these lines with line l are the ones to be found. 8.85. Let us draw line l 1 parallel to line O A at a distance of a. On l, take an arbitrary point B. Let B 1 be the intersection point of lines OB and l 1 . Through point B 1 draw the line parallel to AB; this line intersects line OA at point A 1 . Now, let us draw through points O and A 1 a pair of parallel lines the distance between which is equal to a. There could be two pairs of such lines. Let X and X 1 be the intersection points of the line that passes through point O with lines l and l 1 . Since OA 1 = OX 1 and △OA 1 X 1 ∼ △OAX, point X is the one to be found. 8.86. Let us erect perpendiculars to line O 1 O 2 at points O 1 and O 2 and on the perpen- diculars mark segments O 1 B 1 = O 2 A 2 and O 2 B 2 = O 1 A 1 . Let us construct the midpoint M of segment B 1 B 2 and erect the perpendicular to B 1 B 2 at point M. This perpendicu- lar intersects line O 1 O 2 at point N. Then O 1 N 2 + O 1 B 2 1 = O 2 N 2 + O 2 B 2 2 and, therefore, O 1 N 2 −O 1 A 2 1 = O 2 N 2 −O 2 A 2 2 , i.e., point N lies on the radical axis. It remains to erect the perpendicular to O 1 O 2 at point N. 8.87. First, let us construct an arbitrary line l 1 perpendicular to line l and then draw through point A the line perpendicular to l 1 . 8.88. a) Let us draw through points A and B lines AB and BQ perpendicular to line AB and then draw an arbitrary perpendicular to line AP . As a result we get a rectangle. It remains to drop from the intersection point of its diagonals the perpendicular to line AB. b) Let us raise from point B perpendicular l to line AB and draw through point A two perpendicular lines; they intersect line l at points M and N. Let us complement right triangle MAN to rectangle MANR. The base of the perpendicular dropped from point R to line AB is point C to be found. 8.89. a) Let us drop perpendicular AP from point A to line OB and construct segment AC whose midpoint is points P . Then angle ∠AOC is the one to be found. b) On line OB, take points B and B 1 such that OB = OB 1 . Let us place the right angle so that its sides would pass through points B and B 1 and the vertex would lie on ray OA. If A is the vertex of the right angle, then angle ∠AB 1 B is th e one to be found. 8.90. Let us draw through point O line l ′ parallel to line l. Let us drop perpendiculars BP and BQ from point B to lines l ′ and OA, respectively, and then drop perpendicular OX from point O to line P Q. Then line XO is the desired one (cf. Problem 2.3); if point Y is symmetric to point X through line l ′ , then line Y O is also the one to be found. 8.91. Let us complement triangle OAB to parallelogram OABC and then construct segment CC 1 whose midpoint is point O. Let us place the right angle so that its legs pass through points C and C 1 and the vertex lies on line l. Then the vertex of the right angle coincides with point X to be found. SOLUTIONS 203 8.92. Let us construct segment AB whose midpoint is point O and place the right angle so that its legs p asses through points A and B and the vertex lies on line l. Then the vertex of the right angle coincides with the point to be found. Chapter 9. GEOMETRIC INEQUALITIES Background 1) For elements of a triangle the following notations are used: a, b, c are the lengths of sides BC, CA, AB, respectively; α, β, γ the values of the angles at vertices A, B, C, respectively; m a , m b , m c are the lengths of the medians drawn from vertices A, B, C, respectively; h a , h b , h c are the lengths of the heights dropped from vertices A, B, C, respectively; l a , l b , l c are the lengths of the bisectors drawn from vertices A, B, C, respectively; r and R are the radii of the inscribed and circumscribed circles, respectively. 2) If A, B, C are arbitrary points, then AB ≤ AC + CB and the equality takes place only if point C lies on segment AB (the triangle inequality). 3) The median of a triangle is shorter than a half sum of the sides that confine it: m a < 1 2(b+c) (Problem 9.1). 4) If one convex polygon lies inside another one, then the perimeter of the outer polygon is greater than the perimeter of the inner one (Problem 9.27 b). 5) The sum of the lengths of the diagonals of a convex quadrilateral is greater than the sum of the length of any pair of the opposite sides of the quadrilateral (Problem 9.14). 6) The longer side of a triangle subtends the greater angle (Problem 10.59). 7) The length of the segment that lies inside a convex polygon does not exceed either that of its longest side or that of its longest diagonal (Problem 10.64). Remark. While solving certain problems of this chapter we have to know various alge- braic inequalities. The data on these inequalities and their proof are given in an appendix to this chapter; one should acquaint oneself with them but it should be taken into account that these inequalities are only needed in the solution of comparatively complicated problems; in order to solve simple problems we will only need the inequality √ ab ≤ 1 2 a + b and its corollaries. Introductory problems 1. Prove that S ABC ≤ 1 2 AB · BC. 2. Prove that S ABCD ≤ 1 2 (AB · BC + AD ·DC). 3. Prove that ∠ABC > 90 ◦ if and only if point B lies inside the circle with diameter AC. 4. The radii of two circles are equal to R and r and the distance between the centers of the circles is equal to d. Prove that these circles intersect if and only if |R −r| < d < R + r. 5. Prove that any diagonal of a quadrilateral is shorter than the quadrilateral’s semiperime- ter. §1. A median of a triangle 9.1. Prove that 1 2 (a + b − c) < m c < 1 2 (a + b). 205 206 CHAPTER 9. GEOMETRIC INEQUALITIES 9.2. Prove that in any triangle the sum of the medians is greater than 3 4 of the perimeter but less than the perimeter. 9.3. Given n points A 1 , . . . , A n and a unit circle, prove that it is possible to find a point M on the circle so that MA 1 + ···+ MA n ≥ n. 9.4. Points A 1 , . . . , A n do not lie on one line. Let two distinct points P and Q have the following property A 1 P + ···+ A n P = A 1 Q + ···+ A n Q = s. Prove that A 1 K + ··· + A n K < s for a point K. 9.5. On a table lies 50 working watches (old style, with hands); all work correctly. Prove that at a certain moment the sum of the distances from the center of the table to the endpoints of the minute’s hands becomes greater than the sum of the distances from the center of the table to the centers of watches. (We assume that each watch is of the form of a disk.) §2. Algebraic problems on the triangle inequality In problems of this section a, b and c are the lengths of the sides of an arbitrary triangle. 9.6. Prove that a = y + z, b = x + z and c = x + y, where x, y and z are positive numbers. 9.7. Prove that a 2 + b 2 + c 2 < 2(ab + bc + ca). 9.8. For any positive integer n, a triangle can be composed of segments whose lengths are a n , b n and c n . Prove that among numbers a, b and c two are equal. 9.9. Prove that a(b − c) 2 + b(c − a) 2 + c(a − b) 2 + 4abc > a 3 + b 3 + c 3 . 9.10. Let p = a b + b c + c a and q = a c + c b + b a . Prove that |p −q| < 1. 9.11. Five segments are such that from any three of them a triangle can be constructed. Prove that at least one of these triangles is an acute one. 9.12. Prove that (a + b − c)(a − b + c)(−a + b + c) ≤ abc. 9.13. Prove that a 2 b(a − b) + b 2 c(b − c) + c 2 a(c − a) ≥ 0. §3. The sum of the lengths of quadrilateral’s diagonals 9.14. Let ABCD be a convex quadrilateral. Prove that AB + CD < AC + BD. 9.15. Let ABCD be a convex quadrilateral and AB + BD ≤ AC + CD. Prove that AB < AC. 9.16. Inside a convex quadrilateral the sum of lengths of whose diagonals is equal to d, a convex quadrilateral the sum of lengths of whose diagonals is equal to d ′ is placed. Prove that d ′ < 2d. 9.17. Given closed broken line has the property that any other closed broken line with the same vertices (?) is longer. Prove that the given broken line is not a self-intersecting one. 9.18. How many sides can a convex polygon have if all its diagonals are of equal length? 9.19. In plane, there are n red and n blue dots no three of which lie on one line. Prove that it is possible to draw n segments with the endpoints of distinct colours without common points. 9.20. Prove that the mean arithmetic of the lengths of sides of an arbitrary convex polygon is less than the mean arithmetic of the lengths of all its diagonals. THE AREA OF A TRIANGLE 207 9.21. A convex (2n + 1)-gon A 1 A 3 A 5 . . . A 2n+1 A 2 . . . A 2n is given. Prove that among all the closed broken lines with the vertices in the vertices of the given (2n + 1)-gon the broken line A 1 A 2 A 3 . . . A 2n+1 A 1 is the longest. §4. Miscellaneous problems on the triangle inequality 9.22. In a triangle, the lengths of two sides are equal to 3.14 and 0.67. Find the length of the third side if it is known that it is an integer. 9.23. Prove that the sum of lengths of diagonals of convex pentagon ABCDE is greater than its perimeter but less than the doubled perimeter. 9.24. Prove that if the lengths of a triangle’s sides satisfy the inequality a 2 + b 2 > 5c 2 , then c is the length of the shortest side. 9.25. The lengths of two heights of a triangle are equal to 12 and 20. Prove that the third height is shorter than 30. 9.26. On sides AB, BC, CA of triangle ABC, points C 1 , A 1 , B 1 , respectively, are taken so that BA 1 = λ ·BC, CB 1 = λ ·CA and AC 1 = λ · AB, where 1 2 < λ < 1. Prove that the perimeter P of triangle ABC and the perimeter P 1 of triangle A 1 B 1 C 1 satisfy the following inequality: (2λ −1)P < P 1 < λP . * * * 9.27. a) Prove that under the passage from a nonconvex polygon to its convex hull the perimeter diminishes. (The convex hull of a polygon is the smallest convex polygon that contains the given one.) b) Inside a convex polygon there lies another convex polygon. Prove that the perimeter of the outer polygon is not less than the perimeter of the inner one. 9.28. Inside triangle ABC of perimeter P, a point O is taken. Prove that 1 2 P < AO + BO + CO < P . 9.29. On base AD of trapezoid ABCD, a point E is taken such that the perimeters of triangles ABE, BCE and CDE are equal. Prove that BC = 1 2 AD. See also Problems 13.40, 20.11. §5. The area of a triangle does not exceed a half product of two sides 9.30. Given a triangle of area 1 the lengths of whose sides satisfy a ≤ b ≤ c. Prove that b ≥ √ 2. 9.31. Let E, F , G and H be the midpoints of sides AB, BC, CD and DA of quadrilateral ABCD. Prove that S ABCD ≤ EG · HF ≤ (AB + CD)(AD + BC) 4 . 9.32. The perimeter of a convex quadrilateral is equal to 4. Prove that its area does not exceed 1. 9.33. Inside triangle ABC a point M is taken. Prove that 4S ≤ AM · BC + BM · AC + CM ·AB, where S is the area of triangle ABC. 9.34. In a circle of radius R a polygon of area S is inscribed; the polygon contains the center of the circle and on each of its sides a point is chosen. Prove that the perimeter of the convex polygon with vertices in the chosen points is not less than 2S R . 9.35. Inside a convex quadrilateral ABCD of area S point O is taken such that AO 2 + BO 2 + CO 2 + DO 2 = 2S. Prove that ABCD is a square and O is its center. 208 CHAPTER 9. GEOMETRIC INEQUALITIES §6. Inequalities of areas 9.36. Points M and N lie on sides AB and AC, respectively, of triangle ABC, where AM = CN and AN = BM. Prove that the area of quadrilateral BMNC is at least three times that of triangle AMN. 9.37. Areas of triangles ABC, A 1 B 1 C 1 , A 2 B 2 C 2 are equal to S, S 1 , S 2 , respectively, and AB = A 1 B 1 + A 2 B 2 , AC = A 1 C 1 + A 2 B 2 , BC = B 1 C 1 + B 2 C 2 . Prove that S ≤ 4 √ S 1 S 2 . 9.38. Let ABCD be a convex quadrilateral of area S. The angle between lines AB and CD is equal to α and the angle between AD and BC is equal to β. Prove that AB · CD sin α + AD ·BC sin β ≤ 2S ≤ AB · CD + AD ·BC. 9.39. Through a point inside a triangle three lines parallel to the triangle’s sides are drawn. Figure 94 (9.39) Denote the areas of the parts into which these lines divide the triangle as plotted on Fig. 94. Prove that a α + b β + c γ ≥ 3 2 . 9.40. The areas of triangles ABC and A 1 B 1 C 1 are equal to S and S 1 , respectively, and we know that triangle ABC is not an obtuse one. The greatest of the ratios a 1 a , b 1 b and c 1 c is equal to k. Prove that S 1 ≤ k 2 S. 9.41. a) Points B, C and D divide the (smaller) arc ⌣ AE of a circle into four equal parts. Prove that S ACE < 8S BCD . b) From point A tangents AB and AC to a circle are drawn. Through the midpoint D of the (lesser) arc ⌣ BC the tangent that intersects segments AB and AC at points M and N, respectively is drawn. Prove that S BCD < 2S MAN . 9.42. All sides of a convex polygon are moved outwards at distance h and extended to form a new polygon. Prove that the d ifference of areas of the polygons is more than P h + π h 2 , where P is the perimeter. 9.43. A square is cut into rectangles. Prove that the sum of areas of the disks circum- scribed about all these rectangles is not less than the area of the disk circumscribed about the initial square. 9.44. Prove that the sum of areas of five triangles formed by the pairs of neighbouring sides and the corresponding diagonals of a convex pentagon is greater than the area of the pentagon itself. 9.45. a) Prove that in any convex hexagon of area S there exists a diagonal that cuts off the hexagon a triangle whose area does not exceed 1 6 S. b) Prove that in any convex 8-gon of area S there exists a diagonal that cuts off it a triangle of area not greater than 1 8 S. See also Problem 17.19. §8. BROKEN LINES INSIDE A SQUARE 209 §7. Area. One figure lies inside another 9.46. A convex polygon whose area is greater than 0.5 is placed in a unit square. Prove that inside the polygon one can place a segment of length 0.5 parallel to a side of the square. 9.47. Inside a unit square n points are given. Prove that: a) the area of one of the triangles some of whose vertices are in these points and some in vertices of the square does not exceed 1 2(n+1) ; b) the area of one of the triangles with the vertices in these points does not exceed 1 n−2 . 9.48. a) In a disk of area S a regular n-gon of area S 1 is inscribed and a regular n-gon of area S 2 is circumscribed about the disk. Prove that S 2 > S 1 S 2 . b) In a circle of length L a regular n-gon of perimeter P 1 is inscribed and another regular n-gon of perimeter P 2 is circumscribed about the circle. Prove that L 2 < P 1 P 2 . 9.49. A polygon of area B is inscribed in a circle of area A and circumscribed about a circle of area C. Prove that 2B ≤ A + C. 9.50. In a unit disk two triangles the area of each of which is greater t han 1 are placed. Prove that these triangles intersect. 9.51. a) Prove that inside a convex polygon of area S and perimeter P one can place a disk of radius S P . b) Inside a convex polygon of area S 1 and perimeter P 1 a convex polygon of area S 2 and perimeter P 2 is placed. Prove that 2S 1 P 1 > S 2 P 2 . 9.52. Prove that the area of a parallelogram that lies inside a triangle does not exceed a half area of the triangle. 9.53. Prove that the area of a triangle whose vertices lie on sides of a parallelogram does not exceed a half area of the parallelogram. * * * 9.54. Prove that any acute triangle of area 1 can be placed in a right triangle of area √ 3. 9.55. a) Prove that a convex polygon of area S can be placed in a rectangle of area not greater than 2S. b) Prove that in a convex polygon of area S a parallelogram of area not less than 1 2 S can be inscribed. 9.56. Prove that in any convex polygon of area 1 a triangle whose area is not less than a) 1 4 ; b) 3 8 can be placed. 9.57. A convex n-gon is placed in a unit square. Prove that there are three vertices A, B and C of this n-gon, such that the area of triangle ABC does not exceed a) 8 n 2 ; b) 16π n 3 . See also Problem 15.6. §8. Broken lines inside a square 9.58. Inside a unit square a non-self-intersecting broken line of length 1000 is placed. Prove that there exists a line parallel to one of the sides of the square that intersects this broken line in at least 500 points. 9.59. In a unit square a broken line of length L is placed. It is known that each point of the square is distant from a point of this broken line less than by ε. Prove that L ≥ 1 2ε − 1 2 πε. 9.60. Inside a unit square n 2 points are placed. Prove that there exists a broken line that passes through all these points and whose length does not exceed 2n. 9.61. Inside a square of side 100 a broken line L is placed. This broken line has the following property: the distance from any point of the square to L does not exceed 0.5. [...]... the endpoints of KL lie on sides AB and CD of the quadrilateral Prove that the length of segment KL does not exceed the length of one of the diagonals of the quadrilateral 9.73 Parallelogram P2 is inscribed in parallelogram P1 and parallelogram P3 whose sides are parallel to the corresponding sides of P1 is inscribed in parallelogram P2 Prove that the length of at least one of the sides of P1 does not... parallel lines that contain a pair of sides of the given parallelogram (Fig 107) The area of the given parallelogram does not exceed the sum of areas of the shaded parallelograms which fall in the case considered above If lines that contain a pair of sides of the given parallelogram only intersect two sides of the triangle, then we can restrict ourselves to one shaded parallelogram only SOLUTIONS 225... the length of one of the bases of these trapezoids is greater than 0.5 Suppose that the length of each base of all the trapezoids does not exceed 0.5 Then the area of each trapezoid does not exceed a half height of the strip that confines it Therefore, the area of the polygon, equal to the sum of areas of trapezoids and triangles into which it is cut, does not exceed a half sum of heights of the strips,... case: two vertices A and B of triangle ABC lie on one side P Q of the parallelogram Then AB ≤ P Q and the height dropped to side AB is not longer than the height of the parallelogram Therefore, the area of triangle ABC does not exceed a half area of the parallelogram Figure 108 (Sol 9.53) If the vertices of the triangle lie on distinct sides of the parallelogram, then two of them lie on opposite sides... 228 CHAPTER 9 GEOMETRIC INEQUALITIES Figure 113 (Sol 9.60) Let us consider the segments that connect distinct bands The union of all such segments obtained in both ways is a pair of broken lines such that the sum of the lengths of the horizontal projections of each of them does not exceed 1 Therefore, the sum of the lengths of horizontal projections of the connecting segments for one of these ways... the third vertex of the triangle a line parallel to these sides (Fig 108) This line cuts the parallelogram into two parallelograms and it cuts the triangle into two triangles so that two vertices of each of these triangles lie on sides of the parallelogram We get the case already considered 9.54 Let M be the midpoint of the longest side BC of the given acute triangle ABC The circle of radius M A centered... sum of the lengths of the horizontal projections for connecting links does not exceed 1 and for all the other links it does not exceed (n − 1)(h1 + · · · + hn ), where hi is the width of the i-th strip Clearly, h1 + · · · + hn = 1 The sum of the vertical projections of all links of the broken line does not exceed n As a result we deduce that the sum of the vertical and horizontal projections of all. .. one 9.98 The bases of a trapezoid are equal to a and b and its height is equal to h Prove 2 2 that the length of one of its diagonals is not less than h +(b+a) 4 9.99 The vertices of an n-gon M1 are the midpoints of sides of a convex n-gon M Prove that for n ≥ 3 the perimeter of M1 is not less than the semiperimeter of M and for n ≥ 4 the area of M1 is not less than a half area of M √ 9.100 In a... CHAPTER 9 GEOMETRIC INEQUALITIES Figure 115 (Sol 9.72) then either AA′ ≤ CC ′ (and, therefore, AC > KL) or BB ′ ≥ DD′ (and, therefore, BD > KL) 9.73 Let us introduce the notations as plotted on Fig 116 All the parallelograms considered have a common center (thanks to Problem 1.7) The lengths of the sides of parallelogram P3 are equal to a + a1 and b + b1 and the lengths of the sides of parallelogram... parallel lines Let us shift these lines parallelly until some vertices A and B of the polygon lie on them Then let us perform the same for the strip formed by lines parallel to AB Let the vertices that lie on these new lines be C and D (Fig 110) The initial polygon is confined in a parallelogram and, therefore, the area of this parallelogram is not less than 1 On the other hand, the sum of areas of . from any point of the square to L does not exceed 0.5. Collection of all geometric world champion P3 210 CHAPTER 9. GEOMETRIC INEQUALITIES Prove. Collection of all geometric world champion P3 SOLUTIONS 201 8.77. On one of the given lines take segment AB and