Solutions to the Shortlisted Problems of IMO 1986

4.27 Solutions to the Shortlisted Problems of IMO 1986

1. If w > 2, then setting in (i) x = w−2, y = 2, we get f (w) = f ((w−2) f (w)) f (2) =

0. Thus

f (x) = 0 if and only if x ≥ 2.

Now let 0 ≤ y < 2 and x ≥ 0. The LHS in (i) is zero if and only if x f (y) ≥ 2, while the RHS is zero if and only if x + y ≥ 2. It follows that x ≥ 2/ f (y) if and only if x ≥ 2 − y. Therefore

f (y) = 2 −y for 0 ≤ y < 2;

0 for y ≥ 2.

The confirmation that f satisfies the given conditions is straightforward.

2. No. If a were rational, its decimal expansion would be periodic from some point. Let p be the number of decimals in the period. Since f (10 2p ) has 2np zeros, it contains a full periodic part; hence the period would consist only of zeros, which is impossible.

3. Let E be the point where the boy turned westward, reaching the shore at D. Let the ray DE cut AC at F and the shore again at G. Then EF = AE = x (because AEF is an equilateral triangle) and FG = DE = y. From AE · EB = DE · EG we obtain x (86 − x) = y(x + y). If x is odd, then x(86 − x) is odd, while y(x + y)

is even. Hence x is even, and so y must also be even. Let y = 2y 1 . The above equation can be rewritten as

1 = (43 − y 1 . Since y 1 < 43, we have (2y 1 , 43 − y 1 ) = 1, and thus (|x+ y 1 −43|,2y 1 , 43 −y 1 ) is

a primitive Pythagorean triple. Consequently there exist integers a > b > 0 such that y 1 = ab and 43 − y 1 =a 2 +b 2 . We obtain that a 2 +b 2 + ab = 43, which has the unique solution a = 6, b = 1. Hence y = 12 and x = 2 or x = 72.

Remark. The Diophantine equation x (86 − x) = y(x + y) can be also solved directly. Namely, we have that x

(344 − 3x) = (2y + x) 2 is a square, and since x is even, we have (x, 344 − 3x) = 2 or 4. Consequently x, 344 − 3x are either both

squares or both two times squares. The rest is easy.

4. Let x =p α x ′ ,y=p β y ′ ,z=p γ z ′ with p ∤x ′ y ′ z ′ and α ≥ β ≥ γ . From the given equation it follows that p n

(x + y) = z(xy − p n ) and consequently z ′ | x + y. Since also p γ | x+ y, we have z | x+ y, i.e., x+ y = qz. The given equation together with

the last condition gives us xy =p n (q + 1)

x + y = qz. (1) Conversely, every solution of (1) gives a solution of the given equation.

and

For q = 1 and q = 2 we obtain the following classes of n + 1 solutions each:

492 4 Solutions q = 1 : (x, y, z) = (2p i ,p n −i , 2p i +p n −i )

for i = 0, 1, 2, . . . , n; 3p j +p n q −j = 2 : (x, y, z) = 3p j ,p n −j ,

2 for j = 0, 1, 2, . . . , n. For n = 2k these two classes have a common solution (2p k ,p k , 3p k ); other-

wise, all these solutions are distinct. One further solution is given by (x, y, z) =

1 ,p n (p n + 3)/2, p 2 +2 , not included in the above classes for p > 3. Thus we have found 2 (n + 1) solutions. Another type of solution is obtained if we put q =p k +p n −k . This yields the solutions

(x, y, z) = (p k ,p n +p n −k +p 2n −2k ,p n −k + 1) for k = 0, 1, . . . , n. For k < n these are indeed new solutions. So far, we have found 3(n + 1) − 1 or

3 (n + 1) solutions. One more solution is given by (x, y, z) = (p, p n +p n −1 ,p n −1 + p n −2 + 1).

5. Suppose that for every a , b ∈ {2,5,13,d}, a 6= b, the number ab − 1 is a perfect square. In particular, for some integers x , y, z we have

2d −1=x 2 ,

5d −1=y 2 ,

13d

−1=z 2 . Since x is clearly odd, d = (x 2 + 1)/2 is also odd because 4 ∤ x 2 + 1. It fol-

lows that y and z are even, say y = 2y 1 and z = 2z 1 . Hence (z 1 −y 1 )(z 1 +y 1 )= (z 2 −y 2 )/4 = 2d. But in this case one of the factors z 1 −y 1 ,z 1 +y 1 is odd and the other one is even, which is impossible.

6. There are five such numbers:

7. Let P (x) = (x − x 0 )(x − x 1 ) ···(x − x n )(x − x n +1 ). Then

P (x) = ∑

P ′′ (x) = ∑ ∑

j =0 k 6= j (x − x j )(x − x k ) Therefore

P (x i ) = 2P ′ (x i ) ∑

j 6=i (x i −x j )

for i = 0, 1, . . . , n + 1, and the given condition implies P ′′ (x i ) = 0 for i =

1 , 2, . . . , n. Consequently, x (x − 1)P ′′ (x) = (n + 2)(n + 1)P(x).

4.27 Shortlisted Problems 1986 493 It is easy to observe that there is a unique monic polynomial of degree n +2

satisfying differential equation (1). On the other hand, the polynomial Q (x) = (−1) n P (1 −x) also satisfies this equation, is monic, and degQ = n+2. Therefore (−1) n P (1 − x) = P(x), and the result follows.

8. We shall solve the problem in the alternative formulation. Let L G (v) denote the length of the longest directed chain of edges in the given graph G that begins in a vertex v and is arranged decreasingly relative to the numbering. By the pigeonhole principle it suffices to show that ∑ v L (v) ≥ 2q in every such graph. We do this by induction on q. For q = 1 the claim is obvious. We assume that it is true for q − 1 and consider a graph G with q edges numbered 1 , . . . , q. Let the edge number q connect vertices u and w. Removing this edge, we get a graph G ′ with q − 1 edges. We then have

L G (u) ≥ L G ′ (w) + 1, L G (w) ≥ L G ′ (u) + 1, L G (v) ≥ L G ′ (v) for other v. Since ∑ L G ′ (v) ≥ 2(q − 1) by inductive assumption, it follows that ∑L G (v) ≥

2 (q − 1) + 2 = 2q as desired. Second solution. Let us place a spider at each vertex of the graph. Let us now

interchange the positions of the two spiders at the endpoints of each edge, listing the edges increasingly with respect to the numbering. This way we will move spiders exactly 2q times (two for each edge). Hence there is a spider that will

be moved at least 2q /n times. All that remains is to notice that the path of each spider consists of edges numbered in increasing order.

Remark.

A chain of the stated length having all vertices distinct does not neces- sarily exist. An example is n = 4, q = 6 with the numbering following the order

ab , cd, ac, bd, ad, bc.

9. We shall use induction on the number n of points. The case n = 1 is trivial. Let us suppose that the statement is true for all 1 , 2, . . . , n − 1, and that we are given

a set T of n points. If there exists a point P ∈ T and a line l that is parallel to an axis and contains P and no other points of T , then by the inductive hypothesis we can color the set T \ {P} and then use a suitable color for P. Let us now suppose that whenever a line parallel to an axis contains a point of T , it contains another point of T . It follows

that for an arbitrary point P 0 ∈ T we can choose points P 1 ,P 2 , . . . such that P k P k +1 is parallel to the x-axis for k even, and to the y-axis for k odd. We eventually come to a pair of integers (r, s) of the same parity, 0 ≤ r < s, such that lines P r P r +1 and P s P s +1 coincide. Hence the closed polygonal line P r +1 P r +2 ...P s P r +1 is of even length. Thus we may color the points of this polygonal line alternately and then apply the inductive assumption for the rest of the set T . The induction is complete.

be the points lying on a line l parallel to an axis, going from left to right or from up to down. We draw segments joining P 1 with P 2 ,P 3 with P 4 , and generally P 2i −1 with P 2i . Having this done for every such line l, we obtain a set of segments forming certain polygonal lines. If one

Second solution. Let P 1 ,P 2 ,...,P k

494 4 Solutions of these polygonal lines is closed, then it must have an even number of vertices.

Thus, we can color the vertices on each of the polygonal lines alternately (a point not lying on any of the polygonal lines may be colored arbitrarily). The obtained coloring satisfies the conditions.

10. The set X = {1,...,1986} splits into triads T 1 ,...,T 662 , where T j = {3 j − 2,3 j −

1 , 3 j}. Let F be the family of all k-element subsets P such that |P∩T j | = 1 or 2 for some index j. If j 0 is the smallest such j 0 , we define P ′ to be the k-element set obtained from P by replacing the elements of P ∩T j 0 by the ones following cyclically inside T j 0 . Let s (P) denote the remainder modulo 3 of the sum of elements of P . Then s (P), s(P ′ ), s(P ′′ ) are distinct, and P ′′′ = P. Thus the operator ′ gives us

a bijective correspondence between the sets X ∈ F with s(P) = 0, those with s (P) = 1, and those with s(P) = 2. If 3 ∤ k is not divisible by 3, then each k-element subset of X belongs to F , and the game is fair. If 3 | k, then k-element subsets not belonging to F are those that are unions of several triads. Since every such subset has the sum of elements divisible by 3, it follows that player A has the advantage.

11. Let X be a finite set in the plane and l k a line containing exactly k points of

X (k = 1, . . . , n). Then l n contains n points, l n −1 contains at least n − 2 points not lying on l n ,l n −2 contains at least n − 4 points not lying on l n or l n −1 , etc. It follows that

|X| ≥ g(n) = n + (n − 2) + (n − 4) + ···+ n − 2 n

2 . Hence f (n) ≥ g(n) = n +1 n +2 2 2 , where the last equality is easily proved by

induction. We claim that f (n) = g(n). To prove this, we shall inductively construct a set X n of cardinality g (n) with the required property. For n ≤ 2 a one-point and two- point set satisfy the requirements. Assume that X n is a set of g (n) points and that l k is a line containing exactly k points of X n ,k = 1, . . . , n. Consider any line l not parallel to any of the l k ’s and not containing any point of X n or any intersection point of the l k . Let l intersect l k in a point P k ,k = 1, . . . , n, and let P n +1 ,P n +2 be

two points on l other than P 1 ,...,P n . We define X n +2 =X n ∪ {P 1 ,...,P n +2 }. The set X n +2 consists of g (n)+ (n+ 2) = g(n + 2) points. Since the lines l, l n ,...,l 2 ,l 1 meet X n in n + 2, n + 1, . . ., 3, 2 points respectively (and there clearly exists a line containing only one point of X n +2 ), this set also meets the demands.

12. We define f (x 1 ,...,x 5 5 )=∑ 2 i =1 (x i +1 −x i −1 ) (x 0 =x 5 ,x 6 =x 1 ). Assuming that x 3 < 0, according to the rules the lattice vector X = (x 1 ,x 2 ,x 3 ,x 4 ,x 5 ) changes

into Y = (x 1 ,x 2 +x 3 , −x 3 ,x 4 +x 3 ,x 5 ). Then

f (Y ) − f (X) = (x 2 +x 3 −x 5 ) 2 + (x 1 +x 3 ) 2 + (x 2 −x 4 ) 2 +(x 3 +x 5 ) 2 + (x 1 −x 2 3 −x 4 ) 2 − (x 2 −x 5 ) −(x 3 −x 1 ) 2 − (x 4 −x 2 ) 2 − (x 5 −x 3 ) 2 − (x 1 −x 4 ) 2 = 2x 3 (x 1 +x 2 +x 3 +x 4 +x 5 ) = 2x 3 S < 0.

4.27 Shortlisted Problems 1986 495 Thus f strictly decreases after each step, and since it takes only nonnegative

integer values, the number of steps must be finite. Remark. One could inspect the behavior of g (X ) = ∑ 5 i =1 ∑ 5 j =1 |x i +x i +1 + ··· +

x j −1 | instead. Then g(Y ) − g(X) = |S + x 3 | − |S − x 3 | > 0.

13. Let us consider the infinite integer lattice and assume that having reached a point (x, n) or (n, y), the particle continues moving east and north following the rules of the game. The required probability p k is equal to the probability of getting to one

of the points E 1 (n, n + k), E 2 (n + k, n), but without passing through (n, n + k − 1) or (n + k − 1,n). Thus p is equal to the probability p 1 of getting to E 1 (n, n + k) via D 1 (n − 1,n + k) plus the probability p 2 of getting to E 2 (n + k, n) via D 2 (n +

k , n − 1). Both p 1 and p 2 are easily seen to be equal to 2n +k−1 n −1 2 −2n−k , and therefore p

= 2n +k−1

2 −2n−k+1 .

n −1

14. We shall use the following simple fact. Lemma. If b k is the image of a circle k under an inversion centered at a point Z, and O 1 ,O 2 are centers of k and b k , then O 1 ,O 2 , and Z are collinear. Proof. The result follows immediately from the symmetry with respect to the line ZO 1 . Let I be the center of the inscribed circle i. Since IX ·IA = IE 2 , the inversion with respect to i takes points A into X, and analogously B ,C into Y, Z respectively. It follows from the lemma that the center of circle ABC, the center of circle XY Z, and point I are collinear.

15. (a) This is the same problem as SL82-14. (b) If S is the midpoint of AC, we have B ′ S = AC cos ∠D

2 sin ∠D ,D S = AC 2 sin ∠B ,

′ cos∠B

sin = AC (∠B+∠D)

◦ 2 sin ∠B sin ∠D

or ∠D >

90 ◦ . We similarly obtain that A ′′ C ′′ =B ′ D ′ sin (∠A ′ +∠C ′ ) 2 sin ∠A ′ sin ∠C ′

16. Let Z be the center of the polygon. Suppose that at some moment we have A ∈P i −1 P i and B ∈P i P i +1 , where

Z i P ,P ,P

are adjacent vertices of O P i +1

i −1 i i +1

the polygon. Since ∠AOB = 180 ◦ − Z ∠P i −1 P i P i +1 , the quadrilateral AP i BO

B is cyclic. Hence ∠AP i O = ∠ABO =

∠AP i Z , which means that O ∈P i Z . P i −1 A P i over, from OP i = 2r sin ∠P i AO , where r is the radius of circle AP i BO , we ob-

tain that ZP i ≤ OP i ≤ ZP i /cos( π /n). Thus O traces a segment ZZ i as A and B move along P i −1 P i and P i P i +1 respectively, where Z i is a point on the ray P i Z with P i Z i cos ( π /n) = P i Z . When A , B move along the whole circumference of the polygon, O traces an asterisk consisting of n segments of equal length ema- nating from Z and pointing away from the vertices.

496 4 Solutions

17. We use complex numbers to represent the position of a point in the plane. For convenience, let A 1 ,A 2 ,A 3 ,A 4 ,A 5 , . . . be A, B,C, A, B, . . . respectively, and let P 0

be the origin. After the kth step, the position of P k will be P k =A k + (P k −1 −A k )u,

k = 1, 2, 3, . . ., where u = e 4 πı/3 . We easily obtain

P k = (1 − u)(A k + uA

k −1 +u A k −2 + ··· + u k −1 A 1 ). The condition P 0 ≡P 1986 is equivalent to A 1986 + uA 1985 + ··· + u 1984 A 2 +

u 1985 A 1 = 0, which, having in mind that A 1 =A 4 =A 7 = ··· , A 2 =A 5 =A 8 = ···,

A 3 =A 6 =A 9 = ···, reduces to 662 (A 3 + uA 2 +u 2 A 1 ) = (1 + u 3 + ··· + u 1983 )(A 3 + uA 2 +u 2 A 1 ) = 0. It follows that A 3 −A 1 = u(A 1 −A 2 ), and the assertion follows.

Second solution. Let f P denote the rotation with center P through 120 ◦ clock- wise. Let f 1 =f A . Then f 1 (P 0 )=P 1 . Let B ′ =f 1 (B), C ′ =f 1 (C), and f 2 =f B ′ . Then f 2 (P 1 )=P 2 and f 2 (AB ′ C ′ )=A ′ B ′ C ′′ . Finally, let f 3 =f C ′′ and f 3 (A ′ B ′ C ′′ )=

A ′′ B ′′ C ′′ . Then g =f 3 f 2 f 1 is a translation sending P 0 to P 3 and C to C ′′ . Now P 1986 =P 0 implies that g 662 is the identity, and thus C =C ′′ . Let K be such that ABK is equilateral and positively oriented. We observe that

f 2 f 1 (K) = K; therefore the rotation f 2 f 1 satisfies f 2 f 1 (P) 6= P for P 6= K. Hence

f 2 f 1 (C) = C ′′ = C implies K = C.

18. We shall use the following criterion for a quadrangle to be circumscribable.

Lemma. The quadrangle AY DZ is circumscribable if and only if DB − DC = AB − AC. Proof. Suppose that AY DZ is circumscribable and that the incircle is tangent to AZ , ZD, DY , YA at M, N, P, Q respectively. Then DB − DC = PB − NC = MB −QC = AB− AC. Conversely, assume that DB− DC = AB −AC and let

a tangent from D to the incircle of A the triangle ACZ meet CZ and CA

at D ′

Q 6= Z and Y ′ 6= A respectively.

According to the first part we have Y

D ′ B −D ′ C = AB − AC. It follows

D DD ′ , implying that D ′

that |D ′ B − DB| = |D ′

NP

C − DC| =

≡ D. X B C Let us assume that DZBX and DXCY are circumscribable. Using the lemma

we obtain DC − DA = BC − BA and DA − DB = CA − CB. Adding these two inequalities yields DC − DB = AC − AB, and the statement follows from the lemma.

19. Let M and N be the midpoints of segments AB and CD, respectively. The given conditions imply that △ABD ∼ = △BAC and △CDA ∼ = △DCB; hence MC = MD and NA = NB. It follows that M and N both lie on the perpendicular bisectors of

AB and CD, and consequently MN is the common perpendicular bisector of AB and CD. Points B and C are symmetric to A and D with respect to MN. Now if P

4.27 Shortlisted Problems 1986 497 is a point in space and P ′ the point symmetric to P with respect to MN, we have

BP = AP ′ , CP = DP ′ , and thus f (P) = AP + AP ′ + DP + DP ′ . Let PP ′ intersect MN in Q. Then AP + AP ′ ≥ 2AQ and DP + DP ′ ≥ 2DQ, from which it follows that f (P) ≥ 2(AQ + DQ) = f (Q). It remains to minimize f (Q) with Q moving along the line MN. Let us rotate point D around MN to a point D ′ that belongs to the plane AMN, on the side of MN opposite to A. Then f (Q) = 2(AQ + D ′ Q ) ≥ AD ′ , and equal- ity occurs when Q is the intersection of AD ′ and MN. Thus min f (Q) = AD ′ .

We note that 4MD 2 = 2AD 2 + 2BD 2 − AB 2 = 2a 2 + 2b 2 − AB 2 and 4MN 2 = 4MD 2 − CD 2 = 2a 2 + 2b 2 − AB 2 − CD 2 . Now, AD ′2 = (AM + D ′ N ) 2 + MN 2 , which together with AM +D ′ N = (a + b)/2 gives us

p We conclude that min f (Q) = (a 2 +b 2 +c 2 )/2.

20. If the faces of the tetrahedron ABCD are congruent triangles, we must have AB = CD , AC = BD, and AD = BC. Then the sum of angles at A is ∠BAC + ∠CAD + ∠DAB = ∠BDC + ∠CBD + ∠DCB = 180 ◦ . We now assume that the sum of angles at each vertex is 180 ◦ . Let us construct triangles BCD ′ ,CAD ′′ , ABD ′′′ in the plane ABC, exterior to △ABC, such that △BCD ′ ∼ = △BCD, △CAD ′′ ∼ = △CAD, and △ABD ′′′ ∼ = △ABD. Then by the as-

sumption, A ∈D ′′ D ′′′ ,B ∈D ′′′ D ′ , and C ∈D ′ D ′′ . Since also D ′′ A =D ′′′ A = DA, etc., A , B,C are the midpoints of segments D ′′ D ′′′ ,D ′′′ D ′ ,D ′ D ′′ respectively. Thus the triangles ABC, BCD ′ , CAD ′′ , ABD ′′′ are congruent, and the statement follows.

21. Since the sum of all edges of ABCD is 3, the statement of the problem is an immediate consequence of the following statement:

Lemma. √ Let r be the inradius of a triangle with sides a , b, c. Then a + b + c ≥

6 3 · r, with equality if and only if the triangle is equilateral. Proof. If S and p denotes the area and semiperimeter of the triangle, by Heron’s formula and the AM–GM inequality we have p

pr =S= p (p − a)(p − b)(p − c) s

3 r (p − a) + (p − b) + (p − c)

i.e., p ≥3 3 · r, which is equivalent to the claim.

498 4 Solutions