Solutions to the Shortlisted Problems of IMO 1978

4.20 Solutions to the Shortlisted Problems of IMO 1978

1. There exists an M s that contains at least 2n /k = 2(k 2 + 1) elements. It follows that M s contains either at least k 2 + 1 even numbers or at least k 2 + 1 odd num- bers. In the former case, consider the predecessors of those k 2 + 1 numbers:

among them, at least k 2 k +1 +1 > k, i.e., at least k + 1, belong to the same subset, say M t . Then we choose s ,t. The latter case is similar.

Second solution. For all i , j ∈ {1,2,...,k}, consider the set N ij = {r | 2r ∈ M

i , 2r − 1 ∈ M j }. Then {N ij | i, j} is a partition of {1,2,...,n} into k sub- sets. For n

≥k 3 + 1 one of these subsets contains at least k + 1 elements, and the statement follows. Remark. The statement is not necessarily true when n =k 3 .

2. Consider the transformation φ of the plane defined as the homothety H with center B and coefficient 2 followed by the rotation R about the center O through an angle of 60 ◦ . Being direct, this

O mapping must be a rotational homo-

A thety. We also see that H maps S N B ′

into the point symmetric to S with S respect to OA, and R takes it back

A ′ to S. Hence S is a fixed point, and

is consequently also the center of M φ . Therefore φ is the rotational ho-

B mothety about S with the angle 60 ◦ and coefficient 2. (In fact, this could also be seen from the fact that φ preserves

angles of triangles and maps the segment SR onto SB, where R is the midpoint of AB.) Since φ (M) = B ′ , we conclude that ∠MSB ′ = 60 ◦ and SB ′ /SM = 2. Similarly, ∠NSA ′ = 60 ◦ and SA ′ /SN = 2, so triangles MSB ′ and NSA ′ are indeed similar.

Second solution. Probably the simplest way here is using complex numbers. Put the origin at O and complex numbers a ,a ′ at points A ,A ′ , and denote the

primitive sixth root of 1 by ω . Then the numbers at B, B ′ , S and N are ω a , ω a ′ , (a + ω a )/3, and (a + ω a ′ )/2 respectively. Now it is easy to verify that (n − s) = ω (a ′ − s)/2, i.e., that ∠NSA ′ = 60 ◦ and SA ′ /SN = 2.

3. What we need are m , n for which 1978 m (1978 n −m − 1) is divisible by 1000 =

8 · 125. Since 1978 n −m − 1 is odd, it follows that 1978 m is divisible by 8, so m ≥ 3. Also, 1978 n −m − 1 is divisible by 125, i.e., 1978 n −m ≡ 1 (mod 125). Note that 1978 ≡ −2 (mod 5), and consequently also −2 n −m ≡ 1. Hence 4 | n−m = 4k, k ≥

1. It remains to find the least k such that 1978 4k ≡ 1 (mod 125). Since 1978 4 ≡ (−22) 4 = 484 2 ≡ (−16) 2 = 256 ≡ 6, we reduce it to 6 k ≡ 1. Now 6 k = (1+5) k ≡

1 + 5k + 25 k 2 (mod 125), which reduces to 125 | 5k(5k − 3). But 5k − 3 is not divisible by 5, and so 25 | k. Therefore 100 | n − m, and the desired values are m = 3, n = 103.

438 4 Solutions

be the angles of T 1 and T 2 opposite to c and w respectively. By the cosine theorem, the inequality is transformed into

4. Let γ , ϕ

a 2 (2v 2 − 2uvcos ϕ )+b 2 (2u 2 − 2uvcos ϕ ) +2(a 2 +b 2 − 2ab cos γ )uv cos ϕ ≥ 4abuvsin γ sin ϕ .

This is equivalent to 2 (a 2 v 2 +b 2 u 2 ) − 4abuv(cos γ cos ϕ + sin γ sin ϕ ) ≥ 0, i.e., to

2 (av − bu) 2 + 4abuv(1 − cos( γ − ϕ )) ≥ 0, which is clearly satisfied. Equality holds if and only if γ = ϕ and a /b = u/v, i.e.,

when the triangles are similar, a corresponding to u and b to v.

5. We first explicitly describe the elements of the sets M 1 ,M 2 . x 6∈ M 1 is equivalent to x = a + (a + 1) + ···+ (a + n − 1) = n(2a + n − 1)/2 for some natural numbers n , a, n ≥ 2. Among n and 2a + n − 1, one is odd and the other even, and both are greater than 1; so x has an odd factor ≥ 3. On the other hand, for every x with an odd divisor p > 3 it is easy to see

that there exist corresponding a , n. Therefore M 1 = {2 k | k = 0,1,2,...}. x 6∈ M 2 is equivalent to x = a + (a + 2) + ··· + (a + 2(n − 1)) = n(a + n −

1 ), where n ≥ 2, i.e. to x being composite. Therefore M 2 = {1} ∪ {p | p = prime}. x 6∈ M 3 is equivalent to x = a + (a + 3) + ··· + (a + 3(n − 1)) = n(2a + 3(n −

1 ))/2. It remains to show that every c ∈M 3 can be written as c =2 k p with p prime. Suppose the opposite, that c =2 k pq , where p , q are odd and q ≥ p ≥ 3. Then there exist positive integers a , n (n ≥ 2) such that c = n(2a + 3(n − 1))/2 and

hence c 6∈ M 3 . Indeed, if k = 0, then n = 2 and 2a + 3 = pq work; otherwise, setting n = p one obtains a = 2 k q − 3(p− 1)/2 ≥ 2q− 3(p− 1)/2 ≥ (p+ 3)/2 >

6. For fixed n and the set { ϕ (1), . . . , ϕ (n)}, there are finitely many possibilities for a mapping ϕ on {1,...,n}. Suppose ϕ is the one among these for which

∑ n k =1 ϕ (k)/k 2 is minimal. If i < j and ϕ (i) > ϕ ( j) for some i, j ∈ {1,...,n},

define ψ as ψ (i) = ϕ ( j), ψ ( j) = ϕ (i), and ψ (k) = ϕ (k) for all other k. Then

ϕ (i) ϕ ∑ ( j)

ϕ (k)

ψ (k)

ϕ (i) ϕ ( j)

i +j

= (i − j)( ϕ ( j) − ϕ (i))

i 2 j 2 which contradicts the assumption. This shows that ϕ (1) < ··· < ϕ (n), and con-

sequently ϕ (k) ≥ k for all k. Hence

n ϕ (k)

k =1 k 2

k =1

k =1 k

4.20 Shortlisted Problems 1978 439

7. Let x = OA, y = OB, z = OC, p α = ∠BOC, β = ∠COA, γ = ∠AOB. The conditions yield the equation x +y+ x 2 +y 2 − 2xycos γ = 2p, which transforms to (2p − x − y) 2 =x 2 +y 2 − 2xycos γ , i.e. (p − x)(p − y) = xy(1 − cos γ ). Thus

p −x p −y

= 1 − cos γ ,

and analogously p −y p −z

z · x = 1 − cos β . Setting u = x , v = p −y y ,w = p −z z , the above system becomes

y · z = 1 − cos α , −z −x

−x

uv = 1 − cos γ , vw = 1 − cos α , wu = 1 − cos β .

This system has a unique solution in positive real numbers u, v, w: r

(1 − cos )(1 − cos (1 − cos )(1 − cos =

,v=

1 − cos β s (1 − cos α )(1 − cos β )

1 − cos α

1 − cos γ

Finally, the values of x , y, z are uniquely determined from u, v, w. Remark. It is not necessary that the three lines be in the same plane. Also, there

could be any odd number of lines instead of three.

8. Take the subset {a i } = {1,7,11,13,17,19,23,29,...,30m − 1} of S containing all the elements of S that are not multiples of 3. There are 8m such elements.

Every element in S can be uniquely expressed as 3 t a i for some i and t ≥ 0. In a subset of S with 8m + 1 elements, two of them will have the same a i , hance one will divide the other. On the other hand, for each i = 1, 2, . . . , 8m choose t ≥ 0 such that 10m < b i =

3 t a i < 30m. Then there are 8m b i ’s in the interval (10m, 30m), and the quotient of any two of them is less than 3, so none of them can divide any other. Thus the answer is 8m.

9. Since the nth missing number (gap) is f ( f (n)) + 1 and f ( f (n)) is a member of the sequence, there are exactly n − 1 gaps less than f ( f (n)). This leads to

(1) Since 1 is not a gap, we have f (1) = 1. The first gap is f ( f (1)) + 1 = 2. Two

f ( f (n)) = f (n) + n − 1.

consecutive integers cannot both be gaps (the predecessor of a gap is of the form f ( f (m))). Now we deduce f (2) = 3; a repeated application of the formula above gives f (3) = 3 +1 = 4, f (4) = 4 +2 = 6, f (6) = 9, f (9) = 14, f (14) = 22,

f (22) = 35, f (35) = 56, f (56) = 90, f (90) = 145, f (145) = 234, f (234) = 378. Also, f ( f (35)) + 1 = 91 is a gap, so f (57) = 92. Then by (1), f (92) = 148,

f (148) = 239, f (239) = 386. Finally, here f ( f (148)) + 1 = 387 is a gap, so

f (240) = 388.

440 4 Solutions Second solution. As above, we arrive at formula (1). Then by simple induction

it follows that f (F n + 1) = F n +1 + 1, where F k is the Fibonacci sequence (F 1 =

F 2 = 1). We now prove by induction (on n) that f (F n + x) = F n +1 + f (x) for all x with

1 ≤x≤F n −1 . This is trivially true for n = 0, 1. Supposing that it holds for n − 1, we shall prove it for n: (i) If x = f (y) for some y, then by the inductive assumption and (1)

f (F n + x) = f (F n + f (y)) = f ( f (F n −1 + y))

=F n + f (y) + F n −1 +y−1=F n +1 + f (x). (ii) If x = f ( f (y)) + 1 is a gap, then f (F n + x − 1) + 1 = F n +1 + f (x − 1) + 1 is

a gap also:

F n +1 + f (x) + 1 = F n +1 + f ( f ( f (y))) + 1 = f (F n + f ( f (y))) + 1 = f ( f (F n −1 + f (y))) + 1.

It follows that f (F n + x) = F n +1 + f (x − 1) + 2 = F n +1 + f (x). Now, since we know that each positive integer x is expressible as x =F k 1 +F k 2 + ···+ F k r , where 0 <k r 6= 2, k i ≥k i +1 + 2, we obtain f (x) = F k 1 +1 +F k 2 +1 + ··· +

F k r +1 . Particularly, 240 = 233 + 5 + 2, so f (240) = 377 + 8 + 3 = 388. √ Remark. It can be shown that f (x) = [ α x ], where α = (1 + 5 )/2.

10. Assume the opposite. One of the countries, say A, contains at least 330 members

a 1 ,a 2 ,...,a 330 of the society ( 6 ·329 = 1974). Consider the differences a 330 −a i ,

i = 1, 2, . . . , 329: the members with these numbers are not in A, so at least 66 of them, a 330 −a i 1 ,...,a 330 −a i 66 , belong to the same country, say B. Then the differences (a i 66 −a 330 ) − (a i j −a 330 )=a i 66 −a i j ,j = 1, 2, . . . , 65, are neither in

A nor in B. Continuing this procedure, we find that 17 of these differences are in the same country, say C, then 6 among 16 differences of themselves in a country

D , and 3 among 5 differences of themselves in E; finally, two differences of these

3 differences belong to country F, so that the difference of themselves cannot be in any country. This is a contradiction.

Remark. The following stronger ( [6!e] = 1957) statement can be proved in the same way.

Schur’s lemma. If n is a natural number and e the logarithm base, then for every partition of the set {1,2,...,[en!]} into n subsets one of these subsets contains some two elements and their difference.

11. Set F (x) = f 1 (x) f 2 (x) ··· f n (x): we must prove concavity of F 1 /n . By the as- sumption,

F ( θ x + (1 − θ )y) ≥ ∏ [ θ f i (x) + (1 − θ ) f (y)]

i =1 n

θ ) (1 − n ∑ −k ∑ f i 1 (x) . . . f i k (x) f i k +1 (y) f i n (y),

k =0

4.20 Shortlisted Problems 1978 441 where the second sum goes through all n k k -subsets {i 1 ,...,i k } of {1,...,n}.

The inequality between the arithmetic and geometric means now gives us

F (x) F (y) (n−k)/n .

Inserting this in the above inequality and using the binomial formula, we finally obtain

+ (1 − )y) ≥ ∑ (1 −

n −k F n ( θ x θ θ θ ) F (x) k /n

F (y) (n−k)/n

k =0

= θ F (x) 1 /n + (1 − θ )F(y) 1 /n , which proves the assertion.

12. Let O be the center of the smaller circle, T its contact point with the circumcircle of ABC, and J the midpoint of segment BC. The figure is symmetric with respect to the line through A , O, J, T .

A homothety centered at A taking T into J will take the smaller circle into the incircle of ABC, hence will take O into the incenter I. On the other hand, ∠ABT = ∠ACT = 90 ◦ implies that the quadrilaterals ABTC and APOQ are simi- lar. Hence the above homothety also maps O to the midpoint of PQ. This finishes the proof.

Remark. The assertion is true for a nonisosceles triangle ABC as well, and this (more difficult) case is a matter of SL93-3.

13. Lemma. If MNPQ is a rectangle and O any point in space, then OM 2 + OP 2 = ON 2 + OQ 2 . Proof. Let O 1 be the projection of O onto MNPQ, and m , n, p, q denote the distances of O 1 from MN , NP, PQ, QM, respectively. Then OM 2 = OO 2 1 + q 2 +m 2 , ON 2 = OO 2 +m 2 +n 2 , OP 2 = OO 2 +n 2 +p 2 , OQ 2 = OO 2 +p 2 +

q 2 1 1 , and the lemma follows immediately. 1 Now we return to the problem. Let O be the center of the given sphere S, and

X the point opposite P in the face of the parallelepiped through P , A, B. By the

lemma, we have OP 2 + OQ 2 = OC 2 + OX 2 and OP 2 + OX 2 = OA 2 +OB 2 . Hence 2OP 2 + OQ 2 = OA 2 + OB 2

√ + OC 2 = 3R 2 , i.e. OQ = √ 3R 2 − OP 2 > R.

We claim that the locus of Q is the whole sphere (O, 3R 2 − OP 2 ). Choose any point Q on this sphere. Since OQ > R > OP, the sphere with diameter PQ in- tersects S on a circle. Let C be an arbitrary point on this circle, and X the point

opposite C in the rectangle PCQX . By the lemma, OP 2 + OQ 2 = OC 2 + OX 2 , hence OX 2 = 2R 2 − OP 2 >R 2 . The plane passing through P and perpendicular to PC intersects S in a circle γ ; both P , X belong to this plane, P being inside and X outside the circle, so that the circle with diameter PX intersects γ at some point B. Finally, we choose A to be the point opposite B in the rectangle PBX A:

we deduce that OA 2 + OB 2 = OP 2 + OX 2 , and consequently A ∈ S. By the con- struction, there is a rectangular parallelepiped through P , A, B,C, X, Q.

442 4 Solutions

14. We label the cells of the cube by (a 1 ,a 2 ,a 3 ), a i ∈ {1,2,...,2n + 1}, in a natural way: for example, as Cartesian coordinates of centers of the cells ( (1, 1, 1) is one corner, etc.). Notice that there should be (2n+1) 3 −2n(2n+1)·2(n+1)= 2n+1 void cells, i.e., those not covered by any piece of soap. n = 1. In this case, six pieces of soap 1 × 2 × 2 can be placed on the following positions: [(1, 1, 1), (2, 2, 1)], [(3, 1, 1), (3, 2, 2)], [(2, 3, 1), (3, 3, 2)] and the symmetric ones with respect to the center of the box. (Here [A, B] denotes the rectangle with opposite corners at A , B.)

n is even. Each of the 2n + 1 planes P k = {(a 1 ,a 2 , k) | a i = 1, . . . , 2n + 1} can receive 2n pieces of soap: In fact, P k can be partitioned into four n ×(n+ 1) rectangles at the corners and the central cell, while an n × (n + 1) rectangle can receive n /2 pieces of soap.

n is odd, n > 1. Let us color a cell (a 1 ,a 2 ,a 3 ) blue, red, or yellow if exactly three, two or one a i respectively is equal to n + 1. Thus there are 1 blue, 6n red, and 12n 2 yellow cells. We notice that each piece of soap must contain at least one colored cell (because 2 (n + 1) > 2n + 1). Also, every piece of soap contains an even number (actually, 1 · 2, 1(n + 1), or 2(n + 1)) of cells in P k . On the other hand, 2n + 1 cells are void, i.e., one in each plane. There are several cases for a piece of soap S:

(i) S consists of 1 blue, n + 1 red and n yellow cells; (ii) S consists of 2 red and 2n yellow cells (and no blue cells); (iii) S contains 1 red cell, n + 1 yellow cells, and the rest are uncolored; (iv) S contains 2 yellow cells and no blue or red ones.

From the descriptions of the last three cases, we can deduce that if S con- tains r red cells and no blue, then it contains exactly 2 + (n − 1)r red ones.

(∗) Now, let B 1 ,...,B k

be all boxes put in the cube, with a possible exception for the one covering the blue cell: thus k = 2n(2n + 1) if the blue cell is void, or k = 2n(2n + 1) − 1 otherwise. Let r i and y i respectively be the

numbers of red and yellow cells inside B i . By (∗) we have y 1 + ··· + y k = 2k + (n − 1)(r 1 + ··· + r k ). If the blue cell is void, then r 1 + ··· + r k = 6n and consequently y 1 + ···+y k = 4n(2n + 1)+ 6n(n −1)= 14n 2 −2n, which is impossible because there are only 12n 2 < 14n 2 − 2n yellow cells. Oth- erwise, r 1 + ··· + r k ≥ 5n − 2 (because n + 1 red cells are covered by the box containing the blue cell, and one can be void) and consequently y

1 + ··· + y k ≥ 4n(2n + 1) − 2 + (n − 1)(5n − 2) = 13n − 3n; since there are n more yellow cells in the box containing the blue one, this counts for 13n 2 − 2n > 12n 2 (n ≥ 3), again impossible.

Remark. The following solution of the case n odd is simpler, but does not work for n = 3. For k = 1, 2, 3, let m k

be the number of pieces whose long sides are perpendicular to the plane π k (a k = n + 1). Each of these m k pieces covers exactly

2 cells of π k , while any other piece covers n + 1, 2(n + 1), or none. It follows that 4n 2 + 4n − 2m k is divisible by n + 1, and so is 2m k . This further implies that

4.20 Shortlisted Problems 1978 443 2m 1 + 2m 2 + 2m 3 = 4n(2n + 1) is a multiple of n + 1, which is impossible for

each odd n except n = 1 and n = 3.

15. Let C n = {a 1 ,...,a n } (C 0 = /0) and P n = { f (B) | B ⊆ C n }. We claim that P n contains at least n + 1 distinct elements. First note that P 0 = {0} contains one element. Suppose that P n +1 =P n for some n. Since P n +1 ⊇ {a n +1 +r|r∈P n }, it follows that for each r ∈P n , also r +a n +1 ∈P n . Then obviously 0 ∈P n implies ka n +1 ∈P n for all k; therefore P n = P has at least p ≥ n + 1 elements. Otherwise, if P n +1 ⊃P n for all n, then |P n +1 | ≥ |P n | + 1 and hence |P n | ≥ n + 1, as claimed. Consequently, |P p −1 | ≥ p . (All the operations here are performed modulo p.)

16. Clearly |x| ≤ 1. As x runs over [−1,1], the vector u = (ax,a 1 −x 2 ) runs over all vectors of length a in the plane having a nonnegative vertical component. p Putting v = (by, b 1 −y 2

), w = (cz, c 1 −z 2 ), the system becomes u + v = w, with vectors u , v, w of lengths a, b, c respectively in the upper half-plane. Then

a , b, c are sides of a (possibly degenerate) triangle; i.e, |a − b| ≤ c ≤ a + b is a necessary condition. Conversely, if a , b, c satisfy this condition, one constructs a triangle OMN with

OM = a, ON = b, MN = c. If the vectors −−→ OM , −→ ON have a positive nonnegative component, then so does their sum. For every such triangle, putting u −−→ = OM , v = −→ ON , and w = −−→ OM + −→ ON gives a solution, and every solution is given by one such triangle. This triangle is uniquely determined up to congruence: α = ∠MON = ∠(u, v) and β = ∠(u, w). Therefore, all solutions of the system are

x = cost, y = cos(t + α ), z = y = cos(t + β ), t ∈ [0, π − α ] or x = cost, y = cos(t − α ), z = y = cos(t − β ),

t ∈[ α , π ].

17. Let z 0 ≥ 1 be a positive integer. Supposing that the statement is true for all triples (x, y, z) with z < z 0 , we shall prove that it is true for z =z 0 too. If z 0 = 1, verification is trivial, while x 0 =y 0 is obviously impossible. So let there be given a triple (x 0 ,y 0 ,z 0 ) with z 0 > 1 and x 0 <y 0 , and define another triple (x, y, z) by

z =z 0 −x 0 . Then x, y, z are positive integers. This is clear for x , z, while y = x 0 +y 0 − 2z 0 ≥

2 (√x 0 y 0 −z 0 ) > 2(z 0 −z 2 0 ) = 0. Moreover, xy − z =x 0 (x 0 +y 0 − 2z 0 ) − (z 0 − x 0 ) 2 =x 0 y 0 −z 2 0 = 1 and z < z 0 , so that by the assumption, the statement holds for x , y, z. Thus for some nonnegative integers a, b, c, d we have

z = ac + bd. But then we obtain representations of this sort for x 0 ,y 0 ,z 0 too: x

x =a 2 +b 2 ,

y =c 2 +d 2 ,

0 =a +b , y 0 = (a + c) + (b + d) , z 0 = a(a + c) + b(b + d). For the second part of the problem, we note that for z = (2q)!,

444 4 Solutions z 2 = (2q)!(2q)(2q − 1)···1 ≡ (2q)! · (−(2q + 1))(−(2q + 2))···(−4q)

= (−1) 2q (4q)! ≡ −1 (mod p), by Wilson’s theorem. Hence p |z 2 + 1 = py for some positive integer y > 0. Now

it follows from the first part that there exist integers a , b such that x = p = a 2 +b 2 .

Second solution. Another possibility is using arithmetic of Gaussian integers.

Lemma. Suppose m , n, p, q are elements of Z or any other unique factorization domain, with mn = pq. then there exist elements a, b, c, d such that m = ab, n = cd, p = ac, q = bd.

Proof is direct, for example using factorization of a , b, c, d into primes. We now apply this lemma to the Gaussian integers in our case (because Z[i] has the unique factorization property), having in mind that xy =z 2 + 1 = (z + i)(z −

i ). We obtain (1) x = ab, (2) y = cd, (3) z + i = ac, (4) z − i = bd

for some a , b, c, d ∈ Z[i]. Let a = a 1 +a 2 i , etc. By (3) and (4), gcd (a 1 ,a 2 ) = ··· = gcd (d 1 ,d 2 ). Then (1) and (2) give us b = a, c = d. The statement follows at once: x = ab = aa = a 2 +a 1 2 2 ,y = dd = d 2 1 +d 2 2 and z + i = (a 1 d 1 +a 2 d 2 ) + ı(a 2 d 1 −

a 1 d 2 )⇒z=a 1 d 1 +a 2 d 2 .

4.21 Shortlisted Problems 1979 445