Solutions to the Shortlisted Problems of IMO 1972

4.14 Solutions to the Shortlisted Problems of IMO 1972

1. Suppose that f (x 0 ) 6= 0 and for a given y define the sequence x k by the formula

k + y,

if | f (x k + y)| ≥ | f (x k − y)|;

It follows from (1) that | f (x k +1 )| ≥ | ϕ (y)|| f (x k )|; hence by induction, | f (x k )| ≥

| ϕ (y)| k | f (x 0 )|. Since | f (x k )| ≤ 1 for all k, we obtain | ϕ (y)| ≤ 1. Second solution. Let M = sup | f (x)| ≤ 1, and x k any sequence, possibly constant,

such that | f (x k )| → M, k → ∞. Then for all k, ϕ

| f (x k + y) + f (x k − y)|

2M

| (y)| = k

2 | f (x ≤ k )| 2 → 1, → ∞.

| f (x k )|

2. We use induction. For n = 1 the assertion is obvious. Assume that it is true for

a positive integer n. Let A 1 ,A 2 ,...,A 3n +3

be given 3n + 3 points, and let w.l.o.g.

be their convex hull. Among all the points A i distinct from A 1 ,A 2 , we choose the one, say A k , for which the angle ∠A k A 1 A 2 is minimal (this point is uniquely determined, since no three points are collinear). The line A 1 A k separates the plane into two half- planes, one of which contains A 2 only, and the other one all the remaining 3n points. By the inductive hypothesis, one can construct n disjoint triangles with vertices in these 3n points. Together with the triangle A 1 A 2 A k , they form the required system of disjoint triangles.

A 1 A 2 ...A m

3. We have for each k = 1, 2, . . . , n that m ≤ x k ≤ M, which gives (M −x k )(m−x k )≤

0. It follows directly that

nnn

0 ≥ 2 ∑ (M − x k )(m − x k ) = nmM − (m + M) ∑ x k + ∑ x k .

k =1 k =1 k =1

But ∑ n k =1 x k = 0, implying the required inequality.

4. Choose in E a half-line s beginning at a point O. For every α in the interval [0, 180 ◦ ], denote by s( α ) the line obtained by rotation of s about O by α , and by g ( α ) the oriented line containing s( α ) on which s( α ) defines the positive direction. For each P in M i ,i = 1, 2, let P( α ) be the foot of the perpendicular from P to g ( α ), and l P ( α ) the oriented (positive, negative or zero) distance of P ( α ) from O. Then for i = 1, 2 one can arrange the l P ( α ) (P ∈ M i ) in ascending

order, as l 1 ( α ), l 2 ( α ), . . . , l 2n i ( α ). Call J i ( α ) the interval [l n i ( α ), l n i +1 ( α )]. It is easy to see that any line perpendicular to g ( α ) and passing through the point with the distance l in the interior of J i ( α ) from O, will divide the set M i into two subsets of equal cardinality.

Therefore it remains to show that for some α , the interiors of intervals J 1 ( α ) and J 2 ( α ) have a common point. If this holds for α = 0, then we have finished. Suppose w.l.o.g. that J 1 (0) lies on g(0) to the left of J 2 (0); then J 1 (180 ◦ ) lies to

4.14 Shortlisted Problems 1972 397 the right of J 2 (180 ◦ ). Note that J 1 and J 2 cannot simultaneously degenerate to a

point (otherwise, we would have four collinear points in M 1 ∪M 2 ); also, each of them degenerates to a point for only finitely many values of α . Since J 1 ( α ) and J 2 ( α ) move continuously, there exists a subinterval I of [0, 180 ◦ ] on which they are not disjoint. Thus, at some point of I, they are both nondegenerate and have

a common interior point, as desired.

5. Lemma. If X ,Y, Z, T are points in space, then the lines XZ and Y T are perpen-

dicular if and only if XY 2 + ZT 2 =YZ 2 +TX 2 .

Proof. Consider the plane π through X Z parallel to Y T . If Y ′ ,T ′ are the feet of the perpendiculars to π from Y , T respectively, then

XY 2 + ZT 2 = XY ′2 + ZT ′2 + 2YY ′2 , and

YZ 2 +TX 2 =Y ′ Z 2 +T ′ X 2 + 2YY ′2 . Since by the Pythagorean theorem XY ′2 + ZT ′2 =Y ′ Z 2 +T ′ X 2 , i.e., XY ′2 −

Y ′ Z 2 = XT ′2 −T ′ Z 2 , if and only if Y ′ T ′ ⊥ XZ, the statement follows. Assume that the four altitudes intersect in a point P. Then we have DP ⊥ ABC ⇒ DP ⊥ AB and CP ⊥ ABD ⇒ CP ⊥ AB, which implies that CDP ⊥ AB, and CD ⊥

AB . By the lemma, AC 2 + BD 2 = AD 2 + BC 2 . Using the same procedure we

obtain the relation AD 2 + BC 2 = AB 2 + CD 2 .

Conversely, assume that AB 2 + CD 2 = AC 2 + BD 2 = AD 2 + BC 2 . The lemma implies that AB ⊥ CD, AC ⊥ BD, AD ⊥ BC. Let π

be the plane containing CD that is perpendicular to AB, and let h D be the altitude from D to ABC. Since π ⊥ AB, we have π ⊥ ABC ⇒ h D ⊂ π and π ⊥ ABD ⇒ h C ⊂ π . The altitudes h D and h C are not parallel; thus they have an intersection point P CD . Analogously,

h B ∩h C = {P BC } and h B ∩h D = {P BD }, where both these points belong to π . On the other hand, h B doesn’t belong to π ; otherwise, it would be perpendicular to both ACD and AB ⊂ π , i.e. AB ⊂ ACD, which is impossible. Hence, h B can have at most one common point with π , implying P BD =P CD . Analogously, P AB = P BD =P CD =P ABCD .

6. Let n =2 α 5 β m , where α = 0 or β = 0. These two cases are analogous, and we treat only α = 0, n = 5 β m . The case m = 1 is settled by the following lemma. Lemma. For any integer β ≥ 1 there exists a multiple M β of 5 β with β digits in decimal expansion, all different from 0. Proof. For β = 1, M 1 = 5 works. Assume that the lemma is true for β = k. There is a positive integer C

k ≤ 5 such that C k 2 +m k ≡ 0 (mod 5), where 5 m k = M k , i.e. C k 10 k +M k ≡ 0 (mod 5 k +1 ). Then M k +1 =C k 10 k +M k satisfies the

conditions, and proves the lemma. In the general case, consider, the sequence 1 , 10 β , 10 2 β , . . . . It contains two numbers congruent modulo (10 β − 1)m, and therefore for some k > 0, 10 k β ≡

1 (mod (10 β − 1)m) (this is in fact a consequence of Fermat’s theorem). The number

10 k β −1

M β + ··· + M β −1

M β = 10 (k−1)β M

(k−2)β

398 4 Solutions is a multiple of n =5 β m with the required property.

7. (a) Consider the circumscribing cube R P 1 OQ 1 PR 1 O 1 QP 1 R (that is, the cube

O 1 Q in which the edges of the tetra-

hedron are small diagonals), of √

side b =a 2 /2. The left-hand

side is the sum of squares of the projections of the edges of the

R 1 P tetrahedron onto a perpendicular

l to π . On the other hand, if l forms angles ϕ 1 , ϕ 2 , ϕ 3 with OO 1 , OQ 1 , OR 1 respectively, then the projec- tions of OP and QR onto l have lengths b (cos ϕ 2 + cos ϕ 3 ) and b|cos ϕ 2 − cos ϕ 3 |. Summing up all these expressions, we obtain

4b 2 (cos 2 ϕ 1 + cos 2 ϕ 2 + cos 2 ϕ 3 ) = 4b 2 = 2a 2 . (b) We construct a required tetrahedron of edge length a given in (a). Take O

arbitrarily on π 0 , and let p , q, r be the distances of O from π 1 , π 2 , π 3 . Since

a > p, q, r, |p − q|, we can choose P on π 1 anywhere at distance a from O , and Q at one of the two points on π 2 at distance a from both O and P . Consider the fourth vertex of the tetrahedron: its distance from π 0 will satisfy the equation from (a); i.e., there are two values for this distance;

clearly, one of them is r, putting R on π 3 .

8. Let f (m, n) = (2m)!(2n)! m !n! (m+n)! . Then it is directly shown that

f (m, n) = 4 f (m, n − 1) − f (m + 1,n − 1), and thus n may be successively reduced until one obtains f (m, n) = ∑ r c r f (r, 0).

Now f (r, 0) is a simple binomial coefficient, and the c r ’s are integers. Second solution. For each prime p, the greatest exponents of p that divide the

numerator (2m)!(2n)! and denominator m!n!(m + n)! are respectively 2m

m ∑ +n

hence it suffices to show that the first exponent is not less than the second one for every p. This follows from the fact that for each real x, [2x] + [2y] ≥ [x] + [y] + [x + y], which is straightforward to prove (for example, using [2x] = [x] + [x + 1/2]).

9. Clearly x 1 =x 2 =x 3 =x 4 =x 5 is a solution. We shall show that this describes all solutions. Suppose that not all x i are equal. Then among x 3 ,x 5 ,x 2 ,x 4 ,x 1 two consecutive are distinct: Assume w.l.o.g. that x 3 6= x 5 . Moreover, since (1/x 1 , . . . , 1/x 5 ) is a solution whenever (x 1 ,...,x 5 ) is, we may assume that x 3 <x 5 .

4.14 Shortlisted Problems 1972 399

1 ≤ x 3 x 5 <x 5 and x 2 ≥ √ x

Consider first the case x 1 ≤x 2 . We infer from (i) that x

3 x 5 >x 3 . Then x 2 5 >x 1 x 3 , which together with (iv) gives x 2 4 ≤x 1 x 3 <x 3 x 5 ; but we also have x 2 3 ≤x 5 x 2 ; hence by (iii), x 2 4 ≥x 5 x 2 >x 5 x 3 , a contradiction. Consider next the case x

1 >x 2 . We infer from (i) that x 1 ≥ x 3 x 5 >x 3 and x

2 ≤ x 3 x 5 <x 5 . Then by (ii) and (v),

x 1 x 4 ≤ max(x 2 2 ,x 2 3 )≤x 3 x 5 and

x 2 x 4 ≥ min(x 2 1 ,x 2 5 )≥x 3 x 5 ,

which contradicts the assumption x 1 >x 2 .

Second solution.

0 ≥L 1 = (x 2 1 −x 3 x 5 )(x 2 2 −x 3 x 5 )=x 2 1 x 2 2 2 +x 2 3 x 2 5 2 − (x 1 +x 2 )x 3 x 5

2 2 2 2 1 2 2 2 2 2 2 2 2 ≥x 1 x 2 +x 3 x 5 − (x x 3 +x 1 x 5 +x 2 x 3 2 +x 1 2 x 5 ),

and analogously for L 2 ,...,L 5 . Therefore L 1 +L 2 +L 3 +L 4 +L 5 ≥ 0, with the

only case of equality x 1 =x 2 =x 3 =x 4 =x 5 .

10. Consider first a triangle. It can be decomposed into k = 3 cyclic quadrilaterals by perpendiculars from some interior point of it to the sides; also, it can be decom- posed into a cyclic quadrilateral and a triangle, and it follows by induction that this decomposition is possible for every k. Since every triangle can be cut into two triangles, the required decomposition is possible for each n ≥ 6. It remains to treat the cases n = 4 and n = 5.

n = 4. If the center O of the circumcircle is inside a cyclic quadrilateral ABCD, then the required decomposition is effected by perpendiculars from O to the four sides. Otherwise, let C and D be the vertices of the obtuse angles of the quadrilateral. Draw the perpendiculars at C and D to the lines BC and AD respectively, and choose points P and Q on them such that PQ k

AB . Then the required decomposition is effected by CP , PQ, QD and the perpendiculars from P and Q to AB. n = 5. If ABCD is an isosceles trapezoid with AB k CD and AD = BC, then it is trivially decomposed by lines parallel to AB. Otherwise, ABCD can be decomposed into a cyclic quadrilateral and a trapezoid; this trapezoid can

be cut into an isosceles trapezoid and a triangle, which can further be cut into three cyclic quadrilaterals and an isosceles trapezoid.

Remark. It can be shown that the assertion is not true for n = 2 and n = 3.

11. Let ∠A = 2x, ∠B = 2y, ∠C = 2z. (a) Denote by M i the center of K i ,i = 1, 2, . . . . If N 1 ,N 2 are the projec- tions of M 1 ,M 2 p onto AB, we have AN 1 =r 1 cot x, N 2 B =r 2 cot y, and N

1 N 2 = (r 1 +r 2 ) 2 − (r 1 −r 2 ) 2 =2 r 1 r 2 . The required relation between

r 1 ,r 2 follows from AB = AN 1 +N 1 N 2 +N 2 B .

If this relation is further considered as a quadratic equation in √r 2 , then its discriminant, which equals

∆ = 4 (r(cot x + coty) cot y − r 1 (cotx cot y − 1)),

400 4 Solutions must be nonnegative, and therefore r 1 ≤ r cotycotz. Then t 1 ,t 2 , . . . exist,

and we can assume that t i ∈ [0, π /2]. (b) Substituting r 1 = r cot y cotz sin 2 t 1 ,r 2 = r cot z cot x sin 2 t 2 in the relation of (a) we obtain that sin 2 t 1 + sin 2 t 2 +k 2 + 2k sint 1 sint 2 = 1, where we set k =

√ tan x tan y. It follows that (k + sint 1 sin t 2 ) 2 = (1 − sin 2 t 1 )(1 − sin 2 t 2 )=

cos 2 t 1 cos 2 t 2 , and hence √

cos (t 1 +t 2 ) = cost 1 cost 2 − sint 1 sint 2 =k= tan x tan y , which is constant. Writing the analogous relations for each t i ,t i +1 we con-

clude that t 1 +t 2 =t 4 +t 5 ,t 2 +t 3 =t 5 +t 6 , and t 3 +t 4 =t 6 +t 7 . It follows

that t 1 =t 7 , i.e., K 1 =K 7 .

12. First we observe that it is not essential to require the subsets to be disjoint (if they aren’t, one simply excludes their intersection). There are 2 10 − 1 = 1023 different subsets and at most 990 different sums. By the pigeonhole principle there are two different subsets with equal sums.

4.15 Shortlisted Problems 1973 401