Solutions to the Shortlisted Problems of IMO 1971

4.13 Solutions to the Shortlisted Problems of IMO 1971

1. Assuming that a , b, c in (1) exist, let us find what their values should be. Since P 2 (x) = x 2 −2, equation (1) for n = 1 becomes (x 2 −4) 2 = [a(x 2 −2)+bx+2c] 2 . Therefore, there are two possibilities for (a, b, c): (1, 0, −1) and (−1,0,1). In both cases we must prove that

(x 2 − 4)[P n (x) 2 − 4] = [P n +1 (x) − P n −1 (x)] 2 . (2) It suffices to prove (2) for all x in the interval [−2,2]. In this interval we can set

x = 2 cost for some real t. We prove by induction that

(3) This is trivial for n = 0, 1. Assume (3) holds for some n − 1 and n. Then

P n (x) = 2 cosnt for all n .

P n +1 (x) = 4 cost cos nt − 2cos(n − 1)t = 2cos(n + 1)t by the additive formula for the cosine. This completes the induction. Now (2) reduces to the obviously correct equality

= (2 cos(n + 1)t − 2cos(n − 1)t) 2 . Second solution. If x is fixed, the linear recurrence relation P n +1 (x) + P n −1 (x) =

16 sin 2 t sin 2 nt

xP n (x) can be solved in the standard way. The characteristic polynomial t 2 − xt +

1 has zeros t 1 ,2 with t 1 +t 2 = x and t 1 t 2 = 1; hence, the general P n (x) has the form at n 1 + bt n 2 for some constants a, b. From P 0 = 2 and P 1 = x we obtain that

P n (x) = t n 1 +t n 2 . Plugging in these values and using t 1 t 2 = 1 one easily verifies (2).

2. We will construct such a set S m of 2 m points. Take vectors u 1 ,...,u m in a given plane such that |u i | = 1/2 and 0 6= |c 1 u 1 +

c 2 u 2 + ··· + c n u n | 6= 1/2 for any choice of numbers c i equal to 0 or ±1 (where two or more of the numbers c i are nonzero). Such vectors are easily constructed by induction on m: For u 1 ,...,u m −1 fixed, there are only finitely many vector values u m that violate the upper condition, and we may set u m to be any other vector of length 1 /2. Let S m

be the set of all points M 0 + ε 1 u 1 + ε 2 u 2 + ··· + ε m u m , where M 0 is any fixed point in the plane and ε i = ±1 for i = 1,...,m. Then S m obviously satisfies the condition of the problem.

3. Let x , y, z be a solution of the given system with x 2 +y 2 +z 2 = α < 10. Then

2 2 Furthermore, 3xyz =x 3 +y 3 +z 3 − (x + y + z)(x 2 +y 2 +z 2 − xy − yz− zx), which

gives us xyz = 3(9 − α )/2 − 4. We now have

390 4 Solutions

35 =x 4 +y 4 +z 4 = (x 3 +y 3 +z 3 )(x + y + z) −(x 2 +y 2 +z 2 )(xy + yz + zx) + xyz(x + y + z)

The solutions in α are α = 7 and α = 11. Therefore α = 7, xyz = −1, xy + xz + yz = 1, and

x 5 +y 5 +z 5 = (x 4 +y 4 +z 4 )(x + y + z)

3 +y 3 +z 3 )(xy + xz + yz) + xyz(x 2 +y 2 +z −(x 2 ) = 35 · 3 − 15 · 1 + 7 · (−1) = 83.

4. In the coordinate system in which the x-axis passes through the centers of the circles and the y-axis is their common tangent, the circles have equations

x 2 +y 2 − 2r 2 x = 0. Let p be the desired line with equation y = ax + b. The abscissas of points of

x 2 +y 2 + 2r 1 x = 0,

intersection of p with both circles satisfy one of

(1 + a 2 )x 2 2 )x + b 1 2 + 2(ab − r = 0. Let us denote the lengths of the chords and their projections onto the x-axis by d

(1 + a 2 )x 2 + 2(ab + r )x + b 2 = 0,

and d 1 , respectively. From these equations it follows that

4 (ab + r 1 ) 2 4b 2 2 4 (ab − r 2 ) 2 4b 2

2 2 − . (1) (1 + a )

1 +a 2 Consider the point of intersection of p with the y-axis. This point has equal

+a

(1 + a )

powers with respect to both circles. Hence, if that point divides the segment determined on p by the two circles into two segments of lengths x and y, this power equals x (x + d) = y(y + d), which implies x = y = d/2. Thus each of the equations in (1) has two roots, one of which is thrice the other. This fact gives us

(ab + r 1 ) 2 = 4(1 + a 2 )b 2 /3. We can now use (1) to obtain r

ab 2 −r = 1 , 4b 2 +a 2 b 2 = 3[(ab + r 1 ) 2 2 −a 2 b ] = 3r 1 r 2 ;

(14r 1 r 2 2 2 2

36 (r 1 +r 2 ) 2 Finally, since d 2 =d 2 1 2 (1 + a ), we conclude that

(14r

12 1 r 2 −r 1 −r 2 ), √

√ and that the problem is solvable if and only if 7 −4 3 ≤ r 1 r 2 ≤7+4 3 .

4.13 Shortlisted Problems 1971 391

5. Without loss of generality, we may assume that a ≥ b ≥ c ≥ d ≥ e. Then a − b = −(b − a) ≥ 0, a − c ≥ b − c ≥ 0, a − d ≥ b − d ≥ 0 and a − e ≥ b − e ≥ 0, and hence

(a − b)(a − c)(a − d)(a − e) + (b − a)(b − c)(b − d)(b − e) ≥ 0. Analogously, (d − a)(d − b)(d − c)(d − e) + (e − a)(e − b)(e − c)(e − d) ≥ 0.

Finally, (c − a)(c − b)(c − d)(c − e) ≥ 0 as a product of two nonnegative num- bers, from which the inequality stated in the problem follows.

Remark. The problem in an alternative formulation, accepted for the IMO, asked to prove that the analogous inequality

(a 1 −a 2 )(a 1 −a 2 ) ···(a 1 −a n ) + (a 2 −a 1 )(a 2 −a 3 ) ···(a 2 −a n ) + ··· +(a n −a 1 )(a n −a 2 ) ···(a n −a n −1 )≥0

holds for arbitrary real numbers a i if and only if n = 3 or n = 5. The case n = 3 is analogous to n = 5. For n = 4, a counterexample is a 1 = 0,

a 2 =a 3 =a 4 = 1, while for n > 5 one can take a 1 =a 2 = ··· = a n −4 = 0, a n −3 =

a n −2 =a n −1 = 2, a n = 1 as a counterexample.

6. The proof goes by induction on n. For n = 2, the following labeling satisfies the conditions (i)–(iv): C 1 = 11,C 2 = 12,C 3 = 22,C 4 = 21. Suppose that n > 2, and that the numeration C 1 ,C 2 , . . . ,C 2 n −1 of a regular 2 n −1 - gon, in cyclical order, satisfies (i)–(iv). Then one can assign to the vertices of a

2 n -gon cyclically the following numbers: 1C 1 , 1C 2 , . . . , 1C 2 n −1 , 2C 2 n −1 , . . . , 2C 2 , 2C 1 . The conditions (i), (ii) obviously hold, while (iii) and (iv) follow from the induc-

tive assumption.

7. (a) Suppose that X ,Y, Z are fixed on segments AB, BC,CD. It is proven in a standard way that if ∠AT X 6= ∠ZT D, then ZT + T X can be reduced. It follows that if there exists a broken line XY ZT X of minimal length, then the following conditions hold:

∠DAB = π − ∠ATX − ∠AXT, ∠ABC = π − ∠BXY − ∠BY X = π − ∠AXT − ∠CYZ,

∠BCD = π − ∠CY Z − ∠CZY, ∠CDA = π − ∠DT Z − ∠DZT = π − ∠ATX − ∠CZY.

Thus σ = 0. (b) Now let σ = 0. Let us cut the surface of the tetrahedron along the edges AC , CD, and DB and set it down into a plane. Consider the plane figure S = ACD ′ BD ′′ C ′ thus obtained made up of triangles BCD ′ , ABC, ABD ′′ , and AC ′ D ′′ , with Z ′ ,T ′ ,Z ′′ respectively on CD ′ , AD ′′ ,C ′ D ′′ (here C ′ corre- sponds to C, etc.). Since ∠C ′ D ′′ A + ∠D ′′ AB + ∠ABC + ∠BCD ′ = 0 as an

392 4 Solutions

oriented angle (because σ = 0), the lines CD ′ and C ′ D ′′ are parallel and equally oriented; i.e., CD ′ D ′′ C ′ is a parallelogram. The broken line XY ZT X has minimal length if and only if Z ′′ ,T ′ , X, Y , Z ′ are collinear (where Z ′ Z ′′ k CC ′ ),

and then this length equals Z ′ Z ′′ = A 2

CC ′ = 2AC sin( α /2). There is an in-

Z ′′ finity of such lines, one for every

line Z ′ Z ′′ parallel to CC ′ that meets YX the interiors of all the segments

D ′′ Indeed, the triangles CAB and D ′′ AB are acute-angled, and thus the seg- ment AB has a common interior point with the parallelogram CD ′ D ′′ C ′ . Therefore the desired result follows.

CB , BA, AD ′′ . Such Z ′ Z ′′ exist. D ′

8. Suppose that a , b, c,t satisfy all the conditions. Then abc 6= 0 and

a 2 3 b 3 1 c Multiplying these equations, we obtain x 2 1 x 2 2 x 2 3 = 1, and hence x 1 x 2 x 3 = ε = ±1.

From the above equalities we get x 1 = ε b /a, x 2 = ε c /b, x 3 = ε a /c. Substituting x 1 in the first equation, we get ab 2 /a 2 +t ε b 2 /a + c = 0, which gives us

b 2 (1 + t ε ) = −ac.

(1) Analogously, c 2 (1 +t ε ) = −ab and a 2 (1 +t ε ) = −bc, and therefore (1 +t ε ) 3 =

−1; i.e., 1 + t ε = −1, since it is real. This also implies together with (1) that

b 2 = ac, c 2 = ab, and a 2 = bc, and consequently

a = b = c.

Thus the three equations in the problem are equal, which is impossible. Hence, such a , b, c,t do not exist.

9. We use induction. Since T 1 = 0, T 2 = 1, T 3 = 2, T 4 = 3, T 5 = 5, T 6 = 8, the statement is true for n = 1, 2, 3. Suppose that both formulas from the problem hold for some n ≥ 3. Then

T 2n +1 =1+T 2n +2 n −1

12 +1 2n +2 =1+T 2n +2 n −3 =

2 n −2 +2 n

7 7 Therefore the formulas hold for n + 1, which completes the proof.

10. We use induction. Suppose that every two of the numbers a 1 =2 n 1 − 3,a 2 =

k =2 k − 3, where 2 = n 1 <n 2 < ··· < n k , are coprime. Then one can construct a k +1 =2 n k +1 − 3 in the following way:

2 n 2 − 3,...,a

4.13 Shortlisted Problems 1971 393 Set s =a 1 a 2 ...a k . Among the numbers 2 0 ,2 1 ,...,2 s , two give the same residue

upon division by s, say s |2 α −2 β . Since s is odd, it can be assumed w.l.o.g. that β = 0 (this is actually a direct consequence of Euler’s theorem). Let 2 α −1= qs ,q ∈ N. Since 2 α+2 − 3 = 4qs + 1 is then coprime to s, it is enough to take n k +1 = α + 2. We obviously have n k +1 >n k .

11. We use induction. The statement for n = 1 is trivial. Suppose that it holds for n = k and consider n = k + 1. From the given condition, we have

∑ |a j ,1 x 1 + ··· + a j ,k x k +a j ,k+1 |

j =1

+|a k +1,1 x 1 + ··· + a k +1,k x k +a k +1,k+1 | ≤ M,

∑ |a j ,1 x 1 + ··· + a j ,k x k −a j ,k+1 |

j =1

+|a k +1,1 x 1 + ··· + a k +1,k x k −a k +1,k+1 |≤M for each choice of x i = ±1. Since |a + b| + |a − b| ≥ 2|a| for all a,b, we obtain

2 ∑ |a j 1 x 1 + ··· + a jk x k | + 2|a k +1,k+1 | ≤ 2M, that is,

j =1

∑ |a j 1 x 1 + ··· + a jk x k | ≤ M − |a k +1,k+1 |.

j =1

Now by the inductive assumption ∑ k j =1 |a jj | ≤ M − |a k +1,k+1 |, which is equiva- lent to the desired inequality.

12. Let us start with the case A =A ′ . If the triangles ABC and A ′ B ′ C ′ are oppositely oriented, then they are symmetric with respect to some axis, and the statement is true. Suppose that they are equally oriented. There is a rotation around A by 60 ◦

that maps ABB ′ onto ACC ′ . This rotation also maps the midpoint B 0 of BB ′ onto the midpoint C 0 of CC ′ , hence the triangle AB 0 C 0 is equilateral. In the general case, when A 6= A ′ , let us denote by T the translation that maps A onto A ′ . Let X ′

be the image of a point X under the (unique) isometry mapping ABC onto A ′ B ′ C ′ , and X ′′ the image of X under T . Furthermore, let X 0 ,X ′ 0 be the midpoints of segments X X ′ ,X ′ X ′′ . Then X 0 is the image of X ′ 0 under the translation −(1/2)T. However, since it has already been proven that the triangle

A ′ 0 B ′ 0 C 0 ′ is equilateral, its image A 0 B 0 C 0 under (1/2)T is also equilateral. The statement of the problem is thus proven.

13. Let p be the least of all the sums of elements in one row or column. If p ≥ n/2, then the sum of all elements of the array is s

≥ np ≥ n 2 /2. Now suppose that p < n/2. Without loss of generality, one can assume that the sum of elements in

the first row is p, and that exactly the first q elements of it are different from zero. Then the sum of elements in the last n − q columns is greater than or equal to (n − p)(n− q). Furthermore, the sum of elements in the first q columns is greater than or equal to pq. This implies that the sum of all elements in the array is

394 4 Solutions

n ≥ (n − p)(n − q) + pq = 2

2 2 (n − 2p)(n − 2q) ≥

2 since n ≥ 2p ≥ 2q.

14. Denote by V the figure made by a circle of radius 1 whose center moves along the broken line. From the condition of the problem, V contains the whole 50 ×50 square, and thus the area S (V ) of V is not less than 2500. Let L be the length of the broken line. We shall show that S (V ) ≤ 2L + π , from

which it will follow that L ≥ 1250 − π /2 > 1248. For each segment l i =A i A i +1 of the broken line, consider the figure V i obtained by a circle of radius 1 whose center moves along it, and let V i

be obtained by cutting off the circle of radius

1 with center at the starting point of l i . The area of V i is equal to 2A i A i +1 . It is clear that the union of all the figures V i together with a semicircle with center in

A 1 and a semicircle with center in A n contains V completely. Therefore

S (V ) ≤ π + 2A 1 A 2 + 2A 2 A 3 + ··· + 2A n −1 A n = π + 2L. This completes the proof.

15. Assume the opposite. Then one can numerate the cards 1 to 99, with a number n i written on the card i, so that n 98 6= n 99 . Denote by x i the remainder of n 1 +n 2 + ··· + n i upon division by 100, for i = 1, 2, . . . , 99. All x i must be distinct: Indeed, if x i =x j ,i < j, then n i +1 + ··· + n j is divisible by 100, which is impossible.

Also, no x i can be equal to 0. Thus, the numbers x 1 ,x 2 ,...,x 99 take exactly the values 1 , 2, . . . , 99 in some order. Let x be the remainder of n 1 +n 2 + ··· + n 97 +n 99 upon division by 100. It is not zero; hence it must be equal to x k for some k ∈ {1,2,...,99}. There are three cases:

(i) x =x k ,k ≤ 97. Then n k +1 +n k +2 + ··· + n 97 +n 99 is divisible by 100, a contradiction;

(ii) x =x 98 . Then n 98 =n 99 , a contradiction; (iii) x =x 99 . Then n 98 is divisible by 100, a contradiction. Therefore, all the cards contain the same number.

16. Denote by P ′ the polyhedron defined as the image of P under the homothety with center at A 1 and coefficient of similarity 2. It is easy to see that all P i ,i = 1, . . . , 9, are contained in P ′ (indeed, if M ∈P k , then

for some M ′ ∈ P, and the claim follows from the convexity of P). But the volume of P ′ is exactly 8 times the volume of P, while the volumes of P i add up to 9 times that volume. We conclude that not all P i have disjoint interiors.

17. We use the following obvious consequences of (a + b) 2 ≥ 4ab:

2 (a , 1 +a 2 )(a 3 +a 4 ) (a 1 +a 2 +a 3 +a 4 )

4.13 Shortlisted Problems 1971 395

(a 1 +a 2 +a 3 +a 4 ) 2 Now we have

(a 1 +a 4 )(a 2 +a 3 )

a 1 +a 3 a 2 +a 4 a 3 +a 1 a 4 +a 2

a 1 +a 2 a 2 +a 3 a 3 +a 4 a 4 +a 1

(a 1 +a 3 )(a 1 +a 2 +a 3 +a 4 ) (a 2 +a 4 )(a 1 +a 2 +a 3 +a 4 ) =

(a 1 +a 2 )(a 3 +a 4 )

(a 1 +a 4 )(a 2 +a 3 )

4 (a 1 +a 3 )

4 (a 2 +a 4 )

a 1 +a 2 +a 3 +a 4 a 1 +a 2 +a 3 +a 4

396 4 Solutions