Solutions to the Shortlisted Problems of IMO 1993

4.34 Solutions to the Shortlisted Problems of IMO 1993

1. First we notice that for a rational point O (i.e., with rational coordinates), there exist 1993 rational points in each quadrant of the unit circle centered at O. In fact, it suffices to take

Now consider the set A = {(i/q, j/q) | i, j = 0,1,...,2q}, where q = ∏ 1993 t =1 (t 2 +

1 ). We claim that A gives a solution for the problem. Indeed, for any P ∈ A there is a quarter of the unit circle centered at P that is contained in the square [0, 2] × [0,2]. As explained above, there are 1993 rational points on this quarter circle, and by definition of q they all belong to A.

Remark. Substantially the same problem was proposed by Bulgaria for IMO 71: see (SL71-2), where we give another possible construction of a set A.

2 R . Therefore 3 (1 + r) ≤ 3 1 + 2 = 3 4 . It remains only to show that p

1 1 2 1 1 2. It is well known that r 2 ≤

≤ 1 . We note that p does not exceed one half of

the circumradius of △A ′ B ′ C ′

4 . However, by the theorem on the nine-point circle,

this circumradius is equal to 1 2 R , and the conclusion follows. Second solution. By a well-known relation we have cos A + cos B + cosC =

1 + r R ( = 1 + r when R = 1). Next, recalling that the incenter of △A ′ B ′ C ′ is at the orthocenter of △ABC, we easily obtain p = 2cosAcosBcosC. Cosines of angles of a triangle satisfy the identity cos 2 A + cos 2 B + cos 2 C + 2 cos A cos B cosC = 1 (the proof is straightforward: see (SL81-11)). Thus

1 p + (1 + r) 2 = 2 cosA cos B cosC + 1 (cos A + cosB + cosC) 2

≤ 2cosAcosBcosC + cos 2 A + cos 2 B + cos 2 C = 1.

3. Let O 1 and ρ

be the center and radius of k c . It is clear that C , I, O 1 are collinear and CI /CO 1 = r/ ρ . By Stewart’s theorem applied to △OCO 1 ,

OI = OO 2 1 + 1 −

OC 2

−CI · IO 1 . (1) Since OO 1 =R− ρ , OC = R and by Euler’s formula OI 2 =R 2 −2Rr, substituting

these values in (1) gives CI · IO 1 =r ρ , or equivalently CO 1 · IO 1 = ρ 2 = DO 2 1 . Hence the triangles CO 1 D and DO 1 I are similar, implying ∠DIO 1 = 90 ◦ . Since CD = CE and the line CO 1 bisects the segment DE, it follows that I is the mid- point of DE.

Second solution. Under the inversion with center C and power ab, k c is trans- formed into the excircle of b Ab BC corresponding to C. Thus CD = ab s , where s is

the common semiperimeter of △ABC and △b Ab BC , and consequently the distance from D to BC is ab s sinC = 2S ABC s = 2r. The statement follows immediately.

574 4 Solutions Third solution. We shall prove a stronger statement: Let ABCD be a convex

quadrilateral inscribed in a circle k, and k ′ the circle that is tangent to segments BO , AO at K, L respectively (where O = BD ∩AC), and internally to k at M. Then KL contains the incenters I , J of △ABC and △ABD. Let K ′ ,K ′′ ,L ′ ,L ′′ , N denote the midpoints of arcs BC, BD, AC, AD, AB that don’t contain M; X ′ ,X ′′ the points on k defined by X ′ N = NX ′′ =K ′ K ′′ =L ′ L ′′ (as oriented arcs); and set S = AK ′ ∩ BL ′′ ,M = NS ∩ k, K = K ′′ M ∩ BO, L = L ′ M ∩ AO . It is clear that I = AK ′ ∩BL ′ ,J = AK ′′ ∩BL ′′ . Furthermore, X ′ M contains I (to see this, use the fact that for A , B,C, D, E, F on k, lines AD, BE, CF are concurrent if and only if AB ·CD · EF = BC · DE · FA, and then express AM/MB by applying this rule to AMBK ′ NL ′′ and show that AK ′ , MX ′ , BL ′ are concurrent). Similarly,

X ′′ M contains J. Now the points X ′′ D N X ′

B C , K, I, S, M lie on a circle (∠BKM =

∠BIM = ∠BSM), and points A, L, J, K ′′ O S , M do so as well. Lines IK , JL are

L ′′

parallel to K ′′ L ′ (because ∠MKI = K ′ ∠MBI = ∠MK ′′ L ′ ). On the other hand,

LJI the quadrilateral ABIJ is cyclic, and K S

simple calculation with angles shows that IJ is also parallel to K ′′ L ′ . Hence

A B K , I, J, L are collinear. Finally, K ≡ K,

L ≡ L, and M ≡ M because the homo- thety centered at M that maps k ′ to k

M sends K to K ′′ and L to L ′ (thus M , K, K ′′ , as well as M , L, L ′ , must be collinear). As is seen now, the deciphered picture yields many other interesting properties.

Thus, for example, N , S, M are collinear, i.e., ∠AMS = ∠BMS. Fourth solution. We give an alternative proof of the more general statement in

the third solution. Let W be the foot of the perpendicular from B to AC. We define q = CW , h = BW , t = OL = OK, x = AL, θ = ∡W BO ( θ is negative if B (O,W, A), θ = 0 if W = O), and as usual, a = BC, b = AC, c = AB. Let α = ∡KLC and β = ∡ILC (both angles must be acute). Our goal is to prove α = β . We note that 90 ◦ − θ =2 α . One easily gets

cos 2S θ ABC tan α

= +b+c , tan β = a

2 −x Applying Casey’s theorem to A , B,C, k ′ , we get AC · BK + AL · BC = AB · CL,

1 + sin θ

b +c−a

i.e., b h cos θ −t + xa = c(b − x). Using that t = b − x − q − htan θ we get

b (b + c − q) − bh 1 cos θ + tan θ

a +b+c

Plugging (2) into the second equation of (1) and using bh = 2S

2 ABC and c =

b 2 +a 2 − 2bq, we obtain tan α = tan β , i.e., α = β , which completes our proof.

4.34 Shortlisted Problems 1993 575

4. Let h be the altitude from A and ϕ = ∠BAD. We have BM = 1 2 (BD + AB − AD) and MD = 1 2 (BD − AB + AD), so

2BD sin ϕ =

4BD

2AB · AD(1 − cos ϕ ) 2S ABD (1 − cos ϕ )

2BD sin ϕ

BD · h(1 − cos ϕ ) h tan ϕ 2

It follows that 1 MB + 1 MD depends only on h and ϕ . Specially, 1 1 NC 2 + NE = h tan (ϕ/2) as well.

5. For n = 1 the game is trivially over. If n = 2, it can end, for example, in the following way:

The sequence of moves shown in Fig. 2 enables us to remove three pieces placed in a 1 × 3 rectangle, using one more piece and one more free cell. In that way, for any n ≥ 4 we can reduce an (n + 3) × (n + 3) square to an n × n square ( Fig.

3 ). Therefore the game can end for every n that is not divisible by 3.

Fig. 2 Fig. 3 Suppose now that one can play the game on a 3k × 3k square so that at the

end only one piece remains. Denote the cells by (i, j), i, j ∈ {1,...,3k}, and let S 0 ,S 1 ,S 2 denote the numbers of pieces on those squares (i, j) for which i + j gives remainder 0, 1, 2 respectively upon division by 3. Initially S 0 =S 1 =S 2 = 3k 2 . After each move, two of S 0 ,S 1 ,S 2 diminish and one increases by one. Thus each move reverses the parity of the S i ’s, so that S 0 ,S 1 ,S 2 are always of the same parity. But in the final position one of the S i ’s must be equal to 1 and the other two must be 0, which is impossible.

2 , α 2 n = α n + n for all n ∈ N. We shall show that f (n) = α n + 1 2 (the closest integer to α n ) satisfies the requirements. Observe that f

6. Notice that for α = 1 +

576 4 Solutions

is strictly increasing and f (1) = 2. By the definition of f , | f (n) − α n |≤ 1 2 and

f ( f (n)) − f (n) − n is an integer. On the other hand, | f ( f (n)) − f (n) − n| = | f ( f (n)) − f (n) − α 2 n + α n |

= | f ( f (n)) − α f (n) + α f (n) − α 2 n − f (n) + α n | = |( α − 1)( f (n) − α n ) + ( f ( f (n)) − α f (n))| ≤( α − 1)| f (n) − α n | + | f ( f (n)) − α f (n)|

which implies that f ( f (n)) − f (n) − n = 0.

7. Multiplying by a and c the equation

ax 2 + 2bxy + cy 2 =P k n ,

(1) gives (ax + by) 2 + Py 2 = aP k n and (bx + cy) 2 + Px 2 = cP k n .

It follows immediately that M (n) is finite; moreover, (ax + by) 2 and (bx + cy) 2 are divisible by P, and consequently ax +by, bx + cy are divisible by P because P is not divisible by a square greater than 1. Thus there exist integers X ,Y such that bx + cy = PX, ax+ by = −PY. Then x = −bX −cY and y = aX +bY . Introducing these values into (1) and simplifying the expression obtained we get

(2) Hence (x, y) 7→ (X,Y ) is a bijective correspondence between integral solutions

aX 2 + 2bXY + cY 2 =P k −1 n .

of (1) and (2), so that M (P k n ) = M(P k −1 n ) = ··· = M(n).

8. Suppose that f (n) = 1 for some n > 0. Then f (n + 1) = n + 2, f (n + 2) = 2n + 4,

f (n + 3) = n + 1, f (n + 4) = 2n + 5, f (n + 5) = n, and so by induction f (n + 2k ) = 2n + 3 + k, f (n + 2k − 1) = n + 3 − k for k = 1,2,...,n + 2. Particularly, n ′ = 3n + 3 is the smallest value greater than n for which f (n ′ ) = 1. It follows that

all numbers n with f (n) = 1 are given by n = b i , where b 0 = 1, b n = 3b n −1 + 3. Furthermore, b n = 3 + 3b n −1 =3+3 2 +3 2 b 2 n

n −2 = ··· = 3 + 3 + ···+ 3 +3 = = 1 2 n (5 · 3 − 3). It is seen from above that if n ≤b i , then f (n) ≤ f (b i − 1) = b i + 1. Hence if

f (n) = 1993, then n ≥ b i ≥ 1992 for some i. The smallest such b i is b 7 = 5466, and f (b i + 2k − 1) = b i + 3 − k = 1993 implies k = 3476. Thus the least integer in S is n 1 = 5466 + 2 · 3476 − 1 = 12417. All the elements of S are given by n i =b i +6 + 2k − 1, where b i +6 + 3 − k = 1993, i.e., k =b i +6 − 1990. Therefore n i = 3b i +6 − 3981 = 1 2 (5 · 3 i +7 − 7971). Clearly S is infinite and lim

i →∞ +1

n i = 3.

9. We shall first complete the “multiplication table” for the sets A , B,C. It is clear that this multiplication is commutative and associative, so that we have the fol- lowing relations:

AC = (AB)B = BB = C;

A 2 = AA = (AB)C = BC = A;

C 2 = CC = B(BC) = BA = B.

4.34 Shortlisted Problems 1993 577 (a) Now put 1 in A and distribute the primes arbitrarily in A , B,C. This dis-

tribution uniquely determines the partition of Q + with the stated property. Indeed, if an arbitrary rational number

x =p α 1 α k β 1 β 1 l ··· p k q 1 ··· q l r γ 1 γ 1 m ···r m is given, where p i ∈ A, q i ∈ B, r i ∈ C are primes, it is easy to see that x be-

longs to A , B, or C according as β 1 + ··· + β l +2 γ 1 + ···+ 2 γ m is congruent to 0, 1, or 2 (mod 3). (b) In every such partition, cubes all belong to A. In fact, A 3 =A 2 A = AA = A,

B 3 =B 2 B = CB = A, C 3 =C 2 C = BC = A.

(c) By (b) we have 1 , 8, 27 ∈ A. Then 2 6∈ A, and since the problem is symmetric with respect to B ,C, we can assume 2 ∈ B and consequently 4 ∈ C. Also

7 6∈ A, and also 7 6∈ B (otherwise, 28 = 4 · 7 ∈ A and 27 ∈ A), so 7 ∈ C,

14 ∈ A, 28 ∈ B. Further, we see that 3 6∈ A (since otherwise 9 ∈ A and 8 ∈ A). Put 3 in C. Then 5 6∈ B (otherwise 15 ∈ A and 14 ∈ A), so let 5 ∈ C too. Consequently 6 , 10 ∈ A. Also 13 6∈ A, and 13 6∈ C because 26 6∈ A, so 13 ∈ B. Now it is easy to distribute the remaining primes 11 , 17, 19, 23, 29, 31: one possibility is

A = {1,6,8,10,14,19,23,27,29,31,33,...},

C = {3,4,5,7,18,22,24,26,30,32,34,...},

B = {2,9,11,12,13,15,16,17,20,21,25,28,35,...}. Remark. It can be proved that min {n ∈ N | n ∈ A,n + 1 ∈ A} ≤ 77.

10. (a) Let n = p be a prime and let p | a p −1. By Fermat’s theorem p | a p −1 −1, so that p |a gcd (p,p−1) − 1 = a − 1, i.e., a ≡ 1 (mod p). Since then a i ≡ 1 (mod p ), we obtain p |a p −1 + ··· + a + 1 and hence p 2 |a p − 1 = (a − 1)(a p −1 + ··· + a + 1). (b) Let n =p 1 ··· p k

be a product of distinct primes and let n |a n − 1. Then from p

i |a n − 1 = (a

(n/p

i ) ) p i − 1 and part (a) we conclude that p 2

i |a n − 1. Since this is true for all indices i, we also have n 2 |a n − 1; hence n has the

property P.

11. Due to the extended Eisenstein criterion, f must have an irreducible factor of degree not less than n − 1. Since f has no integral zeros, it must be irreducible.

Second solution. The proposer’s solution was as follows. Suppose that f (x) =

g (x)h(x), where g, h are nonconstant polynomials with integer coefficients. Since | f (0)| = 3, either |g(0)| = 1 or |h(0)| = 1. We may assume |g(0)| = 1 and that

g (x) = (x − n α

1 ) ···(x − α k ). Then | α 1 ··· α k | = 1. Since α −1 i ( α i + 5) = −3, tak- ing the product over i = 1, 2, . . . , k yields |( α 1 + 5) ···( α

k + 5)| = |g(−5)| = 3 . But f (−5) = g(−5)h(−5) = 3, so the only possibility is degg = k = 1. This is

impossible, because f has no integral zeros. Remark. Generalizing this solution, it can be shown that if a , m, n are positive

integers and p < a − 1 is a prime, then F(x) = x m (x + a) n + p is irreducible. The details are left to the reader.

578 4 Solutions

be the elements of S. We use induction on n. The result is trivial for k = 1 or n = k, so assume that it is true for n − 1 numbers. Then there exist m = (k −1)(n−k)+1 distinct sums of k − 1 numbers among x 2 ,...,x n ; call these sums S i ,S 1 <S 2 < ··· < S m . Then x 1 +S 1 ,x 1 +S 2 ,...,x 1 +S m are distinct sums of k of the numbers x 1 ,x 2 ,...,x n . However, the biggest of these sums is

12. Let x 1 <x 2 < ··· < x n

x 1 +S m ≤x 1 +x n −k+2 +x n −k+3 + ··· + x n ; hence we can find n − k sums that are greater and thus not included here:

x 2 +x n −k+2 + ···+ x n ,x 3 +x n −k+2 + ···+x n ,...,x n −k+1 +x n −k+2 + ···+ x n . This counts for k (n − k) + 1 sums in total.

Remark. Equality occurs if S is an arithmetic progression.

13. For an odd integer N > 1, let S N = {(m,n) ∈ S | m+n = N}. If f (m,n) = (m 1 ,n 1 ), then m 1 +n 1 = m + n with m 1 odd and m 1 ≤ n 2 < N 2 <n 1 , so f maps S N to S N . Also f is bijective, since if f (m, n) = (m 1 ,n 1 ), then n is uniquely determined as the even number of the form 2 k m

1 that belongs to the interval [ 2 , N], and this also determines m. Note that S has at most N +1 N 4 elements, with equality if and only if N is prime.

N +1

Thus if (m, n) ∈ S +5 N , there exist s , r with 1 ≤ s < r ≤ N such that f 4 s (m, n) =

f r (m, n). Consequently f t N (m, n) = (m, n), where t = r − s, 0 < t ≤ +1

4 . Suppose that (m, n) ∈ S N and t is the least positive integer with f t (m, n) = (m, n).

m +n+1

We write (m, n) = (m 0 ,n 0 ) and f i (m, n) = (m i ,n i ) for i = 1, . . . ,t. Then there exist positive integers a i such that 2 a i m i =n i −1 ,i = 1, . . . ,t. Since m t =m 0 , multiplying these equalities gives

2 a 1 +a 2 +···+a t m 0 m 1 ···m t −1 =n 0 n 1 ···n t −1 ≡ (−1) (1) t m

0 m 1 ···m t −1 (mod N) . It follows that N

|2 2k k ± 1 and consequently N | 2 − 1, where k = a

1 + ··· + a t . On the other hand, it also follows that 2 k |n 0 n 1 ··· n t −1 | (N − 1)(N − 3)···(N −

2 [N/4]). But since (N − 1)(N − 3)··· N − 2 N

4 2 · 4···(N − 1) N = −1

we conclude that 0 N <k≤ −1

2 , where equality holds if and only if {n 1 ,...,n t } is the set of all even integers from N +1 to N

2 − 1, and consequently t = 4 . Now if N ∤2 h

N +1

−1 for 1 ≤ h < N −1, we must have 2k = N −1. Therefore t = +1 N

14. We first assume that all angles of triangle ABC are less than 120 ◦ . Consider the Torricelli point T of the triangle. It holds that ∠ (AT, EF) = ∠(BT, FD) = ∠ (CT, DE) = θ for some angle θ . Therefore

2S = 2(S AET F +S BFT D +S CDT E ) = (AT · EF + BT · FD +CT · DE)sin θ

(1) = (AT + BT + CT )DE sin θ ≤ (AT + BT +CT )DE.

4.34 Shortlisted Problems 1993 579 On the other hand, by the cosine theorem we get

Adding these four equalities, we obtain 2 (AT + BT + CT ) 2 =a 2 +b 2 +c √ 2 +

4 3S, which together with (1) implies the desired inequality. Assume now that ∠C ≥ 120 ◦ and take T to be the point lying on the same side of AB as C such that ∠BTC = ∠CTA = 60 ◦ (if ∠C = 120 ◦ , take T ≡ C). In this case it is shown as above that

2S ≤ (AT + BT −CT )DE √ and

2 (AT + BT −CT ) 2 =a 2 +b 2 +c 2 +4 3S , and the inequality follows as before.

15. Denote by d (PQR) the diameter of a triangle PQR. It is clear that d(PQR) · m (PQR) = 2S PQR . So if the point X lies inside the triangle ABC or on its bound- ary, we have d (ABX), d(BCX), d(CAX) ≤ d(ABC), which implies

2S BCX 2S CAX (ABX) + m(BCX) + m(CAX) =

2S

ABX

d (ABX) d (BCX) d (CAX) 2S ABX + 2S BCX + 2S CAX

d (ABC)

2S ABC

= m(ABC).

d (ABC)

If X is outside △ABC but inside the angle BAC, consider the point Y of in- tersection of AX and BC. Then m (ABX) + m(BCX ) + m(CAX ) ≥ m(ABY ) + m (BCY ) + m(CAY ) ≥ m(ABC). Also, if X is inside the opposite angle of ∠BAC (i.e., ∠DAE, where B (D, A, B) and B(E, A,C)), then m(ABX) + m(BCX) + m (CAX) ≥ m(BCX) ≥ m(ABC). Since these are essentially all possible different positions of point X, we have finished the proof.

16. Let S n = {A = (a 1 ,...,a n )|0≤a i < i}. For each A = (a 1 ,...,a n ), denote

A ′ = (a 1 ,...,a n −1 ), so we can write A = (A ′ ,a n ). The proof of the statement from the problem will be given by induction on n. For n = 2 there are two pos- sibilities for A 0 , so one directly checks that A 2 =A 0 . Now assume that n ≥3 and that A 0 = (A ′ 0 ,a 0n )∈S n . It is clear that then any A i is in S n too. By the induction hypothesis there exists k ∈ N such that A ′ k =A ′ k +2 =A ′ k +4 = ··· and

A ′ k +1 =A ′ k +3 = ···. Observe that if we increase (decrease) a kn ,a k +1,n will de- crease (respectively increase), and this will also increase (respectively decrease)

a k +2,n . Hence a kn ,a k +2,n ,a k +4,n , . . . is monotonically increasing or decreasing,

580 4 Solutions and since it is bounded (by 0 and n − 1), it follows that we will eventually have

a k +2i,n =a k +2i+2,n = ···. Consequently A k +2i =A k +2i+2 .

17. We introduce the rotation operation Rot to the left by one, so that Step j = Rot −j ◦ Step 0 ◦ Rot j . Now writing Step ∗ = Rot ◦ Step 0 , the problem is trans- formed into the question whether there is an M (n) such that all lamps are on again after M (n) successive applications of Step ∗ .

We operate in the field Z 2 , representing off by 0 and on by 1. So if the status of L j at some moment is given by v j ∈Z 2 , the effect of Step j is that v j is replaced by v j +v j −1 . With the n-tuple v 0 ,...,v n −1 we associate the polynomial

P (x) = v x n −1 +v x n n −1 0 −2 +v 1 x n −3 + ··· + v n −2 . By means of Step ∗ , this polynomial is transformed into the polynomial Q (x) over

Z of degree less than n that satisfies Q(x) ≡ xP(x) (mod x n +x n −1 + 1). From now on, the sign ≡ always stands for congruence with this modulus.

(a) It suffices to show the existence of M (n) with x M (n) ≡ 1. Because the num- ber of residue classes is finite, there are r , q, with r < q, such that x q ≡x r , i.e., x r (x q −r − 1) = 0. One can take M(n) = q − r. (Or simply note that there are only finitely many possible configurations; since each operation is bijective, the configuration that reappears first must be on, on, . . . , on.)

n (b) We shall prove that if n 2 =2 , then x ≡ 1. We have x ≡ (x n −1 + 1) n ≡ x 2 n −n + 1, because all binomial coefficients of order n = 2 k are even, apart

n 2 −1

n 2 n from the first one and the last one. Since also x 2 ≡x −1 +x n 2 −n , this is what we wanted.

n 2 n ≡ 1. We have x +1 −1 ≡ (x ) n −1 ≡ (x + x n ) n −1 ≡x n −1 +x n 2 −n (again by evenness of binomial coefficients of

(c) Now if n =2 k +1, we prove that x n 2 −n+1

order n −1=2 k ). Together with x n 2 ≡x n 2 −1 +x n 2 −n , this leads to x n 2 ≡ x n −1 .

18. Let B n

be the set of sequences with the stated property (S n = |B n |). We shall prove by induction on n that S

n ≥ 2 S n −1 for every n.

i ≥ 2 S i −1 , and consequently S i ≤ 3 S n . Let us consider the 2S n sequences obtained by putting 0 or 1 at the end of any sequence

Suppose that for every i

3 2 n ≤ n, S −i

from B n . If some sequence among them does not belong to B n +1 , then for some k ≥ 1 it can be obtained by extending some sequence from B n +1−6k by a sequence

of k terms repeated six times. The number of such sequences is 2 k S n +1−6k . Hence the number of sequences not satisfying our condition is not greater than

2 6k −1

+1−6k k ≤ ∑ 2 S n = S n

Therefore S n

n +1 is not smaller than 2S n 1 − 2 S n = 3 2 S n . Thus we have S n

19. Let s be the minimum number of nonzero digits that can appear in the b-adic representation of any number divisible by b n − 1. Among all numbers divisible by b n − 1 and having s nonzero digits in base b, we choose the number A with

4.34 Shortlisted Problems 1993 581 the minimum sum of digits. Let A =a 1 b n 1 + ··· + a s b n s , where 0 <a i ≤b−1

and n 1 >n 2 > ··· > n s . First, suppose that n i ≡n j (mod n), i 6= j. Consider the number

B =A−a i b n i −a j b n j

+ (a +kn

i +a j )b n j , with k chosen large enough so that n j + kn > n 1 : this number is divisible by

b n − 1 as well. But if a i +a j < b, then B has s − 1 digits in base b, which is impossible; on the other hand, a i +a j ≥ b is also impossible, for otherwise B would have sum of digits less for b − 1 than that of A (because B would have digits 1 and a i +a j − b in the positions n j + kn + 1, n j + kn). Therefore n i 6≡ n j if i 6= j.

Let n i ≡r i , where r i ∈ {0,1,...,n − 1} are distinct. The number C = a 1 b r 1 + ··· + a s b r s also has s digits and is divisible by b n − 1. But since C < b n , the only possibility is C =b n −1 which has exactly n digits in base b. It follows that s = n.

20. For every real x we shall denote by ⌊x⌋ and ⌈x⌉ the greatest integer less than or equal to x and the smallest integer greater than or equal to x respectively. h

c i The condition c i

i + nk i ∈ [1 − n,n] is equivalent to k i ∈I i = n − 1,1 − n . For every c l i , this interval contains two integers (not necessarily distinct), namely p i = 1 −c i

1 −c i

m −1 c ≤q

i = 1 − n i n . In order to show that there exist integers k i ∈I i with ∑ n k

i =1 i = 0, it is sufficient to show that ∑ i =1 i ≤0≤∑ i =1 q i . Since p i < 1 −c i n , we have

i <1− ∑ ≤ 1,

i =1

i =1 n

and consequently ∑ n i =1 p i ≤ 0 because the p i ’s are integers. On the other hand, q

i >− n implies

∑ q i >−

i =1

i =1 n

which leads to ∑ n i =1 q i ≥ 0. The proof is complete.

21. Assume that S is a circle with center O that cuts S i diametrically in points P i ,Q i ,

i ∈ {A,B,C}, and denote by r i , r the radii of S i and S respectively. Since OA is perpendicular to P A Q A , it follows by Pythagoras’s theorem that OA 2 + AP 2 A = OP 2 A , i.e., r 2 A + OA 2 =r 2 . Analogously r 2 B + OB 2 =r 2 and r 2 C + OC 2 =r 2 . Thus if O A ,O B ,O C are the feet of perpendiculars from O to BC, CA, AB respectively, then O C A 2 −O C B 2 =r 2 B −r 2 A . Since the left-hand side is a monotonic function of O C ∈ AB, the point O C is uniquely determined by the imposed conditions. The same holds for O A and O B . If A , B,C are not collinear, then the positions of O A ,O B ,O C uniquely determine the point O, and therefore the circle S also. On the other hand, if A , B,C are collinear, all one can deduce is that O lies on the lines l A ,l B ,l C through O A ,O B ,O C , perpendicular to BC ,CA, AB respectively.

582 4 Solutions Hence, l ,l ,l are parallel, so O can

be either anywhere on the line if these lines coincide, or nowhere if they don’t C S

coincide. So if there exists more than

one circle S, A , B,C lie on a line and O O C A the foot O ′ of the perpendicular from

A O B O to the line ABC is fixed. If X ,Y are the intersection points of S and

the line ABC, then r 2 = OX 2 = OA 2 +

r 2 A 2 and consequently O ′ X =O ′ A 2 +

A , which implies that X ,Y are fixed.

22. Let M be the point inside ∠ADB that C satisfies DM = DB and DM ⊥ DB.

Then ∠ADM T = ∠ACB and AD/DM = AC /CB. It follows that the triangles

ADM , ACB are similar; hence ∠CAD = ∠BAM (because ∠CAB = ∠DAM) and

D AB /AM = AC/AD. Consequently the

A B triangles CAD, BAM are similar and

M AB AB ·CD BM √

therefore AC = CD = √ CD . Hence

2BD

2. Let CT ,CU be the tangents at C to the circles ACD, BCD respect- ively. Then (in oriented angles) ∠TCU = ∠TCD + ∠DCU = ∠CAD + ∠CBD =

AC ·BD =

90 ◦ , as required. Second solution to the first part. Denote by E , F, G the feet of the perpendiculars

from D to BC ,CA, AB. Consider the pedal triangle EFG. Since FG = AD sin ∠A, from the sine theorem we have FG : GE : EF = (CD · AB) : (BD · AC) : (AD ·

√ ∠CAD = 90 implies that EF : EG =

BC ). Thus EG = FG. On the other hand, ∠EGF = ∠EGD + ∠DGF = ∠CBD +

2 : 1; hence the required ratio is 2. Third solution to the first part. Under inversion centered at C and with power

r 2 = CA · CB, the triangle DAB maps into a right-angled isosceles triangle

D ∗ A ∗ B ∗ , where

Thus D ∗

B ∗ :A ∗ B ∗ = 2 , and this is the required ratio.

23. Let the given numbers be a 1 ,...,a n . Put s =a 1 +···+a n and m = lcm(a 1 ,...,a n ) and write m =2 k r with k ≥ 0 and r odd. Let the binary expansion of r be r =2 k 0 +2 k 1 + ··· + 2 k t , with 0 =k 0 < ··· < k t . Adjoin to the set {a 1 ,...,a n } the numbers 2 k i s ,i = 1, 2, . . . ,t. The sum of the enlarged set is rs. Finally, ad- join rs , 2rs, 2 2 rs ,...,2 l −1 rs for l = max{k,k t }. The resulting set has sum 2 l rs , which is divisible by m and so by each of a j , and also by the 2 i s above and by rs , 2rs, . . . , 2 l −1 rs . Therefore this is a DS-set.

4.34 Shortlisted Problems 1993 583 Second solution. We show by induction that there is a DS-set containing 1 and

n . For n = 2, 3, take {1,2,3}. Assume that {1,n,b 1 ,...,b k } is a DS-set. Then {1,n + 1,n,2(n + 1)n,2(n + 1)b 1 , . . . , 2(n + 1)b k } is a DS-set too. For given a 1 ,...,a n let m be a sufficiently large common multiple of the a i ’s such that u = m − (a 1 + ··· + a n ) 6= a i for all i. There exist b 1 ,...,b k such that {1,u,b 1 ,...,b k } is a DS-set. It is clear that {a 1 ,...,a n , u, mu, mb 1 , . . . , mb k } is a DS -set containing a 1 ,...,a n .

24. By the Cauchy–Schwarz inequality, if x 1 ,x 2 ,...,x n and y 1 ,y 2 ,...,y n are positive numbers, then

Applying this to the numbers a , b, c, d and b + 2c + 3d, c + 2d + 3a, d + 2a + 3b , a + 2b + 3c (here n = 4), we obtain

b + 2c + 3d c + 2d + 3a d + 2a + 3b a + 2b + 3c (a + b + c + d) 2 2

4 (ab + ac + ad + bc + bd + cd) ≥ 3 The last inequality follows, for example, from

2 (a − b) 2 + (a − c) + ··· + (c −

d ) 2 ≥ 0. Equality holds if and only if a = b = c = d. Second solution. Putting A = b + 2c + 3d, B = c + 2d + 3a, C = d + 2a + 3b,

D = a + 2b + 3c, our inequality transforms into −5A + 7B +C+ D −5B + 7C + D + A

≥ 3 This follows from the arithmetic-geometric mean inequality, since B A + C B + D C +

24D

D ≥ 4, etc.

25. We need only consider the case a > 1 (since the case a < −1 is reduced to

a > 1 by taking a ′ = −a, x ′ i = −x i ). Since the left sides of the equations are nonnegative, we have x

i ≥− a > −1, i = 1,...,1000. Suppose w.l.o.g. that x 1 = max

i }. In particular, x 1 ≥x 2 ,x 3 . If x 1 ≥ 0, then we deduce that x 1000 ≥1⇒ x 1000 ≥ 1; further, from this we deduce that x 999 > 1 etc., so either x i > 1 for all

{x 2

i or x i < 0 for all i. (i) x > 1 for every i. Then x 1 2 implies x 2 i 2 ≥x 1 ≥x 2 , so x 2 ≥x 3 . Thus x 1 ≥x 2 ≥

··· ≥ x 1000 ≥x 1 , and consequently x √ 1 = ··· = x 1000 . In this case the only solution is x i = 1 2 (a + a 2 + 4) for all i.

(ii) x i < 0 for every i. Then x 1 3 implies x 2 ≥x 2 1 ≤x 3 ⇒x 2 ≤x 4 . Similarly, this leads to x 3 ≥x 5 , etc. Hence x 1 ≥x 3 ≥x 5 ≥ ··· ≥ x 999 ≥x 1 and x 2 ≤x 4 ≤ ··· ≤ x 2 , so we deduce that x 1 =x 3 = ··· and x 2 =x 4 = ··· . Therefore the

584 4 Solutions system is reduced to x 2 1 = ax 2 +1, x 2 2 = ax 1 +1. Subtracting these equations,

one obtains (x 1 −x 2 )(x 1 +x 2 + a) = 0. There are two possibilities: √ (1) If x 1 =x 2 , then x 1 =x 2 = ··· = 1 2 (a − a 2 + 4). (2) x 1 +x 2 + a = 0 is equivalent to x 2 1 + ax 1 + (a 2 − 1) = 0. The discrim- inant of the last equation is 4 − 3a 2 . Therefore if a > √ 2 3 , this case

yields no solutions, while if a ≤ √ 2 , we obtain x 1 = 1

2 (−a− 4 − 3a 2 ),

x 2 = 1 2 2 (−a + 4 − 3a ), or vice versa.

26. Set 176

f (a, b, c, d) = abc + bcd + cda + dab − abcd

= ab(c + d) + cd a +b−

ab .

27 If a +b− 176 a b ≤ 0, by the arithmetic-geometric inequality we have f (a,b,c,d) ≤

ab (c + d) ≤ 1 27 . On the other hand, if a +b− 176 27 ab > 0, the value of f increases if c, d are replaced by c +d c 2 +d , 2 . Consider now the following fourtuplets:

4 2 2 4 4 4 4 4 From the above considerations we deduce that for i = 0, 1, 2, 3 either f (P i )≤

f (P i +1 ), or directly f (P i ) ≤ 1/27. Since f (P 4 ) = 1/27, in every case we are led to

f (a, b, c, d) = f (P 0 )≤

Equality occurs only in the cases (0, 1/3, 1/3, 1/3) (with permutations) and (1/4, 1/4, 1/4, 1/4).

Remark. Lagrange multipliers also work. On the boundary of the set one of the numbers a , b, c, d is 0, and the inequality immediately follows, while for an extremum point in the interior, among a , b, c, d there are at most two distinct values, in which case one easily verifies the inequality.

4.35 Shortlisted Problems 1994 585