Solutions to the Contest Problems of IMO 1969

4.11 Solutions to the Contest Problems of IMO 1969

1. Set a = 4m 4 , where m ∈ N and m > 1. We then have z = n 4 + 4m 4 = (n 2 + 2m 2 ) 2 − (2mn) 2 = (n 2 + 2m 2 + 2mn)(n 2 + 2m 2 − 2mn). Since n 2 + 2m 2 − 2mn = (n −m) 2 +m 2 ≥m 2 > 1, it follows that z must be composite. Thus we have found infinitely many a that satisfy the condition of the problem.

2. Using cos (a + x) = cos a cos x − sinasinx, we obtain y(x) = Asinx + Bcosx where A = − sina 1 − sin a 2 /2 − ··· − sin a n /2 n −1 and B = cos a 1 + cos a 2 /2 + ··· + cosa n /2 n −1 . Numbers A and B cannot both be equal to 0, for other- wise y would be identically equal to 0, while on the other hand, we have

y (−a 1 ) = cos(a 1 −a 1 ) + cos(a 2 −a 1 )/2 + ··· + cos(a n −a 1 )/2 n −1 ≥ 1 − 1/2 − ··· − 1/2 n −1 = 1/2 n −1 > 0. Setting A = C cos φ and B = C sin φ , where C 6= 0 (such C and φ always exist), we get y (x) = C sin(x + φ ). It follows that the zeros

of y are of the form x 0 ∈− φ + π Z, from which y(x 1 ) = y(x 2 )⇒x 1 −x 2 =m π immediately follows.

3. We have several cases:

1 ◦ k = 1. W.l.o.g. let AB = a and the remaining segments have length 1. Let M √

be the midpoint of CD. Then AM = BM = √ 3 /2 (△CDA and △CDB are √ equilateral) and 0 < AB < AM + BM =

3. It is evident that all values of a within this interval are realizable.

3, i.e., 0 <a<

2 ◦ k = 2. We have two subcases. First, let AC = AD = a. Let M be the midpoint of CD. We have CD = 1, p AM = a 2

√ − 1/4, and BM = 3 /2. Then we have 1 − 3 /2 = AB−BM <

√ p √ AM < AB + BM = 1 + 3 /2, which gives us 2 − 3 <a< 2 + 3.

Second, let AB √ = CD = a. Let M be the midpoint of CD. From △MAB we get a < 2. p

Thus, from 2 − 3 < 2 < 2 + p 3 it follows that the required con- √

3. All values for a in this range are realizable.

dition in this case is 0 <a< 2 +

3 ◦ k = 3. We show that such a tetrahedron exists for all a. Assume a > 1. Assume AB = AC = AD = a. Varying A along the line perpendicular to the plane BCD and through the center of √ √ △BCD we achieve all values of

a > 1/

3. For a ≤ 1/

3 we can observe a similar tetrahedron with three

edges of length 1 /a and three of length 1 and proceed as before.

4 ◦ k = 4. By observing the similar tetrahedron we reduce this case to k = 2 p √ with length 1 /a instead of a. Thus we get a > 2 − 3.

k = 5. We reduce to k = 1 and get a > 1/ 3.

4. Let O be the midpoint of AB, i.e., the center of γ . Let O 1 ,O 2 , and O 3 respec- tively be the centers of γ 1 , γ 2 , and γ 3 and let r 1 ,r 2 ,r 3 respectively be the radii of γ 1 , γ 2 and γ 3 . Let C 1 ,C 2 , and C 3 respectively be the points of tangency of γ 1 , γ 2 and γ 3 with AB. Let D 2 and D 3 respectively be the points of tan- gency of γ 2 and γ 3 with CD. Finally, let G 2 and G 3 respectively be the points of tangency of γ 2 and γ 3 with γ . We have B (G 2 ,O 2 , O), G 2 O 2 =O 2 D 2 , and

4.11 Contest Problems 1969 381

G 2 O = OB. Hence, G 2 ,D 2 , B are collinear. Similarly, G 3 ,D 3 , A are collinear. It follows that AG 2 D 2 D and BG 3 D 3 D are cyclic, since ∠AG 2 D 2 = ∠D 2 DA = ∠D 3 DB = ∠BG 3 D 3 = 90 ◦ . Hence BC 2 2 = BD 2 · BG 2 = BD ·BA = BC 2 ⇒ BC 2 = BC and hence AC 2 = AB − BC. Similarly, AC 3 = AC. We thus have AC 1 =

(AC + AB − BC)/2 = (AC 3 + AC 2 )/2. Hence, C 1 is the midpoint of C 2 C 3 . We also have r 2 +r 3 =C 2 C 3 = AC + BC − AB = 2r 1 , from which it follows that O 1 ,O 2 ,O 3 are collinear.

Second solution. We shall prove the statement for arbitrary points A , B,C on γ . Let us apply the inversion ψ with respect to the circle γ 1 . We denote by b X the

image of an object X under ψ . Also, ψ maps lines BC ,CA, AB onto circles b a ,b b , bc, respectively. Circles b a ,b b , c b pass through the center O 1 of γ 1 and have radii equal to the radius of γ b . Let P , Q, R be the centers of a b ,b b , bcrespectively. The line CD maps onto a circle k through b C and O 1 that is perpendicular to bc. Therefore its center K lies in the intersection of the tangent t to

bc and the line PQ (which bisects b CO 1 ). Let O be a point such that RO 1 KO is a parallelogram and γ ′ , γ ′ the circles centered at O tangent to k. It is easy to see that γ ′ 2 3 2 and γ ′ 3

are also tangent to

bc, since OR and OK have lengths equal to the radii of k and

bc. Hence γ ′ 2 and γ ′ 3 are the images of γ 2 and γ 3 under ψ . Moreover, since Q b AOK and P b BOK are parallelograms and Q , P, K are collinear, it follows that b A ,b B , O are also collinear. Hence the centers of γ 1 , γ 2 , γ 3 are collinear, lying on the line O 1 O , and the statement follows.

Third solution. Moreover, the statement holds for an arbitrary point D ∈ BC. Let

E , F, G, H be the points of tangency of γ 2 with AB ,CD and of γ 3 with AB ,CD, respectively. Let O i

be the center of γ i ,i = 1, 2, 3. As is shown in the third so- lution of (SL93-3), EF and GH meet at O 1 . Hence the problem of proving the collinearity of O 1 ,O 2 ,O 3 reduces to the following simple problem: Let D , E, F, G, H be points such that D ∈ EG, F ∈ DH and DE = DF, DG = DH. Let O 1 ,O 2 ,O 3 be points such that ∠O 2 ED = ∠O 2 FD = 90 ◦ , ∠O 3 GD = ∠O 3 HD = 90 ◦ , and O 1 = EF ∩ GH. Then the points O 1 ,O 2 , and O 3 are collinear. Let K 2 = DO 2 ∩ EF and K 3 = DO 3 ∩ GH. Then O 2 K 2 /O 2 D = DK 3 /DO 3 =

K 2 O 1 /DO 3 and hence by Thales’ theorem O 1 ∈O 2 O 3 .

5. We first prove the following lemma. Lemma. If of five points in a plane no three belong to a single line, then there exist four that are the vertices of a convex quadrilateral. Proof. If the convex hull of the five points A , B,C, D, E is a pentagon or a quadrilateral, the statement automatically holds. If the convex hull is a tri- angle, then w.l.o.g. let △ABC be that triangle and D,E points in its interior. Let the line DE w.l.o.g. intersect [AB] and [AC]. Then B,C, D, E form the desired quadrilateral.

We now observe each quintuplet of points within the set. There are n 5 such quintuplets, and for each of them there is at least one quadruplet of points forming a convex quadrilateral. Each quadruplet, however, will be counted

382 4 Solutions up to n − 4 times. Hence we have found at least 1 n n −4 5 quadruplets. Since

1 n n −3

n −4 5 ≥ 2 ⇔ (n − 5)(n − 6)(n + 8) ≥ 0, which always holds, it follows that we have found at least n −3 2 desired quadruplets of points.

6. Define u 1 = √x 1 y 1 +z 1 ,u 2 = √x 2 y 2 +z 2 ,v 1 = √x 1 y 1 −z 1 , and v 2 = √x 2 y 2 −z 2 . By expanding both sides of the equation we can easily verify (x 1 +x 2 )(y 1 +y 2 )−

1 y 2 − x 2 y 1 ) 2 ≥ (u 1 +u 2 )(v 1 +v 2 ). Since x i y i −z 2 i =u i v i for i = 1, 2, it suffices to prove

(z 1 +z 2 ) 2 = (u 1 +u 2 )(v 1 +v 2 ) + (√x

(u 1 +u 2 )(v 1 +v 2 )

⇔ 8u 1 u 2 v 1 v 2 ≤ (u 1 +u 2 )(v 1 +v 2 )(u 1 v 1 +u 2 v 2 ). This follows from the AM–GM inequalities 2 √

√ u 1 √ u 2 ≤u 1 +u 2 ,2 v 1 v 2 ≤v 1 +v 2

and 2 u 1 v 1 u 2 v 2 ≤u 1 v 1 +u 2 v 2 .

Equality holds if and only if x 1 y 2 =x 2 y 1 ,u 1 =u 2 and v 1 =v 2 , i.e. if and only if

x 1 =x 2 ,y 1 =y 2 and z 1 =z 2 .

Second solution. Let us define f (x, y, z) = 1/(xy − z 2 ). The problem actually states that

x 1 +x 2 y 1 +y 2 z 1 +z 2

2f ,

2 2 2 ≤ f (x 1 ,y 1 ,z 1 ) + f (x 2 ,y 2 ,z 2 ), i.e., that the function f is convex on the set D = {(x,y,z) ∈ R 2 | xy − z 2 > 0}. It

is known that a twice continuously differentiable function f (t 1 ,t 2 , . . . ,t n ) is con- vex if its Hessian [f ′ ′ ij ] n i , j=1 is positive definite, or equivalently (by Sylvester’s

criterion), if its principal minors D k = det[ f ′ ′ ij ] k i , j=1 ,k = 1, 2, . . . , n, are posi- tive. In the case of our f this is directly verified: D 1 = 2y 2 /(xy − z 2 ) 3 ,D 2 =

3xy +z 2 /(xy − z 2 ) 5 ,D 3 = 6/(xy − z 2 ) 6 are obviously positive.

4.12 Shortlisted Problems 1970 383