Solutions to the Shortlisted Problems of IMO 1981

4.22 Solutions to the Shortlisted Problems of IMO 1981

1. Assume that the set {a − n + 1,a − n + 2,...,a} of n consecutive numbers satisfies the condition a

1 p α 2 2 r ...p r be the canonic representation of a, where p 1 <p 2 < ··· < p r are primes and

| lcm[a − n + 1,...,a − 1]. Let a = p α 1 α

α 1 , ··· , α r > 0. Then for each j = 1, 2, . . . , r, there exists m, m = 1, 2, . . . , n − 1,

α such that p j j | a − m, i.e., such that p j | m. Thus p j ≤ n − 1. If r = 1, then

a =p α 1 1 ≤ n − 1, which is impossible. Therefore r ≥ 2. But then there must exist two distinct prime numbers less than n; hence n ≥ 4. For n = 4, we must have p α 1 1 ,p α 2 2 ≤ 3, which leads to p 1 = 2, p 2 = 3, α 1 = α 2 =

1. Therefore a = 6, and {3,4,5,6} is a unique set satisfying the condition of the problem. For every n ≥ 5 there exist at least two such sets. In fact, for n = 5 we easily find two sets: {2,3,4,5,6} and {8,9,10,11,12}. Suppose that n ≥ 6. Let r,s,t be natural numbers such that 2 r ≤ n−1 < 2 r +1 ,3 s ≤ n−1 < 3 s +1 ,5 t ≤ n−1 < 5 t +1 . Taking a =2 r ·3 s and a =2 r ·5 t we obtain two distinct sets with the required property. Thus the answers are (a) n ≥ 4 and (b) n = 4.

2. Lemma. Let E, F, G, H, I, and K be points on edges AB, BC, CD, DA, AC, and BD of a tetrahedron. Then there is a sphere that touches the edges at these points if and only if

AE = AH = AI, BE = BF = BK, CF = CG = CI, DG = DH = DK.

(∗) Proof. The “only if” side of the equivalence is obvious.

We now assume (∗). Denote by

ε , φ , γ , η , ι , and κ planes through

E , F, G, H, I, K perpendicular to G AB , BC, CD, DA, AC and BD re-

spectively. Since the three planes ε , , and are not mutually parallel,

η C ι they intersect in a common point O.

Clearly, △AEO ∼ = △AHO ∼ = △AIO; B

hence OE = OH = OI = r, and the sphere σ (O, r) is tangent to AB, AD, AC. To prove that σ is also tangent to BC ,CD, BD it suffices to show that planes φ , γ , and κ also pass through O. Without loss of generality we can prove this for just φ . By the conditions for E , F, I, these are exactly the points of tangency of the incircle of △ABC and its sides, and if S is the incenter, then SE ⊥ AB, SF ⊥ BC, SI ⊥ AC. Hence ε , ι , and φ all pass through S and are perpendicular to the plane ABC, and consequently all share the line l through S perpendicular to ABC. Since l = ε ∩ ι , the point O will be situated on l, and hence φ will also contain O. This completes our proof of the lemma.

Let AH = AE = x, BE = BF = y, CF = CG = z, and DG = DH = w. If the sphere is also tangent to AC at some point I, then AI = x and IC = z. Using the stated lemma it suffices to prove that if AC = x + z, then BD = y + w.

454 4 Solutions

Let EF = FG = GH = HI = t, ∠BAD = α , ∠ABC = β , ∠BCD = γ , and ∠ADC = δ . We get

t 2 = EH 2 = AE 2 + AH 2 − 2 · AE · AH cos α = 2x 2 (1 − cos α ). We similarly conclude that t 2 = 2y 2 (1 − cos β ) = 2z 2 (1 − cos γ ) = 2w 2 (1 −

cos δ ). Further, using that AB = x + y, BC = y + z, cos β = 1 −t 2 /2y 2 , we obtain

2 2 2 2 2 x z AC = AB + BC − 2AB · BC cos β = (x − z) +t

+1 +1 . y

y Analogously, from the triangle ADC we get AC 2 = (x − z) 2 +t 2 (x/w + 1)(z/w +

1 ), which gives (x/y + 1)(z/y + 1) = (x/w + 1)(z/w + 1). Since f (s) = (x/s +

1 )(z/s + 1) is a decreasing function in s, it follows that y = w; similarly x = z. Hence CF = CG = x and DG = DH = y. Hence AC k EF and AC : t = AC : EF = AB : EB = (x + y) : y; i.e., AC = t(x + y)/y. Similarly, from the triangle ABD , we get that BD = t(x + y)/x. Hence if AC = x + z = 2x, it follows that 2x = t (x + y)/y ⇒ 2xy = t(x + y) ⇒ BD = t(x + y)/x = 2y = y + w. This completes the proof.

Second solution. Without loss of generality, assume that EF = 2. Consider the Cartesian system in which points O , E, F, G, H respectively have coordi- nates (0, 0, 0), (−1,−1,a), (1,−1,a), (1,1,a), (−1,1,a). Line AH is perpen- dicular to OH and AE is perpendicular to OE; hence from Pythagoras’s theo-

rem AO 2 = AH 2 + HO 2 = AE 2 + EO 2 = AE 2 + HO 2 , which implies AH = AE. Therefore the y-coordinate of A is zero; analogously the x-coordinates of B and D and the y-coordinate of C are 0. Let A have coordinates (x 0 , 0, z 1 ): then

−→ EA (x + 1, 1, z

−→ EO

EA −→ 0 −→ 1 − a) ⊥ (1, 1, −a), i.e., · EO =x 0 + 2 + a(a −z 1 ) = 0. Sim- ilarly, for B (0, y 0 ,z 2 ) we have y 0 + 2 + a(a − z 2 ) = 0. This gives us

a a We haven’t used yet that A (x 0 , 0, z 1 ), E(−1,−1,a) and B(0,y 0 ,z 2 ) are collinear,

so let A ′ ,B ′ ,E ′

be the feet of perpendiculars from A , B, E to the plane xy. The line A ′ B ′ , given by y 0 x +x 0 y =x 0 y 0 ,z = 0, contains the point E ′ (−1,−1,0), from which we obtain

(2) In the same way, from the points B and C we get relations similar to (1) and

(x 0 + 1)(y 0 + 1) = 1.

(2) and conclude that C has the coordinates C (−x 0 , 0, z 1 ). Similarly we get

D (0, −y 0 ,z 2 ). The condition that AC is tangent to the sphere √ √ σ (O, OE) is equiv- alent to z √ 1 = a 2 + 2, i.e., to x 0 =a a 2 + 2 − (a 2 + 2). But then (2) implies that y 0 = −a a 2 + 2 − (a 2

+ 2) and z 2 =− a 2 + 2, which means that the sphere σ is tangent to BD as well. This finishes the proof.

3. Denote max (a + b + c, b + c + d, c + d + e, d + e + f , e + f + g) by p. We have

(a + b + c) + (c + d + e) + (e + f + g) = 1 + c + e ≤ 3p,

4.22 Shortlisted Problems 1981 455 which implies that p ≥ 1/3. However, p = 1/3 is achieved by taking (a, b, c, d,

e ,f,g ) = (1/3, 0, 0, 1/3, 0, 0, 1/3). Therefore the answer is 1/3. Remark. In fact, one can prove a more general statement in the same way. Given

positive integers n , k, n ≥ k, if a 1 ,a 2 ,...,a n are nonnegative real numbers with sum 1, then the minimum value of max i =1,...,n−k+1 {a i +a i +1 + ··· + a i +k−1 } is

1 /r, where r is the integer with k(r − 1) < n ≤ kr.

4. We shall use the known formula for the Fibonacci sequence √

− (−1) n α −n ),

where α = . (1)

5 2 (a) Suppose that a f n +bf n +1 =f k n for all n, where k n > 0 is an integer de-

pending on n. By (1), this is equivalent to a( α n − (−1) n α −n ) + b( α n +1 + (−1) n α −n−1 )= α k n − (−1) k n α −k n , i.e.,

α k n −n =a+b α − α −2n (−1) n (a − b α −1 − (− α ) n −k n )→a+b α (2) as n → ∞. Hence, since k n is an integer, k n − n must be constant from some

point on: k n = n + k and α k =a+b α . Then it follows from (2) that α −k =

a −b α −1 , and from (1) we conclude that a f n +b f n +1 =f k +n holds for every n . Putting n = 1 and n = 2 in the previous relation and solving the obtained system of equations we get a =f k −1 ,b =f k . It is easy to verify that such a and b satisfy the conditions.

(b) As in (a), suppose that u f 2 +vf n 2 n +1 =f l n for all n. This leads to √

u +v α 2 5 α l n −2n − n = 2(u − v)(−1) α −2n

l n √ + (−1)

−(u α −4n +v α −4n−2

5 α −l n −2n )

as n → ∞. Thus u+ v α 2 = 5 α l n −2n , and l n − 2n = k is equal to a constant. Putting this into the above equation and multiplying by α 2n we get u −v→

0 as n → ∞, i.e., u = v. Finally, substituting n = 1 and n = 2 in u f 2 n + uf 2 n +1 =f l n we easily get that the only possibility is u = v = 1 and k = 1. It

is easy to verify that such u and v satisfy the conditions.

5. There are four types of small cubes upon disassembling: (1) 8 cubes with three faces, painted black, at one corner; (2) 12 cubes with two black faces, both at one edge; (3) 6 cubes with one black face; (4) 1 completely white cube.

All cubes of type (1) must go to corners, and be placed in a correct way (one of three): for this step we have 3 8 · 8! possibilities. Further, all cubes of type (2) must go in a correct way (one of two) to edges, admitting 2 12 · 12! possibilities; similarly, there are 4 6 · 6! ways for cubes of type (3), and 24 ways for the cube of type (4). Thus the total number of good reassemblings is 3 8 8! ·2 12 12! ·4 6 6! ·

456 4 Solutions

24, while the number of all possible reassemblings is 24 27 · 27!. The desired probability is 3 8 8! ·2 12 12! ·4 6 6! 24 ·24 27 ·27! . It is not necessary to calculate these numbers to

find out that the blind man practically has no chance to reassemble the cube in a right way: in fact, the probability is of order 1 .8 · 10 −37 .

6. Assume w.l.o.g. that n = degP ≥ degQ, and let P 0 = {z 1 ,z 2 ,...,z k }, P 1 = {z k +1 ,z k +2 ,...z k +m }. The polynomials P and Q match at k + m points z 1 ,z 2 , ...,z k +m ; hence if we prove that k + m > n, the result will follow. By the assumption,

P (x) = (x − z 1 ) α 1 ··· (x − z k ) α k = (x − z k +1 ) α k +1 ···(x − z k +m ) α k +m +1 for some positive integers α 1 ,..., α k +m . Let us consider P ′ (x). As we know, it is

divisible by (x − z i ) α i −1 for i = 1, 2, . . . , k + m; i.e.,

k +m

∏ (x − z i ) α i −1 |P ′ (x).

i =1

Therefore 2n k −k −m = deg∏ +m

i =1 (x−z i ) α i −1 ≤ degP ′ = n −1, i.e., k+m ≥ n+1, as we claimed.

7. We immediately find that f (1, 0) = f (0, 1) = 2. Then f (1, y + 1) = f (0, f (1, y)) = f (1, y) + 1; hence f (1, y) = y + 2 for y ≥ 0. Next we find that f (2,0) =

f (1, 1) = 3 and f (2, y + 1) = f (1, f (2, y)) = f (2, y) + 2, from which f (2, y) = 2y + 3. Particularly, f (2, 2) = 7. Further, f (3, 0) = f (2, 1) = 5 and f (3, y + 1) =

f (2, f (3, y)) = 2 f (3, y) + 3. This gives by induction f (3, y) = 2 y +3 − 3. For y = 3, f (3, 3) = 61. Finally, from f (4, 0) = f (3, 1) = 13 and f (4, y + 1) =

f (3, f (4, y)) = 2 f (4,y)+3 − 3, we conclude that

2 .. f .2 (4, y) = 2

(y + 3 twos).

8. Since the number k, k = 1, 2, . . . , n − r + 1, is the minimum in exactly n −k r −1 r - element subsets of {1,2,...,n}, it follows that

f (n, r) = −k n ∑ k .

1 n −r+1 n

r −1

k =1

Using the equality r +j

n −r n j −k r +i−1

n −r+1

r +1 r Therefore f (n, r) = (n + 1)/(r + 1).

j =0

r +1

4.22 Shortlisted Problems 1981 457

9. If we put 1 + 24a n =b 2 n , the given recurrent relation becomes

b n +1 = + n +b n =

3 +b n

i.e., b +1 = , (1)

2 where b 1 = 5. To solve this recurrent equation, we set c n =2 n −1 b n . From (1) we

obtain

c n +1 =c

n +3·2 −1 = ··· = c 1 + 3(1 + 2 + 2 2 + ··· + 2 n −1 ) = 5 + 3(2 n

− 1) = 3 · 2 n + 2.

Therefore b n =3+2 −n+2 and consequently

10. It is easy to see that partitioning into p = 2k squares is possible for k ≥ 2 ( Fig.

1 ). Furthermore, whenever it is possible to partition the square into p squares, there is a partition of the square into p + 3 squares: namely, in the partition into p squares, divide one of them into four new squares.

x -y y

y p =8

y Fig. 1

Fig. 2 This implies that both p = 2k and p = 2k + 3 are possible if k ≥ 2, and therefore

all p ≥ 6 are possible. On the other hand, partitioning the square into 5 squares is not possible. As- suming it is possible, one of its sides would be covered by exactly two squares, which cannot be of the same size ( Fig. 2 ). The rest of the big square cannot be partitioned into three squares. Hence, the answer is n = 6.

11. Let us denote the center of the semicircle by O, and ∠AOB =2 α , ∠BOC =2 β , AC = m, CE = n. We claim that a 2 +b 2 +n 2 + abn = 4. Indeed, since a = 2 sin α ,b = 2 sin β , n = 2 cos( α + β ), we have

a 2 +b 2 +n 2 + abn = 4(sin 2 α + sin 2 β + cos 2 ( α + β ) + 2 sin α sin β cos ( α + β )) cos 2 α cos 2 β

+ cos( α + β ) cos( α

= 4 + 4 (cos( α + β ) cos( α − β ) − cos( α + β ) cos( α − β )) = 4.

458 4 Solutions Analogously, c 2 +d 2 +m 2 + cdm = 4. By adding both equalities and subtracting

m 2 +n 2 = 4 we obtain

a 2 +b 2 +c 2 +d 2 + abn + cdm = 4. Since n > c and m > b, the desired inequality follows.

12. We will solve the contest problem (in which m , n ∈ {1,2,...,1981}). For m = 1, n can be either 1 or 2. If m > 1, then n(n − m) = m 2 ± 1 > 0; hence n − m > 0. Set p = n − m. Since m 2 − mp − p 2 =m 2 − p(m + p) = −(n 2 − nm − m 2 ), we see that (m, n) is a solution of the equation if and only if (p, m) is a solution too. Therefore, all the solutions of the equation are given as two consecutive members of the Fibonacci sequence

So the required maximum is 987 2 + 1597 2 .

13. Lemma. For any polynomial P of degree at most n,

n +1

i n ∑ +1

Proof. We shall use induction on n. For n = 0 it is trivial. Assume that it is true for n = k and suppose that P(x) is a polynomial of degree k + 1. Then P (x) − P(x + 1) clearly has degree at most k; hence (1) gives

i k 0 +1 = ∑ (−1) (P(i) − P(i + 1))

(−1) ∑ i

P (i) +

∑ i (−1)

= ∑ (−1) i

k +2

P (i).

i =0

This completes the proof of the lemma. Now we apply the lemma to obtain the value of P −1 (n + 1). Since P(i) = n +1

for i = 0, 1, . . . , n, we have

0 = , i (−1) P n +1 ∑

It follows that P (n + 1) =

2 ∤ n.

14. We need the following lemma. Lemma. If a convex quadrilateral PQRS satisfies PS = QR and ∠SPQ ≥ ∠RQP, then ∠QRS ≥ ∠PSR.

4.22 Shortlisted Problems 1981 459 Proof. If the lines PS and QR are parallel, then this quadrilateral is a par-

allelogram, and the statement is trivial. Otherwise, let X be the point of intersection of lines PS and QR. Assume that ∠SPQ + ∠RQP > 180 ◦ . Then ∠X PQ ≤ ∠XQP implies that XP ≥ XQ, and consequently XS ≥ XR. Hence, ∠QRS = ∠XRS ≥ ∠XSR = ∠PSR. Similarly, if ∠SPQ + ∠RQP < 180 ◦ , then ∠X PQ ≥ ∠XQP, from which it follows that X P ≤ XQ, and thus XS ≤ XR; hence ∠QRS = 180 ◦ − ∠XRS ≥ 180 ◦ − ∠XSR = ∠PSR.

Now we apply the lemma to the quadrilateral ABCD. Since ∠B ≥ ∠C and AB = CD , it follows that ∠CDA ≥ ∠BAD, which together with ∠EDA = ∠EAD gives ∠D ≥ ∠A. Thus ∠A = ∠B = ∠C = ∠D. Analogously, by applying the lemma to BCDE we obtain ∠E ≥ ∠B, and hence ∠B = ∠C = ∠D = ∠E.

15. Set BC = a, CA = b, AB = c, and denote the area of △ABC by P, and a/PD +

b /PE + c/PF by S. Since a ·PD + b ·PE + c ·PF = 2P, by the Cauchy–Schwarz inequality we have

a b c 2PS = (a · PD + b · PE + c · PF)

≥ (a + b + c) 2 ,

PF with equality if and only if PD = PE = PF, i.e., P is the incenter of △ABC. In

PD PE

that case, S attains its minimum:

16. The sequence {u n } is bounded, whatever u 1 is. Indeed, assume the opposite, and let u m

be the first member of the sequence such that |u m | > max{2,|u 1 |}. Then |u m −1 | = |u 3 m − 15/64| > |u m |, which is impossible. Next, let us see for what values of u m ,u m +1 is greater, equal, or smaller, respec-

tively. If u

m +1 =u m , then u m =u 3 m 3 +1 − 15/64 = u m − 15/64; i.e., u m is a root of x − x − 15/64 = 0. This equation factors as (x + 1/4)(x 2 √ − x/4 − 15/16) = 0, and √ hence u m is equal to x 1 = (1 − 61 )/8, x 2 = −1/4, or x 3 = (1 + 61 )/8, and these are the only possible limits of the sequence. Each of u m +1 >u m ,u m +1 <u m is equivalent to u 3 m −u m − 15/64 < 0 and u 3 m −u m − 15/64 > 0 respectively. Thus the former is satisfied for u m in the inter- val I 1 = (−∞,x 1 ) or I 3 = (x 2 ,x 3 ), while the latter is satisfied for u in I p m 2 = (x 1 ,x 2 ) or I 4 = (x 3 , ∞). Moreover, since the function f (x) = 3 x + 15/64 is strictly in- creasing with fixed points x 1 ,x 2 ,x 3 , it follows that u m will never escape from the interval I 1 ,I 2 ,I 3 , or I 4 to which it belongs initially. Therefore: (1) if u 1 is one of x 1 ,x 2 ,x 3 , the sequence {u m } is constant; (2) if u 1 ∈I 1 , then the sequence is strictly increasing and tends to x 1 ; (3) if u 1 ∈I 2 , then the sequence is strictly decreasing and tends to x 1 ; (4) if u 1 ∈I 3 , then the sequence is strictly increasing and tends to x 3 ; (5) if u 1 ∈I 4 , then the sequence is strictly decreasing and tends to x 3 .

460 4 Solutions

17. Let us denote by S A ,S B ,S C the centers of the given circles, where S A lies on the bisector of ∠A, etc. Then S A S B k AB, S B S C k BC, S C S A k CA, so that the inner bisectors of the angles of triangle ABC are also inner bisectors of the angles of △S A S B S C . These two triangles thus have a common incenter S, which is also the center of the homothety χ mapping △S A S B S C onto △ABC. The point O is the circumcenter of triangle S A S B S C , and so is mapped by χ onto the circumcenter P of ABC. This means that O, P, and the center S of χ are collinear.

18. Let C be the convex hull of the set of the planets: its border consists of parts of planes, parts of cylinders, and parts of the surfaces of some planets. These parts of planets consist exactly of all the invisible points; any point on a planet that is inside C is visible. Thus it remains to show that the areas of all the parts of planets lying on the border of C add up to the area of one planet. As we have seen, an invisible part of a planet is bordered by some main spherical arcs, parallel two by two. Now fix any planet P, and translate these arcs onto arcs on the surface of P. All these arcs partition the surface of P into several parts, each of which corresponds to the invisible part of one of the planets. This correspondence is bijective, and therefore the statement follows.

19. Consider the partition of plane π into regular hexagons, each having inradius 2. Fix one of these hexagons, denoted by γ . For any other hexagon x in the partition, there exists a unique translation τ x taking it onto γ . Define the mapping ϕ : π → γ as follows: If A belongs to the interior of a hexagon x, then ϕ (A) = τ x (A) (if A is on the border of some hexagon, it does not actually matter where its image is). The total area of the images of the union of the given circles equals S, while the √ area of the hexagon γ is 8

3. Thus there exists a point B of γ that is covered at least S 8 √ 3 times, i.e., such that ϕ −1 (B) consists of at least S 8 √ 3 distinct points of the plane that belong to some of the circles. For any of these points, take a circle that contains it. All these circles are disjoint, with total area not less than

8 3 S ≥ 2S/9. Remark. The statement becomes false if the constant 2 /9 is replaced by any number greater than 1 /4. In that case a counterexample is, for example, a set of unit circles inside a circle of radius 2 covering a sufficiently large part of its area.

4.23 Shortlisted Problems 1982 461