SOLUTIONS 401 (here we assume that a point is segment respect to the same axis is a segment of length a+b 2 of zero length) Therefore, the perimeter of H is equal to P1 +P and the number of H’s sides can take any value from the largest — n1 or n2 — to n1 + n2 depending on for how many axes both basic sets of F and G are sides and not vertices simultaneously 22.7 We will prove a more general statement Recall that cardinality of a set is (for a finite set) the number of its element (Ramsey’s theorem.) Let p, q and r be positive integers such that p, q ≥ r Then there exists a number N = N (p, q, r) with the following property: if r-tuples from a set S of cardinality N are divided at random into two nonintersecting families α and β, then either there exists a p-tuple of elements from S all subsets of cardinality r of which are contained in α or there exists a q-tuple all subsets of cardinality r of which are contained in β The desired statement follows easily from Ramsey’s theorem Indeed, let N = N (p, 5, 4) and family α consist of quadruples of elements of an N -element set of points whose convex hulls are quadrilaterals Then there exists a subset of n elements of the given set of points the convex hulls of any its four-elements subset being quadrilaterals because there is no fiveelement subset such that the convex hulls of any four-element subsets of which are triangles (see Problem 22.2) It remains to make use of the result of Problem 22.1 Now, let us prove Ramsey’s theorem It is easy to verify that for N (p, q, 1), N (r, q, r) and N (p, r, r) one can take numbers p + q − 1, q and p, respectively Now, let us prove that if p > r and q > r, then for N (p, q, r) one can take numbers N (p1 , q1 , r − 1) + 1, where p1 = N (p − 1, q, r) and q1 = N (p, q − 1, r) Indeed, let us delete from the N (p, q, r)-element set S one element and divide the (r − 1)-element subsets of the obtained set S ′ into two families: family α′ (resp β ′ ) consists of subsets whose union with the deleted element enters α (resp β) Then either (1) there exists a p1 -element subset of S ′ all (r − 1)-element subsets of which are contained in α′ or (2) there exists a q1 -element subset all whose (r − 1) element subsets are contained in family β ′ Consider case (1) Since p1 = N (p − 1, q, r), it follows that either there exists a q-element subset of S ′ all r-element subsets of which belong to β (then these q elements are the desired one) or there exists a (p − 1)-element subset of S ′ all the r-element subsets of which are contained in α (then these p − elements together with the deleted element are the desired ones) Case (2) is treated similarly Thus, the proof of Ramsey’s theorem can be carried out by induction on r, where in the proof of the inductive step we make use of induction on p + q 22.8 If the polygon is not a triangle or parallelogram, then it has two nonparallel non-neighbouring sides Extending them until they intersect, we get a new polygon which contains the initial one and has fewer number of sides After several such operations we get a triangle or a parallelogram If we have got a triangle, then everything is proved; therefore, let us assume that we have got a parallelogram, ABCD On each of its sides there lies a side of the initial polygon and one of its vertices, say A, does not belong to the initial polygon (Fig 52) Let K be a vertex of the polygon nearest to A and lying on AD; let KL be the side that does not lie on AD Then the polygon is confined inside the triangle formed by lines KL, BC and CD 22.9 The proof will be carried out by induction on n For n = the statement is obvious Let n ≥ By Problem 22.8 there exist lines a, b and c which are extensions of the sides of the given n-gon that constitute triangle T which contains the given n-gon Let line l be the extension of some other side of the given n-gon The extensions of all the sides 402 CHAPTER 22 CONVEX AND NONCONVEX POLYGONS Figure 194 (Sol 22.8) of the n-gon except the side which lies on line l form a convex (n − 1)-gon that lies inside triangle T By the inductive hypothesis for this (n − 1)-gon there exist n − required triangles Moreover, line l and two of the lines a, b and c also form a required triangle Remark If points A2 , , An belong to a circle with center at A1 , where ∠A2 A1 An < 90◦ and the n-gon A1 An is a convex one, then for this n-gon there exist precisely n − triangles required 22.10 Proof will be carried out by induction on n For n = the proof is obvious Now, let us consider n-gons A1 An , where n ≥ Point O lies inside triangle Ap Aq Ar Let Ak be a vertex of the given n-gon distinct from points Ap , Aq and Ar Selecting vertex Ak in n-gon A1 An we get a (n − 1)-gon to which the inductive hypothesis is applicable Moreover, the angles ∠Ak OAp , ∠Ak OAq and ∠Ak OAr cannot all be acute ones because the sum of certain two of them is greater than 180◦ 22.11 Proof will be carried out by induction on n For n = the statement is obvious Let n ≥ Fix one acute triangle Ap Aq Ar and let us discard vertex Ak distinct from the vertices of this triangle The inductive hypothesis is applicable to the obtained (n − 1)gon Moreover, if, for instance, point Ak lies on arc Ap Aq and ∠Ak Ap Ar ≤ ∠Ak Aq Ar , then triangle Ak Ap Ar is an acute one Indeed, ∠Ap Ak Ar = ∠Ap Aq Ar , ∠Ap Ar Ak < ∠Ap Ar Aq and ∠Ak Ap Ar ≤ 90◦ ; hence, ∠Ak Ap Ar < 90◦ 22.12 a) Denote the given figures by M1 , M2 , M3 and M4 Let Ai be the intersection point of all the figures except Mi Two variants of arrangements of points Ai are possible 1) One of the points, for example, A4 lies inside the triangle formed by the remaining points Since points A1 , A2 , A3 belong to the convex figure M4 , all points of A1 A2 A3 also belong to M4 Therefore, point A4 belongs to M4 and it belongs to the other figures by its definition 2) A1 A2 A3 A4 is a convex quadrilateral Let C be the intersection point of diagonals A1 A3 and A2 A4 Let us prove that C belongs to all the given figures Both points A1 and A3 belong to figures M2 and M4 , therefore, segment A1 A3 belongs to these figures Similarly, segment A2 A4 belongs to figures M1 and M3 It follows that the intersection point of segments A1 A3 and A2 A4 belongs to all the given figures b) Proof will be carried out by induction on the number of figures For n = the statement is proved in the preceding problem Let us prove that if the statement holds for n ≥ figures, then it holds also for n + figures Given convex figures Φ1 , , Φn , Φn+1 every three of which have a common point, consider instead of them figures Φ1 , , Φn−1 , Φ′n , where Φ′n is the intersection of Φn and Φn+1 It is clear that Φ′n is also a convex figure Let us prove that any three of the new figures have a common point One can only doubt this for the triple of figures that contain Φ′n but the preceding problem implies that figures SOLUTIONS 403 Φi , Φj , Φn and Φn+1 always have a common point Therefore, by the inductive hypothesis Φ1 , , Φn−1 , Φ′n have a common point; hence, Φ1 , , Φn , Φn+1 have a common point 22.13 A unit disk centered at O covers certain points if and only if unit disks centered at these points contain point O Therefore, our problem admits the following reformulation: Given n points in plane such that any three unit disks centered at these points have a common point, prove that all these disks have a common point This statement clearly follows from Helley’s theorem 22.14 Consider pentagons that remain after deleting pairs of neighbouring vertices of a heptagon It suffices to verify that any three of the pentagons have a common point For three pentagons we delete not more than distinct vertices, i.e., one vertex remains If vertex A is not deleted, then the triangle shaded in Fig 53 belongs to all three pentagons Figure 195 (Sol 22.14) 22.15 Let us introduce the coordinate system with Oy-axis parallel to the given segments For every segment consider the set of all points (a, b) such that the line y = ax + b intersects it It suffices to verify that these sets are convex ones and apply to them Helley’s theorem For the segment with endpoints (x0 , y1 ) and (x0 , y2 ) the considered set is a band between parallel lines ax0 + b = y1 and ax0 + b = y2 22.16 Wrong A counterexample is given on Fig 54 Figure 196 (Sol 22.16) 22.17 The required polygons and points are drawn on Fig 55 Figure 197 (Sol 22.17) 404 CHAPTER 22 CONVEX AND NONCONVEX POLYGONS 22.18 Let the whole contour of polygon A1 An subtend an angle with vertex O Then no other side of the polygon except Ai Ai+1 lies inside angle ∠Ai OAi+1 ; hence, point O lies inside the polygon (Fig 56) Any point X in plane belongs to one of the angles ∠Ai OAi+1 and, therefore, side Ai Ai+1 subtends an angle with vertex in X Figure 198 (Sol 22.18) 22.19 Since all the inner angles of a convex n-gon are smaller than 180◦ and their sum is equal to (n − 2) · 180◦ , the sum of the exterior angles is equal to 360◦ , i.e., for a convex polygon we attain the equality Figure 199 (Sol 22.19) Now, let M be the convex hull of polygon N Each angle of M contains an angle of N smaller than 180◦ and the angle of M can be only greater than the angle of N , i.e., the exterior angle of N is not less than the exterior angle of M (Fig 57) Therefore, even restricting to the angles of N adjacent to the angles of M we will get not less than 360◦ 22.20 a) If the polygon is a convex one, then the statement is proved Now, suppose that the exterior angle of the polygon at vertex A is greater than 180◦ The visible part of the side subtends an angle smaller than 180◦ with vertex at point A, therefore, parts of at least two sides subtend an angle with vertex at A Therefore, there exist rays exiting point A and such that on these rays the change of (parts of) sides visible from A occurs (on Fig 58 all such rays are depicted) Each of such rays determines a diagonal that lies entirely inside the polygon b) On Fig 59 it is plotted how to construct an n-gon with exactly n − diagonals inside it It remains to demonstrate that any n-gon has at least n − diagonals For n = this statement is obvious Suppose the statement holds for all k-gons, where k < n and let us prove it for an n-gon By heading a) it is possible to divide an n-gon by its diagonal into two polygons: a (k + 1)gon and an (n − k + 1)-gon, where k + < n and n − k + < n These parts have at least (k + 1) − and (n − k + 1) − diagonals, respectively, that lie inside these parts Therefore, the n-gon has at least + (k − 2) + (n − k − 2) = n − diagonals that lie inside it 22.21 First, let us prove that if A and B are neighbouring vertices of the n-gon, then either from A or from B it is possible to draw a diagonal The case when the inner angle SOLUTIONS 405 Figure 200 (Sol 22.20 a)) Figure 201 (Sol 22.20 b)) of the polygon at A is greater than 180◦ is considered in the solution of Problem 22.20 a) Now, suppose that the angle at vertex A is smaller than 180◦ Let B and C be vertices neighbouring A If inside triangle ABC there are no other vertices of the polygon, then BC is the diagonal and if P is the nearest to A vertex of the polygon lying inside triangle ABC, then AP is the diagonal Hence, the number of vertices from which it is impossible to draw the diagonal does not exceed [ n2 ] (the integer part of n2 ) On the other hand, there exist n-gons for which this estimate is attained, see Fig 60 Figure 202 (Sol 22.21) 22.22 Let us prove the statement by induction on n For n = it is obvious Let n ≥ Suppose the statement is proved for all k-gons, where k < n; let us prove it for an n-gon Any n-gon can be divided by a diagonal into two polygons (see Problem 22.20 a)) and the number of vertices of every of the smaller polygons is strictly less than n, i.e., they can be divided into triangles by the inductive hypothesis 22.23 Let us prove the statement by induction For n = it is obvious Let n ≥ Suppose it is proved for all k-gons, where k < n, and let us prove it for an n-gon Any n-gon can be divided by a diagonal into two polygons (see Problem 22.20 a)) If the number of sides of one of the smaller polygons is equal to k + 1, then the number of sides of the other one is equal to n − k + and both numbers are smaller than n Therefore, the sum of the 406 CHAPTER 22 CONVEX AND NONCONVEX POLYGONS angles of these polygons are equal to (k − 1) · 180◦ and (n − k − 1) · 180◦ , respectively It is also clear that the sum of the angles of a n-gon is equal to the sum of the angles of these polygons, i.e., it is equal to (k − + n − k − 1) · 180◦ = (n − 2) · 180◦ 22.24 The sum of all the angles of the obtained triangles is equal to the sum of the angles of the polygon, i.e., it is equal to (n − 2) · 180◦ , see Problem 22.23 Therefore, the number of triangles is equal to n − 22.25 Let ki be the number of triangles in the given partition for which precisely i sides are the sides of the polygon We have to prove that k2 ≥ The number of sides of the n-gon is equal to n and the number of the triangles of the partition is equal to n − 2, see Problem 22.24 Therefore, 2k2 + k1 = n and k2 + k1 + k0 = n − Subtracting the second equality from the first one we get k2 = k0 + ≥ 22.26 Suppose that there exists a 13-gon for which on any line that contains its side there lies at least one side Let us draw lines through all the sides of this 13-gon Since the number of sides is equal to 13, it is clear that one of the lines contains an odd number of sides, i.e., one of the lines has at least sides On these sides lie vertices and through each vertex a line passes on which there lie at least sides Therefore, this 13-gon has not less than + · = 15 sides but this is impossible Figure 203 (Sol 22.26) For n even, n ≥ 10, the required example is the contour of a “star” (Fig 61 a)) and an idea of how to construct an example for n odd is illustrated on Fig 61 b) 22.27 Let k be the number of acute angles of the n-gon Then the number of its angles is smaller than k · 90◦ + (n − k) · 360◦ On the other hand, the sum of the angles of an n-gon is equal to (n−2)·180◦ (see Problem 22.23) and, therefore, k ·90◦ +(n−k)·360◦ > (n−2)·180◦ , i.e., 3k < 2n + It follows that k ≤ [ 2n ] + 1, where [x] denotes the largest integer not exceeding x Figure 204 (Sol 22.27) ] + acute angles are given on Fig 62 Examples of n-gons with [ 2n 22.28 Under these operations the vectors of the sides of a polygon remain the same only their order changes (Fig 63) Therefore, there exists only a finite number of polygons that SOLUTIONS 407 Figure 205 (Sol 22.28) may be obtained Moreover, after each operation the area of the polygon strictly increases Hence, the process terminates 22.29 Let us carry out the proof by induction on n For n = the statement is obvious Let n ≥ If one of the numbers αi is equal to π, then the inductive step is obvious and, therefore, we may assume that all the numbers αi are distinct from π If n ≥ 4, then n π 1X (αi + αi+1 ) = 2(n − 2) ≥ π, n i=1 n where the equality is only attained for a quadrilateral Hence, in any case except for a parallelogram (α1 = π − α2 = α3 = π − α4 ), and (?) there exist two neighbouring numbers whose sum is greater than π Moreover, there exist numbers αi and αi+1 such that π < αi + αi+1 < 3π Indeed, if all the given numbers are smaller than π, then we can take the above-mentioned pair of numbers; if αj > π, then we can take numbers αi and αi+1 such that αi < π and αi+1 > π Let αi∗ = αi + αi+1 − Then < αi∗ < 2π and, therefore, by the inductive hypothesis there exists an (n − 1)-gon M with angles α1 , , αi−1 , αi∗ , αi+2 , , αn Three cases might occur: 1) αi∗ < π, 2) αi∗ = π, 3) π < αi∗ < 2π In the first case αi + αi+1 < 2π and, therefore, one of these numbers, say αi , is smaller than π If αi+1 < π, then let us cut from M a triangle with angles π − αi , π − αi+1 , αi∗ (Fig 64 a)) If αi+1 > π, then let us juxtapose to M a triangle with angles αi , αi+1 − π, π − αi∗ (Fig 64 b)) In the second case let us cut from M a trapezoid with the base that belongs to side Ai−1 A∗i Ai+2 (Fig 64 c)) In the third case αi + αi+1 > π and, therefore, one of these numbers, say αi , is greater than π If αi+1 > π, then let us juxtapose to M a triangle with angles αi − π, αi+1 − π, 2π − αi∗ (Fig 64 d)), and if αi+1 < π let us cut off M a triangle with angles 2π − αi , π − αi+1 and αi∗ − π (Fig 64 e)) 408 CHAPTER 22 CONVEX AND NONCONVEX POLYGONS Figure 206 (Sol 22.29) Chapter 23 DIVISIBILITY, INVARIANTS, COLORINGS Background In a number of problems we encounter the following situation A certain system consecutively changes its state and we have to find out something at its final state It might be difficult or impossible to trace the whole intermediate processes but sometimes it is possible to answer the question with the help of a quantity that characterizes the state of the system and is preserved during all the transitions (such a quantity is sometimes called an invariant of the system considered) Clearly, in the final state the value of the invariant is the same as in the initial one, i.e., the system cannot occur in any state with another value of the invariant In practice this method reduces to the following A quantity is calculated in two ways: first, it is simply calculated in the initial and final states and then its variation is studied under consecutive elementary transitions The simplest and most often encountered invariant is the parity of a number; the residue after a division not only by but some other number can also be an invariant In the construction of invariants certain auxiliary colorings are sometimes convenient, i.e., partitions of considered objects into several groups, where each group consists of the objects of the same colour §1 Even and odd 23.1 Can a line intersect (in inner points) all the sides of a nonconvex a) (2n + 1)-gon; b) 2n-gon? 23.2 Given a closed broken plane line with a finite number of links and a line l that intersects it at 1985 points, prove that there exists a line that intersects this broken line in more than 1985 points 23.3 In plane, there lie three pucks A, B and C A hockey player hits one of the pucks so that it passes (along the straight line) between the other two and stands at some point Is it possible that after 25 hits all the pucks return to the original places? 23.4 Is it possible to paint 25 small cells of the graph paper so that each of them has an odd number of painted neighbours? (Riddled cells are called neighbouring if they have a common side) 23.5 A circle is divided by points into 3k arcs so that there are k arcs of length 1, 2, and Prove that there are diametrically opposite division points 23.6 In plane, there is given a non-selfintersecting closed broken line no three vertices of which lie on one line A pair of non-neighbouring links of the broken will be called a singular one if the extension of one of them intersects the other one Prove that the number of singular pairs is always even 23.7 (Sperner’s lemma.) The vertices of a triangle are labeled by figures 0, and This triangle is divided into several triangles so that no vertex of one triangle lies on a side of the other one The vertices of the initial triangle retain their old labels and the additional vertices get labels 0, 1, so that any vertex on a side of the initial triangle should 409 410 CHAPTER 23 DIVISIBILITY, INVARIANTS, COLORINGS be labelled by one of the vertices of this side, see Fig 65 Prove that there exists a triangle in the partition labelled by 0, 1, Figure 207 (23.6) 23.7 The vertices of a regular 2n-gon A1 A2n are divided into n pairs Prove that if n = 4m + or n = 4m + 3, then the two pairs of vertices are the endpoints of equal segments §2 Divisibility 23.9 On Fig 66 there is depicted a hexagon divided into black and white triangles so that any two triangles have either a common side (and then they are painted different colours) or a common vertex, or they have no common points and every side of the hexagon is a side of one of the black triangles Prove that it is impossible to find a similar partition for a 10-gon Figure 208 (23.9) 23.10 A square sheet of graph paper is divided into smaller squares by segments that follow the sides of the small cells Prove that the sum of the lengths of these segments is divisible by (The length of a side of a small cell is equal to 1) §3 Invariants 23.11 Given a chess board, it is allowed to simultaneously repaint into the opposite colour either all the cells of one row or those of a column Can we obtain in this way a board with precisely one black small cell? 23.12 Given a chess board, it is allowed to simultaneously repaint into the opposite colour all the small cells situated inside a 2×2 square Is it possible that after such repaintings there will be exactly one small black cell left? 23.13 Given a convex 2m-gon A1 A2m and point P inside it not belonging to any of the diagonals, prove that P belongs to an even number of triangles with vertices at points A1 , , A2m ... each vertex a line passes on which there lie at least sides Therefore, this 13-gon has not less than + · = 15 sides but this is impossible Figure 203 (Sol 22.26) For n even, n ≥ 10, the required... AB and to each of them assign the number ±l, where l is the length of this side and the sine “plus” is taken if following this side in the direction of ray AB we get inside M and “minus” if we... n ≥ Fix one acute triangle Ap Aq Ar and let us discard vertex Ak distinct from the vertices of this triangle The inductive hypothesis is applicable to the obtained (n − 1)gon Moreover, if, for