Solutions to the Shortlisted Problems of IMO 1976

4.18 Solutions to the Shortlisted Problems of IMO 1976

1. Let r denote the common inradius. Some two of the four triangles with the in- radii ρ have cross angles at M: Suppose these are △AMB 1 and △BMA 1 . We shall show that △AMB 1 ∼ = △BMA 1 . Indeed, the altitudes of these two triangles are both equal to r, the inradius of △ABC, and their interior angles at M are equal to some angle ϕ . If P is the point of tangency of the incircle of △A 1 MB with MB, then r = A 1 M +BM+A 1 ρ B A 1 B , which also implies r −2ρ

A 1 M +BM−A 1 B ρ

2MP 2r cot (ϕ/2)

A 1 B . Since similarly r −2ρ

2r cot

B 1 A , we obtain A 1 B =B 1 A and consequently △AMB 1 ∼ = △BMA 1 . Thus ∠BAC = ∠ABC and CC 1 ⊥ AB. There

are two alternatives for the other two incircles: (i) If the inradii of AMC 1 and AMB 1 are equal to r, it is easy to obtain that △AMC 1 ∼ = △AMB 1 . Hence ∠AB 1 M = ∠AC 1 M = 90 ◦ , and △ABC is equi- lateral. (ii) The inradii of AMB 1 and CMB 1 are equal to r. Put x = ∠MAC 1 = ∠MBC 1 .

In this case ϕ = 2x and ∠B 1 MC = 90 ◦

− x. Now we have AB 1 S CB 1 = AMB1 S CMB1 =

AM +MB 1 +AB 1 = AM +MB 1 −AB 1

CM +MB 1 +CB 1 CM +MB 1 −CB 1 =

cot (45 ◦ −x/2) . On the other hand, we have

cot x

AB 1

CB 1 = AB BC = 2 cos 2x. Thus we have an equation for x: tan(45 ◦ − x/2) =

2 cos 2x tan x, or equivalently x

2 tan 45 ◦ x

2 − 2 Hence sin 3x

2 = 1 − sinx, implying sin3x = 1, i.e., x = 30 ◦ . Therefore △ABC is equilateral.

−sinx = 2sin 2 45 ◦ x −

2. Let us put b i = i(n + 1 − i)/2, and let c i =a i −b i ,i = 0, 1, . . . , n + 1. It is easy to verify that b 0 =b n +1 = 0 and b i −1 − 2b i +b i +1 = −1. Subtracting this inequality from a i −1 − 2a i +a i +1 ≥ −1, we obtain c i −1 − 2c i +c i +1 ≥ 0, i.e., 2c i ≤c i −1 +

c i +1 . We also have c 0 =c n +1 = 0.

Suppose that there exists i ∈ {1,...,n} for which c i > 0, and let c k

be the max- imal such c i . Assuming w.l.o.g. that c k −1 <c k , we obtain c k −1 +c k +1 < 2c k , which is a contradiction. Hence c i ≤ 0 for all i; i.e., a i ≤b i . Similarly, considering the sequence c ′ =a i +b i one can show that c ′ i i ≥ 0, i.e.,

a i ≥ −b i for all i. This completes the proof.

3. (a) Let ABCD be a quadrangle with 16 = d = AB + CD + AC, and let S be its area. Then S ≤ (AC ·AB+AC ·CD)/2 = AC(d −AC)/2 ≤ d 2 /8 = 32, where equality occurs if and only if AB √ ⊥ AC ⊥ CD and AC = AB + CD = 8. In this case BD =8 2.

(b) Let A ′

be the point with DA ′ −→ = AC . The triangular inequality implies AD + BC ≥ AA ′

√ =8 5 . Thus the perimeter attains its minimum for AB = CD =

4. (c) Let us assume w.l.o.g. that CD ≤ AB. Then C lies inside △BDA ′ and hence BC + AD = BC +CA ′ < BD + DA ′ . The maximal value BD + DA ′ of BC + AD is attained when C approaches D, making a degenerate quadrangle.

4.18 Shortlisted Problems 1976 419

4. The first few values are easily verified to be 2 r n +2 −r n , where r 0 = 0, r 1 =r 2 = 1, r 3 = 3, r 4 = 5, r 5 = 11, . . . . Let us put u n =2 r n +2 −r n (we will show that r n exists and is integer for each n). A simple calculation gives us u n (u 2 n −1 − 2) =

2 r n +2r n −1 +2 −r n −2r n −1 +2 r n −2r n −1 +2 −r n +2r n −1 . If an array q n , with q 0 = 0 and

q 1 = 1, is set so as to satisfy the linear recurrence q n +1 =q n + 2q n −1 , then it also

satisfies q n − 2q n = −(q n − 2q n ) = ··· = (−1) n −1 (q 1 − 2q 0 ) = (−1) −1 n −1 −2 −1 . Assuming inductively up to n r i =q i , the expression for u n (u 2 −2) = u n +1 +u 1 reduces to 2 q n +1 +2 −q n

n −1

+1 +u 1 . Therefore, r n +1 =q n +1 . The solution to this linear recurrence with r

3 , and since [u n ]=2 for n ≥ 0, the result follows.

0 = 0, r 1 = 1 is r n =q n =

2 n −(−1) n

Remark. One could simply guess that u =2 r n n +2 −r n for r

n = −(−1) n 3 , and then prove this result by induction.

5. If one substitutes an integer q-tuple (x 1 ,...,x q ) satisfying |x i | ≤ p for all i in an equation of the given system, the absolute value of the right-hand member never exceeds pq. So for the right-hand member of the system there are (2pq + 1) p

possibilities There are (2p +1) q possible q-tuples (x 1 ,...,x q ). Since (2p + 1) q > (2pq + 1) p , there are at least two q-tuples (y 1 ,...,y q ) and (z 1 ,...,z q ) giving the same right-hand members in the given system. The difference (x 1 ,...,x q )=

(y 1 −z 1 ,...,y q −z q ) thus satisfies all the requirements of the problem.

6. Suppose a are the dimensions of the box. If we set b = [a / 1 3 ≤a 2 ≤a 3 i i 2 ], the

condition of the problem is equivalent to a 1 a 2 a 3

b 1 · b 2 · b 3 = 5. We list some values of

a ,b = [a/ 3 2 ] and a/b:

a 23 4 5 6 7 8 9 10

b 12 3 3 4 5 6 7 7

a /b 2 1.5 1.33 1.67 1.5 1.4 1.33 1.29 1.43

We note that if a > 2, then a/b ≤ 5/3, and if a > 5, then a/b ≤ 3/2. If a 1 > 2, then a 1 · a 2 b a 1 b 2 · 3 b 3 < (5/3) 3 < 5, a contradiction. Hence a 1 = 2. If also a 2 = 2, then

a 3 /b 3 = 5/4 < 3 2, which is impossible. Also, if a

2 ≥ 6, then 2 b 2 · 3 b 3 ≤ (1.5) 2 <

2 .5, again a contradiction. We thus have the following cases: (i) a 1 = 2, a 2 = 3, then a 3 /b 3 = 5/3, which holds only if a 3 = 5; (ii) a 1 = 2, a 2 = 4, then a 3 /b 3 = 15/8, which is impossible; (iii) a 1 = 2, a 2 = 5, then a 3 /b 3 = 3/2, which holds only if a 3 = 6. The only possible sizes of the box are therefore (2, 3, 5) and (2, 5, 6).

7. The map T transforms the interval (0, a] onto (1−a,1] and the interval (a,1] onto (0, 1 − a]. Clearly T preserves the measure. Since the measure of the interval

[0, 1] is finite, there exist two positive integers k, l > k such that T k (J) and T l (J) are not disjoint. But the map T is bijective; hence T l −k (J) and J are not disjoint.

8. Every polynomial with real coefficients can be factored as a product of linear and quadratic polynomials with real coefficients. Thus it suffices to prove the result

only for a quadratic polynomial P (x) = x 2 − 2ax + b 2 , with a > 0 and b 2 >a 2 . Using the identity

− (2ax) 2n − 2ax + b −k−1 ∑ (x 2 +b 2 ) k (2ax)

k =0

we have solved the problem if we can choose n such that b 2n 2n n >2 2n a 2n . However, it is is easy to show that 2n 2n n <2 2n ; hence it is enough to take n such that

= 1 < b/a, such an n always exists.

(b/a) 2n > 2n. Since lim

n →∞ (2n)

1 /(2n)

9. The equation P

n (x) = x is of degree 2 , and has at most 2 distinct roots. If x > 2, then by simple induction P n (x) > x for all n. Similarly, if x < −1, then P 1 (x) > 2,

which implies P n (x) > 2 for all n. It follows that all real roots of the equation P n (x) = x lie in the interval [−2,2], and thus have the form x = 2cost.

Observe that P 1 (2 cost) = 4 cos 2 t − 2 = 2cos2t, and in general P n (2 cost) =

2 cos 2 n t . Our equation becomes

cos 2 n t = cost,

which indeed has 2 n different solutions t = 2 n πm (m = 0, 1, . . . , 2 n 2 −1 −1 − 1) and t = 2 2 πm n +1 (m = 1, 2, . . . , 2 n −1 ).

be positive integers whose sum is 1976. Let M denote the maximal value of a 1 a 2 ··· a n . We make the following observations: (1) a 1 = 1 does not yield the maximum, since replacing 1, a 2 by 1 +a 2 in- creases the product. (2) a j −a i ≥ 2 does not yield the maximal value, since replacing a i ,a j by a i +

10. Let a 1 ≤a 2 ≤ ··· ≤ a n

1 ,a j − 1 increases the product. (3) a i ≥ 5 does not yield the maximal value, since 2(a i − 2) = 2a i −4>a i .

Since 4 =2 2 , we may assume that all a i are either 2 or 3, and M =2 k 3 l , where 2k + 3l = 1976. (4) k ≥ 3 does not yield the maximal value, since 2 · 2 · 2 < 3 · 3. Hence k

≤ 2 and 2k ≡ 1976 (mod 3) gives us k = 1, l = 658 and M = 2 · 3 658 .

k for each k = 0, 1, . . . , where q k ∈ N. Indeed, the statement is true for k = 0, and if it holds for some k then

11. We shall show by induction that 5 2 k

−1=2 +2 k q

+1 k 5 2 −1 =2 k +3 d k +1 where d k +1 = 5 2 +1 d k /2 is an integer by the inductive hypothesis. Let us now choose n =2 k + k + 2. We have 5 n = 10 k +2 q k +5 k +2 . It follows from

5 2 k +1 −1= 5 2 k

5 4 < 10 3 that 5 k +2 has at most [3(k + 2)/4] + 2 nonzero digits, while 10 k +2 q k ends in k + 2 zeros. Hence the decimal representation of 5 n contains at least [(k + 2)/4] − 2 consecutive zeros. Now it suffices to take k > 4 · 1978.

12. Suppose the decomposition into k polynomials is possible. The sum of coeffi- cients of each polynomial a 1 x +a 2 x 2 x + ···+ a n n equals 1 + ···+ n = n(n +1)/2 while the sum of coefficients of 1976 (x + x 2 + ···+x n ) is 1976n. Hence we must have 1976n

= kn(n + 1)/2, which reduces to (n + 1) | 3952 = 2 4 ·13·19. In other words, n is of the form n =2 α 13 β 19 γ −1, with 0 ≤ α ≤ 4, 0 ≤ β ≤ 1, 0 ≤ γ ≤ 1.

4.18 Shortlisted Problems 1976 421 We can immediately eliminate the values n = 0 and n = 3951 that correspond to

α = β = γ = 0 and α = 4, β = γ = 1. We claim that all other values n are permitted. There are two cases.

α ≤ 3. In this case k = 3952/(n + 1) is even. The simple choice of the poly- nomials P = x + 2x 2 + ···+ nx n and P ′ = nx + (n − 1)x 2 + ··· + x n suffices,

since k (P + P ′ )/2 = 1976(x + x 2 + ··· + x n ).

α = 4. Then k is odd. Consider (k − 3)/2 pairs (P,P ′ ) of the former case and P 1 = nx + (n − 1)x 3 + ··· + n +1 x n

+ nx 2 4 n + (n − 1)x +3 + ··· + 2 x n −1 . Then P +P 1 +P 2 = 3(n + 1)(x + x 2 + ··· + x n )/2 and therefore (k − 3)(P +

P ′ )/2 + (P + P 1 +P 2 ) = 1976(x + x 2 + ··· + x n ).

It follows that the desired decomposition is possible if and only if 1 < n < 3951 and n + 1 | 2 · 1976.

422 4 Solutions