Solutions to the Shortlisted Problems of IMO 2001

4.42 Solutions to the Shortlisted Problems of IMO 2001

1. First, let us show that there is at most one such function. Suppose that f 1 and f 2 are two such functions, and consider g =f 1 −f 2 . Then g is zero on the boundary and satisfies

6 [g(p + 1, q − 1,r) + ···+ g(p,q − 1,r + 1)], i.e., g (p, q, r) is equal to the average of the values of g at six points (p + 1, q −

g (p, q, r) =

1 , r), . . . that lie in the plane π given by x + y + z = p + q + r. Suppose that (p, q, r) is the point at which g attains its maximum in absolute value on π ∩T. The averaging property of g implies that the values of g at (p + 1, q − 1,r) etc. are all equal to g (p, q, r). Repeating this argument we obtain that g is constant

on the whole of π ∩ T , and hence it equals 0 everywhere. Therefore f 1 ≡f 2 . It remains to guess f . It is natural to try f (p, q, r) = pqr first: it satisfies

6 [ f (p + 1, q − 1,r) + ···+ f (p,q − 1,r + 1)] + 3 . Thus we sim- ply take

f (p, q, r) = +q+r 1 p

p +q+r and directly check that it satisfies the required property. Hence this is the unique

p +q+r

solution.

√ ∈ N, 1 + n ≥ 2, or n 2 ≤1+ 1

2. It follows from Bernoulli’s inequality that for each n

n . Consequently, it will be enough to show that 1 +a n >1+ 1 n a n −1 . Assume the opposite. Then there exists N such that for each n ≥ N,

+a n

a n −1 , i.e.,

≤ +1 n +1

m +1 ≤ −1 N − 1 N +1 + ···+ 1 m +1 . However, it is well known that the sum 1 N 1 +1 + ··· + m +1 can be arbitrarily large for m large enough, so that a m m +1 is eventually negative. This contradiction yields the result.

Summing for n = N, . . . , m yields a m

Second solution. Suppose that 1 +a n ≤ 2a n −1 for all n ≥ N. Set b n =

2 −(1+1/2+···+1/n) and multiply both sides of the above inequality to obtain

b n +b n a n ≤b n −1 a n −1 . Thus

b N a N >b N a N −b n a n ≥b N +b N +1 + ··· + b n . However, it can be shown that ∑ n >N b N diverges: in fact, since 1 + 1 2 + ··· +

2 > 1 2n , and we already know that ∑

1 < 1 + lnn, we have b n >2 −1−lnn = 1 n − ln2

1 n >N 2n diverges. Remark. As can be seen from both solutions, the value 2 in the problem can be increased to e.

678 4 Solutions

3. By the arithmetic–quadratic mean inequality, it suffices to prove that x 2 1 x 2 x 2

2 2 2 (1 + x < 1.

(1 + x 1 + ··· + x n ) Observe that for k ≥ 2 the following holds: x 2 k 2 x k

(1 + x 1 +x 2 )

(1 + x 2 + ··· + x 2 ) 2 (1 + ··· + x 1 2 k k −1 )(1 + ··· + x 2 k )

1 + ··· + x k x 1 For k 2 = 1 we have

(1+x 1 ) 2 ≤1− 1 +x . Summing these inequalities, we obtain 1 2

Second solution. Let a n (k) = sup

k 2 +x 1 2 + ··· + k 2 +x 2 1 and a +···+x 2 n n =a n (1).

We must show that a n < n . Replacing x i by kx i shows that a n (k) = a n /k. Hence 

a n = sup 

 = sup (sin θ cos θ +a n −1 cos θ ), (1)

x 1 1 +x 1 1 +x 2 θ

where tan θ =x 1 . The above supremum can be computed explicitly:

However, the required inequality is weaker and can be proved more easily: if √ √ √ √

a n −1 < n − 1, then by (1) a n < sin θ + n − 1cos θ = n sin ( θ + α √ n )≤ , for α ∈ (0, π /2) with tan α = n .

4. Let (∗) denote the given functional equation. Substituting y = 1 we get f (x) 2 = xf (x) f (1). If f (1) = 0, then f (x) = 0 for all x, which is the trivial solution.

Suppose f (1) = C 6= 0. Let G = {y ∈ R | f (y) 6= 0}. Then

Cx if x

∈ G,

(1) We must determine the structure of G so that the function defined by (1) satisfies

f (x) =

0 otherwise.

(∗). (i) Clearly 1 ∈ G, because f (1) 6= 0.

(ii) If x ∈ G, y 6∈ G, then by (∗) it holds f (xy) f (x) = 0, so xy 6∈ G. (iii) If x , y ∈ G, then x/y ∈ G (otherwise by (ii), y(x/y) = x 6∈ G).

4.42 Shortlisted Problems 2001 679 (iv) If x , y ∈ G, then by (iii) we have x −1 ∈ G, so xy = y/x −1 ∈ G.

Hence G is a set that contains 1, does not contain 0, and is closed under multipli- cation and division. Conversely, it is easy to verify that every such G in (1) gives

a function satisfying (∗).

5. Let a 1 ,a 2 ,...,a n satisfy the conditions of the problem. Then a k >a k −1 , and hence a k ≥ 2 for k = 1,...,n. The inequality (a k +1 − 1)a k −1 ≥a 2 k (a k − 1) can be rewritten as

Summing these inequalities for k equality a n

= i + 1, . . . , n − 1 and using the obvious in-

a n −1 , we obtain a i

a i +1 −1 . Therefore

Consequently, given a 0 ,a 1 ,...,a i , there is at most one possibility for a i +1 . In our case, (1) yields a = 2, a = 5, a = 56, a = 280 2 1 2 3 4 = 78400. These values satisfy the conditions of the problem, so that this is a unique solution.

6. We shall determine a constant k > 0 such that

This inequality is equivalent to (a k +b k +c k ) 2 ≥a 2k −2 (a 2 + 8bc), which further reduces to

−a 2k ≥ 8a −2 bc . On the other hand, the AM–GM inequality yields

(a k +b k +c k ) 2 2k

k /2 b 3k (a /4 +b +c ) −a = (b +c )(2a +b +c ) ≥ 8a c 3k /4 , and therefore k = 4/3 is a good choice. Now we have

4 /3 +c 4 /3 +b + a 4 /3 +b 4 /3 +c 4 /3 = 1. Second solution. The numbers x = √ a and z

+b 4 /3 +c 4 /3

a a 4 /3

2 +8bc ,y = √ a b b 2 +8ca = √

c 2 +8ab

satisfy

2 −1 =8 3 x . y z Our task is to prove x + y + z ≥ 1.

Since f is decreasing on each of the variables x , y, z, this is the same as proving that x , y, z > 0, x + y + z = 1 implies f (x, y, z) ≥ 8 3 . However, since 1 2 −1=

x (2x+y+z)(y+z)

(x+y+z) 2 −x 2

x 2 = x 2 , the inequality f (x, y, z) ≥ 8 becomes

680 4 Solutions

3 x 2 y 2 z 2 ≥8 , which follows immediately by the AM–GM inequality.

(2x + y + z)(x + 2y + z)(x + y + 2z)(y + z)(z + x)(x + y)

Third solution. We shall prove a more general fact: the inequality √ a

+kbc

, b, c > 0 if and only if k ≥ 8. Firstly suppose that k

c 2 +kab ≥ 1 +k

is true for all a

b 2 +kca

≥ 8. Setting x = bc/a 2 ,y = ca/b 2 ,z = ab/c 2 , we reduce the desired inequality to

F (x, y, z) = f (x) + f (y) + f (z) ≥ √ , where f (t) = √ , (2)

1 + kt for x , y, z > 0 such that xyz = 1. We shall prove (2) using the method of Lagrange

1 +k

multipliers. The boundary of the set D = {(x,y,z) ∈ R 3 + | xyz = 1} consists of points (x,y,z) with one of x , y, z being 0 and another one being +∞. If w.l.o.g. x = 0, then √

F (x, y, z) ≥ f (x) = 1 ≥ 3/ 1 + k.

Suppose now that (x, y, z) is a point of local minimum of F on D. There exists λ ∈ R such that (x,y,z) is stationary point of the function F(x,y,z) + λ xyz . Then (x, y, z, λ ) is a solution to the system f ′ (x)+ λ yz =f ′ (y)+ λ xz =f ′ (z)+ λ xy = 0, xyz = 1. Eliminating λ gives us

xf ′ (x) = y f ′ (y) = z f ′ (z), xyz = 1. (3) The function t f ′ (t) =

2 (1+kt) 3 /2 decreases on the interval (0, 2/k] and increases on

−kt

4 (1+kt) 5 /2 . It follows that two of the numbers x , y, z are equal. If x = y = z, then (1, 1, 1) is the only solution to (3). Suppose that x = y 6= z. Since (y f ′ (y)) 2 − (z f ′ (z)) 2 k = 2 (z−y)(k 3 y 2 z 2 4 −3kyz−y−z) (1+ky) 3 (1+kz) 3 , (3) gives

[2/k, +∞) because (t f k (t)) = (kt−2)

us y 2 z = 1 and k 3 y 2 z 2 − 3kyz − y − z = 0. Eliminating z we obtain an equa- tion in y, k 3 /y 2 − 3k/y − y − 1/y 2 = 0, whose only real solution is y = k − 1. Thus

(k − 1,k − 1,1/(k − 1) 2 ) and the cyclic permutations are the only solu- tions to (3) with x , y, z being not all equal. Since F(k − 1,k − 1,1/(k − 1) 2 √ )=

(k + 1)/ k 2 − k + 1 > F(1,1,1) = 1, the inequality (2) follows.

For 0 < k < 8 we have that

If we fix c and let a , b tend to 0, the first two summands will tend to 0 while the third will tend to 1. Hence the inequality cannot be improved.

4.42 Shortlisted Problems 2001 681

7. It is evident that arranging of A in increasing order does not diminish m. Thus we can assume that A is nondecreasing. Assume w.l.o.g. that a 1 = 1, and let b i

be the number of elements of A that are equal to i (1 ≤i≤n=a 2001 ). Then we have b 1 +b 2 + ··· + b n = 2001 and

m =b 1 b 2 b 3 +b 2 b 3 b 4 + ··· + b n −2 b n −1 b n . (1) Now if b i ,b j (i < j) are two largest b’s, we deduce from (1) and the AM–

GM inequality that m ≤b i b j (b 1 + ··· + b i −1 +b i +1 + ··· + b j −1 +b j +1 +b n )≤

3 = 667 3 (b 1 b 2 b 3 ≤b 1 b i b j , etc.). The value 667 3 is attained for b 1 =b 2 =

b 3 = 667 (i.e., a 1 = ··· = a 667 = 1, a 668 = ··· = a 1334 = 2, a 1335 = ··· = a 2001 =

3). Hence the maximum of m is 667 3 .

8. Suppose to the contrary that all the S (a)’s are different modulo n!. Then the sum of S

(n!−1)n! (a)’s over all permutations a satisfies ∑ S n ! a (a) ≡ 0 + 1 + ···+ (n! − 1) =

2 ≡ 2 (mod n!). On the other hand, the coefficient of c i in ∑ a S (a) is equal to

n (n − 1)!(1 + 2 + ···+ n) = +1

2 n ! for all i, from which we obtain

n ∑ +1

a 2 1 + ··· + c n )n! ≡ 0 (mod n!) for odd n. This is a contradiction.

S (a) ≡

(c

9. Consider one such party. The result is trivially true if there is only one 3-clique, so suppose there exist at least two 3-cliques C 1 and C 2 . We distinguish two cases: (i) C 1 = {a,b,c} and C 2 = {a,d,e} for some distinct people a,b,c,d,e. If the departure of a destroys all 3-cliques, then we are done. Otherwise, there is

a third 3-clique C 3 , which has a person in common with each of C 1 ,C 2 and does not include a: say, C 3 = {b,d, f } for some f . We thus obtain another 3-clique C 4 = {a,b,d}, which has two persons in common with C 3 , and the case (ii) is applied. (ii) C 1 = {a,b,c} and C 2 = {a,b,d} for distinct people a,b,c,d. If the depar- ture of a , b leaves no 3-clique, then we are done. Otherwise, for some e there is a clique {c,d,e}. We claim that then the departure of c , d breaks all 3-cliques. Suppose the opposite, that a 3-clique C remains. Since C shares a person with each of the 3-cliques {c,d,a},{c,d,b},{c,d,e}, it must be C = {a,b,e}. However, then {a,b,c,d,e} is a 5-clique, which is assumed to be impossible.

10. For convenience let us write a = 1776, b = 2001, 0 < a < b. There are two types of historic sets:

(1) {x,x + a,x + a + b}

and

(2) {x,x + b,x + a + b}. We construct a sequence of historic sets H 1 ,H 2 ,H 3 , . . . inductively as follows:

(i) H 1 = {0,a,a + b}, and (ii) Let y n

be the least nonnegative integer not occurring in U n =H 1 ∩ ··· ∩ H n . We take H n +1 to be {y n ,y n + a, y n + a + b} if y n + a 6∈ U n , and {y n ,y n +

b ,y n + a + b} otherwise.

682 4 Solutions It remains to show that this construction never fails. Suppose that it failed at the

construction of H n +1 . The element y n + a + b is not contained in U n , since by the construction the smallest elements of H 1 ,...,H n are all less than y n . Hence the reason for the failure must be the fact that both y n + a and y n + b are covered by U n . Further, y n + b must have been the largest element of its set H k , so the smallest element of H k equals y n − a. But since y n is not covered, we conclude that H k is of type (2). This is a contradiction, because y n was free, so by the algorithm we had to choose for H k the set of type (1) (that is, {y n − a,y n ,y n + b}) first.

11. Let (x 0 ,x 1 ,...,x n ) be any such sequence: its terms are clearly nonnegative in- tegers. Also, x 0 = 0 yields a contradiction, so x 0 > 0. Let m be the number of positive terms among x 1 ,...,x n . Since x i counts the terms equal to i, the sum x 1 + ··· + x n counts the total number of positive terms in the sequence, which is known to be m + 1. Therefore among x 1 ,...,x n exactly m − 1 terms are equal to

1, one is equal to 2, and the others are 0. Only x 0 can exceed 2, and consequently at most one of x 3 ,x 4 , . . . can be positive. It follows that m ≤ 3. (i) m = 1: Then x 2 = 2 (since x 1 = 2 is impossible), so x 0 = 2. The resulting sequence is (2, 0, 2, 0). (ii) m = 2: Either x 1 = 2 or x 2 = 2. These cases yield (1, 2, 1, 0) and (2, 1, 2, 0, 0) respectively. (iii) m = 3: This means that x k > 0 for some k > 2. Hence x 0 = k and x k = 1. Fur- ther, x 1 = 1 is impossible, so x 1 = 2 and x 2 = 1; there are no more positive terms in the sequence. The resulting sequence is (p, 2, 1, 0, . . . , 0 , 1, 0, 0, 0).

| {z }

p −3

Remark. The same problem occurred as (LL83-15), proposed by Canada.

12. For each balanced sequence a = (a 1 ,a 2 ,...,a 2n ) denote by f (a) the sum of j’s for which a

j = 1 (for example, f (100101) = 1 + 4 + 6 = 11). Partition the n balanced sequences into n + 1 classes according to the residue of f modulo n +1. Now take S to be a class of minimum size: obviously

2n

1 |S| ≤ 2n n +1 n . We claim that every balanced sequence a is either a member of S or a neighbor of a member

of S. We consider two cases. (i) Let a 1 be 1. It is easy to see that moving this 1 just to the right of the kth 0, we obtain a neighboring balanced sequence b with f (b) = f (a) + k. Thus if a 6∈ S, taking a suitable k ∈ {1,2,...,n} we can achieve that b ∈ S.

(ii) Let a 1 be 0. Taking this 0 just to the right of the kth 1 gives a neighbor b with f (b) = f (a) − k, and the conclusion is similar to that of (i). This justifies our claim.

13. At any moment, let p i

be the number of pebbles in the ith column, i = 1, 2, . . .. The final configuration has obvious properties p 1 ≥p 2 ≥ ··· and p i +1 ∈ {p i ,p i −

1 }. We claim that p i +1 =p i > 0 is possible for at most one i. Assume the opposite. Then the final configuration has the property that for some r and s > r we have p r +1 =p r ,p s +1 =p s > 0 and p r +k =p r +1 − k + 1 for all k =

1 , . . . , s − r. Consider the earliest configuration, say C, with this property. What

4.42 Shortlisted Problems 2001 683 was the last move before C? The only possibilities are moving a pebble either

from the rth or from the sth column; however, in both cases the configuration preceding this last move had the same property, contradicting the assumption that C is the earliest.

Therefore the final configuration looks as follows: p 1 = a ∈ N, and for some r, p i equals a − (i− 1) if i ≤ r, and a− (i−2) otherwise. It is easy to determine a,r:

since n (a+1)(a+2) =p

2 , from which we uniquely find a and then r as well.

1 +p 2 + ··· =

(a+1)(a+2)

2 − r, we get

a (a+1)

2 ≤n<

• The final configuration for n = 13:

14. We say that a problem is difficult for boys if at most two boys solved it, and difficult for girls if at most two girls solved it. Let us estimate the number of pairs boy-girl both of whom solved some problem difficult for boys. Consider any girl. By the condition (ii), among the six prob- lems she solved, at least one was solved by at least 3 boys, and hence at most

5 were difficult for boys. Since each of these problems was solved by at most

2 boys and there are 21 girls, the considered number of pairs does not exceed

5 · 2 · 21 = 210. Similarly, there are at most 210 pairs boy-girl both of whom solved some prob- lem difficult for girls. On the other hand, there are 21 2 > 2 · 210 pairs boy-girl, and each of them solved one problem in common. Thus some problems were difficult neither for girls nor for boys, as claimed.

Remark. The statement can be generalized: if 2 (m − 1)(n − 1) + 1 boys and as many girls participated, and nobody solved more than m problems, then some problem was solved by at least n boys and n girls.

15. Let MNPQ be the square inscribed in △ABC with M ∈ AB, N ∈ AC, P,Q ∈ BC, and let AA 1 meet MN , PQ at K, X respectively. Put MK = PX = m, NK = QX = n , and MN = d. Then

cot β +1 =

d cot β +d

= . XC n

cot γ +1 Similarly, if BB 1 and CC 1 meet AC and BC at Y , Z respectively then CY

XC +n

CQ

d cot γ +d

cot γ+1 YA = cot α+1 and AZ = cot α+1 ZB cot β +1 . Therefore BX CY XC AZ YA ZB = 1, so by Ceva’s theorem, AX, BY,CZ

have a common point. Second solution. Let A 2 be the center of the square constructed over BC out-

side △ABC. Since this square and the inscribed square corresponding to the side BC are homothetic, A, A 1 , and A 2 are collinear. Points B 2 ,C 2 are anal- ogously defined. Denote the angles BAA 2 ,A 2 AC ,CBB 2 ,B 2 BA , ACC 2 ,C 2 CB by α 1 , α 2 , β 1 , β 2 , γ 1 , γ 2 . By the law of sines we have

684 4 Solutions

sin α 1 sin ( β + 45 ◦ ) sin β 1 sin ( γ + 45 ◦ ) sin γ 1 sin ( α + 45 ◦ ) =

= . ) sin β 2 sin ( α + 45 ◦ ) sin γ 2 sin ( β + 45 ◦ )

sin α 2 sin ( γ + 45 ◦

Since the product of these ratios is 1, by the trigonometric Ceva’s theorem AA 2 , BB 2 ,CC 2 are concurrent.

16. Since ∠OCP = 90 ◦ −∠A, we are led to showing that ∠OCP > ∠COP, i.e., OP > CP . By the triangle inequality it suffices to prove CP < 1 2 CO . Let CO = R. The law of sines yields

CP = AC cos γ = 2R sin β cos γ < 2R sin β cos ( β + 30 ◦ ). Finally, we have

2 sin β cos ( β + 30 ◦ ) = sin(2 β + 30 ◦ ) − sin30 ◦ 1 ≤ ,

2 which completes the proof.

17. Let us investigate a more general problem, in which G is any point of the plane such that AG , BG,CG are sides of a triangle. Let F be the point in the plane such that BC : CF : FB = AG : BG : CG and F, A lie on different sides of BC. Then by Ptolemy’s inequality, on BPCF we have

AG · AP + BG · BP + CG ·CP = AG · AP + AG BC (CF · BP + BF ·CP) ≥ AG · AP +

AG

BC BC · PF. Hence AG · AP + BG · BP +CG ·CP ≥ AG · AF,

(1) where equality holds if and only if P lies on the segment AF and on the circle

BCF . Now we return to the case of G the centroid of △ABC. We claim that F is then the point b G C

in which the line AG meets again the F circumcircle of △BGC. Indeed, if M is the midpoint of AB, by the law of sines we have BC

Cb G = sin ∠B b GC sin ∠CB b G = sin ∠BGM sin ∠AGM =

AG BC AG

BG , and similarly Bb G = CG . Thus (1)

implies

A B AG · AP + BG · BP+CG ·CP ≥ AG · A b G .

It is easily seen from the above considerations that equality holds if and only if P ≡ G, and then the (minimum) value of AG · AP + BG · BP+CG ·CP equals

AG 2 + BG +CG 2 +b 2 +c 2 2 = .

Notice that AG −→ −→ −→ −→ −→

Second solution. · AP ≥ AG · AP = AG ·( AG + GP ). Summing this inequality with analogous inequalities for BG · BP and CG · CP gives us

· AP + BG · BP + CG · CP ≥ AG 2 + BG 2 + CG 2 +( AG + BG + CG )· GP = AG 2 + BG 2 + CG 2 = a 2 +b 2 +c 3 2 . Equality holds if and only if P ≡ Q.

AG −→ −→ −→ −→

4.42 Shortlisted Problems 2001 685

18. Let α 1 , β 1 , γ 1 , α 2 , β 2 , γ 2 denote the angles ∠MAB, ∠MBC, ∠MCA, ∠MAC, ∠MBA, ∠MCB respectively. Then

2 = sin α 1 sin α

2 = sin β 1 sin MA β 2 MB , MC 2 = sin γ 1 sin γ 2 ; hence p (M) 2 = sin α 1 sin α 2 sin β 1 sin β 2 sin γ 1 sin γ 2 . Since

1 1 α sin α 1 sin α 2 = (cos( α

1 − α 2 ) − cos( α 1 + α 2 )≤ (1 − cos α ) = sin ,

2 2 2 we conclude that

p (M) ≤ sin sin sin .

Equality occurs when α 1 = α 2 , β 1 = β 2 , and γ 1 = γ 2 , that is, when M is the incenter of △ABC. It is well known that µ (ABC) = sin α 2 sin β 2 sin γ 2 is maximal when △ABC is equilateral (it follows, for example, from Jensen’s inequality applied to ln sin x). Hence max µ (ABC) = 1 8 .

19. The hexagon AEBFCD is obviously

convex. Therefore ∠AEB + ∠BFC + F ∠CDA = 360 ◦ . Using this relation

we obtain that the circles ω 1 , ω 2 ,

ω 3 with centers at D , E, F and radii DA , EB, FC respectively all pass

F ′ O D ′ through a common point O. Indeed, if ω 1 ∩ ω 2 A = {O}, then ∠AOB = B

180 ◦ − ∠AEB/2 and ∠BOC = 180 ◦ − ∠BFC /2. We now have ∠COA =

180 ◦ − ∠CDA/2 as well, i.e., O ∈ ω 3 .

The point O is the reflection of A with E respect to DE. Similarly, it is also the reflection of B with respect to EF, and that

of C with respect to FD. Hence

S DEF Analogously EC =1+ S ODF and FA =1+ S EE ODE ′ S DEF FF ′ S DEF . Adding these relations gives

20. By Ceva’s theorem, we can choose real numbers x , y, z such that −→

BD z CE x

AF y

−→ = , −→ = , and −→ = .

DC y EA z

FB x

686 4 Solutions The point P lies outside the triangle ABC if and only if x , y, z are not all of the

same sign. In what follows, S X will denote the signed area of a figure X. Let us assume that the area S ABC of △ABC is 1. Since S PBC :S PCA :S PAB =x: y : z and S PBD :S PDC = z : y, it follows that S PBD = z

y +z x +y+z . Hence S PBD =

z (z+x) x +y+z ,S PAF = x (x+y) x +y+z . By the condition of the problem we have |S PBD | = |S PCE | = |S PAF |, or

y (y+z) x +y+z PCE

|x(x + y)| = |y(y + z)| = |z(z + x)|. Obviously x , y, z are nonzero, so that we can put w.l.o.g. z = 1. At least two of the

numbers x (x + y), y(y + 1), 1(1 + x) are equal, so we can assume that x(x + y) = y (y + 1). We distinguish two cases:

(i) x (x + y) = y(y + 1) = 1 + x. Then x = y 2 + y − 1, from which we obtain (y 2 2 + y − 1)(y 4 + 2y − 1) = y(y + 1). Simplification gives y + 3y 3 −y 2 − 4y + 1 = 0, or

(y − 1)(y 3 + 4y 2 + 3y − 1) = 0. If y = 1, then also z = x = 1, so P is the centroid of △ABC, which is not an

exterior point. Hence y 3 + 4y 2 + 3y − 1 = 0. Now the signed area of each of the triangles PBD , PCE, PAF equals

yz

S PAF =

(x + y)(x + y + z)

y 3 + 4y 2 + 3y − 2 It is easy to check that not both of x , y are positive, implying that P is indeed

(y 2 + 2y − 1)(y 2 + 2y)

outside △ABC. This is the desired result. (ii) x (x + y) = y(y + 1) = −1 − x. In this case we are led to

f (y) = y 4 + 3y 3 +y 2 − 2y + 1 = 0. We claim that this equation has no real solutions. In fact, assume that y 0 is

a real root of f (y). We must have y 0 < 0, and hence u = −y 0 > 0 satisfies

3u 3 −u 4 = (u + 1) 2 . On the other hand, 3u 3 4 −u 3 =u (3 − u) = 4u

where at least one of the inequalities is strict, a contradiction. Remark. The official solution was incomplete, missing the case (ii).

21. We denote by p (XY Z) the perimeter of a triangle XY Z.

4.42 Shortlisted Problems 2001 687 If O is the circumcenter of △ABC, then A 1 ,B 1 ,C 1 are the midpoints of the corre-

sponding sides of the triangle, and hence p (A 1 B 1 C 1 ) = p(AB 1 C 1 ) = p(A 1 BC 1 )= p (A 1 B 1 C ). Conversely, suppose that p (A 1 B 1 C 1 ) ≥ p(AB 1 C 1 ), p(A 1 BC 1 ), p(A 1 B 1 C ). Let α 1 , α 2 , β 1 , β 2 , γ 1 , γ 2 denote the angles ∠B 1 A 1 C , ∠C 1 A 1 B , ∠C 1 B 1 A , ∠A 1 B 1 C ,

∠A 1 C 1 B , ∠B 1 C 1 A .

C the fourth vertex of the parallel-

Suppose that γ 1 , β 2 ≥ α . If A 2 is

ogram B 1 AC 1 A 2 , then these condi-

A tions imply that A 2

1 is in the interior

or on the border of △B 1 C 1 A 2 , and

therefore p (A 1 B 1 C 1 ) ≤ p(A 2 B 1 C 1 )=

p (AB 1 C 1 ). Moreover, if one of the in- γ 2 γ 1 equalities γ 1 ≥ α , β 2 ≥ α is strict, A C 1 B then p (A 1 B 1 C 1 ) is strictly less than

p (AB 1 C 1 ), contrary to the assumption. Hence

the last two inequalities being obtained analogously to the first one. Because of the symmetry, there is no loss of generality in assuming that γ 1 ≤ α . Then since γ 1 + α 2 = 180 ◦ − β = α + γ , it follows that α 2 ≥ γ . From (1) we deduce β 1 ≤ γ , which further implies γ 2 ≥ β . Similarly, this leads to α 1 ≤ β and β 2 ≥ α . To sum up, γ 1 ≤ α ≤ β 2 , α 1 ≤ β ≤ γ 2 , β 1 ≤ γ ≤ α 2 .

Since OA 1 BC 1 and OB 1 CA 1 are cyclic, we have ∠A 1 OB = γ 1 and ∠A 1 OC = β 2 . Hence BO : CO = cos β 2 : cos γ 1 , hence BO ≤ CO. Analogously, CO ≤ AO and AO ≤ BO. Therefore AO = BO = CO, i.e., O is the circumcenter of ABC.

22. Let S and T respectively be the points on the extensions of AB and AQ over B and Q such that BS = BP and QT = QB. It is given that AS = AB + BP = AQ + QB = AT . Since ∠PAS = ∠PAT , the triangles APS and APT are congruent, from which we deduce that ∠AT P = ∠ASP = β /2 = ∠QBP. Hence ∠QT P = ∠QBP. If P does not lie on BT , then the last equality implies that △QBP and △QT P are congruent, so P lies on the internal bisector of ∠BQT . But P also lies on the internal bisector of ∠QAB; consequently, P is an excenter of △QAB, thus lying on the internal bisector of ∠QBS as well. It follows that ∠PBQ = β /2 = ∠PBS = 180 ◦ − β , so β = 120 ◦ , which is impossible. Therefore P ∈ BT, which means that T ≡ C. Now from QC = QB we conclude that 120 ◦ − β = γ = β /2, i.e., β = 80 ◦ and γ = 40 ◦ .

23. For each positive integer x, define α (x) = x/10 r if r is the positive integer sat- isfying 10 r

r ≤ x < 10 +1 . Observe that if α (x) α (y) < 10 for some x, y ∈ N, then

688 4 Solutions

α (xy) = α (x) α (y). If, as usual, [t] means the integer part of t, then [ α (x)] is actually the leftmost digit of x. Now suppose that n is a positive integer such that k ≤ α ((n + k)!) < k + 1 for k = 1, 2, . . . , 9. We have

α ((n + k)!)

k +1

1 < α (n + k) =

for 2 ≤ k ≤ 9,

α ((n + k − 1)!) k −1

from which we obtain α (n + k + 1) > α (n + k) (the opposite can hold only if α (n + k) ≥ 9). Therefore

1 < α (n + 2) < ··· < α (n + 9) ≤ .

4 On the other hand, this implies that α ((n + 4)!) = α ((n + 1)!) α (n + 2) α (n +

3 ) α (n + 4) < (5/4) 3 α ((n + 1)!) < 4, contradicting the assumption that the left- most digit of (n + 4)! is 4.

24. We shall find the general solution to the system. Squaring both sides of the first equation and subtracting twice the second equation we obtain (x − y) 2 =z 2 +u 2 . Thus (z, u, x − y) is a Pythagorean triple. Then it is well known that there are positive integers t , a, b such that z = t(a 2 −b 2 ), u = 2tab (or vice versa), and x − y = t(a 2 +b 2 ). Using that x + y = z + u we come to the general solution:

z = t(a 2 −b 2 ), u = 2tab. Putting a /b = k we obtain x

x = t(a 2 + ab),

= t(ab − b 2 );

with equality for k √ −1=

2. On the other hand, k can be arbitrarily close to √ √

2. Hence m =3+2 2. Remark. There are several other techniques for solving the given system. The

2, and so x /y can be arbitrarily close to 3 + 2

exact lower bound of m itself can be obtained as follows: by the system x y −

6 x y +1= z −u 2 y √ ≥ 0, so x/y ≥ 3 + 2 2.

25. Define b n = |a n +1 −a n | for n ≥ 1. From the equalities a n +1 =b n −1 +b n −2 , from

a n =b n −2 +b n −3 we obtain b n = |b n −1 −b n −3 |. From this relation we deduce that b m ≤ max(b n ,b n +1 ,b n +2 ) for all m ≥ n, and consequently b n is bounded.

Lemma. If max (b n ,b n +1 ,b n +2 ) = M ≥ 2, then max(b n +6 ,b n +7 ,b n +8 ) ≤ M − 1. Proof. Assume the opposite. Suppose that b j = M, j ∈ {n,n + 1,n + 2}, and let

b j +1 = x and b j +2 = y. Thus b j +3 = M − y. If x,y,M − y are all less than M , then the contradiction is immediate. The remaining cases are these: (i) x = M. Then the sequence has the form M, M, y, M − y,y,... , and since max (y, M − y,y) = M, we must have y = 0 or y = M.

4.42 Shortlisted Problems 2001 689 (ii) y = M. Then the sequence has the form M, x, M, 0, x, M − x,... , and

since max (0, x, M − x) = M, we must have x = 0 or x = M. (iii) y = 0. Then the sequence is M, x, 0, M, M − x,M − x,x,... , and since max (M − x,x,x) = M, we have x = 0 or x = M. In every case M divides both x and y. From the recurrence formula M also divides b i for every i < j. However, b 2 = 12 12 − 11 11 and b 4 = 11 11 are relatively prime, a contradiction. From max (b 1 ,b 2 ,b 3 ) ≤ 13 13 and the lemma we deduce inductively that b n ≤1 for all n ≥ 6 · 13 13 − 5. Hence a n =b n −2 +b n −3 takes only the values 0, 1, 2 for n ≥ 6 · 13 13 − 2. In particular, a 14 14 is 0, 1, or 2. On the other hand, the sequence a n modulo 2 is as follows: 1 , 0, 1, 0, 0, 1, 1; 1, 0, 1, 0, . . .; and therefore it is periodic with period 7. Finally, 14 14 ≡ 0 modulo 7, from which we obtain

a 14 14 ≡a 7 ≡ 1 (mod 2). Therefore a 14 14 = 1.

26. Let C be the set of those a ∈ {1,2,..., p − 1} for which a 2 p −1 ≡ 1 (mod p ). At first, we observe that a , p − a do not both belong to C, regardless of the value

of a. Indeed, by the binomial formula, (p − a) p −1 −a p −1 ≡ −(p − 1)p a p −2 6≡ 0 (mod p 2 ) . As a consequence we deduce that

|C| ≤ p −1

2 . Further, we observe that p −k∈

C ⇔ k ≡ k(p − k) p −1 (mod p 2 ), i.e.,

− (p − 1)p k p −2 )≡k + p (mod p 2 ). (1) Now assume the contrary to the claim, that for every a = 1, . . . , p − 2 one of

p − k ∈ C ⇔ k ≡ k(k p −1

a , a + 1 is in C. In this case it is not possible that a, a + 1 are both in C, for then p − a, p − a − 1 6∈ C. Thus, since 1 ∈ C, we inductively obtain that 2,4,... , p −

1 6∈ C and 1,3,5,..., p − 2 ∈ C. In particular, p − 2, p − 4 ∈ C, which is by (1) equivalent to 2

≡2 2 + p and 4 ≡ 4 + p (mod p ). However, squaring the former equality and subtracting the latter, we obtain

2 p +1 p ≡ p (mod p 2 ), or 4 ≡ 1 (mod p), which is a contradiction unless p = 3. This finishes the proof.

27. The given equality is equivalent to a 2 − ac + c 2 =b 2 + bd + d 2 . Hence (ab +

cd )(ad + bc) = ac(b 2 + bd + d 2 ) + bd(a 2 − ac + c 2 ), or equivalently, (ab + cd)(ad + bc) = (ac + bd)(a 2 − ac + c 2 ).

(1) Now suppose that ab + cd is prime. It follows from a > b > c > d that

ab + cd > ac + bd > ad + bc; (2) hence ac + bd is relatively prime with ab + cd. But then (1) implies that ac + bd

divides ad + bc, which is impossible by (2). Remark. Alternatively, (1) could be obtained by applying the law of cosines

and Ptolemy’s theorem on a quadrilateral XY ZT with XY = a, Y Z = c, ZT = b, TX = d and ∠Y = 60 ◦ , ∠T = 120 ◦ .

690 4 Solutions

28. Yes. The desired result is an immediate consequence of the following fact ap- plied on p = 101.

Lemma. For any odd prime number p, there exist p nonnegative integers less than 2p 2 with all pairwise sums mutually distinct. Proof. We claim that the numbers a n = 2np + (n 2 ) have the desired property, where (x) denotes the remainder of x upon division by p. Suppose that a k +a l =a m +a n . Assume that k ≤ l and m ≤ n. By the con- struction of a i , we have 2p (k + l) ≤ a k +a l < 2p(k + l + 1). Hence we must

have k + l = m + n, and therefore also (k 2 ) + (l 2 ) = (m 2 ) + (n 2 ). Thus k +l≡m+n

k 2 +l 2 ≡m 2 +n 2 (mod p) . But then it holds that

and

(k − l) 2 = 2(k 2 +l 2 2 ) − (k + l) 2 ≡ (m−n) (mod p), so k − l ≡ ±(m − n), which leads to {k,l} = {m,n}. This proves the lemma.

4.43 Shortlisted Problems 2002 691