Solutions to the Shortlisted Problems of IMO 1982

4.23 Solutions to the Shortlisted Problems of IMO 1982

1. From f (1) + f (1) ≤ f (2) = 0 we obtain f (1) = 0. Since 0 < f (3) ≤ f (1) +

f (2) + 1, it follows that f (3) = 1. Note that if f (3n) ≥ n, then f (3n + 3) ≥

f (3n) + f (3) ≥ n + 1. Hence by induction f (3n) ≥ n holds for all n ∈ N. More- over, if the inequality is strict for some n, then it is so for all integers greater than n as well. Since f (9999) = 3333, we deduce that f (3n) = n for all n ≤ 3333. By the given condition, we have 3 f (n) ≤ f (3n) ≤ 3 f (n) + 2. Therefore f (n) = [ f (3n)/3] = [n/3] for n ≤ 3333. In particular, f (1982) = [1982/3] = 660.

2. Since K does not contain a lattice point other than O (0, 0), it is bounded by four lines u , v, w, x that pass through the points U(1, 0), V (0, 1), W (−1,0), X(0,−1) respectively. Let PQRS be the quadrilateral formed by these lines, where U ∈ SP,

V ∈ PQ, W ∈ QR, X ∈ RS. If one of the quadrants, say Q 1 , contains no vertices of PQRS, then K ∩Q 1 is contained in △OUV and hence has area less than 1/2. Consequently the area of K is less than 2. Let us now suppose that P , Q, R, S lie in different quadrants. One of the angles of

PQRS is at least 90 ◦ : let it be ∠P. Then S UPV ≤ PU · PV /2 ≤ (PU 2 + PV 2 )/4 ≤ UV 2 /4 = 1/2, which implies that S K ∩Q 1 <S OUPV ≤ 1. Hence the area of K is less than 4.

3. (a) By the Cauchy–Schwarz inequality we have x 2 0 /x 1 + ··· + x 2 n −1 /x n ·(x 1 + ··· + x n ) ≥ (x 0 + ···+ x n −1 ) 2 . Let us set X n −1 =x 1 +x 2 + ··· + x n −1 . Using x 0 = 1, the last inequality can be rewritten as

1 +x n /X n −1 Since x n ≤x n −1 ≤ ··· ≤ x 1 , it follows that X n −1 ≥ (n− 1)x n . Now (1) yields

X n −1 +x n

X n −1 +x n

1 + ··· + x n −1 /x n ≥ 4(n − 1)/n, which exceeds 3.999 for n > 4000. (b) The sequence x n = 1/2 n obviously satisfies the required condition.

x 2 0 /x

Second solution to part (a). For each n ∈ N, let us find a constant c n such that the inequality x 2 0 /x

1 + ··· + x n −1 /x n ≥c n x 0 holds for any sequence x 0 ≥x 1 ≥ ··· ≥ x n > 0.

For n = 1 we can take c 1 = 1. Assuming that c n exists, we have x 2 x 2 x 2 0 2 1 n x 0 q 2 √

Thus we can take c n +1 = 2√c n . Then inductively c n =2 2 −1/2 n −2 , and since c n →

4 as n → ∞, the result follows. Third solution. Since {x n } is decreasing, there exists lim n →∞ x n = x ≥ 0. If x > 0,

then x 2 n −1 /x n ≥x n ≥ x holds for each n, and the result is trivial. If otherwise x = 0, then we note that x 2 n −1 /x n ≥ 4(x n −1 −x n ) for each n, with equality if and only if x n −1 = 2x n . Hence

462 4 Solutions

lim

x ≥ lim n →∞ ∑ (x k −1 −x k ) = 4x 0 k = 4. =1 k k =1

k −1

Equality holds if and only if x n −1 = 2x n for all n, and consequently x n = 1/2 n .

4. Suppose that a satisfies the requirements of the problem and that x, qx, q 2 x , q 3 x are the roots of the given equation. Then x 6= 0 and we may assume that |q| > 1, so that |x| < |qx| < |q 2 x | < |q 3 x |. Since the equation is symmetric, 1/x is also a root and therefore 1 /x = q 3 x , i.e., q =x −2/3 . It follows that the roots are x ,x 1 /3 ,x −1/3 ,x −1 . Now by Viète’s formula we have x +x 1 /3 +x −1/3 +x −1 =

a /16 and x 4 /3 +x 2 /3 +2+x −2/3 +x −4/3 = (2a + 17)/16. On setting z = x 1 /3 + x −1/3 these equations become

z 3 − 2z = a/16,

(z 2 − 2) 2 +z 2 − 2 = (2a + 17)/16. Substituting a = 16(z 3 4 3 − 2z) in the second equation leads to z 2 − 2z − 3z + 4z +

15 /16 = 0. We observe that this polynomial factors as (z + 3/2)(z − 5/2)(z 2 − z

1 − 1/4). Since |z| = |x /3 +x −1/3 | ≥ 2, the only viable value is z = 5/2. Conse- quently a = 170 and the roots are 1/8, 1/2, 2, 8.

5. Notice that △A 5 B 4 A 4 ∼ = △A 3 B 2 A 2 .

We know that ∠A 5 A 3 A 2 = 90 ◦ and

B 4 that ∠A 2 B 4 A 4 is equal to the sum of

the angles ∠A 2 B 4 A 3 and ∠A 3 B 4 A 4 .

Clearly, ∠A 2 B 4 A 3 = 90 − ∠B 2 A 2 A 3

and ∠A 3 B 4 A 4 = ∠B 4 A 5 A 4 + ∠A 5 A 4 B 4 .

Hence we conclude that ∠A 2 B 4 A 4 =

90 ◦ + ∠B A A

4 A 5 A 4 = 120 ◦ . Hence B 4 be-

longs to the circle with center A √ 3 and radius A 3 A 4 , so A 3 A 4 =A 3 B 4 . Thus λ =

A 3 B 4 /A 3 A 5 =A 3 A 4 /A 3 A 5 = 1/ 3.

6. Denote by d (U,V ) the distance between points or sets of points U and V . For P , Q ∈ L we shall denote by L PQ the part of L between points P and Q and by l PQ the length of this part. Let us denote by S i (i = 1, 2, 3, 4) the vertices of S and

by T i points of L such that S i T i ≤ 1/2 in such a way that l A 0 T 1 is the least of the

l A 0 T i ’s, S 2 and S 4 are neighbors of S 1 , and l A 0 T 2 <l A 0 T 4 .

Now we shall consider the points of the segment S 1 S 4 . Let D and E be the sets of points defined as follows: D = {X ∈ [S 1 S 4 ] | d(X,L A 0 T 2 ) ≤ 1/2} and E = {X ∈ [S 1 S 4 ] | d(X,L T 2 A n ) ≤ 1/2}. Clearly D and E are closed, nonempty (indeed, S 1 ∈ D and S 4 ∈ E) subsets of [S 1 S 4 ]. Since their union is a connected set S 1 S 4 , it follows that they must have a nonempty intersection. Let P ∈ D ∩ E. Then there exist points X ∈L A 0 T 2 and Y ∈L T 2 A n such that d (P, X) ≤ 1/2, d(P,Y ) ≤ 1/2, and consequently d (X ,Y ) ≤ 1. On the other hand, T 2 lies between X and Y on L, and thus L XY =L XT 2 +L T 2 Y ≥ XT 2 +T 2 Y ≥ (PS 2 − XP − S 2 T 2 ) + (PS 2 −YP− S 2 T 2 ) ≥ 99 + 99 = 198.

4.23 Shortlisted Problems 1982 463

7. Let a , b, ab be the roots of the cubic polynomial p(x) = (x − a)(x − b)(x − ab). Observe that

2p (−1) = −2(1 + a)(1 + b)(1 + ab); p (1) + p(−1) − 2(1 + p(0)) = −2(1 + a)(1 + b).

The statement of the problem is trivial if both the expressions are equal to zero. Otherwise, the quotient

p (1)+p(−1)−2(1+p(0)) = 1 + ab is rational and con- sequently ab is rational. But since (ab) 2 = −p(0) is an integer, it follows that ab is also an integer. This completes the proof.

2p (−1)

8. Let F be the given figure. Consider any chord AB of the circumcircle γ that supports F . The other supporting lines to F from A and B intersect γ again at D and C respectively so that ∠DAB = ∠ABC = 90 ◦ . Then ABCD is a rectangle, and hence CD must support F as well, from which it follows that F is inscribed in the rectangle ABCD touching each of its sides. We easily conclude that F is the intersection of all such rectangles. Now, since the center O of γ is the center of symmetry of all these rectangles, it must be so for their intersection F as well.

9. Let X and Y be the midpoints of the segments AP and BP. Then DY PX is a parallelogram. Since X and Y are the

C circumcenters of the triangles APM and BPL, we conclude that X M =

L XP = DY and Y L = Y P = DX . Fur-

P ∠PX M = ∠DXP + 2∠PAM. Simi-

thermore, we have ∠DX M = ∠DXP +

X larly, ∠DY L Y = ∠DY P + 2∠PBL hence ∠DX M = ∠DY L. Therefore, the trian-

gles DX M and LY D are congruent, im- A D B plying DM = DL.

10. If the two balls taken from the box are both white, then the number of white balls decreases by two; otherwise, it remains unchanged. Hence the parity of the number of white balls does not change during the procedure. Therefore if p is even, the last ball cannot be white; the probability is 0. If p is odd, the last ball has to be white; the probability is 1.

11. (a) Suppose {a 1 ,a 2 ,...,a n } is the arrangement that yields the maximal value Q max of Q. Note that the value of Q for the rearrangement {a 1 , ...,a i −1 ,

a j ,a j −1 , ...,a i ,a j +1 , ...,a n } equals Q max − (a i −a j )(a i −1 −a j +1 ), where

1 < i < j < n. Hence (a i −a j )(a i −1 −a j +1 ) ≥ 0 for all 1 < i < j < n.

We may suppose w.l.o.g. that a 1 = 1. Let a i = 2. If 2 < i < n, then (a 2 −

a i )(a 1 −a i +1 ) < 0, which is impossible. Therefore i is either 2 or n; let w.l.o.g. a n = 2. Further, if a j = 3 for 2 < j < n, then (a 1 −a j +1 )(a 2 −

a j ) < 0, which is impossible; therefore a 2 = 3. Continuing this argument we obtain that A = {1,3,5,...,2[(n − 1)/2] + 1,2[n/2],...,4,2}. (b) A similar argument leads to the minimizing rearrangement {1,n,2, n −

1 , . . . , [n/2] + 1}.

464 4 Solutions

12. Let y be the line perpendicular to L passing through the center of C. It can be shown by a continuity argument that there exists a point Y ∈ y such that an

inversion Ψ centered at Y maps C and L onto two concentric circles b C and b L . Let b X denote the image of an object X under Ψ . Then the circles b C i touch b C externally and b L internally, and all have the same radius. Let us now rotate the picture around the common center Z of b C and b L so that c C 3 passes through Y . Applying the inversion Ψ again on the picture thus obtained, b C and b L go back to

C and L, but c C 3 goes to a line C ′ 3 parallel to L, while the images of c C 1 and c C 2 go to two equal circles C ′ 1 and C ′ 2 touching L, C ′ 3 , and C. This way we have achieved that C 3 becomes a line.

C ′ Denote by O 1 ,O 2 , O respectively the

centers of the circles C ′ 1 ,C ′ 2 ,C and by T

xx

O 1 T O 2 C and C ′

the point of tangency of the circles C ′ 1 C ′

2 . If x is the common radius of the

circles C ′ and C ′ 1 2 , then from △O 1 TO

2 +x 2 = (x + 1) (x − 1) 2 , L and thus x = 4. Hence the distance of O from L equals 2x − 1 = 7.

we obtain that

13. The points S 1 ,S 2 ,S 3 clearly lie

on the inscribed circle. Let c XY de- note the oriented arc XY . The arcs

T 1 T d 2 S 1 and d T 1 T 3 are equal, since they

are symmetric with respect to the S 1 bisector of ∠A 1 . Similarly, d

3 T 2 = S d 2 T 1 . Hence d T 3 S 1 =d T 3 T 2 +d T 2 S 1 =

S d 2 T 1 +d T 1 T 3 =d S S

2 T 3 . It follows that

S 1 S 2 is parallel to A 1 A 2 , and con-

sequently S 1 S 2 kM 1 M 2 . Analogously S 1 S 3 kM 1 M 3 and S 2 S 3 kM 2 M 3 . Since the circumcircles of △M 1 M 2 M 3 and △S 1 S 2 S 3 are not equal, these triangles are not congruent and hence they must be homothetic. Then all the lines M i S i pass through the center of homothety.

Second solution. Set the complex plane so that the incenter of △A 1 A 2 A 3 is the unit circle centered at the origin. Let t i ,s i respectively denote the complex numbers of modulus 1 corresponding to T i ,S i . Clearly t 1 t 1 =t 2 t 2 =t 3 t 3 = 1. Since T 2 T 3 and T 1 S 1 are parallel, we obtain t 2 t 3 =t 1 s 1 , or s 1 =t 2 t 3 t 1 . Similarly s 2 =t 1 t 3 t 2 ,s 3 =t 1 t 2 t 3 , from which it follows that s 2 −s 3 =t 1 (t 3 t 2 −t 2 t 3 ). Since the number in parentheses is strictly imaginary, we conclude that OT 1 ⊥S 2 S 3 and consequently S 2 S 3 kA 2 A 3 . We proceed as in the first solution.

14. (a) If any two of A 1 ,B 1 ,C 1 ,D 1 coincide, say A 1 ≡B 1 , then ABCD is inscribed in a circle centered at A 1 and hence all A 1 ,B 1 ,C 1 ,D 1 coincide. Assume now the opposite, and let w.l.o.g. ∠DAB + ∠DCB < 180 ◦ . Then A is outside the circumcircle of △BCD, so A 1 A >A 1 C . Similarly, C 1 C >C 1 A . Hence the perpendicular bisector l AC of AC separates points A 1 and C 1 .

Since B 1 ,D 1 lie on l AC , this means that A 1 and C 1 are on opposite sides

4.23 Shortlisted Problems 1982 465

B 1 D 1 . Similarly one can show that B 1 and D 1 are on opposite sides of

A 1 C 1 . (b) Since A 2 B 2 ⊥C 1 D 1 and C 1 D 1 ⊥ AB, it follows that A 2 B 2 k AB. Similarly

A 2 C 2 k AC, A 2 D 2 k AD, B 2 C 2 k BC, B 2 D 2 k BD, and C 2 D 2 k CD. Hence △A 2 B 2 C 2 ∼ △ABC and △A 2 D 2 C 2 ∼ △ADC, and the result follows.

15. Let a = k/n, where n, k ∈ N, n ≥ k. Putting t n = s, the given inequality becomes

1 −t k

1 −t n ≤ (1 + t n ) k /n−1 , or equivalently k (1 + t + ··· + t −1 ) n (1 + t n ) n −k

n −1 ≤ (1 + t + ··· + t n ) . This is clearly true for k = n. Therefore it is enough to prove that the left-hand

side of the above inequality is an increasing function of k. We are led to show that (1 + t + ··· + t k −1 ) n (1 + t n ) n −k ≤ (1 + t + ··· + t k ) n (1 + t n ) n −k−1 . This is

equivalent to 1 +t n ≤A n , where A = 1 +t+···+t 1 k +t+···+t k −1 . But this easily follows, since

A n −t n = (A − t)(A n −1 +A n −2 t + ··· + t n −1 )

1 + t + ··· + t −1 ≥ (A − t)(1 + t + ··· + t −1 )=

≥ 1. Remark. The original problem asked to prove the inequality for real a.

1 + t + ··· + t k −1

16. It is easy to verify that whenever (x, y) is a solution of the equation x 3 − 3xy 2 + y 3 = n, so are the pairs (y−x,−x) and (−y,x−y). No two of these three solutions are equal unless x = y = n = 0. Observe that 2891 ≡ 2 (mod 9). Since x 3 ,y 3 ≡ 0,±1 (mod 9), x 3 − 3xy 2 +y 3 cannot give the remainder 2 when divided by 9. Hence the above equation for n = 2891 has no integer solutions.

17. Let A be the origin of the Cartesian plane. Suppose that BC : AC = k and that (a, b) and (a 1 ,b 1 ) are coordinates of the points C and C 1 , respectively. Then the coordinates of the point B are (a, b) + k(−b,a) = (a − kb,b + ka), while the coordinates of B 1 are (a 1 ,b 1 ) + k(b 1 , −a 1 ) = (a + kb 1 ,b 1 − ka 1 ). Thus the lines

+kb BC )

1 1 y −b 1 = −(a−kb) and x −a = −(a 1 y 1 −(b+ka) y −b y −(b 1 −ka 1 ) respectively. After multiplying, these equations transform into the forms

and CB are given by the equations x −a 1 x

BC 1 : kax + kby = kaa 1 + kbb 1 + ba 1 − ab 1 − (b − b 1 )x + (a − a 1 )y CB 1 : ka 1 x + kb 1 y = kaa 1 + kbb 1 + ba 1 − ab 1 − (b − b 1 )x + (a − a 1 )y.

The coordinates (x 0 ,y 0 ) of the point M satisfy these equations, from which we deduce that kax

0 + kby 0 = ka 1 x 0 + kb 1 y 0 . This yields 0 y b 0 =− 1 a −b 1 −a , implying that the lines CC 1 and AM are perpendicular.

18. Set the coordinate system with the axes x , y, z along the lines l 1 ,l 2 ,l 3 respectively. The coordinates (a, b, c) of M satisfy a 2 +b 2 +c 2 =R 2 , and so S M is given by the equation

2 2 (x − a) 2 + (y − b) + (z − c) =R 2 . Hence the coordinates of P 1 are (x, 0, 0) with (x − a) 2 +b 2 +c 2 =R 2 , implying that either x = 2a or x = 0.

466 4 Solutions Thus by the definition we obtain x = 2a. Similarly, the coordinates of P 2 and

P 3 are (0, 2b, 0) and (0, 0, 2c) respectively. Now, the centroid of △P 1 P 2 P 3 has the coordinates (2a/3, 2b/3, 2c/3). Therefore the required locus of points is the sphere with center O and radius 2R /3.

19. Let us set x = m/n. Since f (x) = (m + n)/ m 2 +n 2 = (x + 1)/ 1 +x 2 is a con- tinuous function of x, f (x) takes all values between any two values of f ; more- over, the corresponding x can be rational. This completes the proof.

Remark. Since f is increasing for x ≥ 1, 1 ≤ x < z < y implies f (x) < f (z) <

f (y).

20. Since MN is the image of AC under rotation about B for 60 ◦ , we have MN = AC. Similarly, PQ is the image of AC under rotation about D through 60 ◦ , from which it follows that PQ k MN. Hence either M,N,P,Q are collinear or MNPQ is a parallelogram.

4.24 Shortlisted Problems 1983 467