Solutions to the Shortlisted Problems of IMO 2008

4.49 Solutions to the Shortlisted Problems of IMO 2008

) for all x ∈ R + . In particular, f

1. For x = y = z = w the functional equation gives f (x) 2 = f (x 2

(1) = 1. Setting w , x , √y, z in the equation yields

f (w) + f (x)

w +x

, whenever wx = yz.

f (y) + f (z)

y +z

Choosing z (y)+1 = 1, we get w = y/x and f (y/x) + f (x) = y x +x f · y +1 . Now if we place y =x 2 , we get f

f (x) 2 (x) = x · +1 x 2 +1 , which is equivalent to ( f (x) − x)( f (x) −

x ) = 0. Assume that there are x, w ∈ R + \ {1} such that f (x) = x and f (w) = w . √ Choosing y =z= wx and placing in the equation implies 1 w + x = (w + x) ·

√ √ √ √ wx . If f ( wx )= wx then we have w = 1. Otherwise, if f ( wx ) = 1/ wx , we get x = 1, contrary to our assumption. Therefore we either have f (x) = x for all x ∈R + or f (x) = 1 x for all x ∈R + . It is easy to verify that both functions

f ( √ √ wx )

satisfy the original equation.

2. (a) Substituting a = x

x −1 ,b = y −1 ,c = z z −1 , the inequality becomes equivalent to a 2 +b 2 +c 2 ≥ 1, while the constraint becomes a + b + c = ab + bc +

ca + 1. The last equation is equivalent to 2(a + b + c) = (a + b + c) 2 − (a 2 +

b 2 +c 2 ) + 2 = (a + b + c) 2 + 1 − [(a 2 +b 2 +c 2 ) − 1], or [(a + b + c)− 1] 2 = (a 2 +b 2 +c 2 ) − 1, which immediately implies that a 2 +b 2 +c 2 ≥ 1. (b) Equality holds if and only if a + b + c = 1 and ab + bc + ca = 0. Expressing

c 2 = 1 −a− b yields −ab+ a+ b− a 2 −b = 0. It suffices to prove that there are infinitely many rational numbers a for which the quadratic equation

b 2 + (a − 1)b + a(a − 1) = 0 has a rational solution b. This equation has

a rational solution if and only if its discriminant (a − 1) 2 − 4a(a − 1) = (1 −a)(1+3a) is the square of a rational number. We want to find infinitely

1 +3 q is the square of a rational number, which is equivalent to (q − p)(q + 3p) being the square

many rational numbers p q such that 1 p

of an integer. However, for each m , n ∈ N the system q − p = (2m + 1) 2 and q + 3p = (2n + 1) 2 has a solution p =n 2 +n−m 2 − m, q = n 2 +n+ 3m 2 + 3m + 1. Keeping m fixed and increasing n would guarantee that we are getting infinitely many different fractions a = p q .

3. (a) Assume that ( f , g) is a Spanish couple on N. If g(a) > g(b), then a > b (a ≤ b would yield g(a) ≤ g(b)). Let us introduce the notation

g k (x) = g(g(···g (x) ···)).

| {z }

If we assume that g (x) < x for some x ∈ N we get g(g(x)) < g(x) < x, and by induction (g k (x)) ∞ k =1 is an infinite decreasing sequence from N which is impossible. Hence g (x) ≥ x for all x ∈ N. The same holds for f . If for some x ∈ N we had f (x) ≤ g(x), then g( f (x)) ≤ g(g(x)) ≤ f (g(g(x)),

766 4 Solutions which contradicts (ii). Therefore f (x) > g(x) for all x ∈ N. From g( f (x)) >

f (g 2 (x)) > g(g 2 (x)) we conclude that f (x) > g 2 (x), and now easy induc- tion implies f (x) > g n (x) for all n ∈ N. This is impossible if (g n (x)) ∞ n =1 is an infinite increasing sequence. Hence g (x) = x for all x. This can’t be true either, because of (ii). This proves that there is no Spanish couple on N.

(b) The functions f a − 1 b = 3a − 1 b and g (a, b) = a − 1 a +b form a Spanish couple on the given set S.

4. Using the given properties we get f (2) = f (2 0 + 0) = f (2 0 − 3) − f (0) = −1,

f (3) = f (2 + 1) = f (2 − 2) − f (1) = 2, f (4) = −2. For every i ∈ {1,2,3} we have f (2 n − i) = f (2 n −1 +2 n −1 − i) = f (2 n −1 − t(2 n −1 − i)) − f (2 n −1 − i). If

2 | n then 2 n −1 − i ≡ 2 − i ≡ −(i + 1) (mod 3), and if 2 ∤ n then 2 n −1 −i≡ −(i − 1) (mod 3). Set a i k = f (2 2k

k = f (2 2k − i) and b +1 − i). From the previous calculation we get a i =b i +1 k k −1 −b i k −1 and b i

k =a k −a k . This further implies that a i = 2a i

i −1

k =2·3 ,a k =

i −a +1 −1 −a . By induction we now obtain a 1 k −1 2

k k −1 k −1

k −1

a 3 k k = −3 −1 . This further gives b 1 k =a 3 1 k k −a k = −3 ,b 2 k =3 k ,b 3 k = 0. Moreover, for each n ∈ N and each i ∈ {1,2,3}: f (2 n − i) > 0 if and only if 3 | 2 n − i. If

3 |2 n − i, then f (2 n − i) ≥ 3 n /2 . In addition, | f (2 n − i)| ≤ 2 · 3 (n−2)/2 .

Assume that p ≥ 1 is an integer. Let α 1 ,..., α l

be positive integers such that

3p =2 α 1 + ··· + 2 α l . In order to prove f (3p) ≥ 0 let us start with

f (3p) = f (2 α 1 α

α − t (2 l 2 + ··· + 2 )) − f (2 2 − t (2 3 + ··· + 2 ))

+ f (2 α 3 α + ··· + 2 l ).

Since x + y ≡ x − t(y) (mod 3) the first term on the right-hand side is ≥ 3 α 1 /2 , while f (2 α 2 −t(2 α 3 + ···+ 2 α l )) ≤ 0. It suffices to establish f (2 α 3 + ···+ 2 α l )≤

3 α 1 /2 . Let us prove the following: | f (m)| ≤ 3 n /2 for all m, n with m <2 n . The last statement is true for n = 1, 2. Assume that the statement holds for some n and assume that m <2 n +1 . If m <2 n we are done. Otherwise, let m =2 n + j, for some 0

≤j<2 n . Then

| f (m)| = | f (2 n − t( j)) − f ( j)| ≤ | f (2 − t( j)| + | f ( j)|.

Since | f (2 n − t( j))| ≤ 2 · 3 (n−2)/2 and | f ( j)| ≤ 3 n /2 we get | f (m)| ≤ 3 (n+1)/2 .

5. Consider the quantity

a b c d b c d a S =3

Using the inequality between the arithmetic and geometric means, we get

The required inequality now follows immediately.

4.49 Shortlisted Problems 2008 767

6. Assume the contrary, that f is onto. There exists a sequence of numbers a n such that f (a n ) = n. If f (u) = f (v) then f (u +1/n) = f (a n +1/ f (u)) = f (a n + 1/ f (v)) = f (v+1/n) for all n ∈ N, and by induction f (u + m/n) = f (v + m/n) for all m,n ∈ N.

Applying this to u =a 1 ,v =a 1 + 1/ f (a 1 − 1), and n = f (a 1 − 1), we get 1 =

f (a 1 ) = f (a 1 + 1/n) = f (a 1 + 2/n) = ··· = f (a 1 + 1). This further implies that

f (a 1 + q + 1) = f (a 1 + q) for all rational numbers q. For fixed y we have

Particularly, Γ (a 1 ) = N, so we could assume that a 2 ,a 3 , . . . are chosen from Γ (a 1 ). Hence for each n ≥ 2 there exists k n ∈ N such that a n =a 1 + 1/k n . Now we have f (a 1 + 1/n) = f (a 1 + 1/ f (a n )) = f (a n + 1) = f (a 1 + 1 k n + 1) = f (a 1 +

1 /k n ) = f (a n ) = n. Since Γ (a 1 + 1 3 ) = N, there exists d such that f (a 1 + 1 3 + 1 d ) = 1. Assume that

d = q for relatively prime numbers p and q. Since 3 + d 6= 1, we have q > 1. Let k be an integer such that kp ≡ 1 (mod q). Then 1 = f (a 1 + p/q) =

f (a 1 + kp/q) = f (a 1 + 1/q) = q which is a contradiction.

7. The left-hand side can be rewritten as

L = (a − c) + (b − d)

a +b+c b +c+d

2c +a +(a − c)(b − d) ·

2d +b

. (b + c + d)(d + a + b) (a + b + c)(c + d + a)

In order to prove that L ≥ 0, it suffices to establish the following inequality:

2c +a (a − c)(b − d)

2d +b

(b + c + d)(d + a + b) (a + b + c)(c + d + a)

−2|(a − c)(b − d)|

a + b + c· b +c+d

If a = c or b = d, the inequality is obvious. Assume that a > c and b > d. Our goal is to prove that

2d +b

2c +a

√ √ . (b + c + d)(d + a + b) (a + b + c)(c + d + a)

a + b + c· b +c+d Both fractions on the left-hand side are positive hence it is enough to prove

that each of them is smaller than the right-hand side. These two inequalities are analogous, so let us prove the first one. After squaring both sides, cross- multiplying, and subtracting, we get

768 4 Solutions

−(2d + b) 2 (a + b + c) + 4(d + a + b) 2 (b + c + d)

= −(2d + b) 2 (a + b + c) + [(2d + b) + (2a + b)] 2 (b + c + d)

2 2 = −(2d + b) 2 · a + (2d + b) · d + (2a + b) (b + c + d) + 2 · (2d + b)(2a + b)(b + c + d)

> −(2d + b) 2 · a + (2d + b)(2a + b)(2b + 2c + 2d)

2 > −(2d + b) 2 · a + (2d + b) (2a + b) > 0.

Equality holds if and only if a = c and b = d.

8. The largest such n is equal to 6. Six boxes can be placed in a plane as

B 5 shown in the picture.

Let us now prove that n B 1 ≤ 6. Denote by

X i and Y i the projections of the box B i B 3 to the lines Ox and Oy respectively. If

i 6≡ j ± 1 (mod n) then X i ∩X j 6= /0 and B 4 Y i ∩Y j 6= /0. If i ≡ j ± 1 (mod n), then

X i ∩X j = /0 or Y i ∩Y j = /0. We will now prove that there are at most 3 values for i such that X i ∩X i +1 = /0. Assume that X 1 = [a 1 ,b 1 ], . . . , X n = [a n ,b n ]. Without loss of generality we may assume that a 1 = max{a 1 ,a 2 ,...,a n }. If b i <a 1 for some i ∈ {2,3,...,n}, then

X i ∩X 1 = /0, hence i ∈ {2,n}. Therefore a 1 ∈X 3 ∩X 4 ∩ ··· ∩ X n −1 and X 2 ∩X 3 ,

X n ∩X n −1 ,X 1 ∩X 2 , and X 1 ∩X n are the only possible intersections that could

be empty. We will prove that not all of these sets can be empty. Assume the contrary. Then a 3 ∈ (b 2 ,a 1 ), b n ∈ (a 3 ,a 1 ), and a n −1 ∈ (b n ,a 1 ). This implies that

a n −1 >b 2 and X n −1 ∩X 2 = /0, a contradiction.

In a similar way we prove that at most three of the intersections Y 1 ∩Y 2 , ...,Y n ∩ Y 1 are empty. However, there are n empty sets among the intersections X 1 ∩X 2 ,

X 2 ∩X 3 , ...,X n ∩X 1 ,Y 1 ∩Y 2 ,Y 2 ∩Y 3 , ...,Y n ∩Y 1 , yielding n ≤ 3 + 3 = 6.

9. Let us call a permutation nice if it satisfies the stated property. We want to cal- culate the number x n of nice permutations. For n = 1, 2, 3 every permutation is nice hence x n = n! for n ≤ 3.

Assume now that n ≥ 4. From (n − 1) | 2(a 1 + ··· + a n −1 ) = 2[(1 + ··· + n) −

a n ] = n(n + 1) − 2a n = (n + 2)(n − 1) − 2(a n − 1) we conclude that (n − 1) |

2 (a n − 1). If n is even we immediately conclude that a n = n or a n = 1. Let us prove that a n ∈ {1,n} for odd n. Assume the contrary. Then n − 1 =

2 (a n − 1), i.e., a n = n +1 2 . Then n − 2 | 2(a 1 + ··· + a n −2 ) = n(n + 1) − 2a n − 2a n −1 = n(n + 1) − (n + 1) − 2a n −1 = (n − 2)(n + 2) + 3 − 2a n −1 which gives n

− 2 | 2a 2n n −1 − 3. Since n −2 ≤ −3 n −2 =1+ 1 n −2 < 2 we get n − 2 = 2a n −1 − 3. This implies that a

2a n −1 −3

n −1 = 2 =a n , which is a contradiction. Therefore, for n ≥ 4 we must have a n ∈ {1,n}. There are x n −1 nice permuta-

n +1

tions for a n = n, and for a n = 1, the problem reduces to counting the nice per- mutations of the set {2,3,...,n} satisfying the given property. However, since

4.49 Shortlisted Problems 2008 769

2 (a 1 + ··· + a k ) = 2k + 2((a 1 − 1) + ···(a k − 1)), we get k | 2(a 1 + ··· + a k ) if and only if k | 2[(a 1 − 1) + ··· + (a k − 1)]. This provides a bijection between the nice permutations of {2,3,...,n} and the nice permutations of {1,2,...,n − 1}. Thus we have x n = 2x n −1 for n ≥ 4, which implies x n =2 n −3 ·x 3 =6·2 n −3 for n ≥ 4.

10. Since the area of a triangle ABC is equal to 1 2 −→ | AB × AC |, we have that (0,0) and (a, b) are k-friends if and only if there exists a point (x, y) such that ay − bx = ±2k. According to Euclidian algorithm such integers x and y will exist if and only if gcd (a, b) | 2k. Similarly, (a,b) and (c,d) are k-friends if and only if gcd (c − a,d − b) | 2k.

Assume that there exists a k-clique S of size n 2 + 1 for some n ≥ 1. Then there are two elements (a, b), (c, d) ∈ S such that a ≡ c (mod n) and b ≡ d (mod n). This implies n | gcd (a − c,b − d) | 2k, or equivalently, n | 2k. Therefore, for a k-clique of size 200 to exist, we must have n | 2k for all n ∈ {1,2,...,14}. Therefore k ≥ 4 · 9 · 5 · 7 · 11 · 13 = 180180.

It is easy to see that all lattice points from the square [0, 14] 2 are 180180-friends.

11. The number of sequences in which the lamp i is switched (on or off) exactly α i times (i = 1, 2, . . . , 2n) is equal to

α 1 ! ·α 2 ! ···α 2n ! . Therefore

M = k! · ∑

α ! n = k, 2 ∤ α α ! 1 + ··· + α 1 ,...,2∤ α n .

1 ··· n

Similarly, we get

N = k! · ∑

, α 1 ! ··· α n ! · β 1 ! ··· β n !

where the summation is over all possible α 1 ,..., α n , β 1 ,..., β n that satisfy α 1 + ··· + α n + β 1 + ··· + β n = k, 2 ∤ α 1 , ...,2∤ α n ,2 | β 1 , ...,2| β n . We see that the

sum in (1) is equal to the coefficient of X k in the expansion

= sinh n (X ), while the sum in (2) is equal to the coefficient of X k in the expansion

4! + ··· = sinh n

(X ) · cosh (X) = 2 n sinh n (2X). Assume that sinh n (X) = ∑ ∞ i =0 a i X i for some real numbers a 1 ,a 2 , . . . . Then we

have sinh n (2X ) = ∑ ∞

a i (2X) i =∑ ∞

i =0 a i ·2 ·X . Therefore

·a k

=2 n −k .

k ·2 2 k n

1 · k! · a

770 4 Solutions

12. Let m be the average of all elements from S, i.e. m = 1 k +l (x 1 +x 2 + ··· + x k +l ). Define Γ (A) = 1

k ∑ x ∈A x . Clearly, set A is nice if and only if | Γ (A) − m| ≤ . For each permutation π =( π 1 ,..., π k +l ) of S consider the sets A π ,A π , ...,A π 2k 1 2 k +l

defined as A π i ={ π i , π i +1 ,..., π i +k−1 } (indices are modulo k + l). We will prove that at least two of the sets A π i are nice. Let us paint the sets A π i red and green in the following way: A π is green if and only if Γ (A π ) ≥ m. Notice that Γ (A π i i i )−

k ( π i − π i +k ) is of absolute value ≤ k . Therefore | Γ (A i ) − m| ≤ 2k or

Γ (A π i +1 )= 1 1 π

| 1 Γ (A π

i +1 ) − m| ≤ 2k . Hence whenever two consecutive sets in the sequence

A π 1 ,...,A π k +l are of different colors, one of them must be nice. If there are ≥2 sets of each of the colors, it is obvious that at least two of the sets will be nice. Assume that there is only one red set and that it is the only nice set. Without loss of generality assume that A π is red. Then m (k + l) = Γ (A π

)+ Γ (A π 2 ) + ··· + Γ (A π

k +l ) > (m − 2k )+m+ 2k + ···+ m+ 2k ≥ (k +l)m, which is a contradiction. Now we can prove the required statement. To each of the (k + l)! permutations

of S we assign at least two nice sets. Each set is counted (k + l) · k! · l! times, so

2 there are at least (k+l)! 1 k +l · k ! ·l! nice sets.

13. We will prove a stronger result, that for each k we have

n ∑ −1 (|S

i | − (n + 1)) ≤ (2n + 1) · 2 . (1)

i =1

The desired statement follows from (1) because if |S i | ≥ 2n + 2 for each i, then (2n + 1) · 2 n −1 ≥2 n · (n + 1) which is impossible. We will use induction on n to prove (1). First, for n = 1 and any subsets

S 1 ,S 2 ,...,S k of {1,2,3,4} we want to prove |S 1 |+ ···+|S k |≤2 0 ·3+ k(1+ 1) =

3 + 2k. It suffices to verify this only when |S i | ≥ 3 for each i. If there is one set with four elements, then k = 1 and the inequality is satisfied. If all sets have cardinality 3, then k ≤ 3 ({1,3,4} and {2,3,4} can’t both be among the chosen sets), hence 3k ≤ 6 + 2k. Assume now that the statement is true for n − 1. Let us divide the subsets

S 1 ,...,S k of n {1,2,...,2 +1 } into two families: A = {A 1 ,...,A l }, those with all elements greater than 2 n , and B = {B 1 ,...,B k −l }, the remaining subsets. Using

the induction hypothesis we obtain ∑ l

i =1 (|A i |− (n+ 1)) ≤ (2n −1)·2 −2 − l. Let us denote by α i the smallest element of B i ∩ {2 n + 1, . . . , 2 n +1 } if it exists. Let

H i =B i ∩ {1,...,2 n } and G i =B i ∩ {2 n + 1, . . . , 2 n +1 }\{ α i }. If i < j we claim that G i ∩G j = /0. If not, considering z ∈ G i ∩G j and taking y = α i ,x ∈H j , we get a contradiction since x < y < z and x, z ∈ B j ,y ,z∈B i . Sets G i are all disjoint, and the induction hypothesis holds for sets H i . Hence ∑ k −l (|B i | − (n + 1)) = ∑ k −l

nn i =1

i =1 (|H i | − n) + ∑ i =1 |G i | ≤ (2n − 1)2 −2 +2 . Thus

k −l

∑ n (|S

i | − (n + 1)) ≤ (2n − 1) · 2 −2 ·2−l+2 ≤ (2n + 1)2 −1 .

i =1

14. Let A ′ ,B ′ ,C ′

be the midpoints of BC, CA, AB, respectively, and let A ′′ ,B ′′ ,C ′′

be the midpoints of HA, HB, HC respectively. Let O be the circumcenter of

4.49 Shortlisted Problems 2008 771 △ABC and R its circumradius. The Pythagorean theorem implies OA 2 = OA ′2 +

A ′ A 2 = OA ′2 +A ′ H 2 1 1 . Since HA ′ OA ′′ is a parallelogram we have that OA ′2 +

A ′ H 2 = 1 2 ′ ′′2

2 (OH +A A ). However, since A A OA is a parallelogram, we have that A ′ A ′′ = OA = R. Thus OA 2 1 2 1 2 = 2 (R + OH ). Similar relations for OA 2 , OB 1 ,

OB 2 , OC 1 , OC 2 imply that the points A 1 ,A 2 ,B 1 ,B 2 ,C 1 ,C 2 lie on a circle with center O.

15. Assume that the distribution of the points is such that ABEF is a convex quadri- lateral and C belongs to the segment BE (other cases are analogous). Let JF = p, FI = q, IK = r. Then KE = q + r. Let us further define DI = s , IC = t, JA = x, AB = y. Since ABEF

E is cyclic, we have JA

· JB = JF · JE, q +r i.e., x (x + y) = p(p + 2q + 2r). From

r K CD kJB we have q s = and t =

p +2q+2r . The last three equalities im- F q ply that st = q(q + 2r). The quadrilat-

eral ABKI is cyclic if and only if x y (x + x

A y B ) = (p + q)(p + q + r). And JCKD is cyclic if and only if (p + q)r = st. We want to prove that

x (x + y) = (p + q)(p + q + r) ⇔ (p + q)r = st.

Using the equalities we already have, we can eliminate x, x + y, s, and t from the previous equivalence. Hence it suffices to prove that

h p(p+ 2q + 2r)

(p + q)(p + q + r)

q (q + 2r)

⇔ [p(p + 2q + 2r) = (p + q)(p + q + r) ⇔ (p + q)r = q(q + 2r)]. The last equivalence becomes obvious once we multiply out all the terms.

16. Let us first consider the case EQ 6= EP. Assume that EQ < EP and denote by A ′ and D ′ the intersections of EA and ED with the circumcircle of APQD. Then

∠PAA ′ = ∠PAD ′ + ∠D ′ AA ′ = ∠PAD ′ + ∠D ′ DA ′ , while ∠QDD ′ = ∠QDA ′ + ∠A ′ DD ′ hence ∠QDA ′ = ∠PAD ′ . This means that A ′ Q = PD ′ and A ′ D ′ kQP. Therefore ∠DEQ = ∠DD ′ A ′ = ∠DAA ′ ; hence QE is a tangent to the circumcir- cle of

△DAE. Let M be the intersection of AD and PQ . Then ME 2 = MD · MA. Since APQD is cyclic we have that MD

·MA = MQ·MP; hence ME 2 = MQ·MP. Assume that BC intersects PQ at a point N. Then NE 2 = NQ ·NP, and since there

is a unique point X on the line PQ for which X E 2 = XQ · XP, we conclude that M

≡ N. Now from MD · MA = ME 2 = MC · MB we get that ABCD is cyclic. If EQ = EP then it is easy to prove that the perpendicular bisectors of AD, BC,

PQ coincide hence ABCD is an isosceles trapezoid, hence it is cyclic.

17. Let M be the intersection point of QF and PE. We need to prove that ∠QMP = ∠BAC. Since ∠MQP = ∠QAB (QB is a tangent to the circle around △QFA),

772 4 Solutions it is enough to prove that ∠QAB + ∠BAC = ∠QMP + ∠MQP, or equivalently,

∠QAE = ∠EPC. Therefore we need to prove that AQPE is a cyclic quadrilateral. From BQ 2 = BF · BA = BP 2 we get BP = BQ. Adding BF · BA = BP 2 to AF · AB = AE · AC (which holds since BCEF is cyclic) we get AB 2 = AE · AC + BP 2 . From the Pythagorean theorem we have AB 2 = AE 2 + BE 2 = AE 2 + BC 2 −CE 2 ,

from which we get BC 2 −CE 2 = AE · EC + BP 2 . This implies that BC 2 − BP 2 = CE 2 + AE · EC, or equivalently

CE · (CE + AE) = (BC + BP)(BC − BP) = CQ ·CP. Thus CE ·CA = CP ·CQ and QPEA is cyclic.

18. We will use induction on k. The statement is valid for k = 0, since there is at least one point P for which (OP) doesn’t intersect any of the lines from L. Assume that the statement holds for k − 1. Consider the point O and the line (or one of the lines if there are more) whose distance from O is the smallest. Denote this line by l. This line contains n −1 points from I. We will first prove that there are at least k + 1 red points on l. We start by noticing that there exists a point P ∈ l ∩ I such that (OP) doesn’t intersect any of the lines from L. Point P divides the

line l in two rays; assume that one of them contains the points P 1 ,P 2 ,...,P u ∈ I, while the other ray contains the points Q 1 ,...,Q n −2−u ∈ I. Assume that P i ’s are sorted according to their distance from P, and the same holds for Q i ’s. Consider the open segments (OP i ) and (OP i +1 ). Each line not containing any of P i and P i +1 must intersect either both or none of these segments. The line passing through P i (other than l) could intersect (OP i +1 ), and a similar fact holds for the line passing through P i +1 . Hence the numbers of intersections of (OP i ) and (OP i +1 )

with lines from L differ by at most 1. Therefore P 1 ,P 2 , ...,P min {k,u} are all red.

A similar result holds for Q i ’s hence there are at least k + 1 red points on l. If we remove l together with n − 1 points on it, the remaining configuration al- lows us to apply the induction hypothesis. There are at least 1 2 k ·(k + 1) points G from I \{l} for which (OG) intersects at most k − 1 lines from L \{l}. Therefore there are at least 1 2 1 k · (k + 1) + k + 1 = 2 (k + 1)(k + 2) red points.

19. Assume first that there exists a point P inside ABCD with the described property. Let K, L, M, N be the feet of perpendiculars from P to AB, BC, CD, and DA respectively. We have ∠KNM = ∠KNP + ∠PNM = ∠KAP + ∠PDM = 90 ◦ and similarly ∠NKL = ∠KLM = ∠LMN = 90 ◦ ; hence KLMN is a rectangle. Denote by W , X , Y , Z the feet of perpendiculars from P to KL, LM, MN, and NK. From

△PLX ∼ △PCM we get that CM = PW · PM = PM · . Similarly DM = PM PZ

PW . Notice that

hence BD kML. Similarly ACkLK; hence AC ⊥ BD. Conversely, assume that ABCD is a convex quadrilateral for which AC ⊥ BD. Let P ′

be any point in the plane and consider a triangle M ′ P ′ L ′ for which M ′ L ′ kBD,

4.49 Shortlisted Problems 2008 773 P ′ M ′ ⊥ CD, and P ′ L ′ ⊥ CB. Let K ′

be the point for which P ′ K ′ ⊥ AB and K ′ L ′ kAC. Let N ′

be the point such that K ′ L ′ M ′ N ′ is a rectangle. Consider the four lines α k , α l , α m , α n through K ′ ,L ′ ,M ′ ,N ′ perpendicular to P ′ K ′ ,P ′ L ′ ,P ′ M ′ , and P ′ N ′ respectively. Let A ′ = α k ∩ α n ,B ′ = α k ∩ α l ,C ′ = α l ∩ α m , and D ′ = α m ∩ α n .

Using the previously established result we have A ′ C ′ kK ′ L ′ and B ′ D ′ kM ′ L ′ . We also have C ′ D ′ kCD, B ′ C ′ kBC, A ′ B ′ kAB hence △DCB ∼ △D ′ C ′ B ′ and △ABC ∼ △A ′ B ′ C ′ . Thus there exists a homothety that takes A ′ B ′ C ′ D ′ to ABCD, and this homothethy will map P ′ into the point P with the required properties.

20. Let M, N, P, Q be the points of tan- gency of ω with AB, BC, CD, and

DA , respectively. We have that AB + N AD = AB + AQ − QD = AB + AM − DP

= BM − CP + CD = BN − CN + Q CD = BC + CD. Denote by X and Y

the points of tangency of ω

1 and ω 2 Z D X X with AC. Then we have AB = AX +

YY BC − CX and AD = AY + CD − CY .

A B Together with AB +AD = BC +CD, this yields AX −CX = CY −AY. Since AX +

CX = CY + AY , we conclude that AX = CY ; hence Y is the point of tangency of AC and the excircle ω B of △ABC that corresponds to B. Similarly, the excircle ω D corresponding to D of △ADC passes through X. Consider the homothety that maps ω B to ω . Denote by Z the image of Y under this homothety. Then Z belongs to the tangent of ω that is parallel to AC. There- fore Z is the image of X under the homothety with center D that maps ω D to ω . Denote by X ′ and Y ′ the intersections of DX and BY with ω 2 and ω 1 respect- ively. Circles ω 1 and ω B are homothetic with center B; hence Y ′ the image of Y under this homothety. Moreover, Y ′ belongs to the tangent of ω 1 that is parallel to AC. This implies that XY ′ is a diameter of ω 1 . Similarly, X ′ Y is a diameter of ω 2 . This implies that X ′ Y kXY ′ , which means that △ZX ′ Y ∼ △ZXY ′ and Z is the center of the homothety that maps ω 2 to ω 1 . This finishes the proof of the required statement.

21. Assume the contrary. If two of the numbers are the same, then so are all three of them. Let us therefore assume that all of a, b, c are different. The given condi- tions imply that

a n −b n b n −c n c n −a n

a · b · c = −p −b , −c −a which immediately implies that some of the numbers a, b, c have to be negative.

Moreover, n can’t be odd since otherwise each of the fractions would be posi-

a ∤ n −b tive. Assume first that p is odd. Since 2 n

+a b + ··· + b , the numbers a and b have to be of different parity. Similarly, 2 ∤ b − c and 2 ∤ c − a which is not possible. We are left with the case p

= 2. Writing n = 2m, we derive (a

a −b · b −c · c −a = −8. This means that a +b = ±2,

a m −b m = ±(a − b), and analogous equalities hold for the pairs (b,c) and (c,a).

774 4 Solutions If m is even, then |a| = |b| = |c| = 1, which means that at least two of a, b, c have

to be the same. If m is odd, then ±2 = a m +b m is divisible by a + b. Since a m +b m ≡ a + b (mod 2), we conclude that a + b = ±2. Similarly b + c = ±2 and c + a = ±2. At least two of a, b, c have to be the same, which is a contradiction. Remark. The statement of the problem remains valid if we replace the assump- tion that p is prime with the assumption 2 ∤ p or p = 2.

22. Assume the contrary. Without loss of generality we may assume that these num- bers are relatively prime (otherwise we could divide them by their common divi-

sor). We may also assume that a 1 <a 2 < ··· < a n . For each i ∈ {1,2,...,n − 1} there exists j ∈ {1,2,...,n − 1} such that a n +a i | 3a j . This together with

a n +a i >a j implies that a n +a i is divisible by 3 for all i. There exists k ∈ {1,2} such that a n ≡ k (mod 3) and a i ≡ 3 − k (mod 3) for all

i 6= n. For each i ∈ {1,2,...,n − 2} there exists j such that a n −1 +a i | 3a j . Since

a n −1 +a i is not divisible by 3, we must have a n −1 +a i |a j ; hence j = n, and we conclude that a n −1 +a i |a n for all i ∈ {1,2,...,n − 2}. Let l ∈ {1,2,...,n} be such an integer for which a n +a n −1 | 3a l . Adding the inequalities a n +a n −1 ≤ 3a l and a n −1 +a l ≤a n gives that a n −1 ≤a l ; thus either l = n or l = n − 1. In the first case, u (a n −1 +a n ) = 3a n for some u ∈ N. We immediately see that u < 3 and u > 1. Hence u = 2 and 2a n −1 =a n . However, this is impossible, since for each i ∈ {1,2,...,n − 2} the number a n −1 +a i divides a n = 2a n −1 . On the other hand, if a n −1 +a n | 3a n −1 then there exists v ∈ N for which v(a n −1 +

a n ) = 3a n −1 . If v ≥ 2, then 2a n −1 + 2a n ≤ 3a n −1 , which is impossible. Hence v = 1, and we get a n = 2a n −1 . In the same way as in the previous case we get a contradiction.

23. We will use induction on n. Observe that a n ≥ (a n +1 ,a n )>a n −1 . Obviously, a 0 ≥

1, and a 1 ≥a 0 + 1 ≥ 2. From a k +1 −a k ≥ (a k +1 ,a k )≥a k −1 + 1 we get a 2 ≥4 and a 3 ≥ 7. It is impossible to have a 3 = 7, since (a 3 ,a 2 )>a 1 = 2 would imply

a 2 =7=a 3 . Hence we have that the statement is satisfied for n ∈ {0,1,2,3}. Assume that n ≥ 2 and a i ≥2 i for all i ∈ {0,1,...,n}. We need to prove that

a n +1 +1 ≥2 n . Let us define d n = (a n +1 ,a n ). We have d n >a n −1 . Let a n +1 = kd n and a n = ld n . If k ≥ 4 we are done, because a n +1 ≥ 4d n > 4a n −1 ≥4·2 n −1 =

2 n +1 . If l ≥ 3, then a n +1 >a n implies k ≥ 4. If l = 1, then a n +1 ≥ 2a n ≥2 n +1 . Hence the only remaining case to consider is a n = 3d n ,a n −1 = 2d n . Obviously,

d n −1 = (2d n ,a n −1 )>a n −2 . If a n −1 =d n −1 , then from a n −1 <d n and a n −1 | 2d n we get 2d n

3 ≥ 3 and d 3 n ≥ a n −1 ≥ ·2 n −1 . Now a n +1 = 3d n ≥9·2 n a −2 n −1 2 2 >2 n +1 . If a n −1 ≥ 3d n −1 then d n >a n −1 ≥ 3d n −1 . Since d n −1 = (2d n ,a n −1 ), there exists s

∈ N such that 2d n n = sd n −1 . This implies that d n 2d >3· s , which means that s

. It remains to consider the case a n −1 = 2d n −1 . From (2d n , 2d n −1 )=d n −1 we conclude that d n = d n −1 2 w for some odd integer w ≥ 3. From a n −1 <d n we get

n −2 and a +1 = 3d >2 n > 6, or s ≥ 7. Therefore 2d +1

n ≥ 7d n −1 >7·2

nn

2d

2 > 21 · 2 n −3 >2 n +1 ; hence it remains to consider the case w = 5. We now have 2 n −3 ≤a n −3 <

n −1 <d n ; hence w

≥ 5. If w ≥ 7, then a n n +1 ≥3·7· −1

4.49 Shortlisted Problems 2008 775

d n −2 = (2d n −1 ,a n −2 ). If a n −2 ≥ 2d n −2 , then 2d n −1 ≥ 3d n −2 >3·2 n −3 . There- fore a

n +1 =3· −1 2 ≥ 45 · 2 n −4 >2 n . If a n −2 =d n −2 , then from a n −2 <d n −1 we get again 2d n −1 ≥ 3d n −2 and a n +1 ≥2 n +1 .

5d n

24. First we prove that the numbers 2 n −1 k are all odd. Let M be the largest integer for which 2 M divides (2 n − 1)!. Then M = ∑ n −1 n −i 1 n −1 n i −i =1 2 − 2 i =∑ i =1 2 −1 . The largest number N for which 2 N divides k! · (2 n − 1 − k)! satisfies

n −1

i + 2 n ∑ −i − i .

k +1

i =1

Each summand on the right-hand side is equal to 2 n −i

− 1 (write k = q i ·2 +r i , for 0 ≤r i <2 i ). Hence M

= N and 2 −1

is odd. Let us prove that the numbers 2 n −1 k give different remainders modulo 2 n . This

is valid for n n

= 1. Assume that this holds for some n > 1. We claim that the +1 sets A =

− i are the same modulo 2 n +1 for each i = 0, 1, . . . , 2 n −1 − 1. We also claim that that all numbers

from S 2 n −1 i −1 =0 B i are different modulo 2 n +1 . These two claims will imply the

2 n desired result. Let us show that +1 −1

2 n +1 −1

≡− 2i +1 (mod 2 n +1 ) and that one of these two numbers is congruent to 2 n −1 i . The first congruence follows from

(mod 2 n +1 ),

2i +1

while the second is true because

n ≡ (−1) +1 · (mod 2 ).

2 n −1

i It remains to show that S 2 n −1 i −1 =0 B i have all elements different modulo 2 n +1 . The

has a different remainder from 2 −1 j for i

induction hypothesis implies that 2 n −1

6= j. The same holds for 2 n − −1

. From 2 2k −1 +

≡ 0 (mod 2 ) if and only if there exists k such that n {i, j} = {2k,2k + 1} for some k. However, in that case

2k +1 ≡2

2 n −1

n we have that 2 n −1

n +1

2 n −1

n +1

2 −1 + 2 n −1

= 2 n 2 n −1

6≡ 0 (mod 2 ).

ij

2k +1

2k

776 4 Solutions

25. If p is a prime number, then d ( f (p)) has p divisors, and must be a power of a prime. Hence f (p) = q p −1 for some prime number q. Let us show that q = p. Consider first the case p > 2. From f (2p) | (2 − 1) · p 2p −1 · f (2) and f (2p) | (p − 1) · 2 2p −1 · f (p) = (p − 1) · 2 2p −1 ·q p −1 we conclude that f (2p) | (p 2p −1 ·

f (2), (p − 1) · 2 2p −1 ·q p −1 ) = ( f (2), (p − 1) · 2 2p −1 ·q p −1 ). Since f (2p) has 2p divisors and f (2) is prime, this is a contradiction. We also have f (2) = 2. Indeed, this follows from f (6) | 3 6 −1 · f (2), f (6) | 2 · 2 6 −1 ·3 3 −1 , and d ( f (6)) = 6. Assume now that x =p a 1 1 ··· p a n n is a prime factorization of x with p 1 < ··· <

p n . Let f (x) = q b 1 1 1 ··· q b m a a m 1 . From d ( f (x)) = p 1 ··· p n = (b 1 + 1) ···(b m + 1) we conclude that b

a −1

i ≥p 1 −1 for all i. The relation f (x) | (p 1 −1)·(p 1 a n x 1 −1 ··· p n ) ·

f (p 1 ) yields q 1 ,...,q m ∈ {p 1 ,...,p n }. Hence for each prime p and each a ∈ N there is b ∈ N such that f (p a )=p b . From p a = b + 1 we get f (p a )=p p a −1 . Now assume that x ∈ N. There are integers a 1 ,...,a n ,b 1 ,...,b n such that x = p a 1 1 ··· p a n

n and f (x) = p 1 ··· p n . For each i ∈ {1,...,n} we have f (x) | (p i −1)· (x/p a i

p i x ai −1 −1 a

p ai i −1

i . Multiplying this for i = 1, . . . , n we get d( f (x)) = (b

i ) ·p i ; hence p i |p i , which implies b

i i +1≤p

1 + 1) ···(b n + 1) ≤ p 1 ··· p a 1 n = x. Since

p a1 −1 n

i − 1 for all i and f (x) = p 1 ··· p n . It is easy to verify that the function f defined by the previous relation satisfies the required conditions.

d ( f (x)) = x, we must have b =p

1 p n an i −1

26. If p is any prime number of the form p ≡ 1(mod 4); we know that −1 p = 1, and there are exactly two numbers n , m ∈ {0,1,2,..., p − 1} whose square is

congruent to −1 modulo p. Since the sum of these two numbers is equal to p, one of them is smaller than p /2. Assuming that n < p/2; let us set k = p − 2n. It suffices to prove that there exist infinitely many prime numbers p for which

√ k 2 > 2n. From p 2 +1= p −2pk+k |n 2

4 + 1 we conclude that p | k 2 + 4. This implies that k 2 ≥ p−4. It suffices to prove that p−4 > 2n, i.e., 4 < p−2n = k for

infinitely many values of p. However, this will be satisfied, since k √ ≥ p −4>4 for p > 20, and there are infinitely many prime numbers greater than 20 that are congruent to 1 modulo 4.

4.50 Shortlisted Problems 2009 777