Solutions to the Shortlisted Problems of IMO 1984

4.25 Solutions to the Shortlisted Problems of IMO 1984

1. This is the same problem as (SL83-20).

2. (a) For m = t(t −1)/2 and n = t(t + 1)/2 we have 4mn− m− n = (t 2 − 1) 2 − 1. (b) Suppose that 4mn −m−n=p 2 , or equivalently, (4m − 1)(4n − 1) = 4p 2 +

1. The number 4m − 1 has at least one prime divisor, say q, that is of the form 4k + 3. Then 4p 2 ≡ −1 (mod q). However, by Fermat’s theorem we have

1 q −1 = 4p 2 −1

2 (mod q) , which is impossible since (q − 1)/2 = 2k + 1 is odd.

≡ (2p) q 2 ≡ (−1) −1

3. From the equality n =d 2 6 +d 2 7 − 1 we see that d 6 and d 7 are relatively prime and

d 7 2 |d 6 − 1 = (d 6 − 1)(d 6 + 1), d

6 |d 7 − 1 = (d 7 − 1)(d 7 + 1). Suppose that d 6 = ab, d 7 = cd with 1 < a < b, 1 < c < d. Then n has 7 divisors

smaller than d 7 , namely 1 , a, b, c, d, ab, ac, which is impossible. Hence, one of the two numbers d 6 and d 7 is either a prime p or the square p 2 of a prime p 6= 2. Let it be d i , {i, j} = {6,7}; then d i | (d j − 1)(d j + 1) implies that d j ≡ ±1 (mod d i ), and consequently (d 2 i − 1)/d j ≡ ±1 as well. But either d j or (d 2 i − 1)/d j is less than d i , and therefore equals d i − 1 or equals 1. The only nontrivial possibilities are (d 2 i − 1)/d j = 1 and d j =d i ± 1. In the first case we get d i <d j ; hence

d 7 =d 2 6 − 1 = (d 6 − 1)(d 6 + 1); hence d 6 + 1 is a divisor of n that is between d 6 and d 7 . This is impossible. We thus conclude that d 7 =d 6 + 1. Setting d 6 = x,

d 7 = x + 1 we obtain that n = x 2 + (x + 1) 2 − 1 = 2x(x + 1) is even. (i) Assume that one of x , x+1 is a prime p. The other one has at most 6 divisors and hence must be of the form 2 3 ,2 4 ,2 5 , 2q, 2q 2 , 4q, where q is an odd prime. The numbers 2 3 and 2 4 are easily eliminated, while 2 5 yields the solution x = 31, x + 1 = 32, n = 1984. Also, 2q is eliminated because n = 4pq then has only 4 divisors less than x; 2q 2 is eliminated because n = 4pq 2 has at least 6 divisors less than x; 4q is also eliminated because n = 8pq has

6 divisors less than x. (ii) Assume that one of x , x + 1 is p 2 . The other one has at most 5 divisors (p excluded), and hence is of the form 2 3 ,2 4 , 2q, where q is an odd prime. The number 2 3 yields the solution x = 8, x + 1 = 9, n = 144, while 2 4 is easily eliminated. Also, the number 2q is eliminated because n = 4p 2 q has

6 divisors less than x. Thus there are two solutions in total: 144 and 1984.

4. Consider the convex n-gon A 1 A 2 ...A n (the indices are considered modulo n). For any diagonal A i A j we have A i A j +A i +1 A j +1 >A i A i +1 +A j A j +1 . Summing all such n (n − 3)/2 inequalities, we obtain 2d > (n − 3)p, proving the first in- equality.

Let us now prove the second inequality. We notice that for each diagonal A i A i +j (we may assume w.l.o.g. that j ≤ [n/2]) the following relation holds:

A i A i +j <A i A i +1 + ··· + A i + j−1 A i +j . (1)

4.25 Shortlisted Problems 1984 477 If n = 2k + 1, then summing the inequalities (1) for j = 2, 3, . . . , k and i =

1 , 2, . . . , n yields d < (2 + 3 + ···+ k)p = ([n/2][n + 1/2] − 2) p/2. If n = 2k, then summing the inequalities (1) for j = 2, 3, . . . , k − 1, i = 1,2,...,n and for j = k, i = 1, 2, . . . , k again yields d < (2 + 3 + ··· + (k − 1) + k/2)p =

2 ([n/2][n + 1/2] − 2) p.

5. Let f (x, y, z) = xy + yz + zx − 2xyz. The first inequality follows immediately by adding xy ≥ xyz, yz ≥ xyz, and zx ≥ xyz (in fact, a stronger inequality xy + yz + zx − 9xyz ≥ 0 holds).

Assume w.l.o.g. that z is the smallest of x , y, z. Since xy ≤ (x + y) 2 /4 = (1 − z) 2 /4 and z ≤ 1/2, we have

xy + yz + zx − 2xyz = (x + y)z + xy(1 − 2z) (1 − z) 2 (1 − 2z)

≤ (1 − z)z +

7 (1 − 2z)(1 − 3z) 2 7

6. From the given recurrence we infer f n +1 −f n =f n −f n −1 + 2. Consequently,

f n +1 −f n =(f 2 −f 1 ) + 2(n − 1) = c − 1 + 2(n − 1). Summing up for n =

1 , 2, . . . , k − 1 yields the explicit formula

f k =f

1 + (k − 1)(c − 1) + (k − 1)(k − 2) = k + bk − b, where b = c − 4. Now we easily obtain f k f k +1 =k 4 + 2(b + 1)k 3 + (b 2 +b+

1 )k 2 − (b 2 + b)k − b. We are looking for an r for which the last expression equals

f r . Setting r =k 2 + pk + q we get by a straightforward calculation that p = b + 1, q = −b, and r = k 2 + (b + 1)k − b = f k + k. Hence f k f k +1 =f f k +k for all k.

7. It clearly suffices to solve the problem for the remainders modulo 4 (16 of each kind).

(a) The remainders can be placed as shown in Figure 1 , so that they satisfy the conditions.

p 01230123 qrs 10321032

23012301 t 32103210

Fig. 2 (b) Suppose that the required numbering exists. Consider a part of the chess-

Fig. 1

board as in Figure 2 . By the stated condition, all the numbers p +q+r+ s , q + r + s + t, p + q + r + t, p + r + s + t give the same remainder mod- ulo 4, and so do p , q, r, s. We deduce that all numbers on black cells of the

478 4 Solutions board, except possibly the two corner cells, give the same remainder, which

is impossible.

8. Suppose that the statement of the problem is false. Consider two arbitrary circles R = (O, r) and S = (O, s) with 0 < r < s < 1. The point X ∈ R with α (X ) = r (s − r) < 2 π satisfies that C (X ) = S. It follows that the color of the point X does not appear on S. Consequently, the set of colors that appear on R is not the same as the set of colors that appear on S. Hence any two distinct circles with center at O and radii less than 1 have distinct sets of colors. This is a contradiction, since there are infinitely many such circles but only finitely many possible sets of colors.

9. Let us show first that the system has at most one solution. Suppose that (x, y, z) and (x ′ ,y ′ ,z ′ ) are two distinct solutions and that w.l.o.g. x < x ′ . Then the second and third equation imply that y

′ , but then √ √ √

y −a+ √ z −a> y ′ −a+ z ′ − a, which is a contradiction. We shall now prove the existence of at least one solution. Let P be an arbitrary √ point in the plane and K

b , PM , L, M points such that PK = √ = = c , and ∠KPL = ∠LPM = ∠MPK = 120 ◦ . The lines through K , L, M perpendicular respectively to PK , PL, PM form an equilateral triangle ABC, where K ∈ BC, L ∈

√ a , PL

2 AC √ , and M ∈ AB. Since its area equals AB 3 /4 = S △BPC +S △APC +S △APB = √

√ AB a + b + c /2, it follows that AB = 1. Therefore x = PA 2 ,y = PB 2 , and

is a solution of the system (indeed, √ y −a+ z −a= PB 2 − PK 2 + PC 2 − PK 2 = BK +CK = 1, etc.).

z = PC 2

10. Suppose that the product of some five consecutive numbers is a square. It is easily seen that among them at least one, say n, is divisible neither by 2 nor 3. Since n is coprime to the remaining four numbers, it is itself a square of a number m of the form 6k

± 1. Thus n = (6k ± 1) 2 = 24r + 1, where r = k(3k ± 1)/2. Note that neither of the numbers 24r − 1, 24r + 5 is one of our five consecutive

numbers because it is not a square. Hence the five numbers must be 24r , 24r +

1 , . . . , 24r +4. However, the number 24r +4 = (6k ±1) 2 +3 is divisible by 6r +1, which implies that it is a square as well. It follows that these two squares are 1

and 4, which is impossible.

11. Suppose that an integer x satisfies the equation. Then the numbers x −a 1 ,x−

a 2 ,...,x−a 2n are 2n distinct integers whose product is 1 · (−1) · 2 · (−2)··· n · (−n). From here it is obvious that the numbers x −a 1 ,x−a 2 ,...,x−a 2n are some reordering of the numbers −n,−n +1,...,−1,1,...,n −1,n. It follows that their sum is 0, and therefore x = (a 1 +a 2 + ··· + a 2n )/2n. This is the only solution if {a 1 ,a 2 ,...,a 2n } = {x − n,...,x − 1,x + 1,...,x + n} for some x ∈ N. Otherwise there is no solution.

12. By the binomial formula we have (a + b) 7 7 −a 7 −b = 7ab[(a 5 +b 5 ) + 3ab(a 3 +b 3 ) + 5a 2 b 2 (a + b)]

= 7ab(a + b)(a 2 + ab + b 2 ) 2 .

4.25 Shortlisted Problems 1984 479 Thus it will be enough to find a and b such that 7 ∤ a, b and 7 3 |a 2 + ab + b 2 . Such

numbers must satisfy (a + b) 2 >a 2 + ab + b 2 ≥7 3 = 343, implying a + b ≥ 19. Trying a = 1 we easily find the example (a, b) = (1, 18).

13. Let Z be the given cylinder of radius r, altitude h, and volume π r 2 h = 1, k 1 and

k 2 the circles surrounding its bases, and V the volume of an inscribed tetrahedron ABCD . We claim that there is no loss of generality in assuming that A , B,C, D all lie on

k 1 ∪k 2 . Indeed, if the vertices A , B,C are fixed and D moves along a segment EF parallel to the axis of the cylinder (E ∈k 1 ,F ∈k 2 ), the maximum distance of

D from the plane ABC (and consequently the maximum value of V ) is achieved either at E or at F. Hence we shall consider only the following two cases: (i) A ,B∈k 1 and C ,D∈k 2 . Let P , Q be the projections of A, B on the plane of k 2 , and R , S the projections of C, D on the plane of k 1 , respectively. Then

V is one-third of the volume V ′ of the prism ARBSCPDQ with bases ARBS and CPDQ. The area of the quadrilateral ARBS inscribed in k 1 does not exceed the area of the square inscribed therein, which is 2r 2 . Hence 3V =

V ′ ≤ 2r 2 h = 2/ π . (ii) A , B,C ∈ k 1 and D ∈k 2 . The area of the triangle ABC does not exceed the √

area of an equilateral triangle inscribed in k √ √ 1 , which is 3 3r 2 /4. Conse- quently, V

14. Let M and N be the midpoints of AB and CD, and let M ′ ,N ′

be their projections on CD and AB, respectively. We know that MM ′ = AB, and hence

1 1 S ABCD =S AMD +S BMC +S CMD = (S ABD +S ABC )+ AB ·CD.

2 4 The line AB is tangent to the circle with diameter CD if and only if NN ′ = CD/2, or equivalently,

1 1 S ABCD =S AND +S BNC +S ANB = (S BCD +S ACD )+ AB ·CD.

2 4 By (1), this is further equivalent to S ABC +S ABD =S BCD +S ACD . But since S ABC + S ACD =S ABD +S BCD =S ABCD , this reduces to S ABC =S BCD , i.e., to BC k AD.

15. (a) Since rotation by 60 ◦ around A transforms the triangle CAF into △EAB, it follows that ∡ (CF, EB) = 60 ◦ . We similarly deduce that ∡ (EB, AD) = ∡ (AD, FC) = 60 ◦ . Let S be the intersection point of BE and AD. Since ∡CSE = ∡CAE = 60 ◦ , we have that EASC is cyclic. Therefore ∡ (AS, SC) =

60 ◦ = ∡(AD, FC), which implies that S lies on CF as well. (b) A rotation of EASC around E by 60 ◦ transforms A into C and S into a point T for which SE = ST = SC +CT = SC + SA. Summing the equality SE = SC + SA and the analogous equalities SD = SB + SC and SF = SA + SB yields the result.

16. From the first two conditions we can easily conclude that a + d > b + c (indeed, (d + a) 2 − (d − a) 2 = (c + b) 2 − (c − b) 2 = 4ad = 4bc and d − a > c − b > 0). Thus k > m.

480 4 Solutions From d =2 k − a and c = 2 m − b we get a(2 k − a) = b(2 m − b), or equivalently,

(b + a)(b − a) = 2 k (b − 2 −m a ). (1) Since 2 k −m a is even and b is odd, the highest power of 2 that divides the right-

hand side of (1) is m. Hence (b + a)(b − a) is divisible by 2 m but not by 2 m +1 ,

−a=2 m 2 q , where m 1 ,m 2 ≥ 1, m 1 +m 2 = m, and p , q are odd.

which implies b +a=2 m 1 p and b

Furthermore, b = (2 m 1 p +2 m 2 q )/2 and a = (2 m 1 p −2 m 2 q )/2 are odd, so either m 1 = 1 or m 2 = 1. Note that m 1 = 1 is not possible, since it would imply that

b −a = 2 m −1 q ≥2 m −1 , although b +c = 2 m and b < c imply that b < 2 m −1 . Hence m 2 = 1 and m 1 = m − 1. Now since a + b < b + c = 2 m , we obtain a +b=2 m −1 and b −a = 2q, where q is an odd integer. Substituting these into (1) and dividing both sides by 2 m we get

2 k −m a =2 m −2 . Since a is odd and k > m, it follows that a = 1.

q =2 m −2

+q−2 k −m a =⇒

Remark. Now it is not difficult to prove that all quadruples (a, b, c, d) that satisfy the given conditions are of the form (1, 2 m −1 − 1,2 m −1 + 1, 2 2m −2 − 1), where m ∈ N, m ≥ 3.

17. For any m = 0, 1, . . . , n −1, we shall find the number of permutations (x 1 ,x 2 , ..., x n ) with exactly k discordant pairs such that x n = n − m. This x n is a member of exactly m discordant pairs, and hence the permutation (x 1 ,...,x n −1 ) of the set {1,2,...,n} \ {m} must have exactly k − m discordant pairs: there are d(n −

1 , k − m) such permutations. Therefore

d (n, k) = d(n − 1,k) + d(n − 1,k − 1)···+ d(n − 1,k − n + 1) = d(n − 1,k) + d(n,k − 1)

(note that d (n, k) is 0 if k < 0 or k > n 2 ).

We now proceed to calculate d (n, 2) and d(n, 3). Trivially, d(n, 0) = 1. It fol- lows that d (n, 1) = d(n − 1,1)+d(n,0) = d(n −1,1)+1, which yields d(n,1) =

d (1, 1) + n − 1 = n − 1. Further, d (n, 2) = d(n − 1,2) + d(n,1) = d(n − 1,2) + n − 1 = d(2,2) + 2 + 3 + ··· + n − 1 = (n 2 − n − 2)/2. Finally, using the known formula 1 2 +2 2 + ··· + k 2 = k(k + 1)(2k + 1)/6, we have d

(n, 3) = d(n − 1,3) + d(n,2) = d(n − 1,3) + (n 2 − n − 2)/2 = d(2,3) + ∑ n i =3 (n 2 − n − 2)/2 = (n 3 − 7n + 6)/6.

18. Suppose that circles k 1 (O 1 ,r 1 ), k 2 (O 2 ,r 2 ), and k 3 (O 3 ,r 3 ) touch the edges of the angles ∠BAC, ∠ABC, and ∠ACB, respectively. Denote also by O and r the center and radius of the incircle. Let P be the point of tangency of the incircle with AB

and let F be the foot of the perpendicular from O 1 √ to OP. From △O 1 FO we ob- tain cot ( α /2) = 2 rr

1 /(r − r 1 ) and analogously cot( β /2) = 2 rr 2 /(r − r 2 ), cot ( γ /2) = 2√rr 3 /(r − r 3 ). We will now use a well-known trigonometric iden- tity for the angles of a triangle:

4.25 Shortlisted Problems 1984 481

γ cot + cot + cot = cot

2 2 2 2 · cot 2 · cot 2 (This identity follows from tan ( γ /2) = cot( α /2 + β /2) and the formula for the

cotangent of a sum.) Plugging in the obtained cotangents, we get

2√rr 1 2√rr 2 2√rr 3 2√rr 1 2√rr 2 2√rr 3 +

(r − 4)(r − 9) + 2(r − 1)(r − 9) + 3(r − 1)(r − 4) = 24r ⇒ 6(r − 1)(r − 11) = 0. Clearly, r = 11 is the only viable value for r.

19. First, we shall prove that the numbers in the nth row are exactly the numbers

1 1 1 1 n n −1

0 n 1 n 2 n n −1 The proof of this fact can be done by induction. For small n, the statement can

be easily verified. Assuming that the statement is true for some n, we have that the kth element in the (n + 1)st row is, as is directly verified,

Thus (1) is proved. Now the geometric mean of the elements of the nth row becomes:

0 ) + ( 1 ) +···+ ( n n −1

The desired result follows directly from substituting n = 1984.

20. Define the set S =R + r {1}. The given inequality is equivalent to ln b/ln a < ln (b + 1)/ln(a + 1). If b = 1, it is obvious that each a ∈ S satisfies this inequality. Suppose now that

b is also in S. Let us define on S a function f (x) = ln(x + 1)/lnx. Since ln (x + 1) > lnx and

1 /x > 1/x + 1 > 0, we have

f ′ x +1 (x) = −

ln x

ln (x+1)

for all x .

ln 2 x

482 4 Solutions Hence f is always decreasing. We also note that f (x) < 0 for x < 1 and that

f (x) > 0 for x > 1 (at x = 1 there is a discontinuity). Let us assume b > 1. From ln b/ln a < ln (b + 1)/ln (a + 1) we get f (b) > f (a). This holds for b > a or for a < 1. Now let us assume b < 1. This time we get f (b) < f (a). This holds for a < b or for a > 1.

Hence all the solutions to log a b < log a +1 (b + 1) are {b = 1,a ∈ S}, {a > b > 1}, {b > 1 > a}, {a < b < 1}, and {b < 1 < a}.

4.26 Shortlisted Problems 1985 483