Solutions to the Shortlisted Problems of IMO 1974

4.16 Solutions to the Shortlisted Problems of IMO 1974

1. Denote by n the number of exams. We have n (A + B+C) = 20 +10 +9 = 39, and since A , B,C are distinct, their sum is at least 6; therefore n = 3 and A + B +C =

13. Assume w.l.o.g. that A > B > C. Since Betty gained A points in arithmetic, but fewer than 13 points in total, she had C points in both remaining exams (in spelling as well). Furthermore, Carol also gained fewer than 13 points, but with at least B points on two examinations (on which Betty scored C), including spelling. If she had A in spelling, then she would have at least A + B + C = 13 points in total, a contradiction. Hence, Carol scored B and placed second in spelling.

Remark. Moreover, it follows that Alice, Betty, and Carol scored B + A + A,

A +C +C, and C + B + B respectively, and that A = 8, B = 4, C = 1.

i the square with side i . Let us divide the big square into rectangles r

2. We denote by q

i by parallel lines, where the size of r i is 3 2 × 2 i for i = 2, 3, . . . and 3

2 × 1 for i = 1 (this can be done because 1 + ∑ ∞ i =2 2 i = 2 ). In rectan- gle r 1 , one can put the squares q 1 ,q 2 ,q 3 , as is done on the figure. Also, since

2 + ··· + 2 i +1 −1 <2 i 1 · 2 i =1< 3 2 , in each r i ,i ≥ 2, one can put q 2 i ,...,q 2 i +1 −1 . This completes the proof.

q 8 ,...,q 15 q 4 q 5 q 6 q 7

Remark. It can be shown that the squares q 1 ,q 2 cannot fit in any square of side less than 3 2 .

3. For deg (P) ≤ 2 the statement is obvious, since n(P) ≤ deg(P 2 ) = 2 deg(P) ≤ deg (P) + 2. Suppose now that deg (P) ≥ 3 and n(P) > deg(P) + 2. Then there is at least one integer b for which P (b) = −1, and at least one x with P(x) = 1. We may assume w.l.o.g. that b = 0 (if necessary, we consider the polynomial P(x + b) instead). If

k 1 ,...,k m are all integers for which P (k i ) = 1, then P(x) = Q(x)(x − k 1 ) ···(x − k m ) + 1 for some polynomial Q(x) with integer coefficients. Setting x = 0 we obtain

1 ···k m = 1 − P(0) = 2. It follows that k 1 ···k m | 2, and hence m is at most 3. The same holds for the polynomial −P(x), and thus P(x) = −1 also has at most 3 integer solutions. This counts for 6 solutions of P 2 (x) = 1 in

(−1) m Q (0)k

total, implying the statement for deg (P) ≥ 4. It remains to verify the statement for n = 3. If deg(P) = 3 and n(P) = 6, then it

follows from the above consideration that P (x) is either −(x 2 − 1)(x − 2) + 1 or

(x 2 − 1)(x + 2) + 1. It is directly checked that n(P) equals only 4 in both cases.

408 4 Solutions

4. Assume w.l.o.g. that a 1 ≤a 2 ≤a 3 ≤a 4 ≤a 5 . If m is the least value of |a i −a j |,

i 6= j, then a i +1 −a i ≥ m for i = 1,2,...,5, and consequently a i −a j ≥ (i − j)m for any i , j ∈ {1,...,5}, i > j. Then it follows that

(a

) 2 2 ∑ 2 i −a j ≥m ∑ (i − j) = 50m 2 .

i >j

i >j

On the other hand, by the condition of the problem,

(a

) 2 =5 a 2 ) ∑ 2 i −a j ∑ i − (a 1 + ··· + a 5 ≤ 5.

i >j

i =1 Therefore 50m 2 ≤ 5; i.e., m 2 ≤ 1 10 .

5. All the angles are assumed to be oriented and measured modulo 180 ◦ . Denote by α i , β i , γ i the angles of triangle △ i , at A i ,B i ,C i respectively. Let us deter-

mine the angles of △ i +1 . If D i is the intersection of lines B i B i +1 and C i C i +1 , we have ∠B i +1 A i +1 C i +1 = ∠D i B i C i +1 = ∠B i D i C i +1 + ∠D i C i +1 B i = ∠B i D i C i −

∠B i C i +1 C i = −2∠B i A i C i . We conclude that

α i +1 = −2 α i , and analogously β i +1 = −2 β i , γ i +1 = −2 γ i . Therefore α r +t

= (−2) t α r . However, since (−2) 12 ≡ 1 (mod 45) and conse- quently

14 (−2) 2 ≡ (−2) (mod 180), it follows that α 15 = α 3 , since all values are modulo 180 ◦ . Analogously, β 15 = β 3 and γ 15 = γ 3 , and moreover, △ 3 and △ 15

are inscribed in the same circle; hence △ 3 ∼ = △ 15 .

6. We set

2k ∑ +1 =

2n +1

2 3k

8 k ∑ =0 2k +1

Both x and y are positive integers. Also, from the binomial formula we obtain

2n +1 2n +1

y +x 8 =

2n ∑ +1 8 = (1 + 8 ) ,

i =0

and similarly

y −x 8 2n = (1 − +1 8 ) .

√ √ Multiplying these equalities, we get y 2 − 8x 2 = (1 + 8 ) 2n +1 (1 − 8 ) 2n +1 =

−7 +1 2n . Reducing modulo 5 gives us

−y n ≡2 ≡ 2 · (−1) . Now we see that if x is divisible by 5, then y 2 ≡ ±2 (mod 5), which is impossible.

3x 2 2 2n +1

Therefore x is never divisible by 5.

4.16 Shortlisted Problems 1974 409 Second solution. Another standard way is considering recurrent formulas. If we

2k then since a b a = a −1 −1 b + b −1 , it follows that x m +1 =x m +y m and y m +1 =

8x m +y m ; therefore x m +1 = 2x m + 7x m −1 . We need to show that none of x 2n +1 are divisible by 5. Considering the sequence {x m } modulo 5, we get that x m =

0 , 1, 2, 1, 1, 4, 0, 3, 1, 3, 3, 2, 0, 4, 3, 4, 4, 1, . . . . Zeros occur in the initial position of blocks of length 6, where each subsequent block is obtained by multiplying the previous one by 3 (modulo 5). Consequently, x m is divisible by 5 if and only if m is a multiple of 6, which cannot happen if m = 2n + 1.

7. Consider an arbitrary prime number p. If p | m, then there exists b i that is divis- ible by the same power of p as m. Then p divides neither a i m b i nor a i , because (a i ,b i ) = 1. If otherwise p ∤ m, then m b i is not divisible by p for any i, hence p di-

vides a

i and a i b i to the same power. Therefore (a 1 ,...,a k ) and a 1 b 1 ,...,a k b k have the same factorization; hence they are equal.

mmm

Second solution. For k = 2 we can easily verify the formula

m ,m =

b b 2 1 ,a 2 1 ), 2 1 2 1 2 since [b 1 ,b 2 ] · (b 1 ,b 2 )=b 1 b 2 . We proceed by induction: m

a 1 ,...,a k ,a k +1

(a 1 ,...,a k ), a k +1

b k +1

b k +1 = (a 1 ,...,a k ,a k +1 ).

[b 1 ,...,b k ]

8. It is clear that

+ <S

a +b+c+d a +b+c+d a +b+c+d a +b+c+d

a b c d and S <

a +b a +b c +d c +d or equivalently, 1 < S < 2.

On the other hand, all values from (1, 2) are attained. Since S = 1 for (a, b, c, d) = (0, 0, 1, 1) and S = 2 for (a, b, c, d) = (0, 1, 0, 1), due to continuity all the val- ues from (1, 2) are obtained, for example, for (a, b, c, d) = (x(1 − x),x,1 − x,1), where x goes through (0, 1).

Second solution. Set

a 2 +b+d = b +c+d a +b+c a +c+d We may assume without loss of generality that a +b+c+d = 1. Putting a+c = x

and

and b + d = y (then x + y = 1), we obtain that the set of values of

410 4 Solutions

a c 2ac +x−x 2

1 −c 1 −a

ac +1−x is x , 2x 2 −x . Having the analogous result for S 2 in mind, we conclude that the

i values that S

=S 2y

1 +S 2 can take are x + y, 2x 2 −x + 2 −y . Since x + y = 1 and

with equality for xy = 0, the desired set of values for S is (1, 2).

9. There exist real numbers a , b, c with tan a = x, tanb = y, tan c = z. Then using the additive formula for tangents we obtain

x + y + z − xyz

tan (a + b + c) =

1 − xy − xz − yz We are given that xyz = x + y + z. In this case xy + yz + zx = 1 is impossible;

otherwise, x , y, z would be the zeros of a cubic polynomial t 3 − λ t 2 +t− λ = (t 2 + 1)(t − λ ) (where λ = xyz), which has only one real root. It follows that

x + y + z = xyz ⇐⇒ tan(a + b + c) = 0. (1) Hence a +b+c=k π for some k ∈ Z. We note that 3x −x 3 1 −3x 2 actually expresses

tan 3a. Since 3a + 3b + 3c = 3k π , the result follows from (1) for the numbers 3x −x 3 , 3y −y 3 , 3z −z 3

1 −3x 2 1 −3y 2 1 −3z 2 .

10. If we set ∠ACD = γ 1 and ∠BCD = γ 2 for a point D on the segment AB, then by the sine theorem,

CD 2 CD CD sin α sin β

f (D) =

AD AD ·

BD sin γ 1 sin γ 2 The denominator of the last fraction is

· BD

sin γ 1 sin γ 2 = (cos( γ 1 − γ 2 ) − cos( γ 1 + γ 2 ))

− cos

= (cos( γ

2 2 ) − cos γ )≤

= sin .

h Now we deduce that the set of values of f (D) is the interval sin α sin β sin 2 γ , +∞ .

Hence f (D) = 1 (equivalently, CD 2 = AD · BD) is possible if and only if sin α sin β

≤ sin 2γ

2 , i.e.,

sin α sin β ≤ sin .

Second solution. Let E be the second point of intersection of the line CD with the circumcircle k of ABC. Since AD ·BD = CD · ED (power of D with respect to k),

4.16 Shortlisted Problems 1974 411 CD 2 = AD · BD ie equivalent to ED = CD. Clearly the ratio ED CD (D ∈ AB) takes

a maximal value when E is the midpoint of the arc AB not containing C. (This follows from ED : CD =E ′ D :C ′ D when C ′ and E ′ are respectively projections from C and E onto AB.) On the other hand, it is directly shown that in this case

ED sin 2γ

CD sin α sin β

and the assertion follows.

11. First, we notice that a 1 +a 2 + ··· + a p = 32. The numbers a i are distinct, and consequently a i ≥ i and a 1 + ··· + a p ≥ p(p + 1)/2. Therefore p ≤ 7. The number 32 can be represented as a sum of 7 mutually distinct positive inte- gers in the following ways:

The case (1) is eliminated because there is no rectangle with 22 cells on an 8 ×8 chessboard. In the other cases the partitions are realized as below.

12. We say that a word is good if it doesn’t contain any nonallowed word. Let a n be the number of good words of length n. If we prolong any good word of length n by adding one letter to its end (there are 3a n words that can be so obtained), we get either

(i) a good word of length n + 1, or (ii) an (n + 1)-letter word of the form XY , where X is a good word and Y a nonallowed word. The number of words of type (ii) with word Y of length k is exactly a n +1−k ; hence the total number of words of kind (ii) doesn’t exceed a n −1 + ··· + a 1 +a 0 (where a 0 = 1). Hence

a 0 = 1, a 1 = 3. (1) We prove by induction that a n +1 > 2a n for all n. For n = 1 the claim is trivial. If

a n +1 ≥ 3a n − (a n −1 + ··· + a 1 +a 0 ),

it holds for i ≤ n, then a i ≤2 i −n a n ; thus we obtain from (1)

412 4 Solutions

2 2 2 n . Therefore a n ≥2 n for all n (moreover, one can show from (1) that a n ≥ (n +

a n +1 >a n 3 − − 2 − ··· − n > 2a

2 )2 n −1 ); hence there exist good words of length n. Remark. If there are two nonallowed words (instead of one) of each length

greater than 1, the statement of the problem need not remain true.

4.17 Shortlisted Problems 1975 413