Solution The key observation is the following: if row k contains a prime num- ber p > 40, then the same number must be contained by column k, as well.
Therefore, all prime numbers from 1 to 81 must lie on the main diagonal of the table. However, this is impossible, since there are 10 such prime numbers:
41,43,47,53,59,61,67,71,73, and 79.
Problem 3.11 The entries of a matrix are integers. Adding an integer to all entries on a row or on a column is called an operation. It is given that for infinitely many integersNone can obtain, after a finite number of operations, a table with all entries divisible byN. Prove that one can obtain, after a finite number of operations, the zero matrix.
Solution Suppose the matrix hasmrows andncolumns and its entries are denoted byahk. Fix somej,1< j≤n, and consider an arbitraryi,1< i≤m. The expres- sion
Eij=a11+aij−ai1−a1j
is an invariant for our operation. Indeed, addingkto the first row, it becomes (a11+k)+aij−ai1−(a1j+k)=a11+aij−ai1−a1j.
The same happens if we operate on the first column, theith row or thejth column, while operating on any other row or column clearly does not changeEij. From the hypothesis, we deduce thatEij is divisible by infinitely positive integersN, hence Eij=0. We deduce that
a11−a1j=ai1−aij=c,
for all i,1< i ≤m. Adding c to all entries in columnj will make this column identical to the first one.
In the same way we can make all columns identical to the first one. Now, it is not difficult to see that operating on rows we can obtain the zero matrix.
6.2 Functions Defined on Sets of Points
Problem 3.14 LetDbe the union ofn≥1 concentric circles in the plane. Suppose that the functionf :D→Dsatisfies
d
f (A), f (B)
≥d(A, B)
for everyA, B∈D (d(M, N )is the distance between the pointsMandN ).
Prove that
d
f (A), f (B)
=d(A, B) for everyA, B∈D.
Solution LetD1, D2, . . . , Dn be the concentric circles, with radiir1< r2<ã ã ã<
rnand centerO. We will denotef (A)=A, for an arbitrary pointA∈D.
We first notice that ifA, B∈Dnsuch thatABis a diameter, thenAB is also a diameter ofDn. IfCis another point onDn, we have
AC2+BC2≥AC2+BC2=AB2=AB2. BecauseOC is a median of the triangleAB C, it follows that
OC2=1 2
AC2+BC2
−1
4AB2=rn2,
henceC ∈DnandAC =AC, BC =BC. We deduce thatf (Dn)⊂Dnand the restriction off toDnis an isometry. Now takeA, X, Y, ZonDn such thatAX= AY =AZ. It follows that AX =AY =AZ, hence one of the points X, Y coincides withZ. This shows that f (Dn)=Dn and since f is clearly injective it results in the same way thatf (Di)=Di, for all i, 1≤i≤n−1, and that all restrictionsf|Di are isometries.
Next we prove that distances between adjacent circles, sayD1andD2are pre- served. TakeA, B, C, D onD1such thatABCDis a square and letA, B, C, D be the points onD2closest toA, B, C, D, respectively.
ThenAB CD is also a square and the distance fromAtoC is the maximum between any point onD1and any point onD2. Hence the eight points maintain their relative position underf and this shows thatf is an isometry (Fig.6.1).
6.2 Functions Defined on Sets of Points 203 Problem 3.15 LetSbe a set ofn≥4 points in the plane, such that no three of them are collinear and not all of them lie on a circle. Find all functionsf :S→R with the property that for any circleCcontaining at least three points ofS,
P∈C∩S
f (P )=0.
Solution For two distinct pointsA, B ofS we denoteCA,B the set of circles de- termined byA, B and other points ofS. SupposeCA,B haskelements. Since the points ofSare not on the same circle, it follows thatk≥2. Because
P∈C∩S
f (P )=0 for allC∈CA,B, we deduce that
C∈CA,B
P∈C∩S
f (P )=0.
On the other hand, it is not difficult to see that
C∈CA,B
P∈C∩S
f (P )=
P∈S
f (P )+(k−1)
f (A)+f (B) .
Thus the sum
P∈Sf (P )andf (A)+f (B)have opposite signs, for allA, BinS.
If, for instance,
P∈Sf (P )≥0, thenf (A)+f (B)≤0, for all A, B inS. Let S= {A1, A2, . . . , An}. Then
f (A1)+f (A2)≤0, f (A2)+f (A3)≤0, . . . , f (An)+f (A1)≤0, yielding
2
P∈M
f (P )≤0 hence
P∈Mf (P )=0 andf (A)+f (B)=0 for all distinct points A, B. This implies thatf is the zero function. Indeed, letA, B, Cbe three distinct points inS.
It is not difficult to see that the equalities
f (A)+f (B)=0, f (B)+f (C)=0, f (A)+f (C)=0 yield
f (A)=f (B)=f (C)=0 and our claim is proved.
Problem 3.16 LetP be the set of all points in the plane andLbe the set of all lines of the plane. Find, with proof, whether there exists a bijective functionf :P →L such that for any three collinear pointsA, B, C, the linesf (A), f (B)andf (C)are either parallel or concurrent.
Solution LetAi, i=1,2,3 be three distinct points in the plane andli =f (Ai).
We claim that ifl1, l2, l3are concurrent or parallel, then A1, A2, A3 are collinear.
Indeed, suppose thatl1, l2, l3intersect atMand thatA1A2A3is a non-degenerated triangle. Then for any pointB in the plane we can find pointsB2, B3on the lines A1A2, A1A3, respectively, such thatB, B2, B3are collinear (Fig.6.2).
Because A1, B2, A2 are collinear it follows that f (B2) is a line passing throughM. The same is true for f (B1), hence also for f (B). This contradicts the surjectivity off. A similar argument can be given ifl1, l2, l3are parallel.
We conclude that the restriction off to any lineldefines a bijection fromlto a pencil of lines (passing through a point or parallel). Consider two pencilsP1andP2 of parallel lines. The inverse images ofP1, P2are two parallel linesl1, l2(P1andP2
have no common lines, hencel1andl2have no common points). LetP3be a pencil of concurrent lines whose inverse image is a linel, clearly not parallel tol1, l2. Let l be a line parallel tol. Thenf (l)is a pencil of concurrent lines and it follows that there is a line through the points corresponding tolandl whose inverse image would be a point on bothlandl, a contradiction. Hence no such functions exists.
Problem 3.17 LetS be the set of interior points of a sphere andC be the set of interior points of a circle. Find, with proof, whether there exists a functionf:S→ Csuch thatd(A, B)≤d(f (A), f (B)), for any pointsA, B∈S.
Solution No such function exists. Indeed, suppose f :S→C has the enounced property. Consider a cube inscribed in the sphere and assume with no loss of gen- erality that its sides have length 1. Partition the cube inton3smaller cubes and let A1, A2, . . . , A(n+1)3 be their vertices. For alli=j we have
d(Ai, Aj)≥1 n, hence
d
f (Ai), f (Aj)
≥ 1 n.