Solutions to the Shortlisted Problems of IMO 2003

4.44 Solutions to the Shortlisted Problems of IMO 2003

1. Consider the points O (0, 0, 0), P(a 11 ,a 21 ,a 31 ), Q(a 12 ,a 22 ,a 32 ), R(a 13 ,a 23 ,a 33 ) in three-dimensional Euclidean space. It is enough to find a point U (u 1 ,u 2 ,u 3 ) in the interior of the triangle PQR whose coordinates are all positive, all negative, or all zero (indeed, then we have −→ OU

−→ +c 2 OQ +c 3 OR for some c 1 ,c 2 ,c 3 >0

=c 1 OP

with c 1 +c 2 +c 3 = 1). Let P ′ (a 11 ,a 21 , 0), Q ′ (a 12 ,a 22 , 0), and R ′ (a 13 ,a 23 , 0) be the projections of P, Q, and R onto the Oxy plane. We see that P ′ ,Q ′ ,R ′ lie in the fourth, second, and third quadrants, respectively. We have the following two cases:

(i) O is in the exterior of △P ′ Q ′

Set S ′ = OR Q ′ ∩P ′ ′

Q ′ and let S

be the point of the segment PQ O that projects to S ′ . The point S

x has its z coordinate negative (be-

S ′ cause the z coordinates of P and ′ R

Q are negative). Thus any point P ′

of the segment SR sufficiently close to S has all coordinates negative. (ii) O is in the interior or on the boundary of △P ′ Q ′ R ′ . Let T be the point in the plane PQR whose projection is O. If T = O, then all coordinates of T are zero, and we are done. Otherwise O is interior to

△P ′ Q ′ R ′ . Suppose that the z coordinate of T is positive (negative). Since x and y coordinates of T are equal to 0, there is a point U inside PQR close

to T with both x and y coordinates positive (respectively negative), and this point U has all coordinates of the same sign.

2. We can rewrite (ii) as −( f (a) − 1)( f (b) − 1) = f (−(a − 1)(b − 1) + 1) − 1. So putting g (x) = f (x + 1) − 1, this equation becomes −g(a − 1)g(b − 1) =

g (−(a − 1)(b − 1)) for a < 1 < b. Hence −g(x)g(y) = g(−xy) for x < 0 < y,

and g is nondecreasing with g (−1) = −1, g(0) = 0. (1) Conversely, if g satisfies (1), then f is a solution of our problem.

Setting y = 1 in (1) gives −g(−x)g(1) = g(x) for each x > 0, and therefore (1) reduces to g (1)g(yz) = g(y)g(z) for all y, z > 0. We have two cases:

1 ◦ g (1) = 0. By (1) we have g(z) = 0 for all z > 0. Then any nondecreasing function g : R → R with g(−1) = −1 and g(z) = 0 for z ≥ 0 satisfies (1) and gives a solution: f is nondecreasing, f (0) = 0 and f (x) = 1 for every x ≥1

2 (x) ◦ g (1) 6= 0. Then the function h(x) = g

g (1) is nondecreasing and satisfies

h (0) = 0, h(1) = 1, and h(xy) = h(x)h(y). Fix a > 0, and let h(a) = b = a k for some k

∈ R. It follows by induction that h(a q ) = h(a) q = (a q ) k for every rational number q. But h is nondecreasing, so k ≥ 0, and since the

set {a q | q ∈ Q} is dense in R + , we conclude that h (x) = x k for every

4.44 Shortlisted Problems 2003 703 x > 0. Finally, putting g(1) = c, we obtain g(x) = cx k for all x > 0. Then

g (−x) = −x k for all x > 0. This g obviously satisfies (1). Hence 

where c > 0 and k ≥ 0.  1 − (1 − x) k , if x < 1,

if x = 1;

3. (a) Given any sequence c n (in particular, such that C n converges), we shall construct a n and b n such that A n and B n diverge. First, choose n 1 such that n 1 c 1 > 1 and set a 1 =a 2 = ··· = a n 1 =c 1 : this uniquely determines b 2 =c 2 ,...,b n 1 =c n 1 . Next, choose n 2 such that (n 2 −n 1 )c n 1 +1 > 1 and set b n 1 +1 = ··· = b n 2 =c n 1 +1 ; again a n 1 +1 ,...,a n 2 is hereby determined. Then choose n 3 with (n 3 −n 2 )c n 2 +1 > 1 and set

a n 2 +1 = ··· = a n 3 =c n 2 +1 , and so on. It is plain that in this way we con- struct decreasing sequences a n ,b n such that ∑a n and ∑b n diverge, since they contain an infinity of subsums that exceed 1; on the other hand,

c n = min{a n ,b n } and C n is convergent. (b) The answer changes in this situation. Suppose to the contrary that there is such a pair of sequences (a n ) and (b n ). There are infinitely many indices i such that c i =b i (otherwise all but finitely many terms of the sequence (c n ) would be equal to the terms of the sequence (a n ), which has an unbounded

sum). Thus for any n 0 ∈ N there is j ≥ 2n 0 such that c j =b j . Then we have

c k ≥ ∑ c j =(j−n 0 ) ≥ .

2 Hence the sequence (C n ) is unbounded, a contradiction.

k =n 0 k =n 0 j

4. (a) By the Cauchy–Schwarz inequality we have n

(i − j) 2 ∑ (x i −x j ) ∑ 2 ≥ ∑ |i − j| · |x i −x j | . (1)

On the other hand, it is easy to prove (for example by induction) that

n 2 2 2 2 2 (n 2 − 1)

∑ (i − j) = (2n − 2) · 1 + (2n − 4) · 2 + ··· + 2 · (n − 1) =

i , j=1

6 and that

∑ |i − j| · |x i −x j |= 2 ∑ |x i −x j |.

i , j=1

i , j=1

Thus the inequality (1) becomes

n (n − 1)

∑ (x

6 i −x j )

≥ 4 ∑ |x i −x j | ,

i , j=1

i , j=1

which is equivalent to the required one.

704 4 Solutions

(b) Equality holds if and only if it holds in (1), i.e., if and only if there is λ ∈R such that |x i −x j |= λ |i − j| for all i, j. This is equivalent to (x i ) being an arithmetic sequence.

5. Placing x = y = z = 1 in (i) leads to 4 f (1) = f (1) 3 , so by the condition f (1) > 0 we get f (1) = 2. Also, setting x = y = t, z = 1/t yields f (t) = f (1/t); hence

f (t) ≥ f (1) = 2 for each t. Thus f (t) = c + c −1 for some c = c(t) ≥ 1. Now setting (x, y, z) = (ts, t s s , t ) in (i) gives

f (t) f (s) = f (ts) + f

In particular, s = t yields f (t 2 ) = f (t) 2 − 2, so if f (t) = c + c −1 , then f (t 2 )=

c 2 +c −2 . A simple induction shows that f (t n )=c n +c −n for each n ∈ N. Con- sequently, f (t n /m )=c n /m +c −n/m . Now if t > 1 is fixed and λ ∈ R is such that

c =t λ , we obtain f (t q )=t λq +t −λ q , i.e.,

f (x) = x λ +x −λ for x ∈ T = {t q | q ∈ Q}. (1) But since the set T is dense in R + and f is monotone on (0, 1] and [1, ∞), it

follows that (1) holds for all x > 0. It is directly verified that this function satisfies (i) and (ii).

i = X ,y i by y ′ = Y and z i by z ′ i i = √ XY , we may assume that X = Y = 1. It is sufficient to

6. Set X = max{x 1 ,...,x

n } and Y = max{y 1 ,...,y n }. By replacing x i by x ′

prove that M +z 2 + ··· + z 2n ≥x 1 + ··· + x n +y 1 + ··· + y n ,

(1) because this implies the result by the A-G mean inequality.

To prove (1) it is enough to prove that for any r, the number of terms greater than r on the left-hand side of (1) is at least that number on the right-hand side of (1). If r ≥ 1, then there are no terms on the right-hand side greater than r. Suppose that r < 1 and consider the sets A = {i | 1 ≤ i ≤ n, x i > r} and B = {i | 1 ≤ i ≤ n ,y i > r}. Set a = |A| and b = |B|. If x i > r and y j > r, then z i +j ≥ √x i y j > r; hence

C = {k | 2 ≤ k ≤ 2n, z k > r} ⊇ A + B = { α + β | α ∈ A, β ∈ B}. It is easy to verify that |A + B| ≥ |A| + |B| − 1. It follows that the number of z k ’s

greater than r is ≥ a + b − 1. But in that case M > r, implying that at least a + b elements of the left-hand side of (1) is greater than r, which completes the proof.

7. Consider the set D = {x − y | x,y ∈ A}. Obviously, the number of elements of the set D is less than or equal to 101 ·100+ 1. The sets A +t i and A +t j are disjoint if

and only if t i −t j 6∈ D. Now we shall choose inductively 100 elements t 1 , . . . ,t 100 . Let t 1 be any element of the set S \ D (such an element exists, since the number of elements of S is greater than the number of elements of D). Suppose now that we have chosen k (k ≤ 99) elements t 1 , . . . ,t k from D such that the difference of

4.44 Shortlisted Problems 2003 705 any two of the chosen elements does not belong to D. We can select t k +1 to be

an element of S that does not belong to any of the sets t 1 + D,t 2 + D, . . . ,t k +D (this is possible to do, since each of the previous sets has at most 101 · 100 + 1 elements; hence their union has at most 99 (101 · 100 + 1) = 999999 < 1000000 elements).

8. Let S be the disk with the smallest radius, say s, and O the center of that disk. Divide the plane into 7 regions: one bounded by disk s and 6 regions T 1 ,...,T 6 shown in the figure.

Any of the disks different from S, say T 4

D k , has its center in one of the seven P 4

regions. If its center is inside S then D T 3 T 5

contains point O. Hence the number of P 3 P 5 disks different from S having their cen-

O ters in S is at most 2002.

U 6 Consider a disk D that intersects S

P 2 L and whose center is in the region

P 6 T . Let P

be the point such that T 2 T i

OP √ i bisects the region T i and OP i = s T 3. We claim that D 1 K k contains P i .

Divide the region T i by a line l i through P i perpendicular to OP i into two regions U i and V i , where O and U i are on the same side of l i . Let K be the center of D k . Consider two cases:

(i) K ∈U i . Since the disk with the center P i and radius s contains U i , we see that KP i ≤ s. Hence D k contains P i . (ii) K ∈V i . Denote by L the intersection point of the segment KO with the circle s . We want to prove that KL > KP i . It is enough to prove that ∠KP i L > ∠KLP i . However, it is obvious that ∠LP i O ≤ 30 ◦ and ∠LOP i ≤ 30 ◦ , hence ∠KLP i ≤ 60 ◦ , while ∠NP i L = 90 ◦ − ∠LP i O ≥ 60 ◦ . This implies ∠KP i L ≥ ∠NP i L ≥ 60 ◦ ≥ ∠KLP i (N is the point on the edge of T i as shown in the figure). Our claim is thus proved.

Now we see that the number of disks with centers in T i that intersect S is less than or equal to 2003, and the total number of disks that intersect S is not greater than 2002 + 6 · 2003 = 7 · 2003 − 1.

9. Suppose that k of the angles of an n-gon are right. Since the other n −k angles are less than 360 ◦ and the sum of the angles is (n − 2)180 ◦ , we have the inequality k

·90 2n ◦ +4 + (n −k)360 ◦ > (n − 2)180 ◦ , which is equivalent to k <

3 . Since n and

k are integers, it follows that k ≤ 2n 3 + 1.

If n = 5, then 2n 3 + 1 = 4, but if a pentagon has four right angles, the other angle is equal to 180 ◦ , which is impossible. Hence for n = 5, k ≤ 3. It is easy to construct a pentagon with 3 right angles, e.g., as in the picture below. Now we shall show by induction that for n

≥ 6 there is an n-gon with 2n

1 internal right angles. For n = 6, 7, 8 examples are presented in the picture. h Assume that there is a

2 (n−3) i

(n − 3)gon with

3 +1= 2n 3 − 1 internal right

706 4 Solutions angles. Then one of the internal angles, say ∠BAC, is not convex. Interchange

the vertex A with four new vertices A 1 ,A 2 ,A 3 ,A 4 as shown in the picture such

that ∠BA 1 A 2 = ∠A 3 A 4 C = 90 ◦ .

n −3→n 3

10. Denote by b ij the entries of the matrix B. Suppose the contrary, i.e., that there is

a pair (i 0 ,j 0 ) such that a i 0 ,j 0 6= b i 0 ,j 0 . We may assume without loss of generality

that a i 0 ,j 0 = 0 and b i 0 ,j 0 = 1.

Since the sums of elements in the i 0 th rows of the matrices A and B are equal, there is some j 1 for which a i 0 ,j 1 = 1 and b i 0 ,j 1 = 0. Similarly, from the fact that the sums in the j 1 th columns of the matrices A and B are equal, we conclude that there exists i 1 such that a i 1 ,j 1 = 0 and b i 1 ,j 1 = 1. Continuing this procedure, we construct two sequences i k ,j k such that a i k ,j k = 0, b i k ,j k =

1 ,a i k ,j k +1 = 1, b i k ,j k +1 = 0. Since the set of the pairs (i k ,j k ) is finite, there are two different numbers t , s such that (i t ,j t ) = (i s ,j s ). From the given con- dition we have that x i k +y i k < 0 and x i k +1 +y i k +1 ≥ 0. But j t =j s , and hence

0 t −1 t ≤∑ −1 k =s (x i k +y j k +1 )=∑ k =s (x i k +y j k ) < 0, a contradiction.

11. (a) By the pigeonhole principle there are two different integers x √ √ 1 ,x 2 ,x 1 >x 2 , such that |{x 1 3 } − {x 2 3 }| < 0.001. Set a = x √ 1 −x 2 . Consider the equi- lateral triangle with vertices (0, 0), (2a, 0), (a, a 3 ). The points (0, 0) and √ (2a, 0) are lattice points, and we claim that the point (a, a 3 ) is at dis- √ tance less than 0 √ √ .001 from a lattice point. Indeed, since 0.001 > |{x √ √ 1 3 }− {x 2 3 }| = |a √ 3 − ([x 1 3 ] − [x √ 2 3 ])|, we see that the distance between √ the points (a, a √ 3 ) and (a, [x √ 1 3 ] − [x 2 3 ]) is less than 0.001, and the point (a, [x 1 3 ] − [x 2 3 ]) is with integer coefficients. (b) Suppose that P ′ Q ′ R ′ is an equilateral triangle with side length l ≤ 96 such that each of its vertices P ′ ,Q ′ ,R ′ lies in a disk of radius 0 .001 centered at a lattice point. Denote by P , Q, R the centers of these disks. Then we have l − 0.002 ≤ PQ,QR,RP ≤ l + 0.002. Since PQR is not an equilateral triangle, two of its sides are different, say PQ 6= QR. On the other hand,

PQ 2 , QR 2 are integers, so we have 1 ≤ |PQ 2 − QR 2 | = (PQ + QR)|PQ − QR | ≤ 0.004(PQ + QR) ≤ (2l + 0.004) · 0.004 ≤ 2 · 96.002 · 0.004 < 1, which is a contradiction.

12. Denote by a k −1 a k −2 ...a 0 the decimal representation of a number whose digits are a k −1 ,...,a 0 . We will use the following well-known fact:

k −1

a k −1 a k l −2 ...a 0 ≡ i (mod 11) ⇐⇒ ∑ (−1) a l ≡ i (mod 11).

l =0

Let m be a positive integer. Define A as the set of integers n (0 ≤ n < 10 2m ) whose right 2m − 1 digits can be so permuted to yield an integer divisible by 11,

4.44 Shortlisted Problems 2003 707 and B as the set of integers n (0 ≤ n < 10 2m −1 ) whose digits can be permuted so

that the resulting is an integer divisible by 11.

Suppose that a =a 2m −1 ...a 0 ∈ A. Then it satisfies

2m −1

∑ l (−1) a

l ≡ 0 (mod 11).

l =0

The 2m-tuple (a 2m −1 ,...,a 0 ) satisfies (1) if and only if the 2m-tuple (ka 2m −1 + l , . . . , ka 0 + l) satisfies (1), where k, l ∈ Z, 11 ∤ k. Since a 0 + 1 6≡ 0 (mod 11), we can choose k from the set {1,...,10} such that (a 0 + 1)k ≡ 1 (mod 11). Thus there is a permutation of the 2m-tuple ((a 2m −1 +

1 )k −1,...,(a 1 + 1)k −1,0) satisfying (1). Interchanging odd and even positions if necessary, we may assume that this permutation keeps the 0 at the last position. Since (a i + 1)k is not divisible by 11 for any i, there is a unique b i ∈ {0,1,...,9}

such that b i ≡ (a i + 1)k − 1 (mod 11). Hence the number b 2m −1 ...b 1 ∈ B. Thus for fixed a 0 ∈ {0,1,2,...,9}, to each a ∈ A such that the last digit of a is

a 0 we associate a unique b ∈ B. Conversely, having a 0 ∈ {0,1,2,...,9} fixed, from any number b 2m −1 ...b 1 ∈ B we can reconstruct a 2m −1 ...a 1 a 0 ∈ A. Hence |A| = 10|B|, i.e., f (2m) = 10 f (2m − 1).

13. Denote by K and L the intersections of the bisectors of ∠ABC and ∠ADC with

R the line AC, respectively. Since AB :

BC = AK : KC and AD : DC = AL : LC, D we have to prove that

Since the quadrilaterals AQDR and B P C QPCD are cyclic, we deduce that ∠RDQ = ∠BAC and ∠QDP = ∠ACB.

By the law of sines it follows that

AB sin ∠ACB = BC sin ∠BAC

and that QR = AD sin ∠RDQ, QP = CD sin ∠QDP. Now we have

AD · QP =

AB sin ∠ACB

sin ∠QDP

QR ·CD The statement (1) follows directly.

BC sin ∠BAC

sin ∠RDQ

14. Denote by R the intersection point of the bisector of ∠AQC and the line AC. From ∆ ACQ we get

AR

AQ

sin ∠QCA

RC

QC

sin ∠QAC

By the sine version of Ceva’s theorem we have sin ∠APB sin ∠QAC sin ∠QCP sin ∠BPC · sin ∠PAQ · sin ∠QCA = 1, which is equivalent to

708 4 Solutions

sin ∠APB

sin ∠QCA 2

sin ∠BPC

sin ∠QAC

because ∠QCA = ∠PAQ and ∠QAC = ∠QCP. Denote by S(XY Z) the area of a triangle XY Z. Then

sin ∠APB

AP · BP · sin∠APB

S ( ∆ ABP ) AB

= , sin ∠BPC

BP ·CP · sin∠BPC

S ( ∆ BCP ) BC which implies that AR 2 RC = AB BC . Hence R does not depend on Γ .

15. From the given equality we see that 0 = (BP 2 + PE 2 ) − (CP 2 + PF 2 ) = BF 2 − CE 2 , so BF = CE = x for some x. Similarly, there are y and z such that CD = AF = y and BD = AE = z. It is easy to verify that D, E, and F must lie on the segments BC ,CA, AB. Denote by a , b, c the length of the segments BC,CA, AB. It follows that a = z + y,

b = z + x, c = x + y, so D, E, F are the points where the excircles touch the sides

of △ABC. Hence P, D, and I A are collinear and

∠ACB ∠PI A C = ∠DI − ∠ACB A C = 90 ◦ − = .

2 2 In the same way we obtain that ∠PI B C = ∠ACB 2 and PI B = PI A . Analogously, we

get PI C = PI B , which implies that P is the circumcenter of the triangle I A I B I C .

16. Apply an inversion with center at P and radius r; let b X denote the image of X . The circles Γ 1 , Γ 2 , Γ 3 , Γ 4 are transformed into lines b Γ 1 ,b Γ 2 ,b Γ 3 ,b Γ 4 , where b Γ 1 kb Γ 3 and

Γ b 2 kb Γ 4 , and therefore b Ab Bb Cb D is a parallelogram.

Further, we have AB = r 2 2 2 Ab 2 b B , BC = r Bb b C , CD = r Cb b D , DA = r Db b A

Pb A ·P b B Pb B ·P b C Pb C ·P b D Pb D ·Pb A

and PB = Pb B , PD = Pb D . The equality to be proven becomes

Pb D 2 Ab b B ·b Bb C Pb D 2

Pb B 2 Ab b D ·b Db C Pb B 2 which holds because b Ab B =b Cb D and b Bb C =b Db A .

17. The triangles PDE and CFG are ho- C mothetic; hence lines FD, GE, and CP

intersect at one point. Let Q be the in- tersection point of the line CP and the circumcircle of △ABC. The required

F I P G statement will follow if we show that

Q lies on the lines GE and FD. Since ∠CFG = ∠CBA = ∠CQA, the quadri-

D E B lateral AQPF is cyclic. Analogously,

Q BQPG is cyclic. However, the isosce-

les trapezoid BDPG is also cyclic; it

4.44 Shortlisted Problems 2003 709 follows that B , Q, D, P, G lie on a circle. Therefore we get

(1) Since I is the incenter of

∠PQF = ∠PAC, ∠PQD = ∠PBA.

2 ∠CAB = 1 2 ∠CBA = ∠IBA; hence CA is the tangent at A to the circumcircle of △ABI. This implies that

△ABC, we have ∠CAI = 1

∠PAC = ∠PBA, and it follows from (1) that ∠PQF = ∠PQD, i.e., that F, D, Q are also collinear. Similarly, G , E, Q are collinear and the claim is thus proved.

18. Let ABCDEF be the given hexagon. We shall use the following lemma. √ Lemma. If ∠X ZY ≥ 60 ◦ and if M is the midpoint of XY , then MZ

2 XY , with equality if and only if △XY Z is equilateral.

Proof. Let Z ′

be the point such that △XY Z ′ is equilateral. Then Z is inside √ the circle circumscribed about △XY Z ′ . Consequently MZ

2 XY , with equality if and only if Z =Z ′ .

≤ MZ 3 ′ =

Set AD ∩ BE = P, BE ∩CF = Q, and CF ∩ AD = R. Suppose ∠APB = ∠DPE >

60 ◦ , and let K , L be the midpoints of the segments AB and DE respectively. Then by the lemma, √

2 (AB + DE) = KL ≤ PK + PL < 2 which is impossible. Hence ∠APB ≤ 60 ◦ and similarly ∠BQC ≤ 60 ◦ , ∠CRD ≤

(AB + DE),

60 ◦ . But the sum of the angles APB , BQC,CRD is 180 ◦ , from which we con- clude that these angles are all equal to 60 ◦ , and moreover that the triangles APB , BQC,CRD are equilateral. Thus ∠ABC = ∠ABP + ∠QBC = 120 ◦ , and in the same way all angles of the hexagon are equal to 120 ◦ .

D 19. Let D, E, F be the midpoints of ′ BC , CA, AB, respectively. We con-

γ d side △ABC with centers D,E,F and

struct smaller semicircles Γ d , Γ e , Γ f in-

C D B spectively. Since DE = d + e, DF =

radii d = s −a ,e = s −b ,f = s −c 2 2 2 re-

d + f , and EF = e + f , we deduce that Γ d , Γ e , and Γ f touch each other at the

E D 1 F points D 1 ,E 1 ,F 1 of tangency of the in-

circle γ of △DEF with its sides (D 1 ∈

γ e A center O and radius g that lies inside

EF , etc.). Consider the circle Γ g with

△DEF and tangents Γ d , Γ e , Γ f .

Now let OD , OE, OF meet the semicircles Γ d , Γ e , Γ f at D ′ ,E ′ ,F ′ respectively. We have OD ′ = OD + DD ′ =g+d+ a 2 =g+ s 2 and similarly OE ′ = OF ′ =

g + s 2 . It follows that the circle with center O and radius g + s 2 touches all three semicircles, and consequently t =g+ s 2 > s 2 . Now set the coordinate system such that we have the points D 1 (0, 0), E(−e,0), F( f ,0) and such that the y coordinate of D is positive.

710 4 Solutions Apply the inversion with center D 1 and unit radius. This inversion maps the

1 h 1 circles i Γ

e and Γ f to the lines b Γ e x =− 2e and b Γ e x = 2f respectively, and the circle γ goes to the line b γ y = 1 r . The images b Γ d and b Γ g of Γ d , Γ g are the circles that touch the lines b Γ e and b Γ f . Since b Γ d ,b Γ g are perpendicular to γ , they have radii equal to R = 1 4e + 1 4f 1 and centers at − 4e + 1 1 1 1 4f 1 , r and − 4e + 4f , r + 2R respectively. Let p and P be the distances from D 1 (0, 0) to the centers of Γ g and

Γ b respectively. We have that P 2 = 1 g 1 4e − 4f + 1 r + 2R , and that the circles Γ g and b Γ g are homothetic with center of homothety D 1 ; hence p /P = g/R. On

the other hand, b Γ g is the image of Γ g under inversion; hence the product of the tangents from D p √ 1 to these two circles is equal to 1. In other words, we obtain p 2 −g 2 · P 2 −R 2 = 1. Using the relation p/P = g/R we get g = R P 2

√ −R 2 .

The inequality we have to prove is equivalent to (4 + 2 3 )g ≤ r. This can be proved as follows:

r (P 2

−R 2 − (4 + 2 3

)R/r)

4e 4f Remark. One can obtain a symmetric formula for g:

be the remainder when x i is divided by m. Since there are at most m m types of m-consecutive blocks in the sequence (r i ), some type will repeat at least twice. Then since the entire sequence is determined by one m-consecutive block, the entire sequence will be periodic. The formula works both forward and backward; hence using the rule x i =x i +m − ∑ m −1 j =1 x i +j we can define x −1 ,x −2 , . . . . Thus we obtain that

(r −m ,...,r −1 ) = (0, 0, . . . , 0, 1). Hence there are m − 1 consecutive terms in the sequence (x i ) that are divisible

by m. If there were m consecutive terms in the sequence (x i ) divisible by m, then by the recurrence relation all the terms of (x i ) would be divisible by m, which is impossible.

21. Let a be a positive integer for which d (a) = a 2 . Suppose that a has n + 1 digits, n ≥ 0. Denote by s the last digit of a and by f the first digit of c. Then a = ∗ ... ∗ s,

4.44 Shortlisted Problems 2003 711 where ∗ stands for a digit that is not important to us at the moment. We have

∗... ∗ s 2 =a 2 = d = ∗... ∗ f and b 2 = s ∗ ...∗ 2 = c = f ∗ ...∗. We cannot have s = 0, since otherwise c would have at most 2n digits, while a 2 has either 2n + 1 or 2n + 2 digits. The following table gives all possibilities for s and f :

f = last digit of ∗... ∗ s 2 1 4 9 6 5 6 9 4 1

f = first digit of s ∗ ...∗ 2 1 , 2, 3 4 − 8 9,1 1,2 2,3 3,4 4,5,6 6,7,8 8,9 We obtain from the table that s

∈ {1,2,3} and f = s 2 , and consequently c =b 2 and d have exactly 2n + 1 digits each. Put a = 10x + s, where x < 10 n . Then

b = 10 n s + x, c = 10 2n s 2 + 2 ·10 n sx +x 2 , and d

n +1

= 2 ·10 sx + 10x 2 +s 2 , so from

d =a 2 it follows that x = 2s· 10 −1

9 . Thus a =6...6

3, a =4...4 2 or a =2...2 | {z } 1. | {z } | {z }

nn

For n ≥ 1 we see that a cannot be a = 6 ...63 or a = 4...42 (otherwise a 2 would have 2n + 2 digits). Therefore a equals 1, 2, 3 or 2 . . . 2 1 for n | {z } ≥ 0. It is easy to

verify that these numbers have the required property.

22. Let a and b be positive integers for which

2ab 2 −b 3 +1 = k is a positive integer. Since k > 0, it follows that 2ab 2 3 ≥b 2 , so 2a ≥ b. If 2a > b, then from 2ab −

b 3 + 1 > 0 we see that a 2 >b 2 (2a − b) + 1 > b 2 , i.e. a > b. Therefore, if a ≤ b, then a = b/2. We can rewrite the given equation as a quadratic equation in a, a 2 − 2kb 2 a + k (b 3 − 1) = 0, which has two solutions, say a 1 and a 2 , one of which is in N 0 . From a 1 +a 2 = 2kb 2 and a 1 a 2 = k(b 3 − 1) it follows that the other solution is also in N 0 . Suppose w.l.o.g. that a 1 ≥a 2 . Then a 1 ≥ kb 2 and

a 1 ≤ kb 2 < b. By the above considerations we have either a 2 = 0 or a 2 = b/2. If a 2 = 0, then

b 3 − 1 = 0 and hence a 1 = 2k, b = 1. If a 2 = b/2, then b = 2t for some t, and

k =b 2 /4, a 1 =b 4 /2 − b/2. Therefore the only solutions are (a, b) ∈ {(2t,1), (t,2t), (8t 4 − t,2t) | t ∈ N}.

It is easy to show that all of these pairs satisfy the given condition.

23. Assume that b ≥ 6 has the required property. Consider the sequence y n = (b −

1 )x n . From the definition of x n we easily find that y n =b 2n +b n +1 + 3b −5. Then y n y n +1 = (b − 1) 2 x n x n +1 is a perfect square for all n > M. Also, straightforward calculation implies

n +2 n +1

b 2 +b

b n +2

n +1 2

b 2n +1

+b +

3 <y n y −b +1

n +1 < b 2n + +b 3 .

712 4 Solutions Hence for every n > M there is an integer a n such that |a n |<b 3 and

Now considering this equation modulo b n we obtain (3b − 5) 2 ≡a 2 n , so that assuming that n > 3 we get a n = ±(3b − 5). If a n = 3b − 5, then substituting in (1) yields 1 4 b 2n (b 4 − 14b 3 + 45b 2 − 52b +

20 ) = 0, with the unique positive integer solution b = 10. Also, if a n = −3b + 5, we similarly obtain 1 4 b 2n (b 4 − 14b 3 −3b 2 + 28b + 20) − 2b n +1 (3b 2 − 2b− 5) = 0 for each n, which is impossible.

For b = 10 it is easy to show that x n = 10 n +5 2 3 for all n. This proves the statement.

Second solution. In problems of this type, computing z n = √x n asymptotically usually works. From lim 2n

√ →∞ (b−1)x n = 1 we infer that lim n

z n = b − 1. Furthermore, from (bz +z +1 )(bz

→∞ b

+1 =b n +2 + 3b n 2 n n −z n n −x n − 2b − 5 we obtain

Since the z n ’s are integers for all n ≥ M, we conclude that bz n −z n +1 = b b 2 −1 for all n sufficiently large. Hence b − 1 is a perfect square, and moreover b divides 2z n +1 for all large n. It follows that b | 10; hence the only possibility is b = 10.

24. Suppose that m = u + v + w where u, v, w are good integers whose product is a perfect square of an odd integer. Since uvw is an odd perfect square, we have that uvw ≡ 1 (mod 4). Thus either two or none of the numbers u,v,w are congruent to 3 modulo 4. In both cases u + v + w ≡ 3 (mod 4). Hence m ≡ 3 (mod 4). Now we shall prove the converse: every m ≡ 3 (mod 4) has infinitely many representations of the desired type. Let m = 4k + 3. We shall represent m in the form

4k + 3 = xy + yz + zx, for x , y, z odd. (1) The product of the summands is an odd square. Set x = 1 + 2l and y = 1 − 2l. In

order to satisfy (1), z must satisfy z = 2l 2 + 2k + 1. The summands xy, yz, zx are distinct except for finitely many l, so it remains only to prove that for infinitely many integers l, |xy|, |yz|, and |zx| are not perfect squares. First, observe that

|xy| = 4l 2 − 1 is not a perfect square for any l 6= 0. Let p , q > m be fixed different prime numbers. The system of congruences 1 +

2l

2 ) and 1 ≡ p (mod p 2 − 2l ≡ q (mod q ) has infinitely many solutions l by the Chinese remainder theorem. For any such l, the number z = 2l 2 + 2k + 1 is

divisible by neither p nor q, and hence |xz| (respectively |yz|) is divisible by p, but not by p 2 (respectively by q, but not by q 2 ). Thus xz and yz are also good numbers.

4.44 Shortlisted Problems 2003 713

25. Suppose that for every prime q, there exists an n for which n p ≡ p (mod q). Assume that q = kp + 1. By Fermat’s theorem we deduce that p k ≡n kp =n q −1 ≡

1 (mod q), so q |p k − 1. It is known that any prime q such that q p p −1 | p −1 must satisfy q ≡ 1 (mod p).

Indeed, from q |p q −1 − 1 it follows that q | p gcd (p,q−1)

− 1; but q ∤ p − 1 because

p −1 ≡ 1 (mod p − 1), so gcd(p,q − 1) 6= 1. Hence gcd(p,q − 1) = p. Now suppose q is any prime divisor of p p −1 p −1 . Then q | gcd(p k − 1, p p − 1) =

p −1

p gcd (p,k) − 1, which implies that gcd(p,k) > 1, so p | k. Consequently q ≡ 1 (mod p 2 ). However, the number p p −1 p −1 =p p −1 + ··· + p + 1 must have at least one prime divisor that is not congruent to 1 modulo p 2 . Thus we arrived at a contradiction. Remark. Taking q ≡ 1 (mod p) is natural, because for every other q, n p takes all possible residues modulo q (including p too). Indeed, if p ∤ q − 1, then there is an r ∈ N satisfying pr ≡ 1 (mod q − 1); hence for any a the congruence n p ≡a (mod q) has the solution n

≡a r (mod q).

The statement of the problem itself is a special case of the Chebotarev’s theorem.

26. Define the sequence x k of positive reals by a k = cosh x k (cosh is the hyperbolic cosine defined by cosht = e t +e −t 2 ). Since cosh (2x k ) = 2a 2 k − 1 = coshx k +1 , it follows that x k +1 = 2x k and thus x k = λ ·2 k for some λ √ > 0. From the condition

a 0 = 2 we obtain λ = log(2 + 3 ). Therefore

a + (2 − 3 ) n 2 = .

Let p be a prime number such that p |a n . We distinguish the following two cases: (i) There exists an m

∈ Z such that m 2 ≡ 3 (mod p). Then we have

2 + (2 − m) n ≡ 0 (mod p). (1) Since

2 (2 + m) n

(2 + m)(2 − m) = 4 − m 2 ≡ 1 (mod p), multiplying both sides of (1) by (2 + m) 2 n

n gives +1 (2 + m) 2 ≡ −1 (mod p). It follows that the multi- plicative order of (2 + m) modulo p is 2 n +2 , or 2 n +2 | p − 1, which implies

that 2 n +3 | (p − 1)(p + 1) = p 2 − 1.

(ii) m 2 ≡ 3 (mod p) has no integer solutions. We will work in the algebraic √ √ extension Z p ( 3 ) of the field Z p . In this field

√ 3 plays the role of m, so as

= −1; i.e., the order of 2 + 3 in the multiplicative group Z p ( 3 ) ∗ is 2 n +2 . We cannot finish the proof as

in the previous case we obtain (2 + 3 ) 2 n +1

in the previous case: in fact, we would conclude only that 2 n +2 divides the

√ order p − 1 of the group. However, it will be enough to find a u ∈ Z √ p ( 3 )

such that u 2 =2+

3, since then the order of u is equal to 2 n √ +3 √ . Note that (1 + 3 ) 2 = 2(2 + 3 ). Thus it is sufficient to prove that 1

2 is a perfect square in Z p ( 3 ). But we know that in this field a n = 0 = 2a 2 n −1 −1,

and hence 2a 2 1 n 2 −1 = 1 which implies 2 =a n −1 . This completes the proof.

714 4 Solutions

be distinct primes, where r = p − 1. Consider the sets B i = p {p +1

27. Let p 1 ,p 2 ,...,p r

} and B = i r i i =1 B i . Then B has (p − 1) 2 elements and satisfies (i) and (ii). Now suppose that |A| ≥ r 2 + 1 and that A satisfies (i) and (ii). Let {t 1 , . . . ,t r 2 +1 }

i ,p ,...,p (r−1)p+1

1 ·p 2 ··· p r . We shall show that the product of some elements of A is a perfect pth power, i.e., that there exist τ j ∈

be distinct elements of A, where t α

+ 1), not all equal to 0, such that T = t τ 1 2 τ r2 +1 {0,1} (1 ≤ j ≤ r

1 ·t 2 ···t r 2 +1 is a pth power. This is equivalent to the condition that

r 2 +1

∑ α ij τ j ≡ 0 (mod p)

j =1

holds for all i = 1, . . . , r. By Fermat’s theorem it is sufficient to find integers x 1 ,...,x r 2 +1 , not all zero, such that the relation

r 2 +1

∑ α ij x r j ≡ 0 (mod p)

j =1

is satisfied for all i 2 ∈ {1,...,r}. Set F

i =∑ r +1 j =1 α ij x r j . We want to find x 1 ,...,x r such that F 1 ≡F 2 ≡ ··· ≡ F r ≡ 0 (mod p), which is by Fermat’s theorem equiv- alent to

F (x 1 ,...,x r )=F r 1 +F r 2 + ··· + F r r ≡ 0 (mod p). (1) Of course, one solution of (1) is (0, . . . , 0): we are not satisfied with it because it

generates the empty subset of A, but it tells us that (1) has at least one solution. We shall prove that the number of solutions of (1) is divisible by p, which will imply the existence of a nontrivial solution and thus complete the proof. To do

this, consider the sum ∑F(x 1 ,...,x r 2 +1 ) r taken over all elements (x 1 ,...,x r 2 +1 ) in the vector space 2 Z r +1

. Our statement is equivalent to

1 ,...,x r 2 +1 ) ≡ 0 (mod p). (2) Since the degree of F r is r 2 , in each monomial in F r at least one of the variables

∑ r F (x

is missing. Consider any of these monomials, say bx a i 1 1 x a 2 a i k 2 ···x i k . Then the sum

∑ bx 2

i 1 x i 2 ···x i k , taken over the set of all vectors (x 1 ,...,x r 2 +1 )∈Z r +1 p , is equal to

p 2 r +1−u

a 1 a 2 · a ∑ bx k

i 1 x i 2 ···x i ,

(x i1 ,...,x ik )∈Z k

which is divisible by p, so that (2) is proved. Thus the answer is (p − 1) 2 .

4.45 Shortlisted Problems 2004 715