Solutions to the Shortlisted Problems of IMO 1970

4.12 Solutions to the Shortlisted Problems of IMO 1970

1. Denote respectively by R and r the radii of the circumcircle and incircle, by

A 1 ,...,A n ,B 1 ,...,B n ,the vertices of the 2n-gon and by O its center. Let P ′ be the point symmetric to P with respect to O. Then A i P ′ B i P is a parallelogram, and applying cosine theorem on triangles A i B i P and PP ′ B i yields

4R 2 = PA 2 + PB i 2 i − 2PA i · PB i cos a i

4r 2 = PB 2 +P ′ 2 ′

B i − 2PB i ·P B i cos ∠PB i P . Since A i P ′ B i P is a parallelogram, we have that P ′ B i = PA i and ∠PB i P ′ = π −a i .

Subtracting the expression for 4r 2 from the one for 4R 2 yields 4 (R 2 −r 2 )= −4PA i · PB i cos a i = −8S △A i B i P cot a i , hence we conclude that

Denote by M i the foot of the perpendicular from P to A i B i and let m i = PM i . Then S △A i B i P = Rm i . Substituting this into (1) and adding up these relations for

i = 1, 2, . . . , n, we obtain

tan 2 ∑ 4R a i = 2 2 2 ∑ m 2 i .

Note that all the points M i lie on a circle with diameter OP and form a regular n -gon. Denote its center by F. We have that m 2 =k −−→ PM i k 2 i −−→ =k FM −→ i − FP k 2 = −−→

−−→ −→ k FM i 2 k+k FP 2 k − 2h FM i , FP i=r 2 /2 − 2h FM i , FP i. From this it follows that

nn

∑ m 2 i −−→ = 2n(r/2) 2 2 −2 2 ∑ h FM i , −→ FP i = 2n(r/2) − 2h ∑ FM i , −→ FP i = 2n(r/2) ,

n −−→

i =1 i =1 i =1

−−→ → − because ∑ n i =1 FM i = 0 . Thus

4R 2 2 (r/R) 2 cos 2π tan 2 a

∑ 2n

2 (1 − (r/R) ) sin 2n Remark. For n = 1 there is no regular 2-gon. However, if we think of a 2-gon as

i =1

(R −r )

a line segment, the statement will remain true.

2. Suppose that a > b. Consider the polynomial P(X ) = x 1 X n −1 +x 2 X n −2 + ··· + x n −1 X +x n . We have A n = P(a), B n = P(b), A n +1 =x 0 a n + P(a), and B n +1 = x 0 b n + P(b). The inequality A n /A n +1 <B n /B n +1 becomes P (a)/(x 0 a n + P(a)) < P (b)/(x 0 b n + P(b)), i.e.,

b n P (a) < a n P (b).

i b n a ≥x n −i (also, for i ≥1 the inequality is strict). Summing up all these inequalities for i = 1, . . . , n we get

Since a > b, we have that a i >b i and hence x i a n b n −i

a n P (b) > b n P (a), which completes the proof for a > b.

384 4 Solutions On the other hand, for a < b we analogously obtain the opposite inequal-

ity A n /A n +1 >B n /B n +1 , while for a = b we have equality. Thus A n /A n +1 <

B n /B n +1 ⇔ a > b.

3. We shall use the following lemma Lemma. If an altitude of a tetrahedron passes through the orthocenter of the opposite side, then each of the other altitudes possesses the same property. Proof. Denote the tetrahedron by SABC and let a = BC, b = CA, c = AB, m = SA ,n = SB, p = SC. It is enough to prove that an altitude passes through the orthocenter of the opposite side if and only if a 2 +m 2 =b 2 +n 2 =c 2 +p 2 . Suppose that the foot S ′ of the altitude from S is the orthocenter of ABC. Then SS ′ ⊥ ABC ⇒ SB 2 − SC 2 =S ′ B 2 −S ′ C 2 . But from AS ′ ⊥ BC it fol- lows that AB 2 − AC 2 =S ′ B 2 −S ′ C 2 . From these two equalities it can be concluded that n 2 2 2 2 2 2 2 −p 2 =c −b , or equivalently, n +b =c +p . Anal- ogously, a 2 +m 2 =n 2 +b 2 , so we have proved the first part of the equiva- lence. Now suppose that a 2 +m 2 =b 2 +n 2 =c 2 +p 2 . Defining S ′ as before, we get n 2 2 2 2 2 2 2 2 2 −p 2 =S ′ B −S ′ C . From the condition n −p =c −b ( ⇔b +n =

c 2 +p 2 ) we conclude that AS ′ ⊥ BC. In the same way CS ′ ⊥ AB, which proves that S ′ is the orthocenter of △ABC. The lemma is thus proven. Now using the lemma it is easy to see that if one of the angles at S is right, than so are the others. Indeed, suppose that ∠ASB = π /2. From the lemma we have that the altitude from C passes through the orthocenter of △ASB, which is S, so

CS ⊥ ASB and ∠CSA = ∠CSB = π /2. Therefore m 2 +n 2 =c 2 ,n 2 +p 2 =a 2 , and p 2 +m 2 =b 2 , so it follows that m 2 +n 2 +p 2 = (a 2 +b 2 +c 2 )/2. By the inequality between the arithmetic and quadric means, we have that (a 2 +b 2 +c 2 )/2 ≥ 2s 2 /3, where s denotes the semiperimeter of △ABC. It remains to be shown that 2s 2 /3 ≥ 18r 2 . Since S

= sr, this is equivalent to 2s △ABC 4 /3 ≥ 18S 2 ABC = 18s(s − a)(s − b)(s − c) by Heron’s formula. This reduces to s 3 ≥ 27(s − a)(s − b)(s − c), which is an obvious consequence of the AM–GM mean inequality.

Remark. In the place of the lemma one could prove that the opposite edges of the tetrahedron are mutually perpendicular and proceed in the same way.

4. Suppose that n is such a natural number. If a prime number p divides any of the numbers n , n + 1, . . .,n + 5, then it must divide another one of them, so the only possibilities are p = 2, 3, 5. Moreover, n + 1, n + 2, n + 3, n + 4 have no prime divisors other than 2 and 3 (if some prime number greater than 3 divides one of them, then none of the remaining numbers can have that divisor). Since two of these numbers are odd, they must be powers of 3 (greater than 1). However, there are no two powers of 3 whose difference is 2. Therefore there is no such natural number n.

Second solution. Obviously, none of n , n + 1, . . ., n + 5 is divisible by 7; hence they form a reduced system of residues. We deduce that n (n + 1) ···(n + 5) ≡

1 ·2···6 ≡ −1 (mod 7). If {n,...,n + 5} can be partitioned into two subsets with

4.12 Shortlisted Problems 1970 385 the same products, both congruent to, say, p modulo 7, then p 2 ≡ −1 (mod 7),

which is impossible. Remark. Erd ˝os has proved that a set n , n + 1, . . .,n + m of consecutive natural

numbers can never be partitioned into two subsets with equal products of ele- ments.

5. Denote respectively by A 1 ,B 1 ,C 1 and D 1 the points of intersection of the lines AM, BM, CM, and DM with the opposite sides of the tetrahedron. Since

vol(MBCD) = vol(ABCD) MA −−→ −−→ 1 / AA 1 , the relation we have to prove is equivalent to

There exist unique real numbers α , β , γ , and δ such that α + β + γ + δ = 1 and for every point O in space

−−→ OM = α −→ OA + β −→ OB + γ −→ OC + δ −→ OD .

−→ −→ −→ −→ (This follows easily from OM = OA + AM = OA +k AB +l AC +m AD = AB + k −→ −→

) + l( OC − OA ) + m( OD − OA ) for some k, l, m ∈ R.) Further, from the condition that A 1 belongs to the plane BCD we obtain for every O in space the following equality for some β ′ , γ ′ , δ ′ :

OA 1 = β ′ OB + γ ′ OC + δ ′ OD . (3) However, for λ

−−→ = λ OA + (1 − λ ) OA 1 ; hence substituting

= MA

1 / AA 1 ,

(2) and (3) in this expression and equating coefficients for OA −→ we obtain −−→ −−→

−−→ −−→ λ = MA 1 / AA 1 = α . Analogously, β = −−→ MB 1 /−−→ BB 1 , γ = MC 1 / CC 1 , and δ = −−→ MD −−→

1 / DD 1 ; hence (1) follows immediately for O = M. Remark. The statement of the problem actually follows from the fact that M is the center of mass of the system with masses vol(MBCD), vol(MACD), vol(MABD), vol(MABC) at A , B,C, D respectively. Our proof is actually a for- mal verification of this fact.

6. Let F be the midpoint of B ′ C ′ ,A ′ the midpoint of BC, and I the intersection point of the line HF and the circle circumscribed about △BHC ′ . Denote by M the intersection point of the line AA ′ with the circumscribed circle about the

triangle ABC. Triangles HB ′ C ′ and ABC are similar. Since ∠C ′ IF = ∠ABC = ∠A ′ MC , ∠C ′ FI = ∠AA ′ B = ∠MA ′ C , 2C ′ F =C ′ B ′ , and 2A ′ C = CB, it follows that △C ′ IB ′ ∼ △CMB, hence ∠FIB ′ = ∠A ′ MB = ∠ACB. Now one concludes that I belongs to the circumscribed circles of △AB ′ C ′ (since ∠C ′ IB ′ = 180 ◦ − ∠C ′ AB ′ ) and △HCB ′ .

Second Solution. We denote the angles of △ABC by α , β , γ . Evidently △ABC ∼ △HC ′ B ′ . Within △HC ′ B ′ there exists a unique point I such that ∠HIB ′ = 180 ◦ − γ , ∠HIC ′ = 180 ◦ − β , and ∠C ′ IB ′ = 180 ◦ − α , and all three circles must contain

386 4 Solutions this point. Let HI and B ′ C ′ intersect in F. It remains to show that FB ′ = FC ′ .

From ∠HIB ′ + ∠HB ′ F = 180 ◦ we obtain ∠IHB ′ = ∠IB ′ F . Similarly, ∠IHC ′ = ∠IC ′ F . Thus circles around △IHC ′ and △IHB ′ are both tangent to B ′ C ′ , giving us FB ′2 = FI · FH = FC ′2 .

7. For a = 5 one can take n = 10, while for a = 6 one takes n = 11. Now assume

a 6∈ {5,6}. If there exists an integer n such that each digit of n (n + 1)/2 is equal to a, then there is an integer k such that n (n + 1)/2 = (10 k − 1)a/9. After multiplying both

sides of the equation by 72, one obtains 36n 2 + 36n = 8a · 10 k − 8a, which is equivalent to

9 (2n + 1) 2 = 8a · 10 k − 8a + 9. (1) So 8a · 10 k − 8a + 9 is the square of some odd integer. This means that its last

digit is 1, 5, or 9. Therefore a ∈ {1,3,5,6,8}. If a = 3 or a = 8, the number on the RHS of (1) is divisible by 5, but not by 25 (for k ≥ 2), and thus cannot be a square. It remains to check the case a = 1. In that

case, (1) becomes 9(2n + 1) 2 = 8 ·10 k + 1, or equivalently [3(2n +1)−1][3(2n+

1 ) + 1] = 8 ·10 k ⇒ (3n+ 1)(3n+ 2) = 2·10 k . Since the factors 3n + 1, 3n + 2 are relatively prime, this implies that one of them is 2 k +1 and the other one is 5 k . It is directly checked that their difference really equals 1 only for k = 1 and n = 1, which is excluded. Hence, the desired n exists only for a ∈ {5,6}.

8. Let AC = b, BC = a, AM = x, BM = y,CM = l. Denote by I 1 the incenter and by S 1 the center of the excircle of ∆ AMC . Suppose that P 1 and Q 1 are feet of perpen- diculars from I 1 and S 1 , respectively, to the line AC. Then △I 1 CP 1 ∼ △S 1 CQ 1 , hence r 1 / ρ 1 = CP 1 /CQ 1 . We have CP 1 = (AC + MC − AM)/2 = (b + l − x)/2 and CQ 1 = (AC + MC + AM)/2 = (b + l + x)/2. Hence

r 1 b +l−x

ρ 1 b +l+x

We similarly obtain

r 2 b +l−y

a +b−x−y

and

a +b+x+y What we have to prove is now equivalent to (b + l − x)(a + l − y)

ρ 2 b +l+y

a = +b−x−y . (1) (b + l + x)(a + l + y)

a +b+x+y Multiplying both sides of (1) by (a + l + y)(b + l + x)(a + b + x + y) we obtain an

expression that reduces to l 2 x +l 2 y +x 2 y + xy 2 =b 2 y +a 2 x . Dividing both sides by c = x + y, we get that (1) is equivalent to l 2 =b 2 y /(x + y)+ a 2 x /(x + y) − xy,

which is exactly Stewart’s theorem for l. This finally proves the desired result.

9. Let us set a = ∑ n i =1 u 2 i and b = ∑ n i =1 v 2 i . By Minkowski’s inequality (for p =

2) we have ∑ n i =1 (u i +v i ) 2 ≤ (a + b) 2 . Hence the LHS of the desired inequality

4.12 Shortlisted Problems 1970 387 is not greater than 1 + (a + b) 2 , while the RHS is equal to 4 (1 + a 2 )(1 + b 2 )/3.

Now it is sufficient to prove that

3 + 3(a + b) 2 2 )(1 + b ≤ 4(1 + a 2 ). The last inequality can be reduced to the trivial 0 ≤ (a − b) 2 + (2ab − 1) 2 . The

equality in the initial inequality holds if and only if u √ i /v i = c for some c ∈ R and

a = b = 1/ 2.

10. (a) Since a n −1 <a n , we have

a k Summing up all these inequalities for k = 1, 2, . . . , n we obtain

(b) Choose a real number q > 1, and let a k =q k ,k = 1, 2, . . . . Then we deduce (1 − a k −1 /a k ) /√a k = (1 − 1/q)/q k /2 , and consequently

1 n 1 √ q +1

q ∑ k =1 q k /2

q n /2

Since (√q + 1)/q can be arbitrarily close to 2, one can set q such that (√q + 1)/q > b. Then b n ≥ b for all sufficiently large n.

Second solution. (a) Note that

b = ∑ 1 n −1 −

hence b n represents exactly the lower Darboux sum for the function f R R (x) = x −3/2 on the interval

x a −3/2 0 dx < 1 x −3/2 dx = 2. (b) For each b

[a +∞

0 ,a n ]. Then b n ≤

< 2 there exists a number α > 1 such that α −3/2 1 x dx >b+ (2 − b)/2. Now, by Darboux’s theorem, there exists a sequence 1 = a 0 ≤

a 1 ≤ ··· ≤ a n = α such that the corresponding Darboux sums are arbitrarily close to the value of the integral. In particular, there is a sequence a 0 ,...,a n with b n > b.

11. Let S

1 )(x−x 2 ) ···(x−x n ). We have x −x i = (x−x i )( ω x −x i )( ω 2 x − x i ), where ω is a primitive third root of 1. Multiplying these equalities for i =

3 (x) = (x−x 3

1 , . . . , n we obtain

388 4 Solutions

T (x 3 ) = (x 3 −x 3 1 )(x 3 −x 3 2 ) ··· (x 3 −x 3 n ) = S(x)S( ω x )S( ω 2 x ). Since S ( ω x ) = P(x 3 )+ ω xQ (x 3 )+ ω 2 x 2 R (x 3 ) and S( ω 2 x ) = P(x 3 )+ ω 2 xQ (x 3 )+

ω x 2 R (x 3 ), the above expression reduces to T (x 3 )=P 3 (x 3 )+x 3 Q 3 (x 3 )+x 6 R 3 (x 3 ) − 3P(x 3 )Q(x 3 )R(x 3 ). Therefore the zeros of the polynomial T (x) = P 3 (x) + xQ 3 (x) + x 2 R 3 (x) − 3P(x)Q(x)R(x) are exactly x 3 1 ,...,x 3 n . It is easily verified that deg T = deg S = n, and hence T is

the desired polynomial.

12. Lemma. Five points are given in the plane such that no three of them are collinear. Then there are at least three triangles with vertices at these points that are not acute-angled.

Proof. We consider three cases, according to whether the convex hull of these points is a triangle, quadrilateral, or pentagon. (i) Let a triangle ABC be the convex hull and two other points D and E lie inside the triangle. At least two of the triangles ADB , BDC and CDA have obtuse angles at the point D. Similarly, at least two of the triangles AEB , BEC and CEA are obtuse-angled. Thus there are at least four non- acute-angled triangles.

(ii) Suppose that ABCD is the convex hull and that E is a point of its in- terior. At least one angle of the quadrilateral is not acute, determining one non-acute-angled triangle. Also, the point E lies in the interior of either △ABC or △CDA; hence, as in the previous case, it determines another two obtuse-angled triangles.

(iii) It is easy to see that at least two of the angles of the pentagon are not acute. We may assume that these two angles are among the angles corresponding to vertices A, B, and C. Now consider the quadrilateral ACDE . At least one of its angles is not acute. Hence, there are at least three triangles that are not acute-angled.

Now we consider all combinations of 5 points chosen from the given 100. There are 100 5 such combinations, and for each of them there are at least three non- acute-angled triangles with vertices in it. On the other hand, vertices of each of the triangles are counted 97 2 97 times. Hence there are at least 3 100 5 / 2 non- acute-angled triangles with vertices in the given 100 points. Since the number of all triangles with vertices in the given points is 100 3 , the ratio between the num- ber of acute-angled triangles and the number of all triangles cannot be greater than

4.13 Shortlisted Problems 1971 389