Solutions to the Shortlisted Problems of IMO 1995

4.36 Solutions to the Shortlisted Problems of IMO 1995

1. Let x = 1 a ,y = 1 b ,z = 1 c . Then xyz = 1 and

y +z z +x x +y We must prove that S ≥ 3 2 . From the Cauchy–Schwarz inequality,

a 3 (b + c) b 3

(c + a) c (a + b)

2 x +y+z [(y + z) + (z + x) + (x + y)] · S ≥ (x + y + z) ⇒S≥

2 It follows from the A-G mean inequality that x +y+z

2 ≥ 3 2 xyz = 2 ; hence the proof is complete. Equality holds if and only if x = y = z = 1, i.e., a = b = c = 1.

2 2 Remark. 2 After reducing the problem to x y

y +z + z +x + z x +y ≥ 2 , we can solve the problem using Jensen’s inequality applied to the function g (u, v) = u 2 /v. The problem can also be solved using Muirhead’s inequality.

2. We may assume c ≥ 0 (otherwise, we may simply put −y i in the place of y i ). Also, we may assume a ≥ b. If b ≥ c, it is enough to take n = a + b − c, x 1 = ··· = x a = 1, y 1 = ··· = y c =y a +1 = ··· = y a +b−c = 1, and the other x i ’s and y i ’s equal to 0, so we need only consider the case a > c > b. We proceed to prove the statement of the problem by induction on a + b. The case a + b = 1 is trivial. Assume that the statement is true when a + b ≤ N, and let a + b = N + 1. The triple (a + b − 2c,b,c − b) satisfies the condition (since

2 (a + b − 2c)b − (c − b) 2 = ab − c ), so by the induction hypothesis there are n -tuples (x i ) n i =1 and (y i ) n i =1 with the wanted property. It is easy to verify that

(x i +y i ) n i =1 and (y i ) n i =1 give a solution for (a, b, c).

a 2 +a

3. Write A = i i 2 +1 −a 2 i +2

a i +a i +1 −a i +2 . Since 2a i a i +1 ≥ 4(a i +

a i +1 − 2) (which is equivalent to (a i − 2)(a i +1 − 2) ≥ 0), it follows that A i ≤a i +

a i +a i +1 −a i +2 ≤a i +a i +1 +a i +2 −4 1 + 4 , because

i +a i +1 −a i +2 ≤ 4. Therefore A i ≤a i +a i +1 − 2, so ∑ i =1 A i ≤ 2s − 2n as required.

1 ≤a n

4. The second equation is equivalent to 2 a 2

yz + b zx 2 + c abc

xy + xyz = 4. Let x 1 = √ yz ,y 1 = √ b zx ,z 1 = √ c xy . Then x 2 2 1 2 +y 1 +z 1 +x 1 y 1 z 1 = 4, where 0 < x 1 ,y 1 ,z 1 < 2. Regarding

this as a quadratic equation in z , the discriminant

1 (4 − x 1 )(4 − y 1 ) suggests that we let x 1 = 2 sin u, y 1 = 2 sin v, 0 < u, v < π /2. Then it is directly shown

that z 1 will be exactly 2 cos (u + v) as the only positive solution of the quadratic equation. Thus a

= 2√yz sin u, b = 2 xz sin v, c = 2√xy(cos u cosv − sinusinv), so from x + y + z − a − b − c = 0 we obtain

√ ( x cos v − y cos u ) +( x sin v + √ y sin u − z ) 2 = 0,

594 4 Solutions which implies √

Therefore z = a +b . Similarly, x = b 2 +c 2 and y = c +a 2 . It is clear that the triple (x, y, z) = b +c 2 , c +a 2 , a +b 2 is indeed a (unique) solution of the given system of equations. Second solution. Put x = b +c − u, y = c +a − v, z = a 2 +b 2 − w, where u ≤ b 2 +c 2 , v

c ≤ +a ,w a ≤ +b and u + v + w = 0. The equality abc + a 2

2 2 x +b 2 y +c 2 z = 4xyz becomes 2 (au 2 + bv 2 + cw 2 + 2uvw) = 0. Now uvw > 0 is clearly impossible.

On the other hand, if uvw ≤ 0, then two of u,v,w are nonnegative, say u,v ≥ 0. Taking into account w

= −u−v, the above equality reduces to 2[(a+c−2v)u 2 + (b + c − 2u)v 2 + 2cuv] = 0, so u = v = 0.

Third solution. The fact that we are given two equations and three variables suggests that this is essentially a problem on inequalities. Setting f (x, y, z) = 4xyz

2 −a 2 x −b y −c 2 z , we should show that max f (x, y, z) = abc, for 0 < x, y, z, x + y + z = a + b + c, and find when this value is attained. Thus we apply

Lagrange multipliers to F (x, y, z) = f (x, y, z) − λ (x + y + z − a − b − c), and obtain that f takes a maximum at

2 (x, y, z) such that 4yz − a 2 = 4zx − b = 4xy

−c 2 = λ and x + y + z = a + b + c. The only solution of this system is

(x, y, z) = b +c , c +a , a 2 +b 2 2 .

5. Suppose that a function f satisfies the condition, and let c be the least upper bound of { f (x) | x ∈ R}. We have c ≥ 2, since f (2) = f (1 + 1/1 2 ) = f (1) +

f (1) 2 = 2. Also, since c is the least upper bound, for each k = 1, 2, . . . there is an x k ∈ R such that f (x k ) ≥ c − 1/k. Then

x k k On the other hand,

kk It follows that

≥c c −1−

which cannot hold for k sufficiently large. Second solution. Assume that f exists and let n be the least integer such that

f (x) ≤ n 4 for all x. Since f (2) = 2, we have n ≥ 8. Let f (x) > n −1 4 . Then f (1/x) =

f (x + 1/x 2 ) − f (x) < 1/4, so f (1/x) > −1/2. On the other hand, this implies

4 < f (x) = f (1/x + x ) − f (1/x) < 4 + 2 , which is impossible when n ≥ 8.

n −1 2 2 2 n

4.36 Shortlisted Problems 1995 595

6. Let y =x n

(n−1) i

+ ··· + x n =∑ j =2 ( j − 1)x j , and z i = 2 y i − (n − i)Y. Then n −1

i x j − ∑ (n − i)x i Y = ∑ x i y i − ∑ (n − i)x i 2 Y

so it remains to show that ∑ n −1 x z

i =1 i i > 0. Since ∑ −1 i =1 y

i = Y and ∑ −1 i =1 (n − i) =

i = 0. Note that Y < ∑ j =2 ( j − 1)x n = 2 x n , and conse- quently z

n (n−1) n

2 , we have ∑z

(n−1)

n (n−1)

n −1 = 2 x n −Y > 0. Furthermore, we have

2 −i−1 − n −i−1 n −i which means that z 1 n −1 < z 2 z n n −2 < ··· < −1 1 . Therefore there is a k for which

n −i

z 1 ,...,z k ≤ 0 and z k +1 ,...,z n −1 > 0. But then z i (x i −x k ) ≥ 0, i.e., x i z i ≥x k z i for all i, so ∑ n −1

i =1 x i z i >∑ i =1 x k z i = 0 as required. Second solution. Set X

n −1

j =1 (n − j)x j and Y =∑ j =2 ( j − 1)x j . Since 4XY = (X + Y ) 2 − (X −Y ) 2 , the RHS of the inequality becomes

=∑ n −1

1 XY =

4 (n − 1) ∑ i − ∑ (2i − 1 − n)x i i =1 i =1

4 (n − 1) (∑ i =1 x i ) − (n − 1)∑ i <j (x j −x i ) . Since ∑ i =1 (2i−1− n )x i =∑ i <j (x j −x i ) also holds, we must prove that

The LHS is 1 2 n

∑ (x j −x i )

> (n − 1) 2 ∑ (x

Putting x i +1 −x i =d i > 0 (so, x j −x i =d i +d i +1 + ··· + d j −1 ) and expand- ing the obtained expressions, we reduce this inequality to ∑ k 2 k 2 d (n − k) 2

2 k 2∑ + k <l kl (n − k)(n − l)d k d l >∑ k (n − 1)k(n − k)d k +2∑ k <l (n − 1)k(n − l)d k d l ,

which is verified immediately by comparing coefficients. Remark. An inequality significantly stronger than (1) in the second solution has

appeared later, as IMO 03-5.

7. The result is trivial if O coincides with X or Y , so let us assume it does not. From OB · ON = OC · OM = OX · OY we deduce that BCMN is a cyclic quadrilateral. Further, if O lies between X and Y , then ∠MAD + ∠MND = ∠MAD + ∠MNB + ∠BND = ∠MAD + ∠MCA + ∠AMC = 180 ◦ . Similarly, we also have ∠MAD + ∠MND = 180 ◦ if O is not on the segment XY . Therefore ADNM is cyclic. Now let AM and DN intersect at Z and let the line ZX intersect the two circles at Y 1

596 4 Solutions and Y 2 . Then ZX · ZY 1 = ZM · ZA = ZN · ZD = ZX · ZY 2 . Hence Y 1 =Y 2 =Y,

implying that Z lies on XY . Second solution. Let Z 1 ,Z 2 be the points in which AM , DN respectively meet

XY , and P = BC ∩ XY . Then, from △OPC ∼ △APZ 1 , we have PZ 1 = PA ·PC PO =

PX 2 2

PO and analogously PZ 2 = PX PO . Hence, we conclude that Z 1 ≡Z 2 .

be the points symmetric to A , B,C with respect to the midpoints of BC ,CA, AB respectively. From the condition on X we have XB 2 − XC 2 = AC 2 − AB 2 =A ′ B 2 −A ′ C 2 , and hence X must lie on the line through A ′ perpendicular to BC . Similarly, X lies on the line through B ′ perpendicular to CA. It follows that there is a unique position for X , namely the orthocenter of △A ′ B ′ C ′ . It easily follows that this point X satisfies the original equations.

8. Let A ′ ,B ′ ,C ′

9. If EF is parallel to BC, △ABC must be isosceles and E,Y are symmetric to F,Z with respect to AD, so the result follows. Now suppose that EF meets BC at P.

By Menelaus’s theorem, BP CP = BF FA AE · EC = BD DC (since BD = BF, CD = CE, AE = AF ). It follows that the point P depends only on D and not on A. In particular, the same point is obtained as the intersection of ZY with BC. Therefore PE · PF = PD 2 = PY · PZ, from which it follows that EFZY is a cyclic quadrilateral.

Second solution. Since CD = CY = CE and BD = BZ = BF, all angles of EFZY can be calculated in terms of angles of ABC and Y ZBC. In fact, ∠FEY = 1 2 (∠A + ∠C + ∠BCY ) and ∠FZY = 1 2 (180 ◦ + ∠B + ∠BCY ), which gives us ∠FEY + ∠FZY = 180 ◦ .

10. Let the two triangles be X 1 Y 1 Z 1 ,X 2 Y 2 Z 2 , with X 1 = BB 1 ∩CC 1 ,Y 1 = CC 1 ∩ AA 1 , Z 1 =AA 1 ∩ BB 1 ,X 2 = BB 2 ∩CC 2 ,Y 2 = CC 2 ∩ AA 2 ,Z 2 = AA 2 ∩ BB 2 .

First, let us observe that ∠ABB 2 =

∠ACC 1 and ∠ABB 1 = ∠ACC 2 . We

now obtain that ∠BZ 1 A 1 = ∠BAA 1 +

∠ABB 1 = ∠BCC 2 + ∠C 2 CA = ∠C and

similarly ∠AZ 2 B 2 = ∠C, ∠AY 1 C 1 =

∠CY 1

2 A 2 = ∠B. Also the triangles

ABB 2 and ACC 1 are similar; hence AC 1 /AC = AB 2 /AB. From the law of

Y 1 X 2 sines we obtain

AZ 1 AB AB AC AC

sin ∠ABZ 1 sin ∠AZ 1 B sin ∠C

sin ∠B sin ∠AY 2 C

AY 2

sin ∠ACY =⇒ AZ 1 2 2 . Analogously, BX 1 = BZ 2 and CY 1 = CX 2 . Furthermore, again from the sine for-

= AY

mula,

4.36 Shortlisted Problems 1995 597

sin ∠AZ 2 B 2 sin ∠AB 2 Z 2 Hence, AY 1 = AZ 2 and, analogously, BZ 1 = BX 2 and CX 1 = CY 2 . We deduce

that Y 1 Z 2 k BC and Z 2 X 1 k AC, which gives us ∠Y 1 Z 2 X 1 = 180 ◦ − ∠C = 180 ◦ − ∠Y 1 Z 1 X 1 . It follows that Z 2 lies on the circle circumscribed about △X 1 Y 1 Z 1 .

Similarly, so do X 2 and Y 2 .

Second solution. Let H be the orthocenter of △ABC. Triangles AHB, BHC, CHA , ABC have the same circumradius R. Additionally,

∠HAA i = ∠HBB i = ∠HCC i = θ (i = 1, 2).

Since ∠HBX 1 = ∠HCX 1 = θ , BCX 1 H is concyclic and therefore HX 1 = 2R sin θ . The same holds for HY 1 , HZ 1 , HX 2 , HY 2 , HZ 2 . Hence X i ,Y i ,Z i (i = 1, 2) lie on a circle centered at H.

11. Triangles BCD and EFA are equilateral, and hence BE is an axis of symmetry of ABDE. Let C ′ ,F ′ respectively be the points symmetric to C , F with respect to

BE . The points G and H lie on the circumcircles of ABC ′ and DEF ′ respectively (because, for instance, ∠AGB = 120 ◦ = 180 ◦ − ∠AC ′ B ); hence from Ptolemy’s theorem we have AG + GB = C ′ G and DH + HE = HF ′ . Therefore

AG + GB + GH + DH + HE = C ′ G + GH + HF ′ ≥C ′ F ′ = CF,

with equality if and only if G and H both lie on C ′ F ′ .

Remark. Since by Ptolemy’s inequality AG + GB ≥ C ′ G and DH + HE ≥ HF ′ , the result holds without the condition ∠AGB = ∠DHE = 120 ◦ .

12. Let O be the circumcenter and R the circumradius of A 1 A 2 A 3 A 4 . We have OA 2 i = −→

+( OA i − OG )) 2 = OG 2 + GA 2 i +2 OG · GA i . Summing up these equalities for i

= 1, 2, 3, 4 and using that ∑ i =1 GA i = 0 , we obtain

∑ OA 2 i = 4OG 2 + GA 2 2 ∑ 2 i ⇐⇒ ∑ GA 2 i = 4(R − OG ).

Now we have that the potential of G with respect to the sphere equals GA i ·GA ′ i = R 2 − OG 2 . Plugging in these expressions for GA ′ i , we reduce the inequalities we must prove to GA 2 2 1 2 · GA 2 · GA 3 · GA 4 ≤ (R − OG )

and

(R 2 − OG 2 ) ∑ ≥ ∑ GA i .

Inequality (2) immediately follows from (1) and the quadratic-geometric mean inequality for GA i . From the Cauchy–Schwarz inequality we have ∑ 4 i =1 GA 4 i ≥

598 4 Solutions

4 ∑ i =1 GA i and ∑ i =1 GA i ∑ i =1 GA i ≥ 16, hence the inequality (3) fol- lows from (1) and from

∑ GA i ∑

∑ GA i ∑

≥4 ∑ GA i .

13. If O lies on AC, then ABCD, AKON, and OLCM are similar; hence AC √ √ √ = AO +

OC implies S = S 1 + S 2 .

Assume that O does not lie on AC and D that w.l.o.g. it lies inside triangle ADC. Let us denote by T

1 ,T 2 the areas of par-

TM

allelograms KBLO, NOMD respect- A C S 1 S 2 ively. Consider a line through O that

OZY intersects AD, DC, CB, BA respect-

ively at X , Y , Z, W so that OW /OX = K T 1 L

OZ /OY (such a line exists by a conti- nuity argument: the left side is smaller

B when W = X = A, but greater when √ Y = Z = C). The desired inequality is equivalent to T 1 +T 2 ≥2 S 1 S √ . Since 2

triangles W KO, OLZ, W BZ are similar and W O √ + OZ = W Z, we have S √ √ √ W KO + S OLZ = S W BZ √ = S W KO +S OLZ +T 1 , which implies T 1 =2 S W KO S OLZ . Similarly, T 2 =2 S XNO S OMY . Since OW /OZ = OX/OY , we have S W KO /S XNO =S OLZ /S OMY . Therefore we obtain

T 1 +T 2 =2 S W KO S OLZ +2 S XNO S OMY

p =2 (S W KO +S XNO )(S OLZ +S OMY )≥2 S 1 S 2 .

Second solution. Using an affine transformation of the plane one can transform any nondegenerate quadrilateral into a cyclic one, thereby preserving parallel- ness and ratios of areas. Thus we may assume w.l.o.g. that ABCD is cyclic. By a well-known formula, the area of a cyclic quadrilateral with sides a , b, c, d and semiperimeter p is given by

p S = (p − a)(p − b)(p − c)(p − d).

Let us set AK =a 1 , KB =b 1 , BL =a 2 , LC =b 2 , CM =a 3 , MD =b 3 , DN =a 4 , NA =b 4 . Then the sides of quadrilateral AKON are a i , the sides of CLOM are b i , and the sides of ABCD are a i +b i (i = 1, 2, 3, 4). If p and q are the semiperimeters of AKON and CLOM, and x i =p−a i ,y i = q −b i , then we have S 1 = √x 1 x 2 x 3 x 4 p , S 2 = √y 1 y 2 y 3 y 4 , and S = (x 1 +y 1 )(x 2 +y 2 )(x 3 +y 3 )(x 4 +y 4 ) . Thus we need to show that √ 4 x p

4 + 4 y 1 y 2 y 3 y 4 ≤ 4 (x 1 +y 1 )(x 2 +y 2 )(x 3 +y 3 )(x 4 +y 4 ).

4.36 Shortlisted Problems 1995 599 By setting y i =t i x i we reduce this inequality to

1 4 2 3 4 ≤ (1 + t 1 )(1 + t 2 )(1 + t 3 )(1 + t 4 ). One way to prove the last inequality is to apply the simple inequality

1 + uv ≤ (1 + u)(1 + v)

to t 1 t 2 , √ t 3 t 4 and then to t 1 ,t 2 and t 3 ,t 4 .

14. Let BB ′ cut CC ′ at P. Since ∠B ′ BC ′ = ∠B ′ CC ′ , it follows that ∠PBH = ∠PCH. Let D and E be points such that BPCD and HPCE are parallelograms (con- sequently, so is BHED). Triangles BAC and C ′ AB ′ are similar, from which we deduce that △B ′ H ′ C ′ and △BHC

A are similar, as well as

△B ′ PC ′ and C ′ H B ′ △BDC. Hence B ′ PC ′ H ′ and BDCH

are similar, from which we obtain P ∠H ′ PB ′ = ∠HDB. Now ∠CDE =

∠PBH = ∠PCH = ∠CHE implies that H

B HCED C is a cyclic quadrilateral. There- fore ∠BPH = ∠DCE = ∠DHE =

E ∠HDB = ∠H ′ PB ′ ; hence HH ′ also

passes through P. D Second solution. Let us start with observations △HBC ∼ △H ′ B ′ C ′ , ∠PBH =

∠PCH, and ∠PB ′ H ′ = ∠PC ′ H ′ .

By Ceva’s theorem in trigonometric form applied to △BPC and the point H, we have sin ∠BPH = sin ∠HBP sin ∠HCB = sin ∠HCB sin ∠HPC sin ∠HBC · sin ∠HCP sin ∠HBC . Similarly, Ceva’s theorem for

△B ′ ′

PC ′ and point H ′ yields sin ∠B PH ′ = sin ∠H ′ C ′ B ′

sin ∠H ′ PC ′

sin ∠H ′ B ′ C ′ . Thus it follows that

sin ∠B ′ PH ′

sin ∠BPH

sin ∠H ′

PC ′

sin ∠HPC

which finally implies that ∠BPH = ∠B ′ PH ′ .

15. We show by induction on k that there exists a positive integer a k for which a 2 k ≡ −7 (mod 2 k ). The statement of the problem follows, since every a k + r2 k (r =

0 , 1, . . . ) also satisfies this condition. Note that for k = 1, 2, 3 one can take a k = 1. Now suppose that a 2 ≡ −7 (mod

2 k ) for some k > 3. Then either a 2 k +1 ) or a 2 k k ≡ −7 (mod 2 k ≡2 k − 7 (mod 2 k +1 ). In the former case, take a k +1 =a k . In the latter case, set a k +1 =a k +2 k −1 . Then

≡ −7 (mod 2 ) because a k is odd.

16. If A is odd, then every number in M 1 is of the form x (x + A) + B ≡ B (mod 2), while numbers in M 2 are congruent to C modulo 2. Thus it is enough to take

C ≡ B + 1 (mod 2). If A is even, then all numbers in M 2

+B− 4 and are congruent to B − A 4 or B − A 4 + 1 modulo 4, while numbers in M 2 are congruent to C modulo 4. So one can choose any C ≡B− A 4 2 + 2 (mod 4).

1 have the form X + A A 2

600 4 Solutions

17. For n = 4, the vertices of a unit square A 1 A 2 A 3 A 4 and p 1 =p 2 =p 3 =p 4 = 1 6 satisfy the conditions. We claim that there are no solutions for n = 5 (and thus for any n ≥ 5). Suppose to the contrary that points A i and p i ,i = 1, . . . , 5, satisfy the conditions.

Denote the area of △A i A j A k by S i jk =p i +p j +p k ,1 ≤ i < j < k ≤ 5. Ob- serve that all the p i ’s must be distinct. Indeed, if p 4 =p 5 , then S 124 =S 125 and S 234 =S 235 , which implies that A 4 A 5 is parallel to A 1 A 2 and A 2 A 3 , so A 1 ,A 2 ,A 3 are collinear, which is impossible. Also note that if A i A j A k A l is convex, then S i jk +S ikl =S i jl +S jkl gives p i +p k =p j +p l . Now consider the convex hull of

A 1 ,A 2 ,A 3 ,A 4 ,A 5 . There are three cases. (i) The convex hull is the pentagon A 1 A 2 A 3 A 4 A 5 . We deduce that the quadri- laterals A 1 A 2 A 3 A 4 and A 1 A 2 A 3 A 5 are convex, so we have p 1 +p 3 =p 2 +p 4 and p 1 +p 3 =p 2 +p 5 . Hence p 4 =p 5 , a contradiction. (ii) The convex hull is w.l.o.g. the quadrilateral A 1 A 2 A 3 A 4 . Assume that A 5 lies within A 1 A 3 A 4 . Then A 1 A 2 A 3 A 5 is also convex, so as in (1) we get p 4 =p 5 . (iii) The convex hull is w.l.o.g. the triangle A 1 A 2 A 3 . Since S 124 +S 134 +S 234 = S 125 +S 135 +S 235 , we conclude that again p 4 =p 5 .

18. Let x = za and y = zb, where a and b are relatively prime. The given Diophantine

equation becomes a + zb 2 +z 2 =z 2 ab , so a

2 = zc for some c ∈ Z. We obtain

c +b 2 +z=z 2 cb , or c = b z +z 2 b −1 .

(i) If z = 1, then c = b 2 b +1 −1 =b+1+ 2 b −1 , so b = 2 or b = 3. These values yield two solutions: (x, y) = (5, 2) and (x, y) = (5, 3).

(ii) If z = 2, then 16c = 16b 2 4b +32 −1 = 4b + 1 + 33 4b −1 , so b = 1 or b = 3. In this case (x, y) = (4, 2) or (x, y) = (4, 6).

2 z 2 b 2 +z 3 b +z (iii) Let z 3 c = =b+ . Thus b +z ≥ 3. First, we see that z 3

z 2 b −1

z 2 b −1

z 2 b −1 must

be a positive integer, so b +z 3 ≥z 2 b − 1, which implies b ≤ z 2 −z+1 z −1 . It

follows that b ≤ z. But then b 2 +z ≤ z 2 +b < z 2 b

2 −1, with the last inequality because (z 2 b − 1)(b − 1) > 2. Therefore c = +z z 2 b −1 < 1, a contradiction.

The only solutions for (x, y) are (4, 2), (4, 6), (5, 2), (5, 3).

19. For each two people let n be the number of people exchanging greetings with both of them. To determine n in terms of k, we shall count in two ways the number of triples (A, B,C) of people such that A exchanged greetings with both

B and C, but B and C mutually did not. There are 12k possibilities for A, and for each A there are (3k + 6) possibilities for B. Since there are n people who exchanged greetings with both A and B, there are 3k + 5 − n who did so with A but not with B. Thus the number of triples (A, B,C) is 12k(3k + 6)(3k + 5 − n). On the other hand, there are 12k possible choices of B, and 12k − 1 − (3k + 6) = 9k − 7 possible choices of C; for every

B ,C, A can be chosen in n ways, so the number of considered triples equals 12kn (9k − 7).

4.36 Shortlisted Problems 1995 601 Hence

3 (3k + 6)(3k + 5 − n) = n(9k − 7), i.e., n = (k+2)(3k+5)

12k −1

. This gives us that

12k −1 is an integer too. It is directly verified that only k = 3 gives an integer value for n, namely n = 6.

4n

= 2 12k +44k+40 =k+4− 3k −44

3 12k −1

Remark. The solution is complete under the assumption that such a k exists. We give an example of such a party with 36 persons, k = 3. Let the people sit in a

6 × 6 array [P ij ] 6 i , j=1 , and suppose that two persons P ij ,P kl exchanged greetings if and only if i = k or j = l or i − j ≡ k − l (mod 6). Thus each person exchanged greetings with exactly 15 others, and it is easily verified that this party satisfies the conditions.

20. We shall consider the set M = {0,1,...,2p − 1} instead. Let M 1 = {0,1,..., p −

1 } and M 2 = {p, p + 1,...,2p − 1}. We shall denote by |A| and σ (A) the num- ber of elements and the sum of elements of the set A; also, let C p

be the family of all p-element subsets of M. Define the mapping T : C p →C p as T (A) = {x + 1 | x ∈ A ∩ M 1 } ∪ {A ∩ M 2 }, the addition being modulo p. There are exactly two fixed points of T : these are M 1 and M 2 . Now if A is any subset from C p distinct from M 1 ,M 2 , and k = |A ∩ M 1 | with 1 ≤ k ≤ p − 1, then for i = 0, 1, . . . , p − 1, σ (T i (A)) = σ (A) + ik (mod p). Hence subsets

A , T (A), . . . , T p −1 (A) are distinct, and exactly one of them has sum of elements divisible by p. Since σ (M 1 ), σ (M 2 ) are divisible by p and C p \ {M 1 ,M 2 } de- composes into families of the form

{A,T (A),... ,T p −1 (A)}, we conclude that the required number is 1 1 p 2p (|C p | − 2) + 2 = p p −2 + 2. Second solution. Let C k

be the family of all k-element subsets of {1,2,...,2p}. Denote by M k (k = 1, 2, . . . , p) the family of p-element multisets with k distinct elements from {1,2,...,2p}, exactly one of which appears more than once, that have sum of elements divisible by p. It is clear that every subset from C k ,k < p, can be complemented to a multiset from M k ∪M k +1 in exactly two ways, since the equation (p − k)a ≡ 0 (mod p) has exactly two solutions in {1,2,...,2p}. On the other hand, every multiset from M k can be obtained by completing ex- actly one subset from C k . Additionally, a multiset from M k can be obtained from exactly one subset from C k −1 if k < p, and from exactly p subsets from C k −1 if k

= p. Therefore |M 2p k | + |M k +1 | = 2|C k |=2 k for k = 1, 2, . . . , p − 2, and |M 2p

p −1 | + p|M p | = 2|C p −1 |=2 p −1 . Since M 1 = 2p, it is not difficult to show using recursion that

1 |M 2p p |=

p −2 + 2.

i =1 (x − ω ) = (x − 1) = x 2p − 2x p + 1; hence comparing the coefficients at x p , we obtain ∑ ω i 1 +···+i p = ∑ p −1 i =0 a i ω i = 2, where the first sum runs over all p-subsets {i 1 ,...,i p } of the set {1,...,2p}, and a i is the number of such subsets for which i 1 + ···+ i p ≡ i (mod p ). Setting q

p . We have ∏

Third solution. Let ω = cos 2 p π + i sin 2 π

2p

ip 2

i =0 a i x i , we obtain q ( ω j ) = 0 for j = 1, 2, . . . , p − 1. Hence 1 + x + ···+ x p −1 | q(x), and since deg q = p − 1, we have q(x) = −2 + p ∑ −1

p (x) = −2 + ∑ −1

i =0 a i x i = c(1 + x + ···+ x p −1 ) for some constant c. Thus a 0 −2=a 1 = ··· =

−1 2p

, which together with a 0 + ··· + a p −1 = 2p

p yields a 0 = p p −2 + 2.

602 4 Solutions

21. We shall show that there is no such n. Certainly, n = 2 does not work, so suppose n ≥ 3. Let a,b be distinct elements of A 1 , and c any integer greater than −a and −b. We claim that a + c,b + c belong to the same subsets. Suppose to the contrary that a +c∈A 1 and b +c∈A 2 , and take arbitrary elements x i ∈A i ,

i = 3, . . . , n. The number b + x 3 + ··· + x n is in A 2 , so that s = (a + c) + (b + x 3 + ···+x n )+x 4 +···+x n must be in A 3 . On the other hand, a +x 3 + ···+x n ∈A 2 , so s = (a + x 3 + ···+x n ) + (b + c) + x 4 + ···+ x n is in A 1 , a contradiction. Similarly, if a +c∈A 2 and b +c∈A 3 , then s = a + (b + c) + x 4 + ··· + x n belongs to A 2 , but also s = b + (a + c) + x 4 + ··· + x n ∈A 3 , which is impossible. For i = 1, . . . , n choose x i ∈A i ; set s =x 1 + ··· + x n and y i =s−x i . Then y i ∈A i . By what has been proved above, 2x i =x i +x i belongs to the same subset as x i +y i = s does. It follows that all numbers 2x i ,i = 1, . . . , n, are in the same subset. Since we can arbitrarily take x i from each set A i , it follows that all even

numbers belong to the same set, say A 1 . Similarly, 2x i + 1 = (x i + 1) + x i is in the subset to which (x i + 1) + y i = s + 1 belongs for all i = 1, . . . , n; hence all odd numbers greater than 1 are in the same subset, say A 2 . By the above considerations, 3 −2=1∈A 2 also. But then nothing remains in A 3 ,...,A n ,a contradiction. √

22. Let u 2p √ x √ y and v

− u) = 2p − ( − u) = 2p − x − y − 4xy for x , y ∈ N, x ≤ y. Obviously u ≥ 0 if and only if v ≥ 0, and u, v attain minimum positive values simultaneously. Note that v √ √ √ 6= 0. Otherwise u = 0 too,

so y =( 2p − x ) 2 = 2p − x − 2 2px, which implies that 2px is a square, and consequently x is divisible by 2p, which is impossible. √ Now let z be the smallest integer greater than 4xy. We have z 2

− 1 ≥ 4xy, z ≤ 2p − x − y, and z ≤ p because 4xy ≤( x + √y) 2 < 2p. It follows that

Equality holds if and only if z = x + y = p and 4xy = p 2 − 1, which is satisfied only when x = p −1 2 and y = p +1 2 . Hence for these values of x , y, both u and v attain positive minima.

23. By putting F (1) = 0 and F(361) = 1, condition (c) becomes F(F(n 163 )) =

F (F(n)) for n ≥ 2. For n = 2,3,...,360 let F(n) = n, and inductively define

F (n) for n ≥ 362 as follows:

F (m), =m (n) =

∈ N; the least number not in {F(k) | k < n} , otherwise.

F if n 163 ,m

Obviously, (a) each nonnegative integer appears in the sequence because there are infinitely many numbers not of the form m 163 , and (b) each positive integer appears infinitely often because F (m 163 ) = F(m). Since F(n 163 ) = F(n), (c) also holds.

Second solution. Another example of such a sequence is as follows: If n = p α 1 1 p α 2 α 2 k ··· p k is the factorization of n into primes, we put F (n) = α 1 + α 2 +

4.36 Shortlisted Problems 1995 603

··· + α k and F (1) = 0. Conditions (a) and (b) are evidently satisfied for this F, while (c) follows from F (F(n 163 )) = F(163F(n)) = F(F(n)) + 1 (because 163

is a prime) and F (F(361)) = F(F(19 2 )) = F(2) = 1.

24. The given condition is equivalent to (2x i −x i −1 )(x i x i −1 − 1) = 0, so either x i =

=2 k n x e 2 n i −1 i x i −1 ≥ 0, x n 0 for some integer k n , where |k n | ≤ n and e n = (−1) n −k n . Indeed, this is true for

or x = 1 . We shall show by induction on n that for any n

n = 0. If it holds for some n, then x n +1 = 1 x n =2 k n −1 x e 2 n 0 (hence k n +1 =k n −1 and e

x n =2 x 0 (hence k n +1 = −k n and e n +1 = −e n ). Thus x

0 . Note that e 1995 = 1 is impossible, since in that case k

k 1995 0 e =x 1995 =2 x 1995

would be odd, although it should equal 0. Therefore e 1995 1995 = −1, which gives x 2 0 =2 k 1995

0 can have is 2 997 . This value is attained in the case x i =2 997 −i for i = 0, . . . , 997 and x i =2 i −998 for

≤2 1994 , so the maximal value that x

i = 998, . . ., 1995. Second solution. First we show that there is an n, 0 ≤ n ≤ 1995, such that x n = 1.

Suppose the contrary. Then each of x n belongs to one of the intervals I −i−1 = [2 −i−1 ,2 −i ) or I i = (2 i ,2 i +1 ], where i = 0, 1, 2, . . .. Let x n ∈I i n . Note that by the

formula for x n ,i n and i n −1 are of different parity. Hence i 0 and i 1995 are also of

different parity, contradicting x 0 =x 1995 .

It follows that for some n, x n = 1. Now if n ≤ 997, then x 0 ≤2 997 , while if

n ≥ 998, we also have x 0 =x 1995 ≤2 997 .

25. By the definition of q (x), it divides x for all integers x > 0, so f (x) = xp(x)/q(x) is a positive integer too. Let {p 0 ,p 1 ,p 2 , . . . } be all prime numbers in increas- ing order. Since it easily follows by induction that all x n ’s are square-free, we can assign to each of them a unique code according to which primes divide it: if p m is the largest prime dividing x n , the code corresponding to x n will be

. . . 0s m s m −1 ...s 0 , with s i = 1 if p i |x n and s i = 0 otherwise. Let us investigate how f acts on these codes. If the code of x n ends with 0, then x n is odd, so the code of f (x n )=x n +1 is obtained from that of x n by replacing s 0 = 0 by s 0 = 1. Furthermore, if the code of x n ends with 011 . . .1, then the code of x n +1 ends with 100 . . .0 instead. Thus if we consider the codes as binary numbers, f acts on them as an addition of 1. Hence the code of x n is the binary representation of n and thus x n uniquely determines n. Specifically, if x n = 1995 = 3·5·7·19, then its code is 10001110 and corresponds to n = 142.

26. For n = 1 the result is trivial, since x 1 = 1. Suppose now that n ≥ 2 and let

f n (x) = x n n −∑ −1

i =0 x i . Note that x n is the unique positive real root of f n , because

=x−1− 1 − ··· − n x 1 −1 x x −1 is strictly increasing on R + . Consider g n (x) = (x − 1) f n (x) = (x − 2)x n + 1. Obviously g n (x) has no positive roots other than 1 and x

f n (x)

2 for n ≥ 2 (by Bernoulli’s inequality). Since then

n > 1. Observe that 1 − 2 n >1− n 2 n

2 =− 2 n

+1=1− 1 − 2 n +1 > 0,

604 4 Solutions and

we conclude that x n is between 2

2 n −1 and 2 − 2 n , as required.

Remark. Moreover, lim n →∞ 2 n (2 − x n ) = 1.

27. Computing the first few values of f (n), we observe the following pattern:

f (4k) = k, k ≥ 3,

f (8) = 3;

f (4k + 1) = 1, k ≥ 4,

f (5) = f (13) = 2;

f (4k + 2) = k − 3, k ≥ 7,

f (2) = 1, f (6) = f (10) = 2,

f (14) = f (18) = 3, f (26) = 4;

f (4k + 3) = 2. We shall prove these statements simultaneously by induction on n, having veri-

fied them for k ≤ 7. (i) Let n = 4k. Since f (3) = f (7) = ··· = f (4k − 1) = 2, we have f (4k) ≥ k. But f (n) ≤ max m <n f (m) + 1 ≤ (k − 1) + 1, so f (4k) = k. (ii) Let n = 4k + 1, k ≥ 7. Since f (4k) = k and f (m) < k for m < 4k, we deduce that f (4k + 1) = 1. (iii) Let n = 4k +2, k ≥ 7. Since f (17) = f (21) = ··· = f (4k+1) = 1, we obtain

f (4k + 2) ≥ k −3. On the other hand, if f (4k +1) = f (4k +1−d) = 1, then

d ≥ 8, and 4k + 1 − 8(k − 3) < 0. So f (4k + 2) = k − 3. (iv) Let n = 4k + 3, k ≥ 7. We have f (4k + 2) = k − 3 and f (m) = k − 3 for exactly one m < 4k + 2 (namely for m = 4k − 12); hence f (4k + 3) = 2. Therefore, for example, f (4n + 8) = n + 2 for all n; hence we can take a = 4 and

b = 8.

28. Let F (x) = f (x) − 95 for x ≥ 1. Writing k for m + 95, the given condition be- comes

F (k + F(n)) = F(k) + n, k ≥ 96,n ≥ 1. (1) Thus for x , z ≥ 96 and an arbitrary y we have F(x + y) + z = F(x + y + F(z)) =

F (x + F(F(y) + z)) = F(x) + F(y) + z, and consequently F(x + y) = F(x) + F(y) whenever x ≥ 96. Moreover, since then F(x + y) + F(96) = F(x + y + 96) =

F (x) + F(y + 96) = F(x) + F(y) + F(96) for any x, y, we obtain

x , y ∈ N. (2) It follows by induction that F (n) = nc for all n, where F(1) = c. Equation (1)

F (x + y) = F(x) + F(y),

becomes ck +c 2 n = ck + n, and yields c = 1. Hence F(n) = n and f (n) = n + 95 for all n. Finally, ∑ 19 k =1 f (k) = 96 + 97 + ···+ 114 = 1995.

Second solution. First we show that f (n) > 95 for all n. If to the contrary f (n) ≤

95, we have f (m) = n + f (m + 95 − f (n)), so by induction f (m) = kn + f (m + k (95 − f (n))) ≥ kn for all k, which is impossible. Now for m > 95 we have

4.36 Shortlisted Problems 1995 605

f (m + f (n) − 95) = n + f (m), and again by induction f (m + k( f (n) − 95)) = kn + f (m) for all m, n, k. It follows that with n fixed,

n (∀m) lim

f (m + k( f (n) − 95))

; k →∞ m + k( f (n) − 95)

f (n) − 95 hence

f (n)−95 does not depend on n, i.e., f (n) ≡ cn + 95 for some constant c. It is easily checked that only c = 1 is possible.

606 4 Solutions