Solutions to the Shortlisted Problems of IMO 1996

4.37 Solutions to the Shortlisted Problems of IMO 1996

1. We have a 5 +b 5 2 b 2 (a + b) = (a 3 3 )(a 2 2 −a 5 −b −b ) ≥ 0, i.e. a +b 5 ≥a 2 b 2 (a +

b ). Hence

a 2 b 2 c 2 (a + b) + abc 2 a +b+c Now, the left side of the inequality to be proved does not exceed c + a a +b+c a +b+c +

a +b+c = 1. Equality holds if and only if a = b = c.

2. Clearly a 1 > 0, and if p 6= a 1 , we must have a n < 0, |a n | > |a 1 |, and p = −a n . But then for sufficiently large odd k, −a k

n = |a n | > (n − 1)|a 1 | , so that a 1 + ··· + a k n ≤ (n − 1)|a 1 | k − |a n | k < 0, a contradiction. Hence p = a 1 . Now let x >a

kk

1 . From a 1 + ··· + a n ≥ 0 we deduce ∑ j =2 (x − a j ) ≤ (n −

1 )x+ a 1 n −1 , so by the AM–GM inequality,

The last inequality holds because n −1 r ≤ (n − 1) r for all r ≥ 0. Multiplying (1) by (x − a 1 ) yields the desired inequality.

3. Since a 1 > 2, it can be written as a 1 =b+b −1 for some b > 0. Furthermore,

a 2 2 1 2 −2=b +b −2 and hence a 2 = (b +b −2 )(b + b −1 ). We prove that

2 n a −1 =b+b b +b b +b b +b −2 n −1

−2 = b +b . Now we have

by induction. Indeed, a n +1

b +1 (b 2 + 1)(b 4 + 1) + ···

. (b + 1)(b + 1) . . . (b + 1)

Note that 1

2 (a + 2 − a 2 −4)=1+ 1 b ; hence we must prove that the right side in (1) is less than 1 b . This follows from the fact that

(b k 2 + 1)(b 4 + 1) ···(b 2 + 1)

; + 1) ···(b 2 + 1)

+ 1) ···(b

(b 2 + 1)(b 4 2 k −1

+ 1) (b 2 + 1)(b 4 k

hence the right side in (1) equals 1 1 b 1 − (b 2 +1)(b 4 +1)...(b 2n +1) , and this is clearly less than 1 /b .

4.37 Shortlisted Problems 1996 607

4. Consider the function

f (x) =

Since f is strictly decreasing from +∞ to 0 on the interval (0, +∞), there exists exactly one R > 0 for which f (R) = 1. This R is also the only positive real root of the given polynomial. Since ln x is a concave function on (0, +∞), Jensen’s inequality gives us

A R j ≤ ln ∑ A · =1 = ln f (R) = 0. j =1 R j j

ln

Therefore ∑ n j =1 a j (ln A − j ln R) ≤ 0, which is equivalent to AlnA ≤ Bln R, i.e.,

A A ≤R B .

5. Considering the polynomials ±P(±x) we may assume w.l.o.g. that a,b ≥ 0. We have four cases:

(1) c ≥ 0,d ≥ 0. Then |a| + |b| + |c| + |d| = a + b + c + d = P(1) ≤ 1. (2) c ≥ 0,d < 0. Then |a| + |b| + |c| + |d| = a + b + c − d = P(1) − 2P(0) ≤ 3. (3) c < 0, d ≥ 0. Then

|a| + |b| + |c| + |d| = a + b − c + d

3 3 (−1/2) ≤ 7. (4) c < 0, d < 0. Then

3 (1) − 4P(1/2) + 3 (−1/2) ≤ 7. Remark. It can be shown that the maximum of 7 is attained only for P (x) =

±(4x 3 − 3x).

6. Let f (x), g(x) be polynomials with integer coefficients such that

f (x)(x + 1) n + g(x)(x n + 1) = k 0 . (1) Write n =2 r

m for m odd and note that x + 1 = (x + 1)B(x), where B(x) = x

2 r (m−1) 2 r

−x (m−2) + ··· − x 2 r + 1. Moreover, B(−1) = 1; hence B(x) − 1 = (x +

1 )c(x) and thus R

(x)B(x) + 1 = (B(x) − 1) n = (x + 1) n c (x) n (2) for some polynomials c (x) and R(x).

The zeros of the polynomial x 2 r + 1 are ω j , with ω 1 = cos π 2 r + i sin π 2 r , and ω j = ω 2j −1 for 1

≤j≤2 r . We have

608 4 Solutions

( ω 1 + 1)( ω 2 + 1) ···( ω 2 r +1 + 1) = 2. (3) From (1) we also get f ( ω j )( ω j + 1) n =k 0 for j = 1, 2, . . . , 2 r . Since A =

f ( ω 1 )f( ω 2 ) ··· f ( ω 2 r ) is a symmetric polynomial in ω 1 ,..., ω 2 r with integer co- efficients, A is an integer. Consequently, taking the product over j = 1, 2, . . . , 2 r and using (3) we deduce that 2 n

A r =k 2

0 is divisible by 2 n =2 2 m . Hence 2 m |k 0 . Furthermore, since ω j +1=( ω 1 + 1)p r j ( ω 1 ) for some polynomial p j with in- teger coefficients, (3) gives ( ω 1 + 1) 2 p ( ω 1 ) = 2, where p(x) = p 2 (x) ··· p 2 r (x) has integer coefficients. But then the polynomial (x + 1) 2 r r p (x) − 2 has a zero

x = ω 1 , so it is divisible by its minimal polynomial x 2 + 1. Therefore (x + 1) r 2 r p (x) = 2 + (x 2 + 1)q(x)

(4) for some polynomial q

nnm

(x). Raising (4) to the mth power we get (x + 1) p (x) =

2 + (x + 1)Q(x) for some polynomial Q(x) with integer coefficients. Now using (2) we obtain

+ 1)Q(x) = (x r 2 + 1)Q(x) + (x 2 + 1)Q(x)B(x)R(x) = (x + 1) n p (x) n −2 m + (x n + 1)Q(X )R(x). Therefore (x+1) n f (x)+(x n +1)g(x) = 2 m for some polynomials f (x), g(x) with

(x + 1) n c (x) n (x 2 r

integer coefficients, and k 0 =2 m .

7. We are given that f (x + a + b) − f (x + a) = f (x + b) − f (x), where a = 1/6 and

b = 1/7. Summing up these equations for x, x + b, . . . , x + 6b we obtain f (x +

a + 1) − f (x + a) = f (x + 1) − f (x). Summing up the new equations for x,x +

a , . . . , x + 5a we obtain that

f (x + 2) − f (x + 1) = f (x + 1) − f (x). It follows by induction that f (x + n) − f (x) = n[ f (x + 1) − f (x)]. If f (x + 1) 6=

f (x), then f (x + n) − f (x) will exceed in absolute value an arbitrarily large num- ber for a sufficiently large n, contradicting the assumption that f is bounded. Hence f (x + 1) = f (x) for all x.

8. Putting m = n = 0 we obtain f (0) = 0 and consequently f ( f (n)) = f (n) for all n . Thus the given functional equation is equivalent to

f (0) = 0 . Clearly one solution is (∀x) f (x) = 0. Suppose f is not the zero function. We

f (m + f (n)) = f (m) + f (n),

observe that f has nonzero fixed points (for example, any f (n) is a fixed point). Let a be the smallest nonzero fixed point of f . By induction, each ka (k ∈ N) is a fixed point too. We claim that all fixed points of f are of this form. Indeed, suppose that b = ka + i is a fixed point, where i < a. Then

b = f (b) = f (ka + i) = f (i + f (ka)) = f (i) + f (ka) = f (i) + ka; hence f (i) = i. Hence i = 0.

4.37 Shortlisted Problems 1996 609 Since the set of values of f is a set of its fixed points, it follows that for i =

0 , 1, . . . , a − 1, f (i) = an i for some integers n i ≥ 0 with n 0 = 0. Let n = ka + i be any positive integer, 0 ≤ i < a. As before, the functional equa- tion gives us

f (n) = f (ka + i) = f (i) + ka = (n i + k)a.

Besides the zero function, this is the general solution of the given functional equation. To verify this, we plug in m = ka + i, n = la + j and obtain

f (m + f (n)) = f (ka + i + f (la + j)) = f ((k + l + n j )a + i)

= (k + l + n j +n i )a = f (m) + f (n).

9. From the definition of a (n) we obtain

a (n) − a([n/2]) = ≡ 0 or n ≡ 3 (mod 4);

1 if n

−1 if n ≡ 1 or n ≡ 2 (mod 4). Let n =b k b k −1 ...b 1 b 0 be the binary representation of n, where we assume b k =

1. If we define p (n) and q(n) to be the number of indices i = 0, 1, . . . , k − 1 with

b i =b i +1 and the number of i = 0, 1, . . . , k − 1 with b i 6= b i +1 respectively, we get

(1) (a) The maximum value of a (n) for n ≤ 1996 is 9 when p(n) = 9 and q(n) = 0,

a (n) = p(n) − q(n).

i.e., in the case n = 1111111111 2 = 1023.

The minimum value is −10 and is attained when p(n) = 0 and q(n) = 10,

i.e., only for n = 10101010101 2 = 1365.

(b) From (1) we have that a (n) = 0 is equivalent to p(n) = q(n) = k/2. Hence k must be even, and the k /2 indices i for which b i =b i +1 can be chosen in exactly k k ways. Thus the number of positive integers n <2 11 /2 = 2048

with a (n) = 0 is equal to

0 1 2 3 4 5 But five of these numbers exceed 1996: these are 2002 = 11111010010 2 , 2004 = 11111010100 2 , 2006 = 11111010110 2 , 2010 = 11111011010 2 ,

Therefore there are 346 numbers n ≤ 1996 for which a(n) = 0.

10. We first show that H is the common orthocenter of the triangles ABC and AQR.

610 4 Solutions Let G ,G ′ ,H ′

R troid of △ABC, the centroid of △PBC, and the orthocenter of △PBC. Since

be respectively the cen-

A the triangles ABC and PBC have a

X common circumcenter, from the prop-

erties of the Euler line we get HH ′ =

3 ′ GG −→ = AP . But △AQR is exactly the image of △PBC under translation by

−−→ H

B C −→ AP ; hence the orthocenter of AQR co-

incides with H. (Remark: This can be P shown by noting that AHBQ is cyclic.)

Now we have that RH ⊥ AQ; hence ∠AXH = 90 ◦ = ∠AEH. It follows that AX EH is cyclic; hence

− ∠AHE = 180 ◦ − ∠BCA = 180 − ∠BPA = ∠PAQ (as oriented angles). Hence EX k AP.

∠EX Q = 180 ◦

11. Let X ,Y, Z respectively be the feet of the perpendiculars from P to BC, CA, AB. Examining the cyclic quadrilaterals AZPY , BX PZ, CY PX , one can easily see that ∠X ZY = ∠APB − ∠C and XY = PC sin∠C. The first relation gives that XY Z is isosceles with XY = XZ, so from the second relation PB sin ∠B = PC sin ∠C. Hence AB /PB = AC/PC. This implies that the bisectors BD and CD of ∠ABP and ∠ACP divide the segment AP in equal ratios; i.e., they concur with AP.

Second solution. Take that X ,Y, Z are the points of intersection of AP, BP,CP with the circumscribed circle of ABC instead. We similarly obtain XY = XZ. If we write AP · PX = BP · PY = CP · PZ = k, from the similarity of △APC and △ZPX we get

i.e., X Z = k ·AC·BP AP ·BP·CP . It follows again that AC/AB = PC/PB. Third solution. Apply an inversion with center at A and radius r, and denote by

Q the image of any point Q. Then the given condition becomes ∠BCP = ∠CBP, i.e., BP = PC. But

so AC /AB = PC/PB. Remark. Moreover, it follows that the locus of P is an arc of the circle of Apol-

lonius through C.

12. It is easy to see that P lies on the segment AC. Let E be the foot of the altitude BH and Y , Z the midpoints of AC, AB respectively. Draw the perpendicular HR to FP (R ∈ FP). Since Y is the circumcenter of △FCA, we have ∠FYA = 180 ◦ − 2∠A. Also, OFPY is cyclic; hence ∠OPF = ∠OY F = 2∠A − 90 ◦ . Next, △OZF and

△HRF are similar, so OZ/OF = HR/HF.

4.37 Shortlisted Problems 1996 611 This leads to HR · OF = HF · OZ =

2 HF · HC = 1 2 HE · HB = HE · OY. This implies that HR /HE = OY /OF.

Moreover, ∠EHR = ∠FOY ; hence the

triangles EHR and FOY are simi- E H O lar. Consequently ∠HPC = ∠HRE =

∠OY F = 2∠A − 90 ◦ . We finally get

∠FHP = ∠HPC + ∠HCP = ∠A. A F Z B Second solution. As before, ∠HFY = 90 ◦ − ∠A, so it suffices to show that

HP ⊥ FY . The points O,F,P,Y lie on a circle, say Ω 1 with center at the midpoint Q of OP. Furthermore, the points F ,Y lie on the nine-point circle Ω of △ABC with center at the midpoint N of OH. The segment FY is the common chord of

Ω 1 and Ω , from which we deduce that NQ ⊥ FY . However, NQ k HP, and the result follows. Third solution. Let H ′

be the point symmetric to H with respect to AB. Then

H ′ lies on the circumcircle of ABC. Let the line FP meet the circumcircle at U ,V and meet H ′ B at P ′ . Since OF ⊥ UV , F is the midpoint of UV . By the butterfly theorem, F is also the midpoint of PP ′ . Therefore △H ′ FP ′ ∼ = FHP; hence ∠FHP = ∠FH ′ B = ∠A.

Remark. It is possible to solve the problem using trigonometry. For example,

ZO = KP = cosC , where K is on CF with PK ⊥ CF. Then KP = (A−B) cosC + tan A , from which one obtains formulas for KP and KH. Finally, we can calculate

FZ FK sin (A−B) CF sin

tan ∠FHP = KP KH = ··· = tanA. Second remark. Here is what happens when BC ≤ CA. If ∠A > 45 ◦ , then

∠FHP = ∠A. If ∠A = 45 ◦ , the point P escapes to infinity. If ∠A < 45 ◦ , the point P appears on the extension of AC over C, and ∠FHP = 180 ◦ − ∠A.

13. By the law of cosines applied to △CA 1 B 1 , we obtain A 1 B 2 1 =A 1 C 2 +B 1 C 2 −

A 1 C ·B 1 C ≥A 1 C ·B 1 C . Analogously, B 1 C 2 1 ≥B 1 A ·C 1 A and C 1 A 2 1 ≥C 1 B ·A 1 B , so that multiplying these inequalities yields

A B 2 1 2 ·B 1 C 1 2 1 ·C 1 A 1 ≥A 1 B ·A 1 C ·B 1 A ·B 1 C ·C 1 A ·C 1 B . (1) Now, the lines AA 1 , BB 1 ,CC 1 concur, so by Ceva’s theorem, A 1 B ·B 1 C ·C 1 A =

AB 1 · BC 1 · CA 1 , which together with (1) gives the desired inequality. Equality

holds if and only if CA 1 = CB 1 , etc.

14. Let a, b, c, d, e, and f denote the

F E S e lengths of the sides AB, BC, CD, DE,

EF , and FA respectively. Note that f d ∠A = ∠D, ∠B = ∠E, and ∠C = ∠F.

Draw the lines PQ and RS through A D

and D perpendicular to BC and EF cb

612 4 Solutions respectively (P , R ∈ BC, Q,S ∈ EF). Then BF ≥ PQ = RS. Therefore 2BF ≥

PQ + RS, or 2BF ≥ (asinB + f sinC) + (csinC + d sinB),

and similarly, 2BD ≥ (c sinA + bsinB) + (esinB + f sinA),

(1) 2DF ≥ (e sinC + d sinA) + (asinA + bsinC).

Next, we have the following formulas for the considered circumradii: BF BD DF

2 sin E It follows from (1) that

1 sinC sin B

R A +R C +R E ≥ a +

4 sin B sinC + ···

4 sin A sin B

2 (a + b + ···) = 2

with equality if and only if ∠A = ∠B = ∠C = 120 ◦ and FB ⊥ BC etc., i.e., if and only if the hexagon is regular.

Second solution. Let us construct points A ′′ ,C ′′ ,E ′′ such that ABA ′′ F , CDC ′′ B , and EFE ′′ D are parallelograms. It follows that A ′′ ,C ′′ , B are collinear and also

C ′′ ,E ′′ , B and E ′′ ,A ′′ , F. Furthermore, E ′ denote by A ′

be the intersection of the perpendiculars through F and B

E to FA ′′ and BA ′′ , respectively, and let

C and E ′

′ ′′ C ′′

be analogously defined.

Since A FA B is cyclic with the diam- eter being A ′ A ′′ and since △FA ′′ B∼ =

E ′′ A ′′

△BAF, it follows that 2R C

A =A ′ A = A

B C ′ 2R E =E ′ E ′′ = z. We also have AB =

x . Similarly, 2R C =C ′ C ′′

A = y and ′

AB = FA ′′ =y a ,A ′′ B =z a , CD =C ′′ B =z c , CB =C ′′ D =x c , EF =E ′′ D =x e , and ED =E ′′ F =y e . The original inequality we must prove now becomes

x +y+z≥y a +z a +z c +x c +x e +y e . (1) We now follow and generalize the standard proof of the Erd˝os–Mordell in-

equality (for the triangle A ′ C ′ E ′ ), which is what (1) is equivalent to when

A ′′ =C ′′ =E ′′ . We set C ′ E ′ = a, A ′ E ′ = c and A ′ C ′ = e. Let A 1 be the point symmetric to A ′′ with respect to the bisector of ∠E ′ A ′ C ′ . Let F 1 and B 1 be the feet of the perpendiculars from A 1 to A ′ C ′ and A ′ E ′ , respectively. In that case, A 1 F 1 =A ′′ F =y a and A 1 B 1 =

A ′′ B =z a . We have ax =A ′ A 1 ·E ′ C ′ ≥ 2S A ′ E ′ A 1 C ′ = 2S A ′ E ′ A 1 + 2S A ′ C ′ A 1

= cz a + ey a .

4.37 Shortlisted Problems 1996 613

Similarly, cy ≥ ex c + az c and ez ≥ ay e + cx e . Thus

c a e c a e x +y+z≥ z a + z c + x c + x e + y e + y a

1 = −z 2 . We note that △A ′′ C ′′ E ′′ ∼ △A ′ C ′ E ′ and hence a 1 /a = c 1 /c = e 1 /e = k. Thus c − a e 1 + e − c a 1 + a a c c e e − e a c 1 =

k ce − ae + ea − a ca c e + ac c ec e − a = 0. Equation (2) reduces to

a a +z c c e +x c x +y+z≥

e a +y e

Using c /a + a/c, e/c + c/e, a/e + e/a ≥ 2 we finally get x + y + z ≥ y a +z a + z c +x c +x e +y e . Equality holds if and only if a = c = e and A ′′ =C ′′ =E ′′ = center of △A ′ C ′ E ′ , i.e., if and only if ABCDEF is regular.

Remark. From the second proof it is evident that the Erd˝os–Mordell inequality is

a special case of the problem. If P a ,P b ,P c are the feet of the perpendiculars from

a point P inside △ABC to the sides BC,CA,AB, and P a PP b P c ′ ,P b PP c P a ′ ,P c PP a P ′ b parallelograms, we can apply the problem to the hexagon P a P c ′ P b P a ′ P c P ′ b to prove the Erd˝os–Mordell inequality for △ABC and point P.

15. Denote by ABCD and EFGH the two rectangles, where AB = a, BC = b, EF = c, and FG = d. Obviously, the first rectangle can be placed within the second one with the angle α between AB and EF if and only if

a sin α + b cos α ≤ d. (1) Hence ABCD can be placed within EFGH if and only if there is an α ∈ [0, π /2]

a cos α + b sin α ≤ c,

for which (1) holds. The lines l 1 (ax + by = c) and l 2 (bx + ay = d) and the axes x and y bound a region R. By (1), the desired placement of the rectangles is possible if and only if R contains some point (cos α , sin α ) of the unit circle centered at the origin (0, 0).

This in turn holds if and only if the intersection point L of l 1 and l 2 lies outside the unit circle. It is easily computed that L has coordinates bd −ac bc b −ad 2 −a 2 , b 2 −a 2 . Now L being outside the unit circle is exactly equivalent to the inequality we want to

prove. Remark. If equality holds, there is exactly one way of placing. This happens, for

example, when (a, b) = (5, 20) and (c, d) = (16, 19). Second remark. This problem is essentially very similar to (SL89-2).

614 4 Solutions

16. Let A 1 be the point of intersection of OA ′ and BC; similarly define B 1 and C 1 . From the similarity of triangles OBA 1 and OA ′ B we obtain OA 1 · OA ′ =R 2 . Now it is enough to show that 8OA 1 · OB ′ · OC ′ ≤R 3 . Thus we must prove that

λ µν ≤ 1 , where 1 1 OC

1 OA

OB

OC On the other hand, we have

S ABC Simplifying this relation, we get

S ABC S ABC

1 = λµ + µν + νλ +2 λ µν ≥ 3( λ µν ) 2 /3 +2 λ µν , which cannot hold if λ µν > 1 8 . Hence λ µν 1 ≤ 8 , with equality if and only if

λ = µ = ν = 1 2 . This implies that O is the centroid of ABC, and consequently, that the triangle is equilateral.

Second solution. In the official solution, the inequality to be proved is trans- formed into

cos (A − B)cos(B −C)cos(C − A) ≥ 8cosAcosBcosC. Since cos (B−C)

cos (B−C) cos A

=− cos (B+C) = tan B tanC +1 tan B tanC −1 , the last inequality becomes (xy +

1 )(yz + 1)(zx + 1) ≥ 8(xy − 1)(yz − 1)(zx − 1), where we write x, y, z for tanA, tan B, tanC. Using the relation x + y + z = xyz, we can reduce this inequality to

(2x + y + z)(x + 2y + z)(x + y + 2z) ≥ 8(x + y)(y + z)(z + x). This follows from the AM–GM inequality: 2x p + y + z = (x + y) + (x + z) ≥

2 (x + y)(x + z), etc.

17. Let the diagonals AC and BD meet in X. Either ∠AX B or ∠AX D is greater than or equal to 90 ◦ , so we assume w.l.o.g. that ∠AX B ≥ 90 ◦ . Let α , β , α ′ , β ′ denote ∠CAB, ∠ABD, ∠BDC, ∠DCA. These angles are all acute and satisfy α + β = α ′ + β ′ . Furthermore,

AD BC BC AD R A =

2 sin

2 sin α

2 sin α ′

2 sin β ′ Let ∠B + ∠D = 180 ◦ . Then A , B,C, D are concyclic and trivially R A +R C =

R B +R D . Let ∠B + ∠D > 180 ◦ . Then D lies within the circumcircle of ABC, which im- plies that β > β ′ . Similarly α < α ′ , so we obtain R A <R D and R C <R B .

Thus R A +R C <R B +R D .

Let ∠B + ∠D < 180 ◦ . As in the previous case, we deduce that R A >R D and

R C >R B , so R A +R C >R B +R D .

4.37 Shortlisted Problems 1996 615

18. We first prove the result in the simplest case. Given a 2-gon ABA and a point O, let a , b, c, h denote OA, OB, AB, and the distance of O from AB. Then D = a + b, P = 2c, and H = 2h, so we should show that

(1) Indeed, let l be the line through O parallel to AB, and D the point symmetric to B

(a + b) 2 ≥ 4h 2 +c 2 .

with respect to l. Then (a + b) 2 = (OA + OB) 2 = (OA + OD) 2 ≥ AD 2 =c 2 + 4h 2 .

be the polygon F and denote by d i ,p i , and h i respectively OA i ,A i A i +1 , and the distance of O from A i A i +1 (where A n +1 =A q 1 ). By the case proved above, we have for each i, d i +d i +1 ≥

Now we pass to the general case. Let A 1 A 2 ...A n

4h 2 i +p 2 i . Summing these inequalities for i = 1, . . . , n and squaring, we obtain

It remains only to prove that

4h 2 +p 2 ≥

(4h 2 +p 2 )= 4H ∑ 2 i i ∑ i i +D 2 .

i =1

i =1

But this follows immediately from the Minkowski inequality. Equality holds if and only if it holds in (1) and in the Minkowski inequality,

i.e., if and only if d 1 = ··· = d n and h 1 /p 1 = ··· = h n /p n . This means that F is inscribed in a circle with center at O and p 1 = ··· = p n , so F is a regular polygon and O its center.

19. It is easy to check that after 4 steps we will have all a , b, c, d even. Thus |ab −

cd |,|ac − bd|,|ad − bc| remain divisible by 4, and clearly are not prime. The answer is no.

Second solution. After one step we have a + b + c + d = 0. Then ac − bd =

ac + b(a + b + c) = (a + b)(b + c) etc., so |ab − cd| · |ac − bd| · |ad − bc| = (a + b) 2 (a + c) 2 (b + c) 2 . However, the product of three primes cannot be a square, hence the answer is no.

20. Let 15a + 16b = x 2 and 16a − 15b = y 2 , where x , y ∈ N. Then we obtain x 4 +y 4 = (15a + 16b) 2 2 = (15 2 + 16 2 )(a 2 +b 2 ) = 481(a 2 +b + (16a − 15b) 2 ).

In particular, 481 = 13 · 37 | x 4 +y 4 . We have the following lemma. Lemma. Suppose that p 4 |x +y 4 , where x , y ∈ Z and p is an odd prime, where

p 6≡ 1 (mod 8). Then p | x and p | y. Proof. Since p 8 8 p

|x p −y and by Fermat’s theorem p |x −1 −y −1 , we deduce that p d d

|x 4 −y , where d = (p − 1,8). But d 6= 8, so d | 4. Thus p | x 4 −y ,

which implies that p | 2y 4 , i.e., p | y and p | x.

616 4 Solutions In particular, we can conclude that 13 | x,y and 37 | x,y. Hence x and y are divis-

ible by 481. Thus each of them is at least 481. On the other hand, x = y = 481 is possible. It is sufficient to take a = 31 · 481 and b = 481.

Second solution. Note that 15x 2 + 16y 2 = 481a 2 . It can be directly verified that the divisibility of 15x 2 + 16y 2 by 13 and by 37 implies that both x and y are divisible by both primes. Thus 481 | x,y.

21. (a) It clearly suffices to show that for every integer c there exists a quadratic sequence with a 0 = 0 and a n = c, i.e., that c can be expressed as ±1 2 ±

2 2 ± ··· ± n 2 . Since (n + 1) 2 2 − (n + 2) 2 − (n + 3) + (n + 4) 2 = 4, we observe that if our claim is true for c, then it is also true for c ± 4. Thus

it remains only to prove the claim for c = 0, 1, 2, 3. But one immediately finds 1 =1 2 ,2 = −1 2 −2 2 −3 2 +4 2 , and 3 = −1 2 +2 2 , while the case

c = 0 is trivial. (b) We have a 0 = 0 and a n = 1996. Since a n ≤1 2 +2 2 + ··· + n 2 = 1 6 n (n +

1 )(2n + 1), we get a 17 ≤ 1785, so n ≥ 18. On the other hand, a 18 is of the same parity as 1 2 +2 2 + ··· + 18 2 = 2109, so it cannot be equal to 1996. Therefore we must have n ≥ 19. To construct a required sequence with n = 19, we note that 1 2 +2 2 + ···+19 2 = 2470 = 1996 + 2 ·237; hence it is enough to write 237 as a sum of distinct squares. Since 237 = 14 2 +5 2 +4 2 , we finally obtain

22. Let a , b ∈ N satisfy the given equation. It is not possible that a = b (since it leads to a 2 + 2 = 2a), so we assume w.l.o.g. that a > b. Next, for a > b = 1 the equation becomes a 2 = 2a, and one obtains a solution (a, b) = (2, 1). h Let b > 1. If a 2

b = α and b 2

a = β , then we trivially have ab ≥ αβ . Since also

a 2 +b 2

ab ≥ 2, we obtain α + β ≥ αβ + 2, or equivalently ( α −1)( β −1) ≤ −1. But α ≥ 1, and therefore β = 0. It follows that a > b 2 , i.e., a =b h 2 + c for some c > 0.

i h b 4 +2b 2 i

+b + bc, which reduces to

Now the given equation becomes b 3 + 2bc + c 2

If c = 1, then (1) always holds, since both sides are 0. We obtain a family of solutions (a, b) = (n, n 2 + 1) or (a, b) = (n 2 + 1, n). Note that the solution (1, 2) found earlier is obtained for n = 1. 2 If c

b (c+1)+c > 1, then (1) implies that 2

b 3 +bc

≥ (c − 1)b. This simplifies to

c 2 (b 2 − 1) + b 2 (c(b 2 − 2) − (b 2 + 1)) ≤ 0. (2)

4.37 Shortlisted Problems 1996 617 Since c ≥ 2 and b 2 − 2 ≥ 0, the only possibility is b = 2. But then (2) becomes

3c 2 + 8c − 20 ≤ 0, which does not hold for c ≥ 2. Hence the only solutions are (n, n 2 + 1) and (n 2 + 1, n), n ∈ N.

23. We first observe that the given functional equation is equivalent to 4f (3m + 1)(3n + 1) − 1 + 1 = (4 f (m) + 1)(4 f (n) + 1).

3 This gives us the idea of introducing a function g : 3 N 0 + 1 → 4N 0 + 1 defined

as g (x) = 4 f x −1 3 + 1. By the above equality, g will be multiplicative, i.e.,

g (xy) = g(x)g(y) for all x , y ∈ 3N 0 + 1. Conversely, any multiplicative bijection g from 3 N 0 + 1 onto 4N 0 + 1 gives us a

function f with the required property: f

g (x) = (3x+1)−1

4 . It remains to give an example of such a function g. Let P 1 ,P 2 ,Q 1 ,Q 2 be the

sets of primes of the forms 3k + 1, 3k + 2, 4k + 1, and 4k + 3, respectively. It is well known that these sets are infinite. Take any bijection h from P 1 ∪P 2 onto Q 1 ∪Q 2 that maps P 1 bijectively onto Q 1 and P 2 bijectively onto Q 2 . Now define

g as follows: g (1) = 1, and for n = p 1 p 2 ··· p m (p i s need not be different) define

g (n) = h(p 1 )h(p 2 ) ···h(p m ). Note that g is well-defined. Indeed, among the p i s an even number are of the form 3k + 2, and consequently an even number of

h (p i )s are of the form 4k + 3. Hence the product of the h(p i )s is of the form 4k + 1. Also, it is obvious that g is multiplicative. Thus, the defined g satisfies all the required properties.

24. We shall work on the array of lattice points defined by A = {(x,y) ∈ Z 2 |0≤ x ≤ 19, 0 ≤ y ≤ 11}. Our task is to move from (0,0) to (19,0) via the points of A so that each move has the form (x, y) → (x + a,y + b), where a,b ∈ Z and

a 2 +b 2 = r. (a) If r is even, then a + b is even whenever a 2 +b 2 = r (a, b ∈ Z). Thus the parity of x + y does not change after each move, so we cannot reach (19, 0) from (0, 0). If 3 | r, then both a and b are divisible by 3, so if a point (x,y) can be reached from (0, 0), we must have 3 | x. Since 3 ∤ 19, we cannot get to (19,0).

(b) We have r = 73 = 8 2 +3 2 , so each move is either (x, y) → (x ± 8,y ± 3) or (x, y) → (x ± 3,y ± 8). One possible solution is shown in Fig. 1 .

(c) We have 97 =9 2 +4 2 . Let us partition A as B ∪ C , where B = {(x,y) ∈

A | 4 ≤ y ≤ 7}. It is easily seen that moves of the type (x,y) → (x±9,y±4) always take us from the set B to C and vice versa, while the moves (x, y) → (x ± 4,y ± 9) always take us from C to C . Furthermore, each move of the type (x, y) → (x ± 9,y ± 4) changes the parity of x, so to get from (0,0) to (19, 0) we must have an odd number of such moves. On the other hand, with an odd number of such moves, starting from C we can end up only in

B, although the point (19, 0) is not in B. Hence, the answer is no.

618 4 Solutions Remark. Part (c) can also be solved by examining all cells that can be reached

from (0, 0). All these cells are marked in Fig. 2 . • •

25. Let the vertices in the bottom row be assigned an arbitrary coloring, and suppose that some two adjacent vertices receive the same color. The number of such colorings equals 2 n − 2. It is easy to see that then the colors of the remaining

vertices get fixed uniquely in order to satisfy the requirement. So in this case there are 2 n − 2 possible colorings. Next, suppose that the vertices in the bottom row are colored alternately red and blue. There are two such colorings. In this case, the same must hold for every row, and thus we get 2 n possible colorings. It follows that the total number of considered colorings is (2 n

− 2)+2 +1 n =2 n −

26. Denote the required maximum size by M

k (m, n). If m < 2 , then trivially M

n (n+1)

n = k, so from now on we assume that m ≥ (n+1)

First we give a lower bound for M. Let r =r k (m, n) be the largest integer such that r

2 ≤ n (r + 1), so r = m n − −1 n 2 . Clearly no n elements from {r + 1,r + 2,...,k} add up to m, so

n + (r + 1) + ··· + (r + n − 1) ≤ m. This is equivalent to nr ≤ m − (n−1)

M −1

≥k−r k (m, n) = k −

We claim that M is actually equal to k −r k (m, n). To show this, we shall prove by induction on n that if no n elements of a set S ⊆ {1,2,...,k} add up to m, then |S| ≤ k − r k (m, n). For n = 2 the claim is true, because then for each i = 1, . . . , r k (m, 2) = m −1 2 at least one of i and m − i must be excluded from S. Now let us assume that n > 2 and that the result holds for n − 1. Suppose that S ⊆ {1,2,...,k} does not contain n distinct elements with the sum m, and let x be the smallest element of S. We may assume that x ≤r k (m, n), because otherwise the statement is clear. Consider the set S ′ = {y− x | y ∈ S, y 6= x}. Then S ′ is a subset of {1,2,...,k − x} no n − 1 elements of which have the sum m − nx. Also, it is easily checked that n − 1 ≤ m − nx − 1 ≤ k − x, so we may apply the induction hypothesis, which yields that

m −x n |S| ≤ 1 + k − x − r k (m − nx,n − 1) = k − n

4.37 Shortlisted Problems 1996 619 m −x

n (n−1)

On the other hand,

≥ 0 because x ≤ r k (m, n); hence (2) implies |S| ≤ k − r k (m, n) as claimed.

n −1 − 2 −r k (m, n) =

−nx− 2

n (n−1)

27. Suppose that such sets of points A , B exist. First, we observe that there exist five points A , B,C, D, E in A such that their convex hull does not contain any other point of A . Indeed, take any point A ∈

A . Since any two points of A are at distance at least 1, the number of points

X ∈ A with XA ≤ r is finite for every r > 0. Thus it is enough to choose four points B ,C, D, E of A that are closest to A. Now consider the convex hull C of

A , B,C, D, E. Suppose that C is a pentagon, say ABCDE. Then each of the disjoint triangles ABC , ACD, ADE contains a point of B. Denote these points by P, Q, R. Then △PQR contains some point F ∈ A , so F is inside ABCDE, a contradiction. Suppose that C is a quadrilateral, say ABCD, with E lying within ABCD. Then the triangles ABE , BCE,CDE, DAE contain some points P, Q, R, S of B that form two disjoint triangles. It follows that there are two points of A inside ABCD, which is a contradiction. Finally, suppose that C is a triangle with two points of A inside. Then C is the union of five disjoint triangles with vertices in A , so there are at least five points of B inside C . These five points make at least three disjoint triangles containing three points of A . This is again a contradiction. It follows that no such sets A , B exist.

28. Note that w.l.o.g., we can assume that p and q are coprime. Indeed, otherwise it suffices to consider the problem in which all x i ’s and p , q are divided by gcd (p, q). Let k , l be the number of indices i with x i +1 −x i = p and the number of those i

with x i +1 −x i = −q (0 ≤ i < n). From x 0 =x n = 0 we get kp = lq, so for some integer t > 1, k = qt, l = pt, and n = (p + q)t. Consider the sequence y i =x i +p+q −x i ,i = 0, . . . , n− p−q. We claim that at least one of the y i ’s equals zero. We begin by noting that each y i is of the form up −vq, where u +v = p +q; therefore y i = (u + v)p −v(p+q) = (p−v)(p+q) is always divisible by p + q. Moreover, y i +1 −y i = (x i +p+q+1 −x i +p+q ) − (x i +1 −x i ) is 0 or ±(p + q). We conclude that if no y i is 0 then all y i ’s are of the same sign. But

this is in contradiction with the relation y 0 +y p +q + ··· + y n −p−q =x n −x 0 = 0. Consequently some y i is zero, as claimed.

Second solution. As before we assume (p, q) = 1. Let us define a sequence of points A (y ,z ) (i = 0, 1, . . . , n) in N 2 i i i 0 inductively as follows. Set A 0 = (0, 0) and define (y i +1 ,z i +1 ) as (y i ,z i + 1) if x i +1 =x i + p and (y i + 1, z i ) otherwise. The points A

i form a trajectory L in N 0 continuously moving upwards and rightwards by steps of length 1. Clearly, x i = pz i − qy i for all i. Since x n = 0, it follows that

(z n ,y n ) = (kq, kp), k ∈ N. Since y n +z n = n > p + q, it follows that k > 1. We observe that x i =x j if and only if A i A j kA 0 A n . We shall show that such i , j with

i < j and (i, j) 6= (0,n) must exist.

620 4 Solutions If L meets A 0 A n in an interior point, then our statement trivially holds. From

now on we assume the opposite. Let P ij

be the rectangle with sides parallel to the coordinate axes and with vertices at (ip, jq) and ((i + 1)p, ( j + 1)q). Let L ij

be the part of the trajectory L lying inside P ij . We may assume w.l.o.g. that the endpoints of L 00 lie on the vertical sides of P 00 . Then there obviously exists

d ∈ {1,...,k − 1} such that the endpoints of L dd lie on the horizontal sides of P dd . Consider the translate L ′ dd of L dd for the vector −d(p,q). The endpoints of L ′ dd lie on the vertical sides of P 00 . Hence L 00 and L ′ dd have some point X 6= A 0 in common. The translate Y of point X for the vector d (p, q) belongs to L and satisfies XY kA 0 A n .

29. Let the squares be indexed serially by the integers: . . . ,−1,0,1,2,... . When a bean is moved from i to i +1 or from i +1 to i for the first time, we may assign the index i to it. Thereafter, whenever some bean is moved in the opposite direction, we shall assume that it is exactly the one marked by i, and so on. Thus, each pair of neighboring squares has a bean stuck between it, and since the number of beans is finite, there are only finitely pairs of neighboring squares, and thus finitely many squares on which moves are made. Thus we may assume w.l.o.g. that all moves occur between 0 and l ∈ N and that all beans exist at all times within [0, l]. Defining b i to be the number of beans in the ith cell (i ∈ Z) and b the total number

of beans, we define the semi-invariant S =∑ i ∈Z i 2 b i . Since all moves occur above

0, the semi-invariant S increases by 2 with each move, and since we always have S <b·l 2 , it follows that the number of moves must be finite. We now prove the uniqueness of the final configuration and the number of moves for some initial configuration {b i }. Let x i ≥ 0 be the number of moves made in the ith cell (i ∈ Z) during the game. Since the game is finite, only finitely many of x i ’s are nonzero. Also, the number of beans in cell i, denoted as e i , at the end is

(∀i ∈ Z) e i =b i +x i −1 +x i +1 − 2x i ∈ {0,1} . (1) Thus it is enough to show that given b i ≥ 0, the sequence {x i } i ∈Z of nonnegative

integers satisfying (1) is unique. Suppose the assertion is false, i.e., that there exists at least one sequence b i ≥0 for which there exist distinct sequences {x i } and {x ′ i } satisfying (1). We may choose such a {b i } for which min{∑ i ∈Z x i ,∑ i ∈Z x ′ i } is minimal (since ∑ i ∈Z x i is always finite). We choose any index j such that b j > 1. Such an index j exists, since otherwise the game is over. Then one must make at least one move in the

j th cell, which implies that x j ,x ′ j ≥ 1. However, then the sequences {x i } and {x ′ i } with x j and x ′ j decreased by 1 also satisfy (1) for a sequence {b i } where

b j −1 ,b j ,b j +1 is replaced with b j −1 + 1, b j − 2,b j +1 + 1. This contradicts the assumption of minimal min {∑ i ∈Z x i ,∑ i ∈Z x ′ i } for the initial {b i }.

30. For convenience, we shall write f 2 , f g, . . . for the functions f ◦ f , f ◦ g,... . We need two lemmas. Lemma 1. If f (x) ∈ S and g(x) ∈ T , then x ∈ S ∩ T .

4.37 Shortlisted Problems 1996 621 Proof. The given condition means that f 3 (x) = g 2 f (x) and g f g(x) = f g 2 (x).

Since x ∈ S ∪ T = U, we have two cases: x ∈ S. Then f 2 (x) = g 2 (x), which also implies f 3 (x) = f g 2 (x). Therefore gfg (x) = f g 2 (x) = f 3 (x) = g 2 f (x), and since g is a bijection, we ob- tain f g (x) = g f (x), i.e., x ∈ T . x ∈ T . Then f g(x) = g f (x), so g 2 f (x) = g f g(x). It follows that f 3 (x) =

g 2 f (x) = g f g(x) = f g 2 (x), and since f is a bijection, we obtain x ∈ S. Hence x ∈ S ∩T in both cases. Similarly, f (x) ∈ T and g(x) ∈ S again imply x ∈S∩T.

Lemma 2. f (S ∩ T ) = g(S ∩ T ) = S ∩ T . Proof. By symmetry, it is enough to prove f (S ∩ T ) = S ∩ T , or in other words

that f −1 (S ∩T ) = S∩T . Since S∩T is finite, this is equivalent to f (S∩T ) ⊆ S ∩T. Let f (x) ∈ S ∩ T . Then if g(x) ∈ S (since f (x) ∈ T ), Lemma 1 gives x ∈ S ∩ T ; similarly, if g(x) ∈ T , then by Lemma 1, x ∈ S ∩ T .

Now we return to the problem. Assume that f (x) ∈ S. If g(x) 6∈ S, then g(x) ∈ T , so from Lemma 1 we deduce that x ∈ S ∩ T . Then Lemma 2 claims that g(x) ∈ S ∩ T too, a contradiction. Analogously, from g(x) ∈ S we are led to f (x) ∈ S. This finishes the proof.

622 4 Solutions