In this section, we examine some examples that illustrate how mathematical induction can be used, in combinatorial situations, to establish the existence of configurations possessing certain properties. We assume from the reader a thorough acquaintance with the principle of induction, referring to Section 4.1 of [8] for an elementary exposition of the necessary background. In particular, most often we shall not write formal inductive proofs, i.e., sometimes initial cases will not be checked and induction hypotheses will not be explicitly stated.
Example 4.13 (Germany) Givenn ∈ N, prove that one can assemble a square of side length 2nby using one square piece of side length 1 and severalL-triminoes, i.e., pieces of the shape below, where each small square has side length equal to 1:
Proof We shall use induction onnto show that each square of side length 2ncan be assembled as required, with the square piece of side length 1 occupying a corner of the 2n×2nsquare.
Forn=1, the assembling is immediate, as shown in the figure below:
By induction hypothesis, assume that every square of side length 2k can be assembled as prescribed. For the inductive step, take a 2k+1 ×2k+1 square and divide it into four squares of side length 2k, as shown in the figure below. Then, place an L-trimino in the center of the larger square, so that it occupies one square
106 4 Existence of Configurations
of side length 1 in three of the four squares of side length 2k. Finally, place the 1×1 square piece in the corner of the 2k+1×2k+1square that belongs to the square of side length 2kwhich was not intersected by the L-trimino already placed.
2k+1
2k+1
The placement of these two pieces has the effect of letting each of the four squares of side length 2k with exactly one corner occupied by a square piece of side length 1. Hence, induction hypothesis assures that we can finish the assembling of each one of these four squares by using L-triminoes only, and taking these lets us conclude that the same is true for the 2k+1×2k+1square (with the square piece of side length 1 placed in one of its corners).
As one should expect, the coming example shows that, sometimes, strong induction is in force.
Example 4.14 (Leningrad) List all nonempty subsets of the set{1,2, . . . , n}which do not contain consecutive elements. For each one of them, compute the square of the product of its elements. Prove that the sum of all of these numbers is equal to (n+1)! −1.
Proof As was anticipated above, we shall make strong induction onn≥1.
Ifn =1, the only nonempty subset of{1}is itself, so that the required sum is 1=2! −1.
By induction hypothesis, suppose that, for 1≤ n≤ k, the desired sum is equal to(n+1)! −1. Now, notice that the nonempty subsets of{1,2, . . . , k+1}without consecutive elements can be divided in two categories:
(i) the nonempty subsets of{1,2, . . . , k}without consecutive elements;
(ii) the sets of the formA∪{k+1}, whereAis a subset of{1,2, . . . , k−1}without consecutive elements.
The induction hypothesis guarantees that the sum of the squares of the products of the elements of the sets of type (i) equals(k+1)! −1, while that of the sets of type (ii) equals(k+1)2(k! −1)+(k+1)2(the summand(k+1)2corresponding to the set{k+1} = ∅ ∪ {k+1}).
Hence, the sum of the squares of the products of the elements of all desired subsets of{1,2, . . . , k+1}equals
(k+1)! −1+(k+1)2(k! −1)+(k+1)2=
=(k+1)! −1+(k+1)2ãk!
=(k+1)! −1+(k+1)!(k+1)
=(k+1)!(k+2)−1
=(k+2)! −1.
We now give two examples that show that an inductive argument can compose justpartof the analysis of a certain combinatorial situation.
Example 4.15 (Russia) Letn >1 be an odd integer. In open field,nchildren are so positioned that, for each one of them, the distances to the othern−1 children are pairwise distinct. Each child has a water pistol and, at the sound of a whistle, fires at the child closest to him/her. Show that at least one child will remain dry.
Proof Firstly, let us look at the casen = 3. LetA,B andC be the children and assume, without loss of generality, thatAB < BC < AC. Then,Afires atBandB atA, so thatCremains dry.
Now, assume that whenever 2k−1 children (k >1) are positioned in open field as prescribed in the statement of the problem, at least one of them remains dry. We then consider 2k+1 children, also positioned as required by the problem. Since the distances between pairs of children are pairwise distinct, there exist two children, sayAandB, such that the distance between them is the smallest one of all distances between two of the given children. Thus,Afires atBand vice-versa. DiscardingA andB, we are left with 2k−1 children, and there are two distinct possibilities:
(i) One of these 2k−1 children fires atAorB: in this case, at most 2k−2 shots are fired towards some of the remaining 2k−1 children (exceptAandB). By the pigeonhole principle, at least one of them remains dry.
(ii) None of the remaining 2k−1 children fires atAorB: by induction hypothesis, at least one of these 2k−1 children remains dry.
Example 4.16 (IMO Shortlist) We wish to write one of the numbers 0, 1 or 2 in each entry of a table of 19 lines and 86 columns, in such a way that the following conditions are satisfied:
(a) Each column has exactlykzeros.
(b) For any two chosen columns, there exists a line such that the two entries of this line situated at the chosen columns are, in some order, equal to 1 and 2.
Find all values ofkfor which this is possible.
108 4 Existence of Configurations Solution We say that a set ofmlinesdistinguishesncolumns if, for any choice of two of thesencolumns, it is possible to choose one of themlines in such a way that the entries of this line situated in the two chosen columns are equal to 1 and 2, in some order.
We claim thatmlines distinguish at most 2m columns. Indeed, it is clear that a single line can distinguish at most two columns. Now, assume that j lines distinguish at most 2j columns. Then, since a single line distinguishes at most two columns,j +1 lines will distinguish at most two groups of 2j columns (the distinction made by the(j +1)-th line between the two groups of 2j columns is a permutation of the pattern
11. . .1
2j
22. . .2
2j
.
Hence,j +1 lines will distinguish at most 2j+2j =2j+1columns.
Back to the given table, since 86 > 26, the above claim assures that we need at least 7 lines to distinguish all of the 86 columns. On the other hand, 7 lines will suffice, for 86<27. Actually, it suffices to choose 86 distinct sequences of 7 terms, each of which equal to 1 or 2, and fulfill the first 7 lines with the terms of these 86 sequences. The remaining lines can be randomly fulfilled.
Therefore, we must have 0≤k≤19−7=12.
For the coming example, recall that, givenn∈N, we letIn= {1,2, . . . , n}. Also recall that, given setsAandB, their symmetric differenceABis given by
AB=(A∪B)\(A∩B).
Example 4.17 (Argentina) Givenn∈N, letPnbe the family of all subsets ofIn. If f :Pn→Inis any function, show that there exist distinctA, B∈Pnand such that
f (A)=f (B)=max(AB).
Proof Forn=1, we havef : {∅,{1}} → {1}and∅{1} = {1}, so that f (∅)=f ({1})=1=max({1}).
Suppose that the statement is true forn = k, and take a functionf : Pk+1 → Ik+1. There are two cases to consider:
(i) f maps all sets in Pk to elements ofIk: in this case, it suffices to apply the induction hypothesis to the restriction off toIk.
(ii) There existsA∈Pk such thatf (A)=k+1: letF ⊂Pk+1be the family of the elements ofPk+1containingk+1, i.e.,
F = {X∪ {k+1}; X∈Pk}.
We look at two subcases:
• there existsB ∈ F such thatf (B)= k+1: then,k+1 ∈ B\A, so that A=B,k+1∈ABand
f (A)=f (B)=k+1=max(AB).
• fmapsFintoIk: letg:Pk →Ikbe given by settingg(X)=f (X∪{k+1}), for everyX∈Pk. By induction hypothesis, there exist distinctA, B∈Pk, such that
g(A)=g(B)=max(AB).
Since(A∪{k+1})(B∪{k+1})=AB, we conclude thatf (A∪{k+1}) andf (B∪ {k+1})are both equal to max
(A∪ {k+1})(B∪ {k+1}) , withA∪ {k+1} =B∪ {k+1}.
We finish this section by revisiting Problem17, page31, this time with the aid of mathematical induction.
Example 4.18 (Saint Petersburg—Adapted) We are given 2n+1 points on a circle, such that no two of them are the endpoints of a diameter. Prove that, among all triangles having three of the given points as their vertices, at most
1
6n(n+1)(2n+1) of them are acute.
Proof Forn = 1 there is nothing to do, since three points on a circle determine exactly one triangle.
By induction hypothesis, assume that the problem is true whenn= k−1, for some integerk ≥2. Then, letn =k, so that we have 2k+1 points on the circle.
Out of them, take all pairs of points such that the chord joining them leaveskpoints in one arc andk−1 points in the other arc of the circle. Among all such pairs of points, letAandBbe two such thatABis at minimum distance from the centerO of the circle (if there is more than one pair of points with this property, choose any one of them).
Now, there are three kinds of acute triangles having vertices at three of the 2k+1 given points:
(i) Those having AandB as two of their vertices: sinceO must belong to the interior of such a triangle, andABleaveskpoints at one side andk−1 points at the other side, there are at mostksuch triangles.
(ii) Those havingAorB, but not both of them, as one of their vertices: letC and D be the other two vertices of such an acute triangle (so that the third one is eitherAorB). We consider two subcases:
110 4 Existence of Configurations
• CandDare at the same side ofAB: ifACDwas to be acute, thenACor ADwould leaveBandOat different half-planes. In turn, this would assure that the distance fromAC orADtoO would be less than the distance of ABtoO, thus contradicting our choice ofAandB. Therefore,ACDis not acute and, accordingly, the same is true ofBCD.
• CandDare in opposite sides with respect toAB: sinceACBDis convex, at most one of the trianglesACDorBCDcontains the centerO, and hence at most one of such triangles is acute. On the other hand, by the choice ofA andBthe set{C, D}can be chosen in exactlyk(k−1)ways, and this is the largest possible number of acute triangles of this kind.
(iii) Those having neitherAnorBas one of their vertices: by induction hypothesis, there are at most 16(k−1)k(2k−1)such triangles.
Adding the contributions of the three cases above, we conclude that the number of acute triangles having its vertices at three of the 2k+1 given points is at most
k+k(k−1)+1
6(k−1)k(2k−1)=1
6k(k+1)(2k+1).
Problems: Sect. 4.2
1. (India) Prove that, for everyn≥6, every square can be partitioned intonother squares.
2. (Brazil) Given a natural numbern >1, we write a real number of modulus less than 1 in each cell of ann×nchessboard, in such a way that the sum of the numbers written in the four cells of any 2×2 square is equal to 0. Ifnis odd, show that the sum of the numbers written in then2cells is less thann.
3. (TT) Point O is situated in the interior of the convex polygon A1A2. . . An. Consider all of the angles AiOAj, with distinct 1≤ i, j ≤ n. Prove that at leastn−1 of them are not acute.
4. (TT) In a convex polygonP some diagonals were drawn, such that no two of them intersect in the interior ofP. Show that there are at least two vertices ofP such that none of the traced diagonals is incident to none of these two vertices.
5. (Hungary) In ann×nchessboard we have a certain number of towers (at most one tower for each 1×1 square). A tower can move from one square to another if and only if these two squares belong to a single line or column of the board and there are no other towers in between. On the other hand, if two towers belong to a single line or column and there are no other towers in between, then we say that any of these towers canattackthe other. Prove that one can paint the towers with one of three colors in such a way that no tower can attack another tower of the same color.
6. We are givenndistinct lines in the plane. Prove that it is possible to paint the regions into which these lines divide the plane of black or white, so that if two regions share a common edge, then they have distinct colors.
7. (TT) Each square of a chessboard is painted either red or blue. Prove that the squares of one of these two colors are such that a chess queen can move through all of them (possibly passing more than once through one or more of the 1×1 squares of this color), without visiting a single square of the other color. We recall that a chess queen can move an arbitrary number of squares along any line, column or diagonal of the chessboard.
8. We are givennpoints in the plane, not all of which being collinear. Prove that:
(a) There exists at least one line that passes through exactly two of thesen points.
(b) Thenpoints determine at leastndistinct lines.
The result of item (a) is known as theSylvester-Gallai theorem,4while that of item (b) is due to Erdửs and de Bruijn.5
9. In the plane we are given 2npoints ingeneral position, i.e., such that no three of them are collinear. Ifnof the given points are blue and the othernare red, prove that it always possible to drawnline segments satisfying the following conditions:
(a) Each line segment joins a blue point to a red point.
(b) They do not intersect each other.
10. (OCS) Prove that, given a positive integern, there exists another positive integer knwith the following property: given any set ofknpoints in space, no four of them being coplanar, and associating an integer number from 1 tonto each line segment joining two of theseknpoints, one necessarily gets a triangle having vertices at three of thekn points and such that the numbers associated to its sides are all equal.