The algebraic equation x 3 − 3x 2 + 1 = 0 admits three real roots β , γ , a, with √

7. The algebraic equation x 3 − 3x 2 + 1 = 0 admits three real roots β , γ , a, with √

8 < a < 3. Define, for all integers n,

u n = β n + γ n +a n .

It holds that u n +3 = 3u n +2 −u n . Obviously, 0 < β n + γ n < 1 for all n ≥ 2, and we see that u n − 1 = [a n ] for n ≥ 2. It is now a question whether u 1788 − 1 and u 1988 − 1 are divisible by 17.

Working modulo 17, we get u 0 ≡ 3, u 1 ≡ 3, u 2 ≡ 9, u 3 ≡ 7, u 4 ≡ 1,...,u 16 = 3, u 17 = 3, u 18 = 9. Thus, u n is periodic modulo 17, with period 16. Since 1788 =

512 4 Solutions

16 · 111 + 12, 1988 = 16 · 124 + 4, it follows that u 1788 ≡u 12 ≡ 1 and u 1988 ≡ u 4 = 1. So, [a 1788 ] and [a 1988 ] are divisible by 17.

Second solution. The polynomial x 3 − 3x 2 + 1 allows the factorization modulo

17 as (x − 4)(x − 5)(x + 6). Hence it is easily seen that u n ≡4 n +5 n + (−6) n . Fermat’s theorem gives us 4 n ≡5 n ≡ (−6) n ≡ 1 for 16 | n, and the rest follows easily.

Remark. In fact, the roots of x 3 − 3x 2 + 1 = 0 are 1 2 sin 10 ◦ , 1 2 sin 50 1 ◦ , and − 2 sin 70 ◦ .

8. Consider first the case that the vectors are on the same line. Then if e is a unit vector, we can write u 1 =x 1 e ,...,u n =x n e for scalars x i , |x i | ≤ 1, with zero sum. It is now easy to permute x 1 ,x 2 ,...,x n into z 1 ,z 2 ,...z n so that |z 1 | ≤ 1,|z 1 +z 2 |≤

1 , . . . , |z 1 +z 2 + ··· + z n −1 | ≤ 1. Indeed, suppose w.l.o.g. that z 1 =x 1 ≥ 0; then we choose z 2 ,...,z r from the x i ’s to be negative, until we get to the first r with x 1 +x 2 + ··· + x r ≤ 0; we continue successively choosing positive z j ’s from the remaining x i ’s until we get the first partial sum that is positive, and so on. It is easy to verify that |z 1 +z 2 + ··· + z j | ≤ 1 for all j = 1,2,...,n. Now we pass to the general case. Let s be the longest vector that can be obtained by summing a subset of u 1 ,...,u m , and assume w.l.o.g. that s =u 1 + ··· + u p . Further, let δ and δ ′ respectively be the lines through the origin O in the direction of s and perpendicular to s, and e, e ′ respectively the unit vectors on δ and

δ ′ . Put u i =x i e +y i e ′ ,i = 1, 2, . . . , m. By the definition of δ and δ ′ , we have |x i |,|y i | ≤ 1; x 1 + ···+x m =y 1 + ···+y m = 0; y 1 +···+y p =y p +1 + ···+y m = 0; we also have x p +1 ,...,x m ≤ 0 (otherwise, if x i > 0 for some i, then |s + v i |> |s|), and similarly x 1 ,...,x p ≥ 0. Finally, suppose by the one-dimensional case that y 1 ,...,y p and y p +1 ,...,y m are permuted in such a way that all the sums y 1 + ··· + y i and y p +1 + ··· + y p +i are ≤ 1 in absolute value. We apply the construction of the one-dimensional case to x 1 ,...,x m taking, as described above, positive z i ’s from x 1 ,x 2 ,...,x p and negative ones from x p +1 ,...,x m , but so that the order is preserved; this way we get a permutation x σ 1 ,x σ 2 ,...,x σ m . It is then clear that each sum y σ 1 +y σ 2 + ··· + y σ k decomposes into the sum (y 1 +y 2 + ··· + y l ) + (y p +1 + ··· + y p +n ) (because of the preserva- tion of order), and that each of these sums is less than or equal to 1 in absolute value. Thus each sum u σ 1 + ··· + u σ k is composed of a vector of length at most √ 2 and an orthogonal vector of length at most 1, and so is itself of length at most

ab +1 = k ∈ N. We then have a − kab + b = k. Let us assume that k is not an integer square, which implies k ≥ 2. Now we observe the minimal

a 9. Let us assume 2 +b 2 2 2

pair (a, b) such that a 2 − kab + b 2 = k holds. We may assume w.l.o.g. that a ≥ b. For a

= b we get k = (2 − k)a 2 ≤ 0; hence we must have a > b. Let us observe the quadratic equation x 2 − kbx + b 2 − k = 0, which has solutions

a and a 1 . Since a +a 1 = kb, it follows that a 1 ∈ Z. Since a > kb implies k > a +

b 2 > kb and a = kb implies k = b 2 , it follows that a < kb and thus b 2 > k. Since

2 aa 2

1 =b 2 − k > 0 and a > 0, it follows that a 1 ∈ N and a 1 = b −k < a a −1 a < a . We have thus found an integer pair (a 1 , b) with 0 < a 1 < a that satisfies the original

4.29 Shortlisted Problems 1988 513 equation. This is a contradiction of the initial assumption that (a, b) is minimal.

Hence k must be an integer square.

10. We claim that if the family {A 1 ,...,A t } separates the n-set N, then 2 t ≥ n. The proof goes by induction. The case t = 1 is clear, so suppose that the claim holds for t − 1. Since A t does not separate elements of its own or its complement, it

follows that {A 1 ,...,A t −1 } is separating for both A t and N rA t , so that |A t |, |N r A t |≤2 t −1 . Then |N| ≤ 2 · 2 t −1 =2 t , as claimed. Also, if the set N with N =2 t is separated by {A 1 ,...,A t }, then (precisely) one element of N is not covered. To show this, we again use induction. This is trivial for t = 1, so let t ≥ 1. Since A 1 ,...,A t −1 separate both A t and N rA t ,N rA t must have exactly 2 t −1 elements, and thus one of its elements is not covered by

A 1 ,...,A t −1 , and neither is covered by A t . We conclude that a separating and covering family of t subsets can exist only if n ≤2 t − 1. We now construct such subsets for the set N if 2 t −1 ≤n≤2 t − 1, t ≥ 1. For t = 1,

put A 1 = {1}. In the step from t to t +1, let N = N ′ ∪N ′′ ∪{y}, where |N ′ |,|N ′′ |≤

be subsets covering and separating N and A 1 ,...,A t such subsets for N ′′ . Then the subsets A i =A ′ i ∪A ′′ i (i = 1, . . . ,t) and A t +1 =N ′′ ∪ {y} obviously separate and cover N.

2 t −1 ; let A ′ 1 ,...,A ′

′ ′′ ′′ t

The answer: t = [log 2 n ] + 1. Second solution. Suppose that the sets A 1 ,...,A t cover and separate N. Label

each element x ∈ N with a string (x 1 x 2 ...x t ) of 0’s and 1’s, where x i is 1 when x ∈A i , 0 otherwise. Since the A i ’s separate, these strings are distinct; since they cover, the string (00 . . . 0) does not occur. Hence n ≤ 2 t − 1. Conversely, for

2 t −1 ≤n<2 t , represent the elements of N in base 2 as strings of 0’s and 1’s of length t. For 1 ≤ i ≤ t, take A i to be the set of numbers in N whose binary string has a 1 in the ith place. These sets clearly cover and separate.

11. The answer is 32. Write the combinations as triples k = (x, y, z), 0 ≤ x,y,z ≤

7. Define the sets K 1 = {(1,0,0), (0,1,0), (0,0,1), (1,1,1)}, K 2 = {(2,0,0), (0, 2, 0), (0, 0, 2), (2, 2, 2)}, K 3 = {(0,0,0),(4,4,4)}, and K = {k = k 1 +k 2 +k 3 | k i ∈K i , i = 1, 2, 3}. There are 32 combinations in K. We shall prove that these combinations will open the safe in every case.

Let t = (a, b, c) be the right combination. Set k 3 = (0, 0, 0) if at least two of a, b, c are less than 4, and k 3 = (4, 4, 4) otherwise. In either case, the difference t − k 3 contains two nonnegative elements not greater than 3. Choosing a suitable k 2 we can achieve that t −k 3 −k 2 contains two elements that are 0, 1. So, there exists k 1 such that t −k 3 −k 2 −k 1 = t − k contains two zeros, for k ∈ K. This proves that 32 is sufficient. Suppose that K is a set of at most 31 combinations. We say that k ∈ K covers the

combination k 1 if k and k 1 differ in at most one position. One of the eight sets M i = {(i,y,z) | 0 ≤ y,z ≤ 7}, i = 0,1,...,7, contains at most three elements of K. Suppose w.l.o.g. that this is M 0 . Further, among the eight sets N j = {(0, j,z) | 0 ≤ z ≤ 7}, j = 0,...,7, there are at least five, say w.l.o.g. N 0 ,...,N 4 , not containing any of the combinations from K.

514 4 Solutions Of the 40 elements of the set N = {(0,y,z) | 0 ≤ y ≤ 4, 0 ≤ z ≤ 7}, at most 5 ·3 =

15 are covered by K ∩M 0 , and at least 25 aren’t. Consequently, the intersection of K with L = {(x,y,z) | 1 ≤ x ≤ 7,0 ≤ y ≤ 4,0 ≤ z ≤ 7} contains at least 25 elements. So K has at most 31 − 25 = 6 elements in the set P = {(x,y,z) | 0 ≤ x ≤ 7,5 ≤ y ≤ 7,0 ≤ z ≤ 7}. This implies that for some j ∈ {5,6,7}, say w.l.o.g.

j = 7, K contains at most two elements in Q j = {(x,y,z) | 0 ≤ x,z ≤ 7, y = j}; denote them by l 1 ,l 2 . Of the 64 elements of Q 7 , at most 30 are covered by l 1 and l 2 . But then there remain 34 uncovered elements, which must be covered by different elements of K rQ 7 , having itself at most 29 elements. Contradiction.

12. Let E (XY Z) stand for the area of a triangle XY Z. We have

MR AM BK =

E 1 E (AMR) E (AMK) E (ABK)

E E (AMK) E (ABK) E (ABC)

AB BC We similarly obtain

E 1 2 /3 1 KR

BM CK

AB BC Therefore

3 √ 2 /E) √ ≤ 1, i.e., √

E E E . Analogously,

E 3 + E 4 E and ≤ 3 E + 3 E 5 3 6 ≤ E ; hence

13. Let AB = c, AC = b, ∠CBA = β , BC = a, and AD = h. Let r 1 and r 2 be the inradii of ABD and ADC respectively and O 1 and O 2 the centers of the respective incircles.

It is obvious that r √ 1 /r 2 = c/b. We √ also have DO 1 = 2r 1 , DO 2 = 2r 2 ,

and ∠O P

1 DA = ∠O 2 DA = 45 ◦ . Hence

∠O 1 DO 2 = 90 ◦

and DO 1 /DO 2 = c/b

from which it follows that

O 2 △BAC. We now define P as the 1

△O O

1 DO 2 ∼

B D intersection of the circumcircle of C △O 1 DO 2 with DA. From the above

similarity we have ∠DPO 2 = ∠DO 1 O 2 = β = ∠DAC. It follows that PO 2 k AC and from ∠O 1 PO 2 = 90 ◦ it also follows that PO 1 k AB. We also have ∠PO 1 O 2 = ∠PO 2 O 1 = 45 ◦ ; hence ∠LKA = ∠KLA = 45 ◦ , and thus AK = AL. From ∠O 1 KA = ∠O 1 DA = 45 ◦ ,O 1 A =O 1 A , and ∠O 1 KA = ∠O 1 DA we have

△O 1 KA ∼ = △O 1 DA and hence AL = AK = AD = h. Thus

4.29 Shortlisted Problems 1988 515

E ah /2

a a 2 b 2 +c 2

h ah bc ≥2. Remark. It holds that for an arbitrary triangle ABC, AK = AL if and only if AB =

h /2

AC or ∡BAC = 90 ◦ .

14. Consider an array [a ij ] of the given property and denote the sums of the rows and the columns by r i and c j respectively. Among the r i ’s and c j ’s, one element of [−n,n] is missing, so that there are at least n nonnegative and n nonpositive

sums. By permuting rows and columns we can obtain an array in which r 1 ,...,r k and c 1 ,...,c n −k are nonnegative. Clearly

∑ |r i

|+ 2 ∑ |c j |≥ ∑ |r| − n = n .

But on the other hand,

nn

n −k

∑ |r i |+ |c j |= =1 ∑ =1 ∑ r i − ∑ r i + ∑ c j −

ij

i =1

i =k+1

j =1

j =n−k+1

= ∑ a ij − ∑ a ij + ∑ a ij − ∑ a ij =

i ≤k

i >k

j ≤n−k

j >n−k

k n −k

=2 ∑ ∑ a ij −2 ∑ ∑ a ij ≤ 4k(n − k).

i =1 j =1

i =k+1 j =n−k+1

This yields n 2 ≤ 4k(n − k), i.e., (n − 2k) 2 ≤ 0, and thus n must be even. We proceed to show by induction that for all even n an array of the given type exists. For n = 2 the array in Fig. 1 is good. Let such an n × n array be given for

some even n ≥ 2, with c 1 = n, c 2 = −n + 1, c 3 = n − 2,...,c n −1 = 2, c n = −1 and r 1 = n − 1, r 2 = −n + 2,...,r n −1 = 1, r n = 0. Upon enlarging this array as indicated in Fig. 2 , the positive sums are increased by 2, the nonpositive sums are decreased by 2, and the missing sums −1,0,1,2 occur in the new rows and columns, so that the obtained array (n + 2) × (n + 2) is of the same type.

15. Referring to the description of L A , we have ∠AMN = ∠AHN = 90 ◦ − ∠HAC = ∠C, and similarly ∠ANM = ∠B. Since the triangle ABC is acute-angled, the line L A lies inside the angle A. Hence if P =L A ∩ BC and Q = L B ∩ AC, we get ∠BAP = 90 ◦ − ∠C; hence AP passes through the circumcenter O of ∆ ABC .

516 4 Solutions Similarly we prove that L B and L C contains the circumcenter O also. It follows

that L A ,L B and L C intersect at the point O. Remark. Without identifying the point of intersection, one can prove the con-

currence of the three lines using Ceva’s theorem, in usual or trigonometric form.

16. Let f (x) = ∑ 70 k k =1 x −k . For all integers i = 1, . . . , 70 we have that f (x) tends to plus infinity as x tends downward to i, and f (x) tends to minus infinity as x tends upward to i. As x tends to infinity, f (x) tends to 0. Hence it follows that there

exist x 1 ,x 2 ,...,x 70 such that 1 <x 1 <2<x 2 < 3 < ··· < x 69 < 70 < x 70 and

i )= for all i = 1, . . . , 70. Then the solution to the inequality is given by S S 4 = 70 i =1 (i, x i ]. For numbers x for which f (x) is well-defined, the equality f (x) = 5 4 is equivalent to

f (x

p (x) = ∏ (x − j) − ∑ k

5 =1 ∏ j (x − j) = 0. k j =1

j 6=k

The numbers x 1 ,x 2 ,...,x 70 are then the zeros of this polynomial. The sum ∑ 70 i =1 x i is then equal to minus the coefficient of x 69 in p, which is ∑ 70 i =1 i + 4 5 i . Finally,

|S| = ∑ (x i − i) = · ∑ i = ·

17. Let AC and AD meet BE in R , S, respectively. Then by the conditions of the problem,

∠AEB = ∠EBD = ∠BDC = ∠DBC = ∠ADB = ∠EAD = α , ∠ABE = ∠BEC = ∠ECD = ∠CED = ∠ACE = ∠BAC = β ,

∠BCA = ∠CAD = ∠ADE = γ . Since ∠SAE = ∠SEA, it follows that AS = SE, and analogously BR = RA. But

BSDC and REDC are parallelograms; hence BS = CD = RE, giving us BR = SE and AR = AS. Then also AC = AD, because RS k CD. We deduce that 2 β = ∠ACD = ∠ADC = 2 α , i.e., α = β . It will be sufficient to show that α = γ , since that will imply α = β = γ = 36 ◦ . We have that the sum of the interior angles of ACD is 4 α + γ = 180 ◦ . We have

sin γ AE AE AE sin (2 α + γ )

, sin α

sin ( α + γ ) i.e., cos α − cos( α +2 γ ) = 2 sin γ sin ( α + γ ) = 2 sin α sin (2 α + γ ) = cos( α +

DE CD RE

γ ) − cos(3 α + γ ). From 4 α + γ = 180 ◦ we obtain −cos(3 α + γ ) = cos α . Hence

2 α +3 γ cos ( α + γ ) + cos( α +2 γ ) = 2 cos cos

2 2 so that 2 α +3 γ = 180 ◦ . It follows that α = γ .

4.29 Shortlisted Problems 1988 517 Second solution. We have ∠BEC = ∠ECD = ∠DEC = ∠ECA = ∠CAB, and

hence the trapezoid BAEC is cyclic; consequently, AE = BC. Similarly AB = ED , and ABCD is cyclic as well. Thus ABCDE is cyclic and has all sides equal; i.e., it is regular.

18. (a) Define ∠APO =

2 2 p 2 φ and S = AB + AC + BC . We calculate PA = 2r cos φ and PB , PC = R 2 −r 2 cos 2 φ ± r sin φ . We also have AB 2 = PA 2 + PB 2 ,

AC 2 = PA 2 + PC 2 and BC = BP + PC. Combining all these we obtain S = AB 2 + AC 2 + BC 2 = 2(PA 2 + PB 2 + PC 2 + PB · PC)

= 2(4r 2 cos 2 φ + 2(R 2 2 cos 2 φ +r 2 sin −r 2 φ )+R 2 −r 2 )

= 6R 2 + 2r 2 .

Hence it follows that S is constant; i.e., it does not depend on φ . (b) Let B 1 and C 1 respectively be points such that APBB 1 and APCC 1 are rectangles. It is evident that B 1 and C 1 lie on the larger circle and that

1 and PV = 1 2 −−→ 2 PC 1 . It is evident that we can arrange for an arbi- trary point on the larger circle to be B 1 or C 1 . Hence, the locus of U and V is equal to the circle obtained when the larger circle is shrunk by a factor of 1 /2 with respect to point P.

PU −→ = 1 −−→ PB −→

19. We will show that f (n) = n for every n (thus also f (1988) = 1988). Let f (1) = r and f (2) = s. We obtain respectively the following equalities:

f (2r) = f (r + r) = 2; f (2s) = f (s + s) = 4; f (4) = f (2 + 2) = 4r; f (8) =

f (4 + 4) = 4s; f (5r) = f (4r + r) = 5; f (r + s) = 3; f (8) = f (5 + 3) = 6r + s. Then 4s = 6r + s, which means that s = 2r. Now we prove by induction that f (nr) = n and f (n) = nr for every n ≥ 4. First we have that f (5) = f (2 + 3) = 3r + s = 5r, so that the statement is true for n = 4 and n = 5. Suppose that it holds for n − 1 and n. Then f (n + 1) = f (n − 1 + 2) = (n−1)r+2r = (n+1)r, and f ((n+1)r) = f ((n−1)r+2r) = (n−1)+2 = n+1. This completes the induction. Since 4r

≥ 4, we have that f (4r) = 4r 2 , and also f (4r) = 4. Then r = 1, and consequently f (n) = n for every natural number n. Second solution. f ( f (1) + n + m) = f ( f (1) + f ( f (n) + f (m))) = 1 + f (n) +

f (m), so f (n) + f (m) is a function of n + m. Hence f (n + 1) + f (1) = f (n) +

f (2) and f (n + 1) − f (n) = f (2) − f (1), implying that f (n) = An + B for some constants A , B. It is easy to check that A = 1, B = 0 is the only possibility.

20. Suppose that A n = {1,2,...,n} is partitioned into B n and C n , and that neither B n nor C n contains 3 distinct numbers one of which is equal to the product of the other two. If n ≥ 96, then the divisors of 96 must be split up. Let w.l.o.g. 2 ∈ B n . There are four cases.

(i) 3 ∈B n ,4∈B n . Then 6 , 8, 12 ∈ C n ⇒ 48,96 ∈ B n . A contradiction for 96 =

2 · 48. (ii) 3 ∈B n ,4∈C n . Then 6 ∈C n , 24 ∈B n ,8 , 12, 48 ∈ C n . A contradiction for

518 4 Solutions (iii) 3 ∈C n ,4∈B n . Then 8 ∈C n , 24 ∈B n ,6 , 48 ∈ C n . A contradiction for 48 =

6 · 8. (iv) 3 ∈C n ,4∈C n . Then 12 ∈B n ,6 , 24 ∈ C n . A contradiction for 24 = 4 · 6. If n = 95, there is a very large number of ways of partitioning A n . For example,

B n = {1, p, p 2 ,p 3 q 2 ,p 4 q ,p 2 qr | p, q, r are distinct primes}, C n = {p 3 ,p 4 ,p 5 , p 6 , pq, p 2 q ,p 3 q ,p 2 q 2 , pqr | p, q, r are distinct primes}. Then B 95 = {1, 2, 3, 4,

21. Let X be the set of all ordered triples a = (a 1 ,a 2 ,a 3 ) for a i ∈ {0,1,...,7}. Write

a ≺ b if a i ≤b i for i = 1, 2, 3 and a 6= b. Call a subset Y ⊂ X independent if there are no a , b ∈ Y with a ≺ b. We shall prove that an independent set contains at most 48 elements.

For j = 0, 1, . . . , 21 let X j = {(a 1 ,a 2 ,a 3 )∈X|a 1 +a 2 +a 3 = j}. If x ≺ y and x ∈X j ,y ∈X j +1 for some j, then we say that y is a successor of x, and x a predecessor of y.

Lemma. If A is an m-element subset of X j and j ≤ 10, then there are at least m distinct successors of the elements of A. Proof. For k = 0, 1, 2, 3 let X j ,k = {(a 1 ,a 2 ,a 3 )∈X j | min(a 1 ,a 2 ,a 3 ,7−a 1 ,7−

a 2 ,7−a 3 ) = k}. It is easy to verify that every element of X j ,k has at least two successors in X j +1,k and every element of X j +1,k has at most two prede- cessors in X j ,k . Therefore the number of elements of A ∩X j ,k is not greater than the number of their successors. Since X j is a disjoint union of X j ,k , k = 0, 1, 2, 3, the lemma follows.

Similarly, elements of an m-element subset of X j ,j ≥ 11, have at least m prede- cessors. Let Y be an independent set, and let p , q be integers such that p < 10 < q. We can transform Y by replacing all the elements of Y ∩X p with their successors, and all the elements of Y ∩X q with their predecessors. After this transformation

Y will still be independent, and by the lemma its size will not be reduced. Every independent set can be eventually transformed in this way into a subset of X 10 , and X 10 has exactly 48 elements.

22. Set X =∑ p

i =1 x i and w.l.o.g. assume that X ≥ 0 (if (x 1 ,...,x p ) is a solution, then (−x p 1 , . . . , −x p ) is a solution too). Since x 2 ≥ x for all integers x, it follows that ∑ i =1 x 2 i ≥ X. If the last inequality is an equality, then all x i ’s are 0 or 1; then, taking that there are a 1’s, the equation becomes 4p + 1 = 4(a + 1) + 4 a −1 , which forces p = 6 and

a = 5. Otherwise, we have X

i =1 x i = 4p +1 X + 1, so X ≥ p + 1. Also, by the

Cauchy–Schwarz inequality, X 2 p

i =1 x i = 4p +1 X + p, so X ≤ 4p + p and

≤ p∑ 4p 2 2 2 2

X ≤ 2p. Thus 1 ≤ X/p ≤ 2. However,

4.29 Shortlisted Problems 1988 519 p

X 2 2X X 2

i =1

= ∑ x 2 X i 2 −p

p 2 =1−

< 1, p (4p + 1)

and we deduce that −1<x i − X/p < 1 for all i. This finally gives x i ∈ {1,2}. Suppose there are b 2’s. Then 3b + p = 4(b + p) 2 /(4p + 1) + 1, so p = b +

1 /(4b − 3), which leads to p = 2, b = 1. Thus there are no solutions for any p 6∈ {2,6}.

Remark. The condition p = n(n + 1), n ≥ 3, was unnecessary in the official solution, too (its only role was to simplify showing that X 6= p − 1).

23. Denote by R the intersection point of lines AQ and BC. We know that BR : RC =

c : b and AQ : QR = (b + c) : a. By applying Stewart’s theorem to ∆ PBC and ∆ PAR we obtain

a 2 2 · AP 2 + b · BP + c ·CP = aPA 2 + (b + c)PR 2 + (b + c)RB · RC (1) = (a + b + c)QP 2 + (b + c)RB · RC+ (a + b + c)QA · QR.

On the other hand, putting P = Q into (1), we get that

a 2 2 · AQ 2 + b · BQ + c ·CQ = (b + c)RB · RC+ (a + b + c)QA · QR, and the required statement follows.

Second solution. At vertices A , B,C place weights equal to a, b, c in some units respectively, so that Q is the center of gravity of the system. The left side of the equality to be proved is in fact the moment of inertia of the system about the axis through P and perpendicular to the plane ABC. On the other side, the right side expresses the same, due to the parallel axes theorem.

Alternative approach. Analytical geometry. The fact that all the variable seg- ments appear squared usually implies that this is a good approach. Assign co-

ordinates A (x a ,y a ), B(x b ,y b ), C(x c ,y c ), and P(x, y), use that (a + b + c)Q = aA + bB + cC, and calculate. Alternatively, differentiate f (x, y) = a · AP 2 +b·

BP 2 2 + c ·CP 2 − (a + b + c)QP and show that it is constant.

24. The first condition means in fact that a k −a k +1 is decreasing. In particular, if

a k −a k +1 =− δ < 0, then a k −a k +m = (a k −a k +1 ) + ··· + (a k +m−1 −a k +m )< −m δ , which implies that a k +m >a k +m δ , and consequently a k +m > 1 for large enough m, a contradiction. Thus a k −a k +1 ≥ 0 for all k.

Suppose that a

k −a k +1 > 2/k . Then for all i < k, a i −a i +1 > 2/k , so that

a i −a 2

k +1

> 2(k + 1 − i)/k 2 , i.e., a i > 2(k + 1 − i)/k ,i = 1, 2, . . . , k. But this im- plies a 1 +a 2 2 2 2 + ··· + a 2 k > 2/k + 4/k + ··· + 2k/k = k(k + 1)/k , which is im-

possible. Therefore a

k −a k +1 ≤ 2/k for all k.

25. Observe that 1001 = 7 · 143, i.e., 10 3 = −1 + 7a, a = 143. Then by the binomial theorem, 10 21 = (−1 + 7a) 7 = −1 + 7 2 b for some integer b, so that we also have

520 4 Solutions

10 21n ≡ −1 (mod 49) for any odd integer n > 0. Hence N = 9 49 21n (10 + 1) is an integer of 21n digits, and N 2 (10 21n + 1) = 3

7 (10 21n + 1) is a double number that is a perfect square.

26. In the sequel, a 1 a 2 ...a α will be used to representation a number whose binary digits are a 1 , ...,a α . We will show by induction that if n =c k c k −1 ...c 0 = ∑ k i =0 c i 2 i is the binary representation of n (c i ∈ {0,1}), then f (n) = c 0 c 1 ...c k = ∑ k i =0 c i 2 k −i is the number whose binary representation is the palindrome of the binary representation of n. This evidently holds for n ∈ {1,2,3}. Let us as- sume that the claim holds for all numbers up to n − 1 and show it holds for

n =c k c k −1 ...c 0 . We observe three cases: (i) c 0 = 0 ⇒ n = 2m ⇒ f (n) = f (m) = 0c 1 ...c k =c 0 c 1 ...c k . (ii) c 0 = 1, c 1 = 0 ⇒ n = 4m+ 1 ⇒ f (n) = 2 f (2m+ 1)− f (m) = 2·1c 2 ...c k −

c 2 ...c k =2 k +2·c 2 ...c k −c 2 ...c k = 10c 2 ...c k =c 0 c 1 ...c k . (iii) c 0 = 1, c 1 = 1 ⇒ n = 4m+3 ⇒ f (n) = 3 f (2m+1)−2 f (m) = 3·1c 2 ...c k −

2 ·c 2 ...c k =2 k +2 k −1 +3·c 2 ...c k −2·c 2 ...c k = 11c 2 ...c k =c 0 c 1 ...c k . We thus have to find the number of palindromes in binary representation smaller than 1988 = 11111000100. We note that for all m ∈ N the numbers of 2m- and (2m − 1)-digit binary palindromes are both equal to 2 m −1 . We also note that 11111011111 and 11111111111 are the only 11-digit palindromes larger than 1988. Hence we count all palindromes of up to 11 digits and exclude the largest two. The number of n ≤ 1988 such that f (n) = n is thus equal to 1 + 1 + 2 + 2 +

27. Consider a Cartesian system with the x-axis on the line BC and origin at the foot of the perpendicular from A to BC, so that A lies on the y-axis. Let A be (0, α ),