4.36 Shortlisted Problems 1995 601 28. Let F (x)=f(x)− 95 for x ≥ 1. Writing k for m + 95, the given condition becomes F (k + F (n)) = F (k)+n, k ≥ 96,n≥ 1. (1) Thus for x, z ≥ 96 and an arbitrary y we have F (x + y)+z = F (x + y + F (z)) = F(x + F (F (y)+z)) = F (x)+F (y)+z, and consequently F (x + y)=F (x)+F (y) whenever x ≥ 96. Moreover, since then F (x + y)+F (96) = F (x + y + 96) = F (x)+F (y + 96) = F (x)+F (y)+F (96) for any x, y,weobtain F (x + y)=F (x)+F (y),x,y∈ N. (2) It follows by induction that F (n)=nc for all n, where F (1) = c.Equation (1) becomes ck + c 2 n = ck + n, and yields c = 1. Hence F (n)=n and f(n)=n + 95 for all n. Finally,  19 k=1 f(k)=96+97+···+ 114 = 1995. Second solution. First we show that f(n) > 95 for all n. If to the contrary f(n) ≤ 95, we have f(m)=n + f (m +95− f(n)), so by induction f(m)=kn+ f(m +k(95− f(n))) ≥ kn for all k, which is impossible. Now for m>95 we have f(m + f(n)− 95) = n + f (m), and again by induction f(m + k(f(n) − 95)) = kn + f(m) for all m, n, k. It follows that with n fixed, ( ∀m) lim k→∞ f(m + k(f (n)− 95)) m + k(f(n) − 95) = n f(n) − 95 ; hence lim s→∞ f(s) s = n f(n) − 95 . Hence n f(n)−95 does not depend on n, i.e., f(n) ≡ cn+95 for some constant c.Itiseasilycheckedthatonlyc = 1 is possible. 602 4 Solutions 4.37 Solutions to the Shortlisted Problems of IMO 1996 1. We have a 5 + b 5 − a 2 b 2 (a + b)=(a 3 − b 3 )(a 2 − b 2 ) ≥ 0, i.e. a 5 + b 5 ≥ a 2 b 2 (a + b). Hence ab a 5 + b 5 + ab ≤ ab a 2 b 2 (a + b)+ab = abc 2 a 2 b 2 c 2 (a + b)+abc 2 = c a + b + c . Now, the left side of the inequality to be proved does not exceed c a+b+c + a a+b+c + b a+b+c = 1. Equality holds if and only if a = b = c. 2. Clearly a 1 > 0, and if p = a 1 ,wemusthavea n < 0, |a n | > |a 1 |,and p = −a n . But then for sufficiently large odd k, −a k n = |a n | k > (n−1)|a 1 | k , so that a k 1 + ···+ a k n ≤ (n − 1)|a 1 | k −|a n | k < 0, a contradiction. Hence p = a 1 . Now let x>a 1 .Froma 1 + ··· + a n ≥ 0 we deduce  n j=2 (x − a j ) ≤ (n − 1)  x + a 1 n−1  , so by the AM–GM inequality, (x−a 2 )···(x−a n ) ≤  x + a 1 n − 1  n−1 ≤ x n−1 +x n−2 a 1 +···+a n−1 1 . (1) The last inequality holds because  n−1 r  ≤ (n − 1) r for all r ≥ 0. Multi- plying (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 1 − 2=b 2 + b −2 and hence a 2 =(b 2 + b −2 )(b + b −1 ). We prove that a n =  b + b −1  b 2 + b −2  b 4 + b −4  ···  b 2 n−1 + b −2 n−1  by induction. Indeed, a n+1 a n =  a n a n−1  2 − 2=  b 2 n−1 + b −2 n−1  2 − 2= b 2 n + b −2 n . Now we have n  i=1 1 a i =1+ b b 2 +1 + b 3 (b 2 +1)(b 4 +1) + ··· ···+ b 2 n −1 (b 2 +1)(b 4 +1) .(b 2 n +1) . (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 2 k (b 2 +1)(b 4 +1)···(b 2 k +1) = 1 (b 2 +1)(b 4 +1)···(b 2 k−1 +1) − 1 (b 2 +1)(b 4 +1)···(b 2 k +1) ; hence the right side in (1) equals 1 b  1 − 1 (b 2 +1)(b 4 +1) .(b 2 n +1)  , and this is clearly less than 1/b . 4.37 Shortlisted Problems 1996 603 4. Consider the function f(x)= a 1 x + a 2 x 2 + ···+ a n x n . Since f is strictly decreasing from +∞ to 0 on the interval (0, +∞), there exists exactly one R>0forwhichf (R)=1.ThisR 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 n  j=1 a j A  ln A R j  ≤ ln ⎛ ⎝ n  j=1 a j A · A R j ⎞ ⎠ =lnf(R)=0. Therefore  n j=1 a j (ln A − j ln R) ≤ 0, which is equivalent to A ln A ≤ B ln 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 = 4 3 P (1) − 1 3 P (−1)− 8 3 P (1/2) + 8 3 P (−1/2) ≤ 7. (4) c<0,d <0. Then |a| + |b| +|c| + |d| = a + b − c − d = 5 3 P (1) − 4P (1/2) + 4 3 P (−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 . (∗) Write n =2 r m for m odd and note that x n +1=(x 2 r +1)B(x), where B(x)=x 2 r (m−1) − x 2 r (m−2) +···−x 2 r +1.Moreover,B(−1) = 1; hence B(x) − 1=(x +1)c(x)andthus R(x)B(x)+1=(B(x) − 1) n =(x +1) n c(x) n (1) for some polynomials c(x)andR(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 .Wehave 604 4 Solutions (ω 1 +1)(ω 2 +1)···(ω 2 r+1 +1)=2. (2) From (∗)wealsogetf(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 coefficients, A is an integer. Consequently, taking the product over j =1, 2, .,2 r and using (2) we deduce that 2 n A = k 2 r 0 is divisible by 2 n =2 2 r m . Hence 2 m | k 0 . Furthermore, since ω j +1 = (ω 1 +1)p j (ω 1 ) for some polynomial p j with integer coefficients, (2) gives (ω 1 +1) 2 r 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 p(x) − 2 has a zero x = ω 1 , so it is divisible by its minimal poly- nomial x 2 r + 1. Therefore (x +1) 2 r p(x)=2+(x 2 r +1)q(x)(3) for some polynomial q(x). Raising (3) to the mth power we get (x + 1) n p(x) n =2 m +(x 2 r +1)Q(x) for some polynomial Q(x) with integer coefficients. Now using (1) we obtain (x +1) n c(x) n (x 2 r +1)Q(x)= (x 2 r +1)Q(x)+(x 2 r +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 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) = f(x), then f (x + n) − f (x) will exceed in absolute value an arbitrarily large number 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 =0weobtainf (0) = 0 and consequently f (f(n)) = f (n) for all n. Thus the given functional equation is equivalent to f(m + f (n)) = f (m)+f(n),f(0) = 0 . Clearly one solution is (∀x) f(x) = 0. Suppose f is not the zero function. We observe that f has nonzero fixed points (for example, any f(n)isa 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 4.37 Shortlisted Problems 1996 605 b = f(b)=f (ka + i)=f (i + f (ka)) = f (i)+f(ka)=f(i)+ka; hence f (i)=i. Hence i =0. 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 ≥ 0withn 0 =0. Let n = ka+i be any positive integer, 0 ≤ i<a. As before, the functional equation 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 func- tional 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)weobtain a(n) − a([n/2]) =  1ifn ≡ 0orn ≡ 3(mod4); −1ifn ≡ 1orn ≡ 2(mod4). Let n = b k b k−1 .b 1 b 0 be the binary representation of n,whereweas- sume b k = 1. If we define p(n)andq(n) to be the number of indices i =0, 1, .,k− 1withb i = b i+1 and the number of i =0, 1, .,k− 1 with b i = b i+1 respectively, we get a(n)=p(n)− q(n). (1) (a) The maximum value of a(n)forn ≤ 1996 is 9 when p(n)=9and q(n) = 0, i.e., in the case n = 1111111111 2 = 1023. The minimum value is −10 and is attained when p(n)=0andq(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/2  ways. Thus the number of positive integers n<2 11 = 2048 with a(n)=0isequalto  0 0  +  2 1  +  4 2  +  6 3  +  8 4  +  10 5  = 351. But five of these numbers exceed 1996: these are 2002 = 11111010010 2 , 2004 = 11111010100 2 , 2006 = 11111010110 2 , 2010 = 11111011010 2 , 2026 = 11111101010 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. 606 4 Solutions Let G, G  ,H  be respectively the centroid of ABC, the centroid of PBC, and the orthocenter of PBC. Since the triangles ABC and PBC have a common circum- center, from the properties of the Euler line we get −−→ HH  =3 −−→ GG  = −→ AP .ButAQR is exactly the im- age of PBC under translation by −→ AP ; hence the orthocenter of AQR coincides with H.(Remark: This A B C P E H Q R X can be shown by noting that AHBQ is cyclic.) NowwehavethatRH ⊥ AQ; hence ∠AXH =90 ◦ = ∠AEH. It follows that AXEH is cyclic; hence ∠EXQ = 180 ◦ − ∠AHE = 180 ◦ − ∠BCA = 180 ◦ − ∠BPA = ∠PAQ (as oriented angles). Hence EX  AP . 11. Let X, Y, Z respectively be the feet of the perpendiculars from P to BC, CA, AB. Examining the cyclic quadrilaterals AZP Y , BXPZ, CY PX, one can easily see that ∠XZY = ∠AP B − ∠C and XY = PCsin ∠C. The first relation gives that XY Z is isosceles with XY = XZ,sofrom the second relation PBsin ∠B = PCsin ∠C. Hence AB/P B = AC/P C. 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.IfwewriteAP · PX = BP · PY = CP · PZ = k,from the similarity of AP C and ZPX we get AC XZ = AP PZ = AP · CP k , i.e., XZ = 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 PB = r 2 AP · AB PB, so AC/AB = PC/PB. Remark. Moreover, it follows that the locus of P is an arc of the circle of Apollonius through C. 4.37 Shortlisted Problems 1996 607 12. It is easy to see that P lies on the segment AC.LetE 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,wehave∠FYA = 180 ◦ − 2∠A.Also,OF PY is cyclic; hence ∠OPF = ∠OY F =2∠A− 90 ◦ .Next,OZF and HRF are similar, so OZ/OF = HR/HF. This leads to HR · OF = HF · OZ = 1 2 HF · HC = 1 2 HE· HB = HE· OY =⇒ HR/HE = OY/OF. Moreover, ∠EHR = ∠FOY; hence the tri- angles EHR and FOY are similar. Consequently ∠HPC = ∠HRE = ∠OY F =2∠A − 90 ◦ , and finally, ∠FHP = ∠HPC + ∠HCP = ∠A. A B C Y Z O F H E P R Second solution. As before, ∠HFY =90 ◦ −∠A, so it suffices to show that HP ⊥ FY.ThepointsO, F, P, Y lie on a circle, say Ω 1 with center at the midpoint Q of OP.Furthermore,thepointsF, 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 HP, and the result follows. Third solution. Let H  be the point symmetric to H with respect to AB.ThenH  lies on the circumcircle of ABC. Let the line FP meet the circumcircle at U, V and meet H  B at P  .SinceOF ⊥ 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 ex- ample, FZ ZO = FK KP = sin(A−B) cos C , where K is on CF with PK ⊥ CF.Then CF KP = sin(A−B) cos C +tanA, from which one obtains formulas for KP and KH. Finally, we can calculate tan ∠FHP = KP KH = ···=tanA. Second remark. Here is what happens when BC ≤ CA.If∠A>45 ◦ , then ∠FHP = ∠A.If∠A =45 ◦ ,thepointP escapes to infinity. If ∠A<45 ◦ ,thepointP appears on the extension of AC over C,and ∠FHP = 180 ◦ − ∠A. 13. By the law of cosines applied to CA 1 B 1 ,weobtain 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 multi- plying these inequalities yields A 1 B 2 1 · B 1 C 2 1 · C 1 A 2 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 in- equality. Equality holds if and only if CA 1 = CB 1 ,etc. 608 4 Solutions 14. Let a, b, c, d, e,andf denote the lengths of the sides AB, BC, CD, DE, EF,andFA respectively. Note that ∠A = ∠D, ∠B = ∠E, and ∠C = ∠F . Draw the lines PQ and RS through A and D perpen- dicular to BC and EF respectively (P, R ∈ BC, Q, S ∈ EF). Then BF ≥ PQ = RS. Therefore 2BF ≥ PQ+ RS,or A B C D Q EF P R S a b c d e f 2BF ≥ (a sin B + f sin C)+(c sin C + d sin B), and similarly, 2BD ≥ (c sin A + b sin B)+(e sin B + f sin A), 2DF ≥ (e sin C + d sin A)+(a sin A + b sin C). (1) Next, we have the following formulas for the considered circumradii: R A = BF 2sinA ,R C = BD 2sinC ,R E = DF 2sinE . It follows from (1) that R A + R C + R E ≥ 1 4 a  sin B sin A + sin A sin B  + 1 4 b  sin C sin B + sin B sin C  + ··· ≥ 1 2 (a + b + ···)= P 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,andEFE  D are parallelograms. It follows that A  ,C  ,B are collinear and also C  ,E  ,B and E  ,A  ,F. Furthermore, let A  be the intersection of the perpendicu- lars through F and B to FA  and BA  , respectively, and let C  and E  be analogously defined. Since A  FA  B is cyclic with the diameter being A  A  and since FA  B ∼ = BAF, it follows that 2R A = A  A  = x. A B C D E F A  C  E  A  C  E  Similarly, 2R C = C  C  = y and 2R E = E  E  = z. We also have AB = FA  = y a , AF = A  B = z a , CD = C  B = z c , CB = C  D = x c , EF = E  D = x e ,andED = 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) 4.37 Shortlisted Problems 1996 609 We now follow and generalize the standard proof of the Erd˝os–Mordell inequality (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.LetA 1 be the point symmetric to A  with respect to the bisector of ∠E  A  C  .LetF 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 .Wehave 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 . Similarly, cy ≥ ex c + az c and ez ≥ ay e + cx e .Thus x + y + z ≥ c a z a + a c z c + e c x c + c e x e + a e y e + e a y a =  c a + a c   z a + z c 2  +  c a − a c   z a − z c 2  + ··· . (2) Let us set a 1 = x c −x e 2 , c 1 = y e −y a 2 , e 1 = z a −z c 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 − a c  e 1 +  e c − c e  a 1 +  a e − e a  c 1 = k  ce a − ae c + ea c − ca e + ac e − ec a  =0.Equation (2) reduces to x + y + z ≥  c a + a c   z a + z c 2  +  e c + c e   x e + x c 2  +  a e + e a   y a + y e 2  . 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 in- equality 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 prob- lem 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,andFG = 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 cos α + b sin α ≤ c, 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] for which (1) holds. 610 4 Solutions The lines l 1 (ax + by = c)andl 2 (bx + ay = d)andtheaxesx and y bound aregionR. 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 b 2 −a 2 , bc−ad b 2 −a 2  .NowL 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 hap- pens, for example, when (a, b)=(5, 20) and (c, d)=(16, 19). Second remark. This problem is essentially very similar to (SL89-2). 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 λµν ≤ 1 8 , where OA 1 OA = λ, OB 1 OB = µ, OC 1 OC = ν. (1) On the other hand, we have λ 1+λ + µ 1+µ + ν 1+ν = S OBC S ABC + S AOC S ABC + S ABO S ABC =1. Simplifying this relation, we get 1=λµ + µν + νλ +2λµν ≥ 3(λµν) 2/3 +2λµν, which cannot hold if λµν > 1 8 . Hence λµν ≤ 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 transformed into cos(A − B)cos(B − C)cos(C − A) ≥ 8cosA cos B cos C. Since cos(B−C) cos A = − cos(B−C) cos(B+C) = tan B tan C+1 tan B tan C−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 tan A, tan B, tan C. 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+ y + z =(x+ y)+(x + z) ≥ 2  (x + y)(x + z), etc. [...]... + p2 i i 2 n 2 2 It remains only to prove that 4h2 + p2 ≥ i i i=1 i=1 (4hi + pi ) = √ 2 + D 2 But this follows immediately from the Minkowski inequality 4H Equality holds if and only if it holds in (1) and in the Minkowski inequality, i.e., if and only if d1 = · · · = dn and h1 /p1 = · · · = hn /pn This means that F is inscribed in a circle with center at O and p1 = · · · = pn , so F is a regular... 1)b This simplifies to b3 +bc c2 (b2 − 1) + b2 (c(b2 − 2) − (b2 + 1)) ≤ 0 (2) Since c ≥ 2 and b2 − 2 ≥ 0, the only possibility is b = 2 But then (2) becomes 3c2 + 8c − 20 ≤ 0, which does not hold for c ≥ 2 Hence the only solutions are (n, n2 + 1) and (n2 + 1, n), n ∈ N 23 We first observe that the given functional equation is equivalent to 4f (3m + 1)(3n + 1) − 1 3 + 1 = (4f (m) + 1) (4f (n) + 1) This... 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 2n − 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 2n possible colorings It follows that the total number of... largest integer such that r + (r + 1) + · · · + (r + n − 1) ≤ m This is equivalent to nr ≤ m − n(n−1) ≤ n(r + 1), so r = m − n−1 Clearly no n elements 2 n 2 from {r + 1, r + 2, , k} add up to m, so 4.37 Shortlisted Problems 1996 M ≥ k − rk (m, n) = k − m n−1 − n 2 615 (1) We claim that M is actually equal to k − rk (m, n) To show this, we shall prove by induction on n that if no n elements of... below: Q1i Q2i Q3i Q4i Q5i : : : : : 1 1 1 1 1 2 12 3 112 123 4 1112 112123 1234 5 4.38 Shortlisted Problems 1997 619 We claim that Rn is in fact exactly Qn1 Qn2 Qnn Before proving this, we observe that Qni = Qn−1,i This follows by induction, because Qni = Qn−1,i−1 Qn−1,i = Qn−2,i−1 Qn−2,i = Qn−1,i for n ≥ 3, i ≥ 2 (the cases i = 1 and n = 1, 2 are trivial) Now R1 = Q11 and Rn = Rn−1 (n) = Qn−1,1... ⎥ A4 = ⎢ ⎣4 6 1 2⎦ 7431 620 4 Solutions This construction can be generalized Suppose that we are given an n × n coveralls matrix An Let Bn be the matrix obtained from An by adding 2n to each entry, and Cn the matrix obtained from Bn by replacing each diagonal entry (equal to 2n + 1 by induction) with 2n Then the matrix An Bn A2n = Cn An is coveralls To show this, suppose that i ≤ n (the case i > n... i-diagonal For two cells within an i diagonal, x and y, we define x and y to be related if there exists a cross containing both x and y Evidently, for every cell x not on the 0-diagonal there are exactly two other cells related to it The relation thus breaks up each i-diagonal (i > 0) into cycles of length larger than 1 Due to the diagonal translational symmetry (modulo n), all the cycles within a given i-diagonal... BW with Γ Clearly AD = BB But we also have ∠BT C = 2∠BAC = 2∠BB C, which implies that CT = T B It follows that AD = BB = |BT ± T B | = |BT ± CT | Remark This problem is also solved easily using trigonometry 9 For i = 1, 2, 3 (all indices in this problem will be modulo 3) we denote by Oi the center of Ci and by Mi the midpoint of the arc Ai+1 Ai+2 that does not contain Ai First we have that Oi+1... + c(n2 + 3n + 2) xi i!(n − i + 2)! The coefficients Ci of xi appear in the form of a quadratic polynomial in i depending on n We claim that for large enough n this polynomial has negative discriminant, and is thus positive for every i Indeed, this discriminant equals D = ((b + 2c)n + (2b + 3c + 1))2 − 4(b + c + 1)c(n2 + 3n + 2) = (b2 − 4c)n2 − 2U n + V , where U = 2b2 + bc + b − 4c and V = (2b + c +... i=0 624 4 Solutions since p−1 = (p−1)(p−2)···(p−i) ≡ (−1)i (mod p) But this is clearly imi! i possible if f (i) equals 0 or 1 modulo p and f (0) = 0, f (1) = 1 p−1 Remark In proving the essential relation i=0 f (i) ≡ 0 (mod p), it is clearly enough to show that Sk = 1k + 2k + · · · + (p − 1)k is divisible by p for every k ≤ p − 2 This can be shown in two other ways (1) By induction Assume that S0 ≡ . =(n i + k)a. Besides the zero function, this is the general solution of the given func- tional equation. To verify this, we plug in m = ka + i, n = la +. within the second one with the angle α between AB and EF if and only if a cos α + b sin α ≤ c, a sin α + b cos α ≤ d. (1) Hence ABCD can be placed within