Solutions to the Shortlisted Problems of IMO 1979

4.21 Solutions to the Shortlisted Problems of IMO 1979

1. We prove more generally, by induction on n, that any 2n-gon with equal edges and opposite edges parallel to each other can be dissected. For n = 2 the only possible such 2n-gon is a single lozenge, so our theorem holds in this case. We will now show that it holds for general n. Assume by induction that it holds

be an arbitrary 2n-gon with equal edges and opposite edges parallel to each other. Then we can construct points B i for i = 3, 4, . . . , n

for n − 1. Let A 1 A 2 ...A 2n

such that −−→ A i B i −−→ = A 2 A 1 −−−−−−→ = A n +1 A n +2 . We set B 2 =A 2n +1 =A 1 and B n +1 =A n +2 . It follows that A i B i B i +1 A i +1 for i = 2, 3, 4, . . . , n are all lozenges. It also follows that B i B i +1 for i = 2, 3, 4, . . . , n are equal to the edges of A 1 A 2 ...A 2n and parallel to A i A i +1 and hence to A n +i A n +i+1 . Thus B 2 ...B n +1 A n +3 ...A 2n is a 2 (n − 1)- gon with equal edges and opposite sides parallel and hence, by the induction hypothesis, can be dissected into lozenges. We have thus provided a dissection

for A 1 A 2 ...A 2n . This completes the proof.

2. The only way to arrive at the latter alternative is to draw four different socks in the first drawing or to draw only one pair in the first drawing and then draw two different socks in the last drawing. We will call these probabilities respectively

p 1 ,p 2 ,p 3 . We calculate them as follows:

5 2 4 = 4 8 5 4 4 4 p 4 2 2 1 2 10 ,p

21 2 = 10 4 ,p 7 3 = 6 4 . 2 15 We finally calculate the desired probability: P =p 1 +p 2 p 3 = 8 15 .

3. An obvious solution is f (x) = 0. We now look for nonzero solutions. We note that plugging in x = 0 we get f (0) 2 = f (0); hence f (0) = 0 or f (0) = 1. If

f (0) = 0, then f is of the form f (x) = x k g (x), where g(0) 6= 0. Plugging this formula into f (x) f (2x 2 ) = f (2x 3 + x) we get

2 k x 2k g (x)g(2x 2 ) = (2x 2 + 1) k g (2x 3 + x). Plugging in x = 0 gives us g(0) = 0, which is a contradiction. Hence f (0) = 1.

For an arbitrary root α of the polynomial f , 2 α 3 + α must also be a root. Let α

be a root of the largest modulus. If | α | > 1 then |2 α 3 + α | > 2| α | 3 −| α |>| α |, which is impossible. It follows that | α | ≤ 1 and hence all roots of f have modules less than or equal to 1. But the product of all roots of f is | f (0)| = 1, which implies that all the roots have modulus 1. Consequently, for a root α it holds that

| α | = |2 α 3 − α | = 1. This is possible only if α = ±ı. Since the coefficients of

f are real it follows that f must be of the form f (x) = (x 2 + 1) k where k ∈N 0 . These polynomials satisfy the original formula. Hence, the solutions for f are

f (x) = 0 and f (x) = (x 2 + 1) k ,k ∈N 0 .

4. Let us prove first that the edges A 1 A 2 ,A 2 A 3 ,...,A 5 A 1 are of the same color. Assume the contrary, and let w.l.o.g. A 1 A 2 be red and A 2 A 3 be green. Three of the segments A 2 B l (l = 1, 2, 3, 4, 5), say A 2 B i ,A 2 B j ,A 2 B k , have to be of the same

446 4 Solutions color, let it w.l.o.g. be red. Then A 1 B i ,A 1 B j ,A 1 B k must be green. At least one

of the sides of triangle B i B j B k , say B i B j , must be an edge of the prism. Then looking at the triangles A 1 B i B j and A 2 B i B j we deduce that B i B j can be neither green nor red, which is a contradiction. Hence all five edges of the pentagon

A 1 A 2 A 3 A 4 A 5 have the same color. Similarly, all five edges of B 1 B 2 B 3 B 4 B 5 have the same color. We now show that the two colors are the same. Assume otherwise, i.e., that w.l.o.g. the A edges are painted red and the B edges green. Let us call segments

of the form A i B j diagonal (i and j may be equal). We now count the diagonal segments by grouping the red segments based on their A point, and the green segments based on their B point. As above, the assumption that three of A i B j for fixed i are red leads to a contradiction. Hence at most two diagonal segments out of each A i may be red, which counts up to at most 10 red segments. Similarly, at most 10 diagonal segments can be green. But then we can paint at most 20 diagonal segments out of 25, which is a contradiction. Hence all edges in the

pentagons A 1 A 2 A 3 A 4 A 5 and B 1 B 2 B 3 B 4 B 5 have the same color.

5. Let A = {x | (x,y) ∈ M} and B = {y | (x,y) ∈ M. Then A and B are disjoint and hence

(|A| + |B|) 2 n 2 |M| ≤ |A| · |B| ≤

4 ≤ 4 These cardinalities can be achieved for M = {(a,b) | a = 1,2,...,[n/2], b =

[n/2] + 1, . . ., n} .

6. Setting q =x 2 + x − p, the given equation becomes q

(x + 1) 2 − 2q+ (x + 2) 2 −q= (2x + 3) 2 − 3q. (1)

Taking squares of both sides we get 2 ((x + 1) 2 − 2q)((x + 2) 2 − q) = 2(x +

1 )(x + 2). Taking squares again we get

q 2q

2 − 2(x + 2) 2 − (x + 1) = 0. If 2q = 2(x + 2) 2 + (x + 1) 2 , at least one of the expressions under the three square

roots in (1) is negative, and in that case the square root is not well-defined. Thus, we must have q = 0. Now (1) is equivalent to |x + 1| + |x + 2| = |2x + 3|, which holds if and only if

x 6∈ (−2,−1). The number of real solutions x of q = x 2 + x − p = 0 which are not in the interval (−2,−1) is zero if p < −1/4, one if p = −1/4 or 0 < p < 2, and two otherwise. Hence, the answer is −1/4 < p ≤ 0 or p ≥ 2.

7. We denote the sum mentioned above by S. We have the following equalities:

4.21 Shortlisted Problems 1979 447

i =660 i · (1979 − i) Since no term in the sum contains a denominator divisible by 1979 (1979 is a

i =660 i

1979 −i

prime number), it follows that when S is represented as p /q the numerator p will have to be divisible by 1979.

8. By the definition of f , it holds that f (0.b 1 b 2 . . . ) = 3b 1 /4 + f (0.b 2 b 3 . . . )/4 =

0 .b 1 b 1 + f (0.b 2 b 3 . . . )/4. Continuing this argument we obtain

(1) The binary representation of every rational number is eventually periodic. Let

f (0.b 1 b 2 b 3 . . . ) = 0.b 1 b 1 ...b n b n + 2n f (0.b n +1 b

2 n +2 . . . ).

us first determine f (x) for a rational x with the periodic representation x =

0 .b 1 b 2 ...b n n . Using (1) we obtain f (x) = 0.b 1 b 1 ...b n b n + f (x)/2 2n , and hence

f (x) = 2 2 n −1 0 .b 1 b 1 ...b n b n = 0.b 1 b 1 ...b n b n .

be an arbitrary rational number. Then it fol- lows from (1) that

Now let x = 0.a 1 a 2 ...a k b 1 b 2 ...b n

f (x) = 0.a 1 a 1 ...a k a k + 2n f (0.b 1 b ...b ) = 0.a a ...a a b b ...b b .

Hence f (0.b 1 b 2 . . . ) = 0.b 1 b 1 b 2 b 2 . . . for every rational number 0.b 1 b 2 ....

9. Let us number the vertices, starting from S and moving clockwise. In that case S = 1 and F = 5. After an odd number of moves to a neighboring point we can

be only on an even point, and hence it follows that a 2n −1 = 0 for all n ∈ N. Let us define respectively z n and w n as the number of paths from S to S in 2n moves and the number of paths from S to points 3 and 7 in 2n moves. We easily derive the following recurrence relations:

a 2n +2 =w n ,w n +1 = 2w n + 2z n ,z n +1 = 2z n +w n , n = 0, 1, 2, . . . . By subtracting the second equation from the third we get z n +1 =w n +1 −w n . By

plugging this equation into the formula for w n +2 we get w n +2 −4w n +1 +2w √ n = 0.

2 and y =2−

The roots of the characteristic equation r 2 √ − 4r + 2 = 0 are x = 2 +

2. From the conditions w √ 0 = 0 and w 1 = 2 we easily obtain a 2n = w n −1 = (x n −1

−y n −1 )/ 2 .

448 4 Solutions

10. In the cases a − →

= 0,b =

0 , and a k b the inequality is trivial. Otherwise, let us consider a triangle ABC such that −→ CB = a and −→ CA = b. From this point on we

shall refer to α , β , γ as angles of ABC. Since √ |a × b| = |a||b|sin γ , our inequality reduces to

3 |a||b|sin 2 ≤3 |c| /8, which is further reduced to

3 3 sin α sin β sin γ ≤

using the sine law. The last inequality follows immediately from Jensen’s in- equality applied to the function f (x) = ln sin x, which is concave for 0 < x < π because f ′ (x) = cot x is strictly decreasing.

11. Let us define y i =x 2 . We thus have y 1 +y 2 + ··· + y n = 1, y i ≥ 1/n 2 , and P = √

y 1 y 2 ...y n . The upper bound is obtained immediately from the AM–GM inequality: P ≤

1 /n n /2 , where equality holds when x

i = √y i = 1/ n .

For the lower bound, let us assume w.l.o.g. that y 1 ≥y 2 ≥ ··· ≥ y n . We note that if a

2 and s = a + b > 2/n 2 is fixed, then ab = (s 2 ≥ b ≥ 1/n 2 − (a − b) )/4 is minimized when

|a − b| is maximized, i.e., when b = 1/n 2 . Hence y 1 y 2 ···y

is minimal when y √ 2 =y 3 = ··· = y n = 1/n 2 . Then y 1 = (n 2 − n + 1)/n 2 and

therefore P min = n 2 − n + 1/n n .

12. The first criterion ensures that all sets in an S-family are distinct. Since the num- ber of different families of subsets is finite, h has to exist. In fact, we will show that h = 11. First of all, if there exists X ∈ F such that |X| ≥ 5, then by (3) there exists Y ∈ F such that X ∪ Y = R. In this case |F| is at most 2. Similarly, for |X| = 4, for the remaining two elements either there exists a subset in F that contains both, in which case we obtain the previous case, or there exist different

Y and Z containing them, in which case X ∪Y ∪ Z = R, which must not happen. Hence we can assume |X| ≤ 4 for all X ∈ F. Assume |X| = 1 for some X. In that case other sets must not contain that subset and hence must be contained in the remaining 5-element subset. These elements must not be subsets of each other. From elementary combinatorics, the largest

number of subsets of a 5-element set of which none is subset of another is 5 2 =

10. This occurs when we take all 2-element subsets. These subsets also satisfy (2). Hence |F| max = 11 in this case. Otherwise, let us assume |X| = 3 for some X. Let us define the following families of subsets: G = {Z = Y \ X | Y ∈ F} and H = {Z = Y ∩ X | Y ∈ F}. Then no two sets in G must complement each other in R \ X, and G must cover this set. Hence

G contains exactly the sets of each of the remaining 3 elements. For each element of G no two sets in H of which one is a subset of another may be paired with it. There can be only 3 such subsets selected within a 3-element set X . Hence the number of remaining sets is smaller than 3 ·3 = 9. Hence in this case |F| max = 10.

In the remaining case all subsets have two elements. There are 6 2 = 15 of them. But for every three that complement each other one must be discarded; hence the maximal number for F in this case is 2 · 15/3 = 10. It follows that h = 11.

4.21 Shortlisted Problems 1979 449

13. From elementary trigonometry we have sin 3t

= 3 sint − 4sin t . Hence, if we denote y = sin 20 , we have 3 /2 = sin 60 ◦ = 3y − 4y 3 . Obviously 0 <y<

1 /2 = sin 30 ◦ . The function f (x) = 3x − 4x 3 is strictly increasing on [0, 1/2) because f ′ (x) = 3 − 12x 2 > 0 for 0 ≤ x < 1/2. Now the desired inequality 20 60 =

3 < sin 20 ◦ < 21 60 = 7 20 follows from

<f

which is directly verified.

14. Let us assume that a ∈ R\{1} is such that there exist a and x such that x = log a x , or equivalently f (x) := ln x/x = ln a. Then a is a value of the function f (x) for x ∈R + \ {1}, and the converse also holds. First we observe that f (x) tends to −∞ as x → 0 and f (x) tends to 0 as x → 1. Since f (x) > 0 for x > 1, the function f (x) takes its maximum at a point x for

which f ′ (x) = (1 − lnx)/x 2 = 0. Hence max f (x) = f (e) = e 1 /e .

It follows that the set of values of f

1 (x) for x ∈ R /e is the interval (−∞,e ), and consequently the desired set of bases a of logarithms is (0, 1) ∪ (1,e 1 /e ].

15. We note that

i 5 x =a 2 ∑ 2 (a − i i i − 2a ∑ i ∑ i · a − 2a · a +a 3 ∑ = 0.

Since the terms in the sum on the left are all nonnegative, it follows that all the terms have to be 0. Thus, either x i = 0 for all i, in which case a = 0, or a = j 2 for some j and x i = 0 for i 6= j. In this case, x j = a/ j = j. Hence, the only possible values of a are {0,1,4,9,16,25}.

16. Obviously, no two elements of F can be complements of each other. If one of the sets has one element, then the conclusion is trivial. If there exist two different 2-element sets, then they must contain a common element, which in turn must then be contained in all other sets. Thus we can assume that there exists at most one 2-element subset of K in F. Since there can be at most 6 subsets of more than 3 elements of a 5-element set, it follows that at least 9 out of 10 possible 3-element subsets of K belong to F. Let us assume, without loss of generality, that all sets but {c,d,e} belong to F. Then sets {a,b,c}, {a,d,e}, and {b,c,d} have no common element, which is a contradiction. Hence it follows that all sets have a common element.

17. Let K, L, and M be intersections of CQ and BR, AR and CP, and AQ and BP, respectively. Let ∠X denote the angle of the hexagon KQMPLR at the vertex X , where X is one of the six points. By an elementary calculation of angles we get

∠K = 140 ◦ , ∠L = 130 ◦ , ∠M = 150 ◦ , ∠P = 100 ◦ , ∠Q = 95 ◦ , ∠R = 105 ◦ .

450 4 Solutions Since ∠KBC = ∠KCB, it follows that K is on the symmetry line of ABC through

be points symmetric to K with respect to AR and AQ, respectively. Since ∠AK Q Q = ∠AK Q K R = 70 ◦ and ∠AK R R = ∠AK R K Q = 70 ◦ , it fol-

A . Analogous statements hold for L and M. Let K R and K Q

C lows that the points K R , R, Q, and

K o Q are collinear. Hence ∠QRK

= o 25 20 2∠R − 180 ◦ and ∠RQK = 2∠Q −

180 ◦ . In the same way we conclude K R that ∠PRL = 2∠R−180 ◦ , ∠RPL =

RK 2∠P − 180 ◦ , ∠QPM = 2∠P − 180 ◦

L and ∠PQM = 2∠Q −180 ◦ . From these

PQ 20 o formulas we easily get ∠RPQ = 60 ◦ ,

25 o

M K Q 15 A o B ∠RQP = 75 ◦ , and ∠QRP = 45 ◦ .

15 o

18. Let us write all a i in binary representation. For S ⊆ {1,2,...,m} let us define

b (S) as the number in whose binary representation ones appear in exactly the slots where ones appear in all a i where i ⊆ S and don’t appear in any other

a i . Some b (S), including b( /0), will equal 0, and hence there are fewer than 2 m different positive b (S). We note that no two positive b(S 1 ) and b(S 2 ) (S 1 6= S 2 ) have ones in the same decimal places. Hence sums of distinct b (S)’s are distinct. Moreover

a i = ∑ b (S)

i ∈S

and hence the positive b (S) are indeed the numbers b 1 ,...,b n whose existence we had to prove.

19. Let us define i j for two positive integers i and j in the following way: i 1 =i and i j +1 =i i j for all positive integers j. Thus we must find the smallest m such that 100 m >3 100 . Since 100 1 = 100 > 27 = 3 2 , we inductively have 100 j =

10 100 j −1 >3 100 j −1 >3 3 j =3 j +1 and hence m ≤ 99. We now prove that m = 99 by proving 100 98 <3 100 . We note that (100 1 ) 2 = 10 4 < 27 4 =3 12 <3 27 =3 3 . We also note for d > 12 (which trivially holds for all d = 100 i ) that if c >d 2 ,

then we have

3 c >3 d 2 >3 12d = (3 12 ) d > 10000 d = (100 d ) 2 . Hence from 3 3 > (100 1 ) 2 it inductively follows that 3 j > (100 j −2 ) 2 > 100 j −2

and hence that 100 99 >3 100 > 100 98 . Hence m = 99.

20. Let x k = max{x 1 ,x 2 ,...,x n }. Then x i x i +1 ≤x i x k for i = 1, 2, . . . ,k − 1 and x i x i +1 ≤x k x i +1 for i = k, . . . , n − 1. Summing up these inequalities for i =

1 , 2, . . . , n − 1 we obtain

n −1

∑ ≤x k (x 1 + ··· + x k −1 +x k +1 + ··· + x n )=x k (a − x k )≤ .

i =1

4 We note that the value a 2 /4 is attained for x 1 =x 2 = a/2 and x 3 = ··· = x n = 0.

Hence a 2 /4 is the required maximum.

4.21 Shortlisted Problems 1979 451

21. Denote m = 10 6 and let f (n) be the number of different ways n ∈ N can be expressed as x 2 +y 3 with x , y ∈ {0,1,...,m}. Clearly f (n) = 0 for n < 0 or n >m 2 +m 3 . The first equation can be written as x 2 +t 3 =y 2 +z 3 = n, whereas the second equation can be written as x 2 +t 3 = n + 1, y 2 +z 3 = n. Hence we obtain the following formulas for M and N:

m −1

M = ∑ f (i) 2 , N = ∑ f (i) f (i + 1) .

i =0

i =0

Using the AM–GM inequality we get

m −1

N = ∑ f (i) f (i + 1)

+ ∑ f (i) 2 +

<M.

2 The last inequality is strict, since f (0) = 1 > 0. This completes our proof.

i =0

2 2 i =1

22. Let the centers of the two circles be denoted by O and O 1 and their respective radii by r and r M (t)

1 , and let the positions of the points on the circles at time t be

N (t) denoted by M A (t) and N(t). Let Q be

ω φ t φ ω lelogram. We will show that Q is the t O

the point such that OAO 1 Q is a paral-

O 1 point P we are looking for, i.e., that QM (t) = QN(t) for all t. We note that

P =Q OQ =O 1 A =r 1 ,O 1 Q = OA = r and ∠QOA = ∠QO 1 A = φ . Since the two points return to A at the same time, it

follows that ∠M (t)OA = ∠N(t)O 1 A = ω t . Therefore ∠QOM (t) = ∠QO 1 N (t) = φ + ω t , from which it follows that △QOM(t) ∼ = △QO 1 N (t). Hence QM(t) = QN (t), as we claimed.

23. It is easily verified that no solutions exist for n ≤ 8. Let us now assume that n > 8. We note that 2 8 +2 11 +2 n =2 8 · (9 + 2 n −8 ). Hence 9 + 2 n −8 must also

be a square, say 9 +2 n −8 =x 2 ,x ∈ N, i.e., 2 n −8 =x 2 − 9 = (x − 3)(x + 3). Thus x − 3 and x +3 are both powers of 2, which is possible only for x = 5 and n = 12. Hence, n = 12 is the only solution.

24. Clearly O is the midpoint of BC. Let M and N be the points of tangency of the circle with AB and AC, respectively, and let ∠BAC =2 ϕ . Then ∠BOM = ∠CON = ϕ . Let us assume that PQ touches the circle in X. If we set ∠POM = ∠POX = x and ∠QON = ∠QOX = y, then 2x + 2y = ∠MON = 180 ◦ −2 ϕ , i.e., y = 90 ◦ − ϕ − x. It follows that ∠OQC = 180 ◦ −∠QOC−∠OCQ = 180 ◦ −( ϕ +y)−(90 ◦ − ϕ )=

90 ◦ −y=x+ ϕ = ∠BOP. Hence the triangles BOP and CQO are similar, and consequently BP

·CQ = BO ·CO = (BC/2) 2 .

452 4 Solutions

Conversely, let BP ·CQ = (BC/2) 2 and let Q ′

be the point on (AC) such that PQ ′ is tangent to the circle. Then BP ·CQ ′ = (BC/2) 2 , which implies Q ≡Q ′ .

25. Let us first look for such a point R on a ray l in π going through P. Let ∠QPR =

2 θ . Consider a point Q ′ on the extension of l beyond P such that Q ′ P = QP. Then we have

sin ∠QQ ′ R Since ∠QQ ′ R is fixed, the maximum of the expression occurs when ∠Q ′ QR =

QR

QR

90 ◦ , i.e., when PR = PQ. In this case, (QP + PR)/QR = 1/sin θ . Looking at all possible rays l, we see that θ is minimal when l contains the projection of PQ onto π . Hence, if PQ 6⊥ π , the desired point R is the point on the projection of ray PQ onto π such that PR = PQ; otherwise, R is any point of the circle k(P, PQ).

26. Let us assume that f (x + y) = f (x) + f (y) for all reals. In this case we trivially apply the equation to get f (x + y + xy) = f (x + y) + f (xy) = f (x) + f (y) + f (xy). Hence the equivalence is proved in the first direction. Now let us assume that f (x+ y+xy) = f (x)+ f (y)+ f (xy) for all reals. Plugging in x = y = 0 we get f (0) = 0. Plugging in y = −1 we get f (x) = − f (−x). Plugging in y = 1 we get f (2x + 1) = 2 f (x) + f (1) and hence f (2(u + v + uv) +

1 ) = 2 f (u + v + uv) + f (1) = 2 f (uv) + 2 f (u) + 2 f (v) + f (1) for all real u and v. On the other hand, plugging in x = u and y = 2v+1 we get f (2(u+v+uv)+1) =

f (u+ (2v+ 1)+ u(2v+1)) = f (u)+2 f (v)+ f (1)+ f (2uv+ u). Hence it follows that 2 f (uv) + 2 f (u) + 2 f (v) + f (1) = f (u) + 2 f (v) + f (1) + f (2uv + u), i.e.,

(1) Plugging in v = −1/2 we get 0 = 2 f (−u/2) + f (u) = −2 f (u/2) + f (u). Hence,

f (2uv + u) = 2 f (uv) + f (u).

f (u) = 2 f (u/2) and consequently f (2x) = 2 f (x) for all reals. Now (1) reduces to f (2uv + u) = f (2uv) + f (u). Plugging in u = y and x = 2uv, we obtain f (x) +

f (y) = f (x + y) for all nonzero reals x and y. Since f (0) = 0, it trivially holds that f (x + y) = f (x) + f (y) when one of x and y is 0.

Second solution. Assume that f (x + y + xy) = f (x) + f (y) + f (xy) for all x, y. Substituting (x, −y) in the functional equation and adding it to the original equa- tion yields

f (x − t) + f (x + t) = 2 f (x), where t = y(x + 1). (2) Thus (2) holds whenever x 6= −1. Similarly, for t 6= −1 we have f (t − x) + f (x +

t ) = 2 f (t). Summing these two equalities and using f (y) = − f (−y) as shown above we obtain f (x) + f (t) = f (x + t) for all x,t 6= −1. The case x = −1 or

t = −1 is easy to handle with, as f (−1) = 2 f (− 1 2 ).

4.22 Shortlisted Problems 1981 453