Solutions to the Shortlisted Problems of IMO 2007

4.48 Solutions to the Shortlisted Problems of IMO 2007

1. (a) Assume that d =d m for some index m, and let k and l be the indices such that k ≤ m ≤ l such that d m =a k −a l . Then d m =a k −a l ≤ (a k −x k )+ (x l −a l ), and hence a k −x k ≥ d/2 or x l −a l ≥ d/2. The claim follows immediately.

(b) Let M

i = max{a j ≤ j ≤ i} and m i = min{a j ≤ j ≤ n}. Set x i = 2 . Clearly, m i ≤a i ≤M i and both (m i ) and (M i ) are nondecreasing. Fur-

−a i . Similarly x i −a i ≤ 2 ; hence max

thermore, d i = m i −M − i 2 2 =x i −M i ≤x i

−a i i | : 1 ≤ i ≤ n} ≤ max 2 ,1≤i≤n . Thus, the equality holds in (1) for the sequence {x i }.

{|x i

2. Placing n = 1 we get f (m + 1) ≥ f (m)+ f ( f (1))− 1 ≥ f (m); hence the function is nondecreasing. Let n 0 be the smallest integer such that f (n 0 ) > 1. If f (n) = n + k for some k, n ≥ 1, then placing m = 1 gives that f ( f (n)) = f (n + k) ≥

f (k) + f ( f (n))− 1, which implies f (k) = 1. We immediately get k < n 0 . Choose maximal k 0 such that there exists n ∈ N for which f (n) = n + k 0 . Then we have 2n +k 0 ≥ f (2n) ≥ f (n)+ f ( f (n))− 1 = n+k 0 + f (n + k 0 ) − 1 ≥ n+ k 0 + f (n) −

1 = 2n + (2k 0 − 1); hence 2k 0 −1≤k 0 , or k 0 ≤ 1. Therefore f (n) ≤ n + 1 and

f (2007) ≤ 2008. Now we will prove that f (2007) can be any of the numbers 1, 2, . . . , 2008. Define the functions

≤ 2007 − j,

j (n) =

, for j ≤ 2007, and

+ j − 2007, otherwise

∤ n,

2008 (n) = n

+ 1, 2007 | n.

It is easy to verify that the functions f j satisfy the conditions of the problem for j = 1, 2, . . . , 2008.

1 +t 4 < t holds for all t ∈ (0,1) because it is equivalent to 0 < t − t 3 − t + 1 = (1 − t)(1 −t 3 ). Applying it to t = x k and summing over k = 1, . . . , n; we get 2k ∑ n 1 +x

3. The inequality 1 +t 2 1

1 +x 4k <∑ k =1 x k = x n (x−1) = x n (1−x) . Writing the same relation for y and multiplying by this one gives the desired inequality.

k =1

4. Notice that f (x) > x for all x. Indeed, f (x + f (y)) 6= f (x + y) and if f (y) < y for some y, setting x = y − f (y) yields a contradiction. Now we will prove that f (x) − x is injective. If we assume that f (x) − x =

f (y) − y for some x 6= y, we will have x + f (y) = y + f (x), hence f (x + y) +

f (y) = f (x + y) + f (x), implying f (x) = f (y), which is impossible. From the functional equation we conclude that f ( f (x) + f (y)) − ( f (x) + f (y)) = f (x + y), hence f (x)+ f (y) = f (x ′ )+ f (y ′ ) whenever x+ y = x ′ +y ′ . In particular, we have

f x (x) + f (y) = 2 f ( +y 2 ).

Our next goal is to prove that f is injective. If f (x) = f (x + h) for some h > 0 then f (x) + f (x + 2h) = 2 f (x + h) = 2 f (x); hence f (x) = f (x + 2h), and by

4.48 Shortlisted Problems 2007 755 induction f (x + nh) = f (x). Therefore, 0 < f (x + nh) − (x + nh) = f (x)− x − nh

for every n, which is impossible. We now have f

2 + y) and by symmetry f

f ( f (x) + f (y)) = f ( f (x) + y) + f (y) = 2 f ( (x)

f ( f (x) + f (y)) = 2 f ( (y)

f (y)

f (x)

2 + x). Hence 2 +y= 2 + x, and thus

f (x) − 2x = c for some c ∈ R. The functional equation forces c = 0. It is easy to verify that f (x) = 2x satisfies the given relation.

5. Defining a (0) to be 0, the relations in the problem continue to hold. It follows by induction that a (n 1 +n 2 + ··· + n k ) ≤ 2a(n 1 )+2 2 a (n 2 ) + ··· + 2 k a (n k ). We also have a (n 1 +n 2 + ···+ n 2 i ) ≤ 2a(n 1 + ···+ n 2 i −1 ) + 2a(n 2 i −1 +1 + ···+ n 2 i )≤

··· ≤ 2 i (a(n 1 ) + ··· + a(n

)). For integer k ∈ [2 i ,2 i +1 ) we have

a (n 1 + ··· + n k ) = a(n 1 + ··· + n k + 0 + ···+ 0 ) | {z }

2 i +1 −k

i +1 (a(n 1 i ≤2 +1 ) + ··· + a(n k ) + (2 − k)a(0))

≤ 2k(a(n 1 ) + ···+ a(n k )).

be its binary representa- tion (K ∈ N, b i ∈ {0,1}). For each increasing sequence ( τ n ) n ∈N of integers we have

Assume now that N ∈ N is given and let N = ∑ K i =0 b i 2 i

2 ≤ n ∑ · 2( τ n − τ n −1 + 1) ∑ a (2 i )

i =τ n −1 n +1

2 n +1 τ n +1 2 τ

2 2 τ n +1 n − τ

n −1 + 1)

( τ n −1 + 1) c ∑ n ( τ n −1 + 1) c −2

Choosing τ =2 αn

−1, we get a(N) ≤ 2 ·2 n −1−α(c−2)(n−1) . Thus, choos- ing any α > 1 c −2 would give us the sequence τ n for which the last series is

bounded, which proves the required statement.

6. Using the Cauchy–Schwarz inequality we can bound the left-hand side in the fol- lowing way: 1 3 [a 1 (a 2 100 + 2a 1 a 2 )+a 2 (a 2 1 + 2a 2 a 3 ) + ··· + a 100 (a 2 99 + 2a 100 a 1 )] ≤

(the indices are modulo 100). It suffices to show that

(a ∑ 2 k + 2a k +1 a k +2 ) 2 ≤ 2.

k =1

Each term of the last sum can be seen as a 4 k + 4a 2 2 k 2 +1 a k +2 + 4a k (a k +1 ·a k +2 )≤ (a 4 2 2 2 2 2 k 2 + 2a k a k +1 + 2a k a k +2 ) + 4a k +1 a k +2 . The required inequality now follows from ∑ 100 k =1 (a 4 k + 2a 2 2 2 2 2 k 100 a k +1 + 2a k a k +2 ) ≤ (a 1 +···+a 2 100 ) 2 = 1, and ∑ k =1 a 2 k a 2 k +1 ≤ (a 2 +a 2 + ··· + a 2 ) · (a 2 +a 2 + ··· + a 2 1 2 2 2 1 1 3 99 2 4 100 )≤ 4 a 1 + ··· + a 100 = 4 .

756 4 Solutions

7. The union of the planes x = i, y = i, and z = i for 1 ≤ i ≤ n contains S and doesn’t contain 0. Assume now that there exists a collection {a i x +b i y +c i z +d i =0:1≤

i ≤ N} of N < 3n planes with the described properties. Consider the polynomial P (x, y, z) = ∏ N i =1 (a i x +b i y +c i z +d i ).

Let δ 0 = 1, and choose the numbers δ 1 ,..., δ n such that ∑ n

i =0 δ i i = 0 for m =

0 , 1, 2, . . . , n − 1 (here we assume that 0 0 = 1). The choice of these numbers is possible because the given linear system in ( δ 1 ,..., δ n ) has the Vandermonde determinant. Let S =∑ n i =0 ∑ n j =0 ∑ n k =0 δ i δ j δ k P (i, j, k). By the construction of P we know that P (0, 0, 0) 6= 0 and P(i, j,k) = 0 for all other choices of i, j,k ∈ {0,1,...,n}.

Therefore S = δ 3 0 P (0, 0, 0). On the other hand, expanding P as P(x, y, z) = ∑ α+β +γ≤N p α,β ,γ x α y β z γ we get

nn

i α j β k ∑ γ ∑ ∑ ∑ α,β ,γ i =0 j =0 k =0

α+β +γ≤N

∑ δ j j ∑ δ k k = 0,

because for every choice of α , β , γ at least one of them is less than n, making the corresponding sum in the last expression equal to 0. This is a contradiction; hence the required number of planes is 3n.

1 <S n +1 < ··· < S n 2 +1 and since each of these n (i+1)n + 1 numbers belongs to {0,1,...,n}, we have that S

8. Let S m =a k +a k +1 + ··· + a m . Since S n

n 2 +n k

2n

= i. We immediately get a 1 =a 2 = ··· = a n = 0 and a n 2 +1 =a n 2 +2 = ··· = a n 2 +n = 1. For every 0 2 ≤ k ≤ n, consider the sequence l

in +1

k = S k +n k +1 ,S k +2n k +n+1 ,...,S k +n k +n 2 −n+1 . The sequence is strictly increasing and its elements are from the set {0,1,2,...,n}. Let m be the number that doesn’t appear in l k . Define U k to be the total sum

n of elements of l (n+1)

k . Since a 1 + ··· + a

S k 1 +U k +S n 2 k +n +n 2 +1 =U k + n − k, we get m = n − k. Therefore

S k +(s+1)n

< n − k, s + 1, if s ≥ n − k.

if s

k +sn+1 =

Using the previous equality we conclude

S k k +sn+1 =S −1+(s+1)n k −1+sn+1 +a k +(s+1)n −a k −1+sn+1

k +(s+1)n

+a k +(s+1)n −a k +sn , hence for s

k =S −1+(s+1)n

k +sn

+ k < n we have S k k +sn+1 =S −1+(s+1)n k +sn = s, and for s + k ≥ n+ 1 we have S k +(s+1)n

k +(s+1)n

k −1+(s+1)n k +sn+1 =S k +sn = s + 1. Hence a k +(s+1)n =a k +sn if either s +k<n or s

k + k ≥ n + 1. If s + k = n, then S −1+(s+1)n

= s, while S k +sn+1 k + (s + 1)n =

k +sn

4.48 Shortlisted Problems 2007 757 s + 1. Hence a k +(s+1)n = 1 and a k +sn = 0. Now by induction we can easily get

that for 1 ≤ u ≤ n and 0 ≤ v ≤ n,

It is easy to verify that the above sequence satisfies the required properties.

9. Assume the contrary. Consider the minimal such dissection of the square ABCD (i.e., the dissection with the smallest number of rectangles). No two rectangles in this minimal dissection can share an edge. Let AMNP be the rectangle containing the vertex A, and let U BVW be the rectangle containing B. Assume that MN ≤

BV . Let MXY Z be another rectangle containing the point M (this one could be the same as U BVW ). We can have either MN > MZ or MZ > MN. In the first case the rectangle containing the point Z would have to touch the side CD (it can’t touch BC because WU ≥ NM > MZ). The line MN doesn’t intersect any of the interiors of the rectangles. Contradiction. If MZ > MN, consider the rectangle containing the point N. It can’t touch AD because it can’t share the entire side with AMNP. Hence it has to touch CD, and again, MN would be a line that doesn’t intersect any of the interiors. Contradic- tion.

10. Let T = {(x,y,z) ∈ S × S × S : x + y + z is divisible by n}. For any pair (x,y) ∈ S

× S there exists unique z ∈ S such that (x,y,z) ∈ T ; hence |T | = n 2 . Let M ⊆T

be the set of those triples that have all elements of the same color. Denote by R and B the sets of red and blue numbers and assume that the number r of red numbers is not less than n /2. Consider the following function F : T \M → R×B:

If (x, y, z) ∈ T \ M, then F(x,y,z) is defined to be one of the pairs (x,y), (y,z), (z, x) that belongs to R × B (there exists exactly one such pair). For each element (p, q) ∈ R × B there is unique s ∈ S for which n | p + q + s. Then F(p,q,s) =

F (s, p, q) = F(q, s, p) = (p, q). Hence |T \ M| = 3|R × B| = 3r(n − r) and |T | =

n 2 − 3r(n − r) = n 2 − 3rn + 3r 2 .

It remains to solve n 2 − 3nr + 3r 2 = 2007 in the set N × N. First of all, n = 3k for some k ∈ N. Therefore 9k 2 − 9kr + 3r 2 = 2007 and we see that 3 | r. Let r = 3s. The equation becomes k 2 − 3kr + 3r 2 = 223. From our assumption r ≥ n/2 we get 223 =k 2 − 3kr + 3r 2 = (k − r)(k − 2r) + r 2 ≤r 2 . Furthermore, 4 · 223 = (2k − 3r) 2 + 3r 2 ≥ 3r 2 ≥ 3 · 223. Hence r ∈ {15,16,17}. For r = 15 and r = 16,

4 ·223− 3r 2 is not a perfect square, and for r = 17 we get (2k − 3r) 2 = 25, hence 2k − 3 · 17 = ±5. Both k = 28 and k = 23 lead to solutions (n,r) = (84,51) and (n, r) = (69, 51).

11. Denote by a

n k ,1 ,a k ,2 ,...,a k ,n the elements of A k , and let Q k =∑ =1 a i 2 k ,i . Assume the contrary, that ,i

in S

|a k | < n/2 for all k,i. This means that the number of elements k ∈N A k is finite. Hence there are different p , q ∈ N such that A p =A q . For any I ⊆ {1,2,...,n}, define S k (I) = ∑ i ∈I a k ,i . Let (I k ,J k ) be the partition that was chosen in constructing A k +1 from A k . We have

758 4 Solutions

Q k +1 −Q k = ∑ ((a k ,i + 1) 2 − (a k ,i ) 2 )+ ((a k − 1) 2 ∑ ,j − (a k ,j ) 2 )

i ∈I k

j ∈J k

I ,J {|S k (I) − S k (J)|} , where the last minimum is taken over all partitions (I, J) of the set {1,2,...,n}.

= n + 2 (S k (I k )−S k (J k )) = n − 2min

However, for each k, it is easy to inductively build a partition (I ′ ,J ′ ) for which |S k (I ′ )−S k (J ′ )| < n/2. Take I 0 and J 0 to be empty sets, and assume we made the partition I l ,J l of {1,...,l} in such a way that |S k (I l )−S j (I l )| < n/2. Now, take

(I ,J )= l ∪ {l + 1},J l ), if S(I l ) ≤ S(J l ),

(I

l +1 l +1

(I l ,J l ∪ {l + 1}), if S(I l ) > S(J l ). Therefore Q k +1 −Q k >n−2 n 2 = 0 and Q k is increasing. This contradicts the

previously established fact that A p =A q for some p 6= q.

12. Assume that a > b, a = a 1 d ,b =b 1 d , (a 1 ,b 1 ) = 1. There exist p, q ∈ Z such that pa + qb = d. We may assume that the pairs (S n ,S n +a ) and (S n ,S n +b ) are of different colors, since otherwise the statement would follow immediately. By induction we get that (S n ,S n +ua+vb ) are of the same color if and only if u + v is

even. From ab 1 = ba 1 we conclude that S ab 1 =S ba 1 , which means that a 1 and b 1 must be of the same parity; hence they are both odd and a 1 ≥ 3. Furthermore, pa 1 + qb 1 = 1 gives 2 ∤ p + q, which implies that the strips S n +d =S n +pa+qb and S n are of different colors. Now S n and S n +2d have the same color. Consider the rectangle MNPQ such

that MQ = NP = a, MN = PQ =

b and the difference between the x coordinates of M and Q (and con-

P sequently N and P) is 2d. It suf- fices to show that we can choose

a rectangle in such a way that M

and N are of the same color. Simple

2d N x

calculation shows that the difference of the x coordinates of M and N is √ τ =

a 1 b 6∈ Q. Let s be one of the longest single-colored (say red) segments on the x-axis. The translation s ′ of s to the left by τ has noninteger endpoints, hence

a 2 1 −4

it can’t be single-colored. Thus, there is a red point on s ′ ; choose this point to be N . Other points are now easily determined.

Remark. We used the fact a 1 > 2, which is implied by a 6= b. The statement doesn’t hold for squares (a counterexample is the unit square).

13. Consider one of the cliques of maximal size 2n and put its members in a room X . Call these students Π -students . Put the others in the room Y . Let d (X ) and d(Y )

be the maximal sizes of cliques in X and Y at a given moment. If a student moves from X to Y , then d (X) − d(Y) decreases by 1 or 2. Repeating this procedure, we can make this difference 0 or −1. Assume that it is −1 and d(X) = l, d(Y ) = l + 1. If the room Y contains a Π - student that doesn’t belong to a Y -clique of size l + 1, after moving that student

4.48 Shortlisted Problems 2007 759 to X we will manage to have d (X ) = d(Y ). Therefore assume that all 2n − l

Π -students in Y belong to all Y -cliques of size l + 1. Each such clique has to contain 2 (l − n) + 1 ≥ 1 non- Π -students . Take an arbitrary clique in Y of size l + 1 and move a non- Π -student from it to X . Repeat this procedure as long as there are l + 1 cliques in Y . We claim that d(X ) remains l after each such move. If not, consider an (l + 1)-clique in X. All its members would know all of 2n − l Π -students in Y . Together with them they would form a 2n + 1 clique, which is impossible. Hence we will end up with the configuration with d (X ) = d(Y ) = l.

14. Assume that A 1 ,...,A k are disjoint m-element subsets of {1,...,n}. Let S = {S ⊆ {1,...,n} : S ∩ A i 6= /0 for all i} and

T = {T ⊆ {1,...,n} : T ⊇ A i for some i , but T ∩ A j = /0 for some j}. For each A ∈ S and B ∈ T we have A ∩ B 6= /0. It suffices to prove: h

Lemma. For k = k(m) = 2 m · log 3 +

2 and n = mk we have √

2 Proof. For simplicity set ρ

→∞ 2 n

= log 3 2 5 . For every i, each set in S must contain one of 2 m

− 1 nonempty subsets of A k |S | = (2 − 1) . Using the substitution r =2 m , we get

i . Hence

2 In order to calculate |T |, notice that T = U \ (U ∩ S ) where U = {T ⊆

{1,...,n} : T ⊇ A i for some i }. Obviously, |U c | = (2 m − 1) k , which gives us |U | = 2 mk − (2 m − 1) k . Furthermore, |U ∩ S | = |S | − |S \ U |. From S \ U = {T ⊆ {1,...,n} : T ∩ A i 6= 0 and T ∩ A i 6= A i for all i } we get |S \ U | = (2 m − 2) k . Using the substitution r =2 m , we get

15. For each vertex V of P consider all good triangles whose one vertex is V . All the vertices of these triangles belong to the unit circle centered at V . Label them

counterclockwise as V 1 , . . . ,V i . Denote by f (V ) and l(V ) the first and the last of

760 4 Solutions these vertices ( f (V ) = V 1 ,l (V ) = V i ). Denote by T f (V ) the good triangle with

vertices V and f (V ). T l (V ) is defined analogously. We call T f (V ) and T l (V ) the triangles associated with V (they might be the same). Let α

be the total number of associated triangles, and t the total number of good triangles. It is enough to prove that each good triangle is associated with at least three vertices. Indeed this would imply 3t ≤ α ≤ 2n. It suffices to show that for an arbitrary good triangle ABC oriented counterclockwise we have A = l(B) or A = f (C). Assume that A 6= l(B) and A 6= f (C). Then l(B) and f (C) would belong to the half-plane [BC, A. Define A ′ = l(B) if ∠ABl(B) ≤ 60 ◦ . If this angle is bigger than 60 ◦ , define

A ′ to be the third vertex of T l (B). We define A ′′ similarly. We have ∠ABA ′ < 60 ◦ , ∠ACA ′′ < 60 ◦ , hence A belongs to the interior of the rectangle A ′ BCA ′′ and P can’t be convex. This concludes the proof of our claim. Remark. It is easy to refine the proof to show that t

3 (n − 1)]. This result is sharp and the example of a (3k + 1)-gon with 2k good triangles is not hard

to construct: rotate a rhombus ABCD (AB = BC = DA = 1) around A by small angles k times.

16. Let O be the circumcenter of the triangle ABC. We know that ∠CPK = ∠CQL, and hence S RPK S RQL = RP ·PK

RQ ·QL . Since △PKC ∼ △QLC, we have QL = PC QC . Since ROC is isosceles and ∠OPR = ∠OQC, we get △ROQ ∼ = △COP and RQ = PC. This finally implies S RPK S RQL = 1.

PK

17. Let Y be the midpoint of BT . Then MY kCT and TY ⊥ XY ; hence T , Y , M,

X belong to a circle. Thus ∠MT B − ∠CT M = ∠MXY − ∠Y MT = ∠MXY − ∠T XY = ∠MXY − ∠Y XB = ∠MXB = ∠BAM.

18. Let X be the point on the line PQ such that XC kAQ. Then XC : AQ = CP : PA = BC : AD, which implies △BCX ∼ △DAQ. Hence ∠DAQ = ∠BCX and ∠BXC = ∠DQA = ∠BQC. Therefore B, C, Q, X belong to a circle, which implies that ∠BCX = ∠BQX = ∠BQP.

19. Let K and L be the midpoints of FC and CG respectively. Then EK ⊥ CD and EL ⊥ BC hence KL is the Simson’s line of the triangle BCD and intersects BD at point M such that EM ⊥ BD. We also have that KLkl and KL bisects the side CA of △ACG. Hence KL passes through the intersection of the diagonals of ABCD, and thus KL bisects BD. Therefore DM = MB and DEB is isosceles. From △DEB ∼ △KEL we get that EK = EL; hence CF = CG. Thus ∠DAF = ∠FGC = ∠GFC = ∠FAB.

20. Denote by A 0 ,B 0 , and C 0 the given intersection points. Applying Pascal’s theo- rem to the points APCC ′ BA ′ gives us that B 0 ∈C 1 A 1 . Similarly, we get C 0 ∈A 1 B 1 and A

C . From B A we get C 0 0 B ∈B 1 1 0 1 kAB 1 C 0 0 A = C 0 A C 1 0 B 1 . Since BA 1 kB 1 A 0 , we get C 0 A 1 C 0 B . Hence = C 0 B 0 C 0 C B 0 B 1 C 0 A 0 C 0 A = C 0 A 0 or C 0 B 0 ·C 0 A 0 =C 0 A ·C 0 B . Therefore S A 0 B 0 C 0 =S ABC 0 . However, S ABC 0 =S ABB 1 (because A 1 B 1 kAB); hence S A 0 B 0 C 0 =

S ABB 1 = 1 2 S ABC .

21. Let us prove that S 1 ≥ S, i.e., that k ≤ 1.

4.48 Shortlisted Problems 2007 761 Lemma. If X ′ ,Y ′ ,Z ′ are the points on the sides Y Z, ZX , XY of △XY Z, then

S X ′ Y ′ Z ′ ≥ min{S XY ′ Z ′ ,S YZ ′ X ′ ,S ZX ′ Y ′ }. Proof. Denote by X 1 ,Y 1 ,Z 1 the midpoints of Y Z, ZX , XY . If two of X ′ ,Y ′ , Z ′ belong to one of the triangles XY 1 Z 1 ,YZ 1 X 1 , ZX 1 Y 1 , then the statement follows immediately. Indeed, if Y ′ ,Z ′ ∈ XY 1 Z 1 , then Y ′ Z ′ intersects the alti- tude from X to Y Z at the point Q inside △XY 1 Z 1 , which forces d (X , Z ′ Y ′ )≤

d (X , Z 1 Y 1 )= 1 2 d (X,Y Z) ≤ d(X ′ ,Y ′ Z ′ ). Assume now, without loss of gen- erality, that X ′ ∈X 1 Z ,Y ′ ∈Y 1 X ,Z ′ ∈Z 1 Y . Then d (Z ′ ,X ′ Y ′ ) > d(Z 1 ,X ′ Y ′ ), hence S X ′ Y ′ Z ′ >S X ′ Y ′ Z 1 . Similarly S X ′ Y ′ Z 1 >S X ′ Y 1 Z 1 . Since S X ′ Y 1 Z 1 =S X 1 Y 1 Z 1 we have S X ′ Y ′ Z ′ >S X 1 Y 1 Z 1 = 1 4 S XY Z , hence S X ′ Y ′ Z ′ > min{S XY ′ Z ′ ,S YZ ′ X ′ , S ZX ′ Y ′ }. If S A 1 B 1 C 1 ≥ min{S A 1 BB 1 ,S B 1 CC 1 } and S A 1 C 1 D 1 ≥ min{S C 1 DD 1 ,S D 1 AA 1 } then the problem is trivial. The same holds for S A 1 B 1 D 1 and S B 1 C 1 D 1 . Without loss of generality assume that S A 1 B 1 C 1 < min{S A 1 BB 1 ,S B 1 CC 1 } and S A 1 B 1 D 1 < min{S A 1 BB 1 ,S AA 1 D 1 }. Assume also that S A 1 B 1 C 1 ≤S A 1 B 1 D 1 . Then the line C 1 D 1 intersects the ray (BC at some point U. The lines AB and CD can’t intersect at a point that is on the same side of A 1 C 1 as B 1 . Oth- erwise, any line through C 1 that intersects (BA and (BC, say at M and N, would force S C 1 B 1 N >S A 1 B 1 C 1 and S A 1 BB 1 >S A 1 B 1 C 1 . The lemma would imply that S MA 1 C 1 ≤S A 1 B 1 C 1 . This is impossible since we can make S MA 1 C 1 arbitrar- ily large. Therefore C 1 D 1 ∩ (BA = V . Applying the lemma to △V BU, we get

S A 1 B 1 C 1 ≥S VA 1 C 1 >S A 1 C 1 D 1 , a contradiction.

To show that the constant k = 1 is the best possible, consider the cases close to the degenerate one in which ACD is a triangle, D 1 ,C 1 the midpoints of AD and CD , and B =A 1 =B 1 the midpoint of AC.

be the diameter of the circumcircle k of △ABC. Denote by M the intersection of AI with k. Since A ′ M ⊥ AM, we have that K,

22. We will prove that ∠KID = β −γ 2 . Let AA ′

M , and A ′ are collinear. Let A 1 ,B 1 ,

B 1 from I to BC, CA, and AD. Then

E and X be the feet of perpendiculars

X I △XIB 1 ∼ △A ′ MB , since the corre- sponding angles are equal (this is easy

B D F A 1 C to verify). Since MB = MI = MC,

we conclude that IM : KM MA ′

= BM :

1 : IX = XD : IX; hence △KIM ∼ △IDX. Therefore ∠KID = ∠XIM − (∠XID + ∠KIM) = ∠XIM −

= IB K

90 ◦ = 180 ◦ − ∠AA ′ M − 90 ◦ = ∠MAA ′ = β −γ 2 .

Assume now that IE = IF. Since β > γ , we have that A 1 belongs to the segment FC and hence ∠C = ∠DIA 1 + ∠EIB 1 = ∠DIF + 2∠FIA 1 . This is equivalent to 2∠FIA 1 = γ − β −γ 2 ; thus β <3 γ .

23. Let J be the center of circle k tangent to the lines AB, DA, and BC respectively. Denote by a and b the incircles of △ADP and △BCP.

762 4 Solutions First we will prove that F ∈ IJ. Ob-

serve that A is the center of homoth-

C ethy that maps a to k; K is the cen-

ter of negative homothethy that maps ω

a to ω ; and denote by ˆ F the center

of negative homothethy that maps ω

F I L to k. By the consequence of Desar-

E gues’s theorem we have that A, K, and

a F b are collinear. Similarly. we prove B

that ˆ F ∈ BL, and therefore F = ˆ F , and

F ∈ IJ. Now we will prove that E ∈ IJ. Denote by X and Y the centers of inver- sions that map a and b to ω . Comparing the lengths of the tangents from A, B, C,

D to the circles k and a, we get that AP + DC = AD + PC. Hence there exists a circle d inscribed in APCD. Let X be the center of homothethy that maps a to ω . Using the same consequence of Desargues’s theorem, we see that A, C, and X are collinear. Consider the circles a, ω , and k again. We use again that A is the center of homothethy that maps a to k. Notice that X is the center of homothethy that

maps a to ω . Therefore X A contains the center ˆ E of homothethy with positive coefficient that maps ω to k. Similarly ˆ E ∈ BX; hence ˆE = E and E ∈ IJ.

24. k 4 +n 2 must be even because 7 k −3 n is even. If both k and n are odd, then

≡ 2 (mod 4) while 7 n −3 ≡ 7− 3 ≡ 0 (mod 4). Assume that k = 2k ′ and

k 4 +n 2 k

2 +3 n ) and 2(7 k +3 n ) must be a divisor of 2 (8k ′4 + 2n ′2 ); hence 7 k ′

n k′ = 2n ′ . Then 7 k −3 n = 7 −3 n′ · 2(7 k ′

+3 ′ n ≤ 8k ′2 + 2n ′2 . It is easy to prove by induction

that 7 ′ k ′ > 8k ′4 for k ′ ≥ 4 and 3 n > 2n ′2 for n ′ ≥ 1. Therefore k ′ ∈ {1,2,3}. For k ′ = 1 we must have 7 + 3 n ′ | 8 + 2n ′2 . An easy induction gives 7 +3 n ′ >

8 + 2n ′2 for n ′ ≥ 3. Hence n ′ ≤ 2. For n ′ = 1 we get (k, n) = (2, 2), which doesn’t satisfy the given conditions; n ′ = 2 implies (k, n) = (2, 4), which is a solution.

Assume now that k ′ = 2. Then |7 k −3 n | = |7 2 −3 n ′

|·(7 ′ 2 +3 n ) ≥ 22·(49+3 n )>

4 4 + 4n ′2 . This contradiction proves that k ′ 6= 2. If we assume that k ′ = 3, then |7 k

−3 ′ | = |7 3 −3 n ′ | · (7 3 +3 ′ n ) = |343 − 3 n |· (7 3 +3 n ′ ) ≥ 100 · (7 3 +3 n ′ )>6 4 + 4n ′2 . This is again a contradiction.

Thus (k, n) = (2, 4) is the only solution.

25. Assume that b =p α 1 α 1 l ··· p l , where p 1 ,...,p l are prime numbers. Since b −a n b 2 is divisible by b +1 2 we get that p α i α i

∤a b 2 for each i. This implies that n | α i for each i; hence b is a complete nth power.

i |a n

b 2 but p i

26. A set Z of integers will be called good if 47 ∤ a −b+c−d +e for any a,b,c,d,e ∈ Z . Notice that the set G = {−9,−7,−5,−3,−1,1,3,5,7,9} is good. For each integer k ∈ {1,2,...,46} the set G k = {x ∈ X|∃g ∈ G : kx ≡ g (mod 47)} is good

as well. Indeed, if a i ∈G k (1 ≤ i ≤ 5) are such that 47 | a 1 −a 2 +a 3 −a 4 +a 5 , then 47 | ka 1 − ka 2 + ka 3 − ka 4 + ka 5 . There are elements b i ∈ G for which ka i ≡

b i (mod 47) which is impossible. Each element of x is contained in exactly 10

4.48 Shortlisted Problems 2007 763 of the sets G k ; hence 10 |X| = ∑ 46 i =1 |A k |, and therefore |A k | > 2173 > 2007 for at

least one k.

27. The difference of the two binomial coefficients can be written as

2 2 k · (2 k − 1)!!

· P(2 k ),

(2 k )!

for P (x) = (x + 1)(x + 3) ···(x + 2 k − 1) − (x − 1) · (x − 3)···(x − 2 k + 1). The exponent of 2 in (2 k )! is equal to 2 k − 1; hence the exponent of 2 in D is bigger by 1 than the exponent of 2 in P (2 k ). Since P(−x) = −P(x), we get P(x) =

∑ 2 k i −1 =1 c i x 2i −1 . Being the coefficient near x, c 1 satisfies

2 k −1

2 k 1 −1

c 1 = 2 · (2 k − 1)!! ·

i =1 (2i − 1)(2 − 2i + 1) Let a i

be the solution of a i · (2i − 1) ≡ 1 (mod 2 k ). Then

2 k −1

k (2 −1 k − 1)!! 2

≡− ∑ − 1)!! · a i

(2 k

i =1 (2i − 1)(2 − 2i + 1)

i =1

2 k −1

= −(2 2 − 1)!! ∑ (2i − 1)

i =1

2 k −1 (2 k k + 1)(2 k − 1) = −(2 − 1)!!

≡2 k −1 (mod 2 k ).

The exponent of 2 in c 1 has to be 2k − 1. Now P(2 k )=c 1 ·2 k +2 3k Q (2 k ) for some polynomial Q. Clearly, 2 3k −1 | P(2 k ) but 2 3k ∤ P(2 k ). Finally, the exponent of 2 in D is equal to 3k.

28. Fix a prime number p. Let d ∈ N be the smallest number for which p | f (d) (it exists because f is surjective). By induction we have p | f (kd) for every k ∈ N. If p | f (x) but d ∤ x, from the minimality of d we conclude that x = kd + r, where r ∈ {1,2,...,d − 1}. Now we have p | f (kd) + f (r) hence p | f (r), which is impossible. Therefore d | x ⇔ p | f (x). If x ≡ y (mod d), let D > x be some number such that d | D. Then d | (y − x + D) hence p | f (y − x + D), and p | f (y) + f (D − x). Since p | f (x + D − x) we get p | f (x) + f (D − x). We have obtained x ≡ y (mod d) ⇒ f (x) ≡ f (y) (mod p).

764 4 Solutions Now assume that f (x) ≡ f (y) (mod p). Taking the same D as above and assum-

ing that y > x, we get 0 ≡ f (x) + f (D − x) ≡ f (y) + f (D − x) ≡ f (y + D − x) ≡

f (y − x) + f (D) ≡ f (y − x) (mod p). This implies that d | y − x, and thus x ≡ y (mod d) ⇔ f (x) ≡ f (y) (mod p). Now we know that f (1), . . . , f (d) have different residues modulo p hence p ≥

d . Since f is surjective there are numbers x 1 ,...,x p such that f (x 1 ) = 1, . . . ,

f (x p ) = p. They all give different residues modulo p hence x 1 ,...,x p must give p distinct residues modulo d, implying p = d. Now we have p |x ⇔ p| f (x) and x ≡ y (mod p) ⇔ f (x) ≡ f (y) (mod p) for every x , y ∈ N and every prime number p. Since no prime divides 1, we must have

f (1) = 1. We will prove by induction that f (n) = n. Assume that f (k) = k for every k < n. If f (n) > n, then f (n) − n + 1 ≥ 2 will have a prime factor p. This is impossible because f (n) ≡ n − 1 = f (n − 1) (mod p), and hence n ≡ n − 1 (mod p). If f (n) < n, let p be a prime factor of n − f (n) + 1 ≥ 2. Now we have n ≡ f (n) − 1 (mod p) and f (n) ≡ f ( f (n) − 1) = f (n) − 1 (mod p), contradiction. Thus f (n) = n is the only possible solution. It is easy to verify that this f satisfies the given conditions.

29. The statement will follow from the following lemma applied to x = k and y = 2n. Lemma. Given two positive integers x and y, the number 4xy − 1 divides the number (4x 2 − 1) 2 if and only if x = y. Proof. If x = y it is obvious that 4xy − 1 | (4x 2 − 1) 2 . Assume that there is

a pair (x, y) of two distinct positive integers such that 4xy − 1 | (4x 2 − 1). Choose such a pair for which 2x + y is minimal. From (4y 2 − 1) 2 ≡ (4y 2 − (4xy) 2 ) 2 ≡ 16y 2 (4x 2 − 1) 2 ≡ 0 (mod 4xy − 1) we get that (y,x) is such pair as well; hence 2y + x > 2x + y and y > x. Assume that (4x 2 −1) 2 = k ·(4xy− 1). Multiplying 4xy− 1 ≡ −1 (mod 4x) by k, we get (4x 2 − 1) 2 ≡ −k (mod 4x); hence k = 4xl − 1 for some positive integer l. However, this means that 4xl − 1 | (4x 2 − 1) 2 , and since y > x, we must have l < x, implying 2l + y < 2x + y, and this contradicts the mini- mality of (x, y).

Remark: Using the same method one can prove the following more general the- orem: If k > 1 is an integer, then kab − 1 | (ka 2 − 1) 2 ⇔ a = b.

30. Denote by f i (n) the remainder when ν p i (n) is divided by d. Let f (n) = ( f 1 (n), ...,f k (n)). Consider the sequence of integers n j defined inductively as n 1 =1

and n j +1 = (p 1 ··· p k ) n j . Let us first prove that ν p (r + l p m )= ν p (r) + ν p (l p m ) for r <p m . This follows from (r + l p m )! = (l p m )! · (l p m + 1) ···(l p m + r) and for each i <p m , the exponent of p in l p m + i is equal to the exponent of p in i.

If j 1 <j 2 < ··· < j u , the exponent of p i in each of n j 2 ,...,n j u is bigger than n j 1 ; hence f i (n j 1 +n j 2 + ···+n j u )≡f i (n j 1 )+ f i (n j 2 +···+n j u ) (mod d). Continuing by induction, we get f i (n j 1 + ··· + n j u )≡f i (n j 1 ) + ··· + f i (n j u ) (mod d).

Since the range of f has at most (d + 1) k elements, we see that there is an infinite subsequence of n i on which f is constant. Then for any d elements n l 1 , ...,n l d of this subsequence we have f (n l 1 + ··· + n l d ) ≡ f (n l 1 ) + ··· + f (n l d ) ≡ d f (n l 1 )≡ (0, 0, . . . , 0) (mod d).

4.49 Shortlisted Problems 2008 765