Solutions to the Shortlisted Problems of IMO 1968

4.10 Solutions to the Shortlisted Problems of IMO 1968

1. Since the ships are sailing with constant speeds and directions, the second ship is sailing at a constant speed and direction in reference to the first ship. Let A be

the constant position of the first ship in this frame. Let B 1 ,B 2 ,B 3 , and B on line

b defining the trajectory of the ship be positions of the second ship with respect to the first ship at 9:00, 9:35, 9:55, and at the moment the two ships were closest. Then we have the following equations for distances (in miles):

AB 1 = 20, AB 2 = 15, AB 3 = 13,

B 1 B 2 :B 2 B 3 = 7 : 4, AB 2 i = AB 2 + BB 2 i . Since BB 1 > BB 2 > BB 3 , it follows that B (B 3 , B, B 2 ,B 1 ) or B(B, B 3 ,B 2 ,B 1 ).

We get a system of three quadratic equations with three unknowns: AB, BB 3 and

B 3 B 2 (BB 3 being negative if B (B 3 , B, B 1 ,B 2 ), positive otherwise). This can be solved by eliminating AB and then BB 3 . The unique solution ends up being

AB = 12, BB 3 = 5, B 3 B 2 = 4, and consequently, the two ships are closest at 10:20 when they are at a distance

of 12 miles.

2. The sides a , b, c of a triangle ABC with ∠ABC = 2∠BAC satisfy b 2 = a(a + c) (this statement is the lemma in (SL98-7)). Taking into account the remaining condition that a , b, c are consecutive integers with a < b, we obtain three cases:

(i) a = n, b = n+ 1, c = n + 2. We get the equation (n + 1) 2 = n(2n +2), giving us (a, b, c) = (1, 2, 3), which is not a valid triangle. (ii) a = n, b = n + 2, c = n + 1. We get (n +2) 2 = n(2n + 1) ⇒ (n−4)(n+1) =

0, giving us the triangle (a, b, c) = (4, 6, 5). (iii) a = n + 1, b = n + 2, c = n. We get (n + 2) 2 = (n + 1)(2n + 1) ⇒ n 2 −n−

3 = 0, which has no positive integer solutions for n. Hence, the only solution is the triangle with sides of lengths 4, 5, and 6.

3. A triangle cannot be formed out of three lengths if and only if one of them is larger than the sum of the other two. Let us assume this is the case for all triplets of edges out of each vertex in a tetrahedron ABCD. Let w.l.o.g. AB be the largest edge of the tetrahedron. Then AB ≥ AC + AD and AB ≥ BC + BD, from which it follows that 2AB ≥ AC + AD + BC + BD. This implies that either AB ≥ AC + BC or AB ≥ AD + BD, contradicting the triangle inequality. Hence the three edges coming out of at least one of the vertices A and B form a triangle.

Remark. The proof can be generalized to prove that in a polyhedron with only triangular surfaces there is a vertex such that the edges coming out of this vertex form a triangle.

4. We will prove the equivalence in the two directions separately: ( ⇒) Suppose {x 1 ,...,x n } is the unique solution of the equation. Since {x n ,x 1 , x 2 , ...,x n −1 } is also a solution, it follows that x 1 =x 2 = ··· = x n = x and

4.10 Shortlisted Problems 1968 375 the system of equations reduces to a single equation ax 2 + (b − 1)x + c = 0.

For the solution for x to be unique the discriminant (b − 1) 2 − 4ac of this quadratic equation must be 0. ( ⇐) Assume (b − 1) 2 − 4ac = 0. Adding up the equations, we get

∑ f (x i ) = 0, where f (x) = ax 2 + (b − 1)x + c.

i =1

But by the assumed condition, f 2 (x) = a x + b −1 2a . Hence we must have

f (x i ) = 0 for all i, and x

i =− −1 2a , which is indeed a solution.

5. We have h k = r cos( π /k) for all k ∈ N. Using cosx = 1− 2sin 2 (x/2) and cosx =

2 /(1 + tan 2 (x/2))−1 and tanx > x > sinx for all 0 < x < π /2, it suffices to prove

(n + 1) 1 −2

4 (n + 1) −1 2 −n >1 1 + π 2 /(4n 2 )

⇔ 1 + 2n 1 −

1 + π 2 /(4n 2 )

n + π 2 /(4n) n +1

where the last inequality holds because π 2 < 4n. It is also apparent that as n tends to infinity the term in parentheses tends to 0, and hence it is not possible to strengthen the bound. This completes the proof.

6. We define f (x) = a a 1 1 −x + a 2 a a n 2 −x + ··· + a n −x . Let us assume w.l.o.g. a 1 <a 2 < ··· < a n . We note that for all 1 ≤ i < n the function f is continuous in the interval

(a i ,a i +1 ) and satisfies lim x →a i f (x) = −∞ and lim x →a i +1 f (x) = ∞. Hence the equation f (x) = n will have a real solution in each of the n−1 intervals (a i ,a i +1 ).

Remark. In fact, this equation has exactly n solutions, and hence they are all real. Moreover, the solutions are distinct if all a i are of the same sign, since x = 0 is an evident solution.

7. Let r a ,r b ,r c denote the radii of the exscribed circles corresponding to the sides of lengths a, b, c respectively, and R, p and S denote the circumradius, semiperime- ter, and area of the given triangle. It is well-known that r p a (p − a) = r b (p − b) = r c (p − c) = S = √ p (p − a)(p − b)(p − c) = abc √ 4R . Hence, the desired inequality

r a r b r c ≤ 3 8 3 abc reduces to p ≤ 3 2 3 R , which is by the law of sines equivalent to

sin α + sin β + sin γ ≤

This inequality immediately follows from Jensen’s inequality, since the sine is concave on [0, π ]. Equality holds if and only if the triangle is equilateral.

8. Let G be the point such that BCDG is a parallelogram and let H be the midpoint of AG. Obviously HEFD is also a parallelogram, and thus DH = EF = l. If

AD 2 + BC 2 =m 2 is fixed, then from the Stewart theorem we have

4 4 which is fixed.

Thus G and H are fixed points, and from here the locus of D is a circle with center H and radius l. The locus of B is the segment (GI], where I ∈ ∆ is a point in the positive direction such that AI = a. Finally, the locus of C is a region of the plane consisting of a rectangle sandwiched between two semicircles of radius l centered at points H and H ′ , where H ′

−→ is a point such that IH ′ −→ = GH .

9. We note that S a = ad a /2, S b = bd b /2, and S c = cd c /2 are the areas of the tri- angles MBC, MCA, and MAB respectively. The desired inequality now follows from

S a S b +S b S

c +S c S a ≤ (S 3 a +S b +S c ) = .

3 Equality holds if and only if S a =S b =S c , which is equivalent to M being the

centroid of the triangle. √

10. (a) Let us set k = a/b > 1. Then a = kb and c = kb √ , and a > c > b. The segments a , b, c form a triangle if and only if k < √ k + 1, which holds if

and only if 1 <k< 3 + 5 2 .

(b) The triangle is right-angled if and only if a 2 =b 2 +c 2 ⇔k 2 =k+1⇔k=

2 . Also, it is acute-angled if and only if k 2 1 + √ 5 <k+1⇔1<k< 2

and obtuse-angled if 1 +

5 <k< 3 + 5

11. Introducing y i = 1 x i , we transform our equation to

0 =1+y 1 + (1 + y 1 )y 2 + ··· + (1 + y 1 ) ··· (1 + y n −1 )y n

= (1 + y 1 )(1 + y 2 ) ···(1 + y n ). The solutions are n-tuples (y 1 ,...,y n ) with y i 6= 0 for all i and y j = −1 for at least

one index j. Returning to x i , we conclude that the solutions are all the n-tuples (x 1 ,...,x n ) with x i 6= 0 for all i, and x j = −1 for at least one index j.

12. The given inequality is equivalent to (a + b) m /b m + (a + b) m /a m m ≥2 +1 , which can be rewritten as

a +b

Since f (x) = 1/x m is a convex function for every m ∈ Z, the last inequality immediately follows from Jensen’s inequality ( f (a) + f (b))/2 ≥ f ((a + b)/2).

13. Translating one of the triangles if necessary, we may assume w.l.o.g. that

B 1 ≡A 1 . We also assume that B 2 6≡ A 2 and B 3 6≡ A 3 , since the result is obvi- ous otherwise. There exists a plane π through A 1 that is parallel to both A 2 B 2 and A 3 B 3 . Let

A ′ ′ ′ ′ 2 ,A 3 ,B 2 ,B 3 denote the orthogonal projections of A 2 ,A 3 ,B 2 ,B 3 onto π , and let

h 2 ,h 3 denote the distances of A 2 ,B 2 and of A 3 ,B 3 from π . By the Pythagorean theorem, A ′ A ′ 2 2 2 3 =A 2 A 2 3 −(h 2 +h 3 ) 2 =B 2 B 2 3 −(h 2 +h 3 ) 2 =B ′ 2 B ′ 3 , and similarly

4.10 Shortlisted Problems 1968 377

A 1 A ′ =A 1 B ′ and A 1 A 2 ′ 2 3 =A 1 B ′ 3 ; hence △A 1 A ′ 2 A ′ 3 and △A 1 B ′ 2 B ′ 3 are congruent. If these two triangles are equally oriented, then we have finished. Otherwise, they are symmetric with respect to some line a passing through A 1 , and consequently the projections of the triangles A 1 A 2 A 3 and A 1 B 2 B 3 onto the plane through a perpendicular to π coincide.

14. Let O , D, E be the circumcenter of △ABC and the midpoints of AB and AC, and given arbitrary X ∈ AB and Y ∈ AC such that BX = CY , let O 1 ,D 1 ,E 1 be the circumcenter of △AXY and the midpoints of AX and AY , respectively. Since AD = AB/2 and AD 1 = AX/2, it follows that DD 1 = BX/2 and similarly EE 1 = CY /2. Hence O 1 is at the same distance BX /2 = CY /2 from the lines OD and OE and lies on the half-line bisector l of ∠DOE. If we let X ,Y vary along the segments AB and AC, we obtain that the locus of O 1 is the segment OP, where P ∈ l is a point at distance min(AB,AC)/2 from OD and OE.

2 i +1 We prove by induction that f (n) = n. This obviously holds for n = 1. Let us

assume that f (n − 1) = n − 1. Define

. We have that f (n) − f (n + 1) = ∑ ∞ i =0 g (i, n). We also note that g(i, n) = 1 if

2 i +1

2 i +1

and only if 2 i +1 |n+2 i ; otherwise, g (i, n) = 0. The divisibility 2 i +1 |n+2 i is equivalent to 2 i | n and 2 i +1 ∤ n, which for a given n holds for exactly one i ∈ N 0 . Thus it follows that f (n) − f (n − 1) = 1 ⇒ f (n) = n. The proof by induction is now complete.

Second solution. It is easy to show that [x + 1/2] = [2x] − [x] for x ∈ R. Now

f (x) = ([x] − [x/2]) + ([x/2] − [x/4])+ ··· = [x]. Hence, f (n) = n for all n ∈ N.

16. We shall prove the result by induction on k. It trivially holds for k = 0. Assume that the statement is true for some k − 1, and let p(x) be a polynomial of degree

k . Let us set p 1 (x) = p(x + 1) − p(x). Then p 1 (x) is a polynomial of degree k − 1 with leading coefficient ka 0 . Also, m |p 1 (x) for all x ∈ Z and hence by the inductive assumption m | (k − 1)! · ka 0 = k!a 0 , which completes the induction. On the other hand, for any a 0 , k and m | k!a 0 ,p (x) = k!a x 0 k is a polynomial

with leading coefficient a 0 that is divisible by m.

17. Let there be given an equilateral triangle ABC and a point O such that OA = x, OB = y, OC = z. Let X be the point in the plane such that △CXB and △COA are congruent and equally oriented. Then BX = x and the triangle XOC is equilateral, which implies OX = z. Thus we have a triangle OBX with BX = x, BO = y, and OX = z. Conversely, given a triangle OBX with BX = x, BO = y and OX = z it is easy to construct the triangle ABC.

378 4 Solutions

18. The required construction is not feasible. In fact, let us consider the special case ∠BOC = 135 ◦ , ∠AOC = 120 ◦ , ∠AOB = 90 ◦ , where AA ′ ∩ BB ′ ∩ CC ′ = {O}. Denoting OA ′ , OB ′ , OC ′ by a , b, c respectively we obtain the system of equations

√ + 2bc .

a 2 +b 2 =a 2 +c 2 + ac = b 2 +c 2

Assuming w.l.o.g. c = 1 we easily obtain a 3 −a 2 − a − 1 = 0, which is an ir- reducible equation of third degree. By a known theorem, its solution a is not constructible by ruler and compass.

19. We shall denote by d n the shortest curved distance from the initial point to the n th point in the positive direction. The sequence d n goes as follows: 0, 1, 2, 3, 4,

5, 6, 0 .72, 1.72, . . . , 5.72, 0.43, 1.43, . . . , 5.43, 0.15 = d 19 . Hence the required number of points is 20.

20. Let us denote the points A 1 ,A 2 ,...,A n in such a manner that A 1 A n is a diameter of the set of given points, and A 1 A 2 ≤A 1 A 3 ≤ ··· ≤ A 1 A n . Since for each 1 < i < n it holds that A 1 A i <A 1 A n , we have ∠A i A 1 A n < 120 ◦ and hence ∠A i A 1 A n < 60 ◦ (otherwise, all angles in △A 1 A i A n are less than 120 ◦ ). It follows that for all 1 < i < j ≤ n, ∠A i A 1 A j < 120 ◦ . Consequently, the angle in the triangle A 1 A i A j that is at least 120 ◦ must be ∠A 1 A i A j . Moreover, for any 1 <

i < j < k ≤ n it holds that ∠A i A j A k ≥ ∠A 1 A j A k − ∠A 1 A j A i > 120 ◦ − 60 ◦ = 60 ◦ (because ∠A 1 A j A i < 60 ◦ ); hence ∠A i A j A k ≥ 120 ◦ . This proves that the denota- tion is correct.

Remark. It is easy to show that the diameter is unique. Hence the denotation is also unique.

21. The given conditions are equivalent to y −a 0 being divisible by a 0 ,a 0 +a 1 ,a 0 +

a 2 ,...,a 0 +a n , i.e., to y = k[a 0 ,a 0 +a 1 ,...,a 0 +a n ]+a 0 ,k ∈N 0 .

22. It can be shown by induction on the number of digits of x that p (x) ≤ x for all x ∈ N. It follows that x 2 − 10x − 22 ≤ x, which implies x ≤ 12. Since 0 < x 2 − 10x − 22 = (x − 12)(x + 2) + 2, one easily obtains x ≥ 12. Now one can directly check that x = 12 is indeed a solution, and thus the only one.

23. We may assume w.l.o.g. that in all the factors the coefficient of x is 1. Suppose that x + ay + bz is one of the linear factors of p(x, y, z) = x 3 +y 3 +z 3 + mxyz. Then p

(x) is 0 at every point (x, y, z) with z = −ax − by. Hence x 3 +y 3 + (−ax − by ) 3 3 + mxy(−ax − by) = (1 − a 3 )x 3 − (3ab + m)(ax + by)xy + (1 − b 3 )y ≡ 0.

This is obviously equivalent to a 3 =b 3 = 1 and m = −3ab, from which it follows √ that m

2 . Conversely, for each of the three possible values for m there are exactly three possibilities (a, b). Hence −3,−3 2 ω , −3 ω are the desired values.

2 ∈ {−3,−3 3 ω , −3 ω }, where ω = −1+i

24. If the ith digit is 0, then the result is

9 k −j 9! (10− j)! , if i > k − j,

9 k − j−1 9!

(9− j)! , otherwise

4.10 Shortlisted Problems 1968 379 If the ith digit is not 0, then the above results are multiplied by 8.

25. The answer is

+n q .

1 ≤p<q<r≤k

1 ≤p<q≤k

26. (a) We shall show that the period of f is 2a. From ( f (x + a) − 1/2) 2 = f (x) −

f (x) 2 we obtain

f 2 2 (x) − f (x) 1 + f (x + a) − f (x + a) = .

4 Subtracting the above relation for x

+ a in place of x we get f (x) − f (x) 2 =

2 (x+2a)− f (x+2a) 2 ( f (x) − 1/2) = ( f (x + 2a) − 1/2) . Since f (x) ≥ 1/2 holds for all x by the condition of the problem, we con-

f 2 , which implies

clude that f (x + 2a) = f (x). (b) The following function, as is directly verified, satisfies the conditions:

f (x) = /2 if 2n ≤ x < 2n + 1,

for n

1 if 2n

+ 1 ≤ x < 2n + 2,

380 4 Solutions