Springer problem solving strategies 1998

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Polynomials by Edward J Barbeau Problems in Geometry

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Problem-Solving Strategies With 223 Figures

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60054 Frankfurt am Main Germany Mo: artment of ‘Mathematics Santa Clara University Santa Clara, CA 95053 SA Mathematics Subject Classification (1991): O0A07

Library of Congress Cataloging-in-Publication Data Engel, Arthur

Problem solving strategies/Arthur Engel P- — (Problem books in mathematics) Inclu des dex,

0-387-982 19-1 (softcover: alk paper) 1 Problem solving I Title IT Series QA63.E54 1997

510! 76—de21 -10080

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This book is an outgrowth of the training of the German IMO team from a time principles Not only toptes be et also ideas were efficient means of classification

For whom is this book w:

« For trainers and participants of contests of all kinds up to the S highest level of international competitions, including the IMO and the Putnam Competition « For the regular high school teacher, who is conducting a mathematics club and is looking for ideas and problems for his/her club Here, he/she will find ever proposed at any competition « For high school teachers who want to pose the problem of the week, problem

of the month, and research problems of the year This is not so easy Many fail, hile but some persevere, and after a w! ey succeed and generate a c: © atmosphere with continuous discussions of mathematical probl « For the regular high school teacher, who is just lootine fe for ideas to enrich

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tion In this way, it was possible to increase the number of campiesan and probes to over 1300 The reader can increase the effectiveness of the book e' ore by trying to solve the examples

The problems are ve almost ‘exclusively competition problems from all over the world Most of them are from the former USSR, some from Hungary, and some competition problems are usually variations of pr pro oblerns fro Tom journals with prob- If you see a beautiful problem, you firs at the creativity of the problem proposer Later you discover the nen in lier source For this reason, the references to competitions are somewhat sporadic Usually no source is given if I are results that are known to experts in the respective fields

There is a huge literature of mathematical problems But, as a trainer, I kn: that there can never be enough problems You are always in desperate need of new a ”

The problems are arranged in no particular order, and especially not in increasing order of difficulty We do not know ho to rate a problem’s difficulty Even the IMO

ing the difficulty o: b over 40 aining ould like to thank Dr Manfred Grathwohl for his help in imprementing various IATX versions on the workstation at the institute and on my PC at home When difficalties » arose, he was a competent and friendly advisor

There will be some errors in the proofs, for which I take full responsibility, since none of my colleagues has read the manuscript before Readers will miss important strategies So do I, but I have set myself a limit to the size of the book Especially, advanced methods are missing Still, it is probably the most complete

pular in Rus

8

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Preface Abbreviations and Notations

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Abbreviations ARO Allrussian Mathematical Olympiad AuMO Australian Mathematical Olympiad

BMO Balkan Mathematical Olympiad ChNO Chinese National Olympiad HMO Hungarian Mathematical Olympiad (Ktirschak Competition)

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iad (ARO from 1994 on) SPMO St Petersburg Mathematical Olympiad

TT Tournament of the Towns

„ 5 e a

USO US Olympiad Notations for Numerical Sets N or Z the positive integers (natural numbers), i-e., {1,2,3,

mnegative integers, {0,1,2, Z the integers Q the rational numbers @ the positive rational numbers œ the nonnegative rational numbers

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The Invariance Principle

We present our first Higher Problem-Solving Strategy It is extremely useful in solving certain types of difficult problems whic are easily recognizable We will ch it by solving problems which use this tegy In fact, problem solving can be learned only by solving problems Buri it must be supported by strategies provided by the trainer

Our first strategy is the search for invariants, and itis called the Invariance Prin- ciple The principle is Oe to An (games, transformations) Some task is Tepeate dly performed What stays the same? What remains invariant? Here is a saying easy to remeniber

If there is repetition, look for what does not change! In algorithms there is a starting state S and a sequence of legal steps (moves, transformations) One looks for answers to the following questions:

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sequence of points (X», yx) according to the rule Xn + Ya X=4, Y=h, Xp = 7 eH = 2Xn Vn

Xn + Yn Here it is easy to find an invariant From xy41¥a41 = Xn Ya, for all we deduc Xn, = ab for all n This is the invariant we are looking for Initially, we have » < xo This relation also remains invariant Indeed, suppose y, < x, for some Then x,41 is the midpoint of the segment with endpoints y,, x, Moreover, Ya+I < X,41 Since the harmonic mean is strictly less than the arithmetic mean

Thus na — Ya na — Ya — a

atl jm†+†TL — Xn + Ya 2 2

for alln e hav x wi = fab

x? = abo

Here ` invariant ped us very much, bt its s recognition was not yet the solution, although the completion of the solution was trivial E2 Suppose the positive integer odd First Al writes the numbers 1,2, , 2n on the blackboard Then he Dieks an any two numbers a, b, erases them, and w writes, instead, |a — b| Prove that an odd number will remain at the end Solution Suppose S' is the sum of all the numbers still on the blackboard Initially this sumis S = 14+2+4. -+2n = n(2n+4 1), an odd number Each step reduces S by 2 min(a, 5), which is an even number So the parity of S is an invariant During re whole reduction process we have S' = 1 mod 2 Initially the parity is odd So,

will also be odd at the end

E3 A circle is divided into six sectors Then the numbers 1,0, 1,0, 0, 0 are writ- numbers by 1, Is it possible to equalize all numbers by a sequence of such steps?

tially ƒ 2 Th l7=0 t

E4 In the Parliament of Sikinia, each member has at most three enemies Prove that the house can be Be separated into two houses, so that each member has at most one enemy in his own hou

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creases at each step of the algorithm So we know that our all orithm will termi- nate There is no strictly decreasing infini i i strictly an invariant, but decreases &s monotonically until it becomes constant Her: the monotonicity relation is the ant

- Suppose no not all four integers a, by c, a are equal Start with hae b c, oo) and number of the — will nay become arbityarity lovee Solution Let P, = (ay, B

have a) + b, +c, +d, = Oforn > 1 We do not see ee yet how to use this invariant But t geometric interpretation is snot ly helpful A very important function for the space is the square of its distance from the origin (0, 0, 0, 0), which is 42 + b2 + c2 + đ? If we could prove that it has no upper bound, we would be nished

try to find a relation between P,,4, and P,: Gnu + Dau + pay t+ deg, = Gn — ba)” + On = Cn)” + Cn — dn + Gn — On

= 22 + b} + c2 + đ?) apb„ — đy — 2cyd¿ — 2d,an Now we can use a, +b, + + d, = Oor rather it: quare: 0 = (Ga tbntentdn) = (artery +@ntdny +2anb, +2 Adding (1) and (2), for a?,, rh + B,,4+¢,, TH + d?,,, we get

2a, + B+ + a) + (Gn + Cn + On + ary = UG + be + đa + độ `> From this invariant inequality relationship we conclude that, for n > 2,

đ2 + b2 + et Pala t+htcdtad ` (2)

e distance of the points P, from the origin increases without bound, which means that at least one component must become arbitrarily large Can you alw, have equality in (2)?

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X0 _— yn Xn _ 9% aregin V2" 2* — 9 ẽ 2 Yo Yn Ya 2°Yn The right-hand side converges to ye — x6, /y forn — oo Finally, we get y% —% 4 #=Yy=———:

y arccos(xo/ yo) &

It would be pretty la to solve this problem without invariants By the way, this is a hard p m by any competition standard E7 Each of the numbers ay, ., a, is 1 or —1, and we have

ss 1 1 4 as 1 + Gp41a243 = 0 =ÚŨ

Prove that 4|n

Solution This is a number theoretic problem, but it can also be solved by in- changes by +4 Finally, if all four are of the same sign, then S changes by +8

+ 111

¬

ES 2n ambassadors are invited to a banquet Every ambassador has at mostn — enemies Prove that the ambassadors can be seated around a round table, so that nobody sits next to an enemy

Solution First, we seat the ambassadors in any way Let H be the number of neighboring hostile couples We must find an algorithm which reduces this number whenever H > 0 Let a B) be a hostile couple with B sitting to the right of A (Fig 1 3) W must separate them so as to cause as little disturbance as possible

i mains to be shown that

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B A!

A A

Fig 1.3 Invert are A‘B Fig 1.4 Remark This problem is similar to E4, but considerably harder It is the following

‘amiltonian path th ithe sum of the degrees of any two vertices is equal to or larger than n — 1 In our special case, we have proved that there is even a Hamiltonian circuit

To each vertex of a pentagon, we assign an integer x; with sums = >~ x; Ifx, y, z are the numbers anes to three successive vertices and if y < 0, then we ee * y,z) by(x + y,— + z) This step is repeated as long as there Decide if the algorithm always stops (Most difficult problem of IMO 1986 )

Solution The algorithm always stops The key to the proof is (as in Examples 4 and 8) to find an integer-valued, nonnegative function f(x +5) of the vertex labels whose value decreases when the given operation is performed All but one of the eleven students who solved the problem found the same function

5 ŒI, X2, Xã, X4, X5) — Via —XipY, Xe =M1, Xi m2, i=l

0 Th ith

doe loes ns ot stop, w we can find an infinite deen Sequence fo > va > me > - of nonnegative integers Such a sequence does no

Bernard Chazelle (Princeton) asked: How many steps are needed until stop? He considered the infinite multiset S of all sums 'đefned by s@, J) = xy + +++ + xj withl <7 < Sandj >17.A multiset is a set which can have equal elements In this s(4, 5) = x4 changes to —x4 Thus, exactly one negative element of S changes to postive at each step There are only finitely many negative elements in S, since > 0 The number of st eps until stop is equal to the number of negative elements

otha t the x; need not be integers S ¬

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b,c, a Màn integers an the derived sequence bà T(S) = (la — b, |b—el, |e—a|, |d- ab Does the sequence S, S}, Sy = T (81), 53 = T (Sy), - always end up with (0, 0, 0, 0)? Se Let us collect material for solution hints: 021013 4 10.10) (0, 6, 0, 6) +> 6, 6,6, 6) + (0, 0, 0, 0), 5.8 85, 85, 85) > 6,0 0,0, 0.0) (91, 108, 95, 29 203 4 8 104 186), 18 82 182 4 64 100 100) (0.36 0-3 (36, 36, 36, 36) L> (0, 0, 0, 0) Observations:

1 Let max S be the maximal element of S Then max $;4; < max S;, and max Sea < max S; as tone 2s as max S; > 0 Verify these observations This gives a proof of our conjectt 2 Sand rS' have the same life expectancy 3 After four steps at most, all four terms of the sequence become even Indeed,

it is sufficient to calenlie modulo 2 Because of cyclic symmetry, we need most, ¢ each term is divisible by 2, after 8 steps at most, by a » after 4k

+ by 2A 1]

In observation 1, we used another strategy, the Extremal Principle: Pick the maximal element! Chapter 3 is devoted to this principle

In observation 3, we used symmetry You should always think of this strategy, although we did not devote a chapter to this idea

Generalizations: art with four real numbers, e.g.,

” v3 e

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end up with (0, 0, 0, 0) But with + > 1 and S = (1, #, #7, £5) we hav: T(8) = [r — 1,ứ — 1%, — 1), — Dứ? +r+ 1] Tfz3 = 2-+r+1,i.e.,r —= 1.8392867552 , then the process never stops because of the second observation This ¢ is unique up to a transformation f(t) = at + b reach o 0) after 2 ses at most Born = 3, we get, for 011, a pure cycle of lanh 3:011 r> 101 L> 110 r> O11 Eor ø = 5 we get 00011 00101 01111 10001 ++ 10010 be 10111 +> 11000 +> 01001 / 11011 + 01100 10100 L> 11101 r> 00110 r> 01010 Fe 11110 + 00011, which has a pure cycle of length 15

1 Find the periods for n = 6 (n = 7) starting with 000011 (0000011) 2 Prove that, form = 8, the algorithm stops starting with 00000011

3 Prove that, f nd, forn 4 2’, we get

(up to some sreeptions) a vel containing just two vo An boy 0 and evenly a=1.Then| z — b |= z + b mod 2, and we do our calculations in GF(2), ie., the finite field with two elements 0 and 1

-Letn 2 1 be tt le lenott 1 2

some exceptions) 5 Prove that, for odd, S = (0, 0, , 1, 1) always lies ona cycle 6 Algebraization To the Soque ce (ao đụ„ 1), W polynomial

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in finitely many steps:

(a) {4,6, 12}, (b) {x, y, z} with |x — 4], |y — 6], |z — 12| each less than 1/./3? Assume an 8 x 8 chessboard with the usual coloring You may repaint all squares (a) of a row or column (b) of a2 x2 square The goal is to attain just one black square Can you reach the goal

4 \ apply the following algorithm:

b th b @ a, b — a) else b (a, b) <— (a — b, 2b) b b.2b For which starting positions does the algorithm stop? In how many steps does it stop, if it stops? What can you tell about ne and tails? The same que: is, when a, b ate posi teals 5 Around a circle, 5 ones and 4 zeros are arranged in any order Then between any two

equal Gels, you write 0 and between different digits 1 Finally, the original es ed 0 pat HE this proce tepeated indefinitely, you can never get 9 ze: 6 Ther white, b black, and c red chips on a table In one step, you may choo: two chips of different colors and replace them by a chip of the third color Ifjust one chip will remain at the end, its color will not depend on the evolution of the gam Wh len can this final state be reache: hed? 7, There ate a white, & black, and ¢ red chips on a table In one step, you may choo:

two chips of different colors and " cach o one by a chip of the third color Find conditions for all chips to become of the same color Suppose voi have initially 13 white e black and 17 red chips Can a sips become of the color? What states can be reached from these numbers 9 There is a positive integer in each square of a rectangular table In each move, you

may double each number in a row or subtract 1 fromeach number of a column Prove that you can reach a table of zeros by a sequence of these permitted moves 9 Each of the numbers 1 to 10° is hung 1eplaced by its digital sum until we reach

10° one-digit numbers Will these hav e 1’s or 2’s The vertices of an m are labeled by real numbers x), , %, Let a, b, c,d be four successive labels, Sita — ‘Nb —e€@)<0, then we may switch b with c Decide if this switching operation can be performed infinitely often ¬ °

1 In Fig 1.5, you may switch the signs of all numbers of a row, column, or a parallel to one of the diagonals in particular, you may rawiteh the sign of each corner square Prove that at least one —1 will remain in the table

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as

na Under each number a of the first row, there is a positive integer f(a) such at f(a) equals the number of occurrences of a in the first row In the same way, we get the 3rd row from the 2nd row, and so on Prove that, finally, one of the rows is identical to the next row 3 There is an integer in each square of an 8 x 8 chessboard In o e, you may

choose any 4 x 4 or 3 x 3 square and add 1 to each integer of the “chon square Can you always get a table with each entry divisible by (a) 2, (b) 3? We strike the first digit of the number 7°, and then addit to the remaining number This is repeated until a number with 10 digits remains Prove that this number has two equal digits

There is a checker at point (1, 1) of the lattice (x, y) with x, y positive integers It moves as follows At any move it may double one coor or it may subtract the smaller coordinate from the larger Which points of the lattice can the checker teach?

Each term ina sequence 1, 0, 1, 0, 1, 0, starting with the seventh is the sum of the last 6 terms mod 10 Prove that the sequence ,0,1,0,1,0,1, never occurs Star ting with any 35 integers, you may select 23 of them and add 1 to each By tepeating this step, one can make all 35 integers equal Prove this Now replace 35 and 23 by m and n, respectively What condition must m and 7 satisfy to make the equalization still possible? 18 The integers 1, ,2n are arranged in any order on 2n places numbered 1, , 2”

ow we add its place nn number to each integer Prove that there are two among the sums which have the mainder mod 2n 19 The ø holes of a socket are arranged along a circle al (unit) distances and

Imbered 1, n For what n can the prongs ofaplug Những the socket be mumbered such that at Teast one prong in each plug-in goes into a hole of the s: Imber (go numbering)?

A game fo: Than sera, 5) diem 5) We start with x = b and move as follows: fx < y then, sty mm vt if x > y; then set x < x — y and w <_ ‘he game ends with 21 Three integers a, b, c are written on a blackboard Then one of the integer:

and replaced by the sum of the other two diminished by 1 This operation is s repeated many ti wi co the final result 17, 1967, 1983 Could the initial numbers be (a) 2, a — + ¬ n = +4 b d 1 b 2,2 (b 3,

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b ¬ bo a iS) " Trần x (TS ĐỀN HÀ, 2 2

at 7, 72, finally le: ly leads to nonintegral compon: Start with an m x n table of integers In one step, you may change the sign of all

numbers in any row or column Show that you can achieve a nonnegative sum of any

Assume a convex 2m-gon Aj, , Ao In its interior we choose a point P, which does not lie on any diagonal Show that P lies inside an even number of triangles

amo with vertices mg Ai, ,

Three automata J, H, iP print pairs of positive integers on tickets For input (a, 5), I and H give (a+ 1, b+1) and (a/2, b/2), respectively H accepts only even a, b T needs two pairs (a, 6) an ab c)as input and yields output (a, c) Starting with © Hà canyo ou reach the ie « (1, 50) (b) (1, 100)? Initially, we have (a, b), a what 7 is (1, 2) reachal

Three automata J, R, S print pairs of positive integers on tickets For entry (x, y), the automata 1k R, S give tickets (x — y, y), (x + y, y), (y, x), tespectively, as outputs Initially, we have the ticket (1,2) With these automata, can I get the tickets (a) (19, 79) oO tai me Find an invariant What pairs (p, g) can I get starting with (a, b)? Via which pair should I best er 7 numbers are written on a blackboar tep you may e: any two of the

numbers, say a and b, and write, Gost la : bys Repeating than step — 1 times, there is one number left Prove that, initially, if there were m ones on the board, at the end, a number, which is not less than 1/m will remain The following operation is performed with a nonconvex non-self-intersecting poly-

go n P Let A, B be two nonneighboring vertices Suppose P lies on the same side of AB Reflect one part of the polygon connecting A with B at the midpoint O of

afte: fl

AB Prove that the polygon beco: t finitely many such reflections 30 Solve the equat ute? in ey) de? an dyad 31 Let a1, 4 be a permutation of 1,2 n If ø is odd, then the product

P=ứ bai a¿ — 2) (4„ — #) is even Prove this 32 Many handshakes are exchanged at a big international congress We call a person

an odd person if he has ss exchanged an odd number of handshakes Otherwise he will called an even person Show that, at any moment, there is an even number of odd persons

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36 The integers l, , z are afranged in any order In one step you may switch any two neighponne integers Prove that you can never reach the initial order after an odd

ber of steps

37 One step in the preceding oven consists of an interchange of any two integers Prove that the assertion is still tr 38 The integers 1, arranged in order In one step you may take any four integers

me eran ‘te ist with the fourth and the second with the third Prove that, —1) ? is even, then aby means of such steps you may reach the arrangement tifm(m — 1)/2 is odd, you cannot reach this arrangement nn—l,

Consider all lattice squares (x, y) with x integers Assigi lower left corner as a label We shade thes squares ee Đà (1,0), (0, 1), 0 a b, (0, 2) (a) There is a chip on each of the six squares (b) There is only one ch: ‘, , 0)

op: If (x, y) is occupied, but (x + 1, y) and (x, y + 1) are free, you may remove he chip from (x, y) and place a chip on each of ae +1, y) and (x, y +1) The goal is to remove the chips from the shaded squares Is this possible in the cases (a) or (b)? (Kontsevich, TT 1981.) 40 In any way you please, fill up the lattice points below or on the x-axis by s By

nan Tp lì to get one chp n Ko 5) with all other chips cleared st a H nway.) T ceding problem ntsevich might have been suggested by this vroblem

A solitaire jump is a horizontal or vertical jump of any chip over its neighbor point with the chip jumped over removed For instance, with (x, y) and Họ y " 1) occupied and (x, y + 2) free, a jump consists in removing the two chips on (x, y) and (x,y +1) and placing a chip onto (x, y + 2) 41, We may extend a set S of space points by reflecting any point X of S at any space

pointA, A # X a s ” of the 7 vertices of a cube Can you ever get the eight vertex of the cube into S? 42 The following game is played on an infinite chessboard Initially, each cell of an

nxn squate is occupied by a chip A move consists in a jump of a chip over a chip in over (IM! 3 and AUO 19923) 43 Nine 1 x 1 cells of a 10 x are are infected In one time unit, the cells with

at least two infected ighbors hin a common side) become infected Can the infection spread to the whole square 44, Can you get the polynomial h(x) = x from the polynomials f(x) and g(x) by the

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46 + x 51 53 55 56 Bind at L fe yt Start with two numbers 18 and 19 on the blackboard In one step you may add another (IM)?

Ina at (a) pentagon (b) hexagon all diagonals are drawn Initially each vertex and each mint of intersection of the diagonals is labeled by the number 1 In one step io change the signs of all labels to —1 by a seque: ieps (IIM)? 8 In Fig 1.7, two squares are neighbor s if they have a common boundary Consider

same the following operation T: Choose any two neighboring numbers and add the integer to them Can you transform Fig 1.7 into Fig 1.8 by iteration of T?

Fig 1.7 Fig 1.8 There are several signs + and — on a blackboard You may erase two signs and

write, instead, + if they are equal and — if they are unequal Then, the last sign on the board does not depend on the order of erasure There are several letters e, a and b on a blackboard We may replace two e's by one e, two a's by one b, two b's by one a, an a anda b by one e, ana and an e by one a, a b, and an e by one b Prove that the last letter does not depend on the order of A dragon has 100 heads A knight can cut off 15, 17, 20, or 5 heads, respectively,

tly i — 1 othe i’s?

The following operations are permitted with the quadratic polynomial ax? + bx + ¢: (a) wth aandc, (b) replace ce x » x +£ where / is any real By repeating these opera m you transform x* — x — 2 intox? —x—1? veal we have three piles with a, b, and c chips, respectively In In step, you may

transfer on any pile with x chips i Lei

d=y—x-+1.Ifd > 0, the bank pays you d dollars If d < 0, you pay the bank |d| dollars Repeating this step several times “yous observe that the original distribution of chips s has been restored What max: unt can you have gained at this stage? Start with four congruent right triangles In one step you may take any triangle and of congruent triangles (MMO 1995), Starting with a point S(a, b) of the plane with O < a < b, we generate a sequence

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58 Consider any binary word W = It can be transformed by inserting, deleting or appending any word xã mã x Xx being any binary word Our goal is to attained (LMO 1988, oral round)? 59, Seven vertices of a cube are marked by 0 and one by 1 You may repeatedly select

an edge and increase by 1 the numbers at the ends of that edge Your goal is to reach (a) 8 equal numbers, (b) 8 numbers divisible by 3 t» 60 Start with a point S(a, b) of the plane with 0 < b < a, and generate a sequence of

points S,(x,, ¥.) according to the rule m= O = Gs w= xe 2xnYn 0 —,;, 1 be ’ Vent = 2ent1¥n 1 x + Ye

Solutions

= BE one move the number of integers always decreases by one After (4n — 2) steps,

st one integer will be left Initially, there are 27 even integers, which is an even mber If two odd integers are replaced, the number of odd integers decreases by 2 If one of them is odd or both are even, then the number of odd numbers remains

2 (a) (0.6a—0.8b)°+(0.8a+0.6b)? = a?+b* Since a? +b? +c? = 3?4+4°+412? = 137, the point (a, b, c) lies on the sphere around O si radius 13 Because 4?+@?-Ƒ12? = 142, the goal lies on the sphere around O with radius 14 The goal cannot be reached (b) (x — 4) + (y — 6)? +(z — 12)? < 1 The goal cannot be reached (a) Repainting a row or column with b black and 8 — b white squares, you get (8 —

black and b white squares The number of black squares changes by |(8 — b) |8 — 28|, that is an even number The parity of the number of black squares does not change Initially, it was even So, it always remain n One black square is unattainable The reasoning for (b) is similar 4 Heteis a solution valid for natural, rational andirrational numbers With the invariant

a+b=n the algorithm can be reformulated as follows: n/2,teplace a by 2a Ifa >n/2, teplaceabya—-b=a-—(n-—a)=2a-—n=2a (modn)

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(b) The result is nonterminating and periodic ajn = 0.a1a2 apdidy dedidy .dy

le alg

© ite tesult is nonterminating and nonperiodic: ain 0.d,d.d; In this case, orithm will not stop, and the sequence (1) is ot periodic This is a special case of problem E10 on shrinking square 2:0+4+0=14+1=0,14+0=0+1 =1 Let (4, x, , x.) be ginal distribution of zeros and ones around the circle One step consists of the replacement (X15 " 6-65 My) <— (Xì 32, #¿ Ð X:, , X„ + xị) 2, 2T Xã, .ị ).T There are two special distributions i E=(,1, , 1) and/ =(0,0, , 0) Here, we must work backwards Suppose we finally reach i Then the preceding state must be £, and before that an alternating n-tuple (1, 0, 1, 0, ) Since z is odd such an n-tuple does not exist N that 2 i PP The foll Œq, ,; 3u) S— 1 + Xp, Xp +g, 2 + Ky TT XI) < (II + Xa, Xp +X, Ky + Xa) < (XI +12 +1: +34, 32 Ð Xã T X4 Xs, ) — (XI + x5, 824565 ) — shows that, for g = 1, the iteration ends up with J For g > 1, we eventually arrive prove this 14 All three numbers a, b, c change their parity in one step If one of the numbers has 6

different parity from the other two, it will retain this property to the end This will be the one which remains 9 â ơ ° triples (a + 2,6 —1,¢- 1), a—1,b+2,ec— T), (4a — 1, ce+2) In each case, Ï = a — bmod3 is an invariam But b — e = 0 mod 3 and a—c = Omod 3 ate also invariant So monochromatic state

If there are numbers equal to 1 in the first column, then we double the correspondin: a

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a "

ơ â

(a) Let S be the sum of all numbers except the third and sixthrow S mod 2 isinvariant 1S ế 0 a 2) initially, then odd numbers will remain on the chessboard (b) Le e the sum of all numbers, except the fourth and eight row Then J = Ss ned si is an invariant If, initially, 7 a 0 (mea 3) there will always be numbers on the chessboard which are not divisible by We have 7? = 1 mod 9 = 7° = 7! mod 9 This digital sum remains invariant At the endall digits cannot be distinct, else the digital sum would be 0+1+ -+9 = 45, which is 0 mod 9

hed 1.1) iff ocd Ñ.Th Pp itted

re, 1 (x1, x 2x, +4 + 8x4 + 10x5 + 12x5 mod 10 is the invariant Starti th7(1.0.1.0.1 8 th 17(0.1.0.1.0.1 4 t be reached

Suppose gcđớn, z) — 1 Then, in Chapter 4, E5, we prove that øxz — my + 1 has a solution with x and y from {1,2, ,m — nh We rewrite this SN in the fori m(y -1) +m + IN 1 Now we pla lace Hy positive integers x " around a a circle assuming that x, is the smallest number We proceed as follows to around me circle in blocks of m and increase each number of a block by 1 If you de this n uu get around the circle m times, and, in addition, the first number becomes o one more ve then the others In this way, |Xnax — Xmin| decteases by one This is repeated each time placing a minimal element in front until the difference between the maximal and minimal element is reduced to zero Butif ged(x, y) = d > 1, then sucha reduction is not always possible Let one of the m numbers be 2 and all the others be 1 Suppose that, applying the same operation means m+ 1+kn = O0modm But d does not divide m Tim 1 since d > 1 Hence m does not divide m + 1 + kn Contradiction! We proceed by contradiction Suppose all the remainders 0, 1, ,22 — 1 occur The sum of all integers and their place numbers is

5 =2(1+2+ + 2ø) = 2n (2ø + 1) =0 (mod 2n) The sum of all remainders is

S,=O+1+ 4+2n-1l=n(Qn—1)=n (mod 2n)

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21

P and R are obviously invariant We show the invariance of Q Initially, we have ab +ab = = 2a b, and this 3 is spac correct After 0 one step, the left side of 9 b

Initially, if all components are greater than 1, then they will remain greater than 1

b<c, ¬ =a+b—— ], and a backward step yields the triple (2, b,b — a + 1) TÌ we can retrace the last state a sen 1983) mane until the next to last ep (17, 1967, 1983) <— (17, 1967,1951) <— (17 5,1951) <— ++ (17, 15,31) <— (17, 15,3) <— (13, 1 an 5,3,3 The preceding triple should be (1, 3, 3) containing 1, whichis impossible Thus the triple (5,3, hi generated at the first step We can get om (3, 3, 3) to (5, 3, 3) in one step, butn 1n (2, 2, 2)

Let a; be the number of chips on the circle #i We consider the mS = Sia Initially, we have S = S°i #1 = n(@ + 1)/2 ands at at the end, we must have kn for ‘or k € {1,2, ,m} Each move changes S by 0, or n n, thatis, S is invariant mod n At the end, S = 0 mod n Hence, at the bepinning, ¥ we must have S = Omodn se for odd Reaching the goal is trivial in the case of an odd n Ey is is the

Solution 1 Suppose we get only integer n-tuples from (x,, ,x,) Then the dif- axim:

afrive at an integral z-tuple (2, 2, , 2) We will show that we cannot get equal numbers om pairwise different numbers Supppose z;, , Z, ate not all e equ ual, but (21 + 22)/2 = (22 + %3)/2 = (# + Z¡)/2 Then Z¡ = 4 = Z2 ñạ — ñ+ — ñ¿ — s lÊn 1S odd then all z; are equal, oe with our 2ssumpfon For even = 2k3 we must eliminate the case (a, b, ,a, b) witha # b Suppose

8

yỊ T2 WTỊA Yet Ye Ty — 3 Tờ —Ủ

22 23" 3ˆ 2 "

But the sums of the left sides of the two equation chains are equal, i.e., a = ở, that is, _—~ (a,b, , 4, b) with a # b Solution 2 Let ¥ = (x1, ,x,), TX = Y =(n, , yy) Withn +1 =1,

" 1 1 "

yw =| Ver Tên +28 21) < 4 3 G7 Fh tap tx) = doy

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bo + nN ¬ bo a number of steps Another Solution Sketch Try a geometric solution from the fact that the sum of the at each step

Ifyou find a negative sum in any row or column, change the signs of all m umibers it that row or column Then the sum of all numbers in the table strictly increases The nonnegative signs

The diagonals partition the interior of the polygon into convex polygons Consider $ par y; two neighboring polygons P,, P, having a common side ona _ or side XY Then P,, P, both belong or do not belong to the triangles without the side XY Thus, if P goes from P, to P, the number of triangles changes by Tại The — ft; and f are the numbers of vertices of the polygon on the two sides of XY Since tf +t = 2m + 2, the number ft, — f, is also even

t get rid of an odd divisor of the difference b — a, that is, you can reach You cannot

(1, 50) from (5, 19), but not (1, 100) automata leave gcd(x, y) unchanged We can reach (19, 79) from (1, * 27, The thre: rt ot 9 a Wee can a (p, nt tom (a, b) iff ged(p, g) = ged(a, b) = Go from (2, + 1), then, up to (p, 4) hà ` to(1,đ ‘rom the inequality 1/2 + 1/b > si at 6) which is equivalent to (a + b)/2 > 2ab/(a + b), we conclude that the sum ,Š of the inverses of the numbers does not increase Initially, we have S = n Hence, at the end, we have § < n For the last number 1/5, we have 1/5 > 1 The permissible transformations leave the sides of the polygon and their directions

invariant Hence, there are only a finite number of polygons In addition, the area strictly increases after each reflection So the process is finite Remark The corresponding problem for line reflections in AB is considerably harder The theorem is still valid, but the proof is no more elementary The sides still remain the same, but their direction changes So the finiteness of the process cannot be easily deduced (ni the case ofline reflections, there is a conjecture that 27 reflections suffice toreacha x polygon.)

Let f(x) = x? — 3x4 3 We are asked to solve the equation f(f(x)) = x, thatis to find the fixed or invariant points of the function f o of First, let us look at fia) = = “ ie the fixed points of f Every fixed point of f is a fixed point of fo f In

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bets 1 ton C ction!

2 8 S o 3 = 8

32 We partition the participants into the set E of even persons and the set O of odd persons We observe that, during the hand shaking ceremony, the set O cannot shake hands, | O| does not change Since, initially, |O| = 0, the parity of the set is preserved

33 Consider the number U of inversions, computed as follows: Below each 1, write the number of zeros to the right of it, and add up these numbers, Initially V = 0, U doe: not change at all after each move, or it increases or decreases by 2 Thus UV always

But we have U = 1 for the goal Thus, the temains even But we have e goal e goal cannot be reached 34 Consider the tri al f(x has discriminant b* — 4ac The first

transformation changes joo into foe h > + owe +(6+ 2a)x + a with discriminant (+22 -4(a+b+ec = b’ — dac, and, applying the second transformation, get the trinomial cx? an “o_ 2e)x + (a — ? + c) with discriminant b* — 4ac us the discriminant remains invariant But x* + 4x + 3 has discriminant 4, and x? + 10x + 9 has discriminant 64, Hence, one cannot get the second trinomial fro

im h i 2_ x2

35 a = 4) — ay ot (a:

th: đạ — đa

Suppose that a?, a3, a ° đa — đi > đa — đa > đa — ạ nh integers

36 Suppose the integers 1, anged in any order We will say that the numbers 7 and k are out of order if the ‘ager of the two is to the left of the smaller In that y part of the number of im 3 f neighboring integers

38 The number of inversions inn, , 1 is nn 1)/2 Prove that one step does not change the parity of the inversions If (nm — 1)/2 is even, then si the n imlegers into pairs of neighbors (leaving the middle integer unmatched Mã ld 2) Ther

đ, e quadruplets from the first, lạt second, second from behin 39 We assign the weight 1/2”*” to the square with label (x, y) We observe that the total

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1 In (a) the total weight of the shaded squares is 23 The weight of the rest of the board is 13 The total weight of the remaining board is not enough to accommodate the chips on the shaded region In (b) the lone piece has the weight 1 Suppose it is possible to clear the shaded

only a part of them So we have again a contradiction 40 I can get a chip to (0, 4), but not to (0, 5) Indeed, we introduce of a point

hi yes as follows: n(x, y) = |x|+ ở " We oe the weight on that point by a", of w ‘he weight of a set S of chips will s the positive root ned by

W(9= >} ơ pes Cover all the lattice points for y < 0 by chips The weight of the chips with y = 0 is o° + +20°>>).,¢/ = @° + 20 By covering the half plane with y < 0, we have the total weight

5 24 4 (gŠ +2ø#41+z+ø?+- )= “ s =ơ?+2ø2 =1

—Œ We make the following observations: A horizontal solitaire jump toward the y-axi leaves total weight unchanged A vertical j jump up leaves total weight unchansed, Any other jump decreases total weight Total weight al (0, 5) is 1 Thus any distribution of finitely many chips on or below the weight less than 1 Hence, the goal cannot be reached finitely many chips

en given points have coordinates (0,0,0), 41 ‘dinate system s

(oo », oy 0 0, o (q1 " nụ 0 ‘D “OL 1) We observe that a point Posemes the pari nates on reflection Thus, we never get points with all thre coordinsles oh Tem nee we ihe ma 1,1) can never be reached This follows from the pping formula X NAL incoordinates (x, y, Z) +> (2a—x, 2b—y, 2e—z), where A = (a,b,c) and X = in y, z) The invariant, here, is the parity pattern of the coordinates of the points in S 42 Fig 1.10 shows how to reduce an L-tetromino occupied by chips to one square by

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45, We should get x? + y? By looking at a healthy cell wi

Fig 1.9

note the number of occupied cel cells of colors A, B, C by a, b, c, respectively Initially, a = b=c,ie,a mod 2 That is, all three numbers have the ea jump, t se numbers are decreased by 1, and one is the jump, all three numbers change parity, i.e., they still have invariant a = b = c mod 2 This relation is violated if oye one aps temains on the board We can even say more If two chips

n the ust be on squares of the same color ith 2, 3, or 4 infected neighbors, we observe that the perimeter of the contaminated area does not increase, although it may well decrease 4 x 10 = 40 will never be reached

P (FO), (2) =x, ()

whicl d be v: il x In (a) and (b), we specific value of x, for which is is not true ein a Tạo = 9(2) = 6 By repeated application of the three opera ons on 6 we eet again a multiple of 6 But the sight side of (1) is 2 side 1/2 is a fractional number

(f — 8 +2g-3f =x = 1 for all n, but rounding errors corrupt more and mote of the significant digits One gets the table below This is a very robust computation No results In computations involving § muons of operations, one should use double

ion to get single precision

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Rg 1.11 Fig 1.12

and B be the of the numbers on the white and black squares, respectively Application ot Ta does not change the difference W — B For Fig 1.7 and Fig 1.8 the differences are 5 and —1, respectively The goal —1 cannot be reached from 5

49 by —1 1

Obviously, P is an invariant

ece=e, e0a=4, eob=b, acaz=b, bob=a,acb=e o operation is commu since we didnot mention the order It is easy to check

The o tative sin

that it is also associative, i.e.,(pog)or = po(q or) for all letters occurring Thus, the product of all letters is is tndependen ‘nt of the the order in which they are multiplied 52, Replace 1998 by n, and derive a necessary condition for the existence of such an

arrangement Let ø; be the position of the first integer k Then the other k has position p, + k By counting the position numbers twice, we get1+. -+2n = (pit pit) +-+-+ (n+ patn) For P= can 1 Di, We get P = (Gn -+1)/4, and P is an integer form = 0,1 mod 4 Since 1998 = 2 mod 4, this necessary condition

i f đ 8

53, This is an invariance problem As a prime candidate, we think of the discriminant D, The first operation obviously does not ‘change D “The second operation does not change the aitefence of the roots of the polynomial Now, D = b? — 4ac a’((b/ay’ — 4c/a) —b/a = x, + %, andc/a = xx) Hence, D = a 205, — — x))*, Le., the Sond peration does not change D Since the two trinomials have discriminants 9 and 5, the san cannot be reached

= h

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The i sant [

We assume that, at the start, the side lengths are 1, p, ¢, nh , 1 > q Then all succeeding triangles are similar with coefficient p”g" By cu ông ‘such a 'Eiangle 0 of type (m, ), we get two triangles of types (m+ 1, m) and om n-+ 1) We make the following translation Consider the lattice square with nonnegative coordinates We assign the coordinates of its lower left vertex to each square Initially, we place four a chip on square (m, ) by one chip on square (7m + 1, ) and one chip on square (m,n-+ 1) We assign weight 2~"~" to a chip on square (m, 2) Initially, the chips

PP = g az

different square Then the total weight is less than 4 In fact, to get weight 4 we would have to fill the whole plane by single chips This is impossible in a finite number of steps

€ i it! b that x2 2p; invariant If

can show that lim x, = limy, = x, then x ab,orx= ve Because of x, < y, and the arithmeti an-geornetri inequality, yn+1 lies to the left of (%, + yn)/2 and x41 lies to othe left of (x, in wd Thus, x < X41 < n limi ŸWz+‡i < }„ 8D Yzịi — Xạ} 5 Oe Ya — el’ We have, ng it x Actually for large n, say > Sw [Ena © (Yu + Xn)/2 and Yuri — Xng1 © On — ®r)/4

Assign the number /(W) = a,+2a)+3a;+ -+na, to W Deletion or insertion of any word X XX in any place rte Z = bủủ¿ Xã with mis = 1(Z) modulo 3 Since 7(01) = 2 and /(10) = 1, the goal cannot be attait Select four vertices such that no two are joined by an edge Let X be the sum of the numbers at these vertices, and let y be the sum of the numbers at the remaining four vertices Satay f= = +1 A step does not change I So neither (a) nor (b)

n be

1 andt, — 1/y„ Aninvariantis s -L2£, = Hints Consider th

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Coloring Proofs

The problems of this chapter are concerned with the partitioning of a set into a subset by the same color an prototypical example runs as follow

In 1961, the British theoretical physicist M.E Fisher solved a famous and very t problem He showed that tan 8 x 5 chessboard can be covered by 2 x 1 dominoes in 24 x 901? or 12,988,816 uy Now let us cut out two diagonally with 31 domin ) The problem looks even more complicated than the problem solved by Fisher, d the mutilated chessboard has 30s squares of one color and 32 squares of the othe er color

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1 Atectangular flooris covered by 2 x2 and 1 x4 tiles One tile got smashed There isa tile of the other kind available Show that the floor cannot be covered by rearranging the tiles 2

A 10 x 10 chessboard cannot be covered by 25 T-tetrominoes in Fig 2.1 These tiles are called from left to right: straight tetromino, T-tetromino, square tetromino, L-tetromino, and skew tetromino

Fig 2.1 4A 5 Al

6 Consider ann x nm chessboard with the four corners removed For which values of n

can you cover the board with L-tetrominoes as in Ms 22?

er of a (2m + 1) x (2n + 1) chessboard is cut off For which m can you cover the remaining squares by 2 x 1 dominoes, so that half of the dominoes are horizontal?

10 Fig 2.3 shows five heavy boxes which can be displaced only by rolling them about one of their edges Their tops are labeled by the letter T Fig 2.4 shows the same center of the cross?

city exactly once?

Fig 2.2

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N is)

nN tr

Cons:

Every space point is colored with exactly one of the colors red, green, or 0 Every space point is colored either red or biue Show that among the squares with four blue vertices,

Fig 2.6 On one square of a5 x 5 chessboard, we write —1 and on the other 24 squares +1 is to reach +1 on each square On which squares should —1 be to reach the goal? points with any distar

The points of a plane are colored with three colors Show that there exist two points with distance 1 both having the same color All vertices of a convex pentagon are lattice points, and its sides have integral length Show that its perimeter is even nm points (m > 5) of the plane can be colored by two colors so that no line can separate

th le points of one color from those of the other color any 1 x 1 squares You may color their edges with one of four colors and giuei them together along edges of the same color Your aim is to get anm xn tectangle For which m and n is this possible? You have many unit cubes and six colors You may color each cube with 6 colors

and glue together races of the same color Your aim is to set ar <s x £ box, each face having different color For which r, s, ¢ is this possible?

ider three s A = (0,0), B you reach the ‘ourth x vertex D = (1, 1) of the square by reflections at A, B,C or at points previously reflected?

š

e The sets R, G, B consist of the lengths of those segments in space with both endpoints red, green, and blue, respectively Show that at least one of these sets contains all nonnegative real numbers

The Art Gallery Problem An att gallery has the shape of a simple n-gon Find complicated its shape

positions of the 1 x 1 tile? The vertices of aregular 2n-gon A1, , Az, afe partitioned into n pairs Prove that,

2

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Prove that a) + by +++» + as +s £0 Can you pack 53 bricks of dimensions 1 x 1 x 4 into a 6 x 6 x 6 box? The faces of the bricks are parallel to the faces of the box w S

31 Three pucks A, B, C are in a plane An ice hockey player hits the pucks so that any one glides through the other two in a straight line Can all pucks return to their origi

32 A23 x23 square i is completely tiled by 1 x 1,2 x2 and3 x3 tiles What minimum number of 1 x 1 ti eeded (AUO 1989)? 33 The vertices and midpoints of the face: arked on a cube, and all face diagonals

are drawn Tsitpossible to visitall marked points by walking along the face diagonals? 34 There is no closed knight’s tour of a (4 x 35 The plane is colored with two colors Na that there exist three points of the same

color, which are ver of a regular triangle 36 A sphere is colored in two colors Prove that there exist on this sphere three points

of the same color, which are vertices of a regular triangle 37 Given an m xn rectangle, what minimum number of cells (1 x 1 squares) must be

colored, such that there is no place on the remaining cells for an L-tromino? T tegers ate colored black and white The sum of two differently colored numbers is is “hack, and their product is white What is the product of two white numbers? Find all such colorings

Solutions

Color the floor as in Fig 2.7 A 4 x 1 tile always covers 0 or 2 black squares A 2x2 tile * always co covers one black square It follows immediately from this that it is impossible to exchange one tile for a tile of the other kind

‘ig 2.7 2, Any rectangle with 20 squares can be colored like a chessboard with 10 black and 10

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kind On the other hand, one needs 15 tetrominoes for 60 squares Since 15 is odd, a covering is not possible

Color the board diagonally in four colors 0, 1, 2, 3 as shown in Fig 2.10 No matter

squar es with color 1

Fig 2.8 Alternate solution Color the board as shown in Fig 2.9 Each horizontal straight tetromino covers one square of each color Each vertical tetromino covers four a+10,a+10, a, a squares of color 0, 1, 2, 3, respectively Each of these nunbers should be a multiple of 4 But this is impossible since a + 10 and a cannot both be multiples of 4

@ g

There are n? — 4 squares on the board To cover it with tetrominoes n? — 4 must be a ultiple of 4, i.e., 2 mui ‘be even But this is not sufficient To see this, we color the board as in Fig 2.11 An L-tetromino covers three white and one black squares or noes

„ rò 5 œ & 2 5 & œ s 3 Ẵ om 5 3 ° E 8, a oD ° 5 oO § 5 oa a 6 & 2 8 oO 3 a B oO g ° 8 S