Solutions to the Longlisted Problems of IMO 1977

4.19 Solutions to the Longlisted Problems of IMO 1977

1. Let P be the projection of S onto the plane ABCDE. Obviously BS > CS is equivalent to BP > CP. The conditions of the problem imply that PA > PB and PA > PE. The locus of such points P is the region of the plane that is determined by the perpendicular bisectors of segments AB and AE and that contains the point diametrically opposite A. But since AB < DE, the whole of this region lies on one side of the perpendicular bisector of BC. The result follows immediately.

Remark. The assumption BC < CD is redundant.

2. We shall prove by induction on n that f (x) > f (n) whenever x > n. The case n = 0 is trivial. Suppose that n ≥ 1 and that x > k implies f (x) > f (k) for all

k < n. It follows that f (x) ≥ n holds for all x ≥ n. Let f (m) = min x ≥n f (x). If we suppose that m > n, then m − 1 ≥ n and consequently f (m − 1) ≥ n. But in this case the inequality f (m) > f ( f (m − 1)) contradicts the minimality property of m . The inductive proof is thus completed. It follows that f is strictly increasing, so f (n + 1) > f ( f (n)) implies that n + 1 >

f (n). But since f (n) ≥ n we must have f (n) = n.

be k persons who are not acquainted with each other. Let us denote by m the number of acquainted couples and by d j the number of acquain- tances of person v j . Then

3. Let v 1 ,v 2 ,...,v k

k + (n − k) 2 n 2 m ≤d k +1 +d k +2 + ··· + d n ≤ d(n − k) ≤ k(n − k) ≤

4. Consider any vertex v n from which the maximal number d of segments start, and suppose it is not a vertex of a triangle. Let A = {v 1 ,v 2 ,...,v d } be the set of points that are connected to v n , and let B = {v d +1 ,v d +2 ,...,v n } be the set of the other points. Since v n is not a vertex of a triangle, there is no segment both of whose vertices lie in A ; i.e., each segment has an end in B. Thus, if d j denotes the number of segments at v j and m denotes the total number of segments, we

have m ≤d d +1 +d d +2 + ··· + d n ≤ d(n − d) ≤ n 2 /4 = m. This means that each inequality must be equality, implying that each point in B is a vertex of d segments, and each of these segments has the other end in A . Then there is no triangle at all, which is a contradiction.

5. Let us denote by I and E the sets

of interior boundary points and exte- R D CD R C rior boundary points. Let ABCD be

D C the square inscribed in the circle k

with sides parallel to the coordinate axes. Lines AB R , BC,CD, DA divide the DA R

R BC

plane into 9 regions: R, R A ,R B ,R C ,

R ,R A

D AB ,R BC ,R CD ,R DA . There is

a unique pair of lattice points A A B

I ∈ R,

E ∈R A that are opposite vertices of a

A AB B

4.19 Longlisted Problems 1977 423 unit square. We similarly define B I ,C I ,D I ,B E ,C E ,D E . Let us form a graph G by

connecting each point from E lying in R AB (respectively R BC ,R CD ,R DA ) to its upper (respectively left, lower, right) neighbor point (which clearly belongs to

I ). It is easy to see that: (i) All vertices from I other than A I ,B I ,C I ,D I have degree 1. (ii) A E is not in E if and only if A I ∈ I and degA I = 2.

(iii) No other lattice points inside R A belong to E.

Thus if m is the number of edges of the graph G and s is the number of points among A E ,B E ,C E , and D E that are in E, using (i)–(iii) we easily obtain |E| = m + s and |I| = m − (4 − s) = |E| + 4.

6. Let hyi denote the distance from y ∈ R to the closest even integer. We claim that h1 + cosxi ≤ sin x

for all x ∈ [0, π ]. Indeed, if cos x

≥ 0, then h1 + cosxi = 1 − cosx ≤ 1 − cos 2 x = sin 2 x ≤ sin x; the proof is similar if cos x < 0.

We note that hx + yi ≤ hxi + hyi holds for all x,y ∈ R. Therefore

∑ sin x j ≥ ∑ h1 + cosx j i≥ ∑ (1 + cosx j ) = 1.

7. Let us suppose that c 1 ≤c 2 ≤ ··· ≤ c n and that c 1 <0<c n . There exists k,

1 ≤ k < n, such that c k ≤0<c k +1 . Then we have (n − 1)(c 2

1 +c 2 2 2 2 2 2 2 + ··· + c n ) ≥ k(c 1 + ··· + c k ) + (n − k)(c k +1 + ··· + c n ) ) 2 ≥ (c 2 1 + ··· + c k + (c k +1 + ··· + c n )

= (c

1 + ··· + c n )

−2(c 1 + ··· + c k )(c k +1 + ··· + c n ), from which we obtain (c 1 + ··· + c k )(c k +1 + ··· + c n ) ≥ 0, a contradiction.

Second solution. By the given condition and the inequality between arithmetic and quadratic mean we have

(c ) 2 2 2 1 2 + ··· + c n = (n − 1)(c 1 + ··· + c n −1 ) + (n − 1)c n

2 ≥ (c 2

1 + ··· + c n −1 ) + (n − 1)c n , which is equivalent to 2 (c 1 +c 2 + ··· + c n )c n ≥ nc 2 . Similarly, 2 (c 1 +c 2 + ··· +

n i ≥ nc i for all i = 1, . . . , n. Hence all c i are of the same sign.

c )c

8. There is exactly one point satisfying the given condition on each face of the hexahedron. Namely, on the face ABD it is the point that divides the median from D in the ratio 32 : 3.

9. A necessary and sufficient condition for M to be nonempty is that 1 / 10 ≤t≤

424 4 Solutions

10. Integers a , b, q, r satisfy

a 2 +b 2 = (a + b)q + r,

0 ≤ r < a + b,

q 2 + r = 1977. From q 2 ≤ 1977 it follows that q ≤ 44, and consequently a 2 +b 2 < 45(a + b).

Having in mind the inequality (a + b) 2 ≤ 2(a 2 +b 2 ), we get (a + b) 2 < 90(a + b), i.e., a +b < 90 and consequently r < 90. Now from q 2 = 1977−r > 1977−90 = 1887 it follows that q > 43; hence q = 44 and r = 41. It remains to find positive integers a and b satisfying a 2 +b 2 = 44(a + b) + 41, or equivalently

2 (a − 22) 2 + (b − 22) = 1009. The Diophantine equation A 2 +B 2 = 1009 has only two pairs of positive so-

lutions: (15, 28) and (28, 15). Hence (|a − 22|,|b − 22|) ∈ {(15,28), (28,15)}, which implies (a, b) ∈ {(7,50),(37,50),(50,7),(50,37)}.

11. (a) Suppose to the contrary that none of the numbers z 0 ,z 1 ,...,z n −1 is divisible by n. Then two of these numbers, say z k and z l (0 ≤ k < l ≤ n − 1), are congruent modulo n, and thus n |z l −z k =z k +1 z l −k−1 . But since (n, z) = 1, this implies n |z l −k−1 , which is a contradiction.

(b) Again suppose the contrary, that none of z 0 ,z 1 ,...,z n −2 is divisible by n. Since (z − 1,n) = 1, this is equivalent to n ∤ (z − 1)z j , i.e., z k 6≡ 1 (mod n) for all k = 1, 2, . . . , n − 1. But since (z,n) = 1, we also have that z k 6≡ 0 (mod n ). It follows that there exist k , l, 1 ≤ k < l ≤ n − 1 such that z k ≡z l , i.e., z l −k ≡ 1 (mod n), which is a contradiction.

12. According to part (a) of the previous problem we can conclude that T = {n ∈ N | (n, z) = 1}.

13. The figure Φ contains two points A and B having maximum distance. Let h be the semicircle with diameter AB that lies in Φ , and let k be the circle containing

h . Consider any point M inside k. The line passing through M that is orthogonal to AM meets h in some point P (because ∠AMB > 90 ◦ ). Let h ′ and h ′

be the two semicircles with diameter AP, where M ∈h ′ . Since h ′ contains a point C such that BC > AB, it cannot be contained in Φ , implying that h ′ ⊂ Φ . Hence M belongs to Φ . Since Φ contains no points outside the circle k, it must coincide with the disk determined by k. On the other hand, any disk has the required property.

14. We prove by induction on n that independently of the word w 0 , the given algo- rithm generates all words of length n. This is clear for n = 1. Suppose now the statement is true for n − 1, and that we are given a word w 0 =c 1 c 2 ...c n of length n . Obviously, the words w 0 ,w 1 ,...,w 2 n −1 −1 all have the nth digit c n , and by the inductive hypothesis these are all words whose nth digit is c n . Similarly, by the inductive hypothesis w 2 n −1 ,...,w 2 n −1 are all words whose nth digit is 1 −c n , and the induction is complete.

15. Each segment is an edge of at most two squares and a diagonal of at most one square. Therefore p k = 0 for k > 3, and we have to prove that

p 0 =p 2 + 2p 3 .

4.19 Longlisted Problems 1977 425 Let us calculate the number q (n) of considered squares. Each of these squares

is inscribed in a square with integer vertices and sides parallel to the coordinate axes. There are (n − s) 2 squares of side s with integer vertices and sides parallel to the coordinate axes, and each of them circumscribes exactly s of the consid- ered squares. It follows that q (n) = ∑ n −1 s =1 (n − s) 2 s =n 2 (n 2 − 1)/12. Computing the number of edges and diagonals of the considered squares in two ways, we obtain that

(2) On the other hand, the total number of segments with endpoints in the considered

p 1 + 2p 2 + 3p 3 = 6q(n).

integer points is given by

n 2 n 2 (n 2

− 1) = 6q(n). (3)

p 0 +p 1 +p 2 +p 3 =

Now (1) follows immediately from (2) and (3).

16. For i = k and j = l the system is reduced to 1 ≤ i, j ≤ n, and has exactly n 2 solutions. Let us assume that i 6= k or j 6= l. The points A(i, j), B(k,l), C(− j +k + l , i − k + l), D(i− j + l,i+ j − k) are vertices of a negatively oriented square with integer vertices lying inside the square [1, n] × [1,n], and each of these squares corresponds to exactly 4 solutions to the system. By the previous problem there

are exactly q (n) = n 2 (n 2 − 1)/12 such squares. Hence the number of solutions

is equal to n 2 + 4q(n) = n 2 (n 2 + 2)/3.

17. Centers of the balls that are tangent to K are vertices of a regular polyhedron with triangular faces, with edge length 2R and radius of circumscribed sphere r + R. Therefore the number n of these balls is 4, 6, or 20. It is straightforward to obtain that:

√ (i) If n = 4, then r + R = 2R( √ 6 /4), whence R = r(2 + √ 6 ).

(ii) If n = 6, then r + R = 2R( 2 /2), whence R = r(1 + 2 p ).

(iii) If n = 20, then r + R = 2R 5 + 5 hp /8. In this case we can conclude that √

R =r

18. Let U be the midpoint of the segment AB. The point M belongs to CU and √

p√

CM =( 5 − 1)CU/2, r = CU

19. We shall prove the statement by induction on m. For m = 2 it is trivial, since each power of 5 greater than 5 ends in 25. Suppose that the statement is true for some m ≥ 2, and that the last m digits of 5 n alternate in parity. It can be shown

− 1 is 2 consequently the difference 5 −2 −5 n is divisible by 10 m but not by 2

by induction that the maximum power of 2 that divides 5 2 m −2 m , and

· 10 m . It follows that the last m digits of the numbers 5 −2 n +2 and 5 n coincide, but the

n +2 m

digits at the position m +1 have opposite parity. Hence the last m+1 digits of one of these two powers of 5 alternate in parity. The inductive proof is completed.

426 4 Solutions

20. There exist u , v such that a cosx + b sin x = r cos(x − u) and Acos2x + Bsin2x = √

and R = A 2 +B 2 . Then 1 − f (x) = r cos(x− u ) + R cos2(x − v) ≤ 1 holds for all x ∈ R. There exists x ∈ R such that cos(x − u) ≥ 0 and cos2(x − v) = 1 (indeed, either x = v or x = v + π works). It follows that R √ ≤ 1. Similarly, there exists x ∈ R such that cos (x − u) = 1/

R cos 2 (x−v), where r = a 2 +b 2

2 and cos 2 √ (x − v) ≥ 0 (either x = u − π /4 or x = u + π /4 works). It follows that r ≤ 2.

Remark. The proposition of this problem contained as an addendum the follow- ing, more difficult, inequality:

a 2 +b 2

+ A 2 +B 2 ≤ 2.

The proof follows from the existence of x ∈ R such that cos(x − u) ≥ 1/2 and cos 2 (x − v) ≥ 1/2.

21. Let us consider the vectors v 1 = (x 1 ,x 2 ,x 3 ), v 2 = (y 1 ,y 2 ,y 3 ), v 3 = (1, 1, 1) in space. The given equalities express the condition that these three vectors are

x 1 mutually perpendicular. Also, 2

x 2 1 1 2 +x 2 2 +x 2 3 , y 2 1 +y 2 2 +y 2 , and 1 3 /3 are the squares of the projections of the vector (1, 0, 0) onto the directions of v 1 ,v 2 ,v 3 , respectively. The result follows from the fact that the sum of squares of projections of a unit vector on three mutually perpendicular directions is 1.

22. Since the quadrilateral OA 1 BB 1 is cyclic, ∠OA 1 B 1 = ∠OBC. By using the analo- gous equalities we obtain ∠OA 4 B 4 = ∠OB 3 C 3 = ∠OC 2 D 2 = ∠OD 1 A 1 = ∠OAB, and similarly ∠OB 4 A 4 = ∠OBA. Hence △OA 4 B 4 ∼ △OAB. Analogously, we have for the other three pairs of triangles △OB 4 C 4 ∼ △OBC, △OC 4 D 4 ∼ △OCD, △OD 4 A 4 ∼ △ODA, and consequently ABCD ∼ A 4 B 4 C 4 D 4 .

23. Every polynomial q (x 1 ,...,x n ) with integer coefficients can be expressed in the form q =r 1 +x 1 r 2 , where r 1 ,r 2 are polynomials in x 1 ,...,x n with integer co- efficients in which the variable x 1 occurs only with even exponents. Thus if q 1 =r 1 −x 1 r 2 , the polynomial qq 1 =r 2 1 −x 2 1 r 2 2 contains x 1 only with even expo- nents. We can continue inductively constructing polynomials q j ,j = 2, 3, . . . , n, such that qq 1 q 2 ···q j contains each of the variables x 1 ,x 2 ,...,x j only with even exponents. Thus the polynomial qq 1 ··· q n is a polynomial in x 2 1 ,...,x 2 n . The polynomials f and g exist for every n ∈ N. In fact, it suffices to construct q 1 ,...,q n for the polynomial q =x 1 + ··· + x n and take f =q 1 q 2 ··· q n .

24. Setting x = y = 0 gives us f (0) = 0. Let us put g(x) = arctan f (x). The given functional equation becomes tan g (x + y) = tan(g(x) + g(y)); hence

g (x + y) = g(x) + g(y) + k(x, y) π , where k (x, y) is an integer function. But k(x, y) is continuous and k(0, 0) = 0,

therefore k (x, y) = 0. Thus we obtain the classical Cauchy’s functional equation

g (x + y) = g(x) + g(y) on the interval (−1,1), all of whose continuous solutions are of the form g (x) = ax for some real a. Moreover, g(x) ∈ (− π , π ) implies |a| ≤ π /2.

4.19 Longlisted Problems 1977 427

Therefore f (x) = tan ax for some |a| ≤ π /2, and this is indeed a solution to the given equation.

25. Let

f n (z) = z n +a

k ∑ −1 (a − kb) (z + kb) n −k .

k =1 k

We shall prove by induction on n that f n (z) = (z + a) n . This is trivial for n = 1. Suppose that the statement is true for some positive integer n − 1. Then

(z + kb) (n − k)(a − kb) n ∑ −k−1

k −1 (z + kb) (a − kb) n ∑ −k−1

k =1

=nf n −1 (z) = n(z + a) n −1 . It remains to prove that f n (−a) = 0. For z = −a we have by the lemma of (SL81-

n (−a) = (−a)

f +a

∑ (−1) (a − kb) −1

∑ −1 (−1) n −k (a − kb) n = 0.

k =0 k

26. The result is an immediate consequence (for G = {−1,1}) of the following gen- eralization.

(1) Let G be a proper subgroup of Z ∗ n (the multiplicative group of residue classes modulo n coprime to n), and let V be the union of elements of

G . A number m ∈ V is called indecomposable in V if there do not exist numbers p , q ∈ V , p,q 6∈ {−1,1}, such that pq = m. There exists a number r ∈ V that can be expressed as a product of elements indecomposable in V in more than one way.

First proof. We shall start by proving the following lemma. Lemma. There are infinitely many primes not in V that do not divide n. Proof. There is at least one such prime: In fact, any number other than ±1 not in

V must have a prime factor not in V , since V is closed under multiplication. If there were a finite number of such primes, say p 1 ,p 2 ,...,p k , then one of the numbers p 1 p 2 ··· p k + n, p 2 1 p 2 ··· p k + n is not in V and is coprime to n and p 1 ,...,p k , which is a contradiction. [This lemma is actually a direct consequence of Dirichlet’s theorem.] Let us consider two such primes p , q that are congruent modulo n. Let p k

be the least power of p that is in V . Then p k ,q k ,p k −1 q , pq k −1 belong to V and are indecomposable in V . It follows that

r =p k ·q k =p k −1 q · pq k −1

428 4 Solutions has the desired property.

Second proof. Let p be any prime not in V that does not divide n, and let p k be the least power of p that is in V . Obviously p k is indecomposable in V . Then the number

r =p k · (p k −1 + n)(p + n) = p(p k −1 + n) · p k −1 (p + n) has at least two different factorizations into indecomposable factors.

27. The result is a consequence of the generalization from the previous problem for

G = {1}. Remark. There is an explicit example: r = (n − 1) 2 · (2n − 1) 2 = [(n − 1)(2n −

28. The recurrent relations give us that

√ x i +1 =

x i + [n/x i ]

x + n/x

i ≥[ n ].

On the other hand, if x

i >[ n ] for some i, then we have x i +1 <x i . This follows from the fact that x +1 <x is equivalent to x > (x + n/x

i )/2, i.e., to x i > n. Therefore x i =[ n ] holds for at least one i ≤ n − [ n ] + 1.

√ Remark. If n + 1 is a perfect square, then x i =[ n √ ] implies x i +1 =[ n √ ] + 1.

Otherwise, x i =[ n ] implies x i +1 =[ n ].

29. Let us denote the midpoints of segments LM, AN, BL, MN, BK, CM, NK, CL, DN , KL, DM, AK by P 1 ,P 2 ,P 3 ,P 4 ,P 5 ,P 6 ,P 7 ,P 8 ,P 9 ,P 10 ,P 11 ,P 12 , respectively. We shall prove that the dodecagon

D C P 1 P 2 P 3 ...P 11 P 12 is regular. From

BL = BA and ∠ABL = 30 ◦ it fol- lows that ∠BAL = 75 ◦ . Similarly

O ∠DAM = 75 ◦ , and therefore ∠LAM =

60 ◦ , which together with the fact AL

P = AM implies that △ALM is 1 P 2

equilateral. Now, from the triangles M

OLM and ALN, we deduce OP 1 =

LM /2, OP 2 = AL/2 and OP 2 k AL.

Hence OP 1 = OP 2 , ∠P 1 OP 2 = ∠P 1 AL = 30 ◦ and ∠P 2 OM = ∠LAD = 15 ◦ . The desired result follows from symmetry.

30. Suppose ∠SBA = x. By the trigonometric form of Ceva’s theorem we have sin (96 ◦ − x) sin 18 ◦ sin 6 ◦

We claim that x = 12 ◦ is a solution of this equation. To prove this, it is enough to show that sin 84 ◦ sin 6 ◦ sin 18 ◦ = sin 48 ◦ sin 12 ◦ sin 12 ◦ , which is equivalent to sin 18 ◦ = 2 sin 48 ◦ sin 12 ◦ = cos 36 ◦ − cos60 ◦ . The last equality can be checked directly.

4.19 Longlisted Problems 1977 429 Since the equation is equivalent to (sin 96 ◦ cot x

◦ ) sin 6 ◦ sin 18 ◦ = sin 48 ◦ sin 12 − cos96 ◦ , the solution x ∈ [0, π ) is unique. Hence x = 12 ◦ .

Second solution. We know that if a , b, c, a ′ ,b ′ ,c ′ are points on the unit circle in the complex plane, the lines aa ′ , bb ′ , cc ′ are concurrent if and only if

(1) We shall prove that x = 12 ◦ . We may suppose that ABC is the triangle in the

(a − b ′ )(b − c ′ )(c − a ′ ) = (a − c ′ )(b − a ′ )(c − b ′ ).

complex plane with vertices a = 1, b = ε 9 ,c = ε 14 , where ε = cos π 15 + i sin π 15 . If a ′ = ε 12 ,b ′ = ε 28 ,c ′ = ε , our task is the same as proving that lines aa ′ , bb ′ ,

cc ′ are concurrent, or by (1) that ε 28 )( ε 9 − 9 ε )( ε 14 12 12 (1 − 14 − ε ) − (1 − ε )( ε − ε )( ε − ε 28 ) = 0. The last equality holds, since the left-hand side is divisible by the minimum

polynomial of ε :z 8 +z 7 5 4 −z 3 −z −z + z + 1.

31. We obtain from (1) that f (1, c) = f (1, c) f (1, c); hence f (1, c) = 1 and conse- quently f (−1,c) f (−1,c) = f (1,c) = 1, i.e. f (−1,c) = 1. Analogously, f (c,1) =

f (c, −1) = 1. Clearly f (1, 1) = f (−1,1) = f (1,−1) = 1. Now let us assume that a 6= 1. Ob- serve that f (x −1 , y) = f (x, y −1 ) = f (x, y) −1 . Thus by (1) and (2) we get

1 = f (a, 1 − a) f (1/a,1 − 1/a)

−a

1 1 − 1/a = f (a, −a). − 1/a We now have f (a, a) = f (a, −1) f (a,−a) = 1 · 1 = 1 and 1 = f (ab,ab) =

32. It is a known result that among six persons there are 3 mutually acquainted or 3 mutually unacquainted. By the condition of the problem the last case is excluded. If there is a man in the room who is not acquainted with four of the others, then these four men are mutually acquainted. Otherwise, each man is acquainted with at least five others, and since the sum of numbers of acquaintances of all men in the room is even, one of the men is acquainted with at least six men. Among these six there are three mutually acquainted, and they together with the first one make a group of four mutually acquainted men.

33. Let r be the radius of K and s > 2 /r an integer. Consider the points A k (ka 1 − [ka 1 ], ka 2 − [ka 2 ]), where k = 0, 1, 2, . . . , s 2 . Since all these points are in the unit square, two of them, say A p ,A q ,q > p, are in a small square with side 1/s, and √ consequently A p A q ≤ 2 /s < r. Therefore, for n = q − p, m 1 = [qa 1 ] − [pa 1 ] and m 2 = [qa 2 ] − [pa 2 ] the distance between the points n(a 1 ,a 2 ) and (m 1 ,m 2 ) is less then r, i.e., the point (m 1 ,m 2 ) is in the circle K + n(a 1 ,a 2 ).

430 4 Solutions

34. Let A be the set of the 2 n sequences of n terms equal to ±1. Since there are k 2 products ab with a , b ∈ B, by the pigeonhole principle there exists c ∈ A such

that ab = c holds for at most k 2 /2 n pairs (a, b) ∈ B × B. Then cb ∈ B holds for at most k 2 /2 n values b ∈ B, which means that |B ∩ cB| ≤ k 2 /2 n .

35. The solutions are 0 and N k = 10 99 . . .9

89, where k | {z } = 0, 1, 2, . . ..

Remark. If we omit the condition that at most one of the digits is zero, the solutions are numbers of the form N k 1 N k 2 ...N k r , where k 1 =k r ,k 2 =k r −1 etc. The more general problem k ·a 1 a 2 ...a n =a n ...a 2 a 1 has solutions only for k =9 and for k = 4 (namely 0, 2199 . . .978 and combinations as above).

36. It can be shown by simple induction that S m (a 1 ,...,a 2 n ) = (b 1 ,...,b 2 n ), where

b k = ∏ a k +i (assuming that a k +2 n =a k ).

i =0

If we take m =2 n all the binomial coefficients m i apart from i = 0 and i = m

will be even, and thus b k =a k a k +m = 1 for all k.

37. We look for a solution with x A 1 = ··· = x A 1 n n =n A 1 A 2 ···A n x and x n +1 =n y . In order for this to be a solution we must have A 1 A 2 ···A n x +1=A n +1 y . This equation has infinitely many solutions (x, y) in N, since A 1 A 2 ··· A n and A n +1 are coprime.

38. The condition says that the quadratic equation f (x) = 0 has distinct real solu- tions, where

f (x) = 3x 2 ∑ m j − 2x ∑ m j (a j +b j +c j )+ ∑ m j (a j b j +b j c j +c j a j ).

It is easy to verify that the function f is the derivative of

F (x) = ∑ m j (x − a j )(x − b j )(x − c j ).

j =1

Since F (a 1 ) ≤ 0 ≤ F(a n ), F(b 1 ) ≤ 0 ≤ F(b n ) and F(c 1 ) ≤ 0 ≤ F(c n ), F(x) has three distinct real roots, and hence by Rolle’s theorem its derivative f (x) has two distinct real roots.

39. By the pigeonhole principle, we can find 5 distinct points among the given 37 such that their x-coordinates are congruent and their y-coordinates are congruent modulo 3. Now among these 5 points either there exist three with z-coordinates congruent modulo 3, or there exist three whose z-coordinates are congruent to 0,

1, 2 modulo 3. These three points are the desired ones. Remark. The minimum number n such that among any n integer points in space

one can find three points whose barycenter is an integer point is n = 19. Each proof of this result seems to consist in studying a great number of cases.

4.19 Longlisted Problems 1977 431

40. Let us divide the chessboard into 16 squares Q 1 ,Q 2 ,...,Q 16 of size 2 × 2. Let s k

be the sum of numbers in Q k , and let us assume that s 1 ≥s 2 ≥ ··· ≥ s 16 . Since s 4 +s 5 + ··· + s 16 ≥ 1 + 2 + ···+ 52 = 1378, we must have s 4 ≥ 100 and hence s 1 ,s 2 ,s 3 ≥ 100 as well.

41. The considered sums are congruent modulo N to S k =∑ N i =1 (i + k)a i ,k =

0 , 1, . . . , N − 1. Since S k =S 0 + k(a 1 + ··· + a n )=S 0 + k, all these sums give distinct residues modulo N and therefore are distinct.

42. It can be proved by induction on n that

1 + ··· + 3 s n −1 +s n (s i = ±1)}. Thus the result is an immediate consequence of the following lemma.

{a 1 n ,k |1≤k≤2 } = {2 |m=3

n +3 n −1 s

Lemma. Each positive integer s can be uniquely represented in the form

where s i ∈ {−1,0,1}. (1) Proof. Both the existence and the uniqueness can be shown by simple induction

s =3 n +3 n −1 s 1 + ··· + 3 1 s n −1 +s n ,

on s. The statement is trivial for s = 1, while for s > 1 there exist q ∈ N, r ∈ {−1,0,1} such that s = 3q + r, and q has a unique representation of the form (1).

h +p+1

43. Since k

p +2 − p +2 , it fol- lows that

n ∑ (n + 1) ···(n + p + 1)

n +p+1

k (k + 1) ···(k + p) = (p + 1)!

44. Let d (X , σ ) denote the distance from a point X to a plane σ . Let us consider the pair (A, π ) where A ∈ E and π is a plane containing some three points B ,C, D ∈ E such that d (A, π ) is the smallest possible. We may suppose that B,C, D are se- lected such that △BCD contains no other points of E. Let A ′

be the projection of A on π , and let l b ,l c ,l d be lines through B ,C, D parallel to CD, DB, BC re- spectively. If A ′ is in the half-plane determined by l d not containing BC, then

d (D, ABC) ≤ d(A ′ , ABC) < d(A, BCD), which is impossible. Similarly, A ′ lies in the half-planes determined by l b ,l c that contain D, and hence A ′ is inside the triangle bordered by l b ,l c ,l d . The minimality property of (A, π ) and the way in which BCD was selected guarantee that E ∩ T = {A,B,C,D}.

45. As in the previous problem, let us choose the pair (A, π ) such that d(A, π ) is minimal. If π contains only three points of E, we are done. If not, there are four

points in E ∩ P, say A 1 ,A 2 ,A 3 ,A 4 , such that the quadrilateral Q =A 1 A 2 A 3 A 4 contains no other points of E. Suppose Q is not convex, and that w.l.o.g. A 1 is inside the triangle A 2 A 3 A 4 . If A 0 is the projection of A on P, the point A 1 belongs to one of the triangles A 0 A 2 A 3 ,A 0 A 3 A 4 ,A 0 A 4 A 2 , say A 0 A 2 A 3 . Then

d (A 1 , AA 2 A 3 ) ≤ d(A 0 , AA 2 A 3 ) < AA 0 , which is impossible. Hence Q is convex. Also, by the minimality property of (A, π ) the pyramid AA 1 A 2 A 3 A 4 contains no other points of E.

432 4 Solutions

46. We need to consider only the case t > |x|. There is no loss of generality in as- suming x > 0. To obtain the estimate from below, set

2 − f (−(x + t)),

2 Since −(x +t) < x −t and x < (x +t)/2, we have f (x)− f (x −t) ≤ a 1 +a 2 +a 3 .

Since 2 −1 <a j +1 /a j < 2, it follows that

a 4 a 3 /2

g (x,t) >

a 1 +a 2 +a 3 4a 3 + 2a 3 +a 3 To obtain the estimate from above, set

3 If t < 2x, then x − t < −(x + t)/3 and therefore f (x) − f (x − t) ≥ b 1 . If t ≥ 2x,

then (x + t)/3 ≤ x and therefore f (x) − f (x −t) ≥ b 2 . Since 2 −1 <b j +1 /b j < 2, we get

g 2 3 +b 4 b 2 + 2b 2 + 4b (x,t) < 2 < = 14.

47. M lies on AB and N lies on BC. If CQ ≤ 2CD/3, then BM = CQ/2. If CQ > 2CD /3, then N coincides with C.

48. Let a plane cut the edges AB , BC,CD, DA at points K, L, M, N respectively. Let D ′ ,A ′ ,B ′

be distinct points in the plane ABC such that the triangles BCD ′ , CD ′ A ′ ,D ′ A ′ B ′ are equilateral, and

M ′ ∈ [CD ′ ], N ′ ∈ [D ′ A ′ ], and K ′ ∈

[A ′ B ′ ] such that CM ′ = CM, A ′ N ′ =

AN M , and A ′ ′ K ′ = AK. The perimeter

N ′ K ′ P of the quadrilateral KLMN is equal

to the length of the polygonal line KLM ′ ′

A C A ′ N K ′ , which is not less than KK ′ .

It follows that P ≥ 2a. Let us consider all quadrilaterals KLMN that are obtained by intersecting the tetrahedron by a plane parallel to a fixed plane α . The lengths of the segments KL , LM, MN, NK are linear functions in AK, and so is P. Thus P takes its max- imum at an endpoint of the interval, i.e., when the plane KLMN passes through one of the vertices A , B,C, D, and it is easy to see that in this case P ≤ 3a.

4.19 Longlisted Problems 1977 433

49. If one of p , q, say p, is zero, then −q is a perfect square. Conversely, (p,q) = (0, −t 2 ) and (p, q) = (−t 2 , 0) satisfy the conditions for t ∈ Z. We now assume that p , q are nonzero. If the trinomial x 2 + px + q has two integer roots x 1 ,x 2 , then |q| = |x 1 x 2 | ≥ |x 1 | + |x 2 | − 1 ≥ |p| − 1. Similarly, if x 2 + qx + p has integer roots, then |p| ≥ |q| − 1 and q 2 − 4p is a square. Thus we have two cases to investigate: (i) |p| = |q|. Then p 2 − 4q = p 2 ± 4p is a square, so (p,q) = (4,4). (ii) |p| = |q| ± 1. The solutions for (p,q) are (t,−1 − t) for t ∈ Z and (5,6), (6, 5).

50. Suppose that P n (x) = n for x ∈ {x 1 ,x 2 ,...,x n }. Then

P n (x) = (x − x 1 )(x − x 2 ) ···(x − x n ) + n. From P n (0) = 0 we obtain n = |x 1 x 2 ···x n |≥2 n −2 (because at least n − 2 factors

are different from ±1) and therefore n ≥ 2 n −2 . It follows that n ≤ 4. For each positive integer n ≤ 4 there exists a polynomial P n . Here is the list of such polynomials:

51. We shall use the following algorithm: Choose a segment of maximum length (“basic” segment) and put on it un- used segments of the opposite color without overlapping, each time of the maximum possible length, as long as it is possible. Repeat the procedure with remaining segments until all the segments are used.

Let us suppose that the last basic segment is black. Then the length of the used part of any white basic segment is greater than the free part, and consequently at least one-half of the length of the white segments has been used more than once. Therefore all basic segments have total length at most 1 .5 and can be distributed on a segment of length 1 .51. On the other hand, if we are given two white segments of lengths 0 .5 and two black segments of lengths 0 .999 and 0.001, we cannot distribute them on a seg- ment of length less than 1 .499.

52. The maximum and minimum are 2R 4 − 2k 2 and 2R 1 + 1 −k 2 respec- tively.

53. The discriminant of the given equation considered as a quadratic equation in b is 196 − 75a 2 . Thus 75a 2 ≤ 196 and hence −1 ≤ a ≤ 1. Now the integer solutions of the given equation are easily found: (−1,3), (0,0), (1,2).

54. We shall use the following lemma. Lemma. If a real function f is convex on the interval I and x , y, z ∈ I, x ≤ y ≤ z, then (y − z) f (x) + (z − x) f (y) + (x − y) f (z) ≤ 0.

434 4 Solutions Proof. The inequality is obvious for x = y = z. If x < z, then there exist p, r such

that p + r = 1 and y = px + rz. Then by Jensen’s inequality f (px + rz) ≤ pf (x) + r f (z), which is equivalent to the statement of the lemma.

By applying the lemma to the convex function − lnx we obtain x y y z z x ≥y x z y x z for any 0 < x ≤ y ≤ z. Multiplying the inequalities a b b c c a ≥b a c b a c and a c c d d a ≥

c a d c a d we get the desired inequality. Remark. Similarly, for 0 <a

1 ≤a 2 ≤ ··· ≤ a n it holds that a 2 1 a 2 3 ···a n 1 ≥

a a 1 a a 2 a 2 n 3 ···a 1 .

55. The statement is true without the assumption that O ∈ BD. Let BP ∩ DN = {K}. If we denote AB −→

−→ = a, AD

= b and AO = α a + β b for some α , β ∈ R, 1/ α +1/ β 6=

1, by straightforward calculation we obtain that

1 AK −→ =

AO −→ . α + β − αβ

α + β − αβ Hence A , K, O are collinear.

56. See the solution to (LL67-36).

57. Suppose that there exists a sequence of 17 terms a 1 ,a 2 ,...,a 17 satisfying the required conditions. Then the sum of terms in each row of the rectangular array below is positive, while the sum of terms in each column is negative, which is a contradiction.

a 1 a 2 ...a 11

a 2 a 3 ...a 12

a 7 a 8 ...a 17

On the other hand, there exist 16-term sequences with the required property. An example is 5 , 5, −13,5,5,5,−13,5,5,−13,5,5,5,−13,5,5 which can be ob-

tained by solving the system of equations k +10 ∑ i =k a i = 1 (k = 1, 2, . . . , 6) and ∑ l +6 i =l a i = −1 (l = 1,2,...,10). Second solution. We shall prove a stronger statement: If 7 and 11 in the question

are replaced by any positive integers m , n, then the maximum number of terms is m + n − (m,n) − 1.

be a sequence of real numbers, and let us define s 0 = 0 and s k =a 1 + ···+ a k (k = 1, . . . , l). The given conditions are equivalent to s k >s k +m for 0 ≤ k ≤ l − m and s k <s k +n for 0 ≤ k ≤ l − n. Let d = (m, n) and m = m ′ d ,n =n ′ d . Suppose that there exists a sequence (a k ) of length greater than or equal to l = m + n − d satisfying the required conditions.

Let a 1 ,a 2 ,...,a l

−1)d satisfy n inequalities s k +m <s k and m ′ inequalities s k <s k +n . Moreover, each term s kd appears twice in these

Then the m ′ +n ′ numbers s 0 ,s d ,...,s (m ′ +n ′

inequalities: once on the left-hand and once on the right-hand side. It follows that there exists a ring of inequalities s i 1 <s i 2 < ··· < s i k <s i 1 , giving a contradiction. On the other hand, suppose that such a ring of inequalities can be made also for l = m + n − d − 1, say s i 1 <s i 2 < ··· < s i k <s i 1 . If there are p inequalities of

4.19 Longlisted Problems 1977 435 the form a k +m <a k and q inequalities of the form a k +n >a k in the ring, then

qn = rm, which implies m ′ | q, n ′ | p and thus k = p + q ≥ m ′ +n ′ . But since all

i 1 ,i 2 ,...,i k are congruent modulo d, we have k ≤m ′ +n ′ − 1, a contradiction. Hence there exists a sequence of length m + n − d − 1 with the required property.

58. The following inequality (Finsler and Hadwiger, 1938) is sharper than the one we have to prove:

2 2 2 √ + 2bc + 2ca − a −b −c ≥ 4S 3 .

(1) First proof. Let us set 2x = b + c − a, 2y = c + a − b, 2z = a + b − c. Then

2ab

x , y, z > 0 and the inequality (1) becomes y 2 z 2 +z 2 x 2 +x 2 y 2 ≥ xyz(x + y + z), which is equivalent to the obvious inequality

2 (xy − yz) 2 + (yz − zx) + (zx − xy ) 2 ≥ 0. Second proof. Using the known relations for a triangle

a 2 +b 2 +c 2 = 2s 2 − 2r 2 − 8rR,

ab + bc + ca = s 2 +r 2 + 4rR, S = rs,

where r and R are the radii of the incircle and the circumcircle, s the semiperime- ter and S the area, we can transform (1) into

√ s 3 ≤ 4R + r.

The last inequality is a consequence of the inequalities 2r

2 ≤ R and s 2 ≤ 4R + 4Rr + 3r 2 , where the last one follows from the equality HI 2 = 4R 2 + 4Rr + 3r 2 −

s 2 (H and I being the orthocenter and the incenter of the triangle).

59. Let us consider the set R of pairs of coordinates of the points from E reduced modulo 3. If some element of R occurs thrice, then the corresponding points are vertices of a triangle with integer barycenter. Also, no three elements from E can have distinct x-coordinates and distinct y-coordinates. By an easy discussion we can conclude that the set R contains at most four elements. Hence |E| ≤ 8. An example of a set E consisting of 8 points that satisfies the required condition is

E = {(0,0),(1,0),(0,1),(1,1),(3,6),(4,6),(3,7),(4,7)}.

60. By Lagrange’s interpolation formula we have

F (x) = ∑ F (x ) 6= j j

∏ i (x − x j )

∏ i 6= j (x i −x j ) Since the leading coefficient in F (x) is 1, it follows that

j =0

∏ (x i −x j ) = ∏ |x i −x j | ∏ |x i −x j | ≥ j!(n − j)!,

we have

|F(x j )|

n ! ∑ j =0 j

|F(x j )| ≤ max |F(x j )|.

n ! Now the required inequality follows immediately.

j =0

i 6= j (x i −x j )

4.20 Shortlisted Problems 1978 437